An Instructor’s Solutions Manual to Accompany Mechanics of Materials

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An Instructor’s Solutions Manual to Accompany

ISBN-13: 978-0-495-24458-5 ISBN-10: 0-495-24458-9 90000

9 780495 244585

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ISBN-13: 978-0-495-24458-5 ISBN-10: 0-495-24458-9

© 2009, 2004 Cengage Learning ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without the prior written permission of the publisher except as may be permitted by the license terms below. For product information and technology assistance, contact us at Cengage Learning Academic Resource Center, 1-800-423-0563. For permission to use material from this text or product, submit all requests online at www.cengage.com/permissions. Further permissions questions can be emailed to [email protected].

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Contents

1. Tension, Compression, and Shear Normal Stress and Strain 1 Mechanical Properties of Materials 15 Elasticity, Plasticity, and Creep 21 Linear Elasticity, Hooke’s Law, and Poisson’s Ratio 25 Shear Stress and Strain 30 Allowable Stresses and Allowable Loads 51 Design for Axial Loads and Direct Shear 69

2. Axially Loaded Members Changes in Lengths of Axially Loaded Members 89 Changes in Lengths under Nonuniform Conditions 105 Statically Indeterminate Structures 124 Thermal Effects 151 Stresses on Inclined Sections 178 Strain Energy 198 Impact Loading 212 Stress Concentrations 224 Nonlinear Behavior (Changes in Lengths of Bars) 231 Elastoplastic Analysis 237

3. Torsion Torsional Deformations 249 Circular Bars and Tubes 252 Nonuniform Torsion 266 Pure Shear 287 Transmission of Power 294 Statically Indeterminate Torsional Members 302 Strain Energy in Torsion 319 Thin-Walled Tubes 328 Stress Concentrations in Torsion 338

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4. Shear Forces and Bending Moments Shear Forces and Bending Moments 343 Shear-Force and Bending-Moment Diagrams 355

5. Stresses in Beams (Basic Topics) Longitudinal Strains in Beams 389 Normal Stresses in Beams 392 Design of Beams 412 Nonprismatic Beams 431 Fully Stressed Beams 440 Shear Stresses in Rectangular Beams 442 Shear Stresses in Circular Beams 453 Shear Stresses in Beams with Flanges 457 Built-Up Beams 466 Beams with Axial Loads 475 Stress Concentrations 492

6. Stresses in Beams (Advanced Topics) Composite Beams 497 Transformed-Section Method 508 Beams with Inclined Loads 520 Bending of Unsymmetric Beams 529 Shear Stresses in Wide-Flange Beams 541 Shear Centers of Thin-Walled Open Sections 543 Elastoplastic Bending 558

7. Analysis of Stress and Strain Plane Stress 571 Principal Stresses and Maximum Shear Stresses 582 Mohr’s Circle 595 Hooke’s Law for Plane Stress 608 Triaxial Stress 615 Plane Strain 622

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8. Applications of Plane Stress (Pressure Vessels, Beams, and Combined Loadings) Spherical Pressure Vessels 649 Cylindrical Pressure Vessels 655 Maximum Stresses in Beams 664 Combined Loadings 675

9. Deflections of Beams Differential Equations of the Deflection Curve 707 Deflection Formulas 710 Deflections by Integration of the Bending-Moment Equation 714 Deflections by Integration of the Shear Force and Load Equations 722 Method of Superposition 730 Moment-Area Method 745 Nonprismatic Beams 754 Strain Energy 770 Castigliano’s Theorem 775 Deflections Produced by Impact 784 Temperature Effects 790

10. Statically Indeterminate Beams Differential Equations of the Deflection Curve 795 Method of Superposition 809 Temperature Effects 839 Longitudinal Displacements at the Ends of Beams 843

11. Columns Idealized Buckling Models 845 Critical Loads of Columns with Pinned Supports 851 Columns with Other Support Conditions 863 Columns with Eccentric Axial Loads 871 The Secant Formula 880 Design Formulas for Columns 889 Aluminum Columns 903

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12. Review of Centroids and Moments of Inertia Centroids of Plane Ares 913 Centroids of Composite Areas 915 Moment of Inertia of Plane Areas 919 Parallel-Axis Theorem 923 Polar Moments of Inertia 927 Products of Inertia 929 Rotation of Axes 932 Principal Axes, Principal Points, and Principal Moments of Inertia 936

Answers to Problems 944

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1 Tension, Compression, and Shear

Normal Stress and Strain P1

Problem 1.2-1 A hollow circular post ABC (see figure) supports a load

P1  1700 lb acting at the top. A second load P2 is uniformly distributed around the cap plate at B. The diameters and thicknesses of the upper and lower parts of the post are dAB  1.25 in., tAB  0.5 in., dBC  2.25 in., and tBC  0.375 in., respectively.

A tAB dAB P2

(a) Calculate the normal stress AB in the upper part of the post. (b) If it is desired that the lower part of the post have the same compressive stress as the upper part, what should be the magnitude of the load P2? (c) If P1 remains at 1700 lb and P2 is now set at 2260 lb, what new thickness of BC will result in the same compressive stress in both parts?

B dBC tBC C

Solution 1.2-1 PART (a)

PART (b)

P1  1700 dBC  2.25 AAB 

dAB  1.25

tAB  0.5

tBC  0.375

p [ dAB2  (dAB  2 tAB)2] 4

AAB  1.178 sAB  1443 psi

sAB 

P1 AAB

ABC 

p[ dBC2  1dBC  2tBC22]

ABC  2.209

4 P2  ABABC  P1 P2  1488 lbs

CHECK:

;

P1 + P2  1443 psi ABC

;

1

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CHAPTER 1 Tension, Compression, and Shear

Part (c) P2  2260

dBC  2tBC 

P1 + P2  ABC sAB P1 + P2  2.744 sAB

dBC  tBC 

A

dBC2  2

A

dBC 

tBC  0.499 inches

4 P1 + P2 a b p sAB

Problem 1.2-2 A force P of 70 N is applied by a rider to the front hand brake of a bicycle (P is the resultant of an evenly distributed pressure). As the hand brake pivots at A, a tension T develops in the 460-mm long brake cable (Ae  1.075 mm2) which elongates by   0.214 mm. Find normal stress  and strain  in the brake cable.

Brake cable, L = 460 mm

Hand brake pivot A

37.5 mm A P (Resultant of distributed pressure)

m

100

P  70 N L  460 mm

Ae  1.075 mm2

  0.214 mm

Statics: sum moments about A to get T  2P s

T Ae

s  103.2 MPa

â

d L

â  4.65 * 10 4

E

s  1.4 * 105 MPa â

; ;

NOTE: (E for cables is approx. 140 GPa)

;

T

50 m

Solution 1.2-2

4 P1 + P2 b a p sAB

2

(dBC  2tBC)2  dBC2 

4 P1 + P2 b a p sAB

mm

Uniform hand brake pressure

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SECTION 1.2 Normal Stress and Strain

Problem 1.2-3 A bicycle rider would like to compare the effectiveness of cantilever hand brakes [see figure part (a)] versus V brakes [figure part (b)]. (a) Calculate the braking force RB at the wheel rims for each of the bicycle brake systems shown. Assume that all forces act in the plane of the figure and that cable tension T  45 lbs. Also, what is the average compressive normal stress c on the brake pad (A  0.625 in.2)? (b) For each braking system, what is the stress in the brake cable T (assume effective cross-sectional area of 0.00167 in.2)? (HINT: Because of symmetry, you only need to use the right half of each figure in your analysis.)

4 in.

T D TDC = TDE

T

45° TDE

4 in. TDC

TDE 90°

E

C TDCh

5 in.

4.25 in.

RB

A

G Pivot points anchored to frame (a) Cantilever brakes

Solution 1.2-3 Apad  0.625 in.2

Acable  0.00167 in.2 (a) CANTILEVER BRAKES-BRAKING FORCE RB & PAD PRESSURE Statics: sum forces at D to get TDC  T / 2 a MA  0 RB(1)  TDCh(3)  TDCv(1) TDCh  TDCv TDCh  T / 2 RB  90 lbs

;

so RB  2T vs 4.25T for V brakes (below)

HA

B

E RB

1 in.

B

F

RB  2T

T

TDCv

2 in.

T  45 lbs

C

D

1 in.

F

1 in. HA

Pivot points anchored to frame

A

VA

VA (b) V brakes

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CHAPTER 1 Tension, Compression, and Shear

spad 

RB Apad

scable 

spad  144 psi

T Acable

;

scable  26,946 psi

4.25  2.125 2

;

(same for V-brakes (below))

(b) V BRAKES - BRAKING FORCE RB & PAD PRESSURE a MA  0

RB  4.25T spad 

RB Apad

RB  191.3 lbs spad  306 psi

; ;

Problem 1.2-4 A circular aluminum tube of length

L  400 mm is loaded in compression by forces P (see figure). The outside and inside diameters are 60 mm and 50 mm, respectively. A strain gage is placed on the outside of the bar to measure normal strains in the longitudinal direction. (a) If the measured strain in   550 106, what is the shortening  of the bar? (b) If the compressive stress in the bar is intended to be 40 MPa, what should be the load P?

Solution 1.2-4 Aluminum tube in compression   550 106

(b) COMPRESSIVE LOAD P

  40 MPa

L  400 mm d2  60 mm

A

d1  50 mm (a) SHORTENING  OF THE BAR

  L  (550 106)(400 mm)  0.220 mm

;

p 2 p [d  d21]  [160 mm22  150 mm22] 4 2 4

P  A  (40 MPa)(863.9 mm2)  34.6 kN

;

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SECTION 1.2 Normal Stress and Strain

y

Problem 1.2-5

The cross section of a concrete corner column that is loaded uniformly in compression is shown in the figure. (a) Determine the average compressive stress c in the concrete if the load is equal to 3200 k. (b) Determine the coordinates xc and yc of the point where the resultant load must act in order to produce uniform normal stress in the column.

24 in.

16 in. x 8 in.

Solution 1.2-5 P  3200 kips 1 A  (24 + 20)(20 + 16 + 8)  a 82 b  202 2 A  1.504 103 in2 P A

sc  2.13 ksi

;

24  8 b 2 1 2 + (20)(16 + 8)(24 + 10) + 1822a 8b d 2 3

c(24)(20 + 16)(12) + (24  8)(8)a8 + (b) xc 

A

xc  19.22 inches

;

20 + 16 b + (20)(16 + 8) 2 16 + 8 1 2 a b + (24  8)(8)(4) + (82)a 8b d 2 2 3

c(24)(20 + 16)a8 + yc 

yc  19.22 inches

20 in.

20 in.

8 in.

(a) sc 

5

A

;

NOTE: xc & yc are the same as expected due to symmetry about a diagonal

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Problem 1.2-6 A car weighing 130 kN when fully loaded is pulled slowly up a steep inclined track by a steel cable (see figure). The cable has an effective cross-sectional area of 490 mm2, and the angle a of the incline is 30°. Calculate the tensile stress st in the cable.

Solution 1.2-6 Car on inclined track FREE-BODY DIAGRAM OF CAR

TENSILE STRESS IN THE CABLE W  Weight of car T  Tensile force in cable

 Angle of incline A  Effective area of cable R1, R2  Wheel reactions (no friction force between wheels and rails)

st 

Wsin a T  A A

SUBSTITUTE NUMERICAL VALUES: W  130 kN  30° A  490 mm2 st 

(130 kN)(sin 30°) 490 mm2

 133 MPa

;

EQUILIBRIUM IN THE INCLINED DIRECTION ©FT  0 Q + b T  W sin a  0 T  W sin

Problem 1.2-7 Two steel wires support a moveable overhead camera weighing W  25 lb (see figure) used for close-up viewing of field action at sporting events. At some instant, wire 1 is at on angle  20° to the horizontal and wire 2 is at an angle   48°. Both wires have a diameter of 30 mils. (Wire diameters are often expressed in mils; one mil equals 0.001 in.) Determine the tensile stresses 1 and 2 in the two wires.

T2

T1

b

a

W

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SECTION 1.2 Normal Stress and Strain

Solution 1.2-7 NUMERICAL DATA W  25 lb a  20

T1  T2

d  30 103 in.

p 180

b  48

p  radians 180

T1cos( )  T2cos() T1  T2

a Fv  0 T2 a

cos(b) cos (a)

T1sin( )  T2sin()  W

cos (b) sin(a) + sin (b) b  W cos(a)

TENSION IN WIRES T2 

T1  18.042 lb

TENSILE STRESSES IN WIRES A wire 

EQUILIBRIUM EQUATIONS a Fh  0

cos(b) cos (a)

W cos(b) a sin (a) + sin (b)b cos(a)

T2  25.337 lb

Problem 1.2-8 A long retaining wall is braced by wood shores set at an angle of 30° and supported by concrete thrust blocks, as shown in the first part of the figure. The shores are evenly spaced, 3 m apart. For analysis purposes, the wall and shores are idealized as shown in the second part of the figure. Note that the base of the wall and both ends of the shores are assumed to be pinned. The pressure of the soil against the wall is assumed to be triangularly distributed, and the resultant force acting on a 3-meter length of the wall is F  190 kN. If each shore has a 150 mm 150 mm square cross section, what is the compressive stress c in the shores?

p 2 d 4

s1 

T1 A wire

s1  25.5 ksi

;

s2 

T2 A wire

s2  35.8 ksi

;

7

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Solution 1.2-8 Retaining wall braced by wood shores F  190 kN A  area of one shore A  (150 mm)(150 mm)  22,500 mm2  0.0225 m2 SUMMATION OF MOMENTS ABOUT POINT A FREE-BODY DIAGRAM OF WALL AND SHORE

MA  0 哵哴 F(1.5 m)  CV (4.0 m)  CH (0.5 m)  0 or (190 kN)(1.5 m)  C(sin 30°)(4.0 m)  C(cos 30°)(0.5 m)  0 ⬖ C  117.14 kN COMPRESSIVE STRESS IN THE SHORES

C  compressive force in wood shore CH  horizontal component of C

sc 

117.14 kN C  A 0.0225 m2  5.21 MPa

CV  vertical component of C

;

CH  C cos 30° CV  C sin 30°

Problem 1.2-9 A pickup truck tailgate supports a crate

(WC  150 lb), as shown in the figure. The tailgate weighs WT  60 lb and is supported by two cables (only one is shown in the figure). Each cable has an effective crosssectional area Ae  0.017 in2. (a) Find the tensile force T and normal stress  in each cable. (b) If each cable elongates   0.01 in. due to the weight of both the crate and the tailgate, what is the average strain in the cable?

WC = 150 lb

H = 12 in.

dc = 18 in. Ca ble Crate

Truck

Tail gate dT = 14 in. L = 16 in.

WT = 60 lb

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SECTION 1.2 Normal Stress and Strain

Solution 1.2-9 (a) T  2 Tv2 + T h2 T  184.4 lb

Wc  150 lb Ae  0.017 in2

scable 

WT  60

(b) âcable 

  0.01

T Ae d Lc

scable  10.8 ksi cable  5 104

; ; ;

dc  18 dT  14 H  12 L  16 L c  2 L2 + H2 a Mhinge  0

Lc  20

2TvL  Wcdc  WT dT

Tv 

W cd c + W Td T 2L

Th 

L T H v

Tv  110.625 lb T h  147.5

Problem 1.2-10 Solve the preceding problem if the

mass of the tail gate is MT  27 kg and that of the crate is MC  68 kg. Use dimensions H  305 mm, L  406 mm, dC  460 mm, and dT  350 mm. The cable cross-sectional area is Ae  11.0 mm2.

(a) Find the tensile force T and normal stress  in each cable. (b) If each cable elongates   0.25 mm due to the weight of both the crate and the tailgate, what is the average strain in the cable?

MC = 68 kg dc = 460 mm H = 305 mm

Ca

ble

Crate

Truck

Tail gate dT = 350 mm L = 406 mm

MT = 27 kg

9

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Solution 1.2-10 (a) T  2 T2v + T2h

Mc  68 g  9.81

MT  27 kg Wc  Mcg

scable  (b) âcable 

;

T Ae

scable  74.5 MPa

d Lc

cable  4.92 104

;

;

WT  264.87

m

N  kg

s2

  0.25

Ae  11.0 mm2 dc  460

dT  350

H  305

L  406

L c  2 L2 + H2 a Mhinge  0

Lc  507.8 mm

2TvL  Wcdc  WT dT

Wc d c + WT d T 2L

Th 

s2

WT  MTg

Wc  667.08

Tv 

m

T  819 N

L T H v

Tv  492.071 N

Th  655.019 N

Problem ★1.2-11 An L-shaped reinforced concrete slab

12 ft 12 ft (but with a 6 ft 6 ft cutout) and thickness t  9.0 in. is lifted by three cables attached at O, B and D, as shown in the figure. The cables are combined at point Q, which is 7.0 ft above the top of the slab and directly above the center of mass at C. Each cable has an effective crosssectional area of Ae  0.12 in2. (a) Find the tensile force Ti (i  1, 2, 3) in each cable due to the weight W of the concrete slab (ignore weight of cables). (b) Find the average stress i in each cable. (See Table H-1 in Appendix H for the weight density of reinforced concrete.)

F Coordinates of D in ft

Q (5, 5, 7)

T3 1

T1

7

D (5, 12, 0)

1 T2

5 5 z O (0, 0, 0)

y x 6 ft

C (5, 5, 0) 5 7 7

6 ft

W 6 ft B (12, 0, 0) lb Concrete slab g = 150 —3 ft Thickness t, c.g at (5 ft, 5 ft, 0)

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SECTION 1.2 Normal Stress and Strain

11

Solution 1.2-11 CABLE LENGTHS

52  52  72  99 2

L1  299 L2  11.091

L2  25 + 7 + 7 2

2

5  7  7  123 2

2

L2  2123

2

L3  9.899

L3  272 + 72 7  7  98 2

T1 0.484 T2  0.385 P T Q P 0.589 Q 3

L1  9.95

L1  252 + 52 + 72

(a) SOLUTION FOR CABLE FORCES USING STATICS (3 EQU, 3 UNKNOWNS) T1 

7 299 144

T1  0.484

5 2123 T2  144

d1 

5 22 T3  12

T1L1 EA

T2L2 d2  EA

T2  0.385

T3L3 d3  EA

T3  0.589

a Tverti  0 T1

7 299

+ T2

7 2123

+ T3

1 22

 1 CHECK

5

7

299 5

2123 5

299 7

2123 7

22 1

299

2123

22

q

7

0 1

7 2123

+ T3

1 22

1

1. sum about x-axis to get T3v, then T3 2. sum about y-axis to get T2v, then T2 3. sum vertical forces to get T1v, then T1 OR sum forces in x-dir to get T1x in terms of T2x SLAB WEIGHT & C.G. W  150 1122  622 xcg 

0 0 P 1Q

9 12

W  1.215 104

2 A 3 + A (6 + 3) 3A

xcg  5

same for ycg

ycg  xcg

Multiply unit forces by W

-1

r

+ T2

299

T  Tu W

For unit force in Z-direction

T1 T2  PT Q 3

T1

NOTE: preferred solution uses sum of moments about a line as follows –

L3  722

2

check:

T1 Tu  T2 PT Q 3

(b) s 

T 0.12

5877 T  4679 lb P 7159 Q s

;

49.0 ksi 39.0 ksi psi P 60.0 ksi Q

;

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Problem *1.2-12

A round bar ACB of length 2L (see figure) rotates about an axis through the midpoint C with constant angular speed v (radians per second). The material of the bar has weight density g. (a) Derive a formula for the tensile stress sx in the bar as a function of the distance x from the midpoint C. (b) What is the maximum tensile stress smax?

Solution 1.2-12

Rotating Bar Consider an element of mass dM at distance  from the midpoint C. The variable  ranges from x to L. g dM  g A dj

 angular speed (rad/s) A  cross-sectional area

  weight density g  mass density g We wish to find the axial force Fx in the bar at Section D, distance x from the midpoint C. The force Fx equals the inertia force of the part of the rotating bar from D to B.

dF  Inertia force (centrifugal force) of element of mass dM g dF  (dM)(jv2)  g Av2jdj L

g 2 gAv2 2 2 Av jdj  (L x ) 2g LD Lx g (a) TENSILE STRESS IN BAR AT DISTANCE x Fx 

B

sx 

dF 

gv2 2 Fx  (L  x2) A 2g

(b) MAXIMUM TENSILE STRESS x  0 smax 

Problem 1.2-13 Two gondolas on a ski lift are locked in the position shown in the figure while repairs are being made elsewhere. The distance between support towers is L  100 ft. The length of each cable segment under gondola weights WB  450 lb and WC  650 lb are DAB  12 ft, DBC  70 ft, and DCD  20 ft. The cable sag at B is B  3.9 ft and that at C(C) is 7.1 ft. The effective cross-sectional area of the cable is Ae  0.12 in2.

;

gv2L2 2g

;

A

D u1

DB B

u2

DC

u3

C

WB

WC

(a) Find the tension force in each cable segment; neglect the mass of the cable. (b) Find the average stress (s ) in each cable segment. L = 100 ft

Support tower

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13

SECTION 1.2 Normal Stress and Strain

Solution 1.2-13 WB  450

A

Wc  650 lb

D u1

DB

u2

B

¢ B  3.9 ft

DC

u3

C

¢ C  7.1 ft L  100 ft

WB

WC

DAB  12 ft

Support tower

DBC  70 ft DCD  20 ft

L = 100 ft

DAB  DBC  DCD  102 ft Ae  0.12 in2

CONTRAINT EQUATIONS

COMPUTE INITIAL VALUES OF THETA ANGLES (RADIANS)

DAB cos(u1) + DBC cos (u2) + DCD cos(u3)  L

u1  asin a

¢B b DAB

u1  0.331

u2  asin a

¢C  ¢B b DBC

u2  0.046

u3  asin a

¢C b DCD

u3  0.363

DAB sin(u1) + DBC sin (u2)  DCD sin(u3) SOLVE SIMULTANEOUS EQUATIONS NUMERICALLY FOR TENSION FORCE IN EACH CABLE SEGMENT

(a) STATICS AT B & C TAB cos(u1) + TBC cos (u2)  0 TAB sin(u1)  TBC sin(u2)  WB TBC cos(u2) + TCD cos (u3)  0 TBC sin(u2)  TCD sin (u3)  WC

TAB  1620 lb

TCB  1536 lb

TCD  1640 lb

;

CHECK EQUILIBRIUM AT B & C TAB sin(u1)  TBC sin (u2)  450 TBC sin(u2)  TCD sin (u3)  650 (b) COMPUTE STRESSES IN CABLE SEGMENTS sAB 

TAB Ae

sBC 

TBC Ae

sAB  13.5 ksi

sBC  12.8 ksi

sCD  13.67 ksi

;

sCD 

TCD Ae

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CHAPTER 1 Tension, Compression, and Shear

z

Problem 1.2-14 A crane boom of mass 450 kg with its center of mass at C is stabilized by two cables AQ and BQ (Ae  304 mm2 for each cable) as shown in the figure. A load P  20 kN is supported at point D. The crane boom lies in the y–z plane.

P Q y C om

(a) Find the tension forces in each cable: TAQ and TBQ (kN); neglect the mass of the cables, but include the mass of the boom in addition to load P. (b) Find the average stress () in each cable.

D

bo

14

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2 1 B

2

Cr an e

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2m 2m

55°

TBQ

O

5m TAQ

5m

x

5m

A

3m

Solution 1.2-14 Data

Mboom  450 kg

g  9.81

m s2

2TAQZ(3000)  Wboom(5000) + P(9000)

Wboom  Mboom g TAQZ 

Wboom  4415 N P  20 kN Ae  304 mm2

TAQ 

A

22 + 22 + 12 T AQz 2

TAQ  50.5 kN  TBQ

(a) symmetry: TAQ = TBQ (b) s  a Mx  0

Wboom(5000) + P(9000) 2(3000)

TAQ Ae

;

s  166.2 MPa

;

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SECTION 1.3 Mechanical Properties of Materials

Mechanical Properties of Materials Problem 1.3-1 Imagine that a long steel wire hangs vertically from a high-altitude balloon. (a) What is the greatest length (feet) it can have without yielding if the steel yields at 40 ksi? (b) If the same wire hangs from a ship at sea, what is the greatest length? (Obtain the weight densities of steel and sea water from Table H-1, Appendix H.)

Solution 1.3-1

Hanging wire of length L W  total weight of steel wire

S  weight density of steel  490 lb/ft3

w  weight density of sea water

(b) WIRE HANGING IN SEA WATER F  tensile force at top of wire F  (gS  gW)AL smax 

 63.8 lb/ft3 A  cross-sectional area of wire

max  40 ksi (yield strength)



(a) WIRE HANGING IN AIR W  SAL

smax gS gW 40,000 psi (490  63.8) lb/ft3

 13,500 ft

smax 

W  gSL A

Lmax 

smax 40,000 psi  (144 in.2/ft2) gS 490 lb/ft3

 11,800 ft

Lmax 

F  (gS  gW)L A

(144 in.2/ft2)

;

;

Problem 1.3-2 Imagine that a long wire of tungsten hangs vertically from a high-altitude balloon. (a) What is the greatest length (meters) it can have without breaking if the ultimate strength (or breaking strength) is 1500 MPa? (b) If the same wire hangs from a ship at sea, what is the greatest length? (Obtain the weight densities of tungsten and sea water from Table H-1, Appendix H.)

15

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CHAPTER 1 Tension, Compression, and Shear

Solution 1.3-2

Hanging wire of length L (b) WIRE HANGING IN SEA WATER

W  total weight of tungsten wire

F  tensile force at top of wire

T  weight density of tungsten

F  (T  W)AL

 190 kN/m3

W  weight density of sea water  10.0 kN/m3 A  cross-sectional area of wire

max  1500 MPa (breaking strength)

smax 

F  (gT  gW)L A

Lmax 

smax gT  gW



(a) WIRE HANGING IN AIR W  TAL

1500MPa (190  10.0) kN/m3

 8300 m

;

W smax   gTL A Lmax 

smax 1500MPa  gT 190 kN/m3

 7900 m

;

Problem 1.3-3 Three different materials, designated A, B, and C, are tested in tension using test specimens having diameters of 0.505 in. and gage lengths of 2.0 in. (see figure). At failure, the distances between the gage marks are found to be 2.13, 2.48, and 2.78 in., respectively. Also, at the failure cross sections the diameters are found to be 0.484, 0.398, and 0.253 in., respectively. Determine the percent elongation and percent reduction in area of each specimen, and then, using your own judgment, classify each material as brittle or ductile.

P

Gage length

P

Solution 1.3-3 Tensile tests of three materials where L1 is in inches. Percent reduction in area

Percent elongation 

L1 L1  L0 (100)  a  1b 100 L0 L0

L0  2.0 in. Percent elongation  a

L1  1b (100) 2.0

(Eq. 1)

d0  initial diameter

A0  A1 (100) A0 A1  a1  b (100) A0 

d1  final diameter

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17

SECTION 1.3 Mechanical Properties of Materials

A1 d1 2  a b d0  0.505 in. A0 d0 Percent reduction in area  c1  a

d1 2 b d(100) 0.505

(Eq. 2)

Material

L1 (in.)

d1 (in.)

% Elongation (Eq. 1)

% Reduction (Eq. 2)

Brittle or Ductile?

A

2.13

0.484

6.5%

8.1%

Brittle

B

2.48

0.398

24.0%

37.9%

Ductile

C

2.78

0.253

39.0%

74.9%

Ductile

where d1 is in inches.

Problem 1.3-4 The strength-to-weight ratio of a structural material is defined as its load-carrying capacity divided by its weight. For materials in tension, we may use a characteristic tensile stress (as obtained from a stress-strain curve) as a measure of strength. For instance, either the yield stress or the ultimate stress could be used, depending upon the particular application. Thus, the strength-to-weight ratio RS/W for a material in tension is defined as RS/W 

s g

in which  is the characteristic stress and  is the weight density. Note that the ratio has units of length. Using the ultimate stress U as the strength parameter, calculate the strength-to-weight ratio (in units of meters) for each of the following materials: aluminum alloy 6061-T6, Douglas fir (in bending), nylon, structural steel ASTM-A572, and a titanium alloy. (Obtain the material properties from Tables H-1 and H-3 of Appendix H. When a range of values is given in a table, use the average value.)

Solution 1.3-4 Strength-to-weight ratio The ultimate stress U for each material is obtained from Table H-3, Appendix H, and the weight density  is obtained from Table H-1. The strength-to-weight ratio (meters) is RS/W 

sU ( MPa) g( kN/m3)

(103)

Values of U, , and RS/W are listed in the table.

U (MPa)

 (kN/m3)

RS/W (m)

310

26.0

11.9 103

Douglas fir

65

5.1

12.7 103

Nylon

60

9.8

6.1 103

Structural steel ASTM-A572

500

77.0

6.5 103

Titanium alloy

1050

44.0

23.9 103

Aluminum alloy 6061-T6

Titanium has a high strength-to-weight ratio, which is why it is used in space vehicles and high-performance airplanes. Aluminum is higher than steel, which makes it desirable for commercial aircraft. Some woods are also higher than steel, and nylon is about the same as steel.

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CHAPTER 1 Tension, Compression, and Shear

Problem 1.3-5 A symmetrical framework consisting of three pin-connected bars is loaded by a force P (see figure). The angle between the inclined bars and the horizontal is  48°. The axial strain in the middle bar is measured as 0.0713. Determine the tensile stress in the outer bars if they are constructed of aluminum alloy having the stress-strain diagram shown in Fig. 1-13. (Express the stress in USCS units.)

A

B

C a

D P

Solution 1.3-5 Symmetrical framework L  length of bar BD L1  distance BC  L cot  L(cot 48°)  0.9004 L L2  length of bar CD  L csc  L(csc 48°)  1.3456 L Elongation of bar BD  distance DE  BDL BDL  0.0713 L Aluminum alloy

 48°

L3  distance CE L3  2L21 (L âBDL)2

BD  0.0713

 2(0.9004L)2 + L2(1 + 0.0713)2

Use stress-strain diagram of Figure 1-13

 1.3994 L

  elongation of bar CD   L3  L2  0.0538L Strain in bar CD 0.0538L d   0.0400 L2 1.3456L From the stress-strain diagram of Figure 1-13: 

 ⬇ 31 ksi

;

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SECTION 1.3 Mechanical Properties of Materials

Problem 1.3-6 A specimen of a methacrylate plastic is tested in tension at room temperature (see figure), producing the stress-strain data listed in the accompanying table. Plot the stress-strain curve and determine the proportional limit, modulus of elasticity (i.e., the slope of the initial part of the stress-strain curve), and yield stress at 0.2% offset. Is the material ductile or brittle?

19

STRESS-STRAIN DATA FOR PROBLEM 1.3-6

P P

Stress (MPa)

Strain

8.0 17.5 25.6 31.1 39.8 44.0 48.2 53.9 58.1 62.0 62.1

0.0032 0.0073 0.0111 0.0129 0.0163 0.0184 0.0209 0.0260 0.0331 0.0429 Fracture

Solution 1.3-6 Tensile test of a plastic Using the stress-strain data given in the problem statement, plot the stress-strain curve:

PL  proportional limit

PL ⬇ 47 MPa

Modulus of elasticity (slope) ⬇ 2.4 GPa

; ;

Y  yield stress at 0.2% offset Y ⬇ 53 MPa

;

Material is brittle, because the strain after the proportional ; limit is exceeded is relatively small.

Problem 1.3-7 The data shown in the accompanying table were obtained from a tensile test of high-strength steel. The test specimen had a diameter of 0.505 in. and a gage length of 2.00 in. (see figure for Prob. 1.3-3). At fracture, the elongation between the gage marks was 0.12 in. and the minimum diameter was 0.42 in. Plot the conventional stress-strain curve for the steel and determine the proportional limit, modulus of elasticity (i.e., the slope of the initial part of the stress-strain curve), yield stress at 0.1% offset, ultimate stress, percent elongation in 2.00 in., and percent reduction in area.

TENSILE-TEST DATA FOR PROBLEM 1.3-7 Load (lb)

1,000 2,000 6,000 10,000 12,000 12,900 13,400 13,600 13,800 14,000 14,400 15,200 16,800 18,400 20,000 22,400 22,600

Elongation (in.)

0.0002 0.0006 0.0019 0.0033 0.0039 0.0043 0.0047 0.0054 0.0063 0.0090 0.0102 0.0130 0.0230 0.0336 0.0507 0.1108 Fracture

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CHAPTER 1 Tension, Compression, and Shear

Solution 1.3-7 Tensile test of high-strength steel d0  0.505 in. A0 

pd20 4

L0  2.00 in.

ENLARGEMENT OF PART OF THE STRESS-STRAIN CURVE

 0.200 in.2

CONVENTIONAL STRESS AND STRAIN s

P A0

Load P (lb) 1,000 2,000 6,000 10,000 12,000 12,900 13,400 13,600 13,800 14,000 14,400 15,200 16,800 18,400 20,000 22,400 22,600

â

d L0

Elongation  (in.) 0.0002 0.0006 0.0019 0.0033 0.0039 0.0043 0.0047 0.0054 0.0063 0.0090 0.0102 0.0130 0.0230 0.0336 0.0507 0.1108 Fracture

STRESS-STRAIN DIAGRAM

Stress  (psi) 5,000 10,000 30,000 50,000 60,000 64,500 67,000 68,000 69,000 70,000 72,000 76,000 84,000 92,000 100,000 112,000 113,000

Strain  0.00010 0.00030 0.00100 0.00165 0.00195 0.00215 0.00235 0.00270 0.00315 0.00450 0.00510 0.00650 0.01150 0.01680 0.02535 0.05540

RESULTS Proportional limit ⬇ 65,000 psi

;

Modulus of elasticity (slope) ⬇ 30 106 psi Yield stress at 0.1% offset ⬇ 69,000 psi Ultimate stress (maximum stress) ⬇ 113,000 psi

;

Percent elongation in 2.00 in. 

L1  L0 (100) L0



0.12 in. (100)  6% 2.00 in.

;

Percent reduction in area  

A0  A1 (100) A0 0.200 in.2 

 31%

p 4 (0.42

0.200 in.2 ;

in.) 2

(100)

;

;

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SECTION 1.4 Elasticity, Plasticity, and Creep

21

Elasticity, Plasticity, and Creep Problem 1.4-1 A bar made of structural steel having the stressstrain diagram shown in the figure has a length of 48 in. The yield stress of the steel is 42 ksi and the slope of the initial linear part of the stress-strain curve (modulus of elasticity) is 30 103 ksi. The bar is loaded axially until it elongates 0.20 in., and then the load is removed. How does the final length of the bar compare with its original length? (Hint: Use the concepts illustrated in Fig. 1-18b.)

Solution 1.4-1 Steel bar in tension ELASTIC RECOVERY E âE 

sB 42 ksi  0.00140  Slope 30 * 103 ksi

RESIDUAL STRAIN R R  B  E  0.00417  0.00140  0.00277 PERMANENT SET L  48 in. Yield stress Y  42 ksi Slope  30 103 ksi

  0.20 in.

RL  (0.00277)(48 in.)  0.13 in. Final length of bar is 0.13 in. greater than its original ; length.

STRESS AND STRAIN AT POINT B sB  sY  42 ksi âB 

0.20 in. d   0.00417 L 48 in.

Problem 1.4-2 A bar of length 2.0 m is made of a structural steel having the stress-strain diagram shown in the figure. The yield stress of the steel is 250 MPa and the slope of the initial linear part of the stress-strain curve (modulus of elasticity) is 200 GPa. The bar is loaded axially until it elongates 6.5 mm, and then the load is removed. How does the final length of the bar compare with its original length? (Hint: Use the concepts illustrated in Fig. 1-18b.)

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CHAPTER 1 Tension, Compression, and Shear

Solution 1.4-2 Steel bar in tension L  2.0 m  2000 mm Yield stress Y  250 MPa Slope  200 GPa

  6.5 mm

ELASTIC RECOVERY E âE 

sB 250 MPa   0.00125 Slope 200 GPa

RESIDUAL STRAIN R R  B  E  0.00325  0.00125  0.00200 Permanent set  RL  (0.00200)(2000 mm)  4.0 mm

STRESS AND STRAIN AT POINT B sB  sY  250 MPa âB 

6.5 mm d   0.00325 L 2000 mm

Final length of bar is 4.0 mm greater than its original ; length.

Problem 1.4-3 An aluminum bar has length L  5 ft and diameter d  1.25 in. The stress-strain curve for the aluminum is

shown in Fig. 1-13 of Section 1.3. The initial straight-line part of the curve has a slope (modulus of elasticity) of 10 106 psi. The bar is loaded by tensile forces P  39 k and then unloaded. (a) What is the permanent set of the bar? (b) If the bar is reloaded, what is the proportional limit? (Hint: Use the concepts illustrated in Figs. 1-18b and 1-19.)

Solution 1.4-3 RESIDUAL STRAIN

(a) PERMAMENT SET Numerical data L  60 in

âE  âB  âE

d  1.25 in

PERMANENT SET

P  39 kips

STRESS AND STRAIN AT PT B sB 

P p 2 d 4

s B  31.8 ksi

From Figure 1-13 B  0.05 ELASTIC RECOVERY sB 10(10)3

;

(b) PROPORTIONAL LIMIT WHEN RELOADED

B  31.78 ksi

âE 

âRL  2.81 in.

âR  0.047

âE  3.178 * 103

;

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SECTION 1.4 Elasticity, Plasticity, and Creep

Problem 1.4-4 A circular bar of magnesium alloy is 750 mm long. The stressstrain diagram for the material is shown in the figure. The bar is loaded in tension to an elongation of 6.0 mm, and then the load is removed. (a) What is the permanent set of the bar? (b) If the bar is reloaded, what is the proportional limit? (Hint: Use the concepts illustrated in Figs. 1-18b and 1-19.)

23

200 s (MPa) 100

0

0

0.005 e

0.010

Solution 1.4-4 NUMERICAL DATA

L  750 mm

  6 mm

STRESS AND STRAIN AT PT B d B  180 MPa B  8 103 âB  L ELASTIC RECOVERY 178 slope  slope  4.45 104 0.004 sB âE  slope E  4.045 103 RESIDUAL STRAIN R  B  E R  3.955 103

(b) PROPORTIONAL LIMIT WHEN RELOADED sB  180 MPa

;

(a) PERMANENT SET âRL  2.97 mm

;

Problem 1.4-5 A wire of length L  4 ft and diameter d  0.125 in. is stretched by tensile forces P  600 lb. The wire is made of a copper alloy having a stress-strain relationship that may be described mathematically by the following equation: s

18,000P 1 + 300P

0 … P … 0.03 (s  ksi)

in which P is nondimensional and  has units of kips per square inch (ksi). (a) (b) (c) (d)

Construct a stress-strain diagram for the material. Determine the elongation of the wire due to the forces P. If the forces are removed, what is the permanent set of the bar? If the forces are applied again, what is the proportional limit?

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CHAPTER 1 Tension, Compression, and Shear

Solution 1.4-5 Wire stretched by forces P L  4 ft  48 in.

d  0.125 in.

ALTERNATIVE FORM OF THE STRESS-STRAIN RELATIONSHIP

P  600 lb

Solve Eq. (1) for  in terms of :

COPPER ALLOY

â

18,000â s 1 + 300â

0 … â … 0.03 (s  ksi) (Eq. 1)

(a) STRESS-STRAIN DIAGRAM (From Eq. 1)

s 0 … s … 54 ksi (s  ksi) 18,000300s

(Eq. 2)

This equation may also be used when plotting the stressstrain diagram. (b) ELONGATION  OF THE WIRE P 600 lb 48,900 psi  48.9 ksi  p A (0.125 in.)2 4 From Eq. (2) or from the stress-strain diagram: s

  0.0147

  L  (0.0147)(48 in.)  0.71 in. STRESS AND STRAIN AT POINT B (see diagram)

B  48.9 ksi B  0.0147 ELASTIC RECOVERY E âE 

sB 48.9 ksi   0.00272 Slope 18,000 ksi

INITIAL SLOPE OF STRESS-STRAIN CURVE

RESIDUAL STRAIN R

Take the derivative of  with respect to :

R  B  E  0.0147  0.0027  0.0120

(1 + 300â)(18,000)  (18,000)(300)s ds  dâ (1 + 300â)2 

At â  0,

18,000 (1 + 300â)2 ds  18,00 ksi dâ

⬖ Initial slope  18,000 ksi

(c) Permanent set  RL  (0.0120)(48 in.)  0.58 in. ; (d) Proportional limit when reloaded  B

B  49 ksi

;

;

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SECTION 1.5 Linear Elasticity, Hooke’s Law, and Poisson’s Ratio

25

Linear Elasticity, Hooke’s Law, and Poisson’s Ratio When solving the problems for Section 1.5, assume that the material behaves linearly elastically.

Problem 1.5-1 A high-strength steel bar used in a large crane has diameter

d  2.00 in. (see figure). The steel has modulus of elasticity E  29 106 psi and Poisson’s ratio v  0.29. Because of clearance requirements, the diameter of the bar is limited to 2.001 in. when it is compressed by axial forces. What is the largest compressive load Pmax that is permitted?

Solution 1.5-1 Steel bar in compression STEEL BAR d  2.00 in.

AXIAL STRESS Max. d  0.001 in.

E  29 10 psi 6

v  0.29

 50.00 ksi (compression)

LATERAL STRAIN â¿ 

Assume that the yield stress for the high-strength steel is greater than 50 ksi. Therefore, Hooke’s law is valid.

0.001 in. ¢d   0.0005 d 2.00 in.

MAXIMUM COMPRESSIVE LOAD

AXIAL STRAIN â 

  E  (29 106 psi)(0.001724)

0.0005 â¿   0.001724 v 0.29

p Pmax  sA  (50.00 ksi)a b(2.00 in.)2 4  157 k ;

(shortening)

Problem 1.5-2 A round bar of 10 mm diameter is made of aluminum alloy 7075-T6 (see figure). When the bar is stretched by axial forces P, its diameter decreases by 0.016 mm. Find the magnitude of the load P. (Obtain the material properties from Appendix H.)

Solution 1.5-2 d  10 mm

Aluminum bar in tension

d  0.016 mm

AXIAL STRESS

  E  (72 GPa)(0.004848)

(Decrease in diameter)

 349.1 MPa (Tension)

7075-T6 From Table H-2: E  72 GPa

v  0.33

From Table H-3: Yield stress Y  480 MPa LATERAL STRAIN â¿ 

0.016mm ¢d    0.0016 d 10mm

AXIAL STRAIN â¿ 0.0016  v 0.33  0.004848 (Elongation)

â 

Because   Y, Hooke’s law is valid. LOAD P (TENSILE FORCE) p P  sA  (349.1 MPa)a b(10 mm)2 4  27.4 kN ;

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CHAPTER 1 Tension, Compression, and Shear

Problem 1.5-3 A polyethylene bar having diameter d1  4.0 in. is placed inside a

Steel tube

steel tube having inner diameter d2  4.01 in. (see figure). The polyethylene bar is then compressed by an axial force P. At what value of the force P will the space between the nylon bar and the steel tube be closed? (For nylon, assume E  400 ksi and v  0.4.)

d1 d2 Polyethylene bar

Solution 1.5-3 NORMAL STRAIN

NUMERICAL DATA d1  4 in

d2  4.01 in.

v  0.4 p A1  d12 4

d1  0.01 in p A2  d22 4

E  200 ksi

â1 

A1  12.566 in2

AXIAL STRESS

1  1.25 ksi

COMPRESSION FORCE

LATERAL STRAIN 0.01 âp  4

1  6.25 103

v

1  E 1

A2  12.629 in2

¢d1 âp  d1

â p

P  EA11 3

p  2.5 10

P   15.71 kips

Problem 1.5-4 A prismatic bar with a circular cross section is loaded by tensile forces P  65 kN (see figure). The bar has length L  1.75 m and diameter d  32 mm. It is made of aluminum alloy with modulus of elasticity E  75 GPa and Poisson’s ratio   1/3. Find the increase in length of the bar and the percent decrease in its cross-sectional area.

;

d

P L

P

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SECTION 1.5 Linear Elasticity, Hooke’s Law, and Poisson’s Ratio

27

Solution 1.5-4 NUMERICAL DATA P  65 kN

v

LATERAL STRAIN

DECREASE IN DIAMETER

d  32 mm L  1.75(1000) mm

d  ƒpdƒ

E  75 GPa

p 2 d 4

Af 

Ai  804.248 mm2

p 1d  ¢d22 4

Af  803.67 mm2 % decrease in x-sec area 

AXIAL STRAIN â

d  0.011 mm

FINAL AREA OF CROSS SECTION

INITIAL AREA OF CROSS SECTION Ai 

p  3.592 104

p  

1 3

P EA i

  1.078 103

Af  Ai (100) Ai

  0.072

;

;

INCREASE IN LENGTH L  L

¢ L  1.886 mm

;

Problem 1.5-5 A bar of monel metal as in the figure (length L  9 in., diameter d  0.225 in.) is loaded axially by a tensile force P. If the bar elongates by 0.0195 in., what is the decrease in diameter d? What is the magnitude of the load P? Use the data in Table H-2, Appendix H.

Solution 1.5-5 NUMERICAL DATA

DECREASE IN DIAMETER

E  25000 ksi

d  pd

  0.32

¢d   1.56 * 104 in.

L  9 in.

INITIAL CROSS SECTIONAL AREA

  0.0195 in.

Ai 

d  0.225 in.

P  EAi 3

  2.167 10

LATERAL STRAIN p  

Ai  0.04 in.2

MAGNITUDE OF LOAD P

NORMAL STRAIN d â L

p 2 d 4

p  6.933 104

P  2.15 kips

;

;

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CHAPTER 1 Tension, Compression, and Shear

Problem 1.5-6 A tensile test is peformed on a brass specimen 10 mm in diameter using a gage length of 50 mm (see figure). When the tensile load P reaches a value of 20 kN, the distance between the gage marks has increased by 0.122 mm. (a) What is the modulus of elasticity E of the brass? (b) If the diameter decreases by 0.00830 mm, what is Poisson’s ratio?

Solution 1.5-6 Brass specimen in tension d  10 mm Gage length L  50 mm P  20 kN   0.122 mm

d  0.00830 mm

AXIAL STRESS P 20 k s   254.6 MPa p A (10 mm) 2 4 Assume  is below the proportional limit so that Hooke’s law is valid.

(a) MODULUS OF ELASTICITY E

s 254.6 MPa   104 Gpa â 0.002440

;

(b) POISSON’S RATIO   v d  d  vd v

¢d 0.00830 mm   0.34 âd (0.002440)(10 mm)

;

AXIAL STRAIN â

0.122 mm d   0.002440 L 50 mm

Problem 1.5-7 A hollow, brass circular pipe ABC (see figure) supports a load P1  26.5 kips acting at the top. A second load P2  22.0 kips is uniformly distributed around the cap plate at B. The diameters and thicknesses of the upper and lower parts of the pipe are dAB  1.25 in., tAB  0.5 in., dBC  2.25 in., and tAB  0.375 in., respectively. The modulus of elasticity is 14,000 ksi. When both loads are fully applied, the wall thickness of pipe BC increases by 200 3 106 in. (a) Find the increase in the inner diameter of pipe segment BC. (b) Find Poisson’s ratio for the brass. (c) Find the increase in the wall thickness of pipe segment AB and the increase in the inner diameter of AB.

P1 A dAB tAB P2 B Cap plate dBC tBC C

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29

SECTION 1.5 Linear Elasticity, Hooke’s Law, and Poisson’s Ratio

Solution 1.5-7 NUMERICAL DATA P1  26.5 kips P2  22 kips dAB  1.25 in. tAB  0.5 in. dBC  2.25 in. tBC  0.375 in. E  14000 ksi tBC  200 106

(c) INCREASE IN THE WALL THICKNESS OF PIPE SEGMENT AB AND THE INCREASE IN THE INNER DIAMETER OF AB

¢tBC tBC

p c dAB2  1 dAB  2tAB22 d 4

âAB 

P1 EAAB

AB  1.607 103

pAB  brassAB tAB  pABtAB

(a) INCREASE IN THE INNER DIAMETER OF PIPE SEGMENT BC âpBC 

AAB 

pAB  5.464 104

¢tAB  2.73 * 104 in.

;

dABinner  pAB(dAB  2tAB) ¢dABinner  1.366 * 104 inches

pBC  5.333 104

dBCinner  pBC(dBC  2tBC) ¢ dBCinner  8 * 104 inches

;

(b) POISSON’S RATIO FOR THE BRASS ABC 

p c d 2  1 dBC  2tBC22 d 4 BC

ABC  2.209 in.2 â BC 

1P1 + P22

 brass 

1EABC2

âpBC âBC

BC  1.568 103 brass  0.34

(agrees with App. H (Table H-2))

Problem 1.5-8 A brass bar of length 2.25 m with a square cross section of 90 mm on each side is subjected to an axial tensile force of 1500 kN (see figure). Assume that E  110 GPa and   0.34. Determine the increase in volume of the bar.

90 mm

90 mm 1500 kN

1500 kN 2.25 m

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Solution 1.5-8 CHANGE IN DIMENSIONS

NUMERICAL DATA E  110 GPa   0.34 P  1500 kN b  90 mm L  2250 mm

b  pb L  L

INITIAL VOLUME

FINAL LENGTH AND WIDTH

Voli  Lb2 Voli  1.822 107 mm3

Lf  L  L Lf  2.254 103 mm bf  b  b

NORMAL STRESS AND STRAIN P   185 MPa (less than yield so s 2 b Hooke’s Law applies) s â   1.684 103 E LATERAL STRAIN p  

b  0.052 mm L  3.788 mm

bf  89.948 mm

FINAL VOLUME Volf  Lfbf2 Volf  1.823 107 mm3 INCREASE IN VOLUME V  Volf  Vol ¢V  9789 mm3

p  5.724 104

Shear Stress and Strain Problem 1.6-1 An angle bracket having thickness t  0.75 in. is attached to the flange of a column by two 5/8-inch diameter bolts (see figure). A uniformly distributed load from a floor joist acts on the top face of the bracket with a pressure p  275 psi. The top face of the bracket has length L  8 in. and width b  3.0 in. Determine the average bearing pressure b between the angle bracket and the bolts and the average shear stress aver in the bolts. (Disregard friction between the bracket and the column.)

Distributed pressure on angle bracket

P b

Floor slab

L

Floor joist Angle bracket

Angle bracket t

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SECTION 1.6 Shear Stress and Strain

31

Solution 1.6-1 NUMERICAL DATA t  0.75 in.

L  8 in.

b  3. in.

p

275 ksi 1000

Ab  dt d

5 in. 8

Ab  0.469 in.2

BEARING STRESS sb 

F 2Ab

s b  7.04 ksi

;

BEARING FORCE F  pbL

F  6.6 kips

SHEAR AND BEARING AREAS AS 

p 2 d 4

SHEAR STRESS tave 

F 2AS

tave  10.76 ksi

;

AS  0.307 in.2

Roof structure

Problem 1.6-2

Truss members supporting a roof are connected to a 26-mm-thick gusset plate by a 22 mm diameter pin as shown in the figure and photo. The two end plates on the truss members are each 14 mm thick.

Truss member

(a) If the load P  80 kN, what is the largest bearing stress acting on the pin? (b) If the ultimate shear stress for the pin is 190 MPa, what force Pult is required to cause the pin to fail in shear?

P

End plates

(Disregard friction between the plates.)

P

Pin

t = 14 mm

Gusset plate

26 mm

Solution 1.6-2 NUMERICAL DATA

(b) ULTIMATE FORCE IN SHEAR

tep  14 mm

Cross sectional area of pin

tgp  26 mm

Ap 

P  80 kN dp  22 mm

Pult  2t ultAp

(a) BEARING STRESS ON PIN P gusset plate is thinner than dptgp (2 tep) so gusset plate controls

b  139.9 MPa

4

Ap  380.133 mm2

ult  190 MPa

sb 

p d2p

;

Pult  144.4 kN

;

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Problem 1.6-3 The upper deck of a football stadium is supported by braces each of which transfers a load P  160 kips to the base of a column [see figure part (a)]. A cap plate at the bottom of the brace distributes the load P to four flange plates (tf  1 in.) through a pin (dp  2 in.) to two gusset plates (tg  1.5 in.) [see figure parts (b) and (c)]. Determine the following quantities. (a) The average shear stress aver in the pin. (b) The average bearing stress between the flange plates and the pin (bf), and also between the gusset plates and the pin (bg). (Disregard friction between the plates.)

Cap plate Flange plate (tf = 1 in.) Pin (dp = 2 in.) Gusset plate (tg = 1.5 in.) (b) Detail at bottom of brace P P = 160 k Cap plate (a) Stadium brace Pin (dp = 2 in.)

P

Flange plate (tf = 1 in.) Gusset plate (tg = 1.5 in.) P/2

P/2

(c) Section through bottom of brace

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33

SECTION 1.6 Shear Stress and Strain

Solution 1.6-3 (b) BEARING STRESS ON PIN FROM FLANGE PLATE

NUMERICAL DATA P  160 kips

dp  2 in.

tg  1.5 in.

P 4 sbf  dp tf

tf  1 in.

s bf  20 ksi

;

(a) SHEAR STRESS ON PIN

t

V a

p d2p 4

t

b

t  12.73 ksi

BEARING STRESS ON PIN FROM GUSSET PLATE

P 4 a

p d2p 4

P 2 sbg  dp tg

b

sbg  26.7 ksi

;

;

Problem 1.6-4 The inclined ladder AB supports a house painter (82 kg) at C and the self weight (q  36 N/m) of the ladder itself. Each ladder rail (tr  4 mm) is supported by a shoe (ts  5 mm) which is attached to the m) ladder rail by a bolt of diameter dp  8 mm.

B Bx

m) (a) Find support reactions at A and B. (b) Find the resultant force in the shoe bolt at A. = 5 mm) (c) Find maximum average shear () and bearing (b) stresses in the shoe bolt at A.

C H=7m

Typical rung

Shoe bolt at A

36

Ladder rail (tr = 4 mm)

N/

m

tr

q=

01Ch01.qxd

Shoe bolt (dp = 8 mm) Ladder shoe (ts = 5 mm)

A

ts

Ax A —y 2

A —y 2

a = 1.8 m b = 0.7 m Ay Assume no slip at A

Shoe b

Section at base

Solution 1.6-4 (a) SUPPORT REACTIONS

NUMERICAL DATA tr  4 mm dp  8 mm

ts  5 mm

L  2( a + b)2 + H 2

P  82 kg (9.81 m/s ) 2

P  804.42 N a  1.8 m

b  0.7 m

H7m

q  36

N m

L  7.433 m

LAC 

a L a + b

LAC  5.352 m

LCB 

b L a + b

LCB  2.081 m

LAC  LCB  7.433 m

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CHAPTER 1 Tension, Compression, and Shear

SUM MOMENTS ABOUT A

Bx 

Pa + qL a

a + b b 2

H Bx   255 N ; Ax  Bx A y  1072 N

As 

SHEAR AREA

SHEAR STRESS

Ay  P  qL

(b) RESULTANT FORCE IN SHOE BOLT AT A BEARING STRESS

A resultant  2 A 2x + A 2y

ts  5 mm

s bshoe

t  5.48 MPa

tr  4 mm

;

Ab  80 mm2

A resultant 2  Ab

s bshoe  6.89 MPa

;

(c) MAXIMUM SHEAR AND BEARING STRESSES IN SHOE BOLT AT A dp  8 mm

As  50.265 mm2

Aresultant 2 t 2As

BEARING AREA Ab  2dpts

;

Aresultant  1102 N

p 2 d 4 p

;

CHECK BEARING STRESS IN LADDER RAIL Aresultant 2 s brail  dp tr

brail  17.22 MPa

T

Problem 1.6-5 The force in the brake cable of the V-brake system shown in the figure is T  45 lb. The pivot pin at A has diameter dp  0.25 in. and length Lp  5/8 in. Use dimensions show in the figure. Neglect the weight of the brake system. (a) Find the average shear stress aver in the pivot pin where it is anchored to the bicycle frame at B. (b) Find the average bearing stress b,aver in the pivot pin over segment AB.

Lower end of front brake cable D T

T

3.25 in.

Brake pads C

HE

HC

1.0 in. HB

B HF

A VF

Pivot pins anchored to frame (dP)

VB LP

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35

SECTION 1.6 Shear Stress and Strain

Solution 1.6-5 NUMERICAL DATA 5 L  in. 8

dp  0.25 in. BC  1 in.

CD  3.25 in.

AS 

T  45 lb

EQUILIBRIUM - FIND HORIZONTAL FORCES AT B AND C [VERTICAL REACTION VB  0] a MB  0

HC 

HC  191.25 lb HB  T  HC

(a) FIND THE AVE SHEAR STRESS ave IN THE PIVOT PIN WHERE IT IS ANCHORED TO THE BICYCLE FRAME AT B:

T(BC + CD) BC

a FH  0 HB  146.25 lb

tave 

pd2p

As  0.049 in.2

4 | H B| AS

tave  2979 psi

(b) FIND THE AVE BEARING STRESS σb,ave IN THE PIVOT PIN OVER SEGMENT AB. Ab  dpL s b,ave 

Ab  0.156 in.2

| H B| Ab

s b,ave  936 psi

Problem 1.6-6 A steel plate of dimensions 2.5  1.5  0.08 m and weighing 23.1kN is hoisted by steel cables with lengths L1  3.2 m and L2  3.9 m that are each attached to the plate by a clevis and pin (see figure). The pins through the clevises are 18 mm in diameter and are located 2.0 m apart. The orientation angles are measured to be   94.4° and  54.9°. For these conditions, first determine the cable forces T1 and T2, then find the average shear stress aver in both pin 1 and pin 2, and then the average bearing stress b between the steel plate and each pin. Ignore the mass of the cables.

;

;

P

L1 Clevis and pin 1

a = 0.6 m

b 1 b2 u

L2

a 2.0

Clevis and pin 2

m

Center of mass of plate b=

1.0

Steel plate (2.5 × 1.5 × 0.08 m) m

Solution 1.6-6 SOLUTION APPROACH

NUMERICAL DATA L1  3.2 m u  94.4 a a  0.6 m

L2  3.9 m

p b rad. a  54.9a 180

p b rad. 180

d  1.166 m

STEP (1) d  2 a2 + b 2 STEP (2) u1  atan a

a b b

u1

180  30.964 degrees p

STEP (3)-Law of cosines H  2d2 + L12  2dL1cos(u + u1)

b1m

W  77.0(2.5  1.5  0.08)

W  23.1 kN

(77  wt density of steel, kN/m ) 3

H  3.99 m

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STEP (4) b 1  acos a b1

L22 + H2  d2 b 2L1H

180  13.789 degrees p

STEP (5) b 2  acos a

L22 + H2  d2 b 2L2H

180 b2  16.95 degrees p STEP (6) Check (b 1 + b 2 + u + a)

180 p

 180.039 degrees Statics T1sin( 1)  T2sin( 2)

T1  T2 a

sin(b 2) b sin(b 1)

T1  13.18 kN

;

T1cos( 1)  T2cos( 2)  23.1  checks SHEAR & BEARING STRESSES dp  18 mm AS 

p 2 dp 4

T1 2 t1ave  AS T2 2 t2ave  AS

t  100 mm Ab  tdp

t1ave  25.9 MPa

;

t2ave  21.2 MPa

;

sin(b 2) T1  T2 a b sin(b 1)

sb1 

T1 Ab

sb1  7.32 MPa

;

T1cos( 1)  T2cos( 2)  W

s b2 

T2 Ab

s b2  5.99 MPa

;

T2 

W sin(b 2) cos(b 1) + cos(b 2) sin(b 1)

T2  10.77 kN

;

Problem 1.6-7 A special-purpose eye bolt of shank diameter d  0.50 in. passes

y

through a hole in a steel plate of thickness tp  0.75 in. (see figure) and is secured by a nut with thickness t  0.25 in. The hexagonal nut bears directly against the steel plate. The radius of the circumscribed circle for the hexagon is r  0.40 in. (which means that each side of the hexagon has length 0.40 in.). The tensile forces in three cables attached to the eye bolt are T1  800 lb., T2  550 lb., and T3  1241 lb. (a) Find the resultant force acting on the eye bolt. (b) Determine the average bearing stress b between the hexagonal nut on the eye bolt and the plate. (c) Determine the average shear stress aver in the nut and also in the steel plate.

T1 tp

T2

d 30° 2r

x Cables

Nut t

30° Eye bolt

Steel plate

T3

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SECTION 1.6 Shear Stress and Strain

Solution 1.6-7 (c) AVE. SHEAR THROUGH NUT

CABLE FORCES T1  800 lb

T2  550 lb

T3  1241 lb

(a) RESULTANT P  T2

23 + T30.5 2

P  1097 lb

;

d  0.5 in.

t  0.25 in.

Asn  dt

Asn  0

tnut  2793 psi

Ab  0.2194 in.2 P sb  Ab

Aspl  6rtp hexagon (Case 25, App. D)

s b  4999 psi

tpl 

;

P A spl

tp  0.75

r  0.40

Aspl  2 tpl  609 psi

;

b

Problem 1.6-8 An elastomeric bearing pad consisting of two steel plates bonded to a chloroprene elastomer (an artificial rubber) is subjected to a shear force V during a static loading test (see figure). The pad has dimensions a  125 mm and b  240 mm, and the elastomer has thickness t  50 mm. When the force V equals 12 kN, the top plate is found to have displaced laterally 8.0 mm with respect to the bottom plate. What is the shear modulus of elasticity G of the chloroprene?

P A sn

;

SHEAR THROUGH PLATE (b) AVE. BEARING STRESS

tnut 

a V

t

Solution 1.6-8 NUMERICAL DATA V  12 kN b  240 mm

a  125 mm t  50 mm

d  8 mm

AVERAGE SHEAR STRESS tave 

V ab

ave  0.4 MPa

AVERAGE SHEAR STRAIN

g ave 

SHEAR MODULUS G d t

ave  0.16

G

t ave g ave

G  2.5 MPa

;

37

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CHAPTER 1 Tension, Compression, and Shear

Problem 1.6-9 A joint between two concrete slabs A and B is filled with a flexible epoxy that bonds securely to the concrete (see figure). The height of the joint is h  4.0 in., its length is L  40 in., and its thickness is t  0.5 in. Under the action of shear forces V, the slabs displace vertically through the distance d  0.002 in. relative to each other. (a) What is the average shear strain aver in the epoxy? (b) What is the magnitude of the forces V if the shear modulus of elasticity G for the epoxy is 140 ksi?

Solution 1.6-9 Epoxy joint between concrete slabs (a) AVERAGE SHEAR STRAIN gaver 

d  0.004 t

;

(b) SHEAR FORCES V Average shear stress: aver  G aver h  4.0 in.

t  0.5 in.

L  40 in.

d  0.002 in.

G  140 ksi

Problem 1.6-10 A flexible connection consisting of rubber pads

(thickness t  9 mm) bonded to steel plates is shown in the figure. The pads are 160 mm long and 80 mm wide. (a) Find the average shear strain aver in the rubber if the force P  16 kN and the shear modulus for the rubber is G  1250 kPa. (b) Find the relative horizontal displacement  between the interior plate and the outer plates.

V  aver(hL)  G aver(hL)  (140 ksi)(0.004)(4.0 in.)(40 in.)  89.6 k

;

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SECTION 1.6 Shear Stress and Strain

Solution 1.6-10 Rubber pads bonded to steel plates (a) SHEAR STRESS AND STRAIN IN THE RUBBER PADS taver  gaver  Rubber pads: t  9 mm Length L  160 mm

P/2 8kN   625 kPa bL (80 mm)(160 mm)

taver 625 kPa   0.50 G 1250 kPa

;

(b) HORIZONTAL DISPLACEMENT

Width b  80 mm

  avert  (0.50)(9 mm)  4.50 mm

;

G  1250 kPa P  16 kN

Problem 1.6-11 A spherical fiberglass buoy used in an underwater experiment is anchored in shallow water by a chain [see part (a) of the figure]. Because the buoy is positioned just below the surface of the water, it is not expected to collapse from the water pressure. The chain is attached to the buoy by a shackle and pin [see part (b) of the figure]. The diameter of the pin is 0.5 in. and the thickness of the shackle is 0.25 in. The buoy has a diameter of 60 in. and weighs 1800 lb on land (not including the weight of the chain). (a) Determine the average shear stress aver in the pin. (b) Determine the average bearing stress b between the pin and the shackle.

Solution 1.6-11

Submerged buoy d  diameter of buoy  60 in. T  tensile force in chain dp  diameter of pin  0.5 in. t  thickness of shackle  0.25 in. W  weight of buoy  1800 lb

W  weight density of sea water  63.8 lb/ft3 FREE-BODY DIAGRAM OF BUOY FB  buoyant force of water pressure (equals the weight of the displaced sea water) V  volume of buoy pd 3  65.45 ft 3 6 FB  W V  4176 lb 

39

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EQUILIBRIUM T  FB  W  2376 lb (a) AVERAGE SHEAR STRESS IN PIN Ap  area of pin Ap 

p 2 d  0.1963 in.2 4 p

taver 

T  6050 psi 2Ap

;

(b) BEARING STRESS BETWEEN PIN AND SHACKLE Ab  2dpt  0.2500 in.2 sb 

T  9500 psi Ab

;

Problem 1.6-12 The clamp shown in the figure is used to support a load hanging from the lower flange of a steel beam. The clamp consists of two arms (A and B) joined by a pin at C. The pin has diameter d  12 mm. Because arm B straddles arm A, the pin is in double shear. Line 1 in the figure defines the line of action of the resultant horizontal force H acting between the lower flange of the beam and arm B. The vertical distance from this line to the pin is h  250 mm. Line 2 defines the line of action of the resultant vertical force V acting between the flange and arm B. The horizontal distance from this line to the centerline of the beam is c  100 mm. The force conditions between arm A and the lower flange are symmetrical with those given for arm B. Determine the average shear stress in the pin at C when the load P  18 kN.

Solution 1.6-12

Clamp supporting a load P

FREE-BODY DIAGRAM OF CLAMP

h  250 mm c  100 mm P  18 kN From vertical equilibrium: V

P  9 kN 2

d  diameter of pin at C  12 mm

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SECTION 1.6 Shear Stress and Strain

FREE-BODY DIAGRAMS OF ARMS A AND B

41

SHEAR FORCE F IN PIN

F

H 2 P 2 b + a b a A 4 2

 4.847 kN AVERAGE SHEAR STRESS IN THE PIN ©MC  0 哵哴 VC  Hh  0 H

taver 

F F  42.9 MPa  Apin pa 2 4

;

VC Pc   3.6 kN h 2h

FREE-BODY DIAGRAM OF PIN

Problem ★1.6-13 A hitch-mounted bicycle rack is designed to carry up to four 30-lb. bikes mounted on and strapped to two arms GH [see bike loads in the figure part (a)]. The rack is attached to the vehicle at A and is assumed to be like a cantilever beam ABCDGH [figure part (b)]. The weight of fixed segment AB is W1  10 lb, centered 9 in. from A [see the figure part (b)] and the rest of the rack weighs W2  40 lb, centered 19 in. from A. Segment ABCDG is a steel tube, 2  2 in., of thickness t  1/8 in. Segment BCDGH pivots about a bolt at B of diameter dB  0.25 in. to allow access to the rear of the vehicle without removing the hitch rack. When in use, the rack is secured in an upright position by a pin at C (diameter of pin dp  5/16 in.) [see photo and figure part (c)]. The overturning effect of the bikes on the rack is resisted by a force couple Fh at BC. (a) (b) (c) (d)

Find the support reactions at A for the fully loaded rack; Find forces in the bolt at B and the pin at C. Find average shear stresses aver in both the bolt at B and the pin at C. Find average bearing stresses b in the bolt at B and the pin at C.

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Bike loads

y

4 bike loads

19 in. G 27 in.

G

Release pins at C & G 5 (dp = — in.) 16

3 @ 4 in. W2

H a

1 2 in.  2 in.  ( — in.) 8

C Fixed support at A

MA

Ay

6 in.

D

C

D

A

H

F

2.125 in. F

F B Bolt at B 1 (dB = — in.) 4

a h = 7 in.

W1 Ax A

x

B

9 in.

h = 7 in.

F

8 in.

(a)

(b) Pin at C C

Pin at C 2.125 in. D Bolt at B

2  2  1/8 in. tube

(c) Section a–a

Solution *1.6-13 A y  170 lb

NUMERICAL DATA t

1 in. 8

h  7 in. P  30 lb

L1  17  2.125  6

b  2 in. W1  10 lb

W2  40 lb 5 in. dp 16

dB  0.25 in.

(a) REACTIONS AT A Ax  0

L1  25 in.

(dist from A to 1st bike) MA  W1(9)  W2(19)  P(4L1  4  8  12) M A  4585 in.-lb (b) FORCES IN BOLT AT B & PIN AT C Fy  0

;

Ay  W1  W2  4P

;

;

MB  0

By  W2  4P

B y  160 lb

;

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SECTION 1.6 Shear Stress and Strain

RHFB AsC  2

[W2(19  17) + P(6 + 2.125) + P(8.125 + 4) + P(8.125 + 8)

tC 

+ P(8.125 + 12)] Bx  h B x  254 lb

;

Bres  2Bx2 + By2

pd 2B 4

B res tB  A sB

AbB  2tdB sbB 

B res A bB

AbC  2tdp

AsB  0.098 in2 tB  3054 psi

Bx A sC

AsC  0.153 in2 tC  1653 psi

sbC  ;

Cx AbC

AbB  0.063 in2 sbB  4797 psi

sbC  3246 psi

links, each 12 mm long between the centers of the pins (see figure). You might wish to examine a bicycle chain and observe its construction. Note particularly the pins, which we will assume to have a diameter of 2.5 mm. In order to solve this problem, you must now make two measurements on a bicycle (see figure): (1) the length L of the crank arm from main axle to pedal axle, and (2) the radius R of the sprocket (the toothed wheel, sometimes called the chainring). (a) Using your measured dimensions, calculate the tensile force T in the chain due to a force F  800 N applied to one of the pedals. (b) Calculate the average shear stress aver in the pins.

Bicycle chain

F  force applied to pedal  800 N L  length of crank arm

;

AbC  0.078 in2

Problem 1.6-14 A bicycle chain consists of a series of small

Solution 1.6-14

;

t  0.125 in ;

(c) AVERAGE SHEAR STRESSES ave IN BOTH THE BOLT AT B AND THE PIN AT C AsB  2

4

(d) BEARING STRESSES B IN THE BOLT AT B AND THE PIN AT C

Cx  Bx B res  300 lb

pd 2p

R  radius of sprocket

;

43

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CHAPTER 1 Tension, Compression, and Shear

(b) SHEAR STRESS IN PINS

MEASUREMENTS (FOR AUTHOR’S BICYCLE) (1) L  162 mm

(2) R  90 mm

taver 

(a) TENSILE FORCE T IN CHAIN ©M axle  0

FL  TR

T

FL R



Substitute numerical values: T

(800 N)(162 mm)  1440 N 90 mm

2T T/ 2 T   2 Apin pd2 pd 2 (4) 2FL pd 2R

Substitute numerical values: ;

taver 

2(800 N)(162 mm) p(2.5 mm)2(90 mm)

 147 MPa

Problem 1.6-15 A shock mount constructed as shown in the figure is used to support a delicate instrument. The mount consists of an outer steel tube with inside diameter b, a central steel bar of diameter d that supports the load P, and a hollow rubber cylinder (height h) bonded to the tube and bar. (a) Obtain a formula for the shear  in the rubber at a radial distance r from the center of the shock mount. (b) Obtain a formula for the downward displacement  of the central bar due to the load P, assuming that G is the shear modulus of elasticity of the rubber and that the steel tube and bar are rigid.

Solution 1.6-15

Shock mount (a) SHEAR STRESS  AT RADIAL DISTANCE r As  shear area at distance r t

P P  As 2prh

 2prh

;

(b) DOWNWARD DISPLACEMENT  g  shear strain at distance r g

t P  G 2prhG

dd  downward displacement for element dr dd  gdr  d

r  radial distance from center of shock mount to element of thickness dr

L

dd 

Pdr 2prhG b/2

Pdr Ld/2 2prhG

d

b/2 P dr P b/2  [In r]d/2 2phG Ld/2 r 2phG

d

P b ln 2phG d

;

;

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SECTION 1.6 Shear Stress and Strain

45

Problem 1.6-16 The steel plane truss shown in the figure is loaded by three forces P, each of which is 490 kN. The truss mem-

bers each have a cross-sectional area of 3900 mm2 and are connected by pins each with a diameter of dp  18 mm. Members AC and BC each consist of one bar with thickness of tAC  tBC  19 mm. Member AB is composed of two bars [see figure part (b)] each having thickness tAB/2  10 mm and length L  3 m. The roller support at B, is made up of two support plates, each having thickness tsp/2  12 mm. (a) Find support reactions at joints A and B and forces in members AB, BC, and AB. (b) Calculate the largest average shear stress p,max in the pin at joint B, disregarding friction between the members; see figures parts (b) and (c) for sectional views of the joint. (c) Calculate the largest average bearing stress b,max acting against the pin at joint B.

P = 490 kN P

C

a b

A

45°

L=3m Support plate and pin

Ax

b

B

45°

P By

a

Ay (a) Member AB FBC at 45° Member AB

Member BC Support plate

Pin

By — 2

By — 2

(b) Section a–a at joint B (Elevation view)

tAB (2 bars, each — ) 2 FAB ––– 2 Pin P — 2

FBC FAB ––– 2 Support plate

tsp (2 plates, each — ) 2 P — 2

Load P at joint B is applied to the two support plates (c) Section b–b at joint B (Plan view)

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CHAPTER 1 Tension, Compression, and Shear

Solution 1.6-16 (b) MAX. SHEAR STRESS IN PIN AT B

NUMERICAL DATA L  3000 mm

P  490 kN

dp  18 mm

A  3900 mm

2

tAC  19 mm

tBC  tAC

tAB  20 mm

tsp  24 mm

(a) SUPPORT REACTIONS AND MEMBER FORCES

Fx  0

Ax  0

1 L L By  a P  P b L 2 2

a MA  0 By  0

;

;

Fy  0

Ay  P

A y  490 kN

;

METHOD OF JOINTS FAB  P

FBC  0

FAC   22P FAB  490 kN FAC   693 kN

; ;

;

As 

tp max

pd2p 4 FAB 2  As

As  254.469 mm2

tp max  963 MPa

;

(c) MAX. BEARING STRESS IN PIN AT B (tab  tsp SO BEARING STRESS ON AB WILL BE GREATER) Ab  dp

tAB 2

FAB 2 sb max  Ab

sb max  1361 MPa

;

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SECTION 1.6 Shear Stress and Strain

Problem 1.6-17 A spray nozzle for a garden hose requires a force F  5 lb. to open the spring-loaded spray chamber AB.

47

The nozzle hand grip pivots about a pin through a flange at O. Each of the two flanges has thickness t  1/16 in., and the pin has diameter dp  1/8 in. [see figure part (a)]. The spray nozzle is attached to the garden hose with a quick release fitting at B [see figure part (b)]. Three brass balls (diameter db  3/16 in.) hold the spray head in place under water pressure force fp  30 lb. at C [see figure part (c)]. Use dimensions given in figure part (a). (a) Find the force in the pin at O due to applied force F. (b) Find average shear stress aver and bearing stress b in the pin at O. Pin Flange

t

dp

Pin at O

A

F

Top view at O

B

O a = 0.75 in.

Spray nozzle Flange

F

b = 1.5 in. F

F 15°

c = 1.75 in. F

Sprayer hand grip

Water pressure force on nozzle, f p

C (b)

C Quick release fittings Garden hose (c) (a)

3 brass retaining balls at 120°, 3 diameter db = — in. 16

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CHAPTER 1 Tension, Compression, and Shear

Solution 1.6-17 Ox  12.68 lb

NUMERICAL DATA t

F  5 lb

1 in. 16 dN 

fp  30 lb a  0.75 in

dp  5 in. 8

b  1.5 in

1 in. 8

db 

u  15

3 in. 16

p rad. 180

c  1.75 in

(a) FIND THE FORCE IN THE PIN AT O DUE TO APPLIED FORCE F

Mo  0 FAB 

[F cos (u)( b  a)] + F sin (u)(c) a

FAB  7.849 lb a FH  0

Ox  FAB  F cos ()

Oy  1.294 lb

Ores  2 O 2x + O 2y

Ores  12.74 lb

;

(b) FIND AVERAGE SHEAR STRESS tave AND BEARING STRESS sb IN THE PIN AT O As  2

pd2p

Ores As

tO 

4

Ab  2tdp

sbO 

tO  519 psi

Ores Ab

;

sbO  816 psi

;

(c) FIND THE AVERAGE SHEAR STRESS ave IN THE BRASS RETAINING BALLS AT B DUE TO WATER PRESSURE FORCE Fp As  3

pd2b 4

tave 

fp

tave  362 psi

As

;

Oy  F sin () y

Problem 1.6-18

A single steel strut AB with diameter ds  8 mm. supports the vehicle engine hood of mass 20 kg which pivots about hinges at C and D [see figures (a) and (b)]. The strut is bent into a loop at its end and then attached to a bolt at A with diameter db  10 mm. Strut AB lies in a vertical plane.

h = 660 mm W hc = 490 mm C

B

(a) Find the strut force Fs and average normal stress  in the strut. (b) Find the average shear stress aver in the bolt at A. (c) Find the average bearing stress b on the bolt at A.

45° C

x A

30°

D

(a) b = 254 mm c = 506 mm y

a = 760 mm

d = 150 mm

h = 660 mm

Hood

C Hinge

C W

Fs D

z Strut ds = 8 mm

(b)

A

H = 1041 mm

B

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49

SECTION 1.6 Shear Stress and Strain

Solution 1.6-18 NUMERICAL DATA ds  8 mm

db  10 mm

Fsz 

m  20 kg

cd 2 H + ( c  d)2 2

a  760 mm

b  254 mm

c  506 mm

d  150 mm

H

h  660 mm

hc  490 mm

2 H + ( c  d)2

H  h atan a30

Fs

where

2

 0.946

cd

p p b + tan a45 bb 180 180

2 H + ( c  d)2 2

 0.324

H  1041 mm W  m (9.81m/s2)

(a) FIND THE STRUT s IN THE STRUT

W  196.2 N

a + b + c  760 mm 2

M

lineDC

VECTOR rAB

UNIT VECTOR eAB rAB e AB  | rAB| 0 W W P Q 0

hc hc rDC  P b + c Q D

0 rAB  1.041 * 103 P Q 356

AND AVERAGE NORMAL STRESS

Fsy 

|W|hc h

0 0.946 eAB  P 0.324 Q

ƒeAB ƒ  1

rDC 

2 H 2 + ( c  d)2

2 H2 + ( c  d)2

s

p 2 d 4 s

Fs A strut

Astrut  50.265 mm2 s  3.06 MPa

;

db  10 mm As 

490 490 P Q 760

H

Fs  153.9 N

H

(b) FIND THE AVERAGE SHEAR STRESS ave IN THE BOLT AT A

0 W   196.2 P Q 0

p 2 d 4 b

tave 

MD  rDB  Fs eAB  W  rDC

Fsy 

Fsy

Fs 

Astrut 

(ignore force at hinge C since it will vanish with moment about line DC) Fsx  0

0

FS

Fsy  145.664

0 H rAB  P c  dQ

M

FORCE

Fs

Fs As

As  78.54 mm2 tave  1.96 Mpa

;

(c) FIND THE BEARING STRESS b ON THE BOLT AT A Ab  dsdb sb 

Fs Ab

Ab  80 mm2 s b  1.924 MPa

;

;

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CHAPTER 1 Tension, Compression, and Shear

Problem 1.6-19

The top portion of a pole saw used to trim small branches from trees is shown in the figure part (a). The cutting blade BCD [see figure parts (a) and (c)] applies a force P at point D. Ignore the effect of the weak return spring attached to the cutting blade below B. Use properties and dimensions given in the figure.

B Rope, tension = T a

T

Weak return spring

y

2T

x

(a) Find the force P on the cutting blade at D if the tension force in the rope is T  25 lb (see free body diagram in part (b)]. (b) Find force in the pin at C. (c) Find average shear stress ave and bearing stress b in the support pin at C [see Section a–a through cutting blade in figure part (c)].

C

Cutting blade

Collar

Saw blade

D a

P

(a) Top part of pole saw B

T

20°



B 2T 50° BC = 6 in. Cy

70°

C in. D C=1

D

P

Cx

x

20° 20°

Collar 3 (tc = — in.) 8

6 in. C  1 in.

 D

70° (b) Free-body diagram

Cutting blade 3 (tb = — in.) 32

Pin at C 1 (dp = — in.) 8

(c) Section a–a

Solution 1.6-19 NUMERICAL PROPERTIES dp 

1 in 8

T  25 lb dCD  1 in

tb 

3 in 32

SOLVE ABOVE EQUATION FOR P tc 

3 in 8

dBC  6 in a

p rad/deg 180

[T(6 sin (70a)) + 2T cos (20a) P

6 sin (70a))  2T sin (20a)(6 cos (70a))] cos (20a)

P  395 lbs

;

(b) Find force in the pin at C (a) Find the cutting force P on the cutting blade at D if the tension force in the rope is T  25 lb:

Mc  0 MC  T(6 sin(70 ))  2T cos (20 )(6 sin (70 ))  2T sin (20 )(6 cos (70 ))  P cos (20 )(1)

SOLVE FOR FORCES ON PIN AT C

Fx  0

Cx  T  2T cos (20 )  P cos (40 )

Cx  374 lbs

Fy  0

;

Cy  2T sin (20 )  P sin (40 )

Cy   237 lbs

;

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SECTION 1.7 Allowable Stresses and Allowable Loads

RESULTANT AT C

BEARING STRESSES ON PIN ON EACH SIDE OF COLLAR

Cres  2 C 2x + C 2y

Cres  443 lbs

;

(c) Find maximum shear and bearing stresses in the support pin at C (see section a-a through saw).

sbC

Cres 2  dp tc

sbC  4.72 ksi

;

BEARING STRESS ON PIN AT CUTTING BLADE SHEAR STRESS - PIN IN DOUBLE SHEAR p As  d2p 4 tave 

As  0.012 in

Cres 2As

2

sbcb 

Cres dp tb

 bcb  37.8 ksi

;

tave  18.04 ksi

Allowable Stresses and Allowable Loads Problem 1.7.1 A bar of solid circular cross section is loaded in

tension by forces P (see figure). The bar has length L  16.0 in. and diameter d  0.50 in. The material is a magnesium alloy having modulus of elasticity E  6.4  106 psi. The allowable stress in tension is allow  17,000 psi, and the elongation of the bar must not exceed 0.04 in. What is the allowable value of the forces P?

Solution 1.7-1

Magnesium bar in tension p Pmax  smaxA  (16.000 psi)a b (0.50 in.)2 4  3140 lb

L  16.0 in.

d  0.50 in.

E  6.4  106 psi

allow  17,000 psi

max  0.04 in.

MAXIMUM LOAD BASED UPON ELONGATION emax 

dmax 0.04in.  0.00250 L 16 in.

smax  Eâmax  (6.4 * 106 psi)(0.00250)  16,000 psi

51

MAXIMUM LOAD BASED UPON TENSILE STRESS p Pmax  sallowA  (17,000 psi)a b(0.50 in.)2 4  3340 Ib ALLOWABLE LOAD Elongation governs. Pallow  3140 lb

;

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CHAPTER 1 Tension, Compression, and Shear

Problem 1.7-2 A torque T0 is transmitted between two flanged

T0

d

shafts by means of ten 20-mm bolts (see figure and photo). The diameter of the bolt circle is d  250 mm. If the allowable shear stress in the bolts is 90 MPa, what is the maximum permissible torque? (Disregard friction between the flanges.)

T0

T0

Solution 1.7-2 Shafts with flanges NUMERICAL DATA

MAX. PERMISSIBLE TORQUE

r  10 ^ bolts

Tmax  ta As a r

d  250 mm ^ flange

d b 2

Tmax  3.338 * 107 N # mm

As  p r2

Tmax  33.4 kN # m

;

As  314.159 m

2

t a  85 MPa

Problem 1.7-3 A tie-down on the deck of a sailboat consists of a bent bar bolted at both ends, as shown in the figure. The diameter dB of the bar is 1/4 in. , the diameter dW of the washers is 7/8 in. , and the thickness t of the fiberglass deck is 3/8 in. If the allowable shear stress in the fiberglass is 300 psi, and the allowable bearing pressure between the washer and the fiberglass is 550 psi, what is the allowable load Pallow on the tie-down?

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53

SECTION 1.7 Allowable Stresses and Allowable Loads

Solution 1.7-3 Bolts through fiberglass dB 

1 in. 4

P1  309.3 lb 2

dW 

7 in. 8

P1  619 lb

t

3 in. 8

ALLOWABLE LOAD BASED UPON SHEAR STRESS IN FIBERGLASS

allow  300 psi Shear area As  dWt P1  t allow As  t allow (pdWt) 2 7 3  (300 psi)(p)a in. b a in. b 8 8

ALLOWABLE LOAD BASED UPON BEARING PRESSURE

b  550 psi Bearing area Ab 

2 2 P2 p 7 1  sb Ab  (550 psi) a b c a in.b  a in. b d 2 4 8 4

 303.7 lb P2  607 lb ALLOWABLE LOAD Bearing pressure governs. Pallow  607 lb

;

Problem 1.7-4

Two steel tubes are joined at B by four pins (dp  11 mm), as shown in the cross section a–a in the figure. The outer diameters of the tubes are dAB  40 mm and dBC  28 mm. The wall thicknesses are tAB  6 mm and tBC  7 mm. The yield stress in tension for the steel is Y  200 MPa and the ultimate stress in tension is U  340 MPa. The corresponding yield and ultimate values in shear for the pin are 80 MPa and 140 MPa, respectively. Finally, the yield and ultimate values in bearing between the pins and the tubes are 260 MPa and 450 MPa, respectively. Assume that the factors of safety with respect to yield stress and ultimate stress are 4 and 5, respectively.

p 2 (d  d2B) 4 W

a Pin tAB

dAB

A

tBC

B

dBC C P

a

(a) Calculate the allowable tensile force Pallow considering tension in the tubes. (b Recompute Pallow for shear in the pins. (c) Finally, recompute Pallow for bearing between the pins and the tubes. Which is the controlling value of P?

tAB

dp tBC

dAB

dBC

Section a–a

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CHAPTER 1 Tension, Compression, and Shear

Solution 1.7-4 Yield and ultimate stresses (all in MPa)

As 

TUBES:

Y  200

u  340

FSy  4

PIN (SHEAR):

Y  80

u  140

FSu  5

PIN (BEARING):

bY  260

bu  450

tubes and pin dimensions (mm) dAB  40

tAB  6

dBC  dAB  2tAB tBC  7

dBC  28

dp  11

p A netAB  cd2AB  (dAB  2tAB)2  4 dp tAB d 4 AnetAB  433.45 mm

2

p A netBC  cd 2BC  (d BC  2t BC )2  4 dp t BC d 4 AnetAB  219.911 mm2 use smaller sY A FSy netBC

PaT1  11.0 kN PaT2 

su A FSu netBC

p 2 dp 4

PaS1  14As2

As  95.033 mm2 (one pin) tY FSy

PaS1  7.60 kN PaS2  14As2

;

tu FSu

PaT1  1.1  104 N ; PaT2  1.495  104

Problem 1.7-5 A steel pad supporting heavy machinery rests on four short, hollow, cast iron piers (see figure). The ultimate strength of the cast iron in compression is 50 ksi. The outer diameter of the piers is d  4.5 in. and the wall thickness is t  0.40 in. Using a factor of safety of 3.5 with respect to the ultimate strength, determine the total load P that may be supported by the pad.

PaS2  10.64 kN

(c) Pallow CONSIDERING BEARING IN THE PINS AbAB  4dptAB AbAB  264 mm2 AbBC  4dpt BC

(a) Pallow CONSIDERING TENSION IN THE TUBES

PaT1 

(b) Pallow CONSIDERING SHEAR IN THE PINS

Pab1  AbAB a

AbBC  308 mm2

sbY b FSy

Pab1  17.16 kN Pab2  AbAB a

6 smaller controls

sbu b FSu

Pab1  1.716 * 104 ; Pab2  23.8 kN

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55

SECTION 1.7 Allowable Stresses and Allowable Loads

Solution 1.7-5 Cast iron piers in compression Four piers

A

U  50 ksi

 5.152 in.2

n  3.5 s allow

p 2 p (d  d20)  [(4.5 in.)2  (3.7 in.)2] 4 4

sU 50 ksi   14.29 ksi  n 3.5

d  4.5 in.

P1  allowable load on one pier  sallow A  (14.29 ksi)(5.152 in.2)  73.62 k Total load P  4P1  294 k

t  0.4 in.

;

d0  d  2t  3.7 in.

Problem 1.7-6 The rear hatch of a van [BDCF in figure part (a)] is supported by two hinges at B1 and B2 and by two struts A1B1 and A2B2 (diameter ds  10 mm) as shown in figure part (b). The struts are supported at A1 and A2 by pins, each with diameter dp  9 mm and passing through an eyelet of thickness t  8 mm at the end of the strut [figure part (b)]. If a closing force P  50 N is applied at G and the mass of the hatch Mh  43 kg is concentrated at C: (a) What is the force F in each strut? [Use the free-body diagram of one half of the hatch in the figure part (c)] (b) What is the maximum permissible force in the strut, Fallow, if the allowable stresses are as follows: compressive stress in the strut, 70 MPa; shear stress in the pin, 45 MPa; and bearing stress between the pin and the end of the strut, 110 MPa.

127 mm

B2

B1

505 mm

505 mm

F C Mh

D

Bottom part of strut

G P

F

A1

B

710 mm

Mh —g 2

By

ds = 10 mm A2

75 mm Bx

10

460 mm

Eyelet

Pin support

A F

t = 8 mm (c)

(b)

(a)

Solution 1.7-6 (a) FORCE F IN EACH STRUT FROM STATICS (SUM MOMENTS ABOUT B) p FV  F cos1a2 FH  F sin1a2 a  10 180

NUMERICAL DATA Mh  43 kg

a  70 MPa

a  45 MPa

ba  110 MPa

ds  10 mm

dp  9 mm

P  50 N

g  9.81

t  8 mm m s2

G

C

D

g MB  0 FV(127) + FH(75)

P — 2

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CHAPTER 1 Tension, Compression, and Shear



Mh P g (127 + 505) + [127 + 2(505)] 2 2

(b) MAX. PERMISSIBLE FORCE F IN EACH STRUT Fmax IS SMALLEST OF THE FOLLOWING p 2 d 4 s p  ta dp2 4

F (127cos(a) + 75sin(a))

Fa1  sa

Mh P g (127 + 505) + [127 + 2(505)]  2 2

Fa2

Mh P g (127 + 505) + [127 + 2(505)] 2 2 F (127 cos(a) + 75 sin(a)) F  1.171 kN

Fa1  5.50 kN

Fa2  2.86 kN Fa3  sba dp t

;

Fa2  2.445 F

Fa3  7.92 kN

;

Problem 1.7-7 A lifeboat hangs from two ship’s davits, as shown in the

figure. A pin of diameter d  0.80 in. passes through each davit and supports two pulleys, one on each side of the davit. Cables attached to the lifeboat pass over the pulleys and wind around winches that raise and lower the lifeboat. The lower parts of the cables are vertical and the upper parts make an angle  15° with the horizontal. The allowable tensile force in each cable is 1800 lb, and the allowable shear stress in the pins is 4000 psi. If the lifeboat weighs 1500 lb, what is the maximum weight that should be carried in the lifeboat?

Solution 1.7-7 Lifeboat supported by four cables FREE-BODY DIAGRAM OF ONE PULLEY

Pin diameter d  0.80 in. T  tensile force in one cable Tallow  1800 lb

allow  4000 psi W  weight of lifeboat  1500 lb ©Fhoriz  0 ©Fvert  0

RH  T cos 15°  0.9659T RV  T  T sin 15°  0.7412T

V  shear force in pin V  2(RH)2 + (Rv)2  1.2175T

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57

SECTION 1.7 Allowable Stresses and Allowable Loads

ALLOWABLE TENSILE FORCE IN ONE CABLE BASED

MAXIMUM WEIGHT

UPON SHEAR IN THE PINS

Vallow  tallow A pin  2011 lb V  1.2175T

Shear in the pins governs.

p  (4000 psi)a b (0.80 in.)2 4

Tmax  T1  1652 lb Total tensile force in four cables

Vallow  1652 lb T1  1.2175

 4Tmax  6608 lb Wmax  4Tmax  W  6608 lb  1500 lb

ALLOWABLE FORCE IN ONE CABLE BASED UPON TENSION IN THE CABLE

 5110 lb

T2  Tallow  1800 lb

;

Problem 1.7-8 A cable and pulley system in figure part (a) supports a cage of mass 300 kg at B. Assume that this includes the

mass of the cables as well. The thickness of each the three steel pulleys is t  40 mm. The pin diameters are dpA  25 mm, dpB  30 mm and dpC  22 mm [see figure, parts (a) and part (b)].

(a) Find expressions for the resultant forces acting on the pulleys at A, B, and C in terms of cable tension T. (b) What is the maximum weight W that can be added to the cage at B based on the following allowable stresses? Shear stress in the pins is 50 MPa; bearing stress between the pin and the pulley is 110 MPa. a

C

dpA = 25 mm L1 A Cable

Cable Pulley

a

t

L2

dpB tB

Pin

dpC = 22 mm

dp Support bracket

B dpB = 30 mm Cage W

(a)

Section a–a: pulley support detail at A and C

Cage at B

Section a–a: pulley support detail at B (b)

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CHAPTER 1 Tension, Compression, and Shear

Solution 1.7-8 OR check bearing stress

NUMERICAL DATA g  9.81

M  300 kg

m Wmax2 

a  50 MPa

s2 ba  110 MPa

tA  40 mm

tB  40 mm

Wmax2 

tC  50

dpA  25 mm

dpB  30

dpC  22 mm

(a) RESULTANT FORCES F ACTING ON PULLEYS A, B & C FA  22 T FC  T

FB  2T T

Mg Wmax + 2 2

Wmax  2T  M g (b) MAX. LOAD W THAT CAN BE ADDED AT B DUE TO a & ba IN PINS AT A, B & C PULLEY AT A

a sba Ab b  M g

a sba tA dpA b  M g 22  152.6 kN ( bearing at A) 2

Wmax3 

2 1t A 2  M g 2 a s

p Wmax3  cta a 2 dpB2 b d  M g 4

Wmax4 

DOUBLE SHEAR FA  aAs

22 T  t aAs

Mg Wmax ta As +  2 2 22 22 2 22

(shear at B)

2 (s A )  M g 2 ba b

Wmax4  sba t B dpB  M g

PULLEY AT C

2

2T  aAs

PULLEY AT B

Wmax4  129.1 kN

FA tA  As

Wmax1 

22

Wmax3  67.7 kN

From statics at B

Wmax1 

Wmax2

2

ataAs b  M g ata2

p d A2 b  M g 4 p

Wmax1  22.6 Mg Wmax1  66.5 kN ; (shear at A controls)

(bearing at B)

T  ta As

Wmax5  21t a As2  M g Wmax5  c2ta a 2

p 2 d b d  Mg 4 pC

Wmax5  7.3 * 104

Wmax5  73.1 kN

Wmax6  2sbatC dp C  M g Wmax6  239.1 kN

(bearing at C)

(shear at C)

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59

SECTION 1.7 Allowable Stresses and Allowable Loads

Problem 1.7-9 A ship’s spar is attached at the base of a mast by a pin connection

Mast

(see figure). The spar is a steel tube of outer diameter d2  3.5 in. and inner diameter d1  2.8 in. The steel pin has diameter d  1 in., and the two plates connecting the spar to the pin have thickness t  0.5 in. The allowable stresses are as follows: compressive stress in the spar, 10 ksi; shear stress in the pin, 6.5 ksi; and bearing stress between the pin and the connecting plates, 16 ksi. Determine the allowable compressive force Pallow in the spar.

P

Pin

Spar Connecting plate

Solution 1.7-9 COMPRESSIVE STRESS IN SPAR p Pa1  s a 1d22  d122 4

Pa1  34.636 kips

SHEAR STRESS IN PIN Pa2  t a a 2

Pa2  10.21 kips controls

NUMERICAL DATA d2  3.5 in. dp  1 in.

a  10 ksi

d1  2.8 in.

;

^double shear

t  0.5 in.

a  6.5 ksi

p 2 d b 4 p

ba  16 ksi

BEARING STRESS BETWEEN PIN & CONECTING PLATES Pa3  ba(2dpt)

Problem 1.7-10 What is the maximum possible value of the clamping force C in the jaws of the pliers shown in the figure if the ultimate shear stress in the 5-mm diameter pin is 340 MPa? What is the maximum permissible value of the applied load P if a factor of safety of 3.0 with respect to failure of the pin is to be maintained?

Pa3  16 kips

P y

15

mm

90° 38

Rx

50°

x

C

50 mm

90° P

C Pin

10°

mm

Rx 140°

b=

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CHAPTER 1 Tension, Compression, and Shear

Solution 1.7-10 NUMERICAL DATA

u  340 MPa

FS  3 a  40

ta 

p rad 180

ta 

tu ta  FS

As

Pmax 

pin at C in single shear

Rx  C cos ( )

a  163.302 mm

b  38 mm

Rx   P

2 a a cos (a) d + c1 + sin(a) d b b

here

;

a  4.297 a/b  mechanical advantage b

P(a) cos(a) b

R y  Pc1 +

C ult  PmaxFS a a sin(a) d b

2 2 a a c  cos1a2 d + c1 + sin(a) d  ta A s A b b

As 

Ac

FIND MAX. CLAMPING FORCE P(a) C b

a Mpin  0

STATICS

ta A s 2

Pmax  445 N

Ry  P  C sin ( )

a  50 cos ( )  125

t a  113.333 MPa

Find Pmax

d  5 mm

2Rx2 + Ry2

tu FS

a b b

Pult  PmaxFS

C ult  5739 N

;

Pult  1335

C ult  4.297 P ult

p 2 d 4

Problem 1.7-11 A metal bar AB of weight W is suspended by a system of steel

2.0 ft

2.0 ft 7.0 ft

wires arranged as shown in the figure. The diameter of the wires is 5/64 in., and the yield stress of the steel is 65 ksi. Determine the maximum permissible weight Wmax for a factor of safety of 1.9 with respect to yielding.

5.0 ft

5.0 ft W A

B

Solution 1.7-11 NUMERICAL DATA d

5 in. 64

sY sa  FSy

FORCES IN WIRES AC, EC, BD, FD

Y  65 ksi

FSy  1.9

a  34.211 ksi

a FV  0 at A, B, E or F 222 + 52 W * 5 2 Wmax = 0.539 a  A FW 

222 + 52  0.539 10

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SECTION 1.7 Allowable Stresses and Allowable Loads

Wmax  0.539 a

sY p b a d2 b FS y 4

Wmax  0.305 kips

FCD  2a

;

FCD  2c

CHECK ALSO FORCE IN WIRE CD a FH  0

FCD 

at C or D

2 22 + 52 2

2 222 + 52

2 W 5

61

F wb

a

222 + 52 W * bd 5 2

less than FAC so AC controls

Problem 1.7-12 A plane truss is subjected to loads 2P and P at joints B and C, respectively, as shown in the figure

part (a). The truss bars are made of two L102  76  6.4 steel angles [see Table E-5(b): cross sectional area of the two angles, A  2180 mm2, figure part (b)] having an ultimate stress in tension equal to 390 MPa. The angles are connected to an 12 mm-thick gusset plate at C [figure part (c)] with 16-mm diameter rivets; assume each rivet transfers an equal share of the member force to the gusset plate. The ultimate stresses in shear and bearing for the rivet steel are 190 MPa and 550 MPa, respectively. Determine the allowable load Pallow if a safety factor of 2.5 is desired with respect to the ultimate load that can be carried. (Consider tension in the bars, shear in the rivets, bearing between the rivets and the bars, and also bearing between the rivets and the gusset plate. Disregard friction between the plates and the weight of the truss itself.)

F

FCF

G

a

a A

B

a

FCG

Truss bars

C a

a

D

Gusset plate Rivet

C

P

2P

a

FBC

FCD

(a)

P (c) Gusset plate

6.4 mm 12 mm

Rivet (b) Section a–a

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CHAPTER 1 Tension, Compression, and Shear

Solution 1.7-12 PCG  45.8 kN

NUMERICAL DATA

;

A  2180 mm

2

tg  12 mm

dr  16 mm

u  390 MPa

u  190 MPa

bu  550 MPa su FS

sa 

tang  6.4 mm

tu FS

sba 

sbu FS

MEMBER FORCES FROM TRUSS ANALYSIS F BC

5  P 3

4 F CD  P 3

22  0.471 3

22 P FCF  3

4 F CG  P 3

Pallow FOR TENSION ON NET SECTION IN TRUSS BARS Anet  A  2drtang

Anet  1975 mm2

A net  0.906 A Fallow  aAnet

Pallow

3 PCD  2 a b(ta As) 4

PCD  45.8 kN

;

Next, Pallow for bearing of rivets on truss bars Ab  2drtang rivet bears on each angle in two angle pairs

FS  2.5

ta 

< so shear in rivets in CG & CD controls Pallow here

allowable force in a member so BC controls since it has the largest member force for this loading 3 3 Pallow  (sa A net)  F BCmax 5 5

Pallow  184.879 kN

Fmax  sba A b N 3 PBC  3 a b(sba Ab) 5

PBC  81.101 kN

PCF  2 a

PCF  191.156 kN

3 22

b (sba Ab)

3 PCG  2 a b(sba Ab) 4

PCG  67.584 kN

3 PCD  2 a b(sba Ab) 4

PCD  67.584 kN

Finally, Pallow for bearing of rivets on gusset plate Ab  drtg (bearing area for each rivert on gusset plate) tg  12 mm 2tang  12.8 mm so gusset will control over angles

Next, Pallow for shear in rivets (all are in double shear)

3 PBC  3 a b(sba Ab) 5

PBC  76.032 kN

p A s  2 d r2 4

PCF  2 a

PCF  179.209 kN

Fmax  t aA s N

for one rivet in DOUBLE shear N  number of rivets in a particular member (see drawing of conn. detail)

3 22

b (sba Ab)

3 PCG  2 a b(sba Ab) 4

PCG  63.36 kN PCD  63.36 kN

3 PBC  3 a b(ta As) 5

PBC  55.0 kN

3 PCD  2 a b(sba Ab) 4

PCF  2 a

PCF  129.7 kN

So, shear in rivets controls: Pallow = 45.8 kN

3 22

b (ta As)

3 PCG  2 a b(ta As) 4

;

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63

SECTION 1.7 Allowable Stresses and Allowable Loads

Problem 1.7-13

A solid bar of circular cross section (diameter d) has a hole of diameter d/5 drilled laterally through the center of the bar (see figure). The allowable average tensile stress on the net cross section of the bar is allow.

d

d/5

P

d

(a) Obtain a formula for the allowable load Pallow that the bar can carry in tension. (b) Calculate the value of Pallow if the bar is made of brass with diameter d  1.75 in. and allow  12 ksi. (Hint: Use the formulas of Case 15 Appendix D.)

Solution 1.7-13 NUMERICAL DATA

1 1 2 26 b d Pa  sa c d2 a acosa b  2 5 25

a  12 ksi

d  1.75 in

1 2 acosa b  26 5 25

(a) FORMULA FOR PALLOW IN TENSION From Case 15, Appendix D: A  2r2 aa 

ab r2

a a  acosa b r a

b

r

d 2

r  0.875 in.

180  78.463 degrees p

b

d 2 d 2 c a b  a b d A 2 10

b

Aa

Pa  aA

d 10

a  0.175 in.

2

d b  26 5

 0.587

Pa  sa10.587 d22

;

0.587  0.748 0.785 (b) EVALUATE NUMERICAL RESULT d  1.75 in.

b  2r2  a2

6 2 d b 25

a

d/5

P

Pa  21.6 kips

a  12 ksi ;

p  0.785 4

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Problem 1.7-14 A solid steel bar of diameter d1  60 mm has a

hole of diameter d2  32 mm drilled through it (see figure). A steel pin of diameter d2 passes through the hole and is attached to supports. Determine the maximum permissible tensile load Pallow in the bar if the yield stress for shear in the pin is Y  120 MPa, the yield stress for tension in the bar is Y  250 MPa and a factor of safety of 2.0 with respect to yielding is required. (Hint: Use the formulas of Case 15, Appendix D.)

d2 d1 d1 P

Solution 1.7-14 SHEAR AREA (DOUBLE SHEAR)

NUMERICAL DATA d1  60 mm

p As  2a d22 b 4

d2  32 mm

Y  120 MPa

Y  250 MPa

As  1608 mm2

FSy  2

NET AREA IN TENSION (FROM CASE 15, APP. D)

ALLOWABLE STRESSES

Anet  2a

ta 

tY FSy

a  60 MPa

sa 

sY FSy

a  125 MPa

From Case 15, Appendix D: A  2r 2 aa  d2 a 2

b 2

ab r

2 2 d2 c a d1 b  a d2 b d 2 A 2 d2 2 ≥acosa b  ¥ d1 d1 2 a b 2

r

a  arc cos

b  2r2  a2

d1 2 b 2

d1 2 d2 d2/2  arc cos d1/2 d1

Anet  1003 mm2 Pallow in tension: smaller of values based on either shear or tension allowable stress x appropriate area Pa1  aAs Pa2  aAnet

Pa1  96.5kN 6 shear governs Pa2  125.4kN

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SECTION 1.7 Allowable Stresses and Allowable Loads

Resultant of wind pressure

Lh 2

C.P.

F

W

Pipe column

z b 2

D H

Lv

C

A F at each 4 bolt

h

y Overturning moment B about x axis FH x

W at each 4 bolt

(a) W

Pipe column

db dw

FH — = Rh 2 One half of over – turning moment about x axis acts on each bolt pair Base B plate (tbp)

z

A y

Footing

F/4 Tension

h

R

R

Compression

(b) z in.

FH — 2 b= 1

F 4 R W 4

y 2 in

.

B

h

4 =1

D

FH 2

R

F 4

W R 4 (c)

W 4 x

A

four bolts anchored in a concrete footing. Wind pressure p acts normal to the surface of the sign; the resultant of the uniform wind pressure is force F at the center of pressure. The wind force is assumed to create equal shear forces F/4 in the y-direction at each bolt [see figure parts (a) and (c)]. The overturning effect of the wind force also causes an uplift force R at bolts A and C and a downward force (R) at bolts B and D [see figure part (b)]. The resulting effects of the wind, and the associated ultimate stresses for each stress condition, are: normal stress in each bolt (u  60 ksi); shear through the base plate (u  17 ksi); horizontal shear and bearing on each bolt (hu  25 ksi and bu  75 ksi); and bearing on the bottom washer at B (or D) (bw  50 ksi). Find the maximum wind pressure pmax (psf) that can be carried by the bolted support system for the sign if a safety factor of 2.5 is desired with respect to the ultimate wind load that can be carried. Use the following numerical data: bolt db  3⁄4 in.; washer dw  1.5 in.; base plate tbp  1 in.; base plate dimensions h  14 in. and b  12 in.; W  500 lb; H  17 ft; sign dimensions (Lv  10 ft.  Lh  12 ft.); pipe column diameter d  6 in., and pipe column thickness t  3/8 in.

65

Sign (Lv  Lh)

Problem 1.7-15 A sign of weight W is supported at its base by

C

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(1) COMPUTE pmax BASED ON NORMAL STRESS IN EACH BOLT (GREATER AT B & D)

Solution 1.7-15 Numerical Data

u  60 ksi

u  17 ksi

bu  75 ksi db 

3 in. 4

h  14 in.

hu  25 ksi

bw  50 ksi dw  1.5 in.

tbp  1 in.

b  12 in.

W  0.500 kips Lv  10(12)

FSu  2.5

3 in. 8 H  204 in.

d  6 in.

H  17(12) Lh  12(12)

Lv  120 in.

Lh  144 in.

su FSu

a  24

a  6.8 sba 

tha 

sbu FSu

ta 

thu FSu

ba  30

tu FSu

sbwa 

FORCES F AND R IN TERMS OF pmax

R  pmax

LvLhH 2h

p W sa a d b2b  4 4 pmax1  LvLhH 2h pmax1  11.98 psf

FH 2h

;

controls

W 4 t p dw tbp Rmax  ta(p dw tbp) 

sbw FSu pmax2 

R

p W Rmax  sa a db2 b  4 4

p 2 d 4 b

R +

ha  10

bwa  20

F  pmaxLvLh

s

W 4

(2) COMPUTE pmax BASED ON SHEAR THROUGH BASE PLATE (GREATER AT B & D)

ALLOWABLE STRESSES (ksi) sa 

t

R +

W 4

ta a p dw tbp b  Lv Lh H 2h

pmax2  36.5 psf

W 4

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67

SECTION 1.7 Allowable Stresses and Allowable Loads

(3) COMPUTE pmax BASED ON HORIZONTAL SHEAR ON EACH BOLT

th 

F 4

Fmax

p a db2 b 4

pmax3 

p  4tha a db2 b 4

tha(p db2) Lv Lh

pmax3  147.3 psf

(5) COMPUTE pmax BASED ON BEARING UNDER THE TOP WASHER AT A (OR C) AND THE BOTTOM WASHER AT B (OR D) R + sbw 

EACH BOLT

pmax5

F 4 sb  (tbp db) pmax4 

Fmax  4ba(tbpdb)

4sba(tbpdb)

p ad 2  db 2 b 4 w

Rmax  sbwa c

(4) COMPUTE pmax BASED ON HORIZONTAL BEARING ON

W 4

p W adw2  db2b d  4 4

p W sbwa c (dw2  db2) d  4 4  LvLhH 2h

pmax5  30.2 psf So, normal/stress in bolts controls; pmax  11.98 psf

LvLh

pmax4  750 psf

Problem 1.7-16 The piston in an engine is attached to a connecting rod AB, which in turn is connected to a crank arm BC (see figure). The piston slides without friction in a cylinder and is subjected to a force P (assumed to be constant) while moving to the right in the figure. The connecting rod, which has diameter d and length L, is attached at both ends by pins. The crank arm rotates about the axle at C with the pin at B moving in a circle of radius R. The axle at C, which is supported by bearings, exerts a resisting moment M against the crank arm. (a) Obtain a formula for the maximum permissible force Pallow based upon an allowable compressive stress c in the connecting rod. (b) Calculate the force Pallow for the following data: c  160 MPa, d  9.00 mm, and R  0.28L.

Cylinder P

Piston

Connecting rod

A

M

d

C

B L

R

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CHAPTER 1 Tension, Compression, and Shear

Solution 1.7-16 The maximun allowable force P occurs when cos has its smallest value, which means that has its largest value. LARGEST VALUE OF

d  diameter of rod AB FREE-BODY DIAGRAM OF PISTON The largest value of occurs when point B is the farthest distance from line AC. The farthest distance is the radius R of the crank arm.

P  applied force (constant) C  compressive force in connecting rod RP  resultant of reaction forces between cylinder and piston (no friction) : a Fhoriz  0 

; 

P  C cos  0 P  C cos MAXIMUM COMPRESSIVE FORCE C IN CONNECTING ROD Cmax  cAc in which Ac  area of connecting rod pd2 Ac  4 MAXIMUM ALLOWABLE FORCE P P  Cmax cos  c Ac cos

Therefore, — BC  R — Also, AC  2L2 R2 cos a 

R 2 2L2 R2  A 1 a b L L

(a) MAXIMUM ALLOWABLE FORCE P Pallow  c Ac cos  sc a

pd2 4

b

A

R 2 1 a b L

(b) SUBSTITUTE NUMERICAL VALUES c  160 MPa R  0.28L

d  9.00 mm R/L  0.28

Pallow  9.77 kN

;

;

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SECTION 1.8 Design for Axial Loads and Direct Shear

69

Design for Axial Loads and Direct Shear Problem 1.8-1 An aluminum tube is required to transmit an axial tensile force P  33 k [see figure part (a)]. The thickness of the wall of the tube is to be 0.25 in.

d P

(a) What is the minimum required outer diameter dmin if the allowable tensile stress is 12,000 psi? (b) Repeat part (a) if the tube will have a hole of diameter d/10 at mid-length [see figure parts (b) and (c)].

P

(a) Hole of diameter d/10

d

d/10

P

P

d

(b)

(c)

Solution 1.8-1 (b) MIN. DIAMETER OF TUBE (WITH HOLES)

NUMERICAL DATA P  33 kips

t  0.25 in.

a  12 ksi

p d A1  c 3d2 (d  2t)242 a bt d 4 10

(a) MIN. DIAMETER OF TUBE (NO HOLES) p A1  3d2 (d 2 t)24 4

P A2  sa

A2  2.75 in2

t b  pt2 5

equating A1 & A2 and solving for d:

equating A1 & A2 and solving for d: P d + t psat

A1  dapt 

d  3.75 in.

;

P p t2 sa d t p t 5

d  4.01 in.

;

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Problem 1.8-2 A copper alloy pipe having yield stress Y  290

d t =— 8

P

MPa is to carry an axial tensile load P  1500 kN [see figure part (a)]. A factor of safety of 1.8 against yielding is to be used. (a) If the thickness t of the pipe is to be one-eighth of its outer diameter, what is the minimum required outer diameter dmin? (b) Repeat part (a) if the tube has a hole of diameter d/10 drilled through the entire tube as shown in the figure [part (b)].

d

(a)

P

Hole of diameter d/10

d t =— 8

d

(b)

Solution 1.8-2 NUMERICAL DATA

equate A1 & A2 and solve for d:

Y  290 MPa d2 

P  1500 kN FSy  1.8

256 15p

(a) MIN. DIAMETER (NO HOLES) A1 

p 2 d 2 cd  ad b d 4 8

p 15 A1  a d2 b 4 64 P A2  sY FSy

15 A1  p d2 256

A2  9.31  103 mm2

P sY P FS Q y 256 15p

dmin  Q

dmin  225mm

P sY P FSy Q ;

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SECTION 1.8 Design for Axial Loads and Direct Shear

(b) MIN. DIAMETER (WITH HOLES) Redefine A1 - subtract area for two holes - then equate to A2 d p d 2 d A1  c cd2  ad b d  2a b a b d 4 8 10 8 A1 

d2 

a

P sy P FS Q y

15 1 p b 256 40

15 2 1 2 pd  d 256 40

A1  d2 a

15 1 p b 256 40

P sy

15 1 p   0.159 256 40

y

dmin 

a

c Problem 1.8-3 A horizontal beam AB with cross-sectional dimensions (b  0.75 in.)  (h  8.0 in.) is supported by an inclined strut CD and carries a load P  2700 lb at joint B [see figure part (a)]. The strut, which consists of two bars each of thickness 5b/8, is connected to the beam by a bolt passing through the three bars meeting at joint C [see figure part (b)]. (a) If the allowable shear stress in the bolt is 13,000 psi, what is the minimum required diameter dmin of the bolt at C? (b) If the allowable bearing stress in the bolt is 19,000 psi, what is the minimum required diameter dmin of the bolt at C?

P FS Q

15 1 p b 256 40

dmin  242 mm

4 ft

;

5 ft B C

A 3 ft

P

D

(a)

b

Beam AB (b  h)

h — 2

Bolt (dmin)

h — 2

5b — 8

Strut CD (b)

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Solution 1.8-3 NUMERICAL DATA P  2.7 kips a  13 ksi

(b) dmin BASED ON ALLOWABLE BEARING AT JT C

b  0.75 in. ba  19 ksi

h  8 in.

Bearing from beam ACB

(a) dmin BASED ON ALLOWABLE SHEAR - DOUBLE SHEAR

dmin 

IN STRUT

ta 

FDC As

As  2 a

dmin

FDC 

15 P/4 bd

dmin  0.711 inches

;

15 P 4 Bearing from strut DC sb  5 2 bd 8

15 P 4

p 2 d b 4

15 P 4  p t a b a a 2

15 P/4 b sba

sb 

sb  3 dmin  0.704 inches

P bd

(lower than ACB)

;

Problem 1.8-4 Lateral bracing for an elevated pedestrian walkway is shown in the figure part (a). The thickness of the clevis

plate tc  16 mm and the thickness of the gusset plate tg  20 mm [see figure part (b)]. The maximum force in the diagonal bracing is expected to be F  190 kN. If the allowable shear stress in the pin is 90 MPa and the allowable bearing stress between the pin and both the clevis and gusset plates is 150 MPa, what is the minimum required diameter dmin of the pin? Clevis

Gusset plate Gusset plate tc

Pin

tg

Cl

ev is

dmin

Diagonal brace F

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SECTION 1.8 Design for Axial Loads and Direct Shear

Solution 1.8-4 NUMERICAL DATA

(2) dmin BASED ON ALLOW BEARING IN GUSSET & CLEVIS

F  190 kN

a  90 MPa

tg  20 mm

tc  16 mm

ba  150 MPa

(1) dmin BASED ON ALLOW SHEAR - DOUBLE SHEAR IN STRUT

F t As dmin 

p As  2 a d2 b 4

F p t a b Q a 2

dmin  36.7 mm

PLATES

Bearing on gusset plate sb 

F Ab

Ab  tgd

dmin 

dmin  63.3 mm

6 controls

Bearing on clevis

Ab  d(2tc)

dmin 

F 2t csba

dmin  39.6 mm

F tgsba

;

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CHAPTER 1 Tension, Compression, and Shear

Problem 1.8-5 Forces P1  1500 lb and P2  2500 lb are applied at joint C of plane truss ABC

P2

shown in the figure part (a). Member AC has thickness tAC  5/16 in. and member AB is composed of two bars each having thickness tAB/2  3/16 in. [see figure part (b)]. Ignore the effect of the two plates which make up the pin support at A. If the allowable shear stress in the pin is 12,000 psi and the allowable bearing stress in the pin is 20,000 psi, what is the minimum required diameter dmin of the pin?

C

P1 L

A

a B L a

Ax

By

Ay (a) tAC Pin support plates

AC

tAB — 2

AB Pin

A Ay — 2

Ay — 2

Section a–a (b)

Solution 1.8-5 NUMERICAL DATA P1  1.5 kips

P2  2.5 kips

5 tAC  in. 16

3 tAB  2a b in. 16

a  12 ksi

ba  20 ksi

(1) dmin BASED ON ALLOWABLE SHEAR - DOUBLE SHEAR IN STRUT; FIRST CHECK AB (SINGLE SHEAR IN EACH BAR HALF) Force in each bar of AB is P1/2 P1 2 t AS

p A s  a d2 b 4

dmin

P1 2  p t a b a 2 4

dmin  0.282 in.

Next check double shear to AC; force in AC is (P1 + P2)/2 (P1 + P2)/ 2

dmin  Q

ta a

dmin  0.461 inches

p b 4

Finally check RESULTANT force on pin at A R

P1 2 P1 + P2 2 b Aa 2 b + a 2

R  2.136 kips

;

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SECTION 1.8 Design for Axial Loads and Direct Shear

dmin

R 2  p t a b a a 4

dmin  0.476 in.

(2) dmin BASED ON ALLOWABLE BEARING ON PIN member AB bearing on pin dmin 

P1 tABsba

P1 + P2 tACsba

P1 Ab

Ab  tABd

dmin  0.2 in.

member AC bearing on pin dmin 

sb 

Ab  d(tAC)

dmin  0.64 in.

controls

;

Problem 1.8-6 A suspender on a suspension bridge consists of a cable that passes over the main cable (see figure) and supports the bridge deck, which is far below. The suspender is held in position by a metal tie that is prevented from sliding downward by clamps around the suspender cable. Let P represent the load in each part of the suspender cable, and let u represent the angle of the suspender cable just above the tie. Finally, let sallow represent the allowable tensile stress in the metal tie. (a) Obtain a formula for the minimum required cross-sectional area of the tie. (b) Calculate the minimum area if P  130 kN, u  75°, and sallow  80 MPa.

75

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CHAPTER 1 Tension, Compression, and Shear

Solution 1.8-6 Suspender tie on a suspension bridge F  tensile force in cable above tie P  tensile force in cable below tie

allow  allowable tensile stress in the tie

(a) MINIMUM REQUIRED AREA OF TIE Amin 

T Pcotu  sallow sallow

(b) SUBSTITUTE NUMERICAL VALUES: P  130 kN

  75°

allow  80 MPa Amin  435 mm2

FREE-BODY DIAGRAM OF HALF THE TIE Note: Include a small amount of the cable in the free-body diagram T  tensile force in the tie FORCE TRIANGLE cotu 

T P

T  P cot 

Problem 1.8-7 A square steel tube of length L  20 ft and

width b2  10.0 in. is hoisted by a crane (see figure). The tube hangs from a pin of diameter d that is held by the cables at points A and B. The cross section is a hollow square with inner dimension b1  8.5 in. and outer dimension b2  10.0 in. The allowable shear stress in the pin is 8,700 psi, and the allowable bearing stress between the pin and the tube is 13,000 psi. Determine the minimum diameter of the pin in order to support the weight of the tube. (Note: Disregard the rounded corners of the tube when calculating its weight.)

;

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77

SECTION 1.8 Design for Axial Loads and Direct Shear

Solution 1.8-7 Tube hoisted by a crane T  tensile force in cable W  weight of steel tube d  diameter of pin b1  inner dimension of tube  8.5 in. b2  outer dimension of tube  10.0 in.

W  gs AL  (490 lb/ft3)(27.75 in.2)a  1,889 lb DIAMETER OF PIN BASED UPON SHEAR Double shear. 2allow Apin  W 2(8,700 psi)a

L  length of tube  20 ft

allow  8,700 psi b  13,000 psi

1 ft2 b (20 ft) 144 in.

p d2 b  1889 lb 4

d2  0.1382 in.2

d1  0.372 in.

DIAMETER OF PIN BASED UPON BEARING

WEIGHT OF TUBE

b(b2  b1)d  W

s  weight density of steel

(13,000 psi)(10.0 in.  8.5 in.) d  1,889 lb

 490 lb/ft

3

A  area of tube  b 22  b 21  (10.0 in.)2  (8.5 in.)2  27.75 in.

d2  0.097 in. MINIMUM DIAMETER OF PIN Shear governs.

Problem 1.8-8 A cable and pulley system at D is used to bring a 230-kg pole (ACB) to a vertical position as shown in the figure part (a). The cable has tensile force T and is attached at C. The length L of the pole is 6.0 m, the outer diameter is d  140 mm, and the wall thickness t  12 mm. The pole pivots about a pin at A in figure part (b). The allowable shear stress in the pin is 60 MPa and the allowable bearing stress is 90 MPa. Find the minimum diameter of the pin at A in order to support the weight of the pole in the position shown in the figure part (a).

B 1.0 m

dmin  0.372 in.

Pole C Cable 30° Pulley

5.0 m

a

T

A D 4.0 m

a (a) d

ACB Pin support plates

A

Pin Ay — 2

Ay — 2 (b)

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Solution 1.8-8 ALLOWABLE SHEAR & BEARING STRESSES

CHECK SHEAR DUE TO RESULTANT FORCE ON PIN AT A

a  60 MPa

RA  2 A2x + A2y

ba  90 MPa

FIND INCLINATION OF & FORCE IN CABLE, T let  angle between pole & cable at C; use Law of Cosines DC 

A

52 + 42  2(5)(4)cos a120 p b 180 a  acos c

DC  7.81 m a u

180  26.33 degrees p

180  33.67 p

52 + DC2  42 d 2DC(5)

u  60a

p b a 180

< ange between cable & horiz. at D

W  230 kg(9.81 m/s2)

W  2.256  103 N

STATICS TO FIND CABLE FORCE T a MA  0

W(3 sin(30 deg))  TX(5 cos(30 deg))  Ty(5 sin(30 deg))  0

substitute for Tx & Ty in terms of T & solve for T:

T

3 W 2 5 523 sin(u) cos(u) 2 2

T  1.53  103 N Ty  T sin( )

Tx  T cos( )

Tx  1.27  103 N

Ty  846.11 N

(1) dmin BASED ON ALLOWABLE SHEAR - DOUBLE SHEAR AT A Ax  Tx

Ay  Ty  W

dmin

RA  3.35  103 N

RA 2  p t a b a a 4

dmin  5.96 mm 6controls

;

(2) dmin BASED ON ALLOWABLE BEARING ON PIN dpole  140 mm

tpole  12 mm

Lpole  6000 mm

member AB BEARING ON PIN sb 

RA Ab

dmin 

Ab  2tpoled

RA 2tpole sba

dmin  1.55 mm

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SECTION 1.8 Design for Axial Loads and Direct Shear

Problem 1.8-9 A pressurized circular cylinder has a sealed cover plate fastened with steel bolts (see figure). The pressure p of the gas in the cylinder is 290 psi, the inside diameter D of the cylinder is 10.0 in., and the diameter dB of the bolts is 0.50 in. If the allowable tensile stress in the bolts is 10,000 psi, find the number n of bolts needed to fasten the cover.

Solution 1.8-9 Pressurized cylinder P

ppD 2 F  n 4n

Ab  area of one bolt 

p 2 db 4

P  allow Ab sallow  p  290 psi

D  10.0 in.

allow  10,000 psi

db  0.50 in.

n  number of bolts

F  total force acting on the cover plate from the internal pressure F  pa

n

pD 2 b 4

NUMBER OF BOLTS P  tensile force in one bolt

ppD 2 pD 2 P   Ab (4n)(p4 )d 2b nd 2b pD 2 d 2bsallow

SUBSTITUTE NUMERICAL VALUES: n

(290 psi)(10 in.)2 (0.5 in.)2(10,000 psi)

Use 12 bolts

;

 11.6

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Problem 1.8-10 A tubular post of outer diameter d2 is guyed by two cables fitted with turnbuckles (see figure). The cables are tightened by rotating the turnbuckles, thus producing tension in the cables and compression in the post. Both cables are tightened to a tensile force of 110 kN. Also, the angle between the cables and the ground is 60°, and the allowable compressive stress in the post is c  35 MPa. If the wall thickness of the post is 15 mm, what is the minimum permissible value of the outer diameter d2?

Solution 1.8-10

Tubular post with guy cables d2  outer diameter

AREA OF POST

d1  inner diameter

A

t  wall thickness  15 mm T  tensile force in a cable  110 kN

allow  35 MPa P  compressive force in post  2T cos 30° REQUIRED AREA OF POST A

P s allow



2Tcos 30° s allow

p p 2 (d 2  d 21)  [d 22 (d2 2t)2 ] 4 4

 pt (d2  t) EQUATE AREAS AND SOLVE FOR d2: 2T cos 30°  pt (d2  t) sallow d2 

2T cos 30° + t ptsallow

;

SUBSTITUTE NUMERICAL VALUES: (d2)min  131 mm

;

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SECTION 1.8 Design for Axial Loads and Direct Shear

Problem 1.8-11 A large precast concrete panel for a warehouse is being raised to a vertical position using two sets of

81

cables at two lift lines as shown in the figure part (a). Cable 1 has length L1  22 ft and distances along the panel (see figure part (b)) are a  L1/2 and b  L1/4. The cables are attached at lift points B and D and the panel is rotated about its base at A. However, as a worst case, assume that the panel is momentarily lifted off the ground and its total weight must be supported by the cables. Assuming the cable lift forces F at each lift line are about equal, use the simplified model of one half of the panel in figure part (b) to perform your analysis for the lift position shown. The total weight of the panel is W  85 kips. The orientation of the panel is defined by the following angles:  20° and   10°. Find the required cross-sectional area AC of the cable if its breaking stress is 91 ksi and a factor of safety of 4 with respect to failure is desired. F H

F

F

T2 b2

T1 B

a b1

W

D

b — 2

B g

u

y

a C W — 2

D

b

g A

b

A

(a)

x

(b)

Solution 1.8-11 GEOMETRY L1  22 ft

1 a  L1 2

1 b  L1 4

  10 deg

a  2.5b  24.75 ft

 20 deg Using Law of cosines L2  2(a + b)2 + L21  2(a + b)L1 cos(u ) L2  6.425 ft

b  acos c

L21 + L22  ( a + b)2 d 2L1L2

b  26.484 degrees

1   (   ) 2    1

b 1  10 deg

b 2  16.484 deg

SOLUTION APPROACH: FIND T THEN AC  T/(s U/FS)

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STATICS at point H

T1  27.042 kips

g Fx  0

T2  T1

H

SO

T2  T1

g FY  0 H

T1sin(1)  T2sin(2) sin(b 1) sin(b 2)

T1cos(b 1) + T2 cos(b 2)  F

and

F  W/2,

W  85 kips

SO

T1 acos(b 1) +

sin(b 1) sin(b 2)

T2  16.549 kips

COMPUTE REQUIRED CROSS-SECTIONAL AREA

u  91 ksi Ac 

sin(b 1) cos(b 2)b  F sin(b 2)

W 2 T1  sin(b 1) acos(b 1) + cos(b 2)b sin(b 2)

Problem 1.8-12 A steel column of hollow circular cross section is supported on a circular steel base plate and a concrete pedestal (see figure). The column has outside diameter d  250 mm and supports a load P  750 kN. (a) If the allowable stress in the column is 55 MPa, what is the minimum required thickness t? Based upon your result, select a thickness for the column. (Select a thickness that is an even integer, such as 10, 12, 14, . . . , in units of millimeters.) (b) If the allowable bearing stress on the concrete pedestal is 11.5 MPa, what is the minimum required diameter D of the base plate if it is designed for the allowable load Pallow that the column with the selected thickness can support?

T1 su FS

FS  4

su  22.75 ksi FS

Ac  1.189 in2

;

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SECTION 1.8 Design for Axial Loads and Direct Shear

Solution 1.8-12 Hollow circular column SUBSTITUTE NUMERICAL VALUES IN EQ. (1): t 2  250 t +

(750 * 103 N) p(55 N/mm2)

0

(Note: In this eq., t has units of mm.) t2  250t  4,340.6  0 Solve the quadratic eq. for t: t  18.77 mm Use t  20 mm d  250 mm

A  t(d  t) Pallow  allow t(d  t)

b  11.5 MPa (allowable pressure on concrete) (a) THICKNESS t OF THE COLUMN

pt(d  t)  pt  ptd + 2

t 2  td +

P sallow

p (4t)(d  t)  pt(d  t) 4

D2  

P sallow

Pallow pD 2  4 sb

4s allowt(d  t) sb 4(55 MPa)(20 mm)(230 mm) 11.5 MPa

D2  88,000 mm2

0

P 0 psallow

Area of base plate 

sallowpt(d  t) pD 2  4 sb

pd 2 p A  (d  2t)2 4 4 

Pallow  allow A

where A is the area of the column with t  20 mm.

D  diameter of base plate

s allow

;

For the column,

t  thickness of column

A

;

(b) DIAMETER D OF THE BASE PLATE

P  750 kN

allow  55 MPa (compression in column)

P

tmin  18.8 mm

Dmin  297 mm (Eq. 1)

D  296.6 mm ;

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Problem 1.8-13 An elevated jogging track is supported at

intervals by a wood beam AB (L  7.5 ft) which is pinned at A and supported by steel rod BC and a steel washer at B. Both the rod (dBC  3/16 in.) and the washer (dB  1.0 in.) were designed using a rod tension force of TBC  425 lb. The rod was sized using a factor of safety of 3 against reaching the ultimate stress u  60 ksi. An allowable bearing stress ba  565 psi was used to size the washer at B. Now, a small platform HF is to be suspended below a section of the elevated track to support some mechanical and electrical equipment. The equipment load is uniform load q  50 lb/ft and concentrated load WE  175 lb at mid-span of beam HF. The plan is to drill a hole through beam AB at D and install the same rod (dBC) and washer (dB) at both D and F to support beam HF. (a) Use u and ba to check the proposed design for rod DF and washer dF; are they acceptable? (b) Also re-check the normal tensile stress in rod BC and bearing stress at B; if either is inadequate under the additional load from platform HF, redesign them to meet the original design criteria.

Original structure

C Steel rod, 3 dBC = — in. 16

TBC = 425 lb. L — 25

L = 7.5 ft A

Wood beam supporting track

D

B Washer dB = 1.0 in.

3 New steel rod, dDF = — in. 16 WE = 175 lb q = 50 lb/ft

New beam to support equipment

H

L — 2

L — 2 Hx

L — 25 F Washer, dF (same at D above)

Hy

Solution 1.8-13 NUMERICAL DATA L  7.5(12)

u  60 ksi q

50 12

dBC 

L  90 in.

ba  0.565 ksi

FSu  3 q  4.167

3 in. 16

TBC  425 lb

lb in

WE  175 lb

dB  1.0 in

(a) FIND FORCE IN ROD DF AND FORCE ON WASHER AT F

MH  0

L L  qL 2 2 TDF  L aL b 25

NORMAL STRESS IN ROD DF: TDF p 2 d 4 BC

sa 

su FSu

a  20 ksi

BEARING STRESS ON WASHER AT F: sbF 

TDF p 2 2 1d  d BC2 4 B OK - less than ba; washer is ; acceptable

sbF  378 psi

WE

TDF  286.458 lb

s DF 

OK - less than a; rod is ; acceptable

sDF  10.38 ksi

(b) FIND NEW FORCE IN ROD BC - SUM MOMENT ABOUT A FOR UPPER FBD - THEN CHECK NORMAL STRESS IN BC & BEARING STRESS AT B

MA  0 TBC2 

TBCL + TDF a L 

TBC2  700 lb

L

L b 25

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SECTION 1.8 Design for Axial Loads and Direct Shear

REVISED NORMAL STRESS IN ROD BC:

Original structure

TBC2

s BC2 

C Steel rod, 3 dBC = — in. 16

p a dBC2 b 4

sBC2  25.352 ksi

exceeds a = 20 ksi

L — 25

L = 7.5 ft

SO RE-DESIGN ROD BC:

A

Wood beam supporting track

TBC2 dBCreqd  p sa Q4 dBCreqd  0.211 in.

TBC2

B Washer dB = 1.0 in. L — 25

New beam to support equipment

H

RE-CHECK BEARING STRESS IN WASHER AT B:

p c 1dB2 dBC22 d 4

D

3 New steel rod, dDF = — in. 16 WE = 175 lb q = 50 lb/ft

dBCreqd . 16  3.38 in. 1 ^say 4/16  1/4 in. dBC2  in. 4

s bB2 

TBC = 425 lb.

bB2  924 psi ^ exceeds ba = 565 psi

Washer, dF (same at D above)

L — 2

L — 2

F

Hx Hy

SO RE-DESIGN WASHER AT B: TBC2 dBC2 dBreqd  1.281 in. p sba Q4 use 1  5/16 in washer at B: 1 + 5/16  1.312 in. dBreqd 

;

Problem 1.8-14 A flat bar of width b  60 mm and thickness t  10 mm is loaded in tension by a force P (see figure). The bar is attached to a support by a pin of diameter d that passes through a hole of the same size in the bar. The allowable tensile stress on the net cross section of the bar is sT  140 MPa, the allowable shear stress in the pin is tS  80 MPa, and the allowable bearing stress between the pin and the bar is sB  200 MPa. (a) Determine the pin diameter dm for which the load P will be a maximum. (b) Determine the corresponding value Pmax of the load.

d

P

b

t

P

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Solution 1.8-14

Bar with a pin connection SHEAR IN THE PIN PS  2tS Apin  2tS a

pd 2 b 4

p 1  2(80 MPa)a b(d 2)a b 4 1000  0.040 pd 2  0.12566d 2

(Eq. 2)

BEARING BETWEEN PIN AND BAR PB  B td b  60 mm

 (200 MPa)(10 mm)(d )a

t  10 mm

 2.0 d

d  diameter of hole and pin

1 b 1000 (Eq. 3)

GRAPH OF EQS. (1), (2), AND (3)

T  140 MPa S  80 MPa B  200 MPa UNITS USED IN THE FOLLOWING CALCULATIONS: P is in kN

 and  are in N/mm2 (same as MPa) b, t, and d are in mm TENSION IN THE BAR PT  T (Net area)  t(t)(b  d )  (140 MPa)(10 mm) (60 mm  d) a  1.40 (60  d)

1 b 1000 (Eq. 1)

(a) PIN DIAMETER dm PT  PB or 1.40(60  d)  2.0 d 84.0 mm  24.7 mm Solving, dm  3.4 (b) LOAD Pmax Substitute dm into Eq. (1) or Eq. (3): Pmax  49.4 kN

;

;

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SECTION 1.8 Design for Axial Loads and Direct Shear

Problem 1.8-15 Two bars AC and BC of the same material support a vertical load P (see figure). The length L of the horizontal bar is fixed, but the angle  can be varied by moving support A vertically and changing the length of bar AC to correspond with the new position of support A. The allowable stresses in the bars are the same in tension and compression. We observe that when the angle  is reduced, bar AC becomes shorter but the crosssectional areas of both bars increase (because the axial forces are larger). The opposite effects occur if the angle  is increased. Thus, we see that the weight of the structure (which is proportional to the volume) depends upon the angle . Determine the angle  so that the structure has minimum weight without exceeding the allowable stresses in the bars. (Note: The weights of the bars are very small compared to the force P and may be disregarded.)

87

A

θ

B

C L P

Solution 1.8-15 Two bars supporting a load P LENGTHS OF BARS L AC 

L cos u

L BC  L

WEIGHT OF TRUSS g  weight density of material W  g(AAC L AC + ABC L BC)

T  tensile force in bar AC C  compressive force in bar BC a Fvert  0 a Fhoriz  0

T

P sin u

C

P tan u

AREAS OF BARS AAC 

T P  sallow sallow sin u

ABC 

C P  sallow sallow tan u



gPL 1 1 a + b sallow sin u cos u tan u



gPL 1 + cos2u a b sallow sin u cos u

Eq. (1)

, P, L, and allow are constants W varies only with  Let k 

gPL (k has unis of force) sallow

W 1 + cos2u  (Nondimensional) k sin u cos u

Eq. (2)

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GRAPH OF EQ. (2): df  du 

(sin u cos u)(2)(cos u) (  sin u)  (1 + cos2u)( sin2u + cos2 u) sin2u cos2u sin2u cos2 u + sin2u  cos2 u  cos4u sin2 u cos2 u

SET THE NUMERATOR  0 AND SOLVE FOR : sin2 cos2  sin2  cos2  cos4  0 Replace sin2 by 1  cos2: (1  cos2)(cos2)  1  cos2  cos2  cos4  0 Combine terms to simplify the equation: ANGLE  THAT MAKES WA MINIMUM

1  3 cos2  0

Use Eq. (2) Let f 

1 + cos2u sin u cos u

df 0 du

  54.7°

;

cos u 

1 23

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2 Axially Loaded Members Changes in Lengths of Axially Loaded Members Problem 2.2-1 The L-shaped arm ABC shown in the figure lies in a vertical plane and pivots about a horizontal pin at A. The arm has constant cross-sectional area and total weight W. A vertical spring of stiffness k supports the arm at point B. Obtain a formula for the elongation of the spring due to the weight of the arm.

k A

B

C

b

b

b — 2

Solution 2.2-1 Take first moments about A to find c.g.

x⫽

b 2b 2 W(b) + ≥ ¥W(2 b) 5 5 P bQ a bb 2 2

W 6 x⫽ b 5 Find force in spring due to weight of arm

a MA ⫽ 0

Fk ⫽

6 Wa bb 5 b

6 Fk ⫽ W 5

Find elongation of spring due to weight of arm Fk 6W d⫽ d⫽ ; k 5k

89

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Problem 2.2-2 A steel cable with nominal diameter 25 mm (see Table 2-1) is used in a construction yard to lift a bridge section weighing 38 kN, as shown in the figure. The cable has an effective modulus of elasticity E ⫽ 140 GPa. (a) If the cable is 14 m long, how much will it stretch when the load is picked up? (b) If the cable is rated for a maximum load of 70 kN, what is the factor of safety with respect to failure of the cable?

Solution 2.2-2

Bridge section lifted by a cable A ⫽ 304 mm2 (from Table 2-1) W ⫽ 38 kN

(b) FACTOR OF SAFETY PULT ⫽ 406 kN (from Table 2-1) Pmax ⫽ 70 kN

E ⫽ 140 GPa L ⫽ 14 m

n⫽

PULT 406 kN ⫽ ⫽ 5.8 Pmax 70 kN

(a) STRETCH OF CABLE d⫽

(38 kN)(14 m) WL ⫽ EA (140 GPa)(304 mm2)

⫽ 12.5 mm

;

Problem 2.2-3 A steel wire and a copper wire have equal lengths and support equal loads P (see figure). The moduli of elasticity for the steel and copper are Es ⫽ 30,000 ksi and Ec ⫽ 18,000 ksi, respectively. (a) If the wires have the same diameters, what is the ratio of the elongation of the copper wire to the elongation of the steel wire? (b) If the wires stretch the same amount, what is the ratio of the diameter of the copper wire to the diameter of the steel wire?

;

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SECTION 2.2

Solution 2.2-3

91

Changes in Lengths of Axially Loaded Members

Steel wire and copper wire Equal lengths and equal loads Steel: Es ⫽ 30,000 ksi

dc Es 30 ⫽ ⫽ ⫽ 1.67 ds Ec 18

(b) RATIO OF DIAMETERS (EQUAL ELONGATIONS)

Copper: Ec ⫽ 18,000 ksi

dc ⫽ ds

(a) RATIO OF ELONGATIONS (EQUAL DIAMETERS)

dc ⫽

PL EcA

ds ⫽

;

PL EsA

Ec a d2c d2s

Problem 2.2-4 By what distance h does the cage shown in the figure move downward when the weight W is placed inside it? Consider only the effects of the stretching of the cable, which has axial rigidity EA ⫽ 10,700 kN. The pulley at A has diameter dA ⫽ 300 mm and the pulley at B has diameter dB ⫽ 150 mm. Also, the distance L1 ⫽ 4.6 m, the distance L2 ⫽ 10.5 m, and the weight W ⫽ 22 kN. (Note: When calculating the length of the cable, include the parts of the cable that go around the pulleys at A and B.)

PL PL ⫽ or Ec Ac ⫽ Es As Ec Ac Es As

p 2 p bdc ⫽ Es a bd2s 4 4

⫽

Es Ec

dc Es 30 ⫽ ⫽ ⫽ ⫽ 1.29 ds A Ec A 18

;

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Axially Loaded Members

Solution 2.2-4

Cage supported by a cable dA ⫽ 300 mm dB ⫽ 150 mm

LENGTH OF CABLE L ⫽ L1 + 2L2 +

1 1 1pdA2 + (pdB) 4 2

L1 ⫽ 4.6 m

⫽ 4,600 mm + 21,000 mm + 236 mm + 236 mm

L2 ⫽ 10.5 m

⫽ 26,072 mm

EA ⫽ 10,700 kN W ⫽ 22 kN

ELONGATION OF CABLE d⫽

(11 kN)(26,072 mm) TL ⫽ ⫽ 26.8 mm EA (10,700 kN)

LOWERING OF THE CAGE h ⫽ distance the cage moves downward TENSILE FORCE IN CABLE W T⫽ ⫽ 11 kN 2

h⫽

Problem 2.2-5 A safety valve on the top of a tank containing steam under pressure p has a discharge hole of diameter d (see figure). The valve is designed to release the steam when the pressure reaches the value pmax. If the natural length of the spring is L and its stiffness is k, what should be the dimension h of the valve? (Express your result as a formula for h.)

1 d ⫽ 13.4 mm 2

;

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SECTION 2.2

Solution 2.2-5

Changes in Lengths of Axially Loaded Members

Safety valve pmax ⫽ pressure when valve opens L ⫽ natural length of spring (L ⬎ h) k ⫽ stiffness of spring FORCE IN COMPRESSED SPRING F ⫽ k(L ⫺ h) (From Eq. 2-1a) PRESSURE FORCE ON SPRING P ⫽ pmax a

pd2 b 4

EQUATE FORCES AND SOLVE FOR h: h ⫽ height of valve (compressed length of the spring) d ⫽ diameter of discharge hole p ⫽ pressure in tank

F ⫽ P k1L ⫺ h2 ⫽ h⫽L⫺

ppmax d2 4k

ppmaxd2 4 ;

Problem 2.2-6 The device shown in the figure consists of

a pointer ABC supported by a spring of stiffness k ⫽ 800 N/m. The spring is positioned at distance b ⫽ 150 mm from the pinned end A of the pointer. The device is adjusted so that when there is no load P, the pointer reads zero on the angular scale. If the load P ⫽ 8 N, at what distance x should the load be placed so that the pointer will read 3° on the scale?

Solution 2.2-6

Pointer supported by a spring

FREE-BODY DIAGRAM OF POINTER

⌺MA ⫽ 0 哵哴

Px kb Let ␣ ⫽ angle of rotation of pointer Px kb2 d x⫽ tan a tan a ⫽ ⫽ 2 b P kb ⫺ Px + (kd)b ⫽ 0

or d ⫽

SUBSTITUTE NUMERICAL VALUES: P⫽8N k ⫽ 800 N/m b ⫽ 150 mm ␦ ⫽ displacement of spring F ⫽ force in spring ⫽ k␦

a ⫽ 3° (800 N/m)(150 mm)2 tan 3° 8N ⫽ 118 mm ;

x⫽

;

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Axially Loaded Members

Problem 2.2-7 Two rigid bars, AB and CD, rest on a smooth horizontal surface (see figure). Bar AB is pivoted end A, and bar CD is pivoted at end D. The bars are connected to each other by two linearly elastic springs of stiffness k. Before the load P is applied, the lengths of the springs are such that the bars are parallel and the springs are without stress. Derive a formula for the displacement ␦C at point C when the load P is acting near point B as shown. (Assume that the bars rotate through very small angles under the action of the load P.)

b

P

b

b

A B C dC D

Solution 2.2-7 (1) first sum moments about A for the entire structure to get RD then sum vertical forces to get RA a MA ⫽ 0

RD ⫽

DISPLACEMENT DIAGRAMS

1 [ P(2b)] 3b

A

2 RD ⫽ P 3 a FV ⫽ 0

RA ⫽ P ⫺ RD

B

2

UFBD Fk2 ⫽ ⫺RA ⫺ P ⫺4 P 3 ^ spring 2 is in compression

P UFBD

F k2 ⫽

b RA

b Fk1

Fk1 ⫽ RA ⫺ (P ⫹ Fk2)

UFBD P 4 ⫺ P + Pb 3 3

2 F k1 ⫽ P 3 ^ spring 1 is in tension

D

C

(P ⫹ Fk2)b ⫽ ⫺RAb

a Mk1 ⫽ 0

F k1 ⫽ a

dB

dB P RA ⫽ 3

(2) next, cut through both springs & consider equilibrium of upper free body (UFBD) to find forces in springs (assume initially that both springs are in tension)

a FV ⫽ 0

(3) solve displacement equations to find ␦C

Fk2

dC

dC

2

dB Fk1 2P ⫽ ⫽ 2 k 3 k Fk2 dC ⫺4 P elongation of spring 2 ⫽ ⫺ dB ⫽ ⫽ 2 k 3 k multiply 2nd equation above by (⫺1/2) and add to first equation 3 4 P 16 P 16 d ⫽ dC ⫽ ; ⫽ 1.778 4 C 3k 9 k 9 elongation of spring 1 ⫽ dC ⫺

(4) substitute dC into either equation to find ␦B (not a required part of this problem) 4P dB ⫽ 2dC ⫺ 1st equ ⬎ 3k 4P 16 P b ⫺ d dB ⫽ c2a 9 k 3k 20 P 20 dB ⫽ ⫽ 2.222 9 k 9 2nd equ ⬎

dB ⫽

dC 4P + 2 3k

4P 1 16 P dB ⫽ c a b + d 2 9 k 3k

dB ⫽

20 P 9 k

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Changes in Lengths of Axially Loaded Members

Problem 2.2-8 The three-bar truss ABC shown in the figure has a span L ⫽ 3 m and is constructed of steel pipes having cross-sectional area A ⫽ 3900 mm2 and modulus of elasticity E ⫽ 200 GPa. Identical loads P act both vertically and horizontally at joint C, as shown.

P P

C

(a) If P ⫽ 650 kN, what is the horizontal displacement of joint B? (b) What is the maximum permissible load value Pmax if the displacement of joint B is limited to 1.5 mm? 45°

A

45°

B

L

Solution 2.2-8 P P

By

a FV ⫽ 0

Ax Ay

By

Ay ⫽ P ⫺ By

Method of Joints:

NUMERICAL DATA A ⫽ 3900 mm2

E ⫽ 200 GPa

P ⫽ 650 kN

L ⫽ 3000 mm

␦Bmax ⫽ 1.5 mm (a) FIND HORIZ. DISPL. OF JOINT B a MA ⫽ 0

By ⫽

1 L a2P b L 2

By ⫽ P a FH ⫽ 0

Ax ⫽ ⫺P

FAB ⫽ Ax dB ⫽

F ABL EA

Ay ⫽ 0

FACV ⫽ Ay FAC ⫽ 0

FACV ⫽ 0

force in AB is P (tension) so elongation of AB ⫽ horiz. displ. of jt B dB ⫽

PL EA

d B ⫽ 2.5 mm

;

(b) FIND Pmax IF DISPL. OF JOINT B ⫽ d Bmax ⫽ 1.5 mm Pmax ⫽

EA d L Bmax

Pmax ⫽ 390 kN

;

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Problem 2.2-9 An aluminum wire having a diameter d ⫽ 1/10 in. and length L ⫽ 12 ft is subjected to a tensile load P (see figure). The aluminum has modulus of elasticity E ⫽ 10,600 ksi If the maximum permissible elongation of the wire is 1/8 in. and the allowable stress in tension is 10 ksi, what is the allowable load Pmax?

P

d L

Solution 2.2-9 1 in 10 1 d a ⫽ in 8 d⫽

A⫽

pd2 4

L ⫽ 12(12) in

E ⫽ 10600 ⫻ (103) psi

s a ⫽ 10 * (103) psi A ⫽ 7.854 ⫻ 10⫺3 in2 EA ⫽ 8.325 ⫻ 104 lb

Max. load based on elongation EA Pmax1 ⫽ d Pmax1 ⫽ 72.3 lb L a Max. load based on stress Pmax2 ⫽ 78.5 lb Pmax2 ⫽ ␴aA

; controls

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SECTION 2.2

97

Changes in Lengths of Axially Loaded Members

Problem 2.2-10 A uniform bar AB of weight W ⫽ 25 N is supported by two springs, as shown in the figure. The spring on the left has stiffness k1 ⫽ 300 N/m and natural length L1 ⫽ 250 mm. The corresponding quantities for the spring on the right are k2 ⫽ 400 N/m and L2 ⫽ 200 mm. The distance between the springs is L ⫽ 350 mm, and the spring on the right is suspended from a support that is distance h ⫽ 80 mm below the point of support for the spring on the left. Neglect the weight of the springs. (a) At what distance x from the left-hand spring (figure part a) should a load P ⫽ 18 N be placed in order to bring the bar to a horizontal position? (b) If P is now removed, what new value of k1 is required so that the bar (figure part a) will hang in a horizontal position under weight W? (c) If P is removed and k1 ⫽ 300 N/m, what distance b should spring k1 be moved to the right so that the bar (figure part a) will hang in a horizontal position under weight W? (d) If the spring on the left is now replaced by two springs in series (k1 ⫽ 300N/m, k3) with overall natural length L1 ⫽ 250 mm (see figure part b), what value of k3 is required so that the bar will hang in a horizontal position under weight W?

New position of k1 for part (c) only k1 L1

b k2 L2 W

A

B

P x

Load P for part (a) only L (a)

k3 L1 — 2 k1 L1 — 2

h

k2 L2 W

A

B

L (b)

Solution 2.2-10 NUMERICAL DATA W ⫽ 25 N

k1 ⫽ 0.300

N mm

L1 ⫽ 250 mm

N L2 ⫽ 200 mm mm L ⫽ 350 mm h ⫽ 80 mm P ⫽ 18 N k2 ⫽ 0.400

(a) LOCATION OF LOAD P TO BRING BAR TO HORIZ. POSITION

use statics to get forces in both springs a MA ⫽ 0

F2 ⫽ F2 ⫽

1 L aW + Pxb L 2 W x + P 2 L

h

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Axially Loaded Members

a FV ⫽ 0

F1 ⫽ W + P ⫺ F2 F1 ⫽

W x + Pa1 ⫺ b 2 L

use constraint equation to define horiz. position, then solve for location x F1 F2 ⫽ L2 + h + k1 k2

L1 +

substitute expressions for F1 & F2 above into constraint equ. & solve for x ⫺ 2L1 L k1 k2 ⫺ k2WL ⫺ 2k2 P L + 2L2 L k1 k2 + 2 h L k1 k2 + k1W L ⫺2P1k1 + k22 x ⫽ 134.7 mm ; x⫽

(b) NEXT REMOVE P AND FIND NEW VALUE OF SPRING CONSTANT K1 SO THAT BAR IS HORIZ. UNDER WEIGHT W Now, F1 ⫽

W 2

F2 ⫽

W since P ⫽ 0 2

same constraint equation as above but now P ⫽ 0:

Part (c) - continued statics

a Mk1 ⫽ 0

F2 ⫽

wa

L ⫺ bb 2

L⫺b

a FV ⫽ 0

W W a b 2 2 L1 + ⫺ 1L2 + h2 ⫺ ⫽0 k1 k2

F1 ⫽ W ⫺ F2 Wa

solve for k1 k1 ⫽

F1 ⫽ W ⫺

⫺ Wk2 [2k2[L1 ⫺ (L2 + h)]] ⫺ W

N k1 ⫽ 0.204 mm

F1 ⫽

;

(c) USE K1 ⫽ 0.300 N/mm BUT RELOCATE SPRING K1 (x ⫽ b) SO THAT BAR ENDS UP IN HORIZ. POSITION UNDER WEIGHT W L/2 – b F1

b

F2

L/2

L/2

constraint equation - substitute above expressions for F1 & F2 and solve for b F1 F2 ⫺ ( L2 + h) ⫺ ⫽0 k1 k2 use the following data

L1 +

k1 ⫽ 0.300

N mm

L2 ⫽ 200 mm

k2 ⫽ 0.4 L ⫽ 350 mm

L–b FBD

2L1k1k2L + WLk2 ⫺ 2L2k1k2L ⫺ 2hk1k2L ⫺ Wk1L (2L1k1k2) ⫺ 2L2k1k2 ⫺ 2hk1k2 ⫺ 2Wk1

L⫺b

WL 2( L ⫺ b)

W

b⫽

L ⫺ bb 2

b ⫽ 74.1 mm

;

N mm

L1 ⫽ 250 mm

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SECTION 2.2

(d) REPLACE SPRING K1 WITH SPRINGS IN SERIES: K1 ⫽ 0.3N/mm, L1/2 AND K3, L1/2 - FIND K3 SO THAT BAR HANGS IN HORIZ. POSITION statics

F1 ⫽

W 2

k3 ⫽

F2 ⫽

W 2

Changes in Lengths of Axially Loaded Members

new constraint equation; solve for k3 L1 +

F1 F1 F2 + ⫺ ( L2 + h) ⫺ ⫽0 k1 k3 k2

W W W 2 2 2 L1 + + ⫺ ( L2 + h) ⫺ ⫽0 k1 k3 k2

Wk1k2 ⫺2L1k1k2 ⫺ Wk2 + 2L2k1k2 + 2hk1k2 + Wk1

NOTE - equivalent spring constant for series springs k1k3 ke ⫽ k1 + k3

Problem 2.2-11 A hollow, circular, cast-iron pipe (Ec ⫽ 12,000 ksi) supports a brass rod (Eb ⫽ 14,000 ksi) and weight W ⫽ 2 kips, as shown. The outside diameter of the pipe is dc ⫽ 6 in.

(a) If the allowable compressive stress in the pipe is 5000 psi and the allowable shortening of the pipe is 0.02 in., what is the minimum required wall thickness tc,min? (Include the weights of the rod and steel cap in your calculations.) (b) What is the elongation of the brass rod ␦r due to both load W and its own weight? (c) What is the minimum required clearance h?

99

k e ⫽ 0.204

k3 ⫽ 0.638 N mm

;

N mm

;

checks - same as (b) above

Nut & washer 3 dw = — in. 4

(

)

Steel cap (ts = 1 in.) Cast iron pipe (dc = 6 in., tc)

Lr = 3.5 ft

Lc = 4 ft

(

Brass rod 1 dr = — in. 2 h

) W

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Axially Loaded Members

Solution 2.2-11 The figure shows a section cut through the pipe, cap and rod.

LET a ⫽

NUMERICAL DATA

tc2 ⫺ dctc ⫹ ␣ ⫽ 0

Ec ⫽ 12000 ksi W ⫽ 2 kips

Eb ⫽ 14000 ksi

dc ⫽ 6 in

tc ⫽

1 dr ⫽ in. 2

g b ⫽ 3.009 * 10⫺4 Lc ⫽ 48 in

d c ⫺ 2 dc2 ⫺ 4a 2

g s ⫽ 2.836 * 10⫺4

in3

kips

dpipe ⫽

WtLc EcAmin

Amin ⫽

␴a above

Lr ⫽ 42 in ptc( dc ⫺ tc) ⫽

(a) MIN. REQ’D WALL THICKNESS OF CI PIPE, tcmin first check allowable stress then allowable shortening

Wcap ⫽ 8.018 ⫻ 10⫺3 kips

tc ⫽ 0.021 in.

Wt sa

A pipe ⫽

p 2 [ d ⫺ (d c ⫺ 2 t c)2] 4 c

Amin ⫽ 0.402 in2

t c( d c ⫺ t c) ⫽

Wt ps a

WtLc pEc da

dc ⫺ 2 d2c ⫺ 4b 2 ; min. based on da and sa controls

(b) ELONGATION OF ROD DUE TO SELF WEIGHT & ALSO WEIGHT W

Wrod ⫽ 2.482 ⫻ 10⫺3 kips

Amin ⫽

b⫽

␤ ⫽ 0.142 tc ⫽

p ⫽ gb a d2r L r b 4

Wt Lc Ec da

tc2 ⫺ dctc ⫹ ␤ ⫽ 0

p W cap ⫽ g s a d2c t s b 4

Wt ⫽ W ⫹ Wcap ⫹ Wrod

WtLc Ec da

Amin ⫽ 0.447 in2 ⬍ larger than value based on

in3

Apipe ⫽ ␲ tc(dc ⫺ tc)

tc ⫽ 0.021 in

now check allowable shortening requirement kips

ts ⫽ 1 in.

Wrod

␣ ⫽ 0.128

^ min. based on ␴a

␴a ⫽ 5 ksi ␦a ⫽ 0.02 in. unit weights (see Table H-1)

Wt ps a

Wt ⫽ 2.01 kips dr ⫽

aW +

Wrod b Lr 2

p E b a dr 2 b 4

d r ⫽ 0.031 in

(c) MIN. CLEARANCE h hmin ⫽ ␦a ⫹ ␦r

hmin ⫽ 0.051 in.

;

;

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SECTION 2.2

Changes in Lengths of Axially Loaded Members

101

Problem 2.2-12 The horizontal rigid beam ABCD is supported by vertical bars BE and CF and is loaded by vertical forces P1 ⫽ 400 kN and P2 ⫽ 360 kN acting at points A and D, respectively (see figure). Bars BE and CF are made of steel (E ⫽ 200 GPa) and have cross-sectional areas ABE ⫽ 11,100 mm2 and ACF ⫽ 9,280 mm2. The distances between various points on the bars are shown in the figure. Determine the vertical displacements ␦A and ␦D of points A and D, respectively.

Solution 2.2-12 Rigid beam supported by vertical bars

⌺MB ⫽ 0 哵哴 ABE ⫽ 11,100 mm2 ACF ⫽ 9,280 mm2 E ⫽ 200 GPa LBE ⫽ 3.0 m LCF ⫽ 2.4 m P1 ⫽ 400 kN; P2 ⫽ 360 kN

(400 kN)(1.5 m) ⫹ FCF(1.5 m) ⫺ (360 kN)(3.6 m) ⫽ 0 FCF ⫽ 464 kN ⌺MC ⫽ 0 12 (400 kN)(3.0 m) ⫺ FBE(1.5 m) ⫺ (360 kN)(2.1 m) ⫽ 0 FBE ⫽ 296 kN SHORTENING OF BAR BE dBE ⫽

(296 kN)(3.0 m) FBELBE ⫽ EABE (200 GPa)(11,100 mm2) ⫽ 0.400 mm

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Axially Loaded Members

SHORTENING OF BAR CF

␦BE ⫺ ␦A ⫽ ␦CF ⫺ ␦BE or ␦A ⫽ 2␦BE ⫺ ␦CF

(464 kN)(2.4 m) FCFLCF ⫽ dCF ⫽ EACF (200 GPa)(9,280 mm2) ⫽ 0.600 mm

␦A ⫽ 2(0.400 mm) ⫺ 0.600 m ⫽ 0.200 mm ; (Downward)

DISPLACEMENT DIAGRAM

2.1 (d ⫺ dBE) 1.5 CF 12 7 d D ⫽ dCF ⫺ dBE 5 5 12 7 ⫽ (0.600 mm) ⫺ (0.400 mm) 5 5 ⫽ 0.880 mm ; (Downward)

dD ⫺ dCF ⫽ or

bars AB and BC, each having length b (see the first part of the figure). The bars have pin connections at A, B, and C and are joined by a spring of stiffness k. The spring is attached at the midpoints of the bars. The framework has a pin support at A and a roller support at C, and the bars are at an angle ␣ to the hoizontal.

When a vertical load P is applied at joint B (see the second part of the figure) the roller support C moves to the right, the spring is stretched, and the angle of the bars decreases from ␣ to the angle ␪. Determine the angle ␪ and the increase ␦ in the distance between points A and C. (Use the following data; b ⫽ 8.0 in., k ⫽ 16 lb/in., ␣ ⫽ 45°, and P ⫽ 10 lb.)

B

P

Problem 2.2-13 A framework ABC consists of two rigid

b — 2 b — 2

B

b — 2

k a

A

b — 2

a C

A

u

u

C

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SECTION 2.2

Solution 2.2-13

103

Changes in Lengths of Axially Loaded Members

Framework with rigid bars and a spring h ⫽ height from C to B ⫽ b sin ␪ L2 ⫽ bcosu 2 F ⫽ force in spring due to load P ⌺MB ⫽ 0 哵 哴 h P L2 a b ⫺ Fa b ⫽ 0 or Pcosu ⫽ Fsinu 2 2 2

WITH NO LOAD

DETERMINE THE ANGLE ␪

L2 ⫽ span from A to C

⌬S ⫽ elongation of spring

⫽ 2b cos ␪ S1 ⫽ length of spring L1 ⫽ ⫽ bcosa 2

(Eq. 1)

⫽ S2 ⫺ S1 ⫽ b(cos ␪ ⫺ cos ␣) For the spring: F ⫽ k(⌬S) F ⫽ bk(cos ␪ ⫺ cos ␣) Substitute F into Eq. (1): P cos ␪ ⫽ bk(cos ␪ ⫺ cos ␣)(sin ␪) or

P cotu ⫺ cosu + cosa ⫽ 0 bk

;

(Eq. 2)

This equation must be solved numerically for the angle ␪. DETERMINE THE DISTANCE ␦ WITH LOAD P

␦ ⫽ L2 ⫺ L1 ⫽ 2b cos ␪ ⫺ 2b cos ␣ ⫽ 2b(cos ␪ ⫺ cos ␣)

L1 ⫽ span from A to C ⫽ 2b cos ␣ S2 ⫽ length of spring ⫽

L2 ⫽ bcosu 2

FREE-BODY DIAGRAM OF BC

From Eq. (2): cosa ⫽ cosu ⫺

Pcotu bk

Therefore, d ⫽ 2b acosu ⫺ cosu + ⫽

2P cotu k

Pcotu b bk

;

(Eq. 3)

NUMERICAL RESULTS b ⫽ 8.0 in.

k ⫽ 16 lb/in.

␣ ⫽ 45°

P ⫽ 10 lb

Substitute into Eq. (2): 0.078125 cot ␪ ⫺ cos ␪ ⫹ 0.707107 ⫽ 0 Solve Eq. (4) numerically:

␪ ⫽ 35.1°

;

Substitute into Eq. (3):

␦ ⫽ 1.78 in.

;

(Eq. 4)

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Axially Loaded Members

Problem 2.2-14 Solve the preceding problem for the following data: b ⫽ 200 mm, k ⫽ 3.2 kN/m, ␣ ⫽ 45°, and P ⫽ 50 N.

Solution 2.2-14

Framework with rigid bars and a spring

See the solution to the preceding problem. Eq. (2):

P cotu ⫺ cosu + cosa ⫽ 0 bk

Eq. (3):

2P d⫽ cotu k k ⫽ 3.2 kN/m ␣ ⫽ 45°

0.078125 cot ␪ ⫺ cos ␪ ⫹ 0.707107 ⫽ 0 Solve Eq. (4) numerically:

␪ ⫽ 35.1°

;

Substitute into Eq. (3):

NUMERICAL RESULTS b ⫽ 200 mm

Substitute into Eq. (2):

P ⫽ 50 N

␦ ⫽ 44.5 mm

;

(Eq. 4)

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SECTION 2.3

105

Changes in Lengths under Nonuniform Conditions

Changes in Lengths under Nonuniform Conditions Problem 2.3-1 Calculate the elongation of a copper bar of solid circular cross section with tapered ends when it is stretched by axial loads of magnitude 3.0 k (see figure). The length of the end segments is 20 in. and the length of the prismatic middle segment is 50 in. Also, the diameters at cross sections A, B, C, and D are 0.5, 1.0, 1.0, and 0.5 in., respectively, and the modulus of elasticity is 18,000 ksi. (Hint: Use the result of Example 2-4.)

Solution 2.3-1

A

B C

3.0 k

20 in.

50 in.

Bar with tapered ends MIDDLE SEGMENT (L  50 in.) d2 

(3.0 k)(50 in.) PL  EA (18,000 ksi) A p4 B (1.0 in.)2

 0.0106 in. dA  dD  0.5 in.

P  3.0 k

dB  dC  1.0 in.

E  18,000 ksi

END SEGMENT (L  20 in.)

ELONGATION OF BAR d g

From Example 2-4: d d1 

D

20 in.

4PL pE dA dB

NL  2d1 + d2 EA

 2(0.008488 in.) + (0.01061 in.)  0.0276 in.

;

4(3.0 k)(20 in.)  0.008488 in. p(18,000 ksi)(0.5 in.)(1.0 in.)

Problem 2.3-2 A long, rectangular copper bar under a tensile load P hangs from a pin that is supported by two steel posts (see figure). The copper bar has a length of 2.0 m, a cross-sectional area of 4800 mm2, and a modulus of elasticity Ec  120 GPa. Each steel post has a height of 0.5 m, a cross-sectional area of 4500 mm2, and a modulus of elasticity Es  200 GPa.

Steel post

(a) Determine the downward displacement  of the lower end of the copper bar due to a load P  180 kN. (b) What is the maximum permissible load Pmax if the displacement  is limited to 1.0 mm? Copper bar P

3.0 k

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Axially Loaded Members

Solution 2.3-2

Copper bar with a tensile load (a) DOWNWARD DISPLACEMENT  (P  180 kN) dc 

(180 kN)(2.0 m) PLc  Ec Ac (120 GPa)(4800 mm2)

 0.625 mm ds 

(P/2)Ls (90 kN)(0.5 m)  EsAs (200 GPa)(4500 mm2)

 0.050 mm d  dc + ds  0.625 mm + 0.050 mm  0.675 mm

Lc  2.0 m Ac  4800 mm2

;

(b) MAXIMUM LOAD Pmax (max  1.0 mm)

Ec  120 GPa

dmax Pmax  P d

Ls  0.5 m As  4500 mm2

Pmax  Pa

Pmax  (180 kN)a

Es  200 GPa

dmax b d

1.0 mm b  267 kN 0.675 mm

;

Problem 2.3-3 A steel bar AD (see figure) has a cross-sectional

area of 0.40 in.2 and is loaded by forces P1  2700 lb, P2  1800 lb, and P3  1300 lb. The lengths of the segments of the bar are a  60 in., b  24 in., and c  36 in.

P1

(a) Assuming that the modulus of elasticity E  30  10 psi, calculate the change in length  of the bar. Does the bar elongate or shorten? (b) By what amount P should the load P3 be increased so that the bar does not change in length when the three loads are applied? 6

Solution 2.3-3

A

C

B a

P2

b

D c

Steel bar loaded by three forces

A  0.40 in.2

P1  2700 lb

P2  1800 lb

P3  1300 lb

E  30  106 psi

AXIAL FORCES NAB  P1  P2  P3  3200 lb NBC  P2  P3  500 lb NCD  P3  1300 lb

(a) CHANGE IN LENGTH d g 

NiLi EiAi

1 (N L + NBCLBC + NCDLCD) EA AB AB

P3

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SECTION 2.3



1 6

2

The force P must produce a shortening equal to 0.0131 in. in order to have no change in length.

[(3200 lb) (60 in.)

(30 * 10 psi)(0.40 in. ) + (500 lb)(24 in.)  (1300 lb) (36 in.)]

 0.0131 in. (elongation)

107

Changes in Lengths under Nonuniform Conditions

‹ 0.0131 in.  d 

; 

(b) INCREASE IN P3 FOR NO CHANGE IN LENGTH

PL EA P(120 in.)

(30 * 106 psi)(0.40 in.2)

P  1310 lb

;

P  increase in force P3

Problem 2.3-4 A rectangular bar of length L has a slot in the middle half of its length (see figure). The bar has width b, thickness t, and modulus of elasticity E. The slot has width b/4. (a) Obtain a formula for the elongation  of the bar due to the axial loads P. (b) Calculate the elongation of the bar if the material is high-strength steel, the axial stress in the middle region is 160 MPa, the length is 750 mm, and the modulus of elasticity is 210 GPa.

Solution 2.3-4

b — 4

P

t

b L — 4

P L — 2

L — 4

Bar with a slot STRESS IN MIDDLE REGION s

P  A

P 4P  3bt 3 a btb 4

or

P 3s  bt 4

Substitute into the equation for : d

t  thickness L  length of bar



(a) ELONGATION OF BAR d g

P(L/4) P(L/2) P(L/4) NiLi  + + 3 EAi E(bt) E(bt) E A 4 bt B

PL 1 4 1 7PL  a + + b Ebt 4 6 4 6Ebt

;

7L P 7L 3s 7PL  a b  a b 6Ebt 6E bt 6E 4 7sL 8E

(b) SUBSTITUTE NUMERICAL VALUES: s  160 MPa L  750 mm E  210 GPa d

7(160 MPa)(750 mm)  0.500 mm 8(210 GPa)

;

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Axially Loaded Members

Problem 2.3-5 Solve the preceding problem if the axial stress in the middle region is 24,000 psi, the length is 30 in., and the modulus of elasticity is 30  106 psi.

b — 4

P

L — 4

Solution 2.3-5

t

b

P L — 2

L — 4

Bar with a slot STRESS IN MIDDLE REGION P  A

s

P 4P  3bt 3 a btb 4

or

3s P  bt 4

SUBSTITUTE INTO THE EQUATION FOR : t  thickness

L  length of bar

d

(a) ELONGATION OF BAR P(L/4) P(L/4) P(L/2) NiLi + d g  + 3 EAi E(bt) E(bt) E(4bt) 

4 1 7PL PL 1 a + + b Ebt 4 6 4 6Ebt

;



7PL 7L P 7L 3s  a b  a b 6Ebt 6E bt 6E 4 7sL 8E

(B) SUBSTITUTE NUMERICAL VALUES: s  24,000 psi L  30 in. E  30 * 106 psi d

Problem 2.3-6 A two-story building has steel columns AB in the first floor and BC in the second floor, as shown in the figure. The roof load P1 equals 400 kN and the second-floor load P2 equals 720 kN. Each column has length L  3.75 m. The cross-sectional areas of the first- and second-floor columns are 11,000 mm2 and 3,900 mm2, respectively. (a) Assuming that E  206 GPa, determine the total shortening AC of the two columns due to the combined action of the loads P1 and P2. (b) How much additional load P0 can be placed at the top of the column (point C) if the total shortening AC is not to exceed 4.0 mm?

7(24,000 psi)(30 in.) 8(30 * 106 psi)

P1 = 400 kN

 0.0210 in.

;

C

L = 3.75 m P2 = 720 kN

B

L = 3.75 m A

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SECTION 2.3

Solution 2.3-6

Changes in Lengths under Nonuniform Conditions

109

Steel columns in a building (b) ADDITIONAL LOAD P0 AT POINT C (dAC)max  4.0 mm d0  additional shortening of the two columns due to the load P0 d0  (dAC)max  dAC  4.0 mm  3.7206 mm  0.2794 mm Also, d0 

Solve for P0:

(a) SHORTENING AC OF THE TWO COLUMNS

P0 

NiLi NABL NBCL dAC  g  + EiAi EAAB EABC 

Ed0 AAB ABC b a L AAB + ABC

SUBSTITUTE NUMERICAL VALUES:

(1120 kN)(3.75 m)

E  206  109 N/m2

(206 GPa)(11,000 mm2)

L  3.75 m

(400 kN)(3.75 m) +

P0L P0L P0L 1 1 +  + b a EAAB EABC E AAB ABC

0  0.2794  103 m

AAB  11,000  106 m2

ABC  3,900  106 m2

2

(206 GPa)(3,900 mm )

P0  44,200 N  44.2 kN

 1.8535 mm + 1.8671 mm  3.7206 mm dAC  3.72 mm ;

;

Problem 2.3-7 A steel bar 8.0 ft long has a circular cross section of diameter d1  0.75 in. over one-half of its length and diameter d2  0.5 in. over the other half (see figure). The modulus of elasticity E  30  106 psi. (a) How much will the bar elongate under a tensile load P  5000 lb? (b) If the same volume of material is made into a bar of constant diameter d and length 8.0 ft, what will be the elongation under the same load P?

Solution 2.3-7

d1 = 0.75 in.

d2 = 0.50 in. P = 5000 lb

P 4.0 ft

4.0 ft

Bar in tension (a) ELONGATION OF NONPRISMATIC BAR d g

P  5000 lb E  30  106 psi L  4 ft  48 in.

d

NiLi PL 1  g Ei Ai E Ai

(5000 lb)(48 in.) 30 * 106 psi

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Axially Loaded Members

J 4 (0.75 in) 1

*

p

2

 0.0589 in.

1 +

p 4 (0.50

;

in.)2 K

Ap   d

(b) ELONGATION OF PRISMATIC BAR OF SAME VOLUME Original bar: Vo  A1L  A2L  L(A1  A2)

p [(0.75 in.)2 + (0.50 in.)2]  0.3191 in.2 8 (5000 lb)(2)(48 in.) P(2L)  EAp (30 * 106 psi)(0.3191 in.2)

 0.0501 in.

Prismatic bar: Vp  Ap(2L) L(A1  A2)  Ap(2L)

Problem 2.3-8 A bar ABC of length L consists of two parts of equal lengths but different diameters. Segment AB has diameter d1  100 mm, and segment BC has diameter d2  60 mm. Both segments have length L/2  0.6 m. A longitudinal hole of diameter d is drilled through segment AB for one-half of its length (distance L/4  0.3 m). The bar is made of plastic having modulus of elasticity E  4.0 GPa. Compressive loads P  110 kN act at the ends of the bar.

dmax

A

B

d2 C

d1

P L — 4

(a) If the shortening of the bar is limited to 8.0 mm, what is the maximum allowable diameter dmax of the hole? (See figure part a.) (b) Now, if dmax is instead set at d2/2, at what distance b from end C should load P be applied to limit the bar shortening to 8.0 mm? (See figure part b.) (c) Finally, if loads P are applied at the ends and dmax  d2/2, what is the permissible length x of the hole if shortening is to be limited to 8.0 mm? (See figure part c.)

;

NOTE: A prismatic bar of the same volume will always have a smaller change in length than will a nonprismatic bar, provided the constant axial load P, modulus E, and total length L are the same.

Equate volumes and solve for Ap: Vo  Vp

A1 + A2 1 p  a b(d21 + d22) 2 2 4

P

L — 4

L — 2 (a)

d dmax = —2 2

A

B

P

d2

C d1

P L — 4

L — 4

L — 2

b

(b) d dmax = —2 2

A

B

d2 C

d1

P

x

P L — 2

L — x 2 (c)

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SECTION 2.3

Changes in Lengths under Nonuniform Conditions

Solution 2.3-8 b c

NUMERICAL DATA d1  100 mm

d2  60 mm

L  1200 mm

E  4.0 GPa

P  110 kN

(a) find dmax if shortening is limited to a

d

p 2 d 4 1

A2 

p 2 d 4 2

L 4

P ≥ E p 1d 2  dmax 22 4 1

L L 4 2 + ¥ + A1 A2

Eda p d1 2 d2 2  2PLd2 2  2PLd1 2 9 Edapd1 2d2 2  PLd2 2  2PLd1 2

dmax  23.9 mm

;

(b) Now, if dmax is instead set at d2 2, at what distance b from end C should load P be applied to limit the bar shortening to a  8.0 mm? d2 2 p 2 c d1  a b d 4 2 p 2 p A1  d1 A2  d2 2 4 4 A0 

L P L + + d J E 4A0 4A1

a

d

L  bb 2 A2

;

K

no axial force in segment at end of length b; set   a & solve for b

P x J + E A0

a

a

L  xb 2

L b 2

+ A1

A2

set   a & solve for x

x

set  to a and solve for dmax dmax  d1

b  4.16 mm

(c) Finally if loads P are applied at the ends and dmax  d2 2, what is the permissible length x of the hole if shortening is to be limited to a  8.0 mm?

a  8.0 mm

A1 

Eda L L L + bdd  A2 c  a 2 P 4A0 4A1

c A0 A1a

Ed a L 1  b d  A0 L P 2 A2 2

x  183.3 mm

A1  A0 ;

K

111

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Axially Loaded Members

Problem 2.3-9 A wood pile, driven into the earth, supports a load P entirely

P

by friction along its sides (see figure). The friction force f per unit length of pile is assumed to be uniformly distributed over the surface of the pile. The pile has length L, cross-sectional area A, and modulus of elasticity E. (a) Derive a formula for the shortening  of the pile in terms of P, L, E, and A. (b) Draw a diagram showing how the compressive stress c varies throughout the length of the pile.

Solution 2.3-9

Wood pile with friction

FROM FREE-BODY DIAGRAM OF PILE: Fvert  0 *uarr* *darr* fL  P  0 f 

P (Eq. 1) L

(a) SHORTENING  OF PILE: At distance y from the base: N(y)  axial force N(y)  fy dd 

f

N(y)dy fy dy  EA EA

f L fL2 PL L ydy   d  10 dd  1 0 EA 2EA 2EA

d

PL 2EA

(b) COMPRESSIVE STRESS c IN PILE sc 

(Eq. 2)

;

N(y) fy Py   A A AL

;

At the base (y  0): c  0 At the top(y  L): sc  See the diagram above.

P A

L

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SECTION 2.3

Changes in Lengths under Nonuniform Conditions

113

Problem 2.3-10 Consider the copper tubes joined below using a “sweated” joint. Use the properties and dimensions given. (a) Find the total elongation of segment 2-3-4 (2-4) for an applied tensile force of P  5 kN. Use Ec  120 GPa. (b) If the yield strength in shear of the tin-lead solder is y  30 MPa and the tensile yield strength of the copper is y  200 MPa, what is the maximum load Pmax that can be applied to the joint if the desired factor of safety in shear is FS  2 and in tension is FS  1.7? (c) Find the value of L2 at which tube and solder capacities are equal. Sweated joint P

Segment number

Solder joints

1

2

3

4

L2

L3

L4

5

P

d0 = 18.9 mm t = 1.25 mm

d0 = 22.2 mm t = 1.65 mm L3 = 40 mm L2 = L4 = 18 mm

Tin-lead solder in space between copper tubes; assume thickness of solder equal zero

Solution 2.3-10 NUMERICAL DATA P  5 kN

Ec  120 GPa

L2  18 mm

L4  L2

L3  40 mm do3  22.2 mm

t3  1.65 mm

do5  18.9 mm

t5  1.25 mm

Y  30 MPa Y  200 MPa FS  2 tY ta  FSt sa 

sY FSs

FS  1.7

a  15 MPa

(a) ELONGATION OF SEGMENT 2-3-4 p A2  [d2o3  (d o5  2 t5)2] 4 p 2 A3  [d o3  1d o3  2t322] 4 A2  175.835 mm2 A3  106.524 mm2 d24 

L3 P L2 + L4 a + b Ec A2 A3

d24  0.024 mm (b) MAXIMUM

LOAD

;

Pmax

THAT CAN BE APPLIED TO THE

JOINT

a  117.6 MPa

FIRST CHECK NORMAL STRESS

A1 

p 2 [ d o5  1 d o5  2 t522] 4

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Axially Loaded Members

A1  69.311 mm2 smallest cross-sectional area controls normal stress Pmax  aA1 Pmaxs  8.15 kN ; smaller than Pmax based on shear below so normal stress controls

Pmax  taAsh

(c) FIND THE VALUE OF L2 AT WHICH TUBE AND SOLDER CAPACITIES ARE EQUAL

set Pmax based on shear strength equal to Pmax based on tensile strength & solve for L2

next check shear stress in solder joint Ash  do5L2

Pmaxt  16.03 kN

Ash  1.069  103 mm2

L2 

s aA1 ta1pd o52

L2  9.16 mm

Segment 1

;

Segment 2

Problem 2.3-11 The nonprismatic cantilever circular bar shown has an internal cylindrical hole of diameter d/2 from 0 to x, so the net area of the cross section for Segment 1 is (3/4)A. Load P is applied at x, and load P/2 is applied at x  L. Assume that E is constant. (a) Find reaction force R1. (b) Find internal axial forces Ni in segments 1 and 2. (c) Find x required to obtain axial displacement at joint 3 of 3  PL/EA. (d) In (c), what is the displacement at joint 2, 2? (e) If P acts at x  2L/3 and P/2 at joint 3 is replaced by P, find  so that 3  PL/EA. (f) Draw the axial force (AFD: N(x), 0 x L) and axial displacement (ADD: (x), 0 x L) diagrams using results from (b) through (d) above.

3 —A 4

d R1

A

d — 2 x

3

2 L–x

3P — 2

P — 2 0

AFD 0

δ3 δ2 ADD 0

0

Solution 2.3-11 (a) STATICS a FH  0

R1   P  R1 

3 P 2

P 2 ;

(b) DRAW FBD’S CUTTING THROUGH SEGMENT 1 & AGAIN THROUGH SEGMENT 2 3P P N1  6 tension N2  6 tension 2 2 (c) FIND x REQUIRED TO OBTAIN AXIAL DISPLACEMENT AT JOINT 3 OF 3  PL/EA add axial deformations of segments 1 & 2 then set to 3; solve for x N2( L  x) N1x PL  + EA EA 3 E A 4

P — 2

P

3P P x ( L  x) 2 2 PL +  3 EA EA E A 4 3 L L x x ; 2 2 3 (d) WHAT IS THE DISPLACEMENT AT JOINT 2, 2?

d2 

N1x 3 E A 4

d2 

2 PL 3 EA

d2 

a

3P L b 2 3 3 E A 4

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SECTION 2.3

(e) IF x  2L/3 AND P/2 AT JOINT 3 IS REPLACED BY P, FIND  SO THAT 3  PL/EA 2L 3 substitute in axial deformation expression above & solve for  N1  (1  )P

[(1 + b)P]

2L 3

N2  P

bPaL 

x

2L b 3

+ 3 E A 4

EA



PL EA

115

Changes in Lengths under Nonuniform Conditions

1 8 + 11b PL PL  9 EA EA (8  11)  9 1 ; 11   0.091 b

(f) Draw AFD, ADD - see plots above for x 

L 3

Problem 2.3-12 A prismatic bar AB of length L, cross-sectional area A, modulus of elasticity E, A

and weight W hangs vertically under its own weight (see figure). (a) Derive a formula for the downward displacement C of point C, located at distance h from the lower end of the bar.

C

(b) What is the elongation B of the entire bar? (c) What is the ratio  of the elongation of the upper half of the bar to the elongation of the lower half of the bar?

h B

Solution 2.3-12 Prismatic bar hanging vertically W  Weight of bar

A dy C

y

L

(a) DOWNWARD DISPLACEMENT C Consider an element at distance y from the lower end.

B

Wy N(y)dy Wydy N(y)  dd   L EA EAL W L L Wydy dC  1h dd  1h  (L2  h2) EAL 2EAL W (L2  h2) 2EAL

dB 

WL 2EA

;

(c) RATIO OF ELONGATIONS Elongation of upper half of bar a h 

h

dC 

(b) ELONGATION OF BAR (h  0)

;

L b: 2

3WL 8EA Elongation of lower half of bar: d upper 

d lower  dB  d upper  b

d upper d lower



3/8 3 1/8

WL 3WL WL   2EA 8EA 8EA ;

L

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Axially Loaded Members

Problem 2.3-13 A flat bar of rectangular cross section, b2

length L, and constant thickness t is subjected to tension by forces P (see figure). The width of the bar varies linearly from b1 at the smaller end to b2 at the larger end. Assume that the angle of taper is small.

t

(a) Derive the following formula for the elongation of the bar: d

b2 PL ln Et(b2  b1) b1

P

b1 L

P

(b) Calculate the elongation, assuming L  5 ft, t  1.0 in., P  25 k, b1  4.0 in., b2  6.0 in., and E  30  106 psi.

Solution 2.3-13 Tapered bar (rectangular cross section)

t  thickness (constant) L0 + L x b  b1 a b b2  b1 a b L0 L0 x A(x)  bt  b1 t a b L0

From Eq. (1): (Eq. 1)

Solve Eq. (3) for L0: L0  La

PL0 dx Pdx  EA(x) Eb1 tx L0L

d

LL0

d

b1 b b2  b1

b2 PL ln Et (b2  b1) b1

(Eq. 4)

(Eq. 5)

(b) SUBSTITUTE NUMERICAL VALUES:

PL0 L0L dx dd  Eb1 t LL0 x

L0L PL0 L0 + L PL0 ln x `  ln  Eb1 t Eb1 t L0 L0

(Eq. 3)

Substitute Eqs. (3) and (4) into Eq. (2):

(a) ELONGATION OF THE BAR dd 

L0 + L b2  L0 b1

L  5 ft  60 in.

(Eq. 2)

t  10 in.

P  25 k

b1  4.0 in.

b2  6.0 in.

E  30  106 psi

From Eq. (5):   0.010 in.

;

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SECTION 2.3

Changes in Lengths under Nonuniform Conditions

Problem 2.3-14 A post AB supporting equipment in a laboratory is tapered uniformly throughout its height H (see figure). The cross sections of the post are square, with dimensions b  b at the top and 1.5b  1.5b at the base. Derive a formula for the shortening  of the post due to the compressive load P acting at the top. (Assume that the angle of taper is small and disregard the weight of the post itself.)

117

P

A

A

b

b H

B B 1.5b

Solution 2.3-14

Tapered post Ay  cross sectional area at distance y  1by22 

b2 H2

(H + 0.5y)2

SHORTENING OF ELEMENT dy dd 

Pdy  EAy

Pdy 2

Ea

b

H2

b 1H + 0.5y22

SHORTENING OF ENTIRE POST d

Square cross sections b  width at A 1.5b  width at B by  width at distance y y  b + (1.5b  b) H b  1H + 0.5y2 H

L

dd 

PH2

 

dy

Eb2 L0 (H + 0.5y)2

From Appendix C: d

H

dx L (a + bx)2

 

PH2

H 1 c  d (0.5)(H + 0.5y) 0 Eb2

PH2 2

c

Eb 2PH

3Eb2

1 1 + d (0.5)(1.5H ) 0.5H ;

1 b(a + bx)

1.5b

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Axially Loaded Members

Problem 2.3-15 A long, slender bar in the shape of a right circular cone

d

with length L and base diameter d hangs vertically under the action of its own weight (see figure). The weight of the cone is W and the modulus of elasticity of the material is E. Derive a formula for the increase  in the length of the bar due to its own weight. (Assume that the angle of taper of the cone is small.) L

Solution 2.3-15

Conical bar hanging vertically ELEMENT OF BAR

W  weight of cone

TERMINOLOGY

ELONGATION OF ELEMENT dy Ny dy Wy dy 4W y dy   dd  E Ay E ABL pd2 EL

Ny  axial force acting on element dy

ELONGATION OF CONICAL BAR

Ay  cross-sectional area at element dy AB  cross-sectional area at base of cone 

pd2 4

1  ABL 3

d

L

dd 

4W pd2 EL L0

L

y dy 

2WL

;

pd2 E

V  volume of cone Vy  volume of cone below element dy

1  Ay y Wy  weight of cone below element dy 3 Ay yW Vy Ny  Wy  (W )  V AB L

x P

Problem 2.3-16 A uniformly tapered plastic tube AB of circular

dA

B

A

P L

cross section and length L is shown in the figure. The average diameters at the ends are dA and dB  2dA. Assume E is constant. Find the elongation  of the tube when it is subjected to loads P acting at the ends. Use the following numerial data: dA  35 mm, L  300 mm, E  2.1 GPa, P  25 kN. Consider two cases as follows: (a) A hole of constant diameter dA is drilled from B toward A to form a hollow section of length x  L/2 (see figure part a).

dA

dB (a)

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SECTION 2.3

119

Changes in Lengths under Nonuniform Conditions

(b) A hole of variable diameter d(x) is drilled from B toward A to form a hollow section of length x  L/2 and constant thickness t (see figure part b). (Assume that t  dA/20.)

x P

B

A

P

dA

L d(x) t constant dB (b)

Solution 2.3-16 (a)

ELONGATION

 FOR CASE OF CONSTANT DIAMETER HOLE

d()  dA a1 +

d

1 P a d b E L A()

P d E

d

d

 b L

J

p d()2

solid portion of length L-x 4 p

hollow portion of length x A()  (d()2  dA 2) 4

A() 

d

P c E L0

 2 p c cdA a1 + b d d 4 L

L0

P L2 4 + E ( 2  x)pdA 2

J

L

4 pd()2

d +

JJ

4

d +

L pdA2

LLx L

+

LLx

4

LLx p1 d()2  d A 22

L

1

Lx

Lx

d d

1 c

 2 p c cdA a 1 + b d  dA 2 d d 4 L

d

1 d  2 p c c cdA a 1 + b d  dA 2 d d K K K 4 L

ln(Lx) + ln(3Lx ln(3) L L2 P a4 2L + 2L bd c4 2 + 2 2 E ( 2  x)pdA pdA pdA pdA 2

if x  L/2

d

ln(3) P 4 L ±  2L 2  2L E 3 pd2A pdA

Substitute numerical data d  2.18 mm

;

K

5 1  lna Lb + lna Lb 2 2 p d2A

≤

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(b)

Axially Loaded Members

ELONGATION

 FOR CASE OF VARIABLE DIAMETER HOLE BUT CONSTANT WALL THICKNESS t  dA/20 OVER SEGMENT x

d()  dA a1 +

d

d

d

Page 120

 b L

P 1 a d b E L A()

P ≥ E L0

Lx

A() 

p d()2 4

A() 

dA 2 p cd()2  a d()  2 b d 4 20

d

Lx

 L

bd

d +

hollow portion of length x

L

4 pd()2

d +

LLx

L

4 pcdA a1 +

P ≥ E L0

solid portion of length L-x

4 dA 2 pc d()  a d()  2 b d 20 2

4

LLx

pc c dA a 1 +

  dA 2 b d  cdA a 1 + b  2 d d L L 20 2

ln(3) + ln(13) + 2ln( dA) + ln( L) L2 P L  20L c4 + 4 2 2 E ( 2L + x)pdA pdA pdA 2  20L

2ln( dA) + ln (39L  20x) pdA 2

d

if x  L/2 d

ln(3) + ln(13) + 2ln( dA) + ln( L) 2ln( dA) + ln(29L) P 4 L + 20L b a  20L 2 2 E 3 pdA pdA pdA 2

Substitute numerical data d  6.74 mm

;

d¥

d ¥

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SECTION 2.3

121

Changes in Lengths under Nonuniform Conditions

Problem 2.3-17 The main cables of a suspension bridge [see part (a) of the figure] follow a curve that is nearly parabolic because the primary load on the cables is the weight of the bridge deck, which is uniform in intensity along the horizontal. Therefore, let us represent the central region AOB of one of the main cables [see part (b) of the figure] as a parabolic cable supported at points A and B and carrying a uniform load of intensity q along the horizontal. The span of the cable is L, the sag is h, the axial rigidity is EA, and the origin of coordinates is at midspan.

(a) y A

(a) Derive the following formula for the elongation of cable AOB shown in part (b) of the figure: d

L — 2

L — 2

B h

qL3 16h2 (1 + ) 8hEA 3L2

O

q

(b) Calculate the elongation  of the central span of one of the main cables of the Golden Gate Bridge, for which the dimensions and properties are L  4200 ft, h  470 ft, q  12,700 lb/ft, and E  28,800,000 psi. The cable consists of 27,572 parallel wires of diameter 0.196 in.

x

(b)

Hint: Determine the tensile force T at any point in the cable from a free-body diagram of part of the cable; then determine the elongation of an element of the cable of length ds; finally, integrate along the curve of the cable to obtain an equation for the elongation .

Solution 2.3-17

Cable of a suspension bridge dy 8hx  2 dx L FREE-BODY DIAGRAM OF HALF OF CABLE MB  0 哵哴  Hh + H

qL L a b 0 2 4

qL2 8h

Fhorizontal  0 HB  H 

qL2 8h

(Eq. 1)

Fvertical  0 VB  Equation of parabolic curve: y

4hx2 L2

qL 2

(Eq. 2)

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Axially Loaded Members

FREE-BODY DIAGRAM OF SEGMENT DB OF CABLE

dd 

Tds EA

ds  2(dx)2 + (dy)2  dx 1 + a

A

 dx 1 + a

A

 dx 1 +

8hx L2

b

dy 2 b dx

2

64h2x2

A

(Eq. 6)

L4

(a) ELONGATION  OF CABLE AOB d ©F horiz  0

TH  HB

©F vert  0 VB  Tv  q a Tv  VB  q a



qL2 8h

(Eq. 3)

L  xb  0 2



qL2 1 64h 2x 2 a1 + b dx EA L 8h L4 For both halves of cable: (Eq. 4)

qL2 2 b + (qx)2 A 8h a

qL2 64h2x2 1 + 8h A L4

ELONGATION d OF AN ELEMENT OF LENGTH ds

T ds L EA

Substitute for T from Eq. (5) and for ds from Eq. (6):

(Eq. 5)

L/2

qL2 64h2x2 a1 + b dx 8h L4

d

2 EA L0

d

qL3 16h2 a1 + b 8hEA 3L4

TENSILE FORCE T IN CABLE T  2T2H + T2v 

dd 

d

qL qL L  xb   + qx 2 2 2

 qx

L

;

(b) GOLDEN GATE BRIDGE CABLE L  4200 ft

h  470 ft

q  12,700 lb/ft E  28,800,000 psi 27,572 wires of diameter d  0.196 in. p A  (27,572)a b(0.196 in.)2  831.90 in.2 4 Substitute into Eq. (7):

  133.7 in  11.14 ft

;

(Eq. 7)

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SECTION 2.3

123

Changes in Lengths under Nonuniform Conditions

Problem 2.3-18 A bar ABC revolves in a horizontal plane about a

A

vertical axis at the midpoint C (see figure). The bar, which has length 2L and cross-sectional area A, revolves at constant angular speed . Each half of the bar (AC and BC) has weight W1 and supports a weight W2 at its end. Derive the following formula for the elongation of one-half of the bar (that is, the elongation of either AC or BC):

C

v

W1

W2

B W1

L

W2

L

L22 (W + 3W2) 3gEA 1 in which E is the modulus of elasticity of the material of the bar and g is the acceleration of gravity. d

Solution 2.3-18 Rotating bar Centrifugal force produced by weight W2  a

W2 b(L2) g

AXIAL FORCE F(x) F(x) 

  angular speed A  cross-sectional area



E  modulus of elasticity g  acceleration of gravity F(x)  axial force in bar at distance x from point C Consider an element of length dx at distance x from point C. To find the force F(x) acting on this element, we must find the inertia force of the part of the bar from distance x to distance L, plus the inertia force of the weight W2. Since the inertia force varies with distance from point C, we now must consider an element of length d at distance , where  varies from x to L. d W1 b Mass of element d  a L g Acceleration of element  2 Centrifugal force produced by element  ( mass)( acceleration) 

W12 d gL

L

Lx

W12 W2L2 d + gL g

W12 2 W2L2 (L  x2) + 2gL g

ELONGATION OF BAR BC L

F(x) dx L0 EA L L W12 2 W2L2dx (L  x2)dx +  gEA L0 L0 2gL L L 2 W2L2dx L W1L 2 2 c L dx  x dx d + dx  2gLEA L0 gEA L0 L0

d



W2L22 W1L22 + 3gEA gEA



L22 + (W1 + 3W2) 3gEA

;

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Statically Indeterminate Structures Problem 2.4-1 The assembly shown in the figure consists of a brass

core (diameter d1  0.25 in.) surrounded by a steel shell (inner diameter d2  0.28 in., outer diameter d3  0.35 in.). A load P compresses the core and shell, which have length L  4.0 in. The moduli of elasticity of the brass and steel are Eb  15  106 psi and Es  30  106 psi, respectively. (a) What load P will compress the assembly by 0.003 in.? (b) If the allowable stress in the steel is 22 ksi and the allowable stress in the brass is 16 ksi, what is the allowable compressive load Pallow? (Suggestion: Use the equations derived in Example 2-5.)

Solution 2.4-1

Cylindrical assembly in compression Substitute numerical values: Es As + Eb Ab  (30 * 106 psi)(0.03464 in.2) + (15 * 106 psi)(0.04909 in.2)  1.776 * 106 lb P  (1.776 * 106 lb)a  1330 lb

0.003 in. b 4.0 in.

;

(b) ALLOWABLE LOAD

d1  0.25 in.

Eb  15  106 psi

d2  0.28 in.

Es  30  106 psi

d3  0.35 in.

As 

L  4.0 in.

p 2 (d3  d22)  0.03464 in.2 4

p Ab  d21  0.04909 in.2 4

(a) DECREASE IN LENGTH (  0.003 in.) Use Eq. (2-13) of Example 2-5. d

PL Es As + Eb Ab

or

d P  (Es As + Es Ab)a b L

s  22 ksi b  16 ksi Use Eqs. (2-12a and b) of Example 2-5. For steel: ss 

PEs Es As + Eb Ab

Ps  (Es As + Eb Ab)

Ps  (1.776 * 106 lb)a

22 ksi 30 * 106 psi

ss Es

b  1300 lb

For brass: sb 

PEb Es As + Eb Ab

Ps  (Es As + Eb Ab)

Ps  (1.776 * 106 lb)a Steel governs.

16 ksi 15 * 106 psi

Pallow  1300 lb

sb Eb

b  1890 lb

;

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SECTION 2.4

125

Statically Indeterminate Structures

Problem 2.4-2 A cylindrical assembly consisting of a brass core and an aluminum collar is compressed by a load P (see figure). The length of the aluminum collar and brass core is 350 mm, the diameter of the core is 25 mm, and the outside diameter of the collar is 40 mm. Also, the moduli of elasticity of the aluminum and brass are 72 GPa and 100 GPa, respectively. (a) If the length of the assembly decreases by 0.1% when the load P is applied, what is the magnitude of the load? (b) What is the maximum permissible load Pmax if the allowable stresses in the aluminum and brass are 80 MPa and 120 MPa, respectively? (Suggestion: Use the equations derived in Example 2-5.)

Solution 2.4-2

Cylindrical assembly in compression d

PL Ea Aa + Eb Ab

or

d P  (Ea Aa + Eb Ab)a b L Substitute numerical values: Ea Aa + Eb Ab  (72 GPa)(765.8 mm2) (100 GPa)(490.9 mm2)  55.135 MN + 49.090 MN  104.23 MN P  (104.23 MN)a  104.2 kN

A  aluminum

0.350 mm b 350 mm

;

B  brass

(b) ALLOWABLE LOAD

L  350 mm

a  80 MPa b  120 MPa

da  40 mm

Use Eqs. (2-12a and b) of Example 2-5.

db  25 mm

For aluminum:

p Aa  (d2a  d2b) 4

sa 

 765.8 mm2 Ea  72 GPa

Eb  100 GPa

 490.9 mm2

p Ab  d2b 4

(a) DECREASE IN LENGTH (  0.1% of L  0.350 mm) Use Eq. (2-13) of Example 2-5.

PEa Ea Aa + Eb Ab

Pa  (104.23 MN)a

Pa  (Ea Aa + Eb Ab) a

sa b Ea

80 MPa b  115.8 kN 72 GPa

For brass: sb 

PEb Ea Aa + Eb Ab

Pb  (104.23 MN)a

Pb  (Ea Aa + Eb Ab) a 120 MPa b  125.1 kN 100 GPa

Aluminum governs. Pmax  116 kN

;

sb b Eb

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Problem 2.4-3 Three prismatic bars, two of material A and one of material B, transmit a tensile load P (see figure). The two outer bars (material A) are identical. The cross-sectional area of the middle bar (material B) is 50% larger than the cross-sectional area of one of the outer bars. Also, the modulus of elasticity of material A is twice that of material B. (a) What fraction of the load P is transmitted by the middle bar? (b) What is the ratio of the stress in the middle bar to the stress in the outer bars? (c) What is the ratio of the strain in the middle bar to the strain in the outer bars?

Solution 2.4-3

Prismatic bars in tension

FREE-BODY DIAGRAM OF END PLATE

STRESSES:

EQUATION OF EQUILIBRIUM Fhoriz  0

PA  PB  P  0

PA EA P  AA EA AA + EB AB

sB 

PB EB P  AB EA AA + EB AB

(a) LOAD IN MIDDLE BAR

(2)

PB EB AB 1   EA AA P EA AA + EB AB + 1 EB AB Given:

FORCE-DISPLACEMENT RELATIONS AA  total area of both outer bars

‹ PA L dA  EA Ak

PB L dB  EB AB

(3)

Substitute into Eq. (2):

EA 2 EB

PB  P

AA 4 1 + 1   AB 1.5 3

1 3 1   8 11 EA AA + 1 a ba b + 1 3 EB AB

(b) RATIO OF STRESSES

PA L PB L  EA AA EB AB

(4)

sB EB 1   sA EA 2

;

SOLUTION OF THE EQUATIONS (c) RATIO OF STRAINS

Solve simultaneously Eqs. (1) and (4): EA AAP PA  EA AA + EB AB

EB AB P PB  EA AA + EB AB

All bars have the same strain (5)

Substitute into Eq. (3): d  dA  dB 

PL EA AA + EB AB

(7)

(1)

EQUATION OF COMPATIBILITY

A  B

sA 

(6)

Ratio  1

;

;

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SECTION 2.4

Problem 2.4-4 A circular bar ACB of diameter d having a cylindrical hole of length x and diameter d/2 from A to C is held between rigid supports at A and B. A load P acts at L/2 from ends A and B. Assume E is constant. (a) Obtain formulas for the reactions RA and RB at supports A and B, respectively, due to the load P (see figure part a). (b) Obtain a formula for the displacement  at the point of load application (see figure part a). (c) For what value of x is RB  (6/5) RA? (See figure part a.) (d) Repeat (a) if the bar is now tapered linearly from A to B as shown in figure part b and x  L/2. (e) Repeat (a) if the bar is now rotated to a vertical position, load P is removed, and the bar is hanging under its own weight (assume mass density  ). (See figure part c.) Assume that x  L/2

127

Statically Indeterminate Structures

P, d

L — 2 d — 2

d RA

RB B

C

A x

L–x (a)

d dB = — 2

d — 2

dA = d RA

RB

C P, d

A x

L–x L — 2

L — 2 (b)

RB B L–x

C d — 2

x

d A RA

(c)

B P applied L at — 2

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Solution 2.4-4 (a) reactions at A & B due to load P at L/2 AAC 

p 2 d 2 cd  a b d 4 2

ACB 

p 2 d 4

AAC 

3 2 pd 16

select RB as the redundant; use superposition and a compatibility equation at B

Px  B1a  + EA AC

if x L/2

Pa

L  xb 2

L x P x 2 ≤ + B1a  ± p 2 E 3 d pd 2 4 16

ACB

2 2x + 3L B1a  P 3 Epd2 P B1b 

if x  L/2

L 2

EA AC

P  B1b  Ea

L 2

3 pd 2 b 16

B1b 

8 PL 3 Epd2

the following expression for  B2 is good for all x

B2 

RB x Lx + b a E AAC ACB

B2 

RB 16 x Lx + 4 b a 2 E 3 pd pd 2

B2 

RB E

x Lx + p 2 3 P pd2 d Q 4 16

(a.1) solve for RB and RA assuming that x L/2

compatibility:

B1a  B2  0

2 2x + 3L a P b 3 pd 2 RBa  16 x Lx a + 4 b 3 pd 2 pd 2

RBa 

 1 2x + 3L P 2 x + 3L

;

^ check  if x  0, RB  P/2 statics:

RAa  P  RBa

RAa  P 

1 2x + 3L P 2 x + 3L

RAa 

3 L P 2 x + 3L

;

^ check  if x  0, RAa  P/2

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SECTION 2.4

Statically Indeterminate Structures

129

(a.2) solve for RB and RA assuming that x  L/2

B1b  B2  0

compatibility:

RBb

8 PL 3 pd 2  16 x Lx a + 4 b 2 3 pd pd 2

RBb 

2PL x + 3L

;

^ check  if x  L, RB  P/2 RAb   P  a

RAb  P  RBb

statics:

2PL b x + 3L

RAb  P

x + L x + 3L

;

axial force for segment 0 to L/2  RA &   elongation of this segment

(b) find  at point of load application; (b.1) assume that x L/2

da 

RAa x E P AAC

da  PL

L x 2 + ACB Q

2x + 3L 2

(x + 3L)Epd

da 

a

for x  L/2

3 L P b 2 x + 3L E

da 

P 8 L 7 Epd2

L x 2 ± ≤ + p 2 3 d pd2 4 16 x

;

(b.2) assume that x  L/2

b 

1 RAb2 EAAC

L 2

b 

aP

x + L L b x + 3L 2

Ea

3 pd 2 b 16

b 

for x  L/2

8 x + L L Pa b 3 x + 3L Epd 2

8 L b  P 7 Epd 2

;

same as a above (OK)

(c) For what value of x is R B  (6/5) RA? Guess that x L/2 here & use RBa expression above to find x 1 2x + 3L 6 3 L P  a P b 0 2 x + 3L 5 2 x + 3L

 1 10x  3L P 0 10 x + 3L

x

Now try RBb  (6/5)RAb assuming that x L/2 2PL 6 x + L  a P b 0 x + 3L 5 x + 3L So, there are two solutions for x.

2 2L + 3x P 0 5 x + 3L

x

2 L 3

;

3L 10

;

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(d) repeat (a) above for tapered bar & x  L/2 outer diameter d(x)  da1 

AAC 

x b 2L

p d 2 c d(x)2  a b d 4 2

0 … x …

ACB 

p d(x)2 4

ACB 

1 d 2 p a b (4L2  4Lx + x2) 16 L

L … x … L 2

L 2

AAC 

p x 2 d 2 c c da 1  bd  a b d 4 2L 2

AAC 

1 d 2 pa b (3L2  4Lx + x2) 16 L

x 2 p c da 1  bd 4 2L

ACB 

As in (a), use superposition and compatibility to find redundant RB & then RA d B1 

 B1 

P L2 1 d E L0 AAC

 8PL Ep d

2

d B1 

(  ln(5) + ln(3))

P L2 E L0

1 1 d 2 c pa b (3L2  4L + 2) d 16 L

d

PL

 B1  1.301

Ed 2

L

d B2 

L RB 1 2 1 d + d b a L E L0 AAC A L CB 2

L

L

2 RB 1 1 ≥ d  d¥ d B2  2 2 E L d LL 1 2 2 0 1 p a d b 13L2  4L + 22 pa b 14L  4L +  2 2 16 L 16 L

 B2 

8L

RB ( 3 ln(5) + 3 ln(3)  2) 31p d 2 E 2

compatibility: B1  B2  0

statics:

RA  P  RB

RB 

 B2  2.998

 a1.301

PL

Ed2 L a 2.998 b Ed2

b

RA  (P  0.434P)

RBL Ed 2

RB  0.434P

;

RA  0.566P

;

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131

(e) Find reactions if the bar is now rotated to a vertical position, load P is removed, and the bar is hanging under its own weight (assume mass density  ). Assume that x  L/2. AAC 

3 pd2 16

ACB 

p 2 d 4

select RB as the redundant; use superposition and a compatibility equation at B from (a) above: dB2 

compatibility: B1  B2  0

RB x Lx a + b E AAC ACB

 B1 

for x  L/2, dB2 

RB 14 L a b E 3 pd 2

L 2

L NAC NCB d  d L EA EA AC CB L0 L2

WAC  rgAAC

where axial forces in bar due to self weight are: (assume  is measured upward from A) NAC   crgACB

L L + rgAAC a  b d 2 2

AAC 

3 pd2 16

L 2

WCB  rgACB

ACB 

p 2 d 4

NCB  [ gACB(L  )] NAC 

 B1 

1 3 1 rgp d2 L  rgp d2 a L  b 8 16 2 L 2

1 3 1 rgpd2L  rgpd2 a L  b 8 16 2

L0

 B1  a

3 E a pd2 b 16

L2 1 L2  11 rg + rg b 24 E 8 E

 B1 

1 NCB   c rgp d2( L  ) d 4 L

d +

L2 7 rg 12 E

compatibility: B1  B2  0

RB 

statics:

a

7 L2 rg b 12 E

1 rgpd 2L 8

;

RA  (WAC  WCB)  RB

RA  c crga RA 

RB 

14 L a b 3 Epd2

3 L p L 1 pd2 b + rga d2 b d  rgpd2L d 16 2 4 2 8

3 rgp d 2 L 32

;

LL2

1  c rgpd2 (L) d 4 Ea p4 d2 b 7  0.583 12

d

L 2

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Problem 2.4-5 Three steel cables jointly support a load of 12 k (see figure). The diameter of the middle cable is 3/4 in. and the diameter of each outer cable is 1/2 in. The tensions in the cables are adjusted so that each cable carries one-third of the load (i.e., 4 k). Later, the load is increased by 9 k to a total load of 21 k. (a) What percent of the total load is now carried by the middle cable? (b) What are the stresses M and O in the middle and outer cables, respectively? (NOTE: See Table 2-1 in Section 2.2 for properties of cables.)

Solution 2.4-5

Three cables in tension SECOND LOADING P2  9 k (additional load)

AREAS OF CABLES (from Table 2-1) Middle cable: AM  0.268 in.2

EQUATION OF EQUILIBRIUM Fvert  0

2PO  PM  P2  0

Outer cables: AO  0.119 in.2

M  O

(for each cable)

FORCE-DISPLACEMENT RELATIONS

FIRST LOADING P1 P1  12 k aEach cable carries or 4 k.b 3

(1)

EQUATION OF COMPATIBILITY

dM 

PML EAM

(2)

dO 

Po L EAo

(3, 4)

SUBSTITUTE INTO COMPATIBILITY EQUATION: PML POL  EAM EAO

PM PO  AM AO

(5)

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SECTION 2.4

SOLVE SIMULTANEOUSLY EQS. (1) AND (5): PM  P2 a

133

(a) PERCENT OF TOTAL LOAD CARRIED BY MIDDLE CABLE 2

AM 0.268 in. b  (9 k)a b AM + 2AO 0.506 in.2

 4.767 k Po  P2 a

Statically Indeterminate Structures

AM

Percent 

8.767 k (100%)  41.7% 21 k

(b) STRESSES IN CABLES (  P/A) Ao 0.119 in.2 b  (9 k)a b + 2AO 0.506 in.2

 2.117 k

Middle cable: sM 

Outer cables: sO 

FORCES IN CABLES

8.767 k 0.268 in.2 6.117 k 0.119 in.2

 32.7 ksi

 51.4 ksi

Middle cable: Force  4 k  4.767 k  8.767 k Outer cables: Force  4 k  2.117 k  6.117 k (for each cable)

Problem 2.4-6 A plastic rod AB of length L  0.5 m has a

diameter d1  30 mm (see figure). A plastic sleeve CD of length c  0.3 m and outer diameter d2  45 mm is securely bonded to the rod so that no slippage can occur between the rod and the sleeve. The rod is made of an acrylic with modulus of elasticity E1  3.1 GPa and the sleeve is made of a polyamide with E2  2.5 GPa. (a) Calculate the elongation  of the rod when it is pulled by axial forces P  12 kN. (b) If the sleeve is extended for the full length of the rod, what is the elongation? (c) If the sleeve is removed, what is the elongation?

Solution 2.4-6

;

Plastic rod with sleeve

P  12 kN

d1  30 mm

b  100 mm

L  500 mm

d2  45 mm

c  300 mm

Rod: E1  3.1 GPa Sleeve: E2  2.5 GPa

Rod: A1 

pd21  706.86 mm2 4

Sleeve: A2 

p 2 (d  d12)  883.57 mm2 4 2

E1A1  E2A2  4.400 MN

;

;

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Axially Loaded Members

(b) SLEEVE AT FULL LENGTH

(a) ELONGATION OF ROD Part AC: dAC 

Pb  0.5476 mm E1A1

Part CD: d CD 

L 500 mm d  dCD a b  (0.81815 mm)a b c 300 mm  1.36 mm

PC E1A1E2A2

(c) SLEEVE REMOVED PL d  2.74 mm E1A1

 0.81815 mm (From Eq. 2-13 of Example 2-5)

  2AC  CD  1.91 mm

;

;

;

Problem 2.4-7 The axially loaded bar ABCD shown in the figure is held between P

(a) Derive formulas for the reactions RA and RD at the ends of the bar. (b) Determine the displacements B and C at points B and C, respectively. (c) Draw a diagram in which the abscissa is the distance from the left-hand support to any point in the bar and the ordinate is the horizontal displacement  at that point.

Solution 2.4-7

2A1

A1

rigid supports. The bar has cross-sectional area A1 from A to C and 2A1 from C to D. A

B L — 4

C L — 4

D L — 2

Bar with fixed ends

FREE-BODY DIAGRAM OF BAR

(a) REACTIONS Solve simultaneously Eqs. (1) and (6): RA 

2P 3

RD 

P 3

;

(b) DISPLACEMENTS AT POINTS B AND C dB  dAB 

EQUATION OF EQUILIBRIUM Fhoriz  0

RA  RD  P

(Eq. 1) dC  |dCD| 

EQUATION OF COMPATIBILITY

AB  BC  CD  0



FORCE-DISPLACEMENT EQUATIONS

dCD  

(RA  P)(L / 4) dBC  EA1

RD(L/2) E(2A1)

PL (To the right) 12EA1

;

(c) AXIAL DISPLACEMENT DIAGRAM (ADD) (Eqs. 3, 4) (Eq. 5)

SOLUTION OF EQUATIONS Substitute Eqs. (3), (4), and (5) into Eq. (2): (RA  P)(L) RAL RDL +  0 4EA1 4EA1 4EA1

RDL 4EA1

(Eq. 2)

Positive means elongation. RA(L/4) dAB  EA1

RAL PL  (To the right) 4EA1 6EA1

(Eq. 6)

;

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135

Statically Indeterminate Structures

Problem 2.4-8 The fixed-end bar ABCD consists of three prismatic segments, as shown in the figure. The end segments have cross-sectional area A1  840 mm2 and length L1  200 mm. The middle segment has cross-sectional area A2  1260 mm2 and length L2  250 mm. Loads PB and PC are equal to 25.5 kN and 17.0 kN, respectively. (a) Determine the reactions RA and RD at the fixed supports. (b) Determine the compressive axial force FBC in the middle segment of the bar.

Solution 2.4-8

Bar with three segments PB  25.5 kN

PC  17.0 kN

L1  200 mm

L2  250 mm

A1  840 mm

A2  1260 mm2

2

m  meter SOLUTION OF EQUATIONS Substitute Eqs. (3), (4), and (5) into Eq. (2):

FREE-BODY DIAGRAM

RA RA 1 1 a 238.095 b + a 198.413 b E m E m  EQUATION OF EQUILIBRIUM Fhoriz  0 : 

Simplify and substitute PB  25.5 kN:

; 

RA a 436.508

PB  RD  PC  RA  0 or RA  RD  PB  PC  8.5 kN

 5,059.53

AD  elongation of entire bar (Eq. 2)

FORCE-DISPLACEMENT RELATIONS dAB 

RAL1 RA 1  a238.05 b EA1 E m

dBC 

(RA  PB)L2 EA2

RA PB 1 1  a198.413 b  a198.413 b E m E m

1 1 b + RD a 238.095 b m m

(Eq. 1)

EQUATION OF COMPATIBILITY

AD  AB  BC  CD  0

RD PB 1 1 a 198.413 b + a 238.095 b  0 E m E m

kN m

(Eq. 6)

(a) REACTIONS RA AND RD Solve simultaneously Eqs. (1) and (6). From (1): RD  RA  8.5 kN

(Eq. 3)

Substitute into (6) and solve for RA: RA a 674.603

1 kN b  7083.34 m m

RA  10.5 kN (Eq. 4)

;

RD  RA  8.5 kN  2.0 kN

;

(b) COMPRESSIVE AXIAL FORCE FBC dCD 

RDL1 RD 1 a238.095 b  EA1 E m

(Eq. 5)

FBC  PB  RA  PC  RD  15.0 kN

;

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Problem 2.4-9 The aluminum and steel pipes shown in the figure are fastened to rigid supports at ends A and B and to a rigid plate C at their junction. The aluminum pipe is twice as long as the steel pipe. Two equal and symmetrically placed loads P act on the plate at C. (a) Obtain formulas for the axial stresses a and s in the aluminum and steel pipes, respectively. (b) Calculate the stresses for the following data: P  12 k, cross-sectional area of aluminum pipe Aa  8.92 in.2, cross-sectional area of steel pipe As  1.03 in.2, modulus of elasticity of aluminum Ea  10  106 psi, and modulus of elasticity of steel Es  29  106 psi.

Solution 2.4-9

Pipes with intermediate loads SOLUTION OF EQUATIONS Substitute Eqs. (3) and (4) into Eq. (2): RB(2L) RAL  0 Es As EaAa

(Eq. 5)

Solve simultaneously Eqs. (1) and (5): RA 

4Es As P EaAa + 2EsAs

RB 

2EaAaP EaAa + 2EsAs (Eqs. 6, 7)

(a) AXIAL STRESSES Aluminum: sa 

RB 2EaP  Aa EaAa + 2EsAs

; (Eq. 8)

Pipe 1 is steel. Pipe 2 is aluminum.

(compression) Steel: ss 

EQUATION OF EQUILIBRIUM Fvert  0

RA  RB  2P

(Eq. 1)

EQUATION OF COMPATIBILITY

AB  AC  CB  0

dBC  

RB(2L) EaAa

(Eq. 9)

(tension) P  12 k

Aa  8.92 in.2

Ea  10  106 psi

FORCE-DISPLACEMENT RELATIONS RAL EsAs

;

(b) NUMERICAL RESULTS (Eq. 2)

(A positive value of  means elongation.)

dAC 

RA 4EsP  As EaAa + 2EsAs

As  1.03 in.2

Es  29  106 psi

Substitute into Eqs. (8) and (9): (Eqs. 3, 4))

a  1,610 psi (compression) s  9,350 psi (tension)

;

;

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Statically Indeterminate Structures

137

Problem 2.4-10 A nonprismatic bar ABC is composed of two segments: AB of length L1 and cross-sectional area A1; and BC of length L2 and cross-sectional area A2. The modulus of elasticity E, mass density , and acceleration of gravity g are constants. Initially, bar ABC is horizontal and then is restrained at A and C and rotated to a vertical position. The bar then hangs vertically under its own weight (see figure). Let A1  2A2  A and L1  53 L, L2  25 L. (a) Obtain formulas for the reactions RA and RC at supports A and C, respectively, due to gravity. (b) Derive a formula for the downward displacement B of point B. (c) Find expressions for the axial stresses a small distance above points B and C, respectively.

RA A A1 L1

B Stress elements L2 A2 C

RC

Solution 2.4-10 (a) find reactions in 1-degree statically indeterminate structure use superposition; select RA as the redundant compatibility: A1  A2  0 segment weights:

WAB  gA1L1 WBC  gA2L2

find axial forces in each segment; use variable  measured from C toward A NAB   gA1(L1  L2   )

L2  L1  L2

NBC  [WAB  gA2(L2  )]

0  L2

displacement at A in released structure due to self weight L2

d A1 

L0 L2

d A1 

L0

dA1  c

L  L2

1 NBC d + E A2 LL2

NAB d E A1

L1  L2 rgA 1 L + L  2  [rgA1L1 + rgA21 L2  2] 1 1 2  d + d EA2 E A LL2 1

2L1  L2 L12 2L1L2  L22 2A1L1 A2L2 1 1 1 rgL2 rg  rgL2 bd  a 2 EA2 2 E 2 E

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d A1 

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Axially Loaded Members

 rg 12L2A1L1 + A2L22 + A2L122 2( EA2)

Next, displacement at A in released structure due to redundant RA

A2  RA(fAB  fBC)

d A2  RA a

L1 L2 + b EA1 EA2

enforce compatibility: A1  A2  0

solve for RA

rg (2L2A1L1 + A2L22 + A2L12) 2(EA2) RA  fAB  f B 2

RA 

A2L12 + 2A1L1L2 + A2L2 1 rgA1 2 L1A2 + L2A1

statics:

RC  WAB  WBC  RA

RC  crgA1L1 + rgA2L2 

;

A1 1 rg(2L2A1L1 + A2L22 + A2L12) d 2 L1A2 + L2A1 2

A1L1 2 + 2A2L1L2 + A1L2 1 rgA2 2 L1A2 + L2A1

RC 

A1  A

For

RA 

1 rgA 2

RC 

A2 

A 2

L1 

; 3L 5

L2 

A 3L 2 3L 2L A 2L 2 a b + 2A + a b 2 5 5 5 2 5 3L A 2L  A 5 2 5

A 1 rg 2 2

Aa

3L 2 2L 2 A 3L 2L b +2 + Aa b 5 2 5 5 5 3L A 2L + A 5 2 5

(b) use superposition to find displacement at point B due to RA L2

d B1 

d B1 

L 0

NBC d EA2

due to shortening of BC

 rgL2 (2A1L1 + A2L2) 2(EA2)

B2  RA(fBC)

dB2  RA a

L2 b EA 2

2L 5

RA 

37 rgAL 70

RC 

19 rgLA 70

B  B1  B2

37  0.529 70

;

;

19  0.271 70

where B1 is due to gravity and B2 is

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SECTION 2.4

dB  c

Statically Indeterminate Structures

2

A2L12 + 2A1L1L2 + A2L2 L2 1  rgL2 (2A1L1 + A2L2) + rgA1 a bd 2(EA2) 2 L1A2 + L2A1 EA2

A L  A 2L 2 dB   1 r g L2 L1 1 1 2 (L1A2  L2A1) E For

A1  A

A2 

1 2L 3L dB  rg 2 5 5

A 2

L1 

3L 5

3L A 2L + 5 2 5 3L A 2L a + Ab E 5 2 5

L2 

A

dB 

2L 5

L2  24 rg 175 E

;

(c) expressions for the average axial stresses a small distance above points B and C NB  axial force near B  RA  WAB 2

A2L12 + 2A1L1L2 + A2L2 1 b  r gA1L1 NB  a rgA1 2 L1A2 + L2A1 NB 

3L 37 rgAL  rgA 70 5

sB 

NB A

NC  RC

sB 

sC 

NB 

1 rgL 14 a

;

19 rgLAb 70 A 2

1 rgAL 14

sC 

 19 rgL 35

;

139

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Axially Loaded Members

Problem 2.4-11 A bimetallic bar (or composite bar) of square cross

section with dimensions 2b  2b is constructed of two different metals having moduli of elasticity E1 and E2 (see figure). The two parts of the bar have the same cross-sectional dimensions. The bar is compressed by forces P acting through rigid end plates. The line of action of the loads has an eccentricity e of such magnitude that each part of the bar is stressed uniformly in compression. (a) Determine the axial forces P1 and P2 in the two parts of the bar. (b) Determine the eccentricity e of the loads. (c) Determine the ratio 1/2 of the stresses in the two parts of the bar.

Solution 2.4-11

Bimetallic bar in compression

FREE-BODY DIAGRAM

(a) AXIAL FORCES

(Plate at right-hand end)

Solve simultaneously Eqs. (1) and (3): P1 

PE1 E1 + E2

P2 

PE2 E1 + E2

;

(b ECCENTRICITY OF LOAD P Substitute P1 and P2 into Eq. (2) and solve for e: e 

EQUATIONS OF EQUILIBRIUM F  0 P1  P2  P b b Pe + P1 a b  P2 a b  0 2 2

M  0 哵哴

(c) RATIO OF STRESSES (Eq. 2)

2  1 or

P2 P1  E2 E1

;

(Eq. 1)

EQUATION OF COMPATIBILITY

P1L P2L  E2A E1A

b(E2 E1) 2(E2  E1)

(Eq. 3)

s1 

P1 A

s2 

P2 A

s1 P1 E1   s2 P2 E2

;

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SECTION 2.4

Statically Indeterminate Structures

141

Problem 2.4-12 A rigid bar of weight W  800 N hangs from three equally

spaced vertical wires (length L  150 mm, spacing a  50 mm): two of steel and one of aluminum. The wires also support a load P acting on the bar. The diameter of the steel wires is ds  2 mm, and the diameter of the aluminum wire is da  4 mm. Assume Es  210 GPa and Ea  70 GPa.

a L

S

a A

S Rigid bar of weight W

(a) What load Pallow can be supported at the midpoint of the bar (x  a) if the allowable stress in the steel wires is 220 MPa and in the aluminum wire is 80 MPa? (See figure part a.) (b) What is Pallow if the load is positioned at x  a/2? (See figure part a.) (c) Repeat (b) above if the second and third wires are switched as shown in figure part b.

x P (a)

a L

S

a S

A Rigid bar of weight W

x P (b)

Solution 2.4-12 numerical data W  800 N a  50 mm dA  4 mm

L  150 mm dS  2 mm ES  210 GPa

EA  70 GPa

Sa  220 MPa AA 

p 2 d 4 A

AA  13 mm2

Aa  80 MPa AS 

p 2 d 4 S

AS  3 mm2

(a) Pallow at center of bar 1-degree stat-indet - use reaction (RA) at top of aluminum bar as the redundant compatibility: 1  2  0 d1 

P +W L b a 2 ESAS

statics:

2RS  RA  P  W

downward displacement due to elongation of each steel wire under P  W if alum. wire is cut at top

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d2  RA a

Page 142

Axially Loaded Members

L L + b 2E SAS EAAA

upward displ. due to shortening of steel wires & elongation of alum. wire under redundant RA enforce compatibility & then solve for RA

1  2

RA 

so

P +W L a b 2 ESAS

RA  ( P + W)

L L + 2ESAS EAAA

EAAA EAAA + 2ESAS

and

s Aa 

RA AA

now use statics to find RS P + W (P + W)

P + W  RA RS  2

RS 

EAAA EAAA + 2ESAS

2

RS  (P + W) and

ESAS EAAA + 2ESAS RS s Sa  AS

compute stresses & apply allowable stress values s Aa  ( P + W)

EA EAAA + 2ESAS

s Sa  ( P + W)

ES EAAA + 2ESAS

solve for allowable load P PAa  s Aa a

EAAA + 2ESAS b W EA

PAa  1713 N

PSa  sSa a

EAAA + 2ESAS b W ES

PSa  1504 N

lower value of P controls

; Pallow is controlled by steel wires

(b) Pallow if load P at x  a/2 again, cut aluminum wire at top, then compute elongations of left & right steel wires d 1L  a d1 

3P W L b + ba 4 2 ESAS

d 1L + d 1R 2

d1 

d 1R  a

P W L b + ba 4 2 ESAS

P +W L a b where 1  displ. at x  a 2 ESAS

Use 2 from (a) above d2  RA a

L L + b 2ESAS EAAA

so equating 1 & 2, solve for RA RA  ( P + W)

EAAA EAAA + 2ESAS

same as in (a) RSL 

RSL 

RA 3P W +  4 2 2 3P W +  4 2

stress in left steel wire exceeds that in right steel wire

(P + W)

EAAA EAAA + 2ESAS 2

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SECTION 2.4

RSL 

PEAAA + 6PESAS + 4WESAS 4EAAA + 8ESAS

s Sa 

Statically Indeterminate Structures

143

PEAAA + 6PESAS + 4WESAS 1 a b 4EAAA + 8ESAS AS

solve for Pallow based on allowable stresses in steel & alum. sSa(4ASEAAA + 8ESAS2 )  (4WESAS) EAAA + 6ESAS

PSa 

PSa  820 N

;

PAa  1713 N

same as in (a)

steel controls

(c) Pallow if wires are switched as shown & x  a/2 select RA as the redundant statics on the two released structures (1) cut alum. wire - apply P & W, compute forces in left & right steel wires, then compute displacements at each steel wire RSL  d1L 

P 2

RSR 

L P a b 2 ESAS

P +W 2

d1R  a

L P + Wb a b 2 ESAS

by geometry,  at alum. wire location at far right is

d1  a

P L b + 2Wb a 2 ESAS

(2) next apply redundant RA at right wire, compute wire force & displ. at alum. wire RSL  RA

RSR  2RA

d2  RA a

5L L + b ESAS EAAA

(3) compatibility equate 1, 2 and solve for RA then Pallow for alum. wire

RA 

a

P L + 2Wb a b 2 ES AS 5L L + ESAS EAAA

RA 

EAAAP + 4EAAAW 10EAAA + 2ESAS

sAa 

EAP + 4EAW 10EAAA + 2ESAS

PAa 

sAa(10EAAA + 2ESAS)  4EAW EA

s Aa 

RA AA

PAa  1713 N

(4) statics or superposition - find forces in steel wires then Pallow for steel wires RSL 

P + RA 2

RSL 

EAAAP + 4EAAAW P + 2 10EAAA + 2ESAS

RSL 

6EAAAP + PESAS + 4EAAAW 10EAAA + 2ESAS

larger than RSR below so use in allow. stress calcs

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RSR 

sSa 

Page 144

Axially Loaded Members

P + W  2RA 2

RSL AS

RSR 

EAAAP + 4EAAAW P + W 2 5EAAA + ES AS

RSR 

3EAAAP + PESAS + 2EAAAW + 2WESAS 10EAAA + 2ESAS

PSa  sSaAS a

10EAAA + 2ESAS 4EAAAW b 6EAAA + ESAS 6EAAA + ESAS 2

10sSaASEAAA + 2sSaA S E S  4EAAAW 6EAAA + ESAS

PSa 

PSa  703 N ^ steel controls

;

Problem 2.4-13 A horizontal rigid bar of weight W  7200 lb is supported by three slender circular rods that are equally spaced (see figure). The two outer rods are made of aluminum (E1  10  106 psi) with diameter d1  0.4 in. and length L1  40 in. The inner rod is magnesium (E2  6.5  106 psi) with diameter d2 and length L2. The allowable stresses in the aluminum and magnesium are 24,000 psi and 13,000 psi, respectively. If it is desired to have all three rods loaded to their maximum allowable values, what should be the diameter d2 and length L2 of the middle rod?

Solution 2.4-13

Bar supported by three rods BAR 1 ALUMINUM E1  10  106 psi

FREE-BODY DIAGRAM OF RIGID BAR EQUATION OF EQUILIBRIUM Fvert  0

d1  0.4 in.

2F1  F2  W  0

L1  40 in.

(Eq. 1)

FULLY STRESSED RODS

1  24,000 psi

F1  1A1

F2  2A2

BAR 2 MAGNESIUM E2  6.5  106 psi d2  ?

L2  ?

2  13,000 psi

A1  Substitute into Eq. (1): 2s1 a

pd21 4

b + s2 a

pd22 4

b W

pd21 4

A2 

pd22 4

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SECTION 2.4

Diameter d1 is known; solve for d2: d22

4W   ps2

2 2s1d1

s2

;

d2  (Eq. 2)

SUBSTITUTE NUMERICAL VALUES: d22 

2(24,000 psi)(0.4 in.)2 4(7200 lb)  p(13,000 psi) 13,000 psi

 0.70518 in2.  0.59077 in.2  0.11441 in.2 d2  0.338 in.

;

FORCE-DISPLACEMENT RELATIONS F1L1 L1  s1 a b d1  E1A1 E1

F2L2 L2  s2 a b E2A2 E2

(Eq. 5)

Substitute (4) and (5) into Eq. (3): L1 L2 s1 a b  s2 a b E1 E2 Length L1 is known; solve for L2: L2  L1 a

s1E2 b s2E1

;

(Eq. 6)

SUBSTITUTE NUMERICAL VALUES:

EQUATION OF COMPATIBILITY

1  2

145

Statically Indeterminate Structures

(Eq. 3)

L2  (40 in.) a

24,000 psi 6.5 * 106 psi ba b 13,000 psi 10 * 106 psi

 48.0 in. (Eq. 4)

Problem 2.4-14 A circular steel bar ABC (E  200 GPa) has cross-sectional area A1 from A to B and cross-sectional area A2 from B to C (see figure). The bar is supported rigidly at end A and is subjected to a load P equal to 40 kN at end C. A circular steel collar BD having cross-sectional area A3 supports the bar at B. The collar fits snugly at B and D when there is no load. Determine the elongation AC of the bar due to the load P. (Assume L1  2L3  250 mm, L2  225 mm, A1  2A3  960 mm2, and A2  300 mm2.)

A A1 L1 B L3

A3 D

A2

C P

L2

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Axially Loaded Members

Solution 2.4-14

Bar supported by a collar

FREE-BODY DIAGRAM OF BAR ABC AND COLLAR BD

SOLVE SIMULTANEOUSLY EQS. (1) AND (3): PL3A1 PL1A3 RA  RD  L1A3 + L3A1 L1A3 + L3A1 CHANGES IN LENGTHS (Elongation is positive) RAL1 PL1L3 PL2  dBC  dAB  EA1 E(L1A3 + L3A1) EA2 ELONGATION OF BAR ABC

AC  AB  AC SUBSTITUTE NUMERICAL VALUES: P  40 kN

E  200 GPa

L1  250 mm L2  225 mm L3  125 mm EQUILIBRIUM OF BAR ABC Fvert  0

RA  RD  P  0

A1  960 mm2 (Eq. 1)

COMPATIBILITY (distance AD does not change)

AB(bar)  BD(collar)  0

(Eq. 2)

A3  480 mm2 RESULTS:

(Elongation is positive.)

RA  RD  20 kN

FORCE-DISPLACEMENT RELATIONS RAL1 RDL3 dBD   dAB  EA1 EA3 Substitute into Eq. (2): RDL3 RAL1  0 EA1 EA3

A2  300 mm2

AB  0.02604 mm BC  0.15000 mm AC  AB  AC  0.176 mm

;

(Eq. 3)

Problem 2.4-15 A rigid bar AB of length L  66 in. is hinged to a support at A and supported by two vertical wires attached at points C and D (see figure). Both wires have the same cross-sectional area (A  0.0272 in.2) and are made of the same material (modulus E  30  106 psi). The wire at C has length h  18 in. and the wire at D has length twice that amount. The horizontal distances are c  20 in. and d  50 in. (a) Determine the tensile stresses C and D in the wires due to the load P  340 lb acting at end B of the bar. (b) Find the downward displacement B at end B of the bar.

2h h

A

C

D

B

c d

P L

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SECTION 2.4

Solution 2.4-15

147

Statically Indeterminate Structures

Bar supported by two wires EQUATION OF COMPATIBILITY dc dD  c d FORCE-DISPLACEMENT RELATIONS dC 

TCh EA

TD(2h) EA

dD 

(Eq. 2)

(Eqs. 3, 4)

SOLUTION OF EQUATIONS Substitute (3) and (4) into Eq. (2): TD(2h) TCh TC 2TD or   cEA dEA c d TENSILE FORCES IN THE WIRES

h  18 in.

(Eq. 5)

Solve simultaneously Eqs. (1) and (5):

2h  36 in.

TC 

c  20 in. d  50 in.

2cPL 2

2

2c + d

dPL

TD 

2

2c + d2

(Eqs. 6, 7)

TENSILE STRESSES IN THE WIRES TC 2cPL  sC  A A(2c2 + d2)

L  66 in. E  30  106 psi A  0.0272 in.2

sD 

P  340 lb FREE-BODY DIAGRAM

TD dPL  A A(2c2 + d2)

(Eq. 8)

(Eq. 9)

DISPLACEMENT AT END OF BAR 2hTD L L 2hPL2 dB  dD a b  a b  d EA d EA(2c2 + d 2)

(Eq. 10)

SUBSTITUTE NUMERICAL VALUES 2c2  d2  2(20 in.)2  (50 in.)2  3300 in.2 (a) sC 

2cPL 2

2

A(2c + d )



 10,000 psi DISPLACEMENT DIAGRAM

sD 

2

A(2c + d )



 12,500 psi (b) dB 

 EQUATION OF EQUILIBRIUM MA  0 哵 哴 TC (c)  TD(d)  PL

(Eq. 1)

(0.0272 in.2)(3300 in.2)

;

dPL 2

2(20 in.)(340 lb)(66 in.)

(50 in.)(340 lb)(66 in.) (0.0272 in.2)(3300 in.2)

;

2hPL2 EA(2c2 + d2) 2(18 in.)(340 lb)(66 in.)2 (30 * 106 psi)(0.0272 in.2)(3300 in.2)

 0.0198 in.

;

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Axially Loaded Members

Problem 2.4-16 A rigid bar ABCD is pinned at point B and supported by springs at A and D (see figure). The springs at A and D have stiffnesses k1  10 kN/m and k2  25 kN/m, respectively, and the dimensions a, b, and c are 250 mm, 500 mm, and 200 mm, respectively. A load P acts at point C. If the angle of rotation of the bar due to the action of the load P is limited to 3°, what is the maximum permissible load Pmax?

a = 250 mm A

b = 500 mm

B

C

D

P c = 200 mm

k 2 = 25 kN/m

k1 = 10 kN/m

Solution 2.4-16

Rigid bar supported by springs EQUATION OF EQUILIBRIUM MB  0   FA(a)  P(c)  FD(b)  0

NUMERICAL DATA a  250 mm

EQUATION OF COMPATIBILITY dA dD  a b FORCE-DISPLACEMENT RELATIONS FA FD dD  dA  k1 k2

b  500 mm

SOLUTION OF EQUATIONS

c  200 mm

Substitute (3) and (4) into Eq. (2): FA FD  ak1 bk2

k1  10 kN/m k2  25 kN/m p umax  3°  rad 60 FREE-BODY DIAGRAM AND DISPLACEMENT DIAGRAM

(Eq. 1)

(Eq. 2)

(Eqs. 3, 4)

(Eq. 5)

SOLVE SIMULTANEOUSLY EQS. (1) AND (5): ack1P bck2P FA  2 FD  2 2 a k1 + b k2 a k1 + b2k2 ANGLE OF ROTATION FD dD bcP cP dD   2  2 u k2 b a k1 + b2k2 a k1 + b2k2 MAXIMUM LOAD u P  (a2k1 + b2k2) c Pmax 

umax 2 (a k1 + b2k2) c

;

SUBSTITUTE NUMERICAL VALUES: Pmax 

p/60 rad [(250 mm)2(10 kN/m) 200 mm + (500 mm)2(25 kN/m)]

 1800 N

;

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SECTION 2.4

Problem 2.4-17 A trimetallic bar is uniformly compressed by an

axial force P  9 kips applied through a rigid end plate (see figure). The bar consists of a circular steel core surrounded by brass and copper tubes. The steel core has diameter 1.25 in., the brass tube has outer diameter 1.75 in., and the copper tube has outer diameter 2.25 in. The corresponding moduli of elasticity are Es  30,0000 ksi, Eb  16,000 ksi, and Ec  18,000 ksi. Calculate the compressive stresses ss, sb, and sc in the steel, brass, and copper, respectively, due to the force P.

Statically Indeterminate Structures

P=9k

Copper tube

149

Brass tube Steel core

1.25 in. 1.75 in. 2.25 in.

Solution 2.4-17 numerical properties (kips, inches) dc  2.25 in.

db  1.75 in.

Ec  18000 ksi

ds  1.25 in.

Eb  16000 ksi

Es  30000 ksi P  9 kips EQUATION OF EQUILIBRIUM Fvert  0

Ps  Pb  Pc  P

(Eq. 1)

EQUATIONS OF COMPATIBILITY

s  b

c  s

(Eqs. 2)

FORCE-DISPLACEMENT RELATIONS ds 

PsL PbL PcL db  dc  EsAs EbAb EcAc

(Eqs. 3, 4, 5)

SOLUTION OF EQUATIONS Substitute (3), (4), and (5) into Eqs. (2): Pb  Ps

EbAb EcAc P  Ps EsAs c EsAs

(Eqs. 6, 7)

As 

p 2 d 4 s

Ab 

p 2 1d  ds22 4 b

Ac 

p 1dc2  db22 4

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Axially Loaded Members

SOLVE SIMULTANEOUSLY EQS. (1), (6), AND (7): Ps  P

Es As  4 kips Es As + Eb Ab + Ec Ac

Pb  P

EbAb  2 kips Es As + Eb Ab + Ec Ac

Pc  P

Ec Ac Es As + Eb Ab + Ec Ac

Ps  Pb  Pc  9

 3 kips

statics check

COMPRESSIVE STRESSES Let EA  EsAs  EbAb  EcAc ss 

Ps PEs  As ©EA

sc 

Pc PEc  Ac ©EA

sb 

Pb PEb  Ab ©EA

compressive stresses

ss 

Ps As

s  3 ksi

;

sb 

Pb Ab

b  2 ksi

;

sc 

Pc Ac

c  2 ksi

;

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SECTION 2.5

Thermal Effects

151

Thermal Effects Problem 2.5-1 The rails of a railroad track are welded together at their ends (to form continuous rails and thus eliminate the clacking sound of the wheels) when the temperature is 60°F. What compressive stress  is produced in the rails when they are heated by the sun to 120°F if the coefficient of thermal expansion   6.5  106/°F and the modulus of elasticity E  30  106 psi?

Solution 2.5-1

Expansion of railroad rails

The rails are prevented from expanding because of their great length and lack of expansion joints. Therefore, each rail is in the same condition as a bar with fixed ends (see Example 2-7). The compressive stress in the rails may be calculated from Eq. (2-18).

T  120°F  60°F  60°F

  E(T)  (30  106 psi)(6.5  106/°F)(60°F)   11,700 psi ;

Problem 2.5-2 An aluminum pipe has a length of 60 m at a temperature of 10°C. An adjacent steel pipe at the same temperature is 5 mm longer than the aluminum pipe. At what temperature (degrees Celsius) will the aluminum pipe be 15 mm longer than the steel pipe? (Assume that the coefficients of thermal expansion of aluminum and steel are a  23  106/°C and s  12  106/°C, respectively.)

Solution 2.5-2

Aluminum and steel pipes

INITIAL CONDITIONS

or, a(T )La La  L s(T)Ls Ls

La  60 m

T0  10°C

Ls  60.005 m

T0  10°C

a  23  106/°C

s  12  106/°C

Solve for T: ¢T 

¢L + (Ls  La) aaLa  asLs

;

FINAL CONDITIONS

Substitute numerical values:

Aluminum pipe is longer than the steel pipe by the amount L  15 mm.

aLa  sLs  659.9  106 m/°C

T  increase in temperature

¢T 

a  a(T )La

s  s(T )Ls

15 mm + 5 mm 659.9 * 106 m/° C  30.31° C

T  T0 + ¢T  10°C + 30.31°C  40.3°C

From the figure above:

a La  L s Ls

;

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Axially Loaded Members

Problem 2.5-3 A rigid bar of weight W  750 lb hangs from three equally spaced wires, two of steel and one of aluminum (see figure). The diameter of the wires is 1/8 in . Before they were loaded, all three wires had the same length. What temperature increase T in all three wires will result in the entire load being carried by the steel wires? (Assume Es  30  106 psi, s  6.5  106/°F, and a  12  106/°F.)

Solution 2.5-3

Bar supported by three wires 1  increase in length of a steel wire due to temperature increase T S

A

 s (T)L

S

2  increase in length of a steel wire due to load W/2 

WL 2EsAs

3  increase in length of aluminum wire due to temperature increase T

W = 750 lb

 a(T)L S  steel

A  aluminum

W  750 lb

For no load in the aluminum wire:

1 2  3

d

1 in. 8

as(¢T)L +

As 

pd2  0.012272 in.2 4

or

Es  30  106 psi EsAs  368,155 lb

¢T 

6

a  12  10 /°F L  Initial length of wires

W 2EsAs(aa  as)

;

Substitute numerical values:

6

s  6.5  10 /°F

WL  aa(¢T)L 2EsAs

¢T 

750 lb

(2)(368,155 lb)(5.5 * 106/° F) ;  185°F

NOTE: If the temperature increase is larger than T, the aluminum wire would be in compression, which is not possible. Therefore, the steel wires continue to carry all of the load. If the temperature increase is less than T, the aluminum wire will be in tension and carry part of the load.

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SECTION 2.5

Problem 2.5-4 A steel rod of 15-mm diameter is held snugly (but without any initial stresses) between rigid walls by the arrangement shown in the figure. (For the steel rod, use   12  106/°C and E  200 GPa.) (a) Calculate the temperature drop T (degrees Celsius) at which the average shear stress in the 12-mm diameter bolt becomes 45 MPa.

Washer, dw = 20 mm

12-mm diameter bolt ∆T

B

A

18 mm

15 mm

Clevis, tc = 10 mm

(b) What are the average bearing stresses in the bolt and clevis at A and the washer (dw  20 mm) and wall (t wall  18mm) at B?

153

Thermal Effects

Solution 2.5-4 numerical properties dr  15 mm

db  12 mm

dw  20 mm

tc  10 mm

b  45 MPa

twall  18 mm 6

  12  (10 )

solve for T

E  200 GPa

¢T 

(a) TEMPERATURE DROP RESULTING IN BOLT SHEAR STRESS   T

  ET

T  24°C

p rod force  P  ( Ea¢T) d2r and bolt in double 4 P 2 shear with shear stress t  AS

t

P p 2 2 db 4

;

P  ( E a ¢T)

p 2 d 4 r

P  10.18 kN

(b) BEARING STRESSES

p so t b  c( Ea¢T) dr2 d 2 4 pdb 2

tb 

2t b db 2 a b E a dr

Ea¢T dr 2 a b 2 db

P 2 bolt and clevis s bc  bc  42.4 MPa dbtc washer at wall s bw 

bw  74.1 MPa

P p 1d 2  d r2 2 4 w

;

(a) Derive a formula for the compressive stress c in the bar. (Assume that the material has modulus of elasticity E and coefficient of thermal expansion ).

∆TB

∆T

Problem 2.5-5 A bar AB of length L is held between rigid supports and heated nonuniformly in such a manner that the temperature increase T at distance x from end A is given by the expression T  TBx3/L3, where TB is the increase in temperature at end B of the bar (see figure part a).

;

0 A

B x L (a)

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Axially Loaded Members

(b) Now modify the formula in (a) if the rigid support at A is replaced by an elastic support at A having a spring constant k (see figure part b). Assume that only bar AB is subject to the temperature increase.

∆TB

∆T 0 k

A

B x L (b)

Solution 2.5-5 (a) one degree statically indeterminate - use superposition select reaction RB as the redundant; follow procedure below Bar with nonuniform temperature change

COMPRESSIVE FORCE P REQUIRED TO SHORTEN THE BAR BY THE AMOUNT  EAd 1  EAa(¢TB) L 4

P

COMPRESSIVE STRESS IN THE BAR Ea(¢TB) P ;  A 4 (b) one degree statically indeterminate - use superposition select reaction RB as the redundant then compute bar elongations due to T & due to RB L due to temp. from above dB1  a¢TB 4 ac 

At distance x: ¢T  ¢TB a

x3 3

L

dB2  RB a

b

REMOVE THE SUPPORT AT THE END B OF THE BAR:

compatibility: solve for RB

RB 

Consider an element dx at a distance x from end A. d  Elongation of element dx dd  a(¢T)dx  a(¢TB)a

x3 L3

bdx

d  elongation of bar L

L

1 dd  a(¢TB) a 3 bdx  a(¢TB)L d 4 L L0 L0

 aa¢TB a

B1 B2  0

L b 4

1 L + b k EA

RB  a¢TB

EA EA J 4a kL + 1 b K

so compressive stress in bar is: sc 

x3

L 1 + b k EA

RB A

sc 

E a1¢TB2

4a

;

EA + 1b kL

NOTE: sc in (b) is the same as in (a) if spring const. k goes to infinity.

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Problem 2.5-6 A plastic bar ACB having two different solid circular cross sections is held between rigid supports as shown in the figure. The diameters in the left- and right-hand parts are 50 mm and 75 mm, respectively. The corresponding lengths are 225 mm and 300 mm. Also, the modulus of elasticity E is 6.0 GPa, and the coefficient of thermal expansion  is 100  106/°C. The bar is subjected to a uniform temperature increase of 30°C. (a) Calculate the following quantities: (1) the compressive force N in the bar; (2) the maximum compressive stress c; and (3) the displacement C of point C. (b) Repeat (a) if the rigid support at A is replaced by an elastic support having spring constant k  50 MN/m (see figure part b; assume that only the bar ACB is subject to the temperature increase).

SECTION 2.5

Thermal Effects

50 mm C

75 mm

A

225 mm

155

B

300 mm (a)

75 mm

50 mm C

A

k

225 mm

B

300 mm (b)

Solution NUMERICAL DATA d1  50 mm

d2  75 mm

L1  225 mm

L2  300 mm

T  30°C

k  50 MN/m

(a) COMPRESSIVE

FORCE

N,

MAX. COMPRESSIVE STRESS

L1 E A1

C  0.314 mm ; () sign means jt C moves left

  100  (106/°C

E  6.0 GPa

DISPL. OF PT.

d C  a ¢T( L1)  RB

&

(b) COMPRESSIVE FORCE N, MAX. COMPRESSIVE STRESS & DISPL. OF PT. C FOR ELASTIC SUPPORT CASE Use RB as redundant as in (a)

C

p 2 p d1 A2  d22 4 4 one-degree stat-indet - use RB as redundant

B1  T(L1 L2)

B1  T(L1 L2)

^ now add effect of elastic support; equate B1 and B2 then solve for RB

A1 

d B2  RB a

L1 L2 + b E A1 E A2

compatibility: B1  B2, solve for RB RB 

a¢T( L1 + L2) L1 L2 + E A1 E A2

N  RB

N  51.8 kN ; max. compressive stress in AC since it has the smaller area (A1 A2) N cmax  26.4 MPa A1 displacement C of point C  superposition of displacements in two released structures at C scmax 

dB2  RB a

RB 

L1 L2 1 + + b E A1 E A2 k

a¢T1 L1 + L22

L1 L2 1 + + E A1 EA2 k

N  31.2 kN

N  RB

;

N cmax  15.91 MPa A1 super position scmax 

d C  a¢T( L1)  RBa

C  0.546 mm moves left

;

L1 1 + b E A1 k ; () sign means jt C

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Problem 2.5-7 A circular steel rod AB (diameter d1  1.0 in., length L1 

3.0 ft) has a bronze sleeve (outer diameter d2  1.25 in., length L2  1.0 ft) shrunk onto it so that the two parts are securely bonded (see figure). Calculate the total elongation  of the steel bar due to a temperature rise T  500°F. (Material properties are as follows: for steel, Es  30  106 psi and s  6.5  106/°F; for bronze, Eb  15  106 psi and b  11  106/°F.)

Solution 2.5-7

Steel rod with bronze sleeve SUBSTITUTE NUMERICAL VALUES:

s  6.5  106/°F b  11  106/°F Es  30  106 psi

Eb  15  106 psi

d1  1.0 in. L1  36 in.

L2  12 in.

ELONGATION OF THE TWO OUTER PARTS OF THE BAR

1  s(T)(L1  L2) 6

 (6.5  10 /°F)(500°F)(36 in.  12 in.)  0.07800 in. ELONGATION OF THE MIDDLE PART OF THE BAR The steel rod and bronze sleeve lengthen the same amount, so they are in the same condition as the bolt and sleeve of Example 2-8. Thus, we can calculate the elongation from Eq. (2-21): d2 

As 

p 2 d  0.78540 in.2 4 1

d2  1.25 in. Ab 

p (d 2  d12)  0.44179 in.2 4 2

T  500°F

L2  12.0 in.

2  0.04493 in. TOTAL ELONGATION

  1 2  0.123 in.

(as Es As + ab Eb Ab)(¢T)L2 Es As + Eb Ab

Problem 2.5-8 A brass sleeve S is fitted over a steel bolt B (see figure), and the nut is tightened until it is just snug. The bolt has a diameter dB  25 mm, and the sleeve has inside and outside diameters d1  26 mm and d2  36 mm, respectively. Calculate the temperature rise T that is required to produce a compressive stress of 25 MPa in the sleeve. (Use material properties as follows: for the sleeve, S  21  106/°C and ES  100 GPa; for the bolt, B  10  106/°C and EB  200 GPa.) (Suggestion: Use the results of Example 2-8.)

;

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SECTION 2.5

Solution 2.5-8

Thermal Effects

Brass sleeve fitted over a Steel bolt sS ES AS b a1 + ES(aS  aB) EB AB

¢T 

;

SUBSTITUTE NUMERICAL VALUES:

S  25 MPa d2  36 mm Subscript S means “sleeve”. Subscript B means “bolt”.

ES  100 GPa

EQUATION (2-20A): (aS  aB)(¢T)ES EB AB (Compression) ES AS + EB AB SOLVE FOR T: sS 

¢T 

sS(ES AS + EB AB) (aS  aB)ES EB AB

dB  25 mm

EB  200 GPa

6

S  21  10 /°C

Use the results of Example 2-8.

S  compressive force in sleeve

d1  26 mm

B  10  106/°C

AS 

p 2 p (d  d21)  (620 mm2) 4 2 4

AB 

ES AS p p  1.496 (dB)2  (625 mm2) 1 + 4 4 EB AB

¢T 

25 MPa (1.496) (100 GPa)(11 * 106/°C)

T  34°C

;

(Increase in temperature)

or

Problem 2.5-9 Rectangular bars of copper and aluminum are held by pins at their ends, as shown in the figure. Thin spacers provide a separation between the bars. The copper bars have cross-sectional dimensions 0.5 in.  2.0 in., and the aluminum bar has dimensions 1.0 in.  2.0 in. Determine the shear stress in the 7/16 in. diameter pins if the temperature is raised by 100°F. (For copper, Ec  18,000 ksi and c  9.5  106/°F; for aluminum, Ea  10,000 ksi and a  13  106/°F.) Suggestion: Use the results of Example 2-8.

Solution 2.5-9

Rectangular bars held by pins

Diameter of pin: dP  Area of pin: AP 

7 in.  0.4375 in. 16

p 2 d  0.15033 in.2 4 P

Area of two copper bars: Ac  2.0 in.2 Area of aluminum bar: Aa  2.0 in.2 T  100°F

157

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Copper: Ec  18,000 ksi c  9.5  106/°F

SUBSTITUTE NUMERICAL VALUES:

Aluminum: Ea  10,000 ksi

Pa  Pc 

(3.5 * 106/° F)(100°F)(18,000 ksi)(2 in.2) 1 + a

a  13  106/°F Use the results of Example 2-8. Find the forces Pa and Pc in the aluminum bar and copper bar, respectively, from Eq. (2-19).

18 2.0 ba b 10 2.0

 4,500 lb FREE-BODY DIAGRAM OF PIN AT THE LEFT END

Replace the subscript “S” in that equation by “a” (for aluminum) and replace the subscript “B” by “c” (for copper), because  for aluminum is larger than  for copper. Pa  Pc 

(aa  ac)(¢T)Ea Aa Ec Ac Ea Aa + Ec Ac

Note that Pa is the compressive force in the aluminum bar and Pc is the combined tensile force in the two copper bars. Pa  Pc 

(aa  ac)(¢T)Ec Ac Ec Ac 1 + Ea Aa

V  shear force in pin  Pc/2  2,250 lb t  average shear stress on cross section of pin t

2,250 lb V  AP 0.15033 in.2

t  15.0 ksi

;

Problem 2.5-10 A rigid bar ABCD is pinned at end A and supported by two cables at points B and C (see figure). The cable at B has nominal diameter dB  12 mm and the cable at C has nominal diameter dC  20 mm. A load P acts at end D of the bar. What is the allowable load P if the temperature rises by 60°C and each cable is required to have a factor of safety of at least 5 against its ultimate load? (Note: The cables have effective modulus of elasticity E  140 GPa and coefficient of thermal expansion   12  106/°C. Other properties of the cables can be found in Table 2-1, Section 2.2.)

Solution 2.5-10

Rigid bar supported by two cables

FREE-BODY DIAGRAM OF BAR ABCD

From Table 2-1: AB  76.7 mm2 E  140 GPa T  60°C AC  173 mm2   12  106/°C EQUATION OF EQUILIBRIUM

MA  0 哵 哴 TB(2b) TC(4b)  P(5b)  0 (Eq. 1) or 2TB 4TC  5P

TB  force in cable B dB  12 mm

TC  force in cable C

dC  20 mm

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SECTION 2.5

DISPLACEMENT DIAGRAM

COMPATIBILITY:

C  2B

(Eq. 2)

FORCE-DISPLACEMENT AND TEMPERATURE-DISPLACEMENT TBL + a(¢T)L dB  EAB TCL + a(¢T)L dC  EAC

(Eq. 6)

SOLVE SIMULTANEOUSLY EQS. (1) AND (6): TB  0.2494 P  3,480 TC  1.1253 P 1,740 in which P has units of newtons.

(Eq. 7) (Eq. 8)

SOLVE EQS. (7) AND (8) FOR THE LOAD P: PB  4.0096 TB 13,953 PC  0.8887 TC  1,546

(Eq. 3) (Eq. 4)

Factor of safety  5 (TB)allow  20,400 N

(Eq. 9) (Eq. 10)

(TC)ULT  231,000 N (TC)allow  46,200 N

From Eq. (9): PB  (4.0096)(20,400 N) 13,953 N  95,700 N

SUBSTITUTE EQS. (3) AND (4) INTO EQ. (2): TCL 2TBL + a(¢T)L  + 2a(¢T)L EAC EAB

From Eq. (10): PC  (0.8887)(46,200 N)  1546 N  39,500 N Cable C governs.

or 2TBAC  TCAB  E(T)AB AC

SUBSTITUTE NUMERICAL VALUES INTO EQ. (5): TB(346)  TC(76.7)  1,338,000 in which TB and TC have units of newtons.

ALLOWABLE LOADS From Table 2-1: (TB)ULT  102,000 N

RELATIONS

159

Thermal Effects

(Eq. 5)

Pallow  39.5 kN

Problem 2.5-11 A rigid triangular frame is pivoted at C and held by two identical

horizontal wires at points A and B (see figure). Each wire has axial rigidity EA  120 k and coefficient of thermal expansion   12.5  106/°F. (a) If a vertical load P  500 lb acts at point D, what are the tensile forces TA and TB in the wires at A and B, respectively? (b) If, while the load P is acting, both wires have their temperatures raised by 180°F, what are the forces TA and TB? (c) What further increase in temperature will cause the wire at B to become slack?

;

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Solution 2.5-11

Triangular frame held by two wires

FREE-BODY DIAGRAM OF FRAME

(b) LOAD P AND TEMPERATURE INCREASE T Force-displacement and temperature-displacement relations: TAL + a(¢T)L (Eq. 8) dA  EA TBL + a(¢T)L EA Substitute (8) and (9) into Eq. (2): TAL 2TBL + a(¢T)L  + 2a(¢T)L EA EA dB 

(Eq. 9)

TA  2TB  EA(T)

EQUATION OF EQUILIBRIUM

or

MC  0 哵哴

Solve simultaneously Eqs. (1) and (10):

P(2b)  TA(2b)  TB(b)  0 or 2TA TB  2P (Eq. 1)

1 TA  [4P + EAa(¢T)] 5 2 TB  [P  EAa(¢T)] 5

DISPLACEMENT DIAGRAM

(Eq. 10)

(Eq. 11) (Eq. 12)

Substitute numerical values: P  500 lb EA  120,000 lb T  180°F

  12.5  106/°F 1 TA  (2000 lb + 270 lb)  454 lb 5

EQUATION OF COMPATIBILITY

A  2B

(Eq. 2)

(a) LOAD P ONLY Force-displacement relations:

;

(c) WIRE B BECOMES SLACK

TAL TBL dB  dA  EA EA (L  length of wires at A and B.) Substitute (3) and (4) into Eq. (2):

(Eq. 3, 4)

Set TB  0 in Eq. (12): P  EA(T) or 500 lb P  EAa (120,000 lb)(12.5 * 106/°F)  333.3°F

¢T 

TAL 2TBL  EA EA or TA  2TB Solve simultaneously Eqs. (1) and (5): 4P 2P TB  5 5 Numerical values: P  500 lb ⬖TA  400 lb TB  200 lb

2 TB  (500 lb  270 lb)  92 lb 5

;

TA 

(Eq. 5)

Further increase in temperature: (Eqs. 6, 7)

;

T  333.3°F  180°F  153°F

;

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SECTION 2.5

161

Thermal Effects

Misfits and Prestrains Problem 2.5-12 A steel wire AB is stretched between rigid supports (see figure). The initial prestress in the wire is 42 MPa when the temperature is 20°C. (a) What is the stress  in the wire when the temperature drops to 0°C? (b) At what temperature T will the stress in the wire become zero? (Assume   14  106/°C and E  200 GPa.)

Solution 2.5-12

Steel wire with initial prestress From Eq. (2-18): 2  E(T)

  1 2  1 E(T) Initial prestress: 1  42 MPa

 42 MPa (200 GPa)(14  106/°C)(20°C)

Initial temperature: T1  20°C

 42 MPa 56 MPa  98 MPa

E  200 GPa

  14  106/°C (a) STRESS  WHEN TEMPERATURE DROPS TO 0°C T2  0°C

T  20°C

;

(b) TEMPERATURE WHEN STRESS EQUALS ZERO

  1 2  0 1 E(T)  0 ¢T  

s1 Ea

NOTE: Positive T means a decrease in temperature and an increase in the stress in the wire.

(Negative means increase in temp.)

Negative T means an increase in temperature and a decrease in the stress.

¢T  

Stress  equals the initial stress 1 plus the additional stress 2 due to the temperature drop.

42 MPa

(200 GPa)(14 * 106/°C T  20°C 15°C  35°C ;

  15°C

0.008 in.

Problem 2.5-13 A copper bar AB of length 25 in. and diameter 2 in. is placed in position at

A

room temperature with a gap of 0.008 in. between end A and a rigid restraint (see figure). The bar is supported at end B by an elastic spring with spring constant k  1.2  106 lb/in. (a) Calculate the axial compressive stress c in the bar if the temperature rises 50°F. (For copper, use   9.6  106/°F and E  16  106 psi.) (b) What is the force in the spring? (Neglect gravity effects.) (C) Repeat (a) if k : 

25 in.

d = 2 in. B k

C

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Solution 2.5-13 numerical data

compressive stress in bar RA s   957 psi A

L  25 in. d  2 in.   0.008 in. k  1.2  (106) lb/in.

E  16  (106) psi

  9.6  (106)/°F T  50°F p A  d2 A  3.14159 in2 4 (a) one-degree stat.-indet. if gap closes   TL

  0.012 in.

exceeds gap

select RA as redundant & do superposition analysis

A1   d A2  RAa

A1 A2   A2    A1

compatibility RA 

L 1 + b EA k

d ¢ L 1 + EA k

(b) force in spring Fk  RC statics RA RC  0 RC  RA RC  3006 lb ; (c) find compressive stress in bar if k goes to infinity from expression for RA above, 1/k goes to zero, so RA 

d¢ L EA

  2560 psi

RA A

;

RA  3006 lb

2L — 3

Problem 2.5-14 A bar AB having length L and axial rigidity EA is fixed at end A (see figure). At the other end a small gap of dimension s exists between the end of the bar and a rigid surface. A load P acts on the bar at point C, which is two-thirds of the length from the fixed end. If the support reactions produced by the load P are to be equal in magnitude, what should be the size s of the gap?

Solution 2.5-14

s

RA  8042 lb

A

s

L — 3 C

B P

Bar with a gap (load P ) FORCE-DISPLACEMENT RELATIONS

d1 

P A 2L 3B EA

L  length of bar S  size of gap EA  axial rigidity Reactions must be equal; find S.

d2 

RBL EA

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SECTION 2.5

COMPATIBILITY EQUATION

Reactions must be equal.

1  2  S or

⬖ RA  RB P  2RB

RBL 2PL  S 3EA EA

(Eq. 1)

P 2

Substitute for RB in Eq. (1): PL 2PL  S 3EA 2EA

EQUILIBRIUM EQUATION

RB 

163

Thermal Effects

or S 

PL 6EA

;

NOTE: The gap closes when the load reaches the value P/4. When the load reaches the value P, equal to 6EAs/L, the reactions are equal (RA  RB  P/2). When the load is between P/4 and P, RA is greater than RB. If the load exceeds P, RB is greater than RA.

RA  reaction at end A (to the left) RB  reaction at end B (to the left) P  RA RB

Problem 2.5-15 Pipe 2 has been inserted snugly into Pipe 1, but the holes for a connecting pin do not line up: there is a gap s. The user decides to apply either force P1 to Pipe 1 or force P2 to Pipe 2, whichever is smaller. Determine the following using the numerical properties in the box.

Pipe 1 (steel)

Pipe 2 (brass) Gap s

L1

RA

P1

L2

(a) If only P1 is applied, find P1 (kips) required to close gap s; if a pin is then inserted and P1 removed, what are reaction P2 P1 at L1 forces RA and RB for this load case? (b) If only P2 is applied, find P2 (kips) required to close gap s; L P2 at —2 if a pin is inserted and P2 removed, what are reaction 2 forces RA and RB for this load case? Numerical properties (c) What is the maximum shear stress in the pipes, for the E1 = 30,000 ksi, E2 = 14,000 ksi loads in (a) and (b)? a1 = 6.5  10–6/°F, a2 = 11  10–6/°F (d) If a temperature increase T is to be applied to the entire Gap s = 0.05 in. structure to close gap s (instead of applying forces P1 and L1 = 56 in., d1 = 6 in., t1 = 0.5 in., A1 = 8.64 in.2 P2), find the T required to close the gap. If a pin is inserted L2 = 36 in., d2 = 5 in., t2 = 0.25 in., A2 = 3.73 in.2 after the gap has closed, what are reaction forces RA and RB for this case? (e) Finally, if the structure (with pin inserted) then cools to the original ambient temperature, what are reaction forces RA and RB?

RB

Solution 2.5-15 (a) find reactions at A & B for applied force P1: first compute P1, required to close gap P1 

E1A1 s L1

P1  231.4 kips

;

stat-indet analysis with RB as the redundant

B1  s

L1 L2 + b d B2  RB a E1A1 E2 A2

compatibility: B1 B2  0

RB 

s a

L1 L2 + b E1A1 E2A2

RA  RB

RB  55.2 k

;

(b) find reactions at A & B for applied force P2 E2A2 ; s P2  145.1 kips L2 2 analysis after removing P2 is same as in (a) so reaction forces are the same P2 

;

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(c) max. shear stress in pipe 1 or 2 when either P1 or P2 P1 A1 ; is applied t maxa 

maxa  13.39 ksi 2 P2 A2 t maxb  ;

maxb  19.44 ksi 2 (d) required ¢T and reactions at A & B s ¢T reqd  Treqd  65.8°F a1L1 + a2L2

if pin is inserted but temperature remains at T above ambient temp., reactions are zero (e) if temp. returns to original ambient temperature, find reactions at A & B stat-indet analysis with RB as the redundant compatibility: B1 B2  0 analysis is the same as in (a) & (b) above since gap s is the same, so reactions are the same

;

Problem 2.5-16 A nonprismatic bar ABC made up of a, T RA segments AB (length L1, cross-sectional area A1) and BC (length L2, cross-sectional area A2) is fixed at end A and free at A L1, EA1 B end C (see figure). The modulus of elasticity of the bar is E. A small gap of dimension s exists between the end of the bar and an elastic spring of length L3 and spring constant k3. If bar ABC only (not the spring) is subjected to temperature increase T determine the following.

s L2, EA2 C

D L3, k3

(a) Write an expression for reaction forces RA and RD if the elongation of ABC exceeds gap length s. (b) Find expressions for the displacements of points B and C if the elongation of ABC exceeds gap length s.

Solution 2.5-16 With gap s closed due to T, structure is one-degree statically-indeterminate; select internal force (Q) at juncture of bar & spring as the redundant; use superposition of two released structures in the solution

compatibility: rel1 rel2  s rel2  s  rel1

rel1  relative displ. between end of bar at C & end of spring due to T rel1  T(L1 L2) rel1 is greater than gap length s

L1 L2 1 + + E A1 E A2 k3 E A1A2 k3 Q L1A2 k3 + L2A1k3 + EA1A2

rel2  relative displ. between ends of bar & spring due to pair of forces Q, one on end of bar at C & the other on end of spring Q L1 L2 drel2  Q a + b + E A1 E A2 k3 L1 L2 1 + + b drel2  Q a EA1 E A2 k3

rel2  s  T(L1 L2) Q

s  a¢T1 L1 + L22

[ s  a ¢T1 L1 + L22]

(a) REACTIONS AT A & D statics: RA  Q RD  Q RA 

 s + a¢T1 L1 + L22 L1 L2 1 + + E A1 E A2 k3

RD  RA

;

;

RD

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SECTION 2.5

(b) DISPLACEMENTS AT B & C use superposition of displacements in the two released structures d B  a¢T1 L12  RA a

L1 b E A1

d B  a ¢T 1 L12  [  s + a ¢T1 L1 + L22] L1 L2 1 + + E A1 EA2 k3

;

Thermal Effects

d C  a¢T1 L1 + L22  RAa

L1 L2 + b E A1 E A2

;

d C  a ¢T1 L1 + L22 

[  s + a ¢T1 L1 + L22]

a

165

L1 L2 1 + + E A1 EA2 k3

L1 b EA1

a

L1 L2 + b EA1 EA2

Problem 2.5–17 Wires B and C are attached to a support at the left-hand end and to a pin-supported rigid bar at the right-hand end (see figure). Each wire has cross-sectional area A  0.03 in.2 and modulus of elasticity E  30  106 psi. When the bar is in a vertical position, the length of each wire is L  80 in. However, before being attached to the bar, the length of wire B was 79.98 in. and of wire C was 79.95 in. Find the tensile forces TB and TC in the wires under the action of a force P  700 lb acting at the upper end of the bar.

700 lb B

b

C

b b

80 in.

Solution 2.5–17

Wires B and C attached to a bar EQUILIBRIUM EQUATION

Mpin  0

;

TC(b) TB(2b)  P(3b) 2TB TC  3P

P  700 lb A  0.03 in.2 E  30  106 psi LB  79.98 in. LC  79.95 in.

(Eq. 1)

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DISPLACEMENT DIAGRAM

Combine Eqs. (3) and (5):

SB  80 in.  LB  0.02 in.

TCL  SC + d EA

SC  80 in.  LC  0.05 in.

(Eq. 7)

Eliminate  between Eqs. (6) and (7): TB  2TC 

EASB 2EASC  L L

(Eq. 8)

Solve simultaneously Eqs. (1) and (8):

Elongation of wires:

B  SB 2

(Eq. 2)

C  SC 

(Eq. 3)

FORCE-DISPLACEMENT RELATIONS TB L dB  EA

TC L dC  EA

EASB 2EASC 6P +  5 5L 5L

TC 

2EASB 4EASC 3P  + 5 5L 5L

; ;

SUBSTITUTE NUMERICAL VALUES: EA  2250 lb/in. 5L TB  840 lb 45 lb  225 lb  660 lb

(Eqs. 4, 5)

SOLUTION OF EQUATIONS

; ;

TC  420 lb  90 lb 450 lb  780 lb

(Both forces are positive, which means tension, as required for wires.)

Combine Eqs. (2) and (4): TBL  SB + 2d EA

TB 

(Eq. 6)

P

Problem 2.5-18 A rigid steel plate is supported by three posts of high-strength concrete each having an effective cross-sectional area A  40,000 mm2 and length L  2 m (see figure). Before the load P is applied, the middle post is shorter than the others by an amount s  1.0 mm. Determine the maximum allowable load Pallow if the allowable compressive stress in the concrete is allow  20 MPa. (Use E  30 GPa for concrete.)

S s

C

C

C

L

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SECTION 2.5

Solution 2.5-18

Thermal Effects

167

Plate supported by three posts EQUILIBRIUM EQUATION 2P1 P2  P

(Eq. 1)

COMPATIBILITY EQUATION 1  shortening of outer posts 2  shortening of inner post 1  2 s

(Eq. 2)

FORCE-DISPLACEMENT RELATIONS d1  s  size of gap  1.0 mm L  length of posts  2.0 m A  40,000 mm2 allow  20 MPa E  30 GPa C  concrete post DOES THE GAP CLOSE? Stress in the two outer posts when the gap is just closed: s 1.0 mm s  E  Ea b  (30 GPa) a b L 2.0 m

P1 L EA

d2 

(Eqs. 3, 4)

SOLUTION OF EQUATIONS Substitute (3) and (4) into Eq. (2): P1L P2L EAs (Eq. 5)  + s or P1  P2  EA EA L Solve simultaneously Eqs. (1) and (5): P  3P1 

EAs L

By inspection, we know that P1 is larger than P2. Therefore, P1 will control and will be equal to allow A. Pallow  3sallow A 

 15 MPa Since this stress is less than the allowable stress, the allowable force P will close the gap.

P2 L EA

EAs L

 2400 kN  600 kN  1800 kN  1.8 MN

;

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Problem 2.5-19 A capped cast-iron pipe is compressed by a brass rod,

Nut & washer 3 dw = — in. 4

as shown. The nut is turned until it is just snug, then add an additional quarter turn to pre-compress the CI pipe. The pitch of the threads of the bolt is p  52 mils (a mil is one-thousandth of an inch). Use the numerical properties provided.

(

)

Steel cap (tc = 1 in.)

(a) What stresses p and r will be produced in the cast-iron pipe and brass rod, respectively, by the additional quarter turn of the nut? (b) Find the bearing stress b beneath the washer and the shear stress

c in the steel cap.

Cast iron pipe (do = 6 in., di = 5.625 in.) Lci = 4 ft Brass rod 1 dr = — in. 2

)

(

Modulus of elasticity, E: Steel (30,000 ksi) Brass (14,000 ksi) Cast iron (12,000 ksi)

Solution 2.5-19 The figure shows a section through the pipe, cap and rod

Arod  0.196 in2

NUMERICAL PROPERTIES

compatibility equation

Lci  48 in.

Es  30000 ksi

Eb  14000 ksi

1 Ec  12000 ksi tc  1 in. p  52  (103) in. n  4 3 1 dr  in. do  6 in. di  5.625 in. dw  in. 4 2 (a) FORCES & STRESSES IN PIPE & ROD one degree stat-indet - cut rod at cap & use force in rod (Q) as the redundant rel1  relative displ. between cut ends of rod due to 1/4 turn of nut rel1  np ends of rod move apart, not together, so this is () rel2  relative displ. between cut ends of rod due pair of forces Q L + 2tc Lci + b EbArod EcApipe p p Apipe  (do2  di2) Arod  dr2 4 4

Q

Apipe  3.424 in2

rel1 rel2  0

np Lci + 2tc Lci + EbArod EcApipe

Q  0.672 kips

Frod  Q

Fpipe  Q

statics stresses

sc  sb 

Fpipe Apipe Frod Arod

c  0.196 ksi b  3.42 ksi

; ;

(b) BEARING AND SHEAR STRESSES IN STEEL CAP sb 

d rel2  Qa

tc 

Frod p (d 2  dr 2) 4 w Frod pdwtc

b  2.74 ksi

c  0.285 ksi

;

;

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Thermal Effects

169

Problem 2.5-20 A plastic cylinder is held snugly between a rigid plate and a foundation by two steel bolts (see figure). Determine the compressive stress p in the plastic when the nuts on the steel bolts are tightened by one complete turn. Data for the assembly are as follows: length L  200 mm, pitch of the bolt threads p  1.0 mm, modulus of elasticity for steel Es  200 GPa, modulus of elasticity for the plastic Ep  7.5 GPa, cross-sectional area of one bolt As  36.0 mm2, and cross-sectional area of the plastic cylinder Ap  960 mm2.

Solution 2.5-20

Steel bolt

L

Plastic cylinder and two steel bolts COMPATIBILITY EQUATION

L  200 mm P  1.0 mm

s  elongation of steel bolt

Es  200 GPa

p  shortening of plastic cylinder

As  36.0 mm2 (for one bolt)

s p  np

Ep  7.5 GPa

(Eq. 2)

FORCE-DISPLACEMENT RELATIONS

Ap  960 mm2 n  1 (See Eq. 2-22) EQUILIBRIUM EQUATION

ds 

PsL EsAs

dp 

PpL Ep Ap

(Eq. 3, Eq. 4)

SOLUTION OF EQUATIONS Substitute (3) and (4) into Eq. (2): Pp L PsL +  np Es As Ep Ap Solve simultaneously Eqs. (1) and (5): Ps  tensile force in one steel bolt Pp  compressive force in plastic cylinder Pp  2Ps

Pp  (Eq. 1)

2npEs AsEp Ap L(Ep Ap + 2Es As)

(Eq. 5)

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STRESS IN THE PLASTIC CYLINDER sp 

Pp Ap



D  EpAp 2EsAs  21.6  106 N

2np Es As Ep

sp 

L(Ep Ap + 2Es As)

2np N 2(1)(1.0 mm) N a b  a b L D 200 mm D

 25.0 MPa

SUBSTITUTE NUMERICAL VALUES:

;

N  Es As Ep  54.0  10 N /m 15

2

2

Problem 2.5-21 Solve the preceding problem if the data for the assembly are as follows: length L  10 in., pitch of the bolt threads p  0.058 in., modulus of elasticity for steel Es  30  106 psi, modulus of elasticity for the plastic Ep  500 ksi, cross-sectional area of one bolt As  0.06 in.2, and crosssectional area of the plastic cylinder Ap  1.5 in.2

Solution 2.5-21

Steel bolt

L

Plastic cylinder and two steel bolts COMPATIBILITY EQUATION

L  10 in.

s  elongation of steel bolt

p  0.058 in. Es  30  106 psi

p  shortening of plastic cylinder s p  np

(Eq. 2)

As  0.06 in.2 (for one bolt) Ep  500 ksi Ap  1.5 in.2 n  1 (see Eq. 2-22)

FORCE-DISPLACEMENT RELATIONS

EQUILIBRIUM EQUATION Ps  tensile force in one steel bolt

ds 

Pp  compressive force in plastic cylinder Pp  2Ps

(Eq. 1)

Ps L Es As

dp 

Pp L Ep Ap

(Eq. 3, Eq. 4)

SOLUTION OF EQUATIONS Substitute (3) and (4) into Eq. (2): Pp L Ps L +  np Es As Ep Ap

(Eq. 5)

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SECTION 2.5

SUBSTITUTE NUMERICAL VALUES:

Solve simultaneously Eqs. (1) and (5): Pp 

N  Es As Ep  900  109 lb2/in.2

2 np Es As Ep Ap

D  Ep Ap 2Es As  4350  103 lb

L(Ep Ap + 2Es As)

STRESS IN THE PLASTIC CYLINDER sp 

Pp Ap

sp 

2 np Es As Ep



171

Thermal Effects

;

L(Ep Ap + 2Es As)

2np N 2(1)(0.058in.) N a b  a b L D 10 in. D

 2400 psi

;

d = np

Problem 2.5-22 Consider the sleeve made from two copper tubes joined by tin-lead

L1 = 40 mm, d1 = 25 mm, t1 = 4 mm

(a) Find the forces in the sleeve and bolt, Ps and PB, due to both the prestress in the bolt and the temperature increase. For copper, use Ec  120 GPa and c  17  106/°C; for steel, use Es  200 GPa and s  12  106/°C. The pitch of the bolt threads is p  1.0 mm. Assume s  26 mm and bolt diameter db  5 mm. (b) Find the required length of the solder joint, s, if shear stress in the sweated joint cannot exceed the allowable shear stress aj  18.5 MPa. (c) What is the final elongation of the entire assemblage due to both temperature change T and the initial prestress in the bolt?

Brass cap T

S T L2 = 50 mm, d2 = 17 mm, t2 = 3 mm

solder over distance s. The sleeve has brass caps at both ends, which are held in place by a steel bolt and washer with the nut turned just snug at the outset. Then, two “loadings” are applied: n  1/2 turn applied to the nut; at the same time the internal temperature is raised by T  30°C.

Copper sleeve Steel bolt

Solution 2.5-22 p 2 d 4 b

The figure shows a section through the sleeve, cap and bolt

Ab 

NUMERICAL PROPERTIES (SI UNITS)

Ab  19.635 mm2

n

1 2

p  1.0 mm

c  17  (106)/°C

Ec  120 GPa

6

s  12  (10 )/°C

Es  200 GPa

aj  18.5 MPa L1  40 mm d1  25 mm

A2 

T  30°C

s  26 mm

t1  4 mm

db  5 mm

L2  50 mm

d1  2t1  17 mm

t2  3 mm

d2  17 mm

A1 

p 2 [d  1 d1  2 t122] 4 1 A1  263.894 mm2

p [ d 2  1 d2  2t222] 4 2

A2  131.947 mm2

(a) FORCES IN SLEEVE & BOLT one-degree stat-indet - cut bolt & use force in bolt (PB) as redundant (see sketches below)

B1  np sT(L1 L2  s)

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d B2  PB c

Axially Loaded Members

L1 + L2  s L1  s L2  s s + + + d Es Ab EcA1 Ec A2 Ec( A1 + A2)

compatibility PB 

Page 172

B1 B2  0

[ np + a s ¢T( L1 + L2  s)] L1 + L2  s L1  s L2  s s c + + + d EsAb Ec A1 Ec A2 Ec ( A 1 + A2)

PB  25.4 kN

Sketches illustrating superposition procedure for statically-indeterminate analysis δ = np cap

∆T L1

Actual indeterminate structure under load(s)

=

S ∆T L2

1° SI superposition analysis using internal force in bolt as the redundant

sleeve bolt

Two released structures (see below) under: (1)load(s); (2) redundant applied as a load

δ = np

∆T

+ Ps

S

S ∆T

cut bolt

δB1 δB1

relative displacement across cut bolt, δB1 due to both δ and ∆T (positive if pieces move together) relative displacement across cut bolt, δB2 due to Pb (positive if pieces move together)

PB

δB2

PB

δB2

apply redundant internal force Ps & find relative displacement across cut bolt,

δB2

;

Ps  PB

;

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SECTION 2.5

(b) REQUIRED LENGTH OF SOLDER JOINT≈ P t As  d2s As PB sreqd  pd2t aj

sreqd  25.7 mm

d s  Ps c

Thermal Effects

173

L1  s L2  s s + + d Ec A1 Ec A2 Ec (A1 + A2)

s  0.064 mm f  b s

f  0.35 mm

;

(c) FINAL ELONGATION f  net of elongation of bolt (b) & shortening of sleeve (s) d b  PBa

L1 + L2  s b EsAb

b  0.413 mm

Problem 2.5-23 A polyethylene tube (length L) has a cap which when installed compresses a spring (with undeformed length L1  L) by amount   (L1  L). Ignore deformations of the cap and base. Use the force at the base of the spring as the redundant. Use numerical properties in the boxes given. (a) (b) (c) (d)

What is the resulting force in the spring, Fk? What is the resulting force in the tube, Ft? What is the final length of the tube, Lf? What temperature change T inside the tube will result in zero force in the spring?

d = L1 – L Cap (assume rigid) Tube (d0, t, L, at, Et)

Spring (k, L1 > L)

Modulus of elasticity Polyethylene tube (Et = 100 ksi) Coefficients of thermal expansion at = 80  10–6/°F, ak = 6.5  10–6/°F Properties and dimensions 1 d0 = 6 in. t = — in. 8 kip L1 = 12.125 in. > L = 12 in. k = 1.5 ––– in.

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Solution 2.5-23

solve for redundant Q

The figure shows a section through the tube, cap and spring

Q

Properties & dimensions

Fk  0.174 kips

do  6 in. At 

t

1 in. 8

At  2.307 in2 kip k  1.5 in

L1  12.125 in.  L  12 in.

k  6.5(106)/F T  0

  L1  L

(a) Force in spring Fk  redundant Q 1 k

ft 

compressive force in spring (Fk) & also tensile force in tube

;

NOTE: if tube is rigid, Fk  k  0.1875 kips (c) Final length of tube Lf  L c1 c2

t  80  (106)/F



f

(b) Ft  force in tube  Q

  0.125 in.

note that Q result below is for zero temp. (until part(d))

Flexibilities

;

Et  100 ksi

p [ d 2  ( do  2 t)2] 4 o

spring is 1/8 in. longer than tube

 d + ¢T ( a kL1 + a tL)  Fk f + ft

Lf  L  Qft t(T)L

i.e., add displacements for the two released structures to initial tube length L Lf  12.01 in.

;

(d) Set Q  0 to find T required to reduce spring force to zero

L EtAt

2  rel. displ. across cut spring due to redundant  Q(f ft)

¢T reqd 

d (a k L1 + a tL)

Treqd  141.9F

1  rel. displ. across cut spring due to precompression and T   kTL1  tTL

since t  k, a temp. increase is req’d to expand tube so that spring force goes to zero

compatibility: 1 2  0 Steel wires

Problem 2.5-24 Prestressed concrete beams are sometimes manufactured in the following manner. High-strength steel wires are stretched by a jacking mechanism that applies a force Q, as represented schematically in part (a) of the figure. Concrete is then poured around the wires to form a beam, as shown in part (b). After the concrete sets properly, the jacks are released and the force Q is removed [see part (c) of the figure]. Thus, the beam is left in a prestressed condition, with the wires in tension and the concrete in compression. Let us assume that the prestressing force Q produces in the steel wires an initial stress 0  620 MPa. If the moduli of elasticity of the steel and concrete are in the ratio 12:1 and the cross-sectional areas are in the ratio 1:50, what are the final stresses s and c in the two materials?

Q

Q (a) Concrete

Q

Q (b)

(c)

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SECTION 2.5

Solution 2.5-24

Thermal Effects

Prestressed concrete beam L  length

0  initial stress in wires 

Q  620 MPa As

As  total area of steel wires Ac  area of concrete  50 As Es  12 Ec Ps  final tensile force in steel wires Pc  final compressive force in concrete EQUILIBRIUM EQUATION Ps  Pc COMPATIBILITY EQUATION AND FORCE-DISPLACEMENT RELATIONS

STRESSES (Eq. 1)

1  initial elongation of steel wires

sc 

QL s0L   EsAs Es

2  final elongation of steel wires 

Ps  As

s0 EsAs 1 + EcAc

Pc s0  Ac Es Ac + As Ec

0  620 MPa

3  shortening of concrete Pc L  Ec Ac

1  2  3

or

s0L PsL PcL   Es EsAs EcAc Solve simultaneously Eqs. (1) and (3): s0As EsAs 1 + EcAc

;

;

SUBSTITUTE NUMERICAL VALUES:

PsL EsAs

Ps  Pc 

ss 

(Eq. 2, Eq. 3)

Es  12 Ec

As 1  Ac 50

ss 

620 MPa  500 MPa (Tension) 12 1 + 50

sc 

620 MPa  10 MPa (Compression) 50 + 12

;

;

175

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Problem 2.5-25 A polyethylene tube (length L) has a cap which is held in place by a spring (with undeformed length L1 L). After installing the cap, the spring is post-tensioned by turning an adjustment screw by amount . Ignore deformations of the cap and base. Use the force at the base of the spring as the redundant. Use numerical properties in the boxes below. (a) (b) (c) (d)

What is the resulting force in the spring, Fk? What is the resulting force in the tube, Ft? What is the final length of the tube, Lf? What temperature change T inside the tube will result in zero force in the spring?

Cap (assume rigid) Tube (d0, t, L, at, Et)

Spring (k, L1 < L)

d = L – L1

Adjustment screw Modulus of elasticity Polyethylene tube (Et = 100 ksi) Coefficients of thermal expansion at = 80  10–6/°F, ak = 6.5  10–6/°F Properties and dimensions 1 d0 = 6 in. t = — in. 8 kip L = 12 in. L1 = 11.875 in. k = 1.5 ––– in.

Solution 2.5-25 The figure shows a section through the tube, cap and spring Properties & dimensions do  6 in.

1 t  in. 8

L  12 in.  L1  11.875 in.

kip in

k  6.5(106) t  80  (106) p At  [ d2o  1 do  2t22] 4 At  2.307 in2

spring is 1/8 in. shorter than tube

  L  L1   0.125 in. T  0 note that Q result below is for zero temp. (until part (d))

Et  100 ksi k  1.5

Pretension & temperature

Flexibilities

f

1 k

ft 

L EtAt

(a) Force in spring (Fk)  redundant (Q) follow solution procedure outlined in Prob. 2.5-23 solution Q

d + ¢T 1a k L1 + a t L2 f + ft

 Fk

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SECTION 2.5

Fk  0.174 kips

;

also the compressive force in the tube

; (b) force in tube Ft  Q   0.174 k (c) Final length of tube & spring Lf  L c1 c2 Lf  L  Qft t(T)L

Lf  11.99 in.

;

Thermal Effects

177

(d) Set Q  0 to find T required to reduce spring force to zero ¢ Treqd 

d 1a kL1 + a t L2

Treqd  141.6F

since t  k, a temp. drop is req’d to shrink tube so that spring force goes to zero

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Stresses on Inclined Sections Problem 2.6-1 A steel bar of rectangular cross section (1.5 in.  2.0 in.) carries a tensile load P (see figure). The allowable stresses in tension and shear are 14,500 psi and 7,100 psi, respectively. Determine the maximum permissible load Pmax.

2.0 in. P

P

1.5 in.

Solution 2.6-1 MAXIMUM LOAD - tension

2.0 in. P

P

Pmax1  aA

Pmax1  43500 lbs

MAXIMUM LOAD - shear Pmax2  2aA

1.5 in.

Because allow is less than one-half of allow, the shear stress governs.

NUMERICAL DATA A  3 in2

Pmax2  42,600 lbs

a  14500 psi

a  7100 psi

Problem 2.6-2 A circular steel rod of diameter d is subjected to a tensile force P  3.5 kN (see figure). The allowable stresses in tension and shear are 118 MPa and 48 MPa, respectively. What is the minimum permissible diameter dmin of the rod?

d

P

Solution 2.6-2 P

d

P = 3.5 kN

Pmax  2t a a dmin 

P  3.5 kN a  118 MPa a  48 MPa Find Pmax then rod diameter since a is less than 1/2 of a, shear governs NUMERICAL DATA

p dmin2 b 4

2 P pt A a

dmin  6.81 mm

;

P = 3.5 kN

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SECTION 2.6

Problem 2.6-3 A standard brick (dimensions 8 in.  4 in.  2.5 in.) is compressed

P

lengthwise by a force P, as shown in the figure. If the ultimate shear stress for brick is 1200 psi and the ultimate compressive stress is 3600 psi, what force Pmax is required to break the brick? 8 in.

Solution 2.6-3

2.5 in.

Maximum shear stress: t max 

4 in.

4 in.

Standard brick in compression

P

8 in.

179

Stresses on Inclined Sections

2.5 in.

sx P  2 2A

ult  3600 psi

ult  1200 psi

Because ult is less than one-half of ult, the shear stress governs. t max  A  2.5 in.  4.0 in.  10.0 in.2 Maximum normal stress: sx 

P 2A

or P max  2Atult

P max  2(10.0 in.2)(1200 psi)  24,000 lb

;

P A

Problem 2.6-4 A brass wire of diameter d  2.42 mm is stretched tightly

between rigid supports so that the tensile force is T  98 N (see figure). The coefficient of thermal expansion for the wire is 19.5   10-6/°C and the modulus of elasticity is E = 110 GPa (a) What is the maximum permissible temperature drop T if the allowable shear stress in the wire is 60 MPa? (b) At what temperature changes does the wire go slack?

T

d

T

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Axially Loaded Members

Solution 2.6-4

Brass wire in tension d

T

a  60 MPa

T

A

p 2 d 4

T  2 ta A ¢Tmax  Ea

NUMERICAL DATA d  2.42 mm T  98 N  19.5 (106)/°C E  110 GPa

¢Tmax  46°C (drop)

(a) ¢Tmax (DROP IN TEMPERATURE)

(b) ¢T AT WHICH WIRE GOES SLACK

s

T  (E a ¢T) A

ta 

E a ¢T T  2A 2

max 

increase ¢T until s  0

s 2

T E aA ¢T  9.93°C (increase) ¢T 

Problem 2.6-5 A brass wire of diameter d  1/16 in. is stretched between rigid supports with an initial tension T of 37 lb (see figure). Assume that the coefficient of thermal expansion is 10.6  106/°F and the modulus of elasticity is 15  106 psi.)

d

T

T

(a) If the temperature is lowered by 60°F, what is the maximum shear stress max in the wire? (b) If the allowable shear stress is 10,000 psi, what is the maximum permissible temperature drop? (c) At what temperature change T does the wire go slack?

Solution 2.6-5 d

T

T

T  2ta A ¢Tmax  Ea

NUMERICAL DATA d

1 in 16

T  37 lb

 10.6  (106)/°F

Tmax  49.9°F

;

(c) T AT WHICH WIRE GOES SLACK

E  15  (10 ) psi T  60°F p 2 A d 4 (a) max (DUE TO DROP IN TEMPERATURE) 6

tmax 

(b) ¢Tmax FOR ALLOW. SHEAR STRESS

sx 2

T  (E a ¢T) A tmax  2

max  10800 psi

;

increase T until   0 ¢T 

T E aA

T  75.9°F (increase)

;

a  10000 psi

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SECTION 2.6

Problem 2.6-6 A steel bar with diameter d  12 mm is subjected to a tensile load P  9.5 kN (see figure). (a) What is the maximum normal stress max in the bar? (b) What is the maximum shear stress max? (c) Draw a stress element oriented at 45° to the axis of the bar and show all stresses acting on the faces of this element.

Solution 2.6-6

P

Stresses on Inclined Sections

d = 12 mm

181

P = 9.5 kN

Steel bar in tension (b) MAXIMUM SHEAR STRESS The maximum shear stress is on a 45° plane and equals x/2. tmax 

P  9.5 kN (a) MAXIMUM NORMAL STRESS sx 

sx  42.0 MPa 2

;

(c) STRESS ELEMENT AT  45°

P 9.5 kN  84.0 MPa  p A (12 mm)2 4

max  84.0 MPa

;

NOTE: All stresses have units of MPa.

Problem 2.6-7 During a tension test of a mild-steel specimen (see figure), the extensometer shows an elongation of 0.00120 in. with a gage length of 2 in. Assume that the steel is stressed below the proportional limit and that the modulus of elasticity E  30  106 psi. (a) What is the maximum normal stress max in the specimen? (b) What is the maximum shear stress max? (c) Draw a stress element oriented at an angle of 45° to the axis of the bar and show all stresses acting on the faces of this element.

2 in. T

T

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Solution 2.6-7

Page 182

Axially Loaded Members

Tension test (b) MAXIMUM SHEAR STRESS The maximum shear stress is on a 45° plane and equals x/2.

Elongation:   0.00120 in. (2 in. gage length) d 0.00120 in. Strain: â    0.00060 L 2 in.

tmax 

sx  9,000 psi 2

;

(c) STRESS ELEMENT AT  45°

Hooke’s law: x  E  (30  106 psi)(0.00060)  18,000 psi (a) MAXIMUM NORMAL STRESS

x is the maximum normal stress. max  18,000 psi

NOTE: All stresses have units of psi.

;

Problem 2.6-8 A copper bar with a rectangular cross section is held

45∞

without stress between rigid supports (see figure). Subsequently, the temperature of the bar is raised 50°C. Determine the stresses on all faces of the elements A and B, and show these stresses on sketches of the elements. (Assume  17.5  106/°C and E  120 GPa.)

Solution 2.6-8

A

B

Copper bar with rigid supports MAXIMUM SHEAR STRESS tmax 

T  50°C (Increase)

sx  52.5 MPa 2

STRESSES ON ELEMENTS A AND B

 17.5  106/°C E  120 GPa STRESS DUE TO TEMPERATURE INCREASE

x  E (T )

(See Eq. 2-18 of Section 2.5)

 105 MPa (Compression) NOTE: All stresses have units of MPa.

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SECTION 2.6

Problem 2.6-9 The bottom chord AB in a small truss ABC (see figure) is fabricated from a W8  28 wide-flange steel section. The cross-sectional area A  8.25 in.2 (Appendix E, Table E-1 (a)) and each of the three applied loads P  45 k. First, find member force NAB; then, determine the normal and shear stresses acting on all faces of stress elements located in the web of member AB and oriented at (a) an angle  0°, (b) an angle  30°, and (c) an angle  45°. In each case, show the stresses on a sketch of a properly oriented element.

Stresses on Inclined Sections

183

P P = 45 kips

C

9 ft

B

12 ft A

P

Ax Ay

By

NAB

NAB

u

Solution 2.6-9 Statics P  45 kips

MA  0

9 P By  12

By  33.75 k BCV  By

FH  0 at B

degrees in web of AB

x  10.9 ksi (a)  0

12 BCH  BCV 9

BCH  45 k

NAB  BCH  P

NAB  90 kips (compression)

;

Normal and shear stresses on elements at 0, 30 & 45

sx 

 NAB A

A  8.25 in2

;

x  10.91 ksi

;

(b)  30° on x face

  xcos ( )2

  8.18 ksi

  xsin ( ) cos ( )

  4.72 ksi

on y face

uu +

  xcos( )2   xsin( ) cos( )

p 2

  2.73 ksi   4.72 ksi

; ;

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(c)  45 degrees

  xcos( )2

p 2   5.45 ksi

  xsin( )cos( )

  5.45 ksi

uu +

on y face

on x face

  xcos( )

  5.45 ksi

2

  xsin ( ) cos ( )

  5.45 ksi

; ;

Problem 2.6-10 A plastic bar of diameter d  32 mm is

compressed in a testing device by a force P  190 N applied as shown in the figure. (a) Determine the normal and shear stresses acting on all faces of stress elements oriented at (1) an angle  0°, (2) an angle  22.5°, and (3) an angle  45°. In each case, show the stresses on a sketch of a properly oriented element. What are max and max? (b) Find max and max in the plastic bar if a re-centering spring of stiffness k is inserted into the testing device, as shown in the figure. The spring stiffness is 1/6 of the axial stiffness of the plastic bar.

P = 190 N 100 mm

300 mm

200 mm Re-centering spring (Part (b) only)

Plastic bar

u

d = 32 mm

k

Solution NUMERICAL DATA

(2)  22.50 degrees

p 2 d 4 A  804.25 mm2

A

d  32 mm P  190 N

on x face

  xcos( )2   807 kPa

a  100 mm

  xsin( ) cos( )   334 kPa ;

b  300 mm (a) Statics - FIND COMPRESSIVE FORCE F & STRESSES IN PLASTIC BAR

F sx 

P( a + b) a F A

x  0.945 MPa

or

x  945 kPa

(1)  0 degrees

uu +

on x face

  xcos( )2   472 kPa

sx  472 kPa 2

x  945 kPa

  xsin( ) cos( )   334.1 kPa (3)  45 degrees

max  945 kPa

max  472 kPa

on y face

  xcos( )2   138.39 kPa

F  760 N

from (1), (2) & (3) below

max  x

;

;

;

  xsin( ) cos( )   472 kPa ;

p 2

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SECTION 2.6

on y face

  xcos( )2

p 2

uu +

force in plastic bar

  472.49 kPa

  xsin( ) cos( )

  472.49 kPa

δ/3

6k

k

2kδ

kδ

F A

sx 

P

100 mm

185

4P b 5k

8 P 5

F  304 N

normal and shear stresses in plastic bar

IN PLASTIC BAR

200 mm

F  (2k) a F

(b) ADD SPRING - FIND MAX. NORMAL & SHEAR STRESSES

100 mm

Stresses on Inclined Sections

δ

tmax 

x  0.38 max  378 kPa

sx 2

max  189 kPa

; ;

a Mpin  0 P(400)  [2k (100)  k (300)] d

4 P 5k

Problem 2.6-11 A plastic bar of rectangular cross section (b  1.5 in.

L — 2

and h  3 in.) fits snugly between rigid supports at room temperature (68°F) but with no initial stress (see figure). When the temperature of the bar is raised to 160°F, the compressive stress on an inclined plane pq at midspan becomes 1700 psi.

L — 2

L — 4 p

6

u

P

(a) What is the shear stress on plane pq? (Assume  60  10 /°F and E  450  103 psi.) (b) Draw a stress element oriented to plane pq and show the stresses Load P for part (c) only acting on all faces of this element. (c) If the allowable normal stress is 3400 psi and the allowable shear stress is 1650 psi, what is the maximum load P (in x direction) which can be added at the quarter point (in addition to thermal effects above) without exceeding allowable stress values in the bar?

b h

q

Solution 2.6-11 (a)

NUMERICAL DATA b  1.5 in

h  3 in

A  bh

T  92°F

T  (160  68)°F

SHEAR STRESS ON PLANE PQ STAT-INDET ANALYSIS GIVES, FOR REACTION AT RIGHT SUPPORT:

R  EA T

pq  1700 psi

A  4.5 in

2 6

 60  (10 )/°F E  450  (103) psi

sx 

R A

R  11178 lb

x  2484 psi

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Axially Loaded Members

using   xcos( )2 u  acos a

s pq

A sx

cos1u22 

b

s pq

from (a) (for temperature increase T):

sx

RR1  EA T

 34.2°

Stresses in bar (0 to L/4)

pq  xcos( )

pq  1154 psi

;

pq  1700 psi

2

stresses at  /2 (y face) p 2 b 2

s y  s xcosa u +

y  784 psi

tmax 

psi 4 115 ) t (b

 E a¢T 3P + 2 8A 4A Pmax1  12ta + Ea¢ T2 3 Pmax1  34704 lb ta 

3 Pmax1  E a¢T + 2 8A max  1650 psi  check

(b) STRESS ELEMENT FOR PLANE PQ 784

sx 3P tmax  4A 2 set max  a & solve for Pmax1 s x   Ea¢ T +

now with , can find shear stress on plane pq

pq  xsin( )cos( )

RL1  EA T

psi si 0 p θ = 34.2°

170

Par

s x   Ea ¢ T +

x  3300 psi

3Pmax1 4A  less than a

Stresses in bar (L/4 to L) sx P tmax  4A 2 set max  a & solve for Pmax2 s x   E a¢ T 

(c) MAX. LOAD AT QUARTER POINT

a  1650 psi

2 a  3300

a  3400 psi  less than a so shear controls

stat-indet analysis for P at L/4 gives, for reactions: RR2 

P 4

RL2 

3 P 4

(tension for 0 to L/4 & compression for rest of bar)

Pmax2  4A(2a  E T) Pmax2  14688 lb tmax 

;

Pmax2  E a¢ T  2 8A

s x   Ea¢ T 

Pmax2 4A

shear in segment (L/4 to L) controls

max  1650 psi x  3300 psi

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SECTION 2.6

L — 2

Problem 2.6-12 A copper bar of rectangular cross section (b  18 mm

187

Stresses on Inclined Sections

and h  40 mm) is held snugly (but without any initial stress) between rigid supports (see figure). The allowable stresses on the inclined plane pq bL at midspan, for which  55°, are specified as 60 MPa in compression p and 30 MPa in shear. P (a) What is the maximum permissible temperature rise T if the allowable stresses on plane pq are not to be exceeded? (Assume A  17  106/°C and E  120 GPa.) Load for part (c) only (b) If the temperature increases by the maximum permissible amount, what are the stresses on plane pq? (c) If the temperature rise T  28°C, how far to the right of end A (distance L, expressed as a fraction of length L) can load P  15 kN be applied without exceeding allowable stress values in the bar? Assume that a  75 MPa and a  35 MPa.

L — 2

b

u

h

B q

Solution 2.6-12 (b) STRESSES ON PLANE PQ FOR max. TEMP.

x  E Tmax pq  xcos( )2

x  63.85 MPa pq  21.0 MPa

pq  xsin( )cos( ) NUMERICAL DATA p u  55a b 180

(a) FIND Tmax BASED ON ALLOWABLE NORMAL & SHEAR STRESS VALUES ON PLANE pq s x x  E Tmax ¢ Tmax  Ea 2 pq  xcos( ) pq  xsin( )cos( ) ^ set each equal to corresponding allowable & solve for x s pqa s x1  x1  182.38 MPa cos1u22 sin1u2cos1u2

x2  63.85 MPa

lesser value controls so allowable shear stress governs ¢Tmax 

s x2 Ea

;

T  28 DEGREES C P  15 kN from one-degree stat-indet analysis, reactions RA & RB due to load P are:

b  18 mm h  40 mm A  bh A  720 mm2 pqa  60 MPa pqa  30 Mpa 6  17  (10 )/°C E  120 GPa T  20°C P  15 kN

s x2 

pq  30 MPa

(c) ADD LOAD P IN X-DIRECTION TO TEMPERATURE CHANGE & FIND LOCATION OF LOAD

radians

t pqa

;

Tmax  31.3°C

;

RA  (1  )P RB   P now add normal stresses due to P to thermal stresses due to T (tension in segment 0 to L, compression in segment L to L) Stresses in bar (0 to L) RA sx tmax  A 2 shear controls so set max  a & solve for  s x   Ea¢ T +

2t a   E a¢ T + b1

(1  b)P A

A [2 t a + Ea¢ T] P

  5.1 ^ impossible so evaluate segment (L to L)

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Axially Loaded Members

Stresses in bar (L to L)

2t a   Ea¢ T 

RB sx tmax  A 2 set max  a & solve for Pmax2 s x   E a¢ T 

b

bP A

A [ 2 t a + E a ¢T] P

  0.62

;

NAC

Problem 2.6-13 A circular brass bar of diameter d is member AC in truss ABC which has load P  5000 lb applied at joint C. Bar AC is composed of two segments brazed together on a plane pq making an angle  36° with the axis of the bar (see figure). The allowable stresses in the brass are 13,500 psi in tension and 6500 psi in shear. On the brazed joint, the allowable stresses are 6000 psi in tension and 3000 psi in shear. What is the tensile force NAC in bar AC? What is the minimum required diameter dmin of bar AC?

A a p q B

u = 60∞

d C

P

NAC

Solution 2.6-13 NUMERICAL DATA P  5 kips

 36°

a  13.5 ksi

p a 2

P sin(60°)

NAC  5.77 kips

a  6.5 ksi u

NAC 

(tension) ;

min. required diameter of bar AC u  54°

ja  6.0 ksi ja  3.0 ksi tensile force NAC

Method of Joints at C

(1) check tension and shear in bars; a  a/2 so shear sx controls tmax  2 2ta 

NAC A

x  2a= 13 ksi

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SECTION 2.6

Areqd 

NAC 2ta

dmin 

4

Ap

Areqd  0.44 in2

sx 

dmin  0.75 in

Areqd

dreqd 

(2) check tension and shear on brazed joint sx 

NAC A

sX 

NAC p 2 d 4

dreqd 

sja

x  17.37 ksi

cos(u)2 4 NAC

dreqd  0.65 in

A p sx

  xsin( )cos( )

4 NAC

A p sX

sx  `

set equal to ja & solve for x, then dreqd

dreqd 

Problem 2.6-14 Two boards are joined by gluing along a scarf joint, as shown in the figure. For purposes of cutting and gluing, the angle between the plane of the joint and the faces of the boards must be between 10° and 40°. Under a tensile load P, the normal stress in the boards is 4.9 MPa.

tja (sin(u) cos(u)) 4 NAC

`

x  6.31 ksi

dreqd  1.08 in

A p sX

P a

Two boards joined by a scarf joint

 90°   70°   x cos2  (4.9 MPa)(cos 70°)2  0.57 MPa

10°   40° Due to load P: x  4.9 MPa (a) STRESSES ON JOINT WHEN  20°

;

P

(a) What are the normal and shear stresses acting on the glued joint if  20°? (b) If the allowable shear stress on the joint is 2.25 MPa, what is the largest permissible value of the angle ? (c) For what angle will the shear stress on the glued joint be numerically equal to twice the normal stress on the joint?

Solution 2.6-14

189

shear on brazed joint

tension on brazed joint

  xcos( )2

Stresses on Inclined Sections

;

  x sin cos

 (4.9 MPa)(sin 70°)(cos 70°)  1.58 MPa

;

(b) LARGEST ANGLE IF allow  2.25 MPa

allow  x sin cos

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Axially Loaded Members

The shear stress on the joint has a negative sign. Its numerical value cannot exceed allow  2.25 MPa. Therefore,

|  |  2.25 MPa. (c)

2.25 MPa  (4.9 MPa)(sin )(cos ) or sin cos

 0.4592 1 From trigonometry: sin u cos u  sin 2u 2 Therefore: sin 2  2(0.4592)  0.9184 Solving: 2  66.69°

 33.34°

or

 90° 

or

113.31°

56.66° ⬖  56.66°

or

33.34°

Since must be between 10° and 40°, we select

 33.3°

;

NOTE: If is between 10° and 33.3°, |  |  2.25 MPa. If is between 33.3° and 40°,

WHAT IS

if   2 ?

Numerical values only: |  |  x sin cos

`

|  |  x cos2

t0 ` 2 s0

x sin cos  2xcos2

sin  2 cos or

 63.43°  26.6°

tan  2

 90° 

;

NOTE: For  26.6° and  63.4°, we find   0.98 MPa and   1.96 MPa. Thus, `

Problem 2.6-15 Acting on the sides of a stress element cut from a bar in

t0 `  2 as required. s0

5000 psi

uniaxial stress are tensile stresses of 10,000 psi and 5,000 psi, as shown in the figure.

tu tu

su = 10,000 psi u

(a) Determine the angle and the shear stress  and show all stresses on a sketch of the element. (b) Determine the maximum normal stress max and the maximum shear stress max in the material. 10,000 psi

tu tu

5000 psi

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SECTION 2.6

Solution 2.6-15

Stresses on Inclined Sections

191

Bar in uniaxial stress 1 1 tanu  2 12 From Eq. (1) or (2): tan2u 

u  35.26°

;

x  15,000 psi   x sin cos

 (15,000 psi)(sin 35.26°)(cos 35.26°) 7,070 psi

;

Minus sign means that  acts clockwise on the plane for which  35.26°. (a) ANGLE AND SHEAR STRESS 

  x cos2

  10,000 psi sx 

s0



2

cos u

10,000 psi cos2u

(1)

PLANE AT ANGLE  90°

  90°  x[cos(  90°)]2  x[sin ]2

NOTE: All stresses have units of psi.

 x sin2

(b) MAXIMUM NORMAL AND SHEAR STRESSES

  90°  5,000 psi sx 

s 090° sin2u



5,000 psi sin2u

max  x  15,000 psi (2)

tmax 

sx  7,500 psi 2

; ;

Equate (1) and (2): 10,000 psi 2

cos u



5,000 psi sin2u

Problem 2.6-16 A prismatic bar is subjected to an axial force that produces a tensile stress   65 MPa and a shear stress   23 MPa on a certain inclined plane (see figure). Determine the stresses acting on all faces of a stress element oriented at

 30° and show the stresses on a sketch of the element.

65 MPa u 23 MPa

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Axially Loaded Members

Solution 2.6-16 (4754 + 65s x)

0

s 2x sx  x

4754 65

u  acos find & x for stress state shown above

  xcos( )2

cos (u) 

sin (u) 

so

3 18.

Asx A

PA s x Q su

u  19.5°

a

su

1

  65 MPa

 73.1 MPa

MP

a

7 31.

su

MP

a

9 54.

sx

MP

θ = 30°

  xsin( ) cos( ) tu sx a

 

tu sx

2

b 

A

1

su sx

su

su

sx Asx

 a

su sx

b

65 65 2 23 2  a b a b  sx sx sx a

65 2 65 23 2 b  a b + a b 0 sx sx sx

now find  &  for  30°

 1  xcos( )2

 1  54.9 MPa

  xsin( )cos( ) su2  s xcosa u +

Problem 2.6-17 The normal stress on plane pq of a prismatic bar in tension (see figure) is found to be 7500 psi. On plane rs, which makes an angle   30° with plane pq, the stress is found to be 2500 psi. Determine the maximum normal stress max and maximum shear stress max in the bar.

p 2 b 2

;

  31.7 MPa  2  18.3 MPa

; ;

p r b P

P s q

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SECTION 2.6

Solution 2.6-17

193

Stresses on Inclined Sections

Bar in tension SUBSTITUTE NUMERICAL VALUES INTO EQ. (2): cosu1 )

cos(u1 + 30°



7500 psi

A 2500 psi

 23  1.7321

Solve by iteration or a computer program:

1  30°

Eq. (2-29a):

MAXIMUM NORMAL STRESS (FROM EQ. 1)

  xcos2

  30°

smax  sx 

PLANE pq: 1  xcos2 1

1  7500 psi

PLANE rs: 2  xcos2( 1  )

2  2500 psi

s1 cos2u1



s2 cos2(u1 + b)

2

cos u1

 10,000 psi



7500 psi cos2 30°

;

MAXIMUM SHEAR STRESS

Equate x from 1 and 2: sx 

s1

(Eq. 1)

tmax 

sx  5,000 psi 2

;

or cos2u1 2

cos (u1 + b)



cosu1 s1 s1  s2 cos(u1 + b) A s2

(Eq. 2)

Problem 2.6-18 A tension member is to be constructed of two pieces of plastic glued along plane pq (see figure). For purposes of cutting and gluing, the angle must be between 25° and 45°. The allowable stresses on the glued joint in tension and shear are 5.0 MPa and 3.0 MPa, respectively.

u

p

P

P

q

(a) Determine the angle so that the bar will carry the largest load P. (Assume that the strength of the glued joint controls the design.) (b) Determine the maximum allowable load Pmax if the cross-sectional area of the bar is 225 mm2.

Solution 2.6-18

Bar in tension with glued joint ALLOWABLE STRESS x IN TENSION

  xcos2

sx 

su 2

cos u



5.0 MPa cos2u

(1)

  xsin cos

25°   45°

Since the direction of  is immaterial, we can write:  |  xsin cos

A  225 mm2 On glued joint: allow  5.0 MPa

allow  3.0 MPa

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Axially Loaded Members

or sx 

(a) DETERMINE ANGLE  FOR LARGEST LOAD |tu| sin u cosu



3.0 MPa sin u cosu

Point A gives the largest value of x and hence the largest load. To determine the angle corresponding to point A, we equate Eqs. (1) and (2).

(2)

GRAPH OF EQS. (1) AND (2) 5.0 MPa 2

cos u



tan u 

3.0 MPa sin u cos u 3.0 5.0

u  30.96°

;

(b) DETERMINE THE MAXIMUM LOAD From Eq. (1) or Eq. (2): sx 

5.0 MPa 2

cos u



3.0 MPa  6.80 MPa sin u cos u

Pmax  xA  (6.80 MPa)(225 mm2)  1.53 kN

Problem 2.6-19 A nonprismatic bar 1–2–3 of rectangular cross section (cross sectional area A) and two materials is held snugly (but without any initial stress) between rigid supports (see figure). The allowable stresses in compression and in shear are specified as a and a, respectively. Use the following numerical data: (Data: b1  4b2/3  b; A1  2A2  A; E1  3E2/4  E; 1  5 2/4  ; a1  4a2/3  a, a1  2a1/5, a2  3a2/5; let a  11 ksi, P  12 kips, A  6 in.2, b  8 in. E  30,000 ksi,  6.5  10-6/°F; 1  52/3    490 lb/ft3) (a) If load P is applied at joint 2 as shown, find an expression for the maximum permissible temperature rise Tmax so that the allowable stresses are not to be exceeded at either location A or B. (b) If load P is removed and the bar is now rotated to a vertical position where it hangs under its own weight (load intensity  w1 in segment 1–2 and w2 in segment 2–3), find an expression for the maximum permissible temperature rise Tmax so that the allowable stresses are not exceeded at either location 1 or 3. Locations 1 and 3 are each a short distance from the supports at 1 and 3 respectively.

;

b1

b2

1

2

P

A

B E2, A2, a2

E1, A1, a1

(a)

R1 1 W w1 = —1 b1

E1, A1, b1 2

W w2 = —2 b2

E2, A2, b2 3

R3 (b)

3

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SECTION 2.6

Stresses on Inclined Sections

195

Solution 2.6-19 (a) STAT-INDET NONPRISMATIC BAR WITH LOAD P AT jt 2 apply load P and temp. change T - use R3 as redundant & do superposition analysis

¢Tmax 

85s a A + 45 P 64EAa

 3a  Pf12  ( 1b1  2b2)T  3b  R3(f12  f23) f12

b1  E1A1

f23

 Pf12  1a1b1 + a2 b22¢T

R3 

f12 + f23  compression at Location B due to both P and temp. increase  Pf23 + 1a1 b1 + a2b22¢T f12 + f23



aa b + ¢Tmax 

 compression due to temp. increase, tension due to P, at Location A

3 sa 4 9 sa ta2  20

sa2 

2 sa 5 Numerical data ta1 

a  11 ksi A  6 in2 P  12 kips  6.5  (106)/°F steel E  30000 ksi (1) check normal and shear stresses at element A location & solve for Tmax using a1 & a1 sxA  ¢Tmax  f12

¢Tmax 

b1  E1 A1

f23

aa b +

4 3 a bb 5 4

Tmax  67.1°F

R3 A2 [s a2 A2 ( f12 + f23)] + P f12 ¢Tmax  (a1 b1 + a2 b2)  compression due to both temp. rise & load P s xB 

¢Tmax



3 b 3 A b 4 b ≤¥  P ≥ sa ± + 4 2 EA 4 A EA E 3 2

b2  E2 A2

3 3 b b b 4 4 ± ≤ ± ≤ ≥saA + ¥ + P EA 4 A 4 A E E 3 2 3 2

68s aA + 45P 64EAa

(2) check normal and shear stresses at element B location & solve for Tmax using a2 & a2

R1 A1 [s a1 A1 ( f12 + f23)] + P f23 (a1 b1 + a2 b2)

4 3 a bb 5 4

 shear controls for Location A where temp. rise causes compressive stress but ; load P causes tensile stress

numerical data & allowable stresses (normal & shear)

a1  a

sxA 2

3 3 b b b 4 4 2 ≤ + P + 2a s ab A ± 5 EA 4 A 4 A E E 3 2 3 2

R1  P  R3

statics:

compression due to temp. rise but tension due to P

¢Tmax

 3a   3b  0

compatibility:

R1 

max A 

b2  E2A2

Tmax  82.1°F

¢Tmax 

a ab +

4 3 a bb 5 4

255s a A  320P 512EAa

¢ Tmax  21.7°F

;

normal stress controls for Location B where temp. rise & load P both cause compressive stress; as a result, permissible temp. rise is reduced at B compared to Location A where temp. rise effect is offset by load P effect

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tmax B  ¢ Tmax 

Page 196

Axially Loaded Members

sxB 2 [2t a A2(f12 + f23)]  Pf12 (a1b1 + a2b2)

Location A where temp. riseeffect is offset by load P effect

3 b 9 A b 4 b ≤¥  P ≥2 a sab ± + 20 2 EA 4 A EA E 3 2

¢Tmax 

aa b +

¢ Tmax 

4 3 a bb 5 4

153s a A  160 P 256 E Aa

Tmax  27.3°F

(b) STAT-INDET NONPRISMATIC BAR HANGING UNDER ITS OWN WEIGHT (GRAVITY) apply gravity and temp. change T - use R3 as redundant & do superposition analysis d3a 

W1 W2 f12  W2 f12  f23 2 2  (a1b1 + a2 b2) ¢T

3b  R3(f12  f23) f12 

b1 E1A1

f23 

b2 E2 A2

3a  3b  0

compatibility:

R3 

a

W1 W2 f + W2 f12 + f b + (a1b1 + a2 b2) ¢T 2 12 2 23 f12 + f23 ^ compression at Location 3 due to both P and temp. increase

statics:

R1  W1  W2  R3

R1  W1 + W2 

a

W1 W2 f W f + f b + (a1 b1 + a2 b2)¢T 2 12 2 12 2 23 f12 + f23 ^ compression at Location 1 due to temp. increase, tension due to W1 & W2

s x1 

R1 A1

tmax1 

s x1 2

s x3 

R3 A2

tmax3 

s x3 A2

numerical data & allowables stresses (normal & shear)

a1  a a  11 ksi b  8 in.

s a2 

3 s 4 a

A  6 in2 g

0.490 123

t a1 

2 s 5 a

E  30000 ksi k/in3

t a2 

9 s 20 a

 6.5  (106)/°F

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SECTION 2.6

Stresses on Inclined Sections

197

(1) check normal and shear stresses at element 1 location & solve for Tmax using a1 & a1 normal stress

sa1A1  W1 + W2 

¢Tmax 

a

W1 W2 f12 + W2 f12 + f b + (a1b1 + a2 b2)¢T 2 2 23 f12 + f23

g1A1b1f12 + 2(g1A1b1)f23 + g2A2 b2f23  2 s a1A1( f12 + f23) 2(a1b1 + a2 b2) 3 3 3 b b b b 4 3 A 3 4 b 4 ≤ ≥  gAb + 2(g Ab) + g a bb ¥ + 2 salA ± + EA 4 A 5 2 4 4 A EA 4 A E E E 3 2 3 2 3 2

¢Tmax 

2a a b +

1121g b + 1360sa ¢Tmax  1024E a

4 3 a bb 5 4

^ sign difference because gravity offsets effect of temp. rise

Tmax  74.9°F

Next, shear stress 1121g b + 1360a2 ¢Tmax 

2 s b 5 a

Tmax  59.9°F

1024E a

(2) check normal and shear stresses at element 3 location & solve for Tmax using a2 & a2 normal stress

¢ Tmax 

¢Tmax 

s a2A2( f12 + f23) + a

W1 W2 f + W2 f12 + f b 2 12 2 23

a1b1 + a2 b2

same sign because temp. rise & gravity both produce compressive stress at element 3

3 3 A 3 3 b g a bb b A b 4 3 g Ab b 5 2 4 4 3 A 3 b ≥ + ¥ + ≥ a sab + g a bb + + ¥ 4 2 EA 4 A 2 EA 5 2 4 EA 2 4 A E E 3 2 3 2 ab +

¢Tmax 

510sa + 545 g b 1024E a

4 3 a b 5 4

Tmax  28.1°F

shear stress

¢Tmax 

510 a2

9 s b + 545g b 20 a 1024 E a

¢Tmax  25.3°F

;

shear at element 3 location controls

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Axially Loaded Members

Strain Energy When solving the problems for Section 2.7, assume that the material behaves linearly elastically.

Problem 2.7-1 A prismatic bar AD of length L, cross-sectional area A, and modulus of elasticity E is subjected to loads 5P, 3P, and P acting at points B, C, and D, respectively (see figure). Segments AB, BC, and CD have lengths L/6, L/2, and L/3, respectively. (a) Obtain a formula for the strain energy U of the bar. (b) Calculate the strain energy if P ⫽ 6 k, L ⫽ 52 in., A ⫽ 2.76 in.2, and the material is aluminum with E ⫽ 10.4 ⫻ 106 psi.

Solution 2.7-1

Bar with three loads

(a) STRAIN ENERGY OF THE BAR (EQ. 2-40)

P⫽6k L ⫽ 52 in.

U⫽ g

E ⫽ 10.4 ⫻ 10 psi 6

A ⫽ 2.76 in.2

⫽

L L L 1 c(3P)2 a b + (⫺2P)2 a b + (P)2 a b d 2EA 6 2 3

⫽

23P2L P2L 23 a b ⫽ 2EA 6 12EA

INTERNAL AXIAL FORCES NAB ⫽ 3P

NBC ⫽ ⫺2P

NCD ⫽ P

LENGTHS LAB ⫽

L 6

LBC ⫽

L 2

LCD ⫽

L 3

N2i Li 2EiAi

;

(b) SUBSTITUTE NUMERICAL VALUES: U⫽

23(6 k)2(52 in.) 12(10.4 * 106 psi)(2.76 in.2)

⫽ 125 in.-lb

;

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SECTION 2.7

Strain Energy

Problem 2.7-2 A bar of circular cross section having two different diameters d and 2d is shown in the figure. The length of each segment of the bar is L/2 and the modulus of elasticity of the material is E. (a) Obtain a formula for the strain energy U of the bar due to the load P. (b) Calculate the strain energy if the load P ⫽ 27 kN, the length L ⫽ 600 mm, the diameter d ⫽ 40 mm, and the material is brass with E ⫽ 105 GPa.

Solution 2.7-2

Bar with two segments

(b) SUBSTITUTE NUMERICAL VALUES:

(a) STRAIN ENERGY OF THE BAR Add the strain energies of the two segments of the bar (see Eq. 2-40). P2(L/2) N2i Li 1 1 ⫽ cp ⫹p 2 d 2 i⫽1 2 EiAi 2E (2d) (d ) 4 4 2

U⫽ g

1 5P2L P2L 1 a 2 + 2b ⫽ ⫽ pE 4d d 4pEd2

P ⫽ 27 kN

L ⫽ 600 mm

d ⫽ 40 mm

E ⫽ 105 GPa

U⫽

5(27 kN2)(600 mm) 4p(105 GPa)(40 mm)2

; ⫽ 1.036 N # m ⫽ 1.036 J

Problem 2.7-3 A three-story steel column in a building supports roof and floor loads as shown in the figure. The story height H is 10.5 ft, the cross-sectional area A of the column is 15.5 in.2, and the modulus of elasticity E of the steel is 30 ⫻ 106 psi. Calculate the strain energy U of the column assuming P1 ⫽ 40 k and P2 ⫽ P3 ⫽ 60 k.

;

199

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Solution 2.7-3

Page 200

Axially Loaded Members

Three-story column Upper segment: N1 ⫽ ⫺P1 Middle segment: N2 ⫽ ⫺(P1 ⫹ P2) Lower segment: N3 ⫽ ⫺(P1 ⫹ P2 ⫹ P3) STRAIN ENERGY U⫽ g

N2i Li 2EiAi

⫽

H 2 [P + (P1 + P2)2 + (P1 + P2 + P3)2] 2EA 1

⫽

H [Q] 2EA

[Q] ⫽ (40 k)2 + (100 k)2 + (160 k)2 ⫽ 37,200 k2 H ⫽ 10.5 ft

E ⫽ 30 ⫻ 106 psi

A ⫽ 15.5 in.

2

2EA ⫽ 2(30 * 106 psi)(15.5 in.2) ⫽ 930 * 106 lb

P1 ⫽ 40 k

P2 ⫽ P3 ⫽ 60 k To find the strain energy of the column, add the strain energies of the three segments (see Eq. 2-40).

U⫽

(10.5 ft)(12 in./ft) 930 * 106 lb

⫽ 5040 in.-lb

Problem 2.7-4 The bar ABC shown in the figure is loaded by a force P acting at end C and by a force Q acting at the midpoint B. The bar has constant axial rigidity EA. (a) Determine the strain energy U1 of the bar when the force P acts alone (Q ⫽ 0). (b) Determine the strain energy U2 when the force Q acts alone (P ⫽ 0). (c) Determine the strain energy U3 when the forces P and Q act simultaneously upon the bar.

;

[37,200 k2]

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SECTION 2.7

Solution 2.7-4

Strain Energy

201

Bar with two loads (c) FORCES P AND Q ACT SIMULTANEOUSLY

(a) FORCE P ACTS ALONE (Q ⫽ 0) U1 ⫽

P2L 2EA

Segment BC: UBC ⫽

P2(L/2) P2L ⫽ 2EA 4EA

Segment AB: UAB ⫽

(P + Q)2(L/2) 2EA

; ⫽

PQL Q 2L P2L + + 4EA 2EA 4EA

U3 ⫽ UBC + UAB ⫽

PQL Q2L P2L + + 2EA 2EA 4EA

(b) FORCE Q ACTS ALONE (P ⫽ 0) U2 ⫽

Q2(L/2) Q2L ⫽ 2EA 4EA

;

;

(Note that U3 is not equal to U1 ⫹ U2. In this case, U3 ⬎ U1 ⫹ U2. However, if Q is reversed in direction, U3 ⬍ U1 ⫹ U2. Thus, U3 may be larger or smaller than U1 ⫹ U2.)

Problem 2.7-5 Determine the strain energy per unit volume (units of psi) and the strain energy per unit weight (units of in.) that can be stored in each of the materials listed in the accompanying table, assuming that the material is stressed to the proportional limit. DATA FOR PROBLEM 2.7-5

Material

Weight density (lb/in.3)

Modulus of elasticity (ksi)

Proportional limit (psi)

Mild steel Tool steel Aluminum Rubber (soft)

0.284 0.284 0.0984 0.0405

30,000 30,000 10,500 0.300

36,000 75,000 60,000 300

Solution 2.7-5

Strain-energy density STRAIN ENERGY PER UNIT VOLUME

DATA:

Material

Weight density (lb/in.3)

Modulus of elasticity (ksi)

Proportional limit (psi)

Mild steel Tool steel Aluminum Rubber (soft)

0.284 0.284 0.0984 0.0405

30,000 30,000 10,500 0.300

36,000 75,000 60,000 300

U⫽

P2L 2EA

Volume V ⫽ AL Stress s ⫽

u⫽

s2PL U ⫽ V 2E

P A

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Axially Loaded Members

At the proportional limit:

At the proportional limit:

u ⫽ uR ⫽ modulus of resistance

uW ⫽

uR ⫽

s2PL

s2PL 2gE

(Eq. 2)

(Eq. 1)

2E

RESULTS

STRAIN ENERGY PER UNIT WEIGHT U⫽

2

PL 2EA

Weight W ⫽ gAL

␥ ⫽ weight density uW ⫽

Mild steel Tool steel Aluminum Rubber (soft)

uR (psi)

uw (in.)

22 94 171 150

76 330 1740 3700

s2 U ⫽ W 2gE

Problem 2.7-6 The truss ABC shown in the figure is subjected to a horizontal load P at joint B. The two bars are identical with cross-sectional area A and modulus of elasticity E. (a) Determine the strain energy U of the truss if the angle ␤ ⫽ 60°. (b) Determine the horizontal displacement ␦B of joint B by equating the strain energy of the truss to the work done by the load.

Truss subjected to a load P

↓

Solution 2.7-6

↓⫺

␤ ⫽ 60°

⌺Fvert ⫽ 0

LAB ⫽ LBC ⫽ L

⫺FAB sin ␤ ⫹ FBC sin ␤ ⫽ 0

sin b ⫽ 13/2

FAB ⫽ FBC

cos ␤ ⫽ 1/2

⌺Fhoriz ⫽ 0 : ←

FREE-BODY DIAGRAM OF JOINT B

⫺FAB cos ␤ ⫺ FBC cos ␤ ⫹ P ⫽ 0 FAB ⫽ FBC ⫽

⫹

(Eq. 1)

P P ⫽ ⫽P 2 cos b 2(1/2)

(Eq. 2)

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SECTION 2.7

NBC ⫽ ⫺P (compression)

dB ⫽

(a) STRAIN ENERGY OF TRUSS (EQ. 2-40) (NBC)2L (NAB)2L N2i Li P2L ⫽ + ⫽ 2EiAi 2EA 2EA EA

2 P2L 2PL 2U ⫽ a b ⫽ P P EA EA

;

;

Problem 2.7-7 The truss ABC shown in the figure supports a

A

horizontal load P1 ⫽ 300 lb and a vertical load P2 ⫽ 900 lb. Both bars have cross-sectional area A ⫽ 2.4 in.2 and are made of steel with E ⫽ 30 ⫻ 106 psi. (a) Determine the strain energy U1 of the truss when the load P1 acts alone (P2 ⫽ 0). (b) Determine the strain energy U2 when the load P2 acts alone (P1 ⫽ 0). (c) Determine the strain energy U3 when both loads act simultaneously.

Solution 2.7-7

203

(b) HORIZONTAL DISPLACEMENT OF JOINT B (EQ. 2-42)

Axial forces: NAB ⫽ P (tension)

U⫽ g

Strain Energy

30∞

C

B P1 = 300 lb P2 = 900 lb

60 in.

Truss with two loads LAB ⫽

LBC 120 in. ⫽ 69.282 in. ⫽ cos 30° 13

2EA ⫽ 2(30 ⫻ 106 psi)(2.4 in.2) ⫽ 144 ⫻ 106 lb FORCES FAB AND FBC IN THE BARS From equilibrium of joint B: FAB ⫽ 2P2 ⫽ 1800 lb FBC ⫽ P1 ⫺ P2 13 ⫽ 300 lb ⫺ 1558.8 lb P1 ⫽ 300 lb P2 ⫽ 900 lb A ⫽ 2.4 in.2 E ⫽ 30 ⫻ 106 psi LBC ⫽ 60 in.

Force

P1 alone

FAB FBC

0 300 lb

13 cos b ⫽ cos 30° ⫽ 2

P1 and P2

1800 lb ⫺1558.8 lb

1800 lb ⫺1258.8 lb

(a) LOAD P1 ACTS ALONE U1 ⫽

␤ ⫽ 30° 1 sin b ⫽ sin 30° ⫽ 2

P2 alone

(FBC)2LBC (300 lb)2(60 in.) ⫽ 2EA 144 * 106 lb

⫽ 0.0375 in.-lb

;

(b) LOAD P2 ACTS ALONE U2 ⫽

1 c(F )2L + (FBC)2LBC d 2EA AB AB

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⫽

Page 204

Axially Loaded Members

1 c(1800 lb)2(69.282 in.) 2EA

⫽

+ ( ⫺1558.8 lb)2(60 in.) d ⫽

370.265 * 106 lb2-in. 144 * 106 lb

+ (⫺ 1258.8 lb)2(60 in.) d

⫽ 2.57 in.-lb ;

⫽

(c) LOADS P1 AND P2 ACT SIMULTANEOUSLY U3 ⫽

1 c(F )2L + (FBC)2LBC d 2EA AB AB

1 c(1800 lb)2(69.282 in.) 2EA

319.548 * 106 lb2-in. 144 * 106 lb

⫽ 2.22 i n.- lb

;

NOTE: The strain energy U3 is not equal to U1 ⫹ U2.

Problem 2.7-8 The statically indeterminate structure shown in the figure consists of a horizontal rigid bar AB supported by five equally spaced springs. Springs 1, 2, and 3 have stiffnesses 3k, 1.5k, and k, respectively. When unstressed, the lower ends of all five springs lie along a horizontal line. Bar AB, which has weight W, causes the springs to elongate by an amount ␦. (a) Obtain a formula for the total strain energy U of the springs in terms of the downward displacement ␦ of the bar. (b) Obtain a formula for the displacement ␦ by equating the strain energy of the springs to the work done by the weight W. (c) Determine the forces F1, F2, and F3 in the springs. (d) Evaluate the strain energy U, the displacement ␦, and the forces in the springs if W ⫽ 600 N and k ⫽ 7.5 N/mm.

1

3k

k

1.5k 2

3

1.5k 2

A

1

3k B

W

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SECTION 2.7

Solution 2.7-8

205

Strain Energy

Rigid bar supported by springs (c) FORCES IN THE SPRINGS F1 ⫽ 3kd ⫽ F3 ⫽ kd ⫽

3W 3W F2 ⫽ 1.5kd ⫽ 10 20

W 10

;

(d) NUMERICAL VALUES

k1 ⫽ 3k

W ⫽ 600 N k ⫽ 7.5 N/mm ⫽ 7500 N/mm

k2 ⫽ 1.5k k3 ⫽ k

U ⫽ 5kd2 ⫽ 5ka

␦ ⫽ downward displacement of rigid bar kd2 Eq. (2-38b) 2 (a) STRAIN ENERGY U OF ALL SPRINGS

W 2 W2 b ⫽ 10k 20k

⫽ 2.4 N # m ⫽ 2.4 J

For a spring: U ⫽

d⫽

W ⫽ 8.0 mm 10k

F1 ⫽

3W ⫽ 180 N 10

Wd 2

F2 ⫽

3W ⫽ 90 N 20

Strain energy of the springs equals 5k␦2

F3 ⫽

W ⫽ 60 N 10

U ⫽ 2a

2

2

3kd 1.5kd kd b + 2a b + 2 2 2

2

⫽ 5kd2

;

(b) DISPLACEMENT ␦ Work done by the weight W equals

...

;

Wd ⫽ 5kd2 2

and d ⫽

W 10k

;

(a) Determine the strain energy U of the bar. (b) Determine the elongation ␦ of the bar by equating the strain energy to the work done by the force P.

;

;

;

;

NOTE: W ⫽ 2F1 ⫹ 2F2 ⫹ F3 ⫽ 600 N (Check)

Problem 2.7-9 A slightly tapered bar AB of rectangular cross section and length L is acted upon by a force P (see figure). The width of the bar varies uniformly from b2 at end A to b1 at end B. The thickness t is constant.

;

A

B

b2

L

b1 P

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Axially Loaded Members

Solution 2.7-9

Tapered bar of rectangular cross section Apply this integration formula to Eq. (1): U⫽

b(x) ⫽ b2 ⫺

⫽

(b2 ⫺ b1)x L

U⫽

A(x) ⫽ tb(x) ⫽ t cb2 ⫺

(b2 ⫺ b1)x d L

b2 PL 2U ⫽ ln P Et(b2 ⫺ b1) b1

;

NOTE: This result agrees with the formula derived in Prob. 2.3-13. (1)

1 dx ⫽ ln (a + bx) b L a + bx

Problem 2.7-10 A compressive load P is transmitted through a rigid plate to three magnesium-alloy bars that are identical except that initially the middle bar is slightly shorter than the other bars (see figure). The dimensions and properties of the assembly are as follows: length L ⫽ 1.0 m, cross-sectional area of each bar A ⫽ 3000 mm2, modulus of elasticity E ⫽ 45 GPa, and the gap s ⫽ 1.0 mm. (a) (b) (c) (d)

;

L

P2 P2dx dx ⫽ ⫽ x 2Etb(x) 2Et b ⫺ (b L0 L0 2 2 ⫺ b1)L From Appendix C:

b2 P2L ln 2Et(b2 ⫺ b1) b1

d⫽

[N(x)]2dx ( Eq. 2- 41) L 2EA(x) L

P2 ⫺L ⫺L c ln b1 ⫺ ln b2 d 2Et (b2 ⫺ b1) (b2 ⫺ b1)

(b) ELONGATION OF THE BAR (EQ. 2-42)

(a) STRAIN ENERGY OF THE BAR U⫽

(b2 ⫺ b1)x L P2 1 ln cb2 ⫺ c dd 1 2Et ⫺(b2 ⫺ b1)1 2 L 0 L

Calculate the load P1 required to close the gap. Calculate the downward displacement ␦ of the rigid plate when P ⫽ 400 kN. Calculate the total strain energy U of the three bars when P ⫽ 400 kN. Explain why the strain energy U is not equal to P␦/2. (Hint: Draw a load-displacement diagram.)

P s

L

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SECTION 2.7

Solution 2.7-10

Strain Energy

207

Three bars in compression (c) STRAIN ENERGY U FOR P ⫽ 400 kN U⫽ g

EAd2 2L

Outer bars:

␦ ⫽ 1.321 mm

Middle bar:

␦ ⫽ 1.321 mm ⫺ s ⫽ 0.321 mm

U⫽ s ⫽ 1.0 mm L ⫽ 1.0 m

1 ⫽ (135 * 106 N/m)(3.593 mm2) 2

For each bar:

⫽ 243 N # m ⫽ 243 J

A ⫽ 3000 mm2 E ⫽ 45 GPa

;

(d) LOAD-DISPLACEMENT DIAGRAM

EA ⫽ 135 * 106 N/m L

U ⫽ 243 J ⫽ 243 N ⭈ m

(a) LOAD P1 REQUIRED TO CLOSE THE GAP EAd PL In general, d ⫽ and P ⫽ EA L For two bars, we obtain: P1 ⫽ 2 a

EA [2(1.321 mm)2 + (0.321 mm)2] 2L

Pd 1 ⫽ (400 kN)(1.321 mm) ⫽ 264 N # m 2 2 Pd ⫽ because the 2 load-displacement relation is not linear. The strain energy U is not equal to

EAs b ⫽ 2(135 * 106 N/m)(1.0 mm) L

P1 ⫽ 270 kN

;

(b) DISPLACEMENT ␦ FOR P ⫽ 400 kN Since P ⬎ P1, all three bars are compressed. The force P equals P1 plus the additional force required to compress all three bars by the amount ␦ ⫺ s. P ⫽ P1 + 3 a

EA b(d ⫺ s) L

U ⫽ area under line OAB.

or 400 kN ⫽ 270 kN ⫹ 3(135 ⫻ 106 N/m) (␦ ⫺ 0.001 m) Solving, we get ␦ ⫽ 1.321 mm

;

Pd ⫽ area under a straight line from O to B, which is 2 larger than U.

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Axially Loaded Members

Problem 2.7-11 A block B is pushed against three springs by a force P (see figure). The middle spring has stiffness k1 and the outer springs each have stiffness k2. Initially, the springs are unstressed and the middle spring is longer than the outer springs (the difference in length is denoted s). (a) Draw a force-displacement diagram with the force P as ordinate and the displacement x of the block as abscissa. (b) From the diagram, determine the strain energy U1 of the springs when x ⫽ 2s. (c) Explain why the strain energy U1 is not equal to P␦/2, where ␦ ⫽ 2s.

Solution 2.7-11

Block pushed against three springs

Force P0 required to close the gap: P0 ⫽ k1s

(a) FORCE-DISPLACEMENT DIAGRAM (1)

FORCE-DISPLACEMENT RELATION BEFORE GAP IS CLOSED P ⫽ k1x

(0 ⱕ x ⱕ s)(0 ⱕ P ⱕ P0)

(2)

FORCE-DISPLACEMENT RELATION AFTER GAP IS CLOSED All three springs are compressed. Total stiffness equals k1 ⫹ 2k2. Additional displacement equals x ⫺ s. Force P equals P0 plus the force required to compress all three springs by the amount x ⫺ s. P ⫽ P0 ⫹ (k1 ⫹ 2k2)(x ⫺ s) (b) STRAIN ENERGY U1 WHEN x ⫽ 2s

⫽ k1s ⫹ (k1 ⫹ 2k2)x ⫺ k1s ⫺ 2k2s P ⫽ (k1 ⫹ 2k2)x ⫺ 2k2s

(x ⱖ s); (P ⱖ P0)

(3)

P1 ⫽ force P when x ⫽ 2s

⫽

Substitute x ⫽ 2s into Eq. (3): P1 ⫽ 2(k1 ⫹ k2)s

U1 ⫽ Area below force - displacement curve

(4)

⫹

⫹

1 1 1 ⫽ P0s + P0s + (P1 ⫺ P0)s ⫽ P0s + P1s 2 2 2 ⫽ k1s2 + (k1 + k2)s2 U1 ⫽ (2k1 ⫹ k2)s2

;

(5)

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SECTION 2.7

(c) STRAIN ENERGY U1 IS NOT EQUAL TO

Pd 2

Pd 1 ⫽ P1(2 s) ⫽ P1s ⫽ 2(k1 + k2)s2 2 2 (This quantity is greater than U1.) For d ⫽ 2s:

209

Strain Energy

Pd ⫽ area under a straight line from O to B, which 2 is larger than U1. Pd Thus, is not equal to the strain energy because 2 the force-displacement relation is not linear.

U1 ⫽ area under line OAB.

Problem 2.7-12 A bungee cord that behaves linearly

elastically has an unstressed length L0 ⫽ 760 mm and a stiffness k ⫽ 140 N/m.The cord is attached to two pegs, distance b ⫽ 380 mm apart, and pulled at its midpoint by a force P ⫽ 80 N (see figure).

b

A

B

(a) How much strain energy U is stored in the cord? (b) What is the displacement ␦C of the point where the load is applied? (c) Compare the strain energy U with the quantity P␦C/2. (Note: The elongation of the cord is not small compared to its original length.)

Solution 2.7-12

C P

Bungee cord subjected to a load P.

DIMENSIONS BEFORE THE LOAD P IS APPLIED

From triangle ACD: 1 d ⫽ 2L20 ⫺ b2 ⫽ 329.09 mm 2 DIMENSIONS AFTER THE LOAD P IS APPLIED

L0 ⫽ 760 mm

L0 ⫽ 380 mm 2

b ⫽ 380 mm

Let x ⫽ distance CD k ⫽ 140 N/m

Let L1 ⫽ stretched length of bungee cord

(1)

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From triangle ACD:

L1 ⫽ L0 +

or

L1 b 2 ⫽ a b + x2 2 A 2

(2)

L1 ⫽ 2b2 + 4x2

(3)

L0 ⫽ a1 ⫺

P 1 2 2 2 b + 4x2 ⫽ 1b + 4x 4kx

P 1 2 b b + 4x2 4kx

(7)

This equation can be solved for x.

EQUILIBRIUM AT POINT C

SUBSTITUTE NUMERICAL VALUES INTO EQ. (7):

Let F ⫽ tensile force in bungee cord

760 mm ⫽ c1 ⫺

(80 N)(1000 mm/m) d 4(140 N/m)x

* 1(380 mm)2 + 4x2 760 ⫽ a1 ⫺ L1/2 F P L1 1 ⫽ F ⫽ a ba ba b P/2 x 2 2 x

142.857 1 b 144,400 + 4x2 x

(4)

kd2 2 From Eq. (5):

Let ␦ ⫽ elongation of the entire bungee cord F P b2 1 + 2 ⫽ k 2k A 4x

(5)

Final length of bungee cord ⫽ original length ⫹ ␦ P b2 L1 ⫽ L0 + d ⫽ L0 + 1 + 2 2k A 4x SOLUTION OF EQUATIONS Combine Eqs. (6) and (3): L1 ⫽ L0 +

2

P b 1 + 2 ⫽ 1b2 + 4x2 2k A 4x

(a) STRAIN ENERGY U OF THE BUNGEE CORD U⫽

ELONGATION OF BUNGEE CORD

d⫽

(9)

Units: x is in millimeters Solve for x (Use trial & error or a computer program): x ⫽ 497.88 mm

P b 2 ⫽ 1 + a b 2A 2x

(8)

(6)

d⫽

k ⫽ 140 N/m

P ⫽ 80 N

b2 P 1 + 2 ⫽ 305.81 mm 2k A 4x

1 U ⫽ (140 N/m)(305.81 mm)2 ⫽ 6.55 N.m 2 U ⫽ 6.55 J

;

(b) DISPLACEMENT ␦C OF POINT C

␦C ⫽ x ⫺ d ⫽ 497.88 mm ⫺ 329.09 mm ⫽ 168.8 mm

;

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SECTION 2.7

(c) COMPARISON OF STRAIN ENERGY U WITH THE QUANTITY P␦C/2 U ⫽ 6.55 J PdC 1 ⫽ (80 N)(168.8 mm) ⫽ 6.75 J 2 2 The two quantities are not the same. The work done by the load P is not equal to P␦C/2 because the loaddisplacement relation (see below) is non-linear when the displacements are large. (The work done by the load P is equal to the strain energy because the bungee cord behaves elastically and there are no energy losses.) U ⫽ area OAB under the curve OA. PdC ⫽ area of triangle OAB, which is greater than U. 2

Strain Energy

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Impact Loading The problems for Section 2.8 are to be solved on the basis of the assumptions and idealizations described in the text. In particular, assume that the material behaves linearly elastically and no energy is lost during the impact.

Collar

Problem 2.8-1 A sliding collar of weight W ⫽ 150 lb falls from a height

h ⫽ 2.0 in. onto a flange at the bottom of a slender vertical rod (see figure). The rod has length L ⫽ 4.0 ft, cross-sectional area A ⫽ 0.75 in.2, and modulus of elasticity E ⫽ 30 ⫻ 106 psi. Calculate the following quantities: (a) the maximum downward displacement of the flange, (b) the maximum tensile stress in the rod, and (c) the impact factor.

L

Rod h Flange

Probs. 2.8-1, 2.8-2, 2.8-3

Solution 2.8-1

Collar falling onto a flange (a) DOWNWARD DISPLACEMENT OF FLANGE dst ⫽

WL ⫽ 0.00032 in. EA

Eq. of (2-53): dmax ⫽ dst c1 + a1 + ⫽ 0.0361 in.

2h 1/2 b d dst

;

(b) MAXIMUM TENSILE STRESS (EQ. 2-55) smax ⫽

Edmax ⫽ 22,600 psi L

;

(c) IMPACT FACTOR (EQ. 2-61) Impact factor ⫽

W ⫽ 150 lb h ⫽ 2.0 in.

L ⫽ 4.0 ft ⫽ 48 in.

E ⫽ 30 ⫻ 10 psi 6

A ⫽ 0.75 in.

2

dmax 0.0361 in. ⫽ dst 0.00032 in.

⫽ 113

;

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SECTION 2.8

Impact Loading

Problem 2.8-2 Solve the preceding problem if the collar has mass

M ⫽ 80 kg, the height h ⫽ 0.5 m, the length L ⫽ 3.0 m, the cross-sectional area A ⫽ 350 mm2, and the modulus of elasticity E ⫽ 170 GPa.

Solution 2.8-2

Collar falling onto a flange

(a) DOWNWARD DISPLACEMENT OF FLANGE dst ⫽

WL ⫽ 0.03957 mm EA

Eq. (2-53): dmax ⫽ dst c1 + a1 + ⫽ 6.33 mm

2h 1/2 b d dst

;

(b) MAXIMUM TENSILE STRESS (EQ. 2-55) smax ⫽

Edmax ⫽ 359 MPa L

;

(c) IMPACT FACTOR (EQ. 2–61)

M ⫽ 80 kg

Impact factor ⫽

W ⫽ Mg ⫽ (80 kg)(9.81 m/s2) ⫽ 784.8 N h ⫽ 0.5 m

L ⫽ 3.0 m

E ⫽ 170 GPa

A ⫽ 350 mm2

dmax 6.33 mm ⫽ dst 0.03957 mm ⫽ 160

;

Problem 2.8-3 Solve Problem 2.8-1 if the collar has weight W ⫽ 50 lb, the height h ⫽ 2.0 in., the length L ⫽ 3.0 ft, the cross-sectional area A ⫽ 0.25 in.2, and the modulus of elasticity E ⫽ 30,000 ksi.

Solution 2.8-3

Collar falling onto a flange W ⫽ 50 lb

h ⫽ 2.0 in.

L ⫽ 3.0 ft ⫽ 36 in. E ⫽ 30,000 psi

A ⫽ 0.25 in.2

(a) DOWNWARD DISPLACEMENT OF FLANGE dst ⫽

WL ⫽ 0.00024 in. EA 2h 1/2 b d dst ;

Eq. (2 ⫺ 53): dmax ⫽ dst c1 + a1 + ⫽ 0.0312 in.

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(b) MAXIMUM TENSILE STRESS (EQ. 2–55) smax ⫽

Edmax ⫽ 26,000 psi L

(c) IMPACT FACTOR (EQ. 2-61) Impact factor ⫽

;

dmax 0.0312 in. ⫽ dst 0.00024 in. ⫽ 130 ;

Problem 2.8-4 A block weighing W ⫽ 5.0 N drops inside a cylinder

from a height h ⫽ 200 mm onto a spring having stiffness k ⫽ 90 N/m (see figure).

Block

(a) Determine the maximum shortening of the spring due to the impact, and (b) determine the impact factor.

Cylinder

h

k

Prob. 2.8-4 and 2.8-5

Solution 2.8-4

W ⫽ 5.0 N

Block dropping onto a spring

h ⫽ 200 mm

k ⫽ 90 N/m

(a) MAXIMUM SHORTENING OF THE SPRING dst ⫽

W 5.0 N ⫽ ⫽ 55.56 mm k 90 N/m

Eq. (2-53): dmax ⫽ dst c1 + a1 + ⫽ 215 mm

;

2h 1/2 b d dst

(b) IMPACT FACTOR (EQ. 2-61) Impact factor ⫽

dmax 215 mm ⫽ dst 55.56 mm ⫽ 3.9 ;

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SECTION 2.8

Impact Loading

Problem 2.8-5 Solve the preceding problem if the block weighs W ⫽ 1.0 lb, h ⫽ 12 in., and k ⫽ 0.5 lb/in.

Solution 2.8-5

Block dropping onto a spring

(a) MAXIMUM SHORTENING OF THE SPRING dst ⫽

W 1.0 lb ⫽ ⫽ 2.0 in. k 0.5 lb/in.

Eq. (2-53): dmax ⫽ dst c1 + a1 + ⫽ 9.21 in.

2h 1/2 b d dst

;

(b) IMPACT FACTOR (EQ. 2-61) dmax 9.21 in. ⫽ dst 2.0 in. ⫽ 4.6 ;

Impact factor ⫽

W ⫽ 1.0 lb

h ⫽ 12 in.

k ⫽ 0.5 lb/in.

Problem 2.8-6 A small rubber ball (weight W ⫽ 450 mN) is attached by a rubber cord to a wood paddle (see figure). The natural length of the cord is L0 ⫽ 200 mm, its crosssectional area is A ⫽ 1.6 mm2, and its modulus of elasticity is E ⫽ 2.0 MPa. After being struck by the paddle, the ball stretches the cord to a total length L1 ⫽ 900 mm. What was the velocity v of the ball when it left the paddle? (Assume linearly elastic behavior of the rubber cord, and disregard the potential energy due to any change in elevation of the ball.)

Solution 2.8-6

Rubber ball attached to a paddle WHEN THE RUBBER CORD IS FULLY STRETCHED: U⫽

EAd2 EA ⫽ (L ⫺ L0)2 2L0 2L0 1

CONSERVATION OF ENERGY KE ⫽ U

g ⫽ 9.81 m/s

E ⫽ 2.0 MPa

A ⫽ 1.6 mm

L0 ⫽ 200 mm

L1 ⫽ 900 mm

W ⫽ 450 mN

2

2

WHEN THE BALL LEAVES THE PADDLE Wv2 KE ⫽ 2g

v2 ⫽

Wv2 EA (L1 ⫺ L0)2 ⫽ 2g 2L0

gEA (L ⫺ L0)2 WL0 1 gEA A WL0

v ⫽ (L1 ⫺ L0)

;

SUBSTITUTE NUMERICAL VALUES: (9.81 m/s2) (2.0 MPa) (1.6 mm2) A (450 mN) (200 mm) ⫽ 13.1 m/s ;

v ⫽ (700 mm)

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Problem 2.8-7 A weight W ⫽ 4500 lb falls from a height h onto

a vertical wood pole having length L ⫽ 15 ft, diameter d ⫽ 12 in., and modulus of elasticity E ⫽ 1.6 ⫻ 106 psi (see figure). If the allowable stress in the wood under an impact load is 2500 psi, what is the maximum permissible height h?

W = 4,500 lb h d = 12 in.

L = 15 ft

Solution 2.8-7

Weight falling on a wood pole E ⫽ 1.6 ⫻ 106 psi

␴allow ⫽ 2500 psi (⫽ ␴max) Find hmax STATIC STRESS sst ⫽

W 4500 lb ⫽ 39.79 psi ⫽ A 113.10 in.2

MAXIMUM HEIGHT hmax Eq. (2⫺59): smax ⫽ sst c1 + a1 +

2hE 1/2 b d Lsst

or smax 2hE 1/2 ⫺ 1 ⫽ a1 + b sst Lsst Square both sides and solve for h: h ⫽ hmax ⫽ W ⫽ 4500 lb

d ⫽ 12 in.

L ⫽ 15 ft ⫽ 180 in. A⫽

2

pd ⫽ 113.10 in.2 4

Lsmax smax a ⫺ 2b 2E sst

;

SUBSTITUTE NUMERICAL VALUES: hmax ⫽

(180 in.) (2500 psi) 2500 psi ⫺ 2b a 2(1.6 * 106 psi) 39.79 psi

⫽ 8.55 in.

;

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SECTION 2.8

Impact Loading

Problem 2.8-8 A cable with a restrainer at the bottom hangs vertically from its upper end (see figure). The cable has an effective cross-sectional area A ⫽ 40 mm2 and an effective modulus of elasticity E ⫽ 130 GPa. A slider of mass M ⫽ 35 kg drops from a height h ⫽ 1.0 m onto the restrainer. If the allowable stress in the cable under an impact load is 500 MPa, what is the minimum permissible length L of the cable?

Cable

Slider L

h Restrainer

Probs. 2.8-8, 2.8-2, 2.8-9

Solution 2.8-8

Slider on a cable

STATIC STRESS sst ⫽

W 343.4 N ⫽ 8.585 MPa ⫽ A 40 mm2

MINIMUM LENGTH Lmin Eq. (2⫺59): smax ⫽ sst c1 + a1 +

2hE 1/2 b d Lsst

or smax 2hE 1/2 ⫺ 1 ⫽ a1 + b sst Lsst Square both sides and solve for L: L ⫽ Lmin ⫽

2Ehsst smax(smax ⫺ 2sst)

;

SUBSTITUTE NUMERICAL VALUES: W ⫽ Mg ⫽ (35 kg)(9.81 m/s2) ⫽ 343.4 N A ⫽ 40 mm2 h ⫽ 1.0 m

E ⫽ 130 GPa

␴allow ⫽ ␴max ⫽ 500 MPa

Find minimum length Lmin

Lmin ⫽

2(130 GPa) (1.0 m) (8.585 MPa) (500 MPa) [500 MPa ⫺ 2(8.585 MPa)]

⫽ 9.25 mm

;

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Problem 2.8-9 Solve the preceding problem if the slider has

weight W ⫽ 100 lb, h ⫽ 45 in., A ⫽ 0.080 in.2, E ⫽ 21 ⫻ 106 psi, and the allowable stress is 70 ksi. Cable

Slider L

h Restrainer

Solution 2.8-9

Slider on a cable STATIC STRESS sst ⫽

100 lb W ⫽ ⫽ 1250 psi A 0.080 in.2

MINIMUM LENGTH Lmin Eq. (2⫺59): smax ⫽ sst c1 + a1 +

2hE 1/2 b d Lsst

or smax 2hE 1/2 ⫺ 1 ⫽ a1 + b sst Lsst Square both sides and solve for L: L ⫽ Lmin ⫽

2Ehsst smax(smax ⫺ 2sst)

;

SUBSTITUTE NUMERICAL VALUES: Lmin ⫽

W ⫽ 100 lb A ⫽ 0.080 in.2 h ⫽ 45 in

E ⫽ 21 ⫻ 106 psi

␴allow ⫽ ␴max ⫽ 70 ksi

Find minimum length Lmin

2(21 * 106 psi) (45 in.) (1250 psi) (70,000 psi) [70,000 psi ⫺ 2(1250 psi)]

⫽ 500 in.

;

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SECTION 2.8

Impact Loading

Problem 2.8-10 A bumping post at the end of a track in a railway

yard has a spring constant k ⫽ 8.0 MN/m (see figure). The maximum possible displacement d of the end of the striking plate is 450 mm. What is the maximum velocity ␯max that a railway car of weight W ⫽ 545 kN can have without damaging the bumping post when it strikes it?

Solution 2.8-10

Bumping post for a railway car STRAIN ENERGY WHEN SPRING IS COMPRESSED TO THE MAXIMUM ALLOWABLE AMOUNT

U⫽

kd2max kd2 ⫽ 2 2

CONSERVATION OF ENERGY KE ⫽ U k ⫽ 8.0 MN/m

W ⫽ 545 kN

k

v ⫽ vmax ⫽ d

;

A W/g

d ⫽ maximum displacement of spring d ⫽ ␦max ⫽ 450 mm

Wv2 kd2 2 kd2 v ⫽ ⫽ 2g 2 W/g

SUBSTITUTE NUMERICAL VALUES:

Find ␯max KINETIC ENERGY BEFORE IMPACT Mv2 Wv2 KE ⫽ ⫽ 2 2g

8.0 MN/m

vmax ⫽ (450 mm)

A (545 kN)/(9.81 m/s2)

⫽ 5400 mm/s ⫽ 5.4 m/s

Problem 2.8-11 A bumper for a mine car is constructed with

a spring of stiffness k ⫽ 1120 lb/in. (see figure). If a car weighing 3450 lb is traveling at velocity ␯ ⫽ 7 mph when it strikes the spring, what is the maximum shortening of the spring?

;

v k

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Axially Loaded Members

Bumper for a mine car

k ⫽ 1120 lb/in.

W ⫽ 3450 lb

␯ ⫽ 7 mph ⫽ 123.2 in./sec g ⫽ 32.2 ft/sec2 ⫽ 386.4 in./sec2 Find the shortening ␦max of the spring. KINETIC ENERGY JUST BEFORE IMPACT KE ⫽

Mv2 Wv2 ⫽ 2 2g

Conservation of energy KE ⫽ U

kd2max Wv2 ⫽ 2g 2

Solve for ␦max: dmax ⫽

Wv2 A gk

;

SUBSTITUTE NUMERICAL VALUES: dmax ⫽

STRAIN ENERGY WHEN SPRING IS FULLY COMPRESSED kd2max U⫽ 2

Problem 2.8-12 A bungee jumper having a mass of 55 kg leaps from a bridge, braking her fall with a long elastic shock cord having axial rigidity EA ⫽ 2.3 kN (see figure). If the jumpoff point is 60 m above the water, and if it is desired to maintain a clearance of 10 m between the jumper and the water, what length L of cord should be used?

(3450 lb) (123.2 in./sec)2

A (386.4 in./sec2) (1120 lb/in.)

⫽ 11.0 in.

;

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SECTION 2.8

Solution 2.8-12

Impact Loading

Bungee jumper SOLVE QUADRATIC EQUATION FOR ␦max: dmax ⫽ ⫽

WL WL 2 WL 1/2 + ca b + 2L a bd EA EA EA WL 2EA 1/2 c1 + a1 + b d EA W

VERTICAL HEIGHT h ⫽ C + L + dmax h⫺C⫽L + W ⫽ Mg ⫽ (55 kg)(9.81 m/s2) ⫽ 539.55 N

SOLVE FOR L: h⫺C

L⫽

EA ⫽ 2.3 kN

1 +

Height: h ⫽ 60 m

2EA 1/2 WL c1 + a 1 + b d EA W

W 2EA 1/2 c1 + a 1 + b d EA W

Clearance: C ⫽ 10 m

SUBSTITUTE NUMERICAL VALUES:

Find length L of the bungee cord.

W 539.55 N ⫽ ⫽ 0.234587 EA 2.3 kN

P.E. ⫽ Potential energy of the jumper at the top of bridge (with respect to lowest position) U ⫽ strain energy of cord at lowest position EAd2max 2L

or

W(L + dmax) ⫽

d2max ⫺

* c1 + a 1 + ⫽ 1.9586 50 m ⫽ 25.5 m L⫽ 1.9586

CONSERVATION OF ENERGY P.E. ⫽ U

Numerator ⫽ h ⫺ C ⫽ 60 m ⫺ 10 m ⫽ 50 m Denominator ⫽ 1 + (0.234587)

⫽ W(L ⫹ ␦max)

⫽

;

EAd2max 2L

2WL 2WL2 dmax ⫺ ⫽0 EA EA

;

1/2 2 b d 0.234587

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Problem 2.8-13 A weight W rests on top of a wall and is attached to one end of a very flexible cord having cross-sectional area A and modulus of elasticity E (see figure). The other end of the cord is attached securely to the wall. The weight is then pushed off the wall and falls freely the full length of the cord.

W

W

(a) Derive a formula for the impact factor. (b) Evaluate the impact factor if the weight, when hanging statically, elongates the band by 2.5% of its original length.

Solution 2.8-13

Weight falling off a wall CONSERVATION OF ENERGY P.E. ⫽ U or

W(L + dmax) ⫽

d2max ⫺

EAd2max 2L

2WL 2WL2 dmax ⫺ ⫽0 EA EA

SOLVE QUADRATIC EQUATION FOR ␦max: W ⫽ Weight

dmax ⫽

WL 2 WL WL 1/2 + ca b + 2L a bd EA EA EA

Properties of elastic cord: E ⫽ modulus of elasticity

STATIC ELONGATION

A ⫽ cross-sectional area

dst ⫽

L ⫽ original length

␦max ⫽ elongation of elastic cord

WL EA

IMPACT FACTOR

P.E. ⫽ potential energy of weight before fall (with respect to lowest position)

dmax 2EA 1/2 ⫽ 1 + c1 + d dst W

P.E. ⫽ W(L ⫹ ␦max)

NUMERICAL VALUES

Let U ⫽ strain energy of cord at lowest position

␦st ⫽ (2.5%)(L) ⫽ 0.025L

EAd2max U⫽ 2L

dst ⫽

WL EA

W ⫽ 0.025 EA

;

EA ⫽ 40 W

Impact factor ⫽ 1 + [1 + 2(40)]1/2 ⫽ 10

;

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SECTION 2.8

223

Impact Loading

Problem 2.8-14 A rigid bar AB having mass M ⫽ 1.0 kg and

length L ⫽ 0.5 m is hinged at end A and supported at end B by a nylon cord BC (see figure). The cord has cross-sectional area A ⫽ 30 mm2, length b ⫽ 0.25 m, and modulus of elasticity E ⫽ 2.1 GPa. If the bar is raised to its maximum height and then released, what is the maximum stress in the cord?

C b A

B W L

Solution 2.8-14

Falling bar AB GEOMETRY OF BAR AB AND CORD BC

RIGID BAR: W ⫽ Mg ⫽ (1.0 kg)(9.81 m/s2) ⫽ 9.81 N L ⫽ 0.5 m NYLON CORD: A ⫽ 30 mm2

CD ⫽ CB ⫽ b AD ⫽ AB ⫽ L h ⫽ height of center of gravity of raised bar AD

␦max ⫽ elongation of cord From triangle ABC:sin u ⫽ cos u ⫽

b 2b2 + L2 L

E ⫽ 2.1 GPa

2b2 + L2 2h 2h ⫽ From line AD: sin 2 u ⫽ AD L

Find maximum stress ␴max in cord BC.

From Appendix C: sin 2 ␪ ⫽ 2 sin ␪ cos ␪

b ⫽ 0.25 m

L 2bL b 2h ba b ⫽ 2 ⫽ 2a 2 2 2 2 L b + L2 2b + L 2b + L bL2 and h ⫽ 2 (Eq. 1) b + L2 ‹

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Axially Loaded Members

CONSERVATION OF ENERGY P.E. ⫽ potential energy of raised bar AD ⫽ W ah +

Substitute from Eq. (1) into Eq. (3): s2max ⫺

dmax b 2

dmax EAd2max b⫽ 2 2b

smaxb For the cord: dmax ⫽ E Substitute into Eq. (2) and rearrange: s2max ⫺

W 2WhE s ⫺ ⫽0 A max bA

(Eq. 4)

SOLVE FOR ␴max:

EAd2max U ⫽ strain energy of stretched cord ⫽ 2b P.E. ⫽ U W a h +

W 2WL2E ⫽0 smax ⫺ A A(b2 + L2)

(Eq. 2)

smax ⫽

W 8L2EA c1 + 1 + d 2A A W(b2 + L2)

;

SUBSTITUTE NUMERICAL VALUES:

␴max ⫽ 33.3 MPa

;

(Eq. 3)

Stress Concentrations The problems for Section 2.10 are to be solved by considering the stress-concentration factors and assuming linearly elastic behavior.

P

Problem 2.10-1 The flat bars shown in parts (a) and (b) of the figure are

P

d

b

subjected to tensile forces P ⫽ 3.0 k. Each bar has thickness t ⫽ 0.25 in.

(a) For the bar with a circular hole, determine the maximum stresses for hole diameters d ⫽ 1 in. and d ⫽ 2 in. if the width b ⫽ 6.0 in. (b) For the stepped bar with shoulder fillets, determine the maximum stresses for fillet radii R ⫽ 0.25 in. and R ⫽ 0.5 in. if the bar widths are b ⫽ 4.0 in. and c ⫽ 2.5 in.

(a) R P

c

b

(b) Probs. 2.10-1 and 2.10-2

P

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SECTION 2.10

Solution 2.10-1

P ⫽ 3.0 k

225

Flat bars in tension

(b) STEPPED BAR WITH SHOULDER FILLETS

t ⫽ 0.25 in.

(a) BAR WITH CIRCULAR HOLE (b ⫽ 6 in.) FOR d ⫽ 1 in.:

c ⫽ b ⫺ d ⫽ 5 in.

3.0 k P ⫽ 2.40 ksi s nom ⫽ ⫽ ct (5 in.) (0.25 in.) 1 K L 2.60 6

␴max ⫽ k␴nom ⬇ 6.2 ksi

b ⫽ 4.0 in. s nom ⫽

Obtain K from Fig. 2-63

d/b ⫽

Stress Concentrations

c ⫽ 2.5 in.; Obtain k from Fig. 2-64

3.0 k P ⫽ ⫽ 4.80 ksi ct (2.5 in.) (0.25 in.)

FOR R ⫽ 0.25 in.: R/c ⫽ 0.1

b/c ⫽ 1.60

k ⬇ 2.30 ␴max ⫽ K␴nom ⬇ 11.0 ksi FOR R ⫽ 0.5 in.: R/c ⫽ 0.2 K ⬇ 1.87

;

b/c ⫽ 1.60

␴max ⫽ K␴nom ⬇ 9.0 ksi

;

;

FOR d ⫽ 2 in.: c ⫽ b ⫺ d ⫽ 4 in. s nom ⫽ d/b ⫽

P 3.0 k ⫽ ⫽ 3.00 ksi ct (4 in.) (0.25 in.)

1 K L 2.31 3

␴max ⫽ K␴nom ⬇ 6.9 ksi

;

Problem 2.10-2 The flat bars shown in parts (a) and (b) of the figure are subjected to tensile forces P ⫽ 2.5 kN. Each bar has thickness t ⫽ 5.0 mm.

P

(a) For the bar with a circular hole, determine the maximum stresses for hole diameters d ⫽ 12 mm and d ⫽ 20 mm if the width b ⫽ 60 mm. (b) For the stepped bar with shoulder fillets, determine the maximum stresses for fillet radii R ⫽ 6 mm and R ⫽ 10 mm if the bar widths are b ⫽ 60 mm and c ⫽ 40 mm.

P

d

b

(a) R P

c

b

(b)

P

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Solution 2.10-2

P ⫽ 2.5 kN

Page 226

Flat bars in tension

(b) STEPPED BAR WITH SHOULDER FILLETS

t ⫽ 5.0 mm

(a) BAR WITH CIRCULAR HOLE (b ⫽ 60 mm) Obtain K from Fig. 2-63 FOR d ⫽ 12 mm: c ⫽ b ⫺ d ⫽ 48 mm s nom ⫽ d/b ⫽

P 2.5 kN ⫽ ⫽ 10.42 MPa ct (48 mm) (5 mm)

c ⫽ 40 mm;

Obtain K from Fig. 2-64 s nom ⫽

P 2.5 kN ⫽ ⫽ 12.50 MPa ct (40 mm) (5 mm)

FOR R ⫽ 6 mm: R/c ⫽ 0.15 FOR R ⫽ 10 mm: R/c ⫽ 0.25

;

b/c ⫽ 1.5

␴max ⫽ K␴nom ⬇ 25 MPa

K ⬇ 2.00

1 K L 2.51 5

␴max ⫽ K␴nom ⬇ 26 MPa

b ⫽ 60 mm

b/c ⫽ 1.5

␴max ⫽ K␴nom ⬇ 22 MPa

K ⬇ 1.75

;

;

FOR d ⫽ 20 mm: c ⫽ b ⫺ d ⫽ 40 mm s nom ⫽ d/b ⫽

1 3

2.5 kN P ⫽ ⫽ 12.50 MPa ct (40 mm) (5 mm) K L 2.31

␴max ⫽ K␴nom ⬇ 29 MPa

;

Problem 2.10-3 A flat bar of width b and thickness t has a hole of diameter d drilled through it (see figure). The hole may have any diameter that will fit within the bar. What is the maximum permissible tensile load Pmax if the allowable tensile stress in the material is ␴t?

P

b

d

P

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SECTION 2.10

Solution 2.10-3

␴t ⫽ allowable tensile stress Find Pmax Find K from Fig. 2-64 Pmax ⫽ s nom ct ⫽

smax st ct ⫽ (b ⫺ d)t K K

st d bt a1 ⫺ b K b Because ␴t, b, and t are constants, we write: ⫽

P*⫽

stbt

⫽

227

Flat bar in tension

t ⫽ thickness

Pmax

Stress Concentrations

d b

K

P*

0 0.1 0.2 0.3 0.4

3.00 2.73 2.50 2.35 2.24

0.333 0.330 0.320 0.298 0.268

We observe that Pmax decreases as d/b increases. Therefore, the maximum load occurs when the hole becomes very small. d a :0 b Pmax ⫽

and K : 3b stbt 3

;

1 d a1 ⫺ b K b

Problem 2.10-4 A round brass bar of diameter d1 ⫽ 20 mm has

upset ends of diameter d2 ⫽ 26 mm (see figure). The lengths of the segments of the bar are L1 ⫽ 0.3 m and L2 ⫽ 0.1 m. Quarter-circular fillets are used at the shoulders of the bar, and the modulus of elasticity of the brass is E ⫽ 100 GPa. If the bar lengthens by 0.12 mm under a tensile load P, what is the maximum stress ␴max in the bar? Probs. 2.10-4 and 2.10-5

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Axially Loaded Members

Solution 2.10-4

Round brass bar with upset ends

Use Fig. 2-65 for the stress-concentration factor: s nom ⫽

⫽

E ⫽ 100 GPa

␦ ⫽ 0.12 mm

dEA2 P ⫽ ⫽ A1 2L2A1 + L1A2 dE d1 2 2L2 a b + L1 d2

L2 ⫽ 0.1 m

SUBSTITUTE NUMERICAL VALUES:

L1 ⫽ 0.3 m

s nom ⫽

R ⫽ radius of fillets ⫽

26 mm ⫺ 20 mm ⫽ 3 mm 2

PL1 PL2 b + d ⫽ 2a EA2 EA1 Solve for P:

P⫽

dEA1A2 2L2A1 + L1A2

(0.12 mm) (100 GPa) 20 2 2(0.1 m) a b + 0.3 m 26

metal having the following properties: d1 ⫽ 1.0 in., d2 ⫽ 1.4 in., L1 ⫽ 20.0 in., L2 ⫽ 5.0 in., and E ⫽ 25 ⫻ 106 psi. Also, the bar lengthens by 0.0040 in. when the tensile load is applied.

Solution 2.10-5

Use the dashed curve in Fig. 2-65. K ⬇ 1.6

␴max ⫽ K␴nom ⬇ (1.6) (28.68 MPa)

d2

P

;

d2

d1

L1

L2

L2

Round bar with upset ends d ⫽ 2a

PL1 PL2 b + EA2 EA1

Solve for P: P ⫽

s nom ⫽

␦ ⫽ 0.0040 in. L1 ⫽ 20 in. L2 ⫽ 5 in. R ⫽ radius of fillets R ⫽

dEA1A2 2L2A1 + L1A2

Use Fig. 2-65 for the stress-concentration factor.

E ⫽ 25 ⫻ 106 psi

⫽ 0.2 in.

⫽ 28.68 MPa

3 mm R ⫽ ⫽ 0.15 D1 20 mm

⬇ 46 MPa

Problem 2.10-5 Solve the preceding problem for a bar of monel

dE A1 2L2 a b + L1 A2

1.4 in. ⫺ 1.0 in. 2

⫽

dEA2 P ⫽ ⫽ A1 2L2A1 + L1A2 dE d1 2 2L2 a b + L1 d2

dE A1 2L2 a b + L1 A2

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SECTION 2.10

SUBSTITUTE NUMERICAL VALUES: s nom ⫽

(0.0040 in.)(25 * 106 psi) 2(5 in.)a

1.0 2 b + 20 in. 1.4

Stress Concentrations

229

Use the dashed curve in Fig. 2-65. K ⬇ 1.53 ⫽ 3,984 psi

␴max ⫽ K␴nom ⬇ (1.53)(3984 psi) ⬇ 6100 psi

;

0.2 in. R ⫽ ⫽ 0.2 D1 1.0 in.

Problem 2.10-6 A prismatic bar of diameter d0 ⫽ 20 mm is being compared

P1

with a stepped bar of the same diameter (d1 ⫽ 20 mm) that is enlarged in the middle region to a diameter d2 ⫽ 25 mm (see figure). The radius of the fillets in the stepped bar is 2.0 mm.

(a) Does enlarging the bar in the middle region make it stronger than the prismatic bar? Demonstrate your answer by determining the maximum permissible load P1 for the prismatic bar and the maximum permissible load P2 for the enlarged bar, assuming that the allowable stress for the material is 80 MPa. (b) What should be the diameter d0 of the prismatic bar if it is to have the same maximum permissible load as does the stepped bar?

P2 d0

d1

P1

d2 d1

Solution 2.10-6

P2

Prismatic bar and stepped bar Fillet radius: R ⫽ 2 mm Allowable stress: ␴t ⫽ 80 MPa (a) COMPARISON OF BARS Prismatic bar: P1 ⫽ stA0 ⫽ st a

pd20 b 4

p ⫽ (80 MPa)a b(20mm)2 ⫽ 25.1 kN 4

;

Stepped bar: See Fig. 2-65 for the stress-concentration factor.

d0 ⫽ 20 mm d1 ⫽ 20 mm d2 ⫽ 25 mm

R ⫽ 2.0 mm

D1 ⫽ 20 mm

D2 ⫽ 25 mm

R/D1 ⫽ 0.10

D2/D1 ⫽ 1.25

K ⬇ 1.75

s nom ⫽

P2 P2 smax ⫽ s nom ⫽ p 2 A1 K d 4 1

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P2 ⫽ s nom A1 ⫽ ⫽a

Page 230

s max st A1 ⫽ A1 K K

80 MPa p b a b(20 mm)2 1.75 4

L 14.4 kN

(b) DIAMETER OF PRISMATIC BAR FOR THE SAME ALLOWABLE LOAD

d0 ⫽

;

st pd21 pd20 b ⫽ a b 4 K 4

d20 ⫽

20 mm L 15.1 mm 11.75

;

P1 ⫽ P2 st a d1 1K

L

d21 K

Enlarging the bar makes it weaker, not stronger. The ratio of loads is P1/P2 ⫽ K ⫽ 1.75

Problem 2.10-7 A stepped bar with a hole (see figure) has widths b ⫽ 2.4 in. and c ⫽ 1.6 in. The fillets have radii equal to 0.2 in. What is the diameter dmax of the largest hole that can be drilled through the bar without reducing the load-carrying capacity?

Solution 2.10-7

Stepped bar with a hole

b ⫽ 2.4 in.

BASED UPON HOLE (Use Fig. 2-63)

c ⫽ 1.6 in. Fillet radius: R ⫽ 0.2 in.

b ⫽ 2.4 in. c1 ⫽ b ⫺ d

Find dmax

Pmax ⫽ s nom c1t ⫽

smax (b ⫺ d)t K d 1 ⫽ a1 ⫺ bbtsmax K b

BASED UPON FILLETS (Use Fig. 2-64) b ⫽ 2.4 in.

c ⫽ 1.6 in.

R/c ⫽ 0.125

b/c ⫽ 1.5

R ⫽ 0.2 in. K ⬇ 2.10

smax c smax Pmax ⫽ s nomct ⫽ ct ⫽ a b(bt) K K b L 0.317 bt smax

d ⫽ diameter of the hole (in.)

d(in.) 0.3 0.4 0.5 0.6 0.7

d/b

K

Pmax/bt␴max

0.125 0.167 0.208 0.250 0.292

2.66 2.57 2.49 2.41 2.37

0.329 0.324 0.318 0.311 0.299

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SECTION 2.11

Nonlinear Behavior (Changes in Lengths of Bars)

231

Nonlinear Behavior (Changes in Lengths of Bars) A

Problem 2.11-1 A bar AB of length L and weight density ␥ hangs vertically under its own weight (see figure). The stress-strain relation for the material is given by the Ramberg-Osgood equation (Eq. 2-71): P⫽

s0a s m s + a b E E s0

L

Derive the following formula d⫽

gL2 gL m s0aL + a b 2E (m + 1)E s0

B

for the elongation of the bar.

Solution 2.11-1

Bar hanging under its own weight STRAIN AT DISTANCE x Let A ⫽ cross-sectional area Let N ⫽ axial force at distance x N ⫽ ␥Ax s⫽

N ⫽ gx A

␧⫽

s0a s m gx s0 gx m s + a b ⫽ + a b E E s0 E aE s0

ELONGATION OF BAR L

d⫽ ⫽

L0

␧dx ⫽

L

L

gx gx m s0a a b dx dx + E L0 s0 L0 E

gL2 gL m s0aL + a b 2E (m + 1)E s0

Q.E.D.

;

A

B

P1 C

Problem 2.11-2 A prismatic bar of length L ⫽ 1.8 m and cross-sectional

area A ⫽ 480 mm is loaded by forces P1 ⫽ 30 kN and P2 ⫽ 60 kN (see figure). The bar is constructed of magnesium alloy having a stress-strain curve described by the following Ramberg-Osgood equation: 2

P⫽

s 1 s 10 + a b (s ⫽ MPa) 45,000 618 170

in which ␴ has units of megapascals. (a) Calculate the displacement ␦C of the end of the bar when the load P1 acts alone. (b) Calculate the displacement when the load P2 acts alone. (c) Calculate the displacement when both loads act simultaneously.

2L — 3

L — 3

P2

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Axially Loaded Members

Solution 2.11-2

Axially loaded bar

(c) BOTH P1 AND P2 ARE ACTING AB:s ⫽ L ⫽ 1.8 m

␧ ⫽ 0.008477

A ⫽ 480 mm2

P1 ⫽ 30 kN

dAB ⫽ ␧a

P2 ⫽ 60 kN

Ramberg–Osgood Equation: 1 s 10 s + a b (s ⫽ MPa) ␧⫽ 45,000 618 170 Find displacement at end of bar.

P1 30 kN ⫽ 62.5 MPa ⫽ A 480 mm2

␧ ⫽ 0.001389 dc ⫽ ␧ a

2L b ⫽ 1.67 mm 3

BC:s ⫽

2L b ⫽ 10.17 mm 3

P2 60 kN ⫽ 125 MPa ⫽ A 480 mm2

␧ ⫽ 0.002853 L dBC ⫽ ␧a b ⫽ 1.71 mm 3

(a) P1 ACTS ALONE AB: s ⫽

P1 + P2 90 kN ⫽ 187.5 MPa ⫽ A 480 mm2

;

dC ⫽ dAB + dBC ⫽ 11.88 mm

;

(Note that the displacement when both loads act simultaneously is not equal to the sum of the displacements when the loads act separately.)

(b) P2 ACTS ALONE P2 60 kN ⫽ ⫽ 125 MPa A 480 mm2 ␧ ⫽ 0.002853 dc ⫽ ␧L ⫽ 5.13 mm ;

ABC:s ⫽

Problem 2.11-3 A circular bar of length L ⫽ 32 in. and diameter

d ⫽ 0.75 in. is subjected to tension by forces P (see figure). The wire is made of a copper alloy having the following hyperbolic stress-strain relationship: s⫽

18,000P 0 … P … 0.03 (s ⫽ ksi) 1 + 300P

(a) Draw a stress-strain diagram for the material. (b) If the elongation of the wire is limited to 0.25 in. and the maximum stress is limited to 40 ksi, what is the allowable load P?

d

P

P L

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SECTION 2.11

Solution 2.11-3

Nonlinear Behavior (Changes in Lengths of Bars)

233

Copper bar in tension

(b) ALLOWABLE LOAD P Max. elongation ␦max ⫽ 0.25 in. Max. stress ␴max ⫽ 40 ksi Based upon elongation: L ⫽ 32 in. A⫽

d ⫽ 0.75 in.

pd2 ⫽ 0.4418 in.2 4

␧max ⫽

dmax 0.25 in. ⫽ ⫽ 0.007813 L 32 in.

smax ⫽

18,000␧max ⫽ 42.06 ksi 1 + 300␧max

(a) STRESS-STRAIN DIAGRAM s⫽

18,000␧ 0 … ␧ … 0.03 (s ⫽ ksi) 1 + 300␧

BASED UPON STRESS:

␴max ⫽ 40 ksi Stress governs. P ⫽ ␴max A ⫽ (40 ksi)(0.4418 in.2) ⫽ 17.7 k

Problem 2.11-4 A prismatic bar in tension has length L ⫽ 2.0 m

and cross-sectional area A ⫽ 249 mm2. The material of the bar has the stressstrain curve shown in the figure. Determine the elongation ␦ of the bar for each of the following axial loads: P ⫽ 10 kN, 20 kN, 30 kN, 40 kN, and 45 kN. From these results, plot a diagram of load P versus elongation ␦ (load-displacement diagram).

;

200 s (MPa) 100

0

0

0.005 e

0.010

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Axially Loaded Members

Solution 2.11-4

Bar in tension

L ⫽ 2.0 m A ⫽ 249 mm2 STRESS-STRAIN DIAGRAM (See the problem statement for the diagram) LOAD-DISPLACEMENT DIAGRAM P (kN)

␴ ⫽ P/A (MPa)

␧ (from diagram)

␦ ⫽ ␧L (mm)

10 20 30 40 45

40 80 120 161 181

0.0009 0.0018 0.0031 0.0060 0.0081

1.8 3.6 6.2 12.0 16.2

NOTE: The load-displacement curve has the same shape as the stress-strain curve.

Problem 2.11-5 An aluminum bar subjected to tensile forces P has length

L ⫽ 150 in. and cross-sectional area A ⫽ 2.0 in.2 The stress-strain behavior of the aluminum may be represented approximately by the bilinear stress-strain diagram shown in the figure. Calculate the elongation ␦ of the bar for each of the following axial loads: P ⫽ 8 k, 16 k, 24 k, 32 k, and 40 k. From these results, plot a diagram of load P versus elongation ␦ (load-displacement diagram).

s

12,000 psi

E2 = 2.4 106 psi

E1 = 10 106 psi

0

e

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SECTION 2.11

Solution 2.11-5

Nonlinear Behavior (Changes in Lengths of Bars)

Aluminum bar in tension LOAD-DISPLACEMENT DIAGRAM

L ⫽ 150 in. A ⫽ 2.0 in.2 STRESS-STRAIN DIAGRAM

E1 ⫽ 10 ⫻ 106 psi E2 ⫽ 2.4 ⫻ 106 psi

␴1 ⫽ 12,000 psi ␧1 ⫽

12,000 psi s1 ⫽ E1 10 * 106 psi

⫽ 0.0012 For 0 ⱕ ␴ ⱕ ␴1: s s ⫽ (s ⫽ psi) E2 10 * 106psi For ␴ ⱖ ␴1: ␧⫽

␧ ⫽ ␧1 + ⫽

s

Eq. (1)

s ⫺ 12,000 s ⫺ s1 ⫽ 0.0012 + E2 2.4 * 106

2.4 * 106

⫺ 0.0038 (s ⫽ psi)

Eq. (2)

P (k)

␴ ⫽ P/A (psi)

␧ (from Eq. 1 or Eq. 2)

␦ ⫽ ␧L (in.)

8 16 24 32 40

4,000 8,000 12,000 16,000 20,000

0.00040 0.00080 0.00120 0.00287 0.00453

0.060 0.120 0.180 0.430 0.680

235

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Axially Loaded Members

Problem 2.11-6 A rigid bar AB, pinned at end A, is supported by a wire CD and loaded by a force P at end B (see figure). The wire is made of high-strength steel having modulus of elasticity E ⫽ 210 GPa and yield stress ␴Y ⫽ 820 MPa. The length of the wire is L ⫽ 1.0 m and its diameter is d ⫽ 3 mm. The stress-strain diagram for the steel is defined by the modified power law, as follows:

C L A

D

B

s ⫽ EP 0 … s … sY s ⫽ sY a

EP n b s Ú sY sY

P 2b

(a) Assuming n ⫽ 0.2, calculate the displacement ␦B at the end of the bar due to the load P. Take values of P from 2.4 kN to 5.6 kN in increments of 0.8 kN.

b

(b) Plot a load-displacement diagram showing P versus ␦B.

Solution 2.11-6

Rigid bar supported by a wire

sY s 1/n a b E sY 3P Axial force in wire: F ⫽ 2 3P F Stress in wire: s ⫽ ⫽ A 2A PROCEDURE: Assume a value of P Calculate ␴ from Eq. (6) Calculate ␧ from Eq. (4) or (5) Calculate ␦B from Eq. (3) From Eq. (2): ␧ ⫽

Wire: E ⫽ 210 GPa

␴Y ⫽ 820 MPa L ⫽ 1.0 m d ⫽ 3 mm A⫽

pd2 ⫽ 7.0686 mm2 4

STRESS-STRAIN DIAGRAM

␴ (MPa) Eq. (6)

␧ Eq. (4) or (5)

␦B (mm) Eq. (3)

2.4 3.2 4.0 4.8 5.6

509.3 679.1 848.8 1018.6 1188.4

0.002425 0.003234 0.004640 0.01155 0.02497

3.64 4.85 6.96 17.3 37.5

For ␴ ⫽ ␴Y ⫽ 820 MPa: ␧ ⫽ 0.0039048 P ⫽ 3.864 kN

(n ⫽ 0.2)

(2)

(b) LOAD-DISPLACEMENT DIAGRAM

(a) DISPLACEMENT ␦B AT END OF BAR 3 3 ␦ ⫽ elongation of wire dB ⫽ d ⫽ ␧L 2 2 Obtain ␧ from stress-strain equations:

(3)

(0 ⱕ ␴ ⱕ ␴Y)

E␧ n s ⫽ sY a b sY

(␴ ⱖ ␴Y)

From Eq. (1): ␧ ⫽

sE (0 … s … sY)

(4)

(6)

P (kN)

(1)

␴ ⫽ E␧

(5)

␦B ⫽ 5.86 mm

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SECTION 2.12

Elastoplastic Analysis

237

u

C

Elastoplastic Analysis The problems for Section 2.12 are to be solved assuming that the material is elastoplastic with yield stress ␴Y, yield strain ⑀Y, and modulus of elasticity E in the linearly elastic region (see Fig. 2-70).

A

u

Problem 2.12-1 Two identical bars AB and BC support a vertical load P (see figure). The bars are made of steel having a stress-strain curve that may be idealized as elastoplastic with yield stress ␴Y. Each bar has cross-sectional area A. Determine the yield load PY and the plastic load PP.

Solution 2.12-1

B

P

Two bars supporting a load P

JOINT B ⌺Fvert ⫽ 0 Structure is statically determinate. The yield load PY and the plastic lead PP occur at the same time, namely, when both bars reach the yield stress.

(2␴YA) sin ␪ ⫽ P PY ⫽ PP ⫽ 2␴YA sin ␪

Problem 2.12-2 A stepped bar ACB with circular cross sections is held between rigid supports and loaded by an axial force P at midlength (see figure). The diameters for the two parts of the bar are d1 ⫽ 20 mm and d2 ⫽ 25 mm, and the material is elastoplastic with yield stress ␴Y ⫽ 250 MPa. Determine the plastic load PP.

A

d1

;

C

L — 2

d2

P

L — 2

B

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Solution 2.12-2

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Axially Loaded Members

Bar between rigid supports FAC ⫽ ␴YA1

FCB ⫽ ␴YA2

P ⫽ FAC ⫹ FCB PP ⫽ ␴YA1 ⫹ ␴YA2 ⫽ ␴Y(A1 ⫹ A2)

;

SUBSTITUTE NUMERICAL VALUES: d1 ⫽ 20 mm d2 ⫽ 25 mm ␴Y ⫽ 250 MPa DETERMINE THE PLASTIC LOAD PP: At the plastic load, all parts of the bar are stressed to the yield stress.

p PP ⫽ (250 MPa)a b(d21 + d22) 4 p ⫽ (250 MPa)a b[(20 mm)2 + (25 mm)2] 4 ⫽ 201 kN

Point C:

;

Problem 2.12-3 A horizontal rigid bar AB supporting a load P is hung from five symmetrically placed wires, each of cross-sectional area A (see figure). The wires are fastened to a curved surface of radius R.

R

(a) Determine the plastic load PP if the material of the wires is elastoplastic with yield stress ␴Y. (b) How is PP changed if bar AB is flexible instead of rigid? (c) How is PP changed if the radius R is increased? A

B P

Solution 2.12-3

Rigid bar supported by five wires

(b) BAR AB IS FLEXIBLE At the plastic load, each wire is stressed to the yield stress, so the plastic load is not changed. ; (a) PLASTIC LOAD PP At the plastic load, each wire is stressed to the yield stress. ⬖ PP ⫽ 5␴YA ; F ⫽ ␴YA

(c) RADIUS R IS INCREASED Again, the forces in the wires are not changed, so the plastic load is not changed. ;

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SECTION 2.12

Elastoplastic Analysis

239

Problem 2.12-4 A load P acts on a horizontal beam that is supported by four rods arranged in the symmetrical pattern shown in the figure. Each rod has cross-sectional area A and the material is elastoplastic with yield stress ␴Y. Determine the plastic load PP.

a

a

P

Solution 2.12-4

Beam supported by four rods

F ⫽ ␴YA Sum forces in the vertical direction and solve for the load: At the plastic load, all four rods are stressed to the yield stress.

PP ⫽ 2F ⫹ 2F sin ␣ PP ⫽ 2␴YA (1 ⫹ sin ␣)

21 in.

Problem 2.12-5 The symmetric truss ABCDE shown in the figure is constructed of four bars and supports a load P at joint E. Each of the two outer bars has a cross-sectional area of 0.307 in.2, and each of the two inner bars has an area of 0.601 in.2 The material is elastoplastic with yield stress ␴Y ⫽ 36 ksi. Determine the plastic load PP.

A

;

54 in.

21 in. C

B

D

36 in.

E P

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Solution 2.12-5

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Axially Loaded Members

Truss with four bars PLASTIC LOAD PP At the plastic load, all bars are stressed to the yield stress. FAE ⫽ ␴YAAE PP ⫽

FBE ⫽ ␴YABE

6 8 sY AAE + sY ABE 5 5

;

SUBSTITUTE NUMERICAL VALUES: AAE ⫽ 0.307 in.2 ABE ⫽ 0.601 in.2 LAE ⫽ 60 in. JOINT E

LBE ⫽ 45 in.

sY ⫽ 36 ksi 6 8 PP ⫽ (36 ksi) (0.307 in.2) + (36 ksi) (0.601 in.2) 5 5

Equilibrium: 3 4 2FAE a b + 2FBE a b ⫽ P 5 5 or 6 8 P ⫽ FAE + FBE 5 5

⫽ 13.26 k + 34.62 k ⫽ 47.9 k

Problem 2.12-6 Five bars, each having a diameter of 10 mm, support a

b

b

;

b

b

load P as shown in the figure. Determine the plastic load PP if the material is elastoplastic with yield stress ␴Y ⫽ 250 MPa.

2b

P

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SECTION 2.12

Solution 2.12-6

b

Elastoplastic Analysis

241

Truss consisting of five bars

b

b

At the plastic load, all five bars are stressed to the yield stress

b

F ⫽ ␴YA Sum forces in the vertical direction and solve for the load:

2b

PP ⫽ 2Fa P

d ⫽ 10 mm pd2 ⫽ 78.54 mm2 A⫽ 4

␴Y ⫽ 250 MPa

⫽

1 2 b + 2Fa b + F 12 15

sYA (512 + 415 + 5) 5

⫽ 4.2031sYA

;

Substitute numerical values: PP ⫽ (4.2031)(250 MPa)(78.54 mm2) ⫽ 82.5 kN

Problem 2.12-7 A circular steel rod AB of diameter d ⫽ 0.60 in.

;

B

A

is stretched tightly between two supports so that initially the tensile stress in the rod is 10 ksi (see figure). An axial force P is then applied to the rod at an intermediate location C.

d

(a) Determine the plastic load PP if the material is elastoplastic with yield stress ␴Y ⫽ 36 ksi. (b) How is PP changed if the initial tensile stress is doubled to 20 ksi?

Solution 2.12-7

A

P C

Bar held between rigid supports POINT C: sYA

sYA

P

— C ¡ —

d ⫽ 0.6 in.

␴Y ⫽ 36 ksi Initial tensile stress ⫽ 10 ksi (a) PLASTIC LOAD PP The presence of the initial tensile stress does not affect the plastic load. Both parts of the bar must yield in order to reach the plastic load.

p PP ⫽ 2sYA ⫽ (2) (36 ksi)a b(0.60 in.)2 4 ⫽ 20.4 k

;

(B) INITIAL TENSILE STRESS IS DOUBLED PP is not changed.

;

B

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Axially Loaded Members

Problem 2.12-8 A rigid bar ACB is supported on a fulcrum at C and loaded by a force P at end B (see figure). Three identical wires made of an elastoplastic material (yield stress ␴Y and modulus of elasticity E) resist the load P. Each wire has cross-sectional area A and length L. (a) Determine the yield load PY and the corresponding yield displacement ␦Y at point B. (b) Determine the plastic load PP and the corresponding displacement ␦P at point B when the load just reaches the value PP. (c) Draw a load-displacement diagram with the load P as ordinate and the displacement ␦B of point B as abscissa.

Solution 2.12-8

L A

C

B P

L

a

a

a

a

Rigid bar supported by wires (b) PLASTIC LOAD PP

(a) YIELD LOAD PY Yielding occurs when the most highly stressed wire reaches the yield stress ␴Y

At the plastic load, all wires reach the yield stress. ⌺MC ⫽ 0 PP ⫽

4sYA 3

;

At point A: dA ⫽ (sYA)a

sYL L b ⫽ EA E

At point B: dB ⫽ 3dA ⫽ dP ⫽

⌺MC ⫽ 0 PY ⫽ ␴YA At point A:

;

(c) LOAD-DISPLACEMENT DIAGRAM

;

sYA sYL L dA ⫽ a ba b ⫽ 2 EA 2E At point B: dB ⫽ 3dA ⫽ dY ⫽

3sYL E

3sYL 2E

;

4 PP ⫽ PY 3 dP ⫽ 2dY

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SECTION 2.12

243

Elastoplastic Analysis

Problem 2.12-9 The structure shown in the figure consists of a horizontal rigid bar ABCD supported by two steel wires, one of length L and the other of length 3L/4. Both wires have cross-sectional area A and are made of elastoplastic material with yield stress ␴Y and modulus of elasticity E. A vertical load P acts at end D of the bar. (a) Determine the yield load PY and the corresponding yield displacement ␦Y at point D. (b) Determine the plastic load PP and the corresponding displacement ␦P at point D when the load just reaches the value PP. (c) Draw a load-displacement diagram with the load P as ordinate and the displacement ␦D of point D as abscissa.

Solution 2.12-9

L A

3L 4

B

C

D

P 2b

b

b

Rigid bar supported by two wires FREE-BODY DIAGRAM

A ⫽ cross-sectional area

EQUILIBRIUM:

␴Y ⫽ yield stress

⌺MA ⫽ 0 哵哴

E ⫽ modulus of elasticity

FB(2b) ⫹ FC(3b) ⫽ P(4b) 2FB ⫹ 3FC ⫽ 4P

(3)

DISPLACEMENT DIAGRAM

FORCE-DISPLACEMENT RELATIONS FBL dC ⫽ dB ⫽ EA

3 FC a Lb 4 EA

(4, 5)

Substitute into Eq. (1):

COMPATIBILITY: 3 dC ⫽ dB 2

(1)

3FCL 3FBL ⫽ 4EA 2EA

␦D ⫽ 2␦B

(2)

FC ⫽ 2FB

(6)

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Axially Loaded Members

STRESSES

From Eq. (3):

FB FC sC ⫽ sC ⫽ 2sB (7) A A Wire C has the larger stress. Therefore, it will yield first.

2(␴YA) ⫹ 3(␴YA) ⫽ 4P

sB ⫽

(a) YIELD LOAD

␴C ⫽ ␴Y

FB ⫽

(From Eq. 7)

1 s A 2 Y

From Eq. (3): 1 2a sYA b + 3(sYA) ⫽ 4P 2 P ⫽ PY ⫽ ␴YA

;

From Eq. (4):

sC sY ⫽ sB ⫽ 2 2

FC ⫽ ␴Y A

5 P ⫽ PP ⫽ sYA 4

FBL sY L ⫽ EA E From Eq. (2): dB ⫽

dD ⫽ dP ⫽ 2dB ⫽

2sYL E

;

(c) LOAD-DISPLACEMENT DIAGRAM 5 PP ⫽ PY 4

;

From Eq. (4):

␦P ⫽ 2␦Y

FB L sY L ⫽ dB ⫽ EA 2E From Eq. (2): dD ⫽ dY ⫽ 2dB ⫽

sY L E

;

(b) PLASTIC LOAD At the plastic load, both wires yield.

␴B ⫽ ␴Y ⫽ ␴C

FB ⫽ FC ⫽ ␴Y A

Problem 2.12-10 Two cables, each having a length L of approximately 40 m, support a l oaded container of weight W (see figure). The cables, which have effective cross-sectional area A ⫽ 48.0 mm2 and effective modulus of elasticity E ⫽ 160 GPa, are identical except that one cable is longer than the other when they are hanging separately and unloaded. The difference in lengths is d ⫽ 100 mm. The cables are made of steel having an elastoplastic stress-strain diagram with ␴Y ⫽ 500 MPa. Assume that the weight W is initially zero and is slowly increased by the addition of material to the container. L

(a) Determine the weight WY that first produces yielding of the shorter cable. Also, determine the corresponding elongation ␦Y of the shorter cable. (b) Determine the weight WP that produces yielding of both cables. Also, determine the elongation ␦P of the shorter cable when the weight W just reaches the value WP. (c) Construct a load-displacement diagram showing the weight W as ordinate and the elongation ␦ of the shorter cable as abscissa. (Hint: The load displacement diagram is not a single straight line in the region 0 ⱕ W ⱕ WY.) W

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SECTION 2.12

Solution 2.12-10

Elastoplastic Analysis

Two cables supporting a load

L ⫽ 40 m

A ⫽ 48.0 mm2

(b) PLASTIC LOAD WP

E ⫽ 160 GPa

F1 ⫽ ␴YA

d ⫽ difference in length ⫽ 100 mm

WP ⫽ 2␴YA ⫽ 48 kN

␴Y ⫽ 500 MPa INITIAL STRETCHING OF CABLE 1 Initially, cable 1 supports all of the load. Let W1 ⫽ load required to stretch cable 1 to the same length as cable 2 EA W1 ⫽ d ⫽ 19.2 kN L

F2 ⫽ ␴YA ;

␦2P ⫽ elongation of cable 2 ⫽ F2 a

sYL L b ⫽ ⫽ 0.125 mm ⫽ 125 mm EA E

␦1P ⫽ ␦2P ⫹ d ⫽ 225 mm ␦P ⫽ ␦1P ⫽ 225 mm

;

(c) LOAD-DISPLACEMENT DIAGRAM

␦1 ⫽ 100 mm (elongation of cable 1)

s1 ⫽

W1 Ed ⫽ ⫽ 400 MPa (s1 6 sY ‹ 7 OK) A L

(a) YIELD LOAD WY Cable 1 yields first. F1 ⫽ ␴YA ⫽ 24 kN

␦1Y ⫽ total elongation of cable 1 d1Y ⫽ total elongation of cable 1 d1Y ⫽

F1L sY L ⫽ ⫽ 0.125 m ⫽ 125 mm EA E

dY ⫽ d1Y ⫽ 125 mm

;

d2Y ⫽ elongation of cable 2 ⫽ d1Y ⫺ d ⫽ 25 mm EA F2 ⫽ d2Y ⫽ 4.8 kN L WY ⫽ F1 + F2 ⫽ 24 kN + 4.8 kN ⫽ 28.8 kN

;

dY WY ⫽ 1.5 ⫽ 1.25 W1 d1 dP WP ⫽ 1.667 ⫽ 1.8 WY dY 0 ⬍ W ⬍ W1: slope ⫽ 192,000 N/m W1 ⬍ W ⬍ WY: slope ⫽ 384,000 N/m WY ⬍ W ⬍ WP: slope ⫽ 192,000 N/m

245

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Axially Loaded Members

Problem 2.12-11 A hollow circular tube T of length L ⫽ 15 in. is uniformly compressed by a force P acting through a rigid plate (see figure). The outside and inside diameters of the tube are 3.0 and 2.75 in., repectively. A concentric solid circular bar B of 1.5 in. diameter is mounted inside the tube. When no load is present, there is a clearance c ⫽ 0.010 in. between the bar B and the rigid plate. Both bar and tube are made of steel having an elastoplastic stress-strain diagram with E ⫽ 29 ⫻ 103 ksi and ␴Y ⫽ 36 ksi. (a) Determine the yield load PY and the corresponding shortening ␦Y of the tube. (b) Determine the plastic load PP and the corresponding shortening ␦P of the tube. (c) Construct a load-displacement diagram showing the load P as ordinate and the shortening ␦ of the tube as abscissa. (Hint: The load-displacement diagram is not a single straight line in the region 0 ⱕ P ⱕ PY.)

Solution 2.12-11

L ⫽ 15 in. c ⫽ 0.010 in. E ⫽ 29 ⫻ 103 ksi

␴Y ⫽ 36 ksi

P

c T

T

B

T

L

Tube and bar supporting a load

TUBE: d2 ⫽ 3.0 in. d1 ⫽ 2.75 in. AT ⫽

p 2 (d ⫺ d21) ⫽ 1.1290 in.2 4 2

B

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SECTION 2.12

BAR

Elastoplastic Analysis

(b) PLASTIC LOAD PP FT ⫽ ␴YAT

d ⫽ 1.5 in. AB ⫽

pd ⫽ 1.7671 in.2 4

⫽ 104,300 lb

;

␦BP ⫽ shortening of bar

INITIAL SHORTENING OF TUBE T Initially, the tube supports all of the load. Let P1 ⫽ load required to close the clearance

⫽ FB a

sYL L ⫽ 0.018621 in. b ⫽ EAB E

␦TP ⫽ ␦BP ⫹ c ⫽ 0.028621 in.

EAT c ⫽ 21,827 lb L Let ␦1 ⫽ shortening of tube P1 ⫽

P1 ⫽ 19,330 psi s1 ⫽ AT

FB ⫽ ␴YAB

PP ⫽ FT ⫹ FB ⫽ ␴Y(AT ⫹ AB)

2

␦1 ⫽ c ⫽ 0.010 in.

␦P ⫽ ␦TP ⫽ 0.02862 in.

;

(c) LOAD-DISPLACEMENT DIAGRAM

(␴1 ⬍ ␴Y ⬖ OK)

(a) YIELD LOAD PY Because the tube and bar are made of the same material, and because the strain in the tube is larger than the strain in the bar, the tube will yield first. FT ⫽ ␴YAT ⫽ 40,644 lb

␦ TY ⫽ shortening of tube at the yield stress s TY ⫽

FTL sYL ⫽ ⫽ 0.018621 in. EAT E

␦Y ⫽ ␦TY ⫽ 0.018621 in.

;

␦BY ⫽ shortening of bar ⫽ ␦TY ⫺ c ⫽ 0.008621 in.

dY PY ⫽ 3.21 ⫽ 1.86 P1 d1

EAB d ⫽ 29,453 lb L BY

dP PP ⫽ 1.49 ⫽ 1.54 PY dY

FB ⫽

PY ⫽ FT ⫹ FB ⫽ 40,644 lb ⫹ 29,453 lb ⫽ 70,097 lb PY ⫽ 70,100 lb

0 ⬍ P ⬍ P1: slope ⫽ 2180 k/in. P1 ⬍ P ⬍ PY: slope ⫽ 5600 k/in.

;

PY ⬍ P ⬍ PP: slope ⫽ 3420 k/in.

247

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3 Torsion

Torsional Deformations Problem 3.2-1 A copper rod of length L ⫽ 18.0 in. is to be twisted by torques T (see figure) until the angle of rotation between the ends of the rod is 3.0°. If the allowable shear strain in the copper is 0.0006 rad, what is the maximum permissible diameter of the rod?

d T

T

L Probs. 3.2-1 and 3.2-2

Solution 3.2-1

Copper rod in torsion d T

T

L

L ⫽ 18.0 in.

From Eq. (3-3):

p f ⫽ 3.0° ⫽ (3.0)a 180 b rad

␥max ⫽

⫽ 0.05236 rad

␥allow ⫽ 0.0006 rad Find dmax

dmax ⫽

rf df ⫽ L 2L 2L␥ allow f

dmax ⫽ 0.413 in.

⫽

(2)(18.0 in.)(0.0006 rad) 0.05236 rad

;

Problem 3.2-2 A plastic bar of diameter d ⫽ 56 mm is to be twisted by torques T (see figure) until the angle of rotation between the ends of the bar is 4.0°. If the allowable shear strain in the plastic is 0.012 rad, what is the minimum permissible length of the bar?

249

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Torsion

Solution 3.2-2 NUMERICAL DATA d ⫽ 56 mm

␥a ⫽ 0.012 radians f ⫽ 4a

p b radians 180

solution based on Equ. (3-3): Lmin ⫽ 162.9 mm

Lmin ⫽

df 2␥a

;

Problem 3.2-3 A circular aluminum tube subjected to pure torsion by torques T (see figure) has an outer radius r2 equal to 1.5 times the inner radius r1.

T

T

L

(a) If the maximum shear strain in the tube is measured as 400 ⫻ 10⫺6 rad, what is the shear strain ␥1 at the inner surface? (b) If the maximum allowable rate of twist is 0.125 degrees per foot and the maximum shear strain is to be kept at 400 ⫻ 10⫺6 rad by adjusting the torque T, what is the minimum required outer radius (r2)min?

r2 r1 Probs. 3.2-3, 3.2-4, and 3.2-5

Solution 3.2-3 NUMERICAL DATA

(b) MIN. REQUIRED OUTER RADIUS

r2 ⫽ 1.5r1 ␥max ⫽ 400 ⫻ (10⫺6) radians u ⫽ 0.125a

p 1 ba b 180 12

r2min ⫽

␥max u

r2min ⫽

r2min ⫽ 2.2 inches

␥max u

;

␪ ⫽ 1.818 ⫻ 10⫺4 rad /m. (a) SHEAR STRAIN AT INNER SURFACE AT RADIUS r1 ␥1 ⫽

r1 ␥ r2 max

␥1 ⫽

1 ␥ 1.5 max

␥1 ⫽ 267 ⫻ 10⫺6 radians

;

Problem 3.2-4 A circular steel tube of length L ⫽ 1.0 m is loaded in torsion by torques T (see figure). (a) If the inner radius of the tube is r1 ⫽ 45 mm and the measured angle of twist between the ends is 0.5°, what is the shear strain ␥1 (in radians) at the inner surface? (b) If the maximum allowable shear strain is 0.0004 rad and the angle of twist is to be kept at 0.45° by adjusting the torque T, what is the maximum permissible outer radius (r2)max?

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251

SECTION 3.2 Torsional Deformations

Solution 3.2-4 (b) MAX. PERMISSIBLE OUTER RADIUS

NUMERICAL DATA L ⫽ 1000 mm

f ⫽ 0.45a

r1 ⫽ 45 mm f ⫽ 0.5a

p b radians 180

(a) SHEAR STRAIN AT INNER SURFACE f ␥1 ⫽ r1 ␥1 ⫽ 393 ⫻ 10⫺6 radians L

p b radians 180

␥max ⫽ 0.0004 radians r2max ⫽ 50.9 mm

␥max ⫽ r2

f L

r2max ⫽ ␥max

L f

;

;

Problem 3.2-5 Solve the preceding problem if the length L ⫽ 56 in., the inner radius r1 ⫽ 1.25 in., the angle of twist is 0.5°, and the allowable shear strain is 0.0004 rad.

Solution 3.2-5 NUMERICAL DATA

(b) MAXIMUM PERMISSIBLE OUTER RADIUS (r2)max

L ⫽ 56 inches r1 ⫽ 1.25 inches f ⫽ 0.5 a

f ⫽ 0.5 a

p b radians 180

␥max

␥a ⫽ 0.0004 radians (a) SHEAR STRAIN g1 (IN RADIANS) AT THE INNER SURFACE

␥1 ⫽ r1

f L

␥1 ⫽ 195 ⫻ 10⫺6 radians

;

p b radians 180 f ⫽ r2 L

␥a ⫽ 0.0004 radians L r2max ⫽ ␥ a f r2max ⫽ 2.57 inches

;

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Torsion

Circular Bars and Tubes P

Problem 3.3-1 A prospector uses a hand-powered winch (see figure) to raise a bucket of ore in his mine shaft. The axle of the winch is a steel rod of diameter d ⫽ 0.625 in. Also, the distance from the center of the axle to the center of the lifting rope is b ⫽ 4.0 in. If the weight of the loaded bucket is W ⫽ 100 lb, what is the maximum shear stress in the axle due to torsion?

d W b W

Solution 3.3-1

Hand-powered winch d ⫽ 0.625 in.

MAXIMUM SHEAR STRESS IN THE AXLE

b ⫽ 4.0 in.

From Eq. (3-12):

W ⫽ 100 lb

tmax ⫽

Torque T applied to the axle: T ⫽ Wb ⫽ 400 lb-in.

tmax ⫽

16T pd3 (16)(400 lb-in.) p(0.625 in.)3

tmax ⫽ 8,340 psi

;

Problem 3.3-2 When drilling a hole in a table leg, a furniture maker uses a hand-operated drill (see figure) with a bit of diameter d ⫽ 4.0 mm. (a) If the resisting torque supplied by the table leg is equal to 0.3 N⭈m, what is the maximum shear stress in the drill bit? (b) If the shear modulus of elasticity of the steel is G ⫽ 75 GPa, what is the rate of twist of the drill bit (degrees per meter)?

d

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SECTION 3.3

Solution 3.3-2

253

Circular Bars and Tubes

Torsion of a drill bit (b) RATE OF TWIST From Eq. (3-14): u⫽

d ⫽ 4.0 mm

T ⫽ 0.3 N⭈m G ⫽ 75 GPa

(a) MAXIMUM SHEAR STRESS From Eq. (3-12): tmax ⫽

tmax ⫽

16T

u⫽

T GIP 0.3 N # m p (75 GPa)a b(4.0 mm)4 32

u ⫽ 0.1592 rad/m ⫽ 9.12°/m

pd3

;

16(0.3 N # m) p(4.0 mm)3

tmax ⫽ 23.8 MPa

;

Problem 3.3-3 While removing a wheel to change a tire,

a driver applies forces P ⫽ 25 lb at the ends of two of the arms of a lug wrench (see figure). The wrench is made of steel with shear modulus of elasticity G ⫽ 11.4 ⫻ 106 psi. Each arm of the wrench is 9.0 in. long and has a solid circular cross section of diameter d ⫽ 0.5 in. (a) Determine the maximum shear stress in the arm that is turning the lug nut (arm A). (b) Determine the angle of twist (in degrees) of this same arm.

P

9.0

in.

A

9.0

in.

d = 0.5 in. P = 25 lb

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Solution 3.3-3

Page 254

Torsion

Lug wrench P ⫽ 25 lb

(a) MAXIMUM SHEAR STRESS From Eq. (3-12): (16)(450 l b - in.) 16T ⫽ tmax ⫽ 3 pd p(0.5 in.)3

L ⫽ 9.0 in. d ⫽ 0.5 in. G ⫽ 11.4 ⫻ 106 psi T ⫽ torque acting on arm A

tmax ⫽ 18,300 psi

;

(b) ANGLE OF TWIST From Eq. (3-15): (450 lb-in.)(9.0 in.) TL f⫽ ⫽ GIP p (11.4 * 106 psi)a b(0.5 in.)4 32

T ⫽ P(2L) ⫽ 2(25 lb) (9.0 in.) ⫽ 450 lb-in.

f ⫽ 0.05790 rad ⫽ 3.32°

Problem 3.3-4 An aluminum bar of solid circular cross section is twisted by torques T acting at the ends (see figure). The dimensions and shear modulus of elasticity are as follows: L ⫽ 1.4 m, d ⫽ 32 mm, and G ⫽ 28 GPa.

;

d T

T

L

(a) Determine the torsional stiffness of the bar. (b) If the angle of twist of the bar is 5°, what is the maximum shear stress? What is the maximum shear strain (in radians)?

Solution 3.3-4 (a) TORSIONAL STIFFNESS OF BAR d ⫽ 32 mm GI p kT ⫽ L

G ⫽ 28 GPa Ip ⫽

p 4 d 32

Ip ⫽ 1.029 ⫻ 105 mm4

kT ⫽

p 2811092a 0.0324 b 32 1.4

kT ⫽ 2059 N # m

(b) MAX SHEAR STRESS AND STRAIN f ⫽ 5a

T ⫽ kT f

tmax ⫽

␶max ⫽ 27.9 MPa gmax ⫽

;

p b radians 180 d Ta b 2 Ip ;

tmax G

␥max ⫽ 997 ⫻ 10⫺6 radians

;

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SECTION 3.3

Circular Bars and Tubes

255

Problem 3.3-5 A high-strength steel drill rod used for boring a hole in the earth has a diameter of 0.5 in. (see figure). The allowable shear stress in the steel is 40 ksi and the shear modulus of elasticity is 11,600 ksi. What is the minimum required length of the rod so that one end of the rod can be twisted 30° with respect to the other end without exceeding the allowable stress?

Solution 3.3-5

d = 0.5 in.

T

T L

Steel drill rod

T

d = 0.5 in.

T

T⫽

L

G ⫽ 11,600 psi

p b rad ⫽ 0.52360 rad 180

Lmin ⫽

␶allow ⫽ 40 ksi

⫽

MINIMUM LENGTH From Eq. (3-12): tmax ⫽

16T pd3

TL 32TL ⫽ GIP Gpd4

Gpd 4f , substitute Tinto Eq. (1): 32L

tmax ⫽ a

d ⫽ 0.5 in. f ⫽ 30° ⫽ (30°)a

From Eq. (3-15): f ⫽

ba 3

16 pd

Gpd4f Gdf b ⫽ 32L 2L

Gdf 2t allow (11,600 ksi)(0.5 in.)(0.52360 rad) 2(40 ksi)

Lmin ⫽ 38.0 in.

;

(1)

Problem 3.3-6 The steel shaft of a socket wrench has a diameter of 8.0 mm. and a length of 200 mm (see figure). If the allowable stress in shear is 60 MPa, what is the maximum permissible torque Tmax that may be exerted with the wrench? Through what angle ␾ (in degrees) will the shaft twist under the action of the maximum torque? (Assume G ⫽ 78 GPa and disregard any bending of the shaft.)

d = 8.0 mm T L = 200 mm

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Torsion

Solution 3.3-6

Socket wrench ANGLE OF TWIST From Eq. (3-15): f ⫽

d ⫽ 8.0 mm

L ⫽ 200 mm

␶allow ⫽ 60 MPa

G ⫽ 78 GPa

MAXIMUM PERMISSIBLE TORQUE 16T From Eq. (3-12): tmax ⫽ pd3 3 pd tmax Tmax ⫽ 16 3

Tmax ⫽

p(8.0 mm) (60 MPa) 16

Tmax ⫽ 6.03 N # m

;

Problem 3.3-7 A circular tube of aluminum is subjected to torsion by torques T applied at the ends (see figure). The bar is 24 in. long, and the inside and outside diameters are 1.25 in. and 1.75 in., respectively. It is determined by measurement that the angle of twist is 4° when the torque is 6200 lb-in. Calculate the maximum shear stress ␶max in the tube, the shear modulus of elasticity G, and the maximum shear strain ␥max (in radians).

TmaxL GIP

From Eq. (3-12): Tmax ⫽ f⫽ a f⫽

f⫽

pd3t max L ba b 16 GIP

pd3tmaxL(32) 16G(pd4)

⫽

pd3tmax 16 IP ⫽

pd4 32

2tmaxL Gd

2(60 MPa)(200 mm) ⫽ 0.03846 rad (78 GPa)(8.0 mm)

f ⫽ 10.03846 rad2a

180 deg/radb ⫽ 2.20° p

T

;

T

24 in.

1.25 in. 1.75 in.

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Circular Bars and Tubes

257

Solution 3.3-7 NUMERICAL DATA L ⫽ 24 in. f ⫽ 4a

r2 ⫽

1.75 in. 2

p b radians 180

MAX. SHEAR STRESS p Ip ⫽ 1 r24 ⫺ r142 2 tmax ⫽

Tr2 Ip

gmax ⫽

MAX. SHEAR STRAIN 1.25 in. 2

r1 ⫽

T ⫽ 6200 lb-in.

␥max ⫽ 0.00255 radians

r2 f L

;

SHEAR MODULUS OF ELASTICITY

G

G⫽

G ⫽ 3.129 ⫻ 106 psi tmax ⫽

Tr2 Ip

or

G⫽

Ip ⫽ 0.681 in.

4

␶max ⫽ 7965 psi

TL fIp

G ⫽ 3.13 ⫻ 106 psi

tmax gmax

;

;

Problem 3.3-8 A propeller shaft for a small yacht is made of a solid steel bar 104 mm in diameter. The allowable stress in shear is 48 MPa, and the allowable rate of twist is 2.0° in 3.5 meters. Assuming that the shear modulus of elasticity is G ⫽ 80 GPa, determine the maximum torque Tmax that can be applied to the shaft.

d T

T

L

Solution 3.3-8 NUMERICAL DATA d ⫽ 104 mm

FIND MAX. TORQUE BASED ON ALLOWABLE RATE OF TWIST

␶a ⫽ 48 MPa

p 2a b 180 rad u⫽ 3.5 m Ip ⫽

p 4 d 32

f u⫽ L

G ⫽ 80 GPa

Ip ⫽ 1.149 ⫻ 107 mm4

Tmax ⫽

GIpf L

Tmax ⫽ GIp␪

Tmax ⫽ 9164 N # m ^ governs

;

FIND MAX. TORQUE BASED ON ALLOWABLE SHEAR STRESS taIp Tmax ⫽ 10,602 N # m Tmax ⫽ d 2

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Torsion

Problem 3.3-9 Three identical circular disks A, B, and C are welded

P3

to the ends of three identical solid circular bars (see figure). The bars lie in a common plane and the disks lie in planes perpendicular to the axes of the bars. The bars are welded at their intersection D to form a rigid connection. Each bar has diameter d1 ⫽ 0.5 in. and each disk has diameter d2 ⫽ 3.0 in. Forces P1, P2, and P3 act on disks A, B, and C, respectively, thus subjecting the bars to torsion. If P1 ⫽ 28 lb, what is the maximum shear stress ␶max in any of the three bars?

135∞

P1

P3 d1

A D

135∞ P1

90∞

d2

P2 P2

Solution 3.3-9

B

Three circular bars THE THREE TORQUES MUST BE IN EQUILIBRIUM

T3 is the largest torque T3 ⫽ T1 12 ⫽ P1d2 12 MAXIMUM SHEAR STRESS (Eq. 3-12) 16T3 16P1d2 12 16T tmax ⫽ ⫽ ⫽ 3 3 pd pd1 pd31

d1 ⫽ diameter of bars ⫽ 0.5 in.

tmax ⫽

d2 ⫽ diameter of disks ⫽ 3.0 in. P1 ⫽ 28 lb T1 ⫽ P1d2

T2 ⫽ P2d2

T3 ⫽ P3d2

C

16(28 lb)(3.0 in.)12 p(0.5 in.)3

⫽ 4840 psi

;

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SECTION 3.3

Problem 3.3-10 The steel axle of a large winch on an ocean liner is subjected to a

torque of 1.65 kN ⭈ m (see figure). What is the minimum required diameter dmin if the allowable shear stress is 48 MPa and the allowable rate of twist is 0.75°/m? (Assume that the shear modulus of elasticity is 80 GPa.)

Circular Bars and Tubes

T

259

d T

Solution 3.3-10 NUMERICAL DATA T ⫽ 1.65 kN # m u a ⫽ 0.75 a

␶a ⫽ 48 MPa

G ⫽ 80 GPa

p rad b 180 m

32T d ⫽ pGua 4

Ip ⫽

T Gu

;

MIN. REQUIRED DIAMETER OF SHAFT BASED ON ALLOWABLE

MIN. REQUIRED DIAMETER OF SHAFT BASED ON

T GIp

dmin ⫽ 63.3 mm ^ governs

SHEAR STRESS

ALLOWABLE RATE OF TWIST

u⫽

dmin ⫽ 0.063 m

p 4 T d ⫽ 32 Gu

32T dmin ⫽ a b pGua

1 4

t⫽

Td 2Ip

t⫽

Td p 4 2a d b 32

1

16T 3 d dmin ⫽ c pta

dmin ⫽ 0.056 m dmin ⫽ 55.9 mm

Problem 3.3-11 A hollow steel shaft used in a construction auger has

outer diameter d2 ⫽ 6.0 in. and inner diameter d1 ⫽ 4.5 in. (see figure). The steel has shear modulus of elasticity G ⫽ 11.0 ⫻ 106 psi. For an applied torque of 150 k-in., determine the following quantities: (a) shear stress ␶2 at the outer surface of the shaft, (b) shear stress ␶1 at the inner surface, and (c) rate of twist ␪ (degrees per unit of length).

d2

Also, draw a diagram showing how the shear stresses vary in magnitude along a radial line in the cross section.

d1 d2

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Solution 3.3-11 Construction auger d2 ⫽ 6.0 in.

r2 ⫽ 3.0 in.

d1 ⫽ 4.5 in.

r1 ⫽ 2.25 in.

(c) RATE OF TWIST u⫽

G ⫽ 11 ⫻ 106 psi

u ⫽ 157 * 10⫺6 rad/ in. ⫽ 0.00898°/ in.

T ⫽ 150 k-in. IP ⫽

(150 k-in.) T ⫽ GIP (11 * 106 psi)(86.98 in.)4

p 4 (d ⫺ d14) ⫽ 86.98 in.4 32 2

;

(d) SHEAR STRESS DIAGRAM

(a) SHEAR STRESS AT OUTER SURFACE t2 ⫽

(150 k-in.)(3.0 in.) Tr2 ⫽ IP 86.98 in.4 ⫽ 5170 psi

;

(b) SHEAR STRESS AT INNER SURFACE t1 ⫽

Tr1 r1 ⫽ t ⫽ 3880 psi IP r2 2

;

Problem 3.3-12 Solve the preceding problem if the shaft has outer diameter d2 ⫽ 150 mm and inner diameter d1 ⫽ 100 mm. Also, the steel has shear modulus of elasticity G ⫽ 75 GPa and the applied torque is 16 kN ⭈ m.

Solution 3.3-12 Construction auger d2 ⫽ 150 mm

r2 ⫽ 75 mm

d1 ⫽ 100 mm

r1 ⫽ 50 mm

G ⫽ 75 GPa T ⫽ 16 kN # m p 4 IP ⫽ (d2 ⫺ d14) ⫽ 39.88 * 106 mm4 32 (a) SHEAR STRESS AT OUTER SURFACE (16 kN # m)(75 mm)

Tr2 ⫽ IP 39.88 * 106 mm4 ⫽ 30.1 MPa ;

t2 ⫽

(b) SHEAR STRESS AT INNER SURFACE t1 ⫽

Tr1 r1 ⫽ t 2 ⫽ 20.1 MPa IP r2

;

(c) RATE OF TWIST T 16 kN # m u⫽ ⫽ GIP (75 GPa)(39.88 * 106 mm4)

␪ ⫽ 0.005349 rad/m ⫽ 0.306°/m (d) SHEAR STRESS DIAGRAM

;

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SECTION 3.3

Circular Bars and Tubes

Problem 3.3-13 A vertical pole of solid circular cross section is twisted by

c

horizontal forces P ⫽ 1100 lb acting at the ends of a horizontal arm AB (see figure). The distance from the outside of the pole to the line of action of each force is c ⫽ 5.0 in. If the allowable shear stress in the pole is 4500 psi, what is the minimum required diameter dmin of the pole?

Solution 3.3-13

261 P

c B

A P d

Vertical pole tmax ⫽

P(2c + d)d 4

pd /16

⫽

16P(2c + d) pd3

(␲␶max)d3 ⫺ (16P)d ⫺ 32Pc ⫽ 0 P ⫽ 1100 lb

SUBSTITUTE NUMERICAL VALUES:

c ⫽ 5.0 in.

UNITS: Pounds, Inches

␶allow ⫽ 4500 psi

(␲)(4500)d3 ⫺ (16)(1100)d ⫺ 32(1100)(5.0) ⫽ 0

Find dmin

or d3 ⫺ 1.24495d ⫺ 12.4495 ⫽ 0 Solve numerically:

TORSION FORMULA Tr Td ⫽ tmax ⫽ IP 2IP T ⫽ P12c + d2

d ⫽ 2.496 in. dmin ⫽ 2.50 in.

IP ⫽

;

pd 4 32

Problem 3.3-14 Solve the preceding problem if the horizontal forces have magnitude P ⫽ 5.0 kN, the distance c ⫽ 125 mm, and the allowable shear stress is 30 MPa.

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Torsion

Solution 3.3-14

Vertical pole TORSION FORMULA tmax ⫽

Tr Td ⫽ IP 2IP

T ⫽ P(2c + d) tmax ⫽

P(2c + d)d 4

pd /16

IP ⫽ ⫽

pd 4 32

16P (2c + d) pd 3

(␲␶max)d 3 ⫺ (16P)d ⫺ 32Pc ⫽ 0 SUBSTITUTE NUMERICAL VALUES: P ⫽ 5.0 kN

UNITS: Newtons, Meters

c ⫽ 125 mm

(␲)(30 ⫻ 106)d3 ⫺ (16)(5000)d ⫺ 32(5000)(0.125) ⫽ 0

␶allow ⫽ 30 MPa

or

Find dmin

d3 ⫺ 848.826 ⫻ 10⫺6d ⫺ 212.207 ⫻ 10⫺6 ⫽ 0 d ⫽ 0.06438 m

Solve numerically:

dmin ⫽ 64.4 mm

Problem 3.3-15 A solid brass bar of diameter d ⫽ 1.25 in. is subjected to torques T1, as shown in part (a) of the figure. The allowable shear stress in the brass is 12 ksi. (a) What is the maximum permissible value of the torques T1? (b) If a hole of diameter 0.625 in. is drilled longitudinally through the bar, as shown in part (b) of the figure, what is the maximum permissible value of the torques T2? (c) What is the percent decrease in torque and the percent decrease in weight due to the hole?

T1

d

;

T1

(a) d

T2

T2

(b)

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SECTION 3.3

263

Circular Bars and Tubes

Solution 3.3-15

␶a ⫽ 12 ksi

d2 ⫽ 1.25 in. d1 ⫽ 0.625 in. p 1 d24 ⫺ d142 32 T2max ⫽ d2 2 d24 ⫺d12 1 T2max ⫽ tap 16 d2 ta

(a)

MAX. PERMISSIBILE VALUE OF TORQUE

T1max ⫽

taIp

T1 – SOLID BAR

p ta d 4 32 T1max ⫽ d 2

d 2 1 T1max ⫽ tapd3 16 1 T1max ⫽ (12)p (1.25)3 16 ; T1max ⫽ 4.60 in.-k

(b) MAX. PERMISSIBILE VALUE OF TORQUE T2 – HOLLOW BAR

T2max ⫽ 4.31 in.-k (c) PERCENT

;

DECREASE IN TORQUE

IN WEIGHT DUE TO HOLE IN

&

PERCENT DECREASE

(b)

percent decrease in torque T1max ⫺ T2max (100) ⫽ 6.25% T1max

;

percent decrease in weight (weight is proportional to x-sec area) A1 ⫽

d1 d2

p 2 d 4 2

A2 ⫽

p 2 1d ⫺ d122 4 2

A1 ⫺ A2 (100) ⫽ 25 % A1

;

Problem 3.3-16 A hollow aluminum tube used in a roof structure has an outside

diameter d2 ⫽ 104 mm and an inside diameter d1 ⫽ 82 mm (see figure). The tube is 2.75 m long, and the aluminum has shear modulus G ⫽ 28 GPa. (a) If the tube is twisted in pure torsion by torques acting at the ends, what is the angle of twist (in degrees) when the maximum shear stress is 48 MPa? (b) What diameter d is required for a solid shaft (see figure) to resist the same torque with the same maximum stress? (c) What is the ratio of the weight of the hollow tube to the weight of the solid shaft?

d1 d2

d

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CHAPTER 3 Torsion

Solution 3.3-16 Td22

NUMERICAL DATA set tmax expression equal to

d2 ⫽ 104 mm d1 ⫽ 82 mm

⫽

L ⫽ 2.75 ⫻ 103 mm G ⫽ 28 GPa I p ⫽ (p/32)(d 2 4 ⫺ d 14 )

32Td2 pa d24 ⫺ d14 b

d3 ⫽

Ip ⫽ 7.046 ⫻ 106 mm4

p a d42 ⫺ d41 b 32

then solve for d

d24 ⫺ d14 d2

dreqd ⫽ a

1

d24 ⫺ d14 3 b dreqd⫽88.4 mm d2

;

(c) RATIO OF WEIGHTS OF HOLLOW & SOLID SHAFTS (a) FIND ANGLE OF TWIST ␶max ⫽ 48 MPa f⫽

TL GIp

f ⫽ (tmax)

f⫽ a

WEIGHT IS PROPORTIONAL TO CROSS SECTIONAL AREA

p 2 A d ⫺ d12 B 4 2 Ah p A s ⫽ d reqd 2 ⫽ 0.524 4 As

Td2 2L b 2Ip Gd2

Ah ⫽

2L Gd2

So the weight of the tube is 52% of the solid shaft, but they resist the same torque.

␾ ⫽ 0.091 radians ␾ ⫽ 5.19°

;

;

(b) REPLACE HOLLOW SHAFT WITH SOLID SHAFT - FIND DIAMETER

d 2 p

T tmax ⫽

32d 4

tmax ⫽

16T d3p

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SECTION 3.3

265

Circular Bars and Tubes

Problem 3.3-17 A circular tube of inner radius r1 and outer radius r2 is subjected to a torque produced by forces P ⫽ 900 lb (see figure). The forces have their lines of action at a distance b ⫽ 5.5 in. from the outside of the tube.

P

If the allowable shear stress in the tube is 6300 psi and the inner radius r1 ⫽ 1.2 in., what is the minimum permissible outer radius r2? P P r2 r1 P b

Solution 3.3-17

2r2

b

Circular tube in torsion SOLUTION OF EQUATION UNITS: Pounds, Inches Substitute numerical values: 6300 psi ⫽

P ⫽ 900 lb b ⫽ 5.5 in.

␶allow ⫽ 6300 psi r1 ⫽ 1.2 in.

4(900 lb)(5.5 in. + r2)(r2) p[(r42) ⫺ (1.2 in.)4]

or r42 ⫺ 2.07360 ⫺ 0.181891 ⫽ 0 r2(r2 + 5.5) or r42 ⫺ 0.181891 r22 ⫺ 1.000402 r2 ⫺ 2.07360 ⫽ 0

Find minimum permissible radius r2

Solve numerically:

TORSION FORMULA

r2 ⫽ 1.3988 in.

T ⫽ 2P(b ⫹ r2) IP ⫽

p 4 (r ⫺ r41) 2 2

2P(b + r2)r2 4P(b + r2)r2 Tr2 ⫽ ⫽ p 4 IP p (r42 ⫺ r41) (r2 ⫺ r41) 2 All terms in this equation are known except r2. tmax ⫽

MINIMUM PERMISSIBLE RADIUS r2 ⫽ 1.40 in.

;

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Torsion

Nonuniform Torsion T1

Problem 3.4-1 A stepped shaft ABC consisting of two solid

d1

circular segments is subjected to torques T1 and T2 acting in opposite directions, as shown in the figure. The larger segment of the shaft has diameter d1  2.25 in. and length L1  30 in.; the smaller segment has diameter d2  1.75 in. and length L2  20 in. The material is steel with shear modulus G  11  106 psi, and the torques are T1  20,000 lb-in. and T2  8,000 lb-in. Calculate the following quantities: (a) the maximum shear stress max in the shaft, and (b) the angle of twist C (in degrees) at end C.

Solution 3.4-1

d2

T2

B

A

C

L1

L2

Stepped shaft SEGMENT BC TBC  T2  8,000 lb-in.

d1  2.25 in.

L1  30 in.

d2  1.75 in.

L2  20 in.

T2  8,000 lb-in.

` 

TABL1  G(Ip)AB

16(8,000 lb-in.) p(1.75 in.)3

 7602 psi

(8,000 lb-in.)(20 in.) (11 * 106 psi)a

p b(1.75 in.)4 32

16(12,000 lb-in.) p(2.25 in.)3

C  AB  BC  (0.013007  0.015797) rad  5365 psi

(12,000 lb-in.)(30 in.) (11 * 106 psi)a

 0.013007 rad

;

(b) ANGLE OF TWIST AT END C

TAB  T2  T1  12,000 lb-in.

fAB 

TBCL2  G(Ip)BC



max  7600 psi

SEGMENT AB

pd31

fBC 

pd32

(a) MAXIMUM SHEAR STRESS Segment BC has the maximum stress

T1  20,000 lb-in.

tAB  `

16 TBC

 0.015797 rad

G  11  10 psi 6

16 TAB

tBC 

p b(2.25 in.)4 32

C  0.002790 rad  0.16°

;

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SECTION 3.4

Problem 3.4-2 A circular tube of outer diameter d3  70 mm and

inner diameter d2  60 mm is welded at the right-hand end to a fixed plate and at the left-hand end to a rigid end plate (see figure). A solid circular bar of diameter d1  40 mm is inside of, and concentric with, the tube. The bar passes through a hole in the fixed plate and is welded to the rigid end plate. The bar is 1.0 m long and the tube is half as long as the bar. A torque T  1000 N  m acts at end A of the bar. Also, both the bar and tube are made of an aluminum alloy with shear modulus of elasticity G  27 GPa.

267

Nonuniform Torsion

Tube Fixed plate End plate

Bar T A

(a) Determine the maximum shear stresses in both the bar and tube. (b) Determine the angle of twist (in degrees) at end A of the bar.

Tube Bar

d1 d2 d3

Solution 3.4-2

Bar and tube TORQUE T  1000 N  m (a) MAXIMUM SHEAR STRESSES Bar: t bar 

16T

 79.6 MPa ; pd31 T(d3/2)  32.3 MPa Tube: t tube  (Ip) tube

;

TUBE d3  70 mm Ltube  0.5 m (Ip) tube 

(b) ANGLE OF TWIST AT END A

d2  60 mm G  27 GPa

Bar: fbar 

p 4 (d  d24) 32 3

Tube: ftube 

 1.0848 * 106 mm4

A  9.43°

(Ip) bar 

pd14 32

Lbar  1.0 m

TL tube  0.0171 rad G(Ip) tube

A  bar  tube  0.1474  0.0171  0.1645 rad

BAR d1  40 mm

TL bar  0.1474 rad G(Ip) bar

G  27 GPa

 251.3 * 103 mm4

;

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Torsion

Problem 3.4-3 A stepped shaft ABCD consisting of solid circular segments is subjected to three torques, as shown in the figure. The torques have magnitudes 12.5 k-in., 9.8 k-in., and 9.2 k-in. The length of each segment is 25 in. and the diameters of the segments are 3.5 in., 2.75 in., and 2.5 in. The material is steel with shear modulus of elasticity G  11.6  103 ksi.

12.5 k-in. 3.5 in.

25 in.

D

C

B

A

(a) Calculate the maximum shear stress max in the shaft. (b) Calculate the angle of twist D (in degrees) at end D.

9.8 k-in. 9.2 k-in. 2.75 in. 2.5 in.

25 in.

25 in.

Solution 3.4-3 NUMERICAL DATA (INCHES, KIPS) TB  12.5 k-in.

TC  9.8k-in.

TD  9.2 k-in.

tBC 

Check BC:

1TC + TD2 p d 4 32 BC

L  25 in.

dAB  3.5 in.

dBC  2.75 in.

dCD  2.5 in.

G  11.6  (103) ksi

dBC 2

BC  4.65 ksi

; controls

(b) ANGLE OF TWIST AT END D (a) MAX. SHEAR STRESS IN SHAFT

T1  |RA|

torque reaction at A: RA  (TB  TC  TD) RA  31.5 in.-kip |RA| tAB 

dAB 2

p d 4 32 AB

max  3.742 ksi

TD Check CD:

tCD 

dCD 2

p d 4 32 CD

CD  2.999 ksi

T2  TC  TD

T3  TD

IP1 

p d 4 32 AB p IP3  d 4 32 CD

IP2 

p d 4 32 BC

TiLi fD  a GIpi

fD 

T2 T3 L T1 a + + b G IP1 IP2 IP3

D  0.017 radians

fD  0.978 degrees

;

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SECTION 3.4

Problem 3.4-4 A solid circular bar ABC consists of two segments, as shown in the figure. One segment has diameter d1  56 mm and length L1  1.45 m; the other segment has diameter d2  48 mm and length L2  1.2 m. What is the allowable torque Tallow if the shear stress is not to exceed 30 MPa and the angle of twist between the ends of the bar is not to exceed 1.25°? (Assume G  80 GPa.)

269

Nonuniform Torsion

d1

d2

T A

C

B L1

T

L2

Solution 3.4-4 Tallow based on angle of twist

NUMERICAL DATA d1  56 mm

d2  48 mm

L1  1450 mm

L2  1200 mm f a  1.25 a

a  30 MPa

G  80 GPa

fmax 

p b radians 180

Allowable torque

Tallow 

16T d32p

tapd23 Tallow  16

L1

J 32 a

p

d14 b

L2

+

a

Gf a

L1 a

Tallow based on shear stress tmax 

T G

p 4 d1 b 32

+

p 4 d b 32 2 K

L2 a

p 4 d2 b 32

Tallow  459 N # m

;

T1 = T2 = 1000 lb-in. 500 lb-in.

T3 = T4 = 800 lb-in. 500 lb-in.

governs

Tallow  651.441 N # m

Problem 3.4-5 A hollow tube ABCDE constructed of monel metal is subjected to five torques acting in the directions shown in the figure. The magnitudes of the torques are T1  1000 lb-in., T2  T4  500 lb-in., and T3  T5  800 lb-in. The tube has an outside diameter d2  1.0 in. The allowable shear stress is 12,000 psi and the allowable rate of twist is 2.0°/ft. Determine the maximum permissible inside diameter d1 of the tube.

A

B

C

D d2 = 1.0 in.

T5 = 800 lb-in.

E

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Torsion

Solution 3.4-5

Hollow tube of monel metal REQUIRED POLAR MOMENT OF INERTIA BASED UPON ALLOWABLE SHEAR STRESS

tmax  d2  1.0 in.

allow  12,000 psi

allow  2°/ft  0.16667°/in.

Tmaxr Ip

REQUIRED

IP 

T max(d2/2)  0.05417 in.4 tallow

POLAR MOMENT OF INERTIA BASED UPON

ALLOWABLE ANGLE OF TWIST

 0.002909 rad/in. From Table H-2, Appendix H: G  9500 ksi TORQUES

u

Tmax GIP

IP 

Tmax  0.04704 in.4 Gu allow

SHEAR STRESS GOVERNS Required IP  0.05417 in.4 IP 

T1  1000 lb-in.

T2  500 lb-in.

T4  500 lb-in.

T5  800 lb-in.

T3  800 lb-in.

INTERNAL TORQUES

p 4 (d  d41) 32 2

d41  d43 

32(0.05417 in.4) 32IP  (1.0 in.)4  p p

 0.4482 in.4

TAB   T1   1000 lb-in.

d1  0.818 in.

TBC   T1  T2  500 lb-in.

(Maximum permissible inside diameter)

;

TCD   T1  T2  T3   1300 lb-in. TDE   T1  T2  T3  T4   800 lb-in. Largest torque (absolute value only): Tmax  1300 lb-in.

80 mm

Problem 3.4-6 A shaft of solid circular cross section consisting of two segments is shown in the first part of the figure. The left-hand segment has diameter 80 mm and length 1.2 m; the right-hand segment has diameter 60 mm and length 0.9 m. Shown in the second part of the figure is a hollow shaft made of the same material and having the same length. The thickness t of the hollow shaft is d/10, where d is the outer diameter. Both shafts are subjected to the same torque. If the hollow shaft is to have the same torsional stiffness as the solid shaft, what should be its outer diameter d?

1.2 m

60 mm

0.9 m d

2.1 m

d t=— 10

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SECTION 3.4

Solution 3.4-6

Nonuniform Torsion

271

Solid and hollow shafts

SOLID SHAFT CONSISTING OF TWO SEGMENTS

TORSIONAL STIFFNESS kT 

T f

Torque T is the same for both shafts.

⬖ For equal stiffnesses, 1  2 98,741 m3  TLi  f1  g GIPi

T(1.2 m) p Ga b(80 mm)4 32

T(0.9 m) +

p G a b (60 mm)4 32

32T (29,297 m3 + 69,444 m3)  pG 

d4 

3.5569 m d4

3.5569  36.023 * 106 m4 98,741

d  0.0775 m  77.5 mm

;

32T (98,741 m3) pG

HOLLOW SHAFT

d0  inner diameter  0.8d f2 



TL  GIp

T(2.1 m) Ga

p b[d4  (0.8d)4] 32

32T 2.1 m 32T 3.5569 m a b a b 4 pG 0.5904 d pG d4

UNITS: d  meters

Problem 3.4-7 Four gears are attached to a circular shaft and transmit the torques shown in the figure. The allowable shear stress in the shaft is 10,000 psi. 8,000 lb-in. (a) What is the required diameter d of the shaft if it has a solid cross section? (b) What is the required outside diameter d if the shaft is hollow with an inside diameter of 1.0 in.?

19,000 lb-in. 4,000 lb-in. A

7,000 lb-in. B C D

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Torsion

Solution 3.4-7

Shaft with four gears (b) HOLLOW SHAFT Inside diameter d0  1.0 in.

allow  10,000 psi

TBC  11,000 lb-in.

TAB  8000 lb-in.

TCD  7000 lb-in.

Tr tmax  Ip

10,000 psi 

(a) SOLID SHAFT tmax  d3 

t allow 

16T pd3

d Tmax a b 2 Ip

d (11,000 lb-in.)a b 2 a

p b[d4  (1.0 in.)4] 32

UNITS: d  inches

16(11,000 lb-in.) 16Tmax   5.602 in.3 pt allow p(10,000 psi)

Required d  1.78 in.

10,000 

;

56,023 d d4  1

or d 4  5.6023 d  1  0 Solving, d  1.832 Required d  1.83 in.

Problem 3.4-8 A tapered bar AB of solid circular cross section is twisted by torques T (see figure). The diameter of the bar varies linearly from dA at the left-hand end to dB at the right-hand end. For what ratio dB/dA will the angle of twist of the tapered bar be one-half the angle of twist of a prismatic bar of diameter dA? (The prismatic bar is made of the same material, has the same length, and is subjected to the same torque as the tapered bar.) Hint: Use the results of Example 3-5.

T

;

B

A

T

L dA

dB

Problems 3.4-8, 3.4-9 and 3.4-10

Solution 3.4-8

Tapered bar AB

ANGLE OF TWIST

TAPERED BAR (From Eq. 3-27) b2 + b + 1 TL f1  a b G(IP)A 3b 3 PRISMATIC BAR TL f2  G(IP)A

dB b dA

b2 + b + 1

f1 

1 f 2 2

or

3 3  2 2  2  2  0

3b 3

SOLVE NUMERICALLY: b

dB  1.45 dA

;



1 2

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SECTION 3.4 Nonuniform Torsion

273

Problem 3.4-9 A tapered bar AB of solid circular cross section is twisted

by torques T  36,000 lb-in. (see figure). The diameter of the bar varies linearly from dA at the left-hand end to dB at the right-hand end. The bar has length L  4.0 ft and is made of an aluminum alloy having shear modulus of elasticity G  3.9  106 psi. The allowable shear stress in the bar is 15,000 psi and the allowable angle of twist is 3.0°. If the diameter at end B is 1.5 times the diameter at end A, what is the minimum required diameter dA at end A? (Hint: Use the results of Example 3–5).

Solution 3.4-9

Tapered bar

MINIMUM DIAMETER BASED (From Eq. 3-27)

dB  1.5 dA T  36,000 lb-in.

b  dB/dA  1.5

L  4.0 ft  48 in.

b2 + b + 1 TL TL a b  (0.469136) G(IP)A G(IP)A 3b 3 (36,000 lb-in.)(48 in.)  (0.469136) p (3.9 * 106 psi)a bd4A 32

f

G  3.9  106 psi

allow  15,000 psi allow  3.0°  0.0523599 rad MINIMUM

DIAMETER BASED UPON ALLOWABLE SHEAR



STRESS

tmax 

16T pd3A

d3A 

UPON ALLOWABLE ANGLE OF

TWIST

16(36,000 lb-in.) 16 T  ptallow p(15,000 psi)

 12.2231 in.3 dA  2.30 in.

d4A 

2.11728 in.4 d4A 2.11728 in.4 2.11728 in.4  0.0523599 rad fallow

 40.4370 in.4 dA  2.52 in. ANGLE OF TWIST GOVERNS Min. dA  2.52 in.

;

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CHAPTER 3 Torsion

Problem 3.4-10 The bar shown in the figure is tapered linearly from end A to end B and has a solid circular cross section. The diameter at the smaller end of the bar is dA  25 mm and the length is L  300 mm. The bar is made of steel with shear modulus of elasticity G  82 GPa. If the torque T  180 N  m and the allowable angle of twist is 0.3°, what is the minimum allowable diameter dB at the larger end of the bar? (Hint: Use the results of Example 3-5.)

Solution 3.4-10

Tapered bar

dA  25 mm L  300 mm G  82 GPa T  180 N  m

allow  0.3°

p (0.3°)a 180 

rad b degrees (180 N # m)(0.3 m)

p (82 GPa)a b(25 mm)4 32 b2 + b + 1

Find dB

0.304915 

DIAMETER BASED UPON ALLOWABLE ANGLE OF TWIST

0.914745   1  0 3

3b 3 2

(From Eq. 3-27) SOLVE NUMERICALLY: dB b dA f

b2 + b + 1 TL p 4 a b(IP)A  d G(IP)A 32 A 3b 3

 1.94452 Min. dB  dA  48.6 mm

;

a

b2 + b + 1 3b 3

b

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SECTION 3.4

Nonuniform Torsion

Segment 1

Segment 2

275

Problem 3.4-11 The nonprismatic cantilever circular bar shown has an internal cylindrical hole from 0 to x, so the net polar moment of inertia of the cross section for segment 1 is (7/8)Ip. Torque T is applied at x and torque T/2 is applied at x  L. Assume that G is constant. (a) (b) (c) (d) (e)

Find reaction moment R1. Find internal torsional moments Ti in segments 1 & 2. Find x required to obtain twist at joint 3 of 3  TL/GIp What is the rotation at joint 2, 2? Draw the torsional moment (TMD: T(x), 0 x L) and displacement (TDD: (x), 0 x L) diagrams.

x

7 —Ip 8

R1

Ip T

1

2

T — 2 3

x

L–x

T1 T2 TMD 0 φ2

TDD 0

0 φ3

0

Solution 3.4-11 (a) REACTION TORQUE R1 T 3 a Mx  0 R1   a T + 2 b R1  2 T ;

L

1 17 x + L 14 2

x

14 L a b 17 2

(b) INTERNAL MOMENTS IN SEGMENTS 1 & 2 T1  R1

T1  1.5 T

T2 

T 2

(c) FIND X REQUIRED TO OBTAIN TRWIST AT JT 3

TL  GIP

L

T 1x 7 Ga IP b 8

+

3 a Tbx 2 7 Ga IP b 8

3 a bx 2 7 a b 8

+

T2(L  x) GIP T a b(L  x) 2

+

1 (L  x) 2

GIP

;

(d) ROTATION AT JOINT 2 FOR X VALUE IN (C)

f2 

TiLi f3  a GIPi TL  GIP

7 L 17

x

f2 

T1x 7 G a Ip b 8 12TL 17GIP

f2 

3 7 a Tb a Lb 2 17 7 G a Ip b 8

;

(e) TMD & TDD – SEE PLOTS ABOVE TMD is constant - T1 for 0 to x & T2 for x to L; hence TDD is linear - zero at jt 1, 2 at jt 2 & 3 at jt 3

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Torsion

Problem 3.4-12 A uniformly tapered tube AB of hollow circular cross section is shown in the figure. The tube has constant wall thickness t and length L. The average diameters at the ends are dA and dB  2dA. The polar moment of inertia may be represented by the approximate formula IP L d3t/4 (see Eq. 3-18). Derive a formula for the angle of twist  of the tube when it is subjected to torques T acting at the ends.

B

A

T

T

L t

t

dA dB = 2dA

Solution 3.4-12

Tapered tube t  thickness (constant) dA, dB  average diameters at the ends dB  2dA

Ip 

pd3t (approximate formula) 4

ANGLE OF TWIST

Take the origin of coordinates at point O. d(x)  Lp(x) 

x x (d )  dA 2L B L p[d(x)]3t ptd3A 3  x 4 4L3

For element of length dx: df 

Tdx  GIP(x)

Tdx ptd3A G a 3 b x3



4TL3dx pGtdA3 x3

4L

2L

f

LL

df 

4TL3

2L

pGtd3A

LL x3

dx



3TL 2pGtd3A

;

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SECTION 3.4

Problem 3.4-13 A uniformly tapered aluminum-alloy tube AB of circular cross section and length L is shown in the figure. The outside diameters at the ends are dA and dB  2dA. A hollow section of length L/2 and constant thickness t  dA/10 is cast into the tube and extends from B halfway toward A. (a) Find the angle of twist  of the tube when it is subjected to torques T acting at the ends. Use numerical values as follows: dA  2.5 in., L  48 in., G  3.9  106 psi, and T  40,000 in.-lb. (b) Repeat (a) if the hollow section has constant diameter dA. (See figure part b.)

L

dB (a) L — 2

dA

A

from B to centerline outer and inner diameters as function of x 0 … x …

L 2

d0(x)  2d A 

d0(x)  dB  a xdA L

di (x)  (dB  2t) di (x) 

dB  dA bx L

[(2dA  2t) (dA  2t)] x L

1 9L + 5x d 5 A L

solid from centerline to A L … x … L 2

d0(x)  2dA  L

x dA L

L T 32 1 1 2 f a b dx + L 4 dx 4 4 G p P L0 d0  d i L2 d0 Q

dA

B T

L (b)

PART (A) - CONSTANT THICKNESS use x as integration variable measured from B toward A

B T

dA

T

Solution 3.4-13

t constant dB – 2t

L — 2

A

T

277

Nonuniform Torsion

dB

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CHAPTER 3 Torsion

L

T 32 2 f  a b≥ G p L0

f  32

L

1 xdA 4 1  9L + 5x 4 a2dA  b  a dA b L 5 L

dx +

L2 L

1 a 2dA 

xdA 4 b L

T  125 3ln(2) + 2ln(7)  ln(197) 125 2ln(19) + ln(181) 19 a L  L + Lb 4 Gp 2 2 dA dA 4 81dA 4

Simplifying: f 

16TL 81GpdA4

a38 + 10125 ln a

71117 bb 70952

Use numerical properties as follows L  48 in.

a  0.049 radians

fa  2.79°

or

fa  3.868

G  3.9  106 psi

TL GdA4 dA  2.5 in.

;

PART (B) - CONSTANT HOLE DIAMETER 0 … x …

d B  dA bx L

d0( x)  dB  a

L 2

L … x … L 2

d0(x)  2dA  L 2

f

T 32 a b G p

f

2 T 32 a b G p L0

1

4 P L0 d0  di

J

d0(x)  2 dA 

xdA L

xdA L L

dx + 4

1

L d0 4 L 2

L

dx

Q

L

1 1 dx + dx 1 xdA 4 xdA 4 L 2 4 K a2dA  b  dA a 2dA  b L L

3 ln(5) + 2atana b 2 T 1 1 ln(3) + 2atan(2) 19 ± L fb  32  L + L≤ Gp 4 4 dA 4 dA 4 81dA 4 Simplifying, fb  3.057

TL GdA4

Use numerical properties given above

b  0.039 radians fa fb

dx ¥

 1.265

fb  2.21°

;

so tube (a) is more flexible than tube (b)

di (x)  dA

t

dA 10

T  40000 in.-lb

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SECTION 3.4

Problem 3.4-14 For the thin nonprismatic steel pipe of constant thickness t and variable diameter d shown with applied torques at joints 2 and 3, determine the following. (a) Find reaction moment R1. (b) Find an expression for twist rotation 3 at joint 3. Assume that G is constant. (c) Draw the torsional moment diagram (TMD: T(x), 0 x L).

2d t

d

Nonuniform Torsion

t

d T, f3

T/2

R1

2

1

279

L — 2 x

3 L — 2 T

T — 2 0

TMD

Solution 3.4-14 (a) REACTION TORQUE R1 statics: T  0 T R1  + T0 2

L

R1 

T 2

2 2T f3  Gpt L0

;

(b) ROTATION AT JOINT 3

p 3 Ga d12(x) t b 4 L

+

L 2

LL2

T

x 3 bd L

dx

L

Gpd3t LL2

dx

L 2 2T f3  Gpt L0

1 c2d a 1 

x 3 bd L

dx

2TL +

T 2

L 2

L 0

0 … x …

L … x … L 2

d23(x)  d

f3 

x b L

c2da 1 

4T +

d12( x)  2da1 

1

dx

dx

p Ga d23(x)3t b 4 use IP expression for thin walled tubes

f3  f3 

Gpd3t 2TL

3TL 3

8Gp d t 19TL 8Gpd3t

+

Gpd3t ;

(c) TMD TMD is piecewise constant: T(x)  T/2 for segment 1-2 & T(x)  T for segment 2-3 (see plot above)

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Torsion

Problem 3.4-15 A mountain-bike rider going uphill applies

Handlebar extension d01, t01

torque T  Fd (F  15 lb, d  4 in.) to the end of the handlebars ABCD (by pulling on the handlebar extenders DE). Consider the right half of the handlebar assembly only (assume the bars are fixed at the fork at A). Segments AB and CD are prismatic with lengths L1  2 in. and L3  8.5 in., and with outer diameters and thicknesses d01  1.25 in., t01  0.125 in., and d03  0.87 in., t03  0.115 in., respectively as shown. Segment BC of length L2  1.2 in., however, is tapered, and outer diameter and thickness vary linearly between dimensions at B and C.

B A

E

d03, t03 C

L1 L2

T = Fd D

L3

d

Consider torsion effects only. Assume G  4000 ksi is constant. Derive an integral expression for the angle of twist D of half of the handlebar tube when it is subjected to torque T  Fd acting at the end. Evaluate D for the given numerical values.

45∞

Handlebar extension F

D Handlebar

Solution 3.4-15 ASSUME THIN WALLED TUBES Segments AB & CD p p IP1  d01 3t01 IP3  d03 3t03 4 4 Segment BC

0 x L2

d02(x)  d01 a1  d02(x) 

d01L2  d01x + d03x L2

t02(x)  t01 a1  t02(x)  fD 

x x b + d03 a b L2 L2

x x b + t03 a b L2 L2

t01L2  t01x + t03x L2

L2 L3 1 Fd L1 + dx + p G IP1 I L0 P3 P Q d (x)3t02(x) 4 02

fD 

L2 L2 4 L1 4Fd c 3  dx Gp d01 t01 L0 (d01L2  d01x + d03x)3 * (t01L2  t01x + t03x)

+

L3 d03 3t03

d

;

NUMERICAL DATA L1  2 in. L2  1.2 in. L3  8.5 in. t03  0.115 in. d01  1.25 in. t01  0.125 in. F  15 lb d  4 in. d03  0.87 in. G  4  (106) psi f D  0.142°

;

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SECTION 3.4

Nonuniform Torsion

281

Problem 3.4-16 A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant intensity t per unit distance (see figure).

t A

(a) Determine the maximum shear stress max in the bar. (b) Determine the angle of twist  between the ends of the bar.

B

L

Solution 3.4-16

Bar with distributed torque (a) MAXIMUM SHEAR STRESS Tmax  tL

tmax 

16Tmax pd3



16tL pd3

;

(b) ANGLE OF TWIST T(x)  tx df  t  intensity of distributed torque d  diameter

IP 

T(x)dx 32 tx dx  GIP pGd4 L

f

pd4 32

L0

df 

32t pGd4 L0

L

x dx 

16tL2 pGd4

;

G  shear modulus of elasticity

Problem 3.4-17 A prismatic bar AB of solid circular cross section (diameter d) is loaded by a distributed torque (see figure). The intensity of the torque, that is, the torque per unit distance, is denoted t(x) and varies linearly from a maximum value tA at end A to zero at end B. Also, the length of the bar is L and the shear modulus of elasticity of the material is G. (a) Determine the maximum shear stress max in the bar. (b) Determine the angle of twist  between the ends of the bar.

t(x) A

L

B

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Solution 3.4-17

Page 282

Torsion

Bar with linearly varying torque (a) Maximum shear stress tmax 

16Tmax pd

3



16TA pd

3



8tAL pd3

;

(b) ANGLE OF TWIST T(x)  torque at distance x from end B T(x) 

t(x)x tAx2  2 2L

IP 

pd4 32

T(x) dx 16tAx2 dx  GIP pGLd4 L L 16tA 16tA L2 2 df  x dx  ; f pGLd4 L0 3pGd4 L0 df 

t(x)  intensity of distributed torque tA  maximum intensity of torque d  diameter G  shear modulus TA  maximum torque  12 tAL

Problem 3.4-18 A nonprismatic bar ABC of solid circular cross section is loaded by distributed torques (see figure). The intensity of the torques, that is, the torque per unit distance, is denoted t(x) and varies linearly from zero at A to a maximum value T0/L at B. Segment BC has linearly distributed torque of intensity t(x)  T0/3L of opposite sign to that applied along AB. Also, the polar moment of inertia of AB is twice that of BC, and the shear modulus of elasticity of the material is G. (a) Find reaction torque RA. (b) Find internal torsional moments T(x) in segments AB and BC. (c) Find rotation C. (d) Find the maximum shear stress tmax and its location along the bar. (e) Draw the torsional moment diagram (TMD: T(x), 0 x L).

T —0 L

A

T0 — 6

Fc

IP

2Ip

RA

C

B L — 2

T0 — 3L

L — 2

2° 2°

0

TMD –T0 — 12

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SECTION 3.4 Nonuniform Torsion

283

Solution 3.4-18 (a) TORQUE REACTION RA

(d) MAXIMUM SHEAR STRESS ALONG BAR

T  0

STATICS: RA +

1 T0 L 1 T0 L a ba b  a ba b  0 2 L 2 2 3L 2

RA +

1 T 0 6 0

RA 

T0 6

p d 4 32 AB p For BC IP  d 4 32 BC For AB 2IP 

1

;

dBC

1 4  a b dAB 2

(b) INTERNAL TORSIONAL MOMENTS IN AB & BC T0 TAB (x)   6 TAB (x)  a TBC (x) 

T0 x x a b L L 2 P Q 2

T0 x2  2 T0 b 6 L

0 … x …

 (L  x) T0 (L  x) a b L 3L 2 2

TBC (x)   c a

x  L 2 T0 b d L 3

L … x … L 2

L 2

L TAB(x) TBC(x) dx dx + L GIP L0 G(2IP) L2

2 T0 x T0 L  2 2 6 3L fC  dx G(2IP) L0

L

+

c a

LL2

x  L 2 T0 b d L 3 GIP

fC 

T0L T0L  48GIP 72GIP

fC 

T0L 144GIP

;

L 2

;

tmax 

8T0 3pdAB3

tmax

; controls

Just to right of B, T  T0/12 T0 dBC a b 12 2 tmax  p d 4 32 BC T0 0.841dAB a b 12 2 tmax  p (0.841dAB)4 32

;

(c) ROTATION AT C fC 

At A, T  T0/6

T0 dAB 6 2  p d 4 32 AB

dx

tmax 

2.243T0 pdAB 3

(e) TMD  two 2nd degree curves: from T0/6 at A, to T0/12 at B, to zero at C (with zero slopes at A & C since slope on TMD is proportional to ordinate on torsional loading) – see plot of T(x) above

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Torsion

Problem 3.4-19 A magnesium-alloy wire of diameter d  4 mm and length L rotates inside a flexible tube in order to open or close a switch from a remote location (see figure). A torque T is applied manually (either clockwise or counterclockwise) at end B, thus twisting the wire inside the tube. At the other end A, the rotation of the wire operates a handle that opens or closes the switch. A torque T0  0.2 N  m is required to operate the switch. The torsional stiffness of the tube, combined with friction between the tube and the wire, induces a distributed torque of constant intensity t  0.04 N  m/m (torque per unit distance) acting along the entire length of the wire.

T0 = torque

Flexible tube B

d

A t

(a) If the allowable shear stress in the wire is allow  30 MPa, what is the longest permissible length Lmax of the wire? (b) If the wire has length L  4.0 m and the shear modulus of elasticity for the wire is G  15 GPa, what is the angle of twist  (in degrees) between the ends of the wire?

Solution 3.4-19 Wire inside a flexible tube

(b) ANGLE OF TWIST 

d  4 mm T0  0.2 N  m t  0.04 N  m/m (a) MAXIMUM LENGTH Lmax allow  30 MPa Equilibrium: T  tL  T0 16T From Eq. (3-12): tmax  pd3 tL + T0  L

L  4 m G  15 GPa 1  angle of twist due to distributed torque t  T

pGd4

(from problem 3.4-16)

2  angle of twist due to torque T0

pd3tmax 16



pd 3tmax 16

32 T0 L T0 L  (from Eq. 3 -15) GIP pGd4

  total angle of twist  1  2

1 (pd3tmax  16T0) 16t

1 Lmax  (pd3tallow  16T0) 16t

16tL2

f ;

Substitute numerical values: Lmax  4.42 m

;

16L pGd 4

(tL + 2T0)

;

Substitute numerical values:   2.971 rad  170° ;

T

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SECTION 3.4

Problem 3.4-20 Two hollow tubes are connected by a pin at B which is inserted into a hole drilled through both tubes at B (see cross-section view at B). Tube BC fits snugly into tube AB but neglect any friction on the interface. Tube inner and outer diameters di (i  1, 2, 3) and pin diameter dp are labeled in the figure. Torque T0 is applied at joint C. The shear modulus of elasticity of the material is G. Find expressions for the maximum torque T0,max which can be applied at C for each of the following conditions.

B

d3

TA

d2 d1

d2

A

LA

(a) The shear in the connecting pin is less than some allowable value (pin  p,allow). (b) The shear in tube AB or BC is less than some allowable value (tube  t,allow). (c) What is the maximum rotation C for each of cases (a) and (b) above?

Solution 3.4-20

Pin at B is in shear at interface between the two tubes; force couple V # d2  T0 V

T0 d2

tpin 

tpin  T0 d2

a

pdp 2 4

b

V AS

tpin 

T0,max  tp,allow a

ttubeAB 

pd2dp 2

b

p ad3 4  d2 4 b 32

p IPAC  ad 4  d1 4 b 32 2

ttubeAB 

d3 T0 a b 2 IPAB

p (d 4  d2 4) 32 3 16T0d3 p(d3 4  d2 4)

T0,max  tt,allow c

p(d3 4  d2 4) d 16d3

;

and based on tube BC: ;

(b) T0,max BASED ON ALLOWABLE SHEAR IN TUBES AB & BC IPAB 

d3 b 2

so based on tube AB:

4T0

p d2 d2p 4

ttubeAB 

T0 a

ttubeBC 

ttubeBC 

T0 a

d2 b 2

p (d 4  d1 4) 32 2 16T0d2 p(d2 4  d1 4)

T0,max  tt,allow

J

p(d2 4  d1 4) 16d2

K

T0, Fc C

Pin dp LB

Cross-section at B

(a) T0,max BASED ON ALLOWABLE SHEAR IN PIN AT B

285

Nonuniform Torsion

;

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CHAPTER 3 Torsion

(c)

MAX. ROTATION AT SHEAR IN PIN AT

C

B

BASED ON EITHER ALLOWABLE

OR ALLOWABLE SHEAR STRESS IN

TUBES

MAX. ROTATION BASED ON ALLOWABLE SHEAR IN PIN AT B fC 

T0,max

fCmax 

G

a

LA

+

IPAB

t p,allow a

LB IPBC

p d2d2p 4

b

4 4 J 32 (d3  d2 )

p

fCmax  tp, allow a

J

8d2dp 2 G

MAX.

+

LB

p (d 4  d1 4) K 32 2

(d3  d2 ) 4

+

LB (d2  d1 4) 4

K

;

AB

fCmax  tt, allow c

J

2(d3 4  d2 4) d Gd3

LA (d3  d2 ) 4

4

+

LB

K

;

K

;

(d2  d1 4) 4

ROTATION BASED ON ALLOWABLE SHEAR STRESS

IN TUBE

b

LA 4

ROTATION BASED ON ALLOWABLE SHEAR STRESS

IN TUBE

b

G LA

MAX.

BC

fCmax  tt, allow c

J

2(d2 4  d1 4) d Gd2

LA (d3  d2 ) 4

4

+

LB (d2  d1 4) 4

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SECTION 3.5

Pure Shear

287

Pure Shear Problem 3.5-1 A hollow aluminum shaft (see figure) has outside

diameter d2  4.0 in. and inside diameter d1  2.0 in. When twisted by torques T, the shaft has an angle of twist per unit distance equal to 0.54°/ft. The shear modulus of elasticity of the aluminum is G  4.0  106 psi.

d2

T

T

L

(a) Determine the maximum tensile stress max in the shaft. (b) Determine the magnitude of the applied torques T.

d1 d2

Probs. 3.5-1, 3.5-2, and 3.5-3

Solution 3.5-1

d2  4.0 in.

Hollow aluminum shaft

d1  2.0 in.   0.54°/ft

(a) MAXIMUM TENSILE STRESS

G  4.0  106 psi

max occurs on a 45° plane and is equal to max.

MAXIMUM SHEAR STRESS

max  max  6280 psi

max  Gr (from Eq. 3-7a) r  d2 /2  2.0 in. 1 ft prad u  (0.54°/ft)a ba b 12 in. 180 degree  785.40  106 rad/in.

max  (4.0  106 psi)(2.0 in.)(785.40  106 rad/in.)  6283.2 psi

;

(b) APPLIED TORQUE Use the torsion formula tmax  T

tmaxIP r

IP 

p [(4.0 in.)4  (2.0 in.)4] 32

 23.562 in.4 T

(6283.2 psi) (23.562 in.4) 2.0 in.

 74,000 lb-in.

Tr IP

;

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Problem 3.5-2 A hollow steel bar (G  80 GPa) is twisted by torques T (see figure). The twisting of the bar produces a maximum shear strain max  640  106 rad. The bar has outside and inside diameters of 150 mm and 120 mm, respectively. (a) Determine the maximum tensile strain in the bar. (b) Determine the maximum tensile stress in the bar. (c) What is the magnitude of the applied torques T ?

Solution 3.5-2 Hollow steel bar

G  80 GPa max  640  106 rad d2  150 mm IP  

max  Gmax  (80 GPa)(640  106)

d1  120 mm

 51.2 MPa

p 4 (d  d 41) 32 2

max  max  51.2 MPa

p [(150 mm)4  (120 mm)4] 32

Torsion formula: tmax 

(a) MAXIMUM TENSILE STRAIN gmax  320 * 106 2

;

(c) APPLIED TORQUES

 29.343 * 106 mm4

âmax 

(b) MAXIMUM TENSILE STRESS

T ;

Td2 Tr  IP 2IP

2(29.343 * 106 mm4)(51.2 MPa) 2IPtmax  d2 150 mm

 20,030 N # m  20.0 kN # m

;

Problem 3.5-3 A tubular bar with outside diameter d2  4.0 in. is twisted by torques T  70.0 k-in. (see figure). Under the action of these torques, the maximum tensile stress in the bar is found to be 6400 psi. (a) Determine the inside diameter d1 of the bar. (b) If the bar has length L  48.0 in. and is made of aluminum with shear modulus G  4.0  106 psi, what is the angle of twist (in degrees) between the ends of the bar? (c) Determine the maximum shear strain max (in radians)?

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SECTION 3.5

Solution 3.5-3

d2  4.0 in.

T  70.0 k-in.  70,000 lb-in.

f

Torsion formula: tmax 

TL GIp

From torsion formula, T 

(a) INSIDE DIAMETER d1 Td2 Tr  IP 2IP

‹ f

(70.0 k-in.)(4.0 in.) Td2  2tmax 2(6400 psi)



 21.875 in.4 Also, Ip 

289

Tubular bar

max  6400 psi max  max  6400 psi

IP 

Pure Shear

2(48 in.)(6400 psi) (4.0 * 106 psi)(4.0 in.)

 0.03840 rad

;

(c) MAXIMUM SHEAR STRAIN gmax 

Equate formulas: p [256 in.4  d14]  21.875 in.4 32 Solve for d1: d1  2.40 in.

2IPtmax 2Ltmax L a b  d2 GIP Gd2

f  2.20°

p 4 p (d  d 14)  [(4.0 in.)4  d14] 32 2 32

2IP tmax d2

6400 psi tmax  G 4.0 * 106 psi

 1600 * 106 rad

;

;

(b) ANGLE OF TWIST L  48 in.

G  4.0  106 psi

Problem 3.5-4 A solid circular bar of diameter d  50 mm (see figure) is twisted in a testing machine until the applied torque reaches the value T  500 N m. At this value of torque, a strain gage oriented at 45° to the axis of the bar gives a reading P  339  106. What is the shear modulus G of the material?

d = 50 mm

Strain gage

T 45°

T = 500 N·m

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Solution 3.5-4 Bar in a testing machine

Strain gage at 45°:

SHEAR STRESS (FROM EQ. 3-12)

6

max  339  10

tmax 

d  50 mm

16T pd

3



16(500 N # m) p(0.050 m)3

 20.372 MPa

SHEAR MODULUS

T  500 N m

G

SHEAR STRAIN (FROM EQ. 3-33) 6

max  2max  678  10

tmax 20.372 MPa   30.0 GPa gmax 678 * 106

;

Problem 3.5-5 A steel tube (G  11.5  106 psi) has an outer diameter d2  2.0 in. and an inner diameter d1  1.5 in. When twisted by a torque T, the tube develops a maximum normal strain of 170  106. What is the magnitude of the applied torque T ?

Solution 3.5-5

Steel tube

G  11.5  106 psi

d2  2.0 in.

d1  1.5 in.

max  170  106 IP 

p 2 p 1d  d142  [(2.0 in.)4  (1.5 in.)4] 32 2 32

 1.07379 in.

Equate expressions: Td2  Ggmax 2IP SOLVE FOR TORQUE

4

T

SHEAR STRAIN (FROM EQ. 3-33)

2GIPgmax d2 2(11.5 * 106 psi)(1.07379 in.4)(340 * 106) 2.0 in.

max  2max  340  106



SHEAR STRESS (FROM TORSION FORMULA)

 4200 lb-in.

Td2 Tr tmax   IP 2IP Also, max  Gmax

;

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SECTION 3.5

Pure Shear

291

Problem 3.5-6 A solid circular bar of steel (G  78 GPa) transmits a torque T  360 N m. The allowable stresses in tension, compression, and shear are 90 MPa, 70 MPa, and 40 MPa, respectively. Also, the allowable tensile strain is 220  106. Determine the minimum required diameter d of the bar.

Solution 3.5-6

Solid circular bar of steel

T  360 N m

G  78 GPa

DIAMETER BASED UPON ALLOWABLE TENSILE STRAIN

ALLOWABLE STRESSES Tension: 90 MPa Compression: 70 MPa Shear: 40 MPa Allowable tensile strain: max  220  106

max  2max; max  Gmax  2Gmax tmax 

DIAMETER BASED UPON ALLOWABLE STRESS The maximum tensile, compressive, and shear stresses in a bar in pure torsion are numerically equal. Therefore, the lowest allowable stress (shear stress) governs.

allow  40 MPa tmax 

16T 3

pd

d3 

16T 3

pd

d3 

16T 16T  ptmax 2pGâmax

16(360 N # m) 2p(78 GPa)(220 * 106)

 53.423 * 106 m3 d  0.0377 m  37.7 mm TENSILE STRAIN GOVERNS

d3 

16(360 N # m)

16T  pt allow p(40 MPa)

dmin  37.7 mm

;

d  45.837  106 m3 3

d  0.0358 m  35.8 mm

Problem 3.5-7 The normal strain in the 45° direction on the

surface of a circular tube (see figure) is 880  106 when the torque T  750 lb-in. The tube is made of copper alloy with G  6.2  106 psi.

Strain gage T

45°

If the outside diameter d2 of the tube is 0.8 in., what is the inside diameter d1?

Solution 3.5-7

d2  0.80 in.

Circular tube with strain gage

T  750 lb-in. G  6.2  106 psi

Strain gage at 45°: max  880  106

T = 750 lb-in.

d 2 = 0.8 in.

MAXIMUM SHEAR STRAIN

max  2max

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MAXIMUM SHEAR STRESS

INSIDE DIAMETER

tmax  Ggmax  2Gâmax

Substitute numerical values:

tmax  IP 

T(d2/2) IP

IP 

Td2 Td2  2tmax 4Gâmax

8Td2 pGâmax

d14  d24 

8(750 lb-in.) (0.80 in.) p (6.2 * 106 psi) (880 * 10 6)

 0.4096 in.4  0.2800 in.4  0.12956 in.4

Td2 p 4 (d  d14)  32 2 4Gâmax

d24  d14 

d42  (0.8 in.)4 

d1  0.60 in.

;

8Td2 pGâmax

Problem 3.5-8 An aluminium tube has inside diameter d1  50 mm, shear modulus of elasticity G  27 GPa, and torque T  4.0 kN m. The allowable shear stress in the aluminum is 50 MPa and the allowable normal strain is 900  106. Determine the required outside diameter d2.

Solution 3.5-8

d1  50 mm

Aluminum tube

G  27 GPa

T  4.0 kN m allow  50 MPa allow  900  106

NORMAL STRAIN GOVERNS

allow  48.60 MPa

Determine the required diameter d2.

REQUIRED DIAMETER

ALLOWABLE SHEAR STRESS

t

(allow)1  50 MPa

Tr IP

48.6 MPa 

ALLOWABLE SHEAR STRESS BASED ON NORMAL STRAIN âmax 

g t  2 2G

Rearrange and simplify: t  2Gâmax

(allow)2  2Gallow  2(27 GPa)(900  106)  48.6 MPa

(4000 N # m)(d2/2) p 4 [d2  (0.050 m)4] 32

d42  (419.174 * 106)d2  6.25 * 106  0 Solve numerically: d2  0.07927 m d2  79.3 mm

;

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SECTION 3.5

293

Pure Shear

Problem 3.5-9 A solid steel bar (G  11.8  106 psi) of

diameter d  2.0 in. is subjected to torques T  8.0 k-in. acting in the directions shown in the figure.

(a) Determine the maximum shear, tensile, and compressive stresses in the bar and show these stresses on sketches of properly oriented stress elements. (b) Determine the corresponding maximum strains (shear, tensile, and compressive) in the bar and show these strains on sketches of the deformed elements.

Solution 3.5-9

Solid steel bar

T  8.0 k-in.

(b) MAXIMUM STRAINS

G  11.8  10 psi 6

gmax 

(a) MAXIMUM STRESSES tmax 

T = 8.0 k-in.

d = 2.0 in.

T

16T 3

pd



 432 * 106 rad

16(8000 lb-in.)

 5093 psi

3

p(2.0 in.)

âmax 

;

t  5090 psi c  5090 psi

gmax 5093 psi  G 11.8 * 106 psi

Problem 3.5-10 A solid aluminum bar (G  27 GPa) of

(a) Determine the maximum shear, tensile, and compressive stresses in the bar and show these stresses on sketches of properly oriented stress elements. (b) Determine the corresponding maximum strains (shear, tensile, and compressive) in the bar and show these strains on sketches of the deformed elements.

gmax  216 * 106 2

t  216  106 c  216  106

;

diameter d  40 mm is subjected to torques T  300 N m acting in the directions shown in the figure.

;

d = 40 mm T

;

T = 300 N·m

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Page 294

Torsion

Solution 3.5-10

Solid aluminum bar

(b) MAXIMUM STRAINS

(a) MAXIMUM STRESSES tmax 

16T 3

pd



16(300 N #

m)

gmax 

3

p(0.040 m)

 23.87 MPa

tmax 23.87 MPa  G 27 GPa

 884 * 106 rad

;

t  23.9 MPa c  23.9 MPa

;

âmax 

;

gmax  442 * 106 2

t  442  106 c  442  106

;

Transmission of Power Problem 3.7-1 A generator shaft in a small hydroelectric plant turns

120 rpm

at 120 rpm and delivers 50 hp (see figure).

d

(a) If the diameter of the shaft is d  3.0 in., what is the maximum shear stress max in the shaft? (b) If the shear stress is limited to 4000 psi, what is the minimum permissible diameter dmin of the shaft?

Solution 3.7-1 n  120 rpm TORQUE H

Generator shaft H  50 hp

d  diameter

2pnT H  hp n  rpm T  1b-ft 33,000

33,000 H (33,000)(50 hp) T  2pn 2p(120 rpm)  2188 1b-ft  26,260 1b-in. (a) MAXIMUM SHEAR STRESS max d  3.0 in.

50 hp

tmax 

16T 3

pd



16(26,260 1b-in.)

tmax  4950 psi

p (3.0 in.)3 ;

(b) MINIMUM DIAMETER dmin

allow  4000 psi d3 

16(26,260 1b-in.) 16T   33.44 in.3 ptallow p (4000 psi)

dmin  3.22 in.

;

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SECTION 3.7

Transmission of Power

295

Problem 3.7-2 A motor drives a shaft at 12 Hz and delivers 20 kW of power (see figure).

12 Hz d

(a) If the shaft has a diameter of 30 mm, what is the maximum shear stress max in the shaft? (b) If the maximum allowable shear stress is 40 MPa, what is the minimum permissible diameter dmin of the shaft?

Solution 3.7-2 f  12 Hz

20 kW

Motor-driven shaft

P  20 kW  20,000 N m/s

16T

tmax 

pd3

TORQUE P  2 f T

P  watts

T

20,000 W P   265.3 N # m 2pf 2p(12 Hz)

(a) MAXIMUM SHEAR STRESS max

16(265.3 N # m) p(0.030 m)3

 50.0 MPa

f  Hz  s1

T  Newton meters



;

(b) MINIMUM DIAMETER dmin

allow  40 MPa d3 

16(265.3 N # m) 16T  pt allow p(40 MPa)

 33.78  106 m3

d  30 mm

dmin  0.0323 m  32.3 mm

Problem 3.7-3 The propeller shaft of a large ship has outside

;

100 rpm

18 in.

diameter 18 in. and inside diameter 12 in., as shown in the figure. The shaft is rated for a maximum shear stress of 4500 psi. (a) If the shaft is turning at 100 rpm, what is the maximum horsepower that can be transmitted without exceeding the allowable stress?

12 in. 18 in.

(b) If the rotational speed of the shaft is doubled but the power requirements remain unchanged, what happens to the shear stress in the shaft?

Solution 3.7-3

Hollow propeller shaft

d2  18 in. d1  12 in. allow  4500 psi p 4 (d  d42)  8270.2 in.4 IP  32 2 TORQUE T(d2/2) tmax  IP T

2t allowIP T d2

2(4500 psi)(8270.2 in.4) 18 in.

 4.1351 * 106 1b-in.  344,590 1b-ft.

(a) HORSEPOWER n  100 rpm n  rpm H

H

2pnT 33,000

T  lb-ft H  hp

2p(100 rpm)(344,590 lb-ft) 33,000

 6560 hp

;

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Torsion

(b) ROTATIONAL SPEED IS DOUBLED H

2pnT 33,000

If n is doubled but H remains the same, then T is halved. If T is halved, so is the maximum shear stress. ⬖ Shear stress is halved

;

Problem 3.7-4 The drive shaft for a truck (outer diameter 60 mm and inner diameter 40 mm) is running at 2500 rpm (see figure).

2500 rpm 60 mm

(a) If the shaft transmits 150 kW, what is the maximum shear stress in the shaft? (b) If the allowable shear stress is 30 MPa, what is the maximum power that can be transmitted?

40 mm 60 mm

Solution 3.7-4 d2  60 mm IP 

Drive shaft for a truck d1  40 mm

n  2500 rpm

p 4 (d  d14)  1.0210 * 106 m4 32 2

(a) MAXIMUM SHEAR STRESS max P  power (watts) P  150 kW  150,000 W T  torque (newton meters) n  rpm P

2pnT 60

T

60(150,000 W)  572.96 N # m 2p(2500 rpm)

T

60P 2pn

tmax 

(572.96 N # m)(0.060 m) Td2  2 IP 2(1.0210 * 106 m4)

 16.835 MPa tmax  16.8 MPa

;

(b) MAXIMUM POWER Pmax

allow  30 MPa Pmax  P

tallow 30 MPa  (150 kW) a b tmax 16.835 MPa

 267 kW

;

Problem 3.7-5 A hollow circular shaft for use in a pumping station is being designed with an inside diameter equal to 0.75 times the outside diameter. The shaft must transmit 400 hp at 400 rpm without exceeding the allowable shear stress of 6000 psi. Determine the minimum required outside diameter d.

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SECTION 3.7

Solution 3.7-5

H  hp

d0  inside diameter

T

 0.75 d n  400 rpm

allow  6000 psi IP 

n  rpm

T  lb-ft

(33,000)(400 hp) 33,000 H  2pn 2p(400 rpm)

 5252.1 lb-ft  63,025 lb-in. MINIMUM OUTSIDE DIAMETER

p 4 [d  (0.75 d)4]  0.067112 d 4 32

TORQUE H

297

Hollow shaft

d  outside diameter

H  400 hp

Transmission of Power

tmax 

Td 2IP

IP 

0.067112 d 4 

2pnT 33,000

Td Td  2tmax 2t allow

(63,025 lb-in.)(d) 2(6000 psi)

d3  78.259 in.3

dmin  4.28 in.

;

Problem 3.7-6 A tubular shaft being designed for use on a construction site must transmit 120 kW at 1.75 Hz. The inside diameter of the shaft is to be one-half of the outside diameter. If the allowable shear stress in the shaft is 45 MPa, what is the minimum required outside diameter d?

Solution 3.7-6

Tubular shaft

d  outside diameter

T

d0  inside diameter

MINIMUM OUTSIDE DIAMETER

 0.5 d P  120 kW  120,000 W

f  1.75 Hz

allow  45 MPa IP 

p 4 [d  (0.5 d)4]  0.092039 d 4 32

tmax 

Td 2IP

IP 

0.092039 d4 

Td Td  2tmax 2t allow

(10,913.5 N # m)(d) 2(45 MPa)

d3  0.0013175 m3

TORQUE P  2 f T

120,000 W P   10,913.5 N # m 2pf 2p(1.75 Hz)

P  watts

f  Hz

dmin  110 mm

d  0.1096 m ;

T  newton meters

Problem 3.7-7 A propeller shaft of solid circular cross section and diameter d is spliced by a collar of the same material (see figure). The collar is securely bonded to both parts of the shaft. What should be the minimum outer diameter d1 of the collar in order that the splice can transmit the same power as the solid shaft?

d1

d

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Solution 3.7-7 Splice in a propeller shaft

EQUATE TORQUES

SOLID SHAFT tmax 

16 T1 pd3

T1 

pd3tmax 16

For the same power, the torques must be the same. For the same material, both parts can be stressed to the same maximum stress.

HOLLOW COLLAR IP 

T2(d1/2) T2r p 4  (d1  d 4) tmax  32 IP IP

2tmaxIP 2tmax p  a b(d14  d 4) T2  d1 d1 32 ptmax 4  (d1  d 4) 16 d1

‹ T1  T2 or a

pd3tmax ptmax 4  (d  d 4) 16 16d1 1

d1 4 d1 b  10 d d

(Eq. 1)

MINIMUM OUTER DIAMETER Solve Eq. (1) numerically: Min. d1  1.221 d

;

Problem 3.7-8 What is the maximum power that can be delivered by a hollow propeller shaft (outside diameter 50 mm, inside diameter 40 mm, and shear modulus of elasticity 80 GPa) turning at 600 rpm if the allowable shear stress is 100 MPa and the allowable rate of twist is 3.0°/m?

Solution 3.7-8 Hollow propeller shaft d2  50 mm

d1  40 mm

G  80 GPa

n  600 rpm

allow  100 MPa allow  3.0°/m IP 

p 4 (d  d41)  362.3 * 109 m4 32 2

BASED UPON ALLOWABLE SHEAR STRESS tmax 

T1(d2/2) IP

T1 

2t allowIP d2

2(100 MPa)(362.3 * 109 m4) T1  0.050 m  1449 N # m

BASED UPON ALLOWABLE RATE OF TWIST T2 T2  GIPallow u GIP T  (80 GPa) (362.3 * 10 2 * a

9 4 m )(3.0°/m)

p rad /degreeb 180

T2  1517 N m SHEAR STRESS GOVERNS Tallow  T1  1449 N m MAXIMUM POWER 2p(600 rpm)(1449 N # m) 2pnT  P 60 60 P  91,047 W Pmax  91.0 kW

;

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Transmission of Power

299

Problem 3.7-9 A motor delivers 275 hp at 1000 rpm to the end of a shaft (see figure). The gears at B and C take out 125 and 150 hp, respectively. Determine the required diameter d of the shaft if the allowable shear stress is 7500 psi and the angle of twist between the motor and gear C is limited to 1.5°. (Assume G  11.5  106 psi, L1  6 ft, and L2  4 ft.) Motor

C A

d

B

L2

L1

PROBS. 3.7-9 and 3.7-10

Solution 3.7-9

Motor-driven shaft FREE-BODY DIAGRAM

L1  6 ft L2  4 ft

TA  17,332 lb-in.

d  diameter

TC  9454 lb-in.

n  1000 rpm

d  diameter

allow  7500 psi ( AC)allow  1.5°  0.02618 rad G  11.5  106 psi TORQUES ACTING ON THE SHAFT 2pnT H 33,000 T

TB  7878 lb-in. INTERNAL TORQUES TAB  17,332 lb-in. TBC  9454 lb-in.

H  hp n  rpm T  lb-ft

DIAMETER BASED UPON ALLOWABLE SHEAR STRESS The larger torque occurs in segment AB

33,000 H 2pn

At point A: TA 

33,000(275 hp) 2p(1000 rpm)

 1444 lb-ft  17,332 lb-in. At point B: TB  At point C: TC 

tmax  

16TAB

d3 

3

pd

16TAB pt allow

16(17,332 lb-in.)  11.77 in.3 p(7500 psi)

d  2.27 in. DIAMETER BASED UPON ALLOWABLE ANGLE OF TWIST

125 275

TA  7878 lb-in.

150 275

TA  9454 lb-in.

IP 

pd4 32

f

TL 32TL  GIP pGd4

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Segment AB: fAB   fAB 

fBC 

32TAB LAB pGd 4 32(17,330 lb  in.)(6 ft)(12 in./ft) p(11.5 * 106 psi)d 4 1.1052 d



d4

( AC)allow  0.02618 rad 0.02618 

1.5070 d4

d  2.75 in.

32 TBCLBC pGd

1.5070

and

d  2.75 in.

Angle of twist governs

Segment BC: fBC 

d4

From A to C: fAC  fAB + fBC 

⬖

4

0.4018

;

4

32(9450 lb-in.)(4 ft)(12 in./ft) p(11.5 * 106 psi)d 4

Problem 3.7-10 The shaft ABC shown in the figure is driven by a motor that delivers 300 kW at a rotational speed of

32 Hz. The gears at B and C take out 120 and 180 kW, respectively. The lengths of the two parts of the shaft are L1  1.5 m and L2  0.9 m. Determine the required diameter d of the shaft if the allowable shear stress is 50 MPa, the allowable angle of twist between points A and C is 4.0°, and G  75 GPa.

Solution 3.7-10 Motor-driven shaft

L1  1.5 m L2  0.9 m d  diameter f  32 Hz

At point A: TA 

300,000 W  1492 N # m 2p(32 Hz)

At point B: TB 

120 T  596.8 N # m 300 A

At point C: TC 

180 T  895.3 N # m 300 A

FREE-BODY DIAGRAM

allow  50 MPa G  75 GPa ( AC)allow  4°  0.06981 rad TORQUES ACTING ON THE SHAFT P  2 fT P  watts T  newton meters T

P 2pf

f  Hz

TA  1492 N m TB  596.8 N m TC  895.3 N m d  diameter

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SECTION 3.7 Transmission of Power

INTERNAL TORQUES

Segment BC:

TAB  1492 N m

fBC 

TBC  895.3 N m DIAMETER BASED UPON ALLOWABLE SHEAR STRESS

fBC 

32 TBCLBC pGd 4

tmax 

16(1492 N # m) 16 TAB d   pt allow p(50 MPa) 3

pd 3

d3  0.0001520 m3

d  0.0534 m  53.4 mm

DIAMETER BASED UPON ALLOWABLE ANGLE OF TWIST IP 

pd 4 32

f

TL 32TL  GIP pGd 4

fAB 

From A to C: fAC  fAB + fBC  ( AC)allow  0.06981 rad ⬖ 0.06981 

0.1094 * 106 d4

 49.3mm SHEAR STRESS GOVERNS

32 TABLAB pGd

4



0.3039 * 10 d4

32(1492 N #

6

m)(1.5 m)

p(75 GPa)d

4

p(75 GPa)d 4

d4

and d  0.04933 m

Segment AB: fAB 

32(895.3 N # m)(0.9 m)

0.1094 * 106

The larger torque occurs in segment AB 16 TAB



d  53.4mm

;

0.4133 * 106 d4

301

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Torsion

Statically Indeterminate Torsional Members Problem 3.8-1 A solid circular bar ABCD with fixed supports is acted upon by torques T0 and 2T0 at the locations shown in the figure. Obtain a formula for the maximum angle of twist ␾max of the bar. (Hint: Use Eqs. 3-46a and b of Example 3-9 to obtain the reactive torques.)

TA

A

T0

2T0

B

C

3L — 10

3L — 10

D

TD

D

TD

4L — 10 L

Solution 3.8-1

Circular bar with fixed ends ANGLE OF TWIST AT SECTION B

From Eqs. (3-46a and b): TA(3L/10) 9T0 L ⫽ GIP 20GIP

TA ⫽

T0 LB L

fB ⫽ fAB ⫽

TB ⫽

T0 LA L

ANGLE OF TWIST AT SECTION C

APPLY THE ABOVE FORMULAS TO THE GIVEN BAR: TA ⫽ T0 a

15T0 7 4 b + 2T0 a b ⫽ 10 10 10

15T0 3 6 TD ⫽ T0 a b + 2T0 a b ⫽ 10 10 10

fC ⫽ fCD ⫽

TD(4L/10) 3T0 L ⫽ GIP 5GIP

MAXIMUM ANGLE OF TWIST fmax ⫽ fC ⫽

3T0 L 5GIP

Problem 3.8-2 A solid circular bar ABCD with fixed supports at ends A and D is acted upon by two equal and oppositely directed torques T0, as shown in the figure. The torques are applied at points B and C, each of which is located at distance x from one end of the bar. (The distance x may vary from zero to L/2.) (a) For what distance x will the angle of twist at points B and C be a maximum? (b) What is the corresponding angle of twist ␾max? (Hint: Use Eqs. 3-46a and b of Example 3-9 to obtain the reactive torques.)

TA

A

;

T0

T0

B

C

x

x L

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SECTION 3.8

Solution 3.8-2

303

Statically Indeterminate Torsional Members

Circular bar with fixed ends (a) ANGLE OF TWIST AT SECTIONS B AND C φB

φ max

0

From Eqs. (3-46a and b):

fB ⫽ fAB ⫽

L/A

T0 LB L

dfB T0 ⫽ (L ⫺ 4x) dx GIPL

TB ⫽

T0 LA L

dfB ⫽ 0; L ⫺ 4x ⫽ 0 dx or x ⫽

L 4

x

TAx T0 ⫽ (L ⫺ 2x)(x) GIP GIPL

TA ⫽

APPLY THE ABOVE FORMULAS TO THE GIVEN BAR:

L/2

;

(b) MAXIMUM ANGLE OF TWIST fmax ⫽ (fB)max ⫽ (fB)x⫽ L4 ⫽

TA ⫽

T0L 8GIP

;

T0(L ⫺ x) T0x T0 ⫺ ⫽ (L ⫺ 2x) TD ⫽ TA L L L

Problem 3.8-3 A solid circular shaft AB of diameter d is fixed against rotation at both ends (see figure). A circular disk is attached to the shaft at the location shown. What is the largest permissible angle of rotation ␾max of the disk if the allowable shear stress in the shaft is ␶allow? (Assume that a ⬎ b. Also, use Eqs. 3-46a and b of Example 3-9 to obtain the reactive torques.)

Disk A

d

B

a

Solution 3.8-3

b

Shaft fixed at both ends Assume that a torque T0 acts at the disk. The reactive torques can be obtained from Eqs. (3-46a and b): T0b T0a TB ⫽ L L Since a ⬎ b, the larger torque (and hence the larger stress) is in the right hand segment. TA ⫽

L⫽a⫹b a⬎b

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tmax ⫽ T0 ⫽

Page 304

Torsion

ANGLE OF ROTATION OF THE DISK (FROM Eq. 3-49)

TB(d/2) T0 ad ⫽ IP 2LIP

2LIPtmax ad

(T0)max ⫽

2L IPt allow ad

f⫽

T0 ab GLIP (T0)maxab 2bt allow ⫽ GLIp Gd

fmax ⫽

;

Problem 3.8-4 A hollow steel shaft ACB of outside diameter 50 mm and inside diameter 40 mm is held against rotation at ends A and B (see figure). Horizontal forces P are applied at the ends of a vertical arm that is welded to the shaft at point C. Determine the allowable value of the forces P if the maximum permissible shear stress in the shaft is 45 MPa. (Hint: Use Eqs. 3-46a and b of Example 3-9 to obtain the reactive torques.)

200 mm A P

200 mm C B P

600 mm 400 mm

Solution 3.8-4

Hollow shaft with fixed ends

GENERAL FORMULAS:

T0 ⫽ P(400 mm) LB ⫽ 400 mm LA ⫽ 600 mm L ⫽ LA ⫹ LB ⫽ 1000 mm d2 ⫽ 50 mm d1 ⫽ 40 mm

␶allow ⫽ 45 MPa

APPLY THE ABOVE FORMULAS TO THE GIVEN SHAFT

TA ⫽

P(0.4 m)(400 mm) T0 LB ⫽ ⫽ 0.16 P L 1000 mm

TB ⫽

P(0.4 m)(600 mm) T0 LA ⫽ ⫽ 0.24 P L 1000 mm

UNITS: P ⫽ Newtons T ⫽ Newton meters From Eqs. (3-46a and b): TA ⫽

T0 LB L

T0 LA L The larger torque, and hence the larger shear stress, occurs in part CB of the shaft. TB ⫽

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SECTION 3.8

 Tmax ⫽ TB ⫽ 0.24 P Tmax(d/2) IP

Tmax ⫽

2tmaxIP d

0.24P ⫽ (Eq. 1) P⫽

␶max ⫽ 45 ⫻ 10 N/m Ip ⫽

2(45 * 106 N/m2)(362.26 * 10⫺9 m4) 0.05 m

⫽ 652.07 N # m

UNITS: Newtons and meters 6

305

Substitute numerical values into (Eq. 1):

SHEAR STRESS IN PART CB tmax ⫽

Statically Indeterminate Torsional Members

2

652.07 N # m ⫽ 2717 N 0.24 m

Pallow ⫽ 2720 N

p 4 (d ⫺d 4) ⫽ 362.26 * 10⫺9m4 32 2 1

;

d ⫽ d2 ⫽ 0.05 mm

Problem 3.8-5 A stepped shaft ACB having solid circular cross

Solution 3.8-5

B

T0 6.0 in.

15.0 in.

Combine Eqs. (1) and (3) and solve for T0: (T0)AC ⫽

LAIPB 1 pd3 t a1 + b 16 A allow LBIPA

⫽

LAdB4 1 pd3At allow a 1 + b 16 LBdA4

Find (T0)max REACTIVE TORQUES (from Eqs. 3-45a and b) LBIPA b TA ⫽ T0 a LBIPA + LAIPB LAIPB b LBIPA + LAIPB

(1)

ALLOWABLE TORQUE BASED UPON SHEAR STRESS AC

Substitute numerical values:

ALLOWABLE TORQUE BASED UPON SHEAR STRESS IN SEGMENT CB tCB ⫽

16TA

pdA3 1 1 pd3 t ⫽ pd3 t TA ⫽ 16 A AC 16 A allow

(4)

(T0)AC ⫽ 3678 lb-in. (2)

IN SEGMENT

tAC ⫽

C

A

Stepped shaft ACB

dA ⫽ 0.75 in. dB ⫽ 1.50 in. LA ⫽ 6.0 in. LB ⫽ 15.0 in. ␶allow ⫽ 6000 psi

TB ⫽ T0 a

1.50 in.

0.75 in.

sections with two different diameters is held against rotation at the ends (see figure). If the allowable shear stress in the shaft is 6000 psi, what is the maximum torque (T0)max that may be applied at section C? (Hint: Use Eqs. 3-45a and b of Example 3-9 to obtain the reactive torques.)

TB ⫽ (3)

16TB pdB3 1 1 pd 3 t ⫽ pd 3 t 16 B CB 16 B allow

(5)

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Torsion

Combine Eqs. (2) and (5) and solve for T0: (T0)CB ⫽

SEGMENT AC GOVERNS (T0)max ⫽ 3680 lb-in.

LBIPA 1 pd3 t a1 + b 16 B allow LAIPB

LBdA4 1 pd3Bt allow a1 + b ⫽ 16 LAdB4 Substitute numerical values: (T0)CB ⫽ 4597 lb-in.

NOTE: From Eqs. (4) and (6) we find that (6)

(T0)AC LA dB ⫽ a ba b (T0)CB LB dA which can be used as a partial check on the results.

20 mm

Problem 3.8-6 A stepped shaft ACB having solid circular cross sections with two different diameters is held against rotation at the ends (see figure). If the allowable shear stress in the shaft is 43 MPa, what is the maximum torque (T0)max that may be applied at section C? (Hint: Use Eqs. 3-45a and b of Example 3-9 to obtain the reactive torques.)

Solution 3.8-6

;

25 mm B

C

A

T0 225 mm

450 mm

Stepped shaft ACB

1 1 pd 3 t ⫽ pd 3 t 16 A AC 16 A allow

dA

⫽ 20 mm

dB

⫽ 25 mm

LA

⫽ 225 mm

Combine Eqs. (1) and (3) and solve for T0:

LB

⫽ 450 mm

(T0)AC ⫽

TA ⫽

␶allow ⫽ 43 MPa

⫽

Find (T0)max REACTIVE TORQUES (from Eqs. 3-45a and b)

LAIPB 1 pdA3 t allow a 1 + b 16 LBIPA

LAdB4 1 pdA3t allow a 1 + b 16 LBdA4

(1)

(T0)AC ⫽ 150.0 N ⭈ m

TB ⫽ T0 a

(2)

IN SEGMENT

ALLOWABLE TORQUE BASED UPON SHEAR STRESS IN SEGMENT AC tAC ⫽

16TA pd3A

(4)

Substitute numerical values:

LBIPA b TA ⫽ T0 a LBIPA + LAIPB LAIPB b LBIPA + LAIPB

(3)

ALLOWABLE TORQUE BASED UPON SHEAR STRESS CB tCB ⫽ TB ⫽

16TB pd3B 1 1 pd3 t ⫽ pd3 t 16 B CB 16 B allow

(5)

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SECTION 3.8

Combine Eqs. (2) and (5) and solve for T0: (T0)CB ⫽

(T0)max ⫽ 150 N ⭈ m (T0)AC LA dB ⫽ a ba b (T0)CB LB dA

(6)

which can be used as a partial check on the results.

Problem 3.8-7 A stepped shaft ACB is held against rotation at ends

dA

A and B and subjected to a torque T0 acting at section C (see figure). The two segments of the shaft (AC and CB) have diameters dA and dB, respectively, and polar moments of inertia IPA and IPB, respectively. The shaft has length L and segment AC has length a.

C T0

a L

Stepped shaft

SEGMENT AC: dA, IPA LA ⫽ a

or

SEGMENT CB: dB, IPB LB ⫽ L ⫺ a REACTIVE TORQUES (from Eqs. 3-45a and b)

Solve for a/L:

(a)

dB

IPA

A

(a) For what ratio a/L will the maximum shear stresses be the same in both segments of the shaft? (b) For what ratio a/L will the internal torques be the same in both segments of the shaft? (Hint: Use Eqs. 3-45a and b of Example 3-9 to obtain the reactive torques.)

TA ⫽ T0 a

;

NOTE: From Eqs. (4) and (6) we find that

Substitute numerical values: (T0)CB ⫽ 240.0 N ⭈ m

Solution 3.8-7

LBIPA LAIPB b; TB ⫽ T0 a b LBIPA + LAIPB LBIPA + LAIPB

EQUAL SHEAR STRESSES TA(dA/2) TB(dB/2) tCB ⫽ tAC ⫽ IPA IPB

␶AC ⫽ ␶CB or

TAdA TBdB ⫽ IPA IPB

Substitute TA and TB into Eq. (1): LBIPAdA LAIPBdB ⫽ IPA IPB

or

307

SEGMENT AC GOVERNS

LBIPA 1 pd 3 t a1 + b 16 B allow LAIPB

LBdA4 1 b ⫽ pdB3t allow a1 + 16 LAdB4

Statically Indeterminate Torsional Members

LBdA ⫽ LAdB

(L ⫺ a)dA ⫽ adB

(b) EQUAL TORQUES TA ⫽ TB or LBIPA ⫽ LAIPB or

(L ⫺ a)IPA ⫽ aIPB

Solve for a/L: (Eq. 1)

dA a ⫽ L dA + dB

or

IPA a ⫽ L IPA + IPB

dA4 a ⫽ 4 L dA + dB4

;

;

IPB

B

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Torsion

Problem 3.8-8 A circular bar AB of length L is fixed against rotation at the ends and loaded by a distributed torque t(x) that varies linearly in intensity from zero at end A to t0 at end B (see figure). Obtain formulas for the fixed-end torques TA and TB.

t0 t(x) TA

TB A

B x L

Solution 3.8-8

Fixed-end bar with triangular load ELEMENT OF DISTRIBUTED LOAD

dTA ⫽ Elemental reactive torque t(x) ⫽

dTB ⫽ Elemental reactive torque

t0x L

From Eqs. (3-46a and b):

T0 ⫽ Resultant of distributed torque L

L

t0 x t0 L T0 ⫽ t(x)dx ⫽ dx ⫽ 2 L0 L0 L EQUILIBRIUM t0L TA + TB ⫽ T0 ⫽ 2

L⫺x x b dTB ⫽ t(x)dxa b L L REACTIVE TORQUES (FIXED-END TORQUES) dTA ⫽ t(x)dxa

L

t0L x L⫺x TA ⫽ dTA ⫽ a t0 b a bdx ⫽ L L 6 L L0 L t0 L x x TB ⫽ dTB ⫽ a t0 b a b dx ⫽ ; L L 3 L L0 NOTE: TA + TB ⫽

Problem 3.8-9 A circular bar AB with ends fixed against rotation has a hole extending for half of its length (see figure). The outer diameter of the bar is d2 ⫽ 3.0 in. and the diameter of the hole is d1 ⫽ 2.4 in. The total length of the bar is L ⫽ 50 in. At what distance x from the left-hand end of the bar should a torque T0 be applied so that the reactive torques at the supports will be equal?

t0L (check) 2

25 in. A

;

25 in. T0

3.0 in.

B

x

2.4 in.

3.0 in.

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SECTION 3.8

Solution 3.8-9

Statically Indeterminate Torsional Members

309

Bar with a hole IPA ⫽ Polar moment of inertia at left-hand end IPB ⫽ Polar moment of inertia at right-hand end fB ⫽

T0(L/2) GIPA

⫺

L ⫽ 50 in. L/2 ⫽ 25 in.

(2)

Substitute Eq. (1) into Eq. (2) and simplify:

d2 ⫽ outer diameter

fB ⫽

⫽ 3.0 in. d1 ⫽ diameter of hole

T0 L L x L L c + ⫺ + ⫺ d G 4IPB 4IPA IPB 2IPB 2IPA

COMPATIBILITY ␾B ⫽ 0

⫽ 2.4 in.

x

T0 ⫽ Torque applied at distance x

‹ IPB

Find x so that TA ⫽ TB

⫽

3L L ⫺ 4IPB 4IPA

soLVE FOR x:

EQUILIBRIUM TA ⫹ TB ⫽ T0 ‹ TA ⫽ TB ⫽

TB(L/2) TB(L/2) T0(x ⫺ L/2) + ⫺ GIPB GIPA GIPB

T0 2

(1)

REMOVE THE SUPPORT AT END B

x⫽

IPB L a3 ⫺ b 4 IPA

IPB d24 ⫺ d14 d1 4 ⫽ ⫽ 1 ⫺ a b IPA d2 d24 d1 4 L x ⫽ c2 + a b d ; 4 d2 SUBSTITUTE NUMERICAL VALUES: x⫽

2.4 in. 4 50 in. c2 + a b d ⫽ 30.12 in. 4 3.0 in.

;

␾B ⫽ Angle of twist at B

Problem 3.8-10 A solid steel bar of diameter d1 ⫽ 25.0 mm is

enclosed by a steel tube of outer diameter d3 ⫽ 37.5 mm and inner diameter d2 ⫽ 30.0 mm (see figure). Both bar and tube are held rigidly by a support at end A and joined securely to a rigid plate at end B. The composite bar, which has a length L ⫽ 550 mm, is twisted by a torque T ⫽ 400 N ⭈ m acting on the end plate. (a) Determine the maximum shear stresses ␶1 and ␶2 in the bar and tube, respectively. (b) Determine the angle of rotation ␾ (in degrees) of the end plate, assuming that the shear modulus of the steel is G ⫽ 80 GPa. (c) Determine the torsional stiffness kT of the composite bar. (Hint: Use Eqs. 3-44a and b to find the torques in the bar and tube.)

Tube A

B T

Bar

End plate

L

d1 d2 d3

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Torsion

Solution 3.8-10

Bar enclosed in a tube TORQUES IN THE BAR (1) AND TUBE (2) FROM EQS. (3-44A AND B) Bar: T1 ⫽ T a

IP1 b ⫽ 100.2783 N # m IP1 + IP2

Tube: T2 ⫽ T a

IP2 b ⫽ 299.7217 N # m IP1 + IP2

(a) MAXIMUM SHEAR STRESSES Bar: t1 ⫽

T1(d1/2) ⫽ 32.7 MPa IP1

Tube: t2 ⫽ d1 ⫽ 25.0 mm

d2 ⫽ 30.0 mm

d3 ⫽ 37.5 mm

T 2L T1L ⫽ ⫽ 0.017977 rad GIP1 GIP2

f⫽

POLAR MOMENTS OF INERTIA

␾ ⫽ 1.03°

Tube: IP2 ⫽

;

(b) ANGLE OF ROTATION OF END PLATE

G ⫽ 80 GPa

p 4 Bar: IP1 ⫽ d ⫽ 38.3495 * 10⫺9 m4 32 1

T2(d3/2) ⫽ 49.0 MPa IP2

;

;

(c) TORSIONAL STIFFNESS

p 1d 4 ⫺ d242 ⫽ 114.6229 * 10⫺9 m4 32 3

kT ⫽

T ⫽ 22.3 kN # m f

Problem 3.8-11 A solid steel bar of diameter d1 ⫽ 1.50 in. is enclosed

by a steel tube of outer diameter d3 ⫽ 2.25 in. and inner diameter d2 ⫽ 1.75 in. (see figure). Both bar and tube are held rigidly by a support at end A and joined securely to a rigid plate at end B. The composite bar, which has length L ⫽ 30.0 in., is twisted by a torque T ⫽ 5000 lb-in. acting on the end plate. (a) Determine the maximum shear stresses ␶1 and ␶2 in the bar and tube, respectively. (b) Determine the angle of rotation ␾ (in degrees) of the end plate, assuming that the shear modulus of the steel is G ⫽ 11.6 ⫻ 106 psi. (c) Determine the torsional stiffness kT of the composite bar. (Hint: Use Eqs. 3-44a and b to find the torques in the bar and tube.)

;

Tube A

B T

Bar

End plate

L

d1 d2 d3

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SECTION 3.8

Solution 3.8-11

Statically Indeterminate Torsional Members

311

Bar enclosed in a tube TORQUES IN THE BAR (1) AND TUBE (2) FROM EQS. (3-44A AND B) Bar: T1 ⫽ T a

IPI b ⫽ 1187.68 lb-in. IPI + IPI

Tube: T2 ⫽ T a

IP2 b ⫽ 3812.32 lb-in. IP1 + IP2

(a) MAXIMUM SHEAR STRESSES Bar: t1 ⫽

T1(d1/2) ⫽ 1790 psi IP1 T2(d3/2) ⫽ 2690 psi IP2

Tube: t2 ⫽

; ;

(b) ANGLE OF ROTATION OF END PLATE d1 ⫽ 1.50 in.

d2 ⫽ 1.75 in.

d3 ⫽ 2.25 in.

f⫽

G ⫽ 11.6 ⫻ 10 psi 6

␾ ⫽ 0.354°

POLAR MOMENTS OF INERTIA Bar: IP1 ⫽

p 4 d ⫽ 0.497010 in.4 32 1

T 2L T1L ⫽ ⫽ 0.006180015 rad GIP1 GIP2 ;

(c) TORSIONAL STIFFNESS kT ⫽

p Tube: IP2 ⫽ 1d34 ⫺ d242 ⫽ 1.595340 in.4 32

T ⫽ 809 k - in. f

T

Problem 3.8-12 The composite shaft shown in the figure is manufactured by shrink-fitting a steel sleeve over a brass core so that the two parts act as a single solid bar in torsion. The outer diameters of the two parts are d1 ⫽ 40 mm for the brass core and d2 ⫽ 50 mm for the steel sleeve. The shear moduli of elasticity are Gb ⫽ 36 GPa for the brass and Gs ⫽ 80 GPa for the steel.

;

Steel sleeve Brass core T

Assuming that the allowable shear stresses in the brass and steel are ␶b ⫽ 48 MPa and ␶s ⫽ 80 MPa, respectively, determine the maximum permissible torque Tmax that may be applied to the shaft. (Hint: Use Eqs. 3-44a and b to find the torques.)

d1 d2 Probs. 3.8-12 and 3.8-13

Solution 3.8-12

Composite shaft shrink fit d1 ⫽ 40 mm d2 ⫽ 50 mm GB ⫽ 36 GPa

GS ⫽ 80 GPa

Allowable stresses: ␶B ⫽ 48 MPa ␶S ⫽ 80 MPa

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CHAPTER 3 Torsion

BRASS CORE (ONLY)

Eq. (3-44b): TS ⫽ T a

GSIPS b GBIPB + GSIPS

⫽ 0.762082 T T ⫽ TB ⫹ TS (CHECK) ALLOWABLE TORQUE T BASED UPON BRASS CORE IPB ⫽

p 4 d ⫽ 251.327 * 10⫺9 m4 32 1

GBIPB ⫽ 9047.79 N ⭈ m2

TB(d1/2) 2tBIPB TB ⫽ IPB d1 Substitute numerical values: tB ⫽

TB ⫽ 0.237918 T

STEEL SLEEVE (ONLY)

⫽

2(48 MPa)(251.327 * 10⫺9 m4) 40 mm

T ⫽ 2535 N ⭈ m ALLOWABLE TORQUE T BASED UPON STEEL SLEEVE

IPS

p 4 ⫽ (d ⫺ d14) ⫽ 362.265 * 10⫺9 m4 32 2

GSIPS ⫽ 28,981.2 N ⭈ m

2

TORQUES Total torque: T ⫽ TB ⫹ TS Eq. (3-44a): TB ⫽ T a

GBIPB b GBIPB + GS IPS

tS ⫽

TS(d2/2) IPS

TS ⫽

2tSIPS d2

SUBSTITUTE NUMERICAL VALUES: TS ⫽ 0.762082 T ⫽

2(80 MPa)(362.265 * 10⫺9 m4) 50 mm

T ⫽ 1521 N ⭈ m STEEL SLEEVE GOVERNS

Tmax ⫽ 1520 N ⭈ m

;

⫽ 0.237918 T

Problem 3.8-13 The composite shaft shown in the figure is manufactured by shrink-fitting a steel sleeve over a brass core

so that the two parts act as a single solid bar in torsion. The outer diameters of the two parts are d1 ⫽ 1.6 in. for the brass core and d2 ⫽ 2.0 in. for the steel sleeve. The shear moduli of elasticity are Gb ⫽ 5400 ksi for the brass and Gs ⫽ 12,000 ksi for the steel. Assuming that the allowable shear stresses in the brass and steel are ␶b ⫽ 4500 psi and ␶s ⫽ 7500 psi, respectively, determine the maximum permissible torque Tmax that may be applied to the shaft. (Hint: Use Eqs. 3-44a and b to find the torques.)

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SECTION 3.8 Statically Indeterminate Torsional Members

Solution 3.8-13 Composite shaft shrink fit

TORQUES Total torque: T ⫽ TB ⫹ TS Eq. (3-44 a): TB ⫽ T a

GBIPB b GBIPB + GSIPS

⫽ 0.237918 T d1 ⫽ 1.6 in.

Eq. (3-44 b): TS ⫽ T a

d2 ⫽ 2.0 in. GB ⫽ 5,400 psi

GS ⫽ 12,000 psi

GSIPS b GBIPB + GSIPS

⫽ 0.762082 T

Allowable stresses:

T ⫽ TB ⫹ TS (CHECK)

␶B ⫽ 4500 psi ␶S ⫽ 7500 psi

ALLOWABLE TORQUE T BASED UPON BRASS CORE

BRASS CORE (ONLY)

TB(d1/2) 2tBIPB TB ⫽ IPB d1 Substitute numerical values: tB ⫽

TB ⫽ 0.237918 T ⫽ IPB ⫽

p 4 d1 ⫽ 0.643398 in.4 32

GBIPB ⫽ 3.47435 ⫻ 106 lb-in.2 STEEL SLEEVE (ONLY)

2(4500 psi)(0.643398 in.4) 1.6 in.

T ⫽ 15.21 k-in. ALLOWABLE TORQUE T BASED UPON STEEL SLEEVE TS(d2/2) 2tSIPS TS ⫽ IPS d2 Substitute numerical values: tS ⫽

tS ⫽ 0.762082 T ⫽

2(7500 psi)(0.927398 in.4) 2.0 in.

T ⫽ 9.13 k-in. STEEL SLEEVE GOVERNS IPS ⫽

Tmax ⫽ 9.13 k-in.

;

p (d 4 ⫺ d14) ⫽ 0.927398 in.4 32 2

GSIPS ⫽ 11.1288 ⫻ 106 lb-in.2

Problem 3.8-14 A steel shaft (Gs ⫽ 80 GPa) of total length L ⫽ 3.0 m is encased for one-third of its length by a brass sleeve (Gb ⫽ 40 GPa) that is securely bonded to the steel (see figure). The outer diameters of the shaft and sleeve are d1 ⫽ 70 mm and d2 ⫽ 90 mm. respectively.

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(a) Determine the allowable torque T1 that may be applied to the ends of the shaft if the angle of twist between the ends is limited to 8.0°. (b) Determine the allowable torque T2 if the shear stress in the brass is limited to ␶b ⫽ 70 MPa. (c) Determine the allowable torque T3 if the shear stress in the steel is limited to ␶s ⫽ 110 MPa. (d) What is the maximum allowable torque Tmax if all three of the preceding conditions must be satisfied?

Brass sleeve

Steel shaft

d2 = 90 mm

d1 = 70 mm

T

T A

B

1.0 m L = 2.0 m 2 d1

C L = 2.0 m 2

d1 Brass sleeve

d2

d1 Steel shaft

d2

Solution 3.8-14 (a) ALLOWABLE TORQUE T1 BASED ON TWIST AT ENDS OF 8 DEGREES First find torques in steel (Ts) & brass (Tb) in segment in which they are joined - 1 degree stat-indet; use Ts as the internal redundant; see equ. 3-44a in text example

statics Gb IPb b Gs IPs + GbIPb now find twist of 3 segments: Tb ⫽ T1 ⫺ Ts

Tb ⫽ T1 a

L L L Ts T1 4 4 2 ␾⫽ + + Gb IPb Gs IPs Gs IPs T1

Gs IPs Ts ⫽ T1 a b Gs IPs + Gb IPb

For middle term, brass sleeve & steel shaft twist the same so could use Tb(L/4)/(Gb IPb) instead Let ␾a = ␾ allow; substitute expression for Ts then simplifiy; finally, solve for T1, allow Gs IPs L L L T1 a b T1 Gs IPs + Gb IPb 4 4 2 fa ⫽ + + Gb IPb Gs IPs Gs IPs T1

L L L T1 T1 4 4 2 fa ⫽ + + Gb IPb Gs IPs + Gb IPb Gs IPs T1

L 1 2 1 fa ⫽ T1 a + b + 4 Gb IPb GsIPs + Gb IPb Gs IPs T1, allow ⫽

4␾a Gb IPb(GsIPs + Gb IPb)Gs IPs c 2 2 d 2 L Gs IPs + 4 Gb IPb Gs IPs + 2 Gb2 IPb

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SECTION 3.8 Statically Indeterminate Torsional Members

f a ⫽ 8a

NUMERICAL VALUES Gs ⫽ 80 GPa

Gb ⫽ 40 GPa

d1 ⫽ 70 mm

d2 ⫽ 90 mm

IPs ⫽

p 4 d1 32

L ⫽ 3.0 m

IPs ⫽ 2.357 ⫻ 10⫺6 m4

p A d 4 ⫺ d14 B 32 2 T1, allow ⫽ 9.51 kN⭈ m TORQUE

STRESS IN BRASS,

␶b

T2

; BASED ON ALLOWABLE SHEAR

␶b ⫽ 70 MPa First check hollow segment 1 (brass sleeve only) T2 t⫽

d2 2

IPb

T2, allow ⫽

T2, allow ⫽ 6.35 kN⭈m

2tbIPb d2 ;

controls over T2 below also check segment 2 with brass sleeve over steel shaft Tb t⫽

d2 2

IPb

Tb ⫽ T2 a T2, allow

T2, allow ⫽ 13.69 kN⭈m so T2 for hollow segment controls (c) ALLOWABLE

TORQUE

STRESS IN STEEL,

␶s

T3

Ts t⫽

d1 2

where from stat-indet analysis above

IPS

Ts ⫽ T3 a

GsIPS b Gs IPS + Gb IPb

T3, allow ⫽

2ts (Gs IPs + Gb IPb) d1Gs

T3, allow ⫽ 13.83 kN⭈m also check segment 3 with steel shaft alone d1 2 t⫽ IPs T3

T3, allow ⫽

T3, allow ⫽ 7.41 kN⭈m where from stat-indet analysis above

GbIPb b Gs IPS + Gb IPb 2tb(Gs IPS + Gb IPb) ⫽ d2 Gb

BASED ON ALLOWABLE SHEAR

␶s ⫽ 110 MPa First check segment 2 with brass sleeve over steel shaft

IPb ⫽ 4.084 ⫻ 10⫺6 m4

IPb ⫽

(b) ALLOWABLE

p b rad 180

315

2ts IPs d1 ; controls over T3 above

(d) Tmax IF ALL PRECEDING CONDITIONS MUST BE CONSIDERED

from (b) above Tmax ⫽ 6.35 kN⭈m

; max. shear stress in hollow brass sleeve in segment 1 controls overall

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Problem 3.8-15 A uniformly tapered aluminum-alloy tube AB of circular cross section and length L is fixed against rota-

tion at A and B, as shown in the figure. The outside diameters at the ends are dA and dB ⫽ 2dA. A hollow section of length L/2 and constant thickness t ⫽ dA/10 is cast into the tube and extends from B halfway toward A. Torque To is applied at L/2. (a) Find the reactive torques at the supports, TA and TB. Use numerical values as follows: dA ⫽ 2.5 in., L ⫽ 48 in., G ⫽ 3.9 ⫻ 106 psi, T0 ⫽ 40,000 in.-lb. (b) Repeat (a) if the hollow section has constant diameter dA. Fixed against rotation TA

L — 2 A

d(x) t constant

TB

T0

dA

Fixed against rotation

B

dB

L (a)

TA

A

Fixed against rotation

B

L — 2

dA

TB

T0 dA

Fixed against rotation

dB

L (b)

Solution 3.8-15 Solution approach-superposition: select TB as the redundant (1° SI ) L — 2 TA1

A

B

ϕ1(same results for parts a & b)

T0

f1 ⫽

L

+ TA2

L — 2

A

81GpdA

4

f1 ⫽ 2.389

See Prob. 3.4-13 for results for w2 for Parts a & b

B ϕ2 TB

L

608T0L

f2a ⫽ 3.868 f2a ⫽ 3.057

T0 L Gd A 4 T0L Gd A 4

T0L GdA 4

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SECTION 3.8

CONSTANT THICKNESS OF HOLLOW SECTION OF TUBE

CONSTANT DIAMETER OF HOLE

␾1 ⫺ ␾2 ⫽ 0

TB ⫽ a

TB ⫽ 24708 in-lb

TA ⫽ T0 ⫺ TB TA ⫽ 15292 in-1b

ba 4

608T0 L

81GpdA

TB ⫽ 2.45560

GdA4 b b a TB ⫽ a 81G pdA4 3.86804L 608T0 L

T0 TB ⫽ 1.94056 p

317

(b) REACTIVE TORQUES, TA & TB, FOR CASE OF

(a) REACTIVE TORQUES, TA & TB, FOR CASE OF compatibility equation: TB ⫽ redundant T0 ⫽ 40000 in.-lb

Statically Indeterminate Torsional Members

T0 p

GdA4 b 3.05676L

TB ⫽ 31266 in.-lb

TA ⫽ T0 ⫺ TB TA ⫽ 8734 in.-lb

;

;

TA ⫹ TB ⫽ 40,000 in.-1b (check)

; ;

TA ⫹ TB ⫽ 40,000 in.-lb (check)

Problem 3.8-16 A hollow circular tube A (outer diameter dA, wall thickness tA) fits over the end of a circular tube B (dB, tB), as shown in the figure. The far ends of both tubes are fixed. Initially, a hole through tube B makes an angle ␤ with a line through two holes in tube A. Then tube B is twisted until the holes are aligned, and a pin (diameter dp) is placed through the holes. When tube B is released, the system returns to equilibrium. Assume that G is constant. (a) Use superposition to find the reactive torques TA and TB at the supports. (b) Find an expression for the maximum value of ␤ if the shear stress in the pin, ␶p, cannot exceed ␶p,allow. (c) Find an expression for the maximum value of ␤ if the shear stress in the tubes, ␶t, cannot exceed ␶t,allow. (d) Find an expression for the maximum value of ␤ if the bearing stress in the pin at C cannot exceed ␴b,allow.

IPA

IPB

TA Tube A

A

TB

Tube B B

C L

L

b Pin at C Tube A Tube B

Cross-section at C

Solution 3.8-16 (a) SUPERPOSITION

TO FIND TORQUE REACTIONS

-

USE

TB

AS THE REDUNDANT

compatibility: ␾B1 ⫹ ␾B2 ⫽ 0

␾B1 ⫽ ⫺␤ ⬍ joint tubes by pin then release end B fB2 ⫽

TBL 1 1 a + b G IPA IPB

fB2 ⫽

TBL IPB + IPA a b G IPAIPB

TB ⫽

Gb IPAIPB a b L IPA ⫹IPB

TA ⫽ ⫺TB

; statics

(b) ALLOWABLE SHEAR IN PIN RESTRICTS MAGNITUDE OF ␤ TB ⫽ FORCE COUPLE VdB WITH V ⫽ SHEAR IN C TB V V⫽ tp ⫽ dB As

TORQUE PIN AT

TB dB tp, allow ⫽ p 2 d 4 P ;

b max ⫽ tp, allow ca

Gb IPAIPB a b L IPA ⫹ IPB tp, allow ⫽ p dB dp2 4

L 4G

IPB + IPA b dBpdP 2 d IPAIPB

;

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CHAPTER 3 Torsion

(c) ALLOWABLE SHEAR IN TUBES RESTRICTS MAGNITUDE OF ␤ dA TB 2 tmax ⫽ IPA

tmax ⫽

or

tmax

Bearing stresses from tubes A & B are: sbA ⫽

dB TB 2 ⫽ IPB

TB dA ⫺ tA sbA ⫽ dPtA

Gb IPAIPB dA a b L IPA ⫹IPB 2

Gb IPAIPB a b L IPA + IPB

Gb IPAIPB dB a b L IPA ⫹IPB 2

sbA ⫽

IPB

simplifying these two equ., then solving for ␤ gives: tmax ⫽

Gb IPB dA a b L IPA + IPB 2

sbA ⫽ b Gb dB IPA a b L IPA + IPB 2

b max ⫽ tt, allow a

IPA + IPB 2L ba b GdA IPB

sbB ⫽ b ;

or b max

dA ⫺tA dPtA Gb IPAIPB a b L IPA + IPB

IPA + IPB 2L ⫽ tt, allow a ba b GdB IPA

dB ⫺ tB dPtB

sbB ⫽

or tmax ⫽

TB dB ⫺ tB sbB ⫽ dP tB

substitute TB expression from part (a), then simplify ␧ solve for b

IPA

or tmax ⫽

FA FB sbB ⫽ dPtA dPtB

;

where lesser value of ␤ controls (d) ALLOWABLE BEARING STRESS IN PIN RESTRICTS MAGNITUDE OF ␤ Torque TB ⫽ force couple FB (dB ⫺ tB) or FA(dA ⫺ tA), with F ⫽ ave. bearing force on pin at C

L(IPB

IPAIPB G L (IPB + IPA)(dB ⫺ tB)dPtB

b max ⫽ sb, allow c

L G

(IPB + IPA)(dA ⫺ tA)dPtA d IPAIPB

b max ⫽ s b, allow c

GIPAIPB + IPA)(dA ⫺ tA) dP tA

;

L G

(IPB + IPA)(dB ⫺ tB)dPtB d IPAIPB

where lesser value controls

;

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SECTION 3.9

Strain Energy in Torsion

319

Strain Energy in Torsion Problem 3.9-1 A solid circular bar of steel (G ⫽ 11.4 ⫻ 106 psi) with length L ⫽ 30 in. and diameter d ⫽ 1.75 in. is subjected to pure torsion by torques T acting at the ends (see figure).

d

T

T

(a) Calculate the amount of strain energy U stored in the bar when the maximum shear stress is 4500 psi. (b) From the strain energy, calculate the angle of twist ␾ (in degrees).

Solution 3.9-1

L

Steel bar ⫽

pd2Lt2max 16G

(Eq. 2)

Substitute numerical values: U ⫽ 32.0 in.-lb G ⫽ 11.4 ⫻ 106 psi

(b) ANGLE OF TWIST

L ⫽ 30 in.

U⫽

d ⫽ 1.75 in.

␶max ⫽ 4500 psi tmax ⫽

IP ⫽

16 T pd3

Tf 2

f⫽

2U T

Substitute for T and U from Eqs. (1) and (2):

pd3tmax T⫽ 16

pd 4 32

;

f⫽ (Eq. 1)

2Ltmax Gd

(Eq. 3)

Substitute numerical values:

␾ ⫽ 0.013534 rad ⫽ 0.775°

;

(a) STRAIN ENERGY U⫽

pd3tmax 2 L T2L 32 ⫽ a b a b a 4b 2GIP 16 2G pd

Problem 3.9-2 A solid circular bar of copper (G ⫽ 45 GPa) with length L ⫽ 0.75 m and diameter d ⫽ 40 mm is subjected to pure torsion by torques T acting at the ends (see figure). (a) Calculate the amount of strain energy U stored in the bar when the maximum shear stress is 32 MPa. (b) From the strain energy, calculate the angle of twist ␾ (in degrees)

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Torsion

Solution 3.9-2

Copper bar (a) STRAIN ENERGY U⫽

pd3tmax 2 L 32 T2L ⫽ a b a ba b 2GIP 16 2G pd 4

L ⫽ 0.75 m

pd2Lt2max 16G Substitute numerical values:

d ⫽ 40 mm

U ⫽ 5.36 J

⫽

G ⫽ 45 GPa

␶max ⫽ 32 MPa tmax ⫽ IP ⫽

T⫽

3

pd

;

(b) ANGLE OF TWIST 3

16T

pd tmax 16

pd4 32

Tf 2U f⫽ 2 T Substitute for T and U from Eqs. (1) and (2): 2Ltmax f⫽ (Eq. 3) Gd Substitute numerical values: U⫽

(Eq. 1)

␾ ⫽ 0.026667 rad ⫽ 1.53°

Problem 3.9-3 A stepped shaft of solid circular cross sections (see figure) has length L ⫽ 45 in., diameter d2 ⫽ 1.2 in., and diameter d1 ⫽ 1.0 in. The material is brass with G ⫽ 5.6 ⫻ 106 psi. Determine the strain energy U of the shaft if the angle of twist is 3.0°.

d2

;

d1

T

T

L — 2

L — 2

Solution 3.9-3

(Eq. 2)

Stepped shaft ⫽

8T2L 1 1 a 4 + 4b pG d2 d1

Also, U ⫽

(Eq. 1)

Tf 2

(Eq. 2)

Equate U from Eqs. (1) and (2) and solve for T:

d1 ⫽ 1.0 in.

T⫽

d2 ⫽ 1.2 in. L ⫽ 45 in.

pGd14 d24f 16L(d41 + d42) Tf pGf2 d14 d24 ⫽ a b 2 32L d41 + d42

G ⫽ 5.6 ⫻ 10 psi (brass)

U⫽

␾ ⫽ 3.0° ⫽ 0.0523599 rad

SUBSTITUTE NUMERICAL VALUES:

STRAIN ENERGY

U ⫽ 22.6 in.-lb

6

16 T2(L/2) 16 T2(L/2) T2L U⫽ a ⫽ + 4 2GIP pGd2 pGd41

;

f ⫽ radians

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SECTION 3.9

Strain Energy in Torsion

321

Problem 3.9-4 A stepped shaft of solid circular cross sections (see figure) has length L ⫽ 0.80 m, diameter d2 ⫽ 40 mm, and diameter d1 ⫽ 30 mm. The material is steel with G ⫽ 80 GPa. Determine the strain energy U of the shaft if the angle of twist is 1.0°.

Solution 3.9-4

Stepped shaft Also, U ⫽

Tf 2

(Eq. 2)

Equate U from Eqs. (1) and (2) and solve for T: T⫽ d1 ⫽ 30 mm

d2 ⫽ 40 mm

L ⫽ 0.80 m

G ⫽ 80 GPa (steel)

U⫽

pG d14 d24f 16L(d14 + d24) Tf pGf2 d14 d24 ⫽ a b 2 32L d14 + d24

f ⫽ radians

␾ ⫽ 1.0° ⫽ 0.0174533 rad SUBSTITUTE NUMERICAL VALUES: STRAIN ENERGY 2

TL ⫽ U⫽ a 2GIP ⫽

U ⫽ 1.84 J 16T2(L/2) pGd24

8T2L 1 1 a + 4b pG d24 d1

;

16T2(L/2) +

pGd14 (Eq. 1)

Problem 3.9-5 A cantilever bar of circular cross section and length L is fixed at one end and free at the other (see figure). The bar is loaded by a torque T at the free end and by a distributed torque of constant intensity t per unit distance along the length of the bar. (a) What is the strain energy U1 of the bar when the load T acts alone? (b) What is the strain energy U2 when the load t acts alone? (c) What is the strain energy U3 when both loads act simultaneously?

Solution 3.9-5

Cantilever bar with distributed torque G ⫽ shear modulus IP ⫽ polar moment of inertia T ⫽ torque acting at free end t ⫽ torque per unit distance

t

L

T

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(a) LOAD T ACTS ALONE (Eq. 3-51a) U1 ⫽

T2L 2GIP

(c) BOTH LOADS ACT SIMULTANEOUSLY

;

(b) LOAD t ACTS ALONE From Eq. (3-56) of Example 3-11: At distance x from the free end:

t2L3 U2 ⫽ 6GIP

T(x) ⫽ T + tx

;

L

L [T(x)]2 1 dx ⫽ (T + tx)2dx 2GI 2GI P PL L0 0 T2L TtL2 t2L3 ⫽ + + ; 2GIP 2GIP 6GIP

U3 ⫽

NOTE: U3 is not the sum of U1 and U2.

Problem 3.9-6 Obtain a formula for the strain energy U of the statically

2T0

indeterminate circular bar shown in the figure. The bar has fixed supports at ends A and B and is loaded by torques 2T0 and T0 at points C and D, respectively. Hint: Use Eqs. 3-46a and b of Example 3-9, Section 3.8, to obtain the reactive torques.

Solution 3.9-6

3L b 4

L T0 a b 4 +

L

L

⫽

7T0 4

L — 4

D L — 2

⫽

L L L 1 2 2 cT 2 a b + TCD a b + TDB a bd 2GIp AC 4 2 4

⫽

7T0 2 L 1 ca⫺ b a b 2GIP 4 4 + a

INTERNAL TORQUES 7T0 4

C

n Ti2Li U⫽ a i⫽1 2GiIPi

5T0 TB ⫽ 3T0 ⫺ TA ⫽ 4

TAC ⫽ ⫺

B

STRAIN ENERGY (from Eq. 3-53)

From Eq. (3-46a):

TA ⫽

A

Statically indeterminate bar

REACTIVE TORQUES

(2T0)a

T0

TCD ⫽

T0 4

TDB ⫽

5T0 4

U⫽

5T0 2 L T0 2 L b a b + a b a bd 4 2 4 4

19T02L 32GIP

;

L — 4

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SECTION 3.9

Strain Energy in Torsion

323

Problem 3.9-7 A statically indeterminate stepped shaft ACB is fixed at ends A and B and loaded by a torque T0 at point C (see figure). The two segments of the bar are made of the same material, have lengths LA and LB, and have polar moments of inertia IPA and IPB. Determine the angle of rotation ␾ of the cross section at C by using strain energy. Hint: Use Eq. 3-51b to determine the strain energy U in terms of the angle ␾. Then equate the strain energy to the work done by the torque T0. Compare your result with Eq. 3-48 of Example 3-9, Section 3.8.

A

IPA

T0 C

IPB

LA

B

LB

Solution 3.9-7

Statically indeterminate bar WORK DONE BY THE TORQUE T0 W⫽

T0f 2

EQUATE U AND W AND SOLVE FOR ␾ T0f Gf2 IPA IPB a + b ⫽ 2 LA LB 2 f⫽ STRAIN ENERGY (FROM EQ. 3-51B)

;

(This result agrees with Eq. (3-48) of Example 3-9, Section 3.8.)

GIPif2i GIPAf2 GIPBf2 ⫽ + U⫽ a 2LA 2LB i⫽1 2Li n

⫽

T0LALB G(LBIPA + LAIPB)

Gf2 IPA IPB + b a 2 LA LB

Problem 3.9-8 Derive a formula for the strain energy U of the cantilever bar shown in the figure. The bar has circular cross sections and length L. It is subjected to a distributed torque of intensity t per unit distance. The intensity varies linearly from t ⫽ 0 at the free end to a maximum value t ⫽ t0 at the support. t0

t

L

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Solution 3.9-8 Cantilever bar with distributed torque x ⫽ distance from right-hand end of the bar

ELEMENT d␰

STRAIN ENERGY OF ELEMENT dx

Consider a differential element dj at distance j from the right-hand end.

dU ⫽

[T(x)]2dx t0 2 1 ⫽ a b x4dx 2GIP 2GIP 2L ⫽

t20 8L2GIP

x4 dx

STRAIN ENERGY OF ENTIRE BAR L

U⫽

L0

dT ⫽ external torque acting on this element dT ⫽ t(j)dj j ⫽ t0 a bdj L

U⫽

ELEMENT dx AT DISTANCE x

T(x) ⫽ internal torque acting on this element T(x) ⫽ total torque from x ⫽ 0 to x ⫽ x x

T(x) ⫽

⫽

L0 t0x2 2L

dT ⫽

x

L0

t0 a

j bdj L

t20

dU ⫽

t20L3 40GIP

L

x4 dx 8L2GIP L0 t20 L5 ⫽ 2 a b 8L GIP 5 ;

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SECTION 3.9

Problem 3.9-9 A thin-walled hollow tube AB of conical shape has constant thickness t and average diameters dA and dB at the ends (see figure).

B

A

T

T

(a) Determine the strain energy U of the tube when it is subjected to pure torsion by torques T. (b) Determine the angle of twist ␾ of the tube.

L t

Note: Use the approximate formula IP ⬇ ␲d3t/4 for a thin circular ring; see Case 22 of Appendix D.

t

dB

dA

Solution 3.9-9

325

Strain Energy in Torsion

Thin-walled, hollow tube Therefore, L

dx 3 dB ⫺ dA L0 cdA + a bx d L ⫽⫺

t ⫽ thickness dA ⫽ average diameter at end A dB ⫽ average diameter at end B

⫽⫺

d(x) ⫽ average diameter at distance x from end A d(x) ⫽ dA + a

dB ⫺ dA bx L

⫽

pd3t 4

U⫽

3 p[d(x)]3t dB ⫺ dA pt IP(x) ⫽ ⫽ cdA + a bx d 4 4 L

L

T2dx L0 2GIP(x) L dx 2T2 ⫽ 3 pGt L0 dB ⫺ dA cdA + a bx d L

From Appendix C: ⫽ ⫺

L

2(dB ⫺ dA)(dB)

2

+

2(dB ⫺ dA)(dA)2

L(dA + dB) 2dA2 dB2

1 2b(a + bx)2

2T2 L(dA + dB) T2L dA + dB ⫽ a 2 2 b pGt 2dA2dB2 pGt dA dB

Work of the torque T: W ⫽

U⫽

dx

L

(b) ANGLE OF TWIST

(a) STRAIN ENERGY (FROM EQ. 3-54)

L (a + bx)3

2 † 2(dB ⫺ dA) dB ⫺ dA cdA + a bx d L L 0

Substitute this expression for the integral into the equation for U (Eq. 1):

POLAR MOMENT OF INERTIA IP ⫽

L

1

W⫽U (Eq. 1)

Tf T2L(dA + dB) ⫽ 2 pGt d2Ad2B

Solve for ␾: f⫽

Tf 2

2TL(dA + dB) pGt d2A d2B

;

;

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Torsion

Problem 3.9-10 A hollow circular tube A fits over the end of

IPA

a solid circular bar B, as shown in the figure. The far ends of both bars are fixed. Initially, a hole through bar B makes an angle ␤ with a line through two holes in tube A. Then bar B is twisted until the holes are aligned, and a pin is placed through the holes. When bar B is released and the system returns to equilibrium, what is the total strain energy U of the two bars? (Let IPA and IPB represent the polar moments of inertia of bars A and B, respectively. The length L and shear modulus of elasticity G are the same for both bars.)

IPB

Tube A

Bar B

L

L b Tube A Bar B

Solution 3.9-10

Circular tube and bar

TUBE A

COMPATIBILITY

␾A ⫹ ␾B ⫽ ␤ FORCE-DISPLACEMENT RELATIONS fA ⫽

T ⫽ torque acting on the tube

␾A ⫽ angle of twist BAR B

TL GIPA

fB ⫽

TL GIPB

Substitute into the equation of compatibility and solve for T: T⫽

bG IPAIPB a b L IPA + IPB

STRAIN ENERGY U⫽ g ⫽

T 2L T 2L T 2L ⫽ + 2GIP 2GIPA 2GIPB

T2L 1 1 a + b 2G IPA IPB

Substitute for T and simplify: U⫽ T ⫽ torque acting on the bar

␾B ⫽ angle of twist

b 2G IPA IPB a b 2L IPA + IPB

;

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SECTION 3.9

Strain Energy in Torsion

Problem 3.9-11 A heavy flywheel rotating at n revolutions per minute is rigidly attached to the end of a shaft of diameter d (see figure). If the bearing at A suddenly freezes, what will be the maximum angle of twist ␾ of the shaft? What is the corresponding maximum shear stress in the shaft? (Let L ⫽ length of the shaft, G ⫽ shear modulus of elasticity, and Im ⫽ mass moment of inertia of the flywheel about the axis of the shaft. Also, disregard friction in the bearings at B and C and disregard the mass of the shaft.)

A

d

n (rpm)

B C

Hint: Equate the kinetic energy of the rotating flywheel to the strain energy of the shaft.

Solution 3.9-11

Rotating flywheel p 4 d 32

IP ⫽

d ⫽ diameter of shaft U⫽

pGd 4f2 64L

UNITS: d ⫽ diameter n ⫽ rpm

IP ⫽ (length)4

␾ ⫽ radians

KINETIC ENERGY OF FLYWHEEL K.E. ⫽

G ⫽ (force)/(length)2

1 Imv2 2

2pn v⫽ 60 n ⫽ rpm 2pn 2 1 b K.E. ⫽ Im a 2 60

L ⫽ length U ⫽ (length)(force) EQUATE KINETIC ENERGY AND STRAIN ENERGY

Solve for ␾: f⫽

2 2

⫽

p n Im 1800

Im ⫽ (force)(length)(second)2

␻ ⫽ radians per second K.E. ⫽ (length)(force) STRAIN ENERGY OF SHAFT (FROM EQ. 3-51b) U⫽

GIPf 2L

2

2n 2A

15d

2pImL G

;

MAXIMUM SHEAR STRESS t⫽

UNITS:

pGd 4f2 p2n2Im ⫽ 1800 64 L

K.E. ⫽ U

T(d/2) IP

f⫽

TL GIP

Eliminate T: t⫽

Gdf 2L

2pImL 2L15d A G 2pGIm n tmax ⫽ 15d A L tmax ⫽

Gd2n

2

;

327

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Torsion

Thin-Walled Tubes Problem 3.10-1 A hollow circular tube having an inside diameter of 10.0 in.

and a wall thickness of 1.0 in. (see figure) is subjected to a torque T ⫽ 1200 k-in. Determine the maximum shear stress in the tube using (a) the approximate theory of thin-walled tubes, and (b) the exact torsion theory. Does the approximate theory give conservative or nonconservative results?

10.0 in. 1.0 in.

Solution 3.10-1

Hollow circular tube APPROXIMATE THEORY (EQ. 3-63) t1 ⫽

T 2

2pr t

⫽

1200 k-in. 2p(5.5 in.)2(1.0 in.)

␶approx ⫽ 6310 psi

⫽ 6314 psi

;

EXACT THEORY (EQ. 3-11) T ⫽ 1200 k-in. t ⫽ 1.0 in.

t2 ⫽

T(d2/2) ⫽ IP

r ⫽ radius to median line r ⫽ 5.5 in. d2 ⫽ outside diameter ⫽ 12.0 in. d1 ⫽ inside diameter ⫽ 10.0 in.

⫽

Td2 p 2a b 1d24 ⫺ d142 32

16(1200k-in.)(12.0 in.) p[(12.0 in.)4 ⫺ (10.0 in.)4]

⫽ 6831 psi t exact ⫽ 6830 psi

;

Because the approximate theory gives stresses that are too low, it is nonconservative. Therefore, the approximate theory should only be used for very thin tubes.

Problem 3.10-2 A solid circular bar having diameter d is to be replaced by a rectangular tube having cross-sectional dimensions d ⫻ 2d to the median line of the cross section (see figure). Determine the required thickness tmin of the tube so that the maximum shear stress in the tube will not exceed the maximum shear stress in the solid bar.

t t d

d

2d

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SECTION 3.10

Solution 3.10-2

Thin-Walled Tubes

329

Bar and tube

SOLID BAR tmax ⫽

16T pd3

(Eq. 3-12)

Am ⫽ (d)(2d) ⫽ 2d2

(Eq. 3-64)

T T ⫽ 2tAm 4td2

(Eq. 3-61)

tmax ⫽

EQUATE THE MAXIMUM SHEAR STRESSES AND SOLVE FOR t RECTANGULAR TUBE

16T pd3

⫽

T 4td2

tmin ⫽

pd 64

;

If t ⬎ tmin, the shear stress in the tube is less than the shear stress in the bar.

Problem 3.10-3 A thin-walled aluminum tube of rectangular

cross section (see figure) has a centerline dimensions b ⫽ 6.0 in. and h ⫽ 4.0 in. The wall thickness t is constant and equal to 0.25 in.

t h

(a) Determine the shear stress in the tube due to a torque T ⫽ 15 k-in. (b) Determine the angle of twist (in degrees) if the length L of the tube is 50 in. and the shear modulus G is 4.0 ⫻ 106 psi. b

Probs. 3.10-3 and 3.10-4

Solution 3.10-3

Thin-walled tube Eq. (3-64): Am ⫽ bh ⫽ 24.0 in.2 J⫽

Eq. (3-71) with t1 ⫽ t2 ⫽ t: J ⫽ 28.8 in.4 (a) SHEAR STRESS (EQ. 3-61) t⫽ b ⫽ 6.0 in. h ⫽ 4.0 in. t ⫽ 0.25 in. T ⫽ 15 k-in. L ⫽ 50 in. G ⫽ 4.0 ⫻ 106 psi

T ⫽ 1250 psi 2tAm

;

(b) ANGLE OF TWIST (EQ. 3-72) f⫽

TL ⫽ 0.0065104 rad GJ

⫽ 0.373°

;

2b2h2t b + h

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Problem 3.10-4 A thin-walled steel tube of rectangular cross section (see figure) has centerline dimensions b ⫽ 150 mm and h ⫽ 100 mm. The wall thickness t is constant and equal to 6.0 mm.

(a) Determine the shear stress in the tube due to a torque T ⫽ 1650 N ⭈ m. (b) Determine the angle of twist (in degrees) if the length L of the tube is 1.2 m and the shear modulus G is 75 GPa.

Solution 3.10-4

Thin-walled tube b ⫽ 150 mm

(a) SHEAR STRESS (Eq. 3-61)

h ⫽ 100 mm

t⫽

t ⫽ 6.0 mm L ⫽ 1.2 m

f⫽

G ⫽ 75 GPa

TL ⫽ 0.002444 rad GJ

⫽ 0.140°

Eq. (3-64): Am ⫽ bh ⫽ 0.015 m2 J⫽

;

(b) ANGLE OF TWIST (Eq. 3-72)

T ⫽ 1650 N ⭈ m

Eq. (3-71) with t1 ⫽ t2 ⫽ t:

T ⫽ 9.17 MPa 2tAm

;

2b2h2t b + h

J ⫽ 10.8 ⫻ 10⫺6 m4 Tube (1)

Problem 3.10-5 A thin-walled circular tube and a solid circular bar of

Bar (2)

the same material (see figure) are subjected to torsion. The tube and bar have the same cross-sectional area and the same length. What is the ratio of the strain energy U1 in the tube to the strain energy U2 in the solid bar if the maximum shear stresses are the same in both cases? (For the tube, use the approximate theory for thin-walled bars.)

Solution 3.10-5 THIN-WALLED TUBE (1) Am ⫽ ␲r2 tmax ⫽

J ⫽ 2␲r3t

12pr2ttmax22L T 2L ⫽ 2GJ 2G(2pr3t)

prtt2maxL G A But rt ⫽ 2p ⫽

At2max L ‹ U1 ⫽ 2G

SOLID BAR (2)

T T ⫽ 2tAm 2pr 2t

T ⫽ 2␲r 2t␶max U1 ⫽

A ⫽ 2␲rt

A ⫽ pr22

IP ⫽

p 4 r 2 2

Tr2 pr23tmax 2T ⫽ 3 T⫽ IP 2 pr2 3 2 2 2 2 (pr2 tmax) L pr2 tmaxL TL ⫽ U2 ⫽ ⫽ 2GIP 4G p 4 8G a r2 b 2 tmax ⫽

But pr22 ⫽ A RATIO U1 ⫽2 U2

;

‹ U2 ⫽

2 L Atmax 4G

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SECTION 3.10

Thin-Walled Tubes

t = 8 mm

Problem 3.10-6 Calculate the shear stress ␶ and the angle of twist ␾ (in

r = 50 mm

degrees) for a steel tube (G ⫽ 76 GPa) having the cross section shown in the figure. The tube has length L ⫽ 1.5 m and is subjected to a torque T ⫽ 10 kN ⭈ m.

r = 50 mm

b = 100 mm

Solution 3.10-6

Steel tube SHEAR STRESS G ⫽ 76 GPa.

t⫽

10 kN # m T ⫽ 2tAm 2(8 mm)(17,850 mm2)

L ⫽ 1.5 m T ⫽ 10 kN ⭈ m Am ⫽ ␲r2 ⫹ 2br Am ⫽ ␲ (50 mm)2 ⫹ 2(100 mm)(50 mm) ⫽ 17,850 mm2 Lm ⫽ 2b ⫹ 2␲r

⫽ 35.0 MPa

;

ANGLE OF TWIST f⫽

(10 kN # m)(1.5 m) TL ⫽ GJ (76 GPa)(19.83 * 106 mm4) ⫽ 0.00995 rad ⫽ 0.570°

;

⫽ 2(100 mm) ⫹ 2␲(50 mm) ⫽ 514.2 mm J⫽

4(8 mm)(17,850 mm2)2 4tA2m ⫽ Lm 514.2 mm

⫽ 19.83 * 106 mm4

Problem 3.10-7 A thin-walled steel tube having an elliptical cross

331

t

section with constant thickness t (see figure) is subjected to a torque T ⫽ 18 k-in. Determine the shear stress ␶ and the rate of twist ␪ (in degrees per inch) if G ⫽ 12 ⫻ 106 psi, t ⫽ 0.2 in., a ⫽ 3 in., and b ⫽ 2 in. (Note: See Appendix D, Case 16, for the properties of an ellipse.)

2b

2a

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Torsion

Solution 3.10-7

Elliptical tube FROM APPENDIX D, CASE 16: Am ⫽ pab ⫽ p(3.0 in.)(2.0 in.) ⫽ 18.850 in.2 Lm L p[1.5(a + b) ⫺ 1ab] ⫽ p[1.5(5.0 in.) ⫺ 26.0 in.2] ⫽ 15.867 in. J⫽

⫽ 17.92 in.4

T ⫽ 18 k-in.

SHEAR STRESS

G ⫽ 12 ⫻ 106 psi

t⫽

t ⫽ constant t ⫽ 0.2 in

4(0.2 in.)(18.850 in.2)2 4tA2m ⫽ Lm 15.867 in.

a ⫽ 3.0 in.

b ⫽ 2.0 in.

18 k-in. T ⫽ 2tAm 2(0.2 in.)(18.850 in.2)

⫽ 2390 psi

;

ANGLE OF TWIST PER UNIT LENGTH (RATE OF TWIST) u⫽

f T 18 k-in. ⫽ ⫽ L GJ (12 * 106 psi)(17.92 in.)4

u ⫽ 83.73 * 10⫺6 rad/in. ⫽ 0.0048°/in.

Problem 3.10-8 A torque T is applied to a thin-walled tube having a cross section in the shape of a regular hexagon with constant wall thickness t and side length b (see figure). Obtain formulas for the shear stress ␶ and the rate of twist ␪.

;

t

b

Solution 3.10-8

Regular hexagon b ⫽ Length of side t ⫽ Thickness Lm ⫽ 6b FROM APPENDIX D, CASE 25:

␤ ⫽ 60° n ⫽ 6 Am ⫽ ⫽

b nb2 6b2 cot ⫽ cot 30° 4 2 4

3 13b2 2

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SECTION 3.10

SHEAR STRESS T T13 t⫽ ⫽ 2tAm 9b2t

u⫽ ;

Thin-Walled Tubes

2T T 2T ⫽ ⫽ 3 GJ G(9b t) 9Gb3t

333

;

(radians per unit length)

ANGLE OF TWIST PER UNIT LENGTH (RATE OF TWIST) 4A2mt

J⫽

Lm

L0

ds t

⫽

4A2mt 9b3t ⫽ Lm 2

Problem 3.10-9 Compare the angle of twist ␾1 for a thin-walled circular tube (see figure) calculated from the approximate theory for thin-walled bars with the angle of twist ␾2 calculated from the exact theory of torsion for circular bars.

t r

(a) Express the ratio ␾1/␾2 in terms of the nondimensional ratio ␤ ⫽ r/t. (b) Calculate the ratio of angles of twist for ␤ ⫽ 5, 10, and 20. What conclusion about the accuracy of the approximate theory do you draw from these results?

Solution 3.10-9

C

Thin-walled tube (a) RATIO f1 f2

⫽

4r 2 + t 2

Let b ⫽ APPROXIMATE THEORY TL f1 ⫽ GJ

J ⫽ 2␲r 3t

(b) f1 ⫽

TL GIP

2pGr3t

From Eq. (3-17): Ip ⫽

TL 2TL f2 ⫽ ⫽ GIP pGrt(4r 2 + t 2)

r t

f1 f2

t2 4r 2 1

⫽1 +

;

4b 2

␤

␾1/␾2

5 10 20

1.0100 1.0025 1.0006

TL

EXACT THEORY f2 ⫽

4r 2

⫽1 +

prt (4r 2 + t 2) 2

As the tube becomes thinner and ␤ becomes larger, the ratio ␾1/␾2 approaches unity. Thus, the thinner the tube, the more accurate the approximate theory becomes.

Problem 3.10-10 A thin-walled rectangular tube has uniform thickness t and dimensions a ⫻ b to the median line of the cross section (see figure). How does the shear stress in the tube vary with the ratio ␤ ⫽ a/b if the total length Lm of the median line of the cross section and the torque T remain constant? From your results, show that the shear stress is smallest when the tube is square (␤ ⫽ 1).

t

b

a

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Torsion

Solution 3.10-10

Rectangular tube T, t, and Lm are constants. Let k ⫽

2T tL2m

⫽ constant t ⫽ k

(1 + b)2 b

t ⫽ thickness (constant) a, b ⫽ dimensions of the tube b⫽

a b

t a b ⫽4 k min

Lm ⫽ 2(a ⫹ b) ⫽ constant T ⫽ constant

T 2tAm

tL2m

ALTERNATE SOLUTION Am ⫽ ab ⫽ ␤b2

t⫽

Lm ⫽ 2b(1 ⫹ ␤) ⫽ constant

2T (1 + b)2 c d b tL2m

2T b(2)(1 + b) ⫺ (1 + b)2(1) dt ⫽ 2c d ⫽0 db tLm b2

2 Lm d Am ⫽ b c 2(1 + b)

Lm b⫽ 2(1 + b) Am ⫽

8T

From the graph, we see that ␶ is minimum when ␤ ⫽ 1 and the tube is square.

SHEAR STRESS t⫽

tmin ⫽

or 2␤ (1 ⫹ ␤) ⫺ (1 ⫹ ␤)2 ⫽ 0

bL2m

⬖␤ ⫽ 1

2

4(1 + b) T(4)(1 + b)2 2T(1 + b)2 T ⫽ ⫽ t⫽ 2 2tAm 2tbLm tL2m b

;

Thus, the tube is square and ␶ is either a minimum or a maximum. From the graph, we see that ␶ is a minimum.

Problem 3.10-11 A tubular aluminum bar (G ⫽ 4 ⫻ 106 psi) of square

cross section (see figure) with outer dimensions 2 in. ⫻ 2 in. must resist a torque T ⫽ 3000 lb-in. Calculate the minimum required wall thickness tmin if the allowable shear stress is 4500 psi and the allowable rate of twist is 0.01 rad/ft.

t 2 in.

2 in.

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SECTION 3.10

Solution 3.10-11

335

Thin-Walled Tubes

Square aluminum tube THICKNESS t BASED UPON SHEAR STRESS t⫽

T 2tAm

tAm ⫽

T 2t

T 2t

t(b ⫺ t)2 ⫽

b ⫽ in. T ⫽ lb-in. ␶ ⫽ psi

UNITS: t ⫽ in.

t(2.0 in. ⫺ t)2 ⫽

3000 lb-in. 1 ⫽ in.3 2(4500 psi) 3

3t(2 ⫺ t)2 ⫺ 1 ⫽ 0 Solve for t: t ⫽ 0.0915 in.

Outer dimensions: 2.0 in. ⫻ 2.0 in.

THICKNESS t BASED UPON RATE OF TWIST

G ⫽ 4 ⫻ 106 psi T ⫽ 3000 lb-in.

u⫽

␶allow ⫽ 4500 psi u allow ⫽ 0.01 rad/ft ⫽

0.01 rad/in. 12

T T ⫽ GJ Gt(b ⫺ t)3

3000 lb-in

t(2.0 in. ⫺ t)3 ⫽

6

(4 * 10 psi)(0.01/12 rad/in.) 9 ⫽ 10

⫽ 2.0 in. Centerline dimension ⫽ b ⫺ t

10t(2 ⫺ t)3 ⫺ 9 ⫽ 0

Am ⫽ (b ⫺ t)2

Solve for t:

J⫽

Lm

Lm ⫽ 4(b ⫺ t)

4t(b ⫺ t) ⫽ t(b ⫺ t)3 4(b ⫺ t) 4

⫽

T Gu

G ⫽ psi ␪ ⫽ rad/in.

UNITS: t ⫽ in.

Let b ⫽ outer dimension

4tA2m

t(b ⫺ t)3 ⫽

t ⫽ 0.140 in. ANGLE OF TWIST GOVERNS tmin ⫽ 0.140 in.

Problem 3.10-12 A thin tubular shaft of circular cross section (see figure) with inside diameter 100 mm is subjected to a torque of 5000 N ⭈ m. If the allowable shear stress is 42 MPa, determine the required wall thickness t by using (a) the approximate theory for a thin-walled tube, and (b) the exact torsion theory for a circular bar.

;

100 mm t

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CHAPTER 3 Torsion

Solution 3.10-12

Thin tube (b) EXACT THEORY

T ⫽ 5,000 N ⭈ m

d1 ⫽ inner diameter ⫽ 100 mm

t⫽

Tr2 Ip

Ip ⫽

p 4 p (r ⫺ r41) ⫽ [(50 + t)4 ⫺(50)4] 2 2 2

42 MPa ⫽

␶allow ⫽ 42 MPa t is in millimeters.

(50 + t)4 ⫺ (50)4 (5000 N # m)(2) ⫽ 50 + t (p)(42 MPa)

r ⫽ Average radius ⫽ 50 mm +

t 2

⫽

r1 ⫽ Inner radius

t ⫽ 7.02 mm

r2 ⫽ Outer radius ⫽ 50 mm ⫹ t Am ⫽ ␲r2 (a) APPROXIMATE THEORY T T T ⫽ ⫽ 2tAm 2t(pr 2) 2pr 2 t 5,000 N # m 42 MPa ⫽ t 2 2pa50 + b t 2

t⫽

or 5,000 N # m t 2 5 * 106 b ⫽ ⫽ mm3 2 2p(42 MPa) 84p

Solve for t: t ⫽ 6.66 mm

5 * 106 mm3 21p

Solve for t:

⫽ 50 mm

t a50 +

(5,000 N # m)(50 + t) p [(50 + t)4 ⫺ (50)4] 2

;

;

The approximate result is 5% less than the exact result. Thus, the approximate theory is nonconservative and should only be used for thin-walled tubes.

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SECTION 3.10

Problem 3.10-13 A long, thin-walled tapered tube AB of circular cross section (see figure) is subjected to a torque T. The tube has length L and constant wall thickness t. The diameter to the median lines of the cross sections at the ends A and B are dA and dB, respectively. Derive the following formula for the angle of twist of the tube: f⫽

T

T

L t

t

Hint: If the angle of taper is small, we may obtain approximate results by applying the formulas for a thin-walled prismatic tube to a differential element of the tapered tube and then integrating along the axis of the tube.

Solution 3.10-13

B

A

2TL dA + dB a 2 2 b pGt dAdB

337

Thin-Walled Tubes

dB

dA

Thin-walled tapered tube For entire tube: f⫽

4T pGT L0

L

dx 3 dB ⫺ dA cdA + a bx d L

From table of integrals (see Appendix C): 1

t ⫽ thickness

dx (a + bx)3

⫽ ⫺

1 2b(a + bx)2

dA ⫽ average diameter at end A dB ⫽ average diameter at end B T ⫽ torque d(x) ⫽ average diameter at distance x from end A. d(x) ⫽ dA + a J ⫽ 2pr 3t ⫽

dB ⫺ dA bx L 3

pd t 4

3 dB ⫺ dA pt pt J(x) ⫽ [d(x)]3 ⫽ cdA + a bx d 4 4 L

For element of length dx: df ⫽

Tdx ⫽ GJ(x)

4Tdx 3 dB ⫺ dA GptcdA + a bx d L

L

4T f⫽ pGt

⫽ f⫽

J

1

⫺ 2a

2 dB ⫺ dA dB ⫺ dA # xb K0 b adA + L L

4T L L c⫺ + d pGt 2(dB ⫺ dA)d2B 2(dB ⫺ dA)d2A 2TL dA + dB a 2 2 b pGt dAdB

;

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Torsion

Stress Concentrations in Torsion D2

The problems for Section 3.11 are to be solved by considering the stress-concentration factors.

Problem 3.11-1 A stepped shaft consisting of solid circular

D1

T

T

segments having diameters D1 ⫽ 2.0 in. and D2 ⫽ 2.4 in. (see figure) is subjected to torques T. The radius of the fillet is R ⫽ 0.1 in. If the allowable shear stress at the stress concentration is 6000 psi, what is the maximum permissible torque Tmax?

Solution 3.11-1

R

Probs. 3.11-1 through 3.11-5

Stepped shaft in torsion USE FIG. 3-48 FOR THE STRESS-CONCENTRATION FACTOR D2 2.4 in. ⫽ ⫽ 1.2 D1 2.0 in.

R 0.1 in. ⫽ ⫽ 0.05 D1 2.0 in. K ⬇ 1.52

tmax ⫽ Kt nom ⫽ K a

16 Tmax pD31

b

pD31tmax 16K p(2.0 in.)3(6000 psi) ⫽ ⫽ 6200 lb-in. 16(1.52)

D1 ⫽ 2.0 in.

Tmax ⫽

D2 ⫽ 2.4 in. R ⫽ 0.1 in.

␶allow ⫽ 6000 psi

⬖ Tmax ⬇ 6200 lb-in.

;

Problem 3.11-2 A stepped shaft with diameters D1 ⫽ 40 mm and D2 ⫽ 60 mm is loaded by torques T ⫽ 1100 N ⭈ m (see figure). If the allowable shear stress at the stress concentration is 120 MPa, what is the smallest radius Rmin that may be used for the fillet?

Solution 3.11-2

Stepped shaft in torsion USE FIG. 3-48 FOR THE STRESS-CONCENTRATION FACTOR tmax ⫽ Kt nom ⫽ Ka

16T pD31

b

p(40 mm)3(120 MPa) pD31tmax ⫽ ⫽ 1.37 16 T 16(1100 N # m) D2 60 mm ⫽ ⫽ 1.5 D1 40 mm K⫽

D1 ⫽ 40 mm D2 ⫽ 60 mm T ⫽ 1100 N ⭈ m

From Fig. (3-48) with

␶allow ⫽ 120 MPa we get

D2 ⫽ 1.5 and K ⫽ 1.37, D1

R L 0.10 D1

⬖ Rmin ⬇ 0.10(40 mm) ⫽ 4.0 mm

;

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SECTION 3.11 Stress Concentrations in Torsion

Problem 3.11-3 A full quarter-circular fillet is used at the shoulder of a stepped shaft having diameter D2 ⫽ 1.0 in. (see figure). A torque T ⫽ 500 lb-in. acts on the shaft. Determine the shear stress ␶max at the stress concentration for values as follows: D1 5 0.7, 0.8, and 0.9 in. Plot a graph showing ␶max versus D1.

Solution 3.11-3 Stepped shaft in torsion

D1 (in.)

D2/D1

R(in.)

R/D1

K

␶max(psi)

0.7 0.8 0.9

1.43 1.25 1.11

0.15 0.10 0.05

0.214 0.125 0.056

1.20 1.29 1.41

8900 6400 4900

D2 ⫽ 1.0 in. T ⫽ 500 lb-in. D1 ⫽ 0.7, 0.8, and 0.9 in. Full quarter-circular fillet (D2 ⫽ D1 ⫹ 2R) R⫽

D2 ⫺ D1 D1 ⫽ 0.5 in. ⫺ 2 2

USE FIG. 3-48 FOR THE STRESS-CONCENTRATION FACTOR tmax ⫽ Kt nom ⫽ K a ⫽K

16 T pD31

16(500 lb-in.) pD31

b

⫽ 2546

K D31

NOTE that ␶max gets smaller as D1 gets larger, even though K is increasing.

Problem 3.11-4 The stepped shaft shown in the figure is required to transmit 600 kW of power at 400 rpm.

The shaft has a full quarter-circular fillet, and the smaller diameter D1 ⫽ 100 mm. If the allowable shear stress at the stress concentration is 100 MPa, at what diameter D2 will this stress be reached? Is this diameter an upper or a lower limit on the value of D2?

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CHAPTER 3 Torsion

Solution 3.11-4 Stepped shaft in torsion

P ⫽ 600 kW

D1 ⫽ 100 mm

n ⫽ 400 rpm ␶allow ⫽ 100 MPa

Use the dashed line for a full quarter-circular fillet. R L 0.075 D1

R ⬇ 0.075 D1 ⫽ 0.075 (100 mm)

Full quarter-circular fillet

⫽ 7.5 mm

2pnT ( Eq. 3-42 of Section 3.7) POWER P ⫽ 60 P ⫽ watts

n ⫽ rpm

T ⫽ Newton meters

60(600 * 103 W) 60P ⫽ ⫽ 14,320 N # m T⫽ 2pn 2p(400 rpm)

D2 ⫽ D1 ⫹ 2R ⫽ 100 mm ⫹ 2(7.5 mm) ⫽ 115 mm ⬖ D2 ⬇ 115 mm

;

This value of D2 is a lower limit

;

(If D2 is less than 115 mm, R/D1 is smaller, K is larger, and ␶max is larger, which means that the allowable stress is exceeded.)

USE FIG. 3-48 FOR THE STRESS-CONCENTRATION FACTOR tmax ⫽ Kt nom ⫽ K a K⫽ ⫽

16T pD31

b

tmax(pD31) 16T

(100 MPa)(p)(100 mm)3 ⫽ 1.37 16(14,320 N # m)

Problem 3.11-5 A stepped shaft (see figure) has diameter D2 ⫽ 1.5 in. and a full quarter-circular fillet. The allowable shear stress is 15,000 psi and the load T ⫽ 4800 lb-in. What is the smallest permissible diameter D1?

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SECTION 3.11 Stress Concentrations in Torsion

341

Solution 3.11-5 Stepped shaft in torsion

Use trial-and-error. Select trial values of D1

D2 ⫽ 1.5 in.

␶allow ⫽ 15,000 psi T ⫽ 4800 lb-in. Full quarter-circular fillet D2 ⫽ D1 ⫹ 2R R⫽

D1 D2 ⫺ D1 ⫽ 0.75 in. ⫺ 2 2

D1 (in.)

R (in.)

R/D1

K

␶max(psi)

1.30 1.35 1.40

0.100 0.075 0.050

0.077 0.056 0.036

1.38 1.41 1.46

15,400 14,000 13,000

USE FIG. 3-48 FOR THE STRESS-CONCENTRATION FACTOR tmax ⫽ Kt nom ⫽ K a

16T

pD31 K 16(4800 lb-in.) ⫽ 3c d p D1 ⫽ 24,450

b

K D31

From the graph, minimum D1 ⬇ 1.31 in.

;

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4 Shear Forces and Bending Moments

Shear Forces and Bending Moments 800 lb

1600 lb

Problem 4.3-1 Calculate the shear force V and bending moment M at a cross section just to the left of the 1600-1b load acting on the simple beam AB shown in the figure.

A

B 30 in.

50 in. 120 in.

40 in.

Solution 4.3-1 gMA ⫽ 0: RB ⫽

3800 ⫽ 1267 lb 3

3400 gMB ⫽ 0: RA ⫽ ⫽ 1133 lb 3 FREE-BODY DIAGRAM OF SEGMENT DB

gFVERT ⫽ 0: V ⫽ 1600 lb ⫺ 1267 lb ⫽ 333 lb

;

g MD ⫽ 0: M ⫽ 11267 lb2(40 in.) ⫽

152000 # lb in ⫽ 50667 lb # in. 3

;

1600 lb

D B 40 in RB

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Problem 4.3-2 Determine the shear force V and bending moment M at the

6.0 kN

midpoint C of the simple beam AB shown in the figure.

2.0 kN/m

C

A

B

0.5 m 1.0 m 2.0 m 4.0 m

1.0 m

Solution 4.3-2 FREE-BODY DIAGRAM OF SEGMENT AC

g MA ⫽ 0:

RB ⫽ 3.9375 kN

g MB ⫽ 0:

RA ⫽ 5.0625 kN

g FVERT ⫽ 0:

V ⫽ RA ⫺ 6 ⫽ ⫺0.938 kN

g MC ⫽ 0:

M ⫽ RA ⭈ 2 m ⫺ 6 kN ⭈ 1 m ⫽ 4.12 kN⭈m ;

Problem 4.3-3 Determine the shear force V and bending moment M at the

Pb

P

midpoint of the beam with overhangs (see figure). Note that one load acts downward and the other upward. Also clockwise moments Pb are applied at each support.

;

Pb

b

L

P

b

Solution 4.3-3 Pb

Pb

Pb

FREE-BODY DIAGRAM (C IS THE MIDPOINT) 1 (2Pb ⫺ (b + L)P ⫺ Pb) L ⫽ P (upward)

gMB ⫽ 0: RA ⫽ g MA ⫽ 0:

g FVERT ⫽ 0:

V ⫽ RA ⫺ P ⫽ 0

g MC ⫽ 0: M ⫽ ⫺Pa b +

RB ⫽ P (downward) + RA

;

L b 2

L + Pb ⫽ 0 2

;

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SECTION 4.3

Problem 4.3-4 Calculate the shear force V and bending moment M at a cross

4.0 kN

section located 0.5 m from the fixed support of the cantilever beam AB shown in the figure.

B 1.0 m

2.0 m

Cantilever beam

4.0 kN

g FVERT ⫽ 0:

1.5 kN/m

A

B

V ⫽ 4.0 kN ⫹ (1.5 kN/m)(2.0 m) ⫽ 4.0 kN ⫹ 3.0 kN ⫽ 7.0 kN

1.0 m

1.5 kN/m

A

1.0 m

Solution 4.3-4

345

Shear Forces and Bending Moments

1.0 m

2.0 m

g MD ⫽ 0:

;

M ⫽ ⫺(4.0 kN)(0.5 m)

FREE-BODY DIAGRAM OF SEGMENT DB

⫺ (1.5 kN/m)(2.0 m)(2.5 m) ⫽ ⫺2.0 kN ⭈ m ⫺ 7.5 kN ⭈ m

Point D is 0.5 m from support A.

⫽ ⫺9.5 kN ⭈ m

Problem 4.3-5 Determine the shear force V and bending moment M at a cross section located 18 ft from the left-hand end A of the beam with an overhang shown in the figure.

;

400 lb/ft

300 lb/ft B

A 10 ft

10 ft

C 6 ft

6 ft

Solution 4.3-5

FREE-BODY DIAGRAM OF SEGMENT AD

g MB ⫽ 0:

RA ⫽ 2190 lb

g MA ⫽ 0:

RB ⫽ 3610 lb

Point D is 18ft from support A. g FVERT ⫽ 0:

V ⫽ 2190 lb ⫺ (400 lb/ft)(10 ft) ; ⫽ ⫺1810 lb

g Mc ⫽ 0: M ⫽ (2190 lb)(18 ft) ⫺ (400 lb/ft)(10 ft)(13 ft) ⫽ ⫺12580 lb⭈ ft ;

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Problem 4.3-6 The beam ABC shown in the figure is simply supported at A and B and has an overhang from B to C. The loads consist of a horizontal force P1 ⫽ 4.0 kN acting at the end of a vertical arm and a vertical force P2 ⫽ 8.0 kN acting at the end of the overhang. Determine the shear force V and bending moment M at a cross section located 3.0 m from the left-hand support. (Note: Disregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.)

Solution 4.3-6

P1 = 4.0 kN P2 = 8.0 kN 1.0 m A

B

4.0 m

C

1.0 m

Beam with vertical arm FREE-BODY DIAGRAM OF SEGMENT AD Point D is 3.0 m from support A.

g MB ⫽ 0:

RA ⫽ 1.0 kN (downward)

g MA ⫽ 0:

RB ⫽ 9.0 kN (upward)

g FVERT ⫽ 0:

V ⫽ ⫺RA ⫽ ⫺1.0 kN

g MD

M ⫽ ⫺RA(3.0 m) ⫺ 4.0 kN ⭈ m ⫽ ⫺7.0 kN ⭈ m ;

⫽ 0:

Problem 4.3-7 The beam ABCD shown in the figure has overhangs at each end and carries a uniform load of intensity q. For what ratio b/L will the bending moment at the midpoint of the beam be zero?

;

q A

D B b

C L

b

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Solution 4.3-7

347

Shear Forces and Bending Moments

Beam with overhangs FREE-BODY DIAGRAM OF LEFT-HAND HALF OF BEAM: Point E is at the midpoint of the beam.

From symmetry and equilibrium of vertical forces: L RB ⫽ RC ⫽ q ab + b 2

gME ⫽ 0 哵 哴 1 L 2 L ⫺ RB a b + qa b ab + b ⫽ 0 2 2 2 L L 1 L 2 b a b + qa b ab + b ⫽ 0 2 2 2 2

⫺ qa b +

Solve for b/L: b 1 ⫽ L 2

;

Problem 4.3-8 At full draw, an archer applies a pull of 130 N to the bowstring of the bow shown in the figure. Determine the bending moment at the midpoint of the bow.

70° 1400 mm

350 mm

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Solution 4.3-8

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Shear Forces and Bending Moments

Archer’s bow FREE-BODY DIAGRAM OF SEGMENT BC

g MC ⫽ 0 T(cos b)a M⫽Ta

P ⫽ 130 N

␤ ⫽ 70°

⫽

H ⫽ 1400 mm

H b + T(sin b)(b) ⫺ M ⫽ 0 2

H cos b + b sin b b 2

P H a + b tan b b 2 2

SUBSTITUTE NUMERICAL VALUES:

⫽ 1.4 m b ⫽ 350 mm

M⫽

⫽ 0.35 m FREE-BODY DIAGRAM OF POINT A

T ⫽ tensile force in the bowstring g FHORIZ ⫽ 0:

哵哴

2T cos ␤ ⫺ P ⫽ 0 T⫽

P 2 cos b

130 N 1.4 m c + (0.35 m)(tan 70°) d 2 2

M ⫽ 108 N # m

;

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SECTION 4.3

Problem 4.3-9 A curved bar ABC is subjected to loads in the form of two equal and opposite forces P, as shown in the figure. The axis of the bar forms a semicircle of radius r. Determine the axial force N, shear force V, and bending moment M acting at a cross section defined by the angle ␪.

Solution 4.3-9

349

Shear Forces and Bending Moments

M B P A

V

r

u O

N

P

P

C

u A

Curved bar g FN ⫽ 0

Q⫹ b⫺

N ⫺ P sin ␪ ⫽ 0 N ⫽ P sin ␪

g FV ⫽ 0

⫹

R a

⫺

V ⫺ P cos ␪ ⫽ 0 V ⫽ P cos ␪

g MO ⫽ 0

哵哴

;

;

M ⫺ Nr ⫽ 0 M ⫽ Nr ⫽ Pr sin ␪

;

Problem 4.3-10 Under cruising conditions the distributed load acting on the wing of a small airplane has the idealized variation shown in the figure. Calculate the shear force V and bending moment M at the inboard end of the wing.

1600 N/m

2.6 m

Solution 4.3-10

900 N/m

2.6 m

1.0 m

Airplane wing (Minus means the shear force acts opposite to the direction shown in the figure.) LOADING (IN THREE PARTS)

SHEAR FORCE g FVERT ⫽ 0 V + +

c⫹ T⫺

1 (700 N/m)(2.6 m) + (900 N/m)(5.2 m) 2 1 1900 N/m2(1.0 m) ⫽ 0 2

V ⫽ ⫺6040 N ⫽ ⫺6.04 kN

;

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BENDING MOMENT

M ⫽ 788.67 N ⭈ m ⫹ 12,168 N ⭈ m ⫹ 2490 N ⭈ m

g MA ⫽ 0 哵哴

⫽ 15,450 N ⭈ m

1 2.6 m (700 N/m) (2.6 m) a b 2 3 + (900 N/m) (5.2 m) (2.6 m) 1 1.0 m + (900 N/m) (1.0 m) a 5.2 m + b ⫽0 2 3

⫽ 15.45 kN ⭈ m

;

⫺M +

Problem 4.3-11 A beam ABCD with a vertical arm CE is supported as

E

a simple beam at A and D (see figure). A cable passes over a small pulley that is attached to the arm at E. One end of the cable is attached to the beam at point B. What is the force P in the cable if the bending moment in the beam just to the left of point C is equal numerically to 640 lb-ft? (Note: Disregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.)

Cable A

B

8 ft C

6 ft

Solution 4.3-11

6 ft

Beam with a cable FREE-BODY DIAGRAM OF SECTION AC

g MC ⫽ 0 UNITS: P in lb M in lb-ft

P

哵哴

4P 4P (6 ft) + (12 ft) ⫽ 0 5 9 8P M⫽ ⫺ lb-ft 15 M⫺

Numerical value of M equals 640 lb-ft. 8P lb-ft 15 and P ⫽ 1200 lb

‹ 640 lb-ft ⫽

;

D

6 ft

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SECTION 4.3

351

Shear Forces and Bending Moments

Problem 4.3-12 A simply supported beam AB supports a trapezoidally distributed load (see figure). The intensity of the load varies linearly from 50 kN/m at support A to 25 kN/m at support B. Calculate the shear force V and bending moment M at the midpoint of the beam.

50 kN/m 25 kN/m

A

B

4m

Solution 4.3-12 FREE-BODY DIAGRAM OF SECTION CB Point C is at the midpoint of the beam.

g MB ⫽ 0:

⫺RA (4m) + (25 kN/m) (4m) (2m) 2 1 ⫹(25 kN/m)(4 m)a b a 4 m b ⫽ 0 2 3 RA ⫽ 83.33 kN g FVERT ⫽ 0: RA + RB 1 ⫺ (50 kN/m + 25 kN/m)(4 m) ⫽ 0 2 RB ⫽ 66.67 kN

g FVERT ⫽ 0: V ⫺ (25 kN/m)(2 m) ⫺ (12.5 kN/m)(2 m) V ⫽ ⫺4.17 kN gMC ⫽ 0:

1 + RB ⫽ 0 2

; ⫺ M ⫺ (25 kN/m)(2 m)(1 m) ⫺ (12.5 kN/m)(2 m)

1 1 a2 m b 2 3

+ RB (2 m) ⫽ 0 M ⫽ 75 kN ⭈ m

;

Problem 4.3-13 Beam ABCD represents a reinforced-concrete foundation beam

that supports a uniform load of intensity q1 ⫽ 3500 lb/ft (see figure). Assume that the soil pressure on the underside of the beam is uniformly distributed with intensity q2. (a) Find the shear force VB and bending moment MB at point B. (b) Find the shear force Vm and bending moment Mm at the midpoint of the beam.

q1 = 3500 lb/ft B

C

A

D

3.0 ft

q2 8.0 ft

3.0 ft

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Solution 4.3-13

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Shear Forces and Bending Moments

Foundation beam (b) V AND M AT MIDPOINT E

g FVERT ⫽ 0: ‹ q2 ⫽

q2(14 ft) ⫽ q1(8 ft)

8 q ⫽ 2000 lb/ft 14 1

g FVERT ⫽ 0:

(a) V AND M AT POINT B

Vm ⫽ 0 a FVERT

;

g ME ⫽ 0:

⫽ 0:

VB ⫽ 6000 lb g MB ⫽ 0:

Vm ⫽ (2000 lb/ft)(7 ft) ⫺ (3500 lb/ft)(4 ft)

Mm ⫽ (2000 lb/ft)(7 ft)(3.5 ft) ⫺ (3500 lb/ft)(4 ft)(2 ft)

;

Mm ⫽ 21,000 lb-ft MB ⫽ 9000 lb-ft

;

;

Problem 4.3-14 The simply-supported beam ABCD is loaded by a weight W ⫽ 27 kN through the arrangement shown in the figure. The cable passes over a small frictionless pulley at B and is attached at E to the end of the vertical arm. Calculate the axial force N, shear force V, and bending moment M at section C, which is just to the left of the vertical arm. (Note: Disregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.)

E Cable 1.5 m A

B

2.0 m

C

2.0 m

W = 27 kN

D

2.0 m

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SECTION 4.3

Solution 4.3-14

Shear Forces and Bending Moments

353

Beam with cable and weight FREE-BODY DIAGRAM OF PULLEY AT B

RA ⫽ 18 kN

RD ⫽ 9 kN

FREE-BODY DIAGRAM OF SEGMENT ABC OF BEAM

g FHORIZ ⫽ 0: g FVERT ⫽ 0: g MC ⫽ 0:

N ⫽ 21.6 kN (compression) V ⫽ 7.2 kN

;

;

M ⫽ 50.4 kN ⭈ m

;

y

Problem 4.3-15 The centrifuge shown in the figure rotates in a horizontal plane (the xy plane) on a smooth surface about the z axis (which is vertical) with an angular acceleration ␣. Each of the two arms has weight w per unit length and supports a weight W ⫽ 2.0 wL at its end. Derive formulas for the maximum shear force and maximum bending moment in the arms, assuming b ⫽ L/9 and c ⫽ L/10.

c L

b

W

x

W

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Shear Forces and Bending Moments

Solution 4.3-15

Rotating centrifuge

SUBSTITUTE NUMERICAL DATA:

Tangential acceleration ⫽ r␣ Inertial force Mr a ⫽

W ⫽ 2.0 wL b ⫽

W ra g

Maximum V and M occur at x ⫽ b. W (L + b + c)a + g Lb Wa (L + b + c) ⫽ g

L⫹b

Vmax ⫽

+ Mmax ⫽

wLa (L + 2b) 2g

;

Wa (L + b + c)(L + c) g L⫹b

+

wa x(x ⫺ b)dx g

Lb Wa (L + b + c)(L + c) ⫽ g +

wL2a (2L + 3b) 6g

;

wa x dx g

Vmax ⫽

91wL2a 30g

Mmax ⫽

229wL3a 75g

L L c⫽ 9 10 ; ;

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SECTION 4.5

355

Shear-Force and Bending-Moment Diagrams

Shear-Force and Bending-Moment Diagrams When solving the problems for Section 4.5, draw the shear-force and bending-moment diagrams approximately to scale and label all critical ordinates, including the maximum and minimum values. Probs. 4.5-1 through 4.5-10 are symbolic problems and Probs. 4.5-11 through 4.5-24 are numerical problems. The remaining problems (4.5-25 through 4.5-40) involve specialized topics, such as optimization, beams with hinges, and moving loads.

P

a

P

a

A

B

Problem 4.5-1 Draw the shear-force and bending-moment diagrams for a simple beam AB supporting two equal concentrated loads P (see figure). L

Solution 4.5-1

Simple beam

Problem 4.5-2 A simple beam AB is subjected to a counterclockwise couple of moment M0 acting at distance a from the left-hand support (see figure). Draw the shear-force and bending-moment diagrams for this beam.

M0 A

B a L

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Solution 4.5-2

Page 356

Shear Forces and Bending Moments

Simple beam

q

Problem 4.5-3 Draw the shear-force and bending-moment diagrams for

A

a cantilever beam AB carrying a uniform load of intensity q over one-half of its length (see figure).

B L — 2

Solution 4.5-3

Cantilever beam 3qL2 MA = — 8

q A B

RA =

qL 2

L — 2

—

L — 2

qL 2

—

V

0

M

0 qL2 –— 8

3qL2

–— 8

qL 2

—

L — 2

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SECTION 4.5

357

Shear-Force and Bending-Moment Diagrams

Problem 4.5-4 The cantilever beam AB shown in the figure is subjected to a concentrated load P at the midpoint and a counterclockwise couple of moment M1  PL/4 at the free end. Draw the shear-force and bending-moment diagrams for this beam.

Solution 4.5-4

PL M1 = —– 4

P

A

B L — 2

L — 2

Cantilever beam

RA  P MA 

to a concentrated load P and a clockwise couple M1  PL/3 acting at the third points. Draw the shear-force and bending-moment diagrams for this beam.

A

B L — 3

Solution 4.5-5 PL M1 = —– 3

P A

B L — 3

P RA= —– 3

L — 3

L — 3

2P RB= —– 3

P/3 V

0 Vmax = –2P/3

PL/9

Mmax = 2PL/9

M 0 –PL/9

PL M1 = —– 3

P

Problem 4.5-5 The simple beam AB shown in the figure is subjected

PL 4

L — 3

L — 3

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Shear Forces and Bending Moments

Problem 4.5-6 A simple beam AB subjected to couples M1 and 3M1

M1

acting at the third points is shown in the figure. Draw the shear-force and bending-moment diagrams for this beam.

3M1

A

B L — 3

L — 3

L — 3

Solution 4.5-6 M1

3M1

A

B L — 3

L — 3

L — 3

RA

V

RB

2M 1 L 0

7M1 3

5M 1 3

M

0

2M 1 3 2M 1 3

Problem 4.5-7 A simply supported beam ABC is loaded by a vertical load P acting at the end of a bracket BDE (see figure). Draw the shear-force and bending-moment diagrams for beam ABC.

B A

C D

E P

L — 4

L — 4

L — 2 L

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SECTION 4.5

Solution 4.5-7

Beam with bracket

P

Problem 4.5-8 A beam ABC is simply supported at A and B and has an overhang BC (see figure). The beam is loaded by two forces P and a clockwise couple of moment Pa that act through the arrangement shown. Draw the shear-force and bending-moment diagrams for beam ABC.

Solution 4.5-8

359

Shear-Force and Bending-Moment Diagrams

Beam with overhang

P

A

Pa

C

B a

a

a

a

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Shear Forces and Bending Moments

Problem 4.5-9 Beam ABCD is simply supported at B and C and has overhangs at each end (see figure). The span length is L and each overhang has length L/3. A uniform load of intensity q acts along the entire length of the beam. Draw the shear-force and bending-moment diagrams for this beam.

q A

D B L 3

Solution 4.5-9

C L 3

L

Beam with overhangs

x1  L

15  0.3727L 6

q0

Problem 4.5-10 Draw the shear-force and bending-moment diagrams for a cantilever beam AB supporting a linearly varying load of maximum intensity q0 (see figure). A B L

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SECTION 4.5

Solution 4.5-10

Shear-Force and Bending-Moment Diagrams

Cantilever beam

q0 = 10 lb/in.

Problem 4.5-11 The simple beam AB supports a triangular load of maximum intensity q0  10 lb/in. acting over one-half of the span and a concentrated load P  80 lb acting at midspan (see figure). Draw the shear-force and bending-moment diagrams for this beam.

Solution 4.5-11

361

P = 80 lb A

B L = — 40 in. 2

L = — 40 in. 2

Simple beam q0 = 10 lb/in. P = 80 lb

MA  0: RB (80 in.)  (80 lb)(40 in.)

A

1 2  (10 lb/in. )140 in.2(40 + 40 in.)  0 2 3

L = — 40 in. 2

RB  206.7 lb 1 g FVERT  0: RA + RB 80 lb  a10 lb/in. b(40 in.)  0 2 RA  73.3 lb

B L = — 40 in. 2

RA

RB 73.3 lb

V

0 –6.67 lb –207 lb 2933 lb-in

M 0

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Shear Forces and Bending Moments

Problem 4.5-12 The beam AB shown in the figure supports a uniform load

3000 N/m

of intensity 3000 N/m acting over half the length of the beam. The beam rests on a foundation that produces a uniformly distributed load over the entire length. Draw the shear-force and bending-moment diagrams for this beam.

A

B

0.8 m

Solution 4.5-12

1.6 m

0.8 m

Beam with distributed loads

200 lb

Problem 4.5-13 A cantilever beam AB supports a couple and a concentrated load, as shown in the figure. Draw the shear-force and bending-moment diagrams for this beam.

400 lb-ft A

B 5 ft

5 ft

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Solution 4.5-13

Shear-Force and Bending-Moment Diagrams

363

Cantilever beam

Problem 4.5-14 The cantilever beam AB shown in the figure is

2.0 kN/m

subjected to a triangular load acting throughout one-half of its length and a concentrated load acting at the free end. Draw the shear-force and bending-moment diagrams for this beam.

2.5 kN B

A 2m

Solution 4.5-14 4.5 kN V

2.5 kN

2.5 kN

0

M 0

0 –5 kN • m

–11.33 kN • m

2m

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Problem 4.5-15 The uniformly loaded beam ABC has simple supports

25 lb/in.

at A and B and an overhang BC (see figure). A

Draw the shear-force and bending-moment diagrams for this beam.

C B 72 in.

Solution 4.5-15

Beam with an overhang

a uniform load of intensity 12 kN/m and a concentrated moment of magnitude 3 kN # m at C (see figure). Draw the shear-force and bending-moment diagrams for this beam.

A

Beam with an overhang 3 kN • m

12 kN/m

C

B 1.6 m

1.6 m

RA

1.6 m RB

15.34 kN V

0 kN

0 –3.86 kN max 9.80 kN • m 9.18 kN • m

3 kN • m M 0 1.28 m

C

B 1.6 m

A

3 kN • m

12 kN/m

Problem 4.5-16 A beam ABC with an overhang at one end supports

Solution 4.5-16

48 in.

1.6 m

1.6 m

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365

Shear-Force and Bending-Moment Diagrams

Problem 4.5-17 Consider the two beams below; they are loaded the same but have different support conditions. Which beam has the larger maximum moment? First, find support reactions, then plot axial force (N), shear (V) and moment (M) diagrams for both beams. Label all critical N, V & M values and also the distance to points where N, V &/or M is zero. PL L — 2

A

B

L — 2

L — 4

C

P

4 L — 4 3 D

PL Ay

Ax

Cy (a) PL

A

L — 2

B

L — 2

L — 4

P

4 L — 4

3

PL Cy

Dy

Dx

(b)

Solution 4.5-17 BEAM (a): g MA  0: Cy 

0

N 0

1 4 5 a P Lb  P (upward) L 5 4

g FV  0: Ay 

4 P P  Cy   (downward) 5 5

g FH  0: Ax 

3 P (right) 5

–3P/5(compression) 4P/5 0

V 0 –P/5

0

M 0 –PL/10 –11PL/10

–PL/5 –6PL/5

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BEAM (b):

3P/5

g MD  0: Cy 

2 2 4 1 a P Lb  P (upward) L 5 4 5

g FV  0: Dy 

4 2 P  Cy  P (upward) 5 5

N

0

2P/5 V

0

3 g FH  0: Dx  P (right) 5

–2P/5

⬖ The first case has the larger maximum moment 6 a PL b 5

PL/10 M

0

; –PL

Problem 4.5-18 The three beams below are loaded the same and have the same support conditions. However, one has a moment release just to the left of C, the second has a shear release just to the right of C, and the third has an axial release just to the left of C. Which beam has the largest maximum moment? First, find support reactions, then plot axial force (N), shear (V) and moment (M) diagrams for all three beams. Label all critical N, V & M values and also the distance to points where N, V &/or M is zero. PL at C A

L — 2

B

L — 2

L — 4

C

P 3

4 L — 4 D

PL at B Moment release

Ax

Ay

Cy

Dy

(a) PL at C A

L — 2

B

L — 2

L — 4

C

PL at B Ax

Ay

P

4 L — 4 D 3

Shear release

Cy

Dv

(b) PL at C A

L — 2

B PL at B

Ay

Ax

L — 2

Axial force release

(c)

C

L P 4 L — — 4 4 3

Cx Cy

0

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367

Shear-Force and Bending-Moment Diagrams

Solution 4.5-18 BEAM (a): MOMENT RELEASE

N

0

0

Ay  P (upward) Cy   Dy 

–3P/5 (compression)

13 P (downward) 5

12 P (upward) 5

P V

0

–8P/5

3 Ax  P (right) 5

PL/2 M

PL

–12P/5 3PL/5

0

–PL/2

BEAM (b): SHEAR RELEASE Ay 

N

0

0

1 P (upward) 5

1 Cy   P (downward) 5 Dy 

4 P (upward) 5

Ax 

3 P (right) 5

–3P/5 (compression)

P/5 V

0

–4P/5

M

PL/10

0

–9PL/10

BEAM (c): AXIAL RELEASE

N

–3P/5 (compression) 4P/5 V

0

M

0

Ax  0 Cx 

3 P (right) 5

⬖ The third case has the largest maximum moment 6 a PLb 5

;

–4PL/5

0

1 Ay   P (downward) 5 Cy  P (upward)

PL/5

–P/5

–PL/10 –11PL/10

–PL/5 –6PL/5

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Problem 4.5-19 A beam ABCD shown in the figure is simply supported at A and B and has an overhang from B to C. The loads consist of a horizonatal force P1 400 lb acting at the end of the vertical arm and a vertical force P2  900 lb acting at the end of the overhang. Draw the shear-force and bending-moment diagrams for this beam. (Note: Disregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.)

Solution 4.5-19

Beam with vertical arm

Problem 4.5-20 A simple beam AB is loaded by two segments of uniform load and two horizontal forces acting at the ends of a vertical arm (see figure). Draw the shear-force and bending-moment diagrams for this beam.

Solution 4.5-20

Simple beam

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369

Shear-Force and Bending-Moment Diagrams

Problem 4.5-21 The two beams below are loaded the same and have the same support conditions. However, the location of internal axial, shear and moment releases is different for each beam (see figures). Which beam has the larger maximum moment? First, find support reactions, then plot axial force (N), shear (V) and moment (M) diagrams for both beams. Label all critical N, V & M values and also the distance to points where N, V &/or M is zero.

MAz

PL

A

L — 2

B

L — 2

L — 4

C

P 3

4 L — 4 D

Ax PL Axial force release

Ay

Shear release

Moment release

Cy

Dy

Dx

(a) MAz

PL

A

L — 2

B

L — 2

L — 4

C

P 3

4 L — 4 D

Ax PL Ay

Shear release

Axial force release

Moment release

Cy

Dy

Dx

(b)

Solution 4.5-21 Support reactions for both beams: MAz  0, Ax  0, Ay  0 Cy  Dx 

3P/5(tension) N

0

2 2 P ( upward), Dy  P ( upward) 5 5 3 P ( rightward) 5

2P/ 5 V 0 –2P/ 5

⬖ These two cases have the same maximum moment (PL) ; (Both beams have the same N, V and M diagrams)

–PL/10

M 0 –PL

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Shear Forces and Bending Moments

Problem 4.5-22 The beam ABCD shown in the figure has overhangs that extend in both directions for a distance of 4.2 m from the supports at B and C, whiich are 1.2 m apart. Draw the shear-force and bending-moment diagrams for this overhanging beam.

10.6 kN/m 5.1 kN/m

5.1 kN/m

A

D B

C

4.2 m

4.2 m 1.2 m

Solution 4.5-22

Beam with overhangs

Problem 4.5-23 A beam ABCD with a vertical arm CE is supported as a simple beam at A and B (see figure). A cable passes over a small pulley that is attached to the arm at E. One end of the cable is attached to the beam at point B. The tensile force in the cable is 1800 lb. Draw the shear-force and bending-moment diagrams for beam ABCD. (Note: Disregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.)

E

1800 lb

Cable A

B

6 ft

8 ft C

6 ft

D

6 ft

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SECTION 4.5

Solution 4.5-23

Shear-Force and Bending-Moment Diagrams

371

Beam with a cable

Note: All forces have units of pounds.

Problem 4.5-24 Beams ABC and CD are supported

MAz

at A, C and D, and are joined by a hinge (or moment release) just to the left of C and a shear release just to the right of C. The support at A is a sliding support (hence reaction Ay  0 for the loading shown below). Find all support reactions then plot shear (V) and moment (M) diagrams. Label all critical V & M values and also the distance to points where either V &/or M is zero.

W0 = P/L A

L — 2

L — B 2

C

Ax

L — 2

PL Ay

Sliding support

Moment release

Cy

Dy

Solution 4.5-24 MAz  PL (clockwise), Ax  0, Ay  0

;

1 1 P (upward), Dy  P (upward) 12 6

;

Cy 

VMAX 

P MMAX  PL 6

P/12 V 0 0.289L –P/6

PL

; M 0

0.016PL

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Shear Forces and Bending Moments

Problem 4.5-25 The simple beam AB shown in the figure supports a

5k

concentrated load and a segment of uniform load. Draw the shear-force and bending-moment diagrams for this beam.

2.0 k/ft C

A 5 ft

B 10 ft

20 ft

Solution 4.5-25

Simple beam 5k

2.0 k/ft C

A

B

5 ft

10 ft 20 ft 6.25

1.25 V 0 (Kips)

3.125 –13.75 37.5

M 0 (k-ft)

max 47.3

6.25

Problem 4.5-26 The cantilever beam shown in the figure supports a concentrated load and a segment of uniform load. Draw the shear-force and bending-moment diagrams for this cantilever beam.

3 kN

1.0 kN/m

A

0.8 m

B 0.8 m

1.6 m

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SECTION 4.5

Solution 4.5-26

Shear-Force and Bending-Moment Diagrams

373

Cantilever beam

Problem 4.5-27 The simple beam ACB shown in the figure is subjected to

180 lb/ft

a triangular load of maximum intensity 180 lb/ft and a concentrated moment of 300 lb-ft at A. 300 lb-ft Draw the shear-force and bending-moment diagrams for this beam. A

B C 6.0 ft 7.0 ft

Solution 4.5-27

Simple beam 180 lb/ft 300 lb-ft A

B C 6.0 ft 7.0 ft

197.1 V 0 (lb)

0 lb 3.625 ft max 776

300 M 0 (lb-ft)

–343 –433 403

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Shear Forces and Bending Moments

Problem 4.5-28 A beam with simple supports is subjected to a trapezoidally

3.0 kN/m 1.0 kN/m

distributed load (see figure). The intensity of the load varies from 1.0 kN/m at support A to 3.0 kN/m at support B. Draw the shear-force and bending-moment diagrams for this beam. A

B

2.4 m

Solution 4.5-28

V  2.0  x  Set V  0:

Simple beam

x2 2.4

(x  meters; V  kN)

x1  1.2980 m

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SECTION 4.5

Problem 4.5-29 A beam of length L is being designed to support a uniform load

q

of intensity q (see figure). If the supports of the beam are placed at the ends, creating a simple beam, the maximum bending moment in the beam is ql 2/8. However, if the supports of the beam are moved symmetrically toward the middle of the beam (as pictured), the maximum bending moment is reduced. Determine the distance a between the supports so that the maximum bending moment in the beam has the smallest possible numerical value. Draw the shear-force and bending-moment diagrams for this condition.

Solution 4.5-29

A

Beam with overhangs

M1  M2  

a  (2  22) L  0.5858L q (L  a)2 8

qL2 (3  222)  0.02145qL2 8

The maximum bending moment is smallest when M1  M2 (numerically). q(L  a)2 8

qL2 qL a  (2a  L) M2  RA a b  2 8 8 M1  M2

B a L

Solve for a:

M1 

375

Shear-Force and Bending-Moment Diagrams

(L  a)2  L(2a  L)

x1  0.3536 a  0.2071 L

;

;

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Problem 4.5-30 The compound beam ABCDE shown in the figure consists of two beams (AD and DE) joined by a hinged connection at D. The hinge can transmit a shear force but not a bending moment. The loads on the beam consist of a 4-kN force at the end of a bracket attached at point B and a 2-kN force at the midpoint of beam DE. Draw the shear-force and bending-moment diagrams for this compound beam.

4 kN

1m

B

C

1m

A

E

2m

Solution 4.5-30

2 kN D

2m

2m

2m

Compound beam

Problem 4.5-31 The beam shown below has a sliding support at A and an elastic support with spring constant k at B. A distributed load q(x) is applied over the entire beam. Find all support reactions, then plot shear (V) and moment (M) diagrams for beam AB; label all critical V & M values and also the distance to points where any critical ordinates are zero.

MA

y )

A Ax

q(x Linear

q0 B x

L k

By

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SECTION 4.5

Shear-Force and Bending-Moment Diagrams

377

Solution 4.5-31 MA   By 

q0 2 L (clockwise), Ax  0 6

q0 L (upward) 2

V

;

0

–q0L/2

; L2/6

q0 M

Problem 4.5-32 The shear-force diagram for a simple beam is shown in the figure. Determine the loading on the beam and draw the bendingmoment diagram, assuming that no couples act as loads on the beam.

0

12 kN V 0 –12 kN 2.0 m

Solution 4.5-32

Simple beam (V is given)

1.0 m

1.0 m

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Shear Forces and Bending Moments

Problem 4.5-33 The shear-force diagram for a beam is shown in the figure. Assuming that no couples act as loads on the beam, determine the forces acting on the beam and draw the bendingmoment diagram.

652 lb

580 lb

572 lb

500 lb

V 0 –128 lb –448 lb 4 ft

Solution 4.5-33

16 ft

4 ft

Forces on a beam (V is given)

FORCE DIAGRAM

Problem 4.5-34 The compound beam below has an internal moment release just to the left of B and a shear release just to the right of C. Reactions have been computed at A, C and D and are shown in the figure. First, confirm the reaction expressions using statics, then plot shear (V) and moment (M) diagrams. Label all critical V and M values and also the distance to points where either V and/or M is zero.

w0 L2 MA = –––– 12

w0

w0 A

B

L L Ax = 0 — — 2 Moment 2 release w0 L w0 L Ay = –––– Cy = –––– 6 3

C

D L — 2 Shear release –w0 L Dy = –––– 4

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SECTION 4.5

Shear-Force and Bending-Moment Diagrams

379

Solution 4.5-34 FREE-BODY DIAGRAM w 0 L2 –––– 12

w0

0 w0 L –––– 6 w0 L –––– 6 V

w0 L –––– 6

w0 L –––– 6

w0

w0 L2 –––– 24 w0 L2 –––– 24

w0 L –––– 3

w0 L –––– 4 –w0 L –––– 4

0 –w0 L –––– 3

L 6

––––

w0 L2 –––– 72 M 0 L2

–w0 –––– 12

L –––– 3

–w0 L2 –––– 24

Problem 4.5-35 The compound beam below has an shear MA release just to the left of C and a moment release just to the right of C. A plot of the moment diagram is provided below for applied load P at B and triangular distributed loads w(x) on segments BC and CD. Ax First, solve for reactions using statics, then plot axial Ay force (N) and shear (V) diagrams. Confirm that the moment diagram is that shown below. Label all critical N and V & M values and also the distance to points where N, V &/or M is zero.

w0 A

w0

B L — 2 w0 L P = –––– 2

4 3

C

D

L — 2 Shear release

Cy

L — 2 Moment release Dy

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Shear Forces and Bending Moments

Solution 4.5-35 Solve for reactions using statics MA  

7w0 2 L (clockwise), 30

Ax  

3 w L (left) 10 0

Ay  

3 w0L (downward) 20

;

;

Cy 

w0 L (upward) 12

;

Dy 

w0 L (upward) 6

;

;

FREE-BODY DIAGRAM

–7w0L2/60

3w0L/10 3w0L/20

w0L2/24 w0L2/24

w0L/2

w0L/4

w0

w0

w0L/4

w0L/12

3w0L/10 (tension) N 0

w0L/4 w0L/12 V

0 0.289L

–3w0L/20 7w0L2/60

2w0L2/125

M 0 –w0L2/24

–w0L/6

w0L/6

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SECTION 4.5

Shear-Force and Bending-Moment Diagrams

Problem 4.5-36 A simple beam AB supports two connected wheel loads P and 2P that are distance d apart (see figure). The wheels may be placed at any distance x from the left-hand support of the beam.

P

Solution 4.5-36

2P

x

(a) Determine the distance x that will produce the maximum shear force in the beam, and also determine the maximum shear force Vmax. (b) Determine the distance x that will produce the maximum bending moment in the beam, and also draw the corresponding bendingmoment diagram. (Assume P  10 kN, d  2.4 m, and L  12 m.)

381

d

A

B

L

Moving loads on a beam P  10 kN d  2.4 m L  12 m

Reaction at support B: 2P P P x + (x + d)  (2d + 3x) L L L Bending moment at D: RB 

MD  RB (L  x  d)

(a) MAXIMUM SHEAR FORCE By inspection, the maximum shear force occurs at support B when the larger load is placed close to, but not directly over, that support.



P (2d + 3x) (L  x  d) L



P [3x2 + (3L  5d)x + 2d(L  d)] L

dMD P  (6x + 3L  5d)  0 dx L x

Solve for x:

L 5d a3  b  4.0 m 6 L

;

Substitute x into Eq (1): L 2 P 5d 2 c 3a b a 3  b + (3L  5d) L 6 L

Mmax  x  L  d  9.6 m

;

d Vmax  RB  P a3  b  28 kN L

5d L b + 2d(L  d) d * a b a3  6 L

; 

(b) MAXIMUM BENDING MOMENT By inspection, the maximum bending moment occurs at point D, under the larger load 2P.

Note:

PL d 2 a3  b  78.4 kN # m 12 L

RA 

P d a 3 + b  16 kN 2 L

RB 

P d a 3  b  14 kN 2 L

;

Eq.(1)

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Shear Forces and Bending Moments

Problem 4.5-37 The inclined beam below represents the loads applied to a ladder: the weight (W) of the house painter and the self weight (w) of the ladder itself. Find support reactions at A and B, then plot axial force (N), shear (V) and moment (M) diagrams. Label all critical N, V & M values and also the distance to points where any critical ordinates are zero. Plot N, V & M diagrams normal to the inclined ladder.

θ

B

Bx

6f t

W = 150 lb

=2 .5 lb/ ft w

18

ft

θ

θ θ

A

8 ft Ax

θ Ay

θ

Bx sin θ = 47.5 lb

sin

–47.5 lb

–16.79 lb

–30.98 lb

W

W cos θ = 50 lb

Bx cos θ = –16.79 lb

B

θ=

14

1.4

lb

Solution 4.5-37

7.5 lb

–42.5 lb

Ax cos θ + Ay sin θ = 214.8 lb

=0 os θ

wc

ws

in

θ=

2.3 57

.83

lb/

ft

3l b/f

t

–172.4 lb

270 lb·ft

N

A Ay cos θ – Ax sin θ = 22.5 lb

22.5 lb V –214.8 lb

M

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SECTION 4.5

cos u 

8 1 2 12  , sin u  18 + 6 3 3

383

Shear-Force and Bending-Moment Diagrams

(2) Use u to find forces at ends A & B which are along and perpendicular to member AB (see free-body diagram); also resolve forces W and w into components along & perpendicular to member AB

Solution procedure:

(3) Starting at end A, plot N, V and M diagrams (see plots)

(1) Use statics to find reaction forces at A & B g FV  0: Ay  150  2.5 (18  6)  210 lb Ay  210 lb (upward) g MA  0: Bx # 24 sin   150

#

;

6  2.5

Bx  50.38 lb (left) g FH  0; Ax  50.38 lb (right)

# 24 # 4  0

; ;

Problem 4.5-38 Beam ABC is supported by

MD

a tie rod CD as shown (see Prob. 10.4-9). Two configurations are possible: pin support at A and downward triangular load on AB, or pin at B and upward load on AB. Which has the larger maximum moment? First, find all support reactions, then plot axial force (N), shear (V) and moment (M) diagrams for ABC only and label all critical N, V & M values. Label the distance to points where any critical ordinates are zero.

Dy Dx

D Moment releases

q0 at B

y

L — 4

x) ear q(

Lin Ax

A

L

B

PL

(a)

Solution 4.5-38 4q0L/9 q0L/2

7q0L2/9

7q0L2/9 q0L/2

q0L/2 q0L/2

q0L/2 4q0L/9

q0L

q0L2

17q0L/18 4q0L/9

q0L/2

4q0L/9

x

C

L — 2

Ay

FREE-BODY DIAGRAM—BEAM (a)

L — 4P=q L 0

4q0L/9

12:43 PM

Shear Forces and Bending Moments

Use statics to find reactions at A and D for Beam (a)

;

17 Ay  q L (upward) 18 0

4 Dy   q0L (downward) 9

;

MD  0

;

–4q0L/9

D

q0L/2(tension) N

0 A

C 0

B

17q0L/18 V

4q0L/9

0

0

q0L2

7q0L2/9 M

;

(compression)

1 Ax   q0L (left) 2

1 Dx   q0L (left) 2

–q0L/2

CHAPTER 4

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q0L/2

384

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q0L2/4

04Ch04.qxd

0

0

MD

Dy Dx

D Moment releases

q0 at B

y

x) ear q(

Lin A

L — 4 P=q L 0 L — 4

B L

L — 2 By (b)

Bx

x

C PL

;

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SECTION 4.5

Shear-Force and Bending-Moment Diagrams

FREE-BODY DIAGRAM—BEAM (b) 5q0L/3 q0L/2

q0L2/6

q0L2/6

q0L2 q0L/2 q0L/2

q0L/2 5q0L/3

q0L/2

q0L

5q0L/3

5q0L/3

Use statics to find reactions at B and D for Beam (b) 1 q L (right) 2 0

;

1 5 7 By   q0L + q0L  q0L (upward) 2 3 6 1 q L (right) 2 0

;

5 Dy   q0L (downward) 3

; D

N

0

B

A

C 0

q0L/2 (compression) 5q0L/3 q0L/2 V

0

0

q0L2 q0L2/6 M 0

–5q0L/3 (compression)

MD  0

;

q0L/2

Dx 

;

–q0L/2

Bx 

–q0L2/4

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Shear Forces and Bending Moments

Problem 4.5-39 The plane frame below consists of column AB and beam BC which carries a triangular distributed load. Support A is fixed and there is a roller support at C. Column AB has a moment release just below joint B. Find support reactions at A and C, then plot axial force (N), shear (V) and moment (M) diagrams for both members. Label all critical N, V & M values and also the distance to points where any critical ordinates are zero.

q0

B

C L Moment release RCy 2L

A RAx

RAy MA

Solution 4.5-39 Use statics to find reactions at A and C MA  0

;

q0 L (upward) 6

q0 L (upward) 3

RAx  0

; B

B

;

;

B 0

B N –3w0L/20 (compression)

RAy 

RCy 

0

C

w0L/6 0

0

V

0 0.5774L

8w0L2/125 A

A 0

0

A 0

N

V

M

M 0

–w0L/3

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SECTION 4.5

Shear-Force and Bending-Moment Diagrams

Problem 4.5-40 The plane frame shown below is part of an elevated freeway system. Supports at A and D are fixed but there are moment releases at the base of both columns (AB and DE), as well near in column BC and at the end of beam BE. Find all support reactions, then plot axial force (N), shear (V) and moment (M) diagrams for all beam and column members. Label all critical N, V & M values and also the distance to points where any critical ordinates are zero.

387

750 N/m C F

45 kN

Moment release

7m

1500 N/m E B

18 kN

7m

19 m A

D Dx

Ax MA

Ay

MD

Dy

Solution 4.5-40 Solution procedure:

(4) g MB  0 for AB: Ax  0

(1) MA  MD  0 due to moment releases (2) g MA  0: Dy  61,164 N  61.2 kN

(5) g FH  0: Dx  63 kN (6) Draw separate FBD’s of each member (see below) to find N, V and M for each member; plot diagrams (see below)

(3) g Fy  0: Ay  18,414 N  18.41kN

756 kN·m 756 kN·m

FREE-BODY DIAGRAM 750 N/m

C C 32.7 kN

1500 N/m B 32.7 kN B 18.41 kN B

A 18.41 kN

14.25 kN

E

45 kN

45 kN

46.9 kN 61.2 kN

E 14.25 kN

46.9 kN F

F 46.9 kN 441kN·m

32.7 kN

441kN·m

04Ch04.qxd

E

63 kN

D 63 kN

61.2 kN

12:43 PM

Shear Forces and Bending Moments

0 C

–46.9 kN

F

B

E

A

D

–61.2 kN

0

AXIAL FORCE DIAGRAM. () COMPRESSION F

C –32.7 kN

–46.9 kN

45 kN

0

14.25 kN

E –14.25 kN

B

63 kN

0

A

D

SHEAR FORCE DIAGRAM. 756 kN·m F

C 0 0

756 kN·m 67.7 kN·m

B

E

0

CHAPTER 4

Page 388

32.7 kN

388

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18.41 kN

04Ch04.qxd

D

A

BENDING MOMENT DIAGRAM

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5 Stresses in Beams (Basic Topics) d

Longitudinal Strains in Beams Problem 5.4-1 Determine the maximum normal strain âmax produced in a steel wire of diameter d  1/16 in. when it is bent around a cylindrical drum of radius R  24 in. (see figure).

Solution 5.4-1

R

Steel wire R  24 in.

d

1 in. 16

From Eq. (5-4): y âmax  r 

Substitute numerical values: âmax 

1/16 in.  1300 * 106 2(24 in.) + 1/16 in.

d/2 d  R + d/2 2R + d d = diameter

Problem 5.4-2 A copper wire having diameter d  3 mm is bent into a circle and held with the ends just touching (see figure). If the maximum permissible strain in the copper is âmax  0.0024, what is the shortest length L of wire that can be used?

Solution 5.4-2

;

L = length

Copper wire d  3 mm âmax  0.0024 L  2pr r 

L 2p

From Eq. (5-4): âmax 

y d/2 pd   r L/2p L

Lmin 

p(3 mm) pd   3.93 m âmax 0.0024

;

389

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Problem 5.4-3 A 4.5 in. outside diameter polyethylene pipe designed to carry chemical wastes is placed in a trench and bent around a quarter-circular 90° bend (see figure). The bent section of the pipe is 46 ft long. Determine the maximum compressive strain âmax in the pipe. 90°

Solution 5.4-3

Polyethylene pipe Angle equals 90° or p/2 radians, L  length of 90° bend L  46 ft  552 in. d  4.5 in. 2pr pr L  4 2

r  r  radius of curvature r

2L L  p p/2

âmax 

y d/2  r 2L/p

âmax 

pd p 4.5 in.  a b  6400 * 106 4L 4 552 in.

Problem 5.4-4 A cantilever beam AB is loaded by a couple M0 at its free end (see figure.) The length of the beam is L  1.5 m and the longitudinal normal strain at the top surface is 0.001. The distance from the top surface of the beam to the neutral surface is 75 mm. Calculate the radius of curvature r, the curvature k, and the vertical deflection d at the end of the beam.

;

d

A B

M0

L

Solution 5.4-4 Deflection: constant curvature for pure bending so gives a circular arc; assume flat deflection curve (small defl.) so BC  L

NUMERICAL DATA L  2.0 m

âmax  0.0012

c  82.5 mm RADIUS OF CURVATURE c r r  68.8 m âmax

sin(u) 

1 r

L u  asina b r

; u  0.029 radians

CURVATURE k

L r

k  1.455 * 105 m1

;

L  0.029 r

1  cos(u)  4.232 * 104 d  r (1  cos(u))

r  6.875 * 104 mm

d  29.1 mm

;

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SECTION 5.4

Problem 5.4-5 A thin strip of steel of length L  20 in. and thickness

391

Longitudinal Strains in Beams

M0

t  0.2 in. is bent by couples M0 (see figure). The deflection at the midpoint of the strip (measured from a line joining its end points) is found to be 0.20 in. Determine the longitudinal normal strain â at the top surface of the strip.

M0

t

d L — 2

L — 2

Solution 5.4-5 NUMERICAL DATA L  28 inches

solving for r:

t  0.25 inches

r

d  0.20 inches LONGITUDINAL NORMAL STRAIN AT TOP SURFACE t 2 t â â r 2r

d  r (1  cos(u))

L 2 sin(u)  r

L sin(u)  2r

L 2r

d  r a 1  cosa

1  cos a r

L b 2r 0.20

1cos a

14 b r

numerical solution for radius of curvature r gives r  489.719 inches strain at top (compressive): t â  2.552 * 104 â 2r â  255 11062

assume angle is small so that u

insert numerical data:

d

;

L bb 2r

Problem 5.4-6 A bar of rectangular cross section is loaded and supported as shown in the figure. The distance between supports is L  1.5 m and the height of the bar is h  120 mm. The deflection at the midpoint is measured as 3.0 mm. What is the maximum normal strain â at the top and bottom of the bar?

h P

d P

a

L — 2

L — 2

a

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Solution 5.4-6 NUMERICAL DATA

d  r a1  cosa

h  120 mm

L  1.5 m d  3.0 mm

‹ r a 1  cos a

NORMAL STRAIN AT TOP OF BAR: h 2 â r

h 2r

â

L 2 sin(u)  r

u

L b b d  0 2r

numerical solution for radius of curvature r gives r  93.749 m

tensile strain, r  radius of curvature

SMALL DEFLECTION SO SMALL ANGLE

L bb 2r

strain at top (compressive): h â â  640 * 106 2r

;

u

L 2r

Normal Stresses in Beams Problem 5.5-1 A thin strip of hard copper (E  16,000 ksi) having length L  90 in. and thickness t  3/32 in. is bent into a circle and held with the ends just touching (see figure).

3 t = — in. 32

(a) Calculate the maximum bending stress smax in the strip. (b) By what percent does the stress increase or decrease if the thickness of the strip is increased by 1/32 in.?

Solution 5.5-1 smaxnew  69.813 ksi

(a) MAXIMUM BENDING STREES E  16000 ksi

L  90 inches

t 2 sEP Q r

L r 2p

r  14.324 inches

smax  52.4 ksi

smax  ;

(b) % CHANGE IN STRESS tnew 

4 32

smaxnew 

Etnew 2r

Et 2r

t

3 inches 32

smaxnew  smax (100)  33.3 smax

;

33% increase (linear) in max.stress due to increase in t; same as % increase in thickness t 3 4  32 32 (100)  33.3 3 32

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393

Normal Stresses in Beams

Problem 5.5-2 A steel wire (E  200 GPa) of diameter d  1.25 mm is bent around a pulley of radius R0  500 mm (see figure). (a) What is the maximum stress smax in the wire? (b) By what percent does the stress increase or decrease if the radius of the pulley is increased by 25%? R0 d

Solution 5.5-2 (a) MAX. NORMAL STRESS IN WIRE d  1.25 mm

E  200 GPa d 2 s r

E

E

smax 

smax  250 MPa

R0  500 mm

(b) % CHANGE IN MAX. STRESS DUE TO INCREASE IN PULLEY RADIUS BY 25%

d 2

E snew 

d R0 + 2

d 2 d 2

1.25 R0 +

snew  199.8 MPa

snew  smax (100)  20% smax

;

;

L = length

Problem 5.5-3 A thin, high-strength steel rule (E  30  106 psi)

having thickness t  0.175 in. and length L  48 in. is bent by couples M0 into a circular are subtending a central angle a  40° (see figure).

t M0

M0

(a) What is the maximum bending stress smax in the rule? (b) By what percent does the stress increase or decrease if the central angle is increased by 10%?

a

Solution 5.5-3 (b) % CHANGE IN STRESS DUE TO 10% INCREASE IN ANGLE a

(a) MAX. BENDING STRESS a  40 a

p b 180

L  48 inches r

L a

a  0.698 radians t  0.175 in.

E  30 (106) psi

r  68.755 inches

t 2 Et smax  smax  r 2r smax  38.2 ksi ; E

snew 

E t (1.1a) 2L

snew  smax (100)  10% smax linear increase (%)

Eta smax  2L

snew  41997 psi ;

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Problem 5.5-4 A simply supported wood beam AB with span length

q

L  4 m carries a uniform load of intensity q  5.8 kN/m (see figure). (a) Calculate the maximum bending stress smax due to the load q if the beam has a rectangular cross section with width b  140 mm and height h  240 mm. (b) Repeat (a) but use the trapezoidal distuibuted load shown in the figure part (b).

A

h

B

b

L (a) q — 2

q

A

B

L (b)

Solution 5.5-4 (a) MAX. BENDING STRESS DUE TO UNIFORM LOAD q 2

qL 8

Mmax 

S

(b)

MAX. BENDING STRESS DUE TO TRAPEZOIDAL LOAD

RA  c

I h 2

uniform load (q/2) & triang. load (q/2)

3

bh 12 S h 2

smax 

RA 

1 S  bh2 6

L4m

qL2 8

Mmax  11.6 kN # m smax  8.63 MPa

b  140 mm

q 1 x q x a bx  0 2 2 L2

3x2 + 6Lx  4L2  0 6 L  1184 L22 2(3)

x 1  a 1 + 184b L 6 xmax  0.52753 L

h  240 mm Mmax 

RA 

x

3 L2 smax  q 2 4 bh kN q  5.8 m

1 qL 3

find x  location of zero shear

qL2 8 smax  1 a bh2 b 6

Mmax S

1 q 1 q1 a bL + a b Ld 2 2 3 22

;

q

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SECTION 5.5

Mmax  RAxmax 

q xmax2 1 xmax q xmax2  a b 2 2 2 L 2 3

Mmax  9.40376 * 102 qL2 Mmax  8.727 kN # m

smax 

Normal Stresses in Beams

395

Mmax S

smax  6.493 * 103

N m2

smax  6.49 MPa

;

Problem 5.5-5 Each girder of the lift bridge (see figure) is 180 ft long and simply supported at the ends. The design load for each girder is a uniform load of intensity 1.6 k/ft. The girders are fabricated by welding tree steel plates so as to form an I-shaped cross section (see figure) having section modulus S  3600 in.3. What is the maximum bending stress smax in a girder due to the uniform load?

Solution 5.5-5

Bridge girder L  180 ft

q  1.6 k/ft

S  3600 in.

3

Mmax 

qL2 8

smax 

qL2 Mmax  S 8S

smax 

(1.6 k/ft)(180 ft)2(12 in./ft) 8(3600 in.3)

 21.6 ksi

;

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Problem 5.5-6 A freight-car axle AB is loaded approximately as shown in the figure, with the forces P representing the car loads (transmitted to the axle through the axle boxes) and the forces R representing the rail loads (transmitted to the axle through the wheels). The diameter of the axle is d  80 mm, the distance between centers of the rails is L, and the distance between the forces P and is R is b  200 mm. Calculate the maximum bending stress smax in the axle if P  47 kN.

P

P B

A

d

d R b

R L

b

Solution 5.5-6 NUMERICAL DATA d  82 mm

MAX. BENDING STRESS

b  220 mm

P  50 kN I

pd4 64

I  2.219 * 106 m4

Mmax  Pb

smax 

Md 2I

smax  203 MPa

;

Mmax  11 kN # m

Problem 5.5-7 A seesaw weighing 3 lb/ft of length is occupied by two children, each weighing 90 lb (see figure). The center of gravity of each child is 8 ft from the fulcrum. The board is 19 ft long, 8 in. wide, and 1.5 in. thick. What is the maximum bending stress in the board?

Solution 5.5-7

Seesaw b  8 in.

h  1.5 in.

q  3 lb/ft

P  90 lb

d  8.0 ft

L  9.5 ft

2

Mmax  Pd +

qL  720 lb-ft + 135.4 lb-ft 2  855.4 lb-ft  10,264 lb-in.

S

2

bh  3.0 in.3. 6

smax 

10,264 lb-in. M   3420 psi S 3.0 in.3

;

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SECTION 5.5

397

Normal Stresses in Beams

Problem 5.5-8 During construction of a highway bridge, the main girders are cantilevered outward from one pier toward the next (see figure). Each girder has a cantilever length of 48 m and an I-shaped cross section with dimensions shown in the figure. The load on each girder (during construction) is assumed to be 9.5 kN/m, which includes the weight of the girder. Determine the maximum bending stress in a girder due to this load.

52 mm

2600 mm 28 mm

620 mm

Solution 5.5-8 NUMERICAL DATA tf  52 mm h  2600 mm L  48 m I

tw  28 mm bf  620 mm q  9.5

kN m

L Mmax  qL a b 2 Mmax h smax  2I

Mmax  1.094 * 104 kN m

smax  101 MPa

;

1 1 (b ) h3  (b  tw) [ h  2 (tf)]3 12 f 12 f

I  1.41 * 1011 mm4

Problem 5.5-9 The horizontal beam ABC of an oil-well pump has the cross section shown in the figure. If the vertical pumping force acting at end C is 9 k and if the distance from the line of action of that force to point B is 16 ft, what is the maximum bending stress in the beam due to the pumping force?

Horizontal beam transfers loads as part of oil well pump C

B

A

0.875 in.

22 in.

0.625 in.

8.0 in.

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Solution 5.5-9 NUMERICAL DATA FC  9 k

MAX. BENDING STRESS AT B

BC  16 ft

Mmax  F C (BC)

Mmax  144 k-ft

smax 

1 1 I (8) (22)3  (8  0.625) 12 12 * [22  2 (0.875)]

;

4

P a

P a

L

b h

q

Railroad tie (or sleeper)

P  175 kN

b  300 mm

L  1500 mm q

I

smax  9.53 ksi 3

Problem 5.5-10 A railroad tie (or sleeper) is subjected to two rail loads, each of magnitude P  175 kN, acting as shown in the figure. The reaction q of the ballast is assumed to be uniformly distributed over the length of the tie, which has cross-sectional dimensions b  300 mm and h  250 mm. Calculate the maximum bending stress smax in the tie due to the loads P, assuming the distance L  1500 mm and the overhang length a  500 mm.

DATA

22 b 2

I  1.995 * 10 in.

3

Solution 5.5-10

Mmax (12) a

2P L + 2a

h  250 mm

a  500 mm S

bh2  3.125 * 103 m3 6

Substitute numerical values: M1  17,500 N # m

M2  21,875 N # m

Mmax  21,875 N # m MAXIMUM BENDING STRESS

BENDING-MOMENT DIAGRAM

smax 

21,875 N # m Mmax  7.0 MPa  5 3.125 * 103 m3

(Tension on top; compression on bottom)

M1 

qa2 Pa2  2 L + 2a

M2 

2 q L PL a + ab  2 2 2



2 L PL P a + ab  L + 2a 2 2



P (2a  L) 4

;

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Normal Stresses in Beams

399

Problem 5.5-11 A fiberglass pipe is lifted by a sling, as shown in the figure. The outer diameter of the pipe is 6.0 in., its thickness is 0.25 in., and its weight density is 0.053 lb/in.3 The length of the pipe is L  36 ft and the distance between lifting points is s  11 ft. Determine the maximum bending stress in the pipe due to its own weight. s L

Solution 5.5-11

Pipe lifted by a sling

d2  6.0 in.

L  36 ft  432 in. s  11 ft  132 in. g  0.053 lb/in.3

t  0.25 in.

d1  d2  2t  5.5 in. p A  (d22  d21)  4.5160 in.2 4

a  (L  s)/2  150 in. BENDING-MOMENT DIAGRAM

I

p 4 (d  d14)  18.699 in.4 64 2

q  gA  (0.053 lb/in.3)(4.5160 in.2)  0.23935 lb/in. MAXIMUM BENDING STRESS smax  smax 

Mmax c I

M2  

qL L a  sb  2,171.4 lb-in. 4 2

Mmax  2,692.7 lb-in.

d2  3.0 in. 2

(2,692.7 lb-in.)(3.0 in.) 18.699 in.4

(Tension on top) qa2  2,692.7 lb-in. M1   2

c

 432 psi

;

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Problem 5.5-12 A small dam of height h  2.0 m is constructed of vertical wood beams AB of thickness t  120 mm, as shown in the figure. Consider the beams to be simply supported at the top and bottom. Determine the maximum bending stress smax in the beams, assuming that the weight density of water is g  9.81 kN.m3

A

h t

B

Solution 5.5-12

Vertical wood beam MAXIMUM BENDING MOMENT

RA 

q0L 6

q0 x 3 6L q0 Lx q0 x 3   6 6L q0L q0x 2 dM L   0 x dx 6 2L 13 M  RAx 

h  2.0 m t  120 mm g  9.81kN/ m3(water) Let b = width of beam perpendicular to the plane of the figure

Substitute x  L/13 into the equation for M:

Let q0 = maximum intensity of distributed load

Mmax 

q0  gbh S 

bt2 6

q0L q0 q0 L2 L L3 a b  a b  6 6L 313 13 913

For the vertical wood beam: L  h; Mmax 

q0 h 2 9 13

Maximum bending stress smax 

2q0 h2 2gh3 Mmax   S 313 bt2 313 t2

SUBSTITUTE NUMERICAL VALUES: smax  2.10 MPa

;

NOTE: For b  1.0 m, we obtain q0  19,620 N/m, S  0.0024 m3, Mmax  5,034.5 N # m, and smax  Mmax/S  2.10 MPa

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SECTION 5.5

Normal Stresses in Beams

401

y

Problem 5.5-13 Determine the maximum tensile stress st (due to pure bending about a horizontal axis through C by positive bending moments M ) for beams having cross sections as follows (see figure).

x b1 xc

C

(a) A semicircle of diameter d (b) An isosceles trapezoid with bases b1  b and b2  4b/3, and altitude h (c) A circular sector with   p/3 and r  d/2

x

xc

C

h

y

a

xc

C a r

d

b2

O

(a)

(b)

(c)

x

Solution 5.5-13 MAX. TENSILE STRESS DUE TO POSITIVE BENDING MOMENT IS ON BOTTOM OF BEAM CROSS-SECTION

r4 (a + sin (a) cos(a)) 4

Ix 

(a) SEMICIRCLE ybar 

From Appendix D, Case 10: (9p2  64)r4 (9p2  64)d4  72p 1152p

Ic  c

;

A  d2 a

c

2a

p b 12

d b 2

3

From Appendix D, Case 8: h3(b21 + 4b1b2 + b22) 36(b1 + b2)

73bh3  756 c

Mc 360M  Ic 73bh2

Ix 

a

d 4 b 2 4

A  0.2618 d2

p sina b 3 ± ≤ p 3 a

d 2 p b a b 2 3

c  0.276 d

p p p + sina b cos a b b 3 3 3

Ix  0.02313 d 4

h(2b1 + b2) 10h  3(b1 + b2) 21

st 

A a

For a  p/3, r  d/2:

(b) ISOSCELES TRAPEZOID

IC 

c  ybar

d1

2d 4r  3p 3p

Mc 768M M   30.93 3 st  2 3 Ic (9p  64)d d

2r sin (a) a b 3 a

;

(c) CIRCULAR SECTOR WITH a  p/3, r  d/2

IC  Ix  A y2bar IC  cd 4

(4p313) p d 13 2 d 2 a 12 b c a bd d 768 2 p

IC  3.234 * 103 d 4 max. tensile stress st 

From Appendix D, Case 13:

Mc IC

A  r 2 (a)

Problem 5.5-14 Determine the maximum bending stress smax (due to pure bending by a moment M) for a beam having a cross section in the form of a circular core (see figure). The circle has diameter d and the angle b  60°. (Hint: Use the formulas given in Appendix D, Cases 9 and 15.)

C

b b

d

st  85.24

M d3

;

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Solution 5.5-14

Circular core From Appendix D, Cases 9 and 15: Iy  r

4

4

r ab pr 2ab  aa  2 + 4 b 4 2 r r d 2

a

p  b 2

b  radians a  radians a  r sin b Iy 

4

4

4

4

3



d4 p 1 pd 4  a  b + sin 4b b 64 32 2 4



d4 (4b  sin4b) 128

MAXIMUM BENDING STRESS

b  r cos b

pd d p  a  b  sin b cos b + 2 sin b cos3 b b 64 32 2

pd d p   a  b  (sin b cos b)(1  2 cos2 b) b 64 32 2 d4 p 1 pd 4  a  b  a sin 2b b (cos 2b) b  64 32 2 2

smax  smax 

Mc Iy

c  r sin b  64M sin b

;

d (4b  sin 4b) 3

For b  60°  p/3 rad: 576M M smax   10.96 3 3 (8p13 + 9)d d

P

Problem 5.5-15 A simple beam AB of span length L  24 ft is subjected to two wheel loads acting at distance d = 5 ft apart (see figure). Each wheel transmits a load P = 3.0 k, and the carriage may occupy any position on the beam. Determine the maximum bending stress smax due to the wheel loads if the beam is an I-beam having section modulus S  16.2 in.3

Solution 5.5-15

d sin b 2

d

;

P

A

B

C

L

Wheel loads on a beam Substitute x into the equation for M: L  24 ft  288 in. d  5 ft  60 in. P3k S  16.2 in.

3

Mmax 

P d 2 aL  b 2L 2

MAXIMUM BENDING STRESS smax 

Mmax d 2 P aL  b  S 2LS 2

MAXIMUM BENDING MOMENT

Substitute numerical values:

P P P L  x + (L  x  d)  (2L  d  2x) L L L P 2 M  RA x  (2L x  dx  2x ) L P d dM L  (2L  d  4x)  0 x   dx L 2 4

smax 

RA 

3k 2(288 in.)(16.2 in.3)

 21.4 ksi

;

;

(288 in.  30 in.)2

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SECTION 5.5

Problem 5.5-16 Determine the maximum tensile stress st and maximum compressive stress sc due to the load P acting on the simple beam AB (see figure). Data are as follows: P  6.2 kN, L  3.2 m, d  1.25 m, b  80 mm, t  25 mm, h  120 mm, and h1  90 mm.

403

Normal Stresses in Beams

t P

d

A

B

h1

h

L

b

Solution 5.5-16 NUMERICAL DATA

MAX. MOMENT & NORMAL STRESSES

P  6.2 kN

L  3.2 m

d  1.25 m

b  80 mm

t  25 mm

h  120 mm

Mmax 

Beam cross section properties: centroid and moment of inertia

Aw c1 

Af + Aw

c2  h  c1 I

sc 

Mmax c1 I

sc  61.0 MPa

;

MAX. TENSILE STRESS AT BOTTOM (c  c2)

Aw  th1

(h  h1) h1 + Af c h  d 2 2

Mmax  4.7 kN # m

MAX. COMPRESSIVE STRESS AT TOP (c  c1)

h1  90 mm

Af  b (h  h1)

Pd (L  d) L

st  c1  76 mm

Mmax c2 I

st  35.4 MPa

;

c2  44 mm dist. to C from bottom

1 1 t h31 + b (h  h1)3 12 12 (h  h1) 2 h1 2 + Af cc2  d + Aw a c1  b 2 2

I  5879395.2 mm4

250 lb

Problem 5.5-17 A cantilever beam AB, loaded by a uniform load and a concentrated load (see figure), is constructed of a channel section. Find the maximum tensile stress st and maximum compressive stress sc if the cross section has the dimensions indicated and the moment of inertia about the z axis (the neutal axis) is I  3.36 in.4 (Note: The uniform load represents the weight of the beam.)

22.5 lb/ft B

A 5.0 ft

3.0 ft y

z

C

0.617 in. 2.269 in.

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Solution 5.5-17 NUMERICAL DATA

MAXIMUM STRESSES c1  0.617 in.

I  3.36 in.

4

c2  2.269 in. MAmax 

22.5 (8)2 + 250 (5) ft-lb 2

st 

MAmax c1 I

st  4341 psi

sc 

MAmax c2 I

sc  15964 psi

; ;

MAmax  1970 ft-lb MAmax (12)  23640 in.-lb

q

Problem 5.5-18 A cantilever beam AB of isosceles trapezoidal

cross section has length L  0.8 m, dimensions b1  80 mm, b2  90 mm, and height h  110 mm (see figure). The beam is made of brass weighing 85 kN/m3.

b1 C

h

L b2

(a) Determine the maximum tensile stress st and maximum compressive stress sc due to the beam’s own weight. (b) If the width b1 is doubled, what happens to the stresses? (c) If the height h is doubled, what happens to the stresses?

Solution 5.5-18 NUMERICAL DATA

MAX. TENSILE STRESS AT SUPPORT (TOP)

g  85

L  0.8 m b1  80 mm

kN

st 

3

m

b 2  90 mm

(a) MAX. STRESSES DUE TO BEAM’S OWN WEIGHT q L2 2

q  gA

A

1 (b + b 2) h 2 1

A  9.35 * 103 mm2 q  7.9475 * 102

I  h3

h (2b1 b2) 3 (b1  b2)

ybar  53.922 mm

36 (b1 b2)

I  9.417 * 10 mm

4

Mmax ybar I

sc  1.456 MPa

;

(b) DOUBLE b1& RECOMPUTE STRESSES b1  160 mm 1 (b + b2) h 2 1

q  gA

1 b21 4 b1 b2  b222 6

sc 

A

N m

Mmax  254.32 N # m ybar 

st  1.514 MPa

;

MAX. COMPRESSIVE STRESS AT SUPPORT (BOTTOM)

h  110 mm

Mmax 

Mmax (h  ybar) I

A  1.375 * 104 mm2

q  1.169 * 103

N m

qL2 2 Mmax  374 N # m Mmax 

ybar 

h (2 b1 + b2) 3 (b1 + b2)

ybar  60.133 mm

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SECTION 5.5

I  h3

1b21 + 4 b1 b2 + b222

Mmax 

36 (b1 + b2)

I  1.35 * 107 mm4

ybar 

MAX. TENSILE STRESS AT SUPPORT (TOP) st 

Mmax (h  ybar) I

st  1.381 MPa

I  h3

qL2 2

405

Normal Stresses in Beams

Mmax  508.64 N # m

h (2b1 + b2) 3 (b1 + b2)

ybar  107.843 mm

(b12 + 4b1 b2 + b22)

;

36 1b1 + b22

I  7.534 * 107 mm4

MAX. COMPRESSIVE STRESS AT SUPPORT (BOTTOM) sc 

MAX. TENSILE STRESS AT SUPPORT (TOP)

Mmax ybar 2

sc  1.666 MPa

; st 

Mmax (h  ybar) I

st  0.757 MPa

;

(c) DOUBLE h & RECOMPUTE STRESSES MAX. COMPRESSIVE STRESS AT SUPPORT (BOTTOM)

b1  80 mm h  220 mm A

1 (b + b2) h 2 1

q  gA

sc 

A  1.87 * 104 mm2

q  1.589 * 103

Mmax ybar I

sc  0.728 MPa

;

N m 200 lb/ft

Problem 5.5-19 A beam ABC with an overhang from B to C supports a uniform load of 200 lb/ft throughout its length (see figure). The beam is a channel section with dimensions as shown in the figure. The moment of inertia about the z axis (the neutral axis) equals 8.13 in.4 Calculate the maximum tensile stress st and maximum compressive stress sc due to the uniform load.

A

C

B 12 ft

6 ft

y

0.787 in.

z C

2.613 in.

Solution 5.5-19 NUMERICAL DATA

LOCATON OF ZERO SHEAR IN SPAN AB & MAX. (+) MOMENT IN SPAN AB

lb I  8.13 in.4 ft c2  2.613 in. c1  0.787 in.

q  200

xmax 

a MA  0 a Fv  0

RB 

(18)2 2 12

RA  q (18)  RB

xmax  4.5 ft

MmaxAB  RA xmax  q

COMPUTE SUPPORT REACTIONS q

RA q

xmax2 2

MmaxAB  2025 ft-lb RB  2700 lb R A  900 lb

max. () moment at B MB  q

(6)2 2

MB  3600 ft-lb

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MAX. STRESSES IN SPAN AB MmaxAB (12) c1 I MmaxAB (12) c2 st  I st  7810 psi ; sC 

MAX. STRESSES IN SPAN BC MB (12) c2 I sc  13885 psi

sc  2352 psi

sc 

st 

max. tensile stress

MB (12) c1 I

;

st  4182 psi

A

Problem 5.5-20 A frame ABC travels horizontally with an acceleration a0 (see figure). Obtain a formula for the maximum stress smax in the vertical arm AB, which had length L, thickness t , and mass density r.

max. compressive stress

t a0 = acceleration

L B

Solution 5.5-20

Accelerating frame

L  length of vertical arm t  thickness of vertical arm r  mass density a0  acceleration Let b  width of arm perpendicular to the plane of the figure Let q  inertia force per unit distance along vertical arm

TYPICAL UNITS FOR USE

VERTICAL ARM

t  meters (m)

IN THE PRECEDING EQUATION

SI units: r  kg/m3  N # s2/m4 L  meters (m) a0  m/s2 smax  N/m2 (pascals)

q  rbta0 S

bt2 6

USCS units: r  slug/ft3  lb-s2/ft4

qL2 rbta0L2 Mmax   2 2 smax 

3rL2a0 Mmax  S t

L  ft a0  ft/s2

t  ft

smax  lb/ft (Divide by 144 to obtain psi) 2

;

C

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SECTION 5.5

Problem 5.5-21 A beam of T-section is supported and loaded as shown in the figure. The cross section has width b  2 1/2 in., height h  3 in., and thickness t  3/8 in. Determine the maximum tensile and compressive stresses in the beam.

407

Normal Stresses in Beams

3

t=— 8 in.

P = 700 lb q = 100 lb/ft

L1 = 3 ft

3

t=— 8 in.

L2 = 8 ft

h= 3 in.

1

b = 2— 2 in.

L3 = 5 ft

Solution 5.5-21 NUMERICAL DATA L1  3 ft

L2  8 ft q  100

P  700 lb t

3 in. 8

a Fv  0

L3  5 ft

R lf  281 lb

Moment diagram (843.75 ft-lb at load P, 1250 ft-lb at right support)

lb ft

h  3 in.

Rlf  P + qL3  Rrt

8.438E+02

b  2.5 in.

Find centroid of cross section (c2 from bottom, Aw  t (h  t) c1 from top) Af  t b Af c2 

t ht + Aw at + b 2 2

c2  1 in.

Af + Aw

c1  h  c2 check c1  c1  2

–1.250E+03

MP  843.75 ft-lb

c1  2 in. Aw a

Mrt  1250 ft-lb

ht t b + Af ah  b 2 2

MAX. STRESSES IN BEAM at load P

Af + Aw

c1 + c2  3

MP (12) c1 sc  12494 psi I (max. compressive stress)

equals h

sc 

MOMENT OF INERTIA I

1 1 t 2 t (h  t)3 + b t 3 + Af ac2  b 12 12 2 + Aw cc1 

(h  t) 2 d 2

FIND SUPPORT REACTIONS-SUM MOMENTS ABOUT LEFT SUPPORT

a Mlf  0 Rrt  919 lb

Rrt 

L2

MP (12) c2 I

st  5842 psi

at right support

I  2 in.4

PL1 + qL3 aL2 +

st 

;

L3 b 2

Mrt (12) c2 sc  8654 psi I Mrt (12) c1 st  st  18509 psi I (max. tensile stress)

sc 

;

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Stresses in Beams (Basic Topics)

Problem 5.5-22 A cantilever beam AB with a rectangular cross section has a longitudinal hole drilled throughout its length (see figure). The beam supports a load P  600 N. The cross section is 25 mm wide and 50 mm high, and the hole has a diameter of 10 mm. Find the bending stresses at the top of the beam, at the top of the hole, and at the bottom of the beam.

10 mm 50 mm A

B

12.5 mm

37.5 mm

P = 600 N L = 0.4 m 25 mm

Solution 5.5-22

Rectangular beam with a hole MOMENT OF INERTIA ABOUT THE NEUTRAL AXIS (THE z AXIS) All dimensions in millimeters. Rectangle: Iz  Ic + Ad2 1 (25)(50)3 + (25)(50)(25  24.162)2 12

MAXIMUM BENDING MOMENT



M  PL  (600 N)(0.4 m)  240 N # m

 260,420 + 878  261,300 mm4

PROPERTIES OF THE CROSS SECTION

Hole:

A1  area of rectangle

Iz  Ic + Ad2 

 (25 mm)(50 mm)  1250 mm2 A2  area of hole p  (10 mm)2  78.54 mm2 4 A  area of cross section  A1  A2  1171.5 mm Using line B  B as reference axis: ©Aiyi  A1(25 mm)  A2(37.5 mm)  28,305 mm3 Aiyi 28,305 mm3 y a  24.162 mm  A 1171.5 mm2 Distances to the centroid C: c2  y  24.162 mm c1  50 mm  c2  25.838 mm

p (10)4 + (78.54)(37.5  24.162)2 64  490.87 + 13,972  14,460 mm4

Cross-section: I  261,300  14,460  246,800 mm4 STRESS AT THE TOP OF THE BEAM (240 N # m)(25.838 mm) Mc1  s1  I 246,800 mm4  25.1 MPa (tension)

;

STRESS AT THE TOP OF THE HOLE My s2  y  c1  7.5 mm  18.338 mm I (240 N # m)(18.338 mm) s2   17.8 MPa 246,800 mm4 (tension) STRESS AT THE BOTTOM OF THE BEAM (240 N # m)(24.162 mm) Mc2   I 246,800 mm4  23.5 MPa ; (compression)

s3  

;

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SECTION 5.5

Problem 5.5-23 A small dam of height h  6 ft is constructed of

Normal Stresses in Beams

Steel beam

vertical wood beams AB, as shown in the figure. The wood beams, which have thickness t  2.5 in., are simply supported by horizontal steel beams at A and B. Construct a graph showing the maximum bending stress smax in the wood beams versus the depth d of the water above the lower support at B. Plot the stress smax (psi) as the ordinate and the depth d (ft) as the abscissa. (Note: The weight density g of water equals 62.4 lb/ft3.)

A Wood beam t

t Wood beam

Steel beam

h d B

Side view

Solution 5.5-23

Vertical wood beam in a dam h  6 ft t  2.5 in. g  62.4 lb/ft3 Let b  width of beam (perpendicular to the figure) Let q0  intensity of load at depth d q0  gbd

ANALYSIS OF BEAM

L  h  6 ft q0d2 RA  6L q0d d RB  a3  b 6 L

MAXIMUM BENDING STRESS 1 Section modulus: S  bt2 6 Mmax 6 q0d2 d 2d d bd  2c a1  + S 6 L 3L A 3L bt q0  g bd smax 

smax 

smax 

q0d2 d 2d d a1  + b 6 L 3L A 3L

t2

a1 

d 2d d b + L 3L A 3L

(62.4)d3 (2.5)2

a1 

;

d d d b + 6 9 A 18

 0.1849d3(54  9d + d12d )

d A 3L

Mc  RA(L  d) 

gd3

SUBSTITUTE NUMERICAL VALUES: d  depth of water (ft) (Max. d  h  6 ft) L  h  6 ft g  62.4 lb/ ft3 t  2.5 in. smax  psi

x0  d

Mmax 

Top view

q0d2 d a1  b 6 L

d(ft) 0 1 2 3 4 5 6

smax(psi) 0 9 59 171 347 573 830

;

409

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Stresses in Beams (Basic Topics)

Problem 5.5-24 Consider the nonprismatic cantilever beam of circular cross section shown. The beam has an internal cylindrical hole in segment 1; the bar is solid (radius r) in segment 2. The beam is loaded by a downward triangular load with maximum intensity q0 as shown. Find expressions for maximum tensile and compressive flexural stresses at joint 1.

q0

y

Linea

P = q0L/2

r q(x

)

x

M1 1 R1

2 2L — 3 Segment 1 0.5 EI

3 L — 3 Segment 2 EI

Solution 5.5-24 STATICS a Fv  0 R1 

MAX. STRESSES AT JOINT 1 R1 

q0L 1 2L q0 a b  2 3 2

1 qL 6 0

g M1  0 q0 L 1 2L 1 2L b  Ld M1  c q0 a b a 2 3 3 3 2 M1 

23 q0 L2 54

23  0.426 54

MAX. COMPRESSION AT TOP (RADIUS r) 23 q L2 (r) M1 r 54 0 sc  sc  0.5 EI EI 2 sc 

23 q0 L2 r 27 EI

;

23  0.852 27

Max. tensile stress at bottom  same magnitude as compressive stress at top

Problem 5.5-25 A steel post (E  30  106 psi) having thickness t  1/8 in.

and height L  72 in. supports a stop sign (see figure: s  12.5 in.). The height of the post L is measured from the base to the centroid of the sign. The stop sign is subjected to wind pressure p  20 lb/ft2 normal to its surface. Assume that the post is fixed at its base. (a) What is the resultant load on the sign? [See Appendix D, Case 25, for properties of an octagon, n  8]. (b) What is the maximum bending stress smax in the post?

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SECTION 5.5

Normal Stresses in Beams

s

L

y 5/8 in.

Section A–A

z

Circular cutout, d = 0.375 in. Post, t = 0.125 in. c1

1.5 in.

C

c2

Stop sign 0.5 in. 1.0 in.

1.0 in. 0.5 in. Wind load

Numerical properties of post A = 0.578 in.2, c1 = 0.769 in., c2 = 0.731 in., Iy = 0.44867 in.4, Iz = 0.16101 in.4

A

A

Elevation view of post

411

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Stresses in Beams (Basic Topics)

Solution 5.5-25 (a) RESULTANT LOAD F ON SIGN s  12.5 in.

p  20 psf b

(b) MAX. BENDING STRESS IN POST

360 p a b n 180

b ns2 A cot a b 4 2

A  754.442 in.2

F  104.8 lb

;

I Z  0.16101 in.4

c1  0.769 in.

b  0.785 rad

or A  5.239 ft2 F  pA

L  72 in.

n8

c2  0.731 in. Mmax  628.701 ft-lb 12

Mmax  FL sc 

Mmax c1 Iz

sc  36.0 ksi

;

(max. bending stress at base of post) st 

Mmax c2 Iz

st  34.2 ksi

Design of Beams P

Problem 5.6-1 The cross section of a narrow-gage railway

Wood tie

d b Steel girder

(b)

s1 (a)

Railway cross tie Mmax  S

P(s1  s2)  15,000 lb-in. 2

1 5d 2 bd 2  (50 in.)(d 2)  6 6 6

Mmax  s allow S s1  50 in.

P Steel rail

bridge is shown in part (a) of the figure. The bridge is constructed with longitudinal steel girders that support the wood cross ties. The girders are restrained against lateral buckling by diagonal bracing, as indicated by the dashed lines. The spacing of the girders is s1  50 in. and the spacing of the rails is s2  30 in. The load transmitted by each rail to a single tie is P  1500 lb. The cross section of a tie, shown in part (b) of the figure, has width b  5.0 in. and depth d. Determine the minimum value of d based upon an allowable bending stress of 1125 psi in the wood tie. (Disregard the weight of the tie itself.)

Solution 5.6-1

s2

b  5.0 in.

d  depth of tie

s2  30 in.

P  1500 lb

sallow  1125 psi

d  inches

15,000  (1125) a

Solving, d 2  16.0 in.

5d 2 b 6

dmin  4.0 in.

3P(s1  s2) NOTE: Symbolic solution: d 2  bsallow

;

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SECTION 5.6

Problem 5.6-2 A fiberglass bracket ABCD of solid circular cross section has the shape and dimensions shown in the figure. A vertical load p  40 N acts at the free end D. Determine the minimum permissible diameter dmin of the bracket if the allowable bending stress in the material is 30 MPa and b  37 mm. (Note: Disregard the weight of the bracket itself.)

Design of Beams

413

6b A

B

2b D

C 2b

P

Solution 5.6-2

sa 

(3Pb) a a

dmin b 2

pdmin4 64

b

1

dmin3 

96Pb psa

1

96Pb 3 dmin  a b psa

96 (40) (37) 3 dmin  c d p (30)

dmin  11.47 mm

;

P  2750 lb

Problem 5.6-3 A cantilever beam of length L  7.5 ft supports a uniform

load of intensity q  225 lb/ft and a concentrated load P  2750 lb (see figure). Calculate the required section modulus S if sallow  17,000 psi. Then select a suitable wide-flange beam (W shape) from Table E-1(a), Appendix E, and recalculate S taking into account the weight of beam. Select a new beam size if necessary.

q  225 lb/ft

L = 7.5 ft

Solution 5.6-3 sa  17000 psi q  225

lb ft

Mmax1  PL +

P  2750 lb

Mmax2  2.774 * 104 lb-ft smax 

qL2 2

below allowable -OK

Mmax1  2.695 * 104 lb-ft

Mmax1 (12) Sreqd  sa

w  26

Sreqd  19.026 in.

3

try W 8 * 28 (S  24.3 in. ) 3

Check - add weight per ft for beam lb ft

Mmax2  PL +

Sact  24.3 in.3 (q + w) L2 2

smax  13699 psi

Repeat for W14 * 26 which is lighter than W8 * 28

Find Sreqd without beam weight

W  28

Mmax2 (12) Sact

L  7.5 ft

lb ft

Mmax3  PL +

Sact  35.3 in.3 (q + w) L2 2

Mmax3  2.768 * 104 lb-ft smax 

M max3 (12) Sact

smax  9411 psi

well below allowable - OK use W 14 * 26

;

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Problem 5.6-4 A simple beam of length L  5 m carries a uniform load

P = 22.5 kN 1.5 m

kN of intensity q  5.8 and a concentrated load 22.5 kN (see figure). m Assuming sallow  110 MPa, calculate the required section modulus S. Then select an 200 mm wide-flange beam (W shape) from Table E-1(b) Appendix E, and recalculate S taking into account the weight of beam. Select a new 200 mm beam if necessary.

q = 5.8 kN/m

L=5m

Solution 5.6-4 NUMERICAL DATA

RECOMPUTE MAX. MOMENT WITH BEAM MASS INCLUDED &

L  5 m q  5.8

THEN CHECK ALLOWABLE STRESS

kN m

w  a41.7

b  1.5 m

P  22.5 kN

a  L  b a  3.5 m

RA  RB 

qL Pb + 2 L qL Pa + 2 L

qL + P  51.5 kN

N m

w  409.077

sallow  110 MPa statics

kg M b a 9.81 2 b m s

RA  21.25 kN

RA 

aq +

RB  30.25 kN

RA + RB  51.5 kN

Sact  398 * 103 mm3

W bL 1000 + 2

RA  22.273 kN

Pd L

xm 

RA q + W

xm  3.587 m greater than a so max. moment at load pt LOCATE POINT OF ZERO SHEAR xm 

RA q

Mmax  RA a 

xm  3.664 m

Mmax  39.924 kN # m

greater than dist. a to load P so zero shear is at load point Mmax  RA a 

q a2 2

(q + W ) a2 2

Mmax  38.85 kN # m

smax 

Mmax S act

smax  100.311 MPa

OK, less than 110 MPa

FIND REQUIRED SECTION MODULUS Sreqd 

Mmax sallow

Sreqd  353.182 * 103 mm3

select W 200 * 41.7

;

(Sact  398 * 103 mm3)

Calculate the required section modulus S if sallow  17,000 psi, L  28 ft, P  2200 lb, and q  425 lb/ft. Then select a suitable I-beam (S shape) from Table E-2(a), Appendix E, and recalculate S taking into account the weight of the beam. Select a new beam size if necessary.

P

q

Problem 5.6-5 A simple beam AB is loaded as shown in the figure.

q B

A

L — 4

L — 4

L — 4

L — 4

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SECTION 5.6

Design of Beams

415

Solution 5.6-5 NUMERICAL DATA sa  17000 psi

L  28 ft

P  2200 lb

q  425

lb ft

FIND REACTIONS (EQUAL DUE TO SYMMETRY) THEN MAX. MOMENT AT CENTER OF BEAM

RA 

P L + q 2 4

Mmax  RA

RA  4.075 * 103 lb

qL L L 1L  a + b 2 4 4 24

Mmax  2.581 * 104 ft-lb

RECOMPUTE REACTIONS AND MAX. MOMENT THEN CHECK lb MAX. STRESS w  25.4 ft RA 

P L L + q + w 2 4 2

Mmax  RA

RA  4.431 * 103 lb

qL L L 1L L 1L  a + b w a b 2 4 4 24 2 22

Mmax  2.83 * 104 ft-lb smax 

Mmax (12) Sact

smax  13,806 psi less than allowable so OK

Compute Sreqd & then select S shape Sreqd 

Mmax (12) sa

select S 10 * 25.4

Sreqd  18.221 in.3 ;

(Sact  24.6 in. , w  25.4 lb/ft) 3

Problem 5.6-6 A pontoon bridge (see figure) is constructed of two longitudinal wood beams, known an balks, that span between adjacent pontoons and support the transverse floor beams, which are called chesses. For purposes of design, assume that a uniform floor load of 8.0 kPa acts over the chesses. (This load includes an allowance for the weights of the chesses and balks.) Also, assume that the chesses are 2.0 m long and that the balks are simply supported with a span of 3.0 m. The allowable bending stress in the wood is 16 MPa. If the balks have a square cross section, what is their minimum required width bmin?

Solution 5.6-6

Chess Pontoon

Balk

Pontoon bridge FLOOR LOAD: W  8.0 kPa ALLOWABLE STRESS: sallow  16 MPa Lc  length of chesses  2.0 m

Lb  length of balks  3.0 m

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Stresses in Beams (Basic Topics)

LOADING DIAGRAM FOR ONE BALK

Section modulus S  Mmax  S

W  total load

‹

 wLbLc q

b3 6

qL2b (8.0 kN/m)(3.0 m)2   9,000 N # m 8 8

9,000 N # m Mmax   562.5 * 106 m3 sallow 16 MPa b3  562.5 * 106 m3 and b3  3375 * 106 m3 6

Solving, bmin  0.150 m  150 mm

wLc W  2Lb 2

;

(8.0 kPa)(2.0 m) 2  8.0 kN/m 

Problem 5.6-7 A floor system in a small building consists of wood planks supported by 2 in. (nominal width) joists spaced at distance s, measured from center to center (see figure). The span length L of each joist is 10.5 ft, the spacing s of the joists is 16 in., and the allowable bending stress in the wood is 1350 psi. The uniform floor load is 120 lb/ft2, which includes an allowance for the weight of the floor system itself. Calculate the required section modulus S for the joists, and then select a suitable joist size (surfaced lumber) from Appendix F, assuming that each joist may be represented as a simple beam carrying a uniform load.

Planks

s

Joists

Solution 5.6-7

s

L s

Floor joists Mmax 

qL2 1  (13.333 lb/in.)(126 in.)2  26,460 lb-in. 8 8

Required S 

26,460 lb/in. Mmax   19.6 in.3 sallow 1350 psi

From Appendix F: Select 2 * 10 in. joists sallow  1350 psi L  10.5 ft  126 in. w  floor load  120 lb/ft2  0.8333 lb/in.2 s  spacing of joists  16 in. q  ws  13.333 lb/in.

;

;

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Design of Beams

Problem 5.6-8 The wood joists supporting a plank floor (see figure) are 40 mm * 180 mm in cross section (actual dimensions) and have a span length L  4.0 m. The floor load is 3.6 kPa, which includes the weight of the joists and the floor. Calculate the maximum permissible spacing s of the joists if the allowable bending stress is 15 MPa. (Assume that each joist may be represented as a simple beam carrying a uniform load.)

Solution 5.6-8

Spacing of floor joists

L  4.0 m w  floor load  3.6 kPa

sallow  15 MPa

s  spacing of joists

q  ws S

SPACING OF JOISTS

2

bh 6

4 bh2sallow 3wL2

Substitute numerical values:

qL2 wsL2  Mmax  8 8 S

smax 

2

smax  2

Mmax wsL bh   sallow 8sallow 6

4(40 mm)(180 mm)2(15 MPa) 3(3.6 kPa)(4.0 m)2

 0.450 m  450 mm

;

;

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q0

Problem 5.6-9 A beam ABC with an overhang from B to C is

constructed of a C 10  30 channel section (see figure). The beam supports its own weight (30 lb/ft) plus a triangular load of maximum intensity q0 acting on the overhang. The allowable stresses in tension and compression are 20 ksi and 11 ksi, respectively. Determine the allowable triangular load intensity q0,allow if the distance L equals 3.5 ft.

A

C

B L

L

2.384 in. 0.649 in.

C

3.033 in.

10.0 in.

Solution 5.6-9 NUMERICAL DATA w  30

lb ft

check tension on top

sat  20 ksi

sac  11 ksi

L  3.5 ft c1  2.384 in. from Table E-3(a)

MB 

MB c1 I22

q0allow 

c2  0.649 in. I22  3.93 in.

4

MAX. MOMENT IS AT B (TENSION TOP, COMPRESSION BOTTOM) MB  wL

st 

L 1 2 + q0L a Lb 2 2 3

1 1 wL2 + q0L2 2 3

Problem 5.6-10 A so-called “trapeze bar” in a hospital room provides a means for patients to exercise while in bed (see figure). The bar is 2.1 m long and has a cross section in the shape of a regular octagon. The design load is 1.2 kN applied at the midpoint of the bar, and the allowable bending stress is 200 MPa. Determine the minimum height h of the bar. (Assume that the ends of the bar are simply supported and that the weight of the bar is negligible.)

3 L

2

MB  s at csat a

q0allow  628 lb/ft

I22 c1

I22 1 b  wL2 d c1 2 ;

governs

check compression on bottom q0allow 

3 L

2

csac a

q0allow  1314

I22 1 b  wL2 d c2 2

lb ft

C

h

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SECTION 5.6

Solution 5.6-10

Trapeze bar (regular octagon)

P  1.2 kN L  2.1 m

sallow  200 MPa

b  0.41421h

‹ Ic  1.85948(0.41421h)4  0.054738h4

Determine minimum height h.

SECTION MODULUS

MAXIMUM BENDING MOMENT

S

Mmax 

419

Design of Beams

(1.2 kN)(2.1 m) PL   630 N # m 4 4

PROPERTIES OF THE CROSS SECTION Use Appendix D, Case 25, with n  8 b  length of one side b

360° 360°   45° n 8

b b tan  (from triangle) 2 h b h cot  2 b

Ic 0.054738h4   0.109476h3 h/2 h/2

MINIMUM HEIGHT h M s 630 N # m  3.15 * 106 m3 0.109476h3  200 MPa s

M S

S

h3  28.7735 * 106 m3 h  0.030643 m ‹ hmin  30.6 mm

;

ALTERNATIVE SOLUTION (n  8) M

PL 4

b  45° tan

b b  12 1 cot  12 1 2 2

b  (12  1)h h  (12 + 1)b b 45° For b  45°:  tan  0.41421 h 2 45° h  cot  2.41421 b 2

Ic  a S a

11 + 812 4 412  5 4 bb  a bh 12 12

412  5 3 bh 6

3PL 2(412  5)sallow

;

h3  28.7735 * 106 m3 hmin  30.643 mm

;

h3 

Substitute numerical values: MOMENT OF INERTIA 4

Ic 

b b nb acot b a3 cot2 1 b 192 2 2

Ic 

8b4 (2.41421)[3(2.41421)2 1]  1.85948b4 192

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Problem 5.6-11 A two-axle carriage that is part of an overhead traveling crane in a testing laboratory moves slowly across a simple beam AB (see figure). The load transmitted to the beam from the front axle is 2200 lb and from the rear axle is 3800 lb. The weight of the beam itself may be disregarded.

3800 lb

5 ft

2200 lb

A

B

(a) Determine the minimum required section modulus S for the beam if the allowable bending stress is 17.0 ksi, the length of the beam is 18 ft, and the wheelbase of the carriage is 5 ft. (b) Select the most economical I-beam (S shape) from Table E-2(a), Appendix E.

18 ft

Solution 5.6-11 NUMERICAL DATA P1  2200 lb

L  18 ft P2  3800 lb

d  5 ft

sa  17 ksi (a) FIND REACTION RA THEN AN EXPRESSION FOR MOMENT UNDER LARGER LOAD P2; LET X  DIST. FROM A TO LOAD P2 RA  P2 a

L  (x + d ) Lx b + P1 c d L L

M2  RA x M2  x cP2 a

L  (x + d ) Lx b + P1 c dd L L

xP2 LP2 x2 xP1L P1x2 xP1d L Take derivative of MA & set to zero to find max. bending moment at x  x m M2 

xm 

(P1 + P2) L  P1d 2 (P1 + P2)

xm  8.083 ft

L  (xm + d) L  xm b + P1 c d L L RA  2694 lb RA  P2 a

Mmax  xm cP2 a

L(xm  d) L xm b P1 c dd L L

Mmax  21780 ft-lb Sreqd 

Mmax sa

Sreqd  15.37 in.3

;

(b) SELECT MOST ECONOMICAL S SHAPE FROM TABLE E-2(A) select S8 * 23

;

Sact  16.2 in.3

d xP2LP2x2 xP1LP1x2 xP1d a b dx L 

P2L  2P2x + P1L  2P1x  P1d L

P2L  2P2x + P1L  2P1x  P1d  0

Problem 5.6-12 A cantilever beam AB of circular cross section and length

L  450 mm supports a load P  400 N acting at the free end (see figure). The beam is made of steel with an allowable bending stress of 60 MPa. Determine the required diameter dmin of the beam, considering the effect of the beam’s own weight.

A B d P L

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SECTION 5.6

Solution 5.6-12

MINIMUM DIAMETER Mmax  sallow S PL +

 77.0 kN/m

3

pgd 2L2 pd 3  sallow a b 8 32

Rearrange the equation:

WEIGHT OF BEAM PER UNIT LENGTH q  ga

421

Cantilever beam

L  450 mm P  400 N sallow  60 MPa g  weight density of steel

DATA

Design of Beams

sallow d 3  4gL2 d 2 

pd 2 b 4

32 PL 0 p

(Cubic equation with diameter d as unknown.) MAXIMUM BENDING MOMENT

Substitute numerical values (d  meters):

q L2 pgd3L2  PL + Mmax  PL + 2 8

(60 * 106 N/m2)d3  4(77,000 N/m3)(0.45 m)2d2

SECTION MODULUS



pd 3 S 32

32 (400 N)(0.45 m)  0 p

60,000d 3  62.37d 2  1.833465  0 Solve the equation numerically: d  0.031614 m

;

q

Problem 5.6-13 A compound beam ABCD (see figure) is supported at points A, B, and D and has a splice at point C. The distance a  6.25 ft, and the beam is a S 18  70 wide-flange shape with an allowable bending stress of 12,800 psi.

dmin  31.61 mm

A

B

C

D Splice

(a) If the splice is a moment release, find the allowable 4a uniform load qallow that may be placed on top of the beam, taking into account the weight of the beam itself. [See figure part (a).] (b) Repeat assuming now that the splice is a shear release, as in figure part (b).

a

4a

(a) (b) Moment Shear release release

Solution 5.6-13 NUMERICAL DATA lb S  103 in.3 w  70 ft a  6.25 ft sa  12800 psi

MZ 2.000E+00 @ 2.000E+00 MZ 9.453E–01 @ 1.375E+00 ×

×

(a) MOMENT RELEASE AT C-GIVES MAX. MOMENT AT B (SEE MOMENT DIAGRAM)   2.5 q a2 Mmax Mmax  [1qallow + w2 a2 (2.5)] S and Mmax  s a S

sa 

lb S  103 in.3 ft sa  12800 psi a  6.25 ft

w  70

MZ –2.500E+00 @ 4.000E+00

×

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sa S 12 in./ft

qallow 

Page 422

2.5 a2 lb qallow  1055 ft

MZ 8.00E+00

w ;

for moment release

(b) SHEAR RELEASE AT C-GIVES MAX. MOMENT AT C (SEE MOMENT DIAGRAM)  8 q a2 sa S 12 in./ft

qallow 

8a2

qallow  282

;

for shear release

w

Problem 5.6-14 A small balcony constructed of wood is supported by three identical cantilever beams (see figure). Each beam has length L1  2.1 m, width b, and height h  4b/3. The dimensions of the balcon floor are L1 * L2, with L2  2.5 m. The design load is 5.5 kPa acting over the entire floor area. (This load accounts for all loads except the weights of the cantilever beams, which have a weight density g  5.5 kN/m3.) The allowable bending stress in the cantilevers is 15 MPa. Assuming that the middle cantilever supports 50% of the load and each outer cantilever supports 25% of the load, determine the required dimensions b and h. Solution 5.6-14

lb ft

4b h= — 3 L2

b

L1

Compound beam MAXIMUM BENDING MOMENT (q q0)L21 1  (6875 N/m7333b2)(2.1 m)2 Mmax  2 2  15,159 + 16,170b2 (N # m) bh2 8b3  6 27 Mmax  sallow S

L1  2.1 m L2  2.5 m Floor dimensions: L1 * L2 Design load  w  5.5 kPa g  5.5 kN/m3 (weight density of wood beam) sallow  15 MPa

S

MIDDLE BEAM SUPPORTS 50% OF THE LOAD.

Rearrange the equation:

‹ q  wa

L2 2.5 m b  (5.5 kPA)a b  6875 N/m 2 2

WEIGHT OF BEAM q0  gbh 

4gb2 4  (5.5 kN/m2) b2 3 3

 7333b2 (N/m)

(b  meters)

15,159 + 16,170b2  (15 * 106 N/m2)a

8b3 b 27

(120 * 106)b3  436,590b2  409,300  0 SOLVE NUMERICALLY FOR DIMENSION b 4b  0.2023 m h b  0.1517 m 3 REQUIRED DIMENSIONS b  152 mm

h  202 mm

;

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SECTION 5.6

Problem 5.6-15 A beam having a cross section in the form of an unsymmetric wide-flange shape (see figure) is subjected to a negative bending moment acting about the z axis. Determine the width b of the top flange in order that the stresses at the top and bottom of the beam will be in the ratio 4:3, respectively.

423

Design of Beams

y b 1.5 in. 1.25 in. z

12 in.

C

1.5 in. 16 in.

Solution 5.6-15

Unsymmetric wide-flange beam AREAS OF THE CROSS SECTION (in.2) A1  1.5b A2  (12)(1.25)  15 in.2 A3  (16)(1.5)  24 in.2 A  A1 + A2 + A3  39 + 1.5b (in.2) FIRST MOMENT OF THE CROSS-SECTIONAL AREA ABOUT THE LOWER EDGE B-B QBB  gyi Ai  (14.25)(1.5b) + (7.5)(15) + (0.75)(24)

Stresses at top and bottom are in the ratio 4:3. Find b (inches) h  height of beam  15 in. LOCATE CENTROID stop c1 4   sbottom c2 3 4 60  8.57143 in. c1  h  7 7 3 45 c2  h   6.42857 in. 7 7

 130.5 + 21.375b (in.3) DISTANCE c2 FROM LINE B-B TO THE CENTROID C c2 

QBB 130.5 + 21.375b 45   in. A 39 + 1.5b 7

SOLVE FOR b (39 + 1.5b)(45)  (130.5 + 21.375b)(7) 82.125b  841.5 b  10.25 in.

;

y

Problem 5.6-16 A beam having a cross section in the form of a channel (see figure) is subjected to a bending moment acting about the z axis. Calculate the thickness t of the channel in order that the bending stresses at the top and bottom of the beam will be in the ratio 7:3, respectively.

t

z

t

C 152 mm

t

55 mm

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Solution 5.6-16 ratio of top to bottom stresses  c1/c2  7/3

NUMERICAL DATA h  152 mm

b  55 mm

1 11385  186 t + t2 2  131 + t

take 1st moments to find distances c1 & c2 1st moments about base

c2 

J

t b (h  2t) (t) + 2bt a b 2 2 2bt + t (h  2t)



c1  b  c2

c2 

t 55 (152  2t) (t) + 2.55t a b 2 2 t 55 (152  2t) (t) + 2.55t a b 2 2 2.55t + t(152  2t)

 1 11385  186 t + t 2  131 + t

2

111385  186 t + t22  76 t + t2  3025

 7/3

7 a  76 tt2 3025b d  0 t2  109 t + 1298  0 t

109 11092 4 (1298) 2

Problem 5.6-17 Determine the ratios of the weights of three beams that have the same length, are made of the same material, are subjected to the same maximum bending moment, and have the same maximum bending stress if their cross sections are (1) a rectangle with height equal to twice the width, (2) a square, and (3) a circle (see figures).

Solution 5.6-17

K

2.55 t + t (152  2 t)

c3 c  a 11385186 tt2 b d

2.55t + t (152  2t)

c1  55  c1 

55 t (152  2t) (t) + 2.55 t a b 2 2

t  13.61 mm

h = 2b

b

;

a

a

d

Ratio of weights of three beams

Beam 1: Rectangle (h  2b) Beam 2: Square (a  side dimension) Beam 3: Circle (d  diameter) L, g, Mmax, and smax are the same in all three beams. M S  section modulus S  s Since M and s are the same, the section moduli must be the same. bh2 2b3  (1) RECTANGLE: S  6 3 A1  2b2  2 a

3S 1/3 b a b 2

3S 2/3 b  2.6207S2/3 2

(2) SQUARE: S 

a3 6

a  (6S)1/3

A2  a2  (6S)2/3  3.3019 S2/3 (3) CIRCLE: S 

pd 3 32

A3 

d a

32S 1/3 b p

pd 2 p 32S 2/3  a b  3.6905 S2/3 4 4 p

Weights are proportional to the cross-sectional areas (since L and g are the same in all 3 cases). W1 : W2 : W3  A1 : A2 : A3 A1 : A2 : A3  2.6207 : 3.3019 : 3.6905 W1 : W2 : W3  1 : 1.260 : 1.408

;

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425

Design of Beams

t

Problem 5.6-18 A horizontal shelf AD of length L  915 mm, width

b  305 mm, and thickness t  22 mm is supported by brackets at B and C [see part (a) of the figure]. The brackets are adjustable and may be placed in any desired positions between the ends of the shelf. A uniform load of intensity q, which includes the weight of the shelf itself, acts on the shelf [see part (b) of the figure]. Determine the maximum permissible value of the load q if the allowable bending stress in the shelf is sallow  7.5 MPa and the position of the supports is adjusted for maximum load-carrying capacity.

A B

D

C

b

L (a) q A

D B

C L (b)

Solution 5.6-18 NUMERICAL DATA

Substitute x into the equation for either M1 or |M2|: b  305 mm

L  915 mm

t  22 mm

sallow  7.5 MPa

Mmax 

qL2 (3  2 12) 8

Mmax  sallow S  sallow a

MOMENT DIAGRAM

Eq. (1) bt 2 b 6

Eq. (2)

Equate Mmax from Eqs. (1) and (2) and solve for q: qmax 

4bt2sallow 3L2(3  2 12)

Substitute numerical values: For maximum load-carrying capacity, place the supports so that M1  |M2|. Let x  length of overhang M1  ‹

qL (L  4x) 8

|M2| 

qL qx2 (L  4x)  8 2

Solve for x: x 

L (12  1) 2

qx2 2

qmax  10.28 kN/m

;

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Problem 5.6-19 A steel plate (called a cover plate) having cross-sectional dimensions 6.0 in. * 0.5 in. is welded along the full length of the bottom flange of a W 12 * 50 wide-flange beam (see figure, which shows the beam cross section). What is the percent increase in the smaller section modulus (as compared to the wide-flange beam alone)?

W 12  50

6.0  0.5 in. cover plate

Solution 5.6-19 FIND I ABOUT HORIZ. CENTROIDAL AXIS

NUMERICAL PROPERTIES FOR W 12 * 50 (FROM TABEL E-1(a)) 2

A  14.6 in.

d  12.2 in.

c1  c2

c1 

d 2

I  391 in.4

Ih  I + A a c1 

+ (6) (0.5) a c2 

c1 

d 0.5 + (6) (0.5) ad + b 2 2 A + (6) (0.5)

c2  (d + 0.5)  c1

0.5 2 b 2

Ih  491.411in.4

S  64.2 in.3

FIND SMALLER SECTION MODULUS Ih Stop  Stop  68.419 in.3 c1 % increase in smaller section modulus Stop  S ; (100)  6.57% S

FIND CENTROID OF BEAM WITH COVER PLATE (TAKE 1ST MOMENTS ABOUT TOP TO FIND c1 7 c2) A

d 2 1 b + (6) (0.5)3 2 12

c1  7.182 in.

c2  5.518 in.

Problem 5.6-20 A steel beam ABC is simply supported at A and B and has an overhang BC of length L  150 mm (see figure). The beam supports a uniform load of intensity q  4.0 kN/m over its entire span AB and 1.5q over BC. The cross section of the beam is rectangular with width b and height 2b. The allowable bending stress in the steel is sallow  60 MPa, and its weight density is   77.0 kN/m3.

1.5 q q C

A

2b

B 2L

L

(a) Disregarding the weight of the beam, calculate the required width b of the rectangular cross section. (b) Taking into account the weight of the beam, calculate the required width b.

b

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Design of Beams

427

Solution 5.6-20 NUMERICAL DATA L  150 mm sa  60 MPa

(b) NOW MODIFY-INCLUDE BEAM WEIGHT kN q4 m kN g  77 3 m

w  gA

and

and

L2 2

Equate Mmax1 to Mmax2 & solve for bmin at B

Mmax2  s a S

2 3 a sa b b3  1gL22 b2  qL2  0 3 4

2 S  b3 3

Insert numerical values, then solve for b

Equate Mmax1 to Mmax2 & solve for bmin

bmin  11.92 mm

1

9 qL2 3 bmin  a b 8 sa bmin  11.91 mm

L2 2 2 Mmax  s a a b3 b 3

Mmax  (1.5q + w)

(a) IGNORE BEAM SELF WEIGHT-FIND bmin Mmax1  1.5 q

w  g 12b22

;

;

Problem 5.6-21 A retaining wall 5 ft high is constructed of horizontal wood planks 3 in. thick (actual dimension) that are supported by vertical wood piles of 12 in. diameter (actual dimension), as shown in the figure. The lateral earth pressure is p1  100 lb/ft2 at the top of the wall and p2  400 lb/ft2 at the bottom. Assuming that the allowable stress in the wood is 1200 psi, calculate the maximum permissible spacing s of the piles. (Hint: Observe that the spacing of the piles may be governed by the load-carrying capacity of either the planks or the piles. Consider the piles to act as cantilever beams subjected to a trapezoidal distribution of load, and consider the planks to act as simple beams between the piles. To be on the safe side, assume that the pressure on the bottom plank is uniform and equal to the maximum pressure.)

3 in. p1 = 100 lb/ft2

12 in. diam.

12 in. diam.

s

5 ft

3 in.

Top view p2 = 400 lb/ft2 Side view

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Solution 5.6-21

Retaining wall

(1) PLANK AT THE BOTTOM OF THE DAM t  thickness of plank  3 in. b  width of plank (perpendicular to the plane of the figure) p2  maximum soil pressure  400 lb/ft2  2.778 lb/in.2 s  spacing of piles q  p2b sallow  1200 psi S  section modulus 2

Mmax 

qs p2bs  8 8

Mmax  s allow S

2

or

S

bt 6

p2bs 2 bt 2  sallow a b 8 6

Solve for s: s

4sallow t 2

A

3p 2

2

 72.0 in.

q1  p1s q2  p2s d  diameter of pile  12 in. Divide the trapezoidal load into two triangles (see dashed line). Mmax  S

pd 3 32

Mmax  sallow S

or

pd 3 sh 2 (2p1 + p2)  sallow a b 6 32 Solve for s: s

(2) VERTICAL PILE h  5 ft  60 in. p1  soil pressure at the top  100 lb/ft 2  0.6944 lb/in.2

1 sh2 1 2h h (q1) (h)a b  (q2)(h)a b  (2p1  p2) 2 3 2 3 6

3psallow d 3 16h2 (2p1 + p2)

PLANK GOVERNS

 81.4 in.

smax  72.0 in.

Problem 5.6-22 A beam of square cross section (a  length of each side) is bent in the plane of a diagonal (see figure). By removing a small amount of material at the top and bottom corners, as shown by the shaded triangles in the figure, we can increase the section modulus and obtain a stronger beam, even though the area of the cross section is reduced. (a) Determine the ratio b defining the areas that should be removed in order to obtain the strongest cross section in bending. (b) By what percent is the section modulus increased when the areas are removed?

;

y

a z

ba C

a

ba

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SECTION 5.6

Solution 5.6-22 removed

429

Design of Beams

Beam of square cross section with corners RATIO OF SECTION MODULI S  (1 + 3b)(1  b)2 S0

Eq. (1)

GRAPH OF EQ. (1)

a  length of each side ba  amount removed Beam is bent about the z axis. ENTIRE CROSS SECTION (AREA 0) I0 

a4 12

c0 

a 12

S0 

I0 a3 12  c0 12

(a) VALUE OF b

S/S0

d S a b 0 db S0

SQUARE mnpq (AREA 1) I1 

FOR A MAXIMUM VALUE OF

(1  b)4a4 12

Take the derivative and solve this equation for b . b

PARALLELOGRAM mm, n, n (AREA 2) 1 I2  (base)(height)3 3

1 9

;

(b) MAXIMUM VALUE OF S/S0

(1  b)a 3 ba4 1 I2  (ba12)c d  (1  b)3 3 6 12

Substitute b  1/9 into Eq. (1). (S/S0)max  1.0535 The section modulus is increased by 5.35% when ; the triangular areas are removed.

REDUCED CROSS SECTION (AREA qmm, n, p, pq) a4 I  I1 + 2I2  (1 + 3b)(1  b)3 12 c

(1  b) a 12

S

I 12 a3 (1 + 3b)(1  b)2  c 12 b — 9

Problem 5.6-23 The cross section of a rectangular beam having width b and height h is shown in part (a) of the figure. For reasons unknown to the beam designer, it is planned to add structural projections of width b/9 and height d to the top and bottom of the beam [see part (b) of the figure]. For what values of d is the bending-moment capacity of the beam increased? For what values is it decreased?

d

h

b (a)

h

d

b — 9 (b)

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Stresses in Beams (Basic Topics)

Solution 5.6-23

Beam with projections Graph of

S2 d versus S1 h d h 0 0.25 0.50 0.75 1.00

(1) ORIGINAL BEAM I1 

bh3 12

c1 

h 2

S1 

S2 S1 1.000 0.8426 0.8889 1.0500 1.2963

I1 bh2  c1 6

(2) BEAM WITH PROJECTIONS I2 

1 8b 3 1 b a bh + a b(h + 2d)3 12 9 12 9

b [8h3 + (h + 2d)3] 108 1 h c2  + d  (h + 2d) 2 2 

S2 

b[8h3 + (h + 2d)3] I2  c2 54(h + 2d)

RATIO OF SECTION MODULI 3

3

b [8h + (h + 2d) ] S2   S1 9(h + 2d)(bh2)

8 + a1 +

2d 9a 1 + b h

EQUAL SECTION MODULI Set

S2 d  1 and solve numerically for . S1 h

d  0.6861 and h

d 0 h

2d 3 b h

Moment capacity is increased when d 7 0.6861 ; h Moment capacity is decreased when d 6 0.6861 ; h NOTES: S2 2d 3 2d  1 when a1 + b  9a 1 + b + 80 S1 h h or

d  0.6861 and 0 h

3  1 S2 d 14 is minimum when   0.2937 S1 h 2

a

S2 b  0.8399 S1 min

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SECTION 5.7

431

Nonprismatic Beams

Nonprismatic Beams Problem 5.7-1 A tapered cantilever beam AB of length L has square cross sections and supports a concentrated load P at the free end [see figure part (a)]. The width and height of the beam vary linearly from hA at the free end to hB at the fixed end. Determine the distance x from the free end A to the cross section of maximum bending stress if hB  3hA. (a) What is the magnitude smax of the maximum bending stress? What is the ratio of the maximum stress to the largest stress B at the support? (b) Repeat (a) if load P is now applied as a uniform load of intensity q  P/L over the entire beam, A is restrained by a roller support and B is a sliding support [see figure, part (b)].

q = P/L B hA

A B

A

hB x P

Sliding support

x

L

L (a)

(b)

Solution 5.7-1 (a) FIND MAX. BENDING STRESS FOR TAPERED

sB  s(L)

CANTILEVER

h(x)  hA a1 +

2x b L

M(x) s(x)  S(x)

s(x) 

s(x) 

S(x) 

h(x)3 6 2x bd L

3

6PxL3 hA3 (L

2PL 9hA 3

4PL

6(P)(x) chA a1 +

sB 

smax 9hA 3  sB 2PL

smax 2 sB

9hA 3 (b) REPEAT (A) BUT NOW FOR DISTRIBUTED UNIFORM LOAD OF P/L OVER ENTIRE BEAM

+ 2x)3

d s(x)  0 then solve for xmax dx

a Fv  0

6PxL3 d c 3 d 0 dx hA (L + 2x)3

M(x)  c c RA x 

c6P

L3 hA3 (L 2x)3

 L + 4x hA3 (L + 2x)4

0

L smax  s a b 4

smax 

4PL 9hA 3

36Px so

L3 hA3 (L2x)4 x

smax 

;

;

d 0

L 4 L 6P L3 4

L 3 hA3 a L + 2 b 4

M(x)  Px 

M(x) s(x)  S(x)

RA  P P x xa b d d L 2

1 2P x 2 L Px  s(x) 

1 2P x 2 L

c hA a 1 +

s(x)  3xP (2L + x)

2x 3 bd L

6 L2 hA3 (L + 2x)3

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xmax  0.20871 L

d s(x)  0 then solve for xmax dx

smax  s(0.20871 L) PL smax  0.394 3 ; hA

2

d L c  3xP ( 2L x) 3 d 0 dx hA (L2x)3 c 3P (  2L + x)  3xP

L2 hA3(L

sB  s(L) So 3

+ 2x)

L2

sB 

hA3 (L + 2x)3 2

+ 18xP (2L + x)

L hA3 (L

4

+ 2x)

d 0

smax  sB

PL 9hA3

a 0.39385

PL hA3

b

PL 9hA3

smax  3.54 sB

Simplifying

;

L2  5xL + x 2  0 so xmax 5  152  4  L 2

Problem 5.7-2 A tall signboard is supported by two vertical beams consisting of thin-walled, tapered circular tubes [see figure]. For purposes of this analysis, each beam may be represented as a cantilever AB of length L  8.0 m subjected to a lateral load P  2.4 kN at the free end. The tubes have constant thickness t  10.0 mm and average diameters dA  90 mm and dB  270 mm at ends A and B, respectively. Because the thickness is small compared to the diameters, the moment of inertia at any cross section may be obtained from the formula I  pd3t/8 (see Case 22, Appendix D), and therefore, the section modulus may be obtained from the formula S  pd2t/4. (a) At what distance x from the free end does the maximum bending stress occur? What is the magnitude smax of the maximum bending stress? What is the ratio of the maximum stress to the largest stress sB at the support? (b) Repeat (a) if concentrated load P is applied upward at A and downward uniform load q(x)  2P/L is applied over the entire beam as shown. What is the ratio of the maximum stress to the stress at the location of maximum moment? 2P q(x) = — L

P = 2.4 kN Wind load

B

A

t B

A x

P

d

L = 8.0 m t = 10.0 mm

x L = 8.0 m (b)

dA = 90 mm

(a)

dB = 270 mm

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SECTION 5.7

433

Nonprismatic Beams

Solution 5.7-2 (a) FIND MAX. BENDING STRESS FOR TAPERED CANTILEVER d(x)  d A a1 +

2x b L

2

S(x) 

dA  90 mm M(x) s(x)  S(x)

4P s(x)  pt

4P xL2 s(x)  d c 2 pt dA (L + 2x)2

J

c dA a 1 +

xL2 d 4P c c 2 dd 0 dx pt dA (L + 2x)2

xL2 P d 0 pt dA2 (L + 2x)3  L + 2x ptdA2 (L + 2x)3

L 4m 2

d 0

smax 

L 2 aL + 2 b 2

PL

4 P L 9 pt dA2

P x x b L 2

M(x)  a Px  2 M(x)   Pxa

L + x b L

M(x) S(x)

s(x) 

 Px a

L + x b L

pt 2x 2 cdA a 1 + bd 4 L L

ptdA2 (L 2x)2

tension on top, compression on bottom of beam

d L c 4Px (L x) d 0 2 dx ptdA (L 2x)2 ¥

c 4P ( L + x)  4Px

L ptdA2 (L + 2x)2

L ptdA2 (L

+ 2x)2

16Px (L x)

2ptdA2

Stress at support sB  s(L) sB 

(b) REPEAT (A) BUT NOW ADD DISTRIBUTED LOAD

d s(x)  0 then solve for xmax dx

;

L 2 L 2 dA2

2p (0.010) (0.090)2 ; smax  37.7 MPa

s(x)  4Px (L x)

L smax  s a b 2 4P ≥ pt

2x bd L K 2

s(x) 

L2 P c4 2 pt dA (L + 2x)2

smax 

;

(2400) (8)

smax 

x

d s(x)  0 then solve for xmax dx

so xmax 

4 P L b 9 pt dA2

Evaluate using numerical data

dB  270 mm

c4PL2

a

smax 9  sB 8

L  8 m t  10 mm

or

2ptdA2

smax  sB

pd(x) t 4

P  2.4 kN

 16

PL

OR simplifying L 4 xmax  2 m

L ptdA2 (L 2x)3

c 4PL2

so xmax 

;

d 0

 L + 4x ptdA2 (L + 2x)3

d 0

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stress at support

L smax  s a b 4

sB  s(L)

L L smax  4P aL  b 4 4 smax 

J

L p t dA2 aL 2

PL

ptdA2 (L + 2L2) sB  0 so no ratio of smax/sB is possible

L 2 b K 4

MAX. MOMENT AT L/2 SO COMPARE

3 p t dA2 evaluate using numerical data

Stress at location of max. moment L L L sa b  4P a L b 2 2 2

L8m P  2.4 kN d A  90 mm t  10 mm dB  270 mm smax 

L

sB  4PL ( L + L)

L ptdA2 aL 2

L 2 b 2

1 L L sa b  P 2 4 ptdA2

(2400) (8)

3p (0.010) (0.090)2 smax  25.2 MPa ;

PL smax/s(L/2) 

3ptdA2 L 1 a P b 4 ptdA2



4 3

;

Problem 5.7-3 A tapered cantilever beam AB having rectangular cross sections is subjected to a concentrated load P  50 lb and a couple M0  800 lb-in. acting at the free end [see figure part (a)]. The width b of the beam is constant and equal to 1.0 in., but the height varies linearly from hA  2.0 in. at the loaded end to hB  3.0 in. at the support. (a) At what distance x from the free end does the maximum bending stress smax occur? What is the magnitude smax of the maximum bending stress? What is the ratio of the maximum stress to the largest stress sB at the support? (b) Repeat (a) if, in addition to P and M0, a triangular distributed load with peak intensity q0  3P/L acts upward over the entire beam as shown. What is the ratio of the maximum stress to the stress at the location of maximum moment?

P = 50 lb

P = 50 lb A M0 = 800 lb-in. hA = 2.0 in.

B hB = 3.0 in.

x b = 1.0 in.

3P q0 = — L

A M0 = 800 lb-in.

x L = 20 in. (a)

b = 1.0 in.

L = 20 in. (b)

B

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Nonprismatic Beams

435

Solution 5.7-3 (a) FIND MAX. BENDING STRESS FOR TAPERED CANTILEVER FIG. (A) x b h(x)  hA a1 + 2L numerical data

6

24P

+ x)2

L2 bhA2 (2L

+ x)2 L2

0 bhA2 (2L x)3 2PL + Px + 2M0 d 0 OR simplifying c24L2 bhA2 (2L + x)3 2 1PL  M02 so x  P xmax  8 in. ; agrees with plot at left

M(x) S(x)

Evaluate max. stress & stress at B using numerical data

2000

smax  s(8)

1500

sB  s(20) smax  1.042 sB

M(x) (in.-lb) 1000

0

10 x (in.)

20

1260

1240 σ (x) (psi) 1220

20

;

sB  1200 psi ;

h(x)  hA a 1 +

x b 2L

4 PL 5 P q0  3 L

M0  800 in.-lb

I(x)  10 x (in.)

smax  1250 psi

(b) FIND MAX. BENDING STRESS FOR TAPERED CANTILEVER, FIG. (B)

M0 

0

L2 bhA2 (2L

 2 124Px 24M02

6 M(x)  Px + M0

1200

2 x bd 2L

d L2 c24 1Px + M02 d 0 dx bhA2 (2L + x)2

2 x b chA a1 + bd 2L

500

b c hA a 1 +

d s(x)  0 then solve for xmax dx

4 M0  PL M0  800 in.-lb 5 bh(x)3 I(x) I(x)  S(x)  h(x) 12 2 bh(x)2 S(x)  6

s(x) 

Px + M0

s(x)  24 1Px + M02

P  50 lb L  20 in. hA  2 in. hB  3 in. b  1 in.

S(x) 

s(x) 

bh(x)3 12

S(x) 

I(x) h(x) 2

S(x) 

bh(x)2 6

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S(x) 

Page 436

Stresses in Beams (Basic Topics)

b chA a1 +

2 x bd 2L

then solve for xmax

d s(x)  c124PL  12x2 q02 dx

6 1 x x a q0 b x 2 L 3

M(x)  Px + M0 + s(x) 

d s(x)  0 dx

L bhA2 (2L + x)2

M(x) S(x)

 4x3q0) *

1500

 2 (24PxL + 24M0L L

bhA2 (2L

+ x)3

d 0

Simplifying 12PL2 + 6PxL + 6x2 q0L + x3 q0 + 12M0 L  0

M(x) 1000 (in.-lb)

Solve for xmax xmax  4.642 in.

;

Max. stress & stress at B 500

0

10 x (in.)

20

smax  s (xmax) smax  1235 psi

1400

sB  s (20)

; sB  867 psi

FIND MAX. MOMENT AND STRESS AT LOCATION OF MAX. 1200

MOMENT

1000

d M(x)  0 dx

σ (x) (psi)

xm  800

0

10 x (in.)

Px + M0  s(x) 

b chA a1 +

q0 x3 6L

2 x bd 2L

6

s(x)  41 6PxL  6 M0 L + x3 q02 L *

bhA2 (2L

+ x)2

20

P (2L)

A q0

sm  s(xm) smax  1.215 sm

q0x3 d aPx + M0  b 0 dx 6L xm  16.33 in. sm  1017 psi ;

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SECTION 5.7

Nonprismatic Beams

437

Problem 5.7-4 The spokes in a large flywheel are modeled as beams fixed at one end and loaded by a force P and a couple M0 at the other (see figure). The cross sections of the spokes are elliptical with major and minor axes (height and width, respectively) having the lengths shown in the figure part (a). The cross-sectional dimensions vary linearly from end A to end B. Considering only the effects of bending due to the loads P and M0, determine the following quantities. (a) (b) (c) (d) (e)

The largest bending stress sA at end A The largest bending stress sB at end B The distance x to the cross section of maximum bending stress The magnitude smax of the maximum bending stress Repeat (d) if uniform load q(x)  10P/3L is added to loadings P and M0, as shown in the figure part (b). P = 12 kN M0 = 10 kN•m

10P q(x) = — 3L

B

A

P

x M0

L = 1.25 m

A

B

x L = 1.25 m hA = 90 mm

hB = 120 mm (b)

bA = 60 mm bB = 80 mm (a)

Solution 5.7-4 (a-d) FIND MAX. BENDING STRESS FOR TAPERED

30

CANTILEVER

numerical data L  1.25 m bA  60 mm hA  90 mm bB  80 mm hB  120 mm

M(x) 20 (kN•m)

P  12 kn M0  10 kN # m h(x)  hA a1  I(x) 

x b 3L

p b(x) h(x)3 64

b(x)  bA a 1 + S(x) 

p b(x) h(x)2 S(x)  32 S(x) 

p bA hA2 a1 + 32

x 3 b 3L

I(x) h(x) 2

x b 3L

10

0

0.5 x (m)

1

0

0.5 x (m)

1

240 230 σ (x) 220 (MPa) 210 200

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M(x)  Px + M0 s(x) 

s(x) 

M(x) S(x)

I(x) 

p b(x) h(x)3 64

S(x) 

p b(x) h(x)2 32

Px + M0 p bA hA2 a1 +

x 3 b 3L

32 s(x)  864 a

Px + M0

d s (x)  0 dx

p bA hA2

b a

L3 (3L + x)3

b

S(x) 

s(x) 

p bAhA2 (3L + x)3 Px + M0 L3 0  2592 2 p bAhA (3L + x)4

M(x) S(x)

10

5

3PL + 2Px + 3M0 p bAhA2 (3L + x)4

d 0

0

3(PL  M0) so xmax  2P xmax  0.625 m ;

0

σ (x) (MPa)

Evaluate using numerical data smax  231 MPa

;

100

sA  s(0) sB  s(L) smax  1.045 sB

sA  210 MPa sB  221 MPa

; ;

0

(e) FIND MAX. BENDING STRESS INCLUDING UNIFORM LOAD

bB  80 mm P  12 kN

hB  120 mm M0  10 kN # m

x b h(x)  hA a1 + 3L b(x)  bA a1 +

x b 3L

1

0.5 x (m)

1

200

smax  s(xmax)

bA  60 mm

0.5 x (m)

300

agrees with plot above

L  1.25 m

10 P x2 3 L 2

M(x) (kN•m)

OR simplfying c864L3

32

15

L3

P

I(x) h(x) 2

x 3 b 3L

p bA hA2 a 1 +

M(x)  P x + M0 

then solve for xmax

Px + M0 d L3 c864 d 0 2 dx p bAhA (3L + x)3 864

S(x) 

hA  90 mm

0

P x + M0  s(x)  ≥

10 P x2 3 L 2

p bA hA2 a 1 + 32

x 3 b 3L

¥

s(x)  288 13 P x L  3 M0 L L2 + 5 P x22 p bA hA2 (3 L + x)3 d s(x)  0 then solve for xmax dx

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SECTION 5.7

L *

pbAhA2 (3L

+ x)3

9PL2  36PxL + 5Px2  9M0L  0

d 0

Solving for x max: xmax  0.105m

L2 d s (x)  c(864 PL  2880 P x) dx p bA hA2 (3 L x)3  3 1864 P x L 864 M0 L 1440Px22 L2

*

pbAhA2 (3L

+ x)4

439

OR

d c  288 13 Px L  3 M0 L + 5 P x22 dx 2

Nonprismatic Beams

d 0

solution agrees with plot above, evaluate using numerical data smax  s(xmax) sA  s(0) sB  s(L)

smax  214 MPa ; sA  210 MPa ; sB  0 MPa ;

OR simplifying

(288 L2)

c9PL2  36PxL + 5Px2  9M0L d cpbAhA2 (3L + x)4 d

0

Problem 5.7-5 Refer to the tapered cantilever beam of solid circular cross section shown in Fig. 5-24 of Example 5-9. (a) Considering only the bending stresses due to the load P, determine the range of values of the ratio dB/dA for which the maximum normal stress occurs at the support. (b) What is the maximum stress for this range of values?

Solution 5.7-5

Tapered cantilever beam

FROM EQ. (5-32), EXAMPLE 5-9 s1 

32Px

Eq. (1)

x 3 pcdA + (dB  dA)a b d L

After simplification:

FIND THE VALUE OF x THAT MAKES s1 A MAXIMUM Let s1 

u v

ds1  dx

va

du dv b  ua b dx dx 2

v

x 3 N  p cdA + (dB  dA)a b d [32P] L

x 2 1  [32Px][p] [3]cdA + (dB  dA)a b d c (dB  dA) d L L



N D

x 2 x N  32pPcdA + (dB  dA)a b d cdA  2(dB  dA) d L L x 6 D  p 2 cdA + (dB  dA) d L ds1 N   dx D

x 32PcdA  2(dB  dA) d L x 4 pcdA + (dB  dA) a b d L

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ds1 x  0 dA  2(dB  dA)a b  0 dx L dA x ‹   L 2(dB  dA)

1 2a

dB  1b dA

(a) GRAPH OF x/L VERSUS dB/dA (EQ. 2)

Maximum bending stress occurs at the support when 1 …

Eq. (2)

dB … 1.5 dA

;

(b) MAXIMUM STRESS (AT SUPPORT B) Substitute x/L  1 into Eq. (1): smax 

32PL

;

pdB3

Fully Stressed Beams q

Problems 5.7-6 to 5.7-8 pertain to fully stressed beams of rectangular cross section. Consider only the bending stresses obtained from the flexure formula and disregard the weights of the beams.

B

Problem 5.7-6 A cantilever beam AB having rectangular cross sections

A

hx

with constant width b and varying height hx is subjected to a uniform load of intensity q (see figure). How should the height hx vary as a function of x (measured from the free end of the beam) in order to have a fully stressed beam? (Express hx in terms of the height hB at the fixed end of the beam.)

hB

x L

hx

hB b b

Solution 5.7-6

Fully stressed beam with constant width and varying height

hx  height at distance x hB  height at end B b  width (constant) AT DISTANCE x: M  2

3qx M  S bhx2 3q hx  x A bsallow

sallow 

qx 2 2

AT THE FIXED END (x  L): hB  L S

bhx2 6

3q A bsallow

Therefore,

hx x  hB L

hx 

hB x L

;

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SECTION 5.7

Problem 5.7-7 A simple beam ABC having rectangular cross sections with constant height h and varying width bx supports a concentrated load P acting at the midpoint (see figure). How should the width bx vary as a function of x in order to have a fully stressed beam? (Express bx in terms of the width bB at the midpoint of the beam.)

Fully Stressed Beams

P A

h

B

C

x L — 2

L — 2

h

h bx

Solution 5.7-7

441

bB

Fully stressed beam with constant height and varying width

h  height of beam (constant)

L bx  width at distance x from end Aa 0 … x … b 2 bB  width at midpoint B (x  L/2)

Px 1 AT DISTANCE x M  S  b x h2 2 6 3Px M 3Px sallow   bx  2 S bx h sallow h2

AT MIDPOINT B (x  L/2) bB 

3PL 2sallowh2

Therefore,

bx 2bB x 2x  and bx  bb L L

;

NOTE: The equation is valid for 0 … x …

L and the 2

beam is symmetrical about the midpoint.

q

Problem 5.7-8 A cantilever beam AB having rectangular cross sections with varying width bx and varying height hx is subjected to a uniform load of intensity q (see figure). If the width varies linearly with x according to the equation bx  bB x/L, how should the height hx vary as a function of x in order to have a fully stressed beam? (Express hx in terms of the height hB at the fixed end of the beam.)

B hB

hx

A x L

hx

hB bx

bB

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Solution 5.7-8

Fully stressed beam with varying width and varying height

hx  height at distance x hB  height at end B bx  width at distance x bB  width at end B

3qLx

hx 

A bB sallow

AT THE FIXED END (x  L)

x bx  bB a b L

hB 

3qL2 A bB sallow

AT DISTANCE x M

qx 2 2

sallow 

bx h2x bB x  (hx)2 6 6L 3qLx

S

Therefore,

hx x  hB A L

x AL

hx  hB

;

M  S bB h2x

Shear Stresses in Rectangular Beams Problem 5.8-1 The shear stresses t in a rectangular beam are given by Eq. (5-39): t

V h2 a  y21 b 2I 4

in which V is the shear force, I is the moment of inertia of the cross-sectional area, h is the height of the beam, and y1 is the distance from the neutral axis to the point where the shear stress is being determined (Fig. 5-30). By integrating over the cross-sectional area, show that the resultant of the shear stresses is equal to the shear force V.

Solution 5.8-1

Resultant of the shear stresses V  shear force acting on the cross section R  resultant of shear stresses t h/2

R 

Lh/2 12V

h/2

tbdy1  2 h/2

(b)

a

L0

V h2 a  y21 b bdy1 2I 4

2

h  y21 b dy1 4

bh L0 12V 2h3 b V  3 a 24 h I

bh3 12

t

V h2 a  y21 b 2I 4

3

‹ R  V Q.E.D.

;

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SECTION 5.8

443

Shear Stresses in Rectangular Beams

Problem 5.8-2 Calculate the maximum shear stress tmax

22.5 kN/m

and the maximum bending stress smax in a wood beam (see figure) carrying a uniform load of 22.5 kN/m (which includes the weight of the beam) if the length is 1.95 m and the cross section is rectangular with width 150 mm and height 300 mm, and the beam is (a) simply supported as in the figure part (a) and (b) has a sliding support at right as in the figure part (b).

300 mm

150 mm

1.95 m (a)

22.5 kN/m

1.95 m (b)

Solution 5.8-2 q  22

kN m

smax 

b  150 mm

h  300 mm

L  1.95 m

tmax 

tmax  715 kPa

MAXIMUM BENDING STRESS M

qL2 8

S

bh2 6

;

V  qL

A  bh 3V 2A

smax  4.65 MPa

(b) MAXIMUM SHEAR STRESS

(a) MAXIMUM SHEAR STRESS qL V 2

M S

tmax  ;

3V 2A

tmax  1430 kPa

;

MAXIMUM BENDING STRESS M

qL2 2

smax 

M S

Problem 5.8-3 Two wood beams, each of rectangular cross section (3.0 in.  4.0 in., actual dimensions) are glued together to form a 4.0 in. solid beam of dimensions 6.0 in.  4.0 in. (see figure). The beam is simply supported with a span of 8 ft. What is the maximum moment Mmax that may be 6.0 in. applied at the left support if the allowable shear stress in the glued joint is 200 psi? (Include the effects of the beam’s own weight, assuming that the wood weighs 35 lb/ft3.)

smax  18.59 MPa

;

M

8 ft

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Stresses in Beams (Basic Topics)

Solution 5.8-3 L  8 ft

b  4 in.

h  6 in. g  35

t allow  200 psi

Ab#h

lb

tmax 

ft3

q  g A weight of beam per unit distance q  5.833

V

1b ft

Maximum load Mmax

qL 3V 3 M  a + b 2A 2A L 2

qL2 2 AL tmax  3 2 qL2 2 AL tallow  Mmax  3 2 M

Mmax  25.4 k-ft

Problem 5.8-4 A cantilever beam of length L  2 m supports a load P  8.0 kN (see figure). The beam is made of wood with cross-sectional dimensions 120 mm * 200 mm. Calculate the shear stresses due to the load P at points located 25 mm, 75 mm, and 100 mm from the top surface of the beam. from these results, plot a graph showing the distribution of shear stresses from top to bottom of the beam.

Solution 5.8-4

qL M + L 2

;

P = 8.0 kN 200 mm L=2m 120 mm

Shear stresses in a cantilever beam Distance from the top surface (mm)

Eq. (5-39): t 

2

V h a  y21 b 2I 4

h  200 mm (y1  mm) (200)2  y21 d 4 2(80 * 106) 8,000

c

(t  N/mm2  MPa)

t  50 * 106(10,000  y21) (y1  mm; t  MPa)

0

t (kPa)

100

25

75

0.219

219

50

50

0.375

375

75

25

0.469

469

0

0.500

500

GRAPH OF SHEAR STRESS t

bh3 I  80 * 106 mm4 12

t (MPa)

0

100 (N.A.)

V  P  8.0 kN  8,000 N

t

y1 (mm)

0

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SECTION 5.8

Problem 5.8-5 A steel beam of length L  16 in. and cross-

q = 240 lb/in.

sectional dimensions b  0.6 in. and h  2 in. (see figure) supports a uniform load of intensity q  240 lb/in., which includes the weight of the beam. Calculate the shear stresses in the beam (at the cross section of maximum shear force) at points located 1/4 in., 1/2 in., 3/4 in., and 1 in. from the top surface of the beam. From these calculations, plot a graph showing the distribution of shear stresses from top to bottom of the beam.

Solution 5.8-5

h = 2 in.

Shear stresses in a simple beam

V h2 Eq. (5-39): t  a  y21 b 2I 4

y1 (in.)

t (psi)

0

1.00

0

0.25

0.75

1050

0.50

0.50

1800

0.75

0.25

2250

0

2400

qL bh3  1920 lb I   0.4 in.4 2 12

1.00 (N.A.) GRAPH OF SHEAR STRESS t

UNITS: POUNDS AND INCHES t

b = 0.6 in.

L = 16 in.

Distance from the top surface (in.)

V

445

Shear Stresses in Rectangular Beams

1920 (2)2 c  y21  (2400)(1  y21) d 2(0.4) 4

(t  psi; y1  in.)

Problem 5.8-6 A beam of rectangular cross section (width b and height h) supports a uniformly distributed load along its entire length L. The allowable stresses in bending and shear are sallow and tallow, respectively. (a) If the beam is simply supported, what is the span length L0 below which the shear stress governs the allowable load and above which the bending stress governs? (b) If the beam is supported as a cantilever, what is the length L0 below which the shear stress governs the allowable load and above which the bending stress governs?

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Solution 5.8-6 b  width

Page 446

Beam of rectangular cross section

h  height

L  length

(b) CANTILEVER BEAM

q  intensity of load

Uniform load

ALLOWABLE STRESSES

BENDING

sallow and tallow

Mmax 

(a) SIMPLE BEAM smax 

BENDING Mmax 

qL2 8

S

bh2 6

qallow 

qL2 2

3qL Mmax  S 4bh2 4sallow bh2

qallow 

2

3L

qL 2

tmax 

3qL 3V  2A 4bh

Vmax  qL A  bh

(1)

A  bh

tmax 

3qL 3V  2A 2bh

qallow 

2tallow bh 3L

h sallow L0  a b 2 tallow

;

(2)

Equate (1) and (2) and solve for L0: sallow b tallow

(4)

Equate (3) and (4) and solve for L0:

4tallowbh qallow  3L

L 0  ha

(3)

3L2

SHEAR

SHEAR Vmax 

bh2 6

3qL2 Mmax  S bh2 sallowbh2

3

smax 

S

NOTE: If the actual length is less than L 0, the shear stress governs the design. If the length is greater than L0, the bending stress governs.

;

Problem 5.8-7 A laminated wood beam on simple supports is built up by gluing together four 2 in.  4 in. boards (actual dimensions) to form a solid beam 4 in.  8 in. in cross section, as shown in the figure. The allowable shear stress in the glued joints is 65 psi, and the allowable bending stress in the wood is 1800 psi. If the beam is 9 ft long, what is the allowable load P acting at the one-third point along the beam as shown? (Include the effects of the beam’s own weight, assuming that the wood weighs 35 lb/ft3.)

3 ft

P

2 in. 2 in. 2 in. 2 in.

L  9 ft 4 in.

Solution 5.8-7 L  9 ft

b  4 in.

h  8 in.

A  bh

t allow  65 psi

s allow  1800 psi

WEIGHT OF BEAM PER UNIT DISTANCE g  35

lb ft3

q  gA

q  7.778

1b ft

ALLOWABLE LOAD BASED UPON SHEAR STRESS IN THE GLUED JOINTS; MAX. SHEAR STRESS AT NEUTRAL AXIS

t

VQ Ib

tmax 

3V 2A

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SECTION 5.8

VP tmax 

MP

qL 3V 3 2  aP + b 2A 2A 3 2

Pmax  A t allow

447

ALLOWABLE LOAD BASED UPON BENDING STRESS

qL 2 + 3 2

P  aA tmax 

Shear Stresses in Rectangular Beams

S

3 qL b 4

qL q 2 3 ft + 3 ft  (3 ft)2 3 2 2

b h2 6

qL q 2 3 ft + 3 ft  (3 ft)2 3 2 2 M smax   S S q sallow S3 3 qL  a  (3ft)b Pmax  (3 ft) 2 2 2 2 P

3 qL  4

Pmax  2.03 k (governs)

Pmax  3.165 k P allow  2.03 k

Problem 5.8-8 A laminated plastic beam of square cross section is built up by gluing together three strips, each 10 mm  30 mm in cross section (see figure). The beam has a total weight of 3.6 N and is simply supported with span length L  360 mm. Considering the weight of the beam (q) calculate the maximum permissible CCW moment M that may be placed at the right support.

;

M q 10 mm 10 mm 30 mm 10 mm L

30 mm

(a) If the allowable shear stress in the glued joints is 0.3 MPa. (b) If the allowable bending stress in the plastic is 8 MPa.

Solution 5.8-8 (a) FIND M BASED ON ALLOWABLE SHEAR STRESS IN GLUED JOINT

b  30 mm

h  30 mm

W  3.6 N

L  360 mm

q

ta  0.3 MPa

N m

beam distributed weight

MAX. SHEAR ST LEFT SUPPORT Vm  ta 

bh h 3 3

Q

b h2 9

Q 4  Ib 3bh

W L

q  10

Q

qL M + 2 L Vm Q Ib

I

and Vm  t a a 3

bh 12

Ib 

2 3

b h 12

Ib b Q

M  L cta a

qL Ib b  d Q 2

M  L cta a

qL 3bh b  d 4 2

Mmax  72.2 N # M

;

b h2 9

Q  2 3 Ib b h 12

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Stresses in Beams (Basic Topics)

(b) FIND M BASED ON ALLOWABLE BENDING STRESS AT h/2 FROM NA AT LOCATION (xm) OF MAX. BENDING MOMENT, Mm qL qx2 M M(x)  a + bx 2 L 2

Mm  a

d M(x)  0 dx

qL M L M + b a + b 2 L 2 qL



qa

2

use to find location of zero shear where max. moment occurs

simplifying

qx2 M d qL ca + bx d dx 2 L 2

Mm 

 xm 

L M 2 + b 2 qL

2 2 1 1qL + 2 M2 8q L2

bh2 b 6

M 1 qL +  qx  0 2 L

also Mm  sa S

L M + 2 qL

Equating both Mm expressions & solving for M where sa  8 MPa

MAX. MOMENT Mm Mm  a

qL qxm M + b xm  2 L 2

2

M

A

sa a

Mm  sa a

bh2 b a8 qL2 b  qL2 6 2

Mmax  9.01 N # m

Problem 5.8-9 A wood beam AB on simple supports with span length equal to 10 ft is subjected to a uniform load of intensity 125 lb/ft acting along the entire length of the beam, a concentrated load of magnitude 7500 lb acting at a point 3 ft from the right-hand support, and a moment at A of 18,500 ft-lb (see figure). The allowable stresses in bending and shear, respectively, are 2250 psi and 160 psi.

;

7500 lb 18,500 ft-lb 125 lb/ft

3 ft

A

B

(a) From the table in Appendix F, select the lightest beam that will support the loads (disregard the weight of the beam). (b) Taking into account the weight of the beam (weight density 5 35 lb/ft3), verify that the selected beam is satisfactory, or if it is not, select a new beam.

10 ft

Solution 5.8-9 (a) q  125

1b ft

L  10 ft

P  75001b M  18500 ft-b d  3 ft

sAllow  2250 psi

t allow  160 psi

RB  7.725 * 103 1b Vmax  RB

Vmax  7.725 * 103 1b qd2 Mmax  RB d  2

qL d M + P  RA  2 L L

Mmax  2.261 * 104 1b-ft

RA  1.025 * 103 1b

tmax 

RB 

qL Ld M + P + 2 L L

3V 2A

Areq 

Areq  72.422 in.2

3Vmax 2tallow

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SECTION 5.8

smax 

M S

Sreq 

Mmax sallow

Sreq  120.6 in.3

From Appendix F: Select 8 * 12 in. beam (nominal ; dimensions) S  165.3 in.3

A  86.25 in.2

Vmax  RB

Areq 

g  35

ft3

8 * 12 beam is still satisfactory for shear.

Mmax  RB d 

qbeam  g A

qd2 2

Mmax  2.293 * 104 1b-ft Sreq 

RB  7.725 * 103 1b +

qbeam L 2

Mmax sallow

;

Use 8 * 12 in. beam

Problem 5.8-10 A simply supported wood beam of rectangular cross section and span length 1.2 m carries a concentrated load P at midspan in addition to its own weight (see figure). The cross section has width 140 mm and height 240 mm. The weight density of the wood is 5.4 kN/m3. Calculate the maximum permissible value of the load P if (a) the allowable bending stress is 8.5 MPa, and (b) the allowable shear stress is 0.8 MPa.

P 240 mm

0.6 m

0.6 m

140 mm

Simply supported wood beam (a) ALLOWABLE P BASED UPON BENDING STRESS

P 240 mm

0.6 m

0.6 m

h  240 mm

A  bh  33,600 mm2 S

Sreq  122.3 in.3 < S

8 * 12 beam is still satisfactory for moment.

RB  7.83 * 103 1b

b  140 mm

1b ft

qtotal  q + qbeam q total  145.964

1b q beam  20.964 ft

Solution 5.8-10

3Vmax 2 tallow

Areq  73.405 in.2 < A

(b) REPEAT (A) CONSIDERING THE WEIGHT OF THE BEAM 1b

449

Shear Stresses in Rectangular Beams

bh2  1344 * 103 mm3 6

140 mm

sallow  8.5 MPa s  Mmax  +

Mmax S

qL2 P(1.2 m) PL +  4 8 4 (181.44 N/m)(1.2 m)2 8

 0.3 P + 32.66 N # m (P  newtons; M  N # m)

g  5.4 kN/m3

Mmax  Ssallow  (1344 * 103 mm3)(8.5 MPa)

L  1.2 m q  gbh  181.44 N/m

 11,424 N # m Equate values of Mmax and solve for P: 0.3P + 32.66  11,424 P  37,970 N or P  38.0 kN

;

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Stresses in Beams (Basic Topics)

(b) ALLOWABLE LOAD P BASED UPON SHEAR STRESS tallow  0.8 MPa t  V

3V 2A

qL (181.44 N/m)(1.2 m) P P +  + 2 2 2 2

P  + 108.86 (N) 2 2At 2 V  (33,600 mm2)(0.8 MPa)  17,920 N 3 3

Equate values of V and solve for P: P + 108.86  17,920 P  35,622 N 2 or P  35.6 kN

;

NOTE: The shear stress governs and Pallow  35.6 kN

Problem 5.8-11 A square wood platform, 8 ft * 8 ft in area, rests on masonry walls (see figure). The deck of the platform is constructed of 2 in. nominal thickness tongue-and-groove planks (actual thickness 1.5 in.; see Appendix F) supported on two 8-ft long beams. The beams have 4 in. * 6 in. nominal dimensions (actual dimensions 3.5 in. * 5.5 in.). The planks are designed to support a uniformly distributed load w (lb/ft2) acting over the entire top surface of the platform. The allowable bending stress for the planks is 2400 psi and the allowable shear stress is 100 psi. When analyzing the planks, disregard their weights and assume that their reactions are uniformly distributed over the top surfaces of the supporting beams. (a) Determine the allowable platform load w1 (lb/ft2) based upon the bending stress in the planks. (b) Determine the allowable platform load w2 (lb/ft2) based upon the shear stress in the planks. (c) Which of the preceding values becomes the allowable load wallow on the platform? (Hints: Use care in constructing the loading diagram for the planks, noting especially that the reactions are distributed loads instead of concentrated loads. Also, note that the maximum shear forces occur at the inside faces of the supporting beams.)

8 ft

8 ft

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SECTION 5.8

Solution 5.8-11

Shear Stresses in Rectangular Beams

451

Wood platform with a plank deck Load on one plank: q c

w(lb/ft2) 2

2

144 in. / ft

d(b in.) 

Reaction R  qa

wb (lb/in.) 144

96 in. wb wb b  a b (48)  2 144 3

(R  lb; w  lb/ft2; b  in.) Mmax occurs at midspan. Mmax  Ra 

q(48 in.)2 3.5 in. 89 in. + b  2 2 3

wb wb 89 (46.25)  (1152)  wb 3 144 12 (M  lb-in.; w  lb/ft2; b  in.)

Platform: 8 ft * 8 ft t  thickness of planks

Allowable bending moment:

 1.5 in.

Mallow  s allow S  (2400 psi)(0.375 b)

w  uniform load on the deck (lb/ft )

 900 b (lb-in.)

2

sallow  2400 psi

Equate Mmax and Mallow and solve for w:

tallow  100 psi

89 wb  900 b w1  121 lb/ft2 12

2

Find wallow (lb/ft ) (a) ALLOWABLE LOAD BASED UPON BENDING STRESS IN THE

(b) ALLOWABLE

;

LOAD BASED UPON SHEAR STRESS IN THE

PLANKS

PLANKS

Let b  width of one plank (in.)

See the free-body diagram in part (a). A  1.5b (in.2) S

b (1.5 in.)2 6

Vmax occurs at the inside face of the support. Vmax  qa

89 in. b  44.5q 2

 (44.5)a

 0.375b (in.3) Free-body diagram of one plank supported on the beams:

89 wb wb b  144 288

(V  lb; w  lb/ft2; b  in.) Allowable shear force: t 

3V 2A

Vallow 

2Atallow 3

2(1.5 b)(100 psi)  100 b (lb) 3

Equate Vmax and Vallow and solve for w: 89wb  100b w2  324 lb/ft2 288

;

(c) ALLOWABLE LOAD Bending stress governs. wallow  121 lb/ft2

;

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Stresses in Beams (Basic Topics)

Problem 5.8-12 A wood beam ABC with simple supports at A and B and an overhang BC has height h  300 mm (see figure). The length of the main span of the beam is L  3.6 m and the length of the overhang is L/3  1.2 m. The beam supports a concentrated load 3P  18 kN at the midpoint of the main span and a moment PL/2  10.8 kN . m at the free end of the overhang. The wood has weight density g  5.5 kN/m3.

L — 2

3P

PL M = ––– 2

A

h= 300 mm

C

B L — 3

L

b

(a) Determine the required width b of the beam based upon an allowable bending stress of 8.2 MPa. (b) Determine the required width based upon an allowable shear stress of 0.7 MPa.

Solution 5.8-12 Numerical data: L  3.6 m A  bh g  5.5

s

h  300 mm P  6 kN

kN m3

M

PL 2

qbeam  g A

Reactions, max. shear and moment equations RA 

M 4 3P 4  + qbeam L  P  qbeam L 2 L 9 9

RB 

M 8 3P 8 + + qbeam L  2 P + qbeam L 2 L 9 9

Vmax  RB  2 P +

8 L q 9 beam

L L2 PL 17   q L2 MD  RA  qbeam 2 2 2 18 beam MB 

3PL

b  87.8 mm

sallow h2

Vmax  2 P + t  b

8 L q 9 beam

3 Vmax 3 Vmax  2A 2 bh 8 4 3P 3 a2 P + qbeam L b  + gL 2 bh 9 bh 3 3P ha t allow

b  89.1 mm

(a) REQUIRED WIDTH b BASED UPON BENDING STRESS sallow  8.2 MPa PL 2

;

(b) REQUIRED WIDTH b BASED UPON SHEAR STRESS tallow  0.7 MPa

4  gLb 3

b  89.074 mm

Shear stress governs

PL 2

Mmax  MB 

b

6 Mmax Mmax  S bh2

; (governs)

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SECTION 5.9

Shear Stresses in Circular Beams

453

Shear Stresses in Circular Beams Problem 5.9-1 A wood pole of solid circular cross section (d  diameter)

q0 = 20 lb/in.

is subjected to a horizontal force P  450 lb (see figure). The length of the pole is L  6 ft, and the allowable stresses in the wood are 1900 psi in bending and 120 psi in shear. Determine the minimum required diameter of the pole based upon (a) the allowable bending stress, and (b) the allowable shear stress.

d d

L

Solution 5.9-1 q  20

1b in

3

L  6 ft

dmin 

s allow  1900 psi

dmin  5.701 in.

t allow  120 psi Vmax  Mmax 

qL 2

Vmax  720 1b

qL 2 L 2 3

(b) BASED UPON SHEAR STRESS t

Mmax  2.88 * 103 1b-ft

(a) BASED UPON BENDING STRESS s

32 Mmax

A p sallow

32 M M  S pd3

4V 16V  3A 3pd2

dmin 

16 Vmax

A 3p tallow

dmin  3.192 in.

Bending stress governs

dmin  5.70 in.

;

Problem 5.9-2 A simple log bridge in a remote area consists of two parallel logs with planks across them (see figure). The logs are Douglas fir with average diameter 300 mm. A truck moves slowly across the bridge, which spans 2.5 m. Assume that the weight of the truck is equally distributed between the two logs. Because the wheelbase of the truck is greater than 2.5 m, only one set of wheels is on the bridge at a time. Thus, the wheel load on one log is equivalent to a concentrated load W acting at any position along the span. In addition, the weight of one log and the planks it supports is equivalent to a uniform load of 850 N/m acting on the log. Determine the maximum permissible wheel load W based upon (a) an allowable bending stress of 7.0 MPa, and (b) an allowable shear stress of 0.75 MPa.

x

W 850 N/m 300 mm

2.5 m

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Solution 5.9-2

Page 454

Stresses in Beams (Basic Topics)

Log bridge

Diameter d  300 mm sallow  7.0 MPa tallow  0.75 MPa Find allowable load W

(b) BASED UPON SHEAR STRESS Maximum shear force occurs when wheel is adjacent to support (x  0). Vmax  W +

(a) BASED UPON BENDING STRESS Maximum moment occurs when wheel is at midspan (x  L/2). Mmax  

qL WL + 4 8

A

 0.625W + 664.1 (N # m) (W  newtons) S

 W + 1062.5 N (W  newtons)

2

W 1 (2.5 m) + (850 N/m)(2.5 m)2 4 8 pd3  2.651 * 103m3 32

qL 1  W + (850 N/m)(2.5 m) 2 2

pd2  0.070686 m2 4

tmax 

4Vmax 3A

Vmax 

3Atallow 3  (0.070686 m2)(0.75 MPa) 4 4

 39,760 N

Mmax  Ssallow  (2.651 * 103 m3)(7.0 MPa)

‹ W + 1062.5 N  39,760 N

 18,560 N # m

W  38,700 N  38.7 kN

;

‹ 0.625W + 664.1  18,560 W  28,600 N  28.6 kN

;

b

Problem 5.9-3 A sign for an automobile service station is supported by two aluminum poles of hollow circular cross section, as shown in the figure. The poles are being designed to resist a wind pressure of 75 lb/ft2 against the full area of the sign. The dimensions of the poles and sign are h1  20 ft, h2  5 ft, and b  10 ft. To prevent buckling of the walls of the poles, the thickness t is specified as one-tenth the outside diameter d. (a) Determine the minimum required diameter of the poles based upon an allowable bending stress of 7500 psi in the aluminum. (b) Determine the minimum required diameter based upon an allowable shear stress of 2000 psi.

h2

d t=— 10

Wind load

d h1

Probs. 5.9.3 and 5.9.4

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SECTION 5.9

Solution 5.9-3

Wind load on a sign

b  width of sign b  10 ft p  75 lb/ft2 sallow  7500 psi tallow  2000 psi d  diameter t

(b) REQUIRED DIAMETER BASED UPON SHEAR STRESS Vmax  W  1875 lb t

W  wind force on one pole

b W  ph2 a b  1875 lb 2

d 10

(a) REQUIRED DIAMETER BASED UPON BENDING STRESS Mmax  W ah1 +

h2 b  506,250 lb-in. 2

4 p (d24  d24) d2  d d1  d  2t  d 64 5

I

4d 4 pd 4 369 p a b I  cd 4  a b d  64 5 64 625 

369pd 4 (in.4) 40,000

c

d 2

Shear Stresses in Circular Beams

M(d/2) 17.253 M Mc   I 369pd 4/40,000 d3 (17.253)(506,250 lb-in.) 17.253 Mmax  d3  sallow 7500 psi

s

 1164.6 in.

r1 

r2 

d 2

d d d 2d t   2 2 10 5

r22 + r2r1 + r21 r2 2 + r1 2 2d 2 d 2d d 2 a b + a ba b + a b 2 2 5 5 61   2 2 41 d 2d a b + a b 2 5 A

p 2 p 4d 2 9pd2 (d2  d21)  cd2  a b d  4 4 5 100

100 4V 61 V a ba b  7.0160 2 3 41 9pd2 d 7.0160 Vmax d2  tallow t

(d  inches)

3

4V r22 + r2r1 + r12 b a 3A r2 2 + r1 2

d  10.52 in.

;

Problem 5.9-4 Solve the preceding problem for a sign and poles

having the following dimensions: h1  6.0 m, h2  1.5 m, b  3.0 m, and t  d/10. The design wind pressure is 3.6 kPa, and the allowable stresses in the aluminum are 50 MPa in bending and 14 MPa in shear.



(7.0160)(1875 lb)  6.5775 in.2 2000 psi

d  2.56 in.

;

(Bending stress governs.)

455

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Stresses in Beams (Basic Topics)

Solution 5.9-4

Wind load on a sign

b  width of sign

(b) REQUIRED DIAMETER BASED UPON SHEAR STRESS Vmax  W  8.1 kN

b  3.0 m p  3.6 kPa sallow  50 MPa tallow  16 MPa d  diameter

t

4V r2 2 + r1 r2 + r1 2 a b 3A r2 2 + r1 2

r1 

d d d 2d t   2 2 10 5

W  wind force on one pole

(a) REQUIRED DIAMETER BASED UPON BENDING STRESS h2 b  54.675 kN # m Mmax  W ah1 + 2 s

Mc I

I

A

4 d 5



4d 4 p 4 cd  a b d 64 5

I 

d 2

d 2 2d 2 a b + a b 2 5

M(d/2) 17.253 M Mc   s 4 I 369pd /40,000 d3 (17.253)(54.675 kN # m) 17.253Mmax d3  sallow 50 MPa  0.018866 m3 ;



p (d 2  d1 2) 4 2 p 2 4d 2 9pd2 cd  a b d  4 5 100

100 V 4V 61 a ba b  7.0160 2 3 41 9pd2 d (7.0160)(8.1 kN) 7.0160 Vmax d2   tallow 14 MPa  0.004059 m2

(d  meters)

d  0.266 m  266 m

r2 2 + r1 2 d 2d 2d 2 d 2 a b + a ba b + a b 2 5 5 5

t

369pd 4 pd 4 369 a b (m 4) 64 625 40,000

c



p 4 (d  d41) 64 2

d2  d d1  d  2t 

d 2

r2 2 + r1r2 + r1 2

b W  ph2 a b  8.1 kN 2

d t 10

r2 

d  0.06371 m  63.7 mm Bending stress governs

;

61 41

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Page 457

SECTION 5.10

Shear Stresses in Beams with Flanges

Shear Stresses in Beams with Flanges Problem 5.10-1 through 5.10-6 A wide-flange beam (see figure) having

y

the cross section described below is subjected to a shear force V. Using the dimensions of the cross section, calculate the moment of inertia and then determine the following quantites: (a) The maximum shear stress tmax in the web. (b) The minimum shear stress tmin in the web. (c) The average shear stress taver (obtained by dividing the shear force by the area of the web) and the ratio tmax/taver. (d) The shear force Vweb carried in the web and the ratio Vweb /V.

z

O h1

h

t

NOTE: Disregard the fillets at the junctions of the web and flanges and determine all quantities, including the moment of inertia, by considering the cross section to consist of three rectangles.

b Probs 5.10.1through 5.-10.6

Problem 5.10-1 Dimensions of cross section: b  6 in., t  0.5 in., h  12 in., h1  10.5 in., and V  30 k.

Solution 5.10-1

Wide-flange beam (b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 5-48b)

b  6.0 in.

tmin 

t  0.5 in. h  12.0 in.

taver 

V  30 k

1 (bh3  bh31 + th31)  333.4 in.4 12

;

;

(d) SHEAR FORCE IN THE WEB (Eq. 5-49)

(a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-48a) V (bh2  bh21 + th21)  5795 psi 8It

V  5714 psi th1

tmax  1.014 taver

MOMENT OF INERTIA (Eq.5-47)

tmax 

;

(c) AVERAGE SHEAR STREAR IN THE WEB (Eq. 5-50)

h1  10.5 in.

I

Vb 2 (h  h12)  4555 psi 8It

;

Vweb 

th1 (2tmax + tmin)  28.25 k 3

Vweb  0.942 V

;

;

457

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Stresses in Beams (Basic Topics)

Problem 5.10-2 Dimensions of cross section: b  180 mm, t  12 mm, h  420 mm, h1  380 mm, and V  125 kN.

Solution 5.10-2

Wide-flange beam b  180 mm

(b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 5-48b)

t  12 mm

tmin 

h  420 mm

taver 

V  125 kN

V  27.41 MPa th1

tmax  1.037 taver

MOMENT OF INERTIA (Eq. 5-47) 1 (bh3  bh31 + th31)  343.1 * 106 mm4 12

;

;

(d) SHEAR FORCE IN THE WEB (Eq. 5-49) Vweb 

(a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-48a) tmax 

;

(c) AVERAGE SHEAR STRESS IN THE WEB (Eq. 5-50)

h1  380 mm

I

Vb 2 (h  h21)  21.86 MPa 8It

V (bh2  bh21 + th21)  28.43 MPa 8It

;

th1 (2tmax + tmin)  119.7 kN 3

Vweb  0.957 V

;

;

Problem 5.10-3 Wide-flange shape, W 8 * 28 (see Table E-1(a), Appendix E); V  10 k.

Solution 5.10-3

Wide-flange beam (b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 5-48b)

W 8 * 28 b  6.535 in.

tmin 

t  0.285 in. h  8.06 in.

taver 

V  10 k 1 (bh3  bh31 + th31)  96.36 in.4 12

;

;

(d) SHEAR FORCE IN THE WEB (EQ. 5-49)

(a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-48a) V (bh2  bh21 + th21)  4861 psi 8It

V  4921 psi th1

tmax  0.988 taver

MOMENT OF INERTIA (Eq. 5-47)

tmax 

;

(c) AVERAGE SHEAR STRESS IN THE WEB (Eq. 5-50)

h1  7.13 in.

I

Vb 2 (h  h21)  4202 psi 8It

;

Vweb 

th1 (2tmax + tmin)  9.432 k 3

Vweb  0.943 V

;

;

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SECTION 5.10

459

Shear Stresses in Beams with Flanges

Problem 5.10-4 Dimensions of cross section: b  220 mm, t  12 mm, h  600 mm, h1  570 mm, and V  200 kN .

Solution 5.10-4

Wide-flange beam b  220 mm

(c) AVERAGE SHEAR STRESS IN THE WEB (EQ. 5-50)

t  12 mm

taver 

h  600 mm

V  29.24 MPa th1

;

tmax  1.104 taver

h1  570 mm V  200 kN

(d) SHEAR FORCE IN THE WEB (Eq. 5-49) MOMENT OF INERTIA (Eq. 5-47)

Vweb 

1 I (bh3  bh31 + th31)  750.0 * 106 mm4 12 (a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-48a) tmax 

V (bh2  bh21 + th21)  32.28 MPa 8It

th1 (2tmax + tmin)  196.1 kN 3

Vweb  0.981 V

;

;

;

(b) MINIMUM SHEAR STRESS IN THE WEB (EQ. 5-48b) tmin 

Vb 2 (h  h21)  21.45 MPa 8It

;

Problem 5.10-5 Wide-flange shape, W 18 * 71 (see Table E-1(a), Appendix E); V  21 k.

Solution 5.10-5

Wide-flange beam (b) MINIMUM SHEAR STRESS IN THE WEB (EQ. 5-48b)

W 18 * 71 b  7.635 in.

tmin 

t  0.495 in. h  18.47 in.

taver 

V  21 k 1 (bh3  bh31 + th31)  1162 in.4 12

;

;

(d) SHEAR FORCE IN THE WEB (EQ. 5-49)

(a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-48a) V (bh2  bh21 + th21)  2634 psi 8It

V  2518 psi th1

tmax  1.046 taver

MOMENT OF INERTIA (Eq. 5-47)

tmax 

;

(c) AVERAGE SHEAR STRESS IN THE WEB (EQ. 5-50)

h1  16.85 in.

I

Vb 2 (h  h21)  1993 psi 8It

;

Vweb 

th1 (2tmax + tmin)  20.19 k 3

Vweb  0.961 V

;

;

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Page 460

Stresses in Beams (Basic Topics)

Problem 5.10-6 Dimensions of cross section: b  120 mm, t  7 mm, h  350 mm, h1  330 mm, and V  60 kN

Solution 5.10-6

Wide-flange beam (b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 5-48)

b  120 mm

tmin 

t  7 mm h  350 mm

taver 

V  60 kN

V  25.97 MPa th1

tmax  1.093 taver

MOMENT OF INERTIA (Eq. 5-47) 1 (bh3 bh31 + th31)  90.34 * 106 mm4 12

;

;

(d) SHEAR FORCE IN THE WEB (Eq. 5-49) Vweb 

(a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-48a) tmax 

;

(c) AVERAGE SHEAR STRESS IN THE WEB (Eq. 5-50)

h1  330 mm

I

Vb 2 (h  h21)  19.35 MPa 8It

V (bh2  bh21 + th21)  28.40 MPa 8It

th1 (2tmax + tmin)  58.63 kN 3

Vweb  0.977 V

;

;

;

Problem 5.10-7 A cantilever beam AB of length L  6.5 ft supports a trapezoidal distributed load of peak intensity q, and minimum intensity q/2, that includes the weight of the beam (see figure). The beam is a steel W 12  14 wide-flange shape (see Table E-1(a), Appendix E). Calculate the maximum permissible load q based upon (a) an allowable bending stress sallow  18 ksi and (b) an allowable shear stress tallow  7.5 ksi. (Note: Obtain the moment of inertia and section modulus of the beam from Table E-1(a))

q — 2

q

B

A L = 6.5 ft

Solution 5.10-7 b  3.97 in.

I  88.6 # in.4

t  0.2 in.

Vmax 

t f  0.225 in. S  14.9 in.3 h  11.9 in. h1  h  2 tf h1  11.45 in. L  6.5 ft

s allow  18 ksi

t allow  7.5 ksi

a

q + qb L 2 2

Vmax 

Mmax 

1q 2 1 q 2L L + L 22 22 3

Mmax 

5 qL2 12

3 qL 4

W 12  14

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SECTION 5.10

q



12S sallow 5L2

q  1270 lb/ft

q  3210

(b) MAXIMUM LOAD UPON SHEAR STRESS tmax 

Vmax 1 bh2  bh21 + th212 8It

Problem 5.10-8 A bridge girder AB on a simple span of length L  14 m supports a distributed load of maximum intensity q at midspan and minimum intensity q/2 at supports A and B that includes the weight of the girder (see figure). The girder is constructed of three plates welded to form the cross section shown. Determine the maximum permissible load q based upon (a) an allowable bending stress sallow  110 MPa and (b) an allowable shear stress tallow  50 MPa.

461

3 qL 1 bh2  bh21 + th212 32It tallow32It q 3 L1 bh2  bh21 + th212

(a) MAXIMUM LOAD BASED UPON BENDING STRESS 5 2 qL 12 M  s S S

Shear Stresses in Beams with Flanges

lb ft

Shear stress governs

q  1270 lb/ft

q q — 2

q — 2

A

B L = 14 m

;

450 mm 32 mm

16 mm 1800 mm

32 mm 450 mm

Solution 5.10-8 L  14 m

(a) MAXIMUM LOAD BASED UPON BENDING STRESS

h  1864 mm h1  1800 mm b  450 mm I

tf  32 mm tw  16 mm

1 1 bh3  bh31 + tw h312 12

I  3.194 * 10 mm 10

S

2I h

RA  RB 

4

S  3.427 * 107 mm3 qL qL 3 +  qL 22 42 8

sallow  110 MPa qLL qLL 3 L qL   8 2 22 4 24 6

Mmax 

5 qL2 48 5 qL2 Mmax 48 s  S S 

qmax 

sallow S 5 2 L 48

qmax  184.7

kN m

;

;

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Stresses in Beams (Basic Topics)

(b) MAXIMUM LOAD BASED UPON SHEAR STRESS



tallow  50 MPa Vmax  R A  tmax 

3 qL 8

qmax 

3 qL 1 bh2  bh21 + th212 64It 64 tallow Itw

3 L 1bh2 bh12 tw h122

qmax  247 kN/m

Vmax 1 bh2  bh21 + th212 8It

;

‹ Bending stress governs: qmax 184.7 kN/m

Problem 5.10-9 A simple beam with an overhang supports a uniform

load of intensity q  1200 lb/ft and a concentrated load P  3000 lb (see figure). The uniform load includes an allowance for the wight of the beam. The allowable stresses in bending and shear are 18 ksi and 11 ksi, respectively. Select from Table E-2 (a), Appendix E, the lightest I-beam (S shape) A that will support the given loads.

8 ft

q = 1200 lb/ft

C

12 ft

4 ft

Beam with an overhand

sallow  18 ksi q  1200

P = 3000 lb

B

(Hint: Select a beam based upon the bending stress and then calculate the maximum shear stress. If the beam is overstressed in shear, select a heavier beam and repeat.)

Solution 5.10-9

P = 3000 lb

t allow  11 ksi

lb ft

L  12 ft

P  3000 lb

Sum moments about A & Solve for RB

Find moment at D (at Load P between A and B) MD  R A 8 ft  q

(8 ft)2 2

MD  1.28 * 104 lb-ft Mmax  | MB|

Mmax  2.16 * 104 lb-ft

2

RB 

;

4 1 q a Lb + P(8 ft + 16 ft) 3 2 12 ft

RB  1.88 * 104 lb

Sum forces in vertical direction RA  q (16 ft) + 2P  R B R A  6.4 * 103 lb Vmax  R B  (P + q4 ft) Vmax  1.1 * 104 lb at B (4 ft)2 MB   P (4 ft)  q 2 MB  2.16 * 10 lb-ft 4

Required section modulus: S

Mmax sallow

S  14.4 in.3

Lightest beam is S 8 * 23 (from Table E-2(a)) I  64.7 in.4

S  16.2 in.3

b  4.17 in.

t  0.441 in.

t f  0.425 in.

h  8 in.

h1  h  2 tf h1  7.15 in.

Check max. shear stress tmax 

Vmax 1 bh2  bh21 + th212 8 It

tmax  3674 6 11,000 psi so ok for shear Select S 8 * 23 beam

;

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SECTION 5.10

Shear Stresses in Beams with Flanges

Problem 5.10-10 A hollow steel box beam has the rectangular cross section shown in the figure. Determine the maximum allowable shear force V that may act on the beam if the allowable shear stress in 36 Mpa.

463

20 mm

450 10 mm mm

10 mm 20 mm

200 mm

Solution 5.10-10

Rectangular box beam

tallow  36 MPa Find Vallow t Vallow  I

VQ It tallowIt Q 1 1 (200)(450)3  (180)(410)3 12 12

 484.9* 106mm4

Q  (200)a

410 410 450 450 ba b  (180)a ba b 2 4 2 4

 1.280 * 106 mm3 Vallow  

tallow It Q (36 MPa)(484.9 * 106 mm4)(20 mm)

 273 kN

1.280 * 106 mm3 ;

t  2(10 mm)  20 mm

Problem 5.10-11 A hollow aluminum box beam has the square cross section shown in the figure. Calculate the maximum and minimum shear stresses tmax and tmin in the webs of the beam due to a shear force V  28 k.

1.0 in.

1.0 in.

12 in.

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CHAPTER 5

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Stresses in Beams (Basic Topics)

Solution 5.10-11

Square box beam Q a

1  (b3  b31)  91.0 in.3 8

V  28 k  28,000 lb t1  1.0 in . b  12 in.

b21 b1 b2 b ba b  a ba b 2 4 2 4

tmax 

VQ (28,000 lb)(91.0 in.3)  1424 psi  It (894.67 in.4)(2.0 in.)

 1.42 ksi

b1  10 in.

;

MINIMUM SHEAR STRESS IN THE WEB (AT LEVEL A.A) VQ t It

t  2t1  2.0 in .

bt1 b t1 Q  Ay  (bt1)a  b  a b (bt1) 2 2 2

MOMENT OF INERTIA t1 

b  b1 2

MAXIMUM SHEAR STRESS IN THE WEB (AT NEUTRAL AXIS)

Q

(12 in.) [(12 in.)2  (10 in.)2]  66.0 in.3 8

b b2 Q  A1y1 A2y2 A1  ba b  2 2

tmin 

1 4 I (b  b41)  894.67 in.4 12

A2  b1 a

b1 b21 b 2 2

b 1 b y1  a b  2 2 4

b Q  (b2  b21) 8

VQ (28,000 lb)(66.0 in.3)  1033 psi  It (894.67 in4)(2.0 in.)

 1.03 ksi

;

b1 1 b1 y2  a b  2 2 4

y

Problem 5.10-12 The T-beam shown in the figure has cross-sectional dimensions

as follows: b  220 mm, t  15 mm, h  300 mm, and h1  275 mm . The beam is subjected to a shear force V  60 kN. Determine the maximum shear stress tmax in the web of the beam.

t h1

z

C

c b

Probs 5.10.12 and 5.-10.13

Solution 5.10-12 h  300 mm

h1  280 mm

b  210 mm

t  16 mm

t f  h  h1

V  68 kN

t f  20 mm

LOCATION OF NEUTRAL AXIS

c

b1 h  h12 a

c  87.419 mm c1  c

h  h1 h1 b + t h1 a h  b 2 2

b1 h  h12 + t h1

c1  87.419 mm

c2  h  c c2  212.581 mm

h

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SECTION 5.10

MOMENT OF INERTIA ABOUT THE z-AXIS Iweb 

1 3 1 t c + t1 c1  tf 23 3 2 3

Iweb  5.287 * 107 mm4 Iflange 

tf 2 1 b tf3 + b tf ac1  b 12 2

Shear Stresses in Beams with Flanges

Iflange  2.531 * 107 mm4 I  Iweb + Iflange

I  7.818 * 107 mm4

FIRST MOMENT OF AREA ABOVE THE z AXIS c2 Q  tc2 2 VQ tmax  tmax  19.7 MPa ; It

Problem 5.10-13 Calculate the maximum shear stress tmax in the web of the T-beam shown in the figure if b  10 in., t  0.5 in., h  7 in., h1  6.2 in., and the shear force V  5300 lb.

Solution 5.10-13

T-beam

h  7 in.

h1  6.2 in.

b  10 in.

t  0.5 in.

tf  h  h1

tf  0.8 in.

LOCATION OF NEUTRAL AXIS b 1 h  h12 a

c  1.377 in. c1  c

h  h1 h1 b + t h1 a h  b 2 2

b 1 h  h12 + t h1

c1  1.377 in.

c2  h  c

Iweb 

1 1 t c 3 + t1 c1  tf23 3 2 3

Iweb  29.656 in.4

V  5300 lb

c

MOMENT OF INERTIA ABOUT THE z-AXIS

c2  5.623 in.

Iflange 

tf 2 1 btf3 + btf a c1  b 12 2

Iflange  8.07 in.4 I  Iweb + Iflange

I  37.726 in.4

FIRST MOMENT OF AREA ABOVE THE z AXIS c2 Q  tc2 2 VQ tmax  tmax  2221 psi ; It

465

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Stresses in Beams (Basic Topics)

Built-Up Beams Problem 5.11-1 A prefabricated wood I-beam serving as a floor joist

y

has the cross section shown in the figure. The allowable load in shear for the glued joints between the web and the flanges is 65 lb/in. in the longitudinal direction. Determine the maximum allowable shear force Vmax for the beam.

0.75 in.

z 0.625 in.

8 in.

O

0.75 in.

5 in.

Solution 5.11-1

Wood I-beam All dimensions in inches. Find Vmax based upon shear in the glued joints. Allowable load in shear for the glued joints is 65 lb/in. ‹ fallow  65 lb/in. fallow I Q

f

VQ I

I

(b  t)h31 bh3  12 12



Vmax 

;

1 1 (5) (9.5)3  (4.375)(8)3  170.57 in.4 12 12

Q  Qflange  Af df  (5)(0.75)(4.375)  16.406 in.3 Vmax 



fallowI Q (65 lb/in.)(170.57 in.4) 16.406 in.3

 676 lb

;

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SECTION 5.11

Built-Up Beams

Problem 5.11-2 A welded steel girder having the cross section shown in the figure

y

is fabricated of two 300 mm * 25 mm flange plates and a 800 mm * 16 mm web plate. The plates are joined by four fillet welds that run continuously for the length of the girder. Each weld has an allowable load in shear of 920 kN/m. Calculate the maximum allowable shear force Vmax for the girder.

25 mm

z 16 mm

O

800 mm

25 mm 300 mm

Solution 5.11-2 h  850 mm

h1  800 mm

b  300 mm

t  16 mm

Qflange  3.094 * 106 mm3 f allow  920

tf  25 mm I

(b  t)h13 b h3  12 12

f

4

Qflange  A f df Qflange  b t f a

f  2 fallow

(2 welds, one either side of web)

I  3.236 * 10 mm 9

kN m

h  tf 2

b

VQ I

Vmax 

Vmax  1.924 MN

fI Qflange

;

y

Problem 5.11-3 A welded steel girder having the cross section shown in the figure is fabricated of two 20 in. * 1 in. flange plates and a 60 in. * 5/16 in. web plate. The plates are joined by four longitudinal fillet welds that run continuously throughout the length of the girder. If the girder is subjected to a shear force of 280 kips, what force F (per inch of length of weld) must be resisted by each weld?

1 in.

z

O

60 in.

5 — in. 16 1 in. 20 in.

467

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CHAPTER 5

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Stresses in Beams (Basic Topics)

Solution 5.11-3 h  62 in.

h1  60 in.

b  20 in.

5 t in. 16

Qflange  btf a

2

b

Qflange  610 in3

tf  1 in.

V  280 k

(b  t)h13 bh3 I  12 12

F

I  4.284 * 10 in. 4

h  tf

4

f  2F 

VQflange

F  1994 * 103 lb.in.

21

F  1994 lb/in.

Qflange  Af df

VQ I

;

Problem 5.11-4 A box beam of wood is constructed of two

y 25 mm

260 mm * 50 mm boards and two 260 mm * 25 mm boards (see figure). The boards are nailed at a longitudinal spacing s  100 mm. If each nail has a allowable shear force F  1200 N, what is the maximum allowable shear force Vmax? z

O 50 mm

50 mm

260 mm

260 mm

Solution 5.11-4

25 mm

Wood box beam

All dimensions in millimeters. b  260 b1  260  2(50)  160 h  310 h1  260 s  nail spacing  100 mm F  allowable shear force for one nail  1200 N f  shear flow between one flange and both webs 2(1200 N) 2F   24 kN/ m fallow  s 100 mm

fallow I Q

f

VQ I

I

1 (bh3  b1h31)  411.125 * 106 mm4 12

Vmax 

Q  Qflange  Afdf  (260)(25)(142.5)  926.25 * 103 mm4 Vmax 

fallowI (24 kN/ m)(411.25 * 106 mm4) .  Q 926.25 * 103 mm3

 10.7 kN

;

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SECTION 5.11

Problem 5.11-5 A box beam is constructed of four wood

boards as shown in the figure part (a). The webs are 8 in.  1 in. and the flanges are 6 in.  1 in. boards (actual dimensions), joined by screws for which the allowable load in shear is F  250 lb per screw. (a) Calculate the maximum permissible longitudinal spacing smax of the screws if the shear force V is 1200 lb. (b) Repeat (a) if the flanges are attached to the webs using a horizontal arrangement of screws as shown in the figure part (b).

Solution 5.11-5 V  1200 lb

y

z 1 in.

8 in.

1 in. 1 in.

1 in.

6 in.

6 in. 1 in.

(a)

F  250 lb

1 in. (b)

(b) Horizontal screws

h1  8 in. t  1 in.

(b  2t) h13 bh I  12 12 3

Qa  bt (4.5 in.)

I  329.333 in.4

h1  6 in.

b  8 in.

t  1 in.

(b  2t) h13 bh  12 12 3

I  233.333 in.4

Qb  (b  2 t) t (3.5 in.)

Qa  27 in.3

f

Qb  21 in.3

VQ 2F  I s

smax 

2FI VQa

smax  5.08 in.

h  8 in.

I

VQ 2F  I S

smax 

1 in.

Wood box beam

h  10 in.

f

Web 8 in.

O

469

Flange

Flange 1 in. Web

(a) Vertical screws b  6 in.

Built-Up Beams

2FI VQb

smax  4.63 in.

;

;

y

Problem 5.11-6 Two wood box beams (beams A and B) have the same outside dimensions (200 mm * 360 mm) and the same thickness (t  20 mm) throughout, as shown |in the figure on the next page. Both beams are formed by nailing, with each nail having an allowable shear load of 250 N. The beams are designed for a shear force V  3.2 kN.

y

A z

(a) What is the maximum longitudinal spacing SA for the t= nails in beam A? 20 mm (b) What is the maximum longitudinal spacing sB for the nails in beam B? (c) Which beam is more efficient in resisting the shear force?

B O

360 mm

z

O

t= 20 mm 200 mm

200 mm

360 mm

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CHAPTER 5

Solution 5.11-6

Page 470

Stresses in Beams (Basic Topics)

Two wood box beams

Cross-sectional dimensions are the same.

(a) BEAM A

All dimensions in millimeters.

Q  Af df  (bt)a

b  200 b1  200  2(20)  160

 680 * 103 mm3

h  360 h1  360  2(20)  320 t  20

sA 

F  allowable load per nail  250 N V  shear force  3.2 kN I

1 (bh3  b1 h31)  340.69 * 106 mm4 12

‹ smax 

;

(b) BEAM B Q  Afdf  (b  2t)(t)a

f  shear flow between one flange and both webs VQ 2F  s I

(2)(250 N)(340.7 * 106 mm4) 2FI  VQ (3.2 kN)(680 * 103 mm3)

 78.3 mm

s  longitudinal spacing of the nails

f

ht 1 b  (200)(20)a b (340) 2 2

ht b 2

1  (160)(20) (340) 2

2FI VQ

 544 * 103 mm3 sB 

(2)(250 N)(340.7 * 106 mm4) 2FI  VQ (3.2 kN)(544 * 103 mm3)

 97.9 mm

;

(c) BEAM B IS MORE EFFICIENT because the shear flow on the contact surfaces is smaller and therefore fewer ; nails are needed.

3 — in. 16

Problem 5.11-7 A hollow wood beam with plywood webs has the cross-sectional dimensions shown in the figure. The plywood is attached to the flanges by means of small nails. Each nail has an allowable load in shear of 30 lb. Find the maximum allowable spacing s of the nails at cross sections where the shear force V is equal to (a) 200 lb and (b) 300 lb.

3 — in. 16 3 in.

y

z

3 in. 4 8 in.

O 3 in. 4

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SECTION 5.11

Solution 5.11-7

471

Built-Up Beams

Wood beam with plywood webs (a) V  200 lb

All dimensions in inches. b  3.375 b1  3.0

smax 

h  8.0 h1  6.5

 2.77 in.

F  allowable shear force for one nail  30 lb s  longitudinal spacing of the nails f  shear flow between one flange and both webs f

VQ 2F  I s

I

1 (bh3  b1h31)  75.3438 in.4 12

‹ smax 

2FI VQ

2(30 lb)(75.344 in.4) 2FI  VQ (200 lb)(8.1563 in.3) ;

(b) V  300 lb By proportion, smax  (2.77 in.)a

200 b  1.85 in. 300

;

Q  Qflange  Afdf  (3.0)(0.75)(3.625)  8.1563 in.3

y

Problem 5.11-8 A beam of T cross section is formed by nailing together two boards having the dimensions shown in the figure. If the total shear force V acting on the cross section is 1500 N and each nail may carry 760 N in shear, what is the maximum allowable nail spacing s?

240 mm 60 mm z

C 200 mm

60 mm

Solution 5.11-8 V  1500 N

F allow  760 N

h1  200 mm

b  240 mm

t  60 mm

h  260 mm

MOMENT OF INERTIA ABOUT THE NEUTRAL AXIS I

A  bt + h1t A  2.64 * 104 mm2 LOCATION OF NEUTRAL AXIS (z AXIS)

c2 

h1 t bt ah1  b + th1 2 2 A

1 3 1 tc + t1 h1 c223 3 2 3 +

1 3 t 2 bt + bt a c1  b 12 2

I  1.549 * 108 mm4 FIRST MOMENT OF AREA OF FLANGE

c2  170.909 mm

t Q  bt a c1  b 2

c1  h  c2

Q  8.509 * 105 mm3

c1  89.091 mm

MAXIMUM ALLOWABLE SPACING OF NAILS f smax 

VQ F  I s F allowI VQ

smax  92.3 mm

;

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CHAPTER 5

Stresses in Beams (Basic Topics)

Problem 5.11-9 The T-beam shown in the figure is fabricated by welding together two steel plates. If the allowable load for each weld is 1.8 k/in. in the longitudinal direction, what is the maximum allowable shear force V?

y

0.6 in. 5.5 in. z

C

0.5 in.

4.5 in.

Solution 5.11-9 F allow  1.8

T-beam (welded) MOMENT OF INERTIA ABOUT THE NEUTRAL AXIS

k in.

h1  5.5 in.

b  4.5 in.

t1  0.6 in.

t2  0.5 in.

I

+

h  6 in. A  bt2 + h1t1 A  5.55 in.2 LOCATION OF NEUTRAL AXIS (z AXIS)

c2 

t2 h1 bt2 + t1 h1 a + t2 b 2 2

c2  2.034 in. c1  3.966 in.

A c1  h  c 2

1 1 t c 3 + t1 1 c2  t223 3 1 1 3 t2 2 1 b t23 + bt2 ac2  b 12 2

I  20.406 in.4 FIRST MOMENT OF AREA OF FLANGE Q  b t2 ac2 

t2 b 2

Q  4.014 in.3

MAXIMUM ALLOWABLE SHEAR FORCE f

VQ 2F I

Vmax 

2 Fallow I Q

Vmax  18.30 k

Problem 5.11-10 A steel beam is built up from a W 410 * 85 wide-flange beam and two 180 mm * 9 mm cover plates (see figure). The allowable load in shear on each bolt is 9.8 kN. What is the required bolt spacing s in the longitudinal direction if the shear force V = 110kN (Note: Obtain the dimensions and moment of inertia of the W shape from Table E-1(b).)

;

y

z

180 mm  9 mm cover plates

W 410  85 O

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SECTION 5.11

Built-Up Beams

473

Solution 5.11-10 F allow  9.8 kN

V  110 kN

+ Acp a c 

W 410 * 85

I  4.57 * 108 mm4

Aw  10800 mm2 hw  417 mm Iw  310 * 106 mm4

First moment of area of one flange

Acp  (180) (9) (2) mm2 for two plates

Q  180 mm (9 mm)a c 

h  hw + (9 mm) (2)

Maximum allowable spacing of nails

LOCATION OF NEUTRAL AXIS (z AXIS) h 2

f

c  217.5 mm

Moment of inertia about the neutral axis

smax 

3

I  Iw +

9 mm b 2

Q  3.451 * 105 mm3

A  Aw + Acp A  1.404 * 104 mm2

c

9 mm 2 b 2

180 mm (9 mm) (2) 12

VQ 2F  I s 2 Fallow I VQ

smax  236 mm

;

Problem 5.11-11 The three beams shown have approximately the same cross-sectional area. Beam 1 is a W 14  82 with flange plates; Beam 2 consists of a web plate with four angles; and Beam 3 is constructed of 2 C shapes with flange plates. (a) (b) (c) (d)

Which design has the largest moment capacity? Which has the largest shear capacity? Which is the most economical in bending? Which is the most economical in shear?

Assume allowable stress values are: sa  18 ksi and ta  11 ksi. The most economical beam is that having the largest capacityto-weight ratio. Neglect fabrication costs in answering (c) and (d) above. (Note: Obtain the dimensions and properties of all rolled shapes from tables in Appendix E.) 8  0.52

4  0.375 Four angles 1 66— 2

W 14  82

Beam 1

Solution 5.11-11

8  0.52

14  0.675

4  0.375 Beam 2

b1  8 in.

Beam 3

Built-up steel beam

Beam 1: properties and dimensions for W14 * 82 with flange plates AW  24 in.2

C 15  50

hw  14.3 in. t1  0.52 in.

Iw  88l in.4

h1  hw + 2t1

bf1  10.1 in.

tf1  0.855 in.

tw1  0.51 in.

AI  AW + 2b1t1

AI  32.32 in.2

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CHAPTER 5

I1  Iw +

Page 474

Stresses in Beams (Basic Topics)

Q1  b1 t1 a

t1 2 b1 + t31 hw 2 + b1 t1 a + b 2 12 2 2

I1  1.338 * 103 in4 Beam 2: properties and dimensions for L6 * 6 * 1/2 angles with web plate Aa  5.77 in.2

ca  1.67 in.

Ia  19.9 in.4

b2  14 in.

t2  0.675 in.

h2  b2

A2  4Aa + b2t2 I2  4Ia + Aa a

+ tw1

tf1 h1 t1 hw  b + bf1 tf1 a  b 2 2 2 2 a

Q2  2 Aa a

h2  ca b + t2 2

Q3  b3 t3 a

bf3  3.72 in. A3  2Ac + 2b3 t3 I3  Ic 2 +

h3  hc + 2t3 tw3  0.716 in.

A3  32.4 in.2

I3  985.328 in.4 (a) Beam with largest moment capacity; largest section modulus controls Mmax  sallow S 2I1 h1

S1  174.449 in.3

S2 

2I2 h2

S2  127.09 in.3

S3 

2I3 h3

S3  125.121 in.3

largest value

BEAM WITH LARGEST SHEAR CAPACITY: LARGEST

Q3  79.826 in.3

I2 t2  4.964 * 103 mm2 Q2 largest value

Itw Case (3) with maximum has the largest shear Q capacity ; (c) MOST ECONOMICAL BEAM IN BENDING HAS LARGEST BENDING CAPACITY-TO-WEIGHT RATIO S3  3.862 in. A3

6

S2  3.907 in. A2

6

;

Case (1) is the most economical in bending.

Itw/Q

(d) MOST ECONOMICAL BEAM IN SHEAR HAS LARGEST SHEAR CAPACITY-TO-WEIGHT RATIO I1 tw1  0.213 Q1 A1

RATIO CONTROLS

tallow I tw Vmax  O

2

S1  5.398 in. A1

case (1) with maximum S has the largest moment capacity ; (b)

2 hc  tf3 b 2

I2 2tw3  1.14 * 104 mm2 Q3

b3t33 hc t3 2 2 + b3 t3 a + b 2 12 2 2

S1 

a

I1 tw1  4.448 * 103 mm2 Q1

Ic  404 in.4

tf3  0.65 in.

2

tf3 h3 t3 hc  b + 2bf3 tf3 a  b 2 2 2 2

+ 2tw3

Beam 3: properties and dimensions for C15 * 50 with flange plates t3  0.375 in.

b2 2 b 2

Q2  78.046 in.

2 b2 t2 b32  ca b 4 + 2 12

hc  15 in.

a

3

I2  889.627 in.4

b3  4 in.

Q1  98.983 in.3

2

ha  6 in.

A2  32.53 in.2

Ac  14.7 in.2

2 hw  tf1 b 2

6

6

I2 t2  0.237 Q2 A2

I3 tw3  0.273 Q3 A3

Case (3) is the most economical in shear.

;

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SECTION 5.12

Beams with Axial Loads

Problem 5.11-12 Two W 310 * 74 steel wide-flange beams are bolted together to form a built-up beam as shown in the figure. What is the maximum permissible bolt spacing s if the shear force V  80 kN and the allowable load in shear on each bolt is F  13.5 kN (Note: Obtain the dimensions and properties of the W shapes from Table E-1(b).)

W 310  74

W 310  74

Solution 5.11-12 V  80 kN

FIRST MOMENT OF AREA OF FLANGE

W 310 * 74

F allow  13.5 kN hw  310 mm

A w  9420 mm

2 4

Location of neutral axis (z axis) c  hw

Q  Aw

I w  163 * 10 mm 6

c  310 mm hw 2 b d (2) 2

Q  1.46 * 106 mm3

MAXIMUM ALLOWABLE SPACING OF NAILS f

MOMENT OF INERTIA ABOUT THE NEUTRAL AXIS I  c Iw + Aw a

hw 2

VQ 2F  I s

smax 

2Fallow I VQ

smax  180 mm

;

I  7.786 * 108 mm4

Beams with Axial Loads When solving the problems for Section 5.12, assume that the bending moments are not affected by the presence of lateral deflections.

P = 25 lb

Problem 5.12-1 While drilling a hole with a brace and bit, you exert a downward force P  25 lb on the handle of the brace (see figure). The diameter of the crank arm is d  7/16 in. and its lateral offset is b  4-7/8 in. Determine the maximum tensile and compressive stresses st and sc, respectively, in the crank.

7 d= — 16 in. 7

b = 4— 8 in.

475

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CHAPTER 5

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Stresses in Beams (Basic Topics)

Solution 5.12-1

Brace and bit

P  25 lb (compression) M  Pb  (25 lb)(4 7/8 in.)  121.9 lb-in.

MAXIMUM STRESSES st  

 166 psi + 14,828 psi  14,660 psi

d  diameter d  7/16 in.

S

sc  

pd2  0.1503 in.2 4

A

P 121.9 lb-in. M 25 lb + +   2 A S 0.1503 in. 0.008221 in.3 ;

P M   166 psi  14,828 psi A S

 14,990 psi

;

3

pd  0.008221 in.3 32

Problem 5.12-2 An aluminum pole for a street light weights 4600 N and supports an arm that weights 660 N (see figure). The center of gravity of the arm is 1.2 m from the axis of the pole. A wind force of 300 N also acts in the (y) direction at 9 m above the base. The outside diameter of the pole (at its base) is 225 mm, and its thickness is 18 mm. Determine the maximum tensile and compressive stresses st and sc, respectively, in the pole (at its base) due to the weights and the wind force.

W2 = 660 N

1.2 m P1 = 300 N

W1 = 4600 N 18 mm

9m

z

y x y

x

Solution 5.12-2 W1  4600 N

b  1.2 m

Mx  W2 b + P1h

W2  660 N

h9m

Mx  3.492 * 103 N # m

P1  300 N

d1  225 mm t  18 mm

MAXIMUM STRESS

d2  d1  2 t A

p 2 1d1  d222 4

I

A  1.171 * 104 mm2

p 1d1 4  d2 42 64

I  6.317 * 107 mm4

AT BASE OF POLE Pz  W1 + W2 Pz  5.26 * 10 N 3

V y  P1

st  a 

Pz Mx d1 + b A I 2

st  5.77 * 103 kPa  5770 kPa sc  a 

;

Pz Mx d1  b A I 2

sc  6.668 * 103 (Axial force )

V y  300 N

(Shear force)

(Moment)

 6668 kPa

;

;

225 mm

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SECTION 5.12

Problem 5.12-3 A curved bar ABC having a circular axis (radius

h

B

r  12 in.) is loaded by forces P  400 lb (see figure). The cross section of the bar is rectangular with height h and thickness t. If the allowable tensile stress in the bar is 12,000 psi and the height h  1.25 in., what is the minimum required thickness tmin ?

477

Beams with Axial Loads

C

A

P

P 45°

45° r h t

Solution 5.12-3

Curved bar TENSILE STRESS st 

r  radius of curved bar



e  r  r cos 45° tmin 

Pr (2  12) 2

P r c1 + 3(2  12) d hsallow h

SUBSTITUE NUMERICAL VALUES: P  400 lb s allow  12,000 psi

CROSS SECTION h  height t  thickness A  ht S 

P r c1 + 3(2  12) d ht h

MINIMUM THICKNESS

1 b  r a1  12 M  Pe 

3Pr(2  12) P M P +  + A S ht th2

1 2 th 6

r  12 in. h  1.25 in. tmin  0.477 in.

;

B

Problem 5.12-4 A rigid frame ABC is formed by welding two steel pipes at B (see figure). Each pipe has cross-sectional area A  11.31 * 103 mm2, moment of inertia I  46.37 * 106 mm4, and outside diameter d  200 mm. Find the maximum tensile and compressive stresses st and sc, respectively, in the frame due to the load P  8.0 kN if L  H  1.4 m.

d

d

P

H

A

C d L

L

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CHAPTER 5

Solution 5.12-4

Page 478

Stresses in Beams (Basic Topics)

Rigid frame AXIAL FORCE: N  RA sin a 

P sin a 2 PL 2

BENDING MOMENT: M  RAL  TENSILE STRESS st   Load P at midpoint B P REACTIONS: RA  RC  2 BAR AB:

SUBSTITUTE NUMERICAL VALUES P  8.0 kN L  H  1.4 m a  45° sina  1/12 d  200 mm A  11.31 * 103 mm2 I  46.37 * 106 mm4

H tan a  L sin a 

N Mc P sin a PLd +   + A I 2A 4I

st   H

1H2 + L2

2(11.31 * 103 mm2)

(8.0 kN)(1.4 m)(200 mm) + 4(46.37 * 106 mm4)

d  diameter c  d/2

(8.0 kN)(1/12)

 0.250 MPa + 12.08 MPa  11.83 MPa (tension) sc  

;

N Mc   0.250 MPa  12.08 MPa A I

 12.33 MPa (compression)

Problem 5.12-5 A palm tree weighing 1000 lb is inclined at an angle of 60° (see figure). The weight of the tree may be resolved into two resultant forces, a force P1  900 lb acting at a point 12 ft from the base and a force P2  100 lb acting at the top of the tree, which is 30 ft long. The diameter at the base of the tree is 14 in. Calculate the maximum tensile and compressive stresses st and sc, respectively, at the base of the tree due to its weight.

;

P2 = 100 lb

30 ft

12 ft

P1 = 900 lb 60°

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SECTION 5.12

Solution 5.12-5

479

Beams with Axial Loads

Palm tree M  P1L1 cos 60°P2 L2 cos 60°  [(900 lb)(144 in.) + (100 lb)(360 in.)] cos 60°  82,800 lb-in. N  (P1 + P2) sin 60°  (1000 lb) sin 60°  866 lb FREE-BODY DIAGRAM

MAXIMUM TENSILE STRESS

P1  900 lb

st  

P2  100 lb L1  12 ft  144 in. L2  30 ft  360 in. d  14 in. A

pd2  153.94 in.2 4

S

pd3  269.39 in.3 32

82,800 lb-in. N M 866 lb +   + 2 A S 153.94 in. 269.39 in.3

 5.6 psi + 307.4 psi  302 psi MAXIMUM COMPRESSIVE STRESS sc  5.6 psi  307.4 psi  313 psi

Problem 5.12-6 A vertical pole of aluminum is fixed at the base and pulled at the top by a cable having a tensile force T (see figure). The cable is attached at the outer edge of a stiffened cover plate on top of the pole and makes an angle a  20° at the point of attachment. The pole has length L  2.5 m and a hollow circular cross section with outer diameter d2  280 mm and inner diameter d1  220 mm. The circular cover plate has diameter 1.5d2. Determine the allowable tensile force Tallow in the cable if the allowable compressive stress in the aluminum pole is 90 MPa.

T

a L

Solution 5.12-6 d1  220 mm

d2  280 mm t A

d2  d1 2

PN  T cos (a) V  T sin (a)

a  20°

p 1d 2  d1 22 4 2

A  2.356 * 104 mm2

L  2.5 m

I

p 1d 4  d1 42 64 2

I  1.867 * 108 mm4

M  VL + PN a

;

1.5 d2

d2

sallow  90 MPa

;

(Axial force) (Shear force) 1.5 d2 b 2

(Moment).

d1 d2

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Stresses in Beams (Basic Topics)

Allowable Tensile Force sc  



T cos (a) PN M d2    A I 2 A T sin (a) L T cos (a) a I

sallow

Tallow 

1.5 d2 b 2 d2 2

cos (a) + A

sin (a) L + cos (a) a

Tallow  108.6 kN

I

1.5 d2 b 2 d2 2

;

Problem 5.12-7 Because of foundation settlement, a circular tower is leaning at an angle a to the vertical (see figure). The structural core of the tower is a circular cylinder of height h, outer diameter d2, and inner diameter d1. For simplicity in the analysis, assume that the weight of the tower is uniformly distributed along the height. Obtain a formula for the maximum permissible angle a if there is to be no tensile stress in the tower.

h

d1 d2 a

Solution 5.12-7

Leaning tower CROSS SECTION

W  weight of tower a  angle of tilt

A

p 2 (d  d21) 4 2

I

p 4 (d  d41) 64 2



p 2 (d  d21)(d22 + d21) 64 2

d22 + d21 I  A 16 c

d2 2

AT THE BASE OF THE TOWER h N  W cos a M  Wa bsin a 2

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SECTION 5.12

TENSILE STRESS (EQUAL TO ZERO) st  

‹

Mc Wcosa N +   A I A

d2 W h + a sinab a b  0 I 2 2

hd2 sin a cos a  A 4I

MAXIMUM ANGLE a d22 + d12 a  arctan 4hd2

Problem 5.12-8 A steel bar of solid circular cross section and length

Beams with Axial Loads

tan a 

481

d22 + d12 4I  hd2A 4hd2

;

y

L  2.5 m is subjected to an axial tensile force T  24 kN and a bending moment M  3.5 kN m (see figure).

d

(a) Based upon an allowable stress in tension of 110 MPa, determine the required diameter d of the bar; disregard the weight of the bar itself. (b) Repeat (a) including the weight of the bar.

M –z-direction

z

T L

x

Solution 5.12-8 M  3.5 kN # m g steel  77

T  24 kN

kN

L  2.5 m

3

m

p 2 d 4

c

d 2

I

p 4 d 64

(a) DISREGARD WEIGHT OF BAR MAX. TENSILE STRESS AT TOP OF BEAM AT SUPPORT smax 

Md T T M d +  + p 2 p 42 A I 2 d d 4 64

sallow 

4T 2

pd

d  70 mm

d (SUBSTITUTE sallow)

;

(b) INCLUDE WEIGHT OF BAR

sallow  110 MPa A

SOLVE NUMERICALLY FOR

32 M +

p d3

Mmax  M +

Agsteel L2 2

AT TOP OF BEAM AT SUPPORT st  sallow 

Mmax d T + A I 2

SUBSTITUTE MMAX FROM ABOVE, SOLVE FOR d NUMERICALLY d  76.5 mm

;

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Stresses in Beams (Basic Topics)

Problem 5.12-9 A cylindrical brick chimney of height H weighs w  825 lb/ft of height (see figure). The inner and outer diameters are d1  3 ft and d2  4 ft, respectively. The wind pressure against the side of the chimney is p = 10 lb/ft2 of projected area. Determine the maximum height H if there is to be no tension in the brickwork

p w H d1 d2

Solution 5.12-9

Brick Chimney I

d2

H

q

w

p 2 p 4 (d2  d41)  (d  d21) (d22  d21) 64 64 2

I 1 2  (d2 + d21) A 16

d2 2

c

AT BASE OF CHIMNEY M  qH a

N  W  wH V M

TENSILE STRESS (EQUAL TO ZERO) s1  

N

Md2 N + 0 A 2I

p  wind pressure

pd2 H2 d22 + d12  2wH 8d2

q  intensity of load  pd2

SOLVE FOR H

d2  outer diameter

1 H b  pd2 H2 2 2

H

2I M  N Ad2

or

w(d22 + d21)

d1  inner diameter

SUBSTITUTE NUMERICAL VALUES

W  total weight of chimney  wH

w  825 lb/ft

d2  4 ft

CROSS SECTION

q  10 lb/ft

Hmax  32.2 ft

A

p 2 (d2  d21) 4

2

;

4pd22

d1  3 ft ;

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SECTION 5.12

483

Beams with Axial Loads

Problem 5.12-10 A flying buttress transmits a load P  25 kN, acting at an angle of 60° to the horizontal, to the top of a vertical buttress AB (see figure). The vertical buttress has height h  5.0 m and rectangular cross section of thickness t  1.5 m and width b  1.0 m (perpendicular to the plane of the figure). The stone used in the construction weighs y  26 kN/m3. What is the required weight W of the pedestal and statue above the vertical buttress (that is, above section A) to avoid any tensile stresses in the vertical buttress?

Flying buttress P W 60° A

A

—t 2

h t B

Solution 5.12-10

h t

B

Flying buttress

FREE-BODY DIAGRAM OF VERTICAL BUTTRESS

CROSS SECTION A  bt  (1.0 m)(1.5 m)  1.5 m2 1 1 S  bt2  (1.0 m)(1.5 m)2  0.375 m3 6 6 AT THE BASE N  W + WB + P sin 60°  W + 195 kN + (25 kN) sin 60°  W + 216.651 kN M  (Pcos 60°) h  (25 kN) (cos 60°) (5.0 m)  62.5 kN # m TENSILE STRESS (EQUAL TO ZERO)

P  25 kN

st  

h  5.0 m t  1.5 m b  width of buttress perpendicular to the figure g  26 kN/m

3

WB  weight of vertical buttress  195 kN

62.5 kN # m

W + 216.651 kN 2

1.5 m

+

0.375 m3

or W  216.651 kN + 250 kN  0

b  1.0 m

 bthg

 

N M + A S

W  33.3 kN

;

0

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Stresses in Beams (Basic Topics)

Problem 5.12-11 A plain concrete wall (i.e., a wall with no steel

t

reinforcement) rests on a secure foundation and serves as a small dam on a creek (see figure). The height of the wall is h  6.0 ft and the thickness of the wall is t  1.0 ft. (a) Determine the maximum tensile and compressive stresses st and sc, respectively, at the base of the wall when the water level reaches the top (d  h). Assume plain concrete has weight density gc  145 Ib/ft3. (b) Determine the maximum permissible depth dmax of the water if there is to be no tension in the concrete.

Solution 5.12-11

h d

Concrete wall

h  height of wall t  thickness of wall b  width of wall (perpendicular to the figure) gc  width density of concrete gw  weight density of water d  depth of water W  weight of wall

STRESSES AT THE BASE OF THE WALL (d  DEPTH OF WATER) d 3gw W M +  hgc + A S t2 d 3gw W M sc     hgc  2 A S t st  

(a) STRESSES AT THE BASE WHEN d  h

W  bhtgc

h  6.0 ft  72 in. d  72 in.

F  resultant force for the water pressure

t  1.0 ft  12 in.

MAXIMUM WATER PRESSURE = gw d

gc  145 lb/ft3 

F

1 1 (d)(gw d) (b)  bd2gw 2 2

d 1 M  F a b  bd3gw 3 6 1 A  bt S  bt2 6

145 lb/in.3 1728

gw  62.4 Ib/ft3 

62.4 lb/in.3 1728

Eq. (1) Eq.(2)

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SECTION 5.12

Substitute numerical values into Eqs. (1) and (2): st  6.042 psi + 93.600 psi  87.6 psi

d3  (72 in.)(12 in.)2 a

;

sc  6.042 psi  93.600 psi  99.6 psi

dmax  28.9 in.

;

485

Beams with Axial Loads

145 b  24,092 in.3 62.4

;

(b) MAXIMUM DEPTH FOR NO TENSION Set st = 0 in Eq. (1): hgc +

d3gw 2

t

 0 d3  ht2 a

gc b gw

Problem 5.12-12 A circular post, a rectangular post, and a post of cruciform cross section are each compressed by loads that produce a resultant force P acting at the edge of the cross section (see figure). The diameter of the circular post and the depths of the rectangular and cruciform posts are the same. (a) For what width b of the rectangular post will the maximum tensile stresses be the same in the circular and rectangular posts? (b) Repeat (a) for the post with cruciform cross section. (c) Under the conditions described in parts (a) and (b), which post has the largest compressive stress? P

P

P

b 4 — = b 4

x

b d

d

d

Load P here d 4 — = d 4

Solution 5.12-12 (a) EQUAL MAXIMUM TENSILE STRESSES

COMPRESSION sc  

CIRCULAR POST A

p 2 d 4

S

p 3 d 32

M

Pd 2

Tension st  

P M  A S

 

4P pd

2



16 P pd

2



20P pd 2

RECTANGULAR POST

P M 4P 16 P 12 P +   2 +  2 A S pd pd pd2

A  bd TENSION

st  

S

bd2 6

M

Pd 2

P M P 3P 2P +     A S bd bd bd

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Stresses in Beams (Basic Topics)

COMPRESSION s c  

P M P 3P 4P       A S bd bd bd

Equate tensile stress expressions, solve for b 12 P 2

pd



2P bd

6 1  pd b

b

pd 6

;

(b) CRUCIFORM CROSS SECTION A  cbd  a S c

3

bd b d 1 2 3 + a b d  bd 2 2 12 2 2 12 d 32

Pd 16P M  3bd 3 2 a bd2 b 32 TENSION

st    

COMPRESSION

12 P 2

pd



2P 3bd

1 3  pd b

;

substitute expressions for b above & compare compressive stresses

16 P 12 P 4P +  3bd 3bd 3bd

sc  

sc  

M P  A S

20 P pd2

RECTANGULAR POST 4P 24 P   pd pd 2 a bd 6

CRUCIFORM POST 20 P 20 P   sc   pd pd 2 3 d 3 Rectangular post has the largest compressive stress ;

4P 16 P 20 P    3bd 3bd 3bd

Problem 5.12-13 Two cables, each carrying a tensile force P  1200 lb, are bolted to a block of steel (see figure). The block has thickness t  1 in. and width b  3 in.

Steel block loaded by cables

P  1200 lb d  0.25 in. t d +  0.625 in. 2 2

b

P

(a) If the diameter d of the cable is 0.25 in., what are the maximum tensile and compressive stresses st and sc, respectively, in the block? (b) If the diameter of the cable is increased (without changing the force P), what happens to the maximum tensile and compressive stresses?

t  1.0 in. e 

pd 3

(c) THE LARGEST COMPRESSIVE STRESS

sc  

M P + A S

 

Solution 5.12-13

b

CIRCULAR POST

bd bd 22

3

Equate compressive stresses & solve for b

b  width of block  3.0 in.

t

P

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SECTION 5.12

MAXIMUM COMPRESSIVE STRESS (AT BOTTOM OF BLOCK)

CROSS SECTION OF BLOCK A  bt  30 in.2

I

1 3 bt  0.25 in.4 12

t y    0.5 in. 2 sc 

(a) MAMIMUM TENSILE STRESS (AT TOP OF BLOCK) y

t  0.5 in. 2



Pey P st  + A I 

Pey P + A I (1200 lb)(0.625 in.)( 0.5 in.)

1200 lb 3 in.2

+

0.25 in.4

 400 psi  1500 psi  1100 psi

;

(1200 lb)(0.625 in.)(0.5 in.)

1200 lb 3 in.2

+

(b) IF d IS INCREASED, increase the eccentricity e increases and both stresses in magnitude.

0.25 in.4

 400 psi + 1500 psi  1900 psi

;

Problem 5.12-14 A bar AB supports a load P acting at the centroid

b — 2

of the end cross section (see figure). In the middle region of the bar the cross-sectional area is reduced by removing one-half of the bar. A

(a) If the end cross sections of the bar are square with sides of length b, what are the maximum tensile and compressive stresses st and sc, respectively, at cross section mn within the reduced region? (b) If the end cross sections are circular with diameter b, what are the maximum stresses st and sc?

b

b b

b — 2 m

(a) b — 2

n B P b (b)

Solution 5.12-14

Bar with reduced cross section

(a) SQUARE BAR

(b) CIRCULAR BAR

Cross section mn is a rectangle.

Cross section mn is a semicircle

b2 b A  (b)a b  2 2

1 pb2 pb2 A a b   0.3927 b2 2 4 8

b M  Pa b 4

c

I

1 b 3 b4 (b) a b  12 2 96

b 4

STRESSES P Mc 2P 6P 8P +  2 + 2  2 ; A I b b b P Mc 2P 6P 4P sc    2  2   2 ; A I b b b st 

487

Beams with Axial Loads

From Appendix D, Case 10: b 4 I  0.1098a b  0.006860 b4 2 M  Pa

2b b  0.2122 Pb 3p

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Stresses in Beams (Basic Topics)

FOR TENSION

 2.546

FOR COMPRESSION: b 2b   0.2878 b 2 3p



P

(0.2122 Pb)(0.2122 b) Mct P P +  + A I 0.3927 b2 0.006860 b4

Problem 5.12-15 A short column constructed of a W 12 * 35 wide-flange shape is subjected to a resultant compressive load P  12 k having its line of action at the midpoint of one flange (see figure). (a) Determine the maximum tensile and compressive stresses st and sc, respectively, in the column. (b) Locate the neutral axis under this loading condition. (c) Recompute maximum tensile and compressive stresses if a C 10  15.3 is attached to one flange, as shown.

2

0.3927 b

 2.546

STRESSES st 

+ 6.564

2

b Mc P c sc   A I

2b 4r ct    0.2122 b 3p 3p

cc  r  ct 

P

P 2

b



P 2

b

 9.11

P b2

;

(0.2122 Pb)(0.2878 b)

8.903

0.006860 b4 P 2

b

 6.36

P b2

;

y

P = 25 k

C 10  15.3 (Part c only)

z C

2 W 12  35

1

1

2

Solution 5.12-15

Column of wide-flange shape

PROPERTIES OF EACH SHAPE:

sc  

W 12 * 35

C 10 * 15.3

Aw  10.3 in.3

Ac  4.48 in.2

hw  12.5 in.

twc  0.24 in.

tf  0.52 in. Iw  285 in.

y0  

Ic  2.27 in. (2-2 axis) 4

(a) THE MAXIMUM TENSILE AND COMPRESSIVE STRESSES LOCATION OF CENTROID FOR W 12  35 ALONE hw cw  2

cw  6.25 in.

P  25 k

hw tf ew   2 2

st  

Pew P + c Aw Iw w

sc  5711 psi

Iw Aw ew

y0  4.62 in.

h  hw + twc A  Aw + Ac

ew  5.99 in. ;

;

(C) COMBINED COLUMN, W 12 * 35 with C 10 * 15.3 h  12.74 in.

st  857 psi

;

(b) NEUTRAL AXIS (W SHAPE ALONE)

xp  0.634 in.

4

P Pew  c Aw Iw w

A  14.78 in.2

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SECTION 5.12

LOCATION OF CENTROID OF COMBINED SHAPE

c

hw Aw a b + Ac (h  xp) 2 A

I  Iw + Aw ac 

c  8.025 in.

hw 2 b 2

+ Ic + Ac (h  xp  c)2 I  394.334 in.4

(a) Determine the maximum tensile and compressive stresses st and sc, respectively, in the column. (b) Locate the neutral axis under this loading condition. (c) Recompute maximum tensile and compressive stresses if a 120 mm  10 mm cover plate is added to one flange as shown.

2

st  

P Pe + c A I

sc  

P Pe  (h  c) A I

y0  

I Ae

Problem 5.12-16 A short column of wide-flange shape is subjected to a compressive load that produces a resultant force P  55 kN acting at the midpoint of one flange (see figure).

tf

e  hw 

P  25 k

489

Beams with Axial Loads

e  4.215 in.

c

st  453 psi

;

sc  2951 psi

;

y0  6.33 in. (from centroid)

;

y

P = 55 kN z

Cover plate (120 mm  10 mm) (Part c only) y P

C 8 mm z

200 mm

C

12 mm 160 mm

Solution 5.12-16 P  55 kN (a) MAXIMUM TENSILE AND COMPRESSIVE STRESSES FOR W SHAPE ALONE

PROPERTIES AND DIMENSIONS FOR W SHAPE b  160 mm tf  12 mm

d  200 mm tw  8 mm

Aw  bd  (b  tw) (d  2 tf) Aw  5.248 * 103 mm2 (b  tw) (d  2 tf)3 bd3 Iw   12 12 Iw  3.761 * 107 mm4

e

tf d  2 2

e  94 mm

st  

P Pe d + Aw Iw 2

st  3.27 MPa

sc  

P Pe d  Aw Iw 2

sc  24.2 MPa

(b) NEUTRAL AXIS (W SHAPE ALONE) y0  

Iw Aw e

y0  76.2 mm

;

(c) COMBINED COLUMN-W SHAPE & COVER PLATE bp  120 mm tp  10 mm

; ;

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Stresses in Beams (Basic Topics)

h  dtp

I  4.839 * 107 mm4 tf e  74.459 mm ed c 2

h  210 mm A  Aw + bp tp

A  6.448 * 103 mm2

CENTROID OF COMPOSITE SECTION tp d Aw + bp tp ad + b 2 2 c A

st   sc  

c  119.541 mm I  Iw + Aw ac  bp t3p + 12

d b 2

+ bp tp a d +

y0   tp 2

 cb

P Pe  (h  c) Aw Iw

sc  20.3 MPa

;

I Ae

y0  100.8 mm (from centrioid)

2

1

(a) Determine the maximum tensile stress st in the angle section. (b) Recompute the maximum tensile stress if two angels are used and P is applied as shown in the figure part (b).

1 L44— 2 C 1

3 1

1 2L44— 2 C 

P  2

3

P

(a)

(b)

Angle section in tension (b) TWO ANGLES: L 4 * 4 * 1/2

(a) ONE ANGLE: L 4 * 4 * 1/2 AL  3.75 in.2

A  2AL

rmin  0.776 in.

t  0.5 in.

t  0.5 in

c  1.18 in.

c  1.18 in

e ac

t b 12 2

e  1.315 in

P  12.5 k c1  c 12 AL rmin2

M  Pe Mc1 P + AL I3

IL  5.52 in.4 (2-2 axis) e ac

t b 2

e  0.93 in.

P  12.5 k

c1  1.699 in.

I  2IL

I3  2.258 in.

4

M  Pe

M  16.44 k-in.

MAXIMUM TENSILE STRESS OCCURS AT CORNER st 

;

2

L 4  4  2 inch angle section (see Table E-4(a) in Appendix E) is subjected to a tensile load P  12.5 kips that acts through the point where the midlines of the legs intersect [see figure part (a)].

I3 

st  1.587 MPa

NEUTRAL AXIS 2

Problem 5.12-17 A tension member constructed of an

Solution 5.12-17

P Pe + c A I

st  15.48 ksi

;

I  11.04 in.4 M  11.625 k-in.

MAXIMUM TENSILE STRESS OCCURS AT THE LOWER EDGE st 

P Mc + A I

st  2.91 ksi

;

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SECTION 5.12

Beams with Axial Loads

Two L 76  76  6.4 angles

Problem 5.12-18 A short length of a 200 * 17.1 channel is subjected to an axial compressive force P that has its line of action through the midpoint of the web of the channel [(see figure(a)].

y C 200 × 17.1 P



(a) Determine the equation of the neutral axis under this z z C loading condition. (b) If the allowable stresses in tension and compression (a) are 76 MPa and 52 MPa respectively, find the maximum permissible load Pmax. (c) Repeat (a) and (b) if two L 76  76  6.4 angles are added to the channel as shown in the figure part (b). See Table E-3(b) in Appendix E for channel properties and Table E-4(b) for angle properties.

y P





C C 200 × 17.1 (b)

Solution 5.12-18 sc

P 

tw = 5.59mm bf = 57.4

Ac  2170 mm2

1 e  c1 Ac Ic

Pmax  67.3 kN

dc  203 mm c1  14.5 mm

L 76 * 76 * 6.4

Ic  0.545 * 106 mm4 (z-axis)

AL  929 mm2

c2  bf  c1 c2  42.9 mm

cL  21.2 mm A  4.028 * 103 mm2

st  76 MPa s c  52 MPa

A  Ac + 2 AL

ECCENTRICITY OF THE LOAD

h  bf + 76 mm

tw 2

e  11.705 mm

(a) LOCATION OF THE NEUTRAL AXIS (CHANNEL ALONE) Ic y0  Ac # e

y0  21.5 mm

;

Pe P + c A I 2

P  165.025 kN P Pe sc    c A I 1

h  133.4 mm

CENTROID OF COMPOSITE SECTION Ac 1bf  c12 + 2 AL 1bf + cL2 c A c  59.367 mm

I  Ic + Ac 1bf  c1  c22

+ 2 IL + 2 AL 1bf + cL  c22

I  2.845 * 106 mm4

(b) FIND PMAX st  

IL  0.512 * 106 mm4

COMPOSITE SECTION

ALLOWABLE STRESSES

e  c1 

;

(c) COMBINED COLUMN WITH 2-ANGLES

C 200 * 17.1

P

st 1 e  + c2 Ac Ic

491

e  bf 

tw c 2

e  4.762 mm

bf  57.4 mm LOCATION OF THE NEUTRAL AXIS y0  

I Ae

y0  148.3 mm

;



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Stresses in Beams (Basic Topics)

y0  148.3 mm 7 h  133.4 mm

;

P

Thus, this composite section has no tensile stress sc  

P Pe + c A I

sc 1 e  + c A I

Pmax  149.6 kN

;

Stress Concentrations The problems for Section 5.13 are to be solved considering the stress-concentration factors.

M

M h

Problem 5.13-1 The beams shown in the figure are subjected

to bending moments M  2100 lb-in. Each beam has a rectangular cross section with height h  1.5 in. and width b  0.375 in. (perpendicular to the plane of the figure).

d

(a)

(a) For the beam with a hole at midheight, determine the maximum stresses for hole diameters d  0.25, 0.50, 0.75, and 1.00 in. (b) For the beam with two identical notches (inside height h1  1.25 in .), determine the maximum stresses for notch radii R  0.05, 0.10, 0.15, and 0.20 in.

2R M

M h

Probs. 5.13.1 through 5.13-4

h1

(b)

Solution 5.13-1 M  2100 lb-in. h  1.5 in. b  0.375 in.

(b) BEAM WITH NOTCHES

(a) BEAM WITH A HOLE d 1 … h 2

h1  1.25 in.

Eq.(5-57): sc  

1 d Ú h 2

Eq.(5-56): sB  

d (in.)

d h

sc Eq. (1) (psi)

0.25 0.50 0.75 1.00

0.1667 0.3333 0.5000 0.6667

15,000 15,500 17,100 —

6Mh Eq. (5-58)

b(h3  d3) 50,400 3.375  d

3

(1)

snom 

12Md b(h3  d3) 67,200 d 3.375  d3

(2)

sB Eq. (2) (psi) — — 17,100 28,300

h 1.5 in.   1.2 h1 1.25 in.

sm ax (psi) 15,000 15,500 17,100 28,300

NOTE: The larger the hole, the larger the stress.

6M bh21

 21,500 psi

R (in)

R h1

K (Fig. 5-50)

sm ax  Ks nom sm ax (psi)

0.05 0.10 0.15 0.20

0.04 0.08 0.12 0.16

3.0 2.3 2.1 1.9

65,000 49,000 45,000 41,000

NOTE: The larger the notch radius, the smaller the stress.

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SECTION 5.13

Stress Concentrations

493

Problem 5.13-2 The beams shown in the figure are subjected to bending moments M  250 N # m. Each beam has a rectangular cross section with height h  44 mm and width b  10 mm (perpendicular to the plane of the figure). (a) For the beam with a hole at midheight, determine the maximum stresses for hole diameters d  10, 16, 22 and 28 mm. (b) For the beam with two identical notches (inside height h1  40 mm ), determine the maximum stresses for notch radii R  2, 4, 6, and 8 mm.

Solution 5.13-2 M  250 N # m h  44 mm b  10 mm

(b) BEAM WITH NOTCHES

(a) BEAM WITH A HOLE 1 d … h 2 sc 

sB 

d (mm) 10 16 22 28

Eq. (5-57): 6Mh

b(h3  d3)

d 1 Ú h 2

h1  40 mm



Eq. (5-58): snom 

6

6.6 * 10

85,180  d3

MPa

b(h3  d3)

d h 0.227 0.364 0.500 0.636



R (mm) 300 * 103d

MPa 85,180  d3

sB sc Eq. (2) Eq. (1) (MPa) (MPa) 78 81 89 —

— — 89 133

6M bh21

 93.8 MPa

(1)

Eq. (5-56): 12Md

h 44 mm   1.1 h1 40 mm

(2)

sm ax (MPa)

2 4 6 8

R h1

K (Fig. 5-50)

smax  Ks nom smax (MPa)

0.05 0.10 0.15 0.20

2.6 2.1 1.8 1.7

240 200 170 160

NOTE: The larger the notch radius, the smaller the stress.

78 81 89 133

NOTE: The larger the hole, the larger the stress.

Problem 5.13-3 A rectangular beam with semicircular notches, as shown in part (b) of the figure, has dimensions h  0.88 in. and h1  0.80 in. The maximum allowable bending stress in the metal beam is smax  60 ksi, and the bending moment is M  600 lb-in. Determine the minimum permissible width bmin of the beam.

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Solution 5.13-3 h  0.88 in.

Page 494

Stresses in Beams (Basic Topics)

Beam with semicircular notches h1  0.80 in.

smax  60 ksi M  600 lb-in. 1 h  h1 + 2R R  (h  h1)  0.04 in. 2 0.04 in. R   0.05 h1 0.80 in.

smax  Ksnom  Ka 60 ksi  2.57c

6M bh21

b

6(600 lb-in.) b(0.80 in.)2

d

Solve for b: ;

bmin L 0.24 in.

From Fig. 5-50: K L 2.57

Problem 5.13-4 A rectangular beam with semicircular notches, as shown in part (b) of the figure, has dimension h  120 mm and h1  100 mm . The maximum allowable bending stress in the plastic beam is smax  6 MPa, and the bending moment is M  150 N # m. Determine the minimum permissible width bmin of the beam.

Solution 5.13-4

Beam with semicircular notches

h  120 mm

h1  100 mm

smax  6 MPa

M  150 N # m

smax  Ksnom  Ka

1 h  h1 + 2R R  (h  h1)  10 mm 2

6 MPa  2.20c

R 10 mm   0.10 h1 100 mm

Solve for b:

From Fig.5-50: K L 2.20

Problem 5.13-5 A rectangular beam with notches and a hole (see figure) has dimensions h  5.5 in., h1  5 in., and width b  1.6 in. The beam is subjected to a bending moment M  130 k-in., and the maximum allowable bending stress in the material (steel) is smax  42,000 psi. (a) What is the smallest radius Rmin that should be used in the notches? (b) What is the diameter dmax of the largest hole that should be drilled at the midheight of the beam?

6M bh21

b

6(150 N # m) b(100 mm)2

bmin L 33 mm

d

;

2R M

M h1

h

d

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SECTION 5.13

Solution 5.13-5

Beam with notches and a hole

h  5.5 in. h1  5 in. b  1.6 in. M  130 k-in. smax  42,000 psi

(b) LARGEST HOLE DIAMETER Assume

(a) MINIMUM NOTCH RADIUS sB 

5.5 in. h   1.1 h1 5 in. snom  K

6M bh21

Stress Concentrations

12Md b(h3  d3)

42,000 psi 

 19,500 psi

12(130 k-in.)d (1.6 in.)[(5.5 in.)3  d3]

d3 + 23.21d  166.4  0

42,000 psi smax   2.15 snom 19,500 psi

h From Fig. 5-50, with K  2.15 and  1.1, we get h1 R L 0.090 h1 ‹ Rmin L 0.090h1  0.45 in.

1 d 7 and use Eq. (5-56). h 2

;

Solve numerically: dmax  4.13 in.

;

or

495

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Page 497

6 Stresses in Beams (Advanced Topics)

Composite Beams When solving the problems for Section 6.2, assume that the component parts of the beams are securely bonded by adhesives or connected by fasteners. Also, be sure to use the general theory for composite beams described in Sect. 6.2.

y

Problem 6.2-1 A composite beam consisting of fiberglass faces and a core of particle board has the cross section shown in the figure. The width of the beam is 2.0 in., the thickness of the faces is 0.10 in., and the thickness of the core is 0.50 in. The beam is subjected to a bending moment of 250 lb-in. acting about the z axis. Find the maximum bending stresses sface and score in the faces and the core, respectively, if their respective moduli of elasticity are 4 ⫻ 106 psi and 1.5 ⫻ 106 psi.

0.10 in. z

0.50 in.

C

0.10 in. 2.0 in.

Solution 6.2-1 Composite beam b ⫽ 2 in.

h ⫽ 0.7 in.

hc ⫽ 0.5 in.

M ⫽ 250 lb-in.

E1 ⫽ 4 ⫻ 10 psi 6

E2 ⫽ 1.5 ⫻ 106 psi

I1 ⫽

b 3 (h ⫺ h3c) ⫽ 0.03633 in.4 12

I2 ⫽

bh3c ⫽ 0.02083 in.4 12

From Eq. (6-6b): score ⫽ ;

E1I1 + E2I2 ⫽ 176,600 lb-in.2 From Eq. (6-6a): sface ⫽ ;

M(h/2)E1 E1I1 + E2I2

⫽ ; 1980 psi

M(hc / 2)E2 E1I1 + E2I2

⫽ ;531psi

;

;

497

9/24/08

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CHAPTER 6 Stresses in Beams (Advanced Topics)

y

Problem 6.2-2 A wood beam with cross-sectional dimensions 200 mm ⫻ 300 mm is reinforced on its sides by steel plates 12 mm thick (see figure). The moduli of elasticity for the steel and wood are Es ⫽ 190 GPa and Ew ⫽ 11 GPa, respectively. Also, the corresponding allowable stresses are ss ⫽ 110 MPa and sw ⫽ 7.5 MPa. (a) Calculate the maximum permissible bending moment Mmax when the beam is bent about the z axis. (b) Repeat part a if the beam is now bent about its y axis.

z 12 mm

z

C

200 mm

498

5:32 AM

300 mm

06Ch06.qxd

200 mm 12 mm

300 mm

12 mm

12 mm

y

C

(a)

(b)

Solution 6.2-2 MAXIMUM MOMENT BASED UPON THE WOOD

y

Mmax_w ⫽ sallow_w

1

2

J

Ew Iw + Es Is h a b Ew 2

Mmax_w ⫽ 69.1 kN # m

z

300 mm

C

MAXIMUM MOMENT BASED UPON THE STEEL Mmax_s ⫽ sallow_s

200 mm 12 mm

K

J

Ew Iw + Es Is h a b Es 2

Mmax_s ⫽ 58.7 kN # m

12 mm

K

Mmax ⫽ min (Mmax_w, Mmax_s) (a) BENT ABOUT THE Z AXIS b ⫽ 200 mm Ew ⫽ 11 GPa

t ⫽ 12 mm

h ⫽ 300 mm

Es ⫽ 190 GPa

sallow_w ⫽ 7.5 MPa

sallow_s ⫽ 110 MPa

Iw ⫽

bh3 12

Iw ⫽ 4.50 * 108 mm4

Is ⫽

2th3 12

Is ⫽ 5.40 * 107 mm4

EwIw ⫹ EsIs ⫽ 1.52 ⫻ 107 N⭈m2

STEEL GOVERNS.

Mmax ⫽ 58.7 kN # m

(b) BENT ABOUT THE Y AXIS Iw ⫽

b3 h 12

Is ⫽ 2c

Iw ⫽ 2.00 * 108 mm4

t3h b + t 2 + th a b d 12 2

Is ⫽ 8.10 * 107 mm4 Ew Iw + Es Is ⫽ 1.76 * 107 N # m2

;

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SECTION 6.2 Composite Beams

MAXIMUM MOMENT BASED UPON THE STEEL

MAXIMUM MOMENT BASED UPON THE WOOD Mmax_w ⫽ sallow_w

J

Ew Iw + Es Is b a bEw 2

Mmax_w ⫽ 119.9 kN # m

Mmax_s ⫽ sallow_s

K

Ew Iw + Es Is

J a

Mmax_s ⫽ 90.9 kN # m

b + tb Es K 2 ;

Mmax ⫽ min (Mmax_w, Mmax_s) Mmax ⫽ 90.9 kN # m

STEEL GOVERNS.

y

Problem 6.2-3 A hollow box beam is constructed with webs

(a) If the allowable stresses are 2000 psi for the plywood and 1750 psi for the pine, find the allowable bending moment Mmax when the beam is bent about the z axis. (b) Repeat part a if the b.eam is now bent about its y axis.

1.5 in.

z

C

;

z 1.5 in.

1.5 in.

1 in. 3.5 in.

of Douglas-fir plywood and flanges of pine, as shown in the figure in a cross-sectional view. The plywood is 1 in. thick and 12 in. wide; the flanges are 2 in. ⫻ 4 in. (nominal size). The modulus of elasticity for the plywood is 1,800,000 psi and for the pine is 1,400,000 psi.

12 in.

06Ch06.qxd

C

1 in. 1.5 in. 1 in.

12 in. 3.5 in.

1 in. (b)

(a)

Solution 6.2-3 (a) BENT ABOUT THE Z AXIS

t

t 1

I1 ⫽

b(h3 ⫺ h31) 12

I2 ⫽

2th3 12

I1 ⫽ 291 in.4

I2 ⫽ 288 in.4

E1I1 + E2I 2 ⫽ 9.26 * 108 lb # in.2

h1

2

MAXIMUM MOMENT BASED UPON THE WOOD

h

Mmax_1 ⫽ sallow_1 1 2

b ⫽ 3.5 in.

h a b E1 2

Mmax_1 ⫽ 193 k # in.

b t ⫽ 1 in.

J

E1 I1 + E2 I2

;

K

MAXIMUM MOMENT BASED UPON THE PLYWOOD h ⫽ 12 in.

h1 ⫽ 9 in.

E1 ⫽ 1.4 * 106 psi

E2 ⫽ 1.8 * 106 psi

sallow_1 ⫽ 1750 psi

sallow_2 ⫽ 2000 psi

Mmax_2 ⫽ sallow_2

J

E1 I1 + E2 I2

Mmax_2 ⫽ 172 k # in.

h a b E2 2 ;

K

y

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Mmax ⫽ min (Mmax_1, Mmax_2)

MAXIMUM MOMENT BASED UPON THE PLYWOOD

Mmax ⫽ 172 k # in.

PLYWOOD GOVERNS.

;

Mmax_2 ⫽sallow_2

(b) BENT ABOUT THE Y AXIS

Ja

Mmax_2 ⫽ 96 k # in.

b3 (h ⫺ h1) I1 ⫽ 12 I2 ⫽ 2 c

E1 I1 + E2 I2

I1 ⫽ 11 in.

4

3

2

t h b + t + th a b d 12 2

I2 ⫽ 123 in.4

b + tb E2 K 2

Mmax ⫽ min (Mmax_1, Mmax_1) Mmax ⫽ 96 k # in.

PLYWOOD GOVERNS.

;

E1 I1 + E2 I2 ⫽ 2.37 * 108 lb # in.2 MAXIMUM MOMENT BASED UPON THE WOOD Mmax_1 ⫽ sallow_1

J

E1 I1 + E2 I2 b a b E1 2

Mmax_1 ⫽ 170 k # in.

K

Problem 6.2-4 A round steel tube of outside diameter d and an brass core of diameter

S

2d/3 are bonded to form a composite beam, as shown in the figure. Derive a formula for the allowable bending moment M that can be carried by the beam based upon an allowable stress ss in the steel. (Assume that the moduli of elasticity for the steel and brass are Es and Eb, respectively.)

B

2d/3 d

Solution 6.2-4 Core (2): I2 ⫽

B

E1 I1 + E2 I2 ⫽ Es I1 + Eb I2 ⫽

2d/3

Mallow ⫽ ss

d

Tube (1): I1 ⫽

p 2d 4 pd 4 a b ⫽ 64 3 324

S

p 2d 4 65 cd 4 ⫺ a b d ⫽ pd 4 64 3 5184

Mallow ⫽

J

pd 4 (65Es + 16 Eb) 5184

E1I1 + E2 I2 d a b Es 2

K

sspd 3 Eb a 65 + 16 b 2592 Es

;

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SECTION 6.2 Composite Beams

Problem 6.2-5 A beam with a guided support and 10 ft span supports a distributed load of intensity q ⫽ 660 lb/ft over its first half (see figure part a) and a moment M0 ⫽ 300 ft-lb at joint B. The beam consists of a wood member (nominal dimensions 6 in. ⫻ 12 in., actual dimensions 5.5 in. ⫻ 11.5 in. in cross section, as shown in the figure part b) that is reinforced by 0.25-in.-thick steel plates on top and bottom. The moduli of elasticity for the steel and wood are Es ⫽ 30 ⫻ 106 psi and Ew ⫽ 1.5 ⫻ 106 psi, respectively.

y 0.25 in. q

11.5 in.

M0 z A

5 ft

C

C

B

5 ft

0.25 in.

(a) Calculate the maximum bending stresses ss in the steel (a) plates and sw in the wood member due to the applied loads. (b) If the allowable bending stress in the steel plates is sas ⫽ 14,000 psi and that in the wood is saw ⫽ 900 psi, find qmax. (Assume that the moment at B, M0, remains at 300 ft-lb.) (c) If q ⫽ 660 lb/ft and allowable stress values in (b) apply, what is M0,max at B?

5.5 in. (b)

Solution 6.2-5 q ⫽ 660 lb/it

M0 ⫽ 300 lb # ft

L ⫽ 10 ft

(b) MAXIMUM UNIFORM DISTRIBUTED LOAD MAXIMUM MOMENT BASED UPON WOOD

(a) MAXIMUM BENDING STRESSES

sallow_w ⫽ 900 psi

L 3L Mmax ⫽ q a b a b + M0 2 4 Mmax ⫽ 25050 lb # ft b ⫽ 5.5 in.

Wood (1):

From sallow_w ⫽

h1 ⫽ 11.5 in.

I1 ⫽

b 3 1h ⫺ h312 12

sw ⫽

ss ⫽

h1 b Ew 2

Ew I1 + EsI2 h Mmax a b Es 2 Ew I1 + Es I2

Ew I1 + Es I2

sallow_s ⫽ 14000 psi

t ⫽ 0.25 in. Es ⫽ 30 * 106 psi From sallow_s ⫽

I2 ⫽ 94.93 in.4

h Mallow_s a b Es 2 Ew I1 + Es I2

Mallow_s ⫽ 25236 lb-ft

Ew I1 + Es I 2 ⫽ 3.894 * 109 lb # in.2 Mmax a

h1 b Ew 2

MAXIMUM MOMENT BASED UPON STEEL PLATE

I1 ⫽ 697.07 in.4

b ⫽ 5.5 in. h ⫽ 12 in.

Plate (2): I2 ⫽

bh31 12

Mallow_w a

Mallow_w ⫽ 33857 lb-ft

Ew ⫽ 1.5 * 106 psi

MAXIMUM ALLOWABLE MOMENT sw ⫽ 666 psi

;

Mallow ⫽ min (Mallow_s, Mallow_w) STEEL PLATES GOVERN

ss ⫽ 13897 psi

;

501

Mallow ⫽ 25236 lb-ft

;

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CHAPTER 6 Stresses in Beams (Advanced Topics)

MAXIMUM UNIFORM DISTRIBUTED LOAD L 3L From Mallow ⫽ qmax a b a b + M0 2 4 qmax ⫽ 665lb/ft

;

(c) MAXIMUM APPLIED MOMENT L 3L From Mallow ⫽ q a b a b + Mo_max 2 4 M0_max ⫽ 486 lb-ft

;

y

Problem 6.2-6 A plastic-lined steel pipe has the cross-sectional shape shown in the figure. The steel pipe has outer diameter d3 ⫽ 100 mm and inner diameter d2 ⫽ 94 mm. The plastic liner has inner diameter d1 ⫽ 82 mm. The modulus of elasticity of the steel is 75 times the modulus of the plastic. Determine the allowable bending moment Mallow if the allowable stress in the steel is 35 MPa and in the plastic is 600 kPa.

z

C

d1

Solution 6.2-6 Steel pipe with plastic liner MAXIMUM MOMENT BASED UPON THE STEEL (1) From Eq. (6-6a): Mmax ⫽ (s1)allow c ⫽ (s1)allow

(1) Pipe: ds ⫽ 100 mm

d2 ⫽ 94 mm

Es ⫽ E1 ⫽ modulus of elasticity (s1)allow ⫽ 35 MPa (2) Liner: d2 ⫽ 94 mm

d1 ⫽ 32 mm

Ep ⫽ E2 ⫽ modulus of elasticity (s2)allow ⫽ 600 kPa E1 ⫽ 75E2

E1/E2 ⫽ 75

I1 ⫽

p 4 (d ⫺ d 42) ⫽ 1.076 * 10⫺6 m4 64 3

I2 ⫽

p 4 (d ⫺ d 41) ⫽ 1.613 * 10⫺6 m4 64 2

E1I1 + E2I2 d (d3/2)E1

(E1/E2)I1 + I2 ⫽ 768 N # m (d3/2)(E1/E2)

MAXIMUM MOMENT BASED UPON THE PLASTIC (2) From Eq. (6-6b): Mmax ⫽ (s2)allow c ⫽ (s2)allow c STEEL GOVERNS.

E1I1 + E2I2 d (d2/2)E2

(E1/E2)I1 + I2 d ⫽ 1051 N # m (d2/2) Mallow ⫽ 768 N # m

;

d2 d3

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SECTION 6.2 Composite Beams

Problem 6.2-7 The cross section of a sandwich beam consisting of aluminum

y

alloy faces and a foam core is shown in the figure. The width b of the beam is 8.0 in., the thickness t of the faces is 0.25 in., and the height hc of the core is 5.5 in. (total height h ⫽ 6.0 in.). The moduli of elasticity are 10.5 ⫻ 106 psi for the aluminum faces and 12,000 psi for the foam core. A bending moment M ⫽ 40 k-in. acts about the z axis. Determine the maximum stresses in the faces and the core using (a) the general theory for composite beams, and (b) the approximate theory for sand wich beams.

t

z

Probs. 6.2-7 and 6.2-8

Solution 6.2-7

C

hc

b

t

h

Sandwich beam I2 ⫽

bh3c ⫽ 110.92 in.4 12

M ⫽ 40 k.in. E1I1 + E2I2 ⫽ 348.7 * 106 lb-in.2 (a) GENERAL THEORY (EQS. 6-6a AND b) sface ⫽ s1 ⫽

M(h/2)E1 ⫽ 3610 psi E1I1 + E2I2

score ⫽ s2 ⫽

M(hc / 2)E2 ⫽ 4 psi E1I1 + E2I2

;

;

(1) ALUMINUM FACES: b ⫽ 8.0 in.

t ⫽ 0.25 in.

h ⫽ 6.0 in.

E1 ⫽ 10.5 * 106 psi I1 ⫽

I1 ⫽

b 3 (h ⫺ h3c ) ⫽ 33.08 in.4 12

b 3 (h ⫺ h3c ) ⫽ 33.08 in.4 12

sface ⫽

Mh ⫽ 3630 psi 2I1

score ⫽ 0

(2) Foam core: b ⫽ 8.0 in.

(b) APPROXIMATE THEORY (EQS. 6-8 AND 6-9)

hc ⫽ 5.5 in.

;

;

E2 ⫽ 12,000 psi

Problem 6.2-8 The cross section of a sandwich beam consisting of fiberglass faces and a lightweight plastic core is shown in the figure. The width b of the beam is 50 mm, the thickness t of the faces is 4 mm, and the height hc of the core is 92 mm (total height h ⫽ 100 mm). The moduli of elasticity are 75 GPa for the fiberglass and 1.2 GPa for the plastic. A bending moment M ⫽ 275 N # m acts about the z axis. Determine the maximum stresses in the faces and the core using (a) the general theory for composite beams, and (b) the approximate theory for sandwich beams.

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Solution 6.2-8

Sandwich beam (a) GENERAL THEORY (EQS. 6-6a AND b) sface ⫽ s1 ⫽

M(h/2)E1 ⫽ 14.1 MPa E1I1 + E2I2

;

score ⫽ s2 ⫽

M(hc / 2)E2 ⫽ 0.21 MPa E1I1 + E2I2

;

(b) APPROXIMATE THEORY (EQS. 6-8 AND 6-9) I1 ⫽

(1) Fiber glass faces: b ⫽ 50 mm

t ⫽ 4 mm

h ⫽ 100 mm

sface ⫽

E1 ⫽ 75 GPa I1 ⫽

b 3 (h ⫺ h3c ) ⫽ 0.9221 * 106 m4 12 Mh ⫽ 14.9 MPa 2I1

score ⫽ 0

b 3 (h ⫺ h3c ) ⫽ 0.9221 * 10⫺6 m4 12

;

;

(2) Plastic core: hc ⫽ 92 mm

b ⫽ 50 mm I2 ⫽

bh3c 12

E2 ⫽ 1.2 GPa

⫽ 3.245 * 10⫺6 m4

M ⫽ 275 N # m

E1I1 + E2I2 ⫽ 73,050 N # m2

Problem 6.2-9 A bimetallic beam used in a temperature-control switch consists of strips of aluminum and copper bonded together as shown in the figure, which is a cross-sectional view. The width of the beam is 1.0 in., and each strip has a thickness of 1/16 in. Under the action of a bending moment M ⫽ 12 lb-in. acting about the z axis, what are the maximum stresses sa and sc in the aluminum and copper, respectively? (Assume Ea ⫽ 10.5 ⫻ 106 psi and Ec ⫽ 16.8 ⫻ 106 psi.)

y A z O 1.0 in.

Solution 6.2-9 Bimetallic beam NEUTRAL AXIS (EQ. 6-3)

CROSS SECTION

L1

ydA ⫽ y1A1 ⫽ (h1 ⫺ t/2)(bt) ⫽ (h1 ⫺ 1/32)(1)(1/16) in.3

(1) Aluminum E1 ⫽ Ea ⫽ 10.5 ⫻ 106 psi (2) Copper

E2 ⫽ Ec ⫽ 16.8 ⫻ 10 psi 6

M ⫽ 12 lb-in.

L2

1 — in. 16

ydA ⫽ y2A2 ⫽ (h1 ⫺ t ⫺ t/2)(bt) ⫽ (h1 ⫺ 3/32)(1)(1/16) in.3

Eq. (6-3): E1 11 ydA + E2 12 ydA ⫽ 0

C 1 — in. 16

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SECTION 6.2 Composite Beams

(10.5 ⫻ 106)(h1 ⫺ 1/32)(1/16) ⫹ (16.8 ⫻ 106)(h1 ⫺ 3/32)(1/16) ⫽ 0

MAXIMUM STRESSES (EQS. 6-6a AND b) sa ⫽ s1 ⫽

Mh1E1 ⫽ 4120 psi E1I1 + E2I2

;

sc ⫽ s2 ⫽

Mh2E2 ⫽ 5230 psi E1I1 + E2I2

;

Solve for h1: h1 ⫽ 0.06971 in. h2 ⫽ 2(1/16 in.) ⫺ h1 ⫽ 0.05529 in.

505

MOMENTS OF INERTIA (FROM PARALLEL-AXIS THEOREM) I1 ⫽

bt3 + bt(h1 ⫺ t/2)2 ⫽ 0.0001128 in.4 12

I2 ⫽

bt3 + bt(h2 ⫺ t/2)2 ⫽ 0.00005647 in.4 12

E1I1 + E2I2 ⫽ 2133 lb-in.2

Problem 6.2-10 A simply supported composite beam

y

3 m long carries a uniformly distributed load of intensity q ⫽ 3.0 kN/m (see figure). The beam is constructed of a wood member, 100 mm wide by 150 mm deep, reinforced on its lower side by a steel plate 8 mm thick and 100 mm wide. Find the maximum bending stresses sw and ss in the wood and steel, respectively, due to the uniform load if the moduli of elasticity are Ew ⫽ 10 GPa for the wood and Es ⫽ 210 GPa for the steel.

q = 3.0 kN/m 150 mm z

O 8 mm

3m 100 mm

Solution 6.2-10 Simply supported composite beam BEAM: L ⫽ 3 m 2

Mmax ⫽

q ⫽ 3.0 kN/m

qL ⫽ 3375 N # m 8

CROSS SECTION

b ⫽ 100 mm

h ⫽ 150 mm

t ⫽ 8 mm

(1) Wood: E1 ⫽ Ew ⫽ 10 GPa (2) Steel: E2 ⫽ Es ⫽ 210 GPa NEUTRAL AXIS L1

ydA ⫽ y1A1 ⫽ (h1 ⫺ h/2)(bh) ⫽ (h1 ⫺ 75)(100)(150) mm3

L2

ydA ⫽ y2A2 ⫽ ⫺ (h + t/2 ⫺ h1)(bt) ⫽ ⫺ (154 ⫺ h1)(100)(18) mm3

Eq. (6-3): E1 11ydA + E2 12ydA ⫽ 0

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MAXIMUM STRESSES (EQS. 6-6a AND b)

(10 GPa)(h1 ⫺ 75)(100)(150)(10⫺9) ⫹ (210 GPa)(h1 ⫺ 154)(100)(8)(10⫺9) ⫽ 0

sw ⫽ s1 ⫽

Solve for h1: h1 ⫽ 116.74 mm

Mh1E1 E1I1 + E2I2

⫽ 5.1MPa (Compression)

h2 ⫽ h ⫹ t ⫺ h1 ⫽ 41.26 mm MOMENTS OF INERTIA (FROM PARALLEL-AXIS THEOREM) I1 ⫽

bh3 + bh(h1 ⫺ h/2)2 ⫽ 54.26 * 106 mm4 12

I2 ⫽

bt2 + bt(h2 ⫺ t/2)2 ⫽ 1.115 * 106 mm4 12

ss ⫽ s2 ⫽

;

Mh2E2 E1I1 + E2I2

⫽ 37.6MPa (Tension)

;

E1I1 + E2I2 ⫽ 776,750 N # m2

Problem 6.2-11 A simply supported wooden I-beam with a 12 ft span supports

a distributed load of intensity q ⫽ 90 lb/ft over its length (see figure part a). The beam is constructed with a web of Douglas-fir plywood and flanges of pine glued 2 in. to the web as shown in the figure part b. The plywood is 3/8 in. thick; the flanges are 2 in. ⫻ 2 in. (actual size). The modulus of z q 8 in. elasticity for the plywood is 1,600,000 psi and for the pine is 1,200,000 psi. (a) Calculate the maximum bending stresses in the pine flanges and in the plywood web. (b) What is qmax if allowable stresses are 1600 psi in the flanges and 1200 psi in the web?

A

3 — in. plywood 8 (Douglas fir)

2 in.

B

12 ft

y 2 in. ⫻ 2 in. pine flange 1 — in. 2 C

2 in. (b)

(a)

Solution 6.2-11 q ⫽ 90 lb/f

L ⫽ 12ft

I2 ⫽2c

(a) MAXIMUM BENDING STRESSES Mmax

qL2 ⫽ 8

Plywood (1):

+ (b ⫺ t) (h2 ⫺ a) a

Mmax ⫽ 1620 lb # ft t⫽

3 in. 8

I2 ⫽ 65.95 in.4

h1 ⫽7 in.

Eplywood ⫽ 1.6 * 10 psi

Pine (2): b ⫽ 2 in.

I1 ⫽ 10.72 in.4 h2 ⫽ 2 in.

Epine ⫽ 1.2 * 106 psi

h1 h2 ⫺ a 2 ⫺ b d 2 2

;

Eplywood I1 ⫹ EpineI2 ⫽ 96.287 ⫻ 106 lbin.2

6

t h13 I1 ⫽ 12

h1 + a 2 (b ⫺ t)(h2 ⫺ a)3 ba3 + ba a b + 12 2 12

splywood ⫽ a⫽

1 in. 2

Mmax a

h1 b Eplywood 2

Eplywood I1 + Epine I2

splywood ⫽ 1131 psi

;

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SECTION 6.2 Composite Beams

spine ⫽

Mmax a

h1 + a b Epine 2

From sallow_pine ⫽

Eplywood I1 + Epine I2

spine ⫽ 969 psi

h1 + ab Epine 2

Eplywood I1 + Epine I2

Mallow_pine ⫽ 2675 lb # ft

;

(b) MAXIMUM UNIFORM DISTRIBUTED LOAD MAXIMUM MOMENT BASED UPON PLYWOOD

MAXIMUM ALLOWABLE MOMENT Mallow ⫽ min (Mallow_plywood, Mallow_pine) Mallow ⫽ 1719 lb⭈ft

PLYWOOD GOVERNS.

sallow_plywood ⫽ 1200 psi

From sallow_plywood ⫽

Mallow_pine a

Mallow_plywood a

h1 bEplywood 2

Eplywood I1 + Epine I2

;

MAXIMUM UNIFORM DISTRIBUTED LOAD From Mallow ⫽

qmax L2 8

qmax ⫽ 95.5 lb/ft

Mallow_plywood ⫽ 1719 lb # ft

;

MAXIMUM MOMENT BASED UPON PINE sallow_pine ⫽ 1600 psi

6 mm ⫻ 80 mm steel plate

Problem 6.2-12 A simply supported composite beam with a 3.6 m span supports a triangularly distributed load of peak intensity q0 at midspan (see figure part a). The beam is constructed of two wood joists, each 50 mm ⫻ 280 mm, fastened to two steel plates, one of dimensions 6 mm ⫻ 80 mm and the lower plate of dimensions 6 mm ⫻ 120 mm (see figure part b). The modulus of elasticity for the wood is 11 GPa and for the steel is 210 GPa. If the allowable stresses are 7 MPa for the wood and 120 MPa for the steel, find the allowable peak load intensity q0,max when the beam is bent about the z axis. Neglect the weight of the beam.

50 mm ⫻ 280 mm wood joist

y

280 mm

C

z 6 mm ⫻ 120 mm steel plate

q0

A

1.8 m

1.8 m

B

(a)

(b)

Solution 6.2-12 L ⫽ 3.6 m

Steel (2):

t1 ⫽ 50 mm

b1 ⫽ 80 mm

b2 ⫽ 120 mm

DETERMINE NEUTRAL AXIS WOOD (1):

t2 ⫽ 6 mm

h ⫽ 280 mm

Ew ⫽ 11 GPa h y1 dA ⫽ y1 A1 ⫽ a ⫺ h1b (2t1 h) 2 L

L

Es ⫽ 210 GPa

y2 dA ⫽ y2 A 2 ⫽ ah ⫺ h1 ⫺ ⫺ ah1 ⫺

b1 b (t2 b1) 2

b2 b (t 2 b 2) 2

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From E1 Ew a

L

y1 dA + E2

L

MAXIMUM MOMENT BASED UPON WOOD

y2 dA ⫽ 0

sallow_w ⫽ 7 MPa

b1 h ⫺ h1 b (2t1 h) + Es c a h ⫺ h1 ⫺ b 2 2

(t2 b1) ⫺ ah1 ⫺

From

sallow_w ⫽

Mallow_w (h ⫺ h1)Ew Ew I1 + Es I2

Mallow_w ⫽ 18.68 kN # m

b2 b (t2 b 2) d ⫽ 0 2

MAXIMUM MOMENT BASED UPON STEEL

h1 ⫽ 136.4 mm

sallow_s ⫽ 120 MPa

MOMENT OF INERTIA

From sallow_s ⫽

2 t1 h 3 h Wood (1): I1 ⫽ 2 c + (t1 h)a ⫺ h1 b d 12 2

Mallow_s (h ⫺ h1)Es Ew I1 + Es I2

Mallow_s ⫽ 16.78 kN # m

I1 ⫽ 183.30 * 10 mm 6

Steel (2):

I2 ⫽

4

MAXIMUM ALLOWABLE MOMENT

t2 b31 b1 2 + t2 b1 ah ⫺ h1 ⫺ b 12 2 +

t2 b23 12

+ t2 b2 ah1 ⫺

Mallow ⫽ min(Mallow_w,Mallow_s)

b2 2 b 2

Mallow ⫽ 16.78 kN # m

STEEL GOVERNS.

;

MAXIMUM UNIFORM DISTRIBUTED LOAD

I2 ⫽ 10.47 ⫻ 106 mm4 Ew I1 ⫹ Es I2 ⫽ 4.22 ⫻ 1012 N # mm2

From

Mallow ⫽

qomax L2 12

qomax ⫽ 15.53 kN/m

;

Transformed-Section Method When solving the problems for Section 6.3, assume that the component parts of the beams are securely bonded by adhesives or connected by fasteners. Also, be sure to use the transformed-section method in the solutions.

y

Problem 6.3-1 A wood beam 8 in. wide and 12 in. deep

3.5 in.

(nominal dimensions) is reinforced on top and bottom by 0.25-in.-thick steel plates (see figure part a). z

C

z

11.25 in.

0.25 in

11.5 in.

(a) Find the allowable bending moment Mmax about the z axis if the allowable stress in the wood is 1,100 psi and in the steel is 15,000 psi. (Assume that the ratio of the moduli of elasticity of steel and wood is 20.) (b) Compare the moment capacity of the beam in part a with that shown in the figure part b which has two 4 in. ⫻ 12 in. joists (nominal dimensions) attached to a 1/4 in. ⫻ 11.0 in. steel plate.

1 — in. ⫻ 11.0 in. 4 steel plate

y C 4 ⫻ 12 joists

0.25 in 7.5 in. (a)

(b)

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SECTION 6.3 Transformed-Section Method

Solution 6.3-1 Mmax ⫽min(M1, M2)

(a) FIND Mmax b ⫽ 7.5 in.

(1) Wood beam

h1 ⫽ 11.5 in.

Mmax ⫽ 422 k-in.

STELL GOVERNS

;

sallow_w ⫽ 1100 psi b ⫽ 7.5 in.

(2) Steel plates

h2 ⫽ 12 in.

t ⫽ 0.25 in. sallow_s ⫽ 15000 psi TRANSFORMED SECTION (WOOD) n ⫽ 20

IT ⫽

bh31 12

bT ⫽ 150 in. + 2c

b ⫽ 3.5 in.

(1) Wood beam (2) Steel plates

h1 ⫽ 11.25 in.

h2 ⫽ 11 in.

t ⫽ 0.25 in.

WIDTH OF STEEL PLATES bT ⫽ nt bT ⫽ 5 in.

WIDTH OF STEEL PLATES bT ⫽ nb

(b) COMPARE MOMENT CAPACITIES

t3 bT h2 ⫺ t 2 + t bT a b d 12 2

IT ⫽ 2

bh13 bT h23 + 12 12

MAXIMUM MOMENT BASED UPON THE WOOD (1) M1 ⫽

IT ⫽ 3540 in.

4

IT ⫽1385 in.4

sallow_w IT h1 2

M1 ⫽ 271 k # in.

MAXIMUM MOMENT BASED UPON THE WOOD (1) M1 ⫽

sallow_w IT h1 2

M1 ⫽ 677 k # in.

M2 ⫽

MAXIMUM MOMENT BASED UPON THE STEEL (2) M2 ⫽

sallow_s I T h2n 2

MAXIMUM MOMENT BASED UPON THE STEEL (2)

M2 ⫽ 442 k # in.

sallow_s IT h2 n 2

M2 ⫽ 189 k # in.

Mmax ⫽min(M1, M2) STELL GOVERNS.

Mmax ⫽189 k-in.

;

THE MOMENT CAPACITY OF THE BEAM IN (a) IS 2.3 (b)

TIMES MORE THAN THE BEAM IN

y

Problem 6.3-2 A simple beam of span length 3.2 m carries a uniform load of intensity 48 kN/m. The cross section of the beam is a hollow box with wood flanges and steel side plates, as shown in the figure. The wood flanges are 75 mm by 100 mm in cross section, and the steel plates are 300 mm deep. What is the required thickness t of the steel plates if the allowable stresses are 120 MPa for the steel and 6.5 MPa for the wood? (Assume that the moduli of elasticity for the steel and wood are 210 GPa and 10 GPa, respectively, and disregard the weight of the beam.)

75 mm

z

300 mm

C

75 mm 100 mm

t

t

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CHAPTER 6 Stresses in Beams (Advanced Topics)

Solution 6.3-2 Mmax ⫽

Box beam Width of steel plates

qL2 ⫽ 61.44 kN # m 8

SIMPLE BEAM:

L ⫽ 3.2 m

(1) Wood flanges: b ⫽ 100 mm

⫽ nt ⫽ 21t q ⫽ 48 kN/m

All dimensions in millimeters.

h ⫽ 300 mm IT ⫽

h1 ⫽ 150 mm (s1)allow ⫽ 6.5 MPa

1 1 (100 + 42t)(300)3 ⫺ (100)(150)3 12 12

⫽ 196.9 * 106 mm4 + 94.5t * 106 mm4

Ew ⫽ 10 GPa (2) Steel plates:

t ⫽ thickness

h ⫽ 300 mm

(s2)allow ⫽ 120 MPa Es ⫽ 210 GPa TRANSFORMED SECTION (WOOD)

REQUIRED THICKNESS BASED UPON THE WOOD (1) (EQ. 6-15) s1 ⫽

M(h/2) IT

(IT)1 ⫽

Mmax(h/2) (s1)allow

⫽ 1.418 * 109 mm4 Equate IT and (IT)1 and solve for t : t1 ⫽ 12.92 mm REQUIRED THICKNESS BASED UPON THE STEEL (2) (EQ. 6-17) s2 ⫽

M(h/2)n IT

(IT)2 ⫽

Mmax(h/2)n (s2)allow

⫽ 1.612 * 109 mm4 Equate IT and (IT)2 and solve for t : t2 ⫽ 14.97 mm tmin ⫽ 15.0 mm

STEEL GOVERNS.

;

Wood flanges are not changed n⫽

Es ⫽ 21 Ew

y

Problem 6.3-3 A simple beam that is 18 ft long supports a uniform load of intensity q. The beam is constructed of two C 8 ⫻ 11.5 sections (channel sections or C shapes) on either side of a 4 ⫻ 8 (actual dimensions) wood beam (see the cross section shown in the figure part a). The modulus of elasticity of the steel (Es ⫽ 30,000 ksi) is 20 times that of the wood (Ew).

z

C 8 ⫻ 11.5

C

z

y

(a) If the allowable stresses in the steel and wood are 12,000 C psi and 900 psi, respectively, what is the allowable load Wood beam C 8 ⫻ 11.5 Wood beam qallow? (Note: Disregard the weight of the beam, and see Table E-3a of Appendix E for the dimensions and proper(a) (b) ties of the C-shape beam.) (b) If the beam is rotated 90° to bend about its y axis (see figure part b), and uniform load q ⫽ 250 lb/ft is applied, find the maximum stresses ss and sw in the steel and wood, respectively. Include the weight of the beam. (Assume weight densities of 35 lb/ft3 and 490 lb/ft3 for the wood and steel, respectively.)

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SECTION 6.3 Transformed-Section Method

Solution 6.3-3 (b) BENT ABOUT THE Y AXIS (INCLUDING THE WEIGHT OF THE BEAM) q ⫽ 250 lb/ft.

L ⫽ 18 ft

(1) Wood beam

rw ⫽ 35 lb/ft

(1) Wood beam

(a) BENT ABOUT THE Z AXIS b ⫽ 4 in.

qw ⫽ 7.778 Ib/ft.

h ⫽ 8 in.

sallow_w ⫽ 900 psi

qs ⫽ 11.5 lb/ft.

(2) Steel Channels Iz ⫽ 32.5 in.4

(2) Steel Channel h ⫽ 8.0 in. Iy ⫽ 1.31 in.

c ⫽ 0.572 in.

4

qw ⫽ bhrw

sallow_s ⫽ 12000 psi

As ⫽ 3.37 in.2 qtotal ⫽ q + qw + 2qs

bs ⫽ 2.26 in.

qtotal ⫽ 281 lb/ft.

2

Mmax ⫽

qtotal L 8

Mmax ⫽ 11.4 k # ft.

TRANSFORMED SECTION (WOOD) TRANSFORMED SECTION (WOOD)

n ⫽ 20 bh3 IT ⫽ + 2 Iz n 12

IT ⫽

IT ⫽ 1471 in.

4

b3h b 2 + 2n cIy + As a c + b d 12 2

IT ⫽ 987 in.4

MAXIMUM MOMENT BASED UPON THE WOOD (1) M1 ⫽

sallow_w IT h/2

MAXIMUM STRESS IN THE WOOD (1) M1 ⫽ 331 k # in. sw_max ⫽

MAXIMUM MOMENT BASED UPON THE STEEL (2) M2 ⫽

sallow_s IT hn/ 2

M2 ⫽ 221 k # in.

ss_max ⫽

Mmax ⫽ 221 k # in

qallow L2 8

qallow ⫽ 454 lb/ft.

nMmax a

sw_max ⫽ 277 psi

;

b + bsb 2

IT

ss_max ⫽ 11782 psi

ALLOWABLE LOAD ON a 18-FT-LONG SIMPLE BEAM From Mmax ⫽

IT

MAXIMUM MOMENT BASED UPON THE STEEL (2)

Mmax ⫽ min(M1, M2) STEEL GOVERNS.

b Mmax a b 2

;

;

Problem 6.3-4 The composite beam shown in the figure is simply supported and carries a total uniform load of 50 kN/m on a span length of 4.0 m. The beam is built of a wood member having cross-sectional dimensions 150 mm ⫻ 250 mm and two steel plates of cross-sectional dimensions 50 mm ⫻ 150 mm. Determine the maximum stresses ss and sw in the steel and wood, respectively, if the moduli of elasticity are Es ⫽ 209 GPa and Ew ⫽ 11 GPa. (Disregard the weight of the beam.)

y 50 kN/m 50 mm z

C

250 mm 50 mm

4.0 m 150 mm

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Solution 6.3-4 Composite beam SIMPLE BEAM:

L ⫽ 4.0 m

qL2 ⫽ 100 kN # m 8 (1) Wood beam: b ⫽ 150 mm

q ⫽ 50 kN/m

Mmax ⫽

h2 ⫽ 350 mm TRANSFORMED SECTION (WOOD)

⫽ nb ⫽ (19) (150 mm) ⫽ 2850 mm All dimensions in millimeters.

h1 ⫽ 250 mm

Ew ⫽ 11 GPa (2) Steel plates: b ⫽ 150 mm

Width of steel plates

t ⫽ 50 mm Es ⫽ 209 GPa

IT ⫽

1 1 (2850)(350)3 ⫺ (2850 ⫺ 150)(250)3 12 12

⫽ 6.667 * 109 mm4 MAXIMUM STRESS IN THE WOOD (1) (EQ. 6-15) sw ⫽ s1 ⫽

Mmax(h1/2) ⫽ 1.9 MPa IT

;

MAXIMUM STRESS IN THE STEEL (2) (EQ. 6-17) ss ⫽ s2 ⫽

Mmax(h2/2)n ⫽ 49.9 MPa IT

;

Wood beam is not changed. n⫽

Es 209 ⫽ ⫽ 19 Ew 11

Problem 6.3-5 The cross section of a beam made of thin strips of aluminum separated by a lightweight plastic is shown in the figure. The beam has width b ⫽ 3.0 in., the aluminum strips have thickness t ⫽ 0.1 in., and the plastic segments have heights d ⫽ 1.2 in. and 3d ⫽ 3.6 in. The total height of the beam is h ⫽ 6.4 in. The moduli of elasticity for the aluminum and plastic are Ea ⫽ 11 ⫻ 106 psi and Ep ⫽ 440 ⫻ 103 psi, respectively. Determine the maximum stresses sa and sp in the aluminum and plastic, respectively, due to a bending moment of 6.0 k-in.

y t

z

d

C

3d

d

Probs. 6.3-5 and 6.3-6

b

h ⫽ 4t ⫹ 5d

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SECTION 6.3 Transformed-Section Method

513

Solution 6.3-5 Plastic beam with aluminum strips (1) Plastic segments: b ⫽ 3.0 in.

d ⫽ 1.2 in.

3d ⫽ 3.6 in. Ep ⫽ 440 ⫻ 103 psi (2) Aluminum strips: b ⫽ 3.0 in.

All dimensions in inches. Plastic: I1 ⫽ 2c

t ⫽ 0.1 in.

+

Ea ⫽ 11 ⫻ 106 psi h ⫽ 4t ⫹ 5d ⫽ 6.4 in. M ⫽ 6.0 k-in.

1 (3.0)(1.2)3 + (3.0)(1.2)(2.50)2 d 12 1 (3.0)(3.6)3 ⫽ 57.528 in.4 12

Aluminum: I2 ⫽ 2c

TRANSFORMED SECTION (PLASTIC) +

1 (75)(0.1)3 + (75)(0.1)(3.15)2 12 1 (75)(0.1)3 + (75)(0.1)(1.85)2 d 12

⫽ 200.2 in.4 IT ⫽ I1 + I2 ⫽ 257.73 in.4 MAXIMUM STRESS IN THE PLASTIC (1) (EQ. 6-15) sp ⫽ s1 ⫽

M(h/2 ⫺ t) ⫽ 72 psi IT

;

MAXIMUM STRESS IN THE ALUMINUM (2) (EQ. 6-17) Plastic segments are not changed. Ea n⫽ ⫽ 25 Ep

sa ⫽ s2 ⫽

M(h/2)n ⫽ 1860 psi IT

;

Width of aluminum strips ⫽ nb ⫽ (25)(3.0 in.) ⫽ 75 in.

Problem 6.3-6 Consider the preceding problem if the beam has width b ⫽ 75 mm, the aluminum strips have thickness t ⫽ 3 mm, the plastic segments have heights d ⫽ 40 mm and 3d ⫽ 120 mm, and the total height of the beam is h ⫽ 212 mm. Also, the moduli of elasticity are Ea ⫽ 75 GPa and Ep ⫽ 3 GPa, respectively. Determine the maximum stresses sa and sp in the aluminum and plastic, respectively, due to a bending moment of 1.0 kN ⭈ m.

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Solution 6.3-6 Plastic beam with aluminum strips (1) Plastic segments: b ⫽ 75 mm 3d ⫽ 120 mm (2) Aluminum strips: b ⫽ 75 mm

All dimensions in millimeters.

d ⫽ 40 mm Ep ⫽ 3 GPa

Plastic: I1 ⫽ 2c

t ⫽ 3 mm

Ea ⫽ 75 GPa

+

h ⫽ 4t ⫹ 5d ⫽ 212 mm

1 (75)(40)3 + (75)(40)(83)2 d 12 1 (75)(120)3 12

⫽ 52.934 * 106 mm4

M ⫽ 1.0 kN # m

ALUMINUM:

TRANSFORMED SECTION (PLASTIC)

I2 ⫽ 2c +

1 (1875)(3)3 + (1875)(3)(104.5)2 12 1 (1875)(3)3 + (1875)(3)(61.5)2 d 12

⫽ 165.420 * 106 mm4 IT ⫽ I1 + I2 ⫽ 218.35 * 106 mm4 MAXIMUM STRESS IN THE PLASTIC (1) (EQ. 6-15) sp ⫽ s1 ⫽ Plastic segments are not changed.

M(h/2 ⫺ t) ⫽ 0.47 MPa IT

;

MAXIMUM STRESS IN THE ALUMINUM (2) (EQ. 6-17)

Ea n⫽ ⫽ 25 Ep

sa ⫽ s2 ⫽

Width of aluminum strips

M(h/2)n ⫽ 12.14 MPa IT

;

⫽ nb ⫽ (25)(75 mm) ⫽ 1875 mm

Problem 6.3-7 A simple beam that is 18 ft long supports a uniform load of intensity q. The beam is constructed of two angle sections, each L 6 ⫻ 4 ⫻ 1/2, on either side of a 2 in. ⫻ 8 in. (actual dimensions) wood beam (see the cross section shown in the figure part a). The modulus of elasticity of the steel is 20 times that of the wood. (a) If the allowable stresses in the 4 in. steel and wood are 12,000 psi and 900 psi, respectively, what is the allowable load qallow? (Note: Disregard the weight of z the beam, and see Table E-5a 8 in. 6 in. of Appendix E for the dimensions and properties of the Steel angle angles.) (b) Repeat part a if a 1 in. ⫻ 10 in. wood flange (actual dimensions) is added (see figure part b). (a)

4 in.

yC

y

Wood flange C

z

Wood beam 2 in.

Wood beam

Steel angle 2 in. (b)

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SECTION 6.3 Transformed-Section Method

Solution 6.3-7 L ⫽ 18 ft

bf ⫽ 1 in.

(b) ADDITIONAL WOOD FLANGE

hf ⫽ 10 in.

(a) WOOD BEAM AND STEEL ANGLES b ⫽ 2 in.

(1) Wood beam

h ⫽ 8 in.

TRANSFORMED SECTION (WOOD)

sallow_w ⫽ 900 psi

NEUTRAL AXIS

(2) Steel Channel Iz ⫽ 17.3 in.

4

As ⫽ 4.75 in.2

d ⫽ 1.98 in. hs ⫽ 6 in.

sallow_s ⫽ 12000 psi TRANSFORMED SECTION (WOOD) n ⫽ 20

h1_b ⫽ 3.015 in. IT_b ⫽ [IT + (bh + 2nAs)

NEUTRAL AXIS From bh a

L

y1 dA +

L

(h1 + bf ⫺ h1_b)2]

ny2 dA ⫽ 0

+ c

h ⫺ h1 b ⫺ 2nAs(h1 ⫺ d) ⫽ 0 2

12

+ bf hf a h1_b ⫺

bf 2

2

b d

IT_b ⫽ 905 in.

MAXIMUM MOMENT BASED UPON THE WOOD (1)

2 bh3 h + bh a ⫺ h1 b d 12 2

M1 ⫽

+ 2n[Iz + As(h1 ⫺ d) ] IT ⫽ 838 in. 2

b3f hf 4

h1 ⫽ 2.137 in. IT ⫽ c

y1 dA + ny2 dA ⫽ 0 L L (bh + 2nAs) (h1 + bf ⫺ h1_b) bf ⫺ bf hf a h1_b ⫺ b ⫽ 0 2

From

4

sallow_wIT_b h + bf ⫺ h1_b

M1 ⫽ 136.0 k # in.

MAXIMUM MOMENT BASED UPON THE STEEL (2) MAXIMUM MOMENT BASED UPON THE WOOD (1) M1 ⫽

sallow_wIT

M2 ⫽

M1 ⫽ 128.6 k # in.

h ⫺ h1

sallow_sIT

WOOD GOVERNS

M2 ⫽ 130.1 k # in.

(hs ⫺ h1)n

M2 ⫽ 136.2 k # in.

From Mmax ⫽

Mmax ⫽ 128.6 k # in.

WOOD GOVERNS

;

ALLOWABLE LOAD ON A 18-FT-LONG SIMPLE BEAM qallowL2 8

qallow ⫽ 264 lb/ft.

;

Mmax ⫽136.0 k # in

;

ALLOWABLE LOAD ON A 18-FT-LONG SIMPLE BEAM

Mmax ⫽ min (M1, M2)

From Mmax ⫽

(hs + bf ⫺ h1_b)n

Mmax ⫽ min (M1, M2)

MAXIMUM MOMENT BASED UPON THE STEEL (2) M2 ⫽

sallow_sIT_b

qallowL2 8

qallow ⫽ 280 lb/ft

;

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Problem 6.3-8 The cross section of a composite beam made of aluminum

y

and steel is shown in the figure. The moduli of elasticity are Ea ⫽ 75 GPa and Es ⫽ 200 GPa. Under the action of a bending moment that produces a maximum stress of 50 MPa in the aluminum, what is the maximum stress ss in the steel?

Aluminum 40 mm Steel z

O 80 mm

30 mm

Solution 6.3-8 Composite beam of aluminum and steel (1) Aluminum:

b ⫽ 30 mm

ha ⫽ 40 mm

All dimensions in millimeters.

sa ⫽ 50 MPa

Use the base of the cross section as a reference line.

Ea ⫽ 75 GPa (2) Steel: b ⫽ 30 mm Es ⫽ 200 GPa

hs ⫽ 80 mm ss ⫽ ?

TRANSFORMED SECTION (ALUMINUM)

h2 ⫽

©yi Ai (40)(80)(80) + (100)(30)(40) ⫽ ©Ai (80)(80) + (30)(40)

⫽ 49.474 mm h1 ⫽ 120 ⫺ h2 ⫽ 70.526 mm MAXIMUM STRESS IN THE ALUMINUM (1) (EQ. 6-15) sa ⫽ s1 ⫽

Mh1 IT

MAXIMUM STRESS IN THE STEEL (2) (EQ. 6-17) ss ⫽ s2 ⫽ Aluminum part is not changed. Es 200 ⫽ ⫽ 2.667 n⫽ Ea 75

Mh2n IT

(49.474)(2.667) ss h2n ⫽ 1.8707 ⫽ ⫽ sa h1 70.526 ss ⫽ 1.8707 (50 MPa) ⫽ 93.5 MPa

;

Width of steel part ⫽ nb ⫽ (2.667)(30 mm) ⫽ 80 mm

Problem 6.3-9 A beam is constructed of two angle sections, each L 5 ⫻ 3 ⫻ 1/2, which reinforce a 2 ⫻ 8 (actual dimensions) wood plank (see the cross section shown in the figure). The modulus of elasticity for the wood is Ew ⫽ 1.2 ⫻ 106 psi and for the steel is Es ⫽ 30 ⫻ 106 psi. Find the allowable bending moment Mallow for the beam if the allowable stress in the wood is sw ⫽ 1100 psi and in the steel is ss ⫽ 12,000 psi. (Note: Disregard the weight of the beam, and see Table E-5a of Appendix E for the dimensions and properties of the angles.)

2 ⫻ 8 wood plank

C

y

z 5 in. 3 in.

Steel angles

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SECTION 6.3 Transformed-Section Method

Solution 6.3-9 b ⫽ 2 in.

(1) Wood beam

h ⫽ 8 in.

sallow_w ⫽ 1100 psi Iz ⫽ 9.43 in.4

(2) Steel Angle

As ⫽ 3.75 in.

2

sallow_s ⫽ 12000 psi

Ew ⫽ 1.2⭈106 psi

IT ⫽ c

d ⫽ 1.74 in. hs ⫽ 5 in.

Es ⫽ 30⭈10 psi 6

b3h b 2 + bha h1 ⫺ b d 12 2

+ 2n[Iz + As(b + d ⫺ h1)2] IT ⫽ 588 in.4 MAXIMUM MOMENT BASED UPON THE WOOD (1)

TRANSFORMED SECTION (WOOD) Es n⫽ Ew

n ⫽ 25

M1 ⫽

sallow_wIT

M1 ⫽ 183.4 k # in.

h1

MAXIMUM MOMENT BASED UPON THE STEEL (2) NEUTRAL AXIS From

1 y1 dA + 1 ny2 dA ⫽ 0

bha h1 ⫺

b b ⫺ 2nAs(b + d ⫺ h1) ⫽ 0 2

h1 ⫽ 3.525 in.

M2 ⫽

sallow_sIT (hs + b ⫺ h1)n

M2 ⫽ 81.1 k # in.

Mmax ⫽ min (M1, M2) Mmax ⫽ 81.1 k-in.

STEEL GOVERNS

;

y

Problem 6.3-10 The cross section of a bimetallic strip is shown in the figure. Assuming that the moduli of elasticity for metals A and B are EA ⫽ 168 GPa and EB ⫽ 90 GPa, respectively, determine the smaller of the two section moduli for the beam. (Recall that section modulus is equal to bending moment divided by maximum bending stress.) In which material does the maximum stress occur?

z

A

O

B

3 mm 3 mm

10 mm

Solution 6.3-10 Bimetallic strip Metal A: b ⫽ 10 mm

hA ⫽ 3 mm

Metal B does not change.

EA ⫽ 168 GPa Metal B: b ⫽ 10 mm hB ⫽ 3 mm TRANSFORMED SECTION

TRANSFORMED SECTION (METAL B)

n⫽ EB ⫽ 90 GPa

EA 168 ⫽ ⫽ 1.8667 EB 90

Width of metal A ⫽ nb ⫽ (1.8667)(10 mm) ⫽ 18.667 mm All dimensions in millimeters. Use the base of the cross section as a reference line. h2 ⫽

©yi Ai (1.5)(10)(13) + (4.5)(18.667)(3) ⫽ ©Ai (10)(3) + (18.667)(3)

⫽ 3.4535 mm h1 ⫽ 6 ⫺ h2 ⫽ 2.5465 mm

517

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IT ⫽

1 (10)(3)3 + (10)(3)(h2 ⫺ 1.5)2 12 +

1 (18.667)(3)3 + (18.667)(3)(h1 ⫺ 1.5)2 12

MAXIMUM STRESS IN MATERIAL A (EQ. 6-17) sA ⫽ s2 ⫽

Mh1n IT

SA ⫽

IT M ⫽ sA h1n

⫽ 50.6 mm3

⫽ 240.31mm

4

SMALLER SECTION MODULUS MAXIMUM STRESS IN MATERIAL B (EQ. 6-15) Mh2 sB ⫽ s1 ⫽ IT

IT M SB ⫽ ⫽ ⫽ 69.6 mm3 sB h2

SA ⫽ 50.6 mm3

;

‹ Maximum stress occurs in metal A.

;

y

Problem 6.3-11 A W 12 ⫻ 50 steel wide-flange beam and a segment of a 4-inch thick concrete slab (see figure) jointly resist a positive bending moment of 95 k-ft. The beam and slab are joined by shear connectors that are welded to the steel beam. (These connectors resist the horizontal shear at the contact surface.) The moduli of elasticity of the steel and the concrete are in the ratio 12 to 1. Determine the maximum stresses ss and sc in the steel and concrete, respectively. (Note: See Table E-1a of Appendix E for the dimensions and properties of the steel beam.)

30 in. 4 in. z

O

W 12 ⫻ 50

Solution 6.3-11 Steel beam and concrete slab (1) Concrete: b ⫽ 30 in.

t ⫽ 4 in.

(2) Wide-flange beam: W 12 ⫻ 50 d ⫽ 12.19 in.

I ⫽ 394 in.4

A ⫽ 14.7 in.2

M ⫽ 95 k-ft ⫽ 1140 k-in.

TRANSFORMED SECTION (CONCRETE)

All dimensions in inches. Use the base of the cross section as a reference line. nI ⫽ 4728 in.4 h2 ⫽

nA ⫽ 176.4 in.2

©yiAi (12.19/2)(176.4) + (14.19)(30)(4) ⫽ ©Ai 176.4 + (30)(4)

⫽ 9.372 in. h1 ⫽ 16.19 ⫺ h2 ⫽ 6.818 in. IT ⫽

1 (30)(4)3 + (30)(4)(h1 ⫺ 2)2 + 4728 12 + (176.4)(h2 ⫺ 12.19/2)2 ⫽ 9568 in.4

MAXIMUM STRESS IN THE CONCRETE (1) (EQ. 6-15) sc ⫽ s1 ⫽ No change in dimensions of the concrete. n⫽

Es E2 ⫽ ⫽ 12 Ec E1

Width of steel beam is increased by the factor n to transform to concrete.

Mh1 ⫽ 812 psi (Compression) IT

;

MAXIMUM STRESS IN THE STEEL (2) (EQ. 6-17) ss ⫽ s2 ⫽

Mh2n ⫽ 13,400 psi (Tension) IT

;

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SECTION 6.3 Transformed-Section Method

Problem 6.3-12 A wood beam reinforced by an aluminum channel section is

150 mm

shown in the figure. The beam has a cross section of dimensions 150 mm by 250 mm, and the channel has a uniform thickness of 6 mm. If the allowable stresses in the wood and aluminum are 8.0 MPa and 38 MPa, respectively, and if their moduli of elasticity are in the ratio 1 to 6, what is the maximum allowable bending moment for the beam?

y 216 mm 250 mm z

O

40 mm 6 mm 162 mm

Solution 6.3-12

Wood beam and aluminum channel

(1) Wood beam: bw ⫽ 150 mm

hw ⫽ 250 mm

(sw)allow ⫽ 8.0 MPa (2) Aluminum channel: t ⫽ 6 mm

ba ⫽ 162 mm

ha ⫽ 40 mm (sa)allow ⫽ 38 MPa

Use the base of the cross section as a reference line. h2 ⫽

©yiAi ©Ai

Area A1: y1 ⫽ 3

A1 ⫽ (972)(6) ⫽ 5832

y1A1 ⫽ 17,496 mm3 Area A2: y2 ⫽ 23

TRANSFORMED SECTION (WOOD)

A2 ⫽ (36)(34) ⫽ 1224

y2A2 ⫽ 28,152 mm3 Area A3: y3 ⫽ 131

A3 ⫽ (150)(250) ⫽ 37,500

y3A3 ⫽ 4,912,500 mm3 h2 ⫽

y1A1 + 2y2A2 + y3A3 4,986,300 mm3 ⫽ A1 + 2A2 + A3 45,780 mm2

⫽ 108.92 mm h1 ⫽ 256 ⫺ h2 ⫽ 147.08 mm MOMENT OF INERTIA Area A1: I1 ⫽ Wood beam is not changed. Ea n⫽ ⫽6 Ew Width of aluminum channel is increased. nb ⫽ (6)(162 mm) ⫽ 972 mm nt ⫽ (6)(6 mm) ⫽ 36 mm All dimensions in millimeters.

1 (972)(6)3 + (972)(6)(h2 ⫺ 3)2 12

⫽ 65,445,000 mm4 Area A2: I2 ⫽

1 (36)(34)3 12 + (36)(34)(h2 ⫺ 6 ⫺ 17)2

⫽ 9,153,500 mm4

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CHAPTER 6 Stresses in Beams (Advanced Topics)

Area A3: I3 ⫽

MAXIMUM MOMENT BASED UPON ALUMINUM (2) (EQ. 6-17)

1 (150)(250)3 12 + (150)(250)(h1 ⫺ 125)

2

⫽ 213,597,000 mm4

sa ⫽ s2 ⫽

Mh2n IT

WOOD GOVERNS.

M2 ⫽

(sa)allowIT ⫽ 17.3 kN # m h2n

Mallow ⫽ 16.2 kN # m

;

IT ⫽ I1 + 2I2 + I3 ⫽ 297.35 * 106 mm4 MAXIMUM MOMENT BASED UPON THE WOOD (1) (EQ. 6-15) sw ⫽ s1 ⫽

Mh1 IT

M1 ⫽

(sw)allowIT ⫽ 16.2 kN # m h1

Beams with Inclined Loads y

When solving the problems for Section 6.4, be sure to draw a sketch of he cross section showing the orientation of the neutral axis and the locations of the points where the stresses are being found.

Problem 6.4-1 A beam of rectangular cross section supports an z

inclined load P having its line of action along a diagonal of the cross section (see figure). Show that the neutral axis lies along the other diagonal.

h

C

b

P

Solution 6.4-1 Location of neutral axis Iy ⫽

hb3 12

Iz

h2

Iy

⫽

b2

See Figure 6-15b. b ⫽ angle between the z axis and the neutral axis nn u ⫽ angle between the y axis and the load P u ⫽ a ⫹ 180° tan u ⫽ tan (a ⫹ 180°) ⫽ tan a Load P acts along a diagonal. b b/2 ⫽ tan a ⫽ h/2 h Iz ⫽

3

bh 12

(Eq. 6-23):

tan b ⫽

Iz

tan u ⫽

Iy

⫽ a

h2 b2

b2

tan u

b h ba b ⫽ h b

‹ The neutral axis lies along the other diagonal. QED

h2

;

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SECTION 6.4 Beams with Inclined Loads

Problem 6.4-2 A wood beam of rectangular cross section

y

(see figure) is simply supported on a span of length L. The longitudinal axis of the beam is horizontal, and the cross section is tilted at an angle a. The load on the beam is a vertical uniform load of intensity q acting through the centroid C. Determine the orientation of the neutral axis and calculate the maximum tensile stress smax if b  80 mm, h  140 mm, L  1.75 m, a  22.5°, and q  7.5 kN/m.

b h C q

z

a

Probs. 6.4-2 and 6.4-3

Solution 6.4-2 q  7.5 kN/m

L  1.75 m h  140 mm

b  80 mm

a  22.5 deg

Iz 

bh3 12

Iz  18.293 * 106 mm4

NEUTRAL AXIS nn

BENDING MOMENTS 2

My 

qsin(a)L 8

My  1099 N # m

Iz b  a tan a tan(a)b Iy

Mz 

qcos(a)L2 8

Mz  2653 N # m

MAXIMUM TENSILE STRESS (AT POINT A)

MOMENT OF INERTIA hb3 Iy  12

smax 

Iy  5.973 * 10 mm 6

b My a b 2 Iy

4



b  51.8°

Mz a

smax  17.5 MPa

;

h b 2 Iz

;

Problem 6.4-3 Solve the preceding problem for the following data: b  6 in., h  10 in., L  12.0 ft, tan a  1/3, and q  325 lb/ft.

Solution 6.4-3 L  12 ft

q  325 lb/ft

b  6 in.

1 a  a tan a b 3

h  10 in.

BENDING MOMENTS

Mz 

qsin(a)L 8

My  22199 lb-in.

Mz  66598 lb-in.

MOMENT OF INERTIA Iy 

hb3 12

Iy  180 in.4

Iz 

bh3 12

Iz  500 in.4

2

My 

qcos(a)L2 8

521

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MAXIMUM TENSILE STRESS (AT POINT A)

NEUTRAL AXIS nn Iz b  a tan a tan(a)b Iy

b  42.8°

; smax 

b My a b 2 Iy

h b 2

Mz a



Iz

smax  1036 psi

;

y

Problem 6.4-4 A simply supported wide-flange beam of span length L carries a vertical concentrated load P acting through the centroid C at the midpoint of the span (see figure). The beam is attached to supports inclined at an angle a to the horizontal. Determine the orientation of the neutral axis and calculate the maximum stresses at the outside corners of the cross section (points A, B, D, and E ) due to the load P. Data for the beam are as follows: W 250  44.8 section, L  3.5 m, P  18 kN, and a  26.57°. (Note: See Table E-1b of Appendix E for the dimensions and properties of the beam.)

P

n

E D b

C B

z A

n

a

Probs. 6.4-4 and 6.4-5

Solution 6.4-4 L  3.5 m

P  18 kN

W 250  44.8

Wide-flange beam: Iy  6.95  106 mm4 d  267 mm

a  26.57 deg Iz  70.8  106 mm4

b  148 mm

BENDING STRESSES sx(z,y)  POINT A:

Myz Iy



zA 

Mzy Iz b 2

yA 

sA  102 MPa

BENDING MOMENTS My 

Psin(a)L 4

My  7045 N # m

Mz 

Pcos(a)L 4

Mz  14087 N # m

POINT B:

NEUTRAL AXIS NN Iz b  a tan a tan(a)b Iy

b  78.9°

;

d 2

sA  sx(zA, yA)

;

b d yB  2 2 sB   48 MPa sB  sx(zB, yB) zB  

POINT D:

sD   sB

sD  48 MPa

POINT E:

sE   sA

sE   102 MPa

;

; ;

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SECTION 6.4 Beams with Inclined Loads

Problem 6.4-5 Solve the preceding problem using the following data: W 8  21 section, L  84 in., P  4.5 k, and a  22.5°.

Solution 6.4-5 L  84 in.

P  4.5 k

BENDING STRESSES

a  22.5°

sx(z, y) 

WIDE-FLANGE BEAM: W 8  21

Iy  9.77 in.4

d  8.28 in.

Mz 

Iy

Iz  75.3 in.4

b  5.270 in.

POINT A:



zA 

Mz y Iz

b 2

yA 

d 2

sA  sx(zA, yA)

BENDING MOMENTS My 

My z

Psin(a)L 4

My  36164 lb-in.

Pcos(a)L 4

Mz  87307 lb-in.

sA  14554 psi POINT B:

zB  

b 2

;

yB 

sB  sx(zB, yB) sB   4953 psi

NEUTRAL AXIS nn b  a tan a tan(a)b Iy Iz

b  72.6°

d 2

;

;

POINT D:

sD   sB

sD  4953 psi

POINT E:

sE   sA

sE   14554 psi

Problem 6.4-6 A wood cantilever beam of rectangular cross section and length

; ;

y

L supports an inclined load P at its free end (see figure). Determine the orientation of the neutral axis and calculate the maximum tensile stress smax due to the load Pt Data for the beam are as follows: b  80 mm, h  140 mm, L  2.0 m, P  575 N, and a  30°.

h z

C

a P

b

Probs. 6.4-6 and 6.4-7

Solution 6.4-6 L  2.0 m

P  575 N

h  140 mm

b  80 mm

a  30°

BENDING MOMENTS My  Pcos(a)L Mz  Psin(a)L

My  996 N # m Mz  575 N # m

MOMENT OF INERTIA Iy 

hb3 12

Iy  5.973 * 106 mm4

Iz 

bh3 12

Iz  18.293 * 106 mm4

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MAXIMUM TENSILE STRESS (AT POINT A)

NEUTRAL AXIS nn Iz b  a tan a tan(a + 90°)b Iy b   79.3°

smax 

;

b My a b 2 Iy



smax  8.87 MPa

h Mz a b 2 Iz ;

Problem 6.4-7 Solve the preceding problem for a cantilever beam with data as follows: b  4 in., h  9 in., L  10.0 ft, P  325 lb, and a  45°.

Solution 6.4-7 L  10.0 ft h  9 in.

P  325 lb

b  4 in.

Iz b  a tan a tan(a + 90°)b Iy

a  45°

b   78.8 deg

BENDING MOMENTS My  Pcos(a)L Mz  Psin(a)L

My  27577 lb # in. Mz  27577 lb # in.

MOMENT OF INERTIA hb3 Iy  12 Iz 

3

bh 12

NEUTRAL AXIS nn

MAXIMUM TENSILE STRESS (AT POINT A)

smax 

Iy  48.000 in.

;

4

b My a b 2 Iy



smax  1660 psi

h Mz a b 2 Iz ;

Iz  243.000 in.4

Problem 6.4-8 A steel beam of I-section (see figure) is simply supported at the ends. Two equal and oppositely directed bending moments M0 act at the ends of the beam, so that the beam is in pure bending. The moments act in plane mm, which is oriented at an angle a to the xy plane. Determine the orientation of the neutral axis and calculate the maximum tensile stress smax due to the moments M0. Data for the beam are as follows: S 200  27.4 section, M0  4 kN # m., and a  24°. (Note: See Table E-2b of Appendix E for the dimensions and properties of the beam.)

m

y

C

z M0

a

m

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SECTION 6.4 Beams with Inclined Loads

Solution 6.4-8 Mo  4.0 kN # m S 200  27.4

NEUTRAL AXIS nn

a  24° Iy  1.54  106 mm4

Iz  23.9  106 mm4

d  203 mm

b  102 mm

My  1627 N # m

Mz  M0 cos(a)

b   81.8°

;

MAXIMUM TENSILE STRESS (AT POINT A)

BENDING MOMENTS My  M0 sin(a)

Iz b  a tan a tan(a)b Iy

Mz  3654 N # m

smax 

b My a b 2 Iy



d Mz a b 2 Iz

smax  69.4 MPa

;

Problem 6.4-9 A cantilever beam of wide-flange cross section and length

y

L supports an inclined load P at its free end (see figure). Determine the orientation of the neutral axis and calculate the maximum tensile stress smax due to the load P. Data for the beam are as follows: W 10  45 section, L  8.0 ft, P  1.5 k, and a  55°. (Note: See Table E-1a of Appendix E for the dimensions and properties of the beam.)

P a z

C

Probs. 6.4-9 and 6.4-10

Solution 6.4-9 Cantilever beam with inclined load BENDING MOMENTS My  (P cos a)L  82,595 lb-in. Mz  (P sin a)L  117,960 lb-in. NEUTRAL AXIS nn (EQ. 6-23) u  90°  a  35° tan b 

Iz Iy

tan u 

(see Fig. 6-15) 248 tan 35°  3.2519 53.4

b  72.91° P  1.5 k  1500 lb L  8.0 ft  96 in. a  55°

d  10.10 in.

MAXIMUM TENSILE STRESS (POINT A) (EQ. 6-18) zA  b/2  4.01 in. yA   d/2   5.05 in.

W 10  45 Iy  53.4 in.4

;

Iz  248 in.4 b  8.02 in.

smax  sA 

My zA Iy



Mz yA Iz

 8600 psi

;

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CHAPTER 6 Stresses in Beams (Advanced Topics)

Problem 6.4-10 Solve the preceding problem using the following data: W 310  129 section, L  1.8 m, P  9.5 kN, and a  60°. (Note: See Table E-1b of Appendix E for the dimensions and properties of the beam.)

Solution 6.4-10 P  9.5 kN

L  1.8 m

W 310  129

MAXIMUM TENSILE STRESS (AT POINT A)

a  60°

Iy  100  106 mm4

Iz  308  106 mm4

d  318 mm

b  307 mm

smax 

b My a b 2 Iy



d Mz a  b 2 Iz

smax  20.8 MPa

BENDING MOMENTS My  Pcos(a)L

My  8550 N # m

Mz  Psin(a)L

Mz  14809 N # m

;

NEUTRAL AXIS nn Iz b  a tan a tan(90°  a) b Iy b  60.6°

b  60.6°

;

Problem 6.4-11 A cantilever beam of W 12  14 section and

y

length L  9 ft supports a slightly inclined load P  500 lb at the free end (see figure). (a) Plot a graph of the stress sA at point A as a function of the angle of inclination a. (b) Plot a graph of the angle b, which locates the neutral axis nn, as a function of the angle a. (When plotting the graphs, let a vary from 0 to 10°.) (Note: See Table E-1a of Appendix E for the dimensions and properties of the beam.)

A

n

b C

z

n P

a

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SECTION 6.4 Beams with Inclined Loads

Solution 6.4-11

Cantilever beam with inclined load (b) NEUTRAL AXIS nn (EQ. 6-23) u  180° + a tan b  

Iz Iy

(see Fig. 6-15)

tan u 

Iz Iy

tan(180° + a)

88.6 tan(180° + a)  37.54 tan a 2.36

b  arctan(37.54 tan a)

P  500 lb

L  9 ft  108 in.

Iy  2.36 in.

Iz  88.6 in.

d  11.91 in.

b  3.970 in.

4

;

W 12  14

4

BENDING MOMENTS My  (P sin a)L  54,000 sin a Mz  (P cos a)L  54,000 cos a (a) STRESS AT POINT A (EQ. 6-18) zA   b/2   1.985 in. yA  d/2  5.955 in. sA 

MyzA Iy



Mz yA Iz

 45,420 sin a

+ 3629 cos a (psi)

;

y

Problem 6.4-12 A cantilever beam built up from two channel shapes, each

C 200  17.1, and of length L supports an inclined load P at its free end (see figure). Determine the orientation of the neutral axis and calculate the maximum tensile stress smax due to the load P. Data for the beam are as follows: L  4.5 m, P  500 N, and a  30°. z

C a P

C 200  17.1

527

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CHAPTER 6 Stresses in Beams (Advanced Topics)

Solution 6.4-12 L  4.5 m

P  500 N

Iz b  a tan a tan (90 °  a)b Iy

Double C 200  17.1

BUILT UP BEAM:

Icy  0.545  106 mm4

Icz  13.5  106 mm4

c  14.5 mm

bc  57.4 mm

[

]

Iy  2 Icy  Ac(bc  c)2 Iz  2Icz

NEUTRAL AXIS nn

a  30°

Iy  9.08  106 mm4

Iz  27.0  106 mm4

d  203 mm

b  2bc

b  79.0°

;

Ac  2170 mm2 MAXIMUM TEMSILE STRESS sx(z, y) 

b  114.8 mm

POINT A:

BENDING MOMENTS My  Pcos(a)L

My  1949 N # m

Mz  Psin(a)L

Mz  1125 N # m

Myz Iy



Mzy

zA  

Iz b 2

yA 

sA  sx(zA, yA)

d 2 sA  16.6 MPa

;

Problem 6.4-13 A built-up steel beam of I-section with channels attached to the flanges (see figure part a) is simply supported at the ends. Two equal and oppositely directed bending moments M0 act at the ends of the beam, so that the beam is in pure bending. The moments act in plane mm, which is oriented at an angle a to the xy plane. (a) Determine the orientation of the neutral axis and calculate the maximum tensile stress smax due to the moments M0. (b) Repeat part a if the channels are now with their flanges pointing away from the beam flange, as shown in figure part b. Data for the beam are as follows: S 6  12.5 section with C 4  5.4 sections attached to the flanges, M0  45 k-in., and a  40°. (Note: See Tables E-2a and E-3a of Appendix E for the dimensions and properties of the S and C shapes.)

m

y

m

C 4  5.4

y C 4  5.4

z

C M0

S 6  12.5

a

C 4  5.4

C

z

S 6  12.5

a

M0

C 4  5.4

m (a)

m

(b)

Solution 6.4-13 Mo  45 k # in.

a  40°

S 6  12.5:

Isy  1.80 in.4

Isz  22.0 in.4

ds  6.0 in.

bs  3.33 in.

As  3.66 in.2

Iz  Isz + 2cIcy + Ac a

C 4  5.4:

Icy  0.312 in.4

Icz  3.85 in.4

Iz  46.1 in.4

dc  4.0 in.

bc  1.58 in.

Ac  1.58 in.2

d  ds + 2twc

twc  0.184 in.

c  0.457 in.

(a) BUILT UP SECTION:

b  dc

Iy  Isy  2Icz Iy  9.50 in.4 2 ds + twc  cb d 2

d  6.368 in.

b  4.0 in.

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SECTION 6.5 Bending of Unsymmetric Beams

NEUTRAL AXIS nn

BENDING MOMENTS My  Mosin(a)

My  28.9 k # in.

Mz  Mocos(a)

Iz b  a tan a tan(a)b Iy

Mz  34.5 k # in.

b   79.4°

NEUTRAL AXIS nn

MAXIMUM TENSILE STRESS

Iz b  a tan a tan( a) b Iy b   76.2°

sx(z, y) 

;

POINT A:

MAXIMUM TENSILE STRESS sx(z, y)  POINT A:

My z Iy



Mz y

zA  

s A  sx(zA, yA)

;

My z Iy



Mz y Iz

zA  

b 2

sA  sx(zA, yA)

yA  

sA  8704 psi

Iz b 2

yA  

d 2

sA  8469 psi

(b) BUILT UP SECTION: Iy  Isy  2Icz Iz  Isz + 2 cIcy + Ac a

; Iy  9.50 in.4

2 ds + cb d 2

Iz  60.4 in.4 d  ds + 2bc b  dc

d  9.160 in.

b  4.000 in.

Bending of Unsymmetric Beams y

When solving the problems for Section 6.5, be sure to draw a sketch of the cross section showing the orientation of the neutral axis and the locations of the points where the stresses are being found.

Problem 6.5-1 A beam of channel section is subjected to a bending moment M having its vector at an angle u to the z axis (see figure). Determine the orientation of the neutral axis and calculate the maximum tensile stress st and maximum compressive stress sc in the beam. Use the following data: C 8  11.5 section, M  20 k-in., tan u  1/3. (Note: See Table E-3a of Appendix E for the dimensions and properties of the channel section.)

M z

d 2

u

Probs. 6.5-1 and 6.5-2

C

;

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CHAPTER 6 Stresses in Beams (Advanced Topics)

Solution 6.5-1

Channel section NEUTRAL AXIS nn (EQ. 6-40) tan b 

Iz Iy

tan u 

32.6 (1/3)  8.2323 1.32

b  83.07°

;

MAXIMUM TENSILE STRESS (POINT A) (EQ. 6-38) zA  c  0.571 in. st  sA 

yA   d/2   4.00 in.

(M sin u)zA (M cos u)yA  Iy Iz

 5060 psi M  20 k-in.

tan u  1/3

u  18.435°

C 8  11.5 c  0.571 in.

Iy  1.32 in.4

d  8.00 in.

Iz  32.6 in.4

b  2.260 in.

;

MAXIMUM COMPRESSIVE STRESS (POINT B) (EQ. 6-38) zB   (b  c)   (2.260  0.571)   1.689 in. yz  d/2  4.00 in. sc  sB 

(M sin u)zB (M cos u)yB  Iy Iz

  10,420 psi

;

Problem 6.5-2 A beam of channel section is subjected to a bending moment M having its vector at an angle u to the z axis (see figure). Determine the orientation of the neutral axis and calculate the maximum tensile stress st and maximum compressive stress sc in the beam. Use a C 200  20.5 channel section with M  0.75 kN # m and u  20°.

Solution 6.5-2 M  0.75 kN # m C 200  20.5

MAXIMUM TENSILE STRESS (AT POINT A)

u  20° Iy  0.633 # 106 mm4

Iz  15.0  106 mm4

d  203 mm

b  59.4 mm

c  14.1 mm

Iy



d Mz a b 2

smax  10.5 MPa

BENDING MOMENTS My  Msin(u)

My  257 N # m

Mz  Mcos(u)

Mz  705 N # m

NEUTRAL AXIS

smax 

My(c)

smax 

My(b + c) Iy

smax  23.1 MPa b  83.4°

;

MAXIMUM TENSILE STRESS (AT POINT B)

nn

Iz b  atan a tan(u)b Iy

Iz

;



d Mz a b 2 Iz ;

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SECTION 6.5 Bending of Unsymmetric Beams

Problem 6.5-3 An angle section with equal legs is subjected to a

2

bending moment M having its vector directed along the 1-1 axis, as shown in the figure. Determine the orientation of the neutral axis and calculate the maximum tensile stress st and maximum compressive stress sc if the angle is an L 6  6  3/4 section and M  20 k-in. (Note: See Table E-4a of Appendix E for the dimensions and properties of the angle section.)

M

1

Probs. 6.5-3 and 6.5-4

Solution 6.5-3

C

1

2

Angle section with equal legs NEUTRAL AXIS nn (EQ. 6-40) tan b 

Iz Iy

tan u 

44.85 tan 45°  3.8831 11.55

b  75.56°

;

MAXIMUM TENSILE STRESS (POINT A) (EQ. 6-38) zA  c12  2.517 in. st  sA 

(M sin u)zA (M cos u)yA  Iy Iz

 3080 psi M  20 k-in.

L 6  6  3/4 in.

h  b  6 in.

c  1.78 in.

A  8.44 in.2

Iy  Ar2min  11.55 in.4 Iz  I1  I2  Iy  44.85 in.4

;

MAXIMUM COMPRESSIVE STRESS (POINT B) (EQ. 6-38) zB  c12  h/12   1.725 in.

I1  I2  28.2 in.4 u  45°

yA  0

rmin  1.17 in.

yB  h/12  4.243 in. sc  sB 

(M sin u)zB (M cos u)yB  Iy Iz

  3450 psi

;

Problem 6.5-4 An angle section with equal legs is subjected to a bending moment M having its vector directed along the 1–1 axis, as shown in the figure. Determine the orientation of the neutral axis and calculate the maximum tensile stress st and maximum compressive stress sc if the section is an L 152  152  12.7 section and M  2.5 kN # m. (Note: See Table E-4b of Appendix E for the dimensions and properties of the angle section.)

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CHAPTER 6 Stresses in Beams (Advanced Topics)

Solution 6.5-4 M  2.5 kN # m L 152  152  12.7 rmin  30.0 mm Iy  Armin2

MAXIMUM TENSILE STRESS (AT POINT A)

u  45° I1  8.28  106 mm4

I2  I1

A  3720 mm2

h  152 mm

Iz  13.212  106 mm4 bh

Mz  Mcos(u) NEUTRAL AXIS

Iy



smax  My  1768 N # m Mz  1768 N # m

Mz(0) Iz ;

MAXIMUM TENSILE STRESS (AT POINT B)

c  42.4 mm4

BENDING MOMENTS My  Msin(u)

My(c12)

smax  31.7 MPa

Iy  3.348  106 mm4

Iz  I1  I2  Iy

smax 

My a c12 

h b 12

Iy

smax   39.5 MPa



Mz a

h b 12 Iz

;

nn

Iz b  atan a tan(u)b Iy

b  75.8°

;

Problem 6.5-5 A beam made up of two unequal leg angles is subjected to a bending moment M having its vector at an angle u to the z axis (see figure part a). (a) For the position shown in the figure, determine the orientation of the neutral axis and calculate the maximum tensile stress st and maximum compressive stress sc in the beam. Assume that u  30° and M  30 k-in. (b) The two angles are now inverted and attached back-to-back to form a lintel beam which supports two courses of brick façade (see figure part b). Find the new orientation of the neutral axis and calculate the maximum tensile stress st and maximum compressive stress sc in the beam using u  30° and M  30 k-in. y 1 1 L53—  — 2 2

1 1 L53—  — 2 2

y Lintel beam supporting brick facade

z u

C

M

z u

3 — in. 4 (a)

M

C (b)

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SECTION 6.5 Bending of Unsymmetric Beams

533

Solution 6.5-5 M  30 k # in.

u  30°

IL1  10.0 in.4 L5 * 3q1/2 * 1/2 d  1.65 in. c  0.901 in. IL2  4.02 in.4 hL1  5 in. AL  4 in.2

hL2  3.5 in. 3 gap  in. 4

Iz  20.000 in.4

2 gap + cb d 2

Iy  2 cIL2 + AL a Iy  21.065 in.

4

h  hL1

h  5.000 in.

b  gap + 2hL2

b  7.750 in. h1  d

h1  1.650 in.

BENDING MOMENTS My  Msin (u) Mz  Mcos (u)

zA  

st  sx(zA, yA)

gap + t 2

yA   h + h1

st  4263 psi

;

MAXIMUM COMPRESSIVE STRESS 1 t  in. 2

Iz  2IL1

(a) BUILT UP SECTION:

POINT A:

My  1.250 k # ft Mz  2.165 k # ft

POINT B:

zB 

b 2

yB  h1

sc  sx(zB, yB)

sc   4903 psi Iz  2IL1

(b) BUILT UP SECTION: Iy  2(IL2  ALc2) h  hL1

Iz  20.000 in.4

Iy  14.534 in.4

h  5.000 in.

b  7.000 in.

;

b  2hL2

h1  h  d

h1  3.350 in.

NEUTRAL AXIS nn Iz b  a tan a tan(u)b Iy b   38.5 deg

;

MAXIMUM TENSILE STRESS NEUTRAL AXIS nn b  atan a tan( u)b Iy

sx(z, y) 

b   28.7°

POINT A:

Iz

;

MAXIMUM TENSILE STRESS sx(z, y) 

Myz Iy



Mzy Iz

Myz Iy



zA  

st  sx(zA, yA)

Iz b 2

yA   h + h1

st  5756 psi

;

MAXIMUM COMPRESSIVE STRESS POINT B:

zB  t

sc  sx(zB,yB)

yB  h1 sc   4868 psi

y1

Problem 6.5-6 The Z-section of Example 12-7 is subjected to

M  5 kN # m, as shown. Determine the orientation of the neutral axis and calculate the maximum tensile stress st and maximum compressive stress sc in the beam. Use the following numerical data: height h  200 mm, width b  90 mm, constant thickness t  15 mm, and up  19.2°. Use I1  32.6  106 mm4 and I2  2.4  106 mm4 from Example 12-7.

Mzy

;

y

b h — 2 M

t up

x

C h — 2

x1

t t b

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CHAPTER 6 Stresses in Beams (Advanced Topics)

Solution 6.5-6 M  5 kN # m

ypA   91.97 mm

u  19.2°

st 

Z-SECTION Izp  I1

Izp  32.6  10 mm

Iyp  I2

Iyp  2.4  10 mm

6

6

h  200 mm

b  90 mm

4

Myp(zpA) Iyp



st  40.7 MPa

Mzp(ypA) Izp ;

4

t  15 mm

BENDING MOMENTS

MAXIMUM COMPRESSIVE STRESS (AT POINT B) t h zpB  a  b cos(u)  a bsin(u) 2 2

Myp  Msin(u)

Myp  1644 N # m

zpB   39.97 mm

Mzp  Mcos(u)

Mzp  4722 N # m

h t ypB  a  bsin(u) + a bcos(u) 2 2

NEUTRAL AXIS nn b  atan a

Izp Iyp

tan(u)b

ypB  91.97 mm b  78.1°

;

Myp(zpB) Iyp



Mzp(ypB) Izp

sc   40.7 MPa

MAXIMUM TENSILE STRESS (AT POINT A) h t zpA  a bcos(u)  a bsin(u) 2 2

sc 

;

zpA  39.97 mm

h t bcos(u) ypA  a bsin(u) + a 2 2

Problem 6.5-7 The cross section of a steel beam is

constructed of a W 18  71 wide-flange section with a 6 in  1/2 in. cover plate welded to the top flange and a C 10  30 channel section welded to the bottom flange. This beam is subjected to a bending moment M having its vector at an angle u to the z axis (see figure). Determine the orientation of the neutral axis and calculate the maximum tensile stress st and maximum compressive stress sc in the beam. Assume that u  30° and M  75 k-in. (Note: The cross sectional properties of this beam were computed in Examples 12-2 and 12-5.)

Plate 1 6 in.  — in. 2

M

y C1

W 18  71

u z

y1 C2

c

C y3

C 10  30 C3

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SECTION 6.5 Bending of Unsymmetric Beams

535

Solution 6.5-7 M  75 k # in. PLATE: Iplate 

NEUTRAL AXIS nn

u  30° 1 in. 2

bp  bph3p

hp  6 in.

Iplate  9.00 in.4

12

hw  18.47 in.

W SECTION:

bw  7.635 in.

Iz b  atan a tan(u)b Iy

b  82.3°

;

MAXIMUM TENSILE STRESS sx(z, y) 

Myz Iy



Iwy  60.3 in.

Mzy Iz

4

POINT A: hc  10.0 in.

C SECTION:

Icz  103 in.

bc  3.033 in.

zA 

hc 2

yA  

st  sx(zA, yA)

4

st  1397 psi

BUILT-UP SECTION:

MAXIMUM COMPRESSIVE STRESS

cbar  1.80 in.

POINT B:

Iz  2200 in.4

Iy  Iwy  Icz  Iplate

zB  

bw 2

yB 

sc  sx(zB, yB)

Iy  172.3 in.

4

hw  bc + cbar 2 ;

hw + cbar 2

sc   1157 psi

;

BENDING MOMENTS My  Msin(u)

My  3.125 k # ft

Mz  Mcos(u)

Mz  5.413 k # ft

Problem 6.5-8 The cross section of a steel beam is shown in the figure. This beam is subjected to a bending moment M having its vector at an angle u to the z axis. Determine the orientation of the neutral axis and calculate the maximum tensile stress st and maximum compressive stress sc in the beam. Assume that u  22.5° and M  4.5 kN # m. Use cross sectional properties Ix1  93.14  106 mm4, Iy1  152.7  106 mm4, and up  27.3°.

y1

y

180mm

180 mm

x1 30 mm

105 mm

15 mm 30 mm z

up

C u

M

y = 52.5 mm

90 mm

30 mm

90 mm

O 30 mm 120 mm

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CHAPTER 6 Stresses in Beams (Advanced Topics)

Solution 6.5-8 M  4.5 kN # m

u  22.5°

MAXIMUM TENSILE STRESS (AT POINT A) zA  

BUILT-UP SECTION:

t 2

yA  

h  ybar 2

b  120 mm

t  30 mm

zpA  (zA) cos (uP)(yA) sin (uP)

h  180 mm

b1  360 mm

ypA  (zA) sin (uP)(yA) cos (uP)

t + h b 2

ypA   133.51 mm

ybar 

tb1 a

tb1 + [2tb + (h  2t)t]

ybar  52.5 mm

Iyp  152.7 * 106 mm4

BENDING MOMENTS Myp  Msin(uP  u)

Myp  377 N # m

Mzp  Mcos(uP  u)

Mzp  4484 N # m

NEUTRAL AXIS nn

b  2.93°

Izp Iyp

Iyp



Mzp(ypA) Izp ;

MAXIMUM COMPRESSIVE STRESS (AT POINT B)

Iyp  Iy1

Izp  93.14 * 106 mm4

b  atan a

Myp(zpA)

st  6.56 MPa

uP  27.3 deg Izp  Ix1

st 

zpA  52.03 mm

tan(uP  u) b

zB 

b1 2

yB 

h + t  ybar 2

zpB  (zB) cos (uP)(yB) sin (uP)

zpB  128.99 mm

ypB  (zB) sin (uP)(yB) cos (uP)

ypB  142.5 mm

sc 

Myp(zpB) Iyp



sc   6.54 MPa

Mzp(ypB) Izp ;

;

Problem 6.5-9 A beam of semicircular cross section of radius r is subjected to a bending moment M having its vector at an angle u to the z axis (see figure). Derive formulas for the maximum tensile s, and the maximum compressive stress sc in the beam for u  0, 45°, and 90°. (Note: Express the results in the from  M/r3, where  is a numerical value.)

y

M

z

u O

C

r

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SECTION 6.5 Bending of Unsymmetric Beams

Soultion 6.5.-9

537

Semicircle sc  sD 

M(r  c) Iy

  5.244

M

;

r3

FOR u  45°: Eq. (6q40): tan b  tan b 

9p2 9p2  64

Iz Iy

tan u

(1)  3.577897

b  74.3847° 90°  b  15.6153° r  radius c

4r  0.42441r 3p

Iy 

(9p2  64) 4 r 72p

MAXIMUM TENSILE STRESS for u  45° occurs at point A. z A  c  0.42441r From (Eq. 6-38): st  sA 

 0.109757r4 Iz 

pr4 8

(M sin u)zA (M cos u)yA  Iy Iz

 4.535

st  maximum tensile stress

 2.546

Mr 8M  Iz pr 3

z E  c  r cos (90°  )  0.53868r

;

r3

From (Eq. 6-38):

sc  sB   sA   M

Mc Iy

 3.867

M

8M pr

;

r3

FOR u  90°: st  so 

r3

;

r3

y E  r sin (90°  )  0.26918r

M

  2.546

M

MAXIMUM COMPRESSIVE STRESS for u  45° occurs at point E. where the tangent to the circle is parallel to the neutral axis nn.

sc  maximum compressive stress FOR u  0°: st  sA 

yA  r

;

3

sC  sE 

(M sin u)zE (M cos u)yE  Iy Iz

  3.955

M r3

;

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CHAPTER 6 Stresses in Beams (Advanced Topics)

Problem 6.5-10 A built-up beam supporting a condominium balcony is made up of a structural T (one half of a

W 200  31.3) for the top flange and web and two angles (2L 102  76  6.4, long legs back-to-back) for the bottom flange and web, as shown. The beam is subjected to a bending moment M having its vector at an angle u to the z axis (see figure). Determine the orientation of the neutral axis and calculate the maximum tensile stress st and maximum compressive stress sc in the beam. Assume that u  30° and M  15 kN # m. Use the following numerical properties: c1  4.111 mm, c2  4.169 mm, bf  134 mm, Ls  76 mm, A  4144 mm2, Iy  3.88  106 mm4, and Iz  34.18  106 mm4. y bf /2

u, b

bf /2

c1 C

z u

c2

M

Ls

Ls

Solution 6.5-10 M  15 kN # m

MAXIMUM TENSILE STRESS (AT POINT A)

u  30°

zA 

BUILT-UP SECTION: c1  4.111 mm Ls  76 mm

c2  4.169 mm

bf  134 mm

A  4144 mm

2

Iy  3.88  106 mm4

Iz  34.18  106 mm4

BENDING MOMENTS

st 

My  7500 N # m

Mz  Mcos(180°  u)

Mz  12990 N # m

My(zA) Iy



Mz(yA) Iz

st  131.1 MPa

zB  Ls sc 

NEUTRAL AXIS nn Iz b  atan a tan(180°  u) b Iy ;

yA  c1

2

;

MAXIMUM COMPRESSIVE STRESS (AT POINT B)

My  Msin(180°  u)

b   78.9°

bf

My(zB) Iy

yB  c2 

Mz(yB) Iz

sc   148.5 MPa

;

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SECTION 6.5 Bending of Unsymmetric Beams

Problem 6.5-11 A steel post (E  30  106 psi) having

thickness t  1/8 in. and height L  72 in. supports a stop sign (see figure). The stop sign post is subjected to a bending moment M having its vector at an angle u to the z axis. Determine the orientation of the neutral axis and calculate the maximum tensile stress st and maximum compressive stress sc in the beam. Assume that u  30° and M  5.0 k-in. Use the following numerical properties for the post: A  0.578 in2, c1  0.769 in., c2  0.731 in., Iy  0.44867 in4, and Iz  0.16101 in4.

A

Section A–A

5/8 in.

Circular cutout, d = 0.375 in.

y

Post, t = 0.125 in. c1 z

1.5 in.

C

u

Stop sign

c2

M 1.0 in.

0.5 in.

1.0 in.

0.5 in.

A

L

Elevation view of post

Solution 6.5-11 M  5 k # in

MAXIMUM TENSILE STRESS

u  30°

Post: A  0.578 in.2

c1  0.769 in.

sx(z, y) 

c2  0.731 in. Iy  0.44867 in.4

POINT A:

Iz  0.16101 in.4

My z Iy



Mz y Iz

zA   1.5 in.

st  sx(zA, yA)

yA   c2

st  28.0 ksi

;

BENDING MOMENTS My  Msin(u)

My  0.208 k # ft

MAXIMUM COMPRESSIVE STRESS

Mz  Mcos(u)

Mz  0.361 k # ft

POINT B:

Iz b  atan a tan(u)b Iy

zB 

sc  sx(zB, yB)

NEUTRAL AXIS nn b   11.7°

;

539

5 in. 8

yB  c1

sc   24.2 ksi

;

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CHAPTER 6 Stresses in Beams (Advanced Topics)

Problem 6.5-12 A C 200  17.1 channel section has an angle with equal legs attached as shown; the angle serves as a lintel beam. The combined steel section is subjected to a bending moment M having its vector directed along the z axis, as shown in the figure. The centroid C of the combined section is located at distances xc and yc from the centroid (C1) of the channel alone. Principal axes x1 and y1 are also shown in the figure and properties Ix1, Iy1 and up are given below. Find the orientation of the neutral axis and calculate the maximum tensile stress st and maximum compressive stress sc if the angle is an L 76  76  6.4 section and M  3.5 kN # m. Use the following properties for principal axes for the combined section: Ix1  18.49  106 mm4 Iy1  1.602  106 mm4, up  7.448° (CW), xc  10.70 mm, yc  24.07 mm.

y1

y

C

C1

M

z

yc up

xc

x1

L 76  76  6.4 lintel

C 200  17.1

Solution 6.5-12 M  3.5 kN # m ANGLE:

ca  21.2 mm

CHANNEL:

zpA  (zA) cos (up)  (yA) sin (up)

up  7.448°

cc  14.5 mm

zpA   63.18 mm

La  76 mm

ypA  (zA) sin (up) + (yA) cos (up)

dc  203 mm

ypA   69.83 mm

bc  57.4 mm

st 

BUILT-UP SECTION: ybar  24.07 mm Izp  Ix1

xbar  10.70 mm

Iyp



st  31.0 MPa

Iyp  Iy1

Izp  18.49  106 mm4

Myp(zpA)

Iyp  1.602  106 mm4

Izp ;

MAXIMUM COMPRESSIVE STRESS (AT POINT B) zB  xbar  cc

BENDING MOMENTS

Mzp(ypA)

yB 

dc + ybar 2

Myp  Msin(up)

Myp  454 N # m

zpB  (zB) cos (up)  (yB) sin (up)

Mzp  Mcos(up)

Mzp  3470 N # m

zpB  20.05 mm ypB  (zB) sin (up) + (yB) cos (up)

NEUTRAL AXIS nn b  a tan a

Izp Iyp

tan( up)b

ypB  124.0 mm b   56.5°

MAXIMUM TENSILE STRESS (AT POINT A) zA  xbar  cc  bc

dc yA   + ybar 2

; sc 

Myp(zpB) Iyp



sc  29.0 MPa

Mzp(ypB) Izp ;

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SECTION 6.8 Shear Stresses in Wide-Flange Beams

541

Shear Stresses in Wide-Flange Beams When solving the problems for Section 6.8, assume the cross sections are thin-walled. Use centerline dimensions for all calculations and derivations, unless otherwise specified

Problem 6.8-1 A simple beam of W 10  30 wide-flange cross section supports a uniform load of intensity q  3.0 k/ft on a span of length L  12 ft (see figure). The dimensions of the cross section are h  10.5 in., b  5.81 in., tf  0.510 in., and tw  0.300 in.

y b — 2 q

A

(a) Calculate the maximum shear stress tmax on cross section A–A located at distance d  2.5 ft from the end of the beam. (b) Calculate the shear stress t at point B on the cross section. Point B is located at a distance a  1.5 in. from the edge of the lower flange.

A

Probs. 6.8-1 and 6.8-2

L

b — 2 h — 2

tw z

C tf

h — 2

B a

d

Solution 6.8-1 (a) MAXIMUM SHEAR STRESS

SIMPLE BEAM: q  3.0 k/ft R

qL 2

L  12 ft R  18.0 k

V  |R  qd|

d  2.5 ft

V  10.5 k

b  5.81 in.

tf  0.510 in.

tw  0.30 in. 3

tmax  3584 psi

a  1.5 in. t1 

2

twh btfh + Iz  12 2

btf h Vh + b tw 4 2Iz ;

(b) SHEAR STRESS AT POINT B

CROSS SECTION: h  10.5 in.

tmax  a

Iz  192.28 in.4

bhV 4Iz

a tB  (t1) b 2

b  2.9 in. 2 t1  832.8 psi tB  430 psi

;

Problem 6.8-2 Solve the preceding problem for a W 250  44.8 wide-flange shape with the following data: L  3.5 m, q  45 kN/m, h  267 mm, b  148 mm, tf  13 mm, tw  7.62 mm, d  0.5 m, and a  50 mm.

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CHAPTER 6 Stresses in Beams (Advanced Topics)

Solution 6.8-2 (a) MAXIMUM SHEAR STRESS

SIMPLE BEAM: q  45 kN/m R

qL 2

L  3.5 m

tmax  a

R  78.8 kN d  0.5 m

tmax  29.7 MPa

V  56.3 kN

V  |R  qd|

b/2  74.0 mm

a  50 mm

h  267 mm

b  148 mm

tf  13 mm

tw  7.62 mm

;

(b) SHEAR STRESS AT POINT B

CROSS SECTION:

Iz 

btf h Vh + b tw 4 2Iz

twh3 btfh2 + 12 2

t1 

bhV 4Iz

tB 

a (t ) b/2 1

t1  999.1 psi tB  4.65 MPa

;

Iz  80.667 * 106 mm4

Problem 6.8-3 A beam of wide-flange shape, W 8  28, has the cross section

y

shown in the figure. The dimensions are b  6.54 in., h  8.06 in., tw  0.285 in., and tf  0.465 in. The loads on the beam produce a shear force V  7.5 k at the cross section under consideration.

tf

(a) Using centerline dimensions, calculate the maximum shear stress tmax in the web of the beam. (b) Using the more exact analysis of Section 5.10 in Chapter 5, calculate the maximum shear stress in the web of the beam and compare it with the stress obtained in part a.

z

h

C tw

tf

Probs. 6.8-3 and 6.8-4

b

Solution 6.8-3 b  6.54 in.

h  8.06 in.

tf  0.465 in.

V  7.5 k

tw  0.285 in.

(a) CALCULATIONS BASED ON CENTERLINE DIMENSIONS

Maximum shear stress in the web: tmax  a

btf h Vh + b tw 4 2Iz

tmax  3448 psi

;

Moment of inertia: Iz 

twh3 btfh2 + 12 2

Iz  111.216 in.4

(b) CALCULATIONS BASED ON MORE EXACT ANALYSIS h2  h  tf h1  7.6 in.

h2  8.5 in.

h1  h  tf

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SECTION 6.9 Shear Centers of Thin-Walled Open Sections

Moment of inertia: I

Maximum shear stress in the web:

1 (bh32  bh31 + twh31) 12

tmax 

V (bh22  bh21 + twh21) 8Itw

tmax  3446 psi

I  109.295 in.4

;

Problem 6.8-4 Solve the preceding problem for a W 200  41.7 shape with the following data: b  166 mm, h  205 mm, tw  7.24 mm, tf  11.8 mm, and V  38 kN.

Solution 6.8-4 b  166 mm

h  205 mm

tf  11.8 mm

V  38 kN

tw  7.24 mm

(b) CALCULATIONS BASED ON MORE EXACT ANALYSIS h2  h  tf

h2  216.8 mm

h1  h  tf

h1  193.2 mm

(a) CALCULATIONS BASED ON CENTERLINE DIMENSIONS Moment of inertia:

Moment of inertia: Iz 

twh3 btfh2 + 12 2

I

I  45.556 * 106 mm4

Iz  46.357 * 106 mm4 Maximum shear stress in the web: tmax  a

btf h Vh + b tw 4 2Iz

tmax  27.04 MPa

1 (bh32  bh31 + twh31) 12

Maximum shear stress in the web: tmax 

V (bh22  bh21 + twh21) 8Itw

tmax  27.02 MPa

;

;

Shear Centers of Thin-Walled Open Sections When locating the shear centers in the problems for Section 6.9, assume that the cross sections are thin-walled and use centerline dimensions for all calculations and derivations.

y

Problem 6.9-1 Calculate the distance e from the centerline of

the web of a C 15  40 channel section to the shear center S (see figure). (Note: For purposes of analysis, consider the flanges to be rectangles with thickness tf equal to the average flange thickness given in Table E-3a in Appendix E.)

S

z e

Probs. 6.9-1 and 6.9-2

C

543

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CHAPTER 6 Stresses in Beams (Advanced Topics)

Solution 6.9-1 C 15 * 40 bf  3.520 in.

d  15.0 in.

tw  0.520 in.

tf  0.650 in.

b  bf 

h  d  tf tw 2

e

h  14.350 in.

3b2 tf h tw + 6btf

e  1.027 in.

;

b  3.260 in.

Problem 6.9-2 Calculate the distance e from the centerline of the web of a C 310  45 channel section to the shear center S (see figure). (Note: For purposes of analysis, consider the flanges to be rectangles with thickness tf equal to the average flange thickness given in Table E-3b in Appendix E.)

Solution 6.9-2 C 310 * 45 bf  80.5 mm

d  305 mm tf  12.7 mm

tw  13.0 mm b  bf 

tw 2

b  74.0 mm

h  d  tf

h  292.3 mm

2

e

3b tf htw + 6btf

e  22.1 mm

Problem 6.9-3 The cross section of an unbalanced wide-flange

y

beam is shown in the figure. Derive the following formula for the distance h1 from the centerline of one flange to the shear center S: h1 

;

t2

t2b32h z

t1b31 + t2b32

b1

t1 S

Also, check the formula for the special cases of a T-beam (b2  t2  0) and a balanced wide-flange beam (t2  t1 and b2  b1).

h1

h2 h

Solution 6.9-3 Unbalanced wide-flange beam FLANGE 1: t1 

VQ Izt1

Q  (b1/2)(t1)(b1/4)  t1 

b2

C

t1b21 8

Vb21 8Iz

Vt1b31 2 F1  (t1)(b1)(t1)  3 12Iz

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SECTION 6.9 Shear Centers of Thin-Walled Open Sections

Solve Eqs. (1) and (2): h1  FLANGE 2: F2 

Vt2b32 12Iz

t2b32h t1b31 + t2b32

T-BEAM b2  t2  0; ‹ h1  0

Shear force V acts through the shear center S. ‹ a MS  F1h1  F2h2  0

;

WIDE-FLANGE BEAM

or (t1b31 ) h1  (t2b 32) h2

(1)

h1  h2  h

(2)

t2  t1 and b2  b1; ‹ h1  h/2

;

Problem 6.9-4 The cross section of an unbalanced wide-flange beam is

y

tf

shown in the figure. Derive the following formula for the distance e from the centerline of the web to the shear center S: e

;

tw

3tf (b22  b21) htw + 6tf (b1 + b2)

Also, check the formula for the special cases of a channel section (b1  0 and b2  b) and a doubly symmetric beam (b1  b2  b/2).

h — 2 S

z

C

e

tf

b1

h — 2

b2

Solution 6.9-4 Unbalanced wide-flange beam

t1 

VQ b1hV  Itf 2Iz

F1 

b1t1tf b21htfV  2 4Iz

Shear force V acts through the shear center S.

b22htfV

‹ a MS   F3e  F1h + F2h  0

F2 

4Iz

t2 

F3  V

b2hV 2Iz

e

h2tf 2 F2h  F1h  (b  b21) F3 4Iz 2

545

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CHAPTER 6 Stresses in Beams (Advanced Topics)

twh3 h 2 + 2(b1 + b2)(tf)a b 12 2

Iz   e

CHANNEL SECTION (b1  0, b2  b) e

2

h [htw + 6 tf (b1 + b2)] 12 3tf (b22  b21) htw + 6tf (b1 + b2)

3b2tf htw + 6btf

(Eq. 6 - 65)

DOUBLY SYMMETRIC BEAM (b1  b2  b/2) e  0 (Shear center coincides with the centroid)

;

y

Problem 6.9-5 The cross section of a channel beam with double flanges and constant thickness throughout the section is shown in the figure. Derive the following formula for the distance e from the centerline of the web of the shear center S: e

3b2(h21 + h22) h32

+

6b(h21

+

S

z

h22)

C

h1 h2

e

b

Solution 6.9-5 Channel beam with double flanges Shear force V acts through the shear center S. ‹ a MS   F3e + F1h2 + F2h1  0 e

F2h1 + F1h2 b2t 2  (h + h22) F3 4Iz 1

Iz 

th32 + 2 [bt(h2/2)2 + bt(h1/2)2] 12

 e

t  thickness V(bt)a

h2 b 2

tA 

VQA  Izt

F1 

b2h2tV 1 tAbt  2 4Iz

tB 

bh1V 2Iz

F3  V

Izt

F2 



b2h1tV 4Iz

bh2V 2Iz

t 3 [h + 6b(h21 + h22)] 12 2 3b2(h21 + h22) h32 + 6b(h21 + h22)

;

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SECTION 6.9 Shear Centers of Thin-Walled Open Sections

Problem 6.9-6 The cross section of a slit circular tube of

y

constant thickness is shown in the figure. (a) Show that the distance e from the center of the circle to the shear center S is equal to 2r in the figure part a. (b) Find an expression for e if flanges with the same thickness as that of the tube are added, as shown in the figure part b.

Flange (r/2  t)

y

z

z

r

S

r/2 r

S

C

e

C

e

Flange (r/2  t) (b)

(a)

Solution 6.9-6 u

(a) QA 

L

y dA 

L0

for 0 … u 6

(r tsin(f)) df

u

QA 

QA  r t(1  cos(u)) 2

tA 

VQA Vr2(1  cos(u))  Izt Iz

Iz  pr t V(1  cos(u)) prt

TC  moment of shear stresses about center C. 2p

tA 

VQA Vr2(1  cos(u))  Izt Iz

for

3p p … u 6 2 2

Vr p

QA 

Shear force V acts through the shear center S. Moment of the shear force V about any point must be equal to the moment of the shear stresses about that same point. TC  2r e V

(b) Iz  pr3t + 2 J Iz  pr3t +

r 3 ta b 2 12

r 5r 2 + t a b K 2 4

19 3 19 tr  tr3 ap + b 12 12

L

y dA 

L0

(rtsin(f)) df +

QA  r2ta

13  cos(u)b 8

Vr2 a

13  cos(u)b 8

(1  cos(u)) du  2Vr

©MC  Ve  TC

(rtsin(f)) df

L0

u

At point A: dA  rtdu

TC  tAr dA  L L0

L

y dA 

QA  r2t(1  cos(u))

3

tA 

p 2

tA 

r 5r t 2 4

Iz

At point A: dA  rtdu TC  moment of shear stresses about center C.

; TC 

L

tAr dA  2

p 2

Vr4t Iz L 0

(1  cos(u)) du + a

13  cos(u)b du 8

3p 2

Vr4t Iz Lp2

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CHAPTER 6 Stresses in Beams (Advanced Topics)

TC 

L

tAr dA  Vr4t

p 2 Iz 3p 2

+

TC  Vr4t



Vr4t 13 a  cos(u)b du Iz 8 Lp2

p2 1 13p + 16 + Vr4t Iz 8 Iz

21Vr4tp 8Iz

Shear force V acts through the shear center S. Moment of the shear force V about any point must be equal to the moment of the shear stresses about that same point. ©MC  Ve  TC e

e

TC V



21r4tp  8Iz

21r4tp 19 8ctr3 a p + bd 12

63pr  1.745r 24p + 38

;

Problem 6.9-7 The cross section of a slit square tube of constant thickness is

y

shown in the figure. Derive the following formula for the distance e from the corner of the cross section to the shear center S: e

b 212

b

z

S e

Solution 6.9-7 Slit square tube

b  length of each side t  thickness t

VQ Iz t

FROM A TO B: Q

ts2 212

At A: Q  0 tA  0

C

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SECTION 6.9 Shear Centers of Thin-Walled Open Sections

At B: Q  tB  F1 

tb2 212

At B: tB 

b2V 212Iz

F2  tB bt +

tB bt b3tV  3 612Iz

b2V 12Iz

2 5tb3V (tD  tB) bt  3 612Iz

‹ g Ms  0 2(F1/12)(b 12 + e) + 2(F2/ 12) (e)  0

b S b b + st a  b Q  bt a 212 12 212

t

At D: tD 

Shear force V acts through the shear center S.

FROM B TO D:



b2V 212Iz

549

Substitute for F1 and F2 and solve for e: b 212

e

ts tb2 + (2b  s) 212 212

;

VQ V b2 s  c + (2b  s) d Izt Iz 212 212

Problem 6.9-8 The cross section of a slit rectangular tube of

y

constant thickness is shown in the figures. (a) Derive the following formula for the distance e from the centerline of the wall of the tube in figure part (a) to the b(2h + 3b) shear center S: e  2(h + 3b) (b) Find an expression for e if flanges with the same thickness as that of the tube are added as shown in figure part (b).

Flange (h/4  t)

y

h — 2 z

S

z

C

e

h — 2 S C

e h — 2

b — 2

h — 2

b — 2

b — 2

(a)

b — 2 (b)

Solution 6.9-8 (a) FROM A TO B: Q  tA  0 F1 

tB 

ts2 2

h2V 8Iz

tBt h th3V a b  3 2 48Iz

FROM B TO C: tB 

h2V 8Iz

t

VQ s2V  Izt 2Iz

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CHAPTER 6 Stresses in Beams (Advanced Topics)

th h h th a b + bta b  (h + 4b) 2 4 2 8

QC 

h(h + 4b)V 8Iz

tC 

bht(h + 2b)V 1 F2  (tB + tC)bt  2 8Iz ©FVERT  V F3  V a1 +

F3  2 F1  V

In flange: QC_ flange  t sa Fflange 

h 4

s tV h3Vt c (2s h) d ds  Iz 96Iz L0 4

FROM C TO D: Q

th3 b 24Iz

s h ts + b  (2s + h) 2 4 4

FCD 

b h h 3h h th2 + t a b + t a b + t s 8 2 2 4 8 2 b 2

L0

c

th2 b h h 3h t a b t a b  8 2 2 4 8

Shear force V acts through the shear center S. ©Ms  0

(b + e)  0 solve for Iz  2 c

ts2 2

VQ s2V  Izt 2Iz

F1 

tB 

h2V 8Iz

tBt h th3V a b 3 2 48Iz

FROM B TO C: tB  QC  tC 

h2 # V 8Iz

th h b h th a b + t a b  (h + 2b) 2 4 2 2 8 h(h + 2b)V 8Iz

bht(h + b)V 1 b FBC  (tB + tC) t  2 2 16Iz

th3 b 16Iz

Shear force V acts through the shear center S. ©Ms  0

;

t

F3  2F1  2Fflange  V

F3  Va 1 +

b 2h + 3b e a b 2 h + 3b

(b) FROM A TO B: Q  tA  0

©FVERT  V

bh2t(2h + 3b) e 12Iz

1 3 h 2 th2 th + bt a b d  (h + 3b) 12 2 6

Therefore

Vthb h Vt (7h 12b) t s d ds  2 Izt 64Iz

F3e + F2h + 2F1

F3e + (FBC + FCD)h + 2F1 (b + e) + 2Fflange a

e

b2h2t 61bth3 + 192Iz 4Iz

e

th2b (43h + 48b) 192Iz

Iz  2c

b + eb  0 2

1 3 h 2 1 h 3 th + bta b + ta b + 12 2 12 4 th2 h 3h 2 (23h + 48b) ta b a b d  4 8 96

b 43h + 48b b e a 2 23h + 48b

;

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SECTION 6.9 Shear Centers of Thin-Walled Open Sections

Problem 6.9-9 A U-shaped cross section of constant thickness

y

is shown in the figure. Derive the following formula for the distance e from the center of the semicircle to the shear center S: e

551

b r

2(2r2 + b2 + pbr) 4b + pr

S

z

O

Also, plot a graph showing how the distance e (expressed as the nondimensional ratio e/r) varies as a function of the ratio b/r. (Let b/r range from 0 to 2.)

C

e

Solution 6.9-9 U-shaped cross section

At angle u: dA  rtdu r  radius

F1  force in AB

t  thickness

F2  force in EF

p

T0 

trdA  tr2tdu L L0

T0  moment in BDE FROM A TO B: tA  0

F1 

p

tB 

VQ V(btr) Vbr   Izt Izt Iz

2

bttB Vb rt  2 2Iz u

FROM B TO E: Q1 

ydA  (r cos f) rtdf L L0  r2t sin u

QB  btr



Vr3t (b + r sin u)du Iz L0



Vr3t (pb + 2r) Iz

Shear force V acts through the shear center S. Moment of the shear force V about any point must be equal to the moment of the shear stresses about that same point. ‹ a M0  Ve  T0 + F1(2r) e

T0 + 2F1r r2t  (pbr + 2r2 + b2) V Iz

Iz 

pr3t + 2(btr2) 2

Qu  QB + Q1  btr + r2t sin u VQB Vr (b + r sin u) tu   Iz t Iz

e

2(2r 2 + b2 + pbr) 4b + pr

;

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CHAPTER 6 Stresses in Beams (Advanced Topics)

GRAPH

NOTE: When b/r  0,

2(2 + b2/r2 + pb/r) e  r 4b/r + p

e/r 

4 (Eq. 6-73) p

Problem 6.9-10 Derive the following formula for the distance e from the centerline

y

of the wall to the shear center S for the C-section of constant thickness shown in the figure: e

a

3bh2(b + 2a)  8ba3

h — 2

h2(h + 6b + 6a) + 4a2(2a  3h)

S

z

Also, check the formula for the special cases of a channel section (a  0) and a slit rectangular tube (a  h/2).

e

C h — 2 a b

Solution 6.9-10 C-section of constant thickness

a

t  thickness

F1 

FROM A TO B: Q  st a

h s a+ b 2 2

tA  0 tB 

t

a V (h  a) 2 Iz

VQ h s V  sa  a + b Iz t 2 2 Iz



L0

a

ttds 

tV h s s a  a + b ds Iz L0 2 2

a2t(3h  4a)V 12Iz

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SECTION 6.9 Shear Centers of Thin-Walled Open Sections

FROM B TO C: tB 

Substitute for F1, F2, and F3 and solve for e:

V a (h  a) 2 Iz

QC  at a 

a h h  b + bta b 2 2 2

at bht (h  a) + 2 2

a bh V tC  c (h  a) + d 2 2 Iz

e 

bt [3h2(b + 2a)  8a3] 12 Iz

Iz  2a 

1 3 h 2 6 th b + 2bta b  (h  2a)3 12 2 12

t 2 [h (h + 6b + 6a) + 4a2(2a  3h)] 12 3bh2(b + 2a)  8ba3

1 bt V F2  (tB + tC)bt  [2a(h  a) + bh] 2 4 Iz

e 

FROM C TO E:

CHANNEL SECTION (a  0)

g FVERT  V F3  2F1  V

e

a 2 t(3h  4a) F3  V c1 + d 6Iz

h (h + 6b + 6a) + 4a2(2a  3h) 2

3b2 h + 6b

;

(agrees with Eq. 6-65 when tf  tw)

SLIT RECTANGULAR TUBE (a  h/2)

Shear force V acts through the shear center S. ‹ a Ms  0

553

 F3(e) + F2h + 2F1(b + e)  0

e

b(2h + 3b) (agrees with the result of Prob. 6.9-8) 2(h + 3b)

Problem 6.9-11 Derive the following formula for the distance e from the centerline of the

y

wall to the shear center S for the hat section of constant thickness shown in the figure: e

a

3bh2(b + 2a)  8ba3 h2(h + 6b + 6a) + 4a2(2a + 3h)

h — 2

Also, check the formula for the special case of a channel section (a  0). S

z e

C h — 2 a b

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CHAPTER 6 Stresses in Beams (Advanced Topics)

Solution 6.9-11 Hat section of constant thickness

t  thickness FROM A TO B t 

FROM C TO E: Q  st a

s h + a b 2 2

VQ h s V  sa + a  b Izt 2 2 Iz

V a tA  0 tB  (h + a) 2 Iz a

F1  

L0

a

ttds 

s tV h sa + a  bds Iz L0 2 2

a2t (3h + 4a)V 12Iz

FROM B TO C QC  at a

tB 

V a (h + a) 2 Iz

a h at bht h + b + bta b  (h + a) + 2 2 2 2 2

g FVERT  V F3  V c1 

F3 + 2F1  V 2

a t(3h + 4a) d 6Iz

Shear force V acts through the shear center S. ‹ g MS  0

F3e  F2h  2F1(b  e)  0

Substitute for F1, F2, and F3 and solve for e: e

bt[3h2(b + 2a)  8a3] 12Iz

Iz 

1 3 h 2 t 1 th + 2bta b + (h + 2a)3  th3 12 2 12 12

 e

t 2 [h (h + 6b + 6a) + 4a2(2a + 3h)] 12 3bh2(b + 2a)  8ba3 2

h (h + 6b + 6a) + 4a2(2a + 3h)

bh V a tc  c (h + a) + d 2 2 Iz

CHANNEL SECTION (a  0)

1 bt V F2  (tB + tc) bt  [2a(h + a) + bh] 2 4 Iz

e

3b2 h + 6b

;

(agrees with Eq. 6-65 when tf  tw)

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SECTION 6.9 Shear Centers of Thin-Walled Open Sections

Problem 6.9-12 The cross section of a sign post of constant thickness is shown in the figure. Derive the formula below for the distance e from the centerline of the wall of the post to the shear center S. Also, compare this formula with that given in Prob. 6.9-11 for the special case of b  0 here and a  h/2 in both formulas.

y a b

b

b sin(b ) a

S

C

z

a

e b b

b sin(b ) a

Solution 6.9-12 FROM A TO B s QAB  st a2a + bsin (b )  b 2 tAB 

FAB  FAB 

VQAB Iz t

L

tdA 

a V cst a2a bsin (b)

L0

Izt

s bd 2

t ds

Vta2(5a  3bsin (b)) 6Iz

FROM B TO C a s QBC  at a2a + bsin (b)  b + ts a a + bsin (b)  sin (b)b 2 2 3 1 QBC  a2t + atbsin (b) + sta + stbsin (b)  s 2tsin (b) 2 2 tBC 

VQBC Iz t b

FBC 

FBC 

L

tdA 

3 1 V a a2t + atbsin (b) + sta + stbsin (b)  s2tsin (b)b 2 2

L0

Izt

Vtb a9a2 + 6absin (b) + 3ba + 2b2sin (b)b 6Iz

tds

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CHAPTER 6 Stresses in Beams (Advanced Topics)

SHEAR FORCE V ACTS THROUGH THE SHEAR CENTER S. a ME  0 e  2cos (b) e

Ve + 2FABbcos (b)  FBCcos (b)(2a)  0 (FBCa  FABb) V

tbacos (b) a4a2 + 3absin (b) + 3ab + 2b2sin (b)b 3Iz

;

Now, compare this formula with that given in Prob. 6.9-11 for the special case of b  0 here and a = h/2 in both formulas. FIRST MODIFY ABOVE FORMULA FOR b  0 & a = h/2: h tb cos (0) 2 h 2 h h c4a b 3 bsin (0) 3 b2b2sin (0) d e 3Iz 2 2 2

e

bht ah2 

3bh b 2

6Iz

where Iz for the hat section of #6.9-11 is as follows: 3

h 3 ta b 2

2

Iz 

h th + 2bt a b + 2 12 2 12

Iz 

h2t (3b + 4h) 6

h h h 2 + 2t a + b 2 2 4

substituting expression for Iz & simplifying gives: bhtah2  e 6 e

3bh b 2

h2t(3b + 4h) 6

b(3b 2h) 6b  8h

for

b  0 and

a

h 2

;

NOW MODIFY FORMULA FOR e FROM #6.9-11 AND COMPARE TO ABOVE e

e

e

3bh2(b 2a) 8ba3 h2(h  6b 6a)4a2(2a  3h) h h 3 3bh2 ab 2 b  8ba b 2 2 h h 2 h h2 a h 6b  6 b  4a b a2 3hb 2 2 2 b(3b 2h) 6b  8h

same expressions as that above from sign post solution

;

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SECTION 6.9 Shear Centers of Thin-Walled Open Sections

Problem 6.9-13 A cross section in the shape of a circular arc of constant thickness is shown

y

in the figure. Derive the following formula for the distance e from the center of the arc to the shear center S: e

2r(sin b  b cos b) b  sin b cos b

r

in which b is in radians. Also, plot a graph showing how the distance e varies as b varies from 0 to p.

z

b

S

O

C

b

e

Solution 6.9-13

Circular arc Shear force V acts through the shear center S. Moment of the shear force V about any point must be equal to the moment of the shear stresses about that same point. ‹ a M0  Ve  T0 e  T0/V e

t  thickness

r  radius

2r(sin b  b cos b) b  sin b cos b

;

GRAPH

At angle u: b

Q

L

ydA 

Lu

(r sin f) rtdf

 r2t(cos u  cos b) t

Vr2(cos u  cos b) VQ  Izt Iz b

Iz 

L

y2 dA 

L b

(r sin f)2rtdf

 r3t(b  sin b cos b) V(cos u  cos b) t rt(b  sin b cos b)

2(sin b  b cos b) e  r b  sin b cos b SEMICIRCULAR ARC (b  p/2): 4 e  (Eq. 6-73) p r

T0  moment of shear stresses

SLIT CIRCULAR ARC (b  p):

At angle u, dA  rtdu

e  2 (Prob. 6.9-6) r

b

T0  

V(cos u  cos b) trdA  rtdu L L b t (b  sin b cos b 2 Vr (sin b  b cos b) b  sin b cos b

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CHAPTER 6 Stresses in Beams (Advanced Topics)

Elastoplastic Bending

y

The problems for Section 6.10 are to be solved using the assumption that the material is elastoplastic with yield stress sY.

b1

Problem 6.10-1 Determine the shape factor f for a cross section in the shape of a double trapezoid having the dimensions shown in the figure. Also, check your result for the special cases of a rhombus (b1  0) and a rectangle (b1  b2).

h — 2 z

C

b1 b2

Solution 6.10-1

Double trapezoid SHAPE FACTOR f f

(EQ. 6-79)

2(2b1 + b2) Z  S 3b1 + b2

;

SPECIAL CASE – RHOMBUS

Neutral axis passes through the centroid C. Use case 8, Appendix D. SECTION MODULUS S h 3 Iz  2 a b (3b1 + b2)/12 2 h3 (3b1 + b2)  48 c  h/2

S

b1  0

f2

SPECIAL CASE – RECTANGLE

I h2  (3b1 + b2) c 24

PLASTIC MODULUS Z (EQ. 6-78) h h A  2a b(b1 + b2)/2  (b1 + b2) 2 2 1 h 2b1 + b2 b y1  y2  a b a 3 2 b1 + b2 z

h2 A (y1 + y2)  (2b1 + b2) 2 12

b1  b2

f

3 2

h — 2

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559

SECTION 6.10 Elastoplastic Bending

Problem 6.10-2 (a) Determine the shape factor f for a

y

hollow circular cross section having inner radius r1 and outer radius r2 (see figure). (b) If the section is very thin, what is the shape factor? r1 z

C

r2

Solution 6.10-2 Hollow circular cross sections (a) SHAPE FACTOR f (EQ. 6-79) f

16r2(r32  r31) Z  S 3p(r42  r41)

;

(b) THIN SECTION (r1 : r2) Rewrite the expression for the shape factor f. (r23  r13)  (r2  r1)(r22  r1r2  r12) (r24  r14)  (r2  r1)(r2  r1)(r22  r12) Neutral axis passes through the centroid C.

f

Use cases 9 and 10, Appendix D. 

SECTION MODULUS S Iz 

p 4 (r r41 ) c  r2 4 2

S

Iz p 4 (r  r41)  c 4r2 2



gyi Ai g Ai



Let r1  0 f  y

4r 3p

4r2 pr22 4r1 pr21 ba b  a ba b 3p 2 3p 2 p/2(r22  r21)

4 r32  r31 a b 3p r22  r21

y1  y2 Z 

16 3 4 a b  L 1.27 p 3p 4

;

SPECIAL CASE OF A SOLID CIRCULAR CROSS SECTION

A  p(r22  r21 ) For a semicircle,

y1 

16 1 + r1/r2 + (r1/r2)2 d c 3p (1 + r1/r2)(1 + r21/r22)

Let r1/r2 : 1 f 

PLASTIC MODULUS Z (EQ. 6-78)

a

16r2 r22 + r1r2 + r21 d c 3p (r2 + r1)(r22 + r21)

A 4 (y + y2)  (r32  r31) 2 1 3

16 1 16 a b  3p 1 3p

(Eq. 6-90)

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Problem 6.10-3 A propped cantilever beam of length L  54 in.

y

q

with a sliding support supports a uniform load of intensity q (see figure). The beam is made of steel (sY  36 ksi) and has a rectangular cross section of width b  4.5 in. and height h  6.0 in. What load intensity q will produce a fully plastic condition in the beam?

z

C

Sliding support L = 54 in.

h = 6.0 in.

b = 4.5 in.

Solution 6.10-3 L  54 in.

sy  36 ksi

b  4.5 in. Mmax 

MAXIMUM BENDING MOMENT:

h  6.0 in.

qL2 2

Let

Mmax  MP

q

gives

Therefore q  1000 lb/in.

sybh2 2L2

;

2

PLASTIC MOMENT:

MP 

sybh 4

y

Problem 6.10-4 A steel beam of rectangular cross section is 40 mm wide and 80 mm high (see figure). The yield stress of the steel is 210 MPa. (a) What percent of the cross-sectional area is occupied by the elastic core if the beam is subjected to a bending moment of 12.0 kNm acting about the z axis? (b) What is the magnitude of the bending moment that will cause 50% of the cross section to yield?

z

C

40 mm

Solution 6.10-4 sy  210 MPa

b  40 mm

h  80 mm

2e  56.7% h

(a) ELASTIC CORE M  12.0 kN # m

My 

MP 

6

M

is between

(b) ELASTIC CORE e

2

4

MP  13.4 k N # m My

1 3 M eh a  b A2 2 My

and

;

sybh2

My  9.0 kN # m sybh

Percent of cross-sectional area is

MP e  22.678 mm

h 4

M  My a

e  20 mm 3 2e2  2b 2 h

M  12.3 kN # m

;

80 mm

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SECTION 6.10 Elastoplastic Bending

Problem 6.10-5 Calculate the shape factor f for the wideflange beam shown in the figure if h  12.2 in., b  8.08 in., tf  0.64 in., and tw  0.37 in.

y

tf z

h

C tw

tf b

Probs. 6.10-5 and 6.10-6

Solution 6.10-5 h  12.2 in.

b  8.08 in.

PLASTIC MODULUS

tf  0.64 in.

tw  0.37 in.

1 Z  [bh2  (b  tw)(h  2tf)2] 4 Z  70.8 in.3

SECTION MODULUS I

1 3 1 bh  (b  tw)(h  2tf)3 12 12

SHAPE FACTOR

I  386.0 in.4 c

h 2

f

c  6.1 in.

S

I c

Z S

f  1.12

;

S  63.3 in.3

Problem 6.10-6 Solve the preceding problem for a wide-flange beam with h  404 mm, b  140 mm, tf  11.2 mm, and tw  6.99 mm.

Solution 6.10-6 h  404 mm

b  140 mm

PLASTIC MODULUS

tf  11.2 mm

tw  6.99 mm

1 Z  [bh2  (b  tw)(h  2t f)2] 4 Z  870.4 * 103 mm3

SECTION MODULUS 1 3 1 bh  (b  tw)(h  2tf)3 12 12 I  153.4 * 106 mm4 I

h c c  202.0 mm 2 S  759.2 * 103 mm3

I S c

SHAPE FACTOR f

Z S

f  1.15

;

561

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CHAPTER 6 Stresses in Beams (Advanced Topics)

Problem 6.10-7 Determine the plastic modulus Z and shape factor f for a W 12 ⫻ 14 wide-flange beam. (Note: Obtain the cross-sectional dimensions and section modulus of the beam from Table E-1a in Appendix E.)

Solution 6.10-7 W 12 ⫻ 14

SHAPE FACTOR

h ⫽ 11.9 in.

b ⫽ 3.97 in.

tw ⫽ 0.200 in.

S ⫽ 14.9 in.

tf ⫽ 0.225 in. 3

f⫽

Z S

f ⫽ 1.14

;

PLASTIC MODULUS 1 Z ⫽ [bh2 ⫺ (b ⫺ t w)(h ⫺ 2t f)2] 4 Z ⫽ 16.98 in.3

;

Problem 6.10-8 Solve the preceding problem for a W 250 ⫻ 89 wide-flange beam. (Note: Obtain the cross-sectional dimensions and section modulus of the beam from Table E-1b in Appendix E.)

Solution 6.10-8 SHAPE FACTOR

W 250 ⫻ 89 h ⫽ 259 mm

b ⫽ 257 mm

tw ⫽ 10.7 mm

S ⫽ 1090 ⫻ 103 mm3

tf ⫽ 17.3 mm

f⫽

Z S

f ⫽ 1.11

;

PLASTIC MODULUS Z⫽

1 [bh 2 ⫺ (b ⫺ tw)(h ⫺ 2tf)2] 4

Z ⫽ 1.209 * 106 mm3

;

Problem 6.10-9 Determine the yield moment MY, plastic moment MP, and shape factor f for a W 16 ⫻ 100 wide-flange

beam if sY ⫽ 36 ksi. (Note: Obtain the cross-sectional dimensions and section modulus of the beam from Table E-1a in Appendix E.)

Solution 6.10-9 W 16 ⫻ 100

YIELD MOMENT

h ⫽ 17.0 in.

b ⫽ 10.4 in.

tf ⫽ 0.985 in.

tw ⫽ 0.585 in.

S ⫽ 175 in.

sy ⫽ 36 ksi

3

My ⫽ syS

My ⫽ 525 k-f t

;

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SECTION 6.10 Elastoplastic Bending

PLASTIC MODULUS

563

SHAPE FACTOR

1 Z ⫽ [bh2 ⫺(b ⫺ tw)(h ⫺2 tf)2] 4

Z ⫽ 197.1 in.3

f⫽

Z S

f ⫽ 1.13

;

PLASTIC MOMENT MP ⫽ syZ

MP ⫽ 591 k-ft

;

Problem 6.10-10 Solve the preceding problem for a W 410 ⫻ 85 wide-flange beam. Assume that sY ⫽ 250 MPa. (Note: Obtain the cross-sectional dimensions and section modulus of the beam from Table E-1b in Appendix E.)

Solution 6.10-10 W 410 ⫻ 85

PLASTIC MOMENT

h ⫽ 417 mm

b ⫽ 181 mm

tw ⫽ 10.9 mm

S ⫽ 1510⭈10 mm

tf ⫽ 18.2 mm 3

MP ⫽ syZ SHAPE FACTOR f⫽

YIELD MOMENT My ⫽ 378 kN # m

;

3

sy ⫽ 250 MPa

My ⫽ syS

MP ⫽ 427 kN # m

Z S

f ⫽ 1.13

;

;

PLASTIC MODULUS 1 Z ⫽ cbh 2 ⫺(b ⫺ tw) (h ⫺ 2tf )2 d 4 Z ⫽ 1.708 * 106 mm3

Problem 6.10-11 A hollow box beam with height h ⫽ 16 in., width b ⫽ 8 in., and constant wall thickness t ⫽ 0.75 in. is shown in the figure. The beam is constructed of steel with yield stress sY ⫽ 32 ksi. Determine the yield moment MY, plastic moment MP, and shape factor f.

y t

z

C t

Probs. 6.10-11 and 6.10-12

b

h

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Solution 6.10-11 h ⫽ 16 in.

Hollow box beam

b ⫽ 8 in.

t ⫽ 0.75 in.

PLASTIC MODULUS Use (Eq. 6-86) with tw ⫽ 2t and tf ⫽ t:

sY ⫽ 32 ksi

SECTION MODULUS (S ⫽ I/c)

Z⫽

1 1 bh3 ⫺ (b ⫺ 2t)(h ⫺ 2t)3 12 12 ⫽ 1079 in.4

⫽ 170.3 in.3

I⫽

c⫽

h ⫽ 8.0 in. 2

S⫽

1 [bh 2 ⫺(b ⫺2t)(h ⫺ 2t)2] 4

PLASTIC MOMENT (EQ. 6-77)

I ⫽ 134.9 in.3 c

Mp ⫽ sYZ ⫽ 5450 k-in.

;

SHAPE FACTOR (EQ. 6-79) YIELD MOMENT (EQ. 6-74) MY ⫽ sYS ⫽ 4320 k-in.

f⫽

;

MP Z ⫽ ⫽ 1.26 MY S

;

Problem 6.10-12 Solve the preceding problem for a box beam with dimensions h ⫽ 0.5 m, b ⫽ 0.18 m, and t ⫽ 22 mm. The yield stress of the steel is 210 MPa.

Solution 6.10-12 h ⫽ 0.5 m

b ⫽ 0.18 m

t ⫽ 22 mm

PLASTIC MODULUS 1 [bh2 ⫺ (b ⫺ 2t)(h ⫺ 2t)2] 4

sy ⫽ 210 MPa

Z⫽

SECTION MODULUS

Z ⫽ 4.180 * 106 mm3

I⫽

1 3 1 bh ⫺ (b ⫺ 2t)(h ⫺ 2t)3 12 12

PLASTIC MOMENT MP ⫽ syZ

I ⫽ 800.4 * 106 mm4 c⫽

h 2

I S⫽c

SHAPE FACTOR

c ⫽ 250 mm

f⫽ S ⫽ 3.202 * 10 mm 6

3

YIELD MOMENT My ⫽ syS

MP ⫽ 878 kN # m

My ⫽ 672 kN # m

;

Z S

f ⫽ 1.31

;

;

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565

y

Problem 6.10-13 A hollow box beam with height h ⫽ 9.5 in., inside height h1 ⫽ 8.0 in., width b ⫽ 5.25 in., and inside width b1 ⫽ 4.5 in. is shown in the figure. Assuming that the beam is constructed of steel with yield stress sY ⫽ 42 ksi, calculate the yield moment MY, plastic moment MP, and shape factor f.

h1

z

C

h

b1 b

Probs. 6.10-13 through 6.10-16

Solution 6.10-13 h ⫽ 9.5 in.

b ⫽ 5.25 in.

b1 ⫽ 4.5 in.

sy ⫽ 42 ksi

h1 ⫽ 8.0 in.

PLASTIC MODULUS 1 Z ⫽ (bh2 ⫺ b1h12) 4

Z ⫽ 46.5 in.3

SECTION MODULUS PLASTIC MOMENT

1 I ⫽ (bh3 ⫺ b1h13) 12 c⫽

h 2

I ⫽ 183.10 in.

MP ⫽ syZ

I S⫽ c

SHAPE FACTOR

4

c ⫽ 4.75 in.

S ⫽ 38.55 in.3

f⫽ YIELD MOMENT My ⫽ syS

My ⫽ 1619 k-in.

Z S

MP ⫽ 1951 k-in.

f ⫽ 1.21

;

;

;

Problem 6.10-14 Solve the preceding problem for a box beam with dimensions h ⫽ 200 mm, h1 ⫽ 160 mm, b ⫽ 150 mm, and b1 ⫽ 130 mm. Assume that the beam is constructed of steel with yield stress sY ⫽ 220 MPa.

Solution 6.10-14 h ⫽ 200 mm

Hollow box beam b ⫽ 150 mm

h1 ⫽ 160 mm

b1 ⫽ 130 mm

PLASTIC MODULUS sY ⫽ 220 MPa

Use (Eq. 6-86) with b ⫺ tw ⫽ b1 and h ⫺ 2tf ⫽ h1 Z⫽

SECTION MODULUS (S ⫽ I/c) I⫽

1 (bh3 ⫺ b1h31) ⫽ 55.63 * 106 mm4 12

c⫽

h ⫽ 100 mm 2

I S ⫽ c ⫽ 556.3 * 103 mm3

1 (bh 2 ⫺ b1h21) ⫽ 668.0 * 103 mm3 4

PLASTIC MOMENT (EQ. 6-77) MP ⫽ sYZ ⫽ 147 kN # m

;

SHAPE FACTOR (EQ. 6-79) YIELD MOMENT (EQ. 6-74) MY ⫽ sYS ⫽ 122 kN # m

;

f⫽

MP Z ⫽ ⫽ 1.20 MY S

;

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CHAPTER 6 Stresses in Beams (Advanced Topics)

Problem 6.10-15 The hollow box beam shown in the figure is subjected to a bending moment M of such magnitude that the flanges yield but the webs remain linearly elastic. (a) Calculate the magnitude of the moment M if the dimensions of the cross section are h ⫽ 15 in., h1 ⫽ 12.75 in., b ⫽ 9 in., and b1 ⫽ 7.5 in. Also, the yield stress is sY ⫽ 33 ksi.

Solution 6.10-15 h ⫽ 15 in. b1 ⫽ 7.5 in.

b ⫽ 9 in.

h1 ⫽ 12.75 in.

sy ⫽ 33 ksi

1 (b ⫺ b1)h21 6

M1 ⫽ syS1

h + h1 b 2

M2 ⫽ 4636 k-in.

(a) BENDING MOMENT

ELASTIC CORE S1 ⫽

M2 ⫽ Fa

S1 ⫽ 40.64 in.3

M ⫽ M1 + M2

M ⫽ 5977 k-in.

;

(b) PERCENT DUE TO ELASTIC CORE

M1 ⫽ 1341 k-in.

M1 ⫽ 22.4% M

PLASTIC FLANGES

;

F ⫽ force in one flange 1 F ⫽ syba b(h ⫺ h1) 2

F ⫽ 334.1 k

Problem 6.10-16 Solve the preceding problem for a box beam with dimensions h ⫽ 400 mm, h1 ⫽ 360 mm, b ⫽ 200 mm, and b1 ⫽ 160 mm, and with yield stress sY ⫽ 220 MPa.

Solution 6.10-16 h ⫽ 400 mm h1 ⫽ 360 mm

Hollow box beam b ⫽ 200 mm b1 ⫽ 160 mm

(a) BENDING MOMENT sY ⫽ 220 MPa

(see Figure 6-47, Example 6-9) ELASTIC CORE 1 S1 ⫽ (b ⫺ b1)h21 ⫽ 864 * 103 mm3 6 M1 ⫽ sYS1 ⫽ 190.1 kN # m PLASTIC FLANGES F ⫽ force in one flange 1 F ⫽ sYba b(h ⫺h1) ⫽ 880.0 kN 2 M2 ⫽ F a

h + h1 b ⫽ 334.4 kN # m 2

M ⫽ M1 ⫹ M2 ⫽ 524 kN⭈m

;

(b) PERCENT DUE TO ELASTIC CORE Percent ⫽

M1 (100) ⫽ 36% M

;

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567

Problem 6.10-17 A W 10 ⫻ 60 wide-flange beam is subjected to a bending moment M of such magnitude that the flanges yield but the web remains linearly elastic. (a) Calculate the magnitude of the moment M if the yield stress is sY ⫽ 36 ksi. (b) What percent of the moment M is produced by the elastic core?

Solution 6.10-17 W 10 ⫻ 60

(a) BENDING MOMENT

h ⫽ 10.2 in.

b ⫽ 10.1 in.

tw ⫽ 0.420 in.

tf ⫽ 0.680 in.

sy ⫽ 36 ksi

M1 ⫽ syS1

;

(b) PERCENT DUE TO ELASTIC CORE

ELASTIC CORE 1 S1 ⫽ (h ⫺ 2tf ) 2tw 6

M ⫽ 2551 k-in.

M ⫽ M1 + M2

S1 ⫽ 5.47 in.

3

M1 ⫽ 7.7% M

;

M1 ⫽ 196.9 k-in.

PLASTIC FLANGES F ⫽ force in one flange F ⫽ 247.2 k F ⫽ sybtf M2 ⫽ 2354 k-in. M2 ⫽ F(h ⫺ tf) y

Problem 6.10-18 A singly symmetric beam of T-section (see figure) has cross-sectional

dimensions b ⫽ 140 mm, a ⫽ 190.8 mm, tw ⫽ 6.99 mm, and tf ⫽ 11.2 mm. Calculate the plastic modulus Z and the shape factor f.

tw a z

O tf b

Solution 6.10-18 b ⫽ 140 mm

a ⫽ 190.8 mm

tw ⫽ 6.99 mm

tf ⫽ 11.2 mm

c2 ⫽

btf + atw

c1 ⫽ 149.98 mm

1 1 1 t c 3 + bc23 ⫺ (b ⫺ tw)(c2 ⫺ tf)3 3 w 1 3 3 Iz ⫽ 11.41 * 106 mm4 Iz ⫽

ELASTIC BENDING tf a a b btf + a + tfbatw 2 2

c1 ⫽ a + t f ⫺ c2

c2 ⫽ 52.02 mm

S⫽

Iz c1

S ⫽ 76.1 * 103 mm3

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PLASTIC BENDING A ⫽ btf + atw h2 ⫽

A ⫽ 2902 mm

2

A 2b

h2 ⫽ 10.4 mm

h1 ⫽ a + tf ⫺ h2 y2

⫺bar

⫽ h2 /2

h1 ⫽ 191.6 mm y2

⫺bar

⫽ 5.18 mm

1 1 (b ⫺ tw)(tf ⫺ h 2)2 + twh12 2 2 y1 bar ⫽ ⫺ A /2 y1 bar ⫽ 88.50 mm ⫺

Z⫽

A (y + y2 bar) ⫺ 2 1⫺bar

Z ⫽ 136 * 103 mm3 f⫽

Z S

;

f ⫽ 1.79

;

Problem 6.10-19 A wide-flange beam of unbalanced cross section has the dimensions shown in the figure. Determine the plastic moment MP if sY ⫽ 36 ksi.

y 10 in. 0.5 in. z

O 7 in. 0.5 in. 0.5 in. 5 in.

Solution 6.10-19 Unbalanced wide-flange beam NEUTRAL AXIS UNDER FULLY PLASTIC CONDITIONS A ⫽ h1tw + (b1 ⫺ tw)tf 2 from which we get h1 ⫽ 1.50 in. h2 ⫽ d ⫺ h1 ⫽ 8.50 in. PLASTIC MODULUS y1 ⫽ sY ⫽ 36 ksi

b1 ⫽ 10 in.

b2 ⫽ 5 in.

tw ⫽ 0.5 in.

d ⫽ 8 in.

d1 ⫽ 7 in.

tf ⫽ 0.5 in.

A ⫽ b1tf ⫹ b2tf ⫹ d1tw ⫽ 11.0 in.2

⫽

g y i Ai A/2 (h1/ 2)(tw)(h1) + (h1 ⫺ tf / 2)(b1 ⫺ tw)(tf) A/2

⫽ 1.182 in.

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SECTION 6.10 Elastoplastic Bending

y2 ⫽ ⫽

g yi Ai A /2

569

PLASTIC MOMENT MP ⫽ sy Z ⫽ 1120 k-in.

;

(h2/2)(tw)(h2) + (h2 ⫺ tf/2)(b2 ⫺ tw)(tf) A/2

⫽ 4.477 in. Z⫽

A (y + y2) ⫽ 31.12 in.3 2 1

Problem 6.10-20 Determine the plastic moment MP for a beam having

y

the cross section shown in the figure if sY ⫽ 210 MPa.

120 mm z

150 mm

O 250 mm

30 mm

Solution 6.10-20 Cross section of beam sY ⫽ 210 MPa

d2 ⫽ 150 mm

d1 ⫽ 120 mm

NEUTRAL AXIS FOR FULLY PLASTIC CONDITIONS Cross section is divided into two equal areas. A⫽

p [(150 mm) 2 ⫺ (120 mm) 2] 4 + (250 mm) (30 mm) ⫽ 13,862 mm2

A ⫽ 6931 mm2 2

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(h2)(30 mm) ⫽

A ⫽ 6931 mm2 2

h2 ⫽ 231.0 mm

(Dimensions are in millimeters) p Z ⫽ (h1 ⫺ 75)a b(d22 ⫺ d12) 4

h1 ⫽ 150 mm + 250 mm ⫺ h2 ⫽ 169.0 mm

+ ca

PLASTIC MODULUS

+ a

g yi Ai for upper half of cross section y1 ⫽ A/ 2 g yi Ai y2 ⫽ for lower half of cross section A/2 Z⫽

A ( y + y 2) ⫽ (g yi A i)upper + ( g yi Ai)lower 2 1

h1 ⫺ 150 b(30)(h1 ⫺150) d 2

h2 b(30)(h2) 2

⫽ 598,000 + 5,400 + 800,400 ⫽ 1404 * 103 mm3 PLASTIC MOMENT MP ⫽ sPZ ⫽ (210 MPa)(1404 ⫻ 103 mm3) ⫽ 295 kN # m

;

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7 Analysis of Stress and Strain Plane Stress 1200 psi

Problem 7.2-1 An element in plane stress is subjected to stresses sx ⫽ 4750 psi,

sy ⫽ 1200 psi, and txy ⫽ 950 psi, as shown in the figure. Determine the stresses acting on an element oriented at an angle u ⫽ 60˚ from the x axis, where the angle u is positive when counterclockwise. Show these stresses on a sketch of an element oriented at the angle u.

950 psi 4750 psi

Solution 7.2-1 sx ⫽ 4750 psi

sy ⫽ 1200 psi

txy ⫽ 950 psi

u ⫽ 60° sx1 ⫽

sx ⫺ sy

sx + sy + 2

sx1 ⫽ 2910 psi

2 ;

tx1y1 ⫽ ⫺

sx ⫺ sy 2

sin(2u) + txy cos(2u)

tx1y1 ⫽ ⫺2012 psi cos(2u) + txy sin(2u)

;

sy1 ⫽ sx + sy ⫺ sx1 sy1 ⫽ 3040 psi

Problem 7.2-2 Solve the preceding problem for an element in plane stress

subjected to stresses sx ⫽ 100 MPa, sy ⫽ 80 MPa, and txy ⫽ 28 MPa, as shown in the figure. Determine the stresses acting on an element oriented at an angle u ⫽ 30˚ from the x axis, where the angle u is positive when counterclockwise. Show these stresses on a sketch of an element oriented at the angle u.

;

80 MPa

28 MPa 100 MPa

571

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Analysis of Stress and Strain

Solution 7.2-2 sx ⫽ 100 MPa

sy ⫽ 80 MPa

txy ⫽ 28 MPa

u ⫽ 30° sx1 ⫽

sx ⫺ sy

sx + sy + 2

2

sx1 ⫽ 119.2 MPa

tx1y1 ⫽ ⫺

sx ⫺ sy 2

sin(2u) + txy cos(2u)

tx1y1 ⫽ 5.30 MPa cos (2u) + txy sin (2u)

;

sy1 ⫽ sx + sy ⫺ sx1 sy1 ⫽ 60.8 MPa

;

;

Problem 7.2-3 Solve Problem 7.2-1 for an element in plane stress subjected to

2300 psi

stresses sx ⫽ ⫺5700 psi, sy ⫽ ⫺2300 psi, and txy ⫽ 2500 psi, as shown in the figure. Determine the stresses acting on an element oriented at an angle u ⫽ 50˚ from the x axis, where the angle u is positive when counterclockwise. Show these stresses on a sketch of an element oriented at the angle u.

2500 psi 5700 psi

Solution 7.2-3 sx ⫽ ⫺5700 psi

sy ⫽ ⫺2300 psi

txy ⫽ 2500 psi

u ⫽ 50° sx1 ⫽

sx ⫺ sy

sx + sy + 2

sx1 ⫽ ⫺1243 psi

2

tx1y1 ⫽ ⫺

sx ⫺ sy 2

sin (2u) + txy cos (2u)

tx1y1 ⫽ 1240 psi cos (2u) + txy sin (2u)

;

;

sy1 ⫽ sx + sy ⫺ sx1 sy1 ⫽ ⫺6757 psi

;

Problem 7.2-4 The stresses acting on element A in the web of a train rail are found to be 40 MPa tension in the horizontal direction and 160 MPa compression in the vertical direction (see figure). Also, shear stresses of magnitude 54 MPa act in the directions shown. Determine the stresses acting on an element oriented at a counterclockwise angle of 52˚ from the horizontal. Show these stresses on a sketch of an element oriented at this angle.

160 MPa

A A Side View

40 MPa 54 MPa

Cross Section

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SECTION 7.2

Plane Stress

573

Solution 7.2-4 sx ⫽ 40 MPa

sy ⫽ ⫺160 MPa

txy ⫽ ⫺54 MPa

u ⫽ 52° sx1 ⫽

sx ⫺ sy

sx + sy + 2

2

sx1 ⫽ ⫺136.6 MPa

tx1y1 ⫽ ⫺

sx ⫺ sy 2

sin (2u) + txy cos (2u)

tx1y1 ⫽ ⫺84.0 MPa cos(2u) + txy sin(2u)

;

sy1 ⫽ sx + sy ⫺ sx1 sy1 ⫽ 16.6 MPa

;

;

Problem 7.2-5 Solve the preceding problem if the normal and shear stresses acting on element A are 6500 psi, 18,500 psi, and 3800 psi (in the directions shown in the figure). Determine the stresses acting on an element oriented at a counterclockwise angle of 30˚ from the horizontal. Show these stresses on a sketch of an element oriented at this angle.

18,500 psi

6500 psi

A A

3800 psi

Side View

Cross Section

Solution 7.2-5 sx ⫽ 6500 psi

sy ⫽⫺18500 psi

txy ⫽ ⫺3800 psi

u ⫽ 30° sx1 ⫽

sx ⫺ sy

sx + sy + 2

sx1 ⫽ ⫺3041 psi

2 ;

tx1y1 ⫽ ⫺

sx ⫺ sy 2

sin (2u) + txy cos (2u)

tx1y1 ⫽ ⫺12725 psi cos (2u) + txy sin (2u)

;

sy1 ⫽ sx + sy ⫺ sx1 sy1 ⫽ ⫺8959 psi

Problem 7.2-6 An element in plane stress from the fuselage of an airplane is subjected to compressive stresses of magnitude 27 MPa in the horizontal direction and tensile stresses of magnitude 5.5 MPa in the vertical direction (see figure). Also, shear stresses of magnitude 10.5 MPa act in the directions shown. Determine the stresses acting on an element oriented at a clockwise angle of 35˚ from the horizontal. Show these stresses on a sketch of an element oriented at this angle.

;

5.5 MPa

27 MPa 10.5 MPa

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Analysis of Stress and Strain

Solution 7.2-6 sx ⫽ ⫺27 MPa

sy ⫽ 5.5 MPa

txy ⫽ ⫺10.5 MPa

u ⫽ ⫺35° sx1 ⫽

sx ⫺ sy

sx + sy + 2

2

sx1 ⫽ ⫺6.4 MPa

tx1y1 ⫽ ⫺

sx ⫺ sy 2

sin (2u) + txy cos (2u)

tx1y1 ⫽ ⫺18.9 MPa cos (2u) + txy sin (2u)

;

sy1 ⫽ sx ⫹sy ⫺ sx1 sy1 ⫽ ⫺15.1 MPa

;

;

Problem 7.2-7 The stresses acting on element B in the web of a wide-flange beam are found to be 14,000 psi compression in the horizontal direction and 2600 psi compression in the vertical direction (see figure). Also, shear stresses of magnitude 3800 psi act in the directions shown. Determine the stresses acting on an element oriented at a counterclockwise angle of 40˚ from the horizontal. Show these stresses on a sketch of an element oriented at this angle.

2600 psi

B

14,000 psi

B

3800 psi Side View

Cross Section

Solution 7.2-7 sx ⫽ ⫺14000 psi txy ⫽ ⫺3800 psi

sy ⫽ ⫺2600 psi

u ⫽ 40° sx1 ⫽

tx1y1 ⫽ ⫺

sx ⫺ sy 2

tx1y1 ⫽ 4954 psi sx ⫺ sy

sx + sy + 2

sx1 ⫽ ⫺13032 psi

2

cos (2u) + txy sin (2u)

sin (2u) + txy cos (2u) ;

sy1 ⫽ sx + sy ⫺ sx1 sy1 ⫽ ⫺3568 psi

;

;

Problem 7.2-8 Solve the preceding problem if the normal and shear stresses acting on

13 MPa

element B are 46 MPa, 13 MPa, and 21 MPa (in the directions shown in the figure) and the angle is 42.5˚ (clockwise). 21 MPa B

46 MPa

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SECTION 7.2

Plane Stress

Solution 7.2-8 sx ⫽ ⫺46 MPa

sy ⫽ ⫺13 MPa

txy ⫽ 21 MPa

u ⫽ ⫺42.5° sx1 ⫽

sx ⫺ sy

sx + sy + 2

2

sx1 ⫽ ⫺51.9 MPa

tx1y1 ⫽ ⫺

;

;

y 112 psi 30°

350 psi O

x 120 psi

Seam

Plane stress (angle ␪ )

sx ⫽ 350 psi

sy ⫽112 psi

txy ⫽ ⫺120 psi

u ⫽ 30° sx ⫺ sy

sx + sy + 2

2

⫽ 187 psi tx1y1 ⫽ ⫺

sin (2u) + txy cos (2u)

sy1 ⫽ sx + sy ⫺ sx1 sy1 ⫽ ⫺7.1 MPa

;

pond is subjected to stresses sx ⫽ 350 psi, sy ⫽112 psi, and txy ⫽ ⫺120 psi, as shown by the plane-stress element in the first part of the figure. Determine the normal and shear stresses acting on a seam oriented at an angle of 30° to the element, as shown in the second part of the figure. Show these stresses on a sketch of an element having its sides parallel and perpendicular to the seam.

sx1 ⫽

2

tx1y1 ⫽ ⫺14.6 MPa cos (2u) + txy sin (2u)

Problem 7.2-9 The polyethylene liner of a settling

Solution 7.2-9

sx ⫺ sy

sx ⫺ sy 2

⫽ ⫺163 psi

cos 2u + txy sin 2u

; sin 2u + txy cos 2u ;

sy1 ⫽ sx + sy ⫺ sx1 ⫽ 275 psi

;

The normal stress on the seam equals 187 psi ; tension. The shear stress on the seam equals 163 psi, acting clockwise against the seam. ;

575

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Problem 7.2-10 Solve the preceding problem if the normal

y

and shear stresses acting on the element are sx ⫽ 2100 kPa, sy ⫽ 300 kPa, and txy ⫽ ⫺560 kPa, and the seam is oriented at an angle of 22.5° to the element (see figure).

300 kPa

22.5°

2100 kPa x

O

Seam

560 kPa

Solution 7.2-10

Plane stress (angle ␪ ) sx1 ⫽

sx ⫺sy

sx + sy + 2

2

⫽ 1440 kPa tx1 y1 ⫽ ⫺

cos 2u + txy sin 2u

;

sx ⫺ sy 2

⫽ ⫺1030 kPa

sin 2u + txy cos 2u ;

sy1 ⫽ sx + sy ⫺ sx1 ⫽ 960 kPa sx ⫽ 2100 kPa u ⫽ 22.5°

sy ⫽ 300 kPa

txy ⫽⫺560 kPa

;

The normal stress on the seam equals 1440 kPa tension. ; The shear stress on the seam equals 1030 kPa, acting clockwise against the seam. ;

Problem 7.2-11 A rectangular plate of dimensions 3.0 in. * 5.0 in. is formed by welding two triangular plates (see figure). The plate is subjected to a tensile stress of 500 psi in the long direction and a compressive stress of 350 psi in the short direction. Determine the normal stress sw acting perpendicular to the line of the weld and the shear tw acting parallel to the weld. (Assume that the normal stress sw is positive when it acts in tension against the weld and the shear stress tw is positive when it acts counterclockwise against the weld.)

350 psi

ld

We

3 in. 5 in.

500 psi

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SECTION 7.2

Solution 7.2-11

Plane Stress

577

Biaxial stress (welded joint) tx1y1 ⫽ ⫺

sx ⫺ sy 2

sin 2u + txy cos 2u ⫽ ⫺ 375 psi

sy1 ⫽ sx + sy ⫺ sx1 ⫽ ⫺125 psi STRESSES ACTING ON THE WELD

sx ⫽ 500 psi u ⫽ arctan sx1 ⫽

sy ⫽ ⫺350 psi

txy ⫽ 0 sw ⫽ ⫺125 psi

3 in. ⫽ arctan 0.6 ⫽ 30.96° 5 in. sx ⫺ sy

sx + sy + 2

2

tw ⫽ 375 psi

; ;

cos 2u + txy sin 2u

⫽ 275 psi 12.0 MPa

Problem 7.2-12 Solve the preceding problem for a plate of dimensions 100 mm * 250 mm subjected to a compressive stress of 2.5 MPa in the long direction and a tensile stress of 12.0 MPa in the short direction (see figure).

Solution 7.2-12

ld

We

100 mm 250 mm

Biaxial stress (welded joint) tx1y1 ⫽ ⫺

sx ⫺ sy 2

sin 2u + txy cos 2u ⫽ 5.0 MPa

sy1 ⫽ sx + sy ⫺ sx1 ⫽ 10.0 MPa STRESSES ACTING ON THE WELD

sx ⫽ ⫺2.5 MPa u ⫽ arctan sx1 ⫽

sy ⫽ 12.0 MPa

txy ⫽ 0

100 mm ⫽ arctan 0.4 ⫽ 21.80° 250 mm sx ⫺ sy

sx + sy + 2

⫽ ⫺0.5 MPa

2.5 MPa

2

cos 2u + txy sin 2u

sw ⫽ 10.0 MPa

;

tw ⫽ ⫺5.0 MPa

;

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Analysis of Stress and Strain

Problem 7.2-13 At a point on the surface of a machine the material

y

is in biaxial stress with sx ⫽ 3600 psi, and sy ⫽ ⫺1600 psi, as shown in the first part of the figure. The second part of the figure shows an inclined plane aa cut through the same point in the material but oriented at an angle u. Determine the value of the angle u between zero and 90° such that no normal stress acts on plane aa. Sketch a stress element having plane aa as one of its sides and show all stresses acting on the element.

1600 psi

a u

3600 psi O

x a

Solution 7.2-13

Biaxial stress STRESS ELEMENT sx1 ⫽ 0

u ⫽ 56.31°

sy1 ⫽ sx + sy ⫺ sx1 ⫽ 2000 psi

sx ⫽ 3600 psi sy ⫽ ⫺1600 psi txy ⫽ 0

tx1y1 ⫽ ⫺

sx ⫺ sy 2

;

sin 2u + txy cos 2u

⫽ ⫺2400 psi

Find angle u for s ⫽ 0. s ⫽ normal stress on plane a-a sx1 ⫽

sx ⫺ sy

sx + sy + 2

2

cos 2u + txy sin 2u

⫽ 1000 + 2600 cos 2u(psi) For sx1 ⫽ 0, we obtain cos 2u ⫽ ⫺ ‹

2u ⫽112.62° and

1000 2600

u ⫽ 56.31°

Problem 7.2-14 Solve the preceding problem for sx ⫽ 32 MPa

y

and sy ⫽ ⫺50 MPa (see figure).

50 MPa

a u

32 MPa O

x a

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SECTION 7.2

Solution 7.2-14

Plane Stress

579

Biaxial stress STRESS ELEMENT sx1 ⫽ 0

u ⫽ 38.66°

sy1 ⫽ sx + sy ⫺ sx1 ⫽ ⫺18 MPa

sx ⫽ 32 MPa

tx1y1 ⫽ ⫺

sy ⫽ ⫺50 MPa txy ⫽ 0

s x ⫺ sy 2

sin 2u + txy cos 2u

⫽ ⫺40 MPa

Find angles u for s ⫽ 0.

;

;

s ⫽ normal stress on plane a-a sx1 ⫽

sx ⫺ sy

sx + sy + 2

2

cos 2u + txy sin 2u

⫽ ⫺9 + 41 cos 2u ( MPa) For sx1 ⫽ 0, we obtain cos 2u ⫽ ‹

9 41

2u ⫽ 77.32° and u ⫽ 38.66°

;

Problem 7.2-15 An element in plane stress from the frame of a racing car is

y

oriented at a known angle u (see figure). On this inclined element, the normal and shear stresses have the magnitudes and directions shown in the figure. Determine the normal and shear stresses acting on an element whose sides are parallel to the xy axes, that is, determine sx, sy, and txy. Show the results on a sketch of an element oriented at u ⫽ 0˚.

2475 psi 3950 psi

u = 40° 14,900 psi O

Solution 7.2-15 Transform from u ⫽ 40° sx ⫽ ⫺14900 psi txy ⫽ 2475 psi

to

u ⫽ 0°

sy ⫽ ⫺3950 psi

sx ⫺ sy 2

sin (2u) + txy cos (2u)

tx1y1 ⫽ ⫺4962 psi

u ⫽ ⫺40° sx1 ⫽

tx1y1 ⫽ ⫺

;

sy1 ⫽ sx + sy ⫺ sx1 sx ⫺ sy

sx + sy + 2

sx1 ⫽ ⫺12813 psi

2 ;

cos (2u) + txy sin (2u)

sy1 ⫽ ⫺6037 psi

;

x

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Analysis of Stress and Strain

Problem 7.2-16 Solve the preceding problem for the element shown in the

y

figure.

24.3 MPa

62.5 MPa

u = 55°

O

x

24.0 MPa

Solution 7.2-16 Transform from u ⫽ 55° sx ⫽ ⫺24.3 MPa txy ⫽ ⫺24 MPa

to

u ⫽ 0°

sy ⫽ 62.5 MPa

sx ⫺ sy 2

sin (2u) + txy cos (2u)

tx1y1 ⫽ ⫺32.6 MPa

u ⫽ ⫺55° sx1 ⫽

tx1y1 ⫽ ⫺

;

sy1 ⫽ sx + sy ⫺ sx1 sx ⫺ sy

sx + sy + 2

2

sx1 ⫽ 56.5 MPa

cos (2u) + txy sin (2u)

sy1 ⫽ ⫺18.3 MPa

;

;

Problem 7.2-17 A plate in plane stress is subjected to normal stresses sx and sy and shear stress txy, as shown in the figure. At counterclockwise angles u ⫽ 35˚ and u ⫽ 75˚ from the x axis, the normal stress is 4800 psi tension. If the stress sx equals 2200 psi tension, what are the stresses sy and txy?

y sy txy O

sx = 2200 psi x

Solution 7.2-17 sx ⫽ 2200 psi At u ⫽35°

sy unknown

txy unknown

and u ⫽ 75°, sx1 ⫽ 4800 psi

Find sy and txy sx1 ⫽

sx ⫺ sy

sx + sy + 2

2

For

u ⫽ 35°

sx1 ⫽ 4800 psi 4800 psi ⫽

cos (2u) + txy sin (2u) or

2200 psi ⫺ sy

2200 psi + sy +

2 2 * cos (70°) + txy sin (70°)

0.32899 sy + 0.93969 txy ⫽ 3323.8 psi

(1)

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Plane Stress

581

0.93301sy + 0.50000 txy ⫽ 4652.6 psi

(2)

SECTION 7.2

For

u ⫽ 75°:

or

sx1 ⫽ 4800 psi 4800 psi ⫽

Solve Eqs. (1) and (2): 2200 psi ⫺ sy

2200 psi + sy

sy ⫽ 3805 psi

txy ⫽ 2205 psi

;

+ 2 2 * cos (150°) + txy sin (150°)

Problem 7.2-18 The surface of an airplane wing is subjected to plane stress

y

with normal stresses sx and sy and shear stress txy, as shown in the figure. At a counterclockwise angle u ⫽ 32˚ from the x axis, the normal stress is 37 MPa tension, and at an angle u ⫽ 48˚, it is 12 MPa compression. If the stress sx equals 110 MPa tension, what are the stresses sy and txy?

sy txy O

sx = 110 MPa x

Solution 7.2-18 sx ⫽ 110 MPa

sy unknown

At u ⫽ 32°, sx1 ⫽ 37 MPa

sxy unknown

(tension)

At u ⫽ 48°, sx1 ⫽ ⫺12 MPa

(compression)

Find sy and txy sx1 ⫽ For

sx ⫺sy

sx + sy + 2

2

cos(2u) + txy sin (2u)

u ⫽ 32°

37 MPa ⫽

For

0.28081sy + 0.89879txy ⫽ ⫺42.11041 MPa (1) u ⫽ 48°:

sx1 ⫽ ⫺12 MPa 110 MPa + sy 110 MPa ⫺ sy ⫺12 MPa ⫽ + 2 2 * cos (96°) + txy sin (96°) or

0.55226sy + 0.99452txy ⫽ ⫺61.25093 MPa (2)

Solve Eqs. (1) and (2):

sx1 ⫽ 37 MPa 110 MPa + sy

or

sy ⫽ ⫺60.7 MPa 110 MPa ⫺ sy

+ 2 2 * cos (64°) + txy sin (64°)

txy ⫽ ⫺27.9 MPa

;

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Analysis of Stress and Strain

Problem 7.2-19 At a point in a structure subjected to plane stress, the stresses are

y

sx ⫽ ⫺4100 psi, sy ⫽ 2200 psi, and txy ⫽ 2900 psi (the sign convention for these stresses is shown in Fig. 7-1). A stress element located at the same point in the structure (but oriented at a counterclockwise angle u1 with respect to the x axis) is subjected to the stresses shown in the figure (sb, tb, and 1800 psi). Assuming that the angle u1 is between zero and 90˚, calculate the normal stress sb, the shear stress tb, and the angle u1

Solution 7.2-19 sx ⫽ ⫺4100 psi txy ⫽ 2900 psi For

u ⫽ u1:

Find Stress

sy1 ⫽ sb

1800 psi u1 O

x

SOLVE NUMERICALLY:

tx1y1 ⫽ tb

2u1 ⫽ 87.32°

sb, tb, and u1 sb

u1 ⫽ 43.7°

;

Shear Stress tb

sb ⫽ sx + sy ⫺1800 psi

sb ⫽ ⫺3700 psi

; tb ⫽ ⫺

Angle u1 sx1 ⫽

tb

1800 psi ⫽ ⫺950 psi ⫺ 3150 psi cos12u12 + 2900 psi sin 12u12

sy ⫽ 2200 psi

sx1 ⫽ 1800 psi

sb

sx ⫺ sy

sx + sy + 2

2

cos ( 2u) + txy sin ( 2u)

sx ⫺ sy 2

sin 12u12 + txy cos 12u12

tb ⫽ 3282 psi

;

Principal Stresses and Maximum Shear Stresses When solving the problems for Section 7.3, consider only the in-plane stresses (the stresses in the xy plane).

Problem 7.3-1 An element in plane stress is subjected to stresses sx ⫽ 4750 psi, sy ⫽ 1200 psi, and txy ⫽ 950 psi (see the figure for Problem 7.2-1). Determine the principal stresses and show them on a sketch of a properly oriented element.

Solution 7.3-1 sx ⫽ 4750 psi

sy ⫽ 1200 psi

atana up1 ⫽

sx ⫺ sy

up1 ⫽ 14.08°

2

up2 ⫽ up1 + 90° s1 ⫽

PRINCIPAL STRESSES 2 txy

txy ⫽ 950 psi

b

s2 ⫽

up2 ⫽ 104.08° sx ⫺ sy

sx + sy + 2

2 sx ⫺ sy

sx + sy + 2

s1 ⫽ 4988 psi s2 ⫽ 962 psi

2 ; ;

cos 12up12 + txy sin 12up12

cos 12up22 + txy sin 12up22

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SECTION 7.3 Principal Stresses and Maximum Shear Stresses

583

Problem 7.3-2 An element in plane stress is subjected to stresses sx ⫽ 100 MPa, sy ⫽ 80 MPa, and txy ⫽ 28 MPa (see the figure for Problem 7.2-2). Determine the principal stresses and show them on a sketch of a properly oriented element.

Solution 7.3-2 sx ⫽ 100 MPa

sy ⫽ 80 MPa

txy ⫽ 28 MPa

s1 ⫽

PRINCIPAL STRESSES

up1 ⫽

atan a

2 txy sx ⫺ sy

s2 ⫽

b

+ 2

2 sx ⫺ sy

sx + sy + 2

2

s1 ⫽ 120 MPa

2

s2 ⫽ 60 MPa

up1 ⫽ 35.2° up2 ⫽ up1 + 90°

sx ⫺ sy

sx + sy

cos12up12 + txy sin12up12 cos12up22 + txy sin12up22

; ;

up2 ⫽ 125.17°

Problem 7.3-3 An element in plane stress is subjected to stresses sx ⫽ ⫺5700 psi, sy ⫽ ⫺2300 psi, and txy ⫽ 2500 psi (see the figure for Problem 7.2-3). Determine the principal stresses and show them on a sketch of a properly oriented element.

Solution 7.3-3 sx ⫽ ⫺5700 psi

sy ⫽ ⫺2300 psi

txy ⫽ 2500 psi

s1 ⫽

PRINCIPAL STRESSES

up2 ⫽

atan a

2txy sx ⫺ sy

s2 ⫽

b

+ 2

2 sx ⫺ sy

sx + sy + 2

s1 ⫽ ⫺ 977 psi

2

s2 ⫽ ⫺ 7023 psi

up2 ⫽ ⫺27.89° up1 ⫽ up2 + 90°

sx ⫺ sy

sx + sy

2

cos12up12 + txy sin12up12 cos12up22 + txy sin12up22

; ;

up1 ⫽ 62.1°

Problem 7.3-4 The stresses acting on element A in the web of a train rail are found to be 40 MPa tension in the horizontal direction and 160 MPa compression in the vertical direction (see figure). Also, shear stresses of magnitude 54 MPa act in the directions shown (see the figure for Problem 7.2-4). Determine the principal stresses and show them on a sketch of a properly oriented element.

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CHAPTER 7 Analysis of Stress and Strain

Solution 7.3-4 sx ⫽ 40 MPa

sy ⫽ ⫺160 MPa

txy ⫽ ⫺54 MPa

s1 ⫽

PRINCIPAL STRESSES

up1 ⫽

atan a

2txy sx ⫺ sy

s2 ⫽

b

sx ⫺ sy

sx + sy + 2

2 sx ⫺ sy

sx + sy + 2

2

s1 ⫽ 53.6 MPa

2

s2 ⫽ ⫺ 173.6 MPa

up1 ⫽ ⫺ 14.2° up2 ⫽ up1 + 90°

cos 12up12 + txy sin 12up12 cos 12up22 + txy sin 12up22

; ;

up2 ⫽ 75.8°

Problem 7.3-5 The normal and shear stresses acting on element A are 6500 psi, 18,500 psi, and 3800 psi (in the directions shown in the figure) (see the figure for Problem 7.2-5). Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element.

Solution 7.3-5 sx ⫽ 6500 psi

sy ⫽ ⫺18500 psi

txy ⫽ ⫺3800 psi

s2 ⫽ ⫺19065 psi

PRINCIPAL ANGLES

up1 ⫽

atan a

2txy sx ⫺ sy

MAXIMUM SHEAR STRESSES

b

tmax ⫽

2

up1 ⫽ ⫺8.45°

s2 ⫽

up2 ⫽ 81.55° sx ⫺ sy

sx + sy + 2

2 sx ⫺ sy

sx + sy + 2

sx ⫺ sy 2 a b + txy2 A 2

us1 ⫽ up1 ⫺ 45°

up2 ⫽ up1 + 90° s1 ⫽

s1 ⫽ 7065 psi

2

cos 12up12 + txy sin 12up12

saver ⫽

sx + sy 2

tmax ⫽ 13065 psi

us1 ⫽ ⫺53.4°

;

;

saver ⫽ ⫺6000 psi

;

cos 12up22 + txy sin 12up22

Problem 7.3-6 An element in plane stress from the fuselage of an airplane is subjected to compressive stresses of magnitude 27 MPa in the horizontal direction and tensile stresses of magnitude 5.5 MPa in the vertical direction. Also, shear stresses of magnitude 10.5 MPa act in the directions shown (see the figure for Problem 7.2-6). Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element.

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SECTION 7.3 Principal Stresses and Maximum Shear Stresses

585

Solution 7.3-6 sx ⫽ ⫺27 MPa

sy ⫽ 5.5 MPa

txy ⫽ ⫺10.5 MPa

s2 ⫽ ⫺30.1 MPa

PRINCIPAL ANGLES 2txy b atan a sx ⫺ sy up2 ⫽ 2

MAXIMUM SHEAR STRESSES tmax ⫽

up2 ⫽ 16.43° up1 ⫽ up2 + 90° s1 ⫽ s2 ⫽

up1 ⫽ 106.43° sx ⫺ sy

sx + sy + 2

2 sx ⫺ sy

sx + sy + 2

s1 ⫽ 8.6 MPa

2

cos 12up12 + txy sin 12up12

sx ⫺ sy 2 b + txy2 A 2 a

tmax ⫽ 19.3 MPa

;

us1 ⫽ up1 ⫺ 45° saver ⫽

us1 ⫽ 61.4°

sx + sy

saver ⫽ ⫺10.8 MPa

2

;

cos 12up22 + txy sin 12up22

Problem 7.3-7 The stresses acting on element B in the web of a wide-flange beam are found to be 14,000 psi compression in the horizontal direction and 2600 psi compression in the vertical direction. Also, shear stresses of magnitude 3800 psi act in the directions shown (see the figure for Problem 7.2-7). Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element.

Solution 7.3-7 sx ⫽ ⫺14000 psi txy ⫽ ⫺3800 psi

sy ⫽ ⫺2600 psi

up2 ⫽

2txy sx ⫺ sy

b

s1 ⫽

tmax ⫽ up1 ⫽ 106.85° sx ⫺ sy +

2

2

2

cos12up22 + txy sin12up22

MAXIMUM SHEAR STRESSES

up2 ⫽ 16.85° sx + sy

+

s2 ⫽ ⫺15151 psi

2

up1 ⫽ up2 + 90°

sx ⫺ sy

sx + sy

s1 ⫽ ⫺1449 psi

PRINCIPAL ANGLES atan a

s2 ⫽

2

cos12up12 + txy sin12up12

sx ⫺ sy 2 a b + txy2 A 2

us1 ⫽ up1 ⫺ 45° saver ⫽

sx + sy 2

tmax ⫽ 6851 psi

us1 ⫽ 61.8° saver ⫽ ⫺ 8300 psi

;

; ;

Problem 7.3-8 The normal and shear stresses acting on element B are sx ⫽ ⫺46 MPa, sy ⫽ ⫺13 MPa, and txy ⫽ 21 MPa (see figure for Problem 7.2-8). Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element.

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Analysis of Stress and Strain

Solution 7.3-8 sx ⫽ ⫺46 MPa

sy ⫽ ⫺13 MPa

txy ⫽ 21 MPa

s2 ⫽ ⫺56.2 MPa

PRINCIPAL ANGLES

up2 ⫽

atan a

2txy sx ⫺ sy

MAXIMUM SHEAR STRESSES

b

tmax ⫽

2

up1 ⫽ up2 + 90° s1 ⫽ s2 ⫽

up1 ⫽ 64.08° sx ⫺ sy

+ 2

2 sx ⫺ sy

sx + sy

a

sx ⫺ sy

A

2

us1 ⫽ up1 ⫺ 45°

up2 ⫽ ⫺25.92° sx + sy

s1 ⫽ ⫺2.8 MPa

+ 2

2

cos 12up12 + txy sin 12up12

saver ⫽

sx ⫺ sy 2

2

b + txy2

tmax ⫽ 26.7 MPa

us1 ⫽ 19.08°

;

;

saver ⫽ ⫺29.5 MPa

;

cos 12up22 + txy sin 12up22

Problem 7.3-9 A shear wall in a reinforced concrete building is subjected to a vertical uniform load of intensity q and a horizontal force H, as shown in the first part of the figure. (The force H represents the effects of wind and earthquake loads.) As a consequence of these loads, the stresses at point A on the surface of the wall have the values shown in the second part of the figure (compressive stress equal to 1100 psi and shear stress equal to 480 psi).

q

1100 psi H 480 psi

A A

(a) Determine the principal stresses and show them on a sketch of a properly oriented element. (b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element.

Solution 7.3-9 sx ⫽ 0

Shear wall

sy ⫽ ⫺1100 psi

txy ⫽ ⫺480 psi

2txy sx ⫺ sy

⫽ ⫺0.87273

2up ⫽ ⫺41.11° and up ⫽ ⫺20.56° 2up ⫽ 138.89° and up ⫽ 69.44° sx1 ⫽

f

s2 ⫽ ⫺1280 psi and up2 ⫽ 69.44°

(a) PRINCIPAL STRESSES tan 2up ⫽

Therefore, s1 ⫽180 psi and up1 ⫽ ⫺20.56°

sx ⫺ sy

sx + sy + 2

2

cos 2u + txy sin 2u

For 2up ⫽ ⫺41.11°:

sx1 ⫽ 180 psi

For 2up ⫽ 138.89°:

sx1 ⫽ ⫺1280 psi

;

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587

Principal Stresses and Maximum Shear Stresses

(b) MAXIMUM SHEAR STRESSES tmax ⫽

sx ⫺ sy 2 b + t2xy ⫽ 730 psi A 2 a

us1 ⫽ up1⫺ 45° ⫽ ⫺ 65.56° and t ⫽ 730 psi us2 ⫽ up1+ 45° ⫽ 24.44° saver ⫽

sx + sy

and t ⫽ ⫺730 psi

⫽ ⫺550 psi

2

f

;

;

Problem 7.3-10 A propeller shaft subjected to combined torsion and axial thrust is designed to resist a shear stress of 56 MPa and a compressive stress of 85 MPa (see figure). (a) Determine the principal stresses and show them on a sketch of a properly oriented element. (b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element.

85 MPa

56 MPa

Solution 7.3-10 sx ⫽ ⫺85 MPa

sy ⫽ 0 MPa

Txy ⫽ ⫺56 MPa

up2 ⫽

2txy sx ⫺ sy

tmax ⫽

s2 ⫽

up1 ⫽ 116.4° sx ⫺ sy

sx + sy + 2

2 sx ⫺ sy

sx + sy + 2

a

sx ⫺ sy

A 2 tmax ⫽ 70.3 MPa

2

up1 ⫽ up2 + 90°

;

(b) MAXIMUM SHEAR STRESSES

b

up2 ⫽ 26.4°

s1 ⫽

;

s2 ⫽ ⫺112.8 MPa

(a) PRINCIPAL STRESSES atan a

s1 ⫽ 27.8 MPa

2

;

cos 12up12 + txy sin 12up12 cos 12up22 + txy sin 12up22

us1 ⫽ up1 ⫺ 45° saver ⫽

sx + sy 2

2

b + txy2 ; us1 ⫽ 71.4°

;

saver ⫽ ⫺42.5 MPa

;

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y

Problems 7.3-11 sx ⫽ 2500 psi, sy ⫽ 1020 psi, txy ⫽ ⫺900 psi (a) Determine the principal stresses and show them on a sketch of a properly oriented element. (b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element.

sy txy sx O

x

Probs. 7.3-11 through 7.3-16 Solution 7.3-11 sx ⫽ 2500 psi

sy ⫽ 1020 psi

txy ⫽ ⫺900 psi

(a) PRINCIPAL STRESSES

tan(2up) ⫽

2txy

up1 ⫽

sx ⫺ sy

atan a

2txy sx ⫺ sy

b

s1 ⫽

2

s2 ⫽

+ 2

2 sx ⫺ sy

sx + sy + 2

2

s1 ⫽ 2925 psi

For up2 ⫽ 64.7°:

s2 ⫽ 595 psi

tmax ⫽

up2 ⫽ 64.71° sx ⫺ sy

sx + sy

For up1 ⫽ ⫺25.3°:

; ;

(b) MAXIMUM SHEAR STRESSES

up1 ⫽ 25.29° up2 ⫽ 90 ° + up1

Therefore,

A

a

sx ⫺ sy 2 b + txy2 2

tmax ⫽ 1165 psi us1 ⫽ ⫺70.3° and us1 ⫽ up1 ⫺ 45° ; t1 ⫽ 1165 psi

cos 12up12 + txy sin 12up12

us2 ⫽ up1 ⫺ 45° t2 ⫽ ⫺1165 psi

cos 12up22 + txy sin 12up22

saver ⫽

sx + sy

us2 ⫽ 19.71° and ; saver ⫽ 1760 psi

2

;

Problems 7.3-12 sx ⫽ 2150 kPa, sy ⫽ 375 kPa, txy ⫽ ⫺460 kPa (a) Determine the principal stresses and show them on a sketch of a properly oriented element. (b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element.

Solution 7.3-12 sx ⫽ 2150 kPa

sy ⫽ 375 kPa

txy ⫽ ⫺460 kPa

up1 ⫽ ⫺13.70° up2 ⫽ 90° + up1

(a) PRINCIPAL STRESSES

tan(2up) ⫽

2txy sx ⫺ sy

up1 ⫽

atan a

2txy sx ⫺sy 2

b

s1 ⫽ s2 ⫽

sx ⫺ sy

sx + sy + 2

2 sx ⫺ sy

sx + sy + 2

2

up2 ⫽ 76.30°

cos 12up12 + txy sin 12up12 cos 12up22 + txy sin 12up22

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SECTION 7.3 Principal Stresses and Maximum Shear Stresses

tmax ⫽ 145 psi

Therefore, For up1 ⫽ ⫺ 13.70°

s1 ⫽ 2262 kPa

For up2 ⫽ 76.3°

s2 ⫽ 263 kPa

us1 ⫽ ⫺ 58.7° us1 ⫽ up1 ⫺ 45° and t1 ⫽ 1000 kPa

; ;

us2 ⫽ up1 + 45° us2 ⫽ 31.3° ; and t2 ⫽⫺1000 kPa sx + sy saver ⫽ saver ⫽ 1263 kPa 2

(b) MAXIMUM SHEAR STRESSES tmax ⫽

Problems 7.3-13

A

a

sx ⫺ sy 2

;

2

b + txy2

sx ⫽ 14,500 psi, sy ⫽ 1070 psi, txy ⫽ 1900 psi

(a) Determine the principal stresses and show them on a sketch of a properly oriented element. (b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element.

Solution 7.3-13 sx ⫽ 14500 psi

sy ⫽ 1070 psi

txy ⫽ 1900 psi

(a) PRINCIPAL STRESSES

tan(2up) ⫽

2txy

up1 ⫽

sx ⫺ sy

atan a

2txy sx ⫺sy

b

2

up1 ⫽ 7.90° up2 ⫽ 90° + up1 s1 ⫽ s2 ⫽

up2 ⫽ 97.90° sx ⫺ sy

sx + sy + 2

2

sx + sy

sx ⫺ sy +

2

2

cos 12up12 + txy sin 12up12 cos 12up22 + txy sin 12up22

Therefore, For up1 ⫽ 7.90°

s1 ⫽ 14764 psi

;

For up2 ⫽ 97.9°

s2 ⫽ 806 psi

;

(b) MAXIMUM SHEAR STRESSES tmax ⫽

sx ⫺ sy 2

2

b + txy2

tmax ⫽ 6979 psi us1 ⫽ u p1 ⫺ 45° and t1 ⫽ 6979 psi

us1 ⫽ ⫺37.1°

us2 ⫽ 52.9° us2 ⫽ up1 + 45° and t2 ⫽⫺6979 psi saver ⫽

Problems 7.3-14

A

a

sx + sy 2

; ;

saver ⫽ 7785 psi

sx ⫽ 16.5 MPa, sv ⫽ ⫺91 MPa, txy ⫽ ⫺39 MPa

(a) Determine the principal stresses and show them on a sketch of a properly oriented element. (b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element.

589

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Solution 7.3-14 sx ⫽ 16.5 MPa

sy ⫽ ⫺91 MPa

txy ⫽ ⫺39 MPa

(b) MAXIMUM SHEAR STRESSES

(a) PRINCIPAL STRESSES 2txy

tan(2up) ⫽

up1 ⫽

sx ⫺ sy

atan a

2txy sx ⫺sy

tmax ⫽

b

s1 ⫽

us1 ⫽ ⫺63.0° us1 ⫽ up1 ⫺ 45° and t1 ⫽ 66.4 MPa

2

s2 ⫽

+ 2

2 sx ⫺ sy

sx + sy + 2

us2 ⫽ 27.0° us2 ⫽ up1 + 45° and t2 ⫽ ⫺ 66.4 MPa

up2 ⫽72.02°

sx ⫺ sy

sx + sy

2

2

2

b + txy2

tmax ⫽ 9631.7 psi

up1 ⫽ ⫺ 17.98° up2 ⫽ 90° + up1

A

sx ⫺ sy

a

cos 12up12 + txy sin 12up12

saver ⫽

sx + sy

; ;

saver ⫽ ⫺37.3 MPa

2

cos 12up22 + txy sin 12up22

Therefore, For up1 ⫽ ⫺ 17.98° For up2 ⫽ 72.0°

Problems 7.3-15

s1 ⫽ 29.2 MPa s2 ⫽ ⫺ 103.7 MPa

sx ⫽ ⫺3300 psi, sy ⫽ ⫺11,000 psi, txy ⫽ 4500 psi

(a) Determine the principal stresses and show them on a sketch of a properly oriented element. (b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element.

Solution 7.3-15 sy ⫽⫺11000 psi

s ⫽ ⫺3300 psi

txy ⫽ 4500 psi

(a) PRINCIPAL STRESSES

tan(2up) ⫽

2txy sx ⫺ sy

up1 ⫽

atan a

2txy sx ⫺sy

b

2

up1 ⫽ 24.73° up2 ⫽ 90° + up1 s1 ⫽ s2 ⫽

up2 ⫽ 114.73°

sx ⫺ sy

sx + sy + 2

2

sx + sy

sx ⫺ sy +

2

2

cos 12up12 + txy sin 12up12 cos 12up22 + txy sin 12up22

Therefore, For up1 ⫽ 24.7°

s1 ⫽ ⫺1228 psi

For up2 ⫽ 114.7°

s2 ⫽ ⫺13072 psi

(b) MAXIMUM SHEAR STRESSES tmax ⫽

A

a

sx ⫺ sy 2

2

b + txy2

tmax ⫽ 5922 psi us1 ⫽ up1 ⫺ 45° and t1 ⫽ 5922 psi

us1 ⫽ ⫺20.3°

us2 ⫽ up1 + 45° and t2 ⫽ ⫺5922 psi

us2 ⫽ 69.7°

saver ⫽

sx + sy 2

; ;

saver ⫽ ⫺7150 psi

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Principal Stresses and Maximum Shear Stresses

591

Problems 7.3-16 sx ⫽ ⫺108 MPa, sy ⫽ 58 MPa, txy ⫽ ⫺58 MPa (a) Determine the principal stresses and show them on a sketch of a properly oriented element. (b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element.

Solution 7.3-16 sx ⫽ ⫺108 MPa

sy ⫽ 58 MPa

txy ⫽ ⫺58 MPa

(a) PRINCIPAL STRESSES

tan(2up) ⫽

atana

2txy

up2 ⫽

sx ⫺ sy

tmax ⫽

2txy sx ⫺sy

b

s1 ⫽ s2 ⫽

up1 ⫽ 107.47° sx ⫺ sy

sx + sy + 2

2 sx ⫺ sy

sx + sy + 2

2

A

a

sx ⫺ sy 2

2

b + txy2

tmax ⫽ 14686.1 psi us1 ⫽ 62.47° us1 ⫽ up1 ⫺ 45° and t1 ⫽ 101.3 MPa

2

up2 ⫽ 17.47° up1 ⫽ 90° + up2

(b) MAXIMUM SHEAR STRESSES

cos 12up12 + txy sin 12up12

;

us2 ⫽ 152.47° ; us2 ⫽ up1 + 45° and t2 ⫽⫺101.3 MPa sx + sy saver ⫽ saver ⫽ ⫺25.0 MPa 2

cos 12up22 + txy sin 12up22

Therefore, For up1 ⫽ 107.47°

s1 ⫽ 76.3 MPa

For up2 ⫽ 17.47°

s2 ⫽ ⫺126.3 MPa

; ;

Problem 7.3-17 At a point on the surface of a machine component, the stresses

y

acting on the x face of a stress element are sx ⫽ 5900 psi and txy ⫽ 1950 psi (see figure). What is the allowable range of values for the stress sy if the maximum shear stress is limited to t0 ⫽ 2500 psi?

sy txy = 1950 psi O

sx = 5900 psi x

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Solution 7.3-17 sx ⫽ 5900 psi

sy

unknown

txy ⫽ 1950 psi

Therefore, 2771 psi … sy … 9029 psi

Find the allowable range of values for sy if the

From Eq. (1):

maximum allowable shear stresses is tmax ⫽ 2500 psi sx ⫺ sy 2 (1) tmax ⫽ a b + txy2 A 2

t max (sy1) ⫽

A

a

sx ⫺ sy1 2

2

b + txy2

Solve for sy sy ⫽ sx ⫹

J

22tmax 2 ⫺txy2 ⫺ a22tmax2 ⫺txy2 b

K

sy ⫽ a

9029 b psi 2771 2.771 ksi

9.029 ksi

tmax(σy1) 2.5 ksi

σy1

Problem 7.3-18 At a point on the surface of a machine component the stresses

y

acting on the x face of a stress element are sx ⫽ 42 MPa and txy ⫽ 33 MPa (see figure). What is the allowable range of values for the stress sy if the maximum shear stress is limited to t0 ⫽ 35 MPa?

sy txy = 33 MPa O

sx = 42 MPa x

Solution 7.3-18 sx ⫽ 42 MPa

sy

unknown

txy ⫽ 33 MPa

Find the allowable range of values for sy if the maximum allowable shear stresses is tmax ⫽ 35 MPa

tmax ⫽

A

a

sx ⫺ sy 2

2

b + txy2

(1)

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SECTION 7.3

Solve for sy sy ⫽ sx ⫹

593

Principal Stresses and Maximum Shear Stresses

From Eq. (1): 22tmax 2 ⫺txy2

2 2 J ⫺ a22tmax ⫺txy b K

65.3 b MPa sy ⫽ a 18.7

tmax (sy1) ⫽

A

a

sx ⫺ sy1 2

2

b + txy2

Therefore, 18.7 MPa … sy … 65.3 MPa 65.3 MPa

18.7 MPa

35 MPa tmax (σy1)

σy1

An element in plane stress is subjected to stresses sx ⫽ 5700 psi and txy ⫽ ⫺2300 psi (see figure). It is known that one of the principal stresses equals 6700 psi in tension.

Problem 7.3-19

y sy

(a) Determine the stress sy. (b) Determine the other principal stress and the orientation of the principal planes, then show the principal stresses on a sketch of a properly oriented element.

5700 psi O

x 2300 psi

Solution 7.3-19 sx ⫽ 5700 psi

sy unknown txy ⫽ ⫺2300 psi

(a) STRESS sy

s1 ⫽ 6700 psi s1 ⫽

2

;

(b) PRINCIPAL STRESSES

Because sy is smaller than a given principal stress, we know that the given stress is the larger principal stress. sx + sy

sy ⫽ 1410 psi

Solve for sy

sx ⫺ sy 2 + a b + txy2 A 2

tan (2up) ⫽

2txy sx ⫺ sy

atana up1 ⫽

up1 ⫽ ⫺23.50° up2 ⫽ 90° + up1

up2 ⫽ 66.50°

2txy sx ⫺sy 2

b

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Analysis of Stress and Strain

s1 ⫽

s2 ⫽

sx ⫺ sy

sx + sy +

2 2 + txy sin12up12

For up1 ⫽ ⫺23.5° : s1 ⫽ 6700 psi For up2 ⫽ 66.5°

sx ⫺ sy

sx + sy

Therefore,

cos 12up12

+ 2 2 + txy sin 12up22

:

s2 ⫽ 410 psi

; ;

cos 12up22

Problem 7.3-20 An element in plane stress is subjected to stresses sx ⫽ ⫺50 MPa and txy ⫽ 42 MPa (see figure). It is known that one of the principal stresses equals 33 MPa in tension.

y sy

(a) Determine the stress sy. (b) Determine the other principal stress and the orientation of the principal planes, then show the principal stresses on a sketch of a properly oriented element.

42 MPa 50 MPa O

x

Solution 7.3-20 sx ⫽ ⫺50 MPa

sy unknown

txy ⫽ 42 MPa

up2 ⫽ ⫺26.85° up1 ⫽ 90° + up2

(a) STRESS sy

s1 ⫽

Because sy is smaller than a given principal stress, we know that the given stress is the larger principal stress. s1 ⫽ 33 MPa s1 ⫽

sx + sy +

A

2

Solve for sy

a

s x ⫺ sy 2 2

b +

s2 ⫽ txy2

sy ⫽ 11.7 MPa

;

tan(2up) ⫽

2txy

For up1 ⫽ 63.2°

sx ⫺ sy

up2 ⫽

atan a

sx ⫺sy 2

sx ⫺ sy

sx + sy + 2

2 + txy sin 12up12 sx ⫺ sy

sx + sy +

2 2 + txy sin 12up22

cos 12up12

cos 12up22

Therefore,

(b) PRINCIPAL STRESSES 2txy

up1 ⫽ 63.15°

b

:

s1 ⫽ 33.0 MPa

For up2 ⫽ ⫺26.8° : s 2 ⫽ ⫺71.3 MPa

; ;

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SECTION 7.4

595

Mohr’s Circle

Mohr’s Circle y

The problems for Section 7.4 are to be solved using Mohr’s circle. Consider only the in-plane stresses (the stresses in the xy plane).

Problem 7.4-1 An element in uniaxial stress is subjected to tensile stresses sx ⫽ 11,375 psi, as shown in the figure. Using Mohr’s circle, determine:

11,375 psi O

(a) The stresses acting on an element oriented at a counterclockwise angle u ⫽ 24° from the x axis. (b) The maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.

x

Solution 7.4-1 sx ⫽ 11375 psi sy ⫽ 0 psi (a) ELEMENT AT

u ⫽ 24°

2u ⫽ 48° R ⫽

sx 2

txy ⫽ 0 psi ;

sy1 ⫽ 1882 psi (b) MAXIMUM SHEAR STRESSES

R ⫽ 5688 psi

Point C: sc ⫽ R sc ⫽ 5688 psi Point D: sx1 ⫽ R + R cos(2u) sx1 ⫽ 9493 psi

;

;

tx1y1 ⫽ ⫺R sin (2u) tx1y1 ⫽ ⫺4227 psi

Point S1: us1 ⫽ tmax ⫽ R

us1 ⫽⫺45°

tmax ⫽ 5688 psi

Point S2: us2 ⫽ tmax ⫽ ⫺R

;

⫺ 90° 2

saver ⫽ R

90° 2

;

;

us2 ⫽ 45°

;

tmax ⫽ ⫺5688 psi

;

saver ⫽ 5688 psi

;

œ

PointD : sy1 ⫽ R ⫺ R cos (2u)

y

Problem 7.4-2 An element in uniaxial stress is subjected to tensile stresses sx ⫽ 49 MPa, as shown in the figure Using Mohr’s circle, determine:

(a) The stresses acting on an element oriented at an angle u ⫽ ⫺27° from the x axis (minus means clockwise). (b) The maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.

49 MPa O

Solution 7.4-2 sx ⫽ 49 MPa sy ⫽ 0 MPa (a) ELEMENT AT

u ⫽ ⫺27°

2u ⫽ ⫺54.0° R ⫽ Point C:

txy ⫽ 0 MPa

sc ⫽ R

sx 2

R ⫽ 24.5 MPa

sc ⫽ 24.5 MPa

Point D:

sx1 ⫽ R + R cos (|2u|) sx1 ⫽ 38.9 MPa ; tx1y1 ⫽ ⫺R sin (2u) tx1y1 ⫽ 19.8 MPa ; œ

Point D sy1 ⫽ R ⫺ R cos ( |2u| ) ; sy1 ⫽ 10.1 MPa

x

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(b) MAXIMUM SHEAR STRESSES Point S1: tmax ⫽ R

90° 2

Point S2: us2 ⫽

⫺90° us1 ⫽ 2

tmax ⫽ ⫺R

us1 ⫽ ⫺45.0° ; tmax ⫽ 24.5 MPa

saver ⫽ R

us2 ⫽ 45.0°

;

tmax ⫽ ⫺24.5 MPa saver ⫽ 24.5 MPa

; ;

;

Problem 7.4-3 An element in uniaxial stress is subjected to compressive stresses

y

of magnitude 6100 psi, as shown in the figure. Using Mohr’s circle, determine: 1

(a) The stresses acting on an element oriented at a slope of 1 on 2 (see figure). (b) The maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.

2 O 6100 psi

x

Solution 7.4-3 sx ⫽ ⫺6100 psi sy ⫽ 0 psi

txy ⫽ 0 psi

œ

Point D : sy1 ⫽ R ⫺ R cos (2u) sy1 ⫽ ⫺1220 psi

(a) ELEMENT AT A SLOPE OF 1 ON 2 1 u ⫽ atan a b 2

u ⫽ 26.565°

Point S1:

sx 2u ⫽ 53.130° R ⫽ 2 Point C: Point D:

sc ⫽ R

(b) MAXIMUM SHEAR STRESSES ;

R ⫽ ⫺3050 psi

tmax ⫽ ⫺R

sc ⫽ ⫺3050 psi Point S2:

sx1 ⫽ R + R cos (2u) sx1 ⫽ ⫺4880 psi

tx1y1 ⫽ ⫺R sin (2u)

;

tx1y1 ⫽ 2440 psi

us1 ⫽

;

90° 2

us1 ⫽ 45°

tmax ⫽ 3050 psi

us2 ⫽

⫺ 90° 2

; ;

us2 ⫽ ⫺45°

tmax ⫽ R

tmax ⫽ ⫺3050 psi

saver ⫽ R

saver ⫽ ⫺3050 psi

;

; ;

Problem 7.4-4 An element in biaxial stress is subjected to stresses sx ⫽ ⫺48 MPa and sy ⫽ 19 MPa, as shown in the figure. Using Mohr’s circle, determine:

y

(a) The stresses acting on an element oriented at a counterclockwise angle u ⫽ 25° from the x axis. (b) The maximum shear stresses and associated normal stresses.

19 MPa

Show all results on sketches of properly oriented elements. 48 MPa O

x

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SECTION 7.4

597

Mohr’s Circle

Solution 7.4-4 sx ⫽ ⫺48 MPa (a) ELEMENT AT

sy ⫽ 19 MPa u ⫽ 25°

2u ⫽ 50.0 deg R ⫽

txy ⫽ 0 MPa

Point S1: us1 ⫽

;

|sx| + |sy| 2

Point C: sx ⫽ sx + R

(b) MAXIMUM SHEAR STRESSES

us1 ⫽ 45.0°

R ⫽ 33.5 MPa

sc ⫽ ⫺14.5 MPa

tmax ⫽ R

sx1 ⫽ ⫺36.0 MPa

;

tmax ⫽ ⫺R ;

;

tmax ⫽ ⫺33.5 MPa

saver ⫽ sc

Point D œ : sy1 ⫽ sc + R cos(2u)

;

⫺ 90° 2

us2 ⫽ ⫺45.0°

tx1y1 ⫽ ⫺R sin(2u) tx1y1 ⫽ 25.7 MPa

;

tmax ⫽ 33.5 MPa

Point S2: us2 ⫽

Point D: sx1 ⫽ sc ⫺ R cos(2u)

90° 2

saver ⫽ ⫺14.5 MPa

; ;

sy1 ⫽ 7.0 MPa

Problem 7.4-5 An element in biaxial stress is subjected to stresses sx ⫽ 6250 psi

y

and sy ⫽ ⫺1750 psi, as shown in the figure. Using Mohr’s circle, determine:

(a) The stresses acting on an element oriented at a counterclockwise angle u ⫽ 55° from the x axis. (b) The maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.

1750 psi

6250 psi O

x

Solution 7.4-5 sx ⫽ 6250 psi (a) ELEMENT AT

sy ⫽ ⫺1750 psi

txy ⫽ 0 psi

u ⫽ 60°

2u ⫽ 120° R ⫽

|sx| + |sy| 2

Point C: sc ⫽ sx ⫺ R

Point S1: us1 ⫽ R ⫽ 4000 psi

sc ⫽ 2250 psi

Point D: sx1 ⫽ sc + R cos(2u) sx1 ⫽ 250 psi

(b) MAXIMUM SHEAR STRESSES

us1 ⫽ ⫺45° tmax ⫽ R

;

tx1y1 ⫽ ⫺3464 psi Point D œ : sy1 ⫽ sc ⫺ R cos(2u)

; sy1 ⫽ 4250 psi

;

tmax ⫽ 4000 psi

Point S2: us2 ⫽

tx1y1 ⫽ ⫺R sin (2u)

⫺ 90° 2

90° 2

;

us2 ⫽ 45°

;

tmax ⫽ R

tmax ⫽ 4000 psi

;

saver ⫽ sc

saver ⫽ 2250 psi

;

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Analysis of Stress and Strain

y

Problem 7.4-6 An element in biaxial stress is subjected to stresses sx ⫽ ⫺29 MPa and sy ⫽ 57 MPa, as shown in the figure. Using Mohr’s circle, determine:

57 MPa

(a) The stresses acting on an element oriented at a slope of 1 on 2.5 (see figure). (b) The maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.

1 2.5 29 MPa x

O

Solution 7.4-6 sx ⫽ ⫺29 MPa

sy ⫽ 57 MPa

txy ⫽ 0 MPa

(b) MAXIMUM SHEAR STRESSES Point S1: us1 ⫽

(a) ELEMENT AT A SLOPE OF 1 ON 2.5 u ⫽ atan a

1 b 2.5

2u ⫽ 43.603° R ⫽

u ⫽ 21.801° |sx| + |sy| 2

Point C: sc ⫽ sx + R

tmax ⫽ R

;

tmax ⫽ ⫺R

Point D: sx1 ⫽ sc ⫺ R cos (2u) sx1 ⫽ ⫺17.1 MPa tx1y1 ⫽ R sin (2u) Point

Dœ:

saver ⫽ sc

; tx1y1 ⫽ 29.7 MPa

; ;

⫺ 90° 2

us2 ⫽ ⫺45.0°

sc ⫽ 14.0 MPa

us1 ⫽ 45.0°

tmax ⫽ 43.0 MPa

Point S2: us2 ⫽

R ⫽ 43.0 MPa

90° 2

;

tmax ⫽ ⫺43.0 MPa saver ⫽ 14.0 MPa

; ;

;

sy1 ⫽ sc + R cos (2u)

sy1 ⫽ 45.1 MPa

Problem 7.4-7 An element in pure shear is subjected to stresses txy ⫽ 2700 psi, as shown

y

in the figure. Using Mohr’s circle, determine: (a) The stresses acting on an element oriented at a counterclockwise angle u ⫽ 52° from the x axis. (b) The principal stresses. Show all results on sketches of properly oriented elements.

2700 psi O

x

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SECTION 7.4

599

Mohr’s Circle

Solution 7.4-7 sx ⫽ 0 psi

sy ⫽ 0 psi

(a) ELEMENT AT

txy ⫽ 2700 psi

Point P1: up1 ⫽

u ⫽ 52°

2u ⫽ 104.0° R ⫽ txy

R ⫽ 2700 psi

Point D: sx1 ⫽ R cos (2u ⫺ 90°) sx1 ⫽ 2620 psi tx1y1 ⫽ ⫺653 psi Point

90° 2

up1 ⫽ 45°

s1 ⫽ R Point P2: up2 ⫽

;

tx1y1 ⫽ ⫺R sin (2u ⫺ 90°) Dœ

(b) PRINCIPAL STRESSES ;

s1 ⫽ 2700 psi

;

⫺ 90° 2

up2 ⫽ ⫺45°

;

;

: sy1 ⫽ ⫺R cos (2u ⫺ 90°) sy1 ⫽ ⫺2620 psi

s2 ⫽ ⫺R

s2 ⫽ ⫺2700 psi

;

;

Problem 7.4-8 An element in pure shear is subjected to stresses txy ⫽ ⫺14.5 MPa, as shown in the figure. Using Mohr’s circle, determine:

y

(a) The stresses acting on an element oriented at a counterclockwise angle u ⫽ 22.5° from the x axis (b) The principal stresses. Show all results on sketches of properly oriented elements.

O

x 14.5 MPa

Solution 7.4-8 sx ⫽ 0 MPa

sy ⫽ 0 MPa

(a) ELEMENT AT

txy ⫽ ⫺14.5 MPa

u ⫽ 22.5°

Point P1: up1 ⫽

2u ⫽ 45.00° R ⫽ |txy|

270° u ⫽ 135.0° 2 p1

s1 ⫽ R

R ⫽ 14.50 MPa

Point D: sx1 ⫽ ⫺R cos (2u ⫺ 90°) sx1 ⫽ ⫺10.25 MPa

;

tx1y1 ⫽ R sin (2u ⫺ 90°)

Point P2: up2 ⫽

;

s1 ⫽ 14.50 MPa

⫺ 270° 2

up2 ⫽ ⫺135.0°

;

s2 ⫽ ⫺R

tx1y1 ⫽ ⫺10.25 MPa

;

Point D œ : sy1 ⫽ R cos (2u ⫺ 90°) sy1 ⫽ 10.25 MPa

(b) PRINCIPAL STRESSES

;

s2 ⫽ ⫺14.50 MPa

;

;

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Page 600

Analysis of Stress and Strain

Problem 7.4-9 An element in pure shear is subjected to stresses txy ⫽ 3750 psi, as shown in the figure. Using Mohr’s circle, determine:

y

3 4

(a) The stresses acting on an element oriented at a slope of 3 on 4 (see figure). (b) The principal stresses. Show all results on sketches of properly oriented elements.

x

O

3750 psi

Solution 7.4-9 sx ⫽ 0 psi

sy ⫽ 0 psi

txy ⫽ 3750 psi

(b) PRINCIPAL STRESSES

3 u ⫽ atan a b 4

u ⫽ 36.870°

2u ⫽ 73.740°

R ⫽ txy

s1 ⫽ R R ⫽ 3750 psi

Point D: sx1 ⫽ R cos (2u ⫺ 90°) sx1 ⫽ 3600 psi

;

tx1y1 ⫽ ⫺R sin (2u ⫺ 90°) tx1y1 ⫽ 1050 psi

90° 2

Poin P1: up1 ⫽

(a) ELEMENT AT A SLOPE OF 3 ON 4

Point P2: up2 ⫽

up1 ⫽ 45° s1 ⫽ 3750 psi

;

⫺ 90° 2

up2 ⫽ ⫺45° s2 ⫽ ⫺R

;

;

s2 ⫽ ⫺3750 psi

;

;

Point D sy1 ⫽ ⫺R cos (2u ⫺ 90°) œ:

sy1 ⫽ ⫺3600 psi

;

Problem 7.4-10 sx ⫽ 27 MPa, sy ⫽ 14 MPa, txy ⫽ 6 MPa, u ⫽ 40°

y

Using Mohr’s circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these stresses on a sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negative when clockwise.)

sy txy sx O

Probs. 7.4-10 through 7.4-15

x

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SECTION 7.4 Mohr’s Circle

601

Solution 7.4-10 sx ⫽ 27 MPa

sy ⫽ 14 MPa

txy ⫽ 6 MPa

sx + sy

saver ⫽ 20.50 MPa

2

R ⫽ 2(sx ⫺ saver)2 + txy 2 a ⫽ atan a

b ⫽ 37.29°

Point D: sx1 ⫽ saver + R cos (b)

u ⫽ 40° saver ⫽

b ⫽ 2u ⫺ a

txy sx ⫺ saver

R ⫽ 8.8459 MPa

sx1 ⫽ 27.5 MPa

;

tx1y1 ⫽ ⫺ R sin (b) tx1y1 ⫽ ⫺ 5.36 MPa

;

Point D œ : sy1 ⫽ saver ⫺ R cos (b)

b

a ⫽ 42.71°

sy1 ⫽ 13.46 MPa

;

Problem 7.4-11 sx ⫽ 3500 psi, sy ⫽ 12,200 psi, txy ⫽ ⫺3300 psi, u ⫽ ⫺51° Using Mohr’s circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these stresses on a sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negative when clockwise.)

Solution 7.4-11 sx ⫽ 3500 psi u ⫽ ⫺ 51° saver ⫽

sy ⫽ 12200 psi

b ⫽ 180° + 2u ⫺ a

b ⫽ 40.82°

Point D: sx1 ⫽ saver + R cos (b)

;

sx + sy

sx1 ⫽ 11982 psi

saver ⫽ 7850 psi

2

R ⫽ 2(sx ⫺ saver) + txy 2

a ⫽ atan a

txy ⫽ ⫺ 3300 psi

txy sx ⫺ saver

b

tx1y1 ⫽ ⫺ R sin (b)

R ⫽ 5460 psi

2

;

tx1y1 ⫽ ⫺ 3569 psi

;

Point D sy1 ⫽ saver ⫺ R cos (b) œ:

a ⫽ 37.18°

sy1 ⫽ 3718 psi

;

Problem 7.4-12 sx ⫽ ⫺47 MPa, sy ⫽ ⫺186 MPa, txy ⫽ ⫺29 MPa, u ⫽ ⫺33° Using Mohr’s circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these stresses on a sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negative when clockwise.)

Solution 7.4-12 sx ⫽⫺47MPa

sy ⫽⫺186MPa

txy ⫽⫺29MPa

u ⫽ ⫺ 33° saver ⫽

sx1 ⫽ ⫺ 61.7 MPa

sx + sy

saver ⫽ ⫺ 116.50 MPa

2

R ⫽ 2(sx ⫺ saver) + txy 2

a ⫽ atan a `

Point D: sx1 ⫽ saver + R cos (b)

txy sx ⫺ saver

b ⫽ ⫺ 2u ⫺ a

`b

R ⫽ 75.3077 MPa

2

a ⫽ 22.65°

b ⫽ 43.35°

;

tx1y1 ⫽⫺R sin (b) tx1y1 ⫽⫺51.7 MPa

;

Point D œ : sy1 ⫽ saver ⫺ R cos (b) sy1 ⫽ ⫺ 171.3 MPa

;

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CHAPTER 7 Analysis of Stress and Strain

Problem 7.4-13 sx ⫽ ⫺1720 psi, sy ⫽ ⫺680 psi, txy ⫽ 320 psi, u ⫽ 14° Using Mohr’s circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these stresses on a sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negative when clockwise.)

Solution 7.4-13 sx ⫽ ⫺ 1720 psi

sy ⫽ ⫺ 680 psi

txy ⫽ 380 psi

Point D: sx1 ⫽ saver ⫺ R cos(b)

u ⫽ 14° saver ⫽

sx1 ⫽ ⫺ 1481 psi sx + sy

saver ⫽ ⫺ 1200 psi

2

tx1y1 ⫽ 580 psi

;

Point D : sy1 ⫽ saver + R cos (b) œ

R ⫽ 2(sx ⫺ saver) + txy 2

a ⫽ atan a

tx1y1 ⫽ R sin (b)

;

txy |sx ⫺ saver|

b ⫽ 2u + a

b

sy1 ⫽ ⫺ 919 psi

R ⫽ 644.0 psi

2

;

a ⫽ 36.16°

b ⫽ 64.16°

Problem 7.4-14 sx ⫽ 33 MPa, sy ⫽ ⫺9 MPa, txy ⫽ 29 MPa, u ⫽ 35° Using Mohr’s circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these stresses on a sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negative when clockwise.)

Solution 7.4-14 sx ⫽ 33 MPa sy ⫽ ⫺ 9 MPa txy ⫽ 29 MPa

Point D: sx1 ⫽ saver + R cos (b)

u ⫽ 35° saver ⫽

sx1 ⫽ 46.4 MPa sx + sy

R ⫽ 2(sx ⫺ saver) + 2

a ⫽ atan a `

txy sx ⫺ saver

b ⫽ 2u ⫺ a

tx1y1 ⫽ ⫺ R sin (b)

saver ⫽ 12.00 MPa

2

txy2

`b

R ⫽ 35.8050 MPa

;

tx1y1 ⫽ ⫺ 9.81 MPa

;

Point D : sy1 ⫽ saver ⫺ R cos (b)

a ⫽ 54.09°

œ

sy1 ⫽ ⫺22.4 MPa

;

b ⫽ 15.91°

Problem 7.4-15 sx ⫽ ⫺5700 psi, sy ⫽ 950 psi, txy ⫽ ⫺2100 psi, u ⫽ 65° Using Mohr’s circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these stresses on a sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negative when clockwise.)

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SECTION 7.4

603

Mohr’s Circle

Solution 7.4-15 sx ⫽ ⫺5700 psi sy ⫽ 950 psi txy ⫽ ⫺2100 psi u ⫽ 65° saver ⫽

Point D: sx1 ⫽ saver + R cos (b) sx1 ⫽ ⫺1846 psi

sx + sy

R ⫽ 2(sx ⫺ saver) + txy 2

a ⫽ atan a

tx1y1 ⫽ R sin (b)

saver ⫽ ⫺2375 psi

2

|txy| |sx ⫺ saver|

2

b

b ⫽ 180° ⫺ 2u + a

; tx1y1 ⫽ 3897 psi

;

Point D sy1 ⫽ saver ⫺ R cos (b) œ:

sy1 ⫽ ⫺2904 psi

R ⫽ 3933 psi

;

a ⫽ 32.28° b ⫽ 82.28°

Problems 7.4-16 sx ⫽ ⫺29.5 MPa, sy ⫽ 29.5 MPa, txy ⫽ 27 MPa

y

Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.

sy txy sx O

x

Probs. 7.4-16 through 7.4-23

Solution 7.4-16 sx ⫽ ⫺29.5 MPa sy ⫽ 29.5 MPa txy ⫽ 27 MPa saver ⫽

sx + sy 2

a ⫽ atan a `

txy sx ⫺ saver

`b

R ⫽ 39.9906 MPa a ⫽ 42.47°

180° ⫺ a 2

up2 ⫽ up1 ⫺ 90°

90°⫺a 2

s2 ⫽ ⫺40.0 MPa

us1 ⫽ 23.8°

us2 ⫽ 90° + us1

us2 ⫽ 113.8°

Point S1: saver ⫽ 0 MPa up1 ⫽ 68.8° up2 ⫽ ⫺21.2°

tmax ⫽ R

; ;

;

(b) MAXIMUM SHEAR STRESSES us1 ⫽

(a) PRINCIPAL STRESSES up1 ⫽

s1 ⫽ 40.0 MPa

Point P2: s2 ⫽ ⫺R

saver ⫽ 0 MPa

R ⫽ 2(sx ⫺ saver)2 + txy2

Point P1: s1 ⫽ R

; ;

;

tmax ⫽ 40.0 MPa

;

;

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CHAPTER 7 Analysis of Stress and Strain

Problems 7.4-17 sx ⫽ 7300 psi, sy ⫽ 0 psi, txy ⫽ 1300 psi Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.

Solution 7.4-17 sx ⫽ 7300 psi sy ⫽ 0 psi txy ⫽ 1300 psi saver ⫽

sx + sy

s1 ⫽ 7525 psi

saver ⫽ 3650 psi

2

R ⫽ 2(sx ⫺ saver) + 2

a ⫽ atan a `

Point P1: s1 ⫽ R + saver

txy sx ⫺ saver

txy2

`b

Point P2: s2 ⫽ ⫺ R + saver

a 2

up2 ⫽

a + 180° 2

s2 ⫽ ⫺ 225 psi

R ⫽ 3875 psi a ⫽ 19.60°

(b) MAXIMUM SHEAR STRESSES us1 ⫽

(a) PRINCIPAL STRESSES up1 ⫽

up1 ⫽ 9.80°

;

⫺ 90° + a 2

us2 ⫽ 90° + us1

;

us1 ⫽ ⫺ 35.2° us2 ⫽ 54.8°

Point S1: saver ⫽ 3650 psi tmax ⫽ R

up2 ⫽ 99.8°

;

tmax ⫽ 3875 psi

;

Problems 7.4-18 sx ⫽ 0 MPa, sy ⫽ ⫺23.4 MPa, txy ⫽ ⫺9.6 MPa Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.

Solution 7.4-18 sx ⫽ 0 MPa sy ⫽ ⫺ 23.4 MPa txy ⫽ ⫺ 9.6 MPa saver ⫽

sx + sy

R ⫽ 2(sx ⫺ saver)2 + txy2 a ⫽ atan a `

s1 ⫽ 3.43 MPa

saver ⫽ ⫺ 11.70 MPa

2

txy sx ⫺ saver

`b

Point P1: s1 ⫽ R + saver Point P2: s2 ⫽ ⫺ R + saver s2 ⫽ ⫺ 26.8 MPa

R ⫽ 15.1344 MPa

(a) PRINCIPAL STRESSES up1

up2 ⫽ up1 + 90°

⫺ 90° ⫺ a 2

us2 ⫽ 90 ° + us1

up1 ⫽ ⫺ 19.68° up2 ⫽ 70.32°

;

(b) MAXIMUM SHEAR STRESSES

a ⫽ 39.37°

us1 ⫽ ⫺a ⫽ 2

;

;

us1 ⫽ ⫺ 64.7°

;

us2 ⫽ 25.3°

Point S1: saver ⫽ ⫺ 11.70 MPa ;

tmax ⫽ R

tmax ⫽ 15.13 MPa

;

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SECTION 7.4 Mohr’s Circle

605

Problems 7.4-19 sx ⫽ 2050 psi, sy ⫽ 6100 psi, txy ⫽ 2750 psi Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.

Solution 7.4-19 sx ⫽ 2050 psi sy ⫽ 6100 psi txy ⫽ 2750 psi saver ⫽

sx + sy

s1 ⫽ 7490 psi

saver ⫽ 4075 psi

2

R ⫽ 2(sx ⫺ saver) + 2

a ⫽ atan a `

Point P1: s1 ⫽ R + saver

txy sx ⫺ saver

txy2

`b

Point P2: s2 ⫽ ⫺ R + saver

R ⫽ 3415 psi a ⫽ 53.63°

s2 ⫽ 660 psi

;

(b) MAXIMUM SHEAR STRESSES us1 ⫽

(a) PRINCIPAL STRESSES

⫺ 90° + a 2

us2 ⫽ 90° + us1

180° ⫺ a up1 ⫽ 63.2° ; 2 ⫺a ⫽ up2 ⫽ ⫺ 26.8° ; 2

up1 ⫽ up2

;

us1 ⫽ ⫺18.2° us2 ⫽ 71.8°

Point S1: saver ⫽ 4075 psi tmax ⫽ R

;

;

tmax ⫽ 3415 psi

;

Problems 7.4-20 sx ⫽ 2900 kPa, sy ⫽ 9100 kPa, txy ⫽ ⫺3750 kPa Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.

Solution 7.4-20 sx ⫽ 2900 kPa sy ⫽ 9100 kPa txy ⫽ ⫺ 3750 kPa saver ⫽

sx + sy

R ⫽ 2(sx ⫺ saver)2 + txy2 a ⫽ atan a `

s1 ⫽ 10865 KPa

saver ⫽ 6000 kPa

2

txy sx ⫺ saver

`b

Point P1: s1 ⫽ R + saver Point P2: s2 ⫽ ⫺ R + saver s2 ⫽ 1135 kPa

R ⫽ 4865.4393 kPa

(a) PRINCIPAL STRESSES a + 180° 2

up2 ⫽

a 2

90° + a 2

us2 ⫽ 90° + us1 up1 ⫽ 115.2°

up2 ⫽ 25.2°

;

;

(b) MAXIMUM SHEAR STRESSES

a ⫽ 50.42°

us1 ⫽ up1 ⫽

;

;

us1 ⫽ 70.2° us2 ⫽ 160.2°

Point S1: saver ⫽ 6000 kPa tmax ⫽ R

; ;

;

tmax ⫽ 4865 kPa

;

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CHAPTER 7 Analysis of Stress and Strain

Problems 7.4-21 sx ⫽ ⫺11,500 psi, sy ⫽ ⫺18,250 psi, txy ⫽ ⫺7200 psi Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.

Solution 7.4-21 sx ⫽ ⫺ 11500 psi

sy ⫽ ⫺ 18250 psi

Point P1: s1 ⫽ R + saver

txy ⫽ ⫺ 7200 psi saver ⫽

s1 ⫽ ⫺ 6923 psi

sx + sy

R ⫽ 2(sx ⫺ saver)2 + txy2 a ⫽ atan a `

Point P2: s2 ⫽ ⫺ R + saver

saver ⫽ ⫺ 14875 psi

2

txy sx ⫺ saver

`b

s2 ⫽ ⫺ 22827 psi

R ⫽ 7952 psi

⫺a 2

a ⫽ 64.89°

us1 ⫽

up2 ⫽

180° ⫺ a 2

270° ⫺ a 2

us2 ⫽ 90° + us1

up1 ⫽ ⫺ 32.4°

us1 ⫽ 102.6°

tmax ⫽ R

;

us2 ⫽ 192.6°

Point S1: saver ⫽ ⫺ 14875 psi

;

up2 ⫽ 57.6°

;

(b) MAXIMUM SHEAR STRESSES

(a) PRINCIPAL STRESSES up1 ⫽

;

;

tmax ⫽ 7952 psi

;

Problems 7.4-22 sx ⫽ ⫺3.3 MPa, sy ⫽ 8.9 MPa, txy ⫽ ⫺14.1 MPa Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.

Solution 7.4-22 sx ⫽⫺ 3.3 MPa

sy ⫽ 8.9 MPa

txy ⫽⫺ 14.1 MPa saver ⫽

sx + sy

up1 ⫽

a + 180° 2

up2 ⫽

a 2

saver ⫽ 2.8 MPa

2

R ⫽ 2(sx ⫺ saver) + 2

a ⫽ atan a `

(a) PRINCIPAL STRESSES

txy sx ⫺ saver

txy2

`b

R ⫽ 15.4 MPa a ⫽ 66.6°

up1 ⫽ 123.3°

up2 ⫽ 33.3°

Point P1: s1 ⫽ R + saver s1 ⫽ 18.2 MPa Point P2: s2 ⫽ ⫺ R + saver

;

;

;

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Page 607

SECTION 7.4 Mohr’s Circle

s2 ⫽ ⫺ 12.6 MPa

us2 ⫽ 90° + us1

;

Point S1: saver ⫽ 2.8 MPa

(b) MAXIMUM SHEAR STRESSES 90° + a us1 ⫽ 2

us2 ⫽ 168.3°

tmax ⫽ R

;

tmax ⫽ 15.4 MPa

;

us1 ⫽ 78.3°

Problems 7.4-23 sx ⫽ 800 psi, sy ⫽ ⫺2200 psi, txy ⫽ 2900 psi Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.

Solution 7.4-23 sy ⫽ ⫺ 2200 psi txy ⫽ 2900 psi sx ⫽ 800 psi sx + sy saver ⫽ saver ⫽ ⫺ 700 psi 2 R ⫽ 2(sx ⫺ saver)2 + txy2 a ⫽ atan a `

txy sx ⫺ saver

`b

(a) PRINCIPAL STRESSES a up1 ⫽ up1 ⫽ 31.3° 2 up2 ⫽

180° + a 2

R ⫽ 3265 psi a ⫽ 62.65°

Point P1: s1 ⫽ R + saver s1 ⫽ 2565 psi Point P2: s2 ⫽ ⫺ R + saver s2 ⫽ ⫺ 3965 psi

up2 ⫽ 121.3°

⫺ 90° + a 2

us2 ⫽ 90° + us1

us1 ⫽ ⫺ 13.7° us2 ⫽ 76.3°

Point S1: saver ⫽ ⫺ 700 psi ;

;

(b) MAXIMUM SHEAR STRESSES us1 ⫽

;

;

tmax ⫽ R

; ;

;

tmax ⫽ 3265 psi

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Hooke’s Law for Plane Stress When solving the problems for Section 7.5, assume that the material is linearly elastic with modulus of elasticity E and Poisson’s ratio n.

sy

y

Problem 7.5-1 A rectangular steel plate with thickness t  0.25 in. is

subjected to uniform normal stresses x and y, as shown in the figure. Strain gages A and B, oriented in the x and y directions, respectively, are attached to the plate. The gage readings give normal strains ex  0.0010 (elongation) and ey  0.0007 (shortening). Knowing that E  30  106 psi and   0.3, determine the stresses x and y and the change t in the thickness of the plate.

Solution 7.5-1

sx

x

Probs. 7.5-1 and 7.5-2

âz  

SUBSTITUTE NUMERICAL VALUES:

 (s + sy)  128.5 * 106 E x

¢t  âzt  32.1 * 106 in.

Eq. (7-40a): (1  )2

O

Eq. (7-39c):

E  30 * 106 psi   0.3

sx 

A

Rectangular plate in biaxial stress

t  0.25 in. âx  0.0010 ây  0.0007

E

B

;

(Decrease in thickness) (âx + ây)  26,040 psi

;

Eq. (7-40b): sy 

E (1  )2

(ây + âx)  13,190 psi

;

Problem 7.5-2 Solve the preceding problem if the thickness of the steel plate is t  10 mm, the gage readings are ex  480  106 (elongation) and ey  130  106 (elongation), the modulus is E  200 GPa, and Poisson’s ratio is   0.30.

Solution 7.5-2

Rectangular plate in biaxial stress

t  10 mm âx  480 * 106 ây  130 * 10

Eq. (7-40b):

6

sy 

E  200 GPa   0.3

E (1  )2

SUBSTITUTE NUMERICAL VALUES:

Eq. (7-39c):

Eq. (7-40a):

âz  

sx 

E (1  )2

(âx + ây)  114.1 MPa

;

(ây + âx)  60.2 MPa

 (s + sy)  261.4 * 106 E x

¢t  âz t  2610 * 106 mm (Decrease in thickness)

;

;

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SECTION 7.5

Problem 7.5-3 Assume that the normal strains Px and Py for an

Hooke’s Law for Plane Stress

y

element in plane stress (see figure) are measured with strain gages. (a) Obtain a formula for the normal strain Pz in the z direction in terms of Px, Py, and Poisson’s ratio . (b) Obtain a formula for the dilatation e in terms of Px, Py, and Poisson’s ratio .

sy txy sx

O x z

Solution 7.5-3

Plane stress

Given: âx, ây, 

(b) DILATATION Eq. (7-47): e 

(a) NORMAL STRAIN âz Eq. (7-34c): âz   Eq. (7-36a): sx  Eq. (7-36b): sy 

 (s + sy) E x E

(1  2) E (1  2)

(âx + ây)

1  2 (sx + sy) E

Substitute sx and sy from above and simplify: e

1  2 (â + ây) 1 x

;

(ây + âx)

Substitute sx and sy into the first equation and simplify: âz  

 (â + ây) 1 x

;

sy

Problem 7.5-4 A magnesium plate in biaxial stress is subjected to tensile stresses sx 24 MPa and sy 12 MPa (see figure). The corresponding strains in the plate are x  440  106 and y  80  106. Determine Poisson’s ratio  and the modulus of elasticity E for the material.

y O

x

sx

Probs. 7.5-4 through 7.5-7

Solution 7.5-4

Biaxial stress

sx  24 MPa sy  12 MPa âx  440 * 106

ây  80 * 106

POISSON’S RATION AND MODULUS OF ELASTICITY Eq. (7-39a): âx 

1 (s  sy) E x

Eq. (7-39b): ây 

1 (s  sx) E y

Substitute numerical values: E (440 * 106)  24 MPa   (12 MPa) E (80 * 106)  12 MPa   (24 MPa) Solve simultaneously:   0.35 E  45 GPa

;

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Problem 7.5-5 Solve the preceding problem for a steel plate with sx  10,800 psi (tension), sy 5400 psi (compression), ex  420  106 (elongation), and ey  300  106 (shortening).

Solution 7.5-5 Biaxial stress sx  10,800 psi sy   5400 psi âx  420 * 10

6

ây   300 * 10

Substitute numerical values: 6

E (420 * 106)  10,800 psi   ( 5400 psi) E (300 * 106)   5400 psi   (10,800 psi)

POISSON’S RATIO AND MODULUS OF ELASTICITY

Solve simultaneously:   1/3 E  30 * 106 psi

1 Eq. (7-39a): âx  (sx  sy) E Eq. (7-39b): ây 

;

1 (s  sx) E y

Problem 7.5-6 A rectangular plate in biaxial stress (see figure) is subjected to normal stresses x 90 MPa (tension) and y 20 MPa (compression). The plate has dimensions 400 * 800 * 20 mm and is made of steel with E  200 GPa and   0.30. (a) Determine the maximum in-plane shear strain gmax in the plate. (b) Determine the change ¢t in the thickness of the plate. (c) Determine the change ¢V in the volume of the plate.

Solution 7.5-6 Biaxial stress sx  90 MPa sy   20 MPa E  200 GPa   0.30

(b) CHANGE IN THICKNESS

Dimensions of Plate: 400 mm * 800 mm * 20 mm Shear Modulus (Eq. 7-38): E  76.923 GPa G 2(1 + )

Principal stresses: s1  90 MPa s2   20 MPa s1  s2  55.0 MPa 2

Eq. (7-35): gmax 

tmax  715 * 106 G

 (sx + sy)   105 * 106 E

¢t  âz t   2100 * 106 mm

;

(Decrease in thickness) (c) CHANGE IN VOLUME

(a) MAXIMUM IN-PLANE SHEAR STRAIN

Eq. (7-26): tmax 

Eq. (7-39c): âz  

From Eq. (7-47): ¢V  V0 a

1  2 b (sx + sy) E

V0  (400)(800)(20)  6.4 * 106 mm3 ;

Also, a

1  2 b (sx + sy)  140 * 106 E

‹ ¢V  (6.4 * 106 mm3)(140 * 106)  896 mm3 (Increase in volume)

;

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SECTION 7.5

Hooke’s Law for Plane Stress

611

Problem 7.5-7 Solve the preceding problem for an aluminum plate with sx 12,000 psi (tension), sy 3,000 psi (compression), dimensions 20  30  0.5 in., E  10.5  106 psi, and   0.33.

Solution 7.5-7

Biaxial stress

sx  12,000 psi sy  3,000 psi

(b) CHANGE IN THICKNESS

E  10.5 * 10 psi   0.33 6

Eq. (7-39c): âz  

Dimensions of Plate: 20 in. * 30 in. * 0.5 in. Shear Modulus (Eq. 7-38): G

 282.9 * 106 ¢t  âz t  141 * 106 in.

E  3.9474 * 106 psi 2(1 + )

(c) CHANGE IN VOLUME

Principal stresses: s1  12,000 psi

From Eq. (7-47): ¢V  V0 a

s2  3,000 psi s1  s2  7,500 psi 2

tmax Eq. (7-35): gmax   1,900 * 106 G

;

(Decrease in thickness)

(a) MAXIMUM IN-PLANE SHEAR STRAIN

Eq. (7-26): tmax 

 (s + sy) E x

1  2 b (sx + sy) E

V0  (20)(30)(0.5)  300 in.3

;

Also, a

1  2 b (sx + sy)  291.4 * 106 E

‹ ¢V  (300 in.3)(291.4 * 106)  0.0874 in.3

;

(Increase in volume)

Problem 7.5-8 A brass cube 50 mm on each edge is compressed

P = 175 kN

in two perpendicular directions by forces P  175 kN (see figure). Calculate the change V in the volume of the cube and the strain energy U stored in the cube, assuming E  100 GPa and   0.34.

Solution 7.5-8

P = 175 kN

Biaxial stress-cube sx  sy  

P b

2

 

(175 kN) (50 mm)2

 70.0 MPa

CHANGE IN VOLUME Eq. (7-47): e 

1  2 (sx + sy)  448 * 106 E

V0  b 3  (50 mm)3  125 * 103mm3 ¢V  eV0  56 mm3 Side b  50 mm P  175 kN E  100 GPa   0.34 ( Brass)

(Decrease in volume)

;

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U  uV0  (0.03234 MPa)(125 * 103 mm3)

STRAIN ENERGY Eq. (7-50): u 

1 (s2 + s2y  2sxsy) 2E x

 4.04 J

;

 0.03234 MPa

Problem 7.5-9 A 4.0-inch cube of concrete (E  3.0 * 106 psi,   0.1) is compressed in biaxial stress by means of a framework that is loaded as shown in the figure. Assuming that each load F equals 20 k, determine the change ¢V in the volume of the cube and the strain energy U stored in the cube.

F

F

Solution 7.5-9

Biaxial stress – concrete cube CHANGE IN VOLUME

b  4 in.

Eq. (7-47): e 

E  3.0 * 106 psi

V0  b 3  (4 in.)3  64 in.3

  0.1 F  20 kips

1  2 (sx + sy)  0.0009429 E

¢V  eV0  0.0603 in.3

;

(Decrease in volume) STRAIN ENERGY

Joint A:

Eq. (7-50): u 

P  F12  28.28 kips sx  sy  

P  1768 psi b2

1 (s2 + s2y  2sxsy) 2E x

 0.9377 psi U  uV0  60.0 in.-lb

;

Py

Problem 7.5-10 A square plate of width b and thickness t is loaded by normal forces Px and Py, and by shear forces V, as shown in the figure. These forces produce uniformly distributed stresses acting on the side faces of the place. Calculate the change V in the volume of the plate and the strain energy U stored in the plate if the dimensions are b  600 mm and t  40 mm, the plate is made of magnesium with E  45 GPa and v  0.35, and the forces are Px  480 kN, Py  180 kN, and V  120 kN.

t

V y

Px V

V

b O b

x

V Py

Probs. 7.5-10 and 7.5-11

Px

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SECTION 7.5 Hooke’s Law for Plane Stress

613

Solution 7.5-10 Square plate in plane stress b  600 mm

t  40 mm

E  45 GPa

v  0.35 (magnesium) Px  20.0 MPa bt Py sy   7.5 MPa bt V txy   5.0 MPa bt

Px  480 kN

sx 

Py  180 kN V  120 kN CHANGE IN VOLUME

1  2 Eq. (7-47): e  (sx + sy)  183.33 * 106 E

V0  b 2t  14.4 * 106 mm3 ¢V  eV0  2640 mm3

;

(Increase in volume) STRAIN ENERGY Eq. (7-50): u  G

t2xy 1 2 (sx + s2y  2sxsy) + 2E 2G E  16.667 GPa 2(1 + )

Substitute numerical values: u  4653 Pa U  uV0  67.0 N # m  67.0 J

;

Problem 7.5-11 Solve the preceding problem for an aluminum plate with b  12 in., t  1.0 in., E  10,600 ksi,   0.33, Px  90 k, Py  20 k, and V  15 k.

Solution 7.5-11 Square plate in plane stress b  12.0 in.

t  1.0 in.

STRAIN ENERGY

E  10,600 ksi   0.33 (aluminum) Px  7500 psi bt Py sy   1667 psi bt

Px  90 k

sx 

Py  20 k V  15 k

txy 

G

u  2.591 psi

V  1250 psi bt

U  uV0  373 in.-lb

1  2 (sx + sy)  294 * 106 E

V0  b 2t  144 in.3 ¢V  eV0  0.0423 in.3 (Increase in volume)

E  3985 ksi 2(1 + )

Substitute numerical values:

CHANGE IN VOLUME Eq. (7-47): e 

t2xy 1 2 2 Eq. (7-50): u  (s + sy  2sxsy) + 2E x 2G

;

;

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Analysis of Stress and Strain

Problem 7.5-12 A circle of diameter d  200 mm is etched on a

z

brass plate (see figure). The plate has dimensions 400  400  20 mm. Forces are applied to the plate, producing uniformly distributed normal stresses x  42 MPa and y  14 MPa. Calculate the following quantities: (a) the change in length ac of diameter ac; (b) the change in length bd of diameter bd; (c) the change t in the thickness of the plate; (d) the change V in the volume of the plate, and (e) the strain energy U stored in the plate. (Assume E  100 GPa and v  0.34.)

y sy d sx

a

c

sx

b x sy

Solution 7.5-12

Plate in biaxial stress

sx  42 MPa sy  14 MPa

(c) CHANGE IN THICKNESS

 (s + sy) E x  190.4 * 106

Eq. (7-39c): âz  

Dimensions: 400 * 400 * 20 (mm) Diameter of circle: d  200 mm E  100 GPa   0.34 (Brass)

¢t  âz t  0.00381 mm (decrease)

;

(a) CHANGE IN LENGTH OF DIAMETER IN x DIRECTION Eq. (7-39a): âx 

1 (s  sy)  372.4 * 106 E x

¢ac  âx d  0.0745 mm (increase)

;

(b) CHANGE IN LENGTH OF DIAMETER IN y DIRECTION Eq. (7-39b): ây 

1 (s  sx)   2.80 * 106 E y

¢bd  ây d  560 * 106 mm (decrease)

;

(d) CHANGE IN VOLUME Eq. (7-47): e

1  2 (sx + sy)  179.2 * 106 E

V0  (400)(400)(20)  3.2 * 106 mm3 ¢V  eV0  573 mm3 ; (increase) (e) STRAIN ENERGY Eq. (7-50): u 

1 2 (s + s2y  2sx sy) 2E x

 7.801 * 103 MPa U  uV0  25.0 N # m  25.0 J

;

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SECTION 7.6

615

Triaxial Stress

Triaxial Stress When solving the problems for Section 7.6, assume that the material is linearly elastic with modulus of elasticity E and Poisson’s ratio n.

y a c

Problem 7.6-1 An element of aluminum in the form of a rectangular

parallelepiped (see figure) of dimensions a ⫽ 6.0 in., b ⫽ 4.0 in, and c ⫽ 3.0 in. is subjected to triaxial stresses sx ⫽ 12,000 psi, sy ⫽ ⫺ 4,000 psi, and sz ⫽ ⫺1,000 psi acting on the x, y, and z faces, respectively. Determine the following quantities: (a) the maximum shear stress tmax in the material; (b) the changes ¢a, ¢b, and ¢c in the dimensions of the element; (c) the change ¢V in the volume; and (d) the strain energy U stored in the element. (Assume E ⫽ 10,400 ksi and ␯ ⫽ 0.33.)

b O

x

z

Probs. 7.6-1 and 7.6-2

Solution 7.6-1

Triaxial stress

sx ⫽ 12,000 psi sy ⫽ ⫺4,000 psi

¢a ⫽ aâx ⫽ 0.0079 in. ( increase)

sz ⫽ ⫺1,000 psi

¢b ⫽ bây ⫽ ⫺0.0029 in. ( decrease)

a ⫽ 6.0 in. b ⫽ 4.0 in. c ⫽ 3.0 in. E ⫽ 10,400 ksi ␯ ⫽ 0.33 ( aluminum) (a) MAXIMUM SHEAR STRESS s1 ⫽ 12,000 psi s2 ⫽ ⫺1,000 psi

(c) CHANGE IN VOLUME Eq. (7-56): e⫽

s3 ⫽ ⫺4,000 psi tmax ⫽

¢c ⫽ câz ⫽ ⫺0.0011 in. ( decrease)

s1 ⫺ s3 ⫽ 8,000 psi 2

;

M

;

1 ⫺ 2␯ (sx + sy + sz) ⫽ 228.8 * 10⫺6 E

V ⫽ abc ¢V ⫽ e (abc) ⫽ 0.0165 in.3 ( increase)

;

(b) CHANGES IN DIMENSIONS Eq. (7-53 a): âx ⫽

sx ␯ ⫺ (sy + sz) E E

⫽1312.5 * 10⫺6 Eq. (7- 53 b): ây ⫽

sy E

⫺

␯ (s + sx) E z

⫽⫺733.7 * 10⫺6 Eq. (7-53 c): âz ⫽

sz E

⫺

␯ (s + sy) E x

⫽⫺350.0 * 10⫺6

(d) STRAIN ENERGY Eq. (7-57a): u ⫽

1 (s â + sy ây + sz âz) 2 x x

⫽ 9.517 psi U ⫽ u (abc) ⫽ 685 in.-lb

;

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Problem 7.6-2 Solve the preceding problem if the element is steel (E = 200 GPA, ␯ ⫽ 0.30) with dimensions a = 300 mm, b = 150 mm, and c = 150 mm and the stresses are sx ⫽ ⫺60 MPa, sy ⫽ ⫺40 MPa, and sz ⫽ ⫺40 MPa.

Solution 7.6-2

Triaxial stress

sx ⫽ ⫺60 MPa sy ⫽ ⫺40 MPa

¢a ⫽ aâx ⫽ ⫺0.0540 mm (decrease)

sz ⫽ ⫺40 MPa

¢b ⫽ bây ⫽ ⫺0.0075 mm (decrease)

a ⫽ 300 mm

b ⫽ 150 mm

E ⫽ 200 GPa

␯ ⫽ 0.30

c ⫽ 150 mm (steel)

(a) MAXIMUM SHEAR STRESS s1 ⫽ ⫺40 MPa s2 ⫽ ⫺40 MPa s3 ⫽ ⫺60 MPa tmax ⫽

s1 ⫺ s3 ⫽ 10.0 MPa 2

Eq. (7-53 b): ây ⫽ Eq. (7-53 c): âz ⫽

sx ␯ ⫺ (sy + sz) ⫽ ⫺ 180.0 * 10⫺6 E E sy E sz E

(c) CHANGE IN VOLUME Eq. (7-56): e⫽

M

;

1 ⫺ 2␯ (sx + sy + sz) ⫽ ⫺280.0 * 10⫺6 E

V ⫽ abc ;

(b) CHANGES IN DIMENSIONS Eq. (7-53 a): âx ⫽

¢c ⫽ câz ⫽ ⫺0.0075 mm. (decrease)

⫺

␯ (s + sx) ⫽ ⫺50.0 * 10⫺6 E z

⫺

␯ (s + sy) ⫽ ⫺ 50.0 * 10⫺6 E x

¢V ⫽ e(abc) ⫽ ⫺ 1890 mm3 (decrease)

;

(d) STRAIN ENERGY 1 (s â + sy ây + sz âz) 2 x x ⫽ 0.00740 MPa

Eq. (7-57 a): u ⫽

U ⫽ u (abc) ⫽ 50.0 N # m ⫽ 50.0 J

;

y

Problem 7.6-3 A cube of cast iron with sides of length a = 4.0 in. (see figure) is tested in a laboratory under triaxial stress. Gages mounted on the testing machine show that the compressive strains in the material are Px ⫽ ⫺ 225 * 10⫺6 and Py ⫽ Pz ⫽ ⫺37.5 * 10⫺6. Determine the following quantities: (a) the normal stresses sx, sy, and sz acting on the x, y, and z faces of the cube; (b) the maximum shear stress tmax in the material; (c) the change ¢V in the volume of the cube; and (d) the strain energy U stored in the cube. (Assume E = 14,000 ksi and ␯ ⫽ 0.25.)

a a a O

z

Probs. 7.6-3 and 7.6-4

x

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SECTION 7.6 Triaxial Stress

617

Solution 7.6-3 Triaxial stress (cube) âx ⫽ ⫺ 225 * 10⫺6

ây ⫽ ⫺ 37.5 * 10⫺6

âz ⫽ ⫺ 37.5 * 10⫺6

a ⫽ 4.0 in.

(c) CHANGE IN VOLUME Eq. (7-55): e ⫽ âx + ây + âz ⫽ ⫺ 0.000300 V ⫽ a3

E ⫽ 14,000 ksi ␯ ⫽ 0.25 (cast iron)

¢V ⫽ ea 3 ⫽ ⫺ 0.0192 in.3 ( decrease) (a) NORMAL STRESSES Eq. (7-54a):

(d) STRAIN ENERGY

E [(1 ⫺ ␯)âx + ␯(ây + âz)] sx ⫽ (1 + ␯)(1 ⫺ 2␯) ⫽ ⫺ 4200 psi

;

Eq. (7-57a): u ⫽

;

1 (sx âx + sy ây + sz âz) 2

⫽ 0.55125 psi U ⫽ ua ⫽ 35.3 in.-lb 3

In a similar manner, Eqs. (7-54 b and c) give sy ⫽ ⫺ 2100 psi sz ⫽ ⫺ 2100 psi ;

;

(b) MAXIMUM SHEAR STRESS s1 ⫽ ⫺ 2100 psi s2 ⫽ ⫺ 2100 psi s3 ⫽ ⫺ 4200 psi tmax ⫽

s1 ⫺ s3 ⫽ 1050 psi 2

;

Problem 7.6-4 Solve the preceding problem if the cube is granite (E ⫽ 60 GPa, ␯ ⫽ 0.25) with dimensions a = 75 mm and compressive strains Px ⫽ ⫺ 720 * 10⫺6 and Py ⫽ Pz ⫽ ⫺ 270 * 10⫺6.

Solution 7.6-4 Triaxial stress (cube) âx ⫽ ⫺ 720 * 10⫺6

ây ⫽ ⫺ 270 * 10⫺6

âz ⫽ ⫺ 270 * 10⫺6

a ⫽ 75 mm

␯ ⫽ 0.25

E ⫽ 60 GPa

(Granite)

;

Eq. (7-55): e ⫽ âx + ây + âz ⫽ ⫺ 1260 * 10⫺6 V ⫽ a3

E [(1 ⫺ ␯)âx + ␯(âx + âz)] sx ⫽ (1 + ␯)(1 ⫺ 2␯) ;

In a similar manner, Eqs. (7-54 b and c) give sy ⫽ ⫺ 43.2 MPa sz ⫽ ⫺ 43.2 MPa ; (b) MAXIMUM SHEAR STESS s1 ⫽ ⫺ 43.2 MPa s2 ⫽ ⫺ 43.2 MPa s3 ⫽ ⫺ 64.8 MPa

s1 ⫺ s3 ⫽ 10.8 MPa 2

(c) CHANGE IN VOLUME

(a) NORMAL STRESSES Eq.(7-54a):

⫽ ⫺ 64.8 MPa

tmax ⫽

¢V ⫽ ea 3 ⫽ ⫺ 532 mm3 ( decrease)

;

(d) STRAIN ENERGY 1 Eq. (7-57 a): u ⫽ (sxâx + syây + szâz) 2 ⫽ 0.03499 MPa ⫽ 34.99 kPa U ⫽ ua ⫽ 14.8 N # m ⫽ 14.8 J 3

;

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Problem 7.6-5 An element of aluminum in triaxial stress (see figure) is subjected to stresses sx ⫽ 5200 psi (tension), sy ⫽ ⫺4750 psi (compression), and sz ⫽ ⫺3090 psi (compression). It is also known that the normal strains in the x and y directions are Px ⫽ 7138.8 * 10⫺6 (elongation) and Py ⫽ ⫺502.3 * 10⫺6 (shortsx ening). What is the bulk modulus K for the aluminum?

y

sy sz sx

O x

sz sy

z

Probs. 7.6-5 and 7.6-6

Solution 7.6-5

Triaxial stress (bulk modulus)

sx ⫽ 5200 psi sy ⫽ ⫺4750 psi sz ⫽ ⫺3090 psi âx ⫽ 713.8 * 10 ây ⫽ ⫺502.3 * 10

Substitute numerical values and rearrange: ⫺6

⫺6

(713.8 * 10⫺6) E ⫽ 5200 + 7840 ␯ (⫺502.3 * 10

⫺6

) E ⫽ ⫺4750 ⫺ 2110 ␯

Find K.

Units: E = psi

sx ␯ ⫺ (sy + sz) E E sy ␯ Eq. (7-53 b): ây ⫽ ⫺ (sx + sy) E E

Solve simultaneously Eqs. (1) and (2):

Eq. (7-53 a): âx ⫽

(1) (2)

E ⫽ 10.801 * 106 psi ␯ ⫽ 0.3202 Eq. (7-16): K ⫽

E ⫽ 10.0 * 10⫺6 psi 3(1 ⫺ 2␯)

;

Problem 7.6-6

Solve the preceding problem if the material is nylon subjected to compressive stresses sx ⫽ ⫺ 4.5 MPa, sy ⫽ ⫺3.6 MPa, and sz ⫽ ⫺2.1 MPa, and the normal strains are Px ⫽ ⫺740 * 10⫺6 and Py ⫽ ⫺ 320 * 10⫺6 (shortenings).

Solution 7.6-6

Triaxial stress (bulk modulus)

sx ⫽ ⫺4.5 MPa sy ⫽ ⫺3.6 MPa sz ⫽ ⫺2.1 MPa âx ⫽ ⫺740 * 10 ây ⫽ ⫺320 * 10

Substitute numerical values and rearrange: ⫺6

⫺6

Find K. sx ␯ ⫺ (sy + sz) E E sy ␯ Eq. (7-53 b): ây ⫽ ⫺ (sz + sx) E E

Eq. (7-53 a): âx ⫽

(⫺ 740 * 10⫺6) E ⫽ ⫺4.5 + 5.7 ␯ (⫺ 320 * 10

⫺6

(1)

) E ⫽ ⫺3.6 + 6.6 ␯

(2)

Units: E = MPa Solve simultaneously Eqs. (1) and (2): E ⫽ 3,000 MPa ⫽ 3.0 GPa ␯ ⫽ 0.40 Eq. (7-16): K ⫽

E ⫽ 5.0 GPa 3(1 ⫺ 2␯)

;

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SECTION 7.6

Problem 7.6-7 A rubber cylinder R of length L and cross-sectional

F

F

area A is compressed inside a steel cylinder S by a force F that applies a uniformly distributed pressure to the rubber (see figure). (a) Derive a formula for the lateral pressure p between the rubber and the steel. (Disregard friction between the rubber and the steel, and assume that the steel cylinder is rigid when compared to the rubber.) (b) Derive a formula for the shortening d of the rubber cylinder.

Solution 7.6-7

619

Triaxial Stress

R

S

L

S

Rubber cylinder Solve for p: p ⫽

F v a b 1⫺v A

;

(b) SHORTENING Eq. (7-53 b): ây ⫽ sx ⫽ ⫺p sy ⫽ ⫺

F A

sz ⫽ ⫺p âx ⫽ âz ⫽ 0

⫽ ⫺

E

⫺

␯ (s + sx) E z

␯ F ⫺ ( ⫺2p) EA E

Substitute for p and simplify: ây ⫽

F (1 + ␯)(⫺1 + 2␯) EA 1⫺␯

(Positive ây represents an increase in strain, that is, elongation.)

(a) LATERAL PRESSURE Eq. (7-53 a): âx ⫽

sy

sx ␯ ⫺ (sy + sz) E E

F or 0 ⫽ ⫺p ⫺ ␯ a ⫺ ⫺ pb A

d ⫽ ⫺âyL d⫽

(1 + ␯)(1 ⫺ 2␯) FL a b (1 ⫺ ␯) EA

;

(Positive d represents a shortening of the rubber cylinder.)

Problem 7.6-8 A block R of rubber is confined between plane

F

F

parallel walls of a steel block S (see figure). A uniformly distributed pressure p0 is applied to the top of the rubber block by a force F. (a) Derive a formula for the lateral pressure p between the rubber and the steel. (Disregard friction between the rubber and the steel, and assume that the steel block is rigid when compared to the rubber.) (b) Derive a formula for the dilatation e of the rubber. (c) Derive a formula for the strain-energy density u of the rubber.

S S

R

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CHAPTER 7 Analysis of Stress and Strain

Solution 7.6-8 Block of rubber (b) DILATATION Eq. (7-56): e ⫽

1 ⫺ 2␯ (sx + sy + sz) E ⫽

1 ⫺ 2␯ (⫺ p ⫺ p0) E

Substitute for p: e⫽ ⫺

sx ⫽ ⫺ p sy ⫽ ⫺ p0

(a) LATERAL PRESSURE

OR

sx ␯ ⫺ (sy + sz) E E

0 ⫽ ⫺ p ⫺ ␯ ( ⫺p0)

;

sz ⫽ 0

âx ⫽ 0 ây Z 0 âz Z 0

Eq. (7-53 a): âx ⫽

(1 + ␯)(1 ⫺ 2␯)p0 E

(c) STRAIN ENERGY DENSITY Eq. (7-57b): u⫽

‹ p ⫽ ␯p0

;

1 v (s2x + s2y + s2z ) ⫺ (sx sy + sx sz + sy sz) 2E E

Substitute for sx, sy, sz, and p: u⫽

(1 ⫺ ␯2)p 20 2E

;

Problem 7.6-9 A solid spherical ball of brass (E ⫽ 15 * 106 psi ␯ ⫽ 0.34) is lowered into the ocean to a depth of 10,000 ft. The diameter of the ball is 11.0 in. Determine the decrease ¢d in diameter, the decrease ¢V in volume, and the strain energy U of the ball.

Solution 7.6-9 E ⫽ 15 * 10

⫺6

Brass sphere psi ␯ ⫽ 0.34

DECREASE IN VOLUME

Lowered in the ocean to depth h ⫽ 10,000 ft

Eq. (7-60): e ⫽ 3â0 ⫽ 283.6 * 10⫺6 4 4 11.0 in. 3 b ⫽ 696.9 in.3 V0 ⫽ pr 3 ⫽ (p)a 3 3 2

Diameter d ⫽ 11.0 in. Sea water: g ⫽ 63.8 lb/ft3 Pressure: s0 ⫽ gh ⫽ 638,000 lb/ft2 ⫽ 4431 psi DECREASE IN DIAMETER s0 Eq. (7-59): â0 ⫽ (1 ⫺ 2␯) ⫽ 94.53 * 10⫺6 E ¢d ⫽ â0d ⫽ 1.04 * 10⫺3 in. (decrease)

;

¢V ⫽ eV0 ⫽ 0.198 in.3 (decrease)

;

STRAIN ENERGY Use Eq. (7-57 b) with sx ⫽ sy ⫽ sz ⫽ s0: u⫽

3(1 ⫺ 2␯)s20 ⫽ 0.6283 psi 2E

U ⫽ uV0 ⫽ 438 in.-lb

;

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SECTION 7.6 Triaxial Stress

621

Problem 7.6-10 A solid steel sphere (E ⫽ 210 GPa, ␯ = 0.3) is subjected to hydrostatic pressure p such that its volume is reduced by 0.4%. (a) Calculate the pressure p. (b) Calculate the volume modulus of elasticity K for the steel. (c) Calculate the strain energy U stored in the sphere if its diameter is d ⫽ 150 mm

Solution 7.6-10

Steel sphere

E ⫽ 210 GPa ␯ ⫽ 0.3

(b) VOLUME MODULUS OF ELASTICITY

Hydrostatic Pressure. V0 = Initial volume ¢V ⫽ 0.004 V0 Dilatation: e ⫽

¢V ⫽ 0.004 V0

3s0(1 ⫺ 2␯) Eq.(7-60): e ⫽ E s0 ⫽

s0 700 MPa ⫽ ⫽ 175 GPa E 0.004

;

(c) STRAIN ENERGY (d = diameter) d = 150 mm r = 75 mm From Eq. (7-57b) with sx ⫽ sy ⫽ sz ⫽ s0:

(a) PRESSURE

or

Eq. (7-63): K ⫽

Ee ⫽ 700 MPa 3(1 ⫺ 2␯)

Pressure p ⫽ s0 ⫽ 700 MPa

u⫽

3(1 ⫺ 2␯)s20 ⫽ 1.40 MPa 2E

V0 ⫽

4pr 3 ⫽ 1767 * 10⫺6 m3 3

U ⫽ uV0 ⫽ 2470 N # m ⫽ 2470 J

;

;

Problem 7.6-11 A solid bronze sphere (volume modulus of elasticity K ⫽ 14.5 ⫻ 106 psi) is suddenly heated around its outer surface. The tendency of the heated part of the sphere to expand produces uniform tension in all directions at the center of the sphere. If the stress at the center is 12,000 psi, what is the strain? Also, calculate the unit volume change e and the strain-energy density u at the center.

Solution 7.6-11 Bronze sphere (heated) K ⫽ 14.5 * 106 psi s0 ⫽ 12,000 psi (tension at the center) STRAIN AT THE CENTER OF THE SPHERE s0 Eq. (7-59): â0 ⫽ (1 ⫺ 2␯) E E Eq. (7-61): K ⫽ 3(1 ⫺ 2␯) Combine the two equations: â0 ⫽

s0 ⫽ 276 * 10 ⫺6 3K

;

UNIT VOLUME CHANGE AT THE CENTER Eq. (7-62): e ⫽

s0 ⫽ 828 * 10 ⫺6 K

;

STRAIN ENERGY DENSITY AT THE CENTER Eq. (7-57b) with sx ⫽ sy ⫽ sz ⫽ s0: 3(1 ⫺ 2␯)s20 s02 ⫽ 2E 2K u ⫽ 4.97 psi ; u⫽

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Analysis of Stress and Strain

Plane Strain

y sy

When solving the problems for Section 7.7, consider only the in-plane strains (the strains in the xy plane) unless stated otherwise. Use the transformation equations of plane strain except when Mohr’s circle is specified (Problems 7.7-23 through 7.7-28).

sx

h

Problem 7.7-1 A thin rectangular plate in biaxial stress is subjected to stresses sx and sy, as shown in part (a) of the figure on the next page. The width and height of the plate are b  8.0 in. and h  4.0 in., respectively. Measurements show that the normal strains in the x and y directions are Px  195 * 106 and Py  125 * 106, respectively. With reference to part (b) of the figure, which shows a two-dimensional view of the plate, determine the following quantities: (a) the increase ¢d in the length of diagonal Od; (b) the change ¢f in the angle f between diagonal Od and the x axis; and (c) the change ¢c in the angle c between diagonal Od and the y axis.

b x z (a) y d c h f O

x

b (b)

Probs. 7.7-1 and 7.7-2

Solution 7.7-1

Plate in biaxial stress For u  f  26.57°, âx1  130.98 * 106 ¢d  âx1L d  0.00117 in.

;

(b) CHANGE IN ANGLE f Eq. (7-68): a  (âx  ây) sinu cosu  gxy sin2u For u  f  26.57°: a  128.0 * 106 rad b  8.0 in.

h  4.0 in.

ây  125 * 106

âx  195 * 106

gxy  0

Minus sign means line Od rotates clockwise (angle f decreases). ¢f  128 * 106 rad (decrease)

h f  arctan  26.57° b

;

(c) CHANGE IN ANGLE c

L d  1b 2 + h2  8.944 in.

Angle c increases the same amount that f decreases.

(a) INCREASE IN LENGTH OF DIAGONAL

¢c  128 * 106 rad (increase)

âx1 

âx  ây

âx + ây + 2

2

cos 2u +

gxy 2

sin 2u

;

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SECTION 7.7

Plane Strain

623

Problem 7.7-2 Solve the preceding problem if b  160 mm, h  60 mm, Px  410 * 106, and Py  320 * 106.

Solution 7.7-2

Plate in biaxial stress For u  f  20.56°: âx1  319.97 * 10 6 ¢d  âx1 L d  0.0547 mm

;

(b) CHANGE IN ANGLE f Eq. (7-68): a  (âx ây) sin u cos u gxy sin2u For u  f  20.56°: a  240.0 * 106 rad b  160 mm

h  60 mm

ây  320 * 106

Minus sign means line Od rotates clockwise (angle f decreases.)

âx  410 * 106

gxy  0

¢f  240 * 106 rad (decrease)

h f  arctan  20.56° b 2 1 L d  b + h2  170.88 mm

(c) CHANGE IN ANGLE c Angle c increases the same amount that f decreases. ¢c  240 * 106 rad (increase)

(a) INCREASE IN LENGTH OF DIAGONAL âx1 

âx  ây

âx + ây + 2

2

cos 2u +

;

gxy 2

;

sin 2u

Problem 7.7-3 A thin square plate in biaxial stress is subjected to stresses sx and sy, as shown in part (a) of the figure . The width of the plate is b  12.0 in. Measurements show that the normal strains in the x and y directions are Px  427 * 106 and Py  113 * 106, respectively. With reference to part (b) of the figure, which shows a two-dimensional view of the plate, determine the following quantities: (a) the increase ¢d in the length of diagonal Od; (b) the change ¢f in the angle f between diagonal Od and the x axis; and (c) the shear strain g associated with diagonals Od and cf (that is, find the decrease in angle ced).

y sy

y c sx

b

e

b

b

f x

z

d

O

b (b)

(a)

PROBS. 7.7-3 and 7.7-4

f x

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CHAPTER 7 Analysis of Stress and Strain

Solution 7.7-3 Square plate in biaxial stress (b) CHANGE IN ANGLE f Eq. (7-68): a  (âx  ây) sin u cos u  gxy sin2u For u  f  45°: a  157 * 106 rad Minus sign means line Od rotates clockwise (angle f decreases.) ¢f  157 * 106 rad (decrease) (c) SHEAR STRAIN BETWEEN DIAGONALS

âx  427 * 106

b  12.0 in. ây  113 * 106

Eq. (7-71b):

gxy  0

f  45°

;

gx1y1 2



âx ây 2

sin 2u +

gxy 2

cos 2u

For u  f  45°: gx1y1  314 * 106 rad

L d  b12  16.97 in.

(Negative strain means angle ced increases)

(a) INCREASE IN LENGTH OF DIAGONAL

g  314 * 106 rad

âx1 

âx  ây

âx + ây + 2

2

cos 2u +

gxy 2

;

sin 2u

For u  f  45°: âx1  270 * 106 ¢d  âx1L d  0.00458 in.

;

Problem 7.7-4 Solve the preceding problem if b  225 mm, Px  845 * 106 , and Py  211 * 106.

Solution 7.7-4 Square plate in biaxial stress (a) INCREASE IN LENGTH OF DIAGONAL âx1 

âx  ây

âx + ây + 2

2

cos 2u +

gxy 2

sin 2u

For u  f  45°: âx1  528 * 106 ¢d  âx1L d  0.168 mm

;

(b) CHANGE IN ANGLE f Eq. (7-68): a  (âx  ây) sin u cos u  gxy sin2u b  225 mm ây  211 * 106

âx  845 * 106 f  45°

L d  b12  318.2 mm

gxy  0

For u  f  45°: a   317 * 106 rad Minus sign means line Od rotates clockwise (angle f decreases.) ¢f  317 * 106 rad (decrease)

;

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SECTION 7.7

g x1y1 2



âx ây 2

625

For u  f  45°: gx1y1  634 * 106 rad

(c) SHEAR STRAIN BETWEEN DIAGONALS Eq. (7- 71b):

Plane Strain

sin 2u +

gxy 2

cos 2u

(Negative strain means angle ced increases) g  634 * 106 rad

y

Problem 7.7-5 An element of material subjected to plane strain (see

figure) has strains as follows: Px  220 * 106, Py  480 * 106, and gxy  180 * 106. Calculate the strains for an element oriented at an angle u  50° and show these strains on a sketch of a properly oriented element.

ey gxy 1

O

1

ex

x

Probs. 7.7-5 through 7.7-10

Solution 7.7-5

Element in plane strain

âx  220 * 106

ây  480 * 106

gxy  180 * 106 âx1  gx1 y1 2

âx ây

âx + ây

 

+ 2

2

âx  ây 2

cos 2u +

sin 2u +

g xy 2

gxy 2

sin 2u

cos 2u

ây1  âx + ây  âx1 For u  50°: âx1  461 * 106

gx1y1  225 * 106

ây1  239 * 106

Problem 7.7-6 Solve the preceding problem for the following data: Px  420 * 106, Py  170 * 106, gxy 310 * 106,

and u  37.5°.

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CHAPTER 7 Analysis of Stress and Strain

Solution 7.7-6 Element in plane strain âx  420 * 106

ây   170 * 106

gxy  310 * 106 âx1  gx1y1 2

âx  ây

âx + ây + 2



2

âx  ây

sin 2u +

2

cos 2u + gxy 2

gxy 2

sin 2u

cos 2u

ây1  âx + ây  âx1 For u  37.5°: âx1  351 * 106

gx1y1   490 * 106

ây1  101 * 106

Problem 7.7-7 The strains for an element of material in plane strain (see figure) are as follows: Px  480 * 106,

Py  140 * 106, and gxy  350 * 106. Determine the principal strains and maximum shear strains, and show these strains on sketches of properly oriented elements.

Solution 7.7-7 Element in plane strain âx  480 * 106

ây  140 * 106

gxy  350 * 106 PRINCIPAL STRAINS â1, 2 

âx + ây ; 2

âx  ây 2 gxy 2 b + a b A 2 2 a

 310 * 106 ; 244 * 106 â2  66 * 106 â1  554 * 106 gxy  1.0294 tan 2up  âx  ây 2up  45.8° and up  22.9° and

MAXIMUM SHEAR STRAINS âx ây 2 gxy 2 gmax  a b + a b 2 A 2 2

134.2°

 244 * 106

67.1°

gmax  488 * 106

For up   22.9°: âx1 

âx  ây

âx + ây + 2

2

cos 2u +

gxy 2

us1  up1  45°   67.9° or 112.1° sin 2u

up2  67.1°

â1  554 * 106 â2  66 * 106

; ;

;

us2  us1 + 90°  22.1°

 554 * 106 ‹ up1  22.9°

gmax  488 * 106 gmin  488 * 106 âaver 

âx + ây 2

;

 310 * 106

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SECTION 7.7 Plane Strain

Problem 7.7-8 Solve the preceding problem for the following strains: Px  120 * 106, Py   450 * 106, and gxy   360 * 106.

Solution 7.7-8 Element in plane strain âx  120 * 106

ây   450 * 106

MAXIMUM SHEAR STRAINS âx  ây 2 gxy 2 gmax  a b + a b 2 A 2 2

gxy  360 * 106 PRINCIPAL STRAINS â1,2 

âx + ây 2

 337 * 106

âx  ây 2 gxy 2 a ; b + a b A 2 2

gmax  674 * 106 us1  up1  45°  118.9°

  165 * 106 ; 377 * 106 â1  172 * 106 tan 2up 

gxy âx  ây

â2   502 * 106

  0.6316

âaver 

up  163.9° and 73.9° For up  163.9°: âx  ây

âx + ây + 2

2

cos 2u +

gxy 2

sin 2u

 172 * 106 ‹ up1  163.9° up2  73.9°

â1  172 * 106 â2   502 * 10

; 6

;

;

us2  us1  90°  28.9° gmin  674 * 106

2up  327.7° and 147.7°

âx1 

gmax  674 * 106

âx + ây 2

;

  165 * 106

627

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CHAPTER 7 Analysis of Stress and Strain

Problem 7.7-9 A element of material in plane strain (see figure) is subjected to strains Px  480 * 106, Py  70 * 106, and gxy  420 * 106. Determine the following quantities: (a) the strains for an element oriented at an angle u  75°, (b) the principal strains, and (c) the maximum shear strains. Show the results on sketches of properly oriented element.

Solution 7.7-9 Element in plane strain âx  480 * 106

ây  70 * 106

gxy  420 * 106 âx1  gx1y1 2

âx ây

âx + ây + 2



2

âx  ây 2

cos 2u +

sin 2u +

gxy 2

gxy 2

For up  22.85°: âx1 

âx  ây

âx + ây + 2

2

gxy 2

sin 2u

 568 * 106

sin 2u

â1  568 * 106

‹ up1  22.8° cos 2u

cos 2u +

up2  112.8

°

â2   18 * 10

6

; ;

ây1  âx + ây  âx1 For u  75°: âx1  202 * 106

gx1y1  569 * 106

ây1  348 * 106

MAXIMUM SHEAR STRAINS âx  ây 2 gxy 2 gmax  a b + a b  293 * 106 2 A 2 2 gmax  587 * 106 us1  up1  45°   22.2° or 157.8° gmax  587 * 106

us2  us1 + 90°  67.8°

PRINCIPAL STRAINS â1,2 

;

âx + ây ; 2

 275 * 10

âx  ây 2 gxy 2 b + a b A 2 2

6

a

; 293 * 10

6

â2   18 * 106 â1  568 * 106 gxy 1.0244 tan 2up  âx  ây 2up  45.69° and 225.69° up  22.85° and 112.85°

gmin   587 * 106 âaver 

âx + ây 2

;

 275 * 106

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629

SECTION 7.7 Plane Strain

Problem 7.7-10 Solve the preceding problem for the following data: Px  1120 * 106, Py  430 * 106, gxy  780 * 106, and u  45°.

Solution 7.7-10 Element in plane strain âx   1120 * 106 gxy  780 * 10 âx1  gx1y1 2

ây   430 * 106

6

up2  155.7°

âx  ây

âx + ây + 2



2

âx  ây 2

â1   254 * 106

‹ up1  65.7°

sin 2u +

cos 2u + gxy 2

gxy 2

â2   1296 * 10

sin 2u

cos 2u

ây1  âx + ây  âx1 For u  45°: âx1  385 * 106

gx1y1  690 * 106

ây1  1165 * 106

MAXIMUM SHEAR STRAINS âx  ây 2 gxy 2 gmax  a b + a b 2 A 2 2  521 * 106 gmax  1041 * 106 us1  up1  45°  20.7° gmax  1041 * 106

PRINCIPAL STRAINS âx + ây âx  ây 2 gxy 2 â1, 2  ; a b + a b 2 A 2 2   775 * 10

6

; 521 * 10

us2  us1 + 90°  110.7° gmin   1041 * 106

6

6

â2   1296 * 10 â1   254 * 10 gxy   1.1304 tan 2up  âx  ây

6

2up  131.5° and 311.5° up  65.7° and 155.7° For up  65.7°: âx1 

âx  ây

âx + ây + 2

  254 * 106

2

cos 2u +

gxy 2

;

sin 2u

âaver 

âx + ây 2

;

  775 * 106

6

; ;

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Analysis of Stress and Strain

Problem 7.7-11 A steel plate with modulus of elasticity E  30 * 106 psi and Poisson’s ratio   0.30 is loaded in biaxial stress by normal stresses sx and sy (see figure). A strain gage is bonded to the plate at an angle f  30°. If the stress sx is 18,000 psi and the strain measured by the gage is P  407 * 106, what is the maximum in-plane shear stress (tmax)xy and shear strain (gmax)xy? What is the maximum shear strain (gmax)xz in the xz plane? What is the maximum shear strain (gmax)yz in the yz plane?

y sy

sx

f

x z

Probs. 7.7-11 and 7.7-12

Solution 7.7-11

Steel plate in biaxial stress

sx  18,000 psi

gxy  0

E  30 * 106 psi

sy  ?

MAXIMUM IN-PLANE SHEAR STRESS

  0.30

Strain gage: f  30°

â  407 * 10

(tmax)xy 

6

sx  sy 2

 7800 psi

;

UNITS: All stresses in psi.

STRAINS FROM EQS. (1), (2), AND (3)

STRAIN IN BIAXIAL STRESS (EQS. 7-39)

âx  576 * 106

âx 

1 1 (18,000  0.3sy) (s  sy)  E x 30 * 106

(1)

ây 

1 1 (sy  5400) (sy  sx )  E 30 * 106

(2)

âz  

 0.3 (s sy)   (18,000sy) E x 30 * 106

âx  ây

âx + ây + 2

2

cos 2u +

gxy 2

âz  204 * 106 MAXIMUM SHEAR STRAINS (EQ. 7-75) xy plane:

(3)

2



(gmax)xz 2



yz plane:

(gmax) yz 2 gyz  0

(4)

A

a

âx  ây 2

gxy

2

b + a

2

A

a

âx  âz 2

2

b + a

b

2

6

gxz 2

; b

2

(gmax) xz  780 * 106

gxz  0

sin 2u

1 1 407 * 106  a b a b(12,600 + 0.7sy) 2 30 * 106 1 1 + a ba b (23,400 1.3sy) cos 60° 2 30 * 106 sy  2400 psi Solve for sy :

(gmax) xy

gxy  0 (gmax) xy  676 * 10 xz plane:

STRAINS AT ANGLE f  30° (EQ. 7-71a) âx1 

ây  100 * 106



A

a

ây  2

âz 2

b + a

;

gyz 2 2

b

(gmax) yz  104 * 106

;

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SECTION 7.7 Plane Strain

631

Problem 7.7-12 Solve the preceding problem if the plate is made of aluminum with E  72 GPa and   1/3, the stress sx is 86.4 MPa, the angle f is 21° , and the strain P is 946 * 106.

Solution 7.7-12 Aluminum plate in biaxial stress sx  86.4 MPa

gxy  0

E  72 GPa

sy  ?

MAXIMUM IN-PLANE SHEAR STRESS

  1/3 â  946 * 10

Strain gage: f  21°

(tmax) xy 

6

(1)

1 1 (sy  28.8) ây  (sy  sx)  E 72,000

(2)

 1/ 3 (86.4 sy) âz   (sx sy)   E 72,000

946 * 10

2 6

2

cos 2u +

gxy 2

(gmax) xy 2



(gmax)xz 2



âx  ây 2

2

b + a

gxy 2

sin 2u yz plane:

(gmax) yz 2 gyz  0

A

a

âx  âz 2

2

b + a

b

2

6

gxz 2

b

; 2

(gmax) xz  1600 * 106

gxz  0

(4)

A

a

gxy  0 (gmax) xy  1200 * 10 xz plane:

1 4 1 b a115.2  sy b cos 42° + a ba 2 72,000 3 sy  21.55 MPa

xy plane:

(3)

1 1 2  a ba b a 57.6 + sy b 2 72,000 3

Solve for sy:

ây   101 * 106

MAXIMUM SHEAR STRAINS (EQ. 7-75)

STRAINS AT ANGLE f  21° (EQ. 7-71a) +

;

âz   500 * 106

1 1 1 (86.4  sy) âx  (sx   sy)  E 72,000 3

âx1 

 32.4 MPa

âx  1100 * 106

STRAIN IN BIAXIAL STRESS (EQS. 7-39)

âx  ây

2

STRAINS FROM EQS. (1), (2), AND (3)

UNITS: All stresses in MPa.

âx + ây

sx  sy



A

a

ây  2

âz 2

b + a

;

gyz 2 2

b

(gmax) yz  399 * 106

;

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Analysis of Stress and Strain

sy

Problem 7.7-13 An element in plane stress is subjected to stresses sx  8400 psi,

sy  1100 psi, and txy  1700 psi (see figure). The material is aluminum with modulus of elasticity E  10,000 ksi and Poisson’s ratio   0.33. Determine the following quantities: (a) the strains for an element oriented at an angle u  30°, (b) the principal strains, and (c) the maximum shear strains. Show the results on sketches of properly oriented elements.

txy

y O

x

Probs. 7.7-13 and 7.7-14

Solution 7.7-13

Element in plane strain

sx   8400 psi

sy  1100 psi

txy  1700 psi

E  10,000 ksi

PRINCIPAL STRAINS   0.33

HOOKE’S LAW (EQS. 7-34 AND 7-35) âx 

1 (s  sy)  876.3 * 106 E x

ây 

1 (s  sx)  387.2 * 106 E y

gxy 

txy



2txy(1 + )

G FOR u  30°: âx1 

E

+ 2

2

cos 2u +

2

 

âx  ây 2

sin 2u +

gxy 2

; 2

â1  426 * 10 tan 2up 

gxy 2

 756 * 106 gx1y1

âx + ây

A

 245 * 10

 452.2 * 106

âx  ây

âx + ây

â1,2 

âx  ây

a

6

2

; 671 * 10

6

gxy âx  ây

2

b + a

gxy 2

 434 * 106 gx1y1  868 * 106 ây1  âx + ây  âx1  267 * 106

2

6

â2  961 * 106  0.3579

2up  19.7° and 199.7° up  9.8° and 99.8° FOR up  9.8°:

sin 2u

âx1 

âx  ây

âx + ây + 2

2

cos 2u +

gxy 2

 916 * 106 cos 2u

b

‹ up1  99.8° â1  426 * 106 up2  9.8° â2  916 * 10

6

; ;

sin 2u

sx

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SECTION 7.7 Plane Strain

MAXIMUM SHEAR STRAINS âx  ây 2 gxy 2 gmax  a b + a b 2 A 2 2  671 * 106 gmax  1342 * 106 us1  up1  45°  54.8° gmax  1342 * 106

;

us2  us1 + 90°  144.8° gmin   1342 * 106 ; âx + ây âaver    245 * 106 2

Problem 7.7-14 Solve the preceding problem for the following data: sx  150 MPa, sy  210 MPa, txy  16 MPa, and u  50°. The material is brass with E  100 GPa and   0.34.

Solution 7.7-14 Element in plane strain sx   150 MPa sy   210 MPa txy   16 MPa E  100 GPa   0.34 HOOKE’S LAW (EQS. 7-34 AND 7-35) âx 

1 (s  sy)   786 * 106 E x

ây 

1 (s  sx)   1590 * 106 E y

gxy 

txy G



2txy(1 + ) E

  429 * 106

FOR u  50°: âx1 

âx  ây

âx + ây + 2

2

cos 2u +

gxy 2

  1469 * 106 gx1y1 2



âx  ây 2

sin 2u +

gxy 2

cos 2u

  358.5 * 106 gx1y1   717 * 106 ây1  âx + ây  âx1   907 * 106

sin 2u

PRINCIPAL STRAINS âx + ây âx  ây 2 gxy 2 ; b + a b â1,2  a 2 A 2 2   1188 * 106 ; 456 * 106 â1   732 * 106 â2   1644 * 106 gxy   0.5333 tan 2up  âx  ây 2up  151.9° and 331.9° up  76.0° and 166.0°

633

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Analysis of Stress and Strain

FOR up  76.0°: âx1 

MAXIMUM SHEAR STRAINS âx  ây

âx + ây + 2

 1644 * 10

2

cos 2u +

gxy 2

sin 2u

 456 * 106

6

‹ up1  166.0° â1  732 * 106 up2  76.0° â2  1644 * 106

âx  ây 2 gxy 2 gmax  a b + a b 2 A 2 2

; ;

gmax  911 * 106 us1  up1  45°  121.0° gmax  911 * 106

;

us2  us1  90°  31.0° gmin  911 * 106 âaver 

âx + ây 2

;

 1190 * 106

Problem 7.7-15 During a test of an airplane wing, the strain gage readings from a 45° rosette (see figure) are as follows: gage A, 520 * 106; gage B, 360 * 106; and gage C,  80 * 106 . Determine the principal strains and maximum shear strains, and show them on sketches of properly oriented elements.

y 45° B C

45° A

O

Probs. 7.7-15 and 7.7-16

x

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SECTION 7.7 Plane Strain

Solution 7.7-15

45° strain rosette

âA  520 * 106 âC   80 * 10

âB  360 * 106

6

âx  âA  520 * 10

6

ây  âC   80 * 10

gxy  2âB  âA  âC  280 * 10

6

âx + ây ;

;

6

;

MAXIMUM SHEAR STRAINS 6

âx  ây 2 gxy 2 gmax  a b + a b 2 A 2 2  331 * 106

PRINCIPAL STRAINS

A

2

‹ up1  12.5° â1  551 * 106 up2  102.5° â2   111 * 10

FROM EQS. (7-77) AND (7-78) OF EXAMPLE 7-8:

â1,2 

a

âx  ây 2

2

b + a

gxy 2

b

2

 220 * 106 ; 331 * 106 â1  551 * 106

â2   111 * 106

gmax  662 * 106 us1  up1  45°   32.5°or 147.5° gmax  662 * 106

gxy âx  ây

;

us2  us1 + 90°  57.5° gmin   662 * 106 âaver 

tan 2up 

635

âx + ây 2

;

 220 * 106

 0.4667

2up  25.0° and 205.0° up  12.5° and 102.5° For up  12.5°: âx1 

âx  ây

âx + ây + 2

2

cos 2u +

gxy 2

sin 2u

 551 * 106

Problem 7.7-16 A 45°strain rosette (see figure) mounted on the surface of an automobile frame gives the following

readings: gage A, 310 * 106; gage B, 180 * 106; and gage C, 160 * 106. Determine the principal strains and maximum shear strains, and show them on sketches of properly oriented elements.

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CHAPTER 7 Analysis of Stress and Strain

Solution 7.7-16

45° strain rosette

âA  310 * 106

âB  180 * 106

MAXIMUM SHEAR STRAINS âx  ây 2 gxy 2 gmax  a b + a b 2 A 2 2

âC   160 * 106 FROM EQS. (7-77) AND (7-78) OF EXAMPLE 7-8: âx  âA  310 * 10

6

ây  âC   160 * 10

gxy  2âB  âA  âC  210 * 10

6

2

A

a

âx  ây 2

2

b + a

gxy 2

b

2

â1  332 * 106 gxy âx  ây

âaver 

â2   182 * 106  0.4468

2up  24.1° and 204.1° up  12.0° and 102.0° FOR up  12.0°: âx + ây âx  ây gxy âx1  + cos 2u + sin 2u 2 2 2  332 * 106 ‹ up1  12.0° â1  332 * 106 up2  102.0° â2   182 * 10

; 6

;

;

us2  us1 + 90°  57.0° gmin   515 * 106

 75 * 106 ; 257 * 106

tan 2up 

gmax  515 * 106 gmax  515 * 106

âx + ây ;

 257 * 106 us1  up1  45°   33.0° or 147.0°

PRINCIPAL STRAINS â1,2 

6

âx + ây 2

;

 75 * 106

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SECTION 7.7

Problem 7.7-17 A solid circular bar of diameter d  1.5 in. is

Plane Strain

d

subjected to an axial force P and a torque T (see figure). Strain gages A and B mounted on the surface of the bar give reading Pa  100 * 106 and Pb  55 * 106. The bar is made of steel having E  30 * 106 psi and   0.29.

637

T P

C

(a) Determine the axial force P and the torque T. (b) Determine the maximum shear strain gmax and the maximum shear stress tmax in the bar. B 45∞ A C

Solution 7.7-17

Circular bar (plane stress)

Bar is subjected to a torque T and an axial force P. E  30 * 10 psi   0.29

STRAIN AT u  45°

6

âx1 

Diameter d  1.5 in.

+ 2

2

cos 2u +

âx1  âB  55 * 106

STRAIN GAGES At u  0°:

âx  ây

âx + ây

gxy 2

sin 2u

2u  90°

Substitute numerical values into Eq. (1):

âA  âx  100 * 106

55 * 106  35.5 * 106  (0.0649 * 106)T

At u  45°: âB  55 * 106

Solve for T:

T  1390 lb-in

;

ELEMENT IN PLANE STRESS 4P P sx   A pd 2 âx  100 * 106

sy  0

txy  

MAXIMUM SHEAR STRAIN AND MAXIMUM SHEAR STRESS

16T

gxy  (0.1298 * 106)T  180.4 * 106 rad

pd 3

ây  âx  29 * 106

sx 4P  E pd 2E

âx  ây 2 gxy 2 gmax  a b + a b 2 A 2 2  111 * 106 rad

AXIAL FORCE P âx 

Eq.( 7-75):

2

P

pd Eâx  5300 lb 4

SHEAR STRAIN txy 2txy(1 + ) 32T(1 + )    gxy  G E pd 3E  (0.1298 * 106)T (T  lb-in.)

;

gmax  222 * 106 rad tmax  Ggmax  2580 psi

; ;

(1)

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Analysis of Stress and Strain

Problem 7.7-18 A cantilever beam of rectangular cross

h

section (width b  25 mm, height h  100 mm) is loaded by a force P that acts at the midheight of the beam and is inclined at an angle a to the vertical (see figure). Two strain gages are placed at point C, which also is at the midheight of the beam. Gage A measures the strain in the horizontal direction and gage B measures the strain at an angle b = 60° to the horizontal. The measured strains are Pa  125 * 106 and Pb  375 * 106. Determine the force P and the angle a, assuming the material is steel with E  200 GPa and   1/3.

b h

C a

P

b

B b A C

Probs. 7.7-18 and 7.7-19

Solution 7.7-18

Cantilever beam (plane stress)

Beam loaded by a force P acting at an angle a.   1/3

E  200 GPa

b  25 mm

h  100 mm Axial force F  P sin a Shear force V  P cos a (At the neutral axis, the bending moment produces no stresses.)

At u  60°:

âx 

sx P sin a  E bhE

P sin a  bhEâx  62,500 N txy 3(1 + ) P cos a 3P cos a     gxy  G 2bhG bhE  (8.0 * 109) P cos a

(1)

(2)

FOR u  60°:

STRAIN GAGES At u  0°:

HOOKE’S LAW

âA  âx  125 * 10 âB  375 * 10

6

6

ELEMENT IN PLANE STRESS P sin a F sx   A bh

sy  0

âx1 

âx  ây

âx + ây + 2

2

âx1  âB  375 * 106

gxy 2

sin 2u

(3)

2u  120°

Substitute into Eq. (3): 375 * 106  41.67 * 106  41.67 * 106  (3.464 * 109)P cos a

3V 3P cos a txy     2A 2bh

or P cos a  108,260 N

âx  125 * 106

SOLVE EQS. (1) AND (4):

ây  âx  41.67 * 106

cos 2u +

a  30°

tan a  0.5773 P  125 kN

(4) ;

;

Problem 7.7-19 Solve the preceding problem if the cross-sectional dimensions are b  1.0 in. and h  3.0 in., the gage

angle is b = 75°, the measure strains are Pa  171 * 106 and Pb  266 * 106, and the material is a magnesium alloy with modulus E  6.0 * 106 psi and Poisson’s ratio   0.35.

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Plane Strain

639

P sin a  bhEâx  3078 lb txy 3(1 + v)P cos a 3P cos a     gxy  G 2bhG bhE

(1)

SECTION 7.7

Solution 7.7-19

Cantilever beam (plane stress)

Beam loaded by a force P acting at an angle a. E  6.0 * 10 psi 6

  0.35

b  1.0 in.

h  3.0 in. Axial force F  P sin a Shear foce V  P cos a (At the neutral axis, the bending moment produces no stresses.)

HOOKE’S LAW âx 

 (225.0 * 109)P cos a

STRAIN GAGES At u  0°: At u  75°:

sx P sin a  E bhE

âA  âx  171 * 106 âB  266 * 10

6

ELEMENT IN PLANE STRESS F P sin a sx   A bh

(3)

2u  150°

Substitute into Eq. (3):  266 * 106  55.575 * 106  99.961 * 106

3P cos a 3V txy     2A 2bh âx  171 * 106

FOR u  75°: âx + ây âx  ây gxy + cos 2u + sin 2u âx1  2 2 2 âx1  âB  266 * 106

sy  0

(2)

(56.25 * 109)P cos a

ây  âx   59.85 * 106

or P cos a  3939.8 lb

(4)

SOLVE EQS. (1) AND (4): tan a  0.7813 P  5000 lb

a  38°

;

; y

Problem 7.7-20 A 60° strain rosette, or delta rosette, consists of three electrical-resistance strain gages arranged as shown in the figure. Gage A measures the normal strain Pa in the direction of the x axis. Gages B and C measure the strains Pb and Pc in the inclined directions shown. Obtain the equations for the strains Px, Py, and gxy associated with the xy axis.

B

60° O

60°

A

C

60° x

Solution 7.7-20 Delta rosette (60° strain rosette) STRAIN GAGES Gage A at u  0° Gage B at u  60°

Strain  âA Strain  âB

Gage C at u  120° FOR u  0°:

âx  âA

Strain  âC ;

FOR u  60°: âx + ây âx  ây gxy + cos 2u + sin 2u âx1  2 2 2 âA + ây âA  ây gxy + (cos 120°) + 2 (sin 120°) âB  2 2 3ây gxy 13 âA + + âB  (1) 4 4 4

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Analysis of Stress and Strain

FOR u  120°: âx + ây âx  ây gxy + cos 2u + sin 2u âx1  2 2 2 âA + ây âA  ây gxy + (cos 240°) (sin 240°) âC  2 2 2 3ây gxy 13 âA +  âC  (2) 4 4 4

SOLVE EQS. (1) AND (2): ây 

1 (2âB + 2âC  âA) 3

gxy 

2 (âB  âC) 13

; ;

Problem 7.7-21 On the surface of a structural component in a space vehicle, the

y

strainsare monitored by means of three strain gages arranged as shown in the figure. During a certain maneuver, the following strains were recorded: Pa  1100 * 106, Pb  200 * 106, and Pc  200 * 106. Determine the principal strains and principal stresses in the material, which is a magnesium alloy for which E  6000 ksi and   0.35. (Show the principal strains and principal stresses on sketches of properly oriented element.)

B

30° O

Solution 7.7-21

C

x

A

30-60-90° strain rosette

Magnesium alloy: E  6000 ksi   0.35 STRAIN GAGES Gage A at u  0°

PA  1100 * 106

Gage B at u  90°

âB  200 * 106

Gage C at u  150°

âC  200 * 10

6

FOR u  0°:

âx  âA  1100 * 106

FOR u  90°:

ây  âB  200 * 106

FOR u  150°: âx + ây âx  ây gxy âx1  âC  + cos 2u + sin 2u 2 2 2 200 * 106  650 * 106 + 225 * 106  0.43301gxy Solve for gxy: gxy  1558.9 * 106 PRINCIPAL STRAINS âx + ây âx  ây 2 gxy 2 â1,2  ; a b + a b 2 A 2 2  650 * 106 ; 900 * 106 â1  1550 * 106

â2  250 * 106

tan2 up 

gxy

 13  1.7321

âx  ây

2up  60°

up  30°

FOR up  30°: âx + ây âx  ây gxy âx1  + cos 2u + sin 2u 2 2 2  1550 * 106 ‹ up1  30° up2  120°

â1  1550 * 106 â2  250 * 10

6

; ;

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SECTION 7.7

Plane Strain

641

PRINCIPAL STRESSES (see Eqs. 7-36) s1 

E 1

2

(â1 + â2)

s2 

E 1  2

(â2 + â1)

Substitute numerical values: s1  10,000 psi

s2  2,000 psi

;

y

Problem 7.7-22 The strains on the surface of an experimental device made of pure

aluminum (E  70 Gpa,   0.33) and tested in a space shuttle were measured by means of strain gages. The gages were oriented as shown in the figure, and the measured strains were Pa  1100 * 106, Pb  1496 * 106, and Pc  39.44 * 10 * 6. What is the stress sx in the x direction?

B

O

Solution 7.7-22

  0.33

40°

x

FOR u  140°: âx + ây âx  ây gxy + cos 2u + sin 2u âx1  2 2 2

STRAIN GAGES âA  1100 * 106

Gage A at u  0° Gage B at u  40°

âB  1496 * 10

Gage C at u  140°

Substitute âx1  âc  39.44 * 106 and

6

âC  39.44 * 10

âx  1100 * 106; then simplify and rearrange:

6

0.41318ây  0.49240gxy  684.95 * 106

âx  âA  1100 * 106

SOLVE EQS. (1) AND (2): ây  200.3 * 106

FOR u  40°: âx + ây âx  ây gxy + cos 2u + sin 2u âx1  2 2 2 6

sx 

; then simplify and rearrange:

0.41318ây + 0.49240gxy  850.49 * 106

gxy  1559.2 * 106

HOOKE’S LAW

Substitute âx1  âB  1496 * 106 and âx  1100 * 10

A

40-40-100° strain rosette

Pure aluminum: E  70 GPa

FOR u  0°:

40°

C

(1)

E 1  2

(âx + ây)  91.6 MPa

;

(2)

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CHAPTER 7 Analysis of Stress and Strain

Problem 7.7-23 Solve Problem 7.7-5 by using Mohr’s circle for plane strain.

Solution 7.7-23 Element in plane strain

âx  220 * 106 gxy  180 * 106 R  2(130 * 10  158.11 * 10 a  arctan

ây  480 * 106 gxy  90 * 106 u  50° 2

6 2

) + (90 * 10

6

90  34.70° 130

b  180°  a  2u  45.30°

6 2

)

POINT C: âx1  350 * 106 POINT D (u  50° ): âx1  350 * 106 + R cos b  461 * 106 gx1y1  R sin b  112.4 * 106 2 gx1y1  225 * 106 POINT D ¿ (u  140°): âx1  350 * 106  R cos b  239 * 106 gx1y1   R sin b   112.4 * 106 2 gx1y1   225 * 106

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SECTION 7.7 Plane Strain

Problem 7.7-24 Solve Problem 7.7-6 by using Mohr’s circle for plane strain.

Solution 7.7-24 Element in plane strain âx  420 * 106 gxy  310 * 106

ây   170 * 106 gxy  155 * 106 u  37.5° 2

POINT C: âx1  125 * 106 POINT D (u  37.5°): âx1  125 * 106 + R cos b  351 * 106 gx1y1   R sin b   244.8 * 106 2 gx1y1   490 * 106 POINT D œ (u  127.5°): âx1  125 * 106  R cos b   101 * 106 gx1y1  R sin b  244.8 * 106 2 gx1y1  490 * 106

R  2(295 * 106)2 + (155 * 106)2  333.24 * 106 a  arctan

155  27.72° 295

b  2u  a  47.28°

643

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CHAPTER 7 Analysis of Stress and Strain

Problem 7.7-25 Solve Problem 7.7-7 by using Mohr’s circle for plane strain.

Solution 7.7-25 Element in plane strain âx  480 * 106

ây  140 * 106 gxy gxy   350 * 106   175 * 106 2

MAXIMUM SHEAR STRAINS 2us2  90°a  44.17°

us2  22.1°

2us1  2us2 + 180°  224.17° Point S1: âaver  310 * 10

6

gmax  2R  488 * 106 Point S2: âaver  310 * 106 gmin   488 * 106 R  2(175 * 10  243.98 * 10

6 2

) + (170 * 10

6 2

)

6

a  arctan

175  45.83° 170

POINT C:

âx1  310 * 106

PRINCIPAL STRAINS 2up2  180°  a  134.2°

up2  67.1°

2up1  2up2 + 180°  314.2° up1  157.1° Point P1: â1  310 * 106 + R  554 * 106 Point P2: â2  310 * 106  R  66 * 106

us1  112.1°

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SECTION 7.7 Plane Strain

Problem 7.7-26 Solve Problem 7.7-8 by using Mohr’s circle for plane strain.

Solution 7.7-26 Element in plane strain âx  120 * 106

ây   450 * 106 gxy   180 * 106 gxy   360 * 106 2

MAXIMUM SHEAR STRAINS 2us2  90°  a  57.72°

us2  28.9°

2us1  2us2 + 180°  237.72° Point S1: âaver   165 * 10

6

gmax  2R  674 * 106 Point S2: âaver   165 * 106 R  2(285 * 10  337.08 * 10 a  arctan

6 2

) + (180 * 10

6 2

)

6

180  32.28° 285

Point C: âx1   165 * 106 PRINCIPAL STRAINS 2up2  180°  a  147.72°

up2  73.9°

2up1  2up2 + 180°  327.72° Point P1: â1  R  165 * 10

6

up1  163.9°  172 * 106

Point P2: â2   165 * 106  R   502 * 106

gmin   674 * 106

us1  118.9°

645

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CHAPTER 7 Analysis of Stress and Strain

Problem 7.7-27 Solve Problem 7.7-9 by using Mohr’s circle for plane strain.

Solution 7.7-27 Element in plane strain âx  480 * 106 gxy  420 * 106

ây  70 * 106 gxy  210 * 106 2

u  75°

PRINCIPAL STRAINS 2up1  a  45.69° up1  22.8° 2up2  2up1 + 180°  225.69° Point P1: â1  275 * 10

6

up2  112.8°

+ R  568 * 106

Point P2: â2  275 * 106  R   18 * 106

R  2(205 * 106)2 + (210 * 106)2  293.47 * 106 a  arctan

210  45.69° 205

b  a + 180°  2u  75.69° Point C: âx1  275 * 106 Point D (u  75°): âx1  275 * 106  R cos b  202 * 106 gx1y1   R sin b   284.36 * 106 2 gx1y1   569 * 106 Point

Dœ

(u  165°):

âx1  275 * 106 + R cos b  348 * 106 gx1y1  R sin b  284.36 * 106 2 gx1y1  569 * 106

MAXIMUM SHEAR STRAINS 2us2  90° + a  135.69° 2us1  2us2 + 180°  315.69° Point S1: âaver  275 * 10

6

gmax  2R  587 * 106 Point S2: âaver  275 * 106 gmin   587 * 106

us2  67.8° us1  157.8°

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SECTION 7.7 Plane Strain

647

Problem 7.7-28 Solve Problem 7.7-10 by using Mohr’s circle for plane strain.

Solution 7.7-28 Element in plane strain âx   1120 * 106 gxy  780 * 106

ây   430 * 106 gxy 2

 390 * 106

u  45°

PRINCIPAL STRAINS 2up1  180°  a  131.50°

up1  65.7°

2up2  2up1 + 180°  311.50° Point P1: â1   775 * 10

6

up2  155.7°

+ R   254 * 106

Point P2: â2   775 * 106  R   1296 * 106

R  2(345 * 106)2 + (390 * 106)2  520.70 * 106 a  arctan

390  48.50° 345

b  180°  a  2u  41.50° Point C: âx1   775 * 106

MAXIMUM SHEAR STRAINS

Point D: (u  45°):

2us2  2us1 + 180°  221.50°

âx1   775 * 106 + R cos b   385 * 106 gx1y1  R sin b  345 * 106 gx1y1  690 * 106 2 Point D ¿ : (u  135°) âx1   775 * 106  R cos b   1165 * 106 gx1y1   R sin b   345 * 106 2 gx1y1   690 * 106

2us1  90°  a  41.50° Point S1: âaver   775 * 10

us1  20.7° 6

gmax  2R  1041 * 106 Point S2: âaver   775 * 106 gmin   1041 * 106

us2  110.7°

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8 Applications of Plane Stress (Pressure Vessels, Beams, and Combined Loadings)

Spherical Pressure Vessels When solving the problems for Section 8.2, assume that the given radius or diameter is an inside dimension and that all internal pressures are gage pressures.

Problem 8.2-1 A large spherical tank (see figure) contains gas at a pressure of 450 psi. The tank is 42 ft in diameter and is constructed of high-strength steel having a yield stress in tension of 80 ksi. Determine the required thickness (to the nearest 1/4 inch) of the wall of the tank if a factor of safety of 3.5 with respect to yielding is required. Probs. 8.2-1 and 8.2-2

Solution 8.2-1 Radius:

r

1 42 * 12 2

r  252 in.

Internal Pressure: p  450 psi Yield stress:

t

prn 2sY

t  2.481 in.

to nearest 1/4 inch, t min  2.5 in.

;

sY  80 ksi (steel)

Factor of safety: n  3.5 MINIMUM WALL THICKNESS tmin From Eq. (8-1): smax 

pr pr sY  or 2t n 2t

649

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Applications of Plane Stress

Problem 8.2-2 Solve the preceding problem if the internal pressure is 3.75 MPa, the diameter is 19 m, the yield stress is 570 MPa, and the factor of safety is 3.0. Determine the required thickness to the nearest millimeter.

Solution 8.2-2 Radius:

r

1 (19 m) 2

r  9.5  103 mm

Internal Pressure: p  3.75 MPa Yield stress:

t

prn 2sY

t  93.8 mm

Use the next higher millimeter t min  94 mm

sY  570 MPa

Factor of safety: n  3 MINIMUM WALL THICKNESS tmin From Eq. (8-1): smax 

pr pr sY  or 2t n 2t

Problem 8.2-3 A hemispherical window (or viewport) in a decompression chamber (see figure) is subjected to an internal air pressure of 80 psi. The port is attached to the wall of the chamber by 18 bolts. Find the tensile force F in each bolt and the tensile stress s in the viewport if the radius of the hemisphere is 7.0 in. and its thickness is 1.0 in.

Solution 8.2-3

Hemispherical viewport

FREE-BODY DIAGRAM

T  total tensile force in 18 bolts a FHORIZ  T  pA  0

T  pA  p(pr 2)

F  force in one bolt F Radius:

r  7.0 in.

Internal pressure: p  80 psi Wall thickness: 18 bolts

t  1.0 in.

1 T  (ppr 2)  684 lb 18 18

;

TENSILE STRESS IN VIEWPORT (EQ. 8-1) s

pr  280 psi 2t

;

;

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SECTION 8.2

651

Spherical Pressure Vessels

Problem 8.2-4 A rubber ball (see figure) is inflated to a pressure of 60 kPa. At that pressure the diameter of the ball is 230 mm and the wall thickness is 1.2 mm. The rubber has modulus of elasticity E  3.5 MPa and Poisson’s ratio v  0.45. Determine the maximum stress and strain in the ball.

Prob. 8.2-4, 8.2-5

Solution 8.2-4

Rubber ball

CROSS-SECTION

MAXIMUM STRESS (EQ. 8-1) smax 

(60 kPa)(115 mm) pr  2t 2(1.2 mm)

 2.88 MPa

;

MAXIMUM STRAIN (EQ. 8-4) Radius:

r  (230 mm)/2  115 mm

Internal pressure:

p  60 kPa

Wall thickness:

t  1.2 mm

âmax 

pr (60 kPa)(115 mm) (1  v)  (0.55) 2tE 2(1.2 mm)(3.5 MPa)

 0.452

;

Modulus of elasticity: E  3.5 MPa (rubber) v  0.45 (rubber)

Poisson’s ratio:

Problem 8.2-5 Solve the preceding problem if the pressure is 9.0 psi, the diameter is 9.0 in., the wall thickness is 0.05 in., the modulus of elasticity is 500 psi, and Poisson’s ratio is 0.45.

Solution 8.2-5

Rubber ball Modulus of elasticity: E  500 psi (rubber)

CROSS-SECTION

v  0.45 (rubber)

Poisson’s ratio:

MAXIMUM STRESS (EQ. 8-1) smax  1 (9.0 in.)  4.5 in. 2

Radius:

r

Internal pressure:

p  9.0 psi

Wall thickness:

t  0.05 in.

(9.0 psi)(4.5 in.) pr   405 psi 2t 2(0.05 in.)

;

MAXIMUM STRAIN (EQ. 8-4) âmax 

(9.0 psi)(4.5 in.) pr (1  v)  (0.55) 2tE 2(0.05 in.)(500 psi)

 0.446

;

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Applications of Plane Stress

Problem 8.2-6 A spherical steel pressure vessel (diameter 480 mm,

Cracks in coating

thickness 8.0 mm) is coated with brittle lacquer that cracks when the strain reaches 150  106 (see figure). What internal pressure p will cause the lacquer to develop cracks? (Assume E  205 GPa and v  0.30.)

Solution 8.2-6

Spherical vessel with brittle coating Cracks occur when max  150  106 pr (1  v) From Eq. (8-4): âmax  2tE

CROSS-SECTION

‹ P r  240 mm t  8.0 mm

E  205 GPa (steel) v  0.30

P

2tE âmax r(1 v)

2(8.0 mm)(205 GPa)(150 * 106) (240 mm)(0.70)

 2.93 MPa

;

Problem 8.2-7 A spherical tank of diameter 48 in. and wall thickness 1.75 in. contains compressed air at a pressure of 2200 psi. The tank is constructed of two hemispheres joined by a welded seam (see figure).

Weld

(a) What is the tensile load f (lb per in. of length of weld) carried by the weld? (b) What is the maximum shear stress tmax in the wall of the tank? (c) What is the maximum normal strain  in the wall? (For steel, assume E  30  106 psi and v  0.29.)

Probs. 8.2-7 and 8.2-8

Solution 8.2-7 r  24 in. t  1.75 in.

E  30  106 psi v  0.29 (steel)

(b) MAXIMUM SHEAR STRESS IN WALL (EQ. 8-3) tmax 

(2200 psi)(24 in.) pr  7543 psi  4t 4 (1.75 in.)

(a) TENSILE LOAD CARRIED BY WELD T  Total load

f  load per inch

T  pA  ppr 2

c  Circumference of tank  2pr

T p1pr 2 pr (2200 psi)(24 in.) f    c 2pr 2 2 2

 26.4 k/in.

;

(c) MAXIMUM NORMAL STRAIN IN WALL (EQ. 8-4) âmax 

(2200 psi)(24 in.)(0.71) pr (1  v)  2 tE 2(1.75 in.)130106 psi2

 3.57 * 104

;

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653

Spherical Pressure Vessels

Problem 8.2-8 Solve the preceding problem for the following data: diameter 1.0 m, thickness 48 mm, pressure 22 MPa, modulus 210 GPa, and Poisson’s ratio 0.29.

Solution 8.2-8 r  0.5 m

E  210 GPa

t  48 mm

(b) MAXIMUM SHEAR STRESS IN WALL (EQ. 8-3)

v  0.29 (steel)

tmax 

(a) TENSILE LOAD CARRIED BY WELD T  Total load

 57.3 MPa

f  load per inch

T  pA  ppr2

c  Circumference of tank  2pr

p1pr 2 pr (22 MPa)(0.5 m) T    c 2pr 2 2  5.5 MN/m

;

(c) MAXIMUM NORMAL STRAIN IN WALL (EQ. 8-4)

2

f

(22 MPa)(0.5 m) pr  4t 4 (48 mm)

âmax 

pr (1  v) (22 MPa)(0.5 m) 0.71  2 tE 2(48 mm)(210 GPa)

 3.87 * 104

;

;

Problem 8.2-9 A spherical stainless-steel tank having a diameter of 22 in. is used to store propane gas at a pressure of 2450 psi. The properties of the steel are as follows: yield stress in tension, 140,000 psi; yield stress in shear, 65,000 psi; modulus of elasticity, 30  106 psi; and Poisson’s ratio, 0.28. The desired factor of safety with respect to yielding is 2.8. Also, the normal strain must not exceed 1100  106. Determine the minimum permissible thickness tmin of the tank.

Solution 8.2-9 r  11 in.

E  30  106 psi

p  2450 psi

v  0.28 (steel)

sY  140000 psi

t2 

n  2.8 max  1100  106

tY  65000 psi

MIMIMUM WALL THICKNESS t (1) TENSION (EQ. 8-1)

pr sY 2a b n (2450 psi) (11 in.)

t1 



smax 

pr 2 t1

2

2.8

t3  

 0.269 in.

pr 4t2

(2450 psi)(11 in.) pr  0.29 in.  65000 psi tY 4n 4 2.8

(3) STRAIN (EQ. 8-4)

pr  2smax

140000 psi

tmax 

(2) SHEAR (EQ. 8-3)

âmax 

pr (1  v) 2t 3 E

pr (1  v) 2âmax E (2450 psi)(11 in.)

211100 * 1062130 * 106 psi2

0.72

 0.294 in. t3  t2  t1

Thus, tmin  0.294 in.

;

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Problem 8.2-10 Solve the preceding problem if the diameter is 500 mm, the pressure is 18 MPa, the yield stress in tension is 975 MPa, the yield stress in shear is 460 MPa, the factor of safety is 2.5, the modulus of elasticity is 200 GPa, Poisson’s ratio is 0.28, and the normal strain must not exceed 1210  106.

Solution 8.2-10 r  250 mm

E  200 GPa

p  18 MPa

v  0.28 (steel)

(2) SHEAR (EQ. 8-3)

n  2.5

tY  460 MPa

max  1210  106

MINIMUM WALL THICKNESS t

t1 

pr  2smax

smax 

pr 4t 2

(18 MPa)(250 mm) pr   6.114 mm tY 460 MPa 4 4 n 2.5 pr (3) STRAIN (EQ. 8-4) âmax  (1  v) 2t 3E t2 

sY  975 MPa

(1) TENSION (EQ. 8-1)

tmax 

pr 2t 1

pr sY 2a b n

(18 MPa)(250 mm)  5.769 mm  975 MPa 2 2.5

t3  

pr 2âmax E

(1  v)

(18 MPa) (250 mm)

211210 * 1062(200 GPa)

0.72

 6.694 mm t3  t2  t1

Thus, tmin  6.69 mm

Problem 8.2-11 A hollow pressurized sphere having radius r  4.8 in. and

wall thickness t  0.4 in. is lowered into a lake (see figure). The compressed air in the tank is at a pressure of 24 psi (gage pressure when the tank is out of the water). At what depth D0 will the wall of the tank be subjected to a compressive stress of 90 psi?

D0

;

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SECTION 8.3

Solution 8.2-11

Cylindrical Pressure Vessels

655

Pressurized sphere under water

CROSS-SECTION

D0  depth of water (in.)

r  4.8 in.

p1  24 psi

t  0.4 in.

g  density of water  62.4 lb/ft

3

(1) IN AIR: p1  24 psi

p2  gD0  a

( p1  p2)r pr  2t 2t

90 psi 

(2) UNDER WATER: p1  24 psi

1728 in.3/ ft3

b D0  0.036111 D0 ( psi)

Compressive stress in tank wall equals 90 psi. (Note: s is positive in tension.) s

(1) IN AIR

62.4 lb/ft3

s  90 psi

(24 psi  0.03611 D0)(4.8 in.) 2(0.4 in.)

 144  0.21667 D0 Solve for D0: D0 

234 0.21667

 1080 in.  90 ft

(2) UNDER WATER

Cylindrical Pressure Vessels When solving the problems for Section 8.3, assume that the given radius or diameter is an inside dimension and that all internal pressures are gage pressures.

Problem 8.3-1 A scuba tank (see figure) is being designed for an internal pressure of 1600 psi with a factor of safety of 2.0 with respect to yielding. The yield stress of the steel is 35,000 psi in tension and 16,000 psi in shear. If the diameter of the tank is 7.0 in., what is the minimum required wall thickness?

;

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Solution 8.3-1

Page 656

Applications of Plane Stress

Scuba tank sallow 

sY  17,500 psi n

t allow 

tY  8,000 psi n

Find required wall thickness t. (1) BASED ON TENSION (EQ. 8-5) t1 

pr sallow



(1600 psi)(3.5 in.) 17,500 psi

Cylindrical pressure vessel p  1600 psi r  3.5 in.

n  2.0

d  7.0 in.

sY  35,000 psi

Y  16,000 psi

pr t

 0.320 in.

tmax 

(2) BASED ON SHEAR (EQ. 8-10) t2 

smax 

pr 2t

(1600 psi)(3.5 in.) pr   0.350 in. 2tallow 2(8,000 psi)

Shear governs since t2  t1  tmin  0.350 in.

;

Problem 8.3-2 A tall standpipe with an open top (see figure) has diameter d  2.2 m

d

and wall thickness t  20 mm.

(a) What height h of water will produce a circumferential stress of 12 MPa in the wall of the standpipe? (b) What is the axial stress in the wall of the tank due to the water pressure? h

Solution 8.3-2 d  2.2 m

r  1.1 m

t  20 mm

weight density of water g  9.81 kN/m3 height of water

h

water pressure

p  h

(a) HEIGHT OF WATER s1 

pr 0.00981h (1.1 m)  12 MPa  t 20 mm

h

12 (20)  22.2 m 0.00981 (1.1)

(b) AXIAL

STRESS

IN

THE

WALL

DUE

TO

WATER

PRESSURE ALONE

Because the top of the tank is open, the internal pressure of the water produces no axial (longitudinal) stresses in the wall of the tank. Axial stress equals zero. ;

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SECTION 8.3

Problem 8.3-3 An inflatable structure used by a traveling

Cylindrical Pressure Vessels

657

Longitudinal seam

circus has the shape of a half-circular cylinder with closed ends (see figure). The fabric and plastic structure is inflated by a small blower and has a radius of 40 ft when fully inflated. A longitudinal seam runs the entire length of the “ridge” of the structure. If the longitudinal seam along the ridge tears open when it is subjected to a tensile load of 540 pounds per inch of seam, what is the factor of safety n against tearing when the internal pressure is 0.5 psi and the structure is fully inflated?

Solution 8.3-3

Inflatable structure

Half-circular cylinder r  40 ft  480 in. Internal pressure p  0.5 psi T  tensile force per unit length of longitudinal seam Seam tears when T  Tmax  540 lb/in. Find factor of safety against tearing.

CIRCUMFERENTIAL STRESS (EQ. 8-5) s1 

pr where t  thickness of fabric t

Actual value of T due to internal pressure  s1t  T  s1t  pr  (0.5 psi)(480 in.)  240 lb/in. FACTOR OF SAFETY n

Tmax 540 lb/in.   2.25 T 240 lb/in.

;

Problem 8.3-4 A thin-walled cylindrical pressure vessel of radius r is subjected simultaneously to internal gas pressure p and a compressive force F acting at the ends (see figure). What should be the magnitude of the force F in order to produce pure shear in the wall of the cylinder?

Solution 8.3-4

F

Cylindrical pressure vessel STRESSES (SEE EQ. 8-5 AND 8-6):

r  Radius p  Internal pressure

s1 

pr t

s2 

pr pr F F    2t A 2t 2prt

F

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Applications of Plane Stress

FOR PURE SHEAR, the stresses s1 and s2 must be equal in magnitude and opposite in sign (see, e.g., Fig. 7-11 in Section 7.3).  s1  s2 OR

pr pr F  a  b t 2t 2prt

Solve for F: F  3ppr2

;

Problem 8.3-5 A strain gage is installed in the longitudinal direction on the surface of an aluminum beverage can (see figure). The radius-to-thickness ratio of the can is 200. When the lid of the can is popped open, the strain changes by 僆0  170  106. What was the internal pressure p in the can? (Assume E  10  106 psi and v  0.33.)

12 FL OZ (355 mL)

Solution 8.3-5

Aluminum can STRAIN IN LONGITUDINAL DIRECTION (EQ. 8-11a) â2 

pr (1  2v) 2tE

â2  â0

‹

or p  p

2tEâ2 r(1  2v)

2tEâ0 2Eâ0  (r)(1  2v) (r/t) (1  2v)

Substitute numerical values:

r  200 t

E  10  106 psi

v  0.33

0  change in strain when pressure is released  170  106 Find internal pressure p.

p

2(10 * 106 psi)(170 * 106) (200)(1  0.66)

 50 psi

;

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SECTION 8.3

Problem 8.3-6 A circular cylindrical steel tank (see figure) contains a volatile fuel under pressure. A strain gage at point A records the longitudinal strain in the tank and transmits this information to a control room. The ultimate shear stress in the wall of the tank is 84 MPa, and a factor of safety of 2.5 is required. At what value of the strain should the operators take action to reduce the pressure in the tank? (Data for the steel are as follows: modulus of elasticity E  205 GPa and Poisson’s ratio v  0.30.)

659

Cylindrical Pressure Vessels

Pressure relief valve

Cylindrical tank

A

Solution 8.3-6 tULT  84 MPa

E  205 GPa

n  2.5

tULT n

tmax 

v  0.3

tmax  33.6 MPa

Find maximum allowable strain reading at the gage s1 

pr t

s2 

pr 2t

From Eq. (8-11a) â2 

pr (1  2v) 2tE

â2max  âmax 

From Eq. (8-10) tmax 

pr s1  2 2t

Pmax 

pmax r tmax (1  2v)  (1  2v) 2tE E tmax (1  2v) E

max  6.56  105

2ttmax r

Cylinder

Problem 8.3-7 A cylinder filled with oil is under pressure from a piston, as shown in the figure. The diameter d of the piston is 1.80 in. and the compressive force F is 3500 lb. The maximum allowable shear stress tallow in the wall of the cylinder is 5500 psi. What is the minimum permissible thickness tmin of the cylinder wall? (See the figure on the next page.)

F

p Piston

Probs. 8.3-7 and 8.3-8

Solution 8.3-7

Cylinder with internal pressure Maximum shear stress (Eq. 8-10): tmax 

d  1.80 in.

r  0.90 in.

F  3500 lb

tallow  5500 psi

Find minimum thickness tmin. F F Pressure in cylinder: p   A pr 2

pr F  2t 2prt

Minimum thickness: t min 

F 2prtallow

Substitute numerical values: t min 

3500 lb  0.113 in. 2p(0.90 in.)(5500 psi)

;

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Problem 8.3-8 Solve the preceding problem if d  90 mm, F  42 kN, and tallow  40 MPa.

Solution 8.3-8

Cylinder with internal pressure Maximum shear stress (Eq. 8-10): pr F  2t 2prt

tmax 

Minimum thickness: d  90 mm F  42.0 kN

r  45 mm

t min 

tallow  40 MPa

Find minimum thickness tmin. Pressure in cylinder: p 

F 2pr tallow

Substitute numerical values:

F F  A pr 2

t min 

42.0 kN  3.71 mm 2p(45 mm)(40 MPa)

;

Problem 8.3-9 A standpipe in a water-supply system (see figure) is 12 ft in diameter and 6 inches thick. Two horizontal pipes carry water out of the standpipe; each is 2 ft in diameter and 1 inch thick. When the system is shut down and water fills the pipes but is not moving, the hoop stress at the bottom of the standpipe is 130 psi. (a) What is the height h of the water in the standpipe? (b) If the bottoms of the pipes are at the same elevation as the bottom of the standpipe, what is the hoop stress in the pipes?

Solution 8.3-9

Vertical standpipe (a) FIND HEIGHT h OF WATER IN THE STANDPIPE p  pressure at bottom of standpipe  gh From Eq. (8-5): s1 

pr ghr  t t

or h 

Substitute numerical values: h

d  12 ft  144 in. g  62.4 lb/ft3 

r  72 in.

t  6 in.

62.4 lb/in.3 1728

s1  hoop stress at bottom of standpipe  130 psi

(130 psi)(6 in.) 62.4 a lb/in.3 b (72 in.) 1728

 25 ft

;

 300 in.

s1t gr

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SECTION 8.3

HORIZONTAL PIPES d1  2 ft  24 in.

r1  12 in.

t1  1.0 in.

s1 

(b) FIND HOOP STRESS s1 IN THE PIPES Since the pipes are 2 ft in diameter, the depth of water to the center of the pipes is about 24 ft. h1 ⬇ 24 ft  288 in.

p1  gh1



Cylindrical Pressure Vessels

661

p1r1 gh 1r1  t1 t1 a

62.4 lb/in.3 b (288 in.)(12 in.) 1728 1.0 in.

 125 psi Based on the average pressure in the pipes: s1 ⬇ 125 psi

;

Problem 8.3-10 A cylindrical tank with hemispherical heads is constructed of steel sections that are welded circumferentially (see figure). The tank diameter is 1.25 m, the wall thickness is 22 mm, and the internal pressure is 1750 kPa. (a) Determine the maximum tensile stress sh in the heads of the tank. (b) Determine the maximum tensile stress sc in the cylindrical part of the tank. (c) Determine the tensile stress sw acting perpendicular to the welded joints. (d) Determine the maximum shear stress th in the heads of the tank. (e) Determine the maximum shear stress tc in the cylindrical part of the tank.

Welded seams

Probs. 8.3-10 and 8.3-11

Solution 8.3-10 d  1.25 m

r

d 2

t  22 mm

p  1750 kPa

(c) TENSILE STRESS IN WELDS (EQ. 8-6) sw 

pr 2t

sw  24.9 MPa

;

(a) MAXIMUM TENSILE STRESS IN HEMISPHERES (EQ. 8-1) sh 

pr 2t

sh  24.9 MPa

;

(d) MAXIMUM SHEAR STRESS IN HEMISPHERES (EQ. 8-3) th 

pr 4t

th  12.43 MPa

;

(b) MAXIMUM STRESS IN CYLINDER (EQ. 8-5) sc 

pr t

sc  49.7 MPa

;

(e) MAXIMUM SHEAR STRESS IN CYLINDER (EQ. 8-10) tc 

pr 2t

tc  24.9 MPa

;

Problem 8.3-11 A cylindrical tank with diameter d  18 in. is subjected to internal gas pressure p  450 psi. The tank is constructed of steel sections that are welded circumferentially (see figure). The heads of the tank are hemispherical. The allowable tensile and shear stresses are 8200 psi and 3000 psi, respectively. Also, the allowable tensile stress perpendicular to a weld is 6250 psi. Determine the minimum required thickness tmin of (a) the cylindrical part of the tank and (b) the hemispherical heads.

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Solution 8.3-11 d  18 in.

r

d 2

sallow  8200 psi

(tension)

tallow  3000 psi

(shear)

sa  6250 psi

WELD

t min 

p  450 psi

tmax 

pr t min  2tallow s

WELD

tmin  0.324 in.

t min  0.675 in.

TENSION t min 

smax 

pr 2sallow

pr 2t

tmin  0.247 in.

tmax 

SHEAR

pr 2t

t min 

tmin  0.675 in.

t min  0.338 in.

pr 4tallow

pr 4t tmin  0.338 in. ;

pr 2t

Problem *8.3-12 A pressurized steel tank is constructed with a helical weld that makes an angle a  55° with the longitudinal axis (see figure). The tank has radius r  0.6 m, wall thickness t  18 mm, and internal pressure p  2.8 MPa. Also, the steel has modulus of elasticity E  200 GPa and Poisson’s ratio v  0.30. Determine the following quantities for the cylindrical part of the tank. (a) (b) (c) (d)

;

(b) FIND MINIMUM THICKNESS OF HEMISPHERES

(tension)

(a) FIND MINIMUM THICKNESS OF CYLINDER pr TENSION smax  t pr t min  tmin  0.494 in. sallow SHEAR

pr 2sa

Helical weld a

The circumferential and longitudinal stresses. The maximum in-plane and out-of-plane shear stresses. The circumferential and longitudinal strains. The normal and shear stresses acting on planes parallel and perpendicular to the weld (show these stresses on a properly oriented stress element).

Probs. 8.3-12 and 8.3-13

Solution 8.3-12 a  55 deg p  2.8 MPa

r  0.6 m

t  18 mm

E  200 GPa

v  0.3

(a) CIRCUMFERENTIAL STRESS pr s1  t

s1  93.3 MPa

pr 2t

s2  46.7 MPa

t1 

s1  s2 2

t1  23.3 MPa

OUT-OF-PLANE SHEAR STRESS ;

LONGITUDIAL STRESS s2 

(b) IN-PLANE SHEAR STRESS

;

t2 

s1 2

t2  46.7 MPa

;

;

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SECTION 8.3

(c) CIRCUMFERENTIAL STRAIN s1 â1  (2  v) 2E

1  3.97  104

;

For u  35 deg sx  sy sx + sy + cos (2u) sx1  2 2

LONGITUDINAL STRAIN s2 (1  2v) â2  E

+ txy sin (2u) 5

2  9.33  10

;

(d) STRESS ON WELD u  90 deg  a sx  s2

sx1  62.0 MPa ; sx  sy tx1y1   sin (2u) + txy cos (2u) 2 tx1y1  21.9 MPa

u  35 deg

sx  46.667 MPa

sy  93.333 MPa

663

Cylindrical Pressure Vessels

;

sy1  sx sy  sx1

sy  s1

sy1  78.0 MPa

;

txy  0

Problem *8.3-13 Solve the preceding problem for a welded tank with a  62°, r  19 in., t  0.65 in., p  240 psi, E  30  106 psi, and v  0.30.

Solution 8.3-13 a  62 deg

r  19 in.

p  240 psi

E  30  106 psi

LONGITUDINAL STRAIN

t  0.65 in. v  0.3

â2 

(a) CIRCUMFERENTIAL STRESS s1 

pr t

s1  7015 psi

s2 (1  2v) E

2  4.68  105

(d) STRESS ON WELD

;

u  90 deg  a

u  28 deg

LONGITUDIAL STRESS

sx  s2

pr s2  2t

sy  7.015  10 psi

;

s1  s2 2

t1  1754 psi

;

s1 2

t2  3508 psi

sx1  4281 psi ; sx  sy sin (2u) + txy cos (2u) tx1y1   2

;

tx1y1  1454 psi

(c) CIRCUMFERENTIAL STRAIN â1 

s1 (2  v) 2E

txy  0

+ txy sin (2u)

OUT-OF-PLANE SHEAR STRESS t2 

sy  s1

For u  28 deg sx + sy sx  sy + cos (2u) sx1  2 2

(b) IN-PLANE SHEAR STRESS t1 

sx  3.508  103 psi 3

s2  3508 psi

;

1  1.988  104

;

;

sy1  sx sy  sx1

sy1  6242 psi

;

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Applications of Plane Stress

Maximum Stresses in Beams When solving the problems for Section 8.4, consider only the in-plane stresses and disregard the weights of the beams

Problem 8.4-1 A cantilever beam of rectangular cross section is subjected

P

to a concentrated load P ⫽ 17 k acting at the free end (see figure). The beam has width b ⫽ 3 in. and height h ⫽ 12 in. Point A is located at distance c ⫽ 2.5 ft from the free end and distance d ⫽ 9 in. from the bottom of the beam. Calculate the principal stresses s1 and s2 and the maximum shear stress tmax at point A. Show these stresses on sketches of properly oriented elements.

A

h c

b

d

Probs. 8.4-1 and 8.4-2

Solution 8.4-1 P ⫽ 17 k

c ⫽ 2.5 ft

h ⫽ 12 in.

v ⫽ 0.3

b ⫽ 3 in.

d ⫽ 9 in.

For u1 ⫽ up sx1 ⫽

STRESS AT POINT A bh3 I⫽ 12 V⫽P

I ⫽ 432 in.

sx ⫽ ⫺

cos (2 u1)

sx ⫺ sy

2

txy ⫽ t

sy ⫽ 0

2txy 1 b up ⫽ atan a 2 sx ⫺ sy

cos (2 u2)

sx2 ⫽ ⫺77.971 psi Therefore

2txy sx ⫺ sy

up1 ⫽ u2

txy ⫽ 531.25 psi

s2 ⫽ 3620 psi

⫽ 0.3

up ⫽ 8.35 deg

s2 ⫽ sx1

up2 ⫽ u1

up1 ⫽ 98.4 deg

;

;

up2 ⫽ 8.35 deg

;

;

MAXIMUM SHEAR STRESSES tmax ⫽

⫽

2

s1 ⫽ ⫺78.0 psi

PRINCIPAL STRESSES 2txy

sx ⫺ sy +

s1 ⫽ sx2

Q ⫽ 40.5 in.3

t ⫽ 531.25 psi

u2 ⫽ 98.35 deg

+ txy sin (2 u2)

sx ⫽ 3.542 ⫻ 103 psi

sx ⫽ 3.542 ⫻ 103 psi

tan12up2 ⫽

2

sx + sy

sx2 ⫽

yA ⫽ 3 in.

d h Q ⫽ bd a ⫺ b 2 2 VQ t⫽ Ib

2

For u2 ⫽ 90 deg ⫹ up

V ⫽ 1.7 ⫻ 104 lb

MyA I

+

sx1 ⫽ 60.306 Mpa

M ⫽ ⫺5.1 ⫻ 105 lb-in.

h yA ⫽ ⫺ + d 2

sx ⫺ sy

sx + sy

+ txy sin (2u1) 4

M ⫽ ⫺Pc

u1 ⫽ 8.35 deg

A

a

sx ⫺ sy 2

tmax ⫽ 1849 psi

2

2 b + txy

;

us1 ⫽ up1 ⫺ 45 deg

us1 ⫽ 53.4 deg

us2 ⫽ us1 ⫹ 90 deg

us2 ⫽ 143.4 deg

savg ⫽

sx + sy 2

savg ⫽ 1771 psi

; ; ;

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SECTION 8.4

665

Maximum Stresses in Beams

Problem 8.4-2 Solve the preceding problem for the following data: P ⫽ 130 kN, b ⫽ 80 mm, h ⫽ 260 mm, c ⫽ 0.6 m, and d ⫽ 220 mm.

Solution 8.4-2 P ⫽ 130 kN

c ⫽ 0.6 m

b ⫽ 80 mm

d ⫽ 220 mm

h ⫽ 260 mm

v ⫽ 0.3

For u1 ⫽ up sx1 ⫽

STRESS AT POINT A I⫽

bh3 12

M ⫽ ⫺7.8 ⫻ 104 N # m

V⫽P

V ⫽ 1.3 ⫻ 105 N

h yA ⫽ ⫺ + d 2 sx ⫽ ⫺

MyA I

Q ⫽ bd a

2

2

cos (2u1)

Q ⫽ 3.52 ⫻ 105 mm3

t ⫽ 4.882 MPa

txy ⫽ t

sy ⫽ 0

txy ⫽ 4.882 MPa

PRINCIPAL STRESSES 2txy sx ⫺ sy

⫽

2 txy 1 b up ⫽ a tan a 2 sx ⫺ sy

For u2 ⫽ 90 deg ⫹ up

2 txy sx ⫺ sy

u2 ⫽ 94.628 deg

sx ⫺ sy

sx + sy + 2

2

cos (2u2)

+ txy sin (2u 2)

sx ⫽ 59.911 MPa

sx ⫽ 59.911 MPa

tan(2up) ⫽

+

sx1 ⫽ 60.306 MPa

sx2 ⫽

yA ⫽ 90 mm

d h ⫺ b 2 2

VQ t⫽ Ib

sx ⫺ sy

sx + sy

+ txy sin (2u1)

I ⫽ 1.172 ⫻ 108 mm4

M ⫽ ⫺Pc

u1 ⫽ 4.628 deg

⫽ 0.163

up ⫽ 4.628 deg

sx2 ⫽ ⫺0.395 MPa Therefore s1 ⫽ sx1

s1 ⫽ 60.3 MPa

s2 ⫽ sx2

up2 ⫽ u2

up1 ⫽ 4.63 deg

;

s2 ⫽ ⫺0.395 MPa

;

;

up2 ⫽ 94.6 deg

;

MAXIMUM SHEAR STRESSES sx ⫺ sy 2 2 b + t xy tmax ⫽ a A 2 tmax ⫽ 30.4 MPa

;

us1 ⫽ up1 ⫺ 45 deg

us1 ⫽ ⫺40.4 deg

us2 ⫽ us1 ⫹ 90 deg

us2 ⫽ 49.6 deg

savg ⫽

Problem 8.4-3 A simple beam of rectangular cross section (width 4 in., height 10 in.) carries a uniform load of 1200 lb/ft on a span of 12 ft (see figure). Find the principal stresses s1 and s2 and the maximum shear stress tmax at a cross section 2 ft from the left-hand support at each of the following locations: (a) the neutral axis, (b) 2 in. above the neutral axis and (c) the top of the beam. (Disregard the direct compressive stresses produced by the uniform load bearing against the top of the beam.)

up1 ⫽ u1

sx + sy 2

savg ⫽ 30.0 deg

; ; ;

1200 lb/ft 10 in. 4 in. 2 ft 12 ft

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Applications of Plane Stress

Solution 8.4-3 b ⫽ 4 in.

h ⫽ 10 in.

bh3 I⫽ 12

I ⫽ 333.333 in.4 c ⫽ 2 ft

q ⫽ 1200 lb/ft RA ⫽

qL 2

A ⫽ bh

txy ⫽ ⫺ s1 ⫽

L ⫽ 12 ft

s2 ⫽

c2 2

M ⫽ 1.44 ⫻ 105 lb # in.

V ⫽ RA ⫺ qc

sx sx 2 2 + a b + txy 2 A 2

s2 ⫽ ⫺890 psi

V ⫽ 4.8 ⫻ 10 lb

sy ⫽ 0

txy ⫽ ⫺

tmax ⫽

s1 ⫽ 180 psi

sx 2 2 b + txy A 2

tmax ⫽ 458 psi

3V 2A

s2 ⫽ ⫺s1

tmax ⫽ s1

My sx ⫽ ⫺ I

sx ⫽ ⫺2.16 ⫻ 103 psi

I txy ⫽ 0

Uniaxial stress: s1 ⫽ sy s2 ⫽ sx

d ⫽ 3 in. sx ⫽ ⫺864 psi

h d Q ⫽ bd a ⫺ b 2 2

sx ⫽ ⫺

h Ma b 2

sy ⫽ 0

;

(b) 2 IN. ABOVE THE NEUTRAL AXIS y ⫽ 2 in.

;

(c) TOP OF THE BEAM

s2 ⫽ ⫺180 psi

tmax ⫽ 180 psi

;

a

txy ⫽ ⫺180 psi Pure shear: s1 ⫽ ⫺txy

;

sx sx 2 2 ⫺ a b + txy 2 A 2

3

(a) NEUTRAL AXIS sx ⫽ 0

txy ⫽ ⫺151.2 psi

s1 ⫽ 25.7 psi

RA ⫽ 7.2 ⫻ 103 lb

M ⫽ RA c ⫺ q

VQ Ib

sy ⫽ 0

tmax ⫽

s1 ⫽ 0 psi

s2 ⫽ ⫺2.16 ⫻ 103 psi

sx 2 2 b + txy A 2 a

tmax ⫽ 1080 psi

;

Q ⫽ 42 in.

3

Problem 8.4-4 An overhanging beam ABC with a guided support at A is of rectangular cross section and supports concentrated loads P both at A and at the free end C (see figure). The span length from A to B is L, and the length of the overhang is L/2. The cross section has width b and height h. Point D is located midway between the supports at a distance d from the top face of the beam. Knowing that the maximum tensile stress (principal stress) at point D is s1 ⫽ 35 MPa. determine the magnitude of the load P. Data for the beam are as follows: L ⫽ 1.75 m, b ⫽ 50 mm, h ⫽ 200 mm, and d ⫽ 40 mm.

P

P d

A D

C

B L — 2

L — 2

L — 2

Probs. 8.4-4 and 8.4-5

h b

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SECTION 8.4

Maximum Stresses in Beams

Solution 8.4-4 L ⫽ 1.75 m

b ⫽ 50 mm

Q ⫽ bd a

h ⫽ 200 mm

d ⫽ 40 mm

txy ⫽

Maximum principal stress at point D: RB ⫽ 0

M A ⫽ PL + P

L 2

MA ⫽

L 3 M D ⫽ ⫺ PL + P ⫽ ⫺PL 2 2

3 PL 2

bh3 12

Q ⫽ 1.6 ⫻ 105 mm3

VQ ⫽ (96P) N>m2 Ib

PRINCIPAL STRESSES

VD ⫽ P

s1 ⫽

STRESS AT POINT D I⫽

h d ⫺ b 2 2

⫽

I ⫽ 3.333 ⫻ 107 mm4

sx + sy + 2

A

a

y ⫽ 60 mm

P ⫽ 11.10 kN

My ⫺(⫺ PL) y sx ⫽ ⫺ ⫽ ⫽ (3150P) N>m2 I I

2

2

2 b + txy

sx sx 2 2 a b + txy ⫽ (3.153 * 103 ) P + 2 A 2

With s1 ⫽ 35 MPa

h y⫽ ⫺d 2

sx ⫺ sy

P⫽

s1 3.153 * 103

;

sy ⫽ 0

Problem 8.4-5 Solve the preceding problem if the stress and dimensions are as follows: s1 ⫽ 2450 psi, L ⫽ 80 in., b ⫽ 2.5 in., h ⫽ 10 in., and d ⫽ 2.5 in.

Solution 8.4-5 L ⫽ 80 in.

b ⫽ 2.5 in.

h ⫽ 10 in.

d ⫽ 2.5 in.

Maximum principal stress at point D: RB ⫽ 0

M A ⫽ PL + P

L 2

3 L M D ⫽ ⫺ PL + P ⫽ ⫺PL 2 2

MA ⫽

3 PL 2

VD ⫽ P

STRESS AT POINT D I⫽

bh3 12

y⫽

h ⫺d 2

sx ⫽ ⫺ sy ⫽ 0

I ⫽ 208.333 in.4 y ⫽ 2.5 in.

My ⫺(⫺ PL)y lb ⫽ ⫽ (0.96P) I I in.2

Q ⫽ bd a txy ⫽

h d ⫺ b 2 2

Q ⫽ 23.438 in.3

VQ ⫽ (0.045P) lb/in.2 Ib

PRINCIPAL STRESSES s1 ⫽ ⫽

sx + sy + 2

A

a

sx ⫺ sy 2

2

2 b + txy

sx 2 sx 1b 2 + a b + txy ⫽ (0.962P) 2 2 A 2 in.

With s1 ⫽ 2450 psi P ⫽ 2.55 k

;

P⫽

s1 0.962

667

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Applications of Plane Stress

Problem 8.4-6 A beam of wide-flange cross section (see figure) has the following dimensions:

b

b ⫽ 120 mm, t ⫽ 10 mm, h ⫽ 300 mm, and h1 ⫽ 260 mm. The beam is simply supported with span length L ⫽ 3.0 m. A concentrated load P ⫽ 120 kN acts at the midpoint of the span. At a cross section located 1.0 m from the left-hand support, determine the principal stresses s1 and s2 and the maximum shear stress tmax at each of the following locations: (a) the top of the beam, (b) the top of the web, and (c) the neutral axis.

t h1 h

Probs. 8.4-6 and 8.4-7

Solution 8.4-6

Simply supported beam sy ⫽ 0 txy ⫽ 0 Uniaxial stress: s1 ⫽ 0 s2 ⫽ ⫺82.7 MPa tmax ⫽ 41.3 MPa

;

K

(b) TOP OF THE WEB (POINT B) sx ⫽ ⫺ P ⫽ 120 kN M⫽

L ⫽ 3.0 m

Pc ⫽ 60 kN # m 2

V⫽

b ⫽ 120 mm

t ⫽ 10 mm

h ⫽ 300 mm

h1 ⫽ 260 mm

I⫽

c ⫽ 1.0 m P ⫽ 60 kN 2

(b ⫺ t)h 31 bh3 ⫺ ⫽ 108.89 * 106 mm4 12 12

(a) TOP OF THE BEAM (POINT A)

My (60 kN # m)(130 mm) ⫽ ⫺ I 108.89 * 106 mm4

⫽ ⫺71.63 MPa sy ⫽ 0 Q ⫽ (b)a txy ⫽ ⫺

h ⫺ h1 h + h1 ba b ⫽ 336 * 103 mm3 2 4

VQ (60 kN)(336 * 103 mm3) ⫽ ⫺ It (108.89 * 106 mm4)(10 mm)

⫽ ⫺18.51 MPa s1,2 ⫽

sx + sy ; 2

A

a

sx ⫺ sy 2

2

2 b + txy

⫽ ⫺35.82 ; 40.32 MPa s1 ⫽ 4.5 MPa, s2 ⫽ ⫺76.1 MPa tmax ⫽

A

a

sx ⫺ sy 2

;

2

2 b + txy ⫽ 40.3 MPa

(c) NEUTRAL AXIS (POINT C) sx ⫽ 0

sx ⫽

(60 kN # m)(150 mm) Mc ⫽ ⫺ I 108.89 * 106 mm4

⫽ ⫺82.652 MPa

sy ⫽ 0

h1 h1 h h Q ⫽ ba b a b ⫺ (b ⫺ t)a b a b 2 4 2 4 ⫽ 420.5 * 103 mm3

;

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SECTION 8.4

txy ⫽ ⫺

VQ (60 kN)(420.5 * 103 mm3) ⫽ ⫺ It (108.89 * 106 mm4)(10 mm)

⫽ ⫺23.17 MPa

Maximum Stresses in Beams

Pure shear: s1 ⫽ 23.2 MPa, s2 ⫽ ⫺23.2 MPa K tmax ⫽ 23.2 MPa

669

;

Problem 8.4-7 A beam of wide-flange cross section (see figure) has the following dimensions: b ⫽ 5 in., t ⫽ 0.5 in.,

h ⫽ 12 in., and h1 ⫽ 10.5 in. The beam is simply supported with span length L ⫽ 10 ft and supports a uniform load q ⫽ 6 k/ft. Calculate the principal stresses s1 and s2 and the maximum shear stress tmax at a cross section located 3 ft from the left-hand support at each of the following locations: (a) the bottom of the beam, (b) the bottom of the web, and (c) the neutral axis.

Solution 8.4-7

Simply supported beam

q ⫽ 6.0 k/ft

L ⫽ 10 ft ⫽ 120 in. M⫽

c ⫽ 3 ft ⫽ 36 in. V⫽

qL ⫺ qc ⫽ 12,000 lb 2

b ⫽ 5.0 in. I⫽

qc2 qLc ⫺ ⫽ 756,000 lb-in. 2 2

t ⫽ 0.5 in.

(b ⫺ bh ⫺ 12 12 3

t)h 31

sx ⫽ ⫺ h ⫽ 12 in.

h1 ⫽ 10.5 in.

⫽ 285.89 in.4

⫽ 13,883 psi

txy ⫽ ⫺

(756,000 lb-in.)(⫺ 6.0 in.) Mc ⫽ I 285.89 in.4

⫽ 15,866 psi

h ⫺ h1 h + h1 ba b ⫽ 21.094 in.3 2 4

VQ (12,000 lb)(21.094 in.3) ⫽ ⫺ It (285.89 in.4)(0.5 in.)

⫽ ⫺1771 psi s1,2 ⫽

sy ⫽ 0 txy ⫽ 0 Uniaxial stress: s1 ⫽ 15,870 psi, d s2 ⫽ 0 tmax ⫽ 7930 psi

My (756,000 lb-in.)(⫺5.25 in.) ⫽ ⫺ I 285.89 in.4

sy ⫽ 0 Q ⫽ ba

(a) BOTTOM OF THE BEAM (POINT A) sx ⫽ ⫺

(b) BOTTOM OF THE WEB (POINT B)

sx + sy ; 2

A

a

sx ⫺ sy 2

⫽ 6941.5 ; 7163.9 psi ;

2

2 b + txy

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Applications of Plane Stress

s1 ⫽ 14,100 psi, s2 ⫽ ⫺220 psi tmax ⫽

A

a

sx ⫺ sy 2

; txy ⫽ ⫺

2

b + txy2 ⫽ 7160 psi

;

⫽ ⫺2349 psi Pure shear: s1 ⫽ 2350 psi, s2 ⫽ ⫺2350 psi, K tmax ⫽ 2350 psi

(c) NEUTRAL AXIS (POINT C) sx ⫽ 0

VQ (12,000 lb)(27.984 in.3) ⫽ ⫺ It (285.89 in.4)(0.5 in.)

sy ⫽ 0

h1 h1 h h Q ⫽ ba b a b ⫺ (b ⫺ t)a b a b 2 4 2 4

;

⫽ 27.984 in.3

Problem 8.4-8 A W 200 ⫻ 41.7 wide-flange beam (see Table E-1(b).

100 kN

Appendix E) is simply supported with a span length of 2.5 m (see figure). The beam supports a concentrated load of 100 kN at 0.9 m from support B. At a cross section located 0.7 m from the left-hand support, determine the principal stresses s1 and s2 and the maximum shear stress tmax at each of the following locations: (a) the top of the beam, (b) the top of the web, and (c) the neutral axis.

W 200 × 41.7 A

B

D

0.7 m

0.9 m 2.5 m

0.9 m

Solution 8.4-8 RB ⫽ 100 kN a

0.7 + 0.9 b 2.5

RA ⫽ 100 kN ⫺ RB

RB ⫽ 64 kN (upward)

RA ⫽ 36 kN

(upward)

(a) TOP OF THE BEAM (POINT A)

sx ⫽ ⫺

At the point D M ⫽ RA(0.7 m) V ⫽ RA

M ⫽ 25.2 kN # m

V ⫽ 36 kN

sy ⫽ 0

h Ma b 2 I

sx ⫽ ⫺63.309 MPa

txy ⫽ 0

Uniaxial stress: s1 ⫽ sy W200 ⫻ 41.7 I ⫽ 40.8 ⫻ 106 mm4 b ⫽ 166 mm

s1 ⫽ 0

;

s2 ⫽ sx

s2 ⫽ ⫺63.3 MPa

tmax ⫽ 31.7 MPa

;

tw ⫽ 7.24 mm

(b) TOP OF THE WEB (POINT B)

tf ⫽ 11.8 mm

Ma

h ⫽ 205 mm h1 ⫽ h ⫺ 2t f h1 ⫽ 181.4 mm

sx ⫽ ⫺

h1 b 2 I

sx ⫽ ⫺56.021 MPa

sy ⫽ 0 Q ⫽ ba

tmax ⫽ `

h ⫺ h1 h + h1 ba b 2 4

Q ⫽ 1.892 ⫻ 105 mm3

;

sx ` 2

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SECTION 8.4

txy ⫽ ⫺ s1 ⫽

VQ I tw

txy ⫽ ⫺23.061 MPa

sx + sy + 2

A

s1 ⫽ 8.27 MPa s2 ⫽

sx + sy 2

sx ⫺

a

2

A

a

2 b + txy

⫺

A

sx ⫺ sy

a

sx ⫺ sy 2

2

2

2 b + txy

sx ⫽ 0

sy ⫽ 0

h1 h1 h h Q ⫽ b a b a b ⫺ 1 b ⫺ t w2 a b a b 2 4 2 4

txy ⫽ ⫺

VQ I tw

txy ⫽ ⫺26.69 MPa

Pure shear: s1 ⫽ ƒ txy ƒ

; 2

2 b + txy

tmax ⫽ 36.3 MPa

(c) NEUTRAL AXIS (POINT C)

Q ⫽ 2.19 ⫻ 105 mm3

;

s2 ⫽ ⫺64.3 MPa tmax ⫽

sy 2

671

Maximum Stresses in Beams

s1 ⫽ 26.7 Mpa

s2 ⫽ ⫺s1

s2 ⫽ ⫺26.7 Mpa

tmax ⫽ txy

tmax ⫽ ⫺26.7 Mpa

;

; ;

;

Problem 8.4-9 A W 12 ⫻ 14 wide-flange beam (see Table E-1(a), Appendix E)

7.5 k

is simply supported with a span length of 120 in. (see figure). The beam supports two anti-symmetrically placed concentrated loads of 7.5 k each. At a cross section located 20 in. from the right-hand support, determine the principal stresses s1 and s2 and the maximum shear stress tmax at each of the following locations: (a) the top of the beam, (b) the top of the web, and (c) the neutral axis.

7.5 k

W 12 × 14 D

40 in.

40 in.

20 in. 20 in.

120 in.

Solution 8.4-9 RB ⫽

7.5 # 80 ⫺ 7.5 # 40 k 120

RA ⫽ ⫺RB

RA ⫽ ⫺2.5 k

RB ⫽ 2.5 k (upward) (downward)

At Section C-C M ⫽ RB # 20 in. V ⫽ ⫺RB

M ⫽ 50 k # in.

V ⫽ ⫺2.5 k

(a) TOP OF THE BEAM (POINT A)

sx ⫽ ⫺ sy ⫽ 0

h Ma b 2 I

sx ⫽ ⫺3.361 ⫻ 103 psi

txy ⫽ 0

Uniaxial stress: s1 ⫽ sy W12 ⫻ 14 I ⫽ 88.6 in.4 b ⫽ 3.970 in.

s1 ⫽ 0

;

tmax ⫽ 1680 psi

s2 ⫽ ⫺3361 psi

(b) TOP OF THE WEB (POINT B)

tf ⫽ 0.225 in.

Ma

h ⫽ 11.91 in. h1 ⫽ h ⫺ 2tf h1 ⫽ 11.46 in.

sx ⫽ ⫺ sy ⫽ 0

I

tmax ⫽ ` ;

;

tw ⫽ 0.200 in.

h1 b 2

s2 ⫽ sx

sx ⫽ ⫺3.234 ⫻ 103 psi

sx ` 2

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Q ⫽ ba txy ⫽ ⫺ s1 ⫽

Page 672

Applications of Plane Stress

h ⫺ h1 h + h1 ba b 2 4

VQ I tw

sx + sy +

A

2 sx + sy 2

A

a

sx ⫺ sy

a

2

⫺

A

a

sx ⫺ sy 2

Q ⫽ 8.502 in3

2

2 b + txy

txy ⫽ ⫺

2

sy 2

2 b + txy

2

tmax ⫽ 1777 psi

VQ I tw

txy ⫽ 1.2 ⫻ 103 psi

Pure shear:

2 b + txy

;

sx ⫺

sy ⫽ 0

h1 h1 h h Q ⫽ ba b a b ⫺ 1 b ⫺ t w2a b a b 2 4 2 4

;

s2 ⫽ ⫺3393 psi tmax ⫽

sx ⫽ 0

txy ⫽ 736.289 psi

s1 ⫽ 159.8 psi s2 ⫽

(C) NEUTRAL AXIS (POINT C)

Q ⫽ 5.219 in.3

s1 ⫽ ƒ txy ƒ

s1 ⫽ 1200 psi

s2 ⫽ ⫺s1

s2 ⫽ ⫺1200 psi

;

tmax ⫽ txy

tmax ⫽ 1200 psi

;

;

;

Problem *8.4-10 A cantilever beam of T-section is loaded by an inclined force of magnitude 6.5 kN (see figure). The line of action of the force is inclined at an angle of 60° to the horizontal and intersects the top of the beam at the end cross section. The beam is 2.5 m long and the cross section has the dimensions shown. Determine the principal stresses s1 and s2 and the maximum shear stress tmax at points A and B in the web of the beam near the support.

60° B 6.5 kN

A

y

2.5 m

80 80 mm mm 25 mm z

C 160 mm

25 mm

Solution 8.4-10 P ⫽ 6.5 kN

L ⫽ 2.5 m

A ⫽ 8 ⫻ 10 mm 3

Location of centroid C

c2 ⫽

©( yi Ai) A

c2 ⫽

c2 ⫽ 126.25 mm

A ⫽ 2(160 mm)(25 mm)

b ⫽ 160 mm

2

t ⫽ 25 mm

From Eq. (12-7b) in Chapter 12:

(160 mm) (25 mm) a160⫹

c1 ⫽ 185 mm ⫺ c2

25 b mm + (160 mm) (25 mm) (80 mm) 2 A

c1 ⫽ 58.75 mm

MOMENT OF INTERTIA IZ ⫽

1 3 1 1 t c2 + b c31 ⫺ ( b ⫺ t) ( c1 ⫺ t)3 3 3 3

IZ ⫽ 2.585 ⫻ 107 mm4

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SECTION 8.4

EQUIVALENT LOADS AT FREE END OF BEAM

PH ⫽ ⫺P cos (60 deg) Pv ⫽ 5.629 kN

PH ⫽ ⫺3.25 kN

Mc ⫽ PH c1

Pv ⫽ P sin (60 deg)

Mc ⫽ ⫺190.938 N # m

STRESS RESULTANTS AT FIXED END OF BEAM

N0 ⫽ PH

N0 ⫽ ⫺3.25 kN

V ⫽ Pv

V ⫽ 5.629 kN

M ⫽ ⫺Mc ⫺ Pv L

M ⫽ ⫺1.388 ⫻ 104 N # m

Stress at point A (bottom of web) sx ⫽

N0 Mc1 + A IZ

sx ⫽ ⫺31.951 MPa

Uniaxial stress: s1 ⫽ sy s1 ⫽ 0

s2 ⫽ sx

sy ⫽ 0 tmax ⫽ `

s2 ⫽ ⫺32.0 MPa

;

;

txy ⫽ 0

sx ` 2 tmax ⫽ 15.98 MPa

Stress at point B (top of web) sx ⫽

M1c1 ⫺ t2 N0 ⫺ A Iz

Q ⫽ bt ac1 ⫺ txy ⫽ ⫺ s1 ⫽ s2 ⫽

t b 2

VQ Iz t

txy ⫽ ⫺1.611 MPa

+ 2

tmax ⫽

2

A

a

sy ⫽ 0

Q ⫽ 1.85 ⫻ 105 mm3

sx ⫺ sy

A

a

sx ⫺ sy

A

a

sx + sy sx + sy

sx ⫽ 17.715 MPa

⫺

sx ⫺ sy 2

2

2 2

b + txy2

2

b + t2xy 2

b + t2xy

s1 ⫽ 17.86 MPa s2 ⫽ ⫺0.145 MPa

tmax ⫽ 9.00 MPa

;

;

;

;

Maximum Stresses in Beams

673

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Applications of Plane Stress

Problem *8.4-11 A simple beam of rectangular cross section has span length L ⫽ 62 in. and supports a concentrated moment M ⫽ 560 k-in at midspan (see figure). The height of the beam is h ⫽ 6 in. and the width is b ⫽ 2.5 in. Plot graphs of the principal stresses s1 and s2 and the maximum shear stress tmax, showing how they vary over the height of the beam at cross section mn, which is located 24 in. from the left-hand support.

L M at — 2 m h n

b L

Probs. 8.4-11 and 8.4-12

Solution 8.4-11 2 × 103

sx(y) ⫽ ⫺

Prin. stresses (psi)

1 × 103

Q(y) ⫽ ba

σ.1(y) σ.2(y)

0

txy(y) ⫽

τmax (y) –1 × 103

–2.5 –3

M ⫽ 560 k # in. b ⫽ 2.5 in. bh 12

M RA ⫽ L

–2

–1.5

–1

–0.5

0 y

0.5

1

1.5

2

2.5

3

s1(y) ⫽

sx(y) sx(y) 2 a + b + txy(y)2 2 A 2

s2(y) ⫽

sx(y) sx(y) 2 ⫺ a b + txy(y)2 2 A 2

3

L ⫽ 62 in.

c ⫽ 24 in.

h ⫽ 6 in.

RA ⫽ 9.032 k

sx(y) 2 b + txy(y)2 A 2 a

s2(3 in.) ⫽ ⫺14,452 psi (upward)

RB ⫽ ⫺9.032 k

tmax(3 in.) ⫽ 7226 psi s1(0) ⫽ 903 psi

(downward)

At section m-n V ⫽ RA

tmax(y) ⫽

s1(3 in.) ⫽ 0 psi

I ⫽ 45 in4

RB ⫽ ⫺RA M ⫽ RAc

1 h h ⫺ yb a b a + yb 2 2 2

VQ(y) Ib

dist. y (in.)

I⫽

sy ⫽ 0

PRINCIPAL STRESSES

–2 × 103 –3

3

My I

M ⫽ 216.774 k # in. V ⫽ 9.032 k

s2(0) ⫽ ⫺903 psi

tmax(0) ⫽ 903 psi s1(⫺3 in.) ⫽ 14,452 psi s2(⫺3 in.) ⫽ 0 psi tmax(⫺3 in.) ⫽ 7226 psi

Problem *8.4-12 Solve the preceding problem for a cross section mn located 0.18 m from the support if L ⫽ 0.75 m, M ⫽ 65 kN # m, h ⫽ 120 mm, and b ⫽ 20 mm.

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SECTION 8.5

675

Combined Loadings

Solution 8.4-12 sx( y) ⫽ Prin. stresses (MPa)

08Ch08.qxd

Q( y) ⫽ b a txy( y) ⫽

sy ⫽ 0

1 h h ⫺ yb a b a + yb 2 2 2

VQ( y) Ib

PRINCIPAL STRESSES

dist.y(mm)

M ⫽ 65 kN # m b ⫽ 20 mm 3

bh I⫽ 12 RA ⫽

My I

M L

RB ⫽ ⫺RA

L ⫽ 0.75 m

c ⫽ 0.18 m

h ⫽ 120 mm I ⫽ 2.88 * 10 mm

4

V ⫽ RA

s2( y) ⫽

sx( y) sx( y) 2 ⫺ a b + txy( y)2 2 A 2

tmax( y) ⫽

sx( y) 2 b + txy( y)2 A 2 a

s1(0) ⫽ 54.2 MPa (upward)

RB ⫽ ⫺86.667 kN

s2(60 mm) ⫽ ⫺325 MPa

tmax(60 mm) ⫽ 162.5 MPa

(downward)

At section m-n M ⫽ RA c

sx( y) sx( y) 2 a + b + txy( y)2 2 A 2

s1(60 mm) ⫽ 0 MPa

6

RA ⫽ 86.667 kN

s1( y) ⫽

s2(0) ⫽ ⫺54.2 MPa

tmax(0) ⫽ 54.2 MPa s1(⫺60 mm) ⫽ 325 MPa

s2(⫺60 mm) ⫽ 0 MPa

tmax(⫺60 mm) ⫽ 162.5 MPa

M ⫽ 15.6 kN # m V ⫽ 86.667 kN

Combined Loadings y0

The problems for Section 8.5 are to be solved assuming that the structures behave linearly elastically and that the stresses caused by two or more loads may be superimposed to obtain the resultant stresses acting at a point. Consider both in-plane and out-of-plane shear stresses unless otherwise specified.

b2

b1

D

B C

P

Problem 8.5-1 A bracket ABCD having a hollow circular cross section consists of a vertical arm AB, a horizontal arm BC parallel to the x0 axis, and a horizontal arm CD parallel to the z0 axis (see figure). The arms BC and CD have lengths b1 ⫽ 3.6 ft and b2 ⫽ 2.2 ft, respectively. The outer and inner diameters of the bracket are d2 ⫽ 7.5 in. and d1 ⫽ 6.8 in. A vertical load P ⫽ 1400 lb acts at point D. Determine the maximum tensile, compressive, and shear stresses in the vertical arm. A

z0

x0

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Applications of Plane Stress

Solution 8.5-1 b1 ⫽ 3.6 ft

b2 ⫽ 2.2 ft

d2 ⫽ 7.5 in.

P ⫽ 1400 lb

MAXIMUM COMPRESSIVE STRESS

d1 ⫽ 6.8 in.

A⫽

p ad 22 ⫺ d 21 b 4

A ⫽ 7.862 in.2

I⫽

p ad 24 ⫺ d 14 b 64

I ⫽ 50.36 in.4

sc ⫽ ⫺

P ⫺ A

Ma

d2 b 2

I

sc ⫽ ⫺5456 psi

;

MAXIMUM SHEAR STRESS VERTICAL ARM AB M⫽

P 2b 21

+

b 22

Uniaxial stress M ⫽ 7.088 * 10

4

lb # in.

tmax ⫽ |sc|

tmax ⫽ 5456 psi

;

MAXIMUM TENSILE STRESS

st ⫽ ⫺

P + A

Ma

d2 b 2

I

st ⫽ 5100 psi

;

Problem 8.5-2 A gondola on a ski lift is supported by two bent arms, as

W

shown in the figure. Each arm is offset by the distance b ⫽ 180 mm from the line of action of the weight force W. The allowable stresses in the arms are 100 MPa in tension and 50 MPa in shear. If the loaded gondola weighs 12 kN, what is the minimum diameter d of the arms?

d

b

W (a)

(b)

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SECTION 8.5

Solution 8.5-2

Gondola on a ski lift

b ⫽ 180 mm

W⫽

12 kN ⫽ 6 kN 2

sallow ⫽ 100 MPa (tension)

A⫽

pd 4

S⫽

3

pd 32

MAXIMUM TENSILE STRESS W M 4W 32 Wb + + ⫽ 2 A S pd pd 3 pst 3 or a bd ⫺ d ⫺ 8b ⫽ 0 4W st ⫽

677

SUBSTITUTE NUMERICAL VALUES: p(100 Mpa) pst psallow 1 ⫽ ⫽ ⫽ 13,089.97 2 4W 4W 4(6 kN) m 8b ⫽ 1.44 m

Find dmin 2

tallow ⫽ 50 MPa

Combined Loadings

13,090 d3 ⫺ d ⫺ 1.44 ⫽ 0

(d ⫽ meters)

Solve numerically: d ⫽ 0.04845 m ⬖ dmin ⫽ 48.4 mm

;

MAXIMUM SHEAR STRESS st (uniaxial stress) 2 Since tallow is one-half of sallow, the minimum diameter for shear is the same as for tension. tmax ⫽

Problem 8.5-3 The hollow drill pipe for an oil well (see figure) is 6.2 in. in outer diameter and 0.75 in. in thickness. Just above the bit, the compressive force in the pipe (due to the weight of the pipe) is 62 k and the torque (due to drilling) is 185 k-in. Determine the maximum tensile, compressive, and shear stresses in the drill pipe.

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Applications of Plane Stress

Solution 8.5-3 P ⫽ compressive force T ⫽ Torque

txy ⫽

d2 ⫽ outer diameter d1 ⫽ inner diameter P ⫽ 62 k t ⫽ 0.75 in.

T ⫽ 185 k # in. d1 ⫽ d2 ⫺ 2t

p A ⫽ a d 22 ⫺ d 2b 1 4 p Ip ⫽ ad 24 ⫺ d 14 b 32

d2 ⫽ 6.2 in. d1 ⫽ 4.7 in.

Ta

d2 b 2

txy ⫽ 5903 psi

Ip

PRINCIPAL STRESSES s1 ⫽

sx ⫺ sy

A

a

sx ⫺ sy

A

a

sx + sy + 2

2

b + txy2 2

s1 ⫽ 3963 psi

A ⫽ 12.841 in.2 Ip ⫽ 97.16 in.

4

s2 ⫽

sx + sy 2

⫺

2

b + txy2 2

s2 ⫽ ⫺8791 psi

STRESSES AT THE OUTER SURFACE st ⫽ s1

MAXIMUM TENSILE STRESS st ⫽ 3963 psi

;

MAXIMUM COMPRESSIVE STRESS sc ⫽ s2 sc ⫽ ⫺8791 psi

;

MAXIMUM IN-PLANESHEAR STRESS tmax ⫽

sy ⫽ ⫺ sx ⫽ 0

P A

A

a

sx ⫺ sy 2

tmax ⫽ 6377 psi sy ⫽ ⫺4828 psi

b + txy2 ;

NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress.

P

Problem 8.5-4 A segment of a generator shaft is subjected to a torque T and an axial

force P, as shown in the figure. The shaft is hollow (outer diameter d2 ⫽ 300 mm and inner diameter d1 ⫽ 250 mm) and delivers 1800 kW at 4.0 Hz. If the compressive force P ⫽ 540 kN, what are the maximum tensile, compressive, and shear stresses in the shaft?

T

T

Probs. 8.5-4 and 8.5-5

P

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SECTION 8.5

Combined Loadings

679

Solution 8.5-4 P ⫽ Compressive_force P0 ⫽ Power

f ⫽ frequency

T ⫽ torgue ⫽

P0 2p f

d2 ⫽ outer diameter

sy ⫽ ⫺

P ⫽ 540 kN

txy ⫽

P0 ⫽ 1800 kW

P0 2p f

T ⫽ 7.162 ⫻ 104 N # m

d2 ⫽ 300 mm

d1 ⫽ 250 mm

T⫽

A⫽

p ad 22 ⫺ d 21 b 4

A ⫽ 2.16 ⫻ 104 mm2

Ip ⫽

p ad 42 ⫺ d 14 b 32

sy ⫽ ⫺25.002 MPa

sx ⫽ 0 Ta

d2 b 2

txy ⫽ 26.093 MPa

Ip

d1 ⫽ inner diameter f ⫽ 4.0 Hz

P A

PRINCIPAL STRESSES sx + sy sx ⫺ sy 2 2 s1 ⫽ a + b + txy 2 A 2 s1 ⫽ 16.432 MPa s2 ⫽

sx + sy 2

⫺

A

a

sx ⫺ sy 2

s2 ⫽ ⫺41.434 MPa

Ip ⫽ 4.117 ⫻ 108 mm4

STRESSES AT THE OUTER SURFACE

2

2 b + txy

MAXIMUM TENSILE STRESS st ⫽ 16.43 MPa

st ⫽ s1

;

MAXIMUM COMPRESSIVE STRESS sc ⫽ ⫺41.4 MPa

sc ⫽ s2

;

MAXIMUM IN-PLANESHEAR STRESS tmax ⫽

A

a

sx ⫺ sy 2

tmax ⫽ 28.9 MPa

2

2 b + txy

;

NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress.

Problem 8.5-5 A segment of a generator shaft of hollow circular cross section is subjected to a torque T ⫽ 240 k-in. (see figure). The outer and inner diameters of the shaft are 8.0 in. and 6.25 in., respectively. What is the maximum permissible compressive load P that can be applied to the shaft if the allowable in-plane shear stress is tallow ⫽ 6250 psi?

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Applications of Plane Stress

Solution 8.5-5 P ⫽ compressive force d2 ⫽ outer diameter

T ⫽ Torque d1 ⫽ inner diameter

A⫽

P A

sx ⫽ 0

T ⫽ 240 k # in. d2 ⫽ 8 in.

sy ⫽ ⫺

d1 ⫽ 6.25 in.

p ad 22 ⫺ d 21 b 4

tallow ⫽ 6250 psi

A ⫽ 19.586 in.2 Ip ⫽

p ad 42 ⫺ d 14 b 32

txy ⫽

d2 Ta b 2

txy ⫽ 3805 psi

Ip

MAXIMUM IN-PLANESHEAR STRESS a

sx ⫺ sy

2

b + t 2xy

Ip ⫽ 252.321 in.4

tmax ⫽

STRESSES AT THE OUTER SURFACE

2 ⫺ t22 4 sy ⫽ 11tmax xy

A

P ⫽ sy A

2

tmax ⫽ tallow

sy ⫽ 9917 psi

P ⫽ 194.2 k

;

NOTE: The maximum in-plane shear is larger than the maximum out-of-plane shear stress.

Problem 8.5-6 A cylindrical tank subjected to internal pressure p is

simultaneously compressed by an axial force F ⫽ 72 kN (see figure). The cylinder has diameter d ⫽ 100 mm and wall thickness t ⫽ 4 mm. Calculate the maximum allowable internal pressure pmax based upon an allowable shear stress in the wall of the tank of 60 MPa.

Solution 8.5-6

F

F

Cylindrical tank with compressive force CIRCUMFERENTIAL STRESS (TENSION) s1 ⫽

F ⫽ 72 kN

Units: s1 ⫽ MPa

p ⫽ internal pressure d ⫽ 100 mm

pr p(50 mm) ⫽ ⫽ 12.5 p t 4 mm

t ⫽ 4 mm

tallow ⫽ 60 MPa

p ⫽ MPa

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SECTION 8.5

LONGITUDINAL STRESS (TENSION) s2 ⫽

pr pr F F ⫺ ⫽ ⫺ 2t A 2t 2prt

⫽ 6.25p ⫺

72,000 N 2p(50 mm)(4 mm)

⫽ 6.25p ⫺ 57.296 Mpa Units: s2 ⫽ MPa

p ⫽ MPa

Combined Loadings

681

OUT-OF-PLANE SHEAR STRESSES Case 2: tmax ⫽

s1 ⫽ 6.25 p; 60 MPa ⫽ 6.25 p 2

Solving, p2 ⫽ 9.60 MPa Case 3: tmax ⫽

s2 ⫽ 3.125 p ⫺ 28.648 MPa 2

60 MPa ⫽ 3.125 p ⫺ 28.648 MPa Solving, p3 ⫽ 28.37 MPa

BIAXIAL STRESS IN-PLANE SHEAR STRESS (CASE 1)

CASE 2, OUT-OF-PLANE SHEAR STRESS GOVERNS

s1 ⫺ s2 ⫽ 3.125 p + 28.648 Mpa tmax ⫽ 2

pmax ⫽ 9.60 MPa

;

60 MPa ⫽ 3.125 p ⫹ 28.648 MPa Solving, p1 ⫽ 10.03 MPa

Problem 8.5-7 A cylindrical tank having diameter d ⫽ 2.5 in. is

subjected to internal gas pressure p ⫽ 600 psi and an external tensile load T ⫽ 1000 lb (see figure). Determine the minimum thickness t of the wall of the tank based upon an allowable shear stress of 3000 psi.

Solution 8.5-7

T

Cylindrical tank with tensile load CIRCUMFERENTIAL STRESS (TENSION) (600 psi)(1.25 in.) pr 750 ⫽ ⫽ t t t Units: s1 ⫽ psi t ⫽ inches s2 ⫽ psi s1 ⫽

T ⫽ 1000 lb

t ⫽ thickness

p ⫽ 600 psi d ⫽ 2.5 in.

tallow ⫽ 3000 psi

T

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Applications of Plane Stress

LONGITUDINAL STRESS (TENSION) s2 ⫽

pr pr T T + ⫽ + 2t A 2t 2prt

375 1000 lb 375 127.32 502.32 ⫽ + ⫽ + + t 2p(1.25 in.)t t t t BIAXIAL STRESS

3000 psi ⫽

123.84 t

Solving, t 1 ⫽ 0.0413 in. OUT-OF-PLANE SHEAR STRESSES Case 2: tmax ⫽

s1 375 375 ⫽ ; 3000 ⫽ 2 t t

Solving, t2 ⫽ 0.125 in. Case 3: tmax ⫽

s2 251.16 251.16 ⫽ ; 3000 ⫽ 2 t t

Solving, t3 ⫽ 0.0837 in. CASE 2, OUT-OF-PLANE SHEAR STRESS GOVERNS tmin ⫽ 0.125 in.

;

IN-PLANE SHEAR STRESS (CASE 1) tmax ⫽

s1 ⫺ s2 247.68 123.84 ⫽ ⫽ 2 2t t

Problem 8.5-8 The torsional pendulum shown in the figure consists of a horizontal

circular disk of mass M ⫽ 60 kg suspended by a vertical steel wire (G ⫽ 80 GPa) of length L ⫽ 2 m and diameter d ⫽ 4 mm. Calculate the maximum permissible angle of rotation fmax of the disk (that is, the maximum amplitude of torsional vibrations) so that the stresses in the wire do not exceed 100 MPa in tension or 50 MPa in shear.

d = 4 mm

L=2m

fmax M = 60 kg

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SECTION 8.5

Solution 8.5-8

Combined Loadings

683

Torsional pendulum sx ⫽

TENSILE STRESS

W ⫽ 46.839 MPa A

sx ⫽ 0 sy ⫽ st ⫽ 46.839 MPa txy ⫽ ⫺80 fmax (MPa)

L ⫽ 2.0 m

d ⫽ 4.0 mm

M ⫽ 60 kg

G ⫽ 80 GPa

sallow ⫽ 100 MPa A⫽

tallow ⫽ 50 MPa

2

pd ⫽ 12.5664 mm2 4

PRINCIPAL STRESSES

W ⫽ Mg ⫽ (60 kg)(9.81 m/s ) ⫽ 588.6 N 2

s1, 2 ⫽ s1, 2

sx + sy ; 2

A

a

sx ⫺ sy 2

2

b + t 2xy

⫽ 23.420 ; 1123.42022 + 6400f2max ( MPa)

Note that s1 is positive and s2 is negative. Therefore, the maximum in-plane shear stress is greater than the maximum out-of-plane shear stress. MAXIMUM ANGLE OF ROTATION BASED ON TENSILE STRESS s1 ⫽ maximum tensile stress

sallow ⫽ 100 MPa

‹ 100 MPa ⫽ 23.420 ; 1123.42022 + 6400f2max (100 ⫺ 23.420)2 ⫽ 123.42022 + 6400f2max 5316 ⫽ 6400f2max

TORQUE: T ⫽

GIp fmax L

(EQ. 3-15)

fmax ⫽ 0.9114 rad ⫽ 52.2°

MAXIMUM ANGLE OF ROTATION BASED ON IN-PLANE SHEAR STRESS

SHEAR STRESS: t ⫽ t⫽ a

GIpfmax L

t ⫽ 80 fmax

ba

Tr Ip

(EQ. 3-11)

tmax ⫽

Gr fmax r b⫽ ⫽ (80 * 106 Pa)fmax IP L

Units: t ⫽ MPa

fmax ⫽ radians

sx ⫺ sy 2 b + t2xy ⫽ 1(23.420)2 + 6400f2max A 2 a

tallow ⫽ 50 MPa

50 ⫽ 1(23.420)2 + 6400f2max

(50)2 ⫽ (23.420)2 + 6400f2max Solving, fmax ⫽ 0.5522 rad ⫽ 31.6° SHEAR STRESS GOVERNS fmax ⫽ 0.552 rad ⫽ 31.6°

;

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Applications of Plane Stress

Problem 8.5-9 Determine the maximum tensile, compressive,

P = 160 lb

and shear stresses at points A and B on the bicycle pedal crank shown in the figure. The pedal and crank are in a horizontal plane and points A and B are located on the top of the crank. The load P ⫽ 160 lb acts in the vertical direction and the distances (in the horizontal plane) between the line of action of the load and points A and B are b1 ⫽ 5.0 in., b2 ⫽ 2.5 in. and b3 ⫽ 1.0 in. Assume that the crank has a solid circular cross section with diameter d ⫽ 0.6 in.

Crank

d = 0.6 in. A B b3

b1 = 5.0 in.

b3 = 1.0 in. b2 = 2.5 in.

A

B

b3

b3

b2

⫻

P

b1 Top view

Solution 8.5-9 P ⫽ 160 lb

d ⫽ 0.6 in.

STRESS AT POINT A:

b1 ⫽ 5.0 in.

b2 ⫽ 2.5 in.

t⫽

16 TA

s⫽

MA S

b3 ⫽ 1.0 in.

S⫽

pd 32

3

STRESS RESULTANTS on cross section at point A: Torque:

TA ⫽ Pb2

Moment:

MA ⫽ Pb1

Shear Force:

VA ⫽ P

t ⫽ 9.431 ⫻ 103 psi

p d3

s ⫽ 3.773 ⫻ 104 psi

(The shear force V produces no shear stresses at point A) PRINCIPAL STRESSES AND MAXIMUM SHEAR STRESS sx ⫽ 0

sy ⫽ s

txy ⫽ ⫺t

sx + sy

a

STRESS RESULTANTS at point B:

s1 ⫽

TB ⫽ P(b1 ⫹ b3)

s1 ⫽ 3.995 ⫻ 10 psi

MB ⫽ P(b2 ⫹ b3) VB ⫽ P

+

A

2

sx ⫺ sy 2

2

b + t 2xy

4

s2 ⫽

sx + sy 2

⫺

A

a

sx ⫺ sy

s2 ⫽ ⫺2.226 ⫻ 10 psi 3

2

2

b + t 2xy

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SECTION 8.5

MAXIMUM TENSILE STRESS ;

s2 ⫽

sx + sy 2

MAXIMUM COMPRESSIVE STRESS sc ⫽ ⫺2226 psi

sc ⫽ s2

;

MAXIMUM IN-PLANE SHEAR STRESS tmax ⫽

A

a

sx ⫺ sy 2

s⫽

MB S

pd

2

2

b + t2xy

MAXIMUM TENSILE STRESS st ⫽ 39,410 psi

;

s ⫽ 2.641 ⫻ 104 psi

sc ⫽ ⫺13,000 psi

;

MAXIMUM IN-PLANE SHEAR STRESS tmax ⫽

t ⫽ 2.264 ⫻ 104 psi

3

sx ⫺ sy

s2 ⫽ ⫺1.3 ⫻ 10

sc ⫽ s2

STRESS AT POINT B 16TB

A

a

MAXIMUM COMPRESSIVE STRESS

;

NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress.

t⫽

⫺

4

st ⫽ s1

2

b + t 2xy

tmax ⫽ 21,090 psi

A

a

sx ⫺ sy 2

tmax ⫽ 26,210 psi

2

b + txy2 ;

NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress.

(The shear force V produces no shear stresses at point A) PRINCIPAL STRESSES AND MIXIMUM SHEAR STRESS sx ⫽ 0 s1 ⫽

sy ⫽ s

sx + sy + 2

A

txy ⫽ ⫺t a

sx ⫺ sy 2

685

s1 ⫽ 3.941 ⫻ 104 psi

st ⫽ 39,950 psi

st ⫽ s1

Combined Loadings

2

b + t 2xy

Problem 8.5-10 A cylindrical pressure vessel having radius r ⫽ 300 mm and wall thickness t ⫽ 15 mm is subjected to internal pressure p ⫽ 2.5 MPa. In addition, a torque T ⫽ 120 kN # m acts at each end of the cylinder (see figure).

(a) Determine the maximum tensile stress smax and the maximum inplane shear stress tmax in the wall of the cylinder. (b) If the allowable in-plane shear stress is 30 MPa, what is the maximum allowable torque T?

T

T

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Applications of Plane Stress

Solution 8.5-10

Cylindrical pressure vessel s1 ⫽ 56.4 MPa

s2 ⫽ 18.6 MPa

‹ smax ⫽ 56.4 MPa

;

MAXIMUM IN-PLANE SHEAR STRESS tmax ⫽ T ⫽ 120 kN # m t ⫽ 15 mm

A

r ⫽ 300 mm P ⫽ 2.5 MPa

a

sx ⫺ sy 2

2

b + t 2xy ⫽ 18.9 MPa

;

(b) MAXIMUM ALLOWABLE TORQUE T tallow ⫽ 30 MPa (in-plane shear stress)

STRESSES IN THE WALL OF THE VESSEL pr ⫽ 25 MPa sx ⫽ 2t Tr txy ⫽ ⫺ Ip T 2pr 2t

tmax ⫽ sx ⫽

(EQ. 3-11)

Ip ⫽ 2pr 3t txy ⫽ ⫺

pr sy ⫽ ⫽ 50 MPa t

A

a

sx ⫺ sy 2

2

b + t2xy

(1)

pr pr ⫽ 25 MPa sy ⫽ ⫽ 50 MPa 2t t T

⫽ ⫺117.893 * 10⫺6 T

(EQ. 3-18)

txy ⫽ ⫺

⫽ ⫺ 14.147 MPa

Units: txy ⫽ MPa

2pr 2t

T⫽N#m

Substitute into Eq. (1): tmax ⫽ tallow ⫽ 30 MPa ⫽ 2(⫺ 12.5 MPa)2 + (⫺117.893 * 10⫺6 T)2 Square both sides, rearrange, and solve for T: (30)2 ⫽ (12.5)2 ⫹ (117.893 ⫻ 10⫺6)2 T2 T2 ⫽

743.750 13,899 * 10⫺12

⫽ 53,512 * 106 ( N # m)2

T ⫽ 231.3 * 103 N # m Tmax ⫽ 231 kN # m (a) PRINCIPAL STRESSES s1, 2 ⫽

sx + sy ; 2

A

a

sx ⫺ sy

⫽ 37.5 ; 18.878 MPa

2

2

b + t2xy

;

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SECTION 8.5

Problem 8.5-11 An L-shaped bracket lying in a horizontal plane

Combined Loadings

A

supports a load P ⫽ 150 lb (see figure). The bracket has a hollow rectangular cross section with thickness t ⫽ 0.125 in. and outer dimensions b ⫽ 2.0 in. and h ⫽ 3.5 in. The centerline lengths of the arms are b1 ⫽ 20 in. and b2 ⫽ 30 in. Considering only the load P, calculate the maximum tensile stress st, maximum compressive stress sc, and maximum shear stress tmax at point A, which is located on the top of the bracket at the support.

687

t = 0.125 in.

b1 = 20 in.

h = 3.5 in.

b2 = 30 in. b = 2.0 in. P = 150 lb

Solution 8.5-11 P ⫽ 150 lb

L-shaped bracket b1 ⫽ 20 in.

t ⫽ 0.125 in.

b2 ⫽ 30 in.

h ⫽ 3.5 in.

b ⫽ 2.0 in.

STRESSES AT POINT A ON THE TOP OF THE BRACKET t⫽

T 4500 lb-in. ⫽ 2844 psi ⫽ 2tAm 2(0.125 in.)(6.3281 in.2)

s⫽

(3000 lb-in.)(1.75 in.) Mc ⫽ 2454 psi ⫽ I 2.1396 in.4

FREE-BODY DIAGRAM OF BRACKET

(The shear force V produces no stresses at point A.) STRESS ELEMENT AT POINT A (This view is looking downward at the top of the bracket.)

STRESS RESULTANTS AT THE SUPPORT Torque: Moment:

T ⫽ Pb2 ⫽ (150 lb)(30 in.) ⫽ 4500 lb-in. M ⫽ Pb1 ⫽ (150 lb)(20 in.) ⫽ 3000 lb-in.

Shear force:

V ⫽ P ⫽ 150 lb

PROPERTIES OF THE CROSS SECTION For torsion: Am ⫽ (b ⫺ t)(h ⫺ t) ⫽ (1.875 in.)(3.375 in.) ⫽ 6.3281 in.2 For bending: c ⫽ I⫽ ⫽

h ⫽ 1.75 in. 2

1 1 (bh3) ⫺ (b ⫺ 2t)(h ⫺ 2t)3 12 12 1 1 (2.0 in.)(3.5 in.)3 ⫺ (1.75 in.)(3.25 in.)3 12 12

⫽ 2.1396 in.4

sx ⫽ 0

sy ⫽ s ⫽ 2454 psi

txy ⫽ ⫺t ⫽ ⫺2844 psi

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Applications of Plane Stress

PRINCIPAL STRESSES AND MAXIMUM SHEAR STRESS s1, 2 ⫽

sx + sy ;

A

2

a

sx ⫺ sy 2

2

b + t 2xy

⫽ 1227 psi ; 3097 psi s1 ⫽ 4324 psi tmax ⫽

A

a

s2 ⫽ ⫺1870 psi

sx ⫺ sy 2

2

b + t 2xy ⫽ 3097 psi

MAXIMUM COMPRESSIVE STRESS: sc ⫽ ⫺1870 psi

;

MAXIMUM SHEAR STRESS: tmax ⫽ 3100 psi

;

NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear stress is larger than the maximum out-of-plane shear stress.

MAXIMUM TENSILE STRESS: st ⫽ 4320 psi

;

Problem 8.5-12 A semicircular bar AB lying in a horizontal plane is supported at B (see figure). The bar has centerline radius R and weight q per unit of length (total weight of the bar equals pqR). The cross section of the bar is circular with diameter d. Obtain formulas for the maximum tensile stress st, maximum compressive stress sc, and maximum in-plane shear stress tmax at the top of the bar at the support due to the weight of the bar.

Solution 8.5-12

O A

B R

Semicircular bar STRESSES AT THE TOP OF THE BAR AT B 64qR2 M B(d/2) (2qR2)(d/2) ⫽ ⫽ I pd 4/64 pd 3 2 16qR2 TB(d/2) (pqR )(d/2) tB ⫽ ⫽ ⫽ IP pd 4/32 d3 sB ⫽

d ⫽ diameter of bar

R ⫽ radius of bar

q ⫽ weight of bar per unit length W ⫽ weight of bar ⫽ pqR Weight of bar acts at the center of gravity From Case 23, Appendix D, with b ⫽ p/2, we get y⫽

2R p

‹ c⫽

2R p

Bending moment at B: MB ⫽ Wc ⫽ 2qR2 Torque at B: TB ⫽ WR ⫽ pqR2 (Shear force at B produces no shear stress at the top of the bar.)

d

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SECTION 8.5

STRESS ELEMENT AT THE TOP OF THE BAR AT B sx ⫽ 0

MAXIMUM TENSILE STRESS st ⫽ s1 ⫽

sy ⫽ sB txy ⫽ tB

689

Combined Loadings

⫽ 29.15

16qR2 pd 3

(2 + 24 + p2)

qR2

;

d3

PRINCIPAL STRESSES: s1, 2 ⫽

sx + sy ; 2

A

a

sx ⫺ sy 2

MAXIMUM COMPRESSIVE STRESS

2

2 b + txy

sc ⫽ s2 ⫽

sB sB 2 ⫽ ; a ⫺ b + tB2 2 A 2 ⫽ ⫽

32qR 2 pd 3 16qR 2 pd 3

;

A

a

32qR2 pd 3

2

b + a

16qR2 d3

b

pd 3

(2 ⫺ 24 + p2)

⫽ ⫺8.78

2

(2 ; 24 + p2 )

16qR2

qR2 d3

;

MAXIMUM IN-PLANE SHEAR STRESS (EQ. 7-26) tmax ⫽

16qR2 1 24 + p 2 (s1 ⫺ s2) ⫽ 2 pd 3

;

z

Problem 8.5-13 An arm ABC lying in a horizontal plane and supported at A (see figure) is made of two identical solid steel bars AB and BC welded together at a right angle. Each bar is 20 in. long. Knowing that the maximum tensile stress (principal stress) at the top of the bar at support A due solely to the weights of the bars is 932 psi, determine the diameter d of the bars.

y A x B

C

Solution 8.5-13

Horizontal arm ABC L ⫽ length of AB and BC d ⫽ diameter of AB and BC A ⫽ cross-sectional area ⫽ pd2/4 g ⫽ weight density of steel q ⫽ weight per unit length of bars ⫽ gA ⫽ pgd2/4

(1)

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Applications of Plane Stress

RESULTANT FORCES ACTING ON AB

P ⫽ weight of AB and BC P ⫽ qL ⫽ pgLd2/4

(2)

T ⫽ torque due to weight of BC

sx ⫽ 0

sy ⫽ sA

txy ⫽ ⫺tA

(8)

Substitute (8) into (7):

qL2 pgL2d 2 L T ⫽ (qL)a b ⫽ ⫽ 2 2 8

(3)

MA ⫽ bending moment at A MA ⫽ PL ⫹ PL /2 ⫽ 3PL /2 ⫽ 3pgL2d2/8

s1 ⫽

sA sA 2 + a b + t 2A 2 A 2

Substitute from (5) and (6) and simplify: (4) s1 ⫽

STRESSES AT THE TOP OF THE BAR AT A

gL2 2gL2 (6 + 140) ⫽ (3 + 110) d d

sA ⫽ normal stress due to MA

SOLVE FOR d

M(d/2) M(d/2) 12gL2 32M ⫽ ⫽ ⫽ sA ⫽ 4 3 I d pd /64 pd

d⫽

(5)

2gL2 (3 + 110) s1

;

(10)

(11)

SUBSTITUTE NUMERICAL VALUES INTO EQ. (11):

tA ⫽ shear stress due to torque T T(d/2) T(d/2) 2gL2 16T ⫽ tA ⫽ ⫽ ⫽ Ip d pd 4/32 pd 3

(9)

g ⫽ 490 lb/ft3 ⫽ 0.28356 lb/in.3 (6)

L ⫽ 20 in. d ⫽ 1.50 in.

s1 ⫽ 932 psi ;

STRESS ELEMENT ON TOP OF THE BAR AT A s1 ⫽ principal tensile stress (maximum tensile stress) s1 ⫽

sx + sy + 2

A

a

sx ⫺ sy 2

2

b + t 2xy

(7)

Problem 8.5-14 A pressurized cylindrical tank with flat ends is loaded by torques T and tensile forces P (see figure). The tank has radius r ⫽ 50 mm and wall thickness t ⫽ 3 mm. The internal pressure p ⫽ 3.5 MPa and the torque T ⫽ 450 N # m. What is the maximum permissible value of the forces P if the allowable tensile stress in the wall of the cylinder is 72 MPa?

T P

T

P

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SECTION 8.5

Solution 8.5-14 r ⫽ 50 mm T ⫽ 450 N # m

Combined Loadings

691

Cylindrical tank t ⫽ 3.0 mm

p ⫽ 3.5 MPa

sallow ⫽ 72 MPa

Units: sx ⫽ MPa, P ⫽ newtons sy ⫽

pr ⫽ 58.333 MPa t

CROSS SECTION A ⫽ 2prt ⫽ 2p(50 mm)(3.0 mm) ⫽ 942.48 mm2

txy ⫽ ⫺

IP ⫽ 2pr3t ⫽ 2p(50 mm)3(3.0 mm)

(450 N # m)(50 mm) Tr ⫽ ⫺ IP 2.3562 * 106 mm4

⫽ ⫺9.5493 MPa

⫽ 2.3562 ⫻ 106 mm4 MAXIMUM TENSILE STRESS STRESSES IN THE WALL OF THE TANK

smax ⫽ sallow ⫽ 72 MPa ⫽

sx + sy + 2

A

72 ⫽ 43.750 + (530.52 * 10

a

⫺6

sx ⫺ sy 2

2

b + t 2xy

)P

+ 2[⫺ 14.583 + (530.52 * 10⫺6)P]2 + ( ⫺9.5493)2 or 28.250 ⫺ 0.00053052P ⫽ 2( ⫺14.583 + 0.00053052 P)2 + 91.189 Square both sides and simplify: sx ⫽

pr P + 2t A

(3.5 MPa)(50 mm) P + ⫽ 2(3.0 mm) 942.48 mm2

494.21 ⫽ 0.014501 P SOLVE FOR P

P ⫽ 34,080 N

Pmax ⫽ 34.1 kN

OR

;

⫽ 29.167 MPa + 1.0610 * 10⫺3P

z

Problem 8.5-15 A post having a hollow circular cross section supports

a horizontal load P ⫽ 240 lb acting at the end of an arm that is 5 ft long (see figure). The height of the post is 27 ft, and its section modulus is S ⫽ 15 in.3 Assume that outer radius of the post, r2 ⫽ 4.5 in., and inner radius r1 ⫽ 4.243 in. (a) Calculate the maximum tensile stress smax and maximum in-plane shear stress tmax at point A on the outer surface of the post along the x-axis due to the load P. Load P acts in a horizontal plane at an angle of 30° from a line which is parallel to the (⫺x) axis. (b) If the maximum tensile stress and maximum in-plane shear stress at point A are limited to 16,000 psi and 6000 psi, respectively, what is the largest permissible value of the load P?

5 ft 30° P = 240 lb

27 ft

x

A

y

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Applications of Plane Stress

Solution 8.5-15 (a) MAXIMUM

P ⫽ 240 lb b ⫽ 5 ft

length of arm

h ⫽ 27 ft

height of post

r2 ⫽ 4.5 in. A⫽

⫺

s2 ⫽

I ⫽ 67.507 in.4

Vx ⫽ P cos(30deg) Vy ⫽ ⫺P sin(30deg)

+ 2

a

A

sx ⫺

b + t 2xy

2

sx + sy 2

⫺

A

a

sx ⫺ sy 2

2

b + t 2xy

MAXIMUM TENSILE STRESS

Q ⫽ 9.825 in.3

smax ⫽ 4534 psi

smax ⫽ s1

REACTIONS AT THE SUPPORT T ⫽ ⫺P cos(30deg)b

sx + sy

txy ⫽ t

sy 2

s2 ⫽ ⫺44.593 psi

Ip ⫽ 135.014 in.4

M ⫽ P cos(30deg)h

sy ⫽ 4.489 ⫻ 103 psi

s1 ⫽ 4.534 ⫻ 103 psi

A ⫽ 7.059 in.2

2 1 r 32 ⫺ r 312 3

Q⫽

s1 ⫽

t ⫽ 0.257 in. r12 )

p I ⫽ 1 r 42 ⫺ r 412 4 Ip ⫽ 2I

sx ⫽ 0

r1 ⫽ 4.243 in.

t ⫽ r2 ⫺ r1 p(r22

TENSILE STRESS AND MAXIMUM SHEAR

STRESS

;

MAXIMUM SHEAR STRESS

M ⫽ 6.734 ⫻ 104 lb # in T ⫽ ⫺1.247 ⫻ 10

4

lb # in

Vx ⫽ 207.846 lb

tmax ⫽

A

a

sx ⫺ sy 2

tmax ⫽ 2289 psi

2

b + t 2xy ;

Vy ⫽ ⫺120 lb (b) ALLOWABLE LOAD P

STRESSES AT POINT A

sallow ⫽ 16000 psi

tallow ⫽ 6000 psi

The stresses at point A are proportional to the load P. Based on tensile stress: Pallow sallow ⫽ P smax

Pallow ⫽

sallow P smax

Pallow ⫽ 847.01lb Based on shear stress: Pallow tallow ⫽ tmax P Pallow ⫽ 629.07 lb

sx ⫽ 0 sy ⫽ t⫽

Mr2 I

Pallow ⫽

sy ⫽ 4.489 ⫻ 103 psi

Vy Q Tr2 + Ip I 2t

t ⫽ ⫺449.628 psi

(The shear force Vx produces no stress at point A)

Pallow ⫽ 629 lb

;

tallow P tmax

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SECTION 8.5

Problem 8.5-16 A sign is supported by a pipe (see figure) having outer

2.0 m

diameter 110 mm and inner diameter 90 mm. The dimensions of the sign are 2.0 m ⫻ 1.0 m, and its lower edge is 3.0 m above the base. Note that the center of gravity of the sign is 1.05 m from the axis of the pipe. The wind pressure against the sign is 1.5 kPa. Determine the maximum in-plane shear stresses due to the wind pressure on the sign at points A, B, and C, located on the outer surface at the base of the pipe.

Rose’s Editing Co.

1.0 m

1.05 m to c.g.

Pipe

110 mm 3.0 m X

B

X C

Section X-X

Solution 8.5-16 d2 ⫽ 110 mm I⫽

d1 ⫽ 90 mm

p 4 (d ⫺ d 14) 64 2

Ip ⫽ 2I

t ⫽ 10 mm

MAXIMUM SHEAR STRESS AT POINT A

I ⫽ 3.966 ⫻ 106 mm4

Ip ⫽ 7.933 ⫻ 106 mm4

1 1d 3 ⫺ d 312 12 2

Q⫽

Q ⫽ 5.017 * 104 mm3 SIGN:

A ⫽ 2m2 h ⫽ a3 +

Area 1 bm 2

b ⫽ 1.05 m

WIND PRESSURE

horizontal distance from the center of gravity of the sign to the axis of the pipe

pw ⫽ 1.5 kPa P ⫽ pw A

STRESS RESULTANTS AT THE BASE M ⫽ Ph

M ⫽ 10.5 kN # m

T ⫽ Pb

T ⫽ 3.15 kN # m

V⫽P

Height from the base to the center of gravity of the sign

V ⫽ 3 kN

P ⫽ 3 kN

sx ⫽ 0 sy ⫽

Md 2 2I

sy ⫽ 145.603 MPa

txy ⫽

Td 2 2Ip

txy ⫽ 21.84 MPa

tmax ⫽

A

a

sx ⫺ sy 2

tmax ⫽ 76.0 MPa

C A

A

B

PIPE:

693

Combined Loadings

2

b + t 2xy ;

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Applications of Plane Stress

MAXIMUM SHEAR STRESS AT POINT B

MAXIMUM SHEAR STRESS AT POINT C

sx ⫽ 0

sx ⫽ 0

sy ⫽ 0

sy ⫽ 0

txy ⫽

VQ Td2 ⫺ 2 Ip I(2t)

tmax ⫽

A

Pure shear

a

sx ⫺ sy 2

txy ⫽ 19.943 MPa 2

b + t2xy

tmax ⫽ 19.94 MPa

;

VQ Td2 + txy ⫽ 23.738 MPa 2 Ip I(2t) sx ⫺ sy 2 tmax ⫽ a b + t 2xy A 2 txy ⫽

Pure shear

tmax ⫽ 23.7 MPa

Problem 8.5-17 A sign is supported by a pole of hollow circular

8 ft

cross section, as shown in the figure. The outer and inner diameters of the pole are 10.5 in. and 8.5 in., respectively. The pole is 42 ft high and weighs 4.0 k. The sign has dimensions 8 ft ⫻ 3 ft and weighs 500 lb. Note that its center of gravity is 53.25 in. from the axis of the pole. The wind pressure against the sign is 35 lb/ft2. (a) Determine the stresses acting on a stress element at point A, which is on the outer surface of the pole at the “front” of the pole, that is, the part of the pole nearest to the viewer. (b) Determine the maximum tensile, compressive, and shear stresses at point A.

;

Hilda’s Office

3 ft

10.5 in. 8.5 in.

42 ft

X

X A

A Section X-X

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SECTION 8.5

695

Combined Loadings

Solution 8.5-17 PIPE:

d2 ⫽ 10.5 in.

d1 ⫽ 8.5 in.

c⫽

d2 2

(a) STRESSES AT POINT A

p 2 1d ⫺ d 212 A ⫽ 29.845 in.2 4 2 p 4 I ⫽ 340.421 in.4 1d 2 ⫺ d 412 I⫽ 64 Ip ⫽ 2I Ip ⫽ 680.842 in.4 1 3 Q ⫽ 45.292 in.3 1d ⫺ d 31 2 Q⫽ 12 2 A⫽

W1 ⫽ 4000 lb SIGN:

sx ⫽ 0

As ⫽ (8)(3) ft

2

As ⫽ 24 ft

Area

Height from the base to the center of gravity of the sign

sy ⫽ ⫺ sxy ⫽

3 h ⫽ a42 ⫺ b ft 2 Horizontal distance from the center of gravity of the sign to the axis of the pipe b ⫽ a(4)12 +

10.5 b in. 2

s1 ⫽

sy ⫽ 6145 psi

txy ⫽ 345 psi

sx + sy + 2

A

sx ⫺ sy

a

2

sx + sy 2

⫺

A

a

sx ⫺ sy 2

P ⫽ pw As

s2 ⫽ ⫺19.30 psi

P ⫽ 840 lb

M ⫽ 4.082 ⫻ 105 lb # in.

T ⫽ Pb

T ⫽ 4.473 ⫻ 104 lb # in. V ⫽ 840 lb Nz ⫽ 4.5 ⫻ 103 lb

;

;

tmax ⫽

A

a

sx ⫺ sy 2

Max. shear stress

2

b + t 2xy

s1 ⫽ 6164 psi

Max. compressive stress

M ⫽ Ph

Nz ⫽ W1 ⫹ W2

Td2 2Ip

Md2 2I

pw ⫽ 35 lb/ft2

STRESS RESULTANTS AT THE BASE

V⫽P

+ A

Max. tensile stress s2 ⫽

W2 ⫽ 500 lb

Nz

(b) MAXIMUM STRESSES AT POINT A

b ⫽ 53.25 in.

WIND PRESSURE

;

2

;

2

b + t 2xy

; 2

b + t2xy tmax ⫽ 3092 psi

;

NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress.

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Problem 8.5-18 A horizontal bracket ABC consists of two perpendicular arms AB of length 0.5 m, and BC of length of 0.75 m. The bracket has a solid circular cross section with diameter equal to 65 mm. The bracket is inserted in a frictionless sleeve at A (which is slightly larger in diameter) so is free to rotate about the z0 axis at A, and is supported by a pin at C. Moments are applied at point C as follows: M1 ⫽ 1.5 kN # m in the x-direction and M2 ⫽ 1.0 kN # m acts in the (⫺z) direction. Considering only the moments M1 and M2, calculate the maximum tensile stress st, the maximum compressive stress sc, and the maximum in-plane shear stress tmax at point p, which is located at support A on the side of the bracket at midheight.

y0

A 0.75 m

Frictionless sleeve embedded in support

p x0

B z0

y0 C

M2

0.5 m p M1

x0

O

65 mm Cross section at A

Solution 8.5-18 d ⫽ 65 mm

T⫽0

b1 ⫽ 0.5 m

length of arm BC

b2 ⫽ 0.75 m

length of arm AB

Torsional frictionless sleeve at support A (MZ)

Vy ⫽ ⫺

M1 ⫽ 1.5 kN # m

sx ⫽ 0

M2 ⫽ 1.0 kN # m

sy ⫽ 0

PROPERTIES OF THE CROSS SECTION

txy ⫽

d ⫽ 65 mm

r⫽

d 2

M2 b1

Vy Q

tmax ⫽

txy ⫽ ⫺0.804 MPa

Id

A

a

sx ⫺ sy 2

2

b + t 2xy

A⫽

p 2 d 4

A ⫽ 3.318 ⫻ 103 mm2

Pure Shear

I⫽

p 4 d 64

I ⫽ 8.762 ⫻ 105 mm4

STRESSES AT POINT p ON THE SIDE OF THE BRACKET

Ip ⫽ 2 I Q⫽

Ip ⫽ 1.752 ⫻ 106 mm4

2 3 r 3

Q ⫽ 2.289 ⫻ 104 mm3

STRESS RESULTANTS AT SUPPORT A Nz ⫽ 0

Axial force

My ⫽ 0 Mx ⫽ ⫺ M2

b2 + M1 b1

Mx ⫽ 0

tmax ⫽ 0.804 MPa

;

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SECTION 8.5

Problem 8.5-19 A cylindrical pressure vessel with flat ends is

697

Combined Loadings

y0

subjected to a torque T and a bending moment M (see figure). The outer radius is 12.0 in. and the wall thickness is 1.0 in. The loads are as follows: T ⫽ 800 k-in., M ⫽ 1000 k-in., and the internal pressure p ⫽ 900 psi. Determine the maximum tensile stress st, maximum compressive stress sc, and maximum shear stress tmax in the wall of the cylinder.

T M

M T x0

z0

Solution 8.5-19

Cylindrical pressure vessal

Internal pressure: Bending moment:

p ⫽ 900 psi M ⫽ 1000 k-in.

Torque:

T ⫽ 800 k-in.

Outer radius:

r2 ⫽ 12 in.

Wall thickness:

t ⫽ 1.0 in.

Mean radius:

r ⫽ r2 ⫺ t/2 ⫽ 11.5 in.

Outer diameter:

d2 ⫽ 24 in.

Inner diameter:

d1 ⫽ 22 in.

MOMENT OF INERTIA

sx ⫽

⫽ 2668.2 psi sy ⫽

pr ⫽ 10,350 psi t

txy ⫽

Tr2 ⫽ ⫺1002.7 psi Ip

PRINCIPAL STRESSES s1, 2 ⫽

p I ⫽ (d 42 ⫺ d 41) ⫽ 4787.0 in.4 64 Ip ⫽ 2I ⫽ 9574.0 in.4

pr Mr2 ⫺ ⫽ 5175.0 psi ⫺ 2506.8 psi 2t I

sx + sy ; 2

A

a

sx ⫺ sy 2

2

b + t2xy

⫽ 6509.1 psi ; 3969.6 psi s1 ⫽ 10,479 psi

s2 ⫽ 2540 psi

NOTE: Since the stresses due to T and p are the same everywhere in the cylinder, the maximum stresses occur at the top and bottom of the cylinder where the bending stresses are the largest.

MAXIMUM SHEAR STRESSES

PART (a). TOP OF THE CYLINDER

t⫽

Stress element on the top of the cylinder as seen from above.

⬖ tmax ⫽ 5240 psi

In-plane:

t ⫽ 3970 psi

Out-of-plane: s1 2

or

s2 2

t⫽

s1 ⫽ 5240 psi 2

MAXIMUM STRESSES FOR THE TOP OF THE CYLINDER st ⫽ 10,480 psi tmax ⫽ 5240 psi

sc ⫽ 0 (No compressive stresses)

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Applications of Plane Stress

PART (b). BOTTOM OF THE CYLINDER

MAXIMUM SHEAR STRESSES

Stress element on the bottom of the cylinder as seen from below.

In-plane:

t ⫽ 1669 psi

Out-of-plane: t⫽

s1 2

or

s2 2

t⫽

s1 ⫽ 5340 psi 2

⬖ tmax ⫽ 5340 psi MAXIMUM STRESSES FOR THE BOTTOM OF THE CYLINDER st ⫽ 10,680 psi

sc ⫽ 0 (No compressive stresses)

tmax ⫽ 5340 psi PART (c). ENTIRE CYLINDER pr Mr2 sx ⫽ + ⫽ 5175.0 psi + 2506.8 psi 2t I ⫽ 7681.8 psi

st ⫽ 10,680 psi

;

sc ⫽ 0 (No compressive stresses)

pr sy ⫽ ⫽ 10,350 psi t txy ⫽ ⫺

The largest stresses are at the bottom of the cylinder.

tmax ⫽ 5340 psi

;

;

Tr2 ⫽ ⫺1002.7 psi Ip

PRINCIPAL STRESSES s1, 2 ⫽

sx + sy ; 2

A

a

sx ⫺ sy 2

2

b + t2xy

⫽ 9015.9 psi ; 1668.9 psi s1 ⫽ 10,685 psi

s2 ⫽ 7347 psi

y0

Problem 8.5-20 For purposes of analysis, a segment of the crankshaft in a vehicle is represented as shown in the figure. Two loads P act as shown, one parallel to (⫺x0) and another parallel to z0; each load P equals 1.0 kN. The crankshaft dimensions are b1 ⫽ 80 mm, b2 ⫽ 120 mm, and b3 ⫽ 40 mm. The diameter of the upper shaft is d ⫽ 20 mm. (a) Determine the maximum tensile, compressive, and shear stresses at point A, which is located on the surface of the upper shaft at the z0 axis. (b) Determine the maximum tensile, compressive, and shear stresses at point B, which is located on the surface of the shaft at the y0 axis.

b1 = 80 mm B A x0

z0 d = 20 mm b2 = 120 mm

P b3 = 40 mm P = 1.0 kN

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SECTION 8.5

Combined Loadings

699

Solution 8.5-20 P ⫽ 1.0 kN

sx ⫽ ⫺155.972 MPa (compressive)

b1 ⫽ 80 mm

sy ⫽ 0

b2 ⫽ 120 mm

txy ⫽

b3 ⫽ 40 mm

s1 ⫽

PROPERTIES OF THE CROSS SECTION d ⫽ 20 mm

d r⫽ 2

p A ⫽ d2 4

A ⫽ 314.159 mm

I⫽

p 4 d 64

+ 2 sx + sy 2

tmax ⫽

I ⫽ 7.854 ⫻ 103 mm4 4

2 3 r Q ⫽ 666.667 mm3 3

STRESS RESULTANTS AT THE SUPPORT Vx ⫽ P

(Axial force in X-dir.)

Vy ⫽ 0

(Shear force in Y-dir.)

Vz ⫽ P

(Shear force in Z-dir.)

Mx ⫽ Pb2

A

a

⫺

A A

sx ⫺ sy 2

a

sx ⫺ sy

a

sx ⫺ sy

2 2

2

b + t2xy 2

b + t2xy

2

b + t2xy s1 ⫽ 31.2 MPa

MAX. TENSILE STRESS

Ip ⫽ 1.571 ⫻ 10 mm 4

txy ⫽ 76.394 MPa

sx ⫺ sy

2

Ip ⫽ 2I Q⫽

s2 ⫽

M xd 2Ip

MAX. COMPRESSIVE STRESS s2 ⫽ ⫺187.2 MPa

;

MAX. SHEAR STRESS

tmax ⫽ 109.2 MPa

(b) STRESSES AT POINT B sx ⫽ ⫺

Mz Vx + r A I

(Torsional Moment)

Mx ⫽ 120 kN # mm My ⫽ P(b1 ⫹ b3)

(Bending Moment)

My ⫽ 120 kN # mm Mz ⫽ Pb2 (Bending Moment) Mz ⫽ 120 kN # mm (a) STRESSES AT POINT A sx ⫽ 149.606 MPa (tensile) sy ⫽ 0 txy ⫽ s1 ⫽ s2 ⫽ sx ⫽ ⫺

My Vx ⫺ r A I

;

Vz Q Mxd + 2Ip Id

sx ⫺ sy

A

a

sx ⫺ sy

A

a

sx + sy + 2 sx + sy

tmax ⫽

2

A

a

⫺

sx ⫺ sy 2

txy ⫽ 80.639 MPa

2 2 2

b + t2xy

2

b + t2xy 2

b + t2xy

;

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Applications of Plane Stress

s1 ⫽ 184.8 MPa

MAX. TENSILE STRESS

;

MAX. COMPRESSIVE STRESS s2 ⫽ ⫺ 35.2 MPa

tmax ⫽ 110.0 MPa

MAX. SHEAR STRESS

;

NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress.

;

Problem 8.5-21 A moveable steel stand supports an automobile engine weighing W ⫽ 750 lb as shown in figure part (a).

The stand is constructed of 2.5 in. ⫻ 2.5 in. ⫻ 1/8 in. thick steel tubing. Once in position the stand is restrained by pin supports at B and C. Of interest are stresses at point A at the base of the vertical post; point A has coordinates (x ⫽ 1.25, y ⫽ 0, z ⫽ 1.25) (inches). Neglect the weight of the stand. (a) Initially, the engine weight acts in the (⫺z) direction through point Q which has coordinates (24, 0, 1.25); find the maximum tensile, compressive, and shear stresses at point A. (b) Repeat (a) assuming now that, during repair, the engine is rotated about its own longitudinal axis (which is parallel to the x axis) so that W acts through Q⬘ (with coordinates (24, 6, 1.25)) and force Fy ⫽ 200 lb is applied parallel to the y axis at distance d ⫽ 30 in. 17 in. z

17 in.

1.25 in.

O

B W

C y

A 24 in.

d=

. 0 in

Q

3

Q' 12 in.

6 in.

D x

Fy

(b) Top view 2.5 in. ⫻ 2.5 in. ⫻ 1/8 in. A B Q

C

Cx y

24 in.

D x

17 in. 36 in. Dz (a)

Cz

Cy

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SECTION 8.5

701

Combined Loadings

Solution 8.5-21 W ⫽ 750 1b

Fy ⫽ 200 1b

A ⫽ b2 ⫺ (b ⫺ 2t)2 Am ⫽ 6 in.

(a)

A ⫽ 1in.2

1 [ b 4 ⫺ ( b ⫺ 2 t)4] 12

SHEAR

s⫽

I ⫽ 1 in.4

tt ⫽

sx ⫽ 0

sy ⫽ s + 2

sx + sy

s2 ⫽

2

tmax ⫽ s1 ⫽ 0

A

a

⫺

txy ⫽ t sx ⫺ sy

A

a

sx ⫺ sy

A

a

sx ⫺ sy 2

t⫽0

& MAX SHEAR STRESS

sx + sy

s1 ⫽

1 1 b 3 ⫺ b 312 8 Fy Q

SHEAR DUE TO TORSIONAL MOMENT

2 2

t ⫽ tt ⫹ tT

t ⫽ 4633 psi

PRINCIPAL STRESSES

s1 ⫽

2

s2 ⫽

+ 2 sx + sy 2

A

a

⫺

sx ⫺ sy

A

sx ⫺ sy

A

a

sx ⫺ sy 2

s1 ⫽ 988 psi

txy ⫽ t a

sx + sy

tmax ⫽

;

& MAX SHEAR STRESS

sy ⫽ s

2

b + t2xy

2 2 2

b + t2xy

;

s2 ⫽ ⫺20730 psi

;

s2 ⫽ ⫺21719 psi

;

tmax ⫽ 10365 psi

;

tmax ⫽ 11354 psi

;

ENGINE WEIGHT ACTS THROUGH

POINT Q ¿ & FORCE FY

ACTS IN Y-DIR

My ⫽ 18000 in.-1b SHEAR

s⫽

Mx ⫽ Wy1

& NORMAL STRESSES AT A

My c ⫺W ⫺ A I

s ⫽ ⫺20730 psi

same since element A lies on NA for bending about x-axis

tT ⫽

tT ⫽ 4255 psi

2

b + t2xy

MZ USING APPROX.

3-65) Mz ⫽ 6000 in.-1b

sx ⫽ 0 b + t2xy

Q ⫽ 1 in.3

tt ⫽ 378 psi

I (2 t)

Mz ⫽ Fy d

s ⫽ ⫺20730 psi

TRANSVERSE SHEAR

b1 ⫽ 2 in.

THEORY (EQU.

& NORMAL STRESSES AT A

Myc ⫺W ⫺ A I

b1 ⫽ b ⫺ 2t Q⫽

(POINT Q)

MZ (MAX PROB. #5.10-11)

TORSION DUE TO

IN WEB-SEE

b c⫽ . 2

My ⫽ Wx1

PRINCIPAL STRESSES

(b)

Am ⫽ (b ⫺ t)2

ENGINE WEIGHT ACTS THROUGH X-AXIS

Mx ⫽ 0

x1 ⫽ 24 in.

d ⫽ 30 in.

y1 ⫽ 6 in.

2

I⫽

FY &

1 in. 8

t⫽

b ⫽ 2.5 in.

SHEAR STRESS DEPENDS ON TRANSVERSE SHEAR DUE TO

2

b + t2xy 2

b + t2xy

Mz 2 tAm

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Applications of Plane Stress

Problem 8.5-22 A mountain bike rider going uphill applies force P ⫽ 65 N to each end of the handlebars ABCD, made of aluminum alloy 7075-T6, by pulling on the handlebar extenders (DF on right handlebar segment). Consider the right half of the handlebar assembly only (assume the bars are fixed at the fork at A). Segments AB and CD are prismatic with lengths L1 and L3 and with outer diameters and thicknesses d01, t01 and d03, t03, respectively, as shown. Segment BC of length L2, however, is tapered and, outer diameter and thickness vary linearly between dimensions at B and C. Consider shear, torsion, and bending effects only for segment AD; assume DF is rigid. Find maximum tensile, compressive, and shear stresses adjacent to support A. Show where each maximum stress value occurs. y

Handlebar extension d01 = 32 mm F t01 = 3.15 mm d03 = 22 mm t03 = 2.95 mm B A

x

C

F Handlebar extension

d = 100 mm z

D

d03

L3 = 220 mm

P 45°

D

L1 = 50 mm L2 = 30 mm

y

Handlebar

(a)

(b) Section D–F

Solution 8.5-22 z

P ⫽ 65 N L1 ⫽ 50 mm

45°

L2 ⫽ 30 mm

A2

45

A1

L3 ⫽ 220 mm

°

d01 ⫽ 32 mm

ax

y

Vm

d03 ⫽ 22 mm d ⫽ 100 mm

T

M

A3

PROPERTIES OF THE CROSS SECTION AT POINT A r⫽

d01 2

t01 ⫽ 3.15 mm

STRESS RESULTANTS AT POINT A1

p A ⫽ [d 201 ⫺ 1d01 ⫺ t 0122] 4

A ⫽ 150.543 mm2

p 4 [d ⫺ 1d01 ⫺ t 0124] I⫽ 64 01

I ⫽ 1.747 ⫻ 10 mm

Ip ⫽ 2I

Ip ⫽ 3.493 ⫻ 104 mm4

1 Q⫽ [ d 3 ⫺ 1 d01 ⫺ t 0123] 12 01

x

Section at point A

Nx ⫽ 0 Vmax. ⫽ P

4

4

Q ⫽ 729.625 mm

3

(Axial force in X-dir.) (Max. Shear force)

Vmax. ⫽ 0.065 kN T ⫽ Pd

(Torsional Moment)

M ⫽ P(L1 ⫹ L2 ⫹ L3) M ⫽ 19.5 kN # mm

T ⫽ 6.5 kN # mm

(Bending Moment)

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SECTION 8.5

703

Combined Loadings

sx ⫽ 0 sy ⫽ 0 txy ⫽ s1 ⫽ s2 ⫽ sx ⫽ ⫺

M r I

2

A

sx + sy 2

A

a

⫺

sx ⫺ sy

A

a

sx ⫺

2 2

2

b + t2xy 2

b + t2xy

sy 2

b + t2xy

2

s1 ⫽ 3.41 MPa

;

MAX. COMPRESSIVE STRESS

T d01 txy ⫽ 2Ip

+ 2 sx + sy 2

A

a

s2 ⫽ ⫺3.41 MPa

txy ⫽ 2.977 MPa

A

a

sx ⫺ sy sx ⫺ sy

A

a

sx + sy

tmax ⫽

sx ⫺ sy

MAX. TENSILE STRESS

sy ⫽ 0

s2 ⫽

+

txy ⫽ 3.408 MPa

a

sx + sy

tmax ⫽

sx ⫽ ⫺17.863 MPa (compressive stress)

s1 ⫽

Vmax. Q T d01 + 2Ip I 2 t 01

⫺

sx ⫺ sy 2

2 2

; tmax ⫽ 3.41 MPa

MAX. SHEAR STRESS 2

b +

t2xy

NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress.

2

b + t2xy

STRESS RESULTANTS AT POINT A3

2

b + t2xy

MAX. TENSILE STRESS

s1 ⫽ 0.483 MPa

;

MAX. COMPRESSIVE STRESS s2 ⫽ ⫺18.35 MPa

;

MAX. SHEAR STRESS

tmax ⫽ 9.42 MPa

;

NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress. STRESS RESULTANTS AT POINT A2

;

sx ⫽

M r I

sx ⫽ 17.863 MPa (tensile stress)

sy ⫽ 0 txy ⫽ s1 ⫽ s2 ⫽

Td01 2 Ip

txy ⫽ 2.977 MPa sx ⫺ sy

A

a

sx ⫺ sy

A

a

sx + sy + 2 sx + sy 2

⫺

2 2

2

b + t2xy 2

b + t2xy

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tmax ⫽

A

a

Page 704

Applications of Plane Stress

sx ⫺ sy 2

MAX. COMPRESSIVE STRESS

2

b + t2xy

MAX. TENSILE STRESS

s1 ⫽ 18.35 MPa

;

s2 ⫽ ⫺0.483 MPa

;

MAX. SHEAR STRESS

tmax ⫽ 9.42 MPa

;

NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress.

Problem 8.5-23 Determine the maximum tensile, compressive, and shear stresses acting on the cross section of the tube at point A of the hitch bicycle rack shown in the figure. The rack is made up of 2 in. ⫻ 2 in. steel tubing which is 1/8 in. thick. Assume that the weight of each of four bicycles is distributed evenly between the two support arms so that the rack can be represented as a cantilever beam (ABCDEF) in the x-y plane. The overall weight of the rack alone is W ⫽ 60 lb. directed through C, and the weight of each bicycle is B ⫽ 30 lb. B — 2

y

6 in.

Bike loads B

3 @ 4 in. F

E

4 loads, each B

E F

33 in.

B — at each tie down point 2

W MAZ

Ay C

C A Fixed support B

D

1 2 in. ⫻ 2 in. ⫻ — in. steel tube 8

x

Ax

B 17 in.

1 — in. 8 2 in. 2 in.

A

D

7 in.

2 in.

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SECTION 8.5

Combined Loadings

705

Solution 8.5-23 TOP OF THE CROSS SECTION (AT POINT A)

t ⫽ 0.125 in.

Cross section d2 ⫽ 2 in.

Outer width

d1 ⫽ d2 ⫺ 2t

Inner width

A ⫽ d22 ⫺ d12

A ⫽ 0.938 in.2

I⫽

Thickness

d 42 d 41 ⫺ 12 12

I ⫽ 0.552 in.4

Q ⫽ (d2 ⫺ 2t) t a

d2 d2 d2 t ⫺ b + 2 t 2 2 2 4

Q ⫽ 0.33 in.3 The distance between point A and the center of load W

sx ⫽

M Az d2 2I

sx ⫽ 8.591 ⫻ 103 psi (tensile stress) sy ⫽ 0 txy ⫽ 0 s1 ⫽ s2 ⫽

sx ⫺ sy

A

a

sx ⫺ sy

A

a

sx + sy + 2 sx + sy

tmax ⫽

2

A

a

⫺

sx ⫺

2 2

2

b + t2xy 2

b + t2xy

sy 2

2

b + t2xy

b1 ⫽ 17 in. The distance between the point A and the center of a bike load B

MAX. TENSILE STRESS

3.4 b2 ⫽ a17 + 2 + 6 + b in. 2

MAX. COMPRESSIVE STRESS

W ⫽ 60 lb

tmax ⫽ 4295 psi

B ⫽ 30 lb

REACTIONS AT SUPPORT A MAz ⫽ Wb1 ⫹ 4Bb2 Ay ⫽ W ⫹ 4B Ax ⫽ 0

MAz ⫽ 4.74 ⫻ 103 lb # in.

Ay ⫽ 180 lb

s1 ⫽ 8591 psi

; s2 ⫽ 0

;

MAX. SHEAR STRESS ;

NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress.

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Applications of Plane Stress

SIDE OF THE CROSS SECTION (AT POINT A)

BOTTOM OF THE CROSS SECTION (AT POINT A)

sx ⫽ 0

sx ⫽ ⫺

sy ⫽ 0 txy ⫽ s1 ⫽ s2 ⫽

sx ⫽ ⫺8.591 ⫻ 103 psi (compressive stress)

Ay Q

txy ⫽ 430.726 psi

I2 t

sx ⫺ sy

A

a

sx ⫺ sy

A

a

sx + sy + 2 sx + sy

tmax ⫽

2

A

a

M Az d2 2I

⫺

sx ⫺ sy 2

2 2

sy ⫽ 0 txy ⫽ 0

2

b + txy2

s1 ⫽

+ 2

2

b + txy2

s2 ⫽

sx + sy

2

b + txy2 s1 ⫽ 431 psi

MAX. TENSILE STRESS

sx ⫺ sy

A

a

sx ⫺ sy

A

a

sx + sy

tmax ⫽

2

A

a

⫺

sx ⫺

2 2

2

b + txy2

; s1 ⫽ 0

MAX. TENSILE STRESS

s2 ⫽ ⫺431 psi

MAX. COMPRESSIVE STRESS

; tmax ⫽ 431 psi

2

b + txy2

sy 2

MAX. COMPRESSIVE STRESS MAX. SHEAR STRESS

2

b + txy2

;

NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress.

s2 ⫽ ⫺8591 psi

;

;

MAX. SHEAR STRESS tmax ⫽ 4295 psi

;

NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress.

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9 Deflections of Beams

Differential Equations of the Deflection Curve y

The beams described in the problems for Section 9.2 have constant flexural rigidity EI.

Problem 9.2-1 The deflection curve for a simple beam AB (see figure) is

B

A

given by the following equation: ␯⫽ ⫺

q0 x (7L4 ⫺ 10L2x2 + 3x4) 360LEI

x

L

Describe the load acting on the beam.

Probs. 9.2-1 and 9.2-2

Solution 9.2-1

Simple beam

q0 x ␯⫽ ⫺ (7L4 ⫺ 10L2x2 + 3x4) 360LEI Take four consecutive derivatives and obtain: ␯–– ⫽ ⫺

q0 x LEI

From Eq. (9-12c): q ⫽ ⫺ EI␯–– ⫽

q0x L

;

The load is a downward triangular load of maximum ; intensity q0.

707

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Deflections of Beams

Problem 9.2-2 The deflection curve for a simple beam AB (see figure) is given by the following equation: ␯⫽ ⫺

q0 L4 4

p EI

sin

px L

(a) Describe the load acting on the beam. (b) Determine the reactions RA and RB at the supports. (c) Determine the maximum bending moment Mmax.

Solution 9.2-2

Simple beam

4

␯⫽ ⫺ ␯¿ ⫽ ⫺

q0 L 4

p EI q0 L3 3

p EI

sin

(b) REACTIONS (EQ. 9-12b)

px L

cos

V ⫽ EI␯–¿ ⫽

px L

q0 L px cos p L

At x ⫽ 0: V ⫽ RA ⫽

q0 L2

px ␯– ⫽ 2 sin L p EI ␯–¿ ⫽

At x ⫽ L: V ⫽ ⫺RB ⫽ ⫺

q0 L px cos pEI L

␯–– ⫽ ⫺

q0 L p

RB ⫽

q0 px sin EI L

; q0 L ; p

q0 L p

;

(c) MAXIMUM BENDING MOMENT (EQ. 9-12a) M ⫽ EI␯– ⫽

(a) LOAD (EQ. 9-12c) px q ⫽ ⫺EI␯–– ⫽ q0 sin ; L The load has the shape of a sine curve, acts downward, ; and has maximum intensity q0.

q0 L2 2

p

sin

px L

L For maximum moment, x ⫽ ; 2

Problem 9.2-3 The deflection curve for a cantilever beam AB (see figure) is

Mmax ⫽

q0 L2 p2

;

y

given by the following equation: ␯⫽ ⫺

q0 x2 (10L3 ⫺ 10L2x + 5Lx2 ⫺ x3) 120LEI

A

B x

Describe the load acting on the beam.

Probs. 9.2-3 and 9.2.-4

L

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SECTION 9.2 Differential Equations of the Deflection Curve

709

Solution 9.2-3 Cantilever beam ␯⫽ ⫺

q0x2 (10L3 ⫺ 10L2x + 5Lx2 ⫺ x3) 120LEI

Take four consecutive derivatives and obtain: ␯–– ⫽ ⫺

q0 (L ⫺ x) LEI

From Eq. (9-12c): q ⫽ ⫺ EI␯–– ⫽ q0 a1 ⫺

x b L

;

The load is a downward triangular load of maximum ; intensity q0.

Problem 9.2-4 The deflection curve for a cantilever beam AB (see figure) is given by the following equation: ␯⫽ ⫺

q0 x2 360L2EI

(45L4 ⫺ 40L3x + 15L2x2 ⫺ x4)

(a) Describe the load acting on the beam. (b) Determine the reactions RA and MA at the support.

Solution 9.2-4 Cantilever beam

␯– ⫽ ␯– ⫽ ␯–¿ ⫽ ␯–– ⫽

q0x2

(45L4 ⫺ 40L3x + 15L2x2 ⫺ x4) 360L2EI q0 (15L4x ⫺ 20L3x2 + 10L2x2 ⫺ x5) ⫺ 60L2EI q0 (3L4 ⫺ 8 L3x + 6 L2x2 ⫺ x4) ⫺ 12L2EI q0 ⫺ 2 (⫺ 2L3 + 3L2x ⫺ x3) 3L EI q0 ⫺ 2 (L2 ⫺ x2) L EI

␯⫽ ⫺

(a) LOAD (EQ. 9-12c) q ⫽ ⫺ EI␯–– ⫽ q0 a1 ⫺

(b) REACTIONS RA AND MA (EQ. 9-12b AND EQ. 9-12a) V ⫽ EI␯–¿ ⫽ ⫺

x

L2

b

3L2

(⫺2L3 + 3L2x ⫺ x3)

At x ⫽ 0: V ⫽ RA ⫽ M ⫽ EI␯– ⫽ ⫺

2

q0

q0 12L2

2q0 L 3

(3L4 ⫺ 8L3x + 6L2x2 ⫺ x4)

;

The load is a downward parabolic load of maximum intensity q0. ;

;

At x ⫽ 0: M ⫽ MA ⫽ ⫺

q0 L2 4

;

NOTE: Reaction RA is positive upward. Reaction MA is positive clockwise (minus means MA is counterclockwise).

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Deflections of Beams

Deflection Formulas q

Problems 9.3-1 through 9.3-7 require the calculation of deflections using the formulas derived in Examples 9-1, 9-2, and 9-3. All beams have constant flexural rigidity EI.

h

Problem 9.3-1 A wide-flange beam (W 12 ⫻ 35) supports a uniform

load on a simple span of length L ⫽ 14 ft (see figure). Calculate the maximum deflection dmax at the midpoint and the angles of rotation u at the supports if q ⫽ 1.8 k/ft and E ⫽ 30 ⫻ 106 psi. Use the formulas of Example 9-1.

Solution 9.3-1

Simple beam (uniform load)

W 12 ⫻ 35

L ⫽ 14 ft ⫽ 168 in.

q ⫽ 1.8 k/ft ⫽ 150 lb/in. I ⫽ 285 in.

L

Probs. 9.3-1 through 9.3-3

ANGLE OF ROTATION AT THE SUPPORTS (EQs. 9-19 AND 9-20)

E ⫽ 30 ⫻ 10 psi 6

4

u ⫽ uA ⫽ uB ⫽

MAXIMUM DEFLECTION (EQ. 9-18)

⫽

5 qL4 5(150 lb/in.)(168 in.)4 ⫽ 384 EI 384(30 * 106 psi)(285 in.4) ⫽ 0.182 in. ;

dmax ⫽

qL3 24 EI

(150 lb/in.)(168 in.)3 24(30 * 106 psi)(285 in.4)

⫽ 0.003466 rad ⫽ 0.199°

;

Problem 9.3-2 A uniformly loaded steel wide-flange beam with simple supports (see figure) has a downward deflection of 10 mm at the midpoint and angles of rotation equal to 0.01 radians at the ends. Calculate the height h of the beam if the maximum bending stress is 90 MPa and the modulus of elasticity is 200 GPa. (Hint: Use the formulas of Example 9-1.)

Solution 9.3-2

Simple beam (uniform load)

d ⫽ dmax ⫽ 10 mm

u ⫽ uA ⫽ uB ⫽ 0.01 rad

s ⫽ smax ⫽ 90 MPa

Maximum bending moment:

E ⫽ 200 GPa

M⫽

Calculate the height h of the beam. 5qL4 384EId or q ⫽ 384EI 5L4 qL3 24EIu or q ⫽ Eq. (9-19): u ⫽ uA ⫽ 24EI L3 Eq. (9-18): d ⫽ dmax ⫽

16d Equate (1) and (2) and solve for L: L ⫽ 5u Flexure formula: s ⫽

Mc Mh ⫽ I 2I

(1)

qL2 8

‹ s⫽

qL2h 16 I

Solve Eq. (4) for h: h ⫽

(4)

16Is

(5)

qL2

Substitute for q from (2) and for L from (3): (2) h⫽ (3)

32sd 15Eu2

;

Substitute numerical values: h⫽

32(90 MPa)(10 mm) 15(200 GPa)(0.01 rad)2

⫽ 96 mm

;

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SECTION 9.3

Deflection Formulas

711

Problem 9.3-3 What is the span length L of a uniformly loaded simple beam of wide-flange cross section (see figure) if the maximum bending stress is 12,000 psi, the maximum deflection is 0.1 in., the height of the beam is 12 in., and the modulus of elasticity is 30 ⫻ 106 psi? (Use the formulas of Example 9-1.)

Solution 9.3-3

Simple beam (uniform load)

s ⫽ smax ⫽ 12,000 psi h ⫽ 12 in.

d ⫽ dmax ⫽ 0.1 in.

Solve Eq. (2) for q: q ⫽

E ⫽ 30 ⫻ 106 psi

Flexure formula: s ⫽

5qL4 384EId or q ⫽ 384EI 5L4

(1)

qL2 8

‹ s⫽

L2 ⫽

24 Ehd 5s

L⫽

24 Ehd

;

A 5s

Substitute numerical values:

Mh Mc ⫽ I 2I

L2 ⫽

Maximum bending moment: M⫽

(3)

L2h

Equate (1) and (2) and solve for L:

Calculate the span length L. Eq. (9-18): d ⫽ dmax ⫽

16Is

qL2h 16I

(2)

24(30 * 106 psi)(12 in.)(0.1 in.) ⫽ 14,400 in.2 5(12,000 psi)

L ⫽ 120 in. ⫽ 10 ft

Problem 9.3-4 Calculate the maximum deflection dmax of a uniformly loaded simple beam (see figure) if the span length L ⫽ 2.0 m, the intensity of the uniform load q ⫽ 2.0 kN/m, and the maximum bending stress s ⫽ 60 MPa. The cross section of the beam is square, and the material is aluminum having modulus of elasticity E ⫽ 70 GPa. (Use the formulas of Example 9-1.)

;

q = 2.0 kN/m

L = 2.0 m

Solution 9.3-4 L ⫽ 2.0 m

Simple beam (uniform load) q ⫽ 2.0 kN/m

s ⫽ smax ⫽ 60 MPa

Solve for b3: b3 ⫽

E ⫽ 70 GPa

Substitute b into Eq. (2): dmax ⫽

CROSS SECTION (square; b ⫽ width) I⫽

b4 12

S⫽

b3 6

5qL4

32 Eb4 qL2 qL2 M :s⫽ ⫽ Flexure formula with M ⫽ 8 S 8S

5Ls 4Ls 1/3 a b 24E 3q

Substitute numerical values: (1) (2)

3qL

4b3

5(2.0 m)(60 MPa) 5Ls 1 1 ⫽ ⫽ m⫽ mm 24E 24(70 GPa) 2800 2.8 a

4(2.0 m)(60 MPa) 1/3 4 Ls 1/3 b ⫽ c d ⫽ 10(80)1/3 3q 3(2000 N/m)

dmax ⫽

2

Substitute for S: s ⫽

(4)

(The term in parentheses is nondimensional.)

5qL4 Maximum deflection (Eq. 9-18): d ⫽ 384 EI Substitute for I: d ⫽

3qL2 4s

(3)

10(80)1/3 mm ⫽ 15.4 mm 2.8

;

;

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Deflections of Beams

q

Problem 9.3-5 A cantilever beam with a uniform load (see figure) has a height h equal to 1/8 of the length L. The beam is a steel wideflange section with E ⫽ 28 ⫻ 106 psi and an allowable bending stress of 17,500 psi in both tension and compression. Calculate the ratio d/L of the deflection at the free end to the length, assuming that the beam carries the maximum allowable load. (Use the formulas of Example 9-2.)

Solution 9.3-5 1 h ⫽ L 8

L

Cantilever beam (uniform load)

E ⫽ 28 * 106 psi

Solve for q:

s ⫽ 17,500 psi

q⫽

Calculate the ratio d/L. Maximum deflection (Eq. 9-26): dmax ⫽

qL4 8EI

qL3 d ‹ ⫽ L 8EI

(1) (2)

2

4Is

Substitute q from (3) into (2): s L d ⫽ a b L 2E h

2

Problem 9.3-6 A gold-alloy microbeam attached to a silicon

L ⫽ 27.5 ␮m

b ⫽ 4.0 ␮m

qL ⫽ 17.2 ␮N

t ⫽ 0.88 ␮m

dmax ⫽ 2.46 ␮m

b L

Substitute numerical values: Eg ⫽

3(17.2 mN)(27.5 mm)3 2(4.0 mm)(0.88 mm)3(2.46 mm)

⫽ 80.02 * 109 N/m2 or

Determine Eg. 4

bt3 12

t

Gold-alloy microbeam

Cantilever beam with a uniform load.

I⫽

;

q

wafer behaves like a cantilever beam subjected to a uniform load (see figure). The beam has length L ⫽ 27.5 ␮m and rectangular cross section of width b ⫽ 4.0 ␮m and thickness t ⫽ 0.88 ␮m. The total load on the beam is 17.2 ␮N. If the deflection at the end of the beam is 2.46 ␮m, what is the modulus of elasticity Eg of the gold alloy? (Use the formulas of Example 9-2.)

Eq. (9-26): d ⫽

;

17,500 psi 1 d ⫽ (8) ⫽ 6 L 400 2(28 * 10 psi)

qL qL h Mc h ⫽ a ba b ⫽ I 2 2I 4I

Solution 9.3-6

(3)

L2h

Substitute numerical values:

qL2 : Flexure formula with M ⫽ 2 s⫽

h

4

qL qL or Eg ⫽ 8EgI 8Idmax

Eg ⫽

3qL4 2bt3dmax

;

Eg ⫽ 80.0 GPa

;

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SECTION 9.3

Deflection Formulas

Problem 9.3-7 Obtain a formula for the ratio dC /dmax of the deflection at the midpoint to the maximum deflection for a simple beam supporting a concentrated load P (see figure). From the formula, plot a graph of dC/dmax versus the ratio a/L that defines the position of the load (0.5 ⬍ a/L ⬍ 1). What conclusion do you draw from the graph? (Use the formulas of Example 9-3.)

713

P

A

B a

b L

Solution 9.3-7

Simple beam (concentrated load)

Eq. (9-35): dC ⫽

Pb(3L2 ⫺ 4b2) 48EI

Eq. (9-34): dmax ⫽

(a Ú b)

Pb(L2 ⫺ b2)3/2 913LEI

(3 13L)(3L2 ⫺ 4b2) dc ⫽ dmax 16(L2 ⫺ b2)3/2

GRAPH OF dc/dmax VERSUS b ⫽ a/L Because a ⱖ b, the ratio b versus from 0.5 to 1.0.

(a Ú b)

dc b

dmax

0.5 0.6 0.7 0.8 0.9 1.0

1.0 0.996 0.988 0.981 0.976 0.974

(a Ú b)

Replace the distance b by the distance a by substituting L ⫺ a for b: (3 13L)(⫺ L2 + 8aL ⫺ 4a2) dc ⫽ dmax 16(2aL ⫺ a2)3/2 Divide numerator and denominator by L2: dc ⫽ dmax

dc ⫽ dmax

(3 13L)a⫺1 + 8 16L a2

a2 3/2 a ⫺ 2b L L

(3 13L)a⫺ 1 + 8 16a 2

a2 a ⫺ 4 2b L L

a2 a ⫺ 4 2b L L

a2 3/2 a ⫺ 2b L L

; NOTE: The deflection dc at the midpoint of the beam is almost as large as the maximum deflection dmax. The greatest difference is only 2.6% and occurs when the load reaches the end of the beam (b ⫽ 1).

ALTERNATIVE FORM OF THE RATIO Let b ⫽

a L

(3 13)(⫺ 1 + 8b ⫺ 4b 2) dc ⫽ dmax 16(2b ⫺ b 2)3/2

;

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Deflections of Beams

Deflections by Integration of the Bending-Moment Equation Problems 9.3-8 through 9.3-16 are to be solved by integrating the second-order differential equation of the deflection curve (the bending-moment equation). The origin of coordinates is at the left-hand end of each beam, and all beams have constant flexural rigidity EI.

y P A

Problem 9.3-8 Derive the equation of the deflection curve for a cantilever

B

beam AB supporting a load P at the free end (see figure). Also, determine the deflection dB and angle of rotation uB at the free end. (Note: Use the second-order differential equation of the deflection curve.)

Solution 9.3-8

L

Cantilever beam (concentrated load)

BENDING-MOMENT EQUATION (EQ. 9-12a) EI␯– ⫽ M ⫽ ⫺P(L ⫺ x)

␯¿ ⫽ ⫺

Px (2L ⫺ x) 2EI

EI␯¿ ⫽ ⫺PLx +

Px2 + C1 2

dB ⫽ ⫺␯(L) ⫽

n⬘(0) ⫽ 0

⬖ C1 ⫽ 0

uB ⫽ ⫺␯¿(L) ⫽

B.C.

PLx2 Px3 EI␯ ⫽ ⫺ + + C2 2 6 B.C. n (0) ⫽ 0 ⬖ C2 ⫽ 0 ␯⫽ ⫺

Px2 (3L ⫺ x) 6EI

PL3 3EI

;

PL2 2EI

;

(These results agree with Case 4, Table G-1.)

;

y

Problem 9.3-9 Derive the equation of the deflection curve for a simple beam AB loaded by a couple M0 at the left-hand support (see figure). Also, determine the maximum deflection dmax. (Note: Use the second-order differential equation of the deflection curve.)

M0 B

A

L

Solution 9.3-9

Simple beam (couple M0)

BENDING-MOMENT EQUATION (EQ. 9-12a) x EI␯– ⫽ M ⫽ M0 a1 ⫺ b L EI␯¿ ⫽ M0 ax ⫺ EI␯ ⫽ M0 a

x2 b + C1 2L

x2 x3 ⫺ b + C1x + C2 2 6L

B.C.

n(0) ⫽ 0

⬖ C2 ⫽ 0

B.C.

n(L) ⫽ 0

‹ C1 ⫽ ⫺

␯⫽ ⫺

M0 x (2L2 ⫺ 3Lx + x2) 6LEI

M0 L 3 ;

x

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SECTION 9.3

MAXIMUM DEFLECTION ␯¿ ⫽ ⫺

dmax ⫽ ⫺(␯)x⫽x1 ⫽

Set ␯¿ ⫽ 0 and solve for x: 13 b 3

715

Substitute x1 into the equation for n:

M0 (2L2 ⫺ 6Lx + 3x2) 6LEI

x1 ⫽ La 1 ⫺

Deflections by Integration of the Bending-Moment Equation

M0 L2

;

913EI

(These results agree with Case 7, Table G-2.)

;

Problem 9.3-10 A cantilever beam AB supporting a triangularly distributed

y

load of maximum intensity q0 is shown in the figure. Derive the equation of the deflection curve and then obtain formulas for the deflection dB and angle of rotation uB at the free end. (Note: Use the second-order differential equation of the deflection curve.)

q0 x

B

A L

Solution 9.3-10

Cantilever beam (triangular load)

BENDING-MOMENT EQUATION (EQ. 9-12a) EI␯– ⫽ M ⫽ ⫺

q0 (L ⫺ x)3 6L

q0 EI␯¿ ⫽ (L⫺ x)4 + C1 24L 3

B.C.

n⬘(0) ⫽ 0

‹ C1 ⫽ ⫺

q0 L 24 3

EI␯ ⫽ ⫺ B.C.

q0 q0 L x (L ⫺ x)5 ⫺ + C2 120L 24

n(0) ⫽ 0

␯⫽ ⫺

q0 x2 (10L3 ⫺ 10L2x + 5Lx2 ⫺ x3) 120LEI

␯¿ ⫽ ⫺

q0 x (4L3 ⫺ 6L2x + 4Lx2 ⫺ x3) 24LEI

dB ⫽ ⫺␯(L) ⫽ uB ⫽ ⫺␯¿(L) ⫽

q0 L4 30EI q0 L3 24EI

;

;

;

(These results agree with Case 8, Table G-1.)

q0 L4 ‹ C2 ⫽ 120

Problem 9.3-11 A cantilever beam AB is acted upon by a uniformly distributed moment (bending moment, not torque) of intensity m per unit distance along the axis of the beam (see figure). Derive the equation of the deflection curve and then obtain formulas for the deflection dB and angle of rotation uB at the free end. (Note: Use the second-order differential equation of the deflection curve.)

y m B

A L

x

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Deflections of Beams

Solution 9.3-11

Cantilever beam (distributed moment)

BENDING-MOMENT EQUATION (EQ. 9-12a) EI␯– ⫽ M ⫽ ⫺m(L ⫺ x) EI␯¿ ⫽ ⫺m aLx ⫺ B.C.

n⬘(0) ⫽ 0

x2 b + C1 2

⬖ C1 ⫽ 0

x3 Lx2 ⫺ b + C2 EI␯ ⫽ ⫺ma 2 6 B.C.

n(0) ⫽ 0

⬖ C2 ⫽ 0

mx2 (3L ⫺ x) 6 EI mx ␯¿ ⫽ ⫺ (2L ⫺ x) 2EI

␯⫽ ⫺

mL3 3 EI mL2 uB ⫽ ⫺␯¿(L) ⫽ 2EI

dB ⫽ ⫺␯(L) ⫽

Problem 9.3-12 The beam shown in the figure has a guided support

;

; ;

y MA

at A and a spring support at B. The guided support permits vertical movement but no rotation. Derive the equation of the deflection curve and determine the deflection dB at end B due to the uniform load of intensity q. (Note: Use the second-order differential equation of the deflection curve.)

q

A

L

x

B

k = 48EI/L3

RB = kdB

Solution 9.3-12 BENDING-MOMENT EQUATION 2

EI ␯– ⫽ M(x) ⫽

2

qL qx ⫺ 2 2

q L2 x q x3 EI␯¿ ⫽ ⫺ + C1 2 24 2 2

4

qL x qx EI ␯¿ ⫽ ⫺ + C1 x + C2 2 24 B.C.

n⬘(0) ⫽ 0

C1 ⫽ 0

B.C.

␯(L) ⫽

␯(x) ⫽ ⫺

qL qL4 ⫽⫺ k 48EI

C2 ⫽⫺

11qL4 48

q (2 x4 ⫺ 12x2L2 + 11L4) 48EI

dB ⫽ ⫺v(L) ⫽

qL4 48EI

;

;

Note that RB ⫽ kdB ⫽ qL which agrees with a Fvert ⫽ 0

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SECTION 9.3

717

Deflections by Integration of the Bending-Moment Equation

Problem 9.3-13 Derive the equations of the deflection curve for a

y

simple beam AB loaded by a couple M0 acting at distance a from the left-hand support (see figure). Also, determine the deflection d0 at the point where the load is applied. (Note: Use the second-order differential equation of the deflection curve.)

M0 B

A

a

x

b L

Simple beam (couple M0)

Solution 9.3-13

BENDING-MOMENT EQUATION (EQ. 9-12a) M0x EI␯– ⫽ M ⫽ L EI␯¿ ⫽

4 (n)Left ⫽ (n)Right

(0 … x … a)

‹ C4 ⫽ ⫺

M0 x2 + C1 (0 … x … a) 2L

EI␯– ⫽ M ⫽ ⫺ EI␯¿ ⫽ ⫺ B.C.

B.C.

M0 (L ⫺ x) L

(a … x … L)

M0 x2 aLx ⫺ b + C2 L 2

1 (n⬘)Left ⫽ (n⬘)Right

C1 ⫽ ␯⫽ ⫺

(a … x … L) ␯⫽ ⫺

at x ⫽ a

at x ⫽ a

M0a2 2

M0 2 (2L ⫺ 6aL + 3a2) 6L

M0 x (6aL ⫺ 3a2 ⫺ 2L2 ⫺ x2) 6 LEI (0 … x … a) M0 (3a2L ⫺ 3a2x ⫺ 2L2x + 3Lx2 ⫺ x3) 6 LEI

‹ C2 ⫽ C1 + M0a

(a … x … L)

3

EI␯ ⫽ B.C.

M0x + C1x + C3 (0 … x … a) 6L

2 n(0) ⫽ 0 2

d0 ⫽ ⫺␯(a) ⫽

⬖ C3 ⫽ 0 3

M0x M0x EI␯ ⫽ ⫺ + + C1x + M0ax + C4 2 6L (a … x … L) B.C.

3 n(L) ⫽ 0

‹ C4 ⫽ ⫺M0L aa ⫺

⫽

;

;

M0a(L ⫺ a)(2a ⫺ L) 3LEI

M0ab(2a ⫺ L) 3LEI

;

NOTE: d0 is positive downward. The preceding results agree with Case 9, Table G-2.

L b ⫺ C1L 3

Problem 9.3-14 Derive the equations of the deflection curve for a

y

cantilever beam AB carrying a uniform load of intensity q over part of the span (see figure). Also, determine the deflection dB at the end of the beam. (Note: Use the second-order differential equation of the deflection curve.)

q

B

A a

b L

x

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Deflections of Beams

Solution 9.3-14

Cantilever beam (partial uniform load)

BENDING-MOMENT EQUATION (EQ. 9-12a)

B.C.

q q EI␯– ⫽ M ⫽ ⫺ (a ⫺ x)2 ⫽ ⫺ (a2 ⫺ 2ax + x2) 2 2

B.C.

q 2 x3 aa x ⫺ ax2 + b + C1 2 3

1 n⬘(0) ⫽ 0

EI␯¿¿ ⫽ M ⫽ 0

(a ⱕ x ⱕ L)

EI␯¿ ⫽ C2

(a ⱕ x ⱕ L)

B.C.

B.C.

␯ ⫽⫺

2 (v¿)Left ⫽ (v¿)Right at x ⫽ a

q a2x2 ax3 x4 EI␯ ⫽ ⫺ a ⫺ + b + C3 2 2 3 12

‹ C4 ⫽

qx2 (6a2 ⫺ 4ax + x2) (0 … x … a) 24EI

␯⫽ ⫺ qa3 ‹ C2 ⫽ ⫺ 6

qa3x + C4 (a … x … L) 6

4 (n)Left ⫽ (n)Right at x ⫽ a

(0 … x … a)

⬖ C1 ⫽ 0

⬖ C3 ⫽ 0

EI␯ ⫽ C2x + C4 ⫽ ⫺

(0 … x … a) EI␯¿ ⫽ ⫺

3 n(0) ⫽ 0

qa3 (4x ⫺ a) 24EI

dB ⫽ ⫺␯(L) ⫽

(a … x … L)

qa3 (4L ⫺ a) 24EI

;

;

(0 … x … a)

y

q0

MA

beam AB supporting a distributed load of peak intensity q0 acting over one-half of the length (see figure). Also, obtain formulas for the deflections dB and dC at points B and C, respectively. (Note: Use the second-order differential equation of the deflection curve.)

x A

L/2

C

L/2

RA

Solution 9.3-15 L For 0 … x … 2 q0 L x q0 L2 EI␯– ⫽ M(x) ⫽ ⫺ 4 6 q0 L x2 q0L2 x EI␯¿ ⫽ ⫺ + C1 8 6 EI␯ ⫽

q0 L x3 q0 L2 x2 ⫺ + C1 x + C2 24 12

B.C.

n⬘(0) ⫽ 0

B.C.

n(0) ⫽ 0

C1 ⫽ 0 C2 ⫽ 0

5q0 L3 L ␯¿ a b ⫽ ⫺ 2 96EI

;

(These results agree with Case 2, Table G-1.)

Problem 9.3-15 Derive the equations of the deflection curve for a cantilever

BENDING-MOMENT EQUATION

qa4 24

q0 L 3 (x ⫺ 2L x2) 24EI

;

q0 L4 L dC ⫽ ⫺␯a b ⫽ 2 64EI

;

␯(x) ⫽

For

L … x … L 2

EI␯– ⫽ M(x) ⫽

q0L2 q0 q0 L x ⫺ ⫺ (L ⫺ x) 4 6 L ax ⫺

2 q0 L 2 1 b ⫺ cq0 ⫺ 2 2 L

(L ⫺ x)d a x ⫺

L 22 b 2 3

B

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SECTION 9.3

EI␯– ⫽ M(x) ⫽

⫺ q0 (⫺3L2 x + L3 3L

B.C.

5q0 L3 L ␯¿ a b ⫽ ⫺ 2 96EI

B.C.

q0 L4 L ␯a b ⫽ ⫺ 2 64 EI

+ 3L x2 ⫺ x3) q0 ⫺3 2 2 EI␯¿ ⫽⫺ a L x + L3x 3L 2 + L x3 ⫺

␯(x) ⫽

x4 b + C3 4

C3 ⫽ C4 ⫽

5 q L3 192 0

⫺1 q L4 320 0

⫺q0 (⫺160L2 x3 + 160L3 x2 960LEI

+ 80L x4 ⫺ 16 x5 ⫺ 25L4 x

q0 ⫺ 1 2 3 1 1 EI ␯ ⫽⫺ a L x + L3 x2 + L x4 3L 2 2 4 ⫺

719

Deflections by Integration of the Bending-Moment Equation

+ 3 L5)

; 7q0L4 dB ⫽ ⫺␯(L) ⫽ 160EI

1 5 x b + C3 x + C4 20

Problem 9.3-16 Derive the equations of the deflection curve for a simple beam

;

y

AB with a distributed load of peak intensity q0 acting over the left-hand half of the span (see figure). Also, determine the deflection dC at the midpoint of the beam. (Note: Use the second-order differential equation of the deflection curve.)

q0 A

L/2

C

B

L/2

RA

RB

Solution 9.3-16 BENDING-MOMENT EQUATION For 0 … x …

EI␯ ⫽

L 2

+ C 1x + C 2

5q0 L x 2q0 L EI ␯– ⫽ M(x) ⫽ ⫺ a ⫺ xb 24 L 2 a

B.C.

EI␯¿ ⫽

2q0 x 1 b ⫺ cq0 ⫺ 2 2 L

C2 ⫽ 0 2 3

q0 5L x 2x5 a ⫺ x4 L + b 24L 6 5 (2)

+ C 1x For

q0 (5L2 x ⫺12 x2 L + 8 x3) 24L

L … x … L 2

EI␯– ⫽ M(x) ⫽

2 2

q0 5L x a ⫺ 4 x3 L + 2 x4 b 24L 2 + C1

n(0) ⫽ 0

EI␯ ⫽

2

L 2 a ⫺ xb d x x 2 3 EI␯– ⫽

q0 5L2 x3 2 x5 a ⫺ x4 L + b 24L 6 5

(1)

5q0 L x 1 L L ⫺ q0 a x ⫺ b 24 2 2 6

EI␯– ⫽

Lq0 ( ⫺x + L) 24

EI␯¿ ⫽

Lq0 ⫺x2 a + L xb + C3 24 2

(3)

x

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CHAPTER 9

EI␯ ⫽

Page 720

Deflections of Beams

Lq0 ⫺x3 L x2 a + b + C 3x + C 4 24 6 2

(4)

q0 L4 + C3 L + C4 72

(5)

0⫽

B.C.

␯(L) ⫽ 0

B.C.

L L ␯¿ L a b ⫽ ␯¿ R a b 2 2

+ 96 x4 ⫺ 53L4)

1 1 q L3 + C1 ⫽ q L3 + C3 96 0 64 0 B.C.

1 q L4 1920 0 L For 0 … x … 2 q0 x ␯(x) ⫽ (200 x2 L2 ⫺ 240 x3 L 5760LEI C4 ⫽

(6)

L L ␯L a b ⫽ ␯R a b 2 2

For

L … x … L 2

␯(x) ⫽

13 1 5 1 q0 L4 + C1 L ⫽ qo L4 + C 3 L 5760 2 1152 2

⫺ Lq0 (40 x3 ⫺ 120L x2 5760EI

+ 83L2 x ⫺ 3L3)

; 4

(7)

+ C4

;

3q0 L L dC ⫽ ⫺␯a b ⫽ 2 1280EI

;

From (5)–(7) C1 ⫽

⫺ 53 q L3 5760 0

C3 ⫽

⫺83 q L3 5760 0

y

Problem 9.3-17 The beam shown in the figure has a guided support at A and a roller support at B. The guided support permits vertical movement but no rotation. Derive the equation of the deflection curve and determine the deflection dA at end A and also dC at point C due to the uniform load of intensity q ⫽ P/L applied over segment CB and load P at x ⫽ L/3. (Note: Use the secondorder differential equation of the deflection curve.)

MA

L — 3

B

A

C

L — 2

Solution 9.3-17 BENDING-MOMENT EQUATION For 0 … x …

L 3

EI␯– ⫽ M(x) ⫽ EI ␯¿ ⫽ EI␯ ⫽

B.C. ␯¿(0)

For

⫽0

L L … x … 3 2

C1 ⫽ 0

19 PL 24

19 PL x + C1 24

19 PL x2 + C 1 x + C2 48 19 19 EI␯ ⫽ PL x2 + C2 EI ␯¿ ⫽ PL x 24 48

EI␯– ⫽ M(x) ⫽

L 19 PL ⫺ P a x ⫺ b 24 3

P q=— L

P

L — 2

x

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SECTION 9.3 Deflections by Integration of the Bending-Moment Equation

19 PL PL ⫺ P x + 24 3

EI␯– ⫽ M(x) ⫽ EI␯¿ ⫽ EI␯ ⫽ For

B.C.

B.C.

B.C.

B.C.

B.C.

19 Px2 PL x PLx⫺ + + C3 24 2 3

P x3 PL x2 19 PL x2 ⫺ + + C3 x + C4 48 6 6

L … x … L 2

␯(L) ⫽ 0

EI ␯– ⫽ M(x) ⫽

19 P L 21 PL PL ⫺ Px + ⫺ ax ⫺ b 24 3 L 2 2

EI␯– ⫽ M(x) ⫽

19 PL Px PL Px2 PL ⫺ Px + + ⫺ ⫺ 24 3 2L 2 8

EI ␯¿ ⫽

Px2 PL x 19 PL x Px2 P x3 PL x ⫺ + + ⫺ + C5 ⫺ 24 2 3 6L 4 8

EI ␯ ⫽

PL x2 P x4 Px3 PL x2 19 Px3 + PL x2 ⫺ + ⫺ ⫺ + C5 x + C6 48 6 6 24L 12 16

PL4 PL3 PLL2 19 PL3 PLL2 PL L2 ⫺ + ⫺ + ⫺ + C5 L + C6 48 6 6 24L 12 16

0⫽

L L ␯¿L a b ⫽ ␯¿R a b 3 3

0⫽ ⫺

L L ␯L a b ⫽ ␯R a b 3 3

␯L(a) ⫽ ␯R(a)

L 2 Pa b 3

C2 ⫽ ⫺

L L ␯¿L a b ⫽ ␯¿R a b 2 2

+ 2

3

L 3 Pa b 3

C3 ⫽ ⫺

C3

L PL a b 3

+ 6

L 3 Pa b 2

L 2 Pa b 2 +

L + C3 a b + C4 3

⫺

6L

4

L 4 Pa b 2

L 3 Pa b 2

L + C4 ⫽ ⫺ 2 24L

(2)

+ C3

L 2 PL a b 3

6

(1)

+ 12

L PL a b 2 8

⫺

(4)

+ C5

L 2 PL a b 2 16

(3)

L + C5 a b + C6 2

(5)

From (1)–(5) C2 ⫽

721

⫺3565 3 PL 10368

C3 ⫽

⫺1 2 PL 18

C4 ⫽

⫺ 389 PL3 1152

C5 ⫽

⫺5 PL2 144

L 3

␯(x) ⫽

⫺PL (⫺4104 x2 + 3565 L2) 10368 EI

For

L L … x … 3 2

␯(x) ⫽

⫺P (⫺648 L x2 + 192 x3 + 64 L2 x + 389L3) 1152EI

For

L … x … L 2

␯(x) ⫽

⫺P ( ⫺72L2 x2 + 12 x3L + 6 x4 + 5L3 x + 49L4) 144EIL

For 0 … x …

dA ⫽ ⫺␯(0) ⫽

3565PL3 10368EI

;

L 3109PL3 dC ⫽ ⫺␯ a b ⫽ 3 10368EI

C6 ⫽

;

;

; ;

⫺ 49 PL3 144

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CHAPTER 9

Page 722

Deflections of Beams

Deflections by Integration of the Shear Force and Load Equations The beams described in the problems for Section 9.4 have constant flexural rigidity EI. Also, the origin of coordinates is at the left-hand end of each beam.

y

Problem 9.4-1 Derive the equation of the deflection curve for a cantilever

A

M0 B x

beam AB when a couple M0 acts counterclockwise at the free end (see figure). Also, determine the deflection dB and slope uB at the free end. Use the third-order differential equation of the deflection curve (the shear-force equation).

Cantilever beam (couple M0)

Solution 9.4-1

SHEAR-FORCE EQUATION (EQ. 9-12b). EIv–¿ ⫽ V ⫽ 0 EIv– ⫽ M ⫽ M0 ⫽ C1

1 M ⫽ M0

EIv¿ ⫽ C1x + C2 ⫽ M0x + C2 B.C.

B.C.

3 n(0) ⫽ 0

M0 x2 ␯⫽ 2EI

EIv¿¿ ⫽ C1 B.C.

L

2 n⬘(0) ⫽ 0

‹ C2 ⫽ 0

M0x2 EI␯ ⫽ + C3 2

␯¿ ⫽

⬖ C3 ⫽ 0

;

M0 x EI M0 L2 (upward) 2EI

dB ⫽ ␯(L) ⫽

;

M0 L (counterclockwise) EI

uB ⫽ ␯¿(L) ⫽

;

(These results agree with Case 6, Table G-1.)

px q = q0 sin — L

Problem 9.4-2 A simple beam AB is subjected to a distributed load of

intensity q ⫽ q0 sin px/L, where q0 is the maximum intensity of the load (see figure). Derive the equation of the deflection curve, and then determine the deflection dmax at the midpoint of the beam. Use the fourth-order differential equation of the deflection curve (the load equation).

y B

A

L

Solution 9.4-2

Simple beam (sine load)

LOAD EQUATION (EQ. 9-12c). EI␯–– ⫽ ⫺q ⫽ ⫺q0 sin

px L

L px EI␯–¿ ⫽ q0 a b cos + C1 p L 2

L px + C1x + C2 EI␯– ⫽ q0 a b sin p L

EIv–(0) ⫽ 0

B.C.

1 EIv– ⫽ M

B.C.

2 EIv–(L) ⫽ 0

EI␯¿ ⫽ ⫺q0 a EI␯ ⫽ ⫺q0 a

‹ C1 ⫽ 0

3

L px + C3 b cos p L

L 4 px + C3x + C4 b sin p L

‹ C2 ⫽ 0

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SECTION 9.4

B.C.

3 n(0) ⫽ 0

⬖ C4 ⫽ 0

B.C.

4 n(L) ⫽ 0

⬖ C3 ⫽ 0

␯⫽ ⫺

q0 L4 p4EI

q0L4 L dmax ⫽ ⫺␯a b ⫽ 4 2 p EI

;

(These results agree with Case 13, Table G-2.)

px L

sin

723

Deflections by Integration of the Shear Force and Load Equations

;

Problem 9.4-3 The simple beam AB shown in the figure has moments 2M0 and M0 acting at the ends. Derive the equation of the deflection curve, and then determine the maximum deflection dmax. Use the third-order differential equation of the deflection curve (the shear-force equation).

y 2M0 B

A

M0 x

L

Solution 9.4-3

Simple beam with two couples

Reaction at support A: RA ⫽

3M0 L

(downward)

Shear force in beam: V ⫽ ⫺RA ⫽ ⫺

3M0 L

⫽ ⫺

SHEAR-FORCE EQUATION (EQ. 9-12b) EI␯–¿ ⫽ V ⫽ ⫺ EI␯– ⫽ ⫺ B.C.

EI␯¿ ⫽ ⫺ EI␯ ⫽ ⫺

␯¿ ⫽ ⫺

3M0 L

3M0 x2 + 2M0 x + C2 2L

M0 x3 + M0 x2 + C2 x + C3 2L

2 n(0) ⫽ 0

‹ C3 ⫽ 0

B.C.

3 n(L) ⫽ 0

‹ C2 ⫽ ⫺

M0 x (L ⫺ x)2 2LEI

;

M0 (L ⫺ x) (L ⫺ 3x) 2LEI

Set n⬘ ⫽ 0 and solve for x:

EIn⬘⬘ (0) ⫽ 2M0

B.C.

M0 x 2 (L ⫺ 2 Lx + x2) 2LEI

MAXIMUM DEFLECTION

3M0 x + C1 L

1 EIn⬘⬘ ⫽ M

␯⫽ ⫺

M0 L 2

⬖ C1 ⫽ 2M0

x1 ⫽ L and x2 ⫽

L 3

Maximum deflection occurs at x2 ⫽ 2M0 L2 L dmax ⫽ ⫺␯a b ⫽ 3 27EI

L . 3

(downward)

;

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Page 724

Deflections of Beams

Problem 9.4-4 A beam with a uniform load has a guided support at one

y

end and spring support at the other. The spring has stiffness k ⫽ 48EI/L3. Derive the equation of the deflection curve by starting with the third-order differential equation (the shear-force equation). Also, determine the angle of rotation uB at support B.

MA

q

A

L

x

B

k = 48EI/L3

RB = kdB

Solution 9.4-4 SHEAR-FORCE EQUATION

B.C.

n⬘(0) ⫽ 0

B.C.

␯(L) ⫽

EI␯–¿ ⫽ V ⫽ ⫺qx 2

EI␯– ⫽ ⫺

qx + C1 2 2

B.C.

n⬘⬘(L) ⫽ M(L) ⫽ 0 2

EI ␯– ⫽

C1 ⫽

qL 2

2

qL qx ⫺ 2 2

qL2 x q x3 EI␯¿ ⫽ ⫺ + C2 2 6 EI ␯ ⫽

C3 ⫽ ⫺

C2 ⫽ 0

qL4 qL ⫽⫺ k 48EI

11qL4 48

q (2 x4 ⫺12 x2 L2 ⫹11L4) ; 48EI qL3 uB ⫽⫺␯¿(L) ⫽⫺ (Counterclockwise) 3EI ␯(x) ⫽ ⫺

;

q x4 qL2 x2 ⫺ + C2 x + C3 4 24

Problem 9.4-5 The distributed load acting on a cantilever beam AB has an intensity q given by the expression q0 cos px/2L, where q0 is the maximum intensity of the load (see figure). Derive the equation of the deflection curve, and then determine the deflection dB at the free end. Use the fourth-order differential equation of the deflection curve (the load equation).

y q0

px q = q0 cos — 2L

B

A L

x

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SECTION 9.4

Solution 9.4-5

Cantilever beam (cosine load)

LOAD EQUATION (EQ. 9-12 c)

B.C.

px EI␯–– ⫽ ⫺q ⫽ ⫺ q0 cos 2L EI␯–¿ ⫽ ⫺q0 a B.C.

3 n⬘(0) ⫽ 0

EI␯ ⫽ ⫺q0 a

2L px + C1 b sin p 2L

1 EIn⬘⬘⬘ ⫽ V

EI␯– ⫽ q0 a

B.C.

2q0L ‹ C1 ⫽ p

EIn⬘⬘⬘ (L) ⫽ 0

2q0 Lx 2L px + b cos + C2 p p 2L

q0Lx3 q0L2x2 2L px + ⫺ b cos + C4 p p 2L 3p

4 n(0) ⫽ 0

␯⫽ ⫺

2

⬖ C3 ⫽ 0 4

‹ C4 ⫽

2 EIn⬘⬘ ⫽ M

EI␯¿ ⫽ q0 a

EIn⬘⬘(L) ⫽ 0 2

2q0L p

px ⫺ 48L3 + 3p3Lx2 ⫺ p3x3 b 2L

dB ⫽ ⫺␯(L) ⫽

2

q0Lx 2q0L x 2L 3 px + b sin ⫺ + C3 p p p 2L

p4

3p4EI

a48L3 cos

‹ C2 ⫽ ⫺

16q0L4

q0L

2

B.C.

725

Deflections by Integration of the Shear Force and Load Equations

2q0L4 3p4EI

(p3 ⫺ 24)

;

;

(These results agree with Case 10, Table G-1.)

Problem 9.4-6 A cantilever beam AB is subjected to a parabolically

varying load of intensity q ⫽ q0(L2 ⫺ x2)/L2, where q0 is the maximum intensity of the load (see figure). Derive the equation of the deflection curve, and then determine the deflection dB and angle of rotation uB at the free end. Use the fourth-order differential equation of the deflection curve (the load equation).

y L2 ⫺x2 q = q0 — L2

q0

x B

A L

Solution 9.4-6

Cantilever beam (parabolic load)

LOAD EQUATION (EQ. 9-12 c) EI␯–– ⫽ ⫺q ⫽ ⫺ EI␯–¿ ⫽ ⫺ B.C.

q0 2

L

L2

2 EIn⬘⬘ ⫽ M

(L2 ⫺ x2)

aL2x ⫺

1 EIn⬘⬘⬘ ⫽ V

EI␯– ⫽ ⫺

q0

B.C.

EI␯¿ ⫽ ⫺

3

x b + C1 3

EIn⬘⬘⬘(L) ⫽ 0

‹ C1 ⫽

q0 L2x2 2q0L x4 a ⫺ b + x + C2 2 2 12 3 L

2q0L 3

B.C.

EIn⬘⬘(L) ⫽ 0

‹ C2 ⫽ ⫺

q0L2 4

q0 L2x3 q0Lx2 q0L2x x5 a ⫺ b + ⫺ + C3 2 6 60 3 4 L

3 n⬘(0) ⫽ 0

⬖ C3 ⫽ 0

q0 L2x4 q0Lx3 q0L2x2 x6 EI␯ ⫽ ⫺ 2 a ⫺ b + ⫺ + C4 24 360 9 8 L

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CHAPTER 9

B.C.

Page 726

Deflections of Beams

4 n(0) ⫽ 0

⬖ C4 ⫽ 0

␯¿ ⫽ ⫺

2

␯ ⫽⫺

q0 x

360 L2EI

(45L4 ⫺ 40L3x + 15L2x2 ⫺x4)

19q0L4 dB ⫽ ⫺␯(L) ⫽ 360 EI

;

q0 x 60L2EI

(15L4 ⫺ 20L3x + 10L2x2 ⫺ x4)

uB ⫽ ⫺␯¿(L) ⫽

q0L3 15EI

;

;

Problem 9.4-7 A beam on simple supports is subjected to a parabolically

4q0 x (L ⫺ x) q= — L2

distributed load of intensity q ⫽ 4q0 x(L ⫺ x)/L2, where q0 is the maximum intensity of the load (see figure). Derive the equation of the deflection curve, and then determine the maximum deflection dmax. Use the fourthorder differential equation of the deflection curve (the load equation).

y B

A

x

L

Solution 9.4-7

Simple beam (parabolic load)

LOAD EQUATION (EQ. 9-12 c) EI␯–– ⫽ ⫺q ⫽ ⫺ EI␯–¿ ⫽ ⫺ EI␯– ⫽ ⫺ B.C. B.C.

2q0 3L2 q0 3L2

30L2

L

(L ⫺ x) ⫽ ⫺

4q0 L2

(Lx ⫺ x2) EI␯ ⫽ ⫺ B.C.

(2Lx ⫺ x ) + C1x + C2 3

2 EIn⬘⬘(L) ⫽ 0 q0

2

3 (Symmetry)

(3Lx2 ⫺ 2x3) + C1

1 EIn⬘⬘ ⫽ M

EI␯¿ ⫽ ⫺

4q0 x

B.C.

⬖ C2 ⫽ 0

q0L ‹ C1 ⫽ 3

2

30L

a L5x ⫺

4 n(0) ⫽ 0

4

EIn⬘⬘(0) ⫽ 0

q0

␯⫽ ⫺

L ␯¿ a b ⫽ 0 2

q0 x 90L2EI

‹ C3 ⫽ ⫺

qoL3 30

5L3x3 x6 + Lx5 ⫺ b + C4 3 3

⬖ C4 ⫽ 0

(3L5 ⫺ 5L3x2 + 3Lx4 ⫺ x5)

61q0L4 L dmax ⫽ ⫺␯a b ⫽ 2 5760 EI

;

;

( ⫺5L3x2 + 5Lx4 ⫺ 2x5) + C3

Problem 9.4-8 Derive the equation of the deflection curve for beam AB, with guided support at A and roller at B, carrying a triangularly distributed load of maximum intensity q0 (see figure). Also, determine the maximum deflection dmax of the beam. Use the fourth-order differential equation of the deflection curve (the load equation).

MA

y q0 x A

L

B RB

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SECTION 9.4

Deflections by Integration of the Shear Force and Load Equations

727

Solution 9.4-8 LOAD EQUATION EI ␯–– ⫽ ⫺q ⫽ ⫺q0 +

EI ␯¿ ⫽ ⫺

q0 x L

EI ␯ ⫽ ⫺

2

EI ␯–¿ ⫽ ⫺q0 x + B.C.

q0 x + C1 2L

n⬘⬘⬘(0) ⫽ V(0) ⫽ 0

B.C.

C1 ⫽ 0

q0 x3 q0 x2 + + C2 2 6L

n⬘⬘(L) ⫽ M(L) ⫽ 0

EI␯– ⫽ ⫺

q0 x5 q0 L2 x2 q0 x4 + + 24 120L 6

+ C3 x + C4

q0 x2 EI␯–¿ ⫽ ⫺q0 x + 2L EI ␯– ⫽ ⫺

q0 L2 x q0 x4 q0 x3 + + C3 + 6 24L 3

B.C.

n⬘(0) ⫽ 0

C3 ⫽ 0

B.C.

n(L) ⫽ 0

C4 ⫽ ⫺

␯(x) ⫽

C2 ⫽

2q0 L4 15

q0 1⫺ 5x4 L + x5 120EIL

+ 20L3 x2 ⫺ 16L52

q0 L2 3

q0 x2 q0 x3 q0 L2 + + 2 6L 3

;

MAXIMUM DEFLECTION dmax ⫽ ⫺␯(0) ⫽

Problem 9.4-9 Derive the equations of the deflection curve for beam ABC, with guided support at A and roller support at B, supporting a uniform load of intensity q acting on the over-hang portion of the beam (see figure). Also, determine deflection dC and angle of rotation uC. Use the fourth-order differential equation of the deflection curve (the load equation).

2q0 L4 15EI

;

y MA

q x A

L

B

L/2

C

RB

Solution 9.4-9 LOAD EQUATION

B.C.

EI␯–– ⫽ ⫺q ⫽ 0 (0 … x … L)

EI ␯ ⫽ ⫺

EI ␯– ⫽ C1 x + C2 (0 … x … L) n⬘⬘⬘(0) ⫽ V(0) ⫽ 0

B.C.

␯–(0) ⫽ M(0) ⫽ ⫺

EI ␯– ⫽ ⫺ EI ␯¿ ⫽ ⫺

qL2 8

qL2x + C3 8

B.C.

C1 ⫽ 0 2

qL 8

C2 ⫽ ⫺

C3 ⫽ 0

2 2

EI␯–¿ ⫽ C1 (0 … x … L) B.C.

n⬘(0) ⫽ 0

qL2 8

qL x + C4 16

n(L) ⫽ 0

␯(x) ⫽ ⫺

C4 ⫽

qL4 16

qL2 2 1x ⫺ L22 16E I

(0 … x … L)

LOAD EQUATION EI ␯–– ⫽ ⫺q

aL … x …

3L b 2

;

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Deflections of Beams

EI ␯–¿ ⫽ ⫺ qx + C5 aL … x …

3L b 2

⫺ q x2 3L + C5 x + C6 aL … x … b 2 2

EI ␯– ⫽

3L 3L b ⫽ Va b ⫽ 0 2 2

C5 ⫽

3qL 2

3L 3L b ⫽ Ma b ⫽ 0 2 2

C6 ⫽

9qL2 8

B.C.

␯–¿ a

B.C.

␯– a

+ EI␯ ⫽

⫺q x2 3qL x 9qL2 + ⫺ 2 2 8

B.C.

EI ␯¿ ⫽

⫺q x3 3qL x2 9qL2 x + ⫺ + C7 6 4 8

␯(x) ⫽

n⬘L(L) ⫽ n⬘R(L)

qL3 ⫺ qL3 3qL3 9qL3 ⫽ + ⫺ + C7 ⫺ 8 6 4 8 5 qL3 C7 ⫽ 12

5 qL3 12

3qL x3 9 qL2 x2 ⫺q x 4 + ⫺ 24 12 16 5 qL3 x + C8 12

+

EI␯– ⫽

B.C.

⫺qx3 3qL x2 9qL2 x + ⫺ 6 4 8

EI␯¿ ⫽

n(L) ⫽ 0

⫺1 4 qL 16

C8 ⫽

⫺q 1⫺20xL3 + 27L2 x2 48EI aL … x …

⫺12L x3 + 2 x4 + 3L42 dC ⫽ ⫺␯a

9qL4 3L b ⫽ 2 128EI

uC ⫽ ⫺␯¿ a

Problem 9.4-10 Derive the equations of the deflection curve for beam AB, with guided support at A and roller support at B, supporting a distributed load of maximum intensity q0 acting on the right-hand half of the beam (see figure). Also, determine deflection dA, angle of rotation uB, and deflection dC at the midpoint. Use the fourth-order differential equation of the deflection curve (the load equation).

;

;

7qL3 3L b ⫽ 2 48EI

MA

3L b 2

;

(Clockwise)

y

q0 x

A

L/2

C

L/2

B RB

Solution 9.4-10 LOAD EQUATION EI ␯–– ⫽ ⫺q ⫽ 0

B.C.

L a0 … x … b 2

EI ␯–¿ ⫽ C1 a0 … x …

L b 2

EI ␯– ⫽ C1 x + C2 a0 … x …

L b 2

n⬘⬘⬘(0) ⫽ V(0) ⫽ 0

C1 ⫽ 0 2

B.C.

␯–(0) ⫽ M(0) ⫽

EI ␯– ⫽

q0 L2 12

EI ␯¿ ⫽

q0 L2 x + C3 12

q0 L 12

a0 … x …

C2 ⫽

q0 L2 12

L b 2

a0 … x …

L b 2

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729

SECTION 9.4 Deflections by Integration of the Shear Force and Load Equations

B.C.

n⬘(0) ⫽ 0

EI␯ ⫽

C3 ⫽ 0

q0 L2 x2 + C4 24

EI ␯ ⫽ ⫺

a0 … x …

L b 2

q0 x4 q0 x5 q0 L x3 q0 L2 x2 + + ⫺ 12 60L 8 24

+

5q0 L3 x 41 ⫺ q L4 192 960 0

LOAD EQUATION

a

2q0 L L EI ␯–– ⫽ ⫺q0 + ax ⫺ b a … x … Lb L 2 2 B.C.

2 q0 x EI ␯–– ⫽ ⫺ 2 q0 + L 2

EI ␯–¿ ⫽ ⫺ 2q0 x +

a

q0 x + C5 L

L … x … Lb 2

L 2 q0 L2 a b 2 24

+ C4 ⫽ ⫺

L 4 q0 a b 2

B.C.

L L ␯–¿ a b ⫽ V a b ⫽ 0 2 2

+ 60L

B.C.

q0 L2 L L ␯– a b ⫽ Ma b ⫽ 2 2 12

q0 L2 12

q0 x3 3q0 L x q0 L2 EI␯– ⫽ ⫺ q0 x + + ⫺ 3L 4 12 2

EI ␯¿ ⫽⫺

q0 x3 q0 x4 3q0 L x2 + + 3 12 L 8

EI ␯¿ ⫽ ⫺

C7 ⫽

q0 x3 q0 x4 3q0 L x2 + + 3 12L 8

q0 L2 x 5q0 L3 + + 12 192 q0 x4 q0 x5 q0 L x3 EI ␯ ⫽ ⫺ + + 12 60L 8 2 2

B.C.

192 C4 ⫽

3

q0 L x 5q0 L x + + C8 24 192

n(L) ⫽ 0

L 2

⫺

⫺

L 2 q0 L2 a b 2

⫺ 19 q L4 480 0

EI ␯ ⫽

q0 L2 x2 19 ⫺ q L4 24 480 0

␯(x) ⫽ ⫺

⫺41 C8 ⫽ q L4 960 0

5 q L3 192 0

24

41 q L4 960 0

a0 … x …

L b 2

q0 L2 (⫺20x2 + 19L2) 480 EI a 0 … x … Lb

L L ␯¿ L a b ⫽ ␯¿ R a b 2 2

⫺

+

q0 L2 x + C7 12

⫺ B.C.

8 5q0 L3

3q0 L 4

C6 ⫽

L 3 q0 L a b 2

⫹

L … x … Lb 2

C5 ⫽

L 5 q0 a b 2

12

q0 x + C5 x + C6 3L a

;

L L ␯L a b ⫽ ␯R a b 2 2

3

EI␯– ⫽ ⫺ q0 x2 +

L … x … Lb 2

␯(x) ⫽ ⫺

;

q0 (80x4 L ⫺ 16 x5 ⫺120L2 960LEI

x3 + 40 L3 x2 ⫺25L4 x⫹41L5 a dA ⫽ ⫺␯(0) ⫽

19 q L4 480EI 0

uB ⫽ ⫺␯¿(L) ⫽ ⫺

13 q L3 192EI 0

L 7 dC ⫽ ⫺␯ a b ⫽ q L4 2 240EI 0

L … x … Lb 2 ; ; ;

;

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Deflections of Beams

Method of Superposition P

The problems for Section 9.5 are to be solved by the method of superposition. All beams have constant flexural rigidity EI.

P

P

A

B

Problem 9.5-1 A cantilever beam AB carries three equally spaced L — 3

concentrated loads, as shown in the figure. Obtain formulas for the angle of rotation uB and deflection dB at the free end of the beam.

Solution 9.5-1 L 2 Pa b 3

Pa

2L 2 b 3

+ 2EI

2EI

dB 

PL2 7PL2 +  2EI 9EI

L 2 Pa b 3 6EI

; +

P

(a) Determine the deflection d1 at the midpoint of the beam. (b) If the same total load (5P) is distributed as a uniform load on the beam, what is the deflection d2 at the midpoint? (c) Calculate the ratio of d1 to d2.

Solution 9.5-2

2L 2 b 3

6EI

a 3L 

P

P

P

P

A

B

L — 6

d2 

L 2 d1  c3L2  4 a b d 24 EI 6 (c)

L Pa b 3

2

3

L PL c3L2  4 a b d + 24EI 3 48EI ;

2L b 3

;

L — 6

L — 6

L — 6

L — 6

L — 6

(b) Table G-2, Case 1 qL  5P

L Pa b 6

11PL3 144EI

L b + 3

Pa

Simple beam with 5 loads

(a) Table G-2, Cases 4 and 6



a3L 

5PL3 PL3  3EI 9EI

Problem 9.5-2 A simple beam AB supports five equally spaced loads P (see figure).

+

L — 3

Cantilever beam with 3 loads

Table G-1, Cases 4 and 5

uB 

L — 3

5qL4 25PL3  384EI 384EI

;

d1 11 384 88  a b   1.173 d2 144 25 75

;

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SECTION 9.5

Problem 9.5-3 The cantilever beam AB shown in the figure has an extension

L

BCD attached to its free end. A force P acts at the end of the extension.

A

B

(a) Find the ratio a/L so that the vertical deflection of point B will be zero. (b) Find the ratio a/L so that the angle of rotation at point B will be zero.

D C

a P

Solution 9.5-3

Cantilever beam with extension Table G-1, Cases 4 and 6 PL3 PaL2  0 3EI 2EI 2 PL PaL  0 (b) uB  2EI EI (a) dB 

a 2  L 3 a 1  L 2

; ;

Problem 9.5-4 Beam ACB hangs from two springs, as shown in the figure. The springs have stiffnesses k1 and k2 and the beam has flexural rigidity EI. (a) What is the downward displacement of point C, which is at the midpoint of the beam, when the moment M0 is applied? Data for the structure are as follows: M0  10.0 kNm, L  1.8 m, EI  216 kNm2, k1  250 kN/m, and k2  160 kN/m. (b) Repeat (a) but remove M0 and apply uniform load q  3.5 kN/m to the entire beam.

RA = k1 dA

k1 A

RB = k2 dB

k2

M0 L/2

C

L/2

B

q = 3.5 kN/m (for Part (b) only)

Solution 9.5-4 M0  10.0 kN # m k1  250 kN /m

L  1.8 m k2  160 kN / m

q  3.5 kN/ m

EI  216 kN # m2

dA  22.22 mm dB  34.72 mm

Downward Upward

Table G-2, Case 8

(a) RA 

M0 L

RB  

dA 

RA k1

dB 

RB k2

M0 L

731

Method of Superposition

dC  0 +

1# (dA + dB) 2

dC  6.25 mm Upward

;

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CHAPTER 9

(b) RA 

qL 2

RA dA  k1

Page 732

Deflections of Beams

dB  19.69 mm

RB  RA

Table G-2, Case 1 RB dB  k2

dC 

dA  12.60 mm

5qL4 1 + (dA + dB) 384EI 2

dC  18.36 mm Downward

Problem 9.5-5 What must be the equation y  f(x) of the axis of the slightly curved beam AB (see figure) before the load is applied in order that the load P, moving along the bar, always stays at the same level?

;

P

y

B

A

L

Solution 9.5-5

Slightly curved beam

Let x  distance to load P d  downward deflection at load P Table G-2, Case 5: d

Initial upward displacement of the beam must equal d. ‹ y

Px2(L  x)2 3LEI

;

Px2(L  x)2 P(L  x)x 2 [L  (L  x)2  x2]  6LEI 3LEI

Problem 9.5-6 Determine the angle of rotation uB and deflection dB at the free end of a cantilever beam AB having a uniform load of intensity q acting over the middle third of its length (see figure).

q B

A

L — 3

Solution 9.5-6

Cantilever beam (partial uniform load)

q  intensity of uniform load Original load on the beam:

Load No. 1:

L — 3

L — 3

x

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SECTION 9.5

Load No. 2:

Method of Superposition

Table G-1, Case 2 uB 

q 2L 3 q L 3 7qL3 a b  a b  6EI 3 6EI 3 162EI

dB 

q q 2L 3 2L L 3 L a b a 4L  b a b a 4L  b 24EI 3 3 24EI 3 3



SUPERPOSITION:

23qL4 648EI

;

;

Original load  Load No. 1 minus Load No. 2

Problem 9.5-7 The cantilever beam ACB shown in the figure has

A

48 in.

Cantilever beam (two loads) dB    

M0(L/2) L PL3 a 2L  b + 2EI 2 3EI 3M0L2 PL3 + (  downward deflection) 8EI 3EI

SUBSTITUTE NUMERICAL VALUES: EI  2.1  106 k-in.2 M0  35 k-in. P  2.5 k L  96 in. Table G-1, Cases 4, 6, and 7

 

B

C 48 in.

dC  

2.5 k

35 k-in.

flexural rigidity EI  2.1  106 k-in.2 Calculate the downward deflections dC and dB at points C and B, respectively, due to the simultaneous action of the moment of 35 k-in. applied at point C and the concentrated load of 2.5 k applied at the free end B.

Solution 9.5-7

733

P(L/2)2 M0(L/2)2 L + a3L  b 2EI 6EI 2 M0L2 5PL3 + (   downward deflection) 8EI 48EI

dC  0.01920 in.  0.10971 in.  0.0905 in.

;

dB  0.05760 in.  0.35109 in.  0.293 in.

;

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Deflections of Beams

Problem 9.5-8 A beam ABCD consisting of a simple span BD and an

L — 3

L — 2

overhang AB is loaded by a force P acting at the end of the bracket CEF (see figure).

A

(a) Determine the deflection dA at the end of the overhang. (b) Under what conditions is this deflection upward? Under what conditions is it downward?

2L — 3

B

C

F

E

D

P a

Solution 9.5-8

Beam with bracket and overhang

Consider part BD of the beam.

(a) DEFLECTION AT THE END OF THE OVERHANG PL2 L (10L  9a) dA  uB a b  2 324 EI

;

(  upward deflection) a 10 (b) Deflection is upward when 6 and downward L 9 10 a ; when 7 L 9 M0  Pa Table G-2, Cases 5 and 9 uB 

P(L/3)(2L/3)(5L/3) 6LEI +



L2 L2 Pa c6a b  3a b  2L2 d 6LEI 3 9

PL (10L  9a) (  clockwise angle) 162EI

Problem 9.5-9 A horizontal load P acts at end C of the bracket

C

ABC shown in the figure. (a) Determine the deflection dC of point C. (b) Determine the maximum upward deflection dmax of member AB. Note: Assume that the flexural rigidity EI is constant throughout the frame. Also, disregard the effects of axial deformations and consider only the effects of bending due to the load P.

P H

B

A

L

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SECTION 9.5

Method of Superposition

735

Bracket ABC

Solution 9.5-9 BEAM AB

(a) ARM BC

M0  PH

Table G-1, Case 4 3

PH PH3 PH2L + uBH  + 3EI 3EI 3EI

dC 

PH2 (L + H) 3EI



;

(b) MAXIMUM DEFLECTION OF BEAM AB Table G-2, Table G-2, Case 7: uB 

PHL M0L  3EI 3EI

Problem 9.5-10 A beam ABC having flexural rigidity EI  75 kNm2 is loaded by a force P  800 N at end C and tied down at end A by a wire having axial rigidity EA  900 kN (see figure). What is the deflection at point C when the load P is applied?

Solution 9.5-10

Case 7: dmax 

M0L2 913EI



PHL2 913EI

B

A

;

C P = 800 N

0.5 m 0.5 m

0.75 m

D

Beam tied down by a wire CONSIDER AB AS A SIMPLE BEAM M0  PL2 Table G-2, Case 7: u¿B 

M0 L1 PL1L2  3EI 3EI

CONSIDER THE STRETCHING OF WIRE AD d¿A  (Force in AD) a

EI  75 kN # m2 P  800 N

DEFLECTION dC OF POINT C

EA  900 kN H  0.5 m

PL2 PL2H H H b  a ba b  EA L1 EA EAL1

L1  0.5 m

L2  0.75 m



CONSIDER BC AS A CANTILEVER BEAM Table G-1, Case 4: dC¿ 

dc  dc¿ uB¿ (L2)dB¿ a

PL32 3EI

L2 b L1

PL1L22 PL22H PL32 + + 3EI 3EI EAL21

;

SUBSTITUTE NUMERICAL VALUES: dC  1.50 mm  1.00 mm  1.00 mm  3.50 mm

;

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Deflections of Beams

Problem 9.5-11 Determine the angle of rotation uB and deflection dB at the

q0

y

free end of a cantilever beam AB supporting a parabolic load defined by the equation q  q0x2/L2 (see figure).

A

B

x

L

Solution 9.5-11 LOAD: q 

Cantilever beam (parabolic load)

q0x2

qdx  element of load

2

L



L

q0

q0L3 10EI

x4dx 

2EIL2 L0

;

L

(qdx)(x2) (3L  x) 6EI L0 L q0x2 1 a 2 b (x2)(3L  x) dx  6EI L0 L

dB 

TABLE G-1, CASE 5

(Set a equal to x)

L

L

uB 

2

2

(qdx)(x ) q0x 1  a 2 b x2dx 2EI 2EI L L0 L0



q0 6EIL2 L0

Problem 9.5-12 A simple beam AB supports a uniform load of intensity q acting over the middle region of the span (see figure). Determine the angle of rotation uA at the left-hand support and the deflection dmax at the midpoint.

L

(x4)(3Lx) dx 

13q0L4 180EI

;

q B

A

a

a L

Solution 9.5-12

Simple beam (partial uniform load)

LOAD: qdx  element of load

Replace P by qdx

Replace a by x

Integrate x from a to L/2 L/3 qdx L/2 q uA  (x)(L  x)  (xL  x2)dx 2EI La La 2EI q (L3  6a2L + 4a3) ;  24EI TABLE G-2, CASE 6 TABLE G-2, CASE 6

uA 

Pa(L  a) 2EI

Replace P by qdx

Pa (3L2  4a2) 24EI Replace a by x dmax 

Integrate x from a to L/2

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SECTION 9.5

L/2

dmax 

qdx (x)(3L2  4x2) 24EI

dmax 

La L/2 q  (3L2x  4x3)dx 24EI La



q (5L4  24a2L2 + 16a4) 384EI

L 2 L 3  4L(L  a) a b + L a b d 2 2

;



q(L  a)2 qa2 [2L  (L  a)]2  (2L  a)2 24LEI 24LEI

qa2 L L c  La2 + 4L2 a b + a2 a b 24LEI 2 2 L 3 L 2  6La b + 2a b d 2 2

Table G-2, Case 3

q  (L3  6La2 + 4a3) 24EI

q(L/2) c(L  a)4  4L(L  a)3 24LEI L 2 + 4L2(L  a)2 + 2(L  a)2 a b 2

ALTERNATE SOLUTION (not recommended; algebra is extremely lengthy)

uA 

dmax 

;

Problem 9.5-13 The overhanging beam ABCD supports two concentrated loads P and Q (see figure).

q (5L4  24L2a2 + 16a4) 384EI

;

y P

MA

(a) For what ratio P/Q will the deflection at point B be zero? (b) For what ratio will the deflection at point D be zero? (c) If Q is replaced by uniform load with intensity q (on the overhang), repeat (a) and (b) but find ratio P/(qa)

Q

A

C

x

B L — 2

D L — 2

RC q

Solution 9.5-13 (a) DEFLECTION AT POINT B Table G-2 Cases 6 and 10

dB 

L Pa b 2

737

Method of Superposition

L L 2 c3 a b (2L)  3 a b 6EI 2 2 L Qa a b 2 L 2 L a b d  c(2L)  a b d 2 2EI 2

dB  0

9a P  Q 4L

;

a (for Part (c))

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Deflections of Beams

(b) DEFLECTION AT POINT D Table G-2 Case 6; Table G-1 Case 4; Table G-2 Case 10

dD  

L L P a b c(2L)  d 2 2 2E I

Table G-2 Case 6; Table G-1 Case 1; Table G-2 Case 10

(a)

8a (3L + a) P  Q 9L2

dD  

L L P a b c(2L)  d 2 2

;

2EI 4

+ (c.1) DEFLECTION AT POINT B dD  0

Table G-2 Cases 6 and 10

dB 

;

(c.2) DEFLECTION AT POINT D

Qa (2L) Q a3 + (a) + 3EI 2EI dD  0

P 9a  qa 8L

dB  0

L Pa b 2

qa + 8EI

a

qa2 b (2L) 2 2EI

(a)

(a)

a (4L + a) P  qa 3L2

;

L L 2 c3 a b (2L)  3 a b 6EI 2 2 2

L a b d  2

a

qa2 L ba b 2 2 2 EI

L c(2 L) a b d 2

Problem 9.5-14 A thin metal strip of total weight W and length L is placed

d

across the top of a flat table of width L/3 as shown in the figure. What is the clearance d between the strip and the middle of the table? (The strip of metal has flexural rigidity EI.) L — 3

Solution 9.5-14

L — 6

L — 6

Thin metal strip

W  total weight q 

W L

EI  flexural rigidity FREE BODY DIAGRAM (the part of the strip above the table)

TABLE G-2, CASES 1 AND 10 d    

5q M0 L 2 L 4 a b + a b 384EI 3 8EI 3 5qL4 qL4 + 31,104EI 1296EI

19qL4 31,104EI

But q 

W : L

‹ d

19WL3 31,104EI

;

L — 3

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Page 739

SECTION 9.5

Problem 9.5-15 An overhanging beam ABC with flexural rigidity

739

Method of Superposition

y

EI  15 k-in.2 is supported by a guided support at A and by a spring of stiffness k at point B (see figure). Span AB has length L  30 in. and carries a uniform load. The over-hang BC has length b  15 in. For what stiffness k of the spring will the uniform load produce no deflection at the free end C?

q

MA

C A

L

B

k

x

b

RB

Solution 9.5-15 EI  15kip # in2.

L  30 in.

b  15 in.

RB  qL Table G-2, Case 1

3EI

for dC  0

k

Therefore

k  3.33lb/in

bL2 ;

3

dC  uB b  dB 

q (2L) qL (b)  24EI k

Problem 9.5-16 A beam ABCD rests on simple supports at B and C (see figure). The beam has a slight initial curvature so that end A is 18 mm above the elevation of the supports and end D is 12 mm above. What moments M1 and M2, acting at points A and D, respectively, will move points A and D downward to the level of the supports? (The flexural rigidity EI of the beam is 2.5  106 N # m2 and L  2.5 m).

M1

M2 B

18 mm A

C

L

L

12 mm L

D

Solution 9.5-16 EI  2.5106 N # m2

L  2.5 m

dA  18 mm

dD  12 mm Table G-2, Case 7 M2L M 1L + uB  3EI 6 EI

M 2L M 1L + uC  6 EI 3EI

dD 

M 2 L2 M 1L M 2L + a + bL 2EI 6EI 3EI

5M 1 + M 2 

6dA EI

5M 2 + M 1 

6dD EI

(1)

L2

(2)

L2

DEFLECTION AT POINT A AND D SOLVE EQUATION (1) AND (2)

Table G-1, Case 6 dA 

M 1 L2 + uB L 2EI

dA 

M 1 L2 M 1L M 2L + a + bL 2EI 3EI 6EI

dD 

M 2 L2 + uC L 2EI

M1 

EI(5dA  dD)

M2 

2

4L

Therefore M 1  7800 N # m

;

M 2  4200 N # m

;

EI(5dD  dA) 4L2

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Deflections of Beams

Problem 9.5-17 The compound beam ABC shown in the figure has a guided support at A and a fixed support at C. The beam consists of two members joined by a pin connection (i.e., moment release) at B. Find the deflection d under the load P.

P

MA

MC

A

B

3b

C

2b

b

RC

Solution 9.5-17 Table G-1, Case 4 dB 

P(2b)3 3EI

DEFLECTION UNDER THE LOAD P Table G-2, Case 6

d

P(b) c3(b) (8b)  3(b)2  (b)2 d + dB 6EI

d

P (2b)3 P(b) c3(b) (8 b)  3b2  b2 d + 6EI 3EI

d

6 Pb3 EI

Problem 9.5-18 A compound beam ABCDE (see figure) consists of two parts (ABC and CDE) connected by a hinge (i.e., moment release) at C. The elastic support at B has stiffness k  EI/b3 Determine the deflection dE at the free end E due to the load P acting at that point.

;

P D

C

B

A

2b

b

b

Solution 9.5-18 CONSIDER BEAM CDE

CONSIDER BEAM ABC RB 

3P 2

dB 

RB 3P  k 2k

Table G-2, Case 7; Table G-1, Case 4 Upward

Table G-2, Case 7; Table G-1, Case 4 dC   dC 

Pb(2 b) P b3 2b + b b + + dB a b 3EI 3EI 2b Pb(2 b) Pb3 3P 2b + b b + + a b 3EI 3EI 2k 2b P(4b3 k + 9E I) 4EIk

Upward

dE 

E

EI k= — b3

(Pb) (b) (Pb)(b) Pb3 b + + dC  b 3EI 3EI 3EI +

for k  dE 

P(4b3 k + 9EI ) Pb3 + 3EI 4EIk EI b3 47Pb3 12EI

;

b

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Page 741

SECTION 9.5

Method of Superposition

741

Problem 9.5-19 A stekel beam ABC is simply supported at A and held by a high-strength steel wire at B (see figure). A load P  240 lb acts at the free end C. The wire has axial rigidity EA  1500  103 lb, and the beam has flexural rigidity EI  36  106 lb-in.2 What is the deflection dC of point C due to the load P?

Wire 20 in. A

C

20 in.

Solution 9.5-19

P = 240 lb

Beam B

30 in.

Beam supported by a wire (2) ASSUME THAT THE WIRE STRETCHES T  tensile force in the wire 

P (b + c) b

dB 

P  240 lb

b  20 in.

c  30 in.

h  20 in.

Beam: EI  36  106 lb-in.2 Wire: EA  1500  103 lb

Ph(b + c) Th  EA EAb

d– C  dB a

Ph(b + c)2 b + c b  b EAb2

(downward)

(3) DEFLECTION AT POINT C dC  dc¿ dc¿¿  P(bc)c

(1) ASSUME THAT POINT B IS ON A SIMPLE SUPPORT

c2 h(b c) d  3EI EAb2

;

Substitute numerical values: dC  0.10 in. + 0.02 in.  0.12 in.

;

Pc3 Pc3 b uB¿ c  (Pc)a bc 3EI 3EI 3EI Pc2  (b + c) (downward) 3EI

dC¿ 

Problem 9.5-20 The compound beam shown in the figure consists of a cantilever beam AB (length L) that is pin-connected to a simple beam BD (length 2L). After the beam is constructed, a clearance c exists between the beam and a support at C, midway between points B and D. Subsequently, a uniform load is placed along the entire length of the beam. What intensity q of the load is needed to close the gap at C and bring the beam into contact with the support?

q D A

C

B L

Moment release L

c L

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CHAPTER 9

Solution 9.5-20

Page 742

Deflections of Beams

Compound beam dc¿¿  downward displacement of point C due to dB

BEAM BCD WITH A SUPPORT AT B dc¿  

5q(2L)4 384EI

11qL4 1 dc¿¿  dB  2 48EI

5qL4 24EI

DOWNWARD DISPLACEMENT OF POINT C dc  dc¿ dc¿¿ 

BEAM BCD WITH A SUPPORT AT B

5qL4 11qL4 7qL4   24EI 48EI 16EI

c  clearance

CANTILEVER BEAM AB dB 

qL4 (qL)L3 + 8EI 3EI 4



11qL 24EI

(downward)

c  dC 

7qL4 16EI

INTENSITY OF LOAD TO CLOSE THE GAP q

16EIc 7L4

;

Problem 9.5-21 Find the horizontal deflection dh and vertical deflection dn at the free end C of the frame ABC shown in the figure. (The flexural rigidity EI is constant throughout the frame.) Note: Disregard the effects of axial deformations and consider only the effects of bending due to the load P.

P B

C c

b A

Solution 9.5-21

Frame ABC MEMBER BC WITH B FIXED AGAINST ROTATION:

MEMBER AB:

Table G-1, Case 4:

dh  horizontal deflection of point B

dc¿ 

Table G-1, Case 6: dh 

(Pc)b2 Pcb2  2EI 2EI

Pcb uB  EI Since member BC does not change in length, dh is also the horizontal displacement of point C. ‹ dh 

Pcb2 2EI

;

Pc3 3EI

VERTICAL DEFLECTION OF POINT C dc  dv  dc¿ uBC   d 

Pc3 Pcb  (c) 3EI EI

Pc2 (c + 3b) 3EI Pc2 (c3b) 3EI

;

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Page 743

SECTION 9.5

743

Method of Superposition

Problem 9.5-22 The frame ABCD shown in the figure is squeezed by two collinear forces P acting at points A and D. What is the decrease d in the distance between points A and D when the loads P are applied? (The flexural rigidity EI is constant throughout the frame.) Note: Disregard the effects of axial deformations and consider only the effects of bending due to the loads P.

P B

A

a D

C L

Solution 9.5-22

P

Frame ABCD

MEMBER BC:

MEMBER BA:

Table G-1, Case 4: dA 

Table G-2, Case 10: uB 

(PL)a PLa  2EI 2EI

PL3 + uBL 3EI



PLa PL3 + (L) 3EI 2EI



PL2 (2L + 3a) 6EI

DECREASE IN DISTANCE BETWEEN POINTS A AND D d  2dA 

PL2 (2L + 3a) 3EI

Problem 9.5-23 A beam ABCDE has simple supports at B and D and symmetrical overhangs at each end (see figure). The center span has length L and each overhang has length b. A uniform load of intensity q acts on the beam. (a) Determine the ratio b/L so that the deflection dC at the midpoint of the beam is equal to the deflections dA and dE at the ends. (b) For this value of b/L, what is the deflection dC at the midpoint?

Solution 9.5-23 BEAM BCD:

;

q A

E B b

C

D

L

Beam with overhangs Table G-2, Case 1 and Case 10: qL3 qb2 L  a b 24EI 2 2EI qL 2  (L  6b2) (clockwise is positive) 24EI

uB 

b

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CHAPTER 9

dC ⫽

Page 744

Deflections of Beams

qL2 5qL4 qb2 L2 ⫺ a b ⫽ (5L2 ⫺ 24b2) 384EI 2 8EI 384EI (downward is positive)

(1)

Rearrange and simplify the equation: 48b4 ⫹ 96b3L ⫹ 24b2L2 ⫺ 16bL3 ⫺ 5L4 ⫽ 0 or b 3 b 2 b b 4 48a b + 96a b + 24a b ⫺ 16 a b ⫺ 5 ⫽ 0 L L L L

BEAM AB:

(a) RATIO

b L

Solve the preceding equation numerically: Table G-1, Case 1: qb4 qb4 qL 2 ⫺ uBb ⫽ ⫺ (L ⫺ 6b2)b dA ⫽ 8EI 8EI 24EI qb (3b3 + 6b2L ⫺ L3) ⫽ 24EI (downward is positive) DEFLECTION dC EQUALS DEFLECTION dA qL2 qb (5L2 ⫺ 24b2) ⫽ (3b3 + 6b2L ⫺ L3) 384EI 24EI

b ⫽ 0.40301 L

b ⫽ 0.4030 L

Say,

;

(b) DEFLECTION dC (EQ. 1) qL2 (5L2 ⫺ 24b2) 384EI qL2 [5L2 ⫺ 24 (0.40301 L)2] ⫽ 384EI qL4 ⫽ 0.002870 EI

dC ⫽

(downward deflection)

;

Problem 9.5-24 A frame ABC is loaded at point C by a force P acting at an angle a to the horizontal (see figure). Both members of the frame have the same length and the same flexural rigidity. Determine the angle a so that the deflection of point C is in the same direction as the load. (Disregard the effects of axial deformations and consider only the effects of bending due to the load P.) Note: A direction of loading such that the resulting deflection is in the same direction as the load is called a principal direction. For a given load on a planar structure, there are two principal directions, perpendicular to each other.

P

L

a

B

C

L

A

Solution 9.5-24

Principal directions for a frame P1 and P2 are the components of the load P P1 ⫽ P cos a P2 ⫽ P sin a If P1 ACTS ALONE

¿ dH ⫽

P1L3 3EI

dv¿ ⫽ uBL ⫽ a

(to the right) P1L2 P1L3 bL ⫽ 2EI 2EI (downward)

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Page 745

SECTION 9.6

¿¿ dH ⫽

If P2 ACTS ALONE dv¿¿ ⫽

P2L3 2EI

(to the left)

P2L3 P2L3 P2L2 4P2L3 ⫹ uBL ⫽ ⫹a bL ⫽ 3EI 3EI EI 3EI (upward)

DEFLECTIONS DUE TO THE LOAD P dH ⫽

P1L3 P2L3 L3 ⫺ ⫽ (2P1 ⫺ 3P2) 3EI 2EI 6EI (to the right)

P1L3 4P2L3 L3 ⫹ ⫽ (⫺3P1 + 8P2) dv ⫽ ⫺ 2EI 3EI 6EI (upward)

745

Moment-Area Method

dv ⫺3P1 ⫹8P2 ⫽ dH 2P1 ⫺3P2 ⫽

⫺ 3Pcosa + 8Psina ⫺3 + 8 tana ⫽ 2Pcosa ⫺ 3Psina 2 ⫺ 3 tana

PRINCIPAL DIRECTIONS The deflection of point C is in the same direction as the load P. ‹ tan a ⫽

P2 d␯ ⫽ P1 dH

or tan a ⫽

⫺ 3 + 8 tan a 2 ⫺ 3 tan a

Rearrange and simplity: tan2a ⫹ 2 tan a ⫺ 1 ⫽ 0 (quadratic equation) Solving, tan a ⫽ ⫺1 ; 12 a ⫽ 22.5°,

112.5°,

⫺67.5°,

⫺157.5°

;

Moment-Area Method q

The problems for Section 9.6 are to be solved by the moment-area method. All beams have constant flexural rigidity EI.

Problem 9.6-1 A cantilever beam AB is subjected to a uniform load of intensity q

Solution 9.6-1 M EI

B

A

acting throughout its length (see figure). Determine the angle of rotation uB and the deflection dB at the free end.

L

Cantilever beam (uniform load)

DIAGRAM:

uB/A ⫽ uB ⫺ uA ⫽ A1 ⫽ uA ⫽ 0

uB ⫽

qL3 6EI

qL3 6EI

(clockwise)

;

DEFLECTION

ANGLE OF ROTATION Use absolute values of areas. qL2 qL3 1 b ⫽ Appendix D, Case 18: A1 ⫽ (L)a 3 2EI 6EI 3L x⫽ 4

Q1 ⫽ First moment of area A1 with respect to B qL4 qL3 3L ba b ⫽ Q1 ⫽ A1x ⫽ a 6EI 4 8EI 4 qL (Downward) ; dB ⫽ Q1 ⫽ 8EI (These results agree with Case 1, Table G-1.)

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1:33 PM

CHAPTER 9

Page 746

Deflections of Beams

Problem 9.6-2 The load on a cantilever beam AB has a triangular distribution

q0

with maximum intensity q0 (see figure). Determine the angle of rotation uB and the deflection dB at the free end. B

A L

Solution 9.6-2 M EI

Cantilever beam (triangular load) x⫽

DIAGRAM

b(n + 1) 4L ⫽ n + 2 5

uB/A ⫽ uB ⫺ uA ⫽ A1 ⫽ uA ⫽ 0

uB ⫽

q0 L3 24EI

q0 L3 24EI ;

(clockwise)

DEFLECTION ANGLE OF ROTATION

Q1 ⫽ First moment of area A1 with respect to B

Use absolute values of areas.

Q1 ⫽ A1x ⫽ a

Appendix D, Case 20: A1 ⫽

q0 L2 q0 L3 bh 1 ⫽ (L)a b ⫽ n + 1 4 6EI 24EI

dB ⫽ Q1 ⫽

q0 L3 q0 L4 4L ba b ⫽ 24EI 5 30EI

q0 L4 30EI

;

(Downward)

(These results agree with Case 8, Table G-1.)

Problem 9.6-3 A cantilever beam AB is subjected to a concentrated load P and

P

a couple M0 acting at the free end (see figure). Obtain formulas for the angle of rotation uB and the deflection dB at end B.

A

B

M0

L

Solution 9.6-3 M EI

DIAGRAM

Cantilever beam (force P and couple M0) NOTE: A1 is the M/EI diagram for M0 (rectangle). A2 is the M/EI diagram for P (triangle). ANGLE OF ROTATION Use the sign conventions for the moment-area theorems (page 713 of textbook). A1 ⫽

M0 L EI

x1 ⫽

L 2

A2 ⫽ ⫺

PL2 2EI

x2 ⫽

2L 3

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Page 747

SECTION 9.6

A0 ⫽ A1 + A2 ⫽

M0 L PL2 ⫺ EI 2EI

uB/A ⫽ uB ⫺ uA ⫽ A0 uB ⫽ A0 ⫽

tB/A ⫽ Q ⫽ dB

uA ⫽ 0

747

Moment-Area Method

dB ⫽

M0 L2 PL3 ⫺ 2EI 3EI

(dB is positive when upward)

2

M0 L PL ⫺ EI 2EI

FINAL RESULTS

(uB is positive when counterclockwise)

To match the sign conventions for uB and dB used in Appendix G, change the signs as follows.

DEFLECTION

uB ⫽

M0 L PL2 ⫺ 2EI EI

dB ⫽

M0 L2 PL3 ⫺ 3EI 2EI

Q ⫽ first moment of areas A1 and A2 with respect to point B M0 L2 PL3 Q ⫽ A1x1 + A2x2 ⫽ ⫺ 2EI 3EI

(These results agree with Cases 4 and 6, Table G-1.)

q

of a cantilever beam AB with a uniform load of intensity q acting over the middle third of the length (see figure).

A

B L — 3

M EI

;

(positive downward)

Problem 9.6-4 Determine the angle of rotation uB and the deflection dB at the free end

Solution 9.6-4

;

(positive clockwise)

L — 3

L — 3

Cantilever beam with partial uniform load qL3 1 L qL2 b ⫽ A3 ⫽ a b a 2 3 9EI 54EI

DIAGRAM

x3 ⫽

2 L 8L 2L + a b ⫽ 3 3 3 9

A0 ⫽ A1 + A2 + A3 ⫽

7qL3 162EI

uB/A ⫽ uB ⫺ uA ⫽ A0 uA ⫽ 0

ANGLE OF ROTATION Use absolute values of areas. Appendix D, Cases 1, 6, and 18:

(clockwise)

;

DEFLECTION

Q ⫽ A1x1 + A2x2 + A3x3 ⫽

L 3 L 7L + a b⫽ 3 4 3 12

qL2 qL3 L b⫽ A2 ⫽ a b a 3 18EI 54EI

7qL3 162EI

Q ⫽ first moment of area A0 with respect to point B

qL2 qL3 1 L A1 ⫽ a b a b⫽ 3 3 18EI 162EI x1 ⫽

uB ⫽

2L L 5L x2 ⫽ + ⫽ 3 6 6

dB ⫽ Q ⫽

23qL4 648EI

23qL4 648EI

(Downward)

;

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1:33 PM

CHAPTER 9

Page 748

Deflections of Beams

Problem 9.6-5 Calculate the deflections dB and dC at points B and C, respectively, of the cantilever beam ACB shown in the figure. Assume M0 ⫽ 36 k-in., P ⫽ 3.8 k, L ⫽ 8 ft, and EI ⫽ 2.25 ⫻ 109 lb-in.2

A

M0

P

C

B

L — 2

Solution 9.6-5 M EI

L — 2

Cantilever beam (force P and couple M0) DEFLECTION dC

DIAGRAM

QC ⫽ first moment of areas A1 and left-hand part of A2 with respect to point C ⫽ a

M0 L L PL L L ba ba b ⫺ a ba ba b EI 2 4 2EI 2 4

⫺ ⫽

L L 1 PL a ba ba b 2 2EI 2 3

L2 (6M0 ⫺ 5PL) 48EI

NOTE: A1 is the M/EI diagram for M0 (rectangle). A2 is the M/EI diagram for P (triangle).

tC/A ⫽ QC ⫽ dC

Use the sign conventions for the moment-area theorems (page 713 of textbook).

(dC is positive when upward)

DEFLECTION dB

ASSUME DOWNWARD DEFLECTIONS (change the signs of dB and dC)

QB ⫽ first moment of areas A1 and A2 with respect to point B M0 L 3L ba ba b ⫽ A1x1 + A2x2 ⫽ a EI 2 4 2L 1 PL ⫺ a b(L)a b 2 EI 3 L2 (9M0 ⫺ 8PL) ⫽ 24EI tB/A ⫽ QB ⫽ dB

L2 dB ⫽ (9M0 ⫺ 8PL) 24EI

(dB is positive when upward)

dC ⫽

L2 (6M0 ⫺ 5PL) 48EI

dB ⫽

L2 (8PL ⫺ 9M0) 24EI

;

dC ⫽

L2 (5PL ⫺ 6M0) 48EI

;

ARE POSITIVE

SUBSTITUTE NUMERICAL VALUES: M0 ⫽ 36 k-in. L ⫽ 8 ft ⫽ 96 in.

P ⫽ 3.8 k EI ⫽ 2.25 ⫻ 106 k-in.2

dB ⫽ 0.4981 in. ⫺ 0.0553 in. ⫽ 0.443 in.

;

dC ⫽ 0.1556 in. ⫺ 0.0184 in. ⫽ 0.137 in.

;

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Page 749

SECTION 9.6

Problem 9.6-6 A cantilever beam ACB supports two concentrated loads P1 and P2 as shown in the figure. Calculate the deflections dB and dC at points B and C, respectively. Assume P1 ⫽ 10 kN, P2 ⫽ 5 kN, L ⫽ 2.6 m, E ⫽ 200 GPa, and I ⫽ 20.1 ⫻ 106 mm4.

Solution 9.6-6 M EI

749

Moment-Area Method

A

P1

P2

C

B L — 2

L — 2

Cantilever beam (forces P1 and P2) 1 P1L L L L 1 P2L 2L dB ⫽ a ba ba + b + a b(L)a b 2 2EI 2 2 3 2 EI 3

DIAGRAMS

⫽

P2L3 5P1L3 + 48EI 3EI

;

(downward)

DEFLECTION dC dC ⫽ tC/A ⫽ QC ⫽ first moment of areas to the left of point C with respect to point C P2 L L L 1 P1L L L dc ⫽ a ba ba b + a ba ba b 2 2EI 2 3 2EI 2 4 + ⫽ P1 ⫽ 10 kN E ⫽ 200 GPa

P2 ⫽ 5 kN

L ⫽ 2.6 m

I ⫽ 20.1 ⫻ 106 mm4

Use absolute values of areas. DEFLECTION dB dB ⫽ tB/A ⫽ QB ⫽ first moment of areas with respect

1 P2L L L a ba ba b 2 2EI 2 3

5P2L3 P1L3 + 24EI 48EI

;

(downward)

SUBSTITUTE NUMERICAL VALUES: dB ⫽ 4.554 mm + 7.287 mm ⫽ 11.84 mm dC ⫽ 1.822 mm + 2.277 mm ⫽ 4.10 mm

; ;

(deflections are downward)

to point B

Problem 9.6-7 Obtain formulas for the angle of rotation uA at support A and the deflection dmax at the midpoint for a simple beam AB with a uniform load of intensity q (see figure).

q A

B

L

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CHAPTER 9

Solution 9.6-7

Page 750

Deflections of Beams

Simple beam with a uniform load

DEFLECTION CURVE AND

M EI

DIAGRAM

tB/A ⫽ BB1 ⫽ first moment of areas A1 and A2with respect to point B qL4 L ⫽ (A1 + A2)a b ⫽ 2 24EI uA ⫽

qL3 BB1 ⫽ L 24EI

;

(clockwise)

DEFLECTION dmax AT THE MIDPOINT C Distance CC1 ⫽

qL4 1 (BB1) ⫽ 2 48EI

tC2/A ⫽ C2C1 ⫽ first moment of area A1 with respect to point C ⫽ A1x1 ⫽ a dmax ⫽ maximum deflection (distance CC2)

dmax ⫽ CC2 ⫽ CC1 ⫺ C2C1 ⫽

Use absolute values of areas. ANGLE OF ROTATION AT END A Appendix D, Case 17:

qL3 qL4 3L ba b ⫽ 24EI 16 128EI

⫽

5qL4 384EI

qL4 qL4 ⫺ 48EI 128EI ;

(downward)

(These results agree with Case 1 of Table G-2.)

qL3 2 L qL2 b⫽ A1 ⫽ A2 ⫽ a b a 3 2 8EI 24EI 3 L 3L x1 ⫽ a b ⫽ 8 2 16

Problem 9.6-8 A simple beam AB supports two concentrated loads P at the positions shown in the figure. A support C at the midpoint of the beam is positioned at distance d below the beam before the loads are applied. Assuming that d ⫽ 10 mm, L ⫽ 6 m, E ⫽ 200 GPa, and I ⫽ 198 ⫻ 106 mm4, calculate the magnitude of the loads P so that the beam just touches the support at C.

Solution 9.6-8

Simple beam with two equal loads

DEFLECTION CURVE AND

M EI

DIAGRAM

dC ⫽ deflection at the midpoint C

P

d

A

P B

C L — 4

L — 4

L — 4

L — 4

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Page 751

SECTION 9.6

A1 ⫽

PL2 16EI

x1 ⫽

3L 8

A2 ⫽

PL2 32EI

x2 ⫽

L 6

⫽ A1x1 + A2x2 ⫽ ⫽

PL2 3L PL2 L a b + a b 16EI 8 32EI 6

11PL3 384EI

Use absolute values of areas.

d ⫽ gap between the beam and the support at C

DEFLECTION dC AT MIDPOINT OF BEAM

MAGNITUDE OF LOAD TO CLOSE THE GAP

At point C, the deflection curve is horizontal.

d⫽d⫽

dC ⫽ tB/C ⫽ first moment of area between B and C with respect to B

751

Moment-Area Method

11PL3 384EI

P⫽

384EId

;

11L3

SUBSTITUTE NUMERICAL VALUES: d ⫽ 10 mm

L⫽6m

I ⫽ 198 ⫻ 10 mm 6

Problem 9.6-9 A simple beam AB is subjected to a load in the form of a couple M0 acting at end B (see figure). Determine the angles of rotation uA and uB at the supports and the deflection d at the midpoint.

E ⫽ 200 GPa P ⫽ 64 kN

4

;

M0 A

B

L

Solution 9.6-9

Simple beam with a couple M0

DEFLECTION CURVE AND

M EI

DIAGRAM

ANGLE OF ROTATION uA tB/A ⫽ BB1 ⫽ first moment of area between A and B with respect to B M0 L2 1 M0 L ⫽ a b (L)a b ⫽ 2 EI 3 6EI uA ⫽

BB1 M0 L ⫽ L 6EI

(clockwise)

;

ANGLE OF ROTATION uB tA/B ⫽ AA1 ⫽ first moment of area between A and B with respect to A ⫽ d ⫽ deflection at the midpoint C d ⫽ distance CC2 Use absolute values of areas.

uB ⫽

M0 L2 2L 1 M0 a b (L)a b ⫽ 2 EI 3 3EI AA1 M0 L ⫽ L 3EI

(Counterclockwise)

;

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CHAPTER 9

Page 752

Deflections of Beams

DEFLECTION d AT THE MIDPOINT C 2

Distance CC1 ⫽

M0 L 1 (BB1) ⫽ 2 12EI

tC2/A ⫽ C2C1 ⫽ first moment of area between A and C with respect to C

d ⫽ CC1 ⫺ C2C1 ⫽ ⫽

M0 L2 16EI

M0 L2 M0 L2 ⫺ 12EI 48EI

(Downward)

;

(These results agree with Case 7 of Table G-2.)

M0 L2 1 M0 L L ⫽ a ba ba b ⫽ 2 2EI 2 6 48EI

Problem 9.6-10 The simple beam AB shown in the figure supports two

P

P

equal concentrated loads P, one acting downward and the other upward. Determine the angle of rotation uA at the left-hand end, the deflection d1 under the downward load, and the deflection d2 at the midpoint of the beam.

A

B

a

a L

Solution 9.6-10

Simple beam with two loads

Because the beam is symmetric and the load is antisymmetric, the deflection at the midpoint is zero. ‹ d2 ⫽ 0

;

ANGLE OF ROTATION uA AT END A tC/A ⫽ CC1 ⫽ first moment of area between A and C with respect to C ⫽ A1 a ⫽ uA ⫽

a L 2 L ⫺ a + b + A2 a b a ⫺ ab 2 3 3 2

Pa(L ⫺ a)(L ⫺ 2a) 12EI Pa(L ⫺ a)(L ⫺ 2a) CC1 ⫽ L/2 6LEI

(clockwise)

DEFLECTION d1 UNDER THE DOWNWARD LOAD Distance

DD1 ⫽ a ⫽

Pa(L ⫺ 2a) M1 ⫽ EI LEI Pa2(L ⫺ 2a) 1 M1 A1 ⫽ a b (a) ⫽ 2 EI 2LEI Pa(L ⫺ 2a)2 1 M1 L A2 ⫽ a b a ⫺ ab ⫽ 2 EI 2 4LEI

a b(CC1) L/2

Pa2(L ⫺ a)(L ⫺ 2a) 6LEI

tD2/A ⫽ D2D1 ⫽ first moment of area between A and D with respect to D Pa3(L ⫺ 2a) a ⫽ A1 a b ⫽ 3 6LEI d1 ⫽ DD1 ⫺ D2D1 ⫽

Pa2(L ⫺ 2a)2 6LEI

(Downward)

;

;

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Page 753

SECTION 9.6

Moment-Area Method

Problem 9.6-11 A simple beam AB is subjected to couples M0 and 2M0 as

M0

shown in the figure. Determine the angles of rotation uA and uB at the beam and the deflection d at point D where the load M0 is applied.

2M0

A

B D L — 3

Solution 9.6-11

753

E L — 3

L — 3

Simple beam with two couples

DEFLECTION CURVE AND

M EI

ANGLE OF ROTATION uB AT END B

DIAGRAM

tA/B ⫽ AA1 ⫽ first moment of area between A and B with respect to A ⫽ A1 a uB ⫽

2L L 2L L 2L b + A2 a + b + A3 a + b ⫽0 9 3 9 3 9

AA1 ⫽0 L

;

DEFLECTION d AT POINT D M0 L2 1 Distance DD1 ⫽ (BB1) ⫽ 3 18EI tD2/A ⫽ D2D1 ⫽ first moment of area between A and D with respect to D M0L2 L ⫽ A1 a b ⫽ 9 54EI d ⫽ DD1 ⫺ D2D1 ⫽ (downward) M0 L 1 M0 L A1 ⫽ A2 ⫽ a ba b ⫽ 2 EI 3 6EI

A3 ⫽ ⫺

M0 L 6EI

ANGLE OF ROTATION uA AT END A tB/A ⫽ BB1 ⫽ first moment of area between A and B with respect to B ⫽ A1 a ⫽ uA ⫽

L L 2L L 2L + b + A2 a + b ⫹ A3 a b 3 9 3 9 9

M0 L2 6EI

BB1 M0 L ⫽ L 6EI

(clockwise)

;

M0 L2 27EI ;

NOTE: This deflection is also the maximum deflection.

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Deflections of Beams

Nonprismatic Beams Problem 9.7-1 The cantilever beam ACB shown in the figure has moments of

A

inertia I2 and I1 in parts AC and CB, respectively. (a) Using the method of superposition, determine the deflection dB at the free end due to the load P. (b) Determine the ratio r of the deflection dB to the deflection d1 at the free end of a prismatic cantilever with moment of inertia I1 carrying the same load. (c) Plot a graph of the deflection ratio r versus the ratio I2/I1 of the moments of inertia. (Let I2 /I1 vary from 1 to 5.)

Solution 9.7-1

I2

L — 2

P C

I1 B L — 2

Cantilever beam (nonprismatic)

Use the method of superposition.

(3) Total deflection at point B

(a) DEFLECTION dB AT THE FREE END

dB ⫽ (dB)1 + (dB)2 ⫽

(1) Part CB of the beam: (b) PRISMATIC BEAM d1 ⫽ (dB)1 ⫽

P L 3 PL3 a b ⫽ 3EI1 2 24EI1

(2) Part AC of the beam:

3

Ratio: r ⫽

7I1 PL3 a1 + b 24EI1 I2

;

PL3 3 EI1

dB 7I1 1 ⫽ a1 + b d1 8 I2

;

(c) GRAPH OF RATIO

2

3

dC ⫽

P(L/2) (PL/2)(L/2) 5PL + ⫽ 3EI2 2EI2 48EI2

uC ⫽

P(L/2)2 (PL/2)(L/2) 3PL2 + ⫽ 2EI2 EI2 8EI2

L 7PL3 (dB)2 ⫽ dC + uC a b ⫽ 2 24EI2

I2 I1

r

1 2 3 4 5

1.00 0.56 0.42 0.34 0.30

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SECTION 9.7

755

Nonprismatic Beams

Problem 9.7-2 The cantilever beam ACB shown in the figure supports a uniform

q

load of intensity q throughout its length. The beam has moments of inertia I2 and I1 in parts AC and CB, respectively. (a) Using the method of superposition, determine the deflection dB at the free end due to the uniform load. (b) Determine the ratio r of the deflection dB to the deflection d1 at the free end of a prismatic cantilever with moment of inertia I1 carrying the same load. (c) Plot a graph of the deflection ratio r versus the ratio I2/I1 of the moments of inertia. (Let I2 /I1 vary from 1 to 5.)

Solution 9.7-2

A I2 L — 2

dB ⫽ (dB)1 + (dB)2 ⫽

(1) Part CB of the beam: (b) PRISMATIC BEAM d1 ⫽ (dB)1 ⫽

q L 4 qL4 a b ⫽ 8EI1 2 128EI1

Ratio: r ⫽

(c) GRAPH OF RATIO

4

q(L/2) + 8EI2

a

qL b(L/2)3 2 3EI2

2

qL L 2 ba b 8 2

+ 2EI2

17qL4 ⫽ 384EI2

3

⫽

L — 2

q(L/2) (qL/2)(L/2)2 (qL2/8)(L/2) + + 6EI2 2EI2 EI2 7qL3 48EI2

15qL4 L (dB)2 ⫽ dC + uC a b ⫽ 2 128EI2

qL4 15I1 a1 + b 128EI1 I2 qL4 8EI1

dB 15I1 1 ⫽ a1 + b d1 16 I2

(2) Part AC of the beam:

uC ⫽

I1

(3) Total deflection at point B

(a) DEFLECTION dB AT THE FREE END

a

B

Cantilever beam (nonprismatic)

Use the method of superposition

dC ⫽

C

I2 I1

r

1 2 3 4 5

1.00 0.53 0.38 0.30 0.25

;

;

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Deflections of Beams

*Problem 9.7-3 Beam ACB hangs from two springs, as shown in the

RA = k1dA

figure. The springs have stiffnesses k1 and k2 and the beam has flexural rigidity EI. (a) What is the downward displacement of point C, which is at the midpoint of the beam, when the moment M0 is applied? Data for the structure are as follows: M0 ⫽ 7.5 k-ft, L ⫽ 6 ft, EI ⫽ 520 k-ft2, k1 ⫽ 17 k/ft, and k2 ⫽ 11 k/ft. (b) Repeat (a) but remove M0 and, instead, apply uniform load q over the entire beam.

k1

2EI

RB = k2dB M0

k2

EI

B

A L/2

C

L/2

B

q = 250 lb/ft (for Part (b) only)

*Solution 9.7-3 M0 ⫽ 7.5 kip/ft

L ⫽ 6 ft

EI ⫽ 520 kip/ft2

k1 ⫽ 17 kip/ft

k2 ⫽ 11 kip/ft

q ⫽ 250 lb/ft

(a) BENDING-MOMENT EQUATIONS-MOMENT M0 AT C 2EI␯– ⫽ M ⫽

M0x L

a0 … x …

2EI␯¿ ⫽

M0x2 ⫹C1 2L

2EI␯ ⫽

M0x3 + C1x + C 2 6L

B.C.

␯(0) ⫽ 0

a0 … x …

M0 ax ⫺

EI␯¿ ⫽ ⫺ M0x + EI␯ ⫽ ⫺ B.C.

B.C.

B.C.

2EI␯ ⫽ L b 2

L 2

L b 2

a0 … x …

C2 ⫽ 0

M0 + EI␯– ⫽ ⫺ 2

L b 2

M0x + C3 2L

⫺

M0x3 + C 1x 6L

⫽ ⫺M0 + a

M 0x L

a

a0 … x …

L b 2

L … x … Lb 2

L … x … Lb 2

M0x2 M0x3 + + C3x + C4 2 6L

␯(L) ⫽ 0

L b 2

a

L … x … Lb 2

M0L3 M0L2 + + C3L + C4 ⫽ 0 2 6L

L L ␯¿L a b ⫽ ␯¿R a b 2 2 L L ␯L a b ⫽ ␯R a b 2 2

L 2 L 2 M0 a b M0 a b 2 2 1 L J + C1 K ⫽ ⫺M0 + + C3 2 2L 2 2L

L 3 L 2 L 3 M0 a b M0 a b M0 a b 2 2 2 1 L L J + C1 K ⫽ ⫺ + + C3 + C4 2 6L 2 2 6L 2

(1)

(2)

(3)

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SECTION 9.7 Nonprismatic Beams

From (1), (2), and (3) C1 ⫽ 0

C3 ⫽

7 ML 16 0

⫺5 M L2 48 0

C4 ⫽

;

Therefore ␯(x) ⫽

M0x3 12EIL

␯(x) ⫽

M0 (⫺ 24x2L + 8x3 + 21L2x ⫺ 5L3) 48EIL

a0 … x …

L b 2 a

L … x … Lb 2

DEFLECTION AT A AND B RA ⫽

M0 L

dA ⫽

RA k1

RB ⫽ ⫺

M0 L RB k2

dB ⫽

dA ⫽ 0.88 in.

dB ⫽ ⫺1.36 in.

Downward

Upward

DEFLECTION AT POINT C L 1 dC ⫽ ⫺␯ a b + (dA + dB) 2 2

dC ⫽ ⫺

L 3 M0 a b 2 + 12EIL

1 (d + dB) 2 A

dC ⫽ ⫺ 0.31 in. Upward

;

(b) BENDING-MOMENT EQUATIONS-UNIFORM LOAD q qLx qx2 L 2EI␯– ⫽ M ⫽ ⫺ a0 … x … b 2 2 2 2 3 qLx qx L 2EI␯¿ ⫽ ⫺ + C1 a 0 … x … b 4 6 2 2EI␯ ⫽ B.C.

qLx3 qx4 ⫺ + C 1x + C 2 12 24

␯(0) ⫽ 0

C2 ⫽ 0

EI␯– ⫽

qLx qx2 ⫺ 2 2

EI␯¿ ⫽

qLx2 qx3 ⫺ ⫹ C3 4 6

EI␯ ⫽

a

a0 … x … 2EI␯ ⫽

L b 2

qLx3 qx4 ⫺ + C1x 12 24

L … x … Lb 2 a

L … x … Lb 2

qx4 qLx3 ⫺ + C3x + C4 12 24

a

L … x … Lb 2

a0 … x …

L b 2

757

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CHAPTER 9 Deflections of Beams

B.C.

B.C.

B.C.

␯(L) ⫽ 0

qLL3 qL4 + C3L + C4 ⫽ 0 ⫺ 12 24

L L ␯¿L a b ⫽ ␯¿R a b 2 2 L L vL a b ⫽ ␯R a b 2 2

⫽

L 3 qL a b 2 12

(1)

L 2 L 3 L 3 L 2 qL a b qa b qa b qLa b 2 2 2 2 1 J ⫺ + C1 K ⫽ ⫺ + C3 2 4 6 4 6 1 J 2 ⫺

L 3 qLa b 2 12

L 4 qa b 2 24

⫺

+ C3

L 4 qa b 2 24

L + C1 K 2

L + C4 2

(3)

From (1), (2), and (3) ⫺7 3 qL 128

C1 ⫽

C3 ⫽

⫺37 3 qL 768

C4 ⫽

5 qL4 768

Therefore ␯(x) ⫽ ⫺ ␯(x) ⫽

qx (⫺ 32Lx2 + 16x3 + 21L3) 768EI

a0 … x …

q (64Lx3 ⫺ 32x4 ⫺ 37L3x + 5L4) 768EI

a

L b 2

L … x … Lb 2

DEFLECTION AT A AND B RA ⫽ dA ⫽

qL 2

RB ⫽

qL 2

RA k1

dA ⫽ 0.53 in.

dB ⫽ Downward

RB k2

dB ⫽ 0.82 in.

Downward

DEFLECTION AT POINT C L 1 dC ⫽ ⫺␯ a b + (dA + dB) 2 2 L 2 L 2 L 3 1 dC ⫽ c ⫺ 32La b + 16 a b + 21L3 d + (dA + dB) 768EI 2 2 2 q

dC ⫽ 0.75 in. Downward

;

(2)

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SECTION 9.7

759

Nonprismatic Beams

Problem 9.7-4 A simple beam ABCD has moment of inertia I near the supports

q

and moment of inertia 2I in the middle region, as shown in the figure. A uniform load of intensity q acts over the entire length of the beam. Determine the equations of the deflection curve for the left-hand half of the beam. Also, find the angle of rotation uA at the left-hand support and the deflection dmax at the midpoint.

A

B

C

I

D I

2I

L — 4

L — 4 L

Solution 9.7-4

Simple beam (nonprismatic)

Use the bending-moment equation (Eq. 9-12a). B.C.

L 1 Symmetry: ␯¿ a b ⫽ 0 2

REACTIONS, BENDING MOMENT, AND DEFLECTION CURVE From Eq. (4): C2 ⫽ ⫺ 2EI␯¿ ⫽

qL3 24

qL x2 qx3 qL3 ⫺ ⫺ 4 6 24

a

L L … x … b 4 2

(5)

SLOPE AT POINT B (FROM THE RIGHT)

RA ⫽ RB ⫽

qL 2

2

M ⫽ Rx ⫺

Substitute x ⫽

2

qx qLx qx ⫽ ⫺ 2 2 2

EI␯¿B ⫽ ⫺ B.C.

L into Eq. (5): 4

11qL3 768

(6)

2 CONTINUITY OF SLOPES AT POINT B

(vB¿ )Left ⫽ (vB¿ )Right From Eqs. (3) and (6): BENDING-MOMENT

EQUATIONS FOR THE LEFT-HAND HALF

11qL3 qL L 2 q L 3 a b ⫺ a b + C1 ⫽⫺ 4 4 6 4 768

‹C1 ⫽ ⫺

7qL3 256

OF THE BEAM

EI␯– ⫽ M ⫽

qx2 qLx ⫺ 2 2

a0 … x … 2

qLx qx ⫺ E(2I )␯– ⫽ M ⫽ 2 2

L b 4

L L a … x … b 4 2

(1)

qL x2 qx3 ⫺ + C1 4 6 2

2EI␯¿ ⫽

3

qL x qx ⫺ + C2 4 6

a0 … x … a

L b 4

L L … x … b 4 2

EI␯¿ ⫽

7qL3 qL x2 qx3 ⫺ ⫺ 4 6 256

EI␯¿ ⫽

qL3 qL x2 qx3 ⫺ ⫺ 8 12 48

(2)

INTEGRATE EACH EQUATION EI␯¿ ⫽

SLOPE OF THE BEAM (FROM EQS. 3 AND 5)

(3)

a

L b 4

(7)

L L … x … b 4 3

(8)

ANGLE OF ROTATION uA (FROM EQ. 7) uA ⫽ ⫺v¿(0) ⫽

(4)

a0 … x …

7qL3 (positive clockwise) 256EI

;

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Deflections of Beams

INTEGRATE EQS. (7) AND (8) 3

4

3

EI␯ ⫽

7qL x qL x qx ⫺ ⫺ + C3 12 24 256

EI␯ ⫽

qL x3 qx4 qL3x ⫺ ⫺ + C4 24 48 48

B.C.

DEFECTION OF THE BEAM (FROM EQS. 9 AND 10) a0 … x … a

L b 4

(9)

␯⫽ ⫺

qx 121L3 ⫺ 64Lx2 + 32x32 768EI a0 … x …

L L … x … b (10) 4 2 ␯⫽ ⫺

3 Deflection at support A

a

DEFECTION AT POINT B (FROM THE LEFT) L into Eq. (9) with C3 ⫽ 0 4

35 qL4 EI␯B ⫽ ⫺ 6144 B.C.

;

q (13L4 + 256L3x ⫺ 512Lx3 + 256x4) 12,288EI

n(0) ⫽ 0 From Eq. (9): C3 ⫽ 0

Substitute x ⫽

L b 4

L L … x … b 4 2

;

MAXIMUM DEFLECTION (AT THE MIDPOINT E) (From the preceding equation for v.) (11)

dmax ⫽ ⫺va

31qL4 L b ⫽ 2 4096EI

(positive downward)

;

4 Continuity of deflections at point B

(nB)Right ⫽ (nB)Left From Eqs. (10) and (11): q L 4 qL3 L 35qL4 qL L 3 a b ⫺ a b ⫺ a b + C4 ⫽ ⫺ 24 4 48 4 48 4 6144 ‹C4 ⫽ ⫺

13qL4 12,288

Problem 9.7-5 A beam ABC has a rigid segment from A to B and a flexible segment with moment of inertia I from B to C (see figure). A concentrated load P acts at point B. Determine the angle of rotation uA of the rigid segment, the deflection dB at point B, and the maximum deflection dmax.

Rigid

P I

A

B L — 3

2L — 3

C

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761

SECTION 9.7 Nonprismatic Beams

Solution 9.7-5 Simple beam with a rigid segment

EI␯ ⫽

‹ dB ⫽ uA ⫽

3dB x L

a0 … x …

L b 3

(1)

3dB L

a0 … x …

L b 3

(2)

␯¿ ⫽ ⫺

PL2 + 3EIdB 54

8PL3 729EI

dB 8PL2 ⫽ L/3 243EI

EI␯¿ B.C.

(3)

Substitute for dB in Eq. (5) and simplify: ␯⫽

P (7L3 ⫺ 61L2x + 81L x2 ⫺ 27x3) 486EI

1 At x ⫽ L/3, ‹ C1 ⫽ ⫺

a

L … x … Lb 3

(7)

MAXIMUM DEFLECTION v¿ ⫽ 0 gives x1 ⫽

3EIdB Px2 5PL2 PLx ⫺ ⫺ ⫺ EI␯¿ ⫽ 3 6 54 L a EI␯ ⫽

(6)

P (⫺ 61L2 + 162Lx ⫺ 81x2) 486EI

3dB ␯¿ ⫽ ⫺ L

3EIdB 5PL2 ⫺ 54 L

L … x … Lb 3

Also, ␯¿ ⫽

Px2 PLx ⫽ ⫺ + C1 3 6

(5)

;

a Px PL ⫺ 3 3

L … x … Lb 3

;

FROM B TO C EI␯– ⫽ M ⫽

a

L 3 At x ⫽ , (nB)Left ⫽ (nB)Right (Eqs. 1 and 5) 3

B.C.

␯⫽ ⫺

PL3 + 3EIdB 54

3EIdBx Px3 5PL2x PLx2 ⫺ ⫺ ⫺ 6 18 54 L ⫺

FROM A TO B

‹ C2 ⫽ ⫺

2 n(L) ⫽ 0

B.C.

L … x … Lb 3

3EIdB x Px3 5PL2x PLx2 ⫺ ⫺ ⫺ + C2 6 18 54 L a

L … x … Lb 3

L (9 ⫺ 225) ⫽ 0.5031L 9

Substitute x1 in Eq. (6) and simplify: (4)

␯max ⫽ ⫺

4015PL3 6561EI

dmax ⫽ ⫺␯max ⫽

PL3 4015PL3 ⫽ 0.01363 6561EI EI

;

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Deflections of Beams

Problem 9.7-6 A simple beam ABC has moment of inertia 1.5I from A to B and I from

P

B to C (see figure). A concentrated load P acts at point B. Obtain the equations of the deflection curves for both parts of the beam. From the equations, determine the angles of rotation uA and uC at the supports and the deflection dB at point B.

Solution 9.7-6

1.5I

EI␯ ⫽

DEFLECTION CURVE

B.C.

L — 3

2L — 3

PLx2 Px3 ⫺ + C2x + C4 6 18

B.C.

B.C.

2Px 3I b␯– ⫽ M ⫽ 2 3

a0 … x …

Px PL ⫺ EI␯– ⫽ M ⫽ 3 3

4Px2 + C1 18

EI␯¿ ⫽

Px2 PLx ⫺ + C2 3 2

(8)

C4 ⫽ ⫺

PL3 ⫺ C2L 9

L a … x … Lb 3

4 Continuity of deflections at point B

(1)

From Eqs. (6), (8), and (7):

(2)

4P L 3 PL L 2 L P L 3 a b + C1 a b ⫽ a b ⫺ a b 54 3 3 6 3 18 3 L + C2 a b + C4 3

a0 … x … a

L b 3

L … x … Lb 3

10PL3 + C2L + 3C4 243

(3)

C1L ⫽

(4)

SOLVE EQS. (5), (8), (9), AND (10) C1 ⫽ ⫺

(n⬘B)Left ⫽ (n⬘B)Right

38PL2 729

C2 ⫽ ⫺

175PL2 1458

(10)

C3 ⫽ 0

13PL3 1458

From Eqs. (3) and (4):

C4 ⫽

PL L P L 2 4P L 2 a b + C1 ⫽ a b ⫺ a b + C2 18 3 3 3 6 3

SLOPES OF THE BEAM (FROM EQS. 3 AND 4)

11PL2 C2 ⫽ C1 ⫺ 162

(5)

INTEGRATE EQS. (3) AND (4) 4Px3 + C1x + C3 54

(9)

(nB)Left ⫽ (nB)Right

L b 3

1 Continuity of slopes at point B

EI␯ ⫽

(7)

3 Deflection at support C

INTEGRATE EACH EQUATION EI␯¿ ⫽

L … x … Lb 3

C3 ⫽ 0

From Eq. (6):

v(L) ⫽ 0 From Eq. (7): BENDING-MOMENT EQUATIONS

a

2 Deflection at support A

n (0) ⫽ 0

B.C.

C

B

Simple beam (nonprismatic)

Use the bending-moment equation (Eq. 9-12a).

Ea

I

A

a0 … x …

L b 3

(6)

␯¿ ⫽ ⫺

2P L (19L2 ⫺ 81x2) a 0 … x … b 729EI 3

(11)

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SECTION 9.7

␯¿ ⫽ ⫺

DEFLECTIONS OF THE BEAM

P (175L2 ⫺ 486Lx + 243x2) 1458EI a

Substitute C1, C2, C3, and C4 into Eqs. (6) and (7):

L … x … Lb 3

(12)

ANGLE OF ROTATION uA (FROM EQ. 11) uA ⫽ ⫺␯¿(0) ⫽

38PL2 729EI

(positive counterclockwise)

a0 … x …

2Px (19L2 ⫺ 27x2) 729EI

␯⫽ ⫺

P (⫺ 13L3 + 175L2x ⫺ 243Lx2 + 81x3) 1458EI a

;

L b 3

␯⫽ ⫺

;

(positive clockwise)

ANGLE OF ROTATION uC (FROM EQ. 12) 34PL2 uC ⫽ ␯¿(L) ⫽ 729EI

763

Nonprismatic Beams

DEFLECTION AT POINT B a x ⫽ L 32PL3 dB ⫽⫺␯a b ⫽ 3 2187EI

;

L … x … Lb 3

;

L b 3 ;

(positive downward)

Problem 9.7-7 The tapered cantilever beam AB shown in the figure has thin-walled, hollow circular cross sections of constant thickness t. The diameters at the ends A and B are dA and dB ⫽ 2dA, respectively. Thus, the diameter d and moment of inertia I at distance x from the free end are, respectively, dA d⫽ (L + x) L I⫽

P B

A dA

t dB = 2dA d

x

ptd3A IA ptd3 ⫽ (L + x)3 ⫽ 3 (L + x)3 3 8 8L L

L

in which IA is the moment of inertia at end A of the beam. Determine the equation of the deflection curve and the deflection dA at the free end of the beam due to the load P.

Solution 9.7-7 M ⫽ ⫺Px

Tapered cantilever beam EI␯– ⫽ ⫺ Px

I⫽

IA 3

L

(L + x)3

␯¿ ⫽

Px PL3 x ␯– ⫽ ⫺ ⫽ ⫺ c d EI EIA (L + x)3

(1)

From Appendix C: ␯¿ ⫽ B.C.

xdx L (L + x)

3

⫽⫺

3

PL L + 2x c d + C1 EIA 2(L + x)2 1 ␯¿ (L) ⫽ 0

‹ C1 ⫽ ⫺

or ␯¿ ⫽

INTEGRATE EQ. (1) L + 2x

PL3 L + 2x 3PL2 c d ⫺ 2 EIA 2(L + x) 8EIA PL3 L PL3 x 3PL2 c d + c d ⫺ 2 2 EIA 2(L + x) EIA (L + x) 8EIA

INTEGRATE EQ. (2) 2

2(L + x)

From Appendix C: dx L (L + x)

2

2

3PL 8EIA

xdx L (L + x)2

⫽ ⫺ ⫽

1 L + x

L + ln (L + x) L + x

(2)

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⫺

Substitute C2 into Eq. (3).

2

3PL x + C2 8EIA

3x 1 L + x L PL3 ⫺ + + lna bd c EIA 2(L + x) 8L 8 2L

␯⫽

L 3x PL3 c + ln (L + x) ⫺ d + C2 EIA 2(L + x) 8L

B.C.

DEFLECTION OF THE BEAM

PL3 L 1 PL3 L a b a⫺ b+ c + ln (L + x)d EIA 2 L + x EIA L + x

␯⫽

⫽

Page 764

(3)

PL3 1 ‹ C2 ⫽ c ⫺ ln (2L) d EIA 8

2 ␯ (L) ⫽ 0

;

DEFLECTION dA AT END A OF THE BEAM dA ⫽ ⫺␯(0) ⫽ ⫽ 0.06815 Note: ln

PL3 (8 ln 2 ⫺ 5) 8EIA

PL3 (positive downward) EIA

;

1 ⫽ ⫺ln 2 2

Problem 9.7-8 The tapered cantilever beam AB shown in the figure has a solid circular cross section. The diameters at the ends A and B are dA and dB ⫽ 2dA, respectively. Thus, the diameter d and moment of inertia I at distance x from the free end are, respectively,

P dB = 2dA

dA

dA d⫽ (L + x) L pdA4 (L 4

4

I⫽

pd ⫽ 64 64L

+ x)4 ⫽

B

A

x

d L

IA L4

(L + x)4

in which IA is the moment of inertia at end A of the beam. Determine the equation of the deflection curve and the deflection dA at the free end of the beam due to the load P.

Solution 9.7-8 M ⫽ ⫺Px

Tapered cantilever beam EI␯– ⫽ ⫺Px

I⫽

IA

(L + x)4

B.C.

4

L

4

␯– ⫽ ⫺

Px PL x ⫽ ⫺ c d EI EIA (L + x)4

(1)

␯¿ ⫽

‹ C1 ⫽ ⫺

PL2 12EIA

PL2 PL4 L⫹ 3x c d ⫺ EIA 6(L ⫹x)3 12EIA

or

INTEGRATE EQ. (1) xdx

L + 3x

˚L (L + x) ⫽ ⫺ 6(L + x)

From A ppendix C: ␯¿ ⫽

1 n⬘(L) ⫽ 0

PL4 L⫹3x c d ⫹ C1 EIA 6(L⫹ x)3

4

3

␯¿ ⫽

L PL4 x PL4 c d ⫹ c d 3 EIA 6(L ⫹x) EIA 2(L ⫹x)3

⫺

PL2 12EIA

(2)

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SECTION 9.7

INTEGRATE EQ. (2) dx

From Appendix C:

L(L + x)

3

xdx L(L + x)3 ␯⫽

⫽ ⫺ ⫽

B.C.

1

‹ C2 ⫽

PL3 7 a b EIA 24

2(L + x) 2

⫺(L + 2x)

DEFLECTION OF THE BEAM

2(L + x)2

Substitute C2 into Eq. (3)

PL4 L 1 1 2 PL4 1 L + 2x a b a⫺ b a b + a b c⫺ d EIA 6 2 L+x EIA 2 2(L + x)2 ⫺

2 n(L) ⫽ 0

765

Nonprismatic Beams

PL2 x + C2 12EIA

4L (2L + 3x) 2x PL3 c7 ⫺ ⫺ d 24EIA L (L + x)2

␯⫽

;

DEFLECTION dA AT END A OF THE BEAM

L(L + 2x) PL3 L2 x ⫽ c⫺ ⫺ ⫺ d + C2 2 2 EIA 12L 12(L + x) 4(L + x)

(3)

dA ⫽ ⫺␯(0) ⫽

PL3 24EIA

(positive downward)

;

Problem 9.7-9 A tapered cantilever beam AB supports a concentrated load P at the free end (see figure). The cross sections of the beam are rectangular with constant width b, depth dA at support A, and depth dB ⫽ 3dA/2 at the support. Thus, the depth d and moment of inertia I at distance x from the free end are, respectively, dA d⫽ (2L + x) 2L I⫽

P B

A

3dA dB = — 2

dA x

IA bd A3 bd 3 (2L + x)3 ⫽ 3 (2L + x) 3 ⫽ 3 12 96L 8L

d b

L

in which IA is the moment of inertia at end A of the beam. Determine the equation of the deflection curve and the deflection dA at the free end of the beam due to the load P.

Solution 9.7-9 M ⫽ ⫺P x ␯– ⫽ ⫺

Tapered cantilever beam EI␯– ⫽ ⫺P x 3

I⫽

IA 3

8L

(2L + x) 3

x

Px 8PL ⫽⫺ c d EI EIA (2L + x)3

(1)

1 n⬘(L) ⫽ 0

␯¿ ⫽

‹ C1 ⫽ ⫺

16PL2 9EIA

16PL2 8PL3 L⫹ x c d ⫺ EIA (2L ⫹x)2 9EIA

or

INTEGRATE EQ. (1) From Appendix C:

B.C.

xdx L(2L + x)3

8PL3 L⫹x d ⫹ C1 ␯¿ ⫽ EI c 3 A (2L⫹x)

⫽ ⫺

2L + 2x 2(2L + x)2

L 8PL3 x 8PL3 d ⫹ c d ␯¿ ⫽ EI c 2 EIA (2L ⫹x)2 A (2L ⫹x) ⫺

16PL2 9EIA

(2)

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Deflections of Beams

DEFLECTION OF THE BEAM

INTEGRATE EQ. (2) dx

From Appendix C:

L (2L + x)

2

xdx

⫽

L(2L + x)

2

3

⫽ ⫺

Substitute C2 into Eq. (3).

1 2L + x

2L + ln (2L + x) 2L + x

3

8PL L 8PL 2L ␯⫽ a⫺ b + c EIA 2L + x EIA 2L + x

⫽ B.C.

+ lna

2L + x bd 3L

dA ⫽ ⫺␯(0) ⫽

16x PL3 8L c + 8 ln(2L + x) ⫺ d + C2 EIA 2L + x 9L

(3)

8PL3 1 ‹ C2 ⫽ ⫺ c + ln (3L)d EIA 9

2 n(L) ⫽ 0

2x 1 L 8PL3 ⫺ ⫺ c EIA 2L + x 9L 9 ;

DEFLECTION dA AT END A OF THE BEAM

16PL2 x + C2 9EIA

+ ln (2L + x) d ⫺

␯⫽

⫽ 0.1326 NOTE: ln

8PL2 3 7 clna b ⫺ d EIA 2 18

PL3 EIA

(positive downward)

3 2 ⫽ ⫺ln 3 2

Problem 9.7-10 A tapered cantilever beam AB supports a concentrated load P at the free end (see figure). The cross sections of the beam are rectangular tubes with constant width b and outer tube depth dA at A, and outer tube depth dB ⫽ 3dA/2 at support B. The tube thickness is constant, t ⫽ dA/20. IA is the moment of inertia of the outer tube at end A of the beam. If the moment of inertia of the tube is approximated as Ia(x) as defined, find the equation of the deflection curve and the deflection dA at the free end of the beam due to the load P.

(

10x 3 Ia(x) = IA 4 + 27 L

Solution 9.7-10 EI␯ – ⫽ M ⫽ ⫺Px ␯– ⫽

⫺Px ⫽ EIa (x)

⫺ Px EIA a

From Appendix C:

⫺P ␯¿ ⫽ ≥ EIA

␯¿ ⫽

3 10x + b 4 27L

3

x L (a + bx)3

⫽

dx ⫽ ⫺

3 10 ⫹2 x 4 27L 2a

⫺P EIA

2 10 2 3 10 b a ⫹ xb 27L 4 27L

x a

3 10x 3 + b 4 27L

a + 2bx 2b 2 (a + bx)2

¥ ⫹C1

PL3 19683 81L 80x c ⫹ d ⫹C1 EIA 50 (81L⫹40x)2 (81L ⫹40x)2

3

(

IA =

bd 3 12

d b

P dA

dB = 3dA/2 x

BENDING-MOMENT EQUATIONS

;

L

t

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SECTION 9.7

From Appendix C:

1 L (a + bx)2 x L (a + bx)

2

dx ⫽ dx ⫽

Nonprismatic Beams

⫺1 b(a + bx) 1 b

2

a

a + ln(a + bx)b a + bx

␯⫽

PL 19683 ⫺81L 80 81L c ⫹ a + ln (81L + 40x)bd + C1x + C2 EIA 50 40(81L + 40x) 40 2 81L + 40x

␯⫽

19683PL3 81L + 162ln (81L + 40x) L + 80ln (81L + 40x) x a b + C1x + C2 2000EIA 81L + 40 x

B.C.

␯¿(L) ⫽ 0

C1 ⫽

⫺ 3168963 PL2 732050 EIA

B.C.

␯(L) ⫽ 0

C2 ⫽

⫺ 19683PL3 (3361+ 29282 ln (121L)) 29282000EIA

3

␯(x) ⫽

19683PL3 40x 81L 81 6440x 3361 a + 2 ln a + b⫺ ⫺ b 2000EIA 81L + 40x 121 121L 14641L 14641

dA ⫽ ⫺␯(0) ⫽

PL3 19683PL3 11 a⫺2820 + 14641 lna bb ⫽ 0.317 7320500EIA 9 EIA

;

;

P

**Problem 9.7-11 Repeat Problem 9.7-10 but now use the tapered propped cantilever tube AB, with guided support at B, shown in the figure which supports a concentrated load P at the guided end. Find the equation of the deflection curve and the deflection dB at the guided end of the beam due to the load P.

dA

dB = 3dA/2

x

Solution 9.7-11 BENDING-MOMENT EQUATIONS EI␯– ⫽ M ⫽ Px ␯– ⫽

Px ⫽ EIa(x)

Px EIA a

From Appendix C:

␯¿ ⫽

P ≥ EIA

⫽

3

3 10x + b 4 27L x

L (a + bx)3

P EIA

dx ⫽ ⫺

3 10 ⫹2 x 4 27L 10 2 3 10 2 2a xb b a ⫹ 27L 4 27L

¥ ⫹ C1

x a

3 10x 3 + b 4 27L a + 2bx

2b 2(a + bx)2

L

767

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CHAPTER 9

␯¿ ⫽ ⫺

Page 768

Deflections of Beams

80x PL3 19683 81L ⫹ c d ⫹C1 EIA 50 (81L⫹40x)2 (81L⫹40x)2

From Appendix C:

1 2

dx ⫽

2

dx ⫽

L (a + bx) x L (a + bx)

⫺1 b(a + bx) 1 b

2

a

a + ln(a + bx)b a + bx

␯⫽ ⫺

PL 19683 ⫺ 81L 80 81L c ⫹ a + ln (81L + 40x)bd + C1x + C2 EIA 50 40 (81L + 40x) 40 2 81L + 40x

␯⫽ ⫺

19683PL3 81L + 162 ln(81L + 40x)L + 80 ln(81L + 40x)x a b + C1x + C2 2000EIA 81L + 40 x

3

B.C.

␯¿(L) ⫽ 0

C1 ⫽

3168963 PL2 732050 EIA

B.C.

␯(L) ⫽ 0

C2 ⫽

19683PL3 (1+ 2 ln(81L)) 2000EIA

␯(x) ⫽ ⫺

19683PL3 81L 40x 6440x a ⫺1b + 2 ln a1 + b⫺ 2000EIA 81L + 40x 81L 14641L

;

PL3 19683PL3 11 a⫺2820 + 14641 lna bb ⫽ 0.317 7320500EIA 9 EIA

;

dB ⫽ ⫺ ␯(L) ⫽

q

Problem 9.7-12 A simple beam ACB is constructed with square cross sections and a double taper (see figure). The depth of the beam at the supports is dA and at the midpoint is dC ⫽ 2dA. Each half of the beam has length L. Thus, the depth d and moment of inertia I at distance x from the left-hand end are, respectively, dA d ⫽ (L + x) L I⫽

d A4 IA d4 ⫽ (L + x)4 ⫽ 4 (L + x)4 12 12 L4 L

in which IA is the moment of inertia at end A of the beam. (These equations are valid for x between 0 and L, that is, for the left-hand half of the beam.) (a) Obtain equations for the slope and deflection of the left-hand half of the beam due to the uniform load. (b) From those equations obtain formulas for the angle of rotation uA at support A and the deflection dC at the midpoint.

C A

B

d d

x L

L

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SECTION 9.7 Nonprismatic Beams

769

Solution 9.7-12 Simple beam with a double taper L ⫽ length of one-half of the beam I⫽

IA

(L + x) 4 L4

ANGLE OF ROTATION AT SUPPORT A uA ⫽ ⫺␯¿(0) ⫽

(0 … x … L)

(x is measured from the left-hand support A)

qL3 16EIA

qx 2 qx 2 ⫽ qL x ⫺ Bending moment: M ⫽ RAx ⫺ 2 2

From Appendix C:

From Eq. (9-12a):

␯ ⫽⫺

␯– ⫽

qL5x

⫺

EIA(L + x)4

qx 2 2

x 2dx L(L + x)

3

⫽

L(3L + 4 x)

B.C.

+ ln (L + x)

(0 … x … L)

(0 … x … L) (1)

2EIA(L + x)4

2(L + x)2

qL3 8L2(3L + 4x) cx ⫺ 16EIA 2(L + x)2

⫺8L ln (L + x)d + C2

qL4x 2

;

INTEGRATE EQ. (3)

Reactions: RA ⫽ RB ⫽ qL

EI␯– ⫽ M ⫽ qL x ⫺

(positive clockwise)

2 n(0) ⫽ 0

‹ C2 ⫽ ⫺

(4)

qL4 3 a + ln Lb 2EIA 2

INTEGRATE EQ. (1) xdx

From Appendix C:

L (L + x)

4

2

x dx L (L + x)

4

␯¿ ⫽

⫽⫺ ⫽⫺

DEFLECTION OF THE BEAM

L + 3x 3

6(L + x) 2

L + 3L x + 3x

2

3(L + x)3

␯ ⫽⫺

qL5 L⫹3x c⫺ d EIA 6(L⫹x)3 ⫺

qL4 L2 + 3L x + 3x 2 c⫺ d + C1 2EIA 3(L + x)3

B.C.

qL x

2EIA(L + x) 3

+ C1

1 (symmetry) n⬘(L) ⫽ 0

(0 … x … L)

dC ⫽ ⫺␯(L) ⫽ (2)

3

2EIA(L⫹x)

⫽⫺

⫺

qL3 16EIA

qL3 8L x 2 c1 ⫺ d 16EIA (L⫹x)3

qL4 qL4 (3 ⫺ 4 ln 2) ⫽ 0.02843 8EIA EIA (positive downward)

qL3 ‹ C1 ⫽ ⫺ 16EIA

Substitute C1 into Eq. (2). qL4x 2

(0 … x … L)

;

DEFLECTION AT THE MIDPOINT C OF THE BEAM

SLOPE OF THE BEAM

␯¿ ⫽

qL4 (9L2 + 14L x + x 2) x x c ⫺ln a1 + bd 2EIA L 8L(L + x)2 (0 … x … L)

4 2

⫽

Substitute C2 into Eq. (4) and simplify. (The algebra is lengthy.)

(3)

;

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Deflections of Beams

Strain Energy The beams described in the problems for Section 9.8 have constant flexural rigidity EI.

A

Problem 9.8-1 A uniformly loaded simple beam AB (see figure) of span

length L and rectangular cross section (b ⫽ width, h ⫽ height) has a maximum bending stress smax due to the uniform load. Determine the strain energy U stored in the beam.

Solution 9.8-1

b L

Simple beam with a uniform load

Given: L, b, h, smax Bending moment: M ⫽

Find: U (strain energy) qx qLx ⫺ 2 2

Solve for q: q ⫽

2

M 2dx Strain energy (Eq. 9-80a): U ⫽ L0 2EI

U⫽

q 2L5 ⫽ 240EI

smax ⫽

L2h

(1)

16Is2max L 15h2E

Substitute I ⫽

M max c M maxh Maximum stress: smax ⫽ ⫽ I 2I qL2 8

16Ismax

Substitute q into Eq. (1):

L

M max ⫽

bh 3 : 12

U⫽

P

M⫽

Px 2

a0 … x …

Strain energy (Eq. 9-80a):

L0

B

L — 2

L — 2

Simple beam with a concentrated load

(a) BENDING MOMENT

L/2

;

A

(a) Evaluate the strain energy of the beam from the bending moment in the beam. (b) Evaluate the strain energy of the beam from the equation of the deflection curve. (c) From the strain energy, determine the deflection d under the load P.

U⫽2

2 4bhLs max 45E

qL2h 16I

Problem 9.8-2 A simple beam AB of length L supports a concentrated load P at the midpoint (see figure).

Solution 9.8-2

h

B

M 2dx P 2L3 ⫽ 2 EI 96EI

L b 2

(b) DEFLECTION CURVE From Table G-2, Case 4: ␯ ⫽⫺

;

Px (3L2 ⫺ 4x 2) 48EI

d␯ P ⫽ ⫺ (L2 ⫺4x 2) dx 16EI

a0 … x … d 2␯ dx

2

⫽

Px 2EI

L b 2

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SECTION 9.8

Strain energy (Eq. 9-80b): L/2

U⫽2

L0

EI d 2␯ 2 a b dx ⫽ EI 2 dx 2 L0

P 2L3 ⫽ 96EI

Strain Energy

771

(c) DEFLECTION d UNDER THE LOAD P L/2

a

From Eq. (9-82a):

Px 2 b dx 2EI

d⫽

PL3 2U ⫽ P 48EI

;

;

y

Problem 9.8-3 A propped cantilever beam AB of length L, and with guided support at A, supports a uniform load of intensity q (see figure).

q

(a) Evaluate the strain energy of the beam from the bending moment in the beam. (b) Evaluate the strain energy of the beam from the equation of the deflection curve.

x A

B L

Solution 9.8-3 q d ␯⫽ ⫺ (8L3 ⫺12Lx 2 + 4x 3) dx 24EI

(a) BENDING-MOMENT EQUATIONS Measure x from end B

d2

q ( ⫺2Lx + x 2) 2EI

qx 2 M ⫽ qLx ⫺ 2

dx

Strain Energy (Eq. 9-80a):

Strain energy (Eq. 9-80b):

M2 dx L0 2EI qx 2 2 q 2L5 1 aqLx ⫺ b dx ⫽ 2 15EI L0 2EI

(b) DEFLECTION CURVE

2 EI d 2 a 2 ␯b dx L0 2 dx L 2 q EI U⫽ c⫺ (⫺2Lx + x 2 )d dx 2EI L0 2

U⫽

L

⫽

␯⫽ ⫺

L

L

U⫽

2

;

U⫽

q 2L5 15EI

;

Measure x from end B ␯⫽ ⫺

qx (8L3 ⫺ 4Lx 2 + x 3) 24EI

Problem 9.8-4 A simple beam AB of length L is subjected to loads that produce a symmetric deflection curve with maximum deflection d at the midpoint of the span (see figure). How much strain energy U is stored in the beam if the deflection curve is (a) a parabola, and (b) a half wave of a sine curve?

d

A

L — 2

B

L — 2

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Deflections of Beams

Solution 9.8-4

Simple beam (symmetric deflection curve) d ⫽ maximum deflection at midpoint

L, EI, d

GIVEN:

Page 772

Determine the strain energy U. Assume the deflection n is positive downward.

(b) DEFLECTION CURVE IS A SINE CURVE ␯ ⫽ d sin

px L

d 2␯

p 2d

dx (a) DEFLECTION CURVE IS A PARABOLA 4dx

␯⫽ d ␯ 2

dx 2

2

L

d␯ 4d ⫽ 2 (L ⫺2x) dx L

(L ⫺x)

⫽⫺

8d

2

⫽⫺

L

sin

px L

Strain energy (Eq. 9-80b): L

U⫽ ⫽

L2

2

d␯ pd px ⫽ cos dx L L

L

EI d 2␯ 2 EI p 2d 2 px a 2 b dx ⫽ a⫺ 2 b sin2 dx 2 2 L dx L L0 L0 p4EId2 4L3

;

Strain energy (Eq. 9-80b): L

U⫽ ⫽

L

EI d 2␯ 2 EI 8d 2 a⫺ 2 b dx a 2 b dx ⫽ 2 L0 L L0 2 dx 32EId2 L3

;

Problem 9.8-5 A beam ABC with simple supports at A and B and an overhang

P

BC supports a concentrated load P at the free end C (see figure).

B

A

(a) Determine the strain energy U stored in the beam due to the load P. (b) From the strain energy, find the deflection dC under the load P. (c) Calculate the numerical values of U and dC if the length L is 8 ft, the overhang length a is 3 ft, the beam is a W 10 ⫻ 12 steel wide-flange section, and the load P produces a maximum stress of 12,000 psi in the beam. (Use E ⫽ 29 ⫻ 106 psi.)

Solution 9.8-5

a

L

Simple beam with an overhang

(a) STRAIN ENERGY (use Eq. 9-80a)

FROM B TO C: M ⫽ ⫺Px a

UBC ⫽

1 P 2a 3 (⫺Px)2 dx ⫽ 6EI L0 2EI

TOTAL STRAIN ENERGY: U ⫽ UAB + UBC ⫽

P 2a 2 (L + a) 6EI

(b) DEFLECTION dC UNDER THE LOAD P FROM A TO B: M ⫽ ⫺

Pax L

L

UAB ⫽

M 2dx Pax 2 P 2a 2L 1 ⫽ a⫺ b dx ⫽ L 6EI L 2EI L0 2EI

From Eq. (9-82a): dC ⫽

2U Pa 2 ⫽ (L + a) P 3EI

C

;

;

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SECTION 9.8

(c) CALCULATE U AND dc Data: L ⫽ 8 ft ⫽ 96 in. W 10 ⫻ 12

U⫽

a ⫽ 3 ft ⫽ 36 in.

smax ⫽ 12,000 psi I ⫽ 53.8 in.4

c⫽

s2max I(L + a) P 2a 2(L + a) ⫽ 6EI 6c 2E

⫽ 241 in.-lb

E ⫽ 29 ⫻ 106 psi

dc ⫽

9.87 d ⫽ ⫽ 4.935 in. 2 2

773

Strain Energy

;

Pa 2(L + a) smaxa(L + a) ⫽ 3EI 3cE

⫽ 0.133 in.

;

Express load P in terms of maximum stress: smax ⫽

M maxc smax I Mc Pac ⫽ ⫽ ! ! ! ! ‹ P⫽ I I I ac

y

Problem 9.8-6 A simple beam ACB supporting a uniform load q over the first half of the beam and a couple of moment M0 at end B is shown in the figure. Determine the strain energy U stored in the beam due to the load q and the couple M0 acting simultaneously.

q

M0

A

B L — 2

L — 2

Solution 9.8-6 FROM A TO MID-SPAN

STRAIN ENERGY (EQ. 9-80A):

Bending-Moment Equations

U2 ⫽

M⫽ a

L

M0 qx 2 3qL + bx ⫺ 8 L 2

2 L 3qL M0 M 1 dx ⫽ ca + bx 8 L LL 2EI LL 2EI

⫺

STRAIN ENERGY (EQ. 9-80A): U1 ⫽

L 2

L0

U2 ⫽

2

M dx 2 EI

L 2

2 2

3qL M 0 qx 1 ca ⫹ bx ⫺ d dx ⫽ 2EI 8 L 2 L0 U1 ⫽

L a3L4q 2 + 30qL2M 0 + 80M 02b 3840EI

FROM MID-SPAN TO B Bending-Moment Equations M⫽ a

3qL M 0 qL L + b x ⫺ ax ⫺ b 8 L 2 4

2

2

qL L 2 a x⫺ b d dx 2 4

L a L4q 2 + 32qL2M 0 + 448M 20 b 3072EI

STRAIN ENERGY OF THE ENTIRE BEAM U ⫽ U1 + U2 ⫽

L a 17L4q 2 15360EI + 280qL2M 0 + 2560M 20 b

;

x

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Deflections of Beams

Problem 9.8-7 The frame shown in the figure consists of a beam ACB

L

supported by a struct CD. The beam has length 2L and is continuous through joint C. A concentrated load P acts at the free end B. Determine the vertical deflection dB at point B due to the load P. Note: Let EI denote the flexural rigidity of the beam, and let EA denote the axial rigidity of the strut. Disregard axial and shearing effects in the beam, and disregard any bending effects in the strut.

L B

A

C P

L

D

Solution 9.8-7

Frame with beam and strut L CD ⫽ length of strut

BEAM ACB

⫽ 12L F ⫽ axial force in strut ⫽ 2 12P

For part AC of the beam: M ⫽ ⫺Px

USTRUT ⫽

F 2L CD 2EA

USTRUT ⫽

(212P)2(12L) 412P 2L ⫽ 2EA EA

FRAME

U ⫽ UBEAM + USTRUT ⫽

L

1 M 2dx P 2L3 ⫽ (⫺P x) 2dx ⫽ 2EI L0 6EI L 2EI P 2L3 For part CB of the beam: UCB ⫽ UAC ⫽ 6EI UAC ⫽

Entire beam: UBEAM ⫽ UAC SRUCT CD

P 2L3 + UCB ⫽ 3EI

(Eq. 2-37a)

DEFLECTION dB AT POINT B From Eq. (9-82a): dB ⫽

2U 2PL3 812PL ⫽ + P 3EI EA

;

P 2L3 4 12P 2L + 3EI EA

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SECTION 9.9

775

Castigliano’s Theorem

Castigliano’s Theorem M0

The beams described in the problems for Section 9.9 have constant flexural rigidity EI.

A

B

Problem 9.9-1 A simple beam AB of length L is loaded at the left-hand end by a couple of moment M0 (see figure). Determine the angle of rotation uA at support A. (Obtain the solution by determining the strain energy of the beam and then using Castigliano’s theorem.)

Solution 9.9-1

L

Simple beam with couple M0 STRAIN ENERGY U⫽

M 20 L M 20 L x 2 M 2dx ⫽ a 1 ⫺ b dx ⫽ 2EI L0 L 6EI L 2EI

CASTIGLIANO’S THEOREM M0 RA ⫽ L

uA ⫽

(downward)

M ⫽ M 0 ⫺ RAx ⫽ M 0 ⫺ ⫽ M 0 a1 ⫺

M 0x L

M 0L dU ⫽ dM 0 3EI

;

(clockwise)

(This result agrees with Case 7, Table G-2)

x b L

Problem 9.9-2 The simple beam shown in the figure supports a concentrated

P

load P acting at distance a from the left-hand support and distance b from the right-hand support. Determine the deflection dD at point D where the load is applied. (Obtain the solution by determining the strain energy of the beam and then using Castigliano’s theorem.)

A

a

b L

Solution 9.9-2

Simple beam with load P M AD ⫽ RAx ⫽

Pbx L

M DB ⫽ RBx ⫽

Pax L

STRAIN ENERGY RA ⫽

Pb L

RB ⫽

Pa L

a

UAD ⫽

B

D

U⫽

M 2dx L 2EI

1 Pbx 2 P 2a 3b 2 a b dx ⫽ 2EI L0 L 6EIL2

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CHAPTER 9

Page 776

Deflections of Beams

b

UDB ⫽

1 Pax 2 P 2a 2b 3 a b dx ⫽ 2EI L0 L 6EIL2

U ⫽ UAD

P 2a 2b 2 + UDB ⫽ 6LEI

CASTIGLIANO’S THEOREM dD ⫽

Pa 2b 2 dU ⫽ dP 3LEI

;

(downward)

Problem 9.9-3 An overhanging beam ABC supports a concentrated load P at the end of the overhang (see figure). Span AB has length L and the overhang has length a. Determine the deflection dC at the end of the overhang. (Obtain the solution by determining the strain energy of the beam and then using Castigliano’s theorem.)

Solution 9.9-3

Pa L

A

L

U⫽

C

a

M AB ⫽ ⫺RAx ⫽ ⫺

L 1 Pax 2 P 2a 2L a⫺ b dx ⫽ 2EI L0 L 6EI

UCB ⫽

a 1 P 2a 3 (⫺ Px)2dx ⫽ 2EI L0 6EI

U ⫽ UAB + UCB ⫽ Pax L

M 2dx L 2EI

UAB ⫽

(downward)

M CB ⫽ ⫺Px

B

Overhanging beam STRAIN ENERGY

RA ⫽

P

P 2a 2 (L + a) 6EI

CASTIGLIANO’S THEOREM dC ⫽

dU Pa 2 ⫽ (L + a) dP 3EI

Problem 9.9-4 The cantilever beam shown in the figure supports a triangularly

(downward)

;

q0

distributed load of maximum intensity q0. Determine the deflection dB at the free end B. (Obtain the solution by determining the strain energy of the beam and then using Castigliano’s theorem.) B A L

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SECTION 9.9

Solution 9.9-4

Castigliano’s Theorem

777

Cantilever beam with triangular load CASTIGLIANO’S THEOREM dB ⫽

q0L4 0U PL3 ⫽ + 0P 3EI 30EI

(downward)

(This result agrees with Cases 1 and 8 of Table G-1.) SET P ⫽ 0: dB ⫽

q0L4 30EI

;

P ⫽ fictitious load corresponding to deflection dB M ⫽ ⫺Px ⫺

q0x 3 6L

STRAIN ENERGY L q0x 3 2 1 M 2dx a⫺Px ⫺ ⫽ b dx 2EI L0 6L L 2EI Pq0L4 q 20L5 P 2L3 ⫽ + + 6EI 30EI 42EI

U⫽

q

Problem 9.9-5 A simple beam ACB supports a uniform load of intensity q on the left-hand half of the span (see figure). Determine the angle of rotation uB at support B. (Obtain the solution by using the modified form of Castigliano’s theorem.)

C

A

B

L — 2

Solution 9.9-5

L — 2

Simple beam with partial uniform load BENDING MOMENT AND PARTIAL DERIVATIVE FOR AC

SEGMENT

M AC ⫽ RAx ⫺

qx 2 3qL qx 2 M0 ⫽ a + bx ⫺ 2 8 L 2 a0 … x …

M0 ⫽ fictitious load corresponding to angle of rotation uB RA ⫽

3qL M0 + 8 L

RB ⫽

qL M0 ⫺ 8 L

L b 2

0M AC x ⫽ 0M 0 L BENDING MOMENT AND PARTIAL DERIVATIVE FOR SEGMENT CB M CB ⫽ RBx + M 0 ⫽ a

qL M0 ⫺ b x + M0 8 L a0 … x …

L b 2

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Deflections of Beams

SET FICTITIOUS LOAD M0 EQUAL TO ZERO

0M CB x ⫽ ⫺ + 1 0M 0 L

uB ⫽

MODIFIED CASTIGLIANO’S THEOREM (EQ. 9-88) uB ⫽ ⫽

a

M 0M ba bdx EI 0M L 0 1 EI L0

L/2

1 + EI L0

1 EI L0

L/2

a

qx 2 3qLx x ⫺ b a b dx 8 2 L

L/2 qLx x 1 a b a1 ⫺ b dx EI L0 8 L 3 3 qL qL ⫽ + 128EI 96EI

+

ca

3qL qx 2 x M0 + bx⫺ d c d dx 8 L 2 L

L/2

qL M0 x ca ⫺ b x + M 0 d c1 ⫺ d dx 8 L L

⫽

7qL3 384EI

;

(counterclockwise)

(This result agrees with Case 2, Table G-2.)

P1

Problem 9.9-6 A cantilever beam ACB supports two concentrated loads P1 and P2, as shown in the figure. Determine the deflections dC and dB at points C and B, respectively. (Obtain the solution by using the modified form of Castigliano’s theorem.)

A

P2

C

L — 2

B L — 2

Cantilever beam with loads P1 and P2

Solution 9.9-6

MODIFIED CASTIGLIANO’S THEOREM FOR DEFLECTION dC dC ⫽

1 EI L0

L/2

(M CB)a

0M CB b dx 0P1

L

+ BENDING MOMENT AND PARTIAL DERIVATIVES CB

0M AC 1 (M AC)a b dx EI LL/2 0P1

FOR SEGMENT

M CB ⫽ ⫺P2x 0M CB ⫽0 0P1

L

⫽0 +

L a0 … x … b 2 0M CB ⫽ ⫺x 0P2

⫽

1 L L c ⫺P1 a x ⫺ b ⫺ P2x d a ⫺ x b dx EI LL/2 2 2

L3 (2P1 + 5P2) 48EI

;

BENDING MOMENT AND PARTIAL DERIVATIVES FOR AC

MODIFIED CASTIGLIANO’S THEOREM FOR DEFLECTION dB

L M AC ⫽ ⫺P1 ax ⫺ b ⫺ P2x 2

dB ⫽

SEGMENT

0M AC L ⫽ ⫺x 0P1 2

L a … x … Lb 2

0M AC ⫽ ⫺x 0P2

1 EI L0

L/2

(M CB)a L

+

0M CB b dx 0P2

0M AC 1 (M )a b dx EI LL/2 AC 0P2

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SECTION 9.9

⫽

1 EI L0

L/2

(⫺P2x) (⫺x)dx L

L 1 c ⫺P1 ax ⫺ b ⫺ P2x d(⫺x)dx + EI LL/2 2

Castigliano’s Theorem

⫽

P2L3 L3 + (5P1 + 14P2) 24EI 48EI

⫽

L3 (5P1 + 16P2) 48EI

779

;

(These results can be verified with the aid of Cases 4 and 5, Table G-1.)

q

Problem 9.9-7 The cantilever beam ACB shown in the figure is subjected to a uniform load of intensity q acting between points A and C. Determine the angle of rotation uA at the free end A. (Obtain the solution by using the modified form of Castigliano’s theorem.)

C

A L — 2

Solution 9.9-7

B

L — 2

Cantilever beam with partial uniform load MODIFIED CASTIGLIANO’S THEOREM (EQ. 9-88) uA ⫽ ⫽

M0 ⫽ fictitious load corresponding to the angle of rotation uA BENDING MOMENT AND PARTIAL DERIVATIVE FOR AC

SEGMENT

M AC ⫽ ⫺M 0 ⫺

qx 2 2

a0 … x …

L b 2

M 0M a ba b dx L EI 0M 0 1 EI L0 +

L/2

a⫺M 0 ⫺

L qL L 1 c ⫺ M0 ⫺ a x ⫺ b d ( ⫺1)dx EI LL/2 2 4

SET FICTITIOUS LOAD M0 EQUAL TO ZERO uA ⫽

1 EI L0

L/2

L qx 2 qL 1 L a b a x ⫺ b dx dx + 2 EI LL/2 2 4

0M AC ⫽ ⫺1 0M 0

⫽

qL3 qL3 + 48EI 8EI

BENDING MOMENT AND PARTIAL DERIVATIVE FOR SEGMENT CB

⫽

7qL3 48EI

M CB ⫽ ⫺M 0 ⫺ 0M CB ⫽ ⫺1 0M 0

qL L ax ⫺ b 2 4

a

L … x … Lb 2

qx 2 b( ⫺1)dx 2

(counterclockwise)

;

(This result can be verified with the aid of Case 3, Table G-1.)

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Deflections of Beams

Problem 9.9-8 The frame ABC supports a concentrated load P at point C (see figure). Members AB and BC have lengths h and b, respectively. Determine the vertical deflection dC and angle of rotation uC at end C of the frame. (Obtain the solution by using the modified form of Castigliano’s theorem.)

b B

C P

h

A

Solution 9.9-8

Frame with concentrated load MODIFIED CASTIGLIANO’S THEOREM FOR DEFLECTION dC dC ⫽

a

0M M ba bdx EI 0P L h

⫽

b

1 1 (Pb + M 0)(b) dx + (Px + M 0)(x) dx EI L0 EI L0

Set M0 ⫽ 0: h

dC ⫽ P ⫽ concentrated load acting at point C (corresponding to the deflection dC) M0 ⫽ fictitious moment corresponding to the angle of rotation uC BENDING MOMENT AND PARTIAL DERIVATIVES FOR MEMBER AB M AB ⫽ Pb + M 0 (0 … x … h) 0M AB ⫽b 0P

0M AB ⫽1 M0

BENDING MOMENT AND PARTIAL DERIVATIVES FOR BC

MEMBER

⫽

0M BC ⫽1 0M 0

Pb 2 (3h + b) 3EI

;

(downward)

MODIFIED CASTIGLIANO’S THEOREM FOR ANGLE OF uC

ROTATION

uC ⫽

0M M ba b dx L EI 0M 0 a

h

⫽

b

1 1 (Pb + M 0)(1)dx + (Px + M 0)(1) dx EI L0 EI L0

Set M0 ⫽ 0: h

uC ⫽

M BC ⫽ Px + M 0 (0 … x … b) 0M BC ⫽x 0P

b

1 1 Pb 2dx + Px 2dx EI L0 EI L0

⫽

b

1 1 Pb dx + Pxdx EI L0 EI L0 Pb (2h + b) 2EI

(clockwise)

;

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SECTION 9.9

Problem 9.9-9 A simple beam ABCDE supports a uniform load of

q

intensity q (see figure). The moment of inertia in the central part of the beam (BCD) is twice the moment of inertia in the end parts (AB and DE). Find the deflection dC at the midpoint C of the beam. (Obtain the solution by using the modified form of Castigliano’s theorem.)

B

A

C

I

D

L — 4

E I

2I

L — 4

Solution 9.9-9

781

Castigliano’s Theorem

L — 4

L — 4

Nonprismatic beam MODIFIED CASTIGLIANO’S THEOREM (EQ. 9-88) Integrate from A to C and multiply by 2. dC ⫽ 2

a

M AC 0M AC ba b dx 0P L EI

⫽ 2a

1 b EI L0

+ 2a P ⫽ fictitiuous load corresponding to the deflection ␦C at the midpoint qL P + RA ⫽ 2 2

dC ⫽

qLx qx 2 Px ⫺ + 2 2 2

0M AC x ⫽ 0P 2

a0 … x …

L b 2

a0 … x …

L b 2

a

qLx qx 2 Px x ⫺ + b a b dx 2 2 2 2

L/2 qLx qx 2 Px x 1 b a ⫺ + b a b dx 2EI LL/4 2 2 2 2

SET FICTITIOUS LOAD P EQUAL TO ZERO

BENDING MOMENT AND PARTIAL DERIVATIVE FOR THE LEFT-HAND HALF OF THE BEAM (A TO C) M AC ⫽

L/4

2 EI L0 +

⫽ dC ⫽

L/4

a

qLx qx 2 x ⫺ b a bdx 2 2 2

L/2 qLx qx 2 1 x a ⫺ b a bdx EI LL/4 2 2 2

67qL4 13qL4 + 6,144EI 12,288EI 31qL4 4096EI

Problem 9.9-10 An overhanging beam ABC is subjected to a couple MA MA at the free end (see figure). The lengths of the overhang and the main span are a and L, respectively. Determine the angle of rotation uA and deflection dA at end A. (Obtain the solution by using the modified form of Castigliano’s theorem.)

;

(downward)

A

B

a

C

L

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CHAPTER 9

Solution 9.9-10

Page 782

Deflections of Beams

Overhanging beam ABC Set P ⫽ 0: uA ⫽ ⫽

MA ⫽ couple acting at the free end A (corresponding to the angle of rotation uA) P ⫽ fictitious load corresponding to the deflection dA BENDING MOMENT AND PARTIAL DERIVATIVES AB

FOR SEGMENT

M AB ⫽ ⫺M A ⫺ Px (0 … x … a) 0M AB ⫽ ⫺1 0M A

MA (L + 3a) (counterclockwise) 3EI

;

MODIFIED CASTIGLIANO’S THEOREM FOR DEFLECTION dA dA ⫽

a

M 0M ba bdx 0P L EI a

⫽

0M AB ⫽ ⫺x 0P

BENDING MOMENT AND PARTIAL DERIVATIVES BC MA Pa + (downward) Reaction at support C: RC ⫽ L L M Ax Pax ⫺ (0 … x … L) M BC ⫽ ⫺RC x ⫽ ⫺ L L

a L M Ax 1 1 x M Adx + a b a b dx EI L0 EI L0 L L

1 (⫺M A ⫺ Px)(⫺x)dx EI L0 L

+

M Ax Pax ax 1 a⫺ ⫺ b a⫺ bdx EI L0 L L L

Set P ⫽ 0:

FOR SEGMENT

0M BC x ⫽ ⫺ 0M A L

0M BC ax ⫽ ⫺ 0P L

dA ⫽ ⫽

a L M Ax ax 1 1 M Axdx + a b a b dx EI L0 EI L0 L L

M Aa (2L + 3a) (downward) 6EI

;

MODIFIED CASTIGLIANO’S THEOREM FOR ANGLE OF uA

ROTATION

a

M 0M bdx ba EI 0M L A a 1 (⫺ M A ⫺ Px)(⫺1) dx ⫽ EI L0

uA ⫽

+

L M Ax Pax x 1 a⫺ ⫺ b a ⫺ bdx EI L0 L L L

Problem 9.9-11 An overhanging beam ABC rests on a simple support at A and a spring support at B (see figure). A concentrated load P acts at the end of the overhang. Span AB has length L, the overhang has length a, and the spring has stiffness k. Determine the downward displacement dC of the end of the overhang. (Obtain the solution by using the modified form of Castigliano’s theorem.)

P B

A

C

k

L

a

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SECTION 9.9

Solution 9.9-11

783

Castigliano’s Theorem

Beam with spring support TOTAL STRAIN ENERGY U U ⫽ UB + US ⫽

P 2(L + a)2 M 2dx + 2kL2 L 2EI

APPLY CASTIGLIANO’S THEOREM (EQ. 9-87) dC ⫽ RA ⫽ RB ⫽

Pa L

(downward) ⫽

P (L + a) (upward) L

BENDING MOMENT AND PARTIAL DERIVATIVE FOR SEGMENT AB M AB ⫽ ⫺RAx ⫽ ⫺

Pax L

dM AB ax ⫽ ⫺ dP L

dC ⫽

ax 1 Pax a⫺ b a⫺ bdx EI L0 L L +

dM BC ⫽ ⫺x (0 … x … a) dP

P 2(L + a)2 R2B ⫽ US ⫽ 2k 2kL2

P(L + a)2 M dM ba b dx + dP kL2 L EI a

L

⫽

BENDING MOMENT AND PARTIAL DERIVATIVE FOR SEGMENT BC

STRAIN ENERGY OF THE SPRING (EQ. 2-38a)

P(L + a)2 d M 2dx + dP L 2EI kL2

DIFFERENTIATE UNDER THE INTEGRAL SIGN (MODIFIED CASTIGLIANO’S THEOREM)

(0 … x … L)

M BC ⫽ ⫺Px

dU d M 2dx d P 2(L + a)2 d ⫽ + c dP dP L 2EI dP 2kL2

⫽ dC ⫽

a P(L + a)2 1 (⫺Px)(⫺x) dx + EI L0 kL2

P(L + a)2 Pa 3 Pa 2L + + 3EI 3EI kL2 Pa 2(L + a) P(L + a)2 + 3EI kL2

;

STRAIN ENERGY OF THE BEAM (EQ. 9-80a) UB ⫽

M 2dx L 2EI

q

Problem 9.9-12 A symmetric beam ABCD with overhangs at both ends supports a uniform load of intensity q (see figure). Determine the deflection dD at the end of the overhang. (Obtain the solution by using the modified form of Castigliano’s theorem.)

A B L — 4

D

C

L

L — 4

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CHAPTER 9

Solution 9.9-12

Page 784

Deflections of Beams

Beam with overhangs SEGMENT CD M CD ⫽ ⫺

qx 2 ⫺ Px 2

a0 … x …

L b 4

0M CD ⫽ ⫺x 0P MODIFIED CASTIGLIANO’S THEOREM FOR DEFLECTION dD dD ⫽ q ⫽ intensity of uniform load P ⫽ fictitious load corresponding to the deflection dD

⫽

L ⫽ length of segments AB and CD 4 L ⫽ length of span BC RB ⫽

3qL P ⫺ 4 4

RC ⫽

M 0M ba bdx EI 0P L 1 EI L0 +

3qL 5P + 4 4

BENDING MOMENTS AND PARTIAL DERIVATIVES SEGMENT AB qx 2 0M AB L M AB ⫽ ⫺ ⫽ 0 a0 … x … b 2 0P 4 SEGMENT BC M BC ⫽ ⫺ cq ax +

a

3 qL q L 2 P ⫺ bx ⫽ ⫺ ax + b + a 2 4 4 4 (0 … x … L) 0M BC x ⫽ ⫺ 0P 4

a⫺

qx 2 b(0)dx 2

L q 3qL L 2 P 1 c ⫺ ax + b + a ⫺ bx d EI L0 2 4 4 4

1 x * c⫺ ddx + 4 EI L0 SET P ⫽ 0: dD ⫽

L/4

a⫺

qx 2 ⫺ Pxb ( ⫺x)dx 2

L q 3qL L 2 x 1 c ⫺ ax + b + x d c ⫺ d dx EI L0 2 4 4 4

+ L 1 L b d c ax + b d + RBx 4 2 4

L/4

⫽ ⫺

1 EI L0

L/4

a⫺

qx 2 b( ⫺x)dx 2

qL4 37qL4 5qL4 + ⫽ ⫺ 768EI 2048EI 6144EI

(Minus means the deflection is opposite in direction to the fictitious load P.) ‹ dD ⫽

37qL4 6144EI

(upward)

;

Deflections Produced by Impact The beams described in the problems for Section 9.10 have constant flexural rigidity EI. Disregard the weights of the beams themselves, and consider only the effects of the given loads.

W

h

A

Problem 9.10-1 A heavy object of weight W is dropped onto the midpoint of a simple beam AB from a height h (see figure). Obtain a formula for the maximum bending stress smax due to the falling weight in terms of h, sst, and dst, where sst is the maximum bending stressand dst is the deflection at the midpoint when the weight W acts on the beam as a statically applied load. Plot a graph of the ratio smax/sst (that is, the ratio of the dynamic stress to the static stress) versus the ratio h/dst. (Let h/dst vary from 0 to 10.)

B L — 2

L — 2

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SECTION 9.10

Solution 9.10-1

785

Weight W dropping onto a simple beam

MAXIMUM DEFLECTION (EQ. 9-94) dmax ⫽ dst ⫹ (d2st ⫹ 2hdst)1/2 MAXIMUM BENDING STRESS For a linearly elastic beam, the bending stress s is proportional to the deflection d ‹

Deflections Produced by Impact

dmax smax 2h 1/2 ⫽ ⫽ 1 + a1 + b sst dst dst

2h 1/2 smax ⫽ sst c1 + a1 + b d dst

;

NOTE: dst ⫽

h dst

smax sst

0 2.5 5.0 7.5 10.0

2.00 3.45 4.33 5.00 5.58

WL3 for a simple beam with a load at the 48EI

midpoint.

GRAPH OF RATIO smax/sst

W

Problem 9.10-2 An object of weight W is dropped onto the midpoint of a simple beam AB from a height h (see figure). The beam has a rectangular cross section of area A. Assuming that h is very large compared to the deflection of the beam when the weight W is applied statically, obtain a formula for the maximum bending stress smax in the beam due to the falling weight.

h

A

B L — 2

L — 2

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Deflections of Beams

Solution 9.10-2

Weight W dropping onto a simple beam

Height h is very large.

sst ⫽

M WL ⫽ S 4S

dst ⫽

WL3 48EI

MAXIMUM DEFLECTION (EQ. 9-95) dmax ⫽ 12hdst MAXIMUM BENDING STRESS

I⫽

smax dmax 2h ⫽ ⫽ s st dst A dst

s2st 3WEI ⫽ 2 dst S L

bd 3 12

S⫽

bd 2 6

smax ⫽

I S

(1)

18WhE

(2)

2

⫽

3 3 ⫽ bd A

(3)

constructed of a W 8 ⫻ 21 wide-flange section (see figure). A weight W ⫽ 1500 lb falls through a height h ⫽ 0.25 in. onto the end of the beam. Calculate the maximum deflection dmax of the end of the beam and the maximum bending stress smax due to the falling weight. (Assume E ⫽ 30 ⫻ 106 psi.)

;

A AL

W = 1500 lb

Problem 9.10-3 A cantilever beam AB of length L ⫽ 6 ft is

W 8 ⫻ 21

h = 0.25 in.

A

B L = 6 ft

Cantilever beam

DATA: L ⫽ 6 ft ⫽ 72 in. h ⫽ 0.25 in. W 8 ⫻ 21

16S 2

Substitute (2) and (3) into (1):

2hsst2 smax ⫽ A dst

Solution 9.10-3

W 2L2

For a RECTANGULAR BEAM (with b, depth d):

For a linearly elastic beam, the bending stress s is proportional to the deflection d. ‹

s2st ⫽

Equation (9-94):

W ⫽ 1500 lb

dmax ⫽ dst + (d2st + 2hdst)1/2 ⫽ 0.302 in.

E ⫽ 30 ⫻ 106 psi I ⫽ 75.3 in.4

S ⫽ 18.2 in.3

;

MAXIMUM BENDING STRESS

MAXIMUM DEFLECTION (EQ. 9-94)

Consider a cantilever beam with load P at the free end:

Equation (9-94) may be used for any linearly elastic structure by substituting dst ⫽ W/k, where k is the stiffness of the particular structure being considered. For instance: Simple beam with load at midpoint:

smax ⫽

k⫽

Ratio:

M max PL ⫽ S S

Cantilever beam with load at the free end: k ⫽ For the cantilever beam in this problem: dst ⫽

‹ smax ⫽

(1500 lb)(72 in.)3 WL3 ⫽ 3EI 3(30 * 106 psi)(75.3 in.4)

⫽ 0.08261 in.

3 EI L3

PL3 3EI

smax 3 EI ⫽ dmax SL2

48EI L3

dmax ⫽

3EI SL2

dmax ⫽ 21,700 psi

;

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SECTION 9.10

W

Problem 9.10-4 A weight W ⫽ 20 kN falls through a

height h ⫽ 1.0 mm onto the midpoint of a simple beam of A length L ⫽ 3 m (see figure). The beam is made of wood with square cross section (dimension d on each side) and E ⫽ 12 GPa. If the allowable bending stress in the wood is sallow ⫽ 10 MPa, what is the minimum required dimension d?

Solution 9.10-4

787

Deflections Produced by Impact

h d

B d L — 2

L — 2

Simple beam with falling weight W

DATA: W ⫽ 20 kN E ⫽ 12 GPa

h ⫽ 1.0 mm

SUBSTITUTE (2) AND (3) INTO EQ. (1)

L ⫽ 3.0 m

2smaxd 3 8hEd 4 1/2 b ⫽ 1 + a1 + 3WL WL3

sallow ⫽ 10 MPa

CROSS SECTION OF BEAM (SQUARE) SUBSTITUTE NUMERICAL VALUES:

d ⫽ dimension of each side d4 I⫽ 12

2(10 MPa)d 3 8(1.0 mm)(12 GPa)d 4 1/2 d ⫽ 1 + c1 + 3(20 kN)(3.0 m) (20 kN)(3.0 m)3

d3 S⫽ 6

1000 3 1600 4 1/2 (d ⫽ meters) d ⫺ 1 ⫽ c1 + d d 9 9

MAXIMUM DEFLECTION (EQ. 9-94) dmax ⫽ dst ⫹ (d 2st ⫹ 2hdst)1/2

SQUARE BOTH SIDES, REARRANGE, AND SIMPLIFY a

MAXIMUM BENDING STRESS For a linearly elastic beam, the bending stress s is proportional to the deflection d. ‹

smax dmax 2h ⫽ ⫽ 1 + a1 + b s st dst dst

2500d 3 ⫺ 36d ⫺ 45 ⫽ 0 (d ⫽ meters)

1/2

(1)

WL 6 M 3WL s st ⫽ ⫽a b a 3b ⫽ S 4 d 2d 3 WL3 WL3 12 WL3 ⫽ a 4b ⫽ 48 EI 48 E d 4 Ed 4

SOLVE NUMERICALLY d ⫽ 0.2804 m ⫽ 280.4 mm

STATIC TERMS sst AND dst

dst ⫽

1600 2000 1000 2 3 b d ⫺ d⫺ ⫽0 9 9 9

For minimum value, round upward. (2)

‹ d ⫽ 281 mm

;

(3)

W = 4000 lb

Problem 9.10-5 A weight W ⫽ 4000 lb falls through a

height h ⫽ 0.5 in. onto the midpoint of a simple beam of length L ⫽ 10 ft (see figure). Assuming that the allowable bending stress in the beam is sallow ⫽ 18,000 psi and E ⫽ 30 ⫻ 106 psi, select the lightest wide-flange beam listed in Table E-1a in Appendix E that will be satisfactory.

h = 0.5 in. A

B

L — = 5 ft 2

L — = 5 ft 2

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CHAPTER 9

Solution 9.10-5

Page 788

Deflections of Beams

Simple beam of wide-flange shape

DATA: W ⫽ 4000 lb

REQUIRED SECTION MODULUS

h ⫽ 0.5 in.

L ⫽ 10 ft ⫽ 120 in.

S⫽

sallow ⫽ 18,000 psi

E ⫽ 30 ⫻ 106 psi

SUBSTITUTE NUMERICAL VALUES

MAXIMUM DEFLECTION (EQ. 9-94)

S⫽ a

dmax ⫽ dst + (dst2 + 2hdst)1/2 dmax 2h ⫽ 1 + a1 + b dst dst

or

1/2

dmax smax 2h 1/2 ⫽ ⫽ 1 + a1 + b sst dst dst

1. Select a trial beam from Table E-1a. 2. Substitute I into Eq. (4) and calculate required S. 3. Compare with actual S for the beam. 4. Continue until the lightest beam is found.

(1)

STATIC TERMS sst AND dst M WL ⫽ S 4S

dst ⫽

(4)

PROCEDURE

For a linearly elastic beam, the bending stress s is proportional to the deflection d.

s st ⫽

5I 1/2 20 3 in. b c1 + a1 + b d 3 24

(S ⫽ in.3; I ⫽ in.4)

MAXIMUM BENDING STRESS

‹

96hEI 1/2 WL c1 + a1 + b d 4sallow WL3

WL3 48EI

smax 4s allowS 4S ⫽ s allow a b ⫽ sst WL WL

(2)

2h 48EI 96hEI ⫽ 2h a b⫽ dst WL3 WL3

(3)

Trial beam

I

Actual S

Required S

W 8 ⫻ 35 W 10 ⫻ 45 W 10 ⫻ 60 W 12 ⫻ 50 W 14 ⫻ 53 W 16 ⫻ 31

127 248 341 394 541 375

31.2 49.1 66.7 64.7 77.8 47.2

41.6 (NG) 55.0 (NG) 63.3 (OK) 67.4 (NG) 77.8 (OK) 66.0 (NG)

Lightest beam is W 14 ⫻ 53

;

SUBSTITUTE (2) AND (3) INTO EQ. (1): 4s allowS 96hEI 1/2 ⫽ 1 + a1 + b WL WL3

Problem 9.10-6 An overhanging beam ABC of rectangular cross section has the dimensions shown in the figure. A weight W ⫽ 750 N drops onto end C of the beam. If the allowable normal stress in bending is 45 MPa, what is the maximum height h from which the weight may be dropped? (Assume E ⫽ 12 GPa.)

40 mm A

h C

B

1.2 m

W

2.4 m

40 mm 500 mm

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SECTION 9.10 Deflections Produced by Impact

789

Solution 9.10-6 Overhanging beam DATA: W ⫽ 750 N

LAB ⫽ 1.2 m.

E ⫽ 12 GPa

in which a ⫽ LBC and L ⫽ LAB:

LBC ⫽ 2.4 m

sallow ⫽ 45 MPa

dst ⫽

bd 3 1 I⫽ ⫽ (500 mm)(40 mm)3 12 12 ⫽ 2.6667 * 10

(3)

EQUATION (9-94):

⫽ 2.6667 * 106 mm4 ⫺6

W(L 2BC)(L AB + L BC) 3EI

dmax ⫽ dst ⫹ (d2st ⫹ 2hdst)1/2

4

m

or

2

1 bd ⫽ (500 mm)(40 mm)2 S⫽ 6 6

dmax 2h 1/2 ⫽ 1 + a1 + b dst dst

(4)

⫽ 133.33 * 103 mm3 MAXIMUM BENDING STRESS

⫽ 133.33 * 10⫺6 m3

For a linearly elastic beam, the bending stress s is proportional to the deflection d.

DEFLECTION dC AT THE END OF THE OVERHANG

‹

dmax smax 2h 1/2 ⫽ ⫽ 1 + a1 + b sst dst dst

s st ⫽

WL BC M ⫽ S S

(5) (6)

MAXIMUM HEIGHT h P ⫽ load at end C

Solve Eq. (5) for h:

L ⫽ length of span AB

smax 2h 1/2 ⫺ 1 ⫽ a1 + b sst dst

a ⫽ length of overhang BC

a

From the answer to Prob. 9.8-5 or Prob. 9.9-3: dC ⫽

Pa 2(L + a) 3EI

Stiffness of the beam: k ⫽

smax smax 2 2h b ⫺ 2a b + 1⫽1 + sst sst dst

h⫽ P 3EI ⫽ 2 dC a (L + a)

(1)

dst smax smax ba ⫺ 2b a 2 sst sst

Substitute dst from Eq. (3), sst from Eq. (6), and sallow for smax: W(L 2BC)(L AB + L BC) s allowS s allowS ba ⫺ 2b (8) a 6EI WL BC WL BC

MAXIMUM DEFLECTION (EQ. 9-94)

h⫽

Equation (9-94) may be used for any linearly elastic structure by substituting dst ⫽ W/k, where k is the stiffness of the particular structure being considered. For instance:

SUBSTITUTE NUMERICAL VALUES INTO EQ. (8):

Simple beam with load at midpoint: k ⫽

W(L 2BC)(L AB + L BC) ⫽ 0.08100 m 6 EI

48EI L3

Cantilever beam with load at free end: k ⫽

3EI L3

s allowS 10 ⫽ ⫽ 3.3333 WL BC 3

Etc.

For the overhanging beam in this problem (see Eq. 1): dst ⫽

Wa 2(L + a) W ⫽ k 3EI

(2)

(7)

h ⫽ (0.08100 m)a or h ⫽ 360 mm

10 10 ba ⫺ 2b ⫽ 0.36 m 3 3 ;

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CHAPTER 9

Page 790

Deflections of Beams

Problem 9.10-7 A heavy flywheel rotates at an angular speed v (radians per second) around an axle (see figure). The axle is rigidly attached to the end of a simply supported beam of flexural rigidity EI and length L (see figure). The flywheel has mass moment of inertia Im about its axis of rotation. If the flywheel suddenly freezes to the axle, what will be the reaction R at support A of the beam?

Solution 9.10-7

A

R

Im

L

Rotating flywheel

NOTE: We will disregard the mass of the beam and all energy losses due to the sudden stopping of the rotating flywheel. Assume that all of the kinetic energy of the flywheel is transformed into strain energy of the beam. KINETIC ENERGY OF ROTATING FLYWHEEL KE ⫽

v EI

1 Imv 2 2

STRAIN ENERGY OF BEAM

U⫽

CONSERVATION OF ENERGY 1 R2L3 Imv2 ⫽ 2 6 EI

KE ⫽ U R⫽

3 EI Imv2

; L3 NOTE: The moment of inertia Im has units of kg # m2 or N # m # s2

A

M 2dx L 2EI

M ⫽ Rx, where x is measured from support A. L

U⫽

1 R2L3 (Rx)2dx ⫽ 2 EI L0 6 EI

Temperature Effects The beams described in the problems for Section 9.11 have constant flexural rigidity EI. In every problem, the temperature varies linearly between the top and bottom of the beam.

Problem 9.11-1 A simple beam AB of length L and height h undergoes a temperature change such that the bottom of the beam is at temperature T2 and the top of the beam is at temperature T1 (see figure). Determine the equation of the deflection curve of the beam, the angle of rotation uA at the left-hand support, and the deflection dmax at the midpoint.

y A

h

T1

B x

T2 L

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SECTION 9.11

Solution 9.11-1

Simple beam with temperature differential

Eq. (9-147): ␯– ⫽

B.C.

d 2␯ dx

2

⫽

a(T2 ⫺ T1) h

v¿ ⫽ ⫺

a(T2 ⫺T1)(L⫺2x) 2h aL(T2 ⫺T1) 2h

dv a(T2⫺T1)x ⫽ ⫹ C1 dx h

uA ⫽ ⫺v¿(0) ⫽

L 1 (Symmetry) v¿ a b ⫽ 0 2

(positive uA is clockwise rotation)

v¿ ⫽

791

Temperature Effects

aL(T2 ⫺ T1) ‹ C1 ⫽ ⫺ 2h ␯⫽

a(T2 ⫺ T1)x aL(T2 ⫺ T1)x ⫺ + C2 2h 2h

B.C.

2 v(0) ⫽ 0

;

aL2(T2 ⫺ T1) L dmax ⫽ ⫺␯a b ⫽ 2 8h

;

(positive dmax is downward deflection)

2

 C2 ⫽ 0

a(T2 ⫺ T1)(x)(L ⫺ x) ; 2h (positive n is upward deflection) ␯⫽ ⫺

Problem 9.11-2 A cantilever beam AB of length L and height h (see figure) is subjected to a temperature change such that the temperature at the top is T1 and at the bottom is T2. Determine the equation of the deflection curve of the beam, the angle of rotation uB at end B, and the deflection dB at end B.

y T1

A

T2 L

Solution 9.11-2 Eq. (9-147): ␯– ⫽

Cantilever beam with temperature differential d 2␯ dx

2

⫽

a(T2 ⫺ T1) h

dv a(T2 ⫺T1) x ⫹ C1 v¿ ⫽ ⫽ dx h B.C.

1 n⬘(0) ⫽ 0

 C1 ⫽ 0

a(T2 ⫺T1) v¿ ⫽ x h ␯⫽

a(T2 ⫺ T1) x 2 a b + C2 h 2

B.C.

2 n(0) ⫽ 0

␯⫽

a(T2 ⫺ T1)x 2h

 C2 ⫽ 0 2

;

(positive n is upward deflection) uB ⫽ v¿(L) ⫽

aL(T2 ⫺T1) h

;

(positive uB is counterclockwise rotation) dB ⫽ ␯(L) ⫽

aL2(T2 ⫺ T1) 2h

;

(positive dB is upward deflection)

h

B x

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CHAPTER 9

Page 792

Deflections of Beams

Problem 9.11-3 An overhanging beam ABC of height h has a guided

y

support at A and a roller at B. The beam is heated to a temperature T1 on the top and T2 on the bottom (see figure). Determine the equation of the deflection curve of the beam, the angle of rotation uC at end C, and the deflection dC at end C.

A

h B

T1

T1

T2

T2

L

a

C x

Solution 9.11-3 ␯– ⫽

d2 dx

2

␯⫽

a(T 2 ⫺ T 1) h

dC ⫽ ␯(L + a) ⫽

v¿ ⫽

a(T2 ⫺T1) x ⫹C1 h

␯⫽

a (T 2 ⫺ T 1) x 2 + C 1 x + C2 2h

B.C.

n⬘(0) ⫽ 0

C1 ⫽ 0

B.C.

n(L) ⫽ 0

C2 ⫽ ⫺

␯(x) ⫽

⫽

a (T 2 ⫺ T 1) [(L + a)2 ⫺ L2] 2h

a (T 2 ⫺ T 1) (2 L a + a 2) 2h

; Upward

uC ⫽ v ¿(L⫹a) ⫽

a (T2 ⫺T1) (L⫹ a) h

a (T 2 ⫺ T 1) L2 2h

;

Counter Clockwise

a (T 2 ⫺ T 1) (x 2 ⫺ L2) 2h

Problem 9.11-4 A simple beam AB of length L and height h (see figure) is heated in such a manner that the temperature difference T2 ⫺ T1 between the bottom and top of the beam is proportional to the distance from support A; that is, assume the temperature difference varies linearly along the beam:

y A

T1 T2

T2 ⫺ T1 ⫽ T0x in which T0 is a constant having units of temperature (degrees) per unit distance.

L

(a) Determine the maximum deflection dmax of the beam. (b) Repeat for quadratic temperature variation along the beam, T2 ⫺ T1 ⫽ T0 x2.

Solution 9.11-4 (a) (T2 ⫺ T1) ⫽ T0 x

B.C.

n(0) ⫽ 0

C2 ⫽ 0

B.C.

n(L) ⫽ 0

C1 ⫽ ⫺

2

a T0 x d ␯– ⫽ 2 ␯ ⫽ h dx v¿ ⫽

aT0 x 2 ⫹C1 2h

␯(x) ⫽

a T0 (x 3 ⫺ L2 x) 6h

␯⫽

a T0 x 3 + C1 x + C2 6h

v¿(x) ⫽

aT0 2 L2 ax ⫺ b 2h 3

a T0 L2 6h

h B x

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SECTION 9.11

MAXIMUM DEFLECTION Set n⬘(x) ⫽ 0 and solve for x 0⫽

a T0 2 L2 ax ⫺ b 2h 3

dmax ⫽ ⫺␯a

⫽ ⫺ dmax ⫽

x⫽

L 13

L b 13

aT0 c a

a T0 L3

␯(x) ⫽

a T0 (x 4 ⫺ L3 x) 12 h

v¿(x) ⫽

aT0 3 L3 ax ⫺ b 3h 4

MAXIMUM DEFLECTION

L L 3 b ⫺ L2 d 13 13 6h

Downward

913 h

;

Set n⬘(x) ⫽ 0 and solve for x 0⫽

a T0 3 L3 ax ⫺ b 3h 4

dmax ⫽ ⫺␯a

(b) (T2 ⫺ T1) ⫽ T0 x2 ␯– ⫽

d2 dx

␯⫽ 2

aT0 x 2 h

⫽ ⫺

aT0 x 3 v¿ ⫽ ⫹C1 3h

dmax ⫽

␯⫽

a T0 x 4 + C 1 x + C2 12 h

B.C.

n(0) ⫽ 0

C2 ⫽ 0

B.C.

n(L) ⫽ 0

C1 ⫽ ⫺

793

Temperature Effects

x⫽

L 12

L b 12

a T0 c a

L 4 L b ⫺ L3 d 12 12 12h

aT0 L4 (2 12 ⫺ 1) 48h

; Downward

a T0 L3 12 h

Problem 9.11-5 Beam AB, with elastic support kR at A and pin support at B, of length L and height h (see figure) is heated in such a manner that the temperature difference T2 ⫺ T1 between the bottom and top of the beam is proportional to the distance from support A; that is, assume the temperature difference varies linearly along the beam: T2 ⫺ T1 ⫽ T0x in which T0 is a constant having units of temperature (degrees) per unit distance. Assume the spring at A is unaffected by the temperature change. (a) Determine the maximum deflection dmax of the beam. (b) Repeat for quadratic temperature variation along the beam, T2 ⫺ T1 ⫽ T0 x2. (c) What is dmax for (a) and (b) above if kR goes to infinity?

y A

T1 kR

T2 L

h B x

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CHAPTER 9 Deflections of Beams

Solution 9.11-5 (a) (T2 ⫺ T1) ⫽ T0 x ␯– ⫽

d

2

dx

2

␯⫽

(b) (T2 ⫺ T1) ⫽ T0 x2

aT0 x h

␯– ⫽

aT0 x 2 v¿ ⫽ ⫹ C1 2h ␯⫽

a T0 x 3 + C 1 x + C2 6h

B.C.

n⬘(0) ⫽ 0

C1 ⫽ 0

n(L) ⫽ 0

a T0 L3 C2 ⫽ ⫺ 6h

B.C.

d2 dx

␯⫽ 2

a T0 x 2 h

v¿ ⫽

aT0 x 3 + C1 3h

␯⫽

a T0 x 4 + C 1 x + C2 12 h

B.C.

n⬘(0) ⫽ 0

C1 ⫽ 0

B.C.

n(L) ⫽ 0

C2 ⫽ ⫺

␯(x) ⫽

a T0 (x 3 ⫺ L3) 6h

␯(x) ⫽

a T0 (x 4 ⫺ L4) 12 h

v¿(x) ⫽

a T0 x 2 2h

v¿(x) ⫽

a T0 x 3 3h

MAXIMUM DEFLECTION

MAXIMUM DEFLECTION

Set n⬘(x) ⫽ 0 and solve for x a T0 x 2 0⫽ 2h

Set n⬘(x) ⫽ 0 and solve for x

x⫽0

0⫽ 3

dmax ⫽ ⫺␯(0) ⫽

a T0 L 6h

Downward

a T0 L4 12 h

;

a T0 x 3 3h

x⫽0

dmax ⫽ ⫺␯(0) ⫽

a T0 L4 12 h

Downward

;

(c) Changing kR does not change dmax in both cases.

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10 Statically Indeterminate Beams

Differential Equations of the Deflection Curve y

The problems for Section 10.3 are to be solved by integrating the differential equations of the deflection curve. All beams have constant flexural rigidity EI. When drawing shear-force and bending-moment diagrams, be sure to label all critical ordinates, including maximum and minimum values.

M0

A

B

x

MA RA

Problem 10.3-1 A propped cantilever beam AB of length L is loaded by

RB

L

a counterclockwise moment M0 acting at support B (see figure). Beginning with the second-order differential equation of the deflection curve (the bending-moment equation), obtain the reactions, shear forces, bending moments, slopes, and deflections of the beam. Construct the shear-force and bending-moment diagrams, labeling all critical ordinates.

Solution 10.3-1 Propped cantilever beam M 0 ⫽ applied load

EI␯¿ ⫽

Select MA as the redundant reaction.

B.C.

REACTIONS (FROM EQUILIBRIUM) M0 MA RA ⫽ + L L

(1)

RB ⫽ ⫺ RA

(2)

BENDING MOMENT (FROM EQUILIBRIUM) M ⫽ RAx ⫺ M A ⫽

MA M0 x (x ⫺ L) + L L

DIFFERENTIAL EQUATIONS MA M0 x EI␯– ⫽ M ⫽ (x ⫺ L) + L L

M0 x 2 MA x 2 a ⫺ Lxb + + C1 L 2 2L

1 ␯¿(0) ⫽ 0

EI␯ ⫽

(4)

‹ C1 ⫽ 0

M0 x 3 MA x Lx a ⫺ b + + C2 L 6 2 6L 3

2

B.C.

2 ␯(0) ⫽ 0

‹ C2 ⫽ 0

B.C.

3 ␯(L) ⫽ 0

‹ MA ⫽

(5)

M0 2

(3) REACTIONS (SEE EQS. 1 AND 2) MA ⫽

M0 2

RA ⫽

3M 0 2L

RB ⫽ ⫺

3M 0 2L

;

795

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CHAPTER 10 Statically Indeterminate Beams

SHEAR FORCE (FROM EQUILIBRIUM) V ⫽ RA ⫽

3M 0 2L

SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS

;

BENDING MOMENT (FROM EQ. 3) M⫽

M0 (3x ⫺ L) 2L

;

SLOPE (FROM EQ. 4) ␯¿ ⫽ ⫺

M0 x (2L ⫺ 3x) 4LEI

;

DEFLECTION (FROM EQ. 5) ␯⫽ ⫺

M0 x 2 (L ⫺ x) 4LEI

;

Problem 10.3-2 A fixed-end beam AB of length L supports a

y

uniform load of intensity q (see figure). Beginning with the second-order differential equation of the deflection curve (the bending-moment equation), obtain the reactions, shear forces, bending moments, slopes, and deflections of the beam. Construct the shear-force and bending-moment diagrams, labeling all critical ordinates.

q

A

MA

x

B MB L

RA

RB

Solution 10.3-2 Fixed-end beam (uniform load) 1 ␯¿(0) ⫽ 0

Select MA as the redundant reaction.

B.C.

REACTIONS (FROM SYMMETRY AND EQUILIBRIUM)

EI␯ ⫽ ⫺

RA ⫽ RB ⫽

qL 2

MB ⫽ MA

BENDING MOMENT (FROM EQUILIBRIUM) M ⫽ RAx ⫺ M A ⫺

qx 2 q ⫽ ⫺ M A + (Lx ⫺ x 2) 2 2

(1)

EI␯– ⫽ M ⫽ ⫺ M A + EI␯¿ ⫽ ⫺ M Ax +

q (Lx ⫺ x 2) 2

q Lx 2 x3 a ⫺ b + C1 2 2 3

q Lx 3 M Ax 2 x4 + a ⫺ b + C2 2 2 6 12

B.C.

2 ␯(0) ⫽ 0

‹ C2 ⫽ 0

B.C.

3 ␯(L) ⫽ 0

‹ MA ⫽

qL2 12

qL 2

qL2 12

MA ⫽ MB ⫽

SHEAR FORCE (FROM EQUILIBRIUM) (2)

(3)

REACTIONS RA ⫽ RB ⫽

DIFFERENTIAL EQUATIONS

‹ C1 ⫽ 0

q V ⫽ RA ⫺ qx ⫽ (L ⫺ 2x) 2

;

;

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797

SECTION 10.3 Differential Equations of the Deflection Curve

SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS

BENDING MOMENT (FROM EQ. 1) M⫽ ⫺

q 2 (L ⫺ 6Lx + 6x 2) 12

;

SLOPE (FROM EQ. 2) ␯¿ ⫽ ⫺

qx (L2 ⫺ 3Lx + 2x 2) 12EI

;

DEFLECTION (FROM EQ. 3) ␯⫽ ⫺

qx 2 (L ⫺ x)2 24EI

;

qL4 L dmax ⫽ ⫺␯ a b ⫽ 2 384EI

Problem 10.3-3 A cantilever beam AB of length L has a fixed support at A and a roller support at B (see figure). The support at B is moved downward through a distance dB. Using the fourth-order differential equation of the deflection curve (the load equation), determine the reactions of the beam and the equation of the deflection curve. (Note: Express all results in terms of the imposed displacement dB .)

y x A

dB

MA

B

RA L

RB

Solution 10.3-3 Cantilever beam with imposed displacement dB REACTIONS (FROM EQUILIBRIUM) RA ⫽ RB

M A ⫽ RBL

(1)

(2)

EI␯–– ⫽ ⫺ q ⫽ 0

(3)

EI␯–¿ ⫽ V ⫽ C1

(4)

EI␯– ⫽ M ⫽ C1x + C2

(5)

EI␯¿ ⫽ C1x 2/2 + C2x + C3

(6)

EI␯ ⫽ C1x /6 + C2x /2 + C3x + C4

(7)

B.C.

1 ␯(0) ⫽ 0

B.C.

2 ␯¿(0) ⫽ 0

2

‹ C4 ⫽ 0 ‹ C3 ⫽ 0

3 ␯–(L) ⫽ 0

‹ C1L + C2 ⫽ 0

B.C.

4 ␯(L) ⫽ ⫺ dB

‹ C1L + 3C2 ⫽ ⫺ 6EIdB/L2

DIFFERENTIAL EQUATIONS

3

B.C.

(9)

SOLVE EQUATIONS (8) AND (9): C1 ⫽

3EIdB 3

L

C2 ⫽ ⫺

3EIdB L2

SHEAR FORCE (EQ. 4) V⫽

3EIdB 3

L

RA ⫽ V(0) ⫽

(8)

3EIdB L3

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CHAPTER 10 Statically Indeterminate Beams

REACTIONS (EQS. 1 AND 2) RA ⫽ RB ⫽

DEFLECTION (FROM EQ. 7):

3EIdB

M A ⫽ RBL ⫽

␯⫽ ⫺

L3 3EIdB

2L3

(3L ⫺ x)

;

SLOPE (FROM EQ. 6):

;

L2

dBx 2

␯¿ ⫽ ⫺

3dBx 2L3

(2L ⫺ x)

Problem 10.3-4 A cantilever beam of length L and loaded by uniform load

y

of intensity q has a fixed support at A and spring support at B with rotational stiffness kR. A rotation at B, uB , results in a reaction moment MB = kR * uB . Find rotation uB and displacement dB at end B. Use the second-order differential equation of the deflection curve to solve for displacements at end B.

q

A

L

kR x

B

MA RA MB

Solution 10.3-4 q = intensity of uniform load

EIv ⫽ RA

EQUILIBRIUM RA ⫽ qL

(1)

qL2 MA ⫽ ⫺ MB 2

(2)

M B ⫽ k RuB

(3)

qx 2 2

1 v¿(0) ⫽ 0

B.C.

2 v(0) ⫽ 0

B.C.

3 v¿(L) ⫽ uB

EIv– ⫽ M ⫽ RAx ⫺ M A ⫺

qx 3 x2 ⫺ M Ax ⫺ + C1 2 6

‹ C2 ⫽ 0

‹ dB ⫽

qL2 qL3 L2 ⫺ a ⫺ k RuB b L ⫺ 2 2 6

qL3 6(k R L ⫺ EI)

‹ EIdB ⫽ qL qx 2 2

‹ C1 ⫽ 0

Substitute RA and MA from Eqs. (1) and (2):

‹ uB ⫽

DIFFERENTIAL EQUATION

EIv¿ ⫽ RA

B.C.

‹ EIuB ⫽ qL

BENDING MOMENT M ⫽ RA x ⫺ M A ⫺

qx 4 x3 x2 ⫺ MA ⫺ + C1x + C2 6 2 24

;

qL2 qL4 L3 L2 ⫺ a ⫺ k RuB b ⫺ 6 2 2 24

k R qL5 1 1 a ⫺ qL4 + b EI 8 12 (k RL ⫺ EI )

;

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SECTION 10.3 Differential Equations of the Deflection Curve

Problem 10.3-5 A cantilever beam of length L and loaded by a triangularly

q0

y

distributed load of maximum intensity q0 at B. Use the fourth-order differential equation of the deflection curve to solve for reactions at A and B and also the equation of the deflection curve.

A

MA

799

L

x B

RA RB

Solution 10.3-5 Triangular load q ⫽ q0

x L

C2 ⫽ ⫺

DIFFERENTIAL EQUATION

SHEAR FORCE (EQ. 2)

x EI␯–– ⫽ ⫺ q ⫽ ⫺ q0 L

(1)

x2 + C1 2L

(2)

EI␯–¿ ⫽ ⫺ q0

EI␯– ⫽ M ⫽ ⫺ q0 EI␯¿ ⫽ ⫺ q0 EI␯ ⫽ ⫺q0

x + C1x + C2 6L

x4 x2 + C1 + C2 x + C3 24L 2

x3 x2 x5 + C1 + C2 + C3 x + C4 120L 6 2

B.C.

1 ␯–(L) ⫽ 0

B.C.

2 ␯¿(0) ⫽ 0

‹ C3 ⫽ 0

B.C.

3 ␯(0) ⫽ 0 4 ␯(L) ⫽ 0

9 q L 40 0

9 x2 + q L 2L 40 0

REACTIONS RA ⫽ V(0) ⫽

(4)

RB ⫽ ⫺ V(L) ⫽

(5)

;

11 q L 40 0

;

FROM EQUILIBRIUM MA ⫽

(6)

7 q L2 120 0

;

DEFLECTION CURVE (EQ. 5) x5 9 7 x3 x2 + q0 L ⫺ q0 L2 120L 40 6 20 2

or

1 (⫺2 q0 x 5 ⫹9q0 Lx 3 ⫺7q0 L2 x 2) 240LEI

;

EI␯ ⫽ ⫺q0 2

‹ C1

9 q L 40 0

(3)

‹ C4 ⫽ 0

Solve Eqs. (6) and (7): C1 ⫽

V ⫽ ⫺q0

3

L2 ‹ C1L + C2 ⫽ q0 6

B.C.

7 q L2 120 0

L L + C2 ⫽ q0 3 60

(7)

␯⫽

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CHAPTER 10 Statically Indeterminate Beams

Problem 10.3-6 A propped cantilever beam of length L is loaded by a parabolically distributed load with maximum intensity q0 at B.

x2 q0 — L2

( )

y

(a) Use the fourth-order differential equation of the deflection curve to solve for reactions at A and B and also the equation of the deflection curve. (b) Repeat (a) if the parabolic load is replaced by q0 sin (px/2L).

A

MA

q0

x

B

L

RA RB

SOLUTION 10.3-6 q ⫽ q0

(a) Parabolic load

REACTIONS

x2 L2

RA ⫽ V(0) ⫽

DIFFERENTIAL EQUATION EI␯–– ⫽ ⫺ q ⫽ ⫺ q0 EI␯–¿ ⫽ ⫺ q0

2

(1)

+ C1

(2)

3L2

EI␯ ⫽ ⫺q0

x

L

x4 12L2

5

MA ⫽

+ C1

60L

x6 2

360L

x + C2x + C3 2

+ C1

(4)

⫺ (5)

‹ C1L + C2 ⫽ q0

B.C.

2 ␯¿(0) ⫽ 0

‹ C3 ⫽ 0

B.C.

3 ␯(0) ⫽ 0

‹ C4 ⫽ 0

B.C.

4 ␯(L) ⫽ 0

‹ C1L + 3C2 ⫽ q0

Solve Eqs. (6) and (7): C2 ⫽ ⫺

SHEAR FORCE (EQ. 2) x

3L2

+

7 q L 60 0

EI␯ ⫽ ⫺q0

x3 x2 + C2 6 2

1 ␯–(L) ⫽ 0

V ⫽ ⫺ q0

;

;

DEFLECTION CURVE (EQ. 5)

B.C.

3

1 q L2 30 0

2

2

7 q L 60 0

13 q L 60 0

(3)

+ C1x + C2

+ C3x + C4

C1 ⫽

RB ⫽ ⫺ V(L) ⫽

;

FROM EQUILIBRIUM

x3

EI␯– ⫽ M ⫽ ⫺ q0 EI␯¿ ⫽ ⫺ q0

x2

7 q L 60 0

1 q L2 30 0

L2 12

(6)

␯⫽

x6 360L2

x2 1 q0 L2 30 2

(7)

or

q0 360L2EI

1⫺x 6 + 7L3x 3 ⫺ 6q0 L4x 22 (b) Loading q ⫽ q0 sin a

L2 60

7 x3 q0 L 60 6

+

;

px b 2L

DIFFERENTIAL EQUATION EI␯–– ⫽ ⫺ q ⫽ ⫺ q0 sin a EI␯–¿ ⫽ q0

px b 2L

(1)

2L px cos a b + C1 p 2L

(2)

EI␯– ⫽ M ⫽ q0 a

2L 2 px b sin a b + C1 x + C2 p 2L

(3)

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SECTION 10.3 Differential Equations of the Deflection Curve

EI␯¿ ⫽ ⫺ q0 a

REACTIONS

2L 3 px x2 b cos a b + C1 p 2L 2

+ C2x + C3

RA ⫽ V(0) ⫽ 0.31q0 L

2L 4 px x3 x2 EI␯ ⫽ ⫺q0 a b sin a b + C1 + C2 p 2L 6 2 + C3 x + C4 B.C.

⫽ a

(4)

⫽ a6

1 ␯–(L) ⫽ 0

B.C.

2 ␯¿(0) ⫽ 0

B.C.

3 ␯(0) ⫽ 0

B.C.

4 ␯(L) ⫽ 0

4L2 p

(6)

2

+ C1 (7)

Solve Eqs. (6) and (7): C1 ⫽ ⫺ 6 q0 L C2 ⫽ 2 q0 L2

b q0 L

;

From equilibrium

EI␯ ⫽ ⫺q0 a

2L 3 6 b L 2 p L 2L 4 6 b 2 p L

p4

p2 ⫺ 12p + 24

;

p4

DEFLECTION CURVE (EQ. 5)

‹ C4 ⫽ 0

+ q0 a

p2 ⫺ 4p + 8

M A ⫽ ⫺ C2 ⫽ ⫺ 2q0 L2

2L 3 ‹ C3 ⫽ q0 a b p

‹ C1L + 3C2 ⫽ ⫺q0 a

;

RB ⫽ ⫺ V(L) ⫽ 0.327 q0 L (5)

‹ C1L + C2 ⫽ ⫺ q0

2 p2 ⫺ 4p + 8 b q0 L ⫺6 p p4

p ⫺ 4p + 8 2

p4

p2 ⫺ 12p + 24 p4

␯⫽

2L 4 px b sin a b p 2L

x3 x2 + C2 + C3 x + C4, or 6 2

2L 4 px 1 c ⫺q0 a b sin a b p EI 2L ⫺6q0 L

p2 ⫺ 4p + 8 x 3 6 p4

⫹2q0 L2 # ⫹q0 a

p2 ⫺ 12p + 24 x 2 2 p4

2L 3 b xd p

;

SHEAR FORCE (EQ. 2) V ⫽ q0

2L p2 ⫺ 4p + 8 px cos a b ⫺ 6 q0 L p 2L p4

Problem 10.3-7 A fixed-end beam of length L is loaded by distributed

y

load in the form of a cosine curve with maximum intensity q0 at A. (a) Use the fourth-order differential equation of the deflection curve to solve for reactions at A and B and also the equation of the deflection curve. (b) Repeat (a) using the distributed load q0 sin(px/L).

A MA RA

px q0 cos — L

( )

q0

L — 2

L — 2

MB x

B

RB

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CHAPTER 10 Statically Indeterminate Beams

Solution 10.3-7 (a) Loading q ⫽ q0 cos a

REACTIONS

px b L

RA ⫽ V(0) ⫽

24 p4

DIFFERENTIAL EQUATION EI␯–– ⫽ ⫺ q ⫽ ⫺ q0 cos a EI␯–¿ ⫽ ⫺q0

px b L

(1)

L px sin a b + C1 p L

(2)

L 2 px EI␯– ⫽ M ⫽ q0 a b cos a b + C1 x + C2 p L

(3)

L 4 px x3 EI␯ ⫽ ⫺q0 a b cos a b + C1 p L 6

1. ␯¿(0) ⫽ 0

B.C.

2. ␯(0) ⫽ 0

B.C.

3. ␯ œ (L) ⫽ 0

B.C.

4. ␯(L) ⫽ 0

(5)

‹ C1 ‹ C1

p

p2

b q0 L2 ;

12

⫺

1 p2

b q0 L2 ;

24 L 4 px x3 EIv ⫽ ⫺ q0 a b cos a b + 4 q0 L p L 6 p

4

v⫽

12 p

q0 L2

4

1 p4EI

x2 L 4 + q0 a b , or p 2

c ⫺q0 L4 cos a

px b + 4q0 Lx 3 L

⫺ 6q0 L2x 2 + q0 L4 d

L3 L L2 + a⫺C1 b 6 2 2 L 4 b p

1

⫺

4

p4

⫺

L + C2 ⫽ 0 2

⫽ ⫺ 2q0 a

12

(counter-clockwise)

‹ C3 ⫽ 0 L ‹ C4 ⫽ q0 a b p

;

q0 L

DEFLECTION CURVE (EQ. 5)

2

B.C.

MA ⫽ a

(4)

x + C3x + C4 2

24 p4

From equilibrium

MB ⫽ a

x2 + C2 x + C3 2

+ C2

RB ⫽ ⫺V(L) ⫽ ⫺

(counter-clockwise)

L 3 px EI␯¿ ⫽ q0 a b sin a b p L + C1

;

q0 L

;

(b) Loading q ⫽ q0 sin px/L (6)

FROM SYMMETRY: RA ⫽ RB

MA ⫽ MB

DIFFERENTIAL EQUATIONS SOLVE EQS. (6): C1 ⫽

24 p4

C2 ⫽ ⫺

q0 L 12 4

p

q0 L2

SHEAR FORCE (EQ. 2) V⫽ ⫺

q0 L 24 px sin a b + 4 q0 L p L p

Elv–– ⫽ ⫺q ⫽ ⫺ q0 sin px/L EI␯¿– ⫽ V ⫽

q0 L px + C1 cos p L q0 L2

px sin + C1x + C2 L p2 q0 L3 px x2 EI␯¿ ⫽ ⫺ 3 cos + C1 C2x + C3 L 2 p

EI␯– ⫽ M ⫽

(1) (2) (3) (4)

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SECTION 10.3 Differential Equations of the Deflection Curve

EI␯ ⫽ ⫺

q0 L4

sin

p4

BENDING MOMENT (EQ. 3)

px x3 + C1 L 6

M⫽

2

+ C2 B.C.

x + C3x + C4 2

(5)

L 1 From symmetry, V a b ⫽ 0 2 L ‹ C3 ⫽ q0 a b p

B.C.

3 ␯¿(L) ⫽ 0

‹ C2 ⫽ ⫺ 2q0

L2 p

px ⫺ 2b L

2q0 L2 p3

2q0 L2

MB ⫽ MA ⫽

3

;

p3

DEFLECTION CURVE (FROM EQ. 5)

3

EI␯ ⫽ ⫺

‹ C4 ⫽ 0

q 0 L2 p4

sin

q0 L2x 2 q0 L3x px + ⫺ L p3 p3

or

SHEAR FORCE (EQ. 2) q0 L px V⫽ cos p L

a p sin

‹ C1 ⫽ 0

2 ␯¿(0) ⫽ 0

4 v(0) ⫽ 0

p

3

M A ⫽ ⫺M(0) ⫽

B.C.

B.C.

q0 L2

q0 L RA ⫽ V(0) ⫽ p

;

q0 L p

;

RB ⫽ RA ⫽

␯⫽ ⫺

q0 L2 p EI

Problem 10.3-8 A fixed-end beam of length L is loaded by a

y

distributed load in the form of a cosine curve with maximum intensity q0 at A.

q0

(a) Use the fourth-order differential equation of the deflection curve to solve for reactions at A and B and also the equation of the deflection curve. (b) Repeat (a) if the distributed load is now q0 (1 ⫺ x 2/L2) .

a L2 sin

4

px + px 2 ⫺ pLx b L

;

px q0 cos — 2L

( ) MB

A

L

MA

x

B RB

RA

Solution 10.3-8 (a) Loading q ⫽ q0 cos a

px b 2L

EI␯¿ ⫽ q0 a

+ C1

DIFFERENTIAL EQUATION EI␯–– ⫽ ⫺ q ⫽ ⫺ q0 cos a EI␯–¿ ⫽ ⫺ q0

px b 2L

2L px sin a b + C1 p 2L

EI␯– ⫽ M ⫽ q0 a

(1) (2)

2L 2 px b cos a b p 2L

+ C1x + C2

2L 3 px b sin a b p 2L

(3)

x2 + C2x + C3 2

EI␯ ⫽ ⫺q0 a + C1

(4)

2L 4 px b cos a b p 2L

x3 x2 + C2 + C3x + C4 6 2

(5)

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CHAPTER 10 Statically Indeterminate Beams

B.C. B.C. B.C.

1 ␯¿(0) ⫽ 0

v⫽

2L 4 ‹ C4 ⫽ q0 a b p

2 ␯(0) ⫽ 0

L2 2L 3 + C2L ⫽ ⫺ q0 a b p 2

(6)

4 ␯(L) ⫽ 0 (7)

Solve Eqs. (6) and (7): 48(4 ⫺ p) p4

C2 ⫽ ⫺

q0 L

16(6 ⫺ p)

q0 L2

p4

48(4 ⫺ p) 2L px V ⫽ ⫺ q0 sin a b + q0 L p 2L p4

48(4 ⫺ p) 4

p

RB ⫽ ⫺ V(L) ⫽ a

x2 L2

b

2L b + p

32(p ⫺ 3) 4

p

16(6 ⫺ p)

q0 L2

p4

q0 L2

;

b

(1)

b + C1

(2)

x4 x2 ⫺ b + C1x + C2 2 12L2

x2 + C2 x + C3 2

x2 + C3x + C4 2

1 ␯¿(0) ⫽ 0

‹ C3 ⫽ 0

B.C.

2 ␯ (0) ⫽ 0

‹ C4 ⫽ 0

B.C.

3 ␯¿(L) ⫽ 0 ‹ C1L + 2C2 ⫽

B.C.

2

4

x 2L + q0 a b , or p 2

(5)

3 q L2 10 0

(6)

7 q L2 30 0

(7)

4 v( L) ⫽ 0

DEFLECTION CURVE (EQ. 5) 48(4 ⫺ p) 2L 4 px b cos a b + p 2L p4 16(6 ⫺ p) x3 ⫺ * q0 L 6 p4

(4)

x4 x6 x3 b + C ⫺ 1 24 6 360L2

B.C.

;

(3)

x3 x5 b ⫺ 6 60L2

‹ C1L + 3C2 ⫽

EIv ⫽ ⫺ q0 a

L2

EIv– ⫽ M ⫽ ⫺ q0 a

+ C2 ;

x3

x2

3L2

;

q0 L

48(4 ⫺ p) 2 b q0 L ⫺ p p4

2

* q0 L2

;

EIv–¿ ⫽ ⫺ q0 a x ⫺

EIv ⫽ ⫺ q0 a

From equilibrium

MB ⫽ ⫺

px b + 8(4 ⫺ p) q0 L x 3 2L

EIv–– ⫽ ⫺ q ⫽ ⫺ q0 a 1 ⫺

+ C1

REACTIONS

M A ⫽ ⫺ q0 a

(b) Loading q ⫽ q0 a1 ⫺

EIv¿ ⫽ ⫺ q0 a

SHEAR FORCE (EQ. 2)

RA ⫽ V(0) ⫽

c ⫺16 q0 L4 cos a

DIFFERENTIAL EQUATION

L3 L2 2L 4 ‹ C1 + C2 ⫽ ⫺ q0 a b p 6 2

C1 ⫽

1 p4EI

⫺ 8(6 ⫺ p) q0 L2x 2 + 16 q0 L4 d

3 ␯¿(L) ⫽ 0 ‹ C1

B.C.

‹ C3 ⫽ 0

Solve Eqs. (6) and (7): C1 ⫽

13 q L 30 0

C2 ⫽ ⫺

1 q L2 15 0

SHEAR FORCE (EQ. 2) V ⫽ ⫺ q0 a x ⫺

x3 2

3L

b +

13 q L 30 0

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SECTION 10.3 Differential Equations of the Deflection Curve

DEFLECTION CURVE (EQ. 5)

REACTIONS RA ⫽ V(0) ⫽

13 q L 30 0

RB ⫽ ⫺ V(L) ⫽

EIv ⫽ ⫺ q0 a

;

7 q L 30 0

⫺

;

From equilibrium 2

v⫽

4

x x 13 1 b + q Lx ⫺ q L2 M A ⫽ ⫺ q0 a ⫺ 2 30 0 15 0 12L2 1 q0 L2 (counter-clockwise) MA ⫽ 15 MB ⫽ ⫺

1 q L2 (clockwise) 20 0

x6 x4 13 x3 ⫺ q b + L 0 24 30 6 360L2

1 x2 q0 L2 , or 15 2

q0 360L2EI

[x 6 ⫺ 15L2x 4

+ 26L3x 3 ⫺ 12L4x 2]

;

;

;

Problem 10.3-9 A fixed-end beam of length L is loaded by triangularly

q0

y

distributed load of maximum intensity q0 at B. Use the fourth-order differential equation of the deflection curve to solve for reactions at A and B and also the equation of the deflection curve.

MB A

MA

L

x B RB

RA

Solution 10.3-9 Triangular load q ⫽ q0

x L

DIFFERENTIAL EQUATION x L

(1)

x2 + C1 2L

(2)

EIv–– ⫽ ⫺ q ⫽ ⫺ q0 EIv–¿ ⫽ ⫺ q0

x3 EIv– ⫽ M ⫽ ⫺ q0 + C1x + C2 6L EIv¿ ⫽ ⫺ q0 EIv ⫽ ⫺ q0

x4 x2 + C1 + C2x + C3 24L 2 5

3

(3) (4)

3 ␯¿(L) ⫽ 0

B.C.

4 ␯(L) ⫽ 0

(5)

‹ C1L + 2C2 ⫽ ‹ C1L + 3C2 ⫽

Solve Eqs. (6) and (7): C1 ⫽

3 q0 L 20

C2 ⫽ ⫺

1 q L2 30 0

SHEAR FORCE (EQ. 2) V ⫽ ⫺ q0

2

x x x + C1 + C2 + C3x + C4 120L 6 2

B.C.

x2 3 + q L 2L 20 0

B.C.

1 ␯¿(0) ⫽ 0

‹ C3 ⫽ 0

REACTIONS

B.C.

2 v(0) ⫽ 0

‹ C4 ⫽ 0

RA ⫽ V(0) ⫽

3 q L 20 0

;

q0 L2 12 q0 L2 20

(6) (7)

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CHAPTER 10 Statically Indeterminate Beams

7 q L 20 0

RB ⫽ ⫺ V(L) ⫽

DEFLECTION CURVE (EQ. 5)

;

EIv ⫽ ⫺ q0

From equilibrium MA ⫽

1 q L2 30 0

␯⫽

;

Therefore, RA ⫽ ⫺ RB

MA ⫽ ⫺ MB

dC ⫽ 0

Elv– ⫽ M ⫽ RAx ⫺ M A

MA ⫽

x ⫺ M A x + C1 2

x x + C1x + C2 Elv ⫽ RA ⫺ M A 6 2 ‹ C1 ⫽ 0

B.C.

2 v(0) ⫽ 0

B.C.

L 3 va b ⫽ 0 2

‹ MA ⫽

‹ C2 ⫽ 0

RAL ⫺RAL Also, M B ⫽ 6 6

EQUILIBRIUM (OF ENTIRE BEAM) a MB ⫽ 0 or,

A

MA RA

C

x

B

L — 2

MB RB

L — 2

M A + M 0 ⫺ M B ⫺ RAL ⫽ 0

RAL RAL + M0 + ⫺ RAL ⫽ 0 6 6

RAL 6

3M 0 2L

;

‹ MA ⫽ ⫺ MB ⫽

M0 4

;

DEFLECTION CURVE (EQ. 3) (2)

2

1 ␯¿(0) ⫽ 0

M0

(1)

2

B.C.

y

‹ RA ⫽ ⫺ RB ⫽

DIFFERENTIAL EQUATION (0 … x … L/2)

3

;

Fixed-end beam (M0 ⫽ applied load)

Beam is symmetric; load is antisymmetric.

Elv¿ ⫽ RA

or

1 1⫺q0 x 5 + 3q0 L2x 3 ⫺ 2q0 L3x 22 120LEI

Problem 10.3-10 A counterclockwise moment M0 acts at the midpoint of a fixed-end beam ACB of length L (see figure). Beginning with the second-order differential equation of the deflection curve (the bending-moment equation), determine all reactions of the beam and obtain the equation of the deflection curve for the left-hand half of the beam. Then construct the shear-force and bending-moment diagrams for the entire beam, labeling all critical ordinates. Also, draw the deflections curve for the entire beam.

Solution 10.3-10

x5 3 1 x2 x3 + q0 L ⫺ q0 L2 120L 20 6 30 2

M0 x 2 . ␯⫽ ⫺ (L ⫺2x) 8LEI

(3) DIAGRAMS

a0 … x …

L b 2

;

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SECTION 10.3 Differential Equations of the Deflection Curve

dmax ⫽

M 0 L2 216EI

At point of inflection: d ⫽ dmax/2

Problem 10.3-11 A propped cantilever beam of length L is loaded by a concentrated moment M0 at midpoint C. Use the second-order differential equation of the deflection curve to solve for reactions at A and B. Draw shear-force and bending-moment diagrams for the entire beam. Also find the equations of the deflection curves for both halves find the equations of the deflection curves for both halves of the beam, and draw the deflection curve for the entire beam.

y L — 2 MA

A

M0

L — 2

C

x

B

RA

RB

Solution 10.3-11 EQUILIBRIUM RA ⫽ ⫺ RB

(1)

M A ⫽ ⫺ M 0 ⫺ RBL

(2)

EIv¿ ⫽ RB Lx ⫺ RB EIv ⫽ ⫺ RB L

x2 + C3 2

x3 x2 ⫺ RB + C3 x + C4 2 6

BENDING MOMENTS (FROM EQUILIBRIUM) M ⫽ RA x ⫺ M A ⫽ ⫺ RB x + M 0 + RBL a0 … x … M ⫽ RB (L ⫺ x)

L b 2

B.C.

3 v(L) ⫽ 0

B.C.

4 continuity condition at point C

⫺ RB DIFFERENTIAL EQUATIONS (0 … x … L/2) EIv– ⫽ M ⫽ ⫺ RB x + M 0 + RBL EIv¿ ⫽ ⫺ RB EIv ⫽ ⫺ RB

x2 + M 0 x + RBL x + C1 2

x3 x2 x2 + M0 + RBL + C1 x + C2 6 2 2

B.C.

1 ␯¿(0) ⫽ 0

‹ C1 ⫽ 0

B.C.

2 v(0) ⫽ 0

‹ C2 ⫽ 0

(4) (5)

L2 L L + M 0 + RB L 8 2 2

⫽ RB L

(3) C3 ⫽ M 0

L2 L ⫺ RB + C3 2 8

L 2

From eq. (9): C4 ⫽ ⫺ B.C.

RB L3 L2 ⫺ M0 3 2

5 continuity condition at point C

L At x ⫽ : (v )left ⫽ (v )right 2

DIFFERENTIAL EQUATIONS (L/2 … x … L) EIv– ⫽ M ⫽ RB (L ⫺ x)

‹ C3 L + C4 ⫽ ⫺

L At x ⫽ : (␯¿)left ⫽ (␯¿)right 2

L a … x … Lb 2

(6)

(7) (8) RB L3 3

(9)

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EIv ⫽ ⫺ RB ⫺ RB ⫺ RB RB ⫽ ⫺

x3 x2 x2 + M0 + RBL + C1 x + C2 6 2 2

DEFLECTION CURVE FOR (L/2 … x … L)

L3 L2 L2 L2 + M0 + RBL ⫽ RB L 48 8 8 8 3

3

2

L LL L L + M0 ⫺ RB ⫺ M0 48 2 2 3 2

9 M0 8 L

x2 x3 ⫺ RB + C3 x + C4 2 6

EIv ⫽ RB L

x2 x3 L ⫺ RB + M0 x 2 6 2

+ a⫺

; RA ⫽

From eq. (1)

EIv ⫽ RB L

v⫽

9 M0 8 L

;

From eq. (2) M A ⫽ ⫺ M 0 +

9 M0 1 M0 L⫽ 8 L 8 L

RB L3 L2 ⫺ M0 b 3 2

9M 0 2 M0 L M 0 L2 1 9M 0 3 a x ⫺ x + x⫺ b EI 48L 16 2 8 a

;

L … x … Lb 2

DEFLECTION CURVE SHEAR FORCE AND BENDING MOMENT DIAGRAMS 2L /3 0

9M0/8L V

L /3

L /9

7M0 /16

M –M0 /8 –9M0 /16

DEFLECTION CURVE FOR (0 … x … L/2) EIv ⫽ ⫺ RB v⫽

x3 x2 x2 + M0 + RBL + C1x + C2 6 2 2

M0 2 1 9M 0 3 L a x ⫺ x b a0 … x … b EI 48L 16 2

;

δmax = M0L2/72EI

;

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809

SECTION 10.4 Method of Superposition

Method of Superposition P

The problems for Section 10.4 are to be solved by the method of superposition. All beams have constant flexural rigidity EI unless otherwise stated. When drawing shear-force and bending-moment diagrams, be sure to label all critical ordinates, including maximum and minimum values.

B

A

MA RA

Problem 10.4-1 A proposed cantilever beam AB of length L carries a

a

concentrated load P acting at the position shown in the figure. Determine the reactions RA, RB , and MA for this beam. Also, draw the shear-force and bending-moment diagrams, labeling all critical ordinates.

b L

Solution 10.4-1 Propped cantilever beam Select RB as redundant.

OTHER REACTIONS (FROM EQUILIBRIUM)

EQUILIBRIUM

RA 

RA  P  RB

MA  Pa  RBL

RELEASED STRUCTURE AND FORCE-DISPLACEMENT RELATIONS

MA 

Pb 2L3

(3L2  b2)

Pab 2L2

(L + b)

;

SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS

COMPATIBILITY dB  (dB)1  (dB)2  0 dB  RB 

RBL3 Pa2 (3L  a)  0 6EI 3EI Pa2 2L3

(3L  a)

;

RB

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Problem 10.4-2 A beam with a guided support at B is loaded by a

y

uniformly distributed load with intensity q. Use the method of superposition to solve for all reactions. Also draw shear-force and bendingmoment diagrams, labeling all critical ordinates.

q MB x A

MA

L

B

RA

Solution 10.4-2 Select MB as redundant.

FROM EQUILIBRIUM EQS.

EQUILIBRIUM

RA  qL

RA  qL

(1)

MA 

qL2  MB 2

(2)

MA 

qL2 qL2  MB  2 3

SHEAR FORCE V  RA  qx  q(L  x)

RELEASED STRUCTURE AND FORCE OR MOMENT-ROTATION BENDING MOMENTS

RELATIONS

M  RAx  MA 

q

qx2 qx2 qL2  + qLx  2 2 3

SHEAR FORCE AND BENDING MOMENT DIAGRAMS

(θB)1

qL2/3

qL2/6

(θB)2 qL

MB qL V

(uB)1 = Rotation at B due to uniform load q (uB)1 

qL3 6EI

0.423L qL2/6

(uB) 2 = Rotation at B due to Moment MB

2/3

–qL

FROM COMPATIBILITY EQUATION (uB)1  (uB) 2

M 0

MBL (uB) 2  EI

‹ MB 

qL2 6

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SECTION 10.4 Method of Superposition

Another solution by the 2nd order differential equation.

‹ MA 

DIFFERENTIAL EQUATIONS MB 

2

EIv–  M  RA x  MA  EIv¿  RA EIv  RA

qx 2

qx4 x3 x2  MA  + C1 x + C2 6 2 24

1 v¿(0)  0

‹ C1  0

B.C.

2 v¿(0)  0

‹ C2  0

B.C.

3 v¿(L)  0

MA L  qL

;

qL2 qL2 qL2   2 3 6

;

DEFLECTION CURVE

qx3 x2  MAx  + C1 2 6

B.C.

qL2 3

EIv  RA or v 

qx4 qL2 x2 qx4 x3 x2 x3  MA   qL   6 2 24 6 3 2 24

q ( x4 + 4Lx3  4L2x2) 24EI

;

qL3 qL3 L2   2 6 3

Problem 10.4-3 A propped cantilever beam of length 2L with support at B

y

is loaded by a uniformly distributed load with intensity q. Use the method of superposition to solve for all reactions. Also draw shear-force and bendingmoment diagrams, labeling all critical ordinates.

q

MA

x A

L

B

L

C

RA RB

Solution 10.4-3 Select RB as redundant.

(dB) 2  Deflection at B due to Force RB

RELEASED STRUCTURE AND FORCE-DISPLACEMENT RELATIONS

(dB) 2 

q

RBL3 3EI

FROM COMPATIBILITY EQUATION L

( δB ) 1

L

(dB)1  (dB) 2

‹ RB 

17 qL 8

EQUILIBRIUM (δB)2 L

L RB

(dB)1  Deflection at B due to uniform load q (dB)1 

17qL4 24EI

1 RA  2qL  RB   qL 8

(1)

1 MA  2qL2  RBL   qL2 8

(2)

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SHEAR FORCE AND BENDING MOMENT DIAGRAMS

SHEAR FORCE V  RA  qx  

qL  qx 8

V  2qL  qx

(L … x … 2L)

qL

(0 … x … L) L V

L

BENDING MOMENTS

–qL/8

qx2 qLx qL2 x  + M  RAx  MA  q   2 2 8 8 2

–9qL/8 0.3904L

(0 … x … L) M  RAx MA  q

qx x + RB x  2 2

2

2

+ 2qLx 2qL2

qL2/8 M

0

(L … x … 2L) –qL2/2

Problem 10.4-4 Two flat beams AB and CD, lying in horizontal planes, cross

D

P

at right angles and jointly support a vertical load P at their midpoints (see figure). Before the load P is applied, the beams just touch each other. Both beams are made of the same material and have the same widths. Also, the ends of both beams are simply supported. The lengths of beams AB and CD are LAB and LCD, respectively. What should be the ratio tAB/tCD of the thicknesses of the beams if all four reactions are to be the same?

A B tAB C

tCD

Solution 10.4-4 Two beams supporting a load P For all four reactions to be the same, each beam must support one-half of the load P.

MOMENT OF INERTIA IAB 

DEFLECTIONS (P/2) L3AB dAB  48EIAB

(P/2)L3CD dCD  48EICD

COMPATIBILITY dAB  dCD

or

L3AB L3CD  IAB ICD

‹

L3AB t3AB

1 3 bt 12 AB 

L3CD t3CD

ICD 

1 3 bt 12 CD

tAB LAB  tCD LCD

;

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SECTION 10.4 Method of Superposition

Problem 10.4-5 A propped cantilever beam of length 2L is loaded by a

y

uniformly distributed load with intensity q. The beam is supported at B by a linearly elastic spring with stiffness k. Use the method of superposition to solve for all reactions. Also draw shear-force and bending-moment diagrams, labeling all critical ordinates. Let k  6EI/L3.

q0

A

MA

L

x

B

L

k

C

RA RB = kdB

Solution 10.4-5 Select RB as the redundant.

(dB)2  Deflection at B due to Force RB

RELEASED STRUCTURE AND FORCE-DISPLACEMENT RELATIONS

(dB)2 

RBL3 3EI

(dB)3  Shortening in spring due to Force RB

q

(dB)3  L

L (δB)1

RB RBL3  k 6EI

FROM COMPATIBILITY EQUATION (dB)1  (dB)2  (dB)3

(δB)2 RB

17 qL 12

EQUILIBRIUM RA  2qL  RB 

RB (δB)3

‹ RB 

7 qL 12

MA  2qL2  RBL 

(1)

7 qL2 12

(2)

SHEAR FORCE RB

(dB)1  Deflection at B due to uniform load q 17qL4 (dB)1  24EI

V  RA  qx 

7 qL  qx 12

V  RA  qx + RB  2qL  qx

(0 … x … L) (L … x … 2L)

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SHEAR FORCE AND BENDING MOMENT DIAGRAMS

BENDING MOMENTS M  RAx  MA  q

2

2

x 1 7L 7L   qc x2  x + d 2 2 12 12

q 7qL2/12

(0 … x … L) L

x2 M  RAx  MA  q + RB1x  L2 2 1  qc x2  2Lx + 2L2 d 2

L

7qL/12

17qL/12

7L/12

qL

7qL/12

(L … x … 2L)

V 0

–5qL/12

7L/12

M

–.413qL2

–7qL2/12

Problem 10.4-6 A propped cantilever beam of length 2L is loaded by a uniformly distributed load with intensity q. The beam is supported at B by a linearly elastic rotational spring with stiffness kR, which provides a resisting moment MB due to rotation uB . Use the method of superposition to solve for all reactions. Also draw shear-force and bending-moment diagrams, labeling all critical ordinates. Let kR = EI/L.

–qL2/2

y q

MA

B

EI A k =— R L RA L

x C

MB = kRuB L

Solution 10.4-6 Select MB as the redundant.

q

RELEASED STRUCTURE AND FORCE-SLOPE RELATIONS q (θB)1

(uB)1  Slope at B due to uniform load q (uB)1 

(θB)2 MB

(uB)2  Slope at B due to Moment (uB)2 

MB

7qL3 6EI

MBL EI

(uB)3  Spring rotation at B due to Moment (θB)3 MB

(uB)3 

MB MBL  kR EI

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SECTION 10.4 Method of Superposition

SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS

FROM COMPATIBILITY EQUATION (uB)1  (uB)2  (uB)3

‹ MB 

q

7 qL2 12

17qL2/12

L

FROM EQUILIBRIUM EQS. RA  2qL

(1)

17 MA  2qL  MB  qL2 12 2

(2)

L

2qL

7qL2/12

2qL

V

SHEAR FORCE V  RA  qx  2qL  qx

0

(0 … x … 2L)

qL2/12 M

BENDING MOMENTS 1 x2 17L2   qc x2  2Lx + d M  RAx  MA  q 2 2 12

–17qL2/12

0

–qL2/2 2 nd degree curves

(0 … x … L) M  RAx  MA  q

2

x  MB 2

1   q c x2  2Lx + 2L2 d 2

(L … x … 2L)

Problem 10.4-7 Determine the fixed-end moments (MA and MB) and fixed-end forces (RA and RB) for a beam of length L supporting a triangular load of maximum intensity q0 (see figure). Then draw the shear-force and bending-moment diagrams, labeling all critical ordinates.

q0

A

MA RA

B L — 2

L — 2

MB RB

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Solution 10.4-7 Fixed-end beam (triangular load) Select MA and MB as redundants.

COMPATIBILITY uA  (uA)1  (uA)2  0

SYMMETRY MA  MB

Substitute for (uA)1 and (uA) 2 and solve for MA:

RA  RB

EQUILIBRIUM RA  RB  q0 L/4

;

MA  MB 

5q0L2 96

;

RELEASED STRUCTURE AND FORCE-DISPLACEMENT RELATIONS SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS q0L — 4 V O q0L –— 4

M1

M

x0

O

5q0L2 –— 96

M1 

5q0L2 –— 96

q0L2 32

x0  0.2231L

Problem 10.4-8 A continuous beam ABC with two unequal spans, one of length L and one of length 2L, supports a uniform load of intensity q (see figure). Determine the reactions RA, RB, and RC for this beam. Also, draw the shear-force and bending-moment diagrams, labeling all critical ordinates.

q A

RA

C

B

L

RB

2L

RC

Solution 10.4-8 Continuous beam with two spans Select RB as the redundant.

RELEASED STRUCTURE AND FORCE-DISPLACEMENT RELATIONS

EQUILIBRIUM RA 

3qL 2  RB 2 3

Rc 

3qL 1  RB 2 3

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SECTION 10.4 Method of Superposition

(dB)1 

11qL4 12EI

(dB)2 

4RBL3 9EI

SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS

COMPATIBILITY dB  (dB)1  (dB)2  0 11qL4 4RBL3  0 12EI 9EI

RB 

33qL 16

;

OTHER REACTIONS (FROM EQUILIBRIUM) RA 

qL 8

Rc 

13qL 16

;

Problem 10.4-9 Beam ABC is fixed at support A and rests (at point B) upon the midpoint of beam DE (see the first part of the figure). Thus, beam ABC may be represented as a propped cantilever beam with an overhang BC and a linearly elastic support of stiffness k at point B (see the second part of the figure). The distance from A to B is L  10 ft, the distance from B to C is L/2  5 ft, and the length of beam DE is L  10 ft. Both beams have the same flexural rigidity EI. A concentrated load P  1700 lb acts at the free end of beam ABC. Determine the reactions RA, RB, and MA for beam ABC. Also, draw the shear-force and bending-moment diagrams for beam ABC, labeling all critical ordinates.

E

P

A B

C

D

P = 1700 lb MA B

A k RA

RB L = 10 ft

L — = 5 ft 2

C

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Solution 10.4-9 Beam with spring support Select RB as the redundant.

OTHER REACTIONS (FROM EQUILIBRIUM)

EQUILIBRIUM

RA 

RA  RB  P

MA  RBL  3PL/2

11P 17

MA 

5PL 34

;

NUMERICAL VALUES RELEASED STRUCTURE AND FORCE-DISPL. EQS.

P  1700 lb

L  10 ft  120 in.

RA  1100 lb RB  2800 lb f MA  30,000 lb-in.

;

SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS

COMPATIBILITY dB  (dB)1  (dB)2  Beam DE: k 

RB k

48EI L3

RBL3 RBL3 7PL3   12EI 3EI 48EI

RB 

28P 17

; x1 

300 in. 11

 27.27 in.

Problem 10.4-10 A propped cantilever beam has flexural rigidity

y

EI  4.5 MN # m2. When the loads shown are applied to the beam, it settles at joint B by 5 mm. Find the reaction at joint B.

5 kN/m

2 kN x

MA

C

B

A

5 mm settlement RA

4m

1m

2m RB

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SECTION 10.4 Method of Superposition

Solution 10.4-10 EI  4.5 MN # m2  4500 kN # m2 Select RB as the redundant.

(dB)1  Deflection at B due to distributed and concentrated loads

RELEASED STRUCTURE AND FORCE-DISPLACEMENT RELATIONS

(dB)1  

7m

4m

1m

L2m 5m

2m



5 kN/m

5 5 x c (x  1)2  (x  1)3 d dx 2 24 EI

  64.593 * 103   64.593 mm (dB) 2  Deflection at B due to Force RB 5m

(dB) 2  (δB)2 RB

(x  2) dx EI

  (29.63 + 34.963) * 103

2 kN (δB)1

L1 m

2x

L0

RB x

x dx  RB 9.259 * 103 EI

 RB 9.259 mm COMPATIBILITY (SETTLEMENT AT B = 5 mm) (dB)1  (dB) 2  5 mm

Problem 10.4-11 A cantilever beam is supported by a tie rod at B as shown. Both the tie rod and the beam are steel with E  30 * 106 psi. The tie rod is just taut before the distributed load q  200 lb/ft is applied.

‹ RB  6.44 kN

;

D y

1 — in. tie rod 4

H = 3 ft q

(a) Find the tension force in the tie rod. (b) Draw shear-force and bending-moment diagrams for the beam, labeling all critical ordinates. MA

A

S 6  12.5 L1 = 6 ft

RA

x B

C L2 = 2 ft

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Solution 10.4-11 COMPATIBILITY

E  30 * 106 psi A

(dB)1  (dB) 2  (dB)

(1/4)2p  0.0491 in.2 4

I  22.1 in.4

3

0.1282  RD 1.877 * 104  RD 2.445 * 105 ‹ RD  604.3 lb

From S 6 * 12.5

(a) THE TENSION FORCE IN THE TIE ROD  RD  604 lb

Select RD as the redundant.

;

RELEASED STRUCTURE AND FORCE-DISPLACEMENT RELATIONS FROM EQUILIBRIUM EQS. RD

RA  200 # 6 + 200

3 ft RD

(δB)3

MA 

2  RD  795.7 lb 2

(1)

200 # 2 2 62 (6 + ) + 200  RD 6 2 3 2

 1308 lb-ft  1.569 * 104 lb # in.

;

RD (δB)2 6 ft

(b) SHEAR FORCE AND BENDING MOMENT DIAGRAMS

2 ft

2 ft

6 ft 3.98 ft

200 lb/ft 795.7 lb

200 lb V

(δB)1

0 –404.3 lb

(dB)1  Deflection at B due to distributed load q q  200 lb / ft 

200 lb/in. 12

275.3 lb-ft M

–133.3 lb-ft

72

200 2 # 12 2 # 12 x 200 x 2 a + b dx (dB)1  12 2 12 2 3 EI L0  0.1282 in. (dB)2  Deflection at B due to Force RD 72

(dB)2 

RDx 2 dx  RD 1.877 * 104 in. EI L0

(dB) 3  Extension in tie rod due to Force RD RD (3 # 12) RD L (dB) 3   AE 0.0491 # 30 * 106  RD 2.445 * 105 in.

–1307.5 lb-ft

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SECTION 10.4 Method of Superposition

Problem 10.4-12 The figure shows a nonprismatic, propped cantilever beam AB with flexural rigidity 2EI from A to C and EI from C to B. Determine all reactions of the beam due to the uniform load of intensity q. (Hint: Use the results of Problems 9.7-1 and 9.7-2.)

q

MA RA

B

EI

C

2EI

A

L 2

L 2

RB

Solution 10.4-12 Nonprismatic beam Select RB as the redundant.

FORCE-DISPLACEMENT RELATIONS

RELEASED STRUCTURE

From Prob. 9.7-2: dB 

qL4 I1 a1 15 b 128EI1 I2

‹ (dB)1 

I1 : I I2 : 2I

17qL4 256EI

From Prob. 9.7-1: dB  (dB)1  downward deflection of end B due to load q

I1 PL3 a1 7 b 24EI1 I2

‹ (dB) 2 

3RBL3 16EI

COMPATIBILITY dB  (dB)1  (dB) 2  0 or

(dB) 2  upward deflection due to reaction RB

17qL4 3RBL3  0 256EI 16EI

RB 

17qL 48

;

EQUILIBRIUM 31qL 48 qL2 7qL 2 MA   RB L  2 48 RA  qL  RB 

Problem 10.4-13 A beam ABC is fixed at end A and supported by beam DE at point B (see figure). Both beams have the same cross section and are made of the same material. (a) Determine all reactions due to the load P. (b) What is the numerically largest bending moment in either beam?

;

P A

B

C E

D

MA RA L 4

RD L 4

L 4

RE

L 4

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Solution 10.4-13 Beam supported by a beam Let RB  interaction force between beams

COMPATIBILITY

Select RB as the redundant.

(dB)1  (dB) 2  (dB)3

Substitute and solve: RB 

40P 17

;

RELEASED STRUCTURE AND FORCE-DISPL. EQS. SYMMETRY AND EQUILIBRIUM RD  RE 

RB 20P  2 17

RA  PRD RE  

; 23P 17

;

(minus means downward) 3PL L M A  RB a b  PL  2 17

;

BEAM ABC: M max  M B   BEAM DE: M max  M B 

冷 M max 冷 

PL 2

PL 2

5PL 17

;

Problem 10.4-14 A three-span continuous beam ABCD with three equal spans supports a uniform load of intensity q (see figure). Determine all reactions of this beam and drawn the shear-force and bending-moment diagrams, labeling all critical ordinates.

q A

RA

Solution 10.4-14 Three-span continuous beam SELECT RB AND RC AS THE REDUNDANTS. SYMMENTRY AND EQUILIBRIUM RC  RB

RA  RD 

3qL  RB 2

RELEASED STRUCTURE

B

L

RB

D

C

L

L RC

RD

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SECTION 10.4 Method of Superposition

FORCE-DISPLACEMENT RELATIONS 4

(dB)1 

11qL 12EI

OTHER REACTIONS 3

(dB)2 

From symmetry and equilibrium:

5RBL 6EI

RC  RB 

11qL 10

RA  RD 

2qL 5

COMPATIBIITY dB  (dB)1  (dB)2  0

11qL ‹ RB  10

;

;

;

LOADING, SHEAR-FORCE, AND BENDING-MOMENT DIAGRAMS

MB  MC   M max 

2qL2 25

qL2 10

823

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Problem 10.4-15 A beam rests on supports at A and B and is

y

loaded by a distributed load with intensity q as shown. A small gap ¢ exists between the unloaded beam and the support at C. Assume that span length L  40 in. and flexural rigidity of the beam EI  0.4 * 109 lb-in.2 Plot a graph of the bending moment at B as a function of the load intensity q. (HINT: See Example 9-9 for guidance on computing the deflection at C.)

q A

B

L

RA

L

RB

x  = 0.4 in.

C

RC

Solution 10.4-15 (dC)1 

qL4 4EI

RC  2qL  RA  RB

(dC) 2 

2L3 R 3EI C

RELEASED STRUCTURE AND FORCE-DISPLACEMENT RELATIONS

COMPATIBILITY

Select RC as the redundant. EQUILIBRIUM

1) dC  (dC)1

q

for (dC)1 6 0.4 in.

‹ RC (q)  0

2

M B (q)  

2) dC  (dC)1  (dC)2  0.4 in. for (dC)1  0.4 in.

(δC)1

L

L

qL  800q lb # in. for q 6 250 lb/in. 2

‹ RC (q)  15q  3750 M B(q)  RAL 

(δC)2

qL2   200q  150000 lb # in. 2

for q Ú 250 lb/ in.

RC

Moment at B

Moment at B (lb-in.)

0

–1.105 MB(q) –2.105

–3.105

0

100

200

300 q q (lb/in.)

400

500

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SECTION 10.4 Method of Superposition

Problem 10.4-16 A fixed-end beam AB of length L is subjected

M0

to a moment M 0 acting at the position shown in the figure.

A

(a) Determine all reactions for this beam. (b) Draw shear-force and bending-moment diagrams for the special case in which a  b  L/2 .

B MB

MA a

RA

b

RB

L

Solution 10.4-16

Fixed-end beam (M 0 = applied load)

Select RB and MB as redundants.

(uB) 2 

RBL2 2EI

(dB) 2 

RBL3 3EI

(uB)3 

M BL EI

(dB)3 

M BL3 2EI

COMPATIBILITY La + b

dB   (dB)1  (dB) 2 + (dB)3  0

EQUILIBRIUM

or 2RB L3  3MB L2   3M 0 a(a + 2b)

RA   RB

M A  M B  RB L  M 0

RELEASED STRUCTURE AND FORCE-DISPL. EQS.

(1)

uB  (uB)1 + (uB) 2  (uB)3  0 or RBL2  2M BL   2M 0 a

(2)

SOLVE EQS. (1) AND (2): RB  

6M 0 ab

MB  

3

L

M0 a L2

(3b  L)

;

FROM EQUILIBRIUM: RA 

6M 0 ab 3

L

MA 

M0b L2

(3a  L)

;

SPECIAL CASE a  b  L/2 RA   RB 

(uB)1 

M 0a EI

(dB)1 

M0 a (a + 2b) 2EI

3M 0 2L

MA   MB 

M0 4

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CHAPTER 10 Statically Indeterminate Beams

Problem 10.4-17 A temporary wood flume serving as a channel for irrigation water is shown in the figure. The vertical boards forming the sides of the flume are sunk in the ground, which provides a fixed support. The top of the flume is held by tie rods that are tightened so that there is no deflection of the boards at the point. Thus, the vertical boards may be modeled as a beam AB, supported and loaded as shown in the last part of the figure. Assuming that the thickness t of the boards is 1.5 in., the depth d of the water is 40 in., and the height h to the tie rods is 50 in., what is the maximum bending stress s in the boards? (Hint: The numerically largest bending moment occurs at the fixed support.) t = 1.5 in.

B

h= 50 in.

d= 40 in.

A

Solution 10.4-17 Side wall of a wood flume

COMPATIBILITY dB  (dB)1 (dB) 2  0 Select RB as the redundant. Equilibrium: MA 

q0 a 2  RB L 6

RELEASED STRUCTURE AND FORCE-DISPL. EQS.

‹ RB 

q0 a 3(5L a) 40 L3

MAXIMUM BENDING MOMENT 1 M max  M A  q0 a 2 RB L 6 

q0 a 2 120L2

(20 L2 15 aL 3a 2 )

NUMERICAL VALUES a  40 in. From Table G-1, Case B: q0 a 4 q0 a 3 q0 a 3  (L a)  (5L  a) 30EI 24EI 120EI 3 RB L (dB)2  3EI

(dB)1 

L  50 in.

b  width of beam

t  1.5 in.

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SECTION 10.4 Method of Superposition

S

bt 2 6

s

M max S

  62.4 lb/ft3  0.03611 lb/in.3 Pressure p  a

q0  pb  ab

M max  S

ga 3b 120L2

(20 L2 15aL3a 2 )  191.05 b

bt 2  0.3750 b 6

s

Problem 10.4-18 Two identical, simply supported beams AB and CD are placed so that they cross each other at their midpoints (see figure). Before the uniform load is applied, the beams just touch each other at the crossing point. Determine the maximum bending moments (M AB)max and (M CD)max in beams AB and CD, respectively, due to the uniform load it the intensity of the load is q  6.4 kN/m and the length of each beam is L  4 m .

M max  509 psi S

q D

A

B C

Solution 10.4-18 Two beams that cross F  interaction force between the beams

LOWER BEAM

UPPER BEAM

dCD  (dB)1  downward deflection due to q 

5qL4 384EI

(dB) 2  downward deflection due to F 

FL3 48EI

dAB  (dB)1  (dB) 2 

5qL4 FL3  384EI 48EI

FL3 48EI

COMPATIBILITY dAB  dCD 5qL4 FL3 FL3   384EI 48EI 48EI UPPER BEAM

;

‹ F

5qL 16

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CHAPTER 10 Statically Indeterminate Beams

RA 

LOWER BEAM

11qL 32

M max 

5qL2 FL  4 64

(M CD)max  x1 

11L 32

q  6.4 kN/m

121qL 2048

3qL2 M1  64

;

NUMERICAL VALUES 2

M max 

5qL2 64

L4m 121qL2 (M AB)max  2048

(M AB)max  6.05 kN # m

(M CD)max  8.0 kN # m

S 6  12.5 steel I-beam with E  30  106 psi. The simple beam DE is a wood beam 4 in.  12 in. (normal dimension) in cross section with E  1.5  106 psi. A steel rod AC of diameter 0.25 in., length 10 ft, and E  30  106 psi serves as a hanger joining the two beams. The hanger fits snugly between the beam before the uniform load is applied to beam DE. Determine the tensile force F in the hanger and the maximum bending moments MAB and MDE in the two beams due to the uniform load, which has intensity q  400 lb/ft . (Hint: To aid in obtaining the maximum bending moment in beam DE, draw the shear-force and bending-moment diagrams.)

A

S 6  12.5 B 6 ft Steel rod

E

D C Wood beam 10 ft

Solution 10.4-19 Beams joined by a hanger F  tensile force in hanger

(2) HANGER AC

Select F as redundant. (1) CANTILEVER BEAM AB

I1  22.1 in.4

L 1  6 ft  72 in. E 1  30 * 106 psi FL 31  187.66 * 106 F 3E 1I1

e

F  Ib d  in.

10 ft

400 lb/ft

10 ft

(dA)1 

;

;

Problem 10.4-19 The cantilever beam AB shown in the figure is an

S 6 * 12.5

;

L 2  10 ft  120 in.

d  0.25 in.

E 2  30 * 10 psi 6

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SECTION 10.4 Method of Superposition

A2 

pd 2  0.049087 in.2 4

4 in. * 12 in. (nominal) I3  415.28 in.4

¢  elongation of AC ¢

(dC)3 

FL 2  81.488 * 106F E 2A2

5qL 43 FL 33  384E 3I3 48E 3I3

 2.3117 in.462.34 * 106 F

(F  lb, ¢  in.)

COMPATIBILITY

(3) BEAM DCE

(dA)1 + ¢  (dC)3 187.66 * 106 F + 81.488 * 106 F  2.3117  462.34 * 106F F  3160 lb

;

(1) MAX. MOMENT IN AB M AB  FL 1  (3160 lb)(6 ft)  18,960 lb-ft

;

(3) MAX. MOMENT IN DCE

L 3  20 ft  240 in. q  400 lb/ft

RD 

 33.333 lb/in.

qL 3 F   2420 lb 2 2

E 3  1.5 * 106 psi SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS

x 1  6.050 ft

M max  7320 lb-ft M C  4200 lb-ft

M DE  7320 lb-ft

;

e

F  Ib d  in.

829

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CHAPTER 10 Statically Indeterminate Beams

Problem 10.4-20 The beam AB shown in the figure is simply supported at A and B and supported on a spring of stiffness k at its midpoint C. The beam has flexural rigidity EI and length 2L. What should be the stiffness k of the spring in order that the maximum bending moment in the beam (due to the uniform load) will have the smallest possible value?

q A

B

C k L

L

Solution 10.4-20 Beam supported by a spring FOR THE SMALLEST MAXIMUM MOMENT: |M 1|  |M C| or M 1   M C qL2 R2A   RAL + 2q 2 Solve for RA: RA  qL(12  1) EQUILIBRIUM g Fvert  0

2RA + RC  2qL  0 RC  2qL(2  12)

DOWNWARD DEFLECTION OF BEAM qx 2 BENDING MOMENT M  RAx  2

(dC)1 

5qL4 qL4 RCL3   (8 12  11) 24EI 6EI 24EI

DOWNWARD DISPLACEMENT OF SPRING LOCATION OF MAXIMUM POSITIVE MOMENT dM 0 dx

RA  qx  0

RA x1  q

MAXIMUM POSITIVE MOMENT M 1  (M)xx1

R 2A  2q

MAXIMUM NEGATIVE MOMENT M C  (M)xL

qL2  RAL  2

(dC) 2 

2qL RC  (2  12) k k

COMPATIBILITY (dC)1  (dC) 2 Solve for k: k

48EI 7L3

(6 + 512)

 89.63

EI L3

;

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SECTION 10.4 Method of Superposition

Problem 10.4-21 The continuous frame ABC has a fixed support at A,

L

a roller support at C, and a rigid corner connection at B (see figure). Members AB and BC each have length L and flexural rigidity EI. A horizontal force P acts at midheight of member AB.

L — 2

(a) Find all reactions of the frame. (b) What is the largest bending moment M max in the frame?

C

B

P

(Note: Disregard axial deformations in member AB and consider only the effects of bending.)

VC L — 2 A

HA MA

VA

Solution 10.4-21

Frame ABC with fixed support

Slect VC as the redundant. EQUILIBRIUM VA  VC

HA  P

M A  PL/2  VCL

(uB) 2 

VCL2 EI

(dC) 2  (uB) 2L +

VC L3 4VCL3  3EI 3EI

RELEASED STRUCTURE AND FORCE-DISPL. EQS. COMPATIBILITY (dC)1  (dC) 2 Substitute for (dC)1 and (dC) 2 and solve: VC 

3P 32

;

FROM EQUILIBRIUM: VA  PL2 (uB)1  8EI

3P 32

HA  P

MA 

13PL 32

;

REACTIONS AND BENDING MOMENTS

PL3 (dC)1  (uB)1L  8EI

M max 

13PL 32

;

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CHAPTER 10 Statically Indeterminate Beams

Problem 10.4-22 The continuous frame ABC has a pinned support

MC

at A, a guided support at C, and a rigid corner connection at B (see figure). Members AB and BC each have length L and flexural rigidity EI. A horizontal force P acts at midheight of member AB.

L

(a) Find all reactions of the frame. (b) What is the largest bending moment M max in the frame? (Note: Disregard axial deformations in members AB and BC and consider only the effects of bending.)

C

B

L — 2 P L — 2

A HA VA

Solution 10.4-22 Select MC as the redundant.

(uC)1 

PL2 16EI

(uC)1 

4L M 3EI C

EQUILIBRIUM VA  0 HA  

MC P  2 L

HC  

MC P + 2 L

COMPATIBILITY

RELEASED STRUCTURE AND FORCE-DISPLACEMENT RELATIONS (θB)1

MC (θB)2

P

uC  (uC)1  (uC) 2  0

‹ MC 

3 PL 64

FROM EQUILIBRIUM HA  

MC P 35   P 2 L 64

HC  

MC 29 P +  P 2 L 64

M max 

35 PL 128 3PL/64

3PL/64

29P/64 –29P/64 35PL/128 35P/64 35P/64 V 0

0 M

HC

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SECTION 10.4 Method of Superposition

Problem 10.4-23 A wide-flange beam ABC rests on three

P

identical spring supports A, B and C (see figure). The flexural rigidity of the beam is EI  6912  106 1b-in.2 and each spring has stiffness k  62,500 1b/in. The length of the beam is L  16 ft. If the load P is 6000 1b, what are the reactions RA, RB, and RC? Also, draw the shear-force and bending-moment diagrams for the beam, labeling all critical ordinates.

C

A B k

k

L — 4

RA

L — 4

k

L — 2

RB

Solution 10.4-23 Beam on three springs (Case 5, Table G-2) (dB)1 

11PL3 P + 2k 768EI

(dA)2 

RB 2k

(dC)2 

RB 2k

Select RB as the redundant. EQUILIBRIUM RA 

RB 3P  4 2

RC 

RB P  4 2

RELEASED STRUCTURE AND FORCE-DISPL. EQS.

(downward)

RBL3 1 (dB) 2  [(dA) 2 + (dC) 2 ] + 2 48EI 

RBL3 RB + 2k 48EI

(upward)

COMPATIBILITY (dB)1  (dB)2 

RB k

Substitute and solve: RB  Pa

384 EI + 11kL3 b 1152 EI + 16 kL

(dA)1 

3P 4k

Let k * 

(dC)1 

P 4k

RB 

1 (dB)1  [(dA)1 + (dC)1 ] + 2

L L 2 P a b c 3L2  4a b d 4 4 48EI

833

kL3 EI

(nondimensional)

P 384 + 11 k * a b 16 72 + k *

;

;

RC

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CHAPTER 10 Statically Indeterminate Beams

FROM EQUILIBRIUM:

SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS *

RA 

P 1344 + 13 k a b 32 72 + k *

RC 

3P 64  k * a b 32 72 + k *

;

;

NUMERICAL VALUES k  62,500 lb/in.

EI  6912 * 106 lb-in.2 L  16 ft  192 in. k* 

3

kL  64 EI

RA  3000 lb

P  6000 lb

RB  3000 1b RC  0

;

;

Problem 10.4-24 A fixed-end beam AB of length L is subjected to a

q

uniform load of intensity q acting over the middle region of the beam (see figure). (a) Obtain a formula for the fixed-end moments MA and MB in terms of the load q, the length L, and the length b of the loaded part of the beam. (b) Plot a graph of the fixed-end moment MA versus the length b of the loaded part of the beam. For convenience, plot the graph in the following nondimensional form: MA 2

qL /12

versus

B

A

MB

MA RA

a

b L

b L

with the raio b/L varying between its extreme values of 0 and 1. (c) For the special case in which a  b  L/3, draw the shear-force and bending-moment diagrams for the beam, labeling all critical ordinates.

Solution 10.4-24

Fixed-end Beam FROM EXAMPLE 10-4, EQ. (10-25a):

M A  MB RA  RB  Lb a 2

qb 2

MA 

Pa1b 21 L2

a

RB

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835

SECTION 10.4 Method of Superposition

FOR THE PARTIAL UNIFORM LOAD

(b) GRAPH OF FIXED-END MOMENT MA qL2/12

dM A 



 

b b2 a 3 2 b 2L L

(qdx)(x)(L x) 2 L2 ab

MA 



La q

(Lb)/2

dM A 

(c) SPECIAL CASE a  b  L/3

(Lb)/2

RA  RB 

2

x(L x) dx

L2 L(Lb)/2 q

dM A

L(Lb)/2

qL 6

MA  MB 

13qL2 324

(Lb)/2 2

2

3

(L x 2Lx x ) dx

L L(Lb)/2 2

q L2x 2 2Lx 3 x 4 (Lb)/2 c   d 3 4 (Lb)/2 L2 2

. . . (lenghty substitution) . . . qb  (3L2 b 2) 24L (a) M A  M B 

qb (3L2 b 2) 24L

;

Problem 10.4-25 A beam supporting a uniform load of intensity q throughout its length rests on pistons at points A, C and B (see figure). The cylinders are filled with oil and are connected by a tube so that the oil pressure on each piston is the same. The pistons at A and B have diameter d1, and the pistons at C has diameter d2. (a) Determine the ratio of d2 to d1 so that the largest bending moment in the beam is as small as possible. (b) Under these optimum conditions, what is the largest bending moment Mmax in the beam? (c) What is the difference in elevation between point C and the end supports?

L

L

A

B

C p

p

p

d1

d2

d1

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CHAPTER 10 Statically Indeterminate Beams

Solution 10.4-25 Beam supported by pistons Solve for RA: RA  qL(12  1) EQUILIBRIUM g Fvert  0

2 RA + RC  2qL  0 RC  2qL (2  12)

REACTIONS BASED UPON PRESSURE RA  RB  pa (a) ‹

pd 21 b 4

RC  pa

pd 22 b 4

2(2  12) RC d2 4    18 d1 A RA A 12  1  1.682

qx 2 BENDING MOMENT M  RAx  2

(b) M max

 0.08579 qL2

LOCATION OF MAXIMUM POSITIVE MOMENT dM 0 dx

RA  qx  0

x1 

RA q

;

qL2 R2A  M1   (3  212) 2q 2 ;

(c) DIFFERENCE IN ELEVATION By symmetry, beam has zero slope at C.

MAXIMUM POSITIVE MOMENT M 1  (M)xx1 

R 2A 2q

MAXIMUM NEGATIVE MOMENT M C  (M)xL  RA L 

qL2 2

FOR THE SMALLEST MAXIMUM MOMENT: |M 1|  |M C| or M 1   M C qL2 R2A   RAL + 2q 2

dA 

qL4 qL4 RAL3   (8 12  11) 3EI 8EI 24EI

 0.01307 qL4/EI

;

Point C is below points A and B by the amount 0.01307 qL4/EI.

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SECTION 10.4 Method of Superposition

Problem 10.4-26 A thin steel beam AB used in conjunction with an electromagnet in a high-energy physics experiment is securely bolted to rigid supports (see figure). A magnetic field produced by coils C results in a force acting on the beam. The force is trapezoidally distributed with maximum intensity q0  18 kN/m. The length of the beam between supports is L  200 mm and the dimension c of the trapezoidal load is 50 mm. The beam has a rectangular cross section with width b  60 mm and height h  20 mm. Determine the maximum bending stress max and the maximum deflection dmax for the beam. (Disregard any effects of axial deformations and consider only the effects of bending. Use E  200 GPa.)

c

q0

837

c h

A

B

C L

Solution 10.4-26 Fixed-end beam (trapezoidal load) Consider the following beam from Case 6, Table G-2:

FROM SYMMETRY AND EQUILIBRIUM MA  MB

RA  RB 

3q0L 8

SELECT MA AND MB AS REDUNDANTS RELEASED STRUCTURE WITH APPLIED LOAD

u0 

Px(L  x) 2EI

d0 

Px (3L2  4x 2) 24EI

Consider the load P as an element of the distributed load. Replace P by qdx, where 4q0x x from 0 to L/4 L q  q0 x from L/4 to L/2 q

(uA)1 

1 2EI L0

L/2

a

4q0x b(x)(L  x) dx L

L/2

+ 

1 q x(L  x) dx 2EI LL/4 0

11q0 L3 19q0 L3 13q0 L3 +  1536 EI 384 EI 512EI

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CHAPTER 10 Statically Indeterminate Beams

d1 

1 24EI L0

L/4

a

4q0x b(x)(3L2  4x 2)dx L

L/2

+



1 q x(3L2  4x 2)dx 24 EI LL/4 0

19q0 L4 19q0 L4 361q0 L4 +  7680EI 2048EI 30,720EI

BENDING MOMENT AT THE MIDPOINT q0 L2 q0 L2 L  M C  RA a b  M A  2 24 32 

19q0 L2 7q0 L2 31q0 L2 3q0 L L a b    8 2 256 96 768

MAXIMUM BENDING MOMENT MA 7 MC

‹ M max  M A 

19q0 L2 256

NUMERICAL VALUES q0  18 kN/m RELEASED STRUCTURE WITH REDUNDANTS (uA) 2  (uB) 2

MB  MA

From Case 10, Table G-2: (uA)2 

M AL 2EI

d2 

M AL2 8EI

h  20 mm

19q0 L3 MA L  0 512 EI 2 EI

MA 

19q0 L2 256

DEFLECTION AT THE MIDPOINT dmax  d1  d2 

361q0 L4 M A L2  30,720 EI 8 EI



19q0 L2 361q0 L4 L2  a ba b 30,720 EI 256 8 EI



19q0 L4 7680 EI

b  60 mm

E  200 GPa

2

S

bh  4.0 * 106 m3 6

I

bh3  40 * 109 m4 12

M max 

19q0 L2  53.44 N # m 256

s max 

M max  13.4 MPa S

d max 

19q0 L4  0.00891 mm 7680EI

COMPATIBILITY uA  (uA)1  (uA)2  0

L  200 mm

; ;

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SECTION 10.5 Temperature Effects

839

Temperature Effects The beams described in the problems for Section 10.5 have constant flexural rigidity EI.

Problem 10.5-1 A cable CD of length H is attached to the third point of a simple beam AB of length L (see figure). The moment of inertia of the beam is I, and the effective cross-sectional area of the cable is A. The cable is initially taut but without any initial tension.

A

⌬T C

2L —— 3

(a) Obtain a formula for the tensile force S in the cable when the temperature drops uniformly by ¢T degrees, assuming that the beam and cable are made of the same material (modulus of elasticity E and coefficient of thermal expansion a). Use the method of superposition in the solution. (b) Repeat part (a) assuming a wood beam and steel cable.

Cable

L — 3 H

B

D

Solution 10.5-1 ¢T = Decrease in temperature. Use method of superposition. Select tensile force S in the cable as redundant. I ⫽ Moment of inertia of beam A ⫽ Cross-sectional area of cable

COMPATIBILITY (dc)1 ⫽ (dc)2

4SL3 SH ⫽ aH(¢T) ⫺ 243EI EA

SOLVE FOR S: S ⫽

RELEASED STRUCTURE

243EIAHa(¢T) 4AL3 + 243IH

;

(b) WOOD BEAM, STEEL CABLE

S

(dc)1 ⫽

(δC)1 S

(δC)2

CABLE

4SL3 243E W I

(downward)

(dc)2 ⫽ as H(¢T ) ⫺

SH Es A

(downward)

COMPATIBILITY (dc)1 ⫽ (dc)2

S

SOLVE FOR S:

(a) BEAM & CABLE ARE SAME MATERIAL 1dc21 ⫽ CABLE

4SL3 243EI

(downward)

(dc)2 ⫽ aH(¢T) ⫺

S⫽ SH EA

4SL3 SH ⫽ as H (¢T) ⫺ 243E W I ES A

(downward)

243E S E W IAHas(¢T) 4AL3E S + 243IHE W

;

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CHAPTER 10 Statically Indeterminate Beams

Problem 10.5-2 A propped cantilever beam, fixed at the left-hand end A and simply supported at the right-hand end B, is subjected to a temperature differential with temperature T1 on its upper surface and T2 on its lower surface (see figure). (a) Find all reactions for this beam. Use the method of superposition in the solution. Assume the spring support is unaffected by temperature. (b) What are the reactions when k : q ?

y A

T2

MA

h

T1

B x

L k

RA RB = k dB

Probs. 10.5-2 and 10.5-3

Solution 10.5-2 (a) REACTIONS ASSUMING AN ELASTIC SPRING AT B Use the method of superposition. Select RB as the redundant.

RB ⫽

a(T2 ⫺ T1)L2 3EIk b a 2h 3EI + L3k

(downward)

FROM EQUILIBRIUM

RELEASED STRUCTURE

RA ⫽ ⫺ RB ⫽ ⫺

∆T = T2 – T1

(δB)1

M A ⫽ RBL ⫽

L

a(T2 ⫺ T1)L2 3EIk b (upward) a 2h 3EI + L3k

a(T2 ⫺ T1)L3 3EIk b a 2h 3EI + L3k ;

(counter-clockwise)

(b) REACTIONS ASSUMING AN SPRING AT B IS RIGID (δB)2

3EIa(T2 ⫺ T1) 2hL

RB

RB ⫽

RB

RA ⫽ ⫺ RB ⫽ ⫺ (δB)3

M A ⫽ RBL ⫽ RB

(dB)1 ⫽

a(T2 ⫺ T1)L2 2h

(dB)2 ⫽

RBL3 3EI

(dB)3 ⫽ (dB)1 ⫺ (dB)2 ⫽

RB k

(downward)

3EIa(T2 ⫺ T1) 2hL

3EIa(T2 ⫺ T1) 2h

(counter-clockwise)

;

(upward)

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841

SECTION 10.5 Temperature Effects

Problem 10.5-3 Solve the preceding problem by integrating the differential equation of the deflection curve.

Solution 10.5-3 (a) DIFFERENTIAL EQUATION (EQ. 10-39b) EI␯– ⫽ M +

FROM EQUILIBRIUM

aEI(T2 ⫺ T1) h

EI␯– ⫽ ⫺ RB(L ⫺ x) +

RA ⫽ ⫺ RB ⫽ ⫺

aEI(T2 ⫺ T1) h

1 ␯¿(0) ⫽ 0

‹ C1 ⫽ 0

M A ⫽ RBL ⫽

2 ␯(0) ⫽ 0

B.C.

3 ␯(L) ⫽ dB ⫽

‹ RB ⫽

aEI(T2 ⫺ T1)L 3k b a 2h 3EI + L3k ;

(counter-clockwise) (b) SAME REACTIONS AS IN 10.5-2(b) WHEN

aEI(T2 ⫺ T1) 2 x3 x2 EI␯ ⫽ ⫺RBL + RB + x + C2 2 6 2h B.C.

;

(upward) 3

aEI(T2 ⫺ T1) x2 EI␯¿ ⫽ ⫺RBLx + RB + x + C1 2 h B.C.

aEI(T2 ⫺ T1)L2 3k b a 2h 3EI ⫹L3k

k:q

‹ C2 ⫽ 0 RB k

aEI(T2 ⫺ T1)L2 3k b a 2h 3EI + L3k (downward)

;

Problem 10.5-4 A two-span beam with spans of lengths L and L/3 is subjected to a temperature differential with temperature T1 on its upper surface and T2 on its lower surface (see figure). (a) Determine all reactions for this beam. Use the method of superposition in the solution. Assume the spring support is unaffected by temperature. (b) What are the reactions when k : q?

Probs. 10.5-4 and 10.5-5

y A

h

T1 T2

B

L k

RA

C x

L — 3 RC RB = k d B

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CHAPTER 10 Statically Indeterminate Beams

Solution 10.5-4 (a) Use the method of superposition.

RB ⫽ ⫺

Select RB as the redundant. RELEASED STRUCTURE

a(T1 ⫺ T2)L2 6EIk b a h 36EI + L3k

FROM EQUILIBRIUM RA + RB + RC ⫽ 0

∆T = T1 – T2 (δB)1 L /3

L

g M C ⫽ 0: RA ⫽

(δB)2 RB

a(T1 ⫺ T2)L2 3EIk 1 a b RB ⫽ 4 2h 36EI + L3k

a(T1 ⫺ T2)L2 3 9EIk b RB ⫽ a 4 2h 36EI + L3k 6EIa(T1 ⫺ T2) (b) RB ⫽ ⫺ (downward) Lh RC ⫽

RB (δB)3

RB

(dB)1 ⫽

a(T1 ⫺ T2)L2 6h

(dB)2 ⫽

RBL3 36EI

(downward)

RA ⫽

3EIa(T1 ⫺ T2) 2Lh

(upward)

RC ⫽

9EIa(T1 ⫺ T2) 2Lh

(upward)

(upward) (upward)

COMPATIBILITY (dB)3 ⫽ (dB)1 ⫺ (dB)2 ⫽

RB k

Problem 10.5-5 Solve the preceding problem by integrating the differential equation of the deflection curve

Solution 10.5-5 (a) EQUILIBRIUM RA + RB + RC ⫽ 0

EI␯– ⫽ RAx +

aEI(T1 ⫺ T2) h

1 g M C ⫽ 0: RA ⫽ ⫺ RB 4

aEI(T1 ⫺ T2) 1 x + C1 EI␯œ ⫽ RAx 2 + 2 h

3 g M A ⫽ 0: RC ⫽ ⫺ RB 4

EI␯ ⫽

DIFFERENTIAL EQUATION (EQ. 10-39b) For 0 … x … L

B.C.1

aEI(T1 ⫺ T2) 2 1 RAx 3 + x + C1 x + C2 6 2h

␯ (0) ⫽ 0

For L … x …

4 L 3

‹ C2 ⫽ 0

(1) (2)

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SECTION 10.6 Longitudinal Displacements at the Ends of Beams

EI␯– ⫽ RAx +

aEI(T1 ⫺ T2) ⫺ 4RA (x ⫺ L) h

EI␯– ⫽ ⫺ 3RAx +

B.C.

aEI(T1 ⫺ T2) 3 EI␯¿ ⫽ ⫺ RAx 2 + x 2 h

4

4 ␯a Lb ⫽ 0 3

B.C.

5

␯(L)left ⫽ ␯(L)right ⫽

RA ⫽

+ 2RALx 2 + C3x + C4 2

B.C.

(3)

aEI(T1 ⫺ T2) 2 1 x EI␯ ⫽ ⫺ RAx 3 + 2 2h

B.C.

3 continuity condition at point B

At x ⫽ L: (␯)left ⫽ (␯)right

aEI(T1 ⫺ T2) + 4RAL h

+ 4RALx + C3

843

(4)

continuity condition at point B

At x ⫽ L: (␯¿)left ⫽ (␯¿)right C1 ⫽ 2RAL2 + C3

a(T1 ⫺ T2)L2 3EIk b a 2h 36EI + L3k

(upward)

a(T1 ⫺T2)L2 6EIk b a h 36EI + L3k

(downward)

a(T1 ⫺ T2)L2 9EIk b a 2h 36EI + L3k

(upward)

RB ⫽ ⫺ RC ⫽

RB k

(b) SAME REACTIONS AS IN 10.5-4(b) WHEN k : q

Longitudinal Displacements at the Ends of Beams Problem 10.6-1 Assume that the deflected shape of a beam AB with immovable pinned supports (see figure) is given by the equation ␯ ⫽ ⫺d sin px/L, where d is the deflection at the midpoint of the beam and L is the length. Also, assume that the beam has constant axial rigidity EA.

y H

d

A

(a) Obtain formulas for the longitudinal force H at the ends of the beam and the corresponding axial tensile stress st . (b) for an aluminum-alloy beam with E ⫽ 10 * 106 psi, calculate the tensile stress st when the ratio of the deflection d to the length L equals 1/200, 1/400, and 1/600.

B

H

x

L

Solution 10.6-1 Beam with immovable supports Eq. (10-46): s1 ⫽

H p2Ed2 ⫽ A 4L2

;

(b) ALUMINUM ALLOY E ⫽ 10 * 106 psi (a) ␯ ⫽ ⫺ d sin

px L

d␯ pd px ⫽⫺ cos dx L L L

Eq. (10-42): l ⫽ Eq. (10-45): H ⫽

d␯ 2 p2d2 1 a b dx ⫽ 2 L0 dx 4L 2

EAl p EAd ⫽ L 4L2

2

;

s1 ⫽ 24.67 * 106 a

d 2 b (psi) L

d L

1 200

1 400

1 600

st (psi)

617

154

69

NOTE: The axial stress increases as the deflection increases.

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CHAPTER 10 Statically Indeterminate Beams

Problem 10.6-2

q

(a) A simple beam AB with length L and height h supports a uniform load of intensity q (see the first part of the figure). Obtain a formula for the curvature shortening l of this beam. Also, obtain a formula for the maximum bending stress sb in the beam due to the load q. (b) Now assume that the ends of the beam are pinned so that curvature shortening is prevented and a horizontal force H develops at the supports (see the second part of the figure). Obtain a formula for the corresponding axial tensile stress st . (c) Using the formulas in parts (a) and (b), calculate the curvature shortening l, the maximum bending stress sb, and the tensile stress st for the following steel beam: length L = 3 m, height h ⫽ 300 mm, modulus of elasticity E = 200 GPa, and moment of inertia I ⫽ 36 * 106 mm4. Also, the load on the beam has intensity q = 25 kN/m.

A

B h L

q A

B

H

H

h L

Compare the tensile stress st produced by the axial forces with the maximum bending stress sb produced by the uniform load.

Solution 10.6-2 Beam with uniform load (b) IMMOVABLE SUPPORTS

h ⫽ height of beam (a) CURVATURE SHORTENING From Case 1, Table G-2: q d␯ ⫽⫺ (L3 ⫺ 6Lx 2 ⫺ 4x 3 ) dx 24EI

EAl L

Eq. (10-46): st ⫽

17q 2L6 H El ⫽ ⫽ A L 40,320EI 2

(c) NUMERICAL VALUES

L

Eq. (10-42): l ⫽

Eq. (10-45): H ⫽

1 d␯ 2 a b dx 2 L0 dx

L=3m

17q 2L7

⫽

40,320E 2I 2

I ⫽ 36 * 10 mm

sb ⫽

qL2 8

sb ⫽ 117.2 MPa

4

E ⫽ 200 GPa

l ⫽ 0.01112 mm st ⫽ 0.7411 MPa

; ;

The bending stress is much larger than the axial tensile stress due to curvature shortening.

BENDING STRESS M max ⫽

;

q = 25 kN/m

h ⫽300 mm 6

;

c⫽

qhL2 Mc ⫽ I 16I

h 2 ;

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11 Columns

Idealized Buckling Models P

Problem 11.2-1

The figure shows an idealized structure consisting of one or more rigid bars with pinned connections and linearly elastic springs. Rotational stiffness is denoted bR, and translational stiffness is denoted b. Determine the critical load Pcr for the structure.

B L bR A

Solution 11.2-1

Rigid bar AB gM A ⫽ 0 P(uL) ⫺ b R u ⫽ 0 Pcr ⫽

bR L

;

845

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CHAPTER 11 Columns

Problem 11.2-2 The figure shows an idealized structure consisting of one or more rigid bars with pinned connections and linearly elastic springs. Rotational stiffness is denoted bR, and translational stiffness is denoted b. (a) Determine the critical load Pcr for the structure from figure part (a). (b) Find Pcr if another rotational spring is added at B from figure part (b).

P

P

C

C b

B L

bR

L Elastic supports

a

b

B

Elastic supports

a

bR

bR

A

A (a)

(b)

Solution 11.2-2 (a) g M A ⫽ 0

(b) g M A ⫽ 0

PuL ⫺ bua ⫺ b Ru ⫽ 0 2

PuL ⫺ bua 2 ⫺ 2b Ru ⫽ 0

2

Pcr ⫽

ba + b R L

;

Problem 11.2-3 The figure shows an idealized structure consisting of one or more rigid bars with pinned connections and linearly elastic springs. Rotational stiffness is denoted b R and translational stiffiness is denoted b . Determine the critical load Pcr for the structure.

Pcr ⫽

ba 2 + 2b R L

;

P C L — 2 B

bR A

L — 2

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SECTION 11.2 Idealized Buckling Models

Solution 11.2-3

847

Two rigid bars with a pin connection

L g M B ⫽ 0 M B + M C ⫺ Pua b ⫽ 0 2

g M A ⫽ 0 Shows that there are no horizontal reactions at the supports.

b R(2u) + b Ru ⫽ FREE-BODY DIAGRAM OF BAR BC M C ⫽ b Ru

Pcr ⫽

M B ⫽ b R (2u)

6b R L

Problem 11.2-4 The figure shows an idealized structure consisting of bars AB and BC which are connected using a hinge at B and linearly elastic springs at A and B. Rotational stiffness is denoted bR and translational stiffness is denoted b. (a) Determine the critical load Pcr for the structure from figure part (a). (b) Find Pcr if an elastic connection is now used to connect bar segments AB and BC from figure part (b).

PLu 2

;

P

P

C

C

B

Hinge b

L Elastic support

a

L/2 bR B b

Hinge L/2

bR

bR A

A (a)

Elastic connection

(b)

Elastic support

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CHAPTER 11 Columns

Solution 11.2-4 (a) g M A ⫽ 0

(b) g M A ⫽ 0 L 2 bu a b + b Ru ⫺ HL ⫽ 0 2

bua 2 + b R u ⫺ HL ⫽ 0 H⫽

bua 2 + b Ru L

H⫽

buL2 ⫹4b Ru 4L

P

P H

C

H

C

B

F = bua

F = bu L 2

B

u u M = bRu A

M = bRu

A

P

FREE-BODY DIAGRAM OF BAR BC

FREE-BODY DIAGRAM OF BAR BC

P

P

C

H

H

C

u B

B

H

P

M B ⫽ b R(2u)

P

g MB ⫽ 0

H(L ⫺ a) ⫺ P(ua) ⫽ 0

1ba + b R2(L ⫺ a) 2

Pcr ⫽

aL

u ( L2)

;

g MB ⫽ 0

L L H a b ⫺ Pa u b + M B ⫽ 0 2 2

buL2 + 4b Ru L a b 4L 2 L ⫺ Pau b + b R(2u) ⫽ 0 2 Pcr ⫽

bL2 + 20b R 4L

;

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SECTION 11.2 Idealized Buckling Models

Problem 11.2-5 The figure shows an idealized structure consisting of two rigid bars joined by an elastic connection with rotational stiffness b R. Determine the critical load Pcr for the structure.

849

Elastic connection A

B L/2

bR

C

D

L/2

P

L

Solution 11.2-5

θ

C B

D

θL

P

2θ

A

V

©M B ⫽ 0

V⫽0

FREE-BODY DIAGRAM OF BAR CD Moment in elastic connection ⫽ b R * total relative rotation (u ⫹2u)

MC

M C ⫽ b R(3u)

P P

©M C ⫽ 0 PuL ⫺ b R(3u) ⫽ 0 Pcr ⫽

Problem 11.2-6 The figure shows an idealized structure consisting of rigid bars ABC and DEF joined by linearly elastic spring b between C and D. The structure is also supported by translational elastic support b at B and rotational elastic support at b R at E. Determine the critical load Pcr for the structure.

3b R L

A

;

B L/2

Elastic support

C b

L/2 P

b D

bR = (2/5) bL2 L/2 E

F

L/2

Elastic support

Solution 11.2-6

P

θ

RB

∆C

θ'

FREE-BODY DIAGRAM OF DEFORMED STRUCTURE

∆D

ME

œ

M E ⫽ b Ru ⫽

¢ D ⫽ Lu 2 bL2u 5

RB ⫽ b au

œ

L b 2

FREE-BODY DIAGRAM OF MEMBER ABC

RB

ΔC

¢ C ⫽ Lu

θ'

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β (ΔD – ΔC )

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CHAPTER 11 Columns

©M A ⫽ 0 b au

œ

L RB ⫺ b(¢ D ⫺ ¢ C)L ⫽ 0 2

œ L L b ⫺ b(Lu⫺ Lu )L ⫽ 0 2 2

FREE-BODY DIAGRAM OF MEMBER DEF

gMF ⫽ 0

P¢ D ⫺ b1¢ D ⫺ ¢ C2L ⫺ M E ⫽ 0 4 P(Lu) ⫺ b cLu ⫺ L a ub d 5

L⫺

θ

ΔD

β (ΔD – ΔC ) P

ME

4 u 5

œ

u ⫽

2 bL2u ⫽ 0 5

Pcr ⫽

3 bL 5

;

Problem 11.2-7 The figure shown an idealized structure consisting of an L-shaped rigid bar structure supported by linearly elastic springs at A and C. Rotational stiffness in denoted b R and translational stiffness is denoted b . Determine the critical load Pcr for the structure.

P C

B

L/2

L Elastic support

A

Solution 11.2-7 FREE-BODY DIAGRAM OF DEFORMED STRUCTURE P θ θ L/2

θL

βθL/2 θ

3 M A ⫽ b Ru ⫽ bL2u 2 g MA ⫽ 0 L 2 M A ⫺ P(uL) + bu a b ⫽ 0 2 3 2 L 2 bL u ⫺ P(uL) + bu a b ⫽ 0 2 2 7 Pcr ⫽ bL 2

MA

;

bR = 3bL2/2

b

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851

SECTION 11.3 Critical Loads of Columns with Pinned Supports

Critical Loads of Columns with Pinned Supports The problems for Section 11.3 are to be solved using the assumptions of ideal, slender, prismatic, linearly elastic columns (Euler buckling). Buckling occurs in the plane of the figure unless stated otherwise.

2

Problem 11.3-1

Calculate the critical load Pcr for a W 8 * 35 steel column (see figure) having length L ⫽ 24 ft and E ⫽ 30 * 106 psi under the following conditions:

C

1

1

(a) The column buckles by bending about it’s strong axis (axis 1-1), and (b) the column buckles by bending about its weak axis (axis 2-2). In both cases, assume that the column has pinned ends. 2 Problem 11.3.1-3.3

Solution 11.3-1 Column with pinned supports (b) BUCKLING ABOUT WEAK AXIS

W 8 * 35 steel column L ⫽ 24 ft ⫽ 288 in. E ⫽ 30 * 10 psi 6

I1 ⫽ 127 in.4

I2 ⫽ 42.6 in.4

Pcr ⫽

A ⫽ 10.3 in.2

p2EI1 L2

⫽ 453 k

L2

⫽ 152 k

NOTE: scr ⫽

(a) BUCKLING ABOUT STRONG AXIS Pcr ⫽

p2EI2

;

Pcr 453 k ⫽ 44 ksi ⫽ A 10.3 in.2

‹ Solution is satisfactory if sPL Ú 44 ksi

;

Solve the preceding problem for a W 250 ⫻ 89 steel column having length L ⫽ 10 m. Let E ⫽ 200 GPa

Problem 11.3-2

Solution 11.3-2 BUCKLING ABOUT WEAK AXIS

W 250 * 89 E ⫽ 200 GPa

L ⫽ 10.0 m

I2 ⫽ 48.3 * 106 mm4

I1 ⫽ 142 * 10 mm 6

A ⫽ 11400 mm2

p2EI1 L2

Pcr1 ⫽ 2803 kN

Pcr 2 ⫽

p2EI2 L2

Note: scr ⫽

BUCKLING ABOUT STRONG AXIS Pcr1 ⫽

4

;

Problem 11.3-3 Solve Problem 11.3-1 for a W 10 * 45 steel column having length L ⫽ 28 ft.

Pcr1 A

Pcr 2 ⫽ 953 kN

;

scr ⫽ 246 MPa

Solution is satisfactory if sPL Ú 246 MPa

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CHAPTER 11 Columns

Solution 11.3-3 Column with pinned supports (b) BUCKLING ABOUT WEAK AXIS

W 10 * 45 steel column L ⫽ 28 ft ⫽ 336 in. E ⫽ 30 * 10 psi 6

I1 ⫽ 248 in.4

I2 ⫽ 53.4 in.4

A ⫽ 13.3 in.2

p2EI1 L2

⫽ 650 k

p2EI2 L2

⫽ 140 k

NOTE: scr ⫽

(a) BUCKLING ABOUT STRONG AXIS Pcr ⫽

Pcr ⫽

;

;

Pcr 650 k ⫽ ⫽ 49 ksi A 13.3 in.2

‹ Solution is satisfactory if sPL Ú 49 ksi

Problem 11.3-4 A horizontal beam AB is pin-supported at end A and carries a load Q at end B, as shown in the figure. The beam is supported at C by a pinned-end column of length L; the column is restrained laterally at 6.0L from the base at D. Assume the column can only buckle in the plane of the frame. The column is a solid steel bar (E ⫽ 200 GPa) of square cross section having length L ⫽ 2.4 m side dimensions b ⫽ 70 mm. Let dimensions d ⫽ L/2. Based upon the critical load of the column, determine the allowable moment M if the factor of safety with respect to buckling is n ⫽ 2.0.

A

C

B

d

2d Q

L

Solution 11.3-4 COLUMN CD (STEEL) E ⫽ 200 GPa L ⫽ 2.4 m d ⫽ Square cross section: Factor of safety: I⫽

4

b 12

L 2

b ⫽ 70 mm

n ⫽ 2.0

I ⫽ 2.00 * 106 mm4

Pcr ⫽ d ⫽ 1.2 m

p2EI

Pcr ⫽ 1905 kN (0.6 L)2 Pcr Pallow ⫽ 952.3 kN Pallow ⫽ n

Beam ACB

©M A ⫽ 0 M ⫽ Pd

M allow ⫽ Pallow d

Problem 11.3-5 A horizontal beam AB is pin-supported at end A and carries a load Q at joint B, as shown in the figure. The beam is also supported at C by a pinned-end column of length L; the column is restrained laterally at 0.6L from the base at D. Assume the column can only buckle in the plane of the frame. The column is a solid aluminium bar (E ⫽ 10 * 106 psi) of square cross section having length L ⫽ 30 in. and side dimensions b ⫽ 1.5 in. Let dimension d ⫽ L/2. Based upon the critical load of the column, determine the allowable force Q if the factor of safety with respect to buckling is n ⫽ 1.8.

M allow ⫽ 1143 kN # m

A

;

B

C d

2d Q

L 0.6 L D

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853

SECTION 11.3 Critical Loads of Columns with Pinned Supports

Solution 11.3-5 COLUMN CD (STEEL)

d⫽

L 2

Factor of safety: I⫽

b 12

p2EI

Pcr ⫽ 129 k (0.6L)2 Pcr Pallow ⫽ Pallow ⫽ 71 k n

d ⫽ 15 in.

Square cross section: 4

Pcr ⫽

L ⫽ 30 in.

E ⫽ 10 * 10 psi 6

b ⫽ 1.5 in.

Beam ACB

n ⫽ 1.8 Q allow ⫽

I ⫽ 0.422 in.4

©M A ⫽ 0

Pallow 3

Q⫽

P 3

Q allow ⫽ 23.8 k

Problem 11.3-6 A horizontal beam AB is pin-supported at end A and carries a load Q at joind B, as shown in the figure part (a). The beam is also supported at C by a pinned-end column of length L. The column has flexural rigidity EI. (a) For the case of a guided support at A (figure pare (a)), what is the critical load Qcr? (In other words, at what load Qcr? does the system collapse because of Euler buckling of the column DC?) (b) Repeat (a) if the guided support at A is replaced by column AF with length 3/L2 and flexural rigidity EI (see figure part (b)).

;

F

A

C

3L — 2

B

d

2d Q

L

C

B

A d

D

2d Q L

(a)

D

(b)

Solution 11.3-6 (a) Pcr ⫽

p2EI

(b) Q cr BASED UPON Pcr

2

L

FROM FREE-BODY DIAGRAM OF THE SYSTEM ©Fy ⫽ 0

Q⫽P

Therefore Q cr ⫽ Pcr ⫽

p2EI L2

;

Pcr, AF ⫽

p2EI a

3L b 2

2

⫽

IN COLUMN

4p2EI 9L2

AF

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CHAPTER 11 Columns

FROM FREE-BODY DIAGRAM OF BEAM ACB

FROM FREE-BODY DIAGRAM OF BEAM ACB ©M C ⫽ 0

Q⫽

PAF 2

Q cr, AF ⫽

therefore

©M A ⫽ 0

Pcr, AF 2

⫽

2p2EI

therefore Q cr, CD ⫽

9L2

Q cr based upon Pcr in column CD p2EI

Pcr, CD ⫽

Q cr ⫽ Q cr, AF ⫽

;

L2

PCD 3

Q⫽

Pcr, CD 2

2p2EI

⫽

p2EI 3L2

;

9L2

Column AF governs

Problem 11.3-7 A horizontal beam AB has a guided support at end A and carries a load Q at end B, as show in the figure part (a). The beam is supported at C and D by two identical pinned-end columns of length L. Each column has flexural rigidity EI.

A

C

D

d

B

d

C d

2d

L

(a) Find an expression for the critical load Qcr. (In other words, at what load Qcr does the system collapses because of Euler buckling of the columns?) (b) Repeats (a) but assume spin support at A. Find an expression for the critical moment Mcr (i.e., find the moment M at B at which the system collapses because of Euler buckling of the columns).

A

Q

L

D d

2d

L

(a)

B M

L

(b)

Solution 11.3-7 Pcr ⫽

(b) COLLAPSE

p2EI L

(a) COLLAPSE

OCCURS WHEN BOTH COLUMNS REACH THE

CRITICAL LOAD.

A

C d

OCCURS WHEN BOTH COLUMNS REACH THE

CRITICAL LOAD.

2

B

g Fy ⫽ 0 Thus Q cr ⫽

2d Qcr

Pcr Pcr

C d

D d

A

D d

B 2d Mcr

Q cr ⫽ 2Pcr 2p2EI L2

;

Pcr Pcr

g MA ⫽ 0 M cr ⫽ (2d)Pcr + (d)Pcr ⫽ 3dPcr Thus M cr ⫽

3dp2EI L2

;

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855

SECTION 11.3 Critical Loads of Columns with Pinned Supports

Problem 11.3-8 A slender bar AB with pinned ends and length L is held between immovable supports (see figure). What increase ⌬T in the temperature of the bar will produce bucking at the Euler load?

∆T

A

B

L

Solution 11.3-8 Bar with immovable pin supports L ⫽ length A ⫽ cross-sectional area E ⫽ modules of elasticity I ⫽ moment of inertia a ⫽ coefficient of increase in temperature ¢T ⫽ uniform increase in temperature

INCREASE IN TEMPERATURE TO PRODUCE BUCKLING

AXIAL COMPRESSIVE FORCE IN BAR (EQ. 2-17)

P ⫽ Pcr

EULER LOAD Pcr ⫽

p2EI L2

EAa (¢T) ⫽

P ⫽ EAa (¢T)

p2EI L2

¢T ⫽

p2I aAL2

;

P

Problem 11.3-9

A rectangular column with cross-sectional dimensions b and h is pin-supported at ends A and C (see figure). AT midheight, the column is restrained in the plane of the figure but is free to deflect perpendicular to the plane of the figure. Determine the ratio h/b such that the critical load is the same for buckling in the two principal planes of the column.

C

X

X

L — 2

b h

B L — 2

b

Section X-X

A

Solution 11.3-9 Column with restraint at midheight Critical loads for buckling about axes 1-1 and 2-2: P1 ⫽

p2EI1 2

L

P2 ⫽

p2EI2 2

(L/2)

⫽

FOR EQUAL CRITICAL LOADS P1 ⫽ P2

‹ I1 ⫽ 4I2

4p2EI2 L2

I1 ⫽

bh3 12

bh3 ⫽ 4hb 3

I2 ⫽

hb 3 12

h ⫽2 b

;

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CHAPTER 11 Columns

Problem 11.3-10

Three identical, solid circular rods, each of radius r and length L, are placed together to from a compression member (see the cross section shown in the figure). Assuming pinned-end conditions, determine the critical load Pcr as follows: (a) The rods act independently as individual columns, and (b) the rods are bonded by epoxy throughout their lengths so that they function as a single member. What is the effect on the critical load when the rods act as a single member?

2r

Solution 11.3-10 Three solid circular rods (b) RODS ARE BONDED TOGETHER The x and y axes have their origin at the centroid of the cross section. Because there are three different centroidal axes of symmetry, all centroidal axes are principal axes and all centroidal moments of inertia are equal (see Section 12.9). From Case 9, Appendix D: R= Radius

I ⫽ IY ⫽

L= Length

Pcr ⫽

(A) RODS ACT INDEPENDENTLY Pcr ⫽ Pcr ⫽

p2EI L2

pr 4 4

(3) I ⫽

3p3Er 4

pr 4 5pr 4 11pr 4 + 2a b ⫽ 4 4 4

p2EI 2

L

⫽

11p3Er 4

;

4L2

NOTE: Joining the rods so that they act as a single member increases the critical load by a factor of 11/3, ; or 3.67.

;

4L2

Problem 11.3-11

Three pinned-end columns of the same material have the same length and the same cross-sectional area (see figure). The columns are free to buckle in any direction. The columns have cross section as follows: (1) a circle, (2) a square, and (3) an equilateral triangle. Determine the ratios P1 : P2 : P{3} of the critical loads for these columns.

(1)

(2)

(3)

Solution 11.3-11 Three pinned-end columns E,L and A are the same for all three columns. Pcr ⫽

2

p EI L2

‹ P1 : P2 : P3 ⫽ I1 : I2 : I3

(1) CIRCLE Case 9, Appendix D I⫽

pd 4 64

(2) SQUARE I⫽

b4 12

A⫽

pd 2 4

‹ I1 ⫽

Case 1, Appendix D A ⫽ b2

‹ I2 ⫽

A2 12

A2 4p

(3) EQUILATERAL TRIANGLE Case 5, Appendix D I⫽

b 4 13 96

A⫽

b 2 13 4

‹ I3 ⫽

A2 13 18

P1 : P2 : P3 ⫽ I1 : I2 : I3 ⫽ 1 :

p 2p13 : 3 9

⫽ 1.000 : 1.047 : 1.209

;

NOTE: For each of the above cross sections, every centroidal axis has the same moment of inertia (see Section 12.9)

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857

SECTION 11.3 Critical Loads of Columns with Pinned Supports

Problem 11.3-12

A long slender column ABC is pinned at ends A and C and compressed by an axial force P (see figure). At the midpoint B, lateral support is provided to prevent deflection in the plane of the figure. The column is a steel wide-flange section (W 250 * 67) with E = 200 GPa. The distance between lateral supports is L = 5.5 m. Calculate the allowable load P using a factor of safety n = 2.4, taking into account the possibility of Euler buckling about either principal centroidal axis (i.e., axis 1-1 or axis 2-2).

P 2 C

X

W 250 ⫻ 67

L

X

1

1

B L 2 Section X - X

A

Solution 11.3-12 W 250 * 67 E ⫽ 200 GPa L ⫽ 5.5 m

BUCKLING ABOUT AXIS 2-2

I1 ⫽ 103 * 10 mm 6

I2 ⫽ 22.2 * 106 mm4

n ⫽ 2.4

BUCKLING ABOUT AXIS 1-1 Pcr1 ⫽

p2EI1 (2L)2

Pcr1 ⫽ 1680 kN

4

Pcr 2 ⫽

p2EI2 (L)2

Pcr ⫽ Pcr 2

Pcr 2 ⫽ 1449 kN axis 2-2 governs

ALLOWABLE LOAD Pallow ⫽

Pcr n

Pallow ⫽ 604 kN

;

Problem 11.3-13

The roof over a concourse at an airport is supported by the use of pretensioned cables. At a typical joint in the roof structure, a strut AB is compressed by the action of tensile forces F in a cable that makes an angle a ⫽ 75˚ with the strut (see figure and photo). The strut is a circular tube of steel (E ⫽ 30,000 ksi) with outer diameter d2 ⫽ 2.5 in. and inner diameter d1 ⫽ 2.0 in. The strut is 5.75 ft long and is assumed to be pin-connected at both ends. Using a factor of safety n ⫽ 2.5 with respect to the critical load, determine the allowable force F in the cable.

F A

d2 a

Strut

a

B Cable

F

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CHAPTER 11 Columns

Solution 11.3-13 d2 ⫽ 2.5 in.

E ⫽ 30000 ksi

d1 ⫽ 2.0 in.

L ⫽ 5.75 ft n ⫽ 2.5 a ⫽ 75° I⫽

p 1d 24 ⫺ d 142 64

Pcr ⫽

p2EI

I ⫽ 1.132 in.4

Pcr ⫽ 70.40 k

L2

ALLOWABLE LOAD Pallow ⫽

Pcr n

Pallow ⫽ 28.16 k

EQUILIBRIUM OF JOINT B P ⫽ 2F cos (a) Thus

Fallow ⫽

Pallow 2 cos (a)

Problem 11.3-14

The hoisting arrangement for lifting a large pipe is shown in the figure. The spreader is a steel tubular section with outer diameter 70 mm and inner diameter 57 mm. Its length is 2.6 m and its modulus of elasticity is 200 GPa. Based upon a factor of safety of 2.25 with respect ot Euler buckling of the spreader, what is the maximum weight of pipe that can be lifted?(Assume pinned conditions at the ends of the spreader.)

Fallow ⫽ 54.4 k

;

F Cable 7

7

10

10 A

Spreader

B Cable

Pipe

Solution 11.3-14 d2 ⫽ 70 mm

E ⫽ 30000 ksi d1 ⫽ 57 mm n ⫽ 2.25 I⫽

L ⫽ 2.6 m a ⫽ atan a

p 1d 4 ⫺ d1 42 64 2

Pcr ⫽

p2EI L2

7 b 10

I ⫽ 660.4 * 103 mm4

Pcr ⫽ 199 kN

ALLOWABLE LOAD Pallow ⫽

Pcr n

Pallow ⫽ 88.6 kN

EQUILIBRIUM OF JOINT A ©Fhoriz ⫽ 0

⫺P + T cos(a) ⫽ 0 w ©Fvert ⫽ 0 Tsin(a) ⫺ ⫽ 0 2

SOLVE THE EQUATION W ⫽ 2P tan(a) MAXIMUM WEIGHT OF PIPE Wmax ⫽ 2Pallow tan(a)

Wmax ⫽ 124 kN

;

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SECTION 11.3 Critical Loads of Columns with Pinned Supports

A pinned-end strut of aluminium (E ⫽ 10,400 ksi) with length L ⫽ 6 ft is constructed of circular tubing with outside diameter d ⫽ 2 in. (see figure) The strut must resist an axial load P ⫽ 4 kips with a factor of safety n ⫽ 2.0 with respect to the critical load. Determine the required thickness t of the tube.

Problem 11.3-15

t

d = 50 mm

Solution 11.3-15 L ⫽ 6 ft

E ⫽ 10400 ksi d ⫽ 2 in.

n ⫽ 2.0

Pcr ⫽ nP

Pcr ⫽ 8.0 k

Pcr ⫽

2

p EI L2

I⫽

MOMENT OF INERTIA p [d 4 ⫺ (d ⫺ 2t)4] I⫽ 64

P⫽4k

PcrL2 p2E

I ⫽ 0.404 in.

4

d 4 ⫺ (d ⫺ 2t)4 ⫽ I

64 p

t min ⫽ 0.165 in.

Problem 11.3-16

The cross section of a column built up of two steel I-beams (S 150 ⫻ 25.7 sections) is shown in the figure. The beams are connected by spacer bars, or lacing, to ensure that they act together as a single column. (The lacing is represented by dashed lines in the figure.) The column is assumed to have pinned ends and may buckle in any direction. Assuming E = 200 GPa and L = 8.5 m, calculate the critical load Pcr for the column.

S 6 ⫻ 17.25

4 in.

Solution 11.3-16 E ⫽ 200 GPa

S 150 * 25.7 L ⫽ 8.5 m

Buckling occurs about the y axis since

I1 ⫽ 10.9 * 10 mm 6

4

Critical load

I2 ⫽ 0.953 * 10 mm 6

A ⫽ 3260 mm2

4

d⫽

100 mm 2

COMPOSITE COLUMN Ix ⫽ 2I1

Ix ⫽ 21.80 * 106 mm4

Iy ⫽ 21I2 + Ad 22 Iy ⫽ 18.21 * 106 mm4

Pcr ⫽

p2EIy L2

Pcr ⫽ 497 kN

;

Iy 6 Ix

859

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CHAPTER 11 Columns

Problem 11.3-17

The truss ABC shown in the figure supports a vertical load W at joint B. Each member is a slender circular steel pipe (E = 30,000 ksi) with outside diameter 4 in. And wall thickness 0.25 in. The distance between supports is 23 ft. Joint B is restrained against displacement perpendicular to the plane of the truss. Determine the critical value Wcr of the load.

B 100 mm W 40°

55°

A

C

7.0 m

Solution 11.3-17 E ⫽ 30000 ksi

L ⫽ 23 ft

d2 ⫽ 4 in.

t ⫽ 0.25 in.

d1 ⫽ d2 ⫺ 2t I⫽

d1 ⫽ 3.50 in.

p 1d 24 ⫺ d 142 64

u1 ⫽ 40° L AB ⫽ L P

I ⫽ 5.200 in.4 u2 ⫽ 55°

sin au2 b sin(180° ⫺ u1 ⫺ u2)

L AB ⫽ 18.912 ft L BC ⫽ L P

FREE-BODY DIAGRAM OF JOINT B

sin au1 b sin(180° ⫺ u1 ⫺ u2)

L BC ⫽ 14.841 ft Critical loads Pcr– AB ⫽ Pcr– BC ⫽

Problem 11.3-18

p2EI L 2AB 2

p EI L 2BC

©Fhoriz ⫽ 0 FAB cos1u12 ⫺ FBC cos1u22 ⫽ 0

©Fhoriz ⫽ 0 FAB sin1u12 ⫺ FBC sin1u22 ⫺ W ⫽ 0

Q

SOLVE THE TWO EQUATIONS W ⫽ 1.3004 FBC

W ⫽ 1.7361FAB

CRITICAL VALUE OF THE LOAD W BASED ON MEMBER AB:

Q

Wcr– AB ⫽ 1.7361Pcr–AB

Wcr– AB ⫽ 51.90 k

BASED ON MEMBER BC: Pcr– AB ⫽ 29.89 k Pcr–BC ⫽ 48.55 k

A truss ABC supports a load W at joint B, as shown in the figure. The length L1 of member AB is fixed, but the length of strut BC varies as the angle u is changed. Strut BC has a solid circular cross section. Joint B restrained against displacement perpendicular to the plane of the truss. Assuming the collapse occurs by Euler buckling of the strut, determine the angle u for minimum weight of the strut.

Wcr–BC ⫽ 1.3004Pcr⇣– BC

Wcr– BC ⫽ 63.13 k

Wcr ⫽ min(Wcr– AB,Wcr– BC) Wcr ⫽ 51.9 k

;

Member AB governs

A

B u W

C L1

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SECTION 11.3 Critical Loads of Columns with Pinned Supports

861

Solution 11.3-18 Truss ABC (minimum weight) LENGTHS OF MEMBERS

All the terms are constants except cos u and sin u Therefore, we can write VS in the following form:

L AB ⫽ L 1 (a constant) L BC ⫽

VS ⫽

L1 (angle u is variable) cos u

Strut BC may buckle.

k 2

where k is a constant.

cos u2sin u

GRAPH OF

Vs k

FREE-BODY DIGRAM OF JOINT B

a Fvert ⫽ 0

FBC sin u ⫺ W ⫽ 0

STRUT BC (SOLID CIRCULAR BAR) A⫽

pd 2 4

Pcr ⫽

p 2EI L 2BC

I⫽ ⫽

FBC ⫽ Pcr

pd 4 64

‹ I⫽

A2 4p

pEA2 cos2u 4 L 21 or

Solve for area A: A ⫽

W pEA2 cos2 u ⫽ sin u 4 L 21 1/2 2L 1 W a b cos u pE sin u

For minimum weight, the volume VS of the strut must be a minimum. Vs ⫽ AL BC ⫽

1/2 AL 1 2L 21 W a ⫽ b 2 cos u cos u pE sin u

umin ⫽ angle for minimum volume (and minimum weight) For minimum weight, the term cos2u2sin u must be a a maximum. For minimum value, the derivative with respect to u equals zero. Therefore,

d A cos2u2sin u B ⫽ 0 du

Taking the derivative and simplifying, we get cos2 u ⫺ 4 sin2 u ⫽ 0 or 1 ⫺ 4 tan2 u ⫽ 0 and tan u ⫽ ‹ umin ⫽ arctan

1 ⫽ 26.57° 2

;

1 2

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CHAPTER 11 Columns

Problem 11.3-19

An S 6 * 12.5 steel cantilever beam AB is supported by a steel tie rod at B has shown. The tie rod is just taut when a roller support is added at C at a distances S to the left of B, then the distributed load q is applied to beam segment AC. Assume E ⫽ 30 * 106 psi and neglect the self weight of the beam and tie rod. See Table E-2(a) in Appendix E for the properties of the S-Shape beam.

D H

q A

Tie rod, diameter d

S 6 ⫻ 12.5 C

(a) What value of uniform load q will, if exceeded, result in buckling of the tie rod if L1 =6 ft, S = 2 ft, H = 3 ft, d = 0.25 in.? (b) What minimum beam moment of inertia Ib is required to prevent buckling of the tie rod if q ⫽ 200 lb/ft, L 1 ⫽ 6 ft, H ⫽ 3 ft, d ⫽ 0.25 in., S ⫽ 2 ft? (c) For what distances S will the tie rod be just on the verge of buckling if q ⫽ 200 lb/ft, L 1 ⫽ 6 ft, H ⫽ 3 ft, d ⫽ 0.25 in.?

B S

L1

Solution 11.3-19 E ⫽ 30 * 106 psi

L 1 ⫽ 6 ft

q ⫽ 200

H ⫽ 3 ft

Q A

lb ft

Ir ⫽

p 4 d 64

Ir ⫽ 191.7 * 10⫺6 in.4

Ar ⫽

p 2 d 4

Ar ⫽ 0.049 in.2

Pcr ⫽

1 in. 4

d⫽

S L1 = 6 ft

FROM APPENDIX G œœ

p2EIr

dB ⫽

Pcr ⫽ 43.81 lb

H2

B

Qs 3 Qs(L 1 ⫺ s)s + 3EIb 3EIb

S 6 * 12.5 Ib ⫽ 22.0 in.4

Shortening in tie rod

Analyze 1st degree statically-indeterminate beam by letting force in tie rod be a redundant Released beam AB with the uniform load q

Compatibility equation q1L 1 ⫺ s23

⫽

A

B S L1 = 6 ft

Q⫽

⫺ c

Qs1L 1 ⫺ s2s Qs 3 + d 3EIb 3EIb

sqAr1L 1 ⫺ s23

81s 2ArL 1 + 3HIb2

(1)

Q ⫽ Pcr s ⫽ 2.0 ft

From (1)

qmax ⫽

q1L 1 ⫺ s2

3

dB ⫽

œœ

dB ⫺ dB ⫽ dr

QH EAr

(a) FOR Released beam AB with the redundant Q From appendix G

QH EAr

œ

24EIb

q

dr ⫽

8Q

sAr1L 1 ⫺ s23

* 1s 2ArL 1 + 3HIb2

24EIb qmax ⫽ 142.4

lb ft

;

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863

SECTION 11.4 Columns with Other Support Conditions

(b) FOR

q  200

lb ft

Ib– min 

From (1)

1 8QsL 1  qL 31 + 3sqL 1 2  3s 2qL 1 + s 3q sAr 24 QH

l b–min  38.5 in.4 (c) FROM (1)

;

NUMERICALLY SOLVE FOR S WHEN

s  0.264 ft

and s  2.42 ft

Q  Pcr : 2 SOLUTIONS ARE POSSIBLE:

;

Columns with Other Support Conditions The problems for Section 11.4 are to be solved using the assumptions of ideal, slender, prismatic, linearly elastic columns (Euler buckling). Buckling occurs in the plane of the figure unless stated otherwise.

Problem 11.4-1 An aluminum pipe column (E  10,400 ksi) with length L  10.0 ft has inside and outside diameters d1  5.0 in. and d2  6.0 in., respectively (see figure). The column is supported only at the ends and may buckle in any direction. Calculate the critical load Pcr for the following end conditions: (1) pinned-pinned, (2) fixed-free, (3) fixed-pinned, and (4) fixed-fixed.

Solution 11.4-1 d2  6.0 in.

Probs.11.4-1 and 11.4-2

Aluminum pipe column d1  5.0 in.

E  10,400 ksi

p (d 4  d 14 )  32.94 in.4 I 64 2 L  10.0 ft  120 in.

(2) FIXED-FREE

Pcr 

(3) FIXED-PINNED

Pcr 

(4) FIXED-FIXED

Pcr 

(1) PINNED-PINNED Pcr 

p2EI L

2

d2

d1



p2(10,400 ksi) (32.94 in.4)

 235 k

(120 in.)

2

;

Problem 11.4-2 Solve the preceding problem for a steel pipe column (E  210 GPa) with length L  1.2 m, inner diameter d1  36 mm, and outer diameter d2  40 mm.

p2EI 4L2

 58.7 k

2.046p2EI L2 4p2EI L2

;

 480 k

 939 k

;

;

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CHAPTER 11 Columns

Solution 11.4-2

Steel pipe column

d2  40 mm d1  36 mm E  210 GPa I

p 4 (d  d 14)  43.22 * 103 mm4 64 2

(1) PINNED-PINNED

Pcr 

p2 EI 2

L

(2) FIXED-FREE

Pcr 

(3) FIXED-PINNED

Pcr 

(4) FIXED-FIXED

Pcr 

p2 EI 4L2

L  1.2 m

 62.2 kN

;

 15.6 kN

2.046p2EI L2 4p2EI

A wide-flange steel column (E  30  106 psi) of W 12  87 shape (see figure) has length L  28 ft. It is supported only at the ends and may buckle in any direction. Calculate the allowable load Pallow based upon the critical load with a factor of safety n  2.5. Consider the following end conditions: (1) pinned-pinned, (2) fixed-free, (3) fixed-pinned, and (4) fixed-fixed.

L2

1

Pallow 

2.046p2EI2 nL2

 517 k

;

(1) PINNED-PINNED Pallow 

Pcr p2EI2  253 k  n nL2

(4) FIXED-FIXED ; Pallow 

(2) FIXED-FREE Pallow 

p2EI2 4nL2

2

(3) FIXED-PINNED

L  28 ft  336 in. n  2.5 I2  241 in.

 63.2 k

;

Problem 11.4-4 Solve the preceding problem for a W 250  89 shape with length L  7.5 m and L  200 GPa.

;

1

Wide-flange column 4

;

2

Probs. 11.4-3 and 11.4-4

W12 * 87 E  30 * 106 psi

 127 kN

 249 kN

Problem 11.4-3

Solution 11.4-3

;

4p2EI2 nL2

 1011 k

;

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SECTION 11.4 Columns with Other Support Conditions

865

Solution 11.4-4 E  200 GPa

W 250 * 89 L  7.5 m

(3) FIXED-PINNED

n  2.5

Pallow 

I2  48.3 * 106 mm4

p2EI2

;

(4) FIXED-FIED

2

nL

Pallow  678 kN

Pallow 

;

4p2EI2 nL2

Pallow  2712 kN

(2) FIXED-FREE Pallow 

nL2

Pallow  1387 kN

(1) PINNED-PINNED Pallow 

2.046p2EI2

;

p2EI2 4nL2

Pallow  169.5 kN

;

Problem 11.4-5 The upper end of a W 8 * 21 wide-flange steel column (E  30 * 103 ksi) is supported laterally between two pipes (see figure). The pipes are not attached to the column, and friction between the pipes and the column is unreliable. The base of the column provides a fixed support, and the column is 13 ft long. Determine the critical load for the column, considering Euler buckling in the plane of the web and also perpendicular to the plane of the web.

W 8  21

Solution 11.4-5 Wide-flange steel column W 8 * 21

E  30 * 103 ksi

L  13 ft  156 in. I2  9.77 in.4

AXIS 1-1 (FIXED-FREE)

I1  75.3 in.

4

Pcr 

p2 EI1 4L2

 229 k

AXIS 2-2 (FIXED-PINNED) Pcr 

2.046p2 EI2 L2

 243 k

Buckling about axis 1-1 governs. Pcr  229 k

;

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CHAPTER 11 Columns

Problem 11.4-6 A vertical post AB is embedded in a concrete foundation and held at the top by two cables (see figure). The post is a hollow steel tube with modulus of elasticity 200 GPa, outer diameter 40 mm, and thickness 5 mm. The cables are tightened equally by turnbuckles. If a factor of safety of 3.0 against Euler buckling in the plane of the figure is desired, what is the maximum allowable tensile force Tallow in the cables?

B

40 mm

Cable 2.1 m

Steel tube Turnbuckle A

2.0 m

Solution 11.4-6

Steel tube

E  200 GPa

d2  40 mm

d1  30 mm

2.0 m

FREE-BODY DIAGRAM OF JOINT B

L  2.1 m n  3.0 I

p 4 (d  d 41)  85,903 mm4 64 2

Buckling in the plane of the figure means fixed-pinned end conditions. Pcr 

2.046p2EI 2

 78.67 kN

L Pcr 78.67 kN Pallow    26.22 kN n 3.0 DIMENSIONS

T  tensile force in each cable Pallow  compressive force in tube EQUILIBRIUM g Fvert  0 Pallow  2T a

2.1 m b 0 2.9 m

ALLOWAPLE FORCE IN CABLES 1 2.9 m Tallow  (Pallow)a b a b  18.1 kN 2 2.1 m

;

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867

SECTION 11.4 Columns with Other Support Conditions

Problem 11.4-7 The horizontal beam ABC shown in the figure is supported by column BD and CE. The beam is prevented from moving horizontally by the pin support at end A. Each column is pinned at its upper end to the beam, but at the lower ends, support D is a guided support and support E is pinned. Both columns are solid steel bars (E  30 * 106 psi) of square cross section with width equal to 0.625 in. A load Q acts at distance a from column BD.

a

C

B

A

10 in.

40 in. 45 in.

35 in.

(a) If the distance a  12 in., what is the critical value Qcr of the load? (b) If the distance a can be varied between 0 and 40 in., what is the maximum possible value of Qcr? What is the corresponding value of the distance a?

Q

0.625 in.

0.625 in.

D E

Solution E  30 # 106 psi I

b4 12

(a) FIND Q cr if a  12 in.

b  0.625 in.

I  0.01272 in.4

Column BD

L  35 in.

Pcr– BD 

p2EI 4L2

Pcr– BD  768.4 lb Column CE

L  45 in.

Pcr–CE  1859 lb

Pcr–CE 

p2EI L2

The system collapses when both columns buckle. ©M A  0 Pcr– BD(10 in.) + Pcr– CE(50 in.)  Q cr(a + 10in.)  0 Pcr– BD(10 in.) + Pcr– CE(50 in.) a10 in. Q cr  4575 lb ;

Q cr 

(b) Q cr IS MAXIMUM WHEN a  0 in. Q cr 

Pcr– BD(10 in.) + Pcr–CE(50 in.) a + 10 in.

Q cr  10065 lb

Problem 11.4-8 The roof beams of a warehouse are supported by pipe columns (see figure on the next page) having outer diameter d2  100 mm and inner diameter d1  90 mm. The columns have length L  4.0 m, modulus E  210 GPa, and fixed supports at the base. Calculate the critical load Pcr of one of the columns using the following assumptions: (1) the upper end is pinned and the beam prevents horizontal displacement; (2) the upper end is fixed against rotation and the beam prevents horizontal displacement; (3) the upper end is pinned but the beam is free to move horizontally; and (4) the upper end is fixed against rotation but the beam is free to move horizontally.

;

Roof beam

Pipe column d2 L

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CHAPTER 11 Columns

Solution 11.4-8 Pipe column (with fixed base) E  210 GPa L  4.0 m d2  100 mm I 

p 4 (d  d 14)  1688 * 103 mm4 64 2

(3) UPPER END IS PINNED (BUT NO HORIZONTAL RESTRAINT)

d1  90 mm (1) UPPER END IS PINNED (WITH NO HORIZONTAL DISPLACEMENT)

Pcr 

2.046p2EI L2

 447 kN

Pcr 

p2EI 4L2

 54.7 kN

;

; (4) UPPER END IS GUIDED (NO ROTATION; NO HORIZONTAL RESTRAINT)

(2) UPPER END IS FIXED (WITH NO HORIZONTAL DISPLACEMENT)

Pcr 

4p2EI L2

 875 kN

;

The lower half of the column is in the same condition as Case (3) above. Pcr 

Problem 11.4-9

Determine the critical load Pcr and the equation of the buckled shape for an ideal column with ends fixed against rotation (see figure) by solving the differential equation of the deflection curve. (See also Fig. 11-17.)

p2EI 4(L/2)2



p2EI L2

 219 kN

;

P

B

L

A

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SECTION 11.4 Columns with Other Support Conditions

Solution 11.4-9

Fixed-end column

  deflection in the y direction

BUCKLING EQUATION

DIFFERENTIAL EQUATION (EQ.11-3) œœ

El  M  M 0  P 

P k2  EI

M0  + k 2  EI œœ

M0 P

‹ C2  

M0 P

  C1 k cos kx  C2 k sin kx B.C.

2 ¿(0)  0

‹ cos kL  1 k2  a

  C1 sin kx + C2 cos kx + 1 (0)  0

3 (L)  0 0 

M0 (1  cos kL) P

and kL  2p

CRITICAL LOAD

GENERAL SOLUTION

B.C.

B.C.

‹ C1  0

M0  (1  cos kx) P

Pcr 

2p 2 4p2 b  2 L L

4p2EI

;

L2

BUCKLED MODE SHAPE Let d  deflection at midpoint a x 

L b 2

M0 kL L a 1  cos b a b  d  2 P 2 kL p 2

‹ d

M0 (1  cos p) P

2M 0 P

M0 d  P 2

 

Problem 11.4-10 An aluminum tube AB of circular cross section has a guided support at the base and is pinned at the top to a horizontal beam supporting a load Q  200 kN (see figure). Determine the required thickness t of the tube if its outside diameter d is 200 mm and the desired factor of safety with respect to Euler buckling is n  3.0. (Assume E  72 GPa.)

P 4p2  2 EI L

2px d a1  cos b 2 L

;

Q  200 kN B

1.0 m

1.0 m

2.0 m d  200 mm

A

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CHAPTER 11 Columns

Solution 11.4-10 a  1.0 m

E  72 GPa

CRITICAL LOAD

n  3.0

L  2.0 m Q  200 kN

Pcr  P # n

d  200 mm

p2EI

Pcr 

FREE-BODY DIAGRAM OF THE BEAM

I

Pcr  1200 kN

4L2 4PcrL2

I  27.019 * 106 mm4

p2E

MOMENT OF INERTIA I

p cd 4  (d  2t)4 d 64 d

t min  ©M c  0 P  2Q

(a) By solving the differential equation of the deflection curve, derive the following buckling equation for this column: b RL (kL cot kL  1)  k 2L2  0 EI in which L is the length of the column and EI is its flexural rigidity. (b) For the particular case when member BC is identical to member AB, the rotation stiffness b R equals 3EI/L (see Case 7, Table G-2, Appendix G). For this special case, determine the critical load Pcr.

A

d4  I

64 p

2

t min  10.0 mm

P  400 kN

Problem 11.4-11 The frame ABC consists of two members AB and BC that are rigidly connected at joint B, as shown in part (a) of the figure. The frame has pin supports at A and C. A concentrated load P acts at joint B, thereby placing member AB in direct compression. To assist in determining the buckling load for member AB, we represents it as a pinned-end column, as shown in part (b) of the figure. At the top of the column, a rotational spring of stiffness b R represents the restraining action of the horizontal beam BC on the column (note that the horizontal beam provides resistance to rotation of joint B when the column buckles). Also, consider only bending effects in the analysis (i.e., disregard the effects of axial deformations).

4

;

x P P bR

C B

B

L

L

EI

y

A

(a)

A

(b)

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SECTION 11.5 Columns with Eccentric Axial Loads

Solution 11.4-11

Column AB with elastic support at B

FREE-BODY DIAGRAM OF COLUMN

GENERAL SOLUTION   C1 sin kx + C2 cos kx +

b RuB x PL

B.C.

1 (0)  0

‹ C2  0

B.C.

2 (L)  0

‹ C1  

  C1 sin kx +

b RuB Psin kL

b RuB x PL

¿  C1k cos kx +

b RuB PL

(a) BUCKLING EQUATION   deflection in the y direction MB  moment at end B uB  angle of rotation at end B (positive clockwise) MB  bRuB H  horizontal reactions at ends A and B

B.C. 3

‹  uB  

 PL   b R kL cot kL + b R

g M0  g MA  0 MB  HL  0

Substitute P  k 2EI and rearrange:

b RuB MB  L L

b RL (kL cot kL  1)  k 2L2  0 EI

DIFFERENTIAL EQUATION (EQ. 11-3) œœ

EI  M  Hx  P k 2  œœ

 + k 2 

b RuB b RuB (k cos kL)  P sin kL PL

Cancel uB and multiply by PL:

EQUILIBRIUM

H

v¿(L)   uB

P EI

;

(b) CRITICAL LOAD FOR b R  3EI/L

b RuB x LEI

3(kL cot kL  1)  (kL)2  0 Solve numerically for kL : kL  3.7264 Pcr  k 2EI  (kL)2 a

EI L2

b  13.89

EI L2

Columns with Eccentric Axial Loads When solving the problems for Section 11.5, assume that bending occurs in the principal plane containing the eccentric axial load.

Problem 11.5-1

An aluminum bar having a rectangular cross section (2.0 in.  1.0 in.) and length L  30 in. is compressed by axial loads that have a resultant P  2800 lb acting at the midpoint of the long side of the cross section (see figure). Assuming that the modulus of elasticity E is equal to 10  10 6 psi and that the ends of the bar are pinned, calculate the maximum deflection d and the maximum bending moment Mmax.

P = 2800 lb

;

871

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CHAPTER 11 Columns

Solution 11.5-1 Bar with rectangular cross section b  2.0 in. h  1.0 in. L  30 in. P  2800 lb I

e  0.5 in.

bh3  0.1667 in.4 12

E  10 * 106 psi P  1.230 A EI

kL  L

d  e asec

Eq.(11-51):

kL  1b  0.112 in. 2

Eq.(11  56): M max  Pe sec

kL 2

 1710 lb-in.

Problem 11.5-2 A steel bar having a square cross section (50 mm  50 mm 50 mm  50 mm) and length L  2.0 m is compressed by axial loads that have a resultant P  60 kN acting at the midpoint of one side of the cross section (see figure). Assuming that the modulus of elasticity E is equal to 210 GPa and that the ends of the bar are pinned, calculate the maximum deflection d and the maximum bending moment Mmax.

Solution 11.5-2

;

P = 60 kN

Bar with square cross section

b  50 mm. L  2 m. P  60 kN e  25 mm E  210 GPa

;

4

b I  520.8 * 103 mm4 12

P kL  L  1.481 A EI

Eq. (11-51): d  e asec

kL  1b  8.87 mm 2

;

kL  2.03 kN # m 2

;

Eq. (11-56): M max  Pe sec

x

Problem 11.5-3

Determine the bending moment M in the pinned-end column with eccentric axial loads shown in the figure. Then plot the bending-moment diagram for an axial load P  0.3Pcr. Note: Express the moment as a function of the distance x from the end of the column, and plot the diagram in nondimensional form with M/Pe as ordinate and x/L as abscissa.

P

P

e

M0 = Pe

B v L

y

A e P Probs.11.5-3, 11.5-4 and 11.5-5

(a)

P (b)

M0 = Pe

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SECTION 11.5 Columns with Eccentric Axial Loads

Solution 11.5-3

Column with eccentric loads

Column has pinned ends. Use EQ. (11-49):

1.7207 x x M  a tan b asin 1.7207 b + cos 1.7207 Pe 2 L L

   e atan

or

kL sin kx + cos kx  1b 2

‹ M  Pe a tan

M x x  1.162 a sin 1.721 b + cos 1.721 Pe L L

M  Pe  P

From Eq. (11-45):

kL sin kx + cos kx b 2

(NOTE: kL and kx are in radians)

;

BENDING-MOMENT DIAGRAM FOR P  0.3 Pcr

FOR P  0.3 Pcr: From Eq. (11-52): kL  p

P

A Pcr

 p10.3

 1.7207

Problem 11.5-4 Plot the load-deflection diagram for a pinned-end column with eccentric axial loads (see figure) if the eccentricity e of the load is 5 mm and the column has length L  3.6 m, moment of inertia I  9.0 * 106 mm4, and modulus of elasticity E  210 GPa. Note: Plot the axial load as ordinate and the deflection at the midpoint as abscissa.

Solution 11.5-4

Column with eccentric loads

Column has pinned ends.

SOLVE EQ. (2) FOR P:

Use Eq. (11-54) for the deflection at the midpoint (maximum deflection):

P  583.3c arccosa

d  e csec a

p P b  1d 2 A Pcr

(1)

DATA e  5.0 mm L  3.6 m E  210 GPa I  9.0 * 106 mm4 CRITICAL LOAD Pcr 

p2EI L2

 1439.3 kN

MAXIMUM DEFLECTION (FROM EQ. 1) d  (5.0) [sec (0.041404 1P)  1] Units: P  kN d  mm Angles are in radians.

(2)

2 5.0 bd 5.0 + d

LOAD-DEFLECTION DIAGRAM

;

;

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CHAPTER 11 Columns

Problem 11.5-5

Solve the preceding problem for a column with e  0.20 in., L  12 ft, I  21.7 in.4, and E  30 * 106 psi.

Solution 11.5-5

Column with eccentric loads

Column has pinned ends Use Eq. (11-54) for the deflection at the midpoint (maximum deflection): d  e csec a

p P b  1d 2 A Pcr

SOLVE EQ. (2) FOR P: P  125.6carc cos a (1)

2 0.2 bd 0.2 + d

;

LOAD-DEFLECTION DIAGRAM

DATA e  0.20 in. L  12 ft  144 in. E  30 * 106 psi I  21.7 in.4 CRITICAL LOAD Pcr 

p2EI L2

 309.9 k

MAXIMUM DEFLECTION (FROM EQ. 1) d  (0.20) [sec (0.08924 1P)  1]

(2)

Units: P  kips d  inches Angles are in radians.

Problem 11.5-6 A wide-flange member (W 200 * 22.5) is compressed by axial loads that have a resultant P acting at the point shown in the figure. The member has modulus of elasticity E  2000 GPa and pinned conditions at the ends. Lateral supports prevent any bending about the weak axis of the cross section. If the length of the member is 6.2 mm, and the deflection is limited to 6.5 mm, what is the maximum allowable load Pallow?

P

W 200  22.5

Solution 11.5-6 W 200 * 22.5 d  6.5 mm e

d 2

I  20 * 10 mm 6

e  103 mm

CRITICAL LOAD Pcr 

p2EI L2

L  6.2 m

E  200 GPa

Pcr  1027 kN

4

d  206 mm

Mamimum deflection d  easec a Solve for

P

P p b  1b 2 A Pcr

Pallow  49.9 kN

;

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SECTION 11.5 Columns with Eccentric Axial Loads

Problem 11.5-7 A wide-flange member (W 10  30) is compressed by axial loads that have a resultant P  20 k acting at the point shown in the figure. The material is steel with modulus of elasticity E  29,000 ksi. Assuming pinned-end conditions, determine the maximum permissible length Lmax if the deflection is not to exceed 1/400th of the length.

Solution 11.5-7

W 10  30

Column with eccentric axial load

Wide-flange member: W 10 * 30 Pinned-end conditions. Bending occurs about the weak axis (axis 2-2). P  20 k E  29,000 ksi L  length (inches) Maximum allowable deflection  From Table E-1:

P = 20 k

L (  d) 400

I  16.7 in.4

DEFLECTION AT MIDPOINT (EQ. 11-51) d  ea sec

kL  1b 2

L  (2.905 in.) [sec (0.003213 L)  1] 400 Rearrange terms and simplify: L 0 1162 in.

e

5.810 in.  2.905 in. 2

sec (0.003213 L)  1 

k

P  0.006426 in.1 A EI

(NOTE: angles are in radians) Solve the equation numerically for the length L: L  150.5 in. MAXIMUM ALLOWABLE LENGTH L max  150.5 in.  12.5 ft

Problem 11.5-8 Solve the preceding problem (W 250  44.8) if the resultant force P equals 110 kN and E  200 GPa.

Solution 11.5-8 W 250 * 44.8 P  110 kN

E  200 GPa d

L 400

Bending occur about the weak axis (axis 2-2) I  6.95 * 106 mm4 e k

b 2

b  148 mm

e  74 mm P

A EI

k  0.000281 mm1

Deflection at midpoint d  easec a

kL b  1b 2

kL L  easec a b  1b 400 2 Solve for the length L L max  3.14 m

;

;

875

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CHAPTER 11 Columns

Problem 11.5-9 The column shown in the figure is fixed at the base and free at the upper end. A compressive load P acts at the top of the column with an eccentricity e from the axis of the column. Beginning with the differential equation of the deflection curve, derive formulas for the maximum deflection d of the column and the maximum bending moment Mmax in the column.

x P

d

P e

e

B

L

A

y (a)

Solution 11.5-9

(b)

Fixed-free column

e  eccentricity of load P d  deflection at the end of the column   deflection of the column at distance x from base DIFFERENTIAL EQUATION (EQ. 11.3) œœ

EI  M  P(e + d  ) k 2 

P EI

v¿  C1 k cos kx  C2 k sin kx B.C.

1 (0)  0

‹ C2   e  d

2 ¿(0)  0 ‹ C1  0   (e + d)(1  cos kx) B.C. 3 (L)  d ‹ d  (e + d)(1  cos kL) or d  e (sec kL  1) B.C.

d  e(sec kL  1)

;

œœ

MAXIMUM DEFLECTION

œœ

MAXIMUM BENDING MOMENT (AT BASE OF COLUMN)

  k 2(e + d  )  + k 2  k 2(e + d) GENERAL SOLUTION   C1 sin kx + C2 cos kx + e + d

M max  P(e + d)  Pe sec kL NOTE:

;

  (e + d) (1  cos kx)  e(sec kL) (1  cos kx)

Problem 11.5-10

An aluminum box column of square cross section is fixed at the base and free at the top (see figure). The outside dimension b of each side is 100 mm and the thickness t of the wall is 8 mm. The resultant of the compressive loads acting on the top of the column is a force P  50 kN acting at the outer edge of the column at the midpoint of one side. What is the longest permissible length Lmax of the column if the deflection at the top is not to exceed 30 mm? (Assume E  73 GPa.)

P t A

A

L b Section A-A

Probs. 11.5-10 and 11.5-11

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SECTION 11.5 Columns with Eccentric Axial Loads

Solution 11.5-10 Fixed-free column d  deflection at the top Use Eq. (11-51) with L/2 replaced by L: d  e (sec kL  1) (This same equation is obtained in Prob. 11.5-9.)

NUMERICAL DATA (1)

E  73 GPa b  100 mm t  8 mm P  50 kN d  30 mm e 

SOLVE FOR L FROM EQ. (1) I

e + d d sec kL  1 +  e e e e cos kL  kL  arccos e + d e + d 1 e L  arccos k e + d L

EI

AP

arccos

1 4 [b  (b  2t)4]  4.1844 * 106 mm4 12

MAXIMUM ALLOWABLE LENGTH Substitute numerical data into Eq.(2).

P k A EI

e e + d

b  50 mm 2

EI

AP (2)

 2.4717 m

arccos

e  0.625 e + d

e  0.89566 radians e + d

L max  (2.4717 m)(0.89566)  2.21 m

;

Problem 11.5-11

Solve the preceding problem for an aluminum column with b  6.0 in., t  0.5 in., P  30 k, and E  10.6 * 103 ksi. The deflection at the top is limited to 2.0 in.

Solution 11.5-11 Fixed-free column d  deflection at the top Use Eq. (11-51) with L/2 replaced by L: d  e (sec kL  1) (This same equation is obtained in Prob. 11.5-9.) SOLVE FOR L FROM EQ. (1) sec kL  1 + cos kL 

e + d d  e e

e e + d

kL  arccos

1 e L  arccos k e + d L

(1)

E  10.6 * 103 ksi b  6.0 in. t  0.5 in. b P  30 k d  2.0 in. e   3.0 in. 2 1 4 [b  (b  2t)4]  55.917 in.4. I 12 MAXIMUM ALLOWABLE LENGTH

e e + d

P k A EI

EI e arccos AP e + d

NUMERICAL DATA

Substitute numerical data into Eq. (2). EI

e  0.60  140.56 in. e + d e arccos  0.92730 radians e + d L max  (140.56 in.)(0.92730) ;  130.3 in.  10.9 ft

AP

877

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CHAPTER 11 Columns

Problem 11.5-12

A steel post AB of hollow circular cross section is fixed at the base and free at the top (see figure). The inner and outer diameters are d1  96 mm and d2  110 mm, respectively, and the length L  4.0 m. A cable CBD passes through a fitting that is welded to the side of the post. The distance between the plane of the cable (plane CBD) and the axis of the post is e  100 mm, and the angles between the cable and the ground are a  53.13°. The cable is pretensioned by tightening the turnbuckles. If the deflection at the top of the post is limited to d  20 mm, what is the maximum allowable tensile force T in the cable? (Assume E  205 GPa.)

e = 100 mm

B L = 4.0 m Cable d1 d2 d2

a = 53.13°

a = 53.13°

A

C

D

Solution 11.5-12 Fixed-free column d  deflection at the top

MAXIMUM ALLOWABLE COMPRESSIVE FORCE P

P k P  compressive force in post A EI Use Eq. (11-51) with L/2 replaced byL:

Substitute numerical data into Eq. (2).

d  e(sec kL  1) (This same equation is obtained in Prob. 11.5-9.)

Pallow  13,263 N  13.263 kN (1)

MAXIMUM ALLOWABLE TENSILE FORCE T IN THE CABLE Free-body diagram of joint B:

SOLVE FOR P FROM EQ. (1) sec kL  1 + cos kL  kL 

e + d d  e e

e e + d

PL2 A EI

kL  arccos

e e + d

a  53.13° g Fvert  0 P  2T sin a  0 5P P   8289 N T 2 sin a 8

PL2 e  arccos A EI e + d

Square both sides and solve for P: P

2 e a arccos b e + d L2

EI

NUMERICAL DATA E  205 GPa L  4.0 m e  100 mm d  20 mm d2  110 mm d1  96 mm I

p 4 (d  d 41)  3.0177 * 106 mm4 64 2

(2)

‹ Tmax  8.29 kN

;

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SECTION 11.5 Columns with Eccentric Axial Loads

Problem 11.5-13

A frame ABCD is constructed of steel wide-flange members (W 8 * 21; E  30 * 106 psi) and subjected to triangularly distributed loads of maximum intensity q0 acting along the vertical members (see figure). The distance between supports is L  20 ft and the height of the frame is h  4 ft. The members are rigidly connected at B and C.

A

D h

E

B

C q0

q0 E L

(a) Calculate the intensity of load q0 required to produce a maximum bending moment of 80 k-in. in the horizontal member of BC. (b) If the load q0 is reduced to one-half of the value calculated in part (a), what is the maximum bending moment in member BC? What is the ratio of this moment to the moment of 80 k-in. in part (a)?

Section E-E

Solution 11.5-13 Frame with triangular loads (a) LOAD q0 TO PRODUCE M max  80 k-in. Substitute numerical values into Eq. (1). Units: pounds and inches PL2 M max  80,000 lb-in. A 4EI  0.11700932P (radians) 80,000  P(16 in.) [sec (0.00700931P)]

P  resultant force e  eccentricity q0h h P e 2 3

5,000  P sec (0.00700931P) P  5,000 [cos (0.00700931P)]  0 SOLVE EQ. (2) NUMERICALLY

MAXIMUM BENDING MOMENT IN BEAM BC kL From Eq. (11-56): M max  Pe sec 2 k

P

A EI

‹ M max  Pe sec

P  4461.9 lb q0 

2

PL

A 4EI

W 8 * 21 I  I2  9.77 in.4 (from Table E-1a) E  30 * 106 psi L  20 ft  240 in.

2P  186 lb/in.  2230 lb/ft h

;

(1)

NUMERICAL DATA

h  4 ft  48 in. h e   16 in. 3

(2)

(b) LOAD q0 IS REDUCED TO ONE-HALF ITS VALUE ‹ P is reduced to one-half its value. 1 P  (4461.9 lb)  2231.0 lb 2 Substitute numerical values into Eq. (1) and solve for Mmax. M max  37.75 k-in. ; M max 37.7   0.47 Ratio: 80 k-in. 80

;

This result shows that the bending moment varies nonlinearly with the load.

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CHAPTER 11 Columns

The Secant Formula

P e

When solving the problems for Section 11.6, assume that bending occurs in the principal plane containing the eccentric axial load.

Problem 11.6-1

A steel bar has a square cross section of width b  2.0 in. (see figure). The bar has pinned supports at the ends and is 3.0 ft long. The axial forces acting at the end of the bar have a resultant P  20 k located at distance e  0.75 in. from the center of the cross section. Also, the modulus of elasticity of the steel is 29,000 ksi. (a) Determine the maximum compressive stress smax in the bar. (b) If the allowable stress in the steel is 18,000 psi, what is the maximum permissible length L max of the bar? Probs. 11.6-1 through 11.6-3

Solution 11.6-1 Bar with square cross section Pinned supports.

Substitute into Eq. (1):

DATA b  2.0 in. L  3.0 ft  36 in. P  20 k e  0.75 in. E  29,000 ksi

smax  17.3 ksi

(b) MAXIMUM PERMISSIBLE LENGTH sallow  18,000 psi

(a) MAXIMUM COMPRESSIVE STRESS Secant formula (Eq. 11-59): smax 

b4  1.333 in.4 I 12 ec r

2

Solve Eq. (1) for the length L:

P ec L P c1 + 2 sec a bd A 2r A EA r

P P  2  5.0 ksi A b

 2.25

c

;

(1)

b  1.0 in. 2

P(ec/r 2) EI arccos c d AP smax A  P

L2

(2)

Substitute numerical values: L max  46.2 in.

;

I r   0.3333 in.2 A 2

L  62.354 r

P  0.00017241 EA

Problem 11.6-2 A brass bar (E  100 GPa) with a square cross section is subjected to axial force having a resultant P acting at distance e from the center (see figure). The bar is pin supported at the ends and is 0.6 m in length. The side dimension b of the bar is 30 mm and the eccentricity e of the load is 10 mm. If the allowable stress in the brass is 150 MPa, what is the allowable axial force Pallow?

Solution 11.6-2 Bar with square cross section Pinned supports.

SECANT FORMULA (EQ. 11-59):

DATA b  30 mm L  0.6 m sallow  150 MPa e  10 mm E  100 GPa

smax 

P ec L P c1 + 2 sec a bd A 2r A EA r

(1)

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SECTION 11.6 The Secant Formula

Units: Newtons and meters

SUBSTITUTE NUMERICAL VALUES INTO EQ. (1): P [1 + 2sec (0.00365151P)] 150 * 106  900 * 106 or

smax  150 * 106 N/m2 A  b 2  900 * 106 m2 c ec r2

b I b2  0.015 m r 2    75 * 106 m2 2 A 12

 2.0 P  newtons

881

P[1 + 2sec (0.00365151P)]  135,000  0

L P  0.00365151P 2r A EA

(2)

SOLVE EQ. (2) NUMERICALLY: Pallow  37,200 N  37.2 kN

;

Problem 11.6-3 A square aluminum bar with pinned ends carries a load P  25 k acting at distance e  2.0 in. from the center (see figure on the previous page). The bar has length L  54 in. and modulus of elasticity E  10,600 ksi. If the stress in the bar is not exceed 6 ksi, what is the minimum permissible width bmin of the bar?

Solution 11.6-3 Square aluminum bar Pinned ends.

SUBSTITUTE TERMS INTO EQ. (1):

DATA

6,000 

Units: pounds and inches e  2.0 in. P  25 k  25,000 lb

or

L  54 in.

E  10,600 ksi  10,600,000 psi

1 +

smax  6.0 ksi  6,000 psi

P P ec L c1 + 2 sec a bd A 2r A EA r

A  b2 ec r

2



12 b

c

b 2

r2

b

2

c1 +

4.5423 12 sec a bd b b2

4.5423 12 b  0.24 b 2  0 seca b b2

(2)

SOLVE EQ. (2) NUMERICALLY:

SECANT FORMULA (EQ. 11-59): smax 

25,000

(1)

bmin  4.10 in.

;

I b2  A 12

L P 4.5423  2r A EA b2

P

A pinned-end column of length L  2.1 m is constructed of steel pipe (E  210 GPa) having inside diameter d1  60 mm and outside diameter d2  68 mm (see figure). A compressive load P  10 kN acts with eccentricity e  30 mm.

Problem 11.6-4

e

(a) What is the maximum compressive stress smax in the column? (b) If the allowable stress in the steel is 50 MPa, what is the maximum permissible length Lmax of the column?

d1 d2

Probs. 11.6-4 throught 11.6-6

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CHAPTER 11 Columns

Solution 11.6-4

Steel pipe column

Pinned ends. DATA

r2 

Units: Newtons and meters

L  2.1 m

r  22.671 * 103 m

E  210 GPa  210 * 109 N/m2

d1  60 mm  0.06 m

d2  68 mm  0.068 m

P  10 kN  10,000 N

I  513.99 * 106 m2 A

ec

e  30 mm  0.03 m

r2

 1.9845

c

d2  0.034 m 2

L P  0.35638 2r A EA

Substitute into Eq. (1):

TUBULAR CROSS SECTION

smax  38.8 * 106 N/m2  38.8 MPa

p 2 (d 2  d 21)  804.25 * 106 m2 4 p I  (d 24  d 14)  413.38 * 109 m4 64

A

;

(b) MAXIMUM PERMISSIBLE LENGTH sallow  50 MPa Solve Eq. (1) for the length L:

(a) MAXIMUM COMPRESSIVE STRESS

P(ec/r 2) EI arccosc d AP smaxA  P

L2

Secant formula (Eq. 11-59): smax 

ec P P L c1 + 2 sec a bd A 2r A EA r

(1)

(2)

Substitute numerical values: L max  5.03 m

;

P  12.434 * 106 N/m2 A

Problem 11.6-5 A pinned-end strut of length L  5.2 ft is constructed of steel pipe (E  30 * 103 ksi) having inside diameter d1  2.0 in. and outside diameter d2  2.2 in. (see figure). A compressive load P  2.0 k is applied with eccentricity e  1.0 in. (a) What is the maximum compressive stress smax in the strut? (b) What is the allowable load Pallow if a factor of safety n  2 with respect to yielding is required? (Assume that the yield stress sY of the steel is 42 ksi.)

Solution 11.6-5 Pinned-end strut Steel pipe. DATA

(a) MAXIMUM COMPRESSIVE STRESS

Units: kips and inches

L  5.2 ft  62.4 in. E  30 * 103 ksi d1  2.0 in. d2  2.2 in. P  2.0 k

e  1.0 in.

Secant formula (Eq. 11-59): P ec L P c1 + 2 sec a bd A 2r A EA r d2 P  3.0315 ksi c  1.1 in. A 2

smax 

I  0.55250 in.2 A

TUBULAR CROSS SECTION

r2 

p 2 (d  d 21)  0.65973 in.2 4 2 p 4 I (d  d 14)  0.36450 in.4 64 2

r  0.74330 in.

A

ec r2

 1.9910

L P  0.42195 2r A EA

(1)

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SECTION 11.6 The Secant Formula

Substitute into Eq. (1): smax  9.65 ksi

883

Substitute numerical values into Eq. (1):

;

42 

(b) ALLOWABLE LOAD

P [1 + 1.9910sec (0.298361P)] 0.65973

(2)

Solve Eq. (2) numerically: P  PY  7.184 k

sY  42 ksi n  2

Find Pallow

Pallow 

PY  3.59 k n

;

Problem 11.6-6 A circular aluminum tube with pinned ends supports a load P  18 kN acting at distance e  50 mm from the center (see figure). The length of the tube is 3.5m and its modulus of elasticity is 73 GPa. If the maximum permissible stress in the tube is 20 MPa,what is the required outer diameter d2 if the ratio of diameter is to be d1/d2  0.9?

Solution 11.6-6

Aluminum tube

Pinned ends.

c

P  18 kN

DATA

L  3.5 m

e  50 mm

E  73 GPa

smax  20 MPa

d1/d2  0.9

SECANT FORMULA (EQ. 11-59) smax  A

P ec L P bd c1 + 2 sec a A 2r A EA r

p 2 p (d  d 21)  [d 22  (0.9d2)2]  0.14923d 22 4 2 4

(d2  mm;

A  mm2)

18,000 N 120,620 P P   a  MPa b 2 A A 0.14923 d 2 d 22 I

p 4 p 4 (d 2  d 41)  [d 2  (0.9d2)4]  0.016881d 42 64 64

(d2  mm; r2

I  mm ) 4

I  0.11313d 22 A

r  0.33634 d2

(d2  mm; r 2  mm2)

(r  mm)

d2 2

ec r

2



(50 mm)(d2/2) 0.11313

d 22



220.99 d2

5,203.1 L 3500 mm   2r 2(0.33634 d2) d2 18,000 N 1.6524 P   2 2 EA (73,000 N/ mm )(0.14923 d 2) d 22 5,203.1 1.6524 L P 6688.2   2r A EA d2 A d 22 d 22 SUBSTITUTE THE ABOVE EXPRESSIONS INTO EQ. (1): smax  20 MPa   c1

120,620 d 22

6688.2 220.99 sec a bd d2 d 22

SOLVE EQ. (2) NUMERICALLY: d2  131 mm

;

(2)

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CHAPTER 11 Columns

Problem 11.6-7 A steel column (E  30 * 103 ksi) with pinned ends is constructed of W 10 * 60 wide-flange shape (see figure). The column is 24 ft long.The resultant of axial loads acting on the column is a force P acting with eccentricity e  2.0 in.

P

e = 2.0 in.

W 10  60

(a) If P  120 k, determine the maximum compressive stress smax in the column. (b) Determine the allowable load Pallow if the yield stress is Y  42 ksi and the factor of safety with respect to yielding of the material is n  2.5.

Solution 11.6-7

Steel column with pinned ends

E  30 * 10 ksi L  24 ft  288 in. e  2.0 in. W 10  60 wide-flange shape 3

A  17.6 in.2 I  341 in.4 r2 

I  19.38 in.2 A

L  65.42 r

ec r

2

Substitute into Eq. (1):

smax  10.9 ksi

;

d  10.22 in..

r  4.402 in. c 

d  5.11 in. 2

(B) ALLOWABLE LOAD

42 

n  2.5

Find Pallow

P [1 0.5273 sec (0.045021P)] 17.6

Solve numerically: P  PY  399.9 k

Secant formula (Eq. 11-59): ec P P L c1 + 2 sec a bd A 2r A EA r

sY  42 ksi

Substitute into Eq. (1):

 0.5273

(a) MAXIMUM COMPRESSIVE STRESS (P  120 k)

smax 

L P  0.4931 2r A EA

P  6.818 ksi A

Pallow  PY/n  160 k

;

(1)

Problem 11.6-8 A W 410  85 steel column is compressed by a force P  340 kN acting with an eccentricity e  38 mm., as shown in the figure.The column has pinned ends and length L. Also,the steel has modulus of elasticity E  200 GPa and yield stress sY  250 MPa. (a) If the length L  3 m, what is the maximum compressive stress max in the column? (b) If a factor of safety n  2.0 is required with respect to yielding, what is the longest permissible length Lmax of the column?

P = 340 kN

e = 38 mm

W 410  85

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SECTION 11.6 The Secant Formula

885

Solution 11.6-8 W 410 * 85

A  10800 mm2

I  17.9 * 106 mm4

I  I2

b  181 mm c 

c  90.5 mm r

e  38 mm

I

(b) MAXIMUM LENGTH FOR sy  250 MPa

b 2

Py  nP

r  40.711 mm

AA

Py  680 kN

sy 

from

P  340 kN E  200 GPa L  3 m (a) MAXIMUM COMPRESSION STRESS smax 

Py

A P

1

ec r

2

sec a

solve for the length L

P ec L P 1 + 2 sec a b AP 2r A EA Q r

smax  104.5 MPa

n  2.0

L max

Py a

b EI r2 2 acos J K A Py syA  Py

L max  3.66 m

;

Py L b 2r A EA Q

;

P

A steel column (E  30 * 103 ksi) that is fixed at the base and free at the top is constructed of a W 8 * 35 wide-flange member (see figure). The column is 9.0 ft long. The force P acting at the top of the column has an eccentricity e  1.25 in.

Problem 11.6-9

ec

e

e

(a) If P  40 k,what is the maximum compressive stress in the column? (b) If the yield stress is 36 ksi and the required factor of safety with respect to yielding is 2.1,what is allowable load Pallow ?

A

A

P

L Section A

Probs. 11.6-9 and 11.6-10

Solution 11.6-9 Steel column (fixed-free) E  30 * 103 ksi e  1.25 in. L e  2L  2(9.0 ft)  18 ft  216 in. W 8 * 35 WIDE-FLANGE SHAPE A  10.3 in.2 r2  c

I  I2  42.6 in.4

I  4.136 in.2 A

b  4.010 in. 2

smax  9.60 ksi

Substitute into Eq. (1): b  8.020 in. (b) ALLOWABLE LOAD

r  2.034 in. Le  106.2 r

sY  36 ksi ec r

2

36 

Find Pallow

P [1 + 1.212 sec (0.095521P)] 10.3

Solve numerically:

Secant formula (Eq. 11-59): Le P P ec  c1 + 2 sec a bd A 2r A EA r

n  2.1

Substitute into Eq. (1):

 1.212

(a) MAXIMUM COMPRESSIVE STRESS (P  40 k)

smax

Le P  0.6042 2r A EA

P  3.883 ksi A

Pallow  PY/n  53.6 k (1)

P  PY  112.6 k ;

;

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CHAPTER 11 Columns

Problem 11.6-10

A W 310 * 74 wide-flange steel column with length L  3.8 m is fixed at the base and free at the top (see figure). The load P acting on the column is intended to be centrally applied,but because of unavoidable discrepancies in construction, an eccentricity ratio of 0.25 is specified. Also, the following data are supplied: E  200 GPa, sY  290 MPa and P  310 kN. (a) What is the maximum compressive stress smax in the column? (b) What is the factor of safety n with respect to yielding of the steel?

Solution 11.6-10 W 310 * 74

A  9420 mm2

I  23.4 * 106 mm4 ec r2

 0.25

L  3.8 m

r

P  310 kN L e  2L

I  I2 I

AA

(b) FACTOR OF SAFETY WITH RESPECT TO YIELDING sy  290 MPa

r  49.841 mm

E  200 GPa L e  7.6 m

sy 

from

Py A

a1 +

ec r

2

sec a

Py Le bb 2r A EA

solve numerically for Py Py  712 kN

n

Py

n  2.30

P

(a) MAXIMUM COMPRESSION STRESS smax 

Le ec P P a1 + 2 sec a bb A 2r A EA r

smax  47.6 MPa

;

A pinned-end column with length L  18 ft is constructed from a W 12 * 87 wide-flange shape (see figure). The column is subjected to centrally applied load P1  180 k and an eccentrically applied load P2  75 k .The load P2 acts at distance s  5.0 in. from the centroid of the cross section. The properties of the steel are E  29,000 ksi and sY  36 ksi.

Problem 11.6-11

P2

s

P1

Wide-flange column

(a) Calculate the maximum compressive stress in the column. (b) Determine the factor of safety with respect to yielding.

Probs. 11.6.11 and 11.6.12

;

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887

SECTION 11.6 The Secant Formula

Solution 11.6-11

Column with two loads

Pinned-end column.

(a) MAXIMUM COMPRESSIVE STRESS

W 12 * 87

Secant formula (Eq. 11-59):

DATA L  18 ft  216 in. P2  75 k s  5.0 in. P1  180 k sY  36 ksi E  29,000 ksi

I  28.91 in.2 A d c   6.265 in. 2

e

r2 

P  9.961 ksi A

r

2

(1)

smax  13.4 ksi

;

(b) FACTOR OF SAFETY WITH RESPECT TO YIELDING smax  sY  36 ksi

P  PY

Substitute into Eq. (1):

r  5.376 in. ec

P ec L P c1 + 2 sec a bd A 2r A EA r

Substitute into Eq. (1):

P2s  1.471 in. P 4 I  I1  740 in. d  12.53 in.

P  P1 + P2  255 k A  25.6 in.2

smax 

36   0.3188

pY [1 + 0.3188sec(0.023321PY)] 25.6 PY  664.7 k

Solve numerically: P  255 k

L P  0.3723 2r A EA

n

pY 664.7 k   2.61 P 255 k

;

Problem 11.6-12

The wide-flange pinned-end column shown in the figure carries two loads, a force P1  450 kN acting at the centroid and a force P2  270 kN acting at distance s  100 mm, from the centroid. The column is a W 250 * 67 shape with L  4.2 m, E  200 GPa, and sY  290 MPa. (a) What is the maximum compressive stress in the column? (b) If the load P1 remain at 450 kN, what is the largest permissible value of the load P2 in order to maintain a factor of safety of 2.0 with respect to yielding?

Solution 11.6-12 W 250 * 67

L  4.2 m

P1  450 kN

P2  270 kN

s  100 mm

E  200 GPa

sy  290 MPa P  720 kN A  8580 mm2 d  257 mm d c 2

(a) MAXIMUM COMPRESSION STRESS smax 

smax  120.4 MPa

P  P1 + P2 e

P2s P

I  I1 r

I AA

c  128.5 mm

P ec L P a1 + 2 seca bb A 2r A EA r

e  37.5 mm I  103 * 106 mm4 r  109.6 mm

;

(b) LARGEST VALUE OF LOAD P2 WHEN P1  450 kN n  2.0 Py  n(P1 + P2) from sy 

sy 

Py A

n1P1 P22

A Solve for P2

a 1

ec r

2

sec a

Py L bb 2r A EA

L n1P1  P22 dd 2r A EA r P2  387 kN ; * c1 

ec

2

sec c

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CHAPTER 11 Columns

A W 14 * 53 wide-flange column of length L  15 ft is fixed at the base and free at the top (see figure). The column supports a centrally applied load P1  120 k and a load P2  40 k supported on a bracket. The distance from the centroid of the column to the load P2 is s  12 in. Also, the modulus of elasticity is E  29,000 ksi and yield stress is sY  36 ksi.

Problem 11.6-13

P1

(a) Calculate the maximum compressive stress in the column. (b) Determine the factor of safely with respect to yielding.

P2

s

L A

A Section A-A

Probs. 11.6-13 and 11.6-14

Solution 11.6-13

Column with two loads

Fixed-free column.

W 14 * 53

(a) MAXIMUM COMPRESSIVE STRESS Secant formula (Eq. 11-59):

DATA L  15 ft  180 in. L e  2 L  360 in. P1  120 k P2  40 k s  12 in. sY  36 ksi E  29,000 ksi P2s  3.0 in. P 4 I  I1  541 in. d  13.92 in.

P  P1 + P2  160 k A  15.6 in.

2

I  34.68 in.2 A d c   6.960 in. 2

r2 

P  10.26 ksi A

e

r  5.889 in. ec r2

 0.6021

Le P  0.5748 2r A EA

smax 

Le P ec P c1 + 2 sec a bd A 2r A EA r

Substitute into Eq. (1):

(1)

smax  17.6 ksi

;

(b) FACTOR OF SAFETY WITH RESPECT TO YIELDING smax  sY  36 ksi

P  PY

Substitute into Eq. (1): 36 

PY [1 + 0.6021 sec (0.045441PY)] 15.6

Solve numerically: PY  302.6 k P  160 k

Problem 11.6-14

A wide-flange column with a bracket is fixed at the base and free at the top (see figure). The column supports a load P1  340 kN acting at the centroid and a load P2  110 kN acting on the bracket at distance s  250 mm, from the load P1. The column is a W 310 * 52 shape with L  5 m, E  200 GPa, and sY  290 MPa. (a) What is the maximum compressive stress in the column? (b) If the load P1 remains at 340 kN, what is the largest permissible value of the load P2 in order to maintain a factor of safety of 1.8 with respect to yielding?

n

PY 302.6 k   1.89 P 160 k

;

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SECTION 11.9 Design Formulas for Columns

889

Solution 11.6-14 W 310 * 52

L  5.0 m

P1  340 kN

P2  110 kN

s  250 mm

E  200 GPa

sy  290 MPa P  450 kN

d 2

P2 s P

I  I1

;

(b) LARGEST VALUE OF LOAD P2 P1  340 kN n  1.8

I  119 * 106 mm4

WHEN

Py  n (P1 + P2)

r  133.8 mm

sy 

from

c  159.0 mm

Le  2 L

Le P P ec bb a1 + 2 sec a A 2r A EA r

smax  115.2 MPa

e  61.1 mm

I r AA

d  318 mm c

smax 

P  P1 + P2 e

A  6650 mm2

(a) MAXIMUM COMPRESSION STRESS

sy 

L e  10.0 m

Py A

n1P1 P22 A * c1 

Solve for P2

a 1

ec r

2

sec a

L e Py bb 2 r A EA

L e n1P1 P22 dd 2 rA EA r P2  193 kN ec 2

sec c

Design Formulas for Columns

P

The problems for Section 11.9 are to be solved assuming that the axial loads are centrally applied at the ends of the columns. Unless otherwise stated, the columns may buckle in any direction. STEEL COLUMNS L

Problem 11.9-1 Determine the allowable axial load Pallow for a

A

A

W 10  45 steel wide-flange column with pinned ends (see figure) for each of the following lengths: L  8 ft, 16 ft, 24 ft, and 32 ft. (Assume E  29,000 ksi and sY  36 ksi.)

Section A - A

Probs 11.9-1 through 11.9-6

Solution 11.9.1 Steel wide-flange column Pinned ends (K  1). Bucking about axis 2-2 (see Table E-1a). Use AISC formulas. W 10 * 45 E  29,000 ksi

A  13.3 in.2

r2  2.01 in.

sY  36 ksi

L a b  200 r max

2p2 E L Eq. (11  76): a b   126.1 r c A sY L c  126.1 r  253.5 in.  21.1 ft.

L

8 ft

16 ft

24 ft

32 ft

L/r

47.76

95.52

143.3

191.0

n1 (Eq. 11-79)

1.802

1.896

-

-

n2 (Eq. 11-80)

-

-

1.917

1.917

sallow/sY (Eq. 11-81) 0.5152 0.3760

-

-

-

sallow (ksi)

18.55

13.54

7.274

Pallow  A sallow

247 k

180 k

96.7 k 54.4 k

sallow/sY (Eq. 11-82)

0.2020 0.1137 4.091

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CHAPTER 11 Columns

Problem 11.9-2 Determine the allowable axial load Pallow for a W 310  129 steel wide-flange column with pinned ends (see figure) for each of the following lengths: L  3 m, 6 m, 9 m, and 12 m. (Assume E  200 GPa and sY  340 MPa.)

Solution 11.9-2 K1

Pinned ends

Buckling about axis 2-2 W 310 * 129 A  16500 mm2 E  200 GPa L max  15.6 m 3m 6m ≤ L ± 9m 12 m

sallowi  sy

r2  78.0 mm sy  340 MPa Lc  r

2p2 E A sy

r  r2 L max  200 # r

a

L c  8.405 m

2 n 2i a

38.462 L 76.923 ≤  ± r 115.385 153.846

if

K Lc b r

K Lc 2 b r K Li 2 b r

¥ if

K Li K Lc … r r

otherwise

K Li 3a b r K Lc 8a b r



K Li 3 b a r K Lc 3 8a b r

KLi K Lc 23 if 7 12 r r –NA– otherwise

–NA– –NA– ≤ n2  ± 1.917 1.917

2927 3m 2213 6m ≤ kN for ± ≤ Pallow  ± 1276 9m 718 12 m

K Li K Lc 7 r r

1.795 1.889 ≤ n1  ± –NA– –NA– n 2i 

2a

2

Pallowi  A sallowi

–NA–

5 + 3

K Li 2 b r

177.366 134.135 ≤ MPa sallow  ± 77.355 43.512

i  1 .. 4 n 1i 

1 ≥ 1 n li

a

otherwise

;

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SECTION 11.9 Design Formulas for Columns

891

Problem 11.9-3

Determine the allowable axial load Pallow for a W 10 * 60 steel wide-flange column with pinned ends (see figure) for each of the following lengths: L  10 ft, 20 ft, 30 ft, and 40 ft. (Assume E  29,000 ksi and sY  36 ksi.)

Solution 11.9-3 Steel wide-flange column Pinned ends (K  1). Bucking about axis 2-2 (see Table E-1a). Use AISC formulas. A  17.6 in.

2

W 10 * 60

r2  2.57 in.

sY  36 ksi

E  29,000 ksi

L a b  200 r max

2

2p E L  126.1 Eq. (11-76): a b  r c A sY L c  126.1 r  324.in.  27.0 ft

L

10 ft

20 ft

30 ft

40 ft

L/r

46.69

93.39

140.1

186.8

n1 (Eq. 11-79)

1.799

1.894

-

-

n2 (Eq. 11-80)

-

-

1.917

1.917

-

-

sallow/sY (Eq. 11-81) 0.5177 0.3833 -

-

sallow(ksi)

18.64

13.80

7.610

4.281

Pallow  A sallow

328 k

243 k

134 k

75.3 k

sallow/sY (Eq. 11-82)

0.2114 0.1189

Problem 11.9-4 Select a steel wide-flange column of nominal depth 250 mm. (W 250 shape) to support an axial load P  800 kN (see figure). The column has pinned ends and length L  4.25 m. Assume E  200 GPa and sY  250 MPa. (Note: The selection of columns is limited to those listed in Table E-1(b), Appendix E.)

Solution 11.9-4 K1

P  800 kN

L  4.25 m

sy  250 MPa

E  200 GPa

Lc 2p2 E  r A sy

2p2 E  125.664 A sy

(3) TRIAL COLUMN W 250 * 67 A  8580 mm2 r  51.1 mm (4) ALLOWABLE STRESS FOR TRIAL COLUMN

(1) TRIAL VALUE OF sallow Upper limit: sallow– max 

with

L 0 r

sy n1

n1 

5 3

n1 

5 + 3

3a 8a

sallow  110 MPa

KL b r

2p 2 E b A sy

sallow  sy

1 ≥1  n1

(2) TRIAL VALUE OF AREA A

P sallow

A  7273 mm2



a 8a

KL 3 b r

2p2 E 3 b A sy

n 1  1.879

sallow– max  150 MPa Try

Lc L 6 r r

L  83.170 r

sallow  103.9 MPa

a

KL 2 b r

2p2 E 2 2a b A sy

¥

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CHAPTER 11 Columns

(5) ALLOWABLE LOAD FOR TRIAL COLUMN Pallow  sallowA Pallow 7 P

Pallow  891.7 kN

sallow  sy

(OK)

1 ≥1  n1

(W 250 * 67)

n1 

5 + 3

A  5700 mm

2

r  34.8 mm

KL 3a b r 2p 2 E 8a b A sy



Pallow  A sallow Pallow 6 P

Lc L 6 r r

L  122.126 r

KL 2 b r

2p2 E 2 2a b A sy

¥

sallow  68.85 MPa

(6) NEXT SMALLER SIZE COLUMN W 250 * 44.8

a

Pallow  392.4 kN

(Not Satisfactory) ;

‹ Select W250 * 67 KL 3 a b r 8a

2p2 E 3 b A sy

n 1  1.916

Problem 11.9-5 Select a steel wide-flange column of nominal depth 12 in. (W 12 shape) to support an axial load P  175 k (see figure). The column has pinned ends and length L  35 ft. Assume E  29,000 ksi and sY  36 ksi. (Note: The selection of columns is limited to those listed in Table E-1a, Appendix E.)

Solution 11.9-5 Select a column of W 12 shape P  175 k sY  36 ksi

L  35 ft  420 in.

K1

E  29,000 ksi 2

(4) ALLOWABLE STRESS FOR TRIAL COLUMN L L 7 a b r r c

4.20 in. L   136.8 r 3.07 in.

L 2p E Eq. (11-76): a b   126.1 r c A sY

Eqs. (11-80) and (11-82):

(1) TRIAL VALUE OF sallow

sallow  0.2216 sallow  7.979 ksi sY

Upper limit: use Eq. (11-81) with L/r 0 Max. sallow 

sY sY   21.6 ksi n1 5/3

Try sallow  8 ksi (Because column is very long) (2) TRIAL VALUE OF AREA A

175 k P   22 in.2 sallow 8 ksi

(3) TRIAL COLUMN A  25.6 in.2

W 12 * 87 r  3.07 in.

n 2  1.917

(5) ALLOWABLE LOAD FOR TRIAL COLUMN Pallow  sallow A  204 k 7 175 k (ok) (6) NEXT SMALLER SIZE COLUMN W 12 * 50 A  14.7 in.2 r  1.96 in. L  214 Since the maximum permissible value of r L/r is 200, this section is not satisfactory. Select W 12 * 87

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SECTION 11.9 Design Formulas for Columns

Problem 11.9-6 Select a steel wide-flange column of nominal depth 360 mm (W 360 shape) to support an axial load P  1100 kN (see figure). The column has pinned ends and length L  6 m. Assume E  200 GPa and sY  340 MPa. (Note: The selection of columns is limited to those listed in Table E-1 (b), Appendix E.)

Solution 11.9-6 K1

P  1100 kN

sy  340 MPa

L6m

Pallow  sallow A

E  200 GPa

Pallow 6 P

Lc 2p 2 E   107.756 r A sy sallow L with 0 r

sallowmax  Try

sy n1

sallowmax  204 MPa

sallow  110 MPa

P

A  10000 mm2

sallow

(3) TRIAL COLUMN W 360 * 79 r  48.8 mm

A  10100 mm

2

(4) ALLOWABLE STRESS FOR TRIAL COLUMN L  122.951 r

Lc L 7 r r

23 n2  12

n 2  1.917

sallow  sy

a

2p 2 E 2 b A sy

2n 2 a

KL 2 b r

sallow  68.1 MPa

(Not Satisfactory)

(6) NEXT LARGER SIZE COLUMN 5 n1  3

(2) TRIAL VALUE OF AREA A

Pallow  688.1 KN

(W 360 * 79)

(1) TRIAL VALUE OF Upper limit:

(5) ALLOWABLE LOAD FOR TRIAL COLUMN

A  15500 mm2

W 360 * 122

Lc L 6 r r

L  95.238 r

n1 

5 + 3

 63.0 mm

3a

KL b r

2p 2 E 8a b A sy



a 8a

KL 3 b r

2p2 E 3 b A sy

n 1  1.912 sallow  sy

a

1 ≥1  n1

2a

KL 2 b r

2p2 E 2 b A sy

¥

sallow  108.38 MPa Pallow  A sallow Pallow 7 P

Pallow  1679.9 kN

(OK)

‹ Select W 360 * 122

(W 360 * 122)

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CHAPTER 11 Columns

Problem 11.9-7 Determine the allowable axial load Pallow for a steel pipe column with pinned ends having an outside diameter of 4.5 in. and wall thickness of 0.237 in. for each of the following lengths: L  6 ft, 12 ft, 18 ft, and 24 ft. (Assume E  29,000 ksi and sY  36 ksi.)

Solution 11.9.7 Steel pipe column Pinned ends (K  1). Use AISC formulas.

L

d2  4.5 in. t  0.237 in. d1  4.026 in. p A  (d 22  d 12)  3.1740 in.2 4 p I (d 4  d1 4)  7.2326 in.4 64 2 r

I  1.5095 in. AA

L a b  200 r max

E  29,000 ksi sY  36 ksi Eq. (11-76):

6 ft

12 ft

18 ft

24 ft

L/r

47.70

95.39

143.1

190.8

n1 (Eq. 11-79)

1.802

1.896

-

-

n2 (Eq. 11-80)

-

-

1.917

1.917

-

-

sallow/sY (Eq. 11-81) 0.5153 0.3765 -

-

sallow (ksi)

18.55

13.55

Pallow  A sallow

58.9 k 43.0 k 23.1 k 13.0 k

sallow/sY (Eq. 11-82)

0.2026 0.1140 7.293

4.102

2p 2 E L a b   126.1 r c A sY

L c  126.1 r  190.4 in.  15.9 ft

Problem 11.9-8 Determine the allowable axial load Pallow for a steel pipe column with pinned ends having an outside diameter of 220 mm and wall thickness of 12 mm for each of the following lengths: L  2.5 m, 5 m, 7.5 m, and 10 m. (Assume E  200 GPa and sY  250 MPa.)

Solution 11.9.8 Steel pipe column Pinned ends (K  1). Use AISC formulas. d2  220 mm t  12 mm d1  196 mm p A  (d 22  d 21)  7841.4 mm2 4 p 4 I (d  d14)  42.548 mm * 106 mm4 64 2 r

I  73.661 mm AA

L a b  200 r max

E  200 GPa sY  250 MPa Eq. (11-76):

L 2p2 E a b   125.7 r c A sY

L c  125.7 r  9257 mm  9.26 m

L

2.5 m

5.0 m

7.5 m 10.0 m

L/r

33.94

67.88

101.8

135.8

n1 (Eq. 11-79)

1.765

1.850

1.904

-

n2 (Eq. 11-80)

-

-

-

1.917

sallow/sY (Eq. 11-81) 0.5458 0.4618 0.3528 sallow/sY (Eq. 11-82) sallow (MPa) Pallow  A sallow

-

-

-

-

0.2235

136.4

115.5

88.20

55.89

1070 kN 905 kN 692 kN 438 kN

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SECTION 11.9 Design Formulas for Columns

Problem 11.9-9 Determine the allowable axial load Pallow for a steel

895

P

pipe column that is fixed at the base and free at the top (see figure) for each of the following lengths: L  6 ft, 9 ft, 12 ft, and 15 ft. The column has outside diameter d  6.625 in. and wall thickness t  0.280 in. (Assume E  29,000 ksi and sY  36 ksi.)

t A

A

L d Section A-A

Probs. 11.9-9 through 11.9-12

Solution 11.9-9 Steel pipe column Fixed-free column (K  2). Use AISC formulas. d2  6.625 in. t  0.280 in. d1  6.065 in. p A  (d 22  d 21)  5.5814 in.2 4 I r

p 4 (d  d 41)  28.142 in.4 64 2 I

AA

 2.2455

a

KL b  200 r max

E  29,000 ksi sY  36 ksi

Eq. (11-76): L c  126.1

a

2p2 E KL b   126.1 r c A sY

r  141.6 in.  11.8 ft K

L

6 ft

9 ft

12 ft

15 ft

KL/r

64.13

96.19

128.3

160.3

n1 (Eq. 11-79)

1.841

1.897

-

-

n2 (Eq. 11-80)

-

-

1.917

1.917

-

-

sallow/sY (Eq. 11-81) 0.4730 0.3737 -

-

sallow (Ksi)

17.03

13.45

Pallow  A sallow

95.0 k 75.1 k 50.7 k 32.4 k

sallow/sY (Eq.11-82)

Problem 11.9-10 Determine the allowable axial load Pallow for a steel pipe column that is fixed at the base and free at the top (see figure) for each of the following lengths: L  2.6 m, 2.8 m, 3.0 m, and 3.2 m. The column has outside diameter d  140 mm and wall thickness t  7 mm. (Assume E  200 GPa and sY  250 MPa.)

0.2519 0.1614 9.078

5.810

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CHAPTER 11 Columns

Solution 11.9-10 Steel pipe column Fixed-free column (K  2). Use AISC formulas. d2  140 mm t  7.0 mm d1  126 mm p A  (d 22  d 21)  2924.8 mm2 4 p (d 4  d14)  6.4851 * 106 mm4 I 64 2 I  47.09 mm r AA

KL a b  200 r max

E  200 GPa sY  250 MPa

a

Eq. (11-76): L c  125.7

KL 2p2 E b   125.7 r c A sY

r  2959 mm  2.959 m K

L

2.6 m

2.8 m

3.0 m

3.2 m

KL/r

110.4

118.9

127.4

135.9

n1 (Eq. 11-79)

1.911

1.916

-

-

n2 (Eq. 11-80)

-

-

1.917

1.917

-

-

sallow/sY (Eq. 11-81) 0.3212 0.2882 sallow/sY (Eq. 11-82) sallow (MPa) Pallow  A sallow

-

-

0.2537 0.2230

80.29

72.06

63.43

55.75

235 kN 211 kN 186 kN 163 kN

Problem 11.9-11

Determine the maximum permissible length Lmax for a steel pipe column that is fixed at the base and free at the top and must support an axial load P  40 k (see figure). The column has outside diameter d  4.0 in. wall thickness t  0.226 in., E  29,000 ksi, and sY  42 ksi.

Solution 11.9-11 Steel pipe column Fixed-free column (K  2). P  40 k Use AISC formulas. d2  4.0 in. t  0.226 in. d1  3.548 in. p A  (d2 2  d1 2)  2.6795 in. 4 p (d 4  d1 4)  4.7877 in.4 I 64 2 I  1.3367 r AA

KL b a  200 r max

E  29,000 ksi sY  42 ksi Eq. (11-76):

KL 2p2 E a b   116.7 r c A sY

r L c  116.7  78.03 in.  6.502 ft K

Select trial values of the length L and calculate the corresponding values of Pallow (see table). Interpolate between the trial values to obtain the value of L that produces Pallow  P . Note: If L 6 L c, use Eqs.(11-79) and (11-81). If L 7 L c, use Eqs.(11-80) and (11-82). L(ft)

5.20

5.25

5.90

KL/r

93.86

94.26

93.90

n1 (Eq. 11-79)

1.903

1.904

1.903

n2 (Eq. 11-80)

-

-

-

0.3575

0.3541

0.3555

sallow/sY (Eq. 11-81)

-

-

-

sallow (ksi)

15.02

14.87

14.93

Pallow  A sallow

40.2 k

39.8 k

40.0 k

sallow/sY (Eq. 11-82)

For P  40 k,

L max  5.23 ft

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SECTION 11.9 Design Formulas for Columns

897

Problem 11.9-12

Determine the maximum permissible length Lmax for a steel pipe column that is fixed at the base and free at the top and must support an axial load P  500 kN (see figure). The column has outside diameter d  200 mm, wall thickness t  10 mm, E  200 GPa, and sY  250 MPa.

Solution 11.9-12 Steel pipe column Fixed-free column (K  2). P  500 kN Use AISC formulas. d2  200 mm t  10 mm d1  180 mm p A  (d 22  d 21)  5,969.0 mm2 4 p 4 (d 2  d 41)  27.010 * 106 mm4 I 64 r

I  67.27 mm AA

a

KL b  200 r max

E  200 GPa sY  250 GPa Eq. (11-76): L c  125.7

a

2p2 E KL b   125.7 r c A sY

r  4.226 m K

Select trial values of the length L and calculate the corresponding values of Pallow (see table). Interpolate between the trial values to obtain the value of L that produces Pallow  P . Note: If L 6 L c, use Eqs. (11-79) and (11-81). If L 7 L c, use Eqs. (11-80) and (11-82). L(m) KL/r n1 (Eq. 11-79) n2 (Eq. 11-80) sallow/sY (Eq. 11-81) sallow/sY (Eq. 11-82) sallow (MPa) Pallow  A sallow

3.55

3.60

3.59

105.5 1.908 0.3393 84.83 506 kN

107.0 1.909 0.3338 83.46 498 kN

106.7 1.909 0.3349 83.74 500 kN

For P  500 kN, L  3.59 m

;

A steel pipe column with Pinned ends supports an axial load P  21 k. The pipe has outside and inside diameters of 3.5 in. and 2.9 in., respectively. What is the maximum permissible length Lmax of the column if E  29,000 ksi and sY  36 ksi?

Problem 11.9-13

Solution 11.9-13 Steel pipe column Pinned ends (K  1). P  21 k Use AISC formulas. d2  3.5 in. t  0.3 in. d1  2.9 in. p A  (d 22  d 12)  3.0159 in.2 4 p 4 (d  d 41)  3.8943 in.4 I 64 2 r

I  1.1363 in. AA

L a b  200 r max

E  29,000 ksi sY  36 ksi L 2p2 E Eq. (11-76): a b   126.1 r c A sY L c  126.1 r  143.3 in.  11.9 ft

Select trial values of the length L and calculate the corresponding values of Pallow (see table). Interpolate between the trial values to obtain the value of L that produces Pallow  P . Note: If L 6 L c, use Eqs. (11-79) and (11-81). If L 7 L c, use Eqs. (11-80) and (11-82). L(ft) L/r n1 (Eq. 11-79) n2 (Eq. 11-80) sallow/sY (Eq. 11-81) sallow/sY (Eq. 11-82) sallow (ksi) Pallow  A sallow

13.8

13.9

14.0

145.7 1.917 0.1953 7.031 21.2 k

146.8 1.917 0.1925 6.931 20.9 k

147.8 1.917 0.1898 6.832 20.6 k

For P  21 k, L  13.9 ft

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CHAPTER 11 Columns

Problem 11.9-14

A steel column used in a college recreation center are 16.75 m long and are formed by welding three wide-flange sections (see figure). The columns are pin-supported at the ends and may buckle in any direction. Calculate the allowable load Pallow for one column, assuming E  200 GPa and sY  250 MPa.

W 310  129 W 610  241

W 310  129

Solution 11.9-14 L  16.75 m E  200 GPa sy  250 MPa K  1 W 310 * 129 A1  16500 mm I11  308 * 10 mm 6

4

2

d1  318 mm t w  17.9 mm

I12  2150 * 106 mm4 I22  184 * 106 mm4 For built-up column: A  2A1 + A2

A  63800 mm2

Iy  2 I2–1 + I1–2 d1 tw h + 2 2

Iy  2.350 * 109 mm4

h  168 mm

Iz  I2–2 + 2 1I1–1A1h22 Iz  1.731 * 109 mm4 r

Iz

AA

3a

I21  100 * 106 mm4

W 610 * 241 A2  30800 mm2

r  165 mm

Lc 2p2 E  125.664  r A sy

Lc L 6 r r

L  101.694 r

n1 

5 + 3

8a

KL b r

2p 2 E b A sy



a 8a

KL 3 b r

2p2 E 3 b A sy

n 1  1.904 sallow  sy

1 ≥1  n1

a

KL 2 b r

2p2 E 2 2a b A sy

¥

sallow  88.31 MPa Pallow  A sallow

Pallow  5634 kN

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SECTION 11.9 Design Formulas for Columns

899

A W 8  28 steel wide-flange column with pinned ends carries an axial load P. What is the maximum permissible length Lmax of the column if (a) P  50 k, and (b) P  100 k? (Assume E  29,000 ksi and sY  36 ksi.)

Problem 11.9-15

Probs. 11.9-15 and 11.9-16

Solution 11.9-15 Steel wide-flange column (b) P  100 k

Pinned ends (K  1). Buckling about axis 2-2 (see Table E-1a). Use AISC formulas.

L(ft)

W 8 * 28 A  8.25 in.2 r2  1.62 in. E  29,000 ksi sY  36 ksi Eq.(11-76):

L a b  200 r max

L 2p2 E a b   126.1 r c A sY

L c  126.1 r  204.3 in.  17.0 ft For each load P, select trial values of the length L and calculate the corresponding values of Pallow (see table). Interpolate between the trial values to obtain the value of L that produces Pallow  P . Note: If L 6 L c, use Eqs. (11-79) and (11-81). If L 7 L c, use Eqs. (11-80) and (11-82). (a) P  50 k L(ft) L/r n1 (Eq. 11-79) n2 (Eq. 11-80) sallow/sY (Eq. 11-81) sallow/sY (Eq. 11-82) sallow (ksi) Pallow  A sallow

21.0

21.5

21.2

155.6 1.917 0.1714 6.171 50.9 k

159.3 1.917 0.1635 5.888 48.6 k

157.0 1.917 0.1682 6.056 50.0 k

For P  50 k, L max  21.2 ft

;

L/r n1 (Eq. 11-79) n2 (Eq. 11-80) sallow/sY (Eq. 11-81) sallow/sY (Eq. 11-82) sallow (ksi) Pallow  A sallow

14.3

14.4

105.9 106.7 1.908 1.908 0.3393 0.3366 12.21 12.12 100.8 k 100.0 k

For P  100 k, L max  14.4 ft

;

14.5 107.4 1.909 0.3338 12.02 99.2 k

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CHAPTER 11 Columns

Problem 11.9-16

A W 250 * 67 steel wide-flange column with pinned ends carries an axial load P. What is the maximum permissible length Lmax of the column if (a) P  560 kN, and (b) P  890 kN? (Assume E  200 GPa and sY  290 MPa.)

Solution 11.9-16 E  200 GPa sy  290 MPa K  1 W 250 * 67

A  8580 mm

r2  51.1 mm

r  r2

Pallow  A sallow

2p2 E A sy

L c  5.962 m

Lc  116.676 r

For each load, select trial values of the length L and calculate the corresponding values of Pallow

Try

n1 

This solution show only the successful trial. (a) P  560 kN Try

;

L  4.76 m

5 + 3

3a

L  93.151 r

KL b r

2p 2 E 8a b A sy



a 8a

Lc L 6 r r KL 3 b r

2 p2 E 3 b A sy

n 1  1.902 L  125.440 r

L  6.41 m

Lc L 7 r r

23 n2  12

sallow  sy

L max  6.41 m

(b) P  890 kN

L max  200 r Lc  r

Pallow  561.6 kN

Therefore L max  L

2

a

K Lc 2 b r

2n2 a

sallow  sy

a

1 ≥1  n1

2a

KL 2 b r

2 p2 E 2 b A sy

¥

sallow  103.85 MPa Pallow  A sallow

KL 2 b r

Pallow  891 kN

Therefore L max  L L max  4.76 m

sallow  65.45 MPa

Problem 11.9-17

Find the required outside diameter d for a steel pipe column (see figure) of length L  20 ft that is pinned at both ends and must support an axial load P  25 k. Assume that the wall thickness t is equal to d/20. (Use E  29,000 ksi and sY  36 ksi.) Probs. 11.9-17 through 11.9-20

t

d

;

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901

SECTION 11.9 Design Formulas for Columns

Solution 11.9-17

Pipe column

Pinned ends (K  1). L  20 ft  240 in. P  25 k d  outside diameter t  d/20 E  29,000 ksi sY  36 ksi p A  [d 2  (d  2t)2]  0.14923 d 2 4 p 4 [d  (d  2t)4]  0.016881 d 4 I 64 r

I

AA

 0.33634 d 2

2p E L  126.1 L c  (126.1)r a b  r c A sY

d (in.) 2

A (in. ) I (in.4) r (in.) Lc (in.) L/r n2 (Eq. 11-80) sallow/sY (Eq. 11-82) sallow (ksi) Pallow  A sallow

4.80

4.90

5.00

3.438 8.961 1.614 204 148.7 23/12 0.1876 6.754

3.583 9.732 1.648 208 145.6 23/12 0.1957 7.044

3.731 10.551 1.682 212 142.7 23/12 0.2037 7.333

23.2 k

25.2 k

27.4 k

For P  25 k, d  4.89 in.

;

Seltect various values of diameter d until we obtain Pallow  P . If L … L c, Use Eqs. (11-79) and (11-81). If L Ú L c, Use Eqs. (11-80) and (11-82).

Problem 11.9-18

Find the required outside diameter d for a steel pipe column (see figure) of length L  3.5 m that is pinned at both ends and must support an axial load P  130 kN. Assume that the wall thickness t is equal to d/20. (Use E  200 GPa and sY  275 MPa.)

Solution 11.9-18

Pipe column

Pinned ends (K  1). L  3.5 m P  130 kN d  outside diameter t  d/20 E  200 GPa sY  275 MPa A

p 2 [d  (d  2t)2]  0.14923 d 2 4

I

p 4 [d  (d  2t)4]  0.016881 d 4 64

r

I

AA

 0.33634 d

2p2 E L  119.8 L c  (119.8)r a b  r c A sY Select various values of diameter d until we obtain Pallow  P . If L … L c, Use Eqs. (11-79) and (11-81). If L Ú L c, Use Eqs. (11-80) and (11-82).

d (mm)

98

99

2

100

1433 1463 1492 A (mm ) 4 3 3 I (mm ) 1557 * 10 1622 * 10 1688 * 103 r (mm) 32.96 33.30 33.64 Lc (mm) 3950 3989 4030 L/r 106.2 105.1 104.0 n1 (Eq. 11-79) 1.912 1.911 1.910 0.3175 0.3219 0.3263 sallow/sY (Eq. 11-81) 87.32 88.53 89.73 sallow (MPa) Pallow  A sallow 129.5 kN 133.9 kN 125.1 kN For P  130 kN, d  99 mm

;

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CHAPTER 11 Columns

Problem 11.9-19

Find the required outside diameter d for a steel pipe column (see figure) of length L  11.5 ft that is pinned at both ends and must support an axial load P  80 k. Assume that the wall thickness t is 0.30 in. (Use E  29,000 ksi and sY  42 ksi.)

Solution 11.9-19

Pipe column

Pinned ends (K  1). L  11.5 ft  138 in. P  80 k d  outside diameter

t  0.30 in.

E  29,000 ksi sY  42 ksi A

p 2 [d  (d  2t)2] 4

I

I p 4 [d  (d  2t)4] r  64 AA

L 2p 2 E a b   116.7 L c  (116.7)r r c A sY Seltect various values of diameter d until we obtain Pallow  P .

If L … L c, Use Eqs. (11-79) and (11-81). If L Ú L c, Use Eqs. (11-80) and (11-82). d (in.) 2

A (in. ) I (in.4) r (in.) Lc (in.) L/r n1 (Eq. 11-79) sallow/sY (Eq. 11-81) sallow (ksi) Pallow  A sallow

5.20

5.25

5.30

4.618 13.91 1.736 203 79.49 1.883 0.4079 17.13 79.1 k

4.665 14.34 1.753 205 78.72 1.881 0.4107 17.25

4.712 14.78 1.771 207 77.92 1.880 0.4133 17.36

80.5 k

81.8 k

For P  80 k, d  5.23 in.

;

Problem 11.9-20

Find the required outside diameter d for a steel pipe column (see figure) of length L  3.0 m that is pinned at both ends and must support an axial load P  800 kN. Assume that the wall thickness t is 9 mm. (Use E  200 GPa and sY  300 MPa.)

Solution 11.9-20

Pipe column

Pinned ends (K  1). L  3.0 m P  800 kN d  outside diameter t  9.0 mm E  200 GPa sY  300 MPa A

p 2 [d  (d  2t)2] 4

I

I p 4 [d  (d  2t)4] r  64 AA

L 2p2 E a b   114.7 L c  (114.7)r r c A sY Seltect various values of diameter d until we obtain Pallow  P . If L … L c, Use Eqs. (11-79) and (11-81). If L Ú L c, Use Eqs. (11-80) and (11-82).

d (mm)

193

194

2

195

5202 5231 5259 A (mm ) 6 6 I (mm4) 20.08 * 10 22.43 * 10 22.80 * 106 r (mm) 65.13 65.48 65.84 Lc (mm) 7470 7510 7550 L/r 46.06 45.82 45.57 n1 (Eq. 11-79) 1.809 1.809 1.808 0.5082 0.5087 0.5094 sallow/sY (Eq. 11-81) 152.5 152.6 152.8 sallow (MPa) Pallow  A sallow 793.1 kN 798.3 kN 803.8 kN For P  800 kN, d  194 mm

;

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Aluminum Columns Problem 11.9-21

An aluminum pipe column (alloy 2014-T6) with pinned ends has outside diameter d2  5.60 in. and inside diameter d1  4.80 in. (see figure). Determine the allowable axial load Pallow for each of the following lengths: L  6 ft, 8 ft, 10 ft, and 12 ft.

d1 d2

Probs. 11.9-21 through 11.9-24

Solution 11.9-21

Aluminum pipe column

Alloy 2014-T6 Pinned ends (K  1). d2  5.60 in. d1  4.80 in. p 2 (d  d 21)  6.535 in.2 4 2 p I  (d 42  d 41)  22.22 in.4 4 A

r

I

AA

 1.844 in.

Use Eqs. (11-84 a and b): sallow  30.7  0.23 (L/r) ksi L/r … 55 sallow  54,000/(L/r)2 ksi L/r Ú 55 L (ft) L/r sallow (ksi) Pallow  sallow A

6 ft

8 ft

10 ft

12 ft

39.05 21.72

52.06 18.73

65.08 12.75

78.09 8.86

142 k

122 k

83 k

58 k

Problem 11.9-22

An aluminum pipe column (alloy 2014-T6) with pinned ends has outside diameter d2  120 mm and inside diameter d1  110 mm (see figure). Determine the allowable axial load Pallow for each of the following lengths: L  1.0 m, 2.0 m, 3.0 m , and 4.0 m. (Hint: Convert the given data to USCS units, determine the required quantities, and then convert back to SI units.)

Solution 11.9-22

Aluminum pipe column

Alloy 2014-T6 Pinned ends (K  1). d2  120 mm  4.7244 in. d1  110 mm  4.3307 in. p A  (d 22  d 21)  2.800 in.2 4 p 4 (d  d 14)  7.188 in.4 I 64 2 r

I

AA

 40.697 mm  1.6022 in.

Use Eqs. (11-84 a and b): sallow  30.7  0.23 (L/r) ksi L/r … 55 sallow  54,000/(L/r)2 ksi L/r Ú 55 L (m)

1.0 m

2.0 m

3.0 ft

4.0 m

L (in.) L/r sallow (ksi) Pallow  sallow A Pallow(kN)

39.37 24.58 25.05

78.74 49.15 19.40

118.1 73.73 9.934

157.5 98.30 5.588

70.14 k 54.31 k 27.81 k 15.65 k 312 kN 242 kN 124 kN 70 kN

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Problem 11.9-23

An aluminum pipe column (alloy 6061-T6) that is fixed at the base and free at the top has outside diameter d 2  3.25 in. and inside diameter d1  3.00 in. (see figure). Determine the allowable axial load Pallow for each of the following lengths: L  2 ft, 3 ft, 4 ft, and 5 ft.

Solution 11.9-23

Aluminum pipe column

Alloy 6061-T6 Pinned ends (K  2). d2  3.25 in. d1  3.00 in. p A  (d 22  d 21)  1.227 in.2 4 p 4 I (d  d 14)  1.500 in.4 64 2

Use Eqs. (11-85 a and b): sallow  20.2  0.126 (KL/r) ksi KL/r … 66 sallow  51,000/(KL/r)2 ksi KL/r Ú 66 L (ft) KL/r sallow (ksi) Pallow  sallow A

2 ft

3 ft

4 ft

5 ft

43.40 14.73

65.10 12.00

86.80 6.77

108.5 4.33

18.1 k

14.7 k

8.3 k

5.3 k

I  1.106 in. r AA

Problem 11.9-24

An aluminum pipe column (alloy 6061-T6) that is fixed at the base and free at the top has outside diameter d 2  80 mm and inside diameter d1  72 mm (see figure). Determine the allowable axial load Pallow for each of the following lengths: L  0.6 m, 0.8 m, 1.0 m, and 1.2 m. (Hint: Convert the given data to USCS units, determine the required quantities, and then convert back to SI units.)

Solution 11.9-24

Aluminum pipe column

Alloy 6061-T6 Pinned ends (K  2). d2  80 mm  3.1496 in. d1  72 mm  2.8346 in. A

p 2 (d  d 21)  1.480 in.2 4 2

p 4 I (d  d 41)  1.661 in.4 64 2 r

I  26.907 mm  1.059 in. AA

Use Eqs. (11-85 a and b): sallow  20.2  0.126 (L/r) ksi L/r … 66 sallow  51,000/(L/r)2 ksi KL/r Ú 66 L (m) KL (in.) KL/r sallow (ksi) Pallow  sallow A Pallow(kN)

0.6 m

0.8 m

1.0 ft

1.2 m

47.24 62.99 78.74 94.49 44.61 59.48 74.35 89.23 14.58 12.71 9.226 6.405 21.58 k 18.81 k 13.65 k 9.48 k 96 kN 84 kN 61 kN 42 kN

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905

Problem 11.9-25

A solid round bar of aluminum having diameter d (see figure) is compressed by an axial force P  60 k. The bar has pinned supports and is made of alloy 2014-T6. d

(a) If the diameter d  2.0 in., what is the maximum allowable length Lmax of the bar? (b) If the length L  30 in., what is the minimum required diameter dmin?

Solution 11.9-25

Probs. 11.9-25 through 11.9-28

Aluminum bar

Alloy 2014-T6 Pinned ends (K  1). P  60 k

(b) FIND dmin IF L  30 in.

(a) FIND Lmax IF d  2.0 in.

A

A r

pd 4 pd 2  3.142 in.2 I  4 64 I

AA



pd 2 4

sallow 

d  0.5 in. 4

r

d r

L 30 in. 120 in.   r d/4 d

76.39 P 60 k  (ksi)  2 A pd /4 d2

Assume L/r is greater than 55:

P 60 k  19.10 ksi sallow   A 3.142 in.2

Eq.(11-84b): sallow 

Assume L/r is less than 55:

or

Eq. (11-84a): sallow  30.7  0.23(L/r) ksi or 19.10  30.7  0.23(L/r) Solve for L/r:

L  50.43 r

L max  (50.43)r  25.2 in.

76.39 d

2



54,000 ksi (L/r)2

54,000 (120/d)2

d 4  20.37 in.4

dmin  2.12 in.

;

L/r  120/d  120/2.12  56.6 7 55

L 6 55 r

‹ ok

;

Problem 11.9-26

A solid round bar of aluminum having diameter d (see figure) is compressed by an axial force P  175 kN. The bar has pinned supports and is made of alloy 2014-T6.

(a) If the diameter d  40 mm, what is the maximum allowable length Lmax of the bar? (b) If the length L  0.6 m, what is the minimum required diameter dmin? (Hint:Convert the given data to USCS units, determine the required quantities, and then convert back to SI units.)

Solution 11.9-26

Aluminum bar

Alloy 2014-T6 Pinned supports (K  1). P  175 kN  39.34 k

r

(a) FIND Lmax IF d  40 mm  1.575 in.

sallow 

A

2

4

pd pd  1.948 in.2 I  4 64

I d   0.3938 in. AA 4 P 39.34 k   20.20 ksi A 1.948 in.2

‹ ok

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Assume L/r is less than 55:

Assume L/r is greater than 55:

Eq. (11-84a): sallow  30.7  0.23 (L/r) ksi

Eq. (11-84b): sallow 

or 20.20  30.7  0.23 (L/r) Solve for L/r:

L  45.65 r

L 6 55 r

‹ ok

L max  (45.65)r  17.98 in.  457 mm

;

sallow 

d r 4



54,000 (94.48/d)2

d 4  8.280 in.4 ;

L/r  94.48/d  94.48/1.696

L 23.62 in. 94.48 in.   r d/4 d

50.09 39.34 k P   A pd 2/4 d2

d

2

(L/r)2

dmin  1.696 in.  43.1 mm

(b) FIND dmin IF L  0.6 m  23.62 in. pd 2 A 4

50.09

or

54,000 ksi

 55.7 7 55

‹ ok

(ksi)

Problem 11.9-27

A solid round bar of aluminum having diameter d (see figure) is compressed by an axial force P  10 k. The bar has pinned supports and is made of alloy 6061-T6. (a) If the diameter d  1.0 in., what is the maximum allowable length Lmax of the bar? (b) If the length L  20 in., what is the minimum required diameter dmin?

Solution 11.9-27

Aluminum bar

Alloy 6061-T6 Pinned supports (K  1).

(b) FIND dmin IF L  20 in.

P  10 k

A

(a) FIND Lmax IF d 1.0 in. A r

pd 4 pd 2  0.7854 in.2 I  4 64 I

AA



sallow 

d  0.2500 in. 4

or

Eq.(11-85a): sallow  20.2  0.126 (L/r) ksi

Solve For L/r:

12.73  20.2  0.126 (L/r)

L max  (59.29)r  14.8 in.

d 4

20 in. 80 in. L   r d/4 d

12.73 10 k P   (ksi) A pd 2/4 d2

Eq. (11-85b): sallow 

Assume L/r is less than 66:

L  59.29 r

r

Assume L/r is greater than 66:

P 10 k  12.73 ksi sallow   A 0.7854 in.2

or

pd 2 4

L 6 66 r ;

‹ ok

12.73 d

2



51,000 ksi (L/r)2

51,000 (80/d)2

d 4  1.597 in.4

dmin  1.12 in.

L/r  80/d  80/1.12  71 7 66

; ‹ ok

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SECTION 11.9 Aluminum Columns

Problem 11.9-28

A solid round bar of aluminum having diameter d (see figure) is compressed by an axial force P  60 kN. The bar has pinned supports and is made of alloy 6061-T6. (a) If the diameter d  30 mm, what is the maximum allowable length L max of the bar? (b) If the length L  0.6 m, what is the minimum required diameter dmin? (Hint:Convert the given data to USCS units, determine the required quantities, and the convert back to SI units.)

Solution 11.9-28

Aluminum bar

Alloy 6061-T6 Pinned supports (K  1).

P  60 kN  13.49 k

(b) FIND dmin IF L  0.6 m  23.62 in.

(a) FIND Lmax IF d  30 mm  1.181 in. A r

pd 4 pd 2  1.095 in.2 I  4 64 I

AA



L 23.62 in. 94.48 in.   r d/4 d

17.18 13.48 k P   (ksi) A pd 2/4 d2

Eq. (11-85b): sallow  or

Assume L/r is less than 66: Eq. (11-85a): sallow  20.2  0.126 (L/r) ksi 12.32  20.2  0.126 (L/r) L  62.54 r

d 4

r

Assume L/r is greater than 66:

P 13.49 k  12.32 ksi sallow   A 1.095 in.2

Solve For L/r:

pd 2 4

sallow 

d  0.2953 in. 4

or

A

L 6 66 r

L max  (62.54)r  18.47 in.  469 mm

‹ ok

17.18 d

2



51,000 ksi (L/r)2

51,000 (94.48/d)2

d 4  3.007 in.4 dmin  1.317 in.  33.4 mm

;

L/r  94.48/d  94.48/1.317  72 7 66

‹ ok

;

Wood Columns When solving the problems for wood columns, assume that the columns are constructed of sawn lumber (c  0.8 and KcE  0.3) and have pinned-end conditions. Also, buckling may occur about either principal axis of the cross section. h

Problem 11.9-29

A wood post of rectangular cross section (see figure) is constructed of 4 in. * 6 in. structural grade, Douglas fir lumber (Fc  2,000 psi, E  1,800,00 psi). The net cross-sectional dimensions of the post are b  3.5 in. (see Appendix F). Determine the allowable axial load Pallow for each of the following lengths: L  5.0 ft, and 10.0 ft.

b

Probs. 11.9-29 through 11.9-32

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Solution 11.9-29 Wood post (rectangular cross section) Fc  2,000 psi E  1,800,000 psi c  0.8 K cE  0.3 b  3.5 in. h  5.5 in. d  b Find Pallow Eq. (11-94): f 

K cEE Fc(L e/d)2

Eq. (11-95): CP 

Le Le/d f CP Pallow

5 ft

7.5 ft

10.0 ft

17.14 0.9188 0.6610 25.4 k

25.71 0.4083 0.3661 14.1 k

34.29 0.2297 0.2176 8.4 k

1.5 m

2.0 m

2.5 m

15 1.1429 0.7350 154 kN

20 0.6429 0.5261 110 kN

25 0.4114 0.3684 77 kN

;

1 + f 1 + f 2 f  c d  2c A 2c c

Eq. (11-92): Pallow  FcCPA  FcCPbh

Problem 11.9-30

A wood post of rectangular cross section (see figure) is constructed of structural grade, southern pine lumber (Fc  14 MPa, E  12 GPa). The cross-sectional dimensions of the post (actual dimensions) are b  100 mm and h  150 mm. Determine the allowable axial load Pallow for each of the following lengths: L  1.5 m, 2.0 m, and 2.5 m.

Solution 11.9-30 Wood post (rectangular cross section) Fc  14 MPa c  0.8

E  12 GPa

K cE  0.3

b  100 mm h  150 mm d  b Find Pallow Eq. (11-94): f 

K cEE 2

Fc(L e/d)

Eq. (11-95): CP 

1 + f 1 + f 2 f  c d  2c A 2c c

Eq. (11-92): Pallow  FcCPA  FcCPbh

Problem 11.9-31

A wood post column of rectangular cross section (see figure) is constructed of 4 in. * 8 in. construction grade, western hemlock lumber (Fc  1,000 psi, E  1,300,000 psi). The net cross-sectional dimensions of the column are b = 3.5 in. and h  7.25 in. (see Appendix F). Determine the allowable axial load Pallow for each of the following lengths: L  6 ft 8 ft, and 10.0 ft.

Le Le/d f CP Pallow

;

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SECTION 11.9 Aluminum Columns

Solution 11.9-31 Wood column (rectangular cross section) Fc  1,000 psi E  1,300,000 psi c  0.8 K cE  0.3 b  3.5 in. h  7.25 in. d  b Find Pallow Eq. (11-94): f 

K cEE Fc(L e/d)2

Eq. (11-95): CP 

Le

6 ft

8 ft

10 ft

Le/d f CP Pallow

20.57 0.9216 0.6621 16.8 k

27.43 0.5184 0.4464 11.3 k

34.29 0.3318 0.3050 7.7 k

Le

2.5 m

3.5 m

4.5 m

17.86 0.7840 0.6019 212 kN

25.00 0.4000 0.3596 127 kN

32.14 0.2420 0.2284 81 kN

;

1 + f 1 + f 2 f  c d  2c A 2c. c

Eq. (11-92): Pallow  FcCPA  FcCPbh

Problem 11.9-32

A wood column of rectangular cross section (see figure) is constructed of structural grade, Douglas fir lumber (Fc  12 MPa, E  10 GPa). The cross-sectional dimensions of the column (actual dimensions) are b  140 mm and h  210 mm. Determine the allowable axial load Pallow for each of the following lengths: L  2.5 m, 3.5 m, and 4.5 m.

Solution 11.9-32 Wood column (rectangular cross section) Fc  12 MPa E  10 GPa c  0.8 K cE  0.3 b  140 mm h  210 mm d  b Find Pallow Eq. (11-94): Eq. (11-95): Eq. (11-92):

f

K cEE

Fc(L e/d)2 1 + f 1 + f 2 f  c d  CP  2c A 2c c

Le/d f CP Pallow

;

Pallow  FcCPA  FcCPbh

Problem 11.9-33

A square wood column with side dimensions b (see figure) is constructed of a structural grade of Douglas fir for which Fc  1,700 psi and E  1,400,000 psi. An axial force P  40 k acts on the column. (a) If the dimension b  55 in., what is the maximum allowable length Lmax of the column? (b) If the length L  11 ft, what is the minimum required dimension bmin?

b b

Probs. 11.9-33 through 11.9-36

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Solution 11.9-33 Wood column (square cross section) Fc  1,700 psi E  1,400,000 psi c  0.8 K cE  0.3 P  40 k

(b) MINIMUM DIMENSION bmin for L  11 ft Trial and error:

(a) MAXIMUM LENGTH Lmax FOR b  d  5.5 in. From Eq. (11-92):

CP 

P FCb 2

f

 0.77783

From Eq. (11-95):

CP 

1 + f 1 + f 2 f  c d  1.6 A 1.6 0.8

P  FcCPb 2

Trial and error:

f  1.3225

Given load:

K cEE L   13.67 d A fFc

‹ L max  13.67d  (13.67)(5.5 in.)  75.2 in.

Trial b

K cEE Fc(L/d)2

1 + f 1 + f 2 f  c d  1.6 A 1.6 0.8

CP  0.77783 

From Eq. (11-94):

L L  d b

P  40 k

L L  d b

f

CP

(in.) 6.50 6.70 6.71

P (kips)

20.308 19.701 19.672

0.59907 0.63651 0.63841

0.49942 0.52230 0.52343

35.87 39.86 40.06

bmin  6.71 in.

;

Problem 11.9-34

A square wood column with side dimensions b (see figure) is constructed of a structural grade of southern pine for which Fc  10.5 MPa and E  12 GPa. An axial force P  200 kN acts on the column. (a) If the dimension b 150 mm, what is the maximum allowable length Lmax of the column? (b) If the length L  4.0 m, what is the minimum required dimension bmin?

Solution 11.9-34 Wood column(square cross section) Fc  10.5 MPa E  12 GPa c  0.8

Trial and error:

f  1.7807

K cE  0.3 P  200 kN (a) MAXIMUN LENGTH L maxFOR b  d  150 mm From Eq. (11-92):

CP 

P Fcb 2

 0.84656

From Eq. (11-95): CP  0.84656 

1 + f 1 + f 2 f  c d  1.6 A 1.6 0.8

From Eq. (11-94):

K cEE L   13.876 d A fFc

‹ L max  13.876 d  (13.876)(150 mm)  2.08 m

;

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SECTION 11.9 Aluminum Columns

(b) MINIMUM DIMENSION bmin FOR L  4.0 m L L Trial and error:  d b CP 

f

Trial b

K cEE 2

Fc(L/d)

2

1 + f 1 + f f  c d  1.6 A 1.6 0.8

Given load:

P  200 kN

f

CP

(mm) 180 182 183 184

P  FcCPb 2

L L  d b

P (kN)

22.22 21.98 21.86 21.74

0.69429 0.70980 0.71762 0.72549

0.55547 0.56394 0.56814 0.57231

189.0 196.1 199.8 203.5

‹ bmin  184 mm

;

Problem 11.9-35

A square wood column with side dimensions b (see figure) is constructed of a structural grade of spruce for which Fc  900 psi and E  1,500,000 psi. An axial force P  8.0 k acts on the column. (a) If the dimension b  3.5 in., what is the maximum allowable length Lmax of the column? (b) If the length L  10 ft, what is the minimum required dimension bmin?

Solution 11.9-35 Wood column(square cross section) Fc  900 psi E  1,500,000 psi c  0.8 K cE  0.3 P  8.0 k (a) MAXIMUM LENGTH Lmax FOR b  d  3.5 in. From Eq. (11-92): CP 

P Fcb

2

 0.72562

2

1 + f 1 + f f  c d  1.6 A 1.6 0.8

Trial and error: f  1.1094 From Eq. (11-94):

Trial and error. CP 

L L  d b

f

K cEE Fc(L/d)2

1 + f 1 + f 2 f  c d  1.6 A 1.6 0.8

P  FcCPb 2

From Eq. (11-95): CP  0.72562 

(b) MINIMUM DIMENSION bmin FOR L  10 ft

K cEE L   21.23 d A fFc

‹ L max  21.23 d  (21.23)(3.5 in.)  74.3 in. ;

Given load: P  8000 lb Trial b

L L  d b

f

CP

(in.) 4.00 4.20 4.19

P (lb)

30.00 28.57 28.64

0.55556 0.61250 0.60959

0.47145 0.50775 0.50596

6789 8061 7994

‹ bmin  4.20 in.

;

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Problem 11.9-36

A square wood column with side dimensions b (see figure) is constructed of a structural grade of eastern white pine for which Fc  8.0 MPa and E  8.5 GPa. An axial force P  100 kN acts on the column. (a) If the dimension b  120 mm, what is the maximum allowable length Lmax of the column? (b) If the length L  4.0 m, what is the minimum required dimension bmin?

Solution 11.9-36 Wood column (square cross section) Fc  8.0 MPa E  8.5 GPa c  0.8

(b) MINIMUM DIMENSION bmin FOR L  4.0 m

K cE  0.3 P  100 kN

L L Trial and error.  d b

(a) MAXIMUM LENGTH Lmax FOR b  d  120 mm From Eq. (11-92): CP  From Eq. (11.95):

P Fcb

2

Cp 

 0.86806

K cEE L From Eq. (11-94):   12.592 d A fFc ‹ L max  12.592 d  (12.592)(120 mm)  1.51 m

;

K cEE Fc(L/d)2

1 + f 1 + f 2 f  c d  1.6 A 1.6 0.8

P  FcCPb 2

1 + f 1 + f 2 f  c d  Cp  0.86806  1.6 A 1.6 0.8 Trial and error: f  2.0102

f

Given load: P  100 kN Trial b

L L  d b

f

CP

(mm) 160 164 165

P (kN)

25.00 24.39 24.24

0.51000 0.53582 0.54237

0.44060 0.45828 0.46269

90.23 98.61 100.77

‹ bmin  165 mm

;

12 Review of Centroids and Moments of Inertia

Centroids of Plane Areas The problems for Section 12.2 are to be solved by integration.

Problem 12.2-1 Determine the distances
x and
y to the centroid C of a right triangle having base b and altitude h (see Case 6, Appendix D). Solution 12.2-1

Centroid of a right triangle

dA
x dy
b(1  yh) dy A
Qx


h



dA




y dA


y

 b(1  y h) dy
2

bh

y x
b (1  h )

0

h

 yb(1  y h) dy
0

bh2 6

h dy C

y


Qx h
A 3

Similarly, x


y

y

b 3

O

b

x

x

Problem 12.2-2 Determine the distance
y to the centroid C of a trapezoid having bases a and b and altitude h (see Case 8, Appendix D).

Solution 12.2-2

Centroid of a trapezoid a

Qx
dy

h

C

y

y

y b

O

x

dA
[b  (a  b)y h] dy



h

dA


 [b  (a  b)y h] dy 0

 y[b  (a  b) yh]dy 0

2

Width of element
b  (a  b)yh A




h

y dA





h (2a  b) 6 Qx h(2a  b) y

A 3(a  b)


h(a  b) 2

727

728

CHAPTER 12 Review of Centroids and Moments of Inertia

Problem 12.2-3 Determine the distance
y to the centroid C of a semicircle of radius r (see Case 10, Appendix D).

Solution 12.2-3 Centroid of a semicircle dA
2r 2  y 2 dy A




Qx


dA






r

2r 2  y 2 dy




0


r 2 2

r

 2yr

y dA


2

 y 2 dy


0

y


Q x 4r
A 3


2 r2  y2

y

2r 3 3

C

dy r

y

y x

O

Problem 12.2-4 Determine the distances
x and
y to the centroid C of a parabolic spandrel of base b and height h (see Case 18, Appendix D). Solution 12.2-4

Centroid of a parabolic spandrel 2 y
hx2 b

dx

y x

A


0

Qy
h C

 dA


x




Qy A

y O

x

x



0




b

hx 3 b 2h dx
4 b2

3b 4

 y2 dA
 2 0

2

dA
ydx


hx 2 bh dx
3 b2

b

Qx


b

x dA


b

hx dx b2

y


1 hx 2 hx 2 bh2 ¢ 2 ≤ ¢ 2 ≤ dx
10 b b

Q x 3h
A 10

Problem 12.2-5 Determine the distances
x and
y to the centroid C of a semisegment of nth degree having base b and height h (see Case 19, Appendix D). Solution 12.2-5

Centroid of a semisegment of nth degree Qx xn hn dA
y dx
h ¢ 1  n ≤ dx y

b A 2n  1 A


Qy




dA






x dA


b

h ¢1 

0



xn n ≤ n ≤ dx
bh ¢ b n1

b

xh ¢ 1 

0

b(n  1) x

A 2(n  2) Qx




y dA
2



0

1 xn xn h ¢ 1  n ≤ (h) ¢ 1  n ≤ dx 2 b b 2


bh2 B

xn y
h(1  bn ) n0

xn hb 2 n ≤ dx
¢ ≤ n b 2 n2

Qy

b

y

n R (n  1)(2n  1)

h C y x

x

O

dx x b

SECTION 12.3 Centroids of Composite Areas

Centroids of Composite Areas The problems for Section 12.3 are to be solved by using the formulas for composite areas.

Problem 12.3-1 Determine the distance
y to the centroid C of a trapezoid

having bases a and b and altitude h (see Case 8, Appendix D) by dividing the trapezoid into two triangles.

Solution 12.3-1

Centroid of a trapezoid 2h bh A2
3 2 ah bh h A
a Ai

(a  b) 2 2 2

A1
a

y

A1 C

C1 h

C2

y

A2 O

x

b

ah 2

y1


y2


h 3

2h ah h bh h2 Q x
a yi Ai
¢ ≤  ¢ ≤
(2a  b) 3 2 3 2 6 Q x h(2a  b) y

A 3(a  b)

y

Problem 12.3-2 One quarter of a square of side a is removed (see figure).

a — 2

What are the coordinates x
and
y of the centroid C of the remaining area?

a — 2 a — 2 C a — 2

y PROBS. 12.3-2 and 12.5-2

Solution 12.3-2

a2 4 a2 A2
2

3a 4 a y2
4 3a 2 A
a Ai
4 A1


a 2

a 2 O

x

Centroid of a composite area

y

a 2

O

A1 a 2

A2 a

x

y1


3a a 2 a a2 5a 3 Q x
a yi Ai
¢ ≤  ¢ ≤
4 4 4 2 16 Qx 5a
x
y
A 12

x

729

730

CHAPTER 12 Review of Centroids and Moments of Inertia

Problem 12.3-3 Calculate the distance
y to the centroid C of the channel

y

section shown in the figure if a
6 in., b
1 in., and c
2 in.

b

b c C

y

B

b B a — 2

PROBS. 12.3-3, 12.3-4, and 12.5-3

Solution 12.3-3

y

y1
b  c 2
2 in. b y2

0.5 in. A2
ab
6 in.2 2 A
a Ai
2A1  A2
10 in.2

A1

c

y

A2

b

a
6 in.

x

A1
bc
2 in.2

b

C

a — 2

a — 2

Centroid of a channel section

b

A1 c

O

a — 2

O

b
1 in.

Q x
a yi Ai
2y1A1  y2A2
11.0 in.3 Qx y

1.10 in. A

x

c
2 in.

Problem 12.3-4 What must be the relationship between the dimensions a, b, and c of the channel section shown in the figure in order that the centroid C will lie on line BB? Solution 12.3-4

Dimensions of channel section y

b

b

A1 c B

C a — 2

B

A2

y

b

O

a — 2

y1
b  c 2

A2
ab

y2
b 2

A
a Ai
2A1  A2
b(2c  a)

A1

c

A1
bc

x

Q x
a yi Ai
2y1A1  y2A2
b 2(4bc  2c2  ab) y


Q x 4bc  2c2  ab
A 2(2c  a)

Set y
b and solve: 2c2
ab y

Problem 12.3-5 The cross section of a beam constructed of a W 24 

162 wide-flange section with an 8 in.  3/4 in. cover plate welded to the top flange is shown in the figure. Determine the distance
y from the base of the beam to the centroid C of the cross-sectional area.

3 Plate 8 in.  — 4 in.

W 24  162

C y

PROBS. 12.3-5 and 12.5-5

O

x

731

SECTION 12.3 Centroids of Composite Areas

Solution 12.3-5

Centroid of beam cross section W 24  162

A2

A1
47.7 in.2

d
25.00 in.

y1
d 2
12.5 in. PLATE: 8.0  0.75 in.

A2
(8.0)(0.75)
6.0 in.2

y2
25.00  0.75 2
25.375 in. C

A1

A
a Ai
A1  A2
53.70 in.2 Q x
a yi Ai
y1A1  y2A2
748.5 in.3

y

y


x

x

Qx
13.94 in. A

y

Problem 12.3-6 Determine the distance
y to the centroid C of the

180 mm

composite area shown in the figure.

180 mm 105 mm

15 mm

30 mm

30 mm

A1

C y

A2

O

x 30 mm

30 mm PROBS. 12.3-6, 12.5-6 and 12.7-6

Solution 12.3-6

120 mm

Centroid of composite area y

180 mm

105 mm

15 mm

A1
(360)(30)
10,800 mm2 y1
105 mm

180 mm

A2
2(120)(30)  (120)(30)
10,800 mm2 y2
0 A
a Ai
A1  A2
21,600 mm2

30 mm

30 mm

A1

C y

A2

90 mm

O

x 30 mm

30 mm 120 mm

90 mm

90 mm

Q x
a yi Ai
y1A1  y2A2
1.134  106 mm3 Qx y

52.5 mm A

90 mm

732

CHAPTER 12 Review of Centroids and Moments of Inertia

Problem 12.3-7 Determine the coordinates
x and
y of the centroid C of the

y

L-shaped area shown in the figure.

0.5 in. 6 in. C 0.5 in. y O PROBS. 12.3-7, 12.4-7, 12.5-7 and 12.7-7

Solution 12.3-7

4 in.

Centroid of L-shaped area A1
(3.5)(0.5)
1.75 in.2 y1
0.25 in. x 1
2.25 in. A2
(6)(0.5)
3.0 in.2 y2
3.0 in. x 2
0.25 in.

y

A
a Ai
A1  A2
4.75 in.2

A2 6 in.

x

x

Q y
a x i Ai
x 1A1  x 2A2
4.688 in.3

x C y

O 4 in.

x


A1 x


0.99 in. A Q x
a yi Ai
y1A1  y2A2
9.438 in.3 y


Thickness t
0.5 in,

Qy

Qx
1.99 in. A

y 170 mm

Problem 12.3-8 Determine the coordinates
x and
y of the centroid C of the area shown in the figure.

50 mm 50 mm

280 mm

150 mm 80 mm

80 mm

x

O 80 mm

80 mm 300 mm

SECTION 12.4 Moments of Inertia of Plane Areas

Solution 12.3-8

Centroid of composite area A2
12(130)2
8450 mm2 x 2
300  130 3
256.7 mm

y

A2 170

130

A1

y2
280  130 3
236.7 mm 130

A3


280

x 3
80 mm

150 300

O

A3

A4
1963

x

y3
80 mm x 4
220 mm

mm2

Q y
a x i Ai
x 1A1  x 2A2  x 3A3  x 4A4
9.842  106 mm3 x


Qy A




9.842  106
137 mm 71,620

Q x
a yiAi
y1A1  y2A2  y3A3  y4A4
9.446  106 mm3

A1
(280)(300)
84,000 mm2 y1
140 mm

y


Q x 9.446  106

132 mm A 71,620

Moments of Inertia of Plane Areas Problems 12.4-1 through 12.4-4 are to be solved by integration.

Problem 12.4-1 Determine the moment of inertia Ix of a triangle of base b and altitude h with respect to its base (see Case 4, Appendix D).

Solution 12.4-1

Moment of inertia of a triangle Width of element

y dy


b¢ dA
y O

y4
80 mm

A
a Ai
A1  A2  A3  A4
71,620 mm2

A4

A1
large rectangle A2
triangular cutout A3
A4
circular holes All dimensions are in millimeters. Diameter of holes
50 mm Centers of holes are 80 mm from edges. x 1
150 mm


d 2

(50) 2
1963 mm2 4 4

b

h x

Ix


b(h  y) dy h



h

y 2dA


yb 2

0

3




hy ≤ h

bh 12

(h  y) dy h

733

734

CHAPTER 12 Review of Centroids and Moments of Inertia

Problem 12.4-2 Determine the moment of inertia IBB of a trapezoid having bases a and b and altitude h with respect to its base (see Case 8, Appendix D).

Solution 12.4-2

Moment of inertia of a trapezoid Width of element dy

y


a  (b  a) ¢

a

dA
B a  (b  a) ¢

h y B

B

x

b

O

hy ≤ h

IBB




hy ≤ R dy h

h

y 2dA


 y B a  (b  a) 2

0

¢

hy ≤ R dy h

h (3a  b) 12 3




Problem 12.4-3 Determine the moment of inertia Ix of a parabolic spandrel of base b and height h with respect to its base (see Case 18, Appendix D).

Solution 12.4-3

Moment of inertia of a parabolic spandrel Width of element

y dy

y Bh


bx
bb

h

2

y
hx2 b


b(1  y h) y

O

dA
b(1  y h) dy x

b

Ix




y 2dA




h

y 2b (1  y h) dy


0

Problem 12.4-4 Determine the moment of inertia Ix of a circle of radius r with respect to a diameter (see Case 9, Appendix D).

Solution 12.4-4

Moment of inertia of a circle Width of element
2r 2  y 2

y

dA
2r 2  y 2 dy dy y r

C

r

Ix


 y dA
 y (2 r 2

2

r

x





r 4 4

2

 y 2 ) dy

bh3 21

SECTION 12.4 Moments of Inertia of Plane Areas

Problems 12.4-5 through 12.4-9 are to be solved by considering the area to be a composite area.

Problem 12.4-5 Determine the moment of inertia IBB of a rectangle having sides of lengths b and h with respect to a diagonal of the rectangle (see Case 2, Appendix D).

Solution 12.4-5

Moment of inertia of a rectangle with respect to a diagonal L
length of diagonal BB

B

A h1

D

 h

L  B

b

C

L
b 2  h2 h1
distance from A to diagonal BB b Triangle BBC: sin 
L h1 bh h 1
h sin 
Triangle ADB: sin 
L h I1
moment of inertia of triangle ABB with respect to its base BB From Case 4, Appendix D: Lh 31 L bh 3 b 3h3
¢ ≤
12 12 L 12L2 For the rectangle: I1


IBB
2I1


b 3h3 6(b 2  h2 )

Problem 12.4-6 Calculate the moment of inertia Ix for the composite

y

circular area shown in the figure. The origin of the axes is at the center of the concentric circles, and the three diameters are 20, 40, and 60 mm. x

Solution 12.4-6

Moment of inertia of composite area Diameters
20, 40, and 60 mm

y


d 4 (for a circle) 64
Ix
[ (60) 4  (40) 4  (20) 4 ] 64

Ix
x

Ix
518  103 mm4

735

736

CHAPTER 12 Review of Centroids and Moments of Inertia

Problem 12.4-7 Calculate the moments of inertia Ix and Iy with respect to the x and y axes for the L-shaped area shown in the figure for Prob. 12.3-7.

Solution 12.4-7

Moments of inertia of composite area y

y

Ix
I1  I2 1 1
(3.5)(0.5) 3  (0.5)(6) 3 3 3
36.1 in.4 Iy
I3  I4 1 1
(0.5)(4) 3  (5.5)(0.5) 3 3 3 4
10.9 in.

A4

A2 6 in.

6 in.

A3

A1 O

O

x 4 in.

x

4 in.

Thickness t = 0.5 in.

y

Problem 12.4-8 A semicircular area of radius 150 mm has a rectangular cutout of dimensions 50 mm  100 mm (see figure). Calculate the moments of inertia Ix and Iy with respect to the x and y axes. Also, calculate the corresponding radii of gyration rx and ry.

50 mm

O 50 50 mm mm 150 mm

Solution 12.4-8

150 mm

Moments of inertia of composite area All dimensions in millimeters

y

r
150 mm

50 mm

O 50 50 mm mm 150 mm

150 mm

x

b
100 mm

h
50 mm


r4 bh3  Ix
(Ix ) semicircle  (Ix ) rectangle
8 3
194.6  106 mm4 Iy
Ix A



r 2  bh
30.34  103 mm2 2

rx
Ix A
80.1 mm ry
rx

x

SECTION 12.5 Parallel-Axis Theorem

Problem 12.4-9 Calculate the moments of inertia I1 and I2 of a W 16  100 wide-flange section using the cross-sectional dimensions given in Table E-l, Appendix E. (Disregard the cross-sectional areas of the fillets.) Also, calculate the corresponding radii of gyration r1 and r2, respectively.

Solution 12.4-9

Moments of inertia of a wide-flange section All dimensions in inches.

2

1 1 bd3  (b  tw )(d  2tf ) 3 12 12 1 1
(10.425)(16.97) 3  (9.840)(15.00) 3 12 12

I1


tf

C

1


1478 in.4

d

1

say,

I1
1480 in.4

1 1 ≤ tf b3  (d  2tf )t 3w 12 12 1 1
(0.985)(10.425) 3  (15.00)(0.585) 3 6 12

I2
2 ¢

tw 2 b


186.3 in.4 W 16  100 d
16.97 in. tw
tweb
0.585 in. b
10.425 in. tf
tflange
0.985 in.

say,

I2
186 in.4

A
2(btf)  (d  2tf)tw
2(10.425)(0.985)  (15.00)(0.585)
29.31 in.2 r1
I1 A
7.10 in. r2
I2 A
2.52 in. Note that these results are in close agreement with the tabulated values.

Parallel-Axis Theorem Problem 12.5-1 Calculate the moment of inertia Ib of a W 12  50

wide-flange section with respect to its base. (Use data from Table E-l, Appendix E.)

Solution 12.5-1

Moment of inertia W 12  50 I1
394 in.4 d
12.19 in. 1

B

C

A
14.7 in.2

d 2 ≤ 2
394  14.7(6.095)2
940 in.4

Ib
I1  A ¢

1

B

737

738

CHAPTER 12 Review of Centroids and Moments of Inertia

Problem 12.5-2 Determine the moment of inertia Ic with respect to an axis through the centroid C and parallel to the x axis for the geometric figure described in Prob. 12.3-2.

Solution 12.5-2

Moment of inertia From Prob. 12.3-2:

y

A
3a 2 4

yc a — 2

y
5a 12

A1 xc

C

a — 2

A2 O

a — 2

1 a 3 1 a a 3 3a 4 ¢ ≤ (a )  ¢ ≤ ¢ ≤
3 2 3 2 2 16

Ix


Ix
IxC  Ay2

y x

a — 2

Ic
IxC
Ix  Ay2


3a4 3a2 5a 2  ¢ ≤ 16 4 12

11a 4 192




Problem 12.5-3 For the channel section described in Prob. 12.3-3, calculate the moment of inertia Ix with respect to an axis through c the centroid C and parallel to the x axis. Solution 12.5-3

Moment of inertia y

1 in. 2 in. 3 in.

yc

C

From Prob. 12.3-3:

1 in.

y

xc 1 in.

3 in.

O

3 in.

x

A
10.0 in.2 y
1.10 in. Ix
13(4)(1)3  2(13)(1)(3)3
19.33 in.4 Ix
IxC  Ay2 Ixc
Ix  Ay2
19.33  (10.0)(1.10) 2
7.23 in.4

Problem 12.5-4 The moment of inertia with respect to axis 1-1 of the scalene triangle shown in the figure is 90  103 mm4. Calculate its moment of inertia I2 with respect to axis 2-2. 1

1 40 mm

15 mm

2

Solution 12.5-4

Moment of inertia b
40 mm

C 40 mm

1 2

I1
90  103 mm4

I1
bh312

12I1
30 mm B b Ic
bh336
30  103 mm4 h


h h 3 1 15 mm 2

2

3

I2
Ic  Ad 2
Ic  (bh 2) d 2
30  103 1  (40) (30) (25) 2
405  103 mm4 2

SECTION 12.5 Parallel-Axis Theorem

Problem 12.5-5 For the beam cross section described in Prob. 12.3-5, calculate the centroidal moments of inertia Ix and Iy with respect to axes c c through the centroid C such that the xc axis is parallel to the x axis and the yc axis coincides with the y axis. Solution 12.5-5

Moment of inertia

d 2

PLATE

3 8— 4 in.

y, yc

W 24  162 xc

C

1 y

d 2

I–yc
1 12(3 4)(8) 3
32.0 in.4 ENTIRE CROSS SECTION Ixc
I¿xc  I–xc
5269  785
6050 in.4

x

O

I–xc
1 12(8)(3 4) 3  (8)(3 4)(d  3 8  y) 2
0.2813  6(25.00  0.375  13.94)2
0.2813  6(11.44)2
785 in.4

Iyc
I¿yc  I–yc
443  32
475 in.4

From Prob. 12.3-5: y
13.94 in. W 24  162

d
25.00 in.

I1
5170 in.4

d2
12.5 in.

A
47.7 in.2

I2
Iy
443 in.4 I¿xc
I1  A(y  d 2) 2
5170  (47.7)(1.44) 2
5269 in.4 I¿yc
I2
443 in.4

Problem 12.5-6 Calculate the moment of inertia Ixc with respect to an axis

through the centroid C and parallel to the x axis for the composite area shown in the figure for Prob. 12.3-6.

Solution 12.5-6

Moment of inertia From Prob. 12.3-6: y
52.50 mm y

360 mm

A1

30 mm

A2

C 120 mm

y

A4

x

O

30 mm 120 mm

t
30 mm

A
21,600 mm2

A1: Ix
112(360) (30)3  (360) (30) (105)2
119.9  106 mm4 A2: Ix
112(120) (30)3  (120) (30) (75)2
20.52  106 mm4 A3: Ix
112(30) (120)3
4.32  106 mm4 A4: Ix
20.52  106 mm4

A3 ENTIRE AREA: Ix
a Ix
165.26  106 mm4 IxC
Ix  Ay2
165.26  106  (21,600)(52.50) 2 ˇ


106 

106

mm4

739

740

CHAPTER 12 Review of Centroids and Moments of Inertia

Problem 12.5-7 Calculate the centroidal moments of inertia Ixc and Iyc

with respect to axes through the centroid C and parallel to the x and y axes, respectively, for the L-shaped area shown in the figure for Prob. 12.3-7.

Solution 12.5-7

Moments of inertia

y

From Prob. 12.3-7:

A2

yc

t
0.5 in. A
4.75 in.2 y
1.987 in.

0.5 in.

x
0.9869 in. From Problem 12.4-7:

x 6 in. C y

3.5 in. O

Ix
36.15 in.4 Iy
10.90 in.4 Ixc
Ix  Ay2
36.15  (4.75) (1.987) 2

xc

x

4 in.


17.40 in.4 Iyc
Iy  Ax2
10.90  (4.75)(0.9869) 2

A1


6.27 in.4

y

Problem 12.5-8 The wide-flange beam section shown in the figure has a total height of 250 mm and a constant thickness of 15 mm. Determine the flange width b if it is required that the centroidal moments of inertia Ix and Iy be in the ratio 3 to 1, respectively.

15 mm

b

250 mm

C

x 15 mm

15 mm b

Solution 12.5-8

Wide-flange beam y

All dimensions in millimeters. 15 mm

b

250 mm

C

15 mm

b

b
flange width

1 1 3 (220)(15) 3 ≤ (15)(b)  12 12
2.5 b3  61,880 (mm4)

Iy
2 ¢

x

15 mm

t
15 mm

1 1 (b)(250) 3  (b  15)(220) 3 12 12
0.4147  106 b  13.31  106 (mm)4

Ix


Equate Ix to 3Iy and rearrange: 7.5 b3  0.4147  106 b  13.12  10 6
0 Solve numerically: b
250 mm

SECTION 12.6 Polar Moments of Inertia

Polar Moments of Inertia Problem 12.6-1 Determine the polar moment of inertia IP of an isosceles triangle of base b and altitude h with respect to its apex (see Case 5, Appendix D) Solution 12.6-1

Polar moment of inertia y

POINT A (APEX):

A

IP
(IP ) c  A ¢ 2/3 h h

2h 2 ≤ 3

bh bh 2h 2 (4h2  3b 2 )  ¢ ≤ 144 2 3 bh 2 IP
(b  12h2 ) 48


C b

POINT C (CENTROID) FROM CASE 5: (IP ) c


bh (4h2  3b 2 ) 144

Problem 12.6-2 Determine the polar moment of inertia (IP)C with respect to the centroid C for a circular sector (see Case 13, Appendix D).

Solution 12.6-2

Polar moment of inertia A

r 2

y C


y

y

r

POINT C (CENTROID): x

O

(IP ) C
(IP ) O  Ay2


POINT O (ORIGIN) FROM CASE 13: (IP ) o



r 4 2

2r sin
3





(

radians)


r4 2r sin
2 
r2 ¢ ≤ 2 3


r4 (9
2  8 sin2
) 18


Problem 12.6-3 Determine the polar moment of inertia IP for a W 8  21 wide-flange section with respect to one of its outermost corners. Solution 12.6-3

Polar moment of inertia 2

1

C

1

y

W 8  21 I1
75.3 in.4 I2
9.77 in.4 A
6.16 in.2 Depth d
8.28 in. Width b
5.27 in. Ix
I1  A(d2)2
75.3  6.16(4.14)2
180.9 in.4 Iy
I2  A(b2)2
9.77  6.16(2.635)2
52.5 in.4

x

O 2

IP
Ix  Iy
233 in.4

741

742

CHAPTER 12 Review of Centroids and Moments of Inertia

Problem 12.6-4 Obtain a formula for the polar moment of inertia IP with respect to the midpoint of the hypotenuse for a right triangle of base b and height h (see Case 6, Appendix D).

Solution 12.6-4

Polar moment of inertia POINT P:

d = CP

P

h h/2

C h/3 b/3 b/2 b

POINT C FROM CASE 6: (IP ) c


bh 2 (h  b 2 ) 36

IP
(IP ) c  Ad 2 bh A
2 b b 2 h h 2 d2
¢  ≤  ¢  ≤ 2 3 2 3 2 2 2 b h b  h2

36 36 36 bh 2 bh b 2  h2 IP
(h  b 2 )  ¢ ≤ 36 2 36 bh 2
(b  h2 ) 24

Problem 12.6-5 Determine the polar moment of inertia (IP)C with respect to the centroid C for a quarter-circular spandrel (see Case 12, Appendix D).

Solution 12.6-5 y

Polar moment of inertia yc

POINT C (CENTROID): 5 4 ≤r 16 (10  3)r 2   ¢ 1  ≤ (r 2 ) B R 4 3(4  )

Ix c
Ix  Ay2
¢ 1 

x r

C O

y

xc x

COLLECT TERMS AND SIMPLIFY: POINT O FROM CASE 12: 5 4 Ix
¢ 1  ≤r 16 (10  3)r y
3(4  )  A
¢ 1  ≤r 2 4

r4 176  84   9 2 ¢ ≤ 144 4 IyC
IxC (by symmetry) IxC


(IP ) C
2 IxC


r4 176  84  92 ¢ ≤ 72 4

SECTION 12.7 Products of Inertia

Products of Inertia Problem 12.7-1 Using integration, determine the product of inertia Ixy for the parabolic semisegment shown in Fig. 12-5 (see also Case 17 in Appendix D).

Solution 12.7-1 Product of inertia Product of inertia of element dA with respect to axes through its own centroid equals zero. x2 dA
y dx
h ¢ 1  2 ≤ dx b dIxy
product of inertia of element dA with respect to xy axes d1
x d2
y2


dI

xy




h2 2

b


x

¢1 

0

x2 b2

)

dA

h

Parallel-axis theorem applied to element dA: dIxy
0  (dA)(d1d2)
(y dx)(x)(y 2) h2x x2 2
¢ 1  2 ≤ dx 2 b Ixy


(

y
h 1

y

y/2 O

x b

dx

x

x2 2 b 2h2 2 ≤ dx
12 b

Problem 12.7-2 Using integration, determine the product of inertia Ixy for the quarter-circular spandrel shown in Case 12, Appendix D.

Solution 12.7-2

Product of inertia ELEMENT dA:

y

r

dA x

dy y

O

(r  x)/2

EQUATION OF CIRCLE: x 2  ( y  r)2
r 2 or r 2  x 2
(y  r)2

x

d1
distance to its centroid in x direction
(r  x)2 d2
distance to its centroid in y direction
y dA
area of element
(r  x) dy Product of inertia of element dA with respect to axes through its own centroid equals zero. Parallel-axis theorem applied to element dA: rx dIxy
0  (dA)(d1d2 )
(r  x)(dy) ¢ ≤ (y) 2 1 1
(r 2  x 2 ) y dy
(y  r) 2y dy 2 2 r

Ixy
12


y(y  r) dy
24 2

0

r4

743

744

CHAPTER 12 Review of Centroids and Moments of Inertia

Problem 12.7-3 Find the relationship between the radius r and the distance b for the composite area shown in the figure in order that the product of inertia Ixy will be zero.

y

r

x

O b

Solution 12.7-3 yc

Product of inertia

y

C
centroid of semicircle

SEMICIRCLE (CASE 10): Ixy
Ixcyc  Ad1d2

r xc

C

x

O

r 2 2

d1
r

Ixcyc
0

A


Ixy
0  ¢

r 2 4r 2r 4 ≤ (r) ¢  ≤
 2 3 3

b

d2


COMPOSITE AREA (Ixy
0)

TRIANGLE (CASE 7):

Ixy


b 2h2 b 2 (2r) 2 b 2r 2 Ixy


24 24 6

b 2r 2 2r 4 
0 6 3


b
2r

y

Problem 12.7-4 Obtain a formula for the product of inertia Ixy of the symmetrical L-shaped area shown in the figure.

t b t x

O

Solution 12.7-4

Product of inertia

y

AREA 2: (Ixy ) 2
Ixc yc  A2d1d2

t


0  (b  t)(t)(t2) ¢

A1

b

A2




bt t O

b

b

x

t2 2 (b  t 2 ) 4

COMPOSITE AREA: Ixy
(Ixy ) 1  (Ixy ) 2


AREA 1: (Ixy ) 1


t2b2 4

bt ≤ 2

t2 (2b2  t2 ) 4

4r 3

SECTION 12.7 Products of Inertia

Problem 12.7-5 Calculate the product of inertia I12 with respect to the centroidal axes 1-1 and 2-2 for an L 6  6  1 in. angle section (see Table E-4, Appendix E). (Disregard the cross-sectional areas of the fillet and rounded corners.)

Solution 12.7-5

Product of inertia

y

Coordinates of centroid of area A1 with respect to 1–2 axes:

2

d1
 (x  0.5)
 1.3636 in. d2
3.0  y
1.1364 in.

1 in.

A1 1

Product of inertia of area A1 with respect to 1-2 axes:

C

6 in.

I¿12
0  A1d1d2

1 5 in.

y

1 in. 6 in.

O x


(6.0 in.2)(1.3636 in.)(1.1364 in.)
9.2976 in.4 Coordinates of centroid of area A2 with respect to 1–2 axes:

x

A2

d1
3.5  x
1.6364 in. d2
 (y  0.5)
1.3636 in. Product of inertia of area A2 with respect to 1-2 axes:

2

All dimensions in inches. A1
(6)(1)
6.0 in.2 A2
(5)(1)
5.0 in.2 A
A1  A2
11.0 in.2 With respect to the x axis: 6 in. Q 1
(6.0 in.2 ) ¢ ≤
18.0 in.3 2

I–12
0  A2d1d 2
(5.0 in.2)(1.6364 in.)(1.3636 in.)
11.1573 in.4 ANGLE SECTION: I12
I¿12  I–12
20.5 in.4

1.0 in. ≤
2.5 in.3 2 Q 1  Q 2 20.5 in.3 y


1.8636 in. A 11.0 in.2 x
y
1.8636 in.

Q 2
(5.0 in.2 ) ¢

Problem 12.7-6 Calculate the product of inertia Ixy for the composite area shown in Prob. 12.3-6.

Solution 12.7-6

Product of inertia y

AREA A1: (Ixy)1
0

d1

(By symmetry)

AREA A2: (Ixy)2
0  A2 d1d2
(90  30)(60)(75)
12.15  106 mm4

A1

AREA A3: (Ixy)3
0 d2

A2 O

x

A4 A3 All dimensions in millimeters A2
90  30 mm A1
360  30 mm A3
180  30 mm A4
90  30 mm d1
60 mm d2
75 mm

(By symmetry)

AREA A4: (Ixy)4
(Ixy)2
12.15  106 mm4 Ixy
(Ixy)1  (Ixy)2  (Ixy)3  (Ixy)4
(2)(12.15  106 mm4)
24.3  106 mm4

745

746

CHAPTER 12 Review of Centroids and Moments of Inertia

Problem 12.7-7 Determine the product of inertia Ixcyc with respect to centroidal axes xc

and yc parallel to the x and y axes, respectively, for the L-shaped area shown in Prob. 12.3-7.

Solution 12.7-7

Product of inertia With respect to the y axis: Q 1
A1x 1
(3.0 in.2 )(0.25 in.)
0.75 in.3

y yc

t = 0.5 in.

x

Q 2
A2x 2
(1.75 in.2 )(2.25 in.)
3.9375 in.3

C

Q 1  Q 2 4.6875 in.3

0.98684 in. A 4.75 in.2 Product of inertia of area A1 with respect to xy axes:

A1 6.0

xc 3.5

O

x


y

(Ixy)1
(Ixy)centroid  A1 d1 d2
0  (3.0 in.2)(0.25 in.)(3.0 in.)
2.25 in.4 Product of inertia of area A2 with respect to xy axes: (Ixy)2
(Ixy)centroid  A2 d1 d2
0  (1.75 in.2)(2.25 in.)(0.25 in.)
0.98438 in.4

x

4.0

A2 All dimensions in inches. A1
(6.0)(0.5)
3.0 in.2 A2
(3.5)(0.5)
1.75 in.2 A
A1  A2
4.75 in.2

ANGLE SECTION

With respect to the x axis: Q 1
A1 y1
(3.0 in.2 )(3.0 in.)
9.0 in.3 Q 2
A2 y2
(1.75 in.2 )(0.25 in.)
0.4375 in.3 y


Ixy
(Ixy)1  (Ixy)2
3.2344 in.4 CENTROIDAL AXES Ixcyc
Ixy  Ax y

Q 1  Q 2 9.4375 in.3

1.9868 in. A 4.75 in.2


3.2344 in.4  (4.75 in.2)(0.98684 in.)(1.9868 in.)
6.079 in.4

Rotation of Axes

y

y1

The problems for Section 12.8 are to be solved by using the transformation equations for moments and products of inertia.

Problem 12.8-1 Determine the moments of inertia Ix1 and Iy1 and the



b

C

product of inertia Ix y for a square with sides b, as shown in the figure. (Note 1 1 that the x1y1 axes are centroidal axes rotated through an angle  with respect to the xy axes.)

Solution 12.8-1

b

EQ. (12-29): Ix1  Iy1
Ix  Iy 

b

FOR A SQUARE: b4 Ix
Iy
 12

b

EQ. (12-25):




Ix  Iy 2 Ix  Iy 2



Ix  Iy 2

cos 2u  Ixy sin 2u

00



Iy1


b4 12

x1

EQ. (12-27):

x

C

Ix1


x

Rotation of axes

y

y1

x1

b4 12

Ix1y1
Ixy
0

Ix  Iy 2

sin 2u  Ixy cos 2u
0

Since  may be any angle, we see that all moments of inertia are the same and the product of inertia is always zero (for axes through the centroid C).

SECTION 12.8 Rotation of Axes

Problem 12.8-2 Determine the moments and product of inertia with respect to the x1y1 axes for the rectangle shown in the figure. (Note that the x1 axis is a diagonal of the rectangle.)

y

y1

x1

h

x

C

b

Solution 12.8-2

Rotation of axes (rectangle) ANGLE OF ROTATION:

y

y1

x1

cos u


 h

Iy


h b  h2 2

SUBSTITUTE INTO EQS. (12-25), (12-29), AND (12-27)

APPENDIX D, CASE 1: bh3 12

sin u


b 2  h2 b 2  h2 2 bh sin 2
2 sin  cos 
2 b  h2

b

Ix


b  h

2

cos 2
cos2   sin2 


x

C

b 2

AND SIMPLIFY:

hb 3 12

Ixy
0

Ix1


b 3h3 6(b 2  h2 )

Ix1y1


Iy1


bh(b 4  h4 ) 12(b 2  h2 )

b 2h2 (h2  b 2 ) 12(b 2  h2 )

Problem 12.8-3 Calculate the moment of inertia Id for a W 12  50 wide-flange section with respect to a diagonal passing through the centroid and two outside corners of the flanges. (Use the dimensions and properties given in Table E-1.)

Solution 12.8-3

Rotation of axes d 12.19 Tan u


1.509 b 8.080 
56.46º 2
112.92º

y

d

 x

C

EQ. (12-25): Id


b

W 12  50 Ix
394 in.4 4 Iy
56.3 in. Ixy
0 Depth d
12.19 in. Width b
8.080 in.

Ix  Iy



Ix  Iy

cos 2u  Ixy sin 2u 2 2 394  56.3 394  56.3
 cos (112.92)  0 2 2
225 in.4  66 in.4
159 in.4

747

748

CHAPTER 12 Review of Centroids and Moments of Inertia

Problem 12.8-4 Calculate the moments of inertia Ix1 and Iy1 and the product of inertia Ix y with respect to the x1y1 axes for the L-shaped area shown in the 1 1 figure if a
150 mm, b
100 mm, t
15 mm, and 
30°.

y y1 t a x1 t

Solution 12.8-4

b

Rotation of axes 1 Ixy
t 2a 2  Ad1d2 4

y y1

x1 
30° b

A = (bt)(t) bt t
d2
d1
t  2 2

1 Ixy
(15) 2 (150) 2  (85)(15)(57.5)(7.5) 4
1.815  106 mm4

a

O

x

O

Probs. 12.8-4 and 12.9-4



SUBSTITUTE into Eq. (12-25) with 
30º:

x

All dimensions in millimeters. a
150 mm b
100 mm t
15 mm 
30º 1 1 Ix
ta 3  (b  t) t 3 3 3 1 1
(15)(150) 3  (85)(15) 3 3 3
16.971  106 mm4 1 1 Iy
(a  t) t 3  tb 3 3 3 1 1 3
(135)(15)  (15)(100) 3 3 3
5.152  106 mm4

Ix1


Ix  Iy



Ix  Iy

cos 2u  Ixy sin 2u 2 2 6
12.44  10 mm4

SUBSTITUTE into Eq. (12-25) with 
120º: Iy1
9.68  106 mm4 SUBSTITUTE into Eq. (12-27) with 
30º: Ix1y1


Ix  Iy

sin 2u  Ixy cos 2u 2
6.03  106 mm4

y1

Problem 12.8-5 Calculate the moments of inertia Ix1 and Iy1 and the product

y

b

of inertia Ix y with respect to the x1y1 axes for the Z-section shown in the 1 1 figure if b
3 in., h
4 in., t
0.5 in., and 
60°.

h — 2

t x1



x

C h — 2 Probs. 12.8-5, 12.8-6, 12.9-5 and 12.9-6

t

b

t

749

SECTION 12.8 Rotation of Axes

Solution 12.8-5

Rotation of axes y

y1 b h — 2

 C

h — 2

A3

A2

h
4.0 in.

PRODUCT OF INERTIA Ixy Area A1: I¿xy
0  (b  t)(t) ¢ 

t
0.5 in.


60º

b h t ≤¢  ≤ 2 2 2

1
 (bt)(b  t)(h  t)
3.2813 in.4 4 Area A2: I–xy
0 Area A3: I‡ xy
I¿xy 4 Ixy
I¿xy  I–xy  I‡ xy
6.5625 in.

MOMENT OF INERTIA Ix 1 h t 2 (b  t)(t 3 )  (b  t)(t) ¢  ≤ 12 2 2
3.8542 in.4 1 Area A2: I–x
(t)(h3 )
2.6667 in.4 12 Area A3: I‡
I¿x
3.8542 in.4 x I¿x


4 Ix
I¿x  I–x  I‡ x
10.3751 in.

SUBSTITUTE into Eq. (12-25) with 
60º: Ix1


Ix  Iy



Ix  Iy

cos 2u  Ixy sin 2u 2 2
13.50 in.4

SUBSTITUTE into Eq. (12-25) with 
150º: Iy1
3.84 in.4

MOMENT OF INERTIA Iy 1 b 2 (t)(b  t) 3  (b  t)(t) ¢ ≤ 12 2 4
3.4635 in.

I¿y


Area A1:

4 I‡ y
I¿y
3.4635 in.

x

All dimensions in inches.

Area A1:

Area A3:

4 Iy
I¿y  I–y  I‡ y
6.9688 in.

x1

b

b
3.0 in.

I–y


t = thickness

A1

1 (h)(t 3 )
0.0417 in.4 12

Area A2:

SUBSTITUTE into Eq. (12-27) with 
60º: Ix  Iy Ix1y1
sin 2u  Ixy cos 2u
4.76 in.4 2

Problem 12.8-6 Solve the preceding problem if b
80 mm, h
120 mm, t
12 mm, and 
30°.

Solution 12.8-6

Rotation of axes All dimensions in millimeters.

y1

y

t
thickness

b h — 2

A1

 C

h — 2

A2

A3

b

x1 x

b
80 mm h
120 mm t
12 mm 
30º MOMENT OF INERTIA Ix 1 h t 2 (b  t)(t 3 )  (b  t)(t) ¢  ≤ 12 2 2
2.3892  106 mm4 1 Area A2: I–x
(t)(h3 )
1.7280  106 mm4 12 Area A1: I¿x


6 4 Area A3: I‡ x
I¿x
2.3892  10 mm 6 4 Ix
I¿x  I–x  I‡ x
6.5065  10 mm

(Continued)

750

CHAPTER 12 Review of Centroids and Moments of Inertia

SUBSTITUTE into Eq. (12-25) with 
30º:

MOMENT OF INERTIA Iy 1 b 2 (t)(b  t) 3  (b  t)(t) ¢ ≤ 12 2
1.6200  106 mm4 1 Area A2: I–y
(h)(t 3 )
0.01728  106 mm4 12
I¿y
1.6200  106 mm4 Area A3: I‡ y Area A1: I¿y


Ix1


Ix  Iy 2



Ix  Iy 2

cos 2u  Ixy sin 2u


8.75  106 mm4 SUBSTITUTE into Eq. (12-25) with 
120º: Iy1
1.02  106 mm4

6 4 Iy
I¿y  I–y  I‡ y
3.2573  10 mm

SUBSTITUTE into Eq. (12-27) with 
30º: PRODUCT OF INERTIA Ixy b h t Area A1: I¿xy
0  (b  t)(t) ¢  ≤ ¢  ≤ 2 2 2 1
 (bt)(b  t)(h  t)
 1.7626  106 mm4 4 Area A2: I–xy
0

Ix1y1


Ix  Iy 2

sin 2u  Ixy cos 2u


0.356  106 mm4

Area A3: I‡ xy
I¿xy

6 4 Ixy
I¿xy  I–xy  I‡ xy
3.5251  10 mm

Principal Axes, Principal Points, and Principal Moments of Inertia

y

Problem 12.9-1 An ellipse with major axis of length 2a and minor axis of length 2b is shown in the figure. (a) Determine the distance c from the centroid C of the ellipse to the principal points P on the minor axis (y axis). (b) For what ratio a/b do the principal points lie on the circumference of the ellipse? (c) For what ratios do they lie inside the ellipse?

P C

Principal points of an ellipse y

From Case 16: Iy
P1 C

xp

c

x

c

A
ab

Ixp
Ix  Ac2


a

(a) LOCATION OF PRINCIPAL POINTS At a principal point, all moments of inertia are equal. At point P1: Ixp
Iy

ab 3 4

ba 3 4

Parallal-axis theorem:

b

P2 a

Ix


b

Eq. (1)

b

c

b

x

P a

Solution 12.9-1

c

ab 3  abc2 4

Substitute into Eq. (1): ab 3 ba 3  abc2
4 4 1 Solve for c: c
a 2  b 2 2

a

751

SECTION 12.9 Principal Axes, Principal Points, and Principal Moments of Inertia

(b) PRINCIPAL POINTS ON THE CIRCUMFERENCE

(c) PRINCIPAL POINTS INSIDE THE ELLIPSE

1
c
b and b
a 2  b 2 2 a a Solve for ratio :
5 b b


0cb

For c
0:

a
b and a
5 b

For c
b:
1

a 6 5 b y

Problem 12.9-2 Demonstrate that the two points P1 and P2, located as shown in the figure, are the principal points of the isosceles right triangle.

b — 6 b — 6 b — 6

b — 2

P2 x

C P1 b — 2

b — 2

Solution 12.9-2

a
1 b

Principal points of an isosceles right triangle y3

y1 y2

x2 y2 2

yC

xC

x2

P2

3

1

x4

y4

x3

d 4

C x1

P1

CONSIDER POINT P1: Ix1 y1
0 because y1 is an axis of symmetry. Ix2 y2
0 because areas 1 and 2 are symmetrical about the y2 axis and areas 3 and 4 are symmetrical about the x2 axis. Two different sets of principal axes exist at point P1.
P1 is a principal point

CONSIDER POINT P2: Ix3 y3
0 because y3 is an axis of symmetry. Ix2 y2
0 (see above). Parallel-axis theorem: Ix2y2
Ixcyc  Ad1d2

A


b2 4

d
d1
d2


b2 b 2 b4 Ixcyc
 ¢ ≤ ¢ ≤
 4 62 288

b 62

Parallel-axis theorem: Ix4y4
Ixcyc  Ad1d2 4

2

d1
d2
 2

b 62

b b b  ¢ ≤
0 288 4 62 Two different sets of principal axes (x3y3 and x4y4) exist at point P2.
P2 is a principal point Ix4y4


752

CHAPTER 12 Review of Centroids and Moments of Inertia

y

Problem 12.9-3 Determine the angles p1 and p2 defining the orientations

of the principal axes through the origin O for the right triangle shown in the figure if b
6 in. and h
8 in. Also, calculate the corresponding principal moments of inertia I1 and I2.

y1 h x1  O

Solution 12.9-3

x

b

Principal axes y

EQ. (12-30): tan 2up


y1

2p
59.744º h x1  x

O

b

p
29.872º


1.71429 Ix  Iy 120.256º

and and

2Ixy

60.128º

SUBSTITUTE into Eq. (12-25) with 
29.872º: Ix1
311.1 in.4

RIGHT TRIANGLE b
6.0 in.

h
8.0 in.

SUBSTITUTE into Eq. (12-25) with 
60.128º: Ix1
88.9 in.4

CASE 7: Ix


bh3
256 in.4 12

THEREFORE, I1
311.1 in.4 up1
 29.87 I2
88.9 in.4 up2
60.13

hb 3 Iy

144 in.4 12 b 2h2 Ixy

96 in.4 24

}

NOTE: The principal moments of inertia can be verified with Eqs. (12-33a and b) and Eq. (12-29).

Problem 12.9-4 Determine the angles p1 and p2 defining the orientations of the principal axes through the origin O and the corresponding principal moments of inertia I1 and I2 for the L-shaped area described in Prob. 12.8-4 (a
150 mm, b
100 mm, and t
15 mm). Solution 12.9-4

Principal axes ANGLE SECTION y

y1

t
thickness

a
150 mm

b
100 mm

t
15 mm

FROM PROB. 12.8-4: a x1

EQ. (12-30):

 O

b

Ix
16.971  106 mm4 Iy
5.152  106 mm4

x

Ixy
1.815  106 mm4 2 Ixy
 0.3071 tan 2up
 Ix  Iy

2p
17.07º and 162.93º p
8.54º and 81.46º

753

SECTION 12.9 Principal Axes, Principal Points, and Principal Moments of Inertia

SUBSTITUTE into Eq. (12-25) with 
8.54º:

THEREFORE,

Ix1
17.24  106 mm4

I1
17.24  106 mm4 up1
8.54 I2
4.88  106 mm4 up2
81.46

SUBSTITUTE into Eq. (12-25) with 
81.46º:

NOTE: The principal moments of inertia I1 and I2 can be verified with Eqs. (12-33a and b) and Eq. (12-29).

Ix1
4.88  106 mm4

Problem 12.9-5 Determine the angles p1 and p2 defining the orientations of the principal axes through the centroid C and the corresponding principal centroidal moments of inertia I1 and I2 for the Z-section described in Prob. 12.8-5 (b
3 in., h
4 in., and t
0.5 in.).

Solution 12.9-5

Principal axes y

y1

tan 2up


EQ. (12-30): x1

h — 2



Ix  Iy

2p
75.451º

and

255.451º

p
37.726º

and

127.726º


3.8532

SUBSTITUTE into Eq. (12-25) with 
37.726º:

x

C

2 Ixy

h — 2

Ix1
15.452 in.4 SUBSTITUTE into Eq. (12-25) with 
127.726º:

b

Ix1
1.892 in.4

Z-SECTION

THEREFORE, I1
15.45 in.4

t
thickness
0.5 in. b
3.0 in h
4.0 in

I2
1.89

in.4

up1
37.73 up2
127.73

}

FROM PROB. 12.8-5: Ix
10.3751 in.4 Ixy
6.5625 in.4

NOTE: The principal moments of inertia I1 and I2 can be verified with Eqs. (12-33a and b) and Eq. (12-29).

Iy
6.9688 in.4

Problem 12.9-6 Solve the preceding problem for the Z-section described in Prob. 12.8-6 (b
80 mm, h
120 mm, and t
12 mm). Solution 12.9-6

Principal axes y

y1

Z-SECTION x1

h — 2



t
thickness
12 mm b
80 mm h
120 mm

x

C h — 2

FROM PROB. 12.8-6:

b

Iy
3.2573  106 mm4 Ix
6.5065  106 mm4 6 4 Ixy
3.5251  10 mm (Continued)

754

CHAPTER 12 Review of Centroids and Moments of Inertia

tan 2up


Eq. (12-30): 2p
65.257º p
32.628º

and and

2 Ixy

Ix  Iy 245.257º 122.628º


2.1698

THEREFORE, I1
8.76  106 mm4 up1
32.63 I2
1.00  106 mm4 up2
122.63

SUBSTITUTE into EQ. (12-25) with 
32.628º: Ix1
8.763  106 mm4

}

NOTE: The principal moments of inertia I1 and I2 can be verified with Eqs. (12-33a and b) and Eq. (12-29).

SUBSTITUTE into Eq. (12-25) with 
122.628º: Ix1
1.000  106 mm4

y

y1

Problem 12.9-7 Determine the angles p1 and p2 defining the

orientations of the principal axes through the centroid C for the right triangle shown in the figure if h
2b. Also, determine the corresponding principal centroidal moments of inertia I1 and I2.

x1

h

 x

C

b

Solution 12.9-7

Principal axes y

y1

EQ. (12-30):

x1

h
2b



and and

2 Ixy




Ix  Iy 213.6901º 106.8450º

2 3

SUBSTITUTE into Eq. (12-25) with 
16.8450º: x

C

2p
33.6901º p
16.8450º

tan 2up


Ix1
0.23904 b4 SUBSTITUTE into Eq. (12-25) with 
106.8450º:

b

RIGHT TRIANGLE h
2b CASE 6 bh3 2b 4
36 9 hb 3 b 4 Iy

36 18 2 2 b h b4
 Ixy
 72 18

Ix


Ix1
0.03873 b4 THEREFORE, I1
0.2390 b4 up1
16.85 I2
0.0387 b4 up2
106.85

}

NOTE: The principal moments of inertia I1 and I2 can be verified with Eqs. (12-33a and b) and Eq. (12-29).

755

SECTION 12.9 Principal Axes, Principal Points, and Principal Moments of Inertia

Problem 12.9-8 Determine the angles p1 and p2 defining the orientations of the principal centroidal axes and the corresponding principal moments of inertia I1 and I2 for the L-shaped area shown in the figure if a
80 mm, b
150 mm, and t
16 mm.

yc

y1

x1

t a



C

Principal axes (angle section)

y x

yc

t
thickness 

a

xc

C O

1 t 2 1 (a)(t 3 )  A1 ¢ ≤  (t)(b  t 3 ) 12 2 12 bt 2  A2 ¢ ≤ 2 1 1
(80)(16) 3  (1280)(8) 2  (16)(134) 3 12 12 166 2  (2144) ¢ ≤ 2

Iy


x1

A1

y1

xc b

Probs. 12.9-8 and 12.9-9

Solution 12.9-8

t

y

x

b

A2 a
80 mm b
150 mm t
16 mm A1
at
1280 mm2 A2
(b  t)(t)
2144 mm2 A
A1  A2
t (a  b  t)
3424 mm2 LOCATION OF CENTROID C a t Qx
a Ai yi
(at) ¢ ≤  (b  t)(t) ¢ ≤ 2 2
68,352 mm3 Q 68,352 mm3 y
x

19.9626 mm A 3,424 mm2 t bt Qy
a Ai xi
(at) ¢ ≤  (b  t)(t) ¢ ≤ 2 2 3
188,192 mm Qy 188,192 mm3 x


54.9626 mm A 3,424 mm2


18.08738  106 mm4 MOMENTS OF INERTIA (xcyc AXES) Use parallel-axis theorem. IxC
Ix  Ay2
2.91362  106  (3424)(19.9626) 2
1.54914  106 mm4 IyC
Iy  Ax2
18.08738  106  (3424)(54.9626) 2
7.74386  106 mm4 PRODUCT OF INERTIA

Area A1: I¿xCyC
0  A1 B  ¢ x 

t e ≤R B  yR 2 2


(1280)(8  54.9626)(40  19.9626)
 1.20449  106 mm4 Area A2: I–xCyC
0  A2 B

bt t  x R B ¢ y  ≤ R 2 2


(2144)(83  54.9626)(8  19.9626)
 0.71910  106 mm4

MOMENTS OF INERTIA (xy AXES) Use parallel-axis theorem. 1 a 2 1 t 2 Ix
(t)(a 3 )  A1 ¢ ≤  (b  t)(t 3 )  A2 ¢ ≤ 12 2 12 2 1 1
(16)(80) 3  (1280)(40) 2  (134)(16) 3 12 12  (2144)(8)2
2.91362  106 mm4

Ixy
Icentroid  A d1d2

Use parallel-axis theorem:

IxCyC
I¿xCyC  I–xCyC
1.92359  106 mm4 SUMMARY IxC
1.54914  106 mm4

IyC
7.74386  106 mm4

IxCyC
 1.92359  106 mm4

(Continued)

756

CHAPTER 12 Review of Centroids and Moments of Inertia

SUBSTITUTE into Eq. (12-25) with 
74.0790°

PRINCIPAL AXES tan 2up


Eq. (12-30):

2Ixy Ix  Iy


 0.621041

2p
31.8420° and 148.1580° p
15.9210° and 74.0790°

Ix1
8.2926  106 mm4 THEREFORE, I1
8.29  106 mm4 I2
1.00  106 mm4

up1
74.08 up2
15.92

}

SUBSTITUTE into Eq. (12-25) with 
 15.9210° Ix1
1.0004  106 mm4

NOTE: The principal moments of inertia I1 and I2 can be verified with Eqs. (12-33a and b) and Eq. (12-29).

Problem 12.9-9 Solve the preceding problem if a
3 in., b
6 in.,

and t
5/8 in.

Solution 12.9-9

Principal axes (angle section) y x

MOMENTS OF INERTIA (xy AXES)

x1

yc

Use parallel-axis theorem.

A1

y1



a

xc

C O

1 a 2 1 t 2 (t)(a 3 )  A1 ¢ ≤  (b  t)(t 3 )  A2 ¢ ≤ 12 2 12 2 1 5 1 5 3
¢ ≤ (3.0) 3  (1.875)(1.5) 2  (5.375) ¢ ≤ 12 8 12 8 2 5  (3.35938) ¢ ≤ 16
6.06242 in.4

Ix
y

x

b

A2 a
3.0 in. b
6.0 in. t
58 in. A1
at
1.875 in.2 A2
(b  t)(t)
3.35938 in.2 A
A1  A2
t (a  b  t)
5.23438 in.2 LOCATION OF CENTROID C a t Qx
a Aiyi
(at) ¢ ≤  (b  t)(t) ¢ ≤ 2 2
3.86230 in.3 Q x 3.86230 in.3 y

0.73787 in.
A 5.23438 in.2 t bt ≤ Qy
a Ai xi
(at) ¢ ≤  (b  t)(t) ¢ 2 2
11.71387 in.3 Qy 11.71387 in.3 x


2.23787 in. A 5.23438 in.2

1 t 2 1 (a)(t 3 )  A1 ¢ ≤  (t)(b  t 3 ) 12 2 12 bt 2  A2 ¢ ≤ 2 1 5 3 5 2 1 5
(3.0) ¢ ≤  (1.875) ¢ ≤  ¢ ≤ (5.375) 3 12 8 16 12 8 6.625 2  (3.35938) ¢ ≤ 2
45.1933 in.4

Iy


MOMENTS OF INERTIA (xcyc AXES) Use parallel-axis theorem. IxC
Ix  Ay2
6.06242  (5.23438)(0.73787) 2
3.21255 in.4 IyC
Iy  Ax2
45.1933  (5.23438)(2.23787) 2
18.97923 in.4

SECTION 12.9 Principal Axes, Principal Points, and Principal Moments of Inertia

SUBSTITUTE into Eq. (12-25) with 
14.2687°

PRODUCT OF INERTIA Use parallel-axis theorem:

Ixy
Icentroid  A d1d2

t a ≤R B  yR 2 2
(1.875)(1.92537)(0.76213)
2.75134 in.4 bt t  x R B ¢ y ≤ R Area A2: I–xCyC
0  A2 B 2 2 Area A1: I¿xCyC
0  A1 B ¢ x  ˇ

Ix1
2.1223 in.4 SUBSTITUTE into Eq. (12-25) with 
75.7313º Ix1
20.0695 in.4 THEREFORE,

ˇ


(3.35938)(1.07463)(0.42537)
1.53562 in.4 IxCyC
I¿xCyC  I–xCyC
4.28696 in.4 SUMMARY IxC
3.21255 in.4

IyC
18.97923 in.4

IxCyC
 4.28696 in.4 PRINCIPAL AXES EQ. (12-30):

tan 2up


2Ixy Ix  Iy

2p
28.5374° and 151.4626° p
14.2687° and 75.7313°


0.54380

I1
20.07 in.4

up1
75.73

I2
2.12

up2
14.27

in.4

}

NOTE: The principal moments of inertia I1 and I2 can be verified with Eqs. (12-33a and b) and Eq. (12-29).

757