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An Instructor’s Solutions Manual to Accompany
ISBN13: 9780495244585 ISBN10: 0495244589 90000
9 780495 244585
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ISBN13: 9780495244585 ISBN10: 0495244589
© 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, 18004230563. 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
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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 ThinWalled Tubes 328 Stress Concentrations in Torsion 338
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CONTENTS
4. Shear Forces and Bending Moments Shear Forces and Bending Moments 343 ShearForce and BendingMoment 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 BuiltUp Beams 466 Beams with Axial Loads 475 Stress Concentrations 492
6. Stresses in Beams (Advanced Topics) Composite Beams 497 TransformedSection Method 508 Beams with Inclined Loads 520 Bending of Unsymmetric Beams 529 Shear Stresses in WideFlange Beams 541 Shear Centers of ThinWalled 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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CONTENTS
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 BendingMoment Equation 714 Deflections by Integration of the Shear Force and Load Equations 722 Method of Superposition 730 MomentArea 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 ParallelAxis 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 1.21 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 1.21 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 1.22 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 460mm 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 1.22
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 1.23 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 crosssectional 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 1.23 Apad 0.625 in.2
Acable 0.00167 in.2 (a) CANTILEVER BRAKESBRAKING 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 Vbrakes (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 1.24 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 1.24 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 1.25
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 1.25 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 1.26 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 crosssectional area of 490 mm2, and the angle a of the incline is 30°. Calculate the tensile stress st in the cable.
Solution 1.26 Car on inclined track FREEBODY 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 1.27 Two steel wires support a moveable overhead camera weighing W 25 lb (see figure) used for closeup 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 1.27 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 1.28 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 3meter 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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CHAPTER 1 Tension, Compression, and Shear
Solution 1.28 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 FREEBODY 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 1.29 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 1.29 (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 1.210 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 crosssectional 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 1.210 (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 ★1.211 An Lshaped 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 H1 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 1.211 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 xaxis to get T3v, then T3 2. sum about yaxis to get T2v, then T2 3. sum vertical forces to get T1v, then T1 OR sum forces in xdir 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 Zdirection
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 *1.212
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 1.212
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 crosssectional 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 1.213 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 crosssectional 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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SECTION 1.2 Normal Stress and Strain
Solution 1.213 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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z
Problem 1.214 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
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2
Cr an e
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55°
TBQ
O
5m TAQ
5m
x
5m
A
3m
Solution 1.214 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 1.31 Imagine that a long steel wire hangs vertically from a highaltitude 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 H1, Appendix H.)
Solution 1.31
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 crosssectional 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 1.32 Imagine that a long wire of tungsten hangs vertically from a highaltitude 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 H1, Appendix H.)
15
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.32
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 crosssectional 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 1.33 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 1.33 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 1.34 The strengthtoweight ratio of a structural material is defined as its loadcarrying capacity divided by its weight. For materials in tension, we may use a characteristic tensile stress (as obtained from a stressstrain 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 strengthtoweight 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 strengthtoweight ratio (in units of meters) for each of the following materials: aluminum alloy 6061T6, Douglas fir (in bending), nylon, structural steel ASTMA572, and a titanium alloy. (Obtain the material properties from Tables H1 and H3 of Appendix H. When a range of values is given in a table, use the average value.)
Solution 1.34 Strengthtoweight ratio The ultimate stress U for each material is obtained from Table H3, Appendix H, and the weight density is obtained from Table H1. The strengthtoweight 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 ASTMA572
500
77.0
6.5 103
Titanium alloy
1050
44.0
23.9 103
Aluminum alloy 6061T6
Titanium has a high strengthtoweight ratio, which is why it is used in space vehicles and highperformance 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 1.35 A symmetrical framework consisting of three pinconnected 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 stressstrain diagram shown in Fig. 113. (Express the stress in USCS units.)
A
B
C a
D P
Solution 1.35 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 stressstrain diagram of Figure 113
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 stressstrain diagram of Figure 113:
⬇ 31 ksi
;
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SECTION 1.3 Mechanical Properties of Materials
Problem 1.36 A specimen of a methacrylate plastic is tested in tension at room temperature (see figure), producing the stressstrain data listed in the accompanying table. Plot the stressstrain curve and determine the proportional limit, modulus of elasticity (i.e., the slope of the initial part of the stressstrain curve), and yield stress at 0.2% offset. Is the material ductile or brittle?
19
STRESSSTRAIN DATA FOR PROBLEM 1.36
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 1.36 Tensile test of a plastic Using the stressstrain data given in the problem statement, plot the stressstrain 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 1.37 The data shown in the accompanying table were obtained from a tensile test of highstrength steel. The test specimen had a diameter of 0.505 in. and a gage length of 2.00 in. (see figure for Prob. 1.33). At fracture, the elongation between the gage marks was 0.12 in. and the minimum diameter was 0.42 in. Plot the conventional stressstrain curve for the steel and determine the proportional limit, modulus of elasticity (i.e., the slope of the initial part of the stressstrain curve), yield stress at 0.1% offset, ultimate stress, percent elongation in 2.00 in., and percent reduction in area.
TENSILETEST DATA FOR PROBLEM 1.37 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 1.37 Tensile test of highstrength steel d0 0.505 in. A0
pd20 4
L0 2.00 in.
ENLARGEMENT OF PART OF THE STRESSSTRAIN 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
STRESSSTRAIN 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 1.41 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 stressstrain 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. 118b.)
Solution 1.41 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 1.42 A bar of length 2.0 m is made of a structural steel having the stressstrain diagram shown in the figure. The yield stress of the steel is 250 MPa and the slope of the initial linear part of the stressstrain 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. 118b.)
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.42 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 1.43 An aluminum bar has length L 5 ft and diameter d 1.25 in. The stressstrain curve for the aluminum is
shown in Fig. 113 of Section 1.3. The initial straightline 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. 118b and 119.)
Solution 1.43 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 113 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 1.44 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. 118b and 119.)
23
200 s (MPa) 100
0
0
0.005 e
0.010
Solution 1.44 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 1.45 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 stressstrain 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 stressstrain 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 1.45 Wire stretched by forces P L 4 ft 48 in.
d 0.125 in.
ALTERNATIVE FORM OF THE STRESSSTRAIN 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) STRESSSTRAIN 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 stressstrain 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 STRESSSTRAIN 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 1.51 A highstrength 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 1.51 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 highstrength 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 1.52 A round bar of 10 mm diameter is made of aluminum alloy 7075T6 (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 1.52 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)
7075T6 From Table H2: E 72 GPa
v 0.33
From Table H3: 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 1.53 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 1.53 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 1.54 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 crosssectional area.
;
d
P L
P
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SECTION 1.5 Linear Elasticity, Hooke’s Law, and Poisson’s Ratio
27
Solution 1.54 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 xsec 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 1.55 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 H2, Appendix H.
Solution 1.55 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 1.56 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 1.56 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 1.57 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 1.57 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 H2))
Problem 1.58 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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CHAPTER 1 Tension, Compression, and Shear
Solution 1.58 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 1.61 An angle bracket having thickness t 0.75 in. is attached to the flange of a column by two 5/8inch 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 1.61 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 1.62
Truss members supporting a roof are connected to a 26mmthick 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 1.62 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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CHAPTER 1 Tension, Compression, and Shear
Problem 1.63 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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SECTION 1.6 Shear Stress and Strain
Solution 1.63 (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 1.64 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 1.64 (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 1.65 The force in the brake cable of the Vbrake 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 1.65 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 1.66 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 1.66 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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CHAPTER 1 Tension, Compression, and Shear
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 1.67 A specialpurpose 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 1.67 (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 1.68 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 1.68 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 1.69 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 1.69 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 1.610 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 1.610 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 1.611 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 1.611
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 FREEBODY 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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CHAPTER 1 Tension, Compression, and Shear
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 1.612 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 1.612
Clamp supporting a load P
FREEBODY 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
FREEBODY 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
FREEBODY DIAGRAM OF PIN
Problem ★1.613 A hitchmounted bicycle rack is designed to carry up to four 30lb. 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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CHAPTER 1 Tension, Compression, and Shear
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 *1.613 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 1.614 A bicycle chain consists of a series of small
Solution 1.614
;
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 1.615 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 1.615
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 1.616 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 crosssectional 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 1.616 (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 1.617 A spray nozzle for a garden hose requires a force F 5 lb. to open the springloaded 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 1.617 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 1.618
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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SECTION 1.6 Shear Stress and Strain
Solution 1.618 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
Whc 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 1.619
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) Freebody diagram
Cutting blade 3 (tb = — in.) 32
Pin at C 1 (dp = — in.) 8
(c) Section a–a
Solution 1.619 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 aa 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 1.71
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 1.72 A torque T0 is transmitted between two flanged
T0
d
shafts by means of ten 20mm 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 1.72 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 1.73 A tiedown 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 tiedown?
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SECTION 1.7 Allowable Stresses and Allowable Loads
Solution 1.73 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 1.74
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 1.74 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 1.75 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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SECTION 1.7 Allowable Stresses and Allowable Loads
Solution 1.75 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 1.76 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 freebody 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 1.76 (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 1.77 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 1.77 Lifeboat supported by four cables FREEBODY 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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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 1.78 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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Solution 1.78 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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SECTION 1.7 Allowable Stresses and Allowable Loads
Problem 1.79 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 1.79 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 1.710 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 5mm 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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Solution 1.710 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 1.711 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 1.711 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 1.712 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 E5(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 mmthick gusset plate at C [figure part (c)] with 16mm 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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Solution 1.712 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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SECTION 1.7 Allowable Stresses and Allowable Loads
Problem 1.713
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 1.713 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 1.714 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 1.714 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 ydirection 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 1.715 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 1.715 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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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 1.716 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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Solution 1.716 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 FREEBODY 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 1.81 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 midlength [see figure parts (b) and (c)].
P
(a) Hole of diameter d/10
d
d/10
P
P
d
(b)
(c)
Solution 1.81 (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 1.82 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 oneeighth 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 1.82 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 1.83 A horizontal beam AB with crosssectional 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 1.83 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 1.84 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 1.84 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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Problem 1.85 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 1.85 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 1.86 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 crosssectional area of the tie. (b) Calculate the minimum area if P 130 kN, u 75°, and sallow 80 MPa.
75
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Solution 1.86 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
FREEBODY DIAGRAM OF HALF THE TIE Note: Include a small amount of the cable in the freebody diagram T tensile force in the tie FORCE TRIANGLE cotu
T P
T P cot
Problem 1.87 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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SECTION 1.8 Design for Axial Loads and Direct Shear
Solution 1.87 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 1.88 A cable and pulley system at D is used to bring a 230kg 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 1.88 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 1.89 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 1.89 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 1.810 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 1.810
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 1.811 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 crosssectional 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 1.811 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 CROSSSECTIONAL 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 1.812 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 1.812 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 1.813 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 midspan 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 recheck 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 1.813 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 REDESIGN 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
RECHECK 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 REDESIGN 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 1.814 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 1.814
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 1.815 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 1.815 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 2.21 The Lshaped arm ABC shown in the figure lies in a vertical plane and pivots about a horizontal pin at A. The arm has constant crosssectional 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 2.21 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
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Axially Loaded Members
Problem 2.22 A steel cable with nominal diameter 25 mm (see Table 21) 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 2.22
Bridge section lifted by a cable A ⫽ 304 mm2 (from Table 21) W ⫽ 38 kN
(b) FACTOR OF SAFETY PULT ⫽ 406 kN (from Table 21) 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 2.23 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 2.23
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 2.24 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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Solution 2.24
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 2.25 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 2.25
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. 21a) 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 2.26 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 2.26
Pointer supported by a spring
FREEBODY 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 2.27 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 2.27 (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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95
Changes in Lengths of Axially Loaded Members
Problem 2.28 The threebar truss ABC shown in the figure has a span L ⫽ 3 m and is constructed of steel pipes having crosssectional 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 2.28 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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Axially Loaded Members
Problem 2.29 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 2.29 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 2.210 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 lefthand 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 2.210 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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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 2.211 A hollow, circular, castiron 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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Solution 2.211 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 H1)
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 2.212 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 crosssectional 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 2.212 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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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 2.213 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 2.213
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
FREEBODY 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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Problem 2.214 Solve the preceding problem for the following data: b ⫽ 200 mm, k ⫽ 3.2 kN/m, ␣ ⫽ 45°, and P ⫽ 50 N.
Solution 2.214
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 2.31 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 24.)
Solution 2.31
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 24: 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 2.32 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 crosssectional area of 4800 mm2, and a modulus of elasticity Ec 120 GPa. Each steel post has a height of 0.5 m, a crosssectional 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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Solution 2.32
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 2.33 A steel bar AD (see figure) has a crosssectional
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 2.33
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 2.34 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 highstrength 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 2.34
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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Problem 2.35 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 2.35
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 2.36 A twostory 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 secondfloor load P2 equals 720 kN. Each column has length L 3.75 m. The crosssectional areas of the first and secondfloor 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 2.36
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 2.37 A steel bar 8.0 ft long has a circular cross section of diameter d1 0.75 in. over onehalf 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 2.37
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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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 2.38 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 onehalf 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 2.38 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 2.39 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, crosssectional 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 2.39
Wood pile with friction
FROM FREEBODY 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 2.310 Consider the copper tubes joined below using a “sweated” joint. Use the properties and dimensions given. (a) Find the total elongation of segment 234 (24) for an applied tensile force of P 5 kN. Use Ec 120 GPa. (b) If the yield strength in shear of the tinlead 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
Tinlead solder in space between copper tubes; assume thickness of solder equal zero
Solution 2.310 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 234 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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A1 69.311 mm2 smallest crosssectional 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 2.311 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 2.311 (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 2.312 A prismatic bar AB of length L, crosssectional 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 2.312 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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Problem 2.313 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 2.313 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 2.314 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 2.314
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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Problem 2.315 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 2.315
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 crosssectional area at element dy AB crosssectional 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 2.316 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 2.316 (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 Lx 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 Lx
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 2.317 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 freebody 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 2.317
Cable of a suspension bridge dy 8hx 2 dx L FREEBODY 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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FREEBODY 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 2.318 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 crosssectional 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 onehalf 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 2.318 Rotating bar Centrifugal force produced by weight W2 a
W2 b(L2) g
AXIAL FORCE F(x) F(x)
angular speed A crosssectional 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 2.41 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 25.)
Solution 2.41
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. (213) of Example 25. d
PL Es As + Eb Ab
or
d P (Es As + Es Ab)a b L
s 22 ksi b 16 ksi Use Eqs. (212a and b) of Example 25. 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 2.42 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 25.)
Solution 2.42
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. (212a and b) of Example 25.
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. (213) of Example 25.
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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Axially Loaded Members
Problem 2.43 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 crosssectional area of the middle bar (material B) is 50% larger than the crosssectional 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 2.43
Prismatic bars in tension
FREEBODY 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:
FORCEDISPLACEMENT 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 2.44 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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Axially Loaded Members
Solution 2.44 (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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SECTION 2.4
Statically Indeterminate Structures
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 2.45 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 onethird 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 21 in Section 2.2 for properties of cables.)
Solution 2.45
Three cables in tension SECOND LOADING P2 9 k (additional load)
AREAS OF CABLES (from Table 21) 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)
FORCEDISPLACEMENT 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 2.46 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 2.46
;
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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(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. 213 of Example 25)
2AC CD 1.91 mm
;
;
;
Problem 2.47 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 lefthand support to any point in the bar and the ordinate is the horizontal displacement at that point.
Solution 2.47
2A1
A1
rigid supports. The bar has crosssectional 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
FREEBODY 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
FORCEDISPLACEMENT 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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SECTION 2.4
135
Statically Indeterminate Structures
Problem 2.48 The fixedend bar ABCD consists of three prismatic segments, as shown in the figure. The end segments have crosssectional area A1 840 mm2 and length L1 200 mm. The middle segment has crosssectional 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 2.48
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):
FREEBODY 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)
FORCEDISPLACEMENT 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 2.49 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, crosssectional area of aluminum pipe Aa 8.92 in.2, crosssectional 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 2.49
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
FORCEDISPLACEMENT 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 2.410 A nonprismatic bar ABC is composed of two segments: AB of length L1 and crosssectional area A1; and BC of length L2 and crosssectional 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 2.410 (a) find reactions in 1degree 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
Page 138
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 2.411 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 crosssectional 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 2.411
Bimetallic bar in compression
FREEBODY DIAGRAM
(a) AXIAL FORCES
(Plate at righthand 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 2.412 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 2.412 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 1degree statindet  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 2.413 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 2.413
Bar supported by three rods BAR 1 ALUMINUM E1 10 106 psi
FREEBODY 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.
;
FORCEDISPLACEMENT 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 2.414 A circular steel bar ABC (E 200 GPa) has crosssectional area A1 from A to B and crosssectional 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 crosssectional 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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Solution 2.414
Bar supported by a collar
FREEBODY 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
FORCEDISPLACEMENT 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 2.415 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 crosssectional 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 2.415
147
Statically Indeterminate Structures
Bar supported by two wires EQUATION OF COMPATIBILITY dc dD c d FORCEDISPLACEMENT 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 FREEBODY 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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Problem 2.416 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 2.416
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 FORCEDISPLACEMENT 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 FREEBODY 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 2.417 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 2.417 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)
FORCEDISPLACEMENT 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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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 2.51 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 2.51
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 27). The compressive stress in the rails may be calculated from Eq. (218).
T 120°F 60°F 60°F
E(T) (30 106 psi)(6.5 106/°F)(60°F) 11,700 psi ;
Problem 2.52 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 2.52
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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Problem 2.53 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 2.53
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 2.54 A steel rod of 15mm 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 12mm diameter bolt becomes 45 MPa.
Washer, dw = 20 mm
12mm 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 2.54 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 2.55 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 2.55 (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 2.56 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 righthand 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 onedegree statindet  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 2.57 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 2.57
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 28. Thus, we can calculate the elongation from Eq. (221): 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 2.58 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 28.)
;
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SECTION 2.5
Solution 2.58
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 (220A): (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 28.
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 2.59 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 crosssectional 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 28.
Solution 2.59
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 28. Find the forces Pa and Pc in the aluminum bar and copper bar, respectively, from Eq. (219).
18 2.0 ba b 10 2.0
4,500 lb FREEBODY 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 2.510 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 21, Section 2.2.)
Solution 2.510
Rigid bar supported by two cables
FREEBODY DIAGRAM OF BAR ABCD
From Table 21: 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)
FORCEDISPLACEMENT AND TEMPERATUREDISPLACEMENT 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 21: (TB)ULT 102,000 N
RELATIONS
159
Thermal Effects
(Eq. 5)
Pallow 39.5 kN
Problem 2.511 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 2.511
Triangular frame held by two wires
FREEBODY DIAGRAM OF FRAME
(b) LOAD P AND TEMPERATURE INCREASE T Forcedisplacement and temperaturedisplacement 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 Forcedisplacement 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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161
Thermal Effects
Misfits and Prestrains Problem 2.512 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 2.512
Steel wire with initial prestress From Eq. (218): 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 2.513 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 2.513 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) onedegree 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 2.514 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 twothirds 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 2.514
s
RA 8042 lb
A
s
L — 3 C
B P
Bar with a gap (load P ) FORCEDISPLACEMENT 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 2.515 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 2.515 (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
;
statindet 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 statindet 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 2.516 A nonprismatic bar ABC made up of a, T RA segments AB (length L1, crosssectional area A1) and BC (length L2, crosssectional 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 2.516 With gap s closed due to T, structure is onedegree staticallyindeterminate; 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 lefthand end and to a pinsupported rigid bar at the righthand end (see figure). Each wire has crosssectional 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)
FORCEDISPLACEMENT 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 2.518 A rigid steel plate is supported by three posts of highstrength concrete each having an effective crosssectional 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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Solution 2.518
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)
FORCEDISPLACEMENT 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 2.519 A capped castiron 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 precompress the CI pipe. The pitch of the threads of the bolt is p 52 mils (a mil is onethousandth of an inch). Use the numerical properties provided.
(
)
Steel cap (tc = 1 in.)
(a) What stresses p and r will be produced in the castiron 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 2.519 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 statindet  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 2.520 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, crosssectional area of one bolt As 36.0 mm2, and crosssectional area of the plastic cylinder Ap 960 mm2.
Solution 2.520
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)
FORCEDISPLACEMENT RELATIONS
Ap 960 mm2 n 1 (See Eq. 222) 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 2.521 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, crosssectional area of one bolt As 0.06 in.2, and crosssectional area of the plastic cylinder Ap 1.5 in.2
Solution 2.521
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. 222)
FORCEDISPLACEMENT 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 2.522 Consider the sleeve made from two copper tubes joined by tinlead
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 2.522 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 onedegree statindet  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 staticallyindeterminate 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 2.523 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 2.523
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 2.524 Prestressed concrete beams are sometimes manufactured in the following manner. Highstrength 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 crosssectional 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 2.524
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 FORCEDISPLACEMENT 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 2.525 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 posttensioned 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 2.525 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. 2.523 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 2.61 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 2.61 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 onehalf of allow, the shear stress governs.
NUMERICAL DATA A 3 in2
Pmax2 42,600 lbs
a 14500 psi
a 7100 psi
Problem 2.62 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 2.62 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 2.63 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 2.63
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 onehalf 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 2.64 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 19.5 106/°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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Solution 2.64
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 2.65 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 2.65 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 2.66 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 2.66
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 2.67 During a tension test of a mildsteel 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 2.67
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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 2.68 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 2.68
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. 218 of Section 2.5)
105 MPa (Compression) NOTE: All stresses have units of MPa.
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SECTION 2.6
Problem 2.69 The bottom chord AB in a small truss ABC (see figure) is fabricated from a W8 28 wideflange steel section. The crosssectional area A 8.25 in.2 (Appendix E, Table E1 (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 2.69 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 2.610 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 recentering 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 Recentering 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 2.611 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 2.611 (a)
NUMERICAL DATA b 1.5 in
h 3 in
A bh
T 92°F
T (160 68)°F
SHEAR STRESS ON PLANE PQ STATINDET 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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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
statindet 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 2.612 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 2.612 (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 onedegree statindet 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 XDIRECTION 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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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 2.613 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 2.613 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 2.614 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 2.614
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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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 2.615 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 2.615
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 2.616 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 2.616 (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 2.617 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 2.617
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. (229a):
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 2.618 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 crosssectional area of the bar is 225 mm2.
Solution 2.618
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 2.619 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, 6.5 106/°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 2.619 (a) STATINDET 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) STATINDET 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 2.71 A prismatic bar AD of length L, crosssectional 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 2.71
Bar with three loads
(a) STRAIN ENERGY OF THE BAR (EQ. 240)
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 2.72 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 2.72
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. 240). 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 2.73 A threestory steel column in a building supports roof and floor loads as shown in the figure. The story height H is 10.5 ft, the crosssectional 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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CHAPTER 2
Solution 2.73
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Axially Loaded Members
Threestory 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. 240).
U⫽
(10.5 ft)(12 in./ft) 930 * 106 lb
⫽ 5040 in.lb
Problem 2.74 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 2.74
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 2.75 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 2.75
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 2.75
Strainenergy 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 2.76 The truss ABC shown in the figure is subjected to a horizontal load P at joint B. The two bars are identical with crosssectional 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 2.76
↓⫺
 ⫽ 60°
⌺Fvert ⫽ 0
LAB ⫽ LBC ⫽ L
⫺FAB sin  ⫹ FBC sin  ⫽ 0
sin b ⫽ 13/2
FAB ⫽ FBC
cos  ⫽ 1/2
⌺Fhoriz ⫽ 0 : ←
FREEBODY 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. 240) (NBC)2L (NAB)2L N2i Li P2L ⫽ + ⫽ 2EiAi 2EA 2EA EA
2 P2L 2PL 2U ⫽ a b ⫽ P P EA EA
;
;
Problem 2.77 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 crosssectional 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 2.77
203
(b) HORIZONTAL DISPLACEMENT OF JOINT B (EQ. 242)
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 lb2in. 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 lb2in. 144 * 106 lb
⫽ 2.22 i n. lb
;
NOTE: The strain energy U3 is not equal to U1 ⫹ U2.
Problem 2.78 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 2.78
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. (238b) 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 2.79 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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Solution 2.79
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. 2.313. (1)
1 dx ⫽ ln (a + bx) b L a + bx
Problem 2.710 A compressive load P is transmitted through a rigid plate to three magnesiumalloy 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, crosssectional 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. 242)
(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 loaddisplacement diagram.)
P s
L
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SECTION 2.7
Solution 2.710
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) LOADDISPLACEMENT 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 loaddisplacement 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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Problem 2.711 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 forcedisplacement 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 2.711
Block pushed against three springs
Force P0 required to close the gap: P0 ⫽ k1s
(a) FORCEDISPLACEMENT DIAGRAM (1)
FORCEDISPLACEMENT RELATION BEFORE GAP IS CLOSED P ⫽ k1x
(0 ⱕ x ⱕ s)(0 ⱕ P ⱕ P0)
(2)
FORCEDISPLACEMENT 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 forcedisplacement relation is not linear.
U1 ⫽ area under line OAB.
Problem 2.712 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 2.712
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 nonlinear 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 2.81 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, crosssectional 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. 2.81, 2.82, 2.83
Solution 2.81
Collar falling onto a flange (a) DOWNWARD DISPLACEMENT OF FLANGE dst ⫽
WL ⫽ 0.00032 in. EA
Eq. of (253): dmax ⫽ dst c1 + a1 + ⫽ 0.0361 in.
2h 1/2 b d dst
;
(b) MAXIMUM TENSILE STRESS (EQ. 255) smax ⫽
Edmax ⫽ 22,600 psi L
;
(c) IMPACT FACTOR (EQ. 261) 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 2.82 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 crosssectional area A ⫽ 350 mm2, and the modulus of elasticity E ⫽ 170 GPa.
Solution 2.82
Collar falling onto a flange
(a) DOWNWARD DISPLACEMENT OF FLANGE dst ⫽
WL ⫽ 0.03957 mm EA
Eq. (253): dmax ⫽ dst c1 + a1 + ⫽ 6.33 mm
2h 1/2 b d dst
;
(b) MAXIMUM TENSILE STRESS (EQ. 255) 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 2.83 Solve Problem 2.81 if the collar has weight W ⫽ 50 lb, the height h ⫽ 2.0 in., the length L ⫽ 3.0 ft, the crosssectional area A ⫽ 0.25 in.2, and the modulus of elasticity E ⫽ 30,000 ksi.
Solution 2.83
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. 261) Impact factor ⫽
;
dmax 0.0312 in. ⫽ dst 0.00024 in. ⫽ 130 ;
Problem 2.84 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. 2.84 and 2.85
Solution 2.84
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. (253): dmax ⫽ dst c1 + a1 + ⫽ 215 mm
;
2h 1/2 b d dst
(b) IMPACT FACTOR (EQ. 261) Impact factor ⫽
dmax 215 mm ⫽ dst 55.56 mm ⫽ 3.9 ;
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SECTION 2.8
Impact Loading
Problem 2.85 Solve the preceding problem if the block weighs W ⫽ 1.0 lb, h ⫽ 12 in., and k ⫽ 0.5 lb/in.
Solution 2.85
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. (253): dmax ⫽ dst c1 + a1 + ⫽ 9.21 in.
2h 1/2 b d dst
;
(b) IMPACT FACTOR (EQ. 261) 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 2.86 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 2.86
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 2.87 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 2.87
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 2.88 A cable with a restrainer at the bottom hangs vertically from its upper end (see figure). The cable has an effective crosssectional 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. 2.88, 2.82, 2.89
Solution 2.88
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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Axially Loaded Members
Problem 2.89 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 2.89
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 2.810 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 2.810
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 2.811 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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Solution 2.811
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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 2.812 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 2.812
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
221
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Problem 2.813 A weight W rests on top of a wall and is attached to one end of a very flexible cord having crosssectional 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 2.813
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 ⫽ crosssectional 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 2.814 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 crosssectional 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 2.814
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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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 stressconcentration factors and assuming linearly elastic behavior.
P
Problem 2.101 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. 2.101 and 2.102
P
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SECTION 2.10
Solution 2.101
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 ⫽ knom ⬇ 6.2 ksi
b ⫽ 4.0 in. s nom ⫽
Obtain K from Fig. 263
d/b ⫽
Stress Concentrations
c ⫽ 2.5 in.; Obtain k from Fig. 264
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 ⫽ Knom ⬇ 11.0 ksi FOR R ⫽ 0.5 in.: R/c ⫽ 0.2 K ⬇ 1.87
;
b/c ⫽ 1.60
max ⫽ Knom ⬇ 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 ⫽ Knom ⬇ 6.9 ksi
;
Problem 2.102 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 2.102
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. 263 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. 264 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 ⫽ Knom ⬇ 25 MPa
K ⬇ 2.00
1 K L 2.51 5
max ⫽ Knom ⬇ 26 MPa
b ⫽ 60 mm
b/c ⫽ 1.5
max ⫽ Knom ⬇ 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 ⫽ Knom ⬇ 29 MPa
;
Problem 2.103 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 2.103
t ⫽ allowable tensile stress Find Pmax Find K from Fig. 264 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 2.104 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. Quartercircular 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. 2.104 and 2.105
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Solution 2.104
Round brass bar with upset ends
Use Fig. 265 for the stressconcentration 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 2.105
Use the dashed curve in Fig. 265. K ⬇ 1.6
max ⫽ Knom ⬇ (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. 265 for the stressconcentration factor.
E ⫽ 25 ⫻ 106 psi
⫽ 0.2 in.
⫽ 28.68 MPa
3 mm R ⫽ ⫽ 0.15 D1 20 mm
⬇ 46 MPa
Problem 2.105 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. 265. K ⬇ 1.53 ⫽ 3,984 psi
max ⫽ Knom ⬇ (1.53)(3984 psi) ⬇ 6100 psi
;
0.2 in. R ⫽ ⫽ 0.2 D1 1.0 in.
Problem 2.106 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 2.106
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. 265 for the stressconcentration 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 2.107 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 loadcarrying capacity?
Solution 2.107
Stepped bar with a hole
b ⫽ 2.4 in.
BASED UPON HOLE (Use Fig. 263)
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. 264) 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/btmax
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 2.111 A bar AB of length L and weight density ␥ hangs vertically under its own weight (see figure). The stressstrain relation for the material is given by the RambergOsgood equation (Eq. 271): 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 2.111
Bar hanging under its own weight STRAIN AT DISTANCE x Let A ⫽ crosssectional 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 2.112 A prismatic bar of length L ⫽ 1.8 m and crosssectional
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 stressstrain curve described by the following RambergOsgood 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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Solution 2.112
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 2.113 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 stressstrain relationship: s⫽
18,000P 0 … P … 0.03 (s ⫽ ksi) 1 + 300P
(a) Draw a stressstrain 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 2.113
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 ⫽
18,000max ⫽ 42.06 ksi 1 + 300max
(a) STRESSSTRAIN 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 2.114 A prismatic bar in tension has length L ⫽ 2.0 m
and crosssectional 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 ␦ (loaddisplacement diagram).
;
200 s (MPa) 100
0
0
0.005 e
0.010
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Solution 2.114
Bar in tension
L ⫽ 2.0 m A ⫽ 249 mm2 STRESSSTRAIN DIAGRAM (See the problem statement for the diagram) LOADDISPLACEMENT 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 loaddisplacement curve has the same shape as the stressstrain curve.
Problem 2.115 An aluminum bar subjected to tensile forces P has length
L ⫽ 150 in. and crosssectional area A ⫽ 2.0 in.2 The stressstrain behavior of the aluminum may be represented approximately by the bilinear stressstrain 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 ␦ (loaddisplacement diagram).
s
12,000 psi
E2 = 2.4 106 psi
E1 = 10 106 psi
0
e
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SECTION 2.11
Solution 2.115
Nonlinear Behavior (Changes in Lengths of Bars)
Aluminum bar in tension LOADDISPLACEMENT DIAGRAM
L ⫽ 150 in. A ⫽ 2.0 in.2 STRESSSTRAIN 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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Problem 2.116 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 highstrength 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 stressstrain 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 loaddisplacement diagram showing P versus ␦B.
Solution 2.116
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
STRESSSTRAIN 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) LOADDISPLACEMENT DIAGRAM
(a) DISPLACEMENT ␦B AT END OF BAR 3 3 ␦ ⫽ elongation of wire dB ⫽ d ⫽ L 2 2 Obtain from stressstrain 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. 270).
A
u
Problem 2.121 Two identical bars AB and BC support a vertical load P (see figure). The bars are made of steel having a stressstrain curve that may be idealized as elastoplastic with yield stress Y. Each bar has crosssectional area A. Determine the yield load PY and the plastic load PP.
Solution 2.121
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.
(2YA) sin ⫽ P PY ⫽ PP ⫽ 2YA sin
Problem 2.122 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 2.122
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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 2.123 A horizontal rigid bar AB supporting a load P is hung from five symmetrically placed wires, each of crosssectional 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 2.123
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 ⫽ 5YA ; 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 2.124 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 crosssectional area A and the material is elastoplastic with yield stress Y. Determine the plastic load PP.
a
a
P
Solution 2.124
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 ⫽ 2YA (1 ⫹ sin ␣)
21 in.
Problem 2.125 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 crosssectional 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 2.125
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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 2.126 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 2.126
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 2.127 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 2.127
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 2.128 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 crosssectional 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 loaddisplacement diagram with the load P as ordinate and the displacement ␦B of point B as abscissa.
Solution 2.128
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) LOADDISPLACEMENT 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 2.129 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 crosssectional 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 loaddisplacement diagram with the load P as ordinate and the displacement ␦D of point D as abscissa.
Solution 2.129
L A
3L 4
B
C
D
P 2b
b
b
Rigid bar supported by two wires FREEBODY DIAGRAM
A ⫽ crosssectional 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
FORCEDISPLACEMENT 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) LOADDISPLACEMENT 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 2.1210 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 crosssectional 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 stressstrain 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 loaddisplacement 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 2.1210
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 ⫽ 2YA ⫽ 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) LOADDISPLACEMENT 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
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Axially Loaded Members
Problem 2.1211 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 stressstrain 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 loaddisplacement diagram showing the load P as ordinate and the shortening ␦ of the tube as abscissa. (Hint: The loaddisplacement diagram is not a single straight line in the region 0 ⱕ P ⱕ PY.)
Solution 2.1211
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) LOADDISPLACEMENT 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 3.21 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. 3.21 and 3.22
Solution 3.21
Copper rod in torsion d T
T
L
L ⫽ 18.0 in.
From Eq. (33):
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 3.22 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?
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Torsion
Solution 3.22 NUMERICAL DATA d ⫽ 56 mm
␥a ⫽ 0.012 radians f ⫽ 4a
p b radians 180
solution based on Equ. (33): Lmin ⫽ 162.9 mm
Lmin ⫽
df 2␥a
;
Problem 3.23 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. 3.23, 3.24, and 3.25
Solution 3.23 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 3.24 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 3.24 (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 3.25 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 3.25 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 3.31 A prospector uses a handpowered 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 3.31
Handpowered winch d ⫽ 0.625 in.
MAXIMUM SHEAR STRESS IN THE AXLE
b ⫽ 4.0 in.
From Eq. (312):
W ⫽ 100 lb
tmax ⫽
Torque T applied to the axle: T ⫽ Wb ⫽ 400 lbin.
tmax ⫽
16T pd3 (16)(400 lbin.) p(0.625 in.)3
tmax ⫽ 8,340 psi
;
Problem 3.32 When drilling a hole in a table leg, a furniture maker uses a handoperated 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 3.32
253
Circular Bars and Tubes
Torsion of a drill bit (b) RATE OF TWIST From Eq. (314): u⫽
d ⫽ 4.0 mm
T ⫽ 0.3 N⭈m G ⫽ 75 GPa
(a) MAXIMUM SHEAR STRESS From Eq. (312): 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 3.33 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 3.33
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Torsion
Lug wrench P ⫽ 25 lb
(a) MAXIMUM SHEAR STRESS From Eq. (312): (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. (315): (450 lbin.)(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 lbin.
f ⫽ 0.05790 rad ⫽ 3.32°
Problem 3.34 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 3.34 (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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255
Problem 3.35 A highstrength 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 3.35
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. (312): 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. (315): 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 3.36 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 3.36
Socket wrench ANGLE OF TWIST From Eq. (315): f ⫽
d ⫽ 8.0 mm
L ⫽ 200 mm
allow ⫽ 60 MPa
G ⫽ 78 GPa
MAXIMUM PERMISSIBLE TORQUE 16T From Eq. (312): tmax ⫽ pd3 3 pd tmax Tmax ⫽ 16 3
Tmax ⫽
p(8.0 mm) (60 MPa) 16
Tmax ⫽ 6.03 N # m
;
Problem 3.37 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 lbin. 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. (312): 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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SECTION 3.3
Circular Bars and Tubes
257
Solution 3.37 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 lbin.
␥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 3.38 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 3.38 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 3.39 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 3.39
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. 312) 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 3.310 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 3.310 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 3.311 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 kin., 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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CHAPTER 3 Torsion
Solution 3.311 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 kin. IP ⫽
(150 kin.) 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 kin.)(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 3.312 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 3.312 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 3.313 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 3.313
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 3.314 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 3.314
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 3.315 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 3.315
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 xsec area) A1 ⫽
d1 d2
p 2 d 4 2
A2 ⫽
p 2 1d ⫺ d122 4 2
A1 ⫺ A2 (100) ⫽ 25 % A1
;
Problem 3.316 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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Solution 3.316 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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265
Circular Bars and Tubes
Problem 3.317 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 3.317
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 3.41 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 lbin. and T2 8,000 lbin. 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 3.41
d2
T2
B
A
C
L1
L2
Stepped shaft SEGMENT BC TBC T2 8,000 lbin.
d1 2.25 in.
L1 30 in.
d2 1.75 in.
L2 20 in.
T2 8,000 lbin.
`
TABL1 G(Ip)AB
16(8,000 lbin.) p(1.75 in.)3
7602 psi
(8,000 lbin.)(20 in.) (11 * 106 psi)a
p b(1.75 in.)4 32
16(12,000 lbin.) p(2.25 in.)3
C AB BC (0.013007 0.015797) rad 5365 psi
(12,000 lbin.)(30 in.) (11 * 106 psi)a
0.013007 rad
;
(b) ANGLE OF TWIST AT END C
TAB T2 T1 12,000 lbin.
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 lbin.
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 3.42 A circular tube of outer diameter d3 70 mm and
inner diameter d2 60 mm is welded at the righthand end to a fixed plate and at the lefthand 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 3.42
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 3.43 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 kin., 9.8 kin., and 9.2 kin. 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 kin. 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 kin. 9.2 kin. 2.75 in. 2.5 in.
25 in.
25 in.
Solution 3.43 NUMERICAL DATA (INCHES, KIPS) TB 12.5 kin.
TC 9.8kin.
TD 9.2 kin.
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 3.44 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 3.44 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 lbin. 500 lbin.
T3 = T4 = 800 lbin. 500 lbin.
governs
Tallow 651.441 N # m
Problem 3.45 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 lbin., T2 T4 500 lbin., and T3 T5 800 lbin. 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 lbin.
E
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Torsion
Solution 3.45
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 H2, 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 lbin.
T2 500 lbin.
T4 500 lbin.
T5 800 lbin.
T3 800 lbin.
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 lbin.
d1 0.818 in.
TBC T1 T2 500 lbin.
(Maximum permissible inside diameter)
;
TCD T1 T2 T3 1300 lbin. TDE T1 T2 T3 T4 800 lbin. Largest torque (absolute value only): Tmax 1300 lbin.
80 mm
Problem 3.46 A shaft of solid circular cross section consisting of two segments is shown in the first part of the figure. The lefthand segment has diameter 80 mm and length 1.2 m; the righthand 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 3.46
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 3.47 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 lbin. (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 lbin. 4,000 lbin. A
7,000 lbin. B C D
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Torsion
Solution 3.47
Shaft with four gears (b) HOLLOW SHAFT Inside diameter d0 1.0 in.
allow 10,000 psi
TBC 11,000 lbin.
TAB 8000 lbin.
TCD 7000 lbin.
Tr tmax Ip
10,000 psi
(a) SOLID SHAFT tmax d3
t allow
16T pd3
d Tmax a b 2 Ip
d (11,000 lbin.)a b 2 a
p b[d4 (1.0 in.)4] 32
UNITS: d inches
16(11,000 lbin.) 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 3.48 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 lefthand end to dB at the righthand end. For what ratio dB/dA will the angle of twist of the tapered bar be onehalf 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 35.
T
;
B
A
T
L dA
dB
Problems 3.48, 3.49 and 3.410
Solution 3.48
Tapered bar AB
ANGLE OF TWIST
TAPERED BAR (From Eq. 327) 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 3.49 A tapered bar AB of solid circular cross section is twisted
by torques T 36,000 lbin. (see figure). The diameter of the bar varies linearly from dA at the lefthand end to dB at the righthand 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 3.49
Tapered bar
MINIMUM DIAMETER BASED (From Eq. 327)
dB 1.5 dA T 36,000 lbin.
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 lbin.)(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 lbin.) 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 3.410 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 35.)
Solution 3.410
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. 327) 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 3.411 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 3.411 (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 3.412 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. 318). 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 3.412
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 3.413 A uniformly tapered aluminumalloy 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 3.413
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 3.414 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 3.414 (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 12 & T(x) T for segment 23 (see plot above)
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Torsion
Problem 3.415 A mountainbike 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 3.415 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 3.416 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 3.416
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 3.417 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 3.417
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 3.418 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 3.418 (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 3.419 A magnesiumalloy 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 3.419 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. (312): tmax pd3 tL + T0 L
L 4 m G 15 GPa 1 angle of twist due to distributed torque t T
pGd4
(from problem 3.416)
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 3.420 Two hollow tubes are connected by a pin at B which is inserted into a hole drilled through both tubes at B (see crosssection 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 3.420
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
Crosssection at B
(a) T0,max BASED ON ALLOWABLE SHEAR IN PIN AT B
285
Nonuniform Torsion
;
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(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 3.51 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. 3.51, 3.52, and 3.53
Solution 3.51
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. 37a) 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 lbin.
Tr IP
;
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Problem 3.52 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 3.52 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 3.53 A tubular bar with outside diameter d2 4.0 in. is twisted by torques T 70.0 kin. (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 3.53
d2 4.0 in.
T 70.0 kin. 70,000 lbin.
f
Torsion formula: tmax
TL GIp
From torsion formula, T
(a) INSIDE DIAMETER d1 Td2 Tr IP 2IP
‹ f
(70.0 kin.)(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 3.54 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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CHAPTER 3 Torsion
Solution 3.54 Bar in a testing machine
Strain gage at 45°:
SHEAR STRESS (FROM EQ. 312)
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. 333) 6
max 2max 678 10
tmax 20.372 MPa 30.0 GPa gmax 678 * 106
;
Problem 3.55 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 3.55
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. 333)
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 lbin.
Td2 Tr tmax IP 2IP Also, max Gmax
;
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Pure Shear
291
Problem 3.56 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 3.56
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 3.57 The normal strain in the 45° direction on the
surface of a circular tube (see figure) is 880 106 when the torque T 750 lbin. 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 3.57
d2 0.80 in.
Circular tube with strain gage
T 750 lbin. G 6.2 106 psi
Strain gage at 45°: max 880 106
T = 750 lbin.
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 lbin.) (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 3.58 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 3.58
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 3.59 A solid steel bar (G 11.8 106 psi) of
diameter d 2.0 in. is subjected to torques T 8.0 kin. 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 3.59
Solid steel bar
T 8.0 kin.
(b) MAXIMUM STRAINS
G 11.8 10 psi 6
gmax
(a) MAXIMUM STRESSES tmax
T = 8.0 kin.
d = 2.0 in.
T
16T 3
pd
432 * 106 rad
16(8000 lbin.)
5093 psi
3
p(2.0 in.)
âmax
;
t 5090 psi c 5090 psi
gmax 5093 psi G 11.8 * 106 psi
Problem 3.510 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 3.510
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 3.71 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 3.71 n 120 rpm TORQUE H
Generator shaft H 50 hp
d diameter
2pnT H hp n rpm T 1bft 33,000
33,000 H (33,000)(50 hp) T 2pn 2p(120 rpm) 2188 1bft 26,260 1bin. (a) MAXIMUM SHEAR STRESS max d 3.0 in.
50 hp
tmax
16T 3
pd
16(26,260 1bin.)
tmax 4950 psi
p (3.0 in.)3 ;
(b) MINIMUM DIAMETER dmin
allow 4000 psi d3
16(26,260 1bin.) 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 3.72 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 3.72 f 12 Hz
20 kW
Motordriven 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 3.73 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 3.73
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 1bin. 344,590 1bft.
(a) HORSEPOWER n 100 rpm n rpm H
H
2pnT 33,000
T lbft H hp
2p(100 rpm)(344,590 lbft) 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 3.74 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 3.74 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 3.75 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 3.75
H hp
d0 inside diameter
T
0.75 d n 400 rpm
allow 6000 psi IP
n rpm
T lbft
(33,000)(400 hp) 33,000 H 2pn 2p(400 rpm)
5252.1 lbft 63,025 lbin. 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 lbin.)(d) 2(6000 psi)
d3 78.259 in.3
dmin 4.28 in.
;
Problem 3.76 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 onehalf of the outside diameter. If the allowable shear stress in the shaft is 45 MPa, what is the minimum required outside diameter d?
Solution 3.76
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 3.77 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 3.77 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 3.78 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 3.78 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 3.79 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. 3.79 and 3.710
Solution 3.79
Motordriven shaft FREEBODY DIAGRAM
L1 6 ft L2 4 ft
TA 17,332 lbin.
d diameter
TC 9454 lbin.
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 lbin. INTERNAL TORQUES TAB 17,332 lbin. TBC 9454 lbin.
H hp n rpm T lbft
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 lbft 17,332 lbin. At point B: TB At point C: TC
tmax
16TAB
d3
3
pd
16TAB pt allow
16(17,332 lbin.) 11.77 in.3 p(7500 psi)
d 2.27 in. DIAMETER BASED UPON ALLOWABLE ANGLE OF TWIST
125 275
TA 7878 lbin.
150 275
TA 9454 lbin.
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 lbin.)(4 ft)(12 in./ft) p(11.5 * 106 psi)d 4
Problem 3.710 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 3.710 Motordriven 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
FREEBODY 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 3.81 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. 346a and b of Example 39 to obtain the reactive torques.)
TA
A
T0
2T0
B
C
3L — 10
3L — 10
D
TD
D
TD
4L — 10 L
Solution 3.81
Circular bar with fixed ends ANGLE OF TWIST AT SECTION B
From Eqs. (346a 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 3.82 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. 346a and b of Example 39 to obtain the reactive torques.)
TA
A
;
T0
T0
B
C
x
x L
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SECTION 3.8
Solution 3.82
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. (346a 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 3.83 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. 346a and b of Example 39 to obtain the reactive torques.)
Disk A
d
B
a
Solution 3.83
b
Shaft fixed at both ends Assume that a torque T0 acts at the disk. The reactive torques can be obtained from Eqs. (346a 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. 349)
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 3.84 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. 346a and b of Example 39 to obtain the reactive torques.)
200 mm A P
200 mm C B P
600 mm 400 mm
Solution 3.84
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. (346a 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 3.85 A stepped shaft ACB having solid circular cross
Solution 3.85
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. 345a 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 lbin. (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. 345a and b of Example 39 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 lbin.
LBIPA 1 pd3 t a1 + b 16 B allow LAIPB
LBdA4 1 pd3Bt allow a1 + b ⫽ 16 LAdB4 Substitute numerical values: (T0)CB ⫽ 4597 lbin.
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 3.86 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. 345a and b of Example 39 to obtain the reactive torques.)
Solution 3.86
;
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. 345a 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 3.87 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. 345a 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. 345a and b of Example 39 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 3.87
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 3.88 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 fixedend torques TA and TB.
t0 t(x) TA
TB A
B x L
Solution 3.88
Fixedend bar with triangular load ELEMENT OF DISTRIBUTED LOAD
dTA ⫽ Elemental reactive torque t(x) ⫽
dTB ⫽ Elemental reactive torque
t0x L
From Eqs. (346a 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 (FIXEDEND 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 3.89 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 lefthand 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 3.89
Statically Indeterminate Torsional Members
309
Bar with a hole IPA ⫽ Polar moment of inertia at lefthand end IPB ⫽ Polar moment of inertia at righthand 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 3.810 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. 344a 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 3.810
Bar enclosed in a tube TORQUES IN THE BAR (1) AND TUBE (2) FROM EQS. (344A 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 3.811 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 lbin. 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. 344a 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 3.811
Statically Indeterminate Torsional Members
311
Bar enclosed in a tube TORQUES IN THE BAR (1) AND TUBE (2) FROM EQS. (344A AND B) Bar: T1 ⫽ T a
IPI b ⫽ 1187.68 lbin. IPI + IPI
Tube: T2 ⫽ T a
IP2 b ⫽ 3812.32 lbin. 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 3.812 The composite shaft shown in the figure is manufactured by shrinkfitting 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. 344a and b to find the torques.)
d1 d2 Probs. 3.812 and 3.813
Solution 3.812
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. (344b): 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. (344a): 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 3.813 The composite shaft shown in the figure is manufactured by shrinkfitting 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. 344a and b to find the torques.)
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SECTION 3.8 Statically Indeterminate Torsional Members
Solution 3.813 Composite shaft shrink fit
TORQUES Total torque: T ⫽ TB ⫹ TS Eq. (344 a): TB ⫽ T a
GBIPB b GBIPB + GSIPS
⫽ 0.237918 T d1 ⫽ 1.6 in.
Eq. (344 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 lbin.2 STEEL SLEEVE (ONLY)
2(4500 psi)(0.643398 in.4) 1.6 in.
T ⫽ 15.21 kin. 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 kin. STEEL SLEEVE GOVERNS IPS ⫽
Tmax ⫽ 9.13 kin.
;
p (d 4 ⫺ d14) ⫽ 0.927398 in.4 32 2
GSIPS ⫽ 11.1288 ⫻ 106 lbin.2
Problem 3.814 A steel shaft (Gs ⫽ 80 GPa) of total length L ⫽ 3.0 m is encased for onethird 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.
313
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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 3.814 (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 statindet; use Ts as the internal redundant; see equ. 344a 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 ⫽
4a 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 statindet 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 statindet 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 3.815 A uniformly tapered aluminumalloy 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 3.815 Solution approachsuperposition: 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. 3.413 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 inlb
TA ⫽ T0 ⫺ TB TA ⫽ 15292 in1b
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 3.816 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
Crosssection at C
Solution 3.816 (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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(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 3.91 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 3.91
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 3.92 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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Solution 3.92
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 3.93 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 3.93
(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 3.94 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 3.94
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 3.95 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 3.95
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. 351a) U1 ⫽
T2L 2GIP
(c) BOTH LOADS ACT SIMULTANEOUSLY
;
(b) LOAD t ACTS ALONE From Eq. (356) of Example 311: 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 3.96 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. 346a and b of Example 39, Section 3.8, to obtain the reactive torques.
Solution 3.96
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. 353)
From Eq. (346a):
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 3.97 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. 351b 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. 348 of Example 39, Section 3.8.
A
IPA
T0 C
IPB
LA
B
LB
Solution 3.97
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. 351B)
;
(This result agrees with Eq. (348) of Example 39, 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 3.98 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 3.98 Cantilever bar with distributed torque x ⫽ distance from righthand end of the bar
ELEMENT d
STRAIN ENERGY OF ELEMENT dx
Consider a differential element dj at distance j from the righthand 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 3.99 A thinwalled 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 3.99
325
Strain Energy in Torsion
Thinwalled, 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. 354)
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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Problem 3.910 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 3.910
Circular tube and bar
TUBE A
COMPATIBILITY
A ⫹ B ⫽  FORCEDISPLACEMENT 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 3.911 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 3.911
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. 351b) 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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ThinWalled Tubes Problem 3.101 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 kin. Determine the maximum shear stress in the tube using (a) the approximate theory of thinwalled tubes, and (b) the exact torsion theory. Does the approximate theory give conservative or nonconservative results?
10.0 in. 1.0 in.
Solution 3.101
Hollow circular tube APPROXIMATE THEORY (EQ. 363) t1 ⫽
T 2
2pr t
⫽
1200 kin. 2p(5.5 in.)2(1.0 in.)
approx ⫽ 6310 psi
⫽ 6314 psi
;
EXACT THEORY (EQ. 311) T ⫽ 1200 kin. 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(1200kin.)(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 3.102 A solid circular bar having diameter d is to be replaced by a rectangular tube having crosssectional 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 3.102
ThinWalled Tubes
329
Bar and tube
SOLID BAR tmax ⫽
16T pd3
(Eq. 312)
Am ⫽ (d)(2d) ⫽ 2d2
(Eq. 364)
T T ⫽ 2tAm 4td2
(Eq. 361)
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 3.103 A thinwalled 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 kin. (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. 3.103 and 3.104
Solution 3.103
Thinwalled tube Eq. (364): Am ⫽ bh ⫽ 24.0 in.2 J⫽
Eq. (371) with t1 ⫽ t2 ⫽ t: J ⫽ 28.8 in.4 (a) SHEAR STRESS (EQ. 361) t⫽ b ⫽ 6.0 in. h ⫽ 4.0 in. t ⫽ 0.25 in. T ⫽ 15 kin. L ⫽ 50 in. G ⫽ 4.0 ⫻ 106 psi
T ⫽ 1250 psi 2tAm
;
(b) ANGLE OF TWIST (EQ. 372) f⫽
TL ⫽ 0.0065104 rad GJ
⫽ 0.373°
;
2b2h2t b + h
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Problem 3.104 A thinwalled 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 3.104
Thinwalled tube b ⫽ 150 mm
(a) SHEAR STRESS (Eq. 361)
h ⫽ 100 mm
t⫽
t ⫽ 6.0 mm L ⫽ 1.2 m
f⫽
G ⫽ 75 GPa
TL ⫽ 0.002444 rad GJ
⫽ 0.140°
Eq. (364): Am ⫽ bh ⫽ 0.015 m2 J⫽
;
(b) ANGLE OF TWIST (Eq. 372)
T ⫽ 1650 N ⭈ m
Eq. (371) with t1 ⫽ t2 ⫽ t:
T ⫽ 9.17 MPa 2tAm
;
2b2h2t b + h
J ⫽ 10.8 ⫻ 10⫺6 m4 Tube (1)
Problem 3.105 A thinwalled 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 crosssectional 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 thinwalled bars.)
Solution 3.105 THINWALLED TUBE (1) Am ⫽ r2 tmax ⫽
J ⫽ 2r3t
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 ⫽ 2r 2tmax U1 ⫽
A ⫽ 2rt
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
ThinWalled Tubes
t = 8 mm
Problem 3.106 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 3.106
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 ⫹ 2r
⫽ 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 3.107 A thinwalled steel tube having an elliptical cross
331
t
section with constant thickness t (see figure) is subjected to a torque T ⫽ 18 kin. 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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Solution 3.107
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 kin.
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 kin. 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 kin. ⫽ ⫽ L GJ (12 * 106 psi)(17.92 in.)4
u ⫽ 83.73 * 10⫺6 rad/in. ⫽ 0.0048°/in.
Problem 3.108 A torque T is applied to a thinwalled 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 3.108
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⫽ ;
ThinWalled 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 3.109 Compare the angle of twist 1 for a thinwalled circular tube (see figure) calculated from the approximate theory for thinwalled 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 3.109
C
Thinwalled tube (a) RATIO f1 f2
⫽
4r 2 + t 2
Let b ⫽ APPROXIMATE THEORY TL f1 ⫽ GJ
J ⫽ 2r 3t
(b) f1 ⫽
TL GIP
2pGr3t
From Eq. (317): 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 3.1010 A thinwalled 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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Solution 3.1010
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 3.1011 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 lbin. 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 3.1011
335
ThinWalled 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 ⫽ lbin. ⫽ psi
UNITS: t ⫽ in.
t(2.0 in. ⫺ t)2 ⫽
3000 lbin. 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 lbin.
u⫽
allow ⫽ 4500 psi u allow ⫽ 0.01 rad/ft ⫽
0.01 rad/in. 12
T T ⫽ GJ Gt(b ⫺ t)3
3000 lbin
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 3.1012 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 thinwalled tube, and (b) the exact torsion theory for a circular bar.
;
100 mm t
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CHAPTER 3 Torsion
Solution 3.1012
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 thinwalled tubes.
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SECTION 3.10
Problem 3.1013 A long, thinwalled 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 thinwalled prismatic tube to a differential element of the tapered tube and then integrating along the axis of the tube.
Solution 3.1013
B
A
2TL dA + dB a 2 2 b pGt dAdB
337
ThinWalled Tubes
dB
dA
Thinwalled 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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Stress Concentrations in Torsion D2
The problems for Section 3.11 are to be solved by considering the stressconcentration factors.
Problem 3.111 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 3.111
R
Probs. 3.111 through 3.115
Stepped shaft in torsion USE FIG. 348 FOR THE STRESSCONCENTRATION 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 lbin. 16(1.52)
D1 ⫽ 2.0 in.
Tmax ⫽
D2 ⫽ 2.4 in. R ⫽ 0.1 in.
allow ⫽ 6000 psi
⬖ Tmax ⬇ 6200 lbin.
;
Problem 3.112 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 3.112
Stepped shaft in torsion USE FIG. 348 FOR THE STRESSCONCENTRATION 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. (348) 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 3.113 A full quartercircular fillet is used at the shoulder of a stepped shaft having diameter D2 ⫽ 1.0 in. (see figure). A torque T ⫽ 500 lbin. 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 3.113 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 lbin. D1 ⫽ 0.7, 0.8, and 0.9 in. Full quartercircular fillet (D2 ⫽ D1 ⫹ 2R) R⫽
D2 ⫺ D1 D1 ⫽ 0.5 in. ⫺ 2 2
USE FIG. 348 FOR THE STRESSCONCENTRATION FACTOR tmax ⫽ Kt nom ⫽ K a ⫽K
16 T pD31
16(500 lbin.) pD31
b
⫽ 2546
K D31
NOTE that max gets smaller as D1 gets larger, even though K is increasing.
Problem 3.114 The stepped shaft shown in the figure is required to transmit 600 kW of power at 400 rpm.
The shaft has a full quartercircular 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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Solution 3.114 Stepped shaft in torsion
P ⫽ 600 kW
D1 ⫽ 100 mm
n ⫽ 400 rpm allow ⫽ 100 MPa
Use the dashed line for a full quartercircular fillet. R L 0.075 D1
R ⬇ 0.075 D1 ⫽ 0.075 (100 mm)
Full quartercircular fillet
⫽ 7.5 mm
2pnT ( Eq. 342 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. 348 FOR THE STRESSCONCENTRATION 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 3.115 A stepped shaft (see figure) has diameter D2 ⫽ 1.5 in. and a full quartercircular fillet. The allowable shear stress is 15,000 psi and the load T ⫽ 4800 lbin. What is the smallest permissible diameter D1?
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SECTION 3.11 Stress Concentrations in Torsion
341
Solution 3.115 Stepped shaft in torsion
Use trialanderror. Select trial values of D1
D2 ⫽ 1.5 in.
allow ⫽ 15,000 psi T ⫽ 4800 lbin. Full quartercircular 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. 348 FOR THE STRESSCONCENTRATION FACTOR tmax ⫽ Kt nom ⫽ K a
16T
pD31 K 16(4800 lbin.) ⫽ 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 4.31 Calculate the shear force V and bending moment M at a cross section just to the left of the 16001b load acting on the simple beam AB shown in the figure.
A
B 30 in.
50 in. 120 in.
40 in.
Solution 4.31 gMA ⫽ 0: RB ⫽
3800 ⫽ 1267 lb 3
3400 gMB ⫽ 0: RA ⫽ ⫽ 1133 lb 3 FREEBODY 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
343
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Shear Forces and Bending Moments
Problem 4.32 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 4.32 FREEBODY 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 4.33 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 4.33 Pb
Pb
Pb
FREEBODY 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 4.34 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 4.34
345
Shear Forces and Bending Moments
1.0 m
2.0 m
g MD ⫽ 0:
;
M ⫽ ⫺(4.0 kN)(0.5 m)
FREEBODY 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 4.35 Determine the shear force V and bending moment M at a cross section located 18 ft from the lefthand 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 4.35
FREEBODY 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 4.36 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 lefthand support. (Note: Disregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.)
Solution 4.36
P1 = 4.0 kN P2 = 8.0 kN 1.0 m A
B
4.0 m
C
1.0 m
Beam with vertical arm FREEBODY 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 4.37 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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SECTION 4.3
Solution 4.37
347
Shear Forces and Bending Moments
Beam with overhangs FREEBODY DIAGRAM OF LEFTHAND 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 4.38 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 4.38
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Shear Forces and Bending Moments
Archer’s bow FREEBODY 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 FREEBODY 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 4.39 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 4.39
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 4.310 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 4.310
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 4.311 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 lbft? (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 4.311
6 ft
Beam with a cable FREEBODY DIAGRAM OF SECTION AC
g MC ⫽ 0 UNITS: P in lb M in lbft
P
哵哴
4P 4P (6 ft) + (12 ft) ⫽ 0 5 9 8P M⫽ ⫺ lbft 15 M⫺
Numerical value of M equals 640 lbft. 8P lbft 15 and P ⫽ 1200 lb
‹ 640 lbft ⫽
;
D
6 ft
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SECTION 4.3
351
Shear Forces and Bending Moments
Problem 4.312 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 4.312 FREEBODY 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 4.313 Beam ABCD represents a reinforcedconcrete 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 4.313
Page 352
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 lbft MB ⫽ 9000 lbft
;
;
Problem 4.314 The simplysupported 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 4.314
Shear Forces and Bending Moments
353
Beam with cable and weight FREEBODY DIAGRAM OF PULLEY AT B
RA ⫽ 18 kN
RD ⫽ 9 kN
FREEBODY 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 4.315 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 4.315
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
ShearForce and BendingMoment Diagrams
ShearForce and BendingMoment Diagrams When solving the problems for Section 4.5, draw the shearforce and bendingmoment diagrams approximately to scale and label all critical ordinates, including the maximum and minimum values. Probs. 4.51 through 4.510 are symbolic problems and Probs. 4.511 through 4.524 are numerical problems. The remaining problems (4.525 through 4.540) involve specialized topics, such as optimization, beams with hinges, and moving loads.
P
a
P
a
A
B
Problem 4.51 Draw the shearforce and bendingmoment diagrams for a simple beam AB supporting two equal concentrated loads P (see figure). L
Solution 4.51
Simple beam
Problem 4.52 A simple beam AB is subjected to a counterclockwise couple of moment M0 acting at distance a from the lefthand support (see figure). Draw the shearforce and bendingmoment diagrams for this beam.
M0 A
B a L
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Solution 4.52
Page 356
Shear Forces and Bending Moments
Simple beam
q
Problem 4.53 Draw the shearforce and bendingmoment diagrams for
A
a cantilever beam AB carrying a uniform load of intensity q over onehalf of its length (see figure).
B L — 2
Solution 4.53
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
ShearForce and BendingMoment Diagrams
Problem 4.54 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 shearforce and bendingmoment diagrams for this beam.
Solution 4.54
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 shearforce and bendingmoment diagrams for this beam.
A
B L — 3
Solution 4.55 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 4.55 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 4.56 A simple beam AB subjected to couples M1 and 3M1
M1
acting at the third points is shown in the figure. Draw the shearforce and bendingmoment diagrams for this beam.
3M1
A
B L — 3
L — 3
L — 3
Solution 4.56 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 4.57 A simply supported beam ABC is loaded by a vertical load P acting at the end of a bracket BDE (see figure). Draw the shearforce and bendingmoment 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 4.57
Beam with bracket
P
Problem 4.58 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 shearforce and bendingmoment diagrams for beam ABC.
Solution 4.58
359
ShearForce and BendingMoment Diagrams
Beam with overhang
P
A
Pa
C
B a
a
a
a
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Shear Forces and Bending Moments
Problem 4.59 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 shearforce and bendingmoment diagrams for this beam.
q A
D B L 3
Solution 4.59
C L 3
L
Beam with overhangs
x1 L
15 0.3727L 6
q0
Problem 4.510 Draw the shearforce and bendingmoment 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 4.510
ShearForce and BendingMoment Diagrams
Cantilever beam
q0 = 10 lb/in.
Problem 4.511 The simple beam AB supports a triangular load of maximum intensity q0 10 lb/in. acting over onehalf of the span and a concentrated load P 80 lb acting at midspan (see figure). Draw the shearforce and bendingmoment diagrams for this beam.
Solution 4.511
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 lbin
M 0
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Shear Forces and Bending Moments
Problem 4.512 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 shearforce and bendingmoment diagrams for this beam.
A
B
0.8 m
Solution 4.512
1.6 m
0.8 m
Beam with distributed loads
200 lb
Problem 4.513 A cantilever beam AB supports a couple and a concentrated load, as shown in the figure. Draw the shearforce and bendingmoment diagrams for this beam.
400 lbft A
B 5 ft
5 ft
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SECTION 4.5
Solution 4.513
ShearForce and BendingMoment Diagrams
363
Cantilever beam
Problem 4.514 The cantilever beam AB shown in the figure is
2.0 kN/m
subjected to a triangular load acting throughout onehalf of its length and a concentrated load acting at the free end. Draw the shearforce and bendingmoment diagrams for this beam.
2.5 kN B
A 2m
Solution 4.514 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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Shear Forces and Bending Moments
Problem 4.515 The uniformly loaded beam ABC has simple supports
25 lb/in.
at A and B and an overhang BC (see figure). A
Draw the shearforce and bendingmoment diagrams for this beam.
C B 72 in.
Solution 4.515
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 shearforce and bendingmoment 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 4.516 A beam ABC with an overhang at one end supports
Solution 4.516
48 in.
1.6 m
1.6 m
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SECTION 4.5
365
ShearForce and BendingMoment Diagrams
Problem 4.517 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 4.517 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 4.518 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
ShearForce and BendingMoment Diagrams
Solution 4.518 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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Shear Forces and Bending Moments
Problem 4.519 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 shearforce and bendingmoment diagrams for this beam. (Note: Disregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.)
Solution 4.519
Beam with vertical arm
Problem 4.520 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 shearforce and bendingmoment diagrams for this beam.
Solution 4.520
Simple beam
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SECTION 4.5
369
ShearForce and BendingMoment Diagrams
Problem 4.521 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 4.521 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 4.522 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 shearforce and bendingmoment 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 4.522
Beam with overhangs
Problem 4.523 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 shearforce and bendingmoment 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 4.523
ShearForce and BendingMoment Diagrams
371
Beam with a cable
Note: All forces have units of pounds.
Problem 4.524 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 4.524 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 4.525 The simple beam AB shown in the figure supports a
5k
concentrated load and a segment of uniform load. Draw the shearforce and bendingmoment diagrams for this beam.
2.0 k/ft C
A 5 ft
B 10 ft
20 ft
Solution 4.525
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 (kft)
max 47.3
6.25
Problem 4.526 The cantilever beam shown in the figure supports a concentrated load and a segment of uniform load. Draw the shearforce and bendingmoment 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 4.526
ShearForce and BendingMoment Diagrams
373
Cantilever beam
Problem 4.527 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 lbft at A. 300 lbft Draw the shearforce and bendingmoment diagrams for this beam. A
B C 6.0 ft 7.0 ft
Solution 4.527
Simple beam 180 lb/ft 300 lbft A
B C 6.0 ft 7.0 ft
197.1 V 0 (lb)
0 lb 3.625 ft max 776
300 M 0 (lbft)
–343 –433 403
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Shear Forces and Bending Moments
Problem 4.528 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 shearforce and bendingmoment diagrams for this beam. A
B
2.4 m
Solution 4.528
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 4.529 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 shearforce and bendingmoment diagrams for this condition.
Solution 4.529
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
ShearForce and BendingMoment Diagrams
(L a)2 L(2a L)
x1 0.3536 a 0.2071 L
;
;
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Shear Forces and Bending Moments
Problem 4.530 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 4kN force at the end of a bracket attached at point B and a 2kN force at the midpoint of beam DE. Draw the shearforce and bendingmoment diagrams for this compound beam.
4 kN
1m
B
C
1m
A
E
2m
Solution 4.530
2 kN D
2m
2m
2m
Compound beam
Problem 4.531 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
ShearForce and BendingMoment Diagrams
377
Solution 4.531 MA By
q0 2 L (clockwise), Ax 0 6
q0 L (upward) 2
V
;
0
–q0L/2
; L2/6
q0 M
Problem 4.532 The shearforce 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 4.532
Simple beam (V is given)
1.0 m
1.0 m
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Shear Forces and Bending Moments
Problem 4.533 The shearforce 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 4.533
16 ft
4 ft
Forces on a beam (V is given)
FORCE DIAGRAM
Problem 4.534 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
ShearForce and BendingMoment Diagrams
379
Solution 4.534 FREEBODY 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 4.535 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 4.535 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
;
;
FREEBODY 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
ShearForce and BendingMoment Diagrams
Problem 4.536 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 lefthand support of the beam.
P
Solution 4.536
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 4.537 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 4.537
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
ShearForce and BendingMoment Diagrams
(2) Use u to find forces at ends A & B which are along and perpendicular to member AB (see freebody 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 4.538 Beam ABC is supported by
MD
a tie rod CD as shown (see Prob. 10.49). 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 4.538 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
FREEBODY 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
Page 384
q0L/2
384
9/25/08
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
ShearForce and BendingMoment Diagrams
FREEBODY 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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0
385
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Shear Forces and Bending Moments
Problem 4.539 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 4.539 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
ShearForce and BendingMoment Diagrams
Problem 4.540 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 4.540 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
FREEBODY 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
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Page 388
32.7 kN
388
9/25/08
18.41 kN
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D
A
BENDING MOMENT DIAGRAM
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5 Stresses in Beams (Basic Topics) d
Longitudinal Strains in Beams Problem 5.41 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 5.41
R
Steel wire R 24 in.
d
1 in. 16
From Eq. (54): 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 5.42 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 5.42
;
L = length
Copper wire d 3 mm âmax 0.0024 L 2pr r
L 2p
From Eq. (54): 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 5.43 A 4.5 in. outside diameter polyethylene pipe designed to carry chemical wastes is placed in a trench and bent around a quartercircular 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 5.43
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 5.44 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 5.44 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 5.45 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 5.45 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 5.46 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 5.46 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 5.51 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 5.51 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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Problem 5.52 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 5.52 (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 5.53 A thin, highstrength 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 5.53 (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 5.54 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 5.54 (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 5.55 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 Ishaped 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 5.55
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 5.56 A freightcar 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 5.56 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 5.57 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 5.57
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 lbft + 135.4 lbft 2 855.4 lbft 10,264 lbin.
S
2
bh 3.0 in.3. 6
smax
10,264 lbin. M 3420 psi S 3.0 in.3
;
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Normal Stresses in Beams
Problem 5.58 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 Ishaped 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 5.58 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 5.59 The horizontal beam ABC of an oilwell 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 5.59 NUMERICAL DATA FC 9 k
MAX. BENDING STRESS AT B
BC 16 ft
Mmax F C (BC)
Mmax 144 kft
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 5.510 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 crosssectional 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 5.510
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
BENDINGMOMENT 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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399
Problem 5.511 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 5.511
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. BENDINGMOMENT 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 lbin. 4 2
Mmax 2,692.7 lbin.
d2 3.0 in. 2
(2,692.7 lbin.)(3.0 in.) 18.699 in.4
(Tension on top) qa2 2,692.7 lbin. M1 2
c
432 psi
;
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Problem 5.512 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 5.512
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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Normal Stresses in Beams
401
y
Problem 5.513 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 5.513 MAX. TENSILE STRESS DUE TO POSITIVE BENDING MOMENT IS ON BOTTOM OF BEAM CROSSSECTION
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 5.514 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 5.514
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 5.515 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 Ibeam having section modulus S 16.2 in.3
Solution 5.515
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 5.516 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 5.516 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 5.517 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 5.517 NUMERICAL DATA
MAXIMUM STRESSES c1 0.617 in.
I 3.36 in.
4
c2 2.269 in. MAmax
22.5 (8)2 + 250 (5) ftlb 2
st
MAmax c1 I
st 4341 psi
sc
MAmax c2 I
sc 15964 psi
; ;
MAmax 1970 ftlb MAmax (12) 23640 in.lb
q
Problem 5.518 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 5.518 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 5.519 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 5.519 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 ftlb RB 2700 lb R A 900 lb
max. () moment at B MB q
(6)2 2
MB 3600 ftlb
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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 5.520 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 5.520
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 lbs2/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 5.521 A beam of Tsection 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 5.521 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 ftlb at load P, 1250 ftlb 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 ftlb
c1 2 in. Aw a
Mrt 1250 ftlb
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 REACTIONSSUM 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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Problem 5.522 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 5.522
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
Crosssection: 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 5.523 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 5.523
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
;
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Stresses in Beams (Basic Topics)
Problem 5.524 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 5.524 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 5.525 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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Solution 5.525 (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 ftlb 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 5.61 The cross section of a narrowgage railway
Wood tie
d b Steel girder
(b)
s1 (a)
Railway cross tie Mmax S
P(s1 s2) 15,000 lbin. 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 5.61
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 5.62 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 5.62
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 5.63 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 wideflange beam (W shape) from Table E1(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 5.63 sa 17000 psi q 225
lb ft
Mmax1 PL +
P 2750 lb
Mmax2 2.774 * 104 lbft smax
qL2 2
below allowable OK
Mmax1 2.695 * 104 lbft
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 lbft smax
M max3 (12) Sact
smax 9411 psi
well below allowable  OK use W 14 * 26
;
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Problem 5.64 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 wideflange beam (W shape) from Table E1(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 5.64 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 Ibeam (S shape) from Table E2(a), Appendix E, and recalculate S taking into account the weight of the beam. Select a new beam size if necessary.
P
q
Problem 5.65 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 5.65 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 ftlb
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 ftlb 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 5.66 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 5.66
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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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 5.67 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 5.67
s
L s
Floor joists Mmax
qL2 1 (13.333 lb/in.)(126 in.)2 26,460 lbin. 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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SECTION 5.6
Design of Beams
Problem 5.68 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 5.68
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 5.69 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 5.69 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 E3(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 5.610 A socalled “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 5.610
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 5.611 A twoaxle 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 Ibeam (S shape) from Table E2(a), Appendix E.
18 ft
Solution 5.611 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 ftlb Sreqd
Mmax sa
Sreqd 15.37 in.3
;
(b) SELECT MOST ECONOMICAL S SHAPE FROM TABLE E2(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 5.612 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 5.612
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 5.613 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 wideflange 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 5.613 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 CGIVES 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 CGIVES MAX. MOMENT AT C (SEE MOMENT DIAGRAM) 8 q a2 sa S 12 in./ft
qallow
8a2
qallow 282
;
for shear release
w
Problem 5.614 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 5.614
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 5.615 A beam having a cross section in the form of an unsymmetric wideflange 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 5.615
Unsymmetric wideflange 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 CROSSSECTIONAL AREA ABOUT THE LOWER EDGE BB 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 BB 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 5.616 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 5.616 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 5.617 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 5.617
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 crosssectional 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 5.618 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 loadcarrying capacity.
A B
D
C
b
L (a) q A
D B
C L (b)
Solution 5.618 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 loadcarrying 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 5.619 A steel plate (called a cover plate) having crosssectional dimensions 6.0 in. * 0.5 in. is welded along the full length of the bottom flange of a W 12 * 50 wideflange beam (see figure, which shows the beam cross section). What is the percent increase in the smaller section modulus (as compared to the wideflange beam alone)?
W 12 50
6.0 0.5 in. cover plate
Solution 5.619 FIND I ABOUT HORIZ. CENTROIDAL AXIS
NUMERICAL PROPERTIES FOR W 12 * 50 (FROM TABEL E1(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 5.620 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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SECTION 5.6
Design of Beams
427
Solution 5.620 NUMERICAL DATA L 150 mm sa 60 MPa
(b) NOW MODIFYINCLUDE 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 WEIGHTFIND bmin Mmax1 1.5 q
w g 12b22
;
;
Problem 5.621 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 loadcarrying 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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Stresses in Beams (Basic Topics)
Solution 5.621
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 5.622 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 5.622 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 5.623 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 bendingmoment 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 5.623
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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431
Nonprismatic Beams
Nonprismatic Beams Problem 5.71 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 5.71 (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 5.72 A tall signboard is supported by two vertical beams consisting of thinwalled, 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 5.72 (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 5.73 A tapered cantilever beam AB having rectangular cross sections is subjected to a concentrated load P 50 lb and a couple M0 800 lbin. 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 lbin. hA = 2.0 in.
B hB = 3.0 in.
x b = 1.0 in.
3P q0 = — L
A M0 = 800 lbin.
x L = 20 in. (a)
b = 1.0 in.
L = 20 in. (b)
B
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SECTION 5.7
Nonprismatic Beams
435
Solution 5.73 (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)
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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 5.74 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 crosssectional 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 5.74 (ad) 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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Stresses in Beams (Basic Topics)
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 5.75 Refer to the tapered cantilever beam of solid circular cross section shown in Fig. 524 of Example 59. (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 5.75
Tapered cantilever beam
FROM EQ. (532), EXAMPLE 59 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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Stresses in Beams (Basic Topics)
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 5.76 to 5.78 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 5.76 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 5.76
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 5.77 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 5.77
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 5.78 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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Stresses in Beams (Basic Topics)
Solution 5.78
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 5.81 The shear stresses t in a rectangular beam are given by Eq. (539): t
V h2 a y21 b 2I 4
in which V is the shear force, I is the moment of inertia of the crosssectional 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. 530). By integrating over the crosssectional area, show that the resultant of the shear stresses is equal to the shear force V.
Solution 5.81
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 5.82 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 5.82 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 5.83 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 5.83 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 kft
Problem 5.84 A cantilever beam of length L 2 m supports a load P 8.0 kN (see figure). The beam is made of wood with crosssectional 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 5.84
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. (539): 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 5.85 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 5.85
h = 2 in.
Shear stresses in a simple beam
V h2 Eq. (539): 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 5.86 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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Stresses in Beams (Basic Topics)
Solution 5.86 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 5.87 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 onethird 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 5.87 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 5.88 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 5.88 (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 5.89 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 righthand support, and a moment at A of 18,500 ftlb (see figure). The allowable stresses in bending and shear, respectively, are 2250 psi and 160 psi.
;
7500 lb 18,500 ftlb 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 5.89 (a) q 125
1b ft
L 10 ft
P 75001b M 18500 ftb 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 1bft
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 1bft Sreq
RB 7.725 * 103 1b +
qbeam L 2
Mmax sallow
;
Use 8 * 12 in. beam
Problem 5.810 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 5.810
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 5.811 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 tongueandgroove planks (actual thickness 1.5 in.; see Appendix F) supported on two 8ft 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 5.811
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 lbin.; 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 (lbin.)
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 freebody 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) Freebody 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 5.812 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 5.812 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 5.91 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 5.91 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 1bft
(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 5.92 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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CHAPTER 5
Solution 5.92
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 5.93 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 onetenth 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 5.93
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 lbin. 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 lbin.) 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 5.94 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 5.94
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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SECTION 5.10
Shear Stresses in Beams with Flanges
Shear Stresses in Beams with Flanges Problem 5.101 through 5.106 A wideflange 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 5.101 Dimensions of cross section: b 6 in., t 0.5 in., h 12 in., h1 10.5 in., and V 30 k.
Solution 5.101
Wideflange beam (b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 548b)
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. 549)
(a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 548a) V (bh2 bh21 + th21) 5795 psi 8It
V 5714 psi th1
tmax 1.014 taver
MOMENT OF INERTIA (Eq.547)
tmax
;
(c) AVERAGE SHEAR STREAR IN THE WEB (Eq. 550)
h1 10.5 in.
I
Vb 2 (h h12) 4555 psi 8It
;
Vweb
th1 (2tmax + tmin) 28.25 k 3
Vweb 0.942 V
;
;
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Stresses in Beams (Basic Topics)
Problem 5.102 Dimensions of cross section: b 180 mm, t 12 mm, h 420 mm, h1 380 mm, and V 125 kN.
Solution 5.102
Wideflange beam b 180 mm
(b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 548b)
t 12 mm
tmin
h 420 mm
taver
V 125 kN
V 27.41 MPa th1
tmax 1.037 taver
MOMENT OF INERTIA (Eq. 547) 1 (bh3 bh31 + th31) 343.1 * 106 mm4 12
;
;
(d) SHEAR FORCE IN THE WEB (Eq. 549) Vweb
(a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 548a) tmax
;
(c) AVERAGE SHEAR STRESS IN THE WEB (Eq. 550)
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 5.103 Wideflange shape, W 8 * 28 (see Table E1(a), Appendix E); V 10 k.
Solution 5.103
Wideflange beam (b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 548b)
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. 549)
(a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 548a) V (bh2 bh21 + th21) 4861 psi 8It
V 4921 psi th1
tmax 0.988 taver
MOMENT OF INERTIA (Eq. 547)
tmax
;
(c) AVERAGE SHEAR STRESS IN THE WEB (Eq. 550)
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 5.104 Dimensions of cross section: b 220 mm, t 12 mm, h 600 mm, h1 570 mm, and V 200 kN .
Solution 5.104
Wideflange beam b 220 mm
(c) AVERAGE SHEAR STRESS IN THE WEB (EQ. 550)
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. 549) MOMENT OF INERTIA (Eq. 547)
Vweb
1 I (bh3 bh31 + th31) 750.0 * 106 mm4 12 (a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 548a) 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. 548b) tmin
Vb 2 (h h21) 21.45 MPa 8It
;
Problem 5.105 Wideflange shape, W 18 * 71 (see Table E1(a), Appendix E); V 21 k.
Solution 5.105
Wideflange beam (b) MINIMUM SHEAR STRESS IN THE WEB (EQ. 548b)
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. 549)
(a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 548a) V (bh2 bh21 + th21) 2634 psi 8It
V 2518 psi th1
tmax 1.046 taver
MOMENT OF INERTIA (Eq. 547)
tmax
;
(c) AVERAGE SHEAR STRESS IN THE WEB (EQ. 550)
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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Stresses in Beams (Basic Topics)
Problem 5.106 Dimensions of cross section: b 120 mm, t 7 mm, h 350 mm, h1 330 mm, and V 60 kN
Solution 5.106
Wideflange beam (b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 548)
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. 547) 1 (bh3 bh31 + th31) 90.34 * 106 mm4 12
;
;
(d) SHEAR FORCE IN THE WEB (Eq. 549) Vweb
(a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 548a) tmax
;
(c) AVERAGE SHEAR STRESS IN THE WEB (Eq. 550)
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 5.107 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 wideflange shape (see Table E1(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 E1(a))
q — 2
q
B
A L = 6.5 ft
Solution 5.107 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 5.108 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 5.108 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 5.109 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 E2 (a), Appendix E, the lightest Ibeam (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 5.109
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 lbft Mmax  MB
Mmax 2.16 * 104 lbft
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 lbft 4
Required section modulus: S
Mmax sallow
S 14.4 in.3
Lightest beam is S 8 * 23 (from Table E2(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 5.1010 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 5.1010
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 5.1011 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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Stresses in Beams (Basic Topics)
Solution 5.1011
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 5.1012 The Tbeam shown in the figure has crosssectional 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 5.1012 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 zAXIS 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 5.1013 Calculate the maximum shear stress tmax in the web of the Tbeam 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 5.1013
Tbeam
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 zAXIS
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
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Stresses in Beams (Basic Topics)
BuiltUp Beams Problem 5.111 A prefabricated wood Ibeam 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 5.111
Wood Ibeam 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
BuiltUp Beams
Problem 5.112 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 5.112 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 5.113 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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Stresses in Beams (Basic Topics)
Solution 5.113 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 5.114 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 5.114
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 5.115 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 5.115 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.
BuiltUp Beams
2FI VQb
smax 4.63 in.
;
;
y
Problem 5.116 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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Solution 5.116
Page 470
Stresses in Beams (Basic Topics)
Two wood box beams
Crosssectional 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 5.117 A hollow wood beam with plywood webs has the crosssectional 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 5.117
471
BuiltUp 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 5.118 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 5.118 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 5.119 The Tbeam 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 5.119 F allow 1.8
Tbeam (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 5.1110 A steel beam is built up from a W 410 * 85 wideflange 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 E1(b).)
;
y
z
180 mm 9 mm cover plates
W 410 85 O
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SECTION 5.11
BuiltUp Beams
473
Solution 5.1110 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 5.1111 The three beams shown have approximately the same crosssectional 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 capacitytoweight 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 5.1111
8 0.52
14 0.675
4 0.375 Beam 2
b1 8 in.
Beam 3
Builtup 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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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 CAPACITYTOWEIGHT 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 CAPACITYTOWEIGHT 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 5.1112 Two W 310 * 74 steel wideflange beams are bolted together to form a builtup 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 E1(b).)
W 310 74
W 310 74
Solution 5.1112 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 5.121 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 47/8 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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Stresses in Beams (Basic Topics)
Solution 5.121
Brace and bit
P 25 lb (compression) M Pb (25 lb)(4 7/8 in.) 121.9 lbin.
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 lbin. 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 5.122 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 5.122 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 5.123 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 5.123
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 5.124 A rigid frame ABC is formed by welding two steel pipes at B (see figure). Each pipe has crosssectional 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 5.124
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 5.125 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 5.125
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 lbin. N (P1 + P2) sin 60° (1000 lb) sin 60° 866 lb FREEBODY 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 lbin. 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 5.126 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 5.126 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 5.127 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 5.127
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 5.128 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 –zdirection
z
T L
x
Solution 5.128 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 5.129 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 5.129
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 5.1210 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 5.1210
h t
B
Flying buttress
FREEBODY 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 5.1211 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 5.1211
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 5.1212 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 5.1212 (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 5.1213 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 5.1213
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 5.1214 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 crosssectional area is reduced by removing onehalf 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 5.1214
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 5.1215 A short column constructed of a W 12 * 35 wideflange 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 5.1215
Column of wideflange 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. (22 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 5.1216 A short column of wideflange 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 5.1216 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 COLUMNW 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 (22 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 kin.
MAXIMUM TENSILE STRESS OCCURS AT CORNER st
;
2
L 4 4 2 inch angle section (see Table E4(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 5.1217 A tension member constructed of an
Solution 5.1217
P Pe + c A I
st 15.48 ksi
;
I 11.04 in.4 M 11.625 kin.
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 5.1218 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 E3(b) in Appendix E for channel properties and Table E4(b) for angle properties.
y P
C C 200 × 17.1 (b)
Solution 5.1218 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 (zaxis)
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 2ANGLES
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 stressconcentration factors.
M
M h
Problem 5.131 The beams shown in the figure are subjected
to bending moments M 2100 lbin. 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 5.134
h1
(b)
Solution 5.131 M 2100 lbin. 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.(557): sc
1 d Ú h 2
Eq.(556): 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. (558)
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. 550)
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 5.132 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 5.132 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. (557): 6Mh
b(h3 d3)
d 1 Ú h 2
h1 40 mm
Eq. (558): 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. (556): 12Md
h 44 mm 1.1 h1 40 mm
(2)
sm ax (MPa)
2 4 6 8
R h1
K (Fig. 550)
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 5.133 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 lbin. Determine the minimum permissible width bmin of the beam.
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Solution 5.133 h 0.88 in.
Page 494
Stresses in Beams (Basic Topics)
Beam with semicircular notches h1 0.80 in.
smax 60 ksi M 600 lbin. 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 lbin.) b(0.80 in.)2
d
Solve for b: ;
bmin L 0.24 in.
From Fig. 550: K L 2.57
Problem 5.134 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 5.134
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.550: K L 2.20
Problem 5.135 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 kin., 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 5.135
Beam with notches and a hole
h 5.5 in. h1 5 in. b 1.6 in. M 130 kin. 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 kin.)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. 550, 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. (556). h 2
;
Solve numerically: dmax 4.13 in.
;
or
495
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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 6.21 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 lbin. 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 6.21 Composite beam b ⫽ 2 in.
h ⫽ 0.7 in.
hc ⫽ 0.5 in.
M ⫽ 250 lbin.
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. (66b): score ⫽ ;
E1I1 + E2I2 ⫽ 176,600 lbin.2 From Eq. (66a): sface ⫽ ;
M(h/2)E1 E1I1 + E2I2
⫽ ; 1980 psi
M(hc / 2)E2 E1I1 + E2I2
⫽ ;531psi
;
;
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CHAPTER 6 Stresses in Beams (Advanced Topics)
y
Problem 6.22 A wood beam with crosssectional 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
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300 mm
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300 mm
12 mm
12 mm
y
C
(a)
(b)
Solution 6.22 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 6.23 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 Douglasfir plywood and flanges of pine, as shown in the figure in a crosssectional 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.
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1 in. 1.5 in. 1 in.
12 in. 3.5 in.
1 in. (b)
(a)
Solution 6.23 (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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CHAPTER 6 Stresses in Beams (Advanced Topics)
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 6.24 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 6.24 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 6.25 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 ftlb 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 0.25in.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 ftlb.) (c) If q ⫽ 660 lb/ft and allowable stress values in (b) apply, what is M0,max at B?
5.5 in. (b)
Solution 6.25 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 lbft
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 lbft
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 lbft
;
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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 lbft
;
y
Problem 6.26 A plasticlined steel pipe has the crosssectional 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 6.26 Steel pipe with plastic liner MAXIMUM MOMENT BASED UPON THE STEEL (1) From Eq. (66a): 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. (66b): 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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503
SECTION 6.2 Composite Beams
Problem 6.27 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 kin. 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. 6.27 and 6.28
Solution 6.27
C
hc
b
t
h
Sandwich beam I2 ⫽
bh3c ⫽ 110.92 in.4 12
M ⫽ 40 k.in. E1I1 + E2I2 ⫽ 348.7 * 106 lbin.2 (a) GENERAL THEORY (EQS. 66a 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. 68 AND 69)
hc ⫽ 5.5 in.
;
;
E2 ⫽ 12,000 psi
Problem 6.28 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 6.28
Sandwich beam (a) GENERAL THEORY (EQS. 66a 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. 68 AND 69) 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 6.29 A bimetallic beam used in a temperaturecontrol switch consists of strips of aluminum and copper bonded together as shown in the figure, which is a crosssectional 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 lbin. 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 6.29 Bimetallic beam NEUTRAL AXIS (EQ. 63)
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 lbin.
L2
1 — in. 16
ydA ⫽ y2A2 ⫽ (h1 ⫺ t ⫺ t/2)(bt) ⫽ (h1 ⫺ 3/32)(1)(1/16) in.3
Eq. (63): 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. 66a 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 PARALLELAXIS 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 lbin.2
Problem 6.210 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 6.210 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. (63): E1 11ydA + E2 12ydA ⫽ 0
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MAXIMUM STRESSES (EQS. 66a 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 PARALLELAXIS 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 6.211 A simply supported wooden Ibeam 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 Douglasfir 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 6.211 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 6.212 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 6.212 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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CHAPTER 6 Stresses in Beams (Advanced Topics)
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
;
TransformedSection 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 transformedsection method in the solutions.
y
Problem 6.31 A wood beam 8 in. wide and 12 in. deep
3.5 in.
(nominal dimensions) is reinforced on top and bottom by 0.25in.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 TransformedSection Method
Solution 6.31 Mmax ⫽min(M1, M2)
(a) FIND Mmax b ⫽ 7.5 in.
(1) Wood beam
h1 ⫽ 11.5 in.
Mmax ⫽ 422 kin.
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 kin.
;
THE MOMENT CAPACITY OF THE BEAM IN (a) IS 2.3 (b)
TIMES MORE THAN THE BEAM IN
y
Problem 6.32 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 6.32 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. 615) 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. 617) 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 6.33 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 E3a of Appendix E for the dimensions and proper(a) (b) ties of the Cshape 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 TransformedSection Method
Solution 6.33 (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 18FTLONG SIMPLE BEAM From Mmax ⫽
IT
MAXIMUM MOMENT BASED UPON THE STEEL (2)
Mmax ⫽ min(M1, M2) STEEL GOVERNS.
b Mmax a b 2
;
;
Problem 6.34 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 crosssectional dimensions 150 mm ⫻ 250 mm and two steel plates of crosssectional 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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CHAPTER 6 Stresses in Beams (Advanced Topics)
Solution 6.34 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. 615) sw ⫽ s1 ⫽
Mmax(h1/2) ⫽ 1.9 MPa IT
;
MAXIMUM STRESS IN THE STEEL (2) (EQ. 617) ss ⫽ s2 ⫽
Mmax(h2/2)n ⫽ 49.9 MPa IT
;
Wood beam is not changed. n⫽
Es 209 ⫽ ⫽ 19 Ew 11
Problem 6.35 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 kin.
y t
z
d
C
3d
d
Probs. 6.35 and 6.36
b
h ⫽ 4t ⫹ 5d
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SECTION 6.3 TransformedSection Method
513
Solution 6.35 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 kin.
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. 615) sp ⫽ s1 ⫽
M(h/2 ⫺ t) ⫽ 72 psi IT
;
MAXIMUM STRESS IN THE ALUMINUM (2) (EQ. 617) 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 6.36 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 6.36 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. 615) sp ⫽ s1 ⫽ Plastic segments are not changed.
M(h/2 ⫺ t) ⫽ 0.47 MPa IT
;
MAXIMUM STRESS IN THE ALUMINUM (2) (EQ. 617)
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 6.37 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 E5a 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 TransformedSection Method
Solution 6.37 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 18FTLONG SIMPLE BEAM qallowL2 8
qallow ⫽ 264 lb/ft.
;
Mmax ⫽136.0 k # in
;
ALLOWABLE LOAD ON A 18FTLONG 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 6.38 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 6.38 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. 615) sa ⫽ s1 ⫽
Mh1 IT
MAXIMUM STRESS IN THE STEEL (2) (EQ. 617) 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 6.39 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 E5a 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 TransformedSection Method
Solution 6.39 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 kin.
STEEL GOVERNS
;
y
Problem 6.310 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 6.310 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. 617) sA ⫽ s2 ⫽
Mh1n IT
SA ⫽
IT M ⫽ sA h1n
⫽ 50.6 mm3
⫽ 240.31mm
4
SMALLER SECTION MODULUS MAXIMUM STRESS IN MATERIAL B (EQ. 615) Mh2 sB ⫽ s1 ⫽ IT
IT M SB ⫽ ⫽ ⫽ 69.6 mm3 sB h2
SA ⫽ 50.6 mm3
;
‹ Maximum stress occurs in metal A.
;
y
Problem 6.311 A W 12 ⫻ 50 steel wideflange beam and a segment of a 4inch thick concrete slab (see figure) jointly resist a positive bending moment of 95 kft. 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 E1a of Appendix E for the dimensions and properties of the steel beam.)
30 in. 4 in. z
O
W 12 ⫻ 50
Solution 6.311 Steel beam and concrete slab (1) Concrete: b ⫽ 30 in.
t ⫽ 4 in.
(2) Wideflange beam: W 12 ⫻ 50 d ⫽ 12.19 in.
I ⫽ 394 in.4
A ⫽ 14.7 in.2
M ⫽ 95 kft ⫽ 1140 kin.
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. 615) 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. 617) ss ⫽ s2 ⫽
Mh2n ⫽ 13,400 psi (Tension) IT
;
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SECTION 6.3 TransformedSection Method
Problem 6.312 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 6.312
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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Area A3: I3 ⫽
MAXIMUM MOMENT BASED UPON ALUMINUM (2) (EQ. 617)
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. 615) 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 6.41 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 6.41 Location of neutral axis Iy ⫽
hb3 12
Iz
h2
Iy
⫽
b2
See Figure 615b. 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. 623):
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 6.42 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. 6.42 and 6.43
Solution 6.42 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 6.43 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 6.43 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 lbin.
Mz 66598 lbin.
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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CHAPTER 6 Stresses in Beams (Advanced Topics)
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 6.44 A simply supported wideflange 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 E1b of Appendix E for the dimensions and properties of the beam.)
P
n
E D b
C B
z A
n
a
Probs. 6.44 and 6.45
Solution 6.44 L 3.5 m
P 18 kN
W 250 44.8
Wideflange 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 6.45 Solve the preceding problem using the following data: W 8 21 section, L 84 in., P 4.5 k, and a 22.5°.
Solution 6.45 L 84 in.
P 4.5 k
BENDING STRESSES
a 22.5°
sx(z, y)
WIDEFLANGE 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 lbin.
Pcos(a)L 4
Mz 87307 lbin.
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 6.46 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. 6.46 and 6.47
Solution 6.46 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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CHAPTER 6 Stresses in Beams (Advanced Topics)
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 6.47 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 6.47 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 6.48 A steel beam of Isection (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 E2b 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 6.48 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 6.49 A cantilever beam of wideflange 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 E1a of Appendix E for the dimensions and properties of the beam.)
P a z
C
Probs. 6.49 and 6.410
Solution 6.49 Cantilever beam with inclined load BENDING MOMENTS My (P cos a)L 82,595 lbin. Mz (P sin a)L 117,960 lbin. NEUTRAL AXIS nn (EQ. 623) u 90° a 35° tan b
Iz Iy
tan u
(see Fig. 615) 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. 618) 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 6.410 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 E1b of Appendix E for the dimensions and properties of the beam.)
Solution 6.410 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 6.411 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 E1a 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 6.411
Cantilever beam with inclined load (b) NEUTRAL AXIS nn (EQ. 623) u 180° + a tan b
Iz Iy
(see Fig. 615)
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. 618) 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 6.412 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 6.412 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 6.413 A builtup steel beam of Isection 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 kin., and a 40°. (Note: See Tables E2a and E3a 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 6.413 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 6.51 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 kin., tan u 1/3. (Note: See Table E3a of Appendix E for the dimensions and properties of the channel section.)
M z
d 2
u
Probs. 6.51 and 6.52
C
;
529
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Solution 6.51
Channel section NEUTRAL AXIS nn (EQ. 640) tan b
Iz Iy
tan u
32.6 (1/3) 8.2323 1.32
b 83.07°
;
MAXIMUM TENSILE STRESS (POINT A) (EQ. 638) 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 kin.
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. 638) 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 6.52 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 6.52 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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531
SECTION 6.5 Bending of Unsymmetric Beams
Problem 6.53 An angle section with equal legs is subjected to a
2
bending moment M having its vector directed along the 11 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 kin. (Note: See Table E4a of Appendix E for the dimensions and properties of the angle section.)
M
1
Probs. 6.53 and 6.54
Solution 6.53
C
1
2
Angle section with equal legs NEUTRAL AXIS nn (EQ. 640) tan b
Iz Iy
tan u
44.85 tan 45° 3.8831 11.55
b 75.56°
;
MAXIMUM TENSILE STRESS (POINT A) (EQ. 638) zA c12 2.517 in. st sA
(M sin u)zA (M cos u)yA Iy Iz
3080 psi M 20 kin.
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. 638) 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 6.54 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 E4b of Appendix E for the dimensions and properties of the angle section.)
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Solution 6.54 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 6.55 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 kin. (b) The two angles are now inverted and attached backtoback 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 kin. 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 6.55 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 6.56 The Zsection of Example 127 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 127.
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 6.56 M 5 kN # m
ypA 91.97 mm
u 19.2°
st
ZSECTION 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 6.57 The cross section of a steel beam is
constructed of a W 18 71 wideflange 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 kin. (Note: The cross sectional properties of this beam were computed in Examples 122 and 125.)
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 6.57 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
BUILTUP 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 6.58 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 6.58 M 4.5 kN # m
u 22.5°
MAXIMUM TENSILE STRESS (AT POINT A) zA
BUILTUP 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 6.59 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. 638): 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. 638):
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 6.510 A builtup 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 backtoback) 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 6.510 M 15 kN # m
MAXIMUM TENSILE STRESS (AT POINT A)
u 30°
zA
BUILTUP 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 6.511 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 kin. 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 6.511 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 6.512 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 6.512 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
BUILTUP 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 WideFlange Beams
541
Shear Stresses in WideFlange Beams When solving the problems for Section 6.8, assume the cross sections are thinwalled. Use centerline dimensions for all calculations and derivations, unless otherwise specified
Problem 6.81 A simple beam of W 10 30 wideflange 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. 6.81 and 6.82
L
b — 2 h — 2
tw z
C tf
h — 2
B a
d
Solution 6.81 (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 6.82 Solve the preceding problem for a W 250 44.8 wideflange 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 6.82 (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 6.83 A beam of wideflange 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. 6.83 and 6.84
b
Solution 6.83 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 ThinWalled 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 6.84 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 6.84 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 ThinWalled Open Sections When locating the shear centers in the problems for Section 6.9, assume that the cross sections are thinwalled and use centerline dimensions for all calculations and derivations.
y
Problem 6.91 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 E3a in Appendix E.)
S
z e
Probs. 6.91 and 6.92
C
543
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Solution 6.91 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 6.92 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 E3b in Appendix E.)
Solution 6.92 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 6.93 The cross section of an unbalanced wideflange
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 Tbeam (b2 t2 0) and a balanced wideflange beam (t2 t1 and b2 b1).
h1
h2 h
Solution 6.93 Unbalanced wideflange 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 ThinWalled Open Sections
Solve Eqs. (1) and (2): h1 FLANGE 2: F2
Vt2b32 12Iz
t2b32h t1b31 + t2b32
TBEAM b2 t2 0; ‹ h1 0
Shear force V acts through the shear center S. ‹ a MS F1h1 F2h2 0
;
WIDEFLANGE BEAM
or (t1b31 ) h1 (t2b 32) h2
(1)
h1 h2 h
(2)
t2 t1 and b2 b1; ‹ h1 h/2
;
Problem 6.94 The cross section of an unbalanced wideflange 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 6.94 Unbalanced wideflange 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 6.95 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 6.95 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 ThinWalled Open Sections
Problem 6.96 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 6.96 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 6.97 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 6.97 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 ThinWalled 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 6.98 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 6.98 (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 ThinWalled Open Sections
Problem 6.99 A Ushaped 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 6.99 Ushaped 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. 673) p
Problem 6.910 Derive the following formula for the distance e from the centerline
y
of the wall to the shear center S for the Csection 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 6.910 Csection 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 ThinWalled 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. 665 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. 6.98) 2(h + 3b)
Problem 6.911 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 6.911 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. 665 when tf tw)
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SECTION 6.9 Shear Centers of ThinWalled Open Sections
Problem 6.912 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. 6.911 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 6.912 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. 6.911 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 #6.911 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 #6.911 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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557
SECTION 6.9 Shear Centers of ThinWalled Open Sections
Problem 6.913 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 6.913
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. 673) p r
T0 moment of shear stresses
SLIT CIRCULAR ARC (b p):
At angle u, dA rtdu
e 2 (Prob. 6.96) 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 6.101 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 6.101
Double trapezoid SHAPE FACTOR f f
(EQ. 679)
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. 678) 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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SECTION 6.10 Elastoplastic Bending
Problem 6.102 (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 6.102 Hollow circular cross sections (a) SHAPE FACTOR f (EQ. 679) 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. 678)
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. 690)
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Problem 6.103 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 6.103 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 6.104 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 crosssectional 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 6.104 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 crosssectional 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 6.105 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. 6.105 and 6.106
Solution 6.105 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 6.106 Solve the preceding problem for a wideflange beam with h 404 mm, b 140 mm, tf 11.2 mm, and tw 6.99 mm.
Solution 6.106 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