Engineering Mechanics Statics Eleventh Edition- R - Instructor's Solutions Manual

Engineering Mechanics - Statics Chapter 1 Problem 1-1 Represent each of the following combinations of units in the cor...

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Engineering Mechanics - Statics

Chapter 1

Problem 1-1 Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) m/ms (b) μkm (c) ks/mg (d) km⋅ μN Units Used: μN = 10

−6

μkm = 10

N

−6

km

9

Gs = 10 s 3

ks = 10 s mN = 10

−3

−3

ms = 10

N s

Solution: ( a)

m 3m = 1 × 10 ms s m km =1 ms s

( b)

μkm = 1 × 10

−3

m

μkm = 1 mm ( c)

ks 9 s = 1 × 10 mg kg ks Gs =1 mg kg

( d)

−3

km⋅ μN = 1 × 10

mN

km⋅ μN = 1 mm⋅ N

1

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


Chapter 1

Problem 1-2 Wood has a density d. What is its density expressed in SI units? Units Used: Mg = 1000 kg Given: d = 4.70

slug ft

3

Solution: 1slug = 14.594 kg d = 2.42

Mg m

3

Problem 1-3 Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) Mg/mm (b) mN/μs (c) μm⋅ Mg Solution: 3

( a)

6

Mg 10 kg 10 kg Gg = = = −3 mm m m 10 m Mg Gg = mm m

( b)

mN μs mN μs

( c)

=

10

−3

10 =

N

−6

3

=

s

10 N kN = s s

kN s

(

−6

μm⋅ Mg = 10

)(

3

)

m 10 kg = 10

−3

m⋅ kg

μm⋅ Mg = mm⋅ kg

2



Chapter 1

Problem 1-4 Represent each of the following combinations of units in the correct SI form: (a) Mg/ms, (b) N/mm, (c) mN/( kg⋅ μs). Solution: 3

( a)

6

Mg 10 kg 10 kg Gg = = = −3 ms s s 10 s Mg Gg = ms s N = mm

( b)

1N 10

−3

= 10

3N

m

m

=

kN m

N kN = mm m mN

( c)

kg⋅ μs

10

=

mN kg⋅ μs

−3

−6

10 =

N

=

kg⋅ s

kN kg⋅ s

kN kg⋅ s

Problem 1-5 Represent each of the following with SI units having an appropriate prefix: (a) S1, (b) S2, (c) S3. Units Used: kg = 1000 g

−3

ms = 10

s

3

kN = 10 N

Given: S1 = 8653 ms S2 = 8368 N S3 = 0.893 kg Solution: ( a)

S1 = 8.653 s

3



( b)

S2 = 8.368 kN

( c)

S3 = 893 g

Chapter 1

Problem 1-6 Represent each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) x, (b) y, and (c) z. Units Used: 6

MN = 10 N μg = 1 × 10

−6

gm

3

kN = 10 N Given: x = 45320 kN

(

y = 568 × 10

5

) mm

z = 0.00563 mg Solution: ( a)

x = 45.3 MN

( b)

y = 56.8 km

( c)

z = 5.63 μg

Problem 1-7 Evaluate ( a⋅ b)/c to three significant figures and express the answer in SI units using an appropriate prefix. Units Used: μm = 10

−6

m

Given: a = ( 204 mm) 4



Chapter 1

b = ( 0.00457 kg) c = ( 34.6 N) Solution: l =

ab

l = 26.945

c

μm⋅ kg N

Problem 1-8 If a car is traveling at speed v, determine its speed in kilometers per hour and meters per second. Given: v = 55

mi hr

Solution: v = 88.514

km hr

m

v = 24.6

s

Problem 1-9 Convert: (a) S1 to N ⋅ m , (b) S2 to kN/m3, (c) S3 to mm/s. Express the result to three significant figures. Use an appropriate prefix. Units Used: 3

kN = 10 N Given: S1 = 200g lb⋅ ft S2 = 350g

lb ft

S3 = 8

3

ft hr

5



Chapter 1

Solution: ( a)

S1 = 271 N⋅ m

( b)

S2 = 55.0

kN m

S3 = 0.677

( c)

3

mm s

Problem 1-10 What is the weight in newtons of an object that has a mass of: (a) m1, (b) m2, (c) m3? Express the result to three significant figures. Use an appropriate prefix. Units Used: 3

Mg = 10 kg mN = 10

−3

N

3

kN = 10 N Given: m1 = 10 kg m2 = 0.5 gm m3 = 4.50 Mg Solution: ( a)

W = m1 g W = 98.1 N

( b)

W = m2 g W = 4.90 mN

( c)

W = m3 g W = 44.1 kN

6



Chapter 1

Problem 1-11 If an object has mass m, determine its mass in kilograms. Given: m = 40 slug Solution: m = 584 kg

Problem 1-12 The specific weight (wt./vol.) of brass is ρ. Determine its density (mass/vol.) in SI units. Use an appropriate prefix. Units Used: 3

Mg = 10 kg Given: lb

ρ = 520

ft

3

Solution: Mg

ρ = 8.33

m

3

Problem 1-13 A concrete column has diameter d and length L. If the density (mass/volume) of concrete is ρ, determine the weight of the column in pounds. Units Used: 3

Mg = 10 kg 3

kip = 10 lb Given: d = 350 mm L = 2m

7



ρ = 2.45

Chapter 1

Mg m

3

Solution: 2

d V = π ⎛⎜ ⎟⎞ L ⎝ 2⎠ W = ρ V

V = 192.423 L W = 1.04 kip

Problem 1-14 The density (mass/volume) of aluminum is ρ. Determine its density in SI units. Use an appropriate prefix. Units Used: Mg = 1000 kg Given:

ρ = 5.26

slug ft

3

Solution:

ρ = 2.17

Mg m

3

Problem 1-15 Determine your own mass in kilograms, your weight in newtons, and your height in meters. Solution: Example W = 150 lb m = W

m = 68.039 kg W g = 667.233 N

h = 72 in

h = 1.829 m

8



Chapter 1

Problem 1-16 Two particles have masses m1 and m2, respectively. If they are a distance d apart, determine the force of gravity acting between them. Compare this result with the weight of each particle. Units Used: G = 66.73 × 10

− 12

m

3 2

kg⋅ s nN = 10

−9

N

Given: m1 = 8 kg m2 = 12 kg d = 800 mm Solution: F =

G m1 m2 d

2

F = 10.0 nN W1 = m1 g

W1 = 78.5 N

W2 = m2 g

W2 = 118 N

W1 F

W2 F

= 7.85 × 10

9

= 1.18 × 10

10

Problem 1-17 Using the base units of the SI system, show that F = G(m1m2)/r2 is a dimensionally homogeneous equation which gives F in newtons. Compute the gravitational force acting between two identical spheres that are touching each other. The mass of each sphere is m1, and the radius is r. Units Used: μN = 10

−6

N

G = 66.73 10

− 12

m

3 2

kg⋅ s

9



Chapter 1

Given: m1 = 150 kg r = 275 mm Solution: F =

G m1 ( 2r)

2

2

F = 4.96 μN Since the force F is measured in Newtons, then the equation is dimensionally homogeneous.

Problem 1-18 Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) x, (b) y, (c) z. Units Used: 6

MN = 10 N 3

kN = 10 N μm = 10

−6

m

Given: x = ( 200 kN)

2

y = ( 0.005 mm) z = ( 400 m)

2

3

Solution: 2

( a)

x = 0.040 MN

( b)

y = 25.0 μm

( c)

z = 0.0640 km

2

3

10



Chapter 1

Problem 1-19 Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) a 1/b1, (b) a2b2/c2, (c) a3b3. Units Used: μm = 10

−6

6

m

Mm = 10 m

Mg = 10 gm

kg = 10 gm

6

−3

ms = 10

3

s

Given: a1 = 684 μm b1 = 43 ms a2 = 28 ms b2 = 0.0458 Mm c2 = 348 mg a3 = 2.68 mm b3 = 426 Mg Solution: a1

( a)

b1

= 15.9

a2 b2

( b)

c2

mm s

= 3.69 Mm

s kg

a3 b3 = 1.14 km⋅ kg

( c)

Problem 1-20 Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) a1/b12 (b) a22b23. Units Used: 6

Mm = 10 m

11



Chapter 1

Given: a1 = 0.631 Mm b1 = 8.60 kg a2 = 35 mm b2 = 48 kg Solution:

( a)

a1 b1

( b)

2

2

= 8.532

km kg

3

2

3

a2 b2 = 135.48 kg ⋅ m

2

12



Chapter 2

Problem 2-1 Determine the magnitude of the resultant force FR = F1 + F 2 and its direction, measured counterclockwise from the positive x axis. Given: F 1 = 600 N F 2 = 800 N F 3 = 450 N

α = 45 deg β = 60 deg γ = 75 deg

Solution:

ψ = 90 deg − β + α FR =

F1 + F 2 − 2 F1 F 2 cos ( ψ) 2

2

F R = 867 N FR

sin ( ψ)

=

F2

sin ( θ )

⎛ ⎝

θ = asin ⎜F 2 θ = 63.05 deg

sin ( ψ) ⎞ ⎟ FR ⎠

φ = θ+α φ = 108 deg

Problem 2-2 Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis. Given: F 1 = 80 lb F 2 = 60 lb 13



Chapter 2

θ = 120 deg Solution: FR =

F1 + F 2 − 2 F1 F 2 cos ( 180 deg − θ ) 2

2

F R = 72.1 lb

⎛ ⎝

β = asin ⎜ F1

sin ( 180 deg − θ ) ⎞ ⎟ FR ⎠

β = 73.9 deg

Problem 2-3 Determine the magnitude of the resultant force F R = F1 + F 2 and its direction, measured counterclockwise from the positive x axis. Given: F 1 = 250 lb F 2 = 375 lb

θ = 30 deg φ = 45 deg Solution: FR =

F1 + F 2 − 2 F1 F 2 cos ( 90 deg + θ − φ ) 2

2

F R = 178 kg FR

sin ( 90 deg + θ − φ )

⎛ F1

β = asin ⎜

⎝ FR

=

F1

sin ( β )

⎞

sin ( 90 deg + θ − φ )⎟

⎠

β = 37.89 deg 14



Chapter 2

Angle measured ccw from x axis 360 deg − φ + β = 353 deg

Problem 2-4 Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured counterclockwise from the positive u axis. Given: F 1 = 300 N F 2 = 500 N

α = 30 deg β = 45 deg γ = 70 deg Solution: FR =

F1 + F 2 − 2 F1 F 2 cos ( 180 deg − β − γ + α ) 2

2

F R = 605 N FR

sin ( 180 deg − β − γ + α )

⎛ ⎝

θ = asin ⎜F 2

=

F2

sin ( θ )

sin ( 180 deg − β − γ + α ) ⎞ ⎟ FR ⎠

θ = 55.40 deg φ = θ+α φ = 85.4 deg Problem 2-5 Resolve the force F 1 into components acting along the u and v axes and determine the magnitudes of the components. Given: F 1 = 300 N

α = 30 deg 15



F 2 = 500 N

Chapter 2

β = 45 deg

γ = 70 deg Solution: F1u

=

sin ( γ − α )

F1

sin ( 180 deg − γ ) sin ( γ − α )

F 1u = F1

sin ( 180 deg − γ )

F 1u = 205 N F 1v

sin ( α )

=

F 1v = F 1

F1

sin ( 180 deg − γ ) sin ( α )

sin ( 180 deg − γ )

F 1v = 160 N

Problem 2-6 Resolve the force F2 into components acting along the u and v axes and determine the magnitudes of the components. Given: F 1 = 300 N F 2 = 500 N

α = 30 deg β = 45 deg γ = 70 deg Solution:

⎛ sin ( β ) ⎞ ⎟ ⎝ sin ( γ ) ⎠

F 2u = F2 ⎜

16



Chapter 2

F 2u = 376.2 N

⎡ sin ⎡⎣180 deg − ( β + γ )⎤⎦⎤ ⎥ sin ( γ ) ⎣ ⎦

F 2v = F 2 ⎢

F 2v = 482.2 N

Problem 2-7 Determine the magnitude of the resultant force F R = F 1 + F 2 and its direction measured counterclockwise from the positive u axis. Given: F 1 = 25 lb F 2 = 50 lb

θ 1 = 30 deg θ 2 = 30 deg θ 3 = 45 deg Solution:

α = 180 deg − ( θ 3 + θ 1 ) FR =

F2 + F 1 − 2 F1 F 2 cos ( α ) 2

2

F R = 61.4 lb sin ( θ' ) sin ( α ) = F2 FR

⎛

F2 ⎞

⎝

FR ⎠

θ' = asin ⎜ sin ( α )

⎟

θ' = 51.8 deg θ = θ' − θ 3 θ = 6.8 deg

Problem 2-8 Resolve the force F 1 into components acting along the u and v axes and determine the components. 17



Chapter 2

Given: F 1 = 25 lb F 2 = 50 lb

θ 1 = 30 deg θ 2 = 30 deg θ 3 = 45 deg Solution: −F u

=

sin ( θ 3 − θ 2 )

Fu =

F1

sin ( θ 2 )

−F1 sin ( θ 3 − θ 2 ) sin ( θ 2 )

F u = −12.9 lb Fv

sin ( 180 deg − θ 3 )

Fv =

=

F1

sin ( θ 2 )

F 1 sin ( 180 deg − θ 3 ) sin ( θ 2 )

F v = 35.4 lb

Problem 2-9 Resolve the force F2 into components acting along the u and v axes and determine the components. Given: F 1 = 25 lb F 2 = 50 lb

θ 1 = 30 deg 18



Chapter 2

θ 2 = 30 deg θ 3 = 45 deg Solution: Fu

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦

=

F2

sin ( θ 2 )

F2 sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦

Fu =

sin ( θ 2 )

F u = 86.6 lb −Fv

=

sin ( θ 1 ) Fv =

F2

sin ( θ 2 )

−F 2 sin ( θ 1 ) sin ( θ 2 )

F v = −50 lb

Problem 2-10 Determine the components of the F force acting along the u and v axes. Given:

θ 1 = 70 deg θ 2 = 45 deg θ 3 = 60 deg F = 250 N Solution: Fu

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦

Fu =

=

F sin ( θ 2 )

F sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦ sin ( θ 2 )

F u = 320 N

19



Fv

sin ( θ 1 )

Fv =

=

Chapter 2

F sin ( θ 2 )

F sin ( θ 1 )

F v = 332 N

sin ( θ 2 )

Problem 2-11 The force F acts on the gear tooth. Resolve this force into two components acting along the lines aa and bb. Given: F = 20 lb

θ 1 = 80 deg θ 2 = 60 deg Solution: F

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦

Fa =

Fa

sin ( θ 1 )

F sin ( θ 1 )

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦ F

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦

Fb =

=

F sin ( θ 2 )

=

F a = 30.6 lb

Fb

sin ( θ 2 )

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦

F b = 26.9 lb

Problem 2-12 The component of force F acting along line aa is required to be Fa. Determine the magnitude of F and its component along line bb.

20



Chapter 2

Given: F a = 30 lb

θ 1 = 80 deg θ 2 = 60 deg

Solution: Fa

=

sin ( θ 1 )

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦

⎛ sin ( 180 deg − θ 1 − θ 2) ⎞ ⎜ ⎟ sin ( θ 1 ) ⎝ ⎠

F = Fa

Fa

=

sin ( θ 1 ) Fb =

F

F = 19.6 lb

Fb

sin ( θ 2 )

Fa sin ( θ 2 ) sin ( θ 1 )

F b = 26.4 lb

Problem 2-13 A resultant force F is necessary to hold the ballon in place. Resolve this force into components along the tether lines AB and AC, and compute the magnitude of each component.

Given: F = 350 lb

θ 1 = 30 deg θ 2 = 40 deg Solution: F AB

sin ( θ 1 )

=

F

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦

21



Chapter 2

sin ( θ 1 ) ⎡ ⎤ ⎢ ⎥ ⎣ sin ⎡⎣180 deg − ( θ 1 + θ 2)⎤⎦⎦

F AB = F

F AB = 186 lb F AC

sin ( θ 2 )

F

=

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦ sin ( θ 2 ) ⎡ ⎤ ⎢ ⎥ ⎣sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦ ⎦

F AC = F

F AC = 239 lb

Problem 2-14 The post is to be pulled out of the ground using two ropes A and B. Rope A is subjected to force F 1 and is directed at angle θ1 from the horizontal. If the resultant force acting on the post is to be F R, vertically upward, determine the force T in rope B and the corresponding angle θ. Given: F R = 1200 lb F 1 = 600 lb

θ 1 = 60 deg Solution: T =

F 1 + FR − 2 F1 F R cos ( 90 deg − θ 1 ) 2

2

T = 744 lb sin ( 90 − θ 1 ) sin ( θ ) = FR T

⎛ ⎝

θ = asin ⎜ sin ( 90 deg − θ 1 )

F1 ⎞ T

⎟ ⎠

θ = 23.8 deg

22



Chapter 2

Problem 2-15 Resolve the force F 1 into components acting along the u and v axes and determine the magnitudes of the components. Given: F 1 = 250 N F 2 = 150 N

θ 1 = 30 deg θ 2 = 30 deg θ 3 = 105 deg Solution: F 1v

sin ( θ 1 )

=

F1

sin ( θ 3 )

⎛ sin ( θ 1 ) ⎞ ⎜ ⎟ ⎝ sin ( θ 3 ) ⎠

F 1v = F 1

F 1v = 129 N F 1u

sin ( 180 deg − θ 1 − θ 3 ) F 1u = F1

=

F1

sin ( θ 3 )

⎛ sin ( 180 deg − θ 1 − θ 3 ) ⎞ ⎜ ⎟ sin ( θ 3 ) ⎝ ⎠

F 1u = 183 N

Problem 2-16 Resolve the force F 2 into components acting along the u and v axes and determine the magnitudes of the components. Given: F 1 = 250 N 23



Chapter 2

F 2 = 150 N

θ 1 = 30 deg θ 2 = 30 deg θ 3 = 105 deg Solution: F 1v

sin ( θ 1 )

F2

=

F 1v = F 2

sin ( 180 deg − θ 3 ) sin ( θ 1 ) ⎛ ⎞ ⎜ ⎟ ⎝ sin ( 180 deg − θ 3 ) ⎠

F 1v = 77.6 N F2u

sin ( 180 deg − θ 3 ) F 2u = F2

=

F2

sin ( 180 deg − θ 3 )

⎛ sin ( 180 deg − θ 3) ⎞ ⎜ ⎟ ⎝ sin ( 180 deg − θ 3) ⎠

F 2u = 150 N

Problem 2-17 Determine the magnitude and direction of the resultant force F R. Express the result in terms of the magnitudes of the components F 1 and F 2 and the angle φ.

Solution: F R = F 1 + F2 − 2F1 F 2 cos ( 180 deg − φ ) 2

2

2

24



Chapter 2

Since cos ( 180 deg − φ ) = −cos ( φ ), F 1 + F2 + 2 F 1 F2 cos ( φ ) 2

FR =

2

From the figure, tan ( θ ) =

F1 sin ( φ )

F 2 + F 1 cos ( φ )

Problem 2-18 If the tension in the cable is F1, determine the magnitude and direction of the resultant force acting on the pulley. This angle defines the same angle θ of line AB on the tailboard block. Given: F 1 = 400 N

θ 1 = 30 deg

Solution: FR =

F1 + F 1 − 2F 1 F 1 cos ( 90 deg − θ 1 ) 2

2

F R = 400 N sin ( θ 1 ) sin ( 90 deg − θ ) = FR F1

⎛ FR

θ = 90 deg − asin ⎜

⎝ F1

⎞

sin ( θ 1 )⎟

⎠

θ = 60 deg

25



Chapter 2

Problem 2-19 The riveted bracket supports two forces. Determine the angle θ so that the resultant force is directed along the negative x axis. What is the magnitude of this resultant force? Given: F 1 = 60 lb F 2 = 70 lb

θ 1 = 30 deg Solution: sin ( θ 1 ) sin ( θ ) = F1 F2

⎛

F1 ⎞

⎝

F2 ⎠

θ = asin ⎜ sin ( θ 1 )

⎟

θ = 25.4 deg φ = 180 deg − θ − θ 1 φ = 124.6 deg F1 + F 2 − 2F 1 F 2 cos ( φ ) 2

R =

2

R = 115 lb

Problem 2-20 The plate is subjected to the forces acting on members A and B as shown. Determine the magnitude of the resultant of these forces and its direction measured clockwise from the positive x axis.

Given: F A = 400 lb F B = 500 lb

26



Chapter 2

θ 1 = 30 deg θ = 60 deg Solution: Cosine law: FR =

FB + FA − 2F B FA cos ( 90 deg − θ + θ 1 ) 2

2

F R = 458 lb Sine law: sin ( 90 deg − θ + θ 1 ) FR

=

sin ( θ − α ) FA

⎛

FA ⎞

⎝

FR ⎠

α = θ − asin ⎜ sin ( 90 deg − θ + θ 1 )

⎟

α = 10.9 deg

Problem 2-21 Determine the angle θ for connecting member B to the plate so that the resultant of FA and FB is directed along the positive x axis. What is the magnitude of the resultant force? Given: F A = 400 lb F B = 500 lb

θ 1 = 30 deg 27



Chapter 2

Solution: Sine law: sin ( 90 deg − θ 1 ) sin ( θ ) = FA FB

⎛

FA ⎞

⎝

FB ⎠

θ = asin ⎜ sin ( 90 deg − θ 1 )

⎟

θ = 43.9 deg FR

sin ( 90 deg + θ 1 − θ )

FR = FA

=

FA

sin ( θ )

sin ( 90 deg − θ + θ 1 ) sin ( θ )

F R = 561 lb

Problem 2-22 Determine the magnitude and direction of the resultant FR = F 1 + F2 + F 3 of the three forces by first finding the resultant F' = F1 + F 2 and then forming F R = F' + F 3. Given: F 1 = 30 N F 2 = 20 N F 3 = 50 N

θ = 20 deg c = 3 d = 4

28



Chapter 2

Solution: c α = atan ⎛⎜ ⎟⎞

⎝ d⎠

F' =

F1 + F 2 − 2F 1 F 2 cos ( 90 deg + θ − α ) 2

2

F' = 30.9 N F'

sin ( ( 90 deg − θ + α ) )

=

⎛ ⎝

F1

sin ( 90 deg − θ − β )

β = 90 deg − θ − asin ⎜F 1

sin ( 90 deg − θ + α ) ⎞ ⎟ F' ⎠

β = 1.5 deg Now add in force F3. FR =

F' + F3 − 2F' F3 cos ( β ) 2

2

F R = 19.2 N FR

sin ( β )

F'

=

sin ( φ )

⎛ ⎝

φ = asin ⎜F'

sin ( β ) ⎞ ⎟ FR ⎠

φ = 2.4 deg

Problem 2-23 Determine the magnitude and direction of the resultant FR = F1 + F2 + F3 of the three forces by first finding the resultant F' = F 2 + F 3 and then forming F R = F' + F 1. Given: F 1 = 30 N F 2 = 20 N 29



Chapter 2

F 3 = 50 N

θ = 20 deg c = 3 d = 4 Solution: F' =

F2 + F 3 − 2F 2 F 3 cos ( ( 90 deg − θ ) ) 2

2

F' = 47.07 N F2

sin ( β )

F'

=

sin ( 90 deg − θ )

⎛ ⎝

β = asin ⎜ F2

sin ( 90 deg − θ ) ⎞ ⎟ F' ⎠

β = 23.53 deg c α = atan ⎛⎜ ⎟⎞

⎝ d⎠

γ = α−β FR =

F' + F1 − 2F' F 1 cos ( γ ) 2

2

F R = 19.2 N FR

sin ( γ )

=

F1

sin ( φ )

⎛ ⎝

φ = asin ⎜F 1

sin ( γ ) ⎞ ⎟ FR ⎠

φ = 21.16 deg ψ = β−φ ψ = 2.37 deg

30



Chapter 2

Problem 2-24 Resolve the force F into components acting along (a) the x and y axes, and (b) the x and y' axes.

Given: F = 50 lb

α = 65 deg β = 45 deg γ = 30 deg Solution: (a)

F x = F cos ( β ) F x = 35.4 lb F y = F sin ( β ) F y = 35.4 lb

(b)

Fx

sin ( 90 deg − β − γ )

=

F

sin ( 90 deg + γ )

sin ( 90 deg − β − γ )

Fx = F

sin ( 90 deg + γ )

F x = 14.9 lb Fy'

sin ( β )

=

F y' = F

F

sin ( 90 deg + γ ) sin ( β )

sin ( 90 deg + γ )

31



Chapter 2

F y' = 40.8 lb

Problem 2-25 The boat is to be pulled onto the shore using two ropes. Determine the magnitudes of forces T and P acting in each rope in order to develop a resultant force F1, directed along the keel axis aa as shown. Given:

θ = 40 deg θ 1 = 30 deg F 1 = 80 lb

Solution: F1

sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦

T = F1

=

T

sin ( θ )

sin ( θ )

sin ( 180 deg − θ − θ 1 )

T = 54.7 lb F1

sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦ P = sin ( θ 1 )

=

P

sin ( θ 1 )

F1

sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦

P = 42.6 lb

32



Chapter 2

Problem 2-26 The boat is to be pulled onto the shore using two ropes. If the resultant force is to be F1, directed along the keel aa as shown, determine the magnitudes of forces T and P acting in each rope and the angle θ of P so that the magnitude of P is a minimum. T acts at θ from the keel as shown.

Given:

θ 1 = 30 deg F 1 = 80 lb

Solution: From the figure, P is minimum when

θ + θ 1 = 90 deg θ = 90 deg − θ 1 θ = 60 deg P

sin ( θ 1 )

P =

F1

=

sin ( 90 deg)

F 1 sin ( θ 1 ) sin ( 90 deg)

P = 40 lb T

sin ( θ ) T = F1

=

F1 sin ( 90 deg)

sin ( θ ) sin ( 90 deg)

T = 69.3 lb

33



Chapter 2

Problem 2-27 The beam is to be hoisted using two chains. Determine the magnitudes of forces FA and FB acting on each chain in order to develop a resultant force T directed along the positive y axis.

Given: T = 600 N

θ 1 = 30 deg θ = 45 deg

Solution: FA

sin ( θ ) FA =

T

=

sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦ T sin ( θ )

sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦

F A = 439 N

FB

sin ( θ 1 )

FB = T

=

T

sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦ sin ( θ 1 )

sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦

F B = 311 N

34



Chapter 2

Problem 2-28 The beam is to be hoisted using two chains. If the resultant force is to be F, directed along the positive y axis, determine the magnitudes of forces FA and F B acting on each chain and the orientation θ of F B so that the magnitude of FB is a minimum. Given: F = 600 N

θ 1 = 30 deg Solution: For minimum FB, require

θ = 90 deg − θ 1 θ = 60 deg F A = F cos ( θ 1 ) F A = 520 N F B = F sin ( θ 1 ) F B = 300 N

Problem 2-29 Three chains act on the bracket such that they create a resultant force having magnitude F R. If two of the chains are subjected to known forces, as shown, determine the orientation θ of the third chain,measured clockwise from the positive x axis, so that the magnitude of force F in this chain is a minimum. All forces lie in the x-y plane.What is the magnitude of F ? Hint: First find the resultant of the two known forces. Force F acts in this direction.

35



Chapter 2

Given: F R = 500 lb F 1 = 200 lb F 2 = 300 lb

φ = 30 deg

Solution: Cosine Law: F R1 =

F 1 + F2 − 2 F 1 F2 cos ( 90 deg − φ ) 2

2

F R1 = 264.6 lb Sine Law:

Make F parallel to FR1

sin ( φ + θ ) sin ( 90 deg − φ ) = F1 F R1

⎛

θ = −φ + asin ⎜ sin ( 90 deg − φ )

⎝

⎞ ⎟ FR1 ⎠ F1

θ = 10.9 deg When F is directed along F R1, F will be minimum to create the resultant forces. F = F R − FR1 F = 235 lb

Problem 2-30 Three cables pull on the pipe such that they create a resultant force having magnitude F R. If two of the cables are subjected to known forces, as shown in the figure, determine the direction θ of the third cable so that the magnitude of force F in this cable is a minimum. All forces lie in the 36



Chapter 2

g x–y plane.What is the magnitude of F ? Hint: First find the resultant of the two known forces.

Given: F R = 900 lb F 1 = 600 lb F 2 = 400 lb

α = 45 deg β = 30 deg

Solution:

F' =

F1 + F 2 − 2F 1 F 2 cos ( 90 deg + α − β ) 2

2

F' = 802.64 lb F = F R − F' F = 97.4 lb sin ( φ ) F1

=

sin ( 90 deg + α − β ) F'

⎛ ⎝

φ = asin ⎜ sin ( 90 deg + α − β )

F1 ⎞ F'

⎟ ⎠

φ = 46.22 deg θ = φ−β θ = 16.2 deg

Problem 2-31 Determine the x and y components of the force F.

37



Chapter 2

Given: F = 800 lb

α = 60 deg β = 40 deg

Solution: F x = F sin ( β ) F y = −F cos ( β ) F x = 514.2 lb F y = −612.8 lb

Problem 2-32 Determine the magnitude of the resultant force and its direction, measured clockwise from the positive x axis. Given: F 1 = 70 N F 2 = 50 N F 3 = 65 N

θ = 30 deg φ = 45 deg Solution: + →

F Rx = ΣFx;

F RX = F 1 + F 2 cos ( θ ) − F 3 cos ( φ )

38



+

↑

FR =

F RY = −F2 sin ( θ ) − F 3 sin ( φ )

F Ry = ΣF y; 2

FRX + FRY

Chapter 2

2

⎛ FRY ⎞ ⎟ ⎝ FRX ⎠

θ = atan ⎜

F R = 97.8 N

θ = 46.5 deg

Problem 2-33 Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis. Given: F 1 = 50 lb F 2 = 35 lb

α = 120 deg β = 25 deg

Solution: + F Rx = ΣF x; →

F Rx = F 1 sin ( α ) − F2 sin ( β ) F Rx = 28.5 lb

+

↑ FRy = ΣFy;

F Ry = −F 1 cos ( α ) − F 2 cos ( β )

39



Chapter 2

F Ry = −6.7 lb FR =

2

FRx + FRy

2

F R = 29.3 lb

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ' = atan ⎜

θ' = 13.3 deg θ = 360 deg − θ' θ = 347 deg

Problem 2-34 Determine the magnitude of the resultant force and its direction , measured counterclockwise from the positive x axis.

Given: F 1 = 850 N F 2 = 625 N F 3 = 750 N

θ = 45 deg φ = 30 deg c = 3 d = 4

Solution: + →

F Rx = SF x;

F RX = F 1

d 2

c +d

2

− F 2 sin ( φ ) − F3 sin ( θ )

40



+

↑

F Ry = SF y;

F RY = −F1

c 2

2

− F2 cos ( φ ) + F3 cos ( θ )

c +d

F RX = −162.8 N FR =

Chapter 2

2

FRX + FRY

F RY = −520.9 N 2

F R = 546 N

⎛ FRY ⎞ ⎟ ⎝ FRX ⎠

α = atan ⎜

α = 72.64 deg β = α + 180 deg β = 252.6 deg

Problem 2-35 Three forces act on the bracket. Determine the magnitude and direction θ of F 1 so that the resultant force is directed along the positive x' axis and has a magnitude of FR. Units Used: 3

kN = 10 N Given: F R = 1 kN F 2 = 450 N F 3 = 200 N

α = 45 deg β = 30 deg Solution: + →

F Rx = SF x;

F R cos ( β ) = F 3 + F 2 cos ( α ) + F 1 cos ( θ + β )

+

F Ry = SF y;

−F R sin ( β ) = F2 sin ( α ) − F 1 sin ( θ + β )

↑

41



Chapter 2

F 1 cos ( θ + β ) = FR cos ( β ) − F3 − F2 cos ( α ) F 1 sin ( θ + β ) = F 2 sin ( α ) + FR sin ( β )

⎛

F2 sin ( α ) + F R sin ( β )

⎞ ⎟−β ⎝ FR cos ( β ) − F3 − F2 cos ( α ) ⎠

θ = atan ⎜

θ = 37 deg

(FR cos (β ) − F3 − F2 cos (α ))2 + (F2 sin (α ) + FR sin (β ))2

F1 =

F 1 = 889 N

Problem 2-36 Determine the magnitude and direction, measured counterclockwise from the x' axis, of the resultant force of the three forces acting on the bracket.

Given: F 1 = 300 N F 2 = 450 N F 3 = 200 N

α = 45 deg β = 30 deg θ = 20 deg Solution: F Rx = F 1 cos ( θ + β ) + F3 + F2 cos ( α )

F Rx = 711.03 N

F Ry = −F 1 sin ( θ + β ) + F 2 sin ( α )

F Ry = 88.38 N

42



FR =

2

FRx + FRy

Chapter 2

2

F R = 717 N

φ (angle from x axis)

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

φ = atan ⎜

φ = 7.1 deg

φ' (angle from x' axis) φ' = β + φ

φ' = 37.1 deg

Problem 2-37 Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

Given: F 1 = 800 N F 2 = 600 N

θ = 40 deg c = 12 d = 5 Solution: + F Rx = ΣF x; →

F Rx = F 1 cos ( θ ) − F 2 ⎜

2

+

F Ry = F 1 sin ( θ ) + F2 ⎜

⎛

d

↑

F Ry = ΣF y;

⎛

c

⎞

2⎟

⎝ c +d ⎠ 2

⎞

2⎟

⎝ c +d ⎠

FR =

2

FRx + FRy

2

F Rx = 59 N

F Ry = 745 N

F R = 747 N

43



Chapter 2

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ = atan ⎜

θ = 85.5 deg

Problem 2-38 Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis. Units Used: 3

kN = 10 N Given: F 1 = 30 kN F 2 = 26 kN

θ = 30 deg c = 5 d = 12

Solution: + F Rx = ΣF x; →

F Rx = −F 1 sin ( θ ) −

c ⎛ ⎞ ⎜ 2 2 ⎟ F2 ⎝ c +d ⎠

F Rx = −25 kN

+

F Ry = −F 1 cos ( θ ) +

⎛ d ⎞F ⎜ 2 2⎟ 2 ⎝ c +d ⎠

F Ry = −2 kN

↑ FRy = ΣFy;

FR =

2

FRx + FRy

2

F R = 25.1 kN

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

φ = atan ⎜

φ = 4.5 deg

β = 180 deg + φ

β = 184.5 deg

44 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


Chapter 2

Problem 2-39 Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis. Given: F 1 = 60 lb F 2 = 70 lb F 3 = 50 lb

θ 1 = 60 deg θ 2 = 45 deg c = 1 d = 1 Solution: d θ 3 = atan ⎛⎜ ⎟⎞

⎝c⎠

F Rx = −F 1 cos ( θ 3 ) − F2 sin ( θ 1 )

F Rx = −103 lb

F Ry = F 1 sin ( θ 3 ) − F 2 cos ( θ 1 ) − F3

F Ry = −42.6 lb

FR =

2

FRx + FRy

2

F R = 111.5 lb

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ = 180 deg + atan ⎜

θ = 202 deg

Problem 2-40 Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured counterclockwise from the positive x axis by summing the rectangular or x, y components of the forces to obtain the resultant force. 45



Chapter 2

Given: F 1 = 600 N F 2 = 800 N F 3 = 450 N

θ 1 = 60 deg θ 2 = 45 deg θ 3 = 75 deg Solution: + F Rx = ΣF x; → +

↑ FRy = ΣFy;

F Rx = F 1 cos ( θ 2 ) − F2 sin ( θ 1 )

F Rx = −268.556 N

F Ry = F 1 sin ( θ 2 ) + F 2 cos ( θ 1 )

F Ry = 824.264 N

FR =

2

FRx + FRy

2

F R = 867 N

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ = 180 deg − atan ⎜

θ = 108 deg

Problem 2-41 Determine the magnitude and direction of the resultant FR = F 1 + F2 + F 3 of the three forces by summing the rectangular or x, y components of the forces to obtain the resultant force. Given: F 1 = 30 N F 2 = 20 N

46



Chapter 2

F 3 = 50 N

θ = 20 deg c = 3 d = 4

Solution: F Rx = −F 1

F Ry = F 1

FR =

⎛ d ⎞ − F ( sin ( θ ) ) + F 2 3 ⎜ 2 2⎟ ⎝ c +d ⎠

F Rx = 19.2 N

c ⎛ ⎞ ⎜ 2 2 ⎟ − F2 cos ( θ ) ⎝ c +d ⎠ 2

FRx + FRy

F Ry = −0.8 N

2

F R = 19.2 N

⎛ −FRy ⎞ ⎟ ⎝ FRx ⎠

θ = atan ⎜

θ = 2.4 deg

Problem 2-42 Determine the magnitude and orientation, measured counterclockwise from the positive y axis, of the resultant force acting on the bracket. Given: F A = 700 N

47



Chapter 2

F B = 600 N

θ = 20 deg φ = 30 deg

Solution: Scalar Notation: Suming the force components algebraically, we have F Rx = ΣF x;

F Rx = F A sin ( φ ) − F B cos ( θ ) F Rx = −213.8 N

F Ry = ΣF y;

F Ry = F A cos ( φ ) + FB sin ( θ ) F Ry = 811.4 N

The magnitude of the resultant force F R is 2

FR =

FRx + FRy

2

F R = 839 N The directional angle θ measured counterclockwise from the positive x axis is

⎛ FRx ⎞ ⎟ ⎝ FRy ⎠

θ = atan ⎜

θ = 14.8 deg

Problem 2-43 Determine the magnitude and direction, measured counterclockwise from the positive x' axis, of the resultant force of the three forces acting on the bracket. Given: F 1 = 300 N

48



Chapter 2

F 2 = 200 N F 3 = 180 N

θ = 10 deg θ 1 = 60 deg c = 5 d = 12

Solution: + →

+

↑

F Rx = ΣF x;

F Ry = ΣF y;

F Rx = F 1 sin ( θ 1 + θ ) −

⎛ d ⎞F ⎜ 2 2⎟ 3 ⎝ c +d ⎠

F Ry = F 1 cos ( θ 1 + θ ) + F2 +

FR =

2

FRx + FRy

2

c ⎛ ⎞ ⎜ 2 2 ⎟ F3 ⎝ c +d ⎠

F Rx = 115.8 N

F Ry = 371.8 N

F R = 389 N

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

φ = atan ⎜

φ = 72.7 deg

ψ = φ − ( 90 deg − θ 1 )

ψ = 42.7 deg

Problem 2-44 Determine the x and y components of F 1 and F 2.

49



Chapter 2

Given: F 1 = 200 N F 2 = 150 N

θ = 45 deg φ = 30 deg Solution: F 1x = F 1 sin ( θ ) F 1x = 141.4 N F 1y = F 1 cos ( θ ) F 1y = 141.4 N F 2x = −F 2 cos ( φ ) F 2x = −129.9 N F 2y = F 2 sin ( φ ) F 2y = 75 N

Problem 2-45 Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis. Given: F 1 = 200 N F 2 = 150 N

θ = 45 deg φ = 30 deg

50



Chapter 2

Solution: + F Rx = ΣF x; → +

↑ FRy = ΣFy;

F Rx = F 1 sin ( θ ) − F2 cos ( φ )

F Rx = 11.5 N

F Ry = F 1 cos ( θ ) + F 2 sin ( φ )

F Ry = 216.4 N

F =

2

FRx + FRy

2

F = 217 N

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

β = atan ⎜

β = 87 deg

Problem 2-46 Determine the x and y components of each force acting on the gusset plate of the bridge truss. Given: F 1 = 200 lb

c = 3

F 2 = 400 lb

d = 4

F 3 = 300 lb

e = 3

F 4 = 300 lb

f = 4

Solution: F 1x = −F 1 F 1x = −200 lb F 1y = 0 lb F 2x = F 2 ⎛ ⎜

d 2

⎞

2⎟

⎝ c +d ⎠

F 2x = 320 lb F 2y = −F 2 ⎛

c ⎞ ⎜ 2 2⎟ ⎝ c +d ⎠

F 2y = −240 lb F 3x = F 3 ⎛ ⎜

e 2

⎞

2⎟

⎝ e +f ⎠ 51



Chapter 2

F 3x = 180 lb F 3y = F 3 ⎛ ⎜

f 2

⎞

2⎟

⎝ e +f ⎠

F 3y = 240 lb F 4x = −F 4 F 4x = −300 lb F 4y = 0 lb

Problem 2-47 Determine the magnitude of the resultant force and its direction measured clockwise from the positive x axis. Units Used: 3

kN = 10 N

Given: F 1 = 20 kN F 2 = 40 kN F 3 = 50 kN

θ = 60 deg c = 1 d = 1 e = 3 f = 4 Solution: + F Rx = ΣF x; →

F Rx = F 3 ⎛ ⎜

⎞ + F ⎛ d ⎞ − F cos ( θ ) 2⎜ 1 2 2⎟ 2 2⎟ ⎝ e +f ⎠ ⎝ c +d ⎠ f

F Rx = 58.28 kN

52



+

Chapter 2

⎛

c ⎞−F ⎛ ⎞ 2 ⎜ 2 2 ⎟ − F1 sin ( θ ) 2 2⎟ ⎝ e +f ⎠ ⎝ c +d ⎠

↑ FRy = ΣFy;

F Ry = F 3 ⎜

e

F Ry = −15.6 kN F =

2

FRx + FRy

2

F = 60.3 kN

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ = atan ⎜

θ = 15 deg

Problem 2-48 Three forces act on the bracket. Determine the magnitude and direction θ of F 1 so that the resultant force is directed along the positive x' axis and has magnitude FR. Given: F 2 = 200 N F 3 = 180 N

θ 1 = 60 deg F R = 800 N c = 5 d = 12

Solution: Initial Guesses:

F 1 = 100 N

θ = 10 deg

Given + →

F Rx = ΣF x;

F R sin ( θ 1 ) = F1 sin ( θ 1 + θ ) −

⎛ d ⎞F ⎜ 2 2⎟ 3 ⎝ c +d ⎠

53



+

↑

F Ry = ΣF y;

Chapter 2

F R( cos ( θ 1 ) ) = F 1 cos ( θ 1 + θ ) + F2 +

c ⎛ ⎞ ⎜ 2 2 ⎟ F3 ⎝ c +d ⎠

⎛ F1 ⎞ ⎜ ⎟ = Find ( F1 , θ ) ⎝θ ⎠ F 1 = 869 N

θ = 21.3 deg

Problem 2-49 Determine the magnitude and direction, measured counterclockwise from the positive x' axis, of the resultant force acting on the bracket. Given: F 1 = 300 N F 2 = 200 N F 3 = 180 N

θ 1 = 60 deg θ = 10 deg c = 5 d = 12

Solution: Guesses

F Rx = 100 N

F Ry = 100 N

Given + →

F Rx = ΣF x;

F Rx = F1 sin ( θ 1 + θ ) −

+

F Ry = ΣF y;

F Ry = F1 cos ( θ 1 + θ ) + F 2 +

↑

⎛ d ⎞F ⎜ 2 2⎟ 3 ⎝ c +d ⎠ c ⎛ ⎞ ⎜ 2 2 ⎟ ( F3 ) ⎝ c +d ⎠

54



Chapter 2

⎛ FRx ⎞ ⎜ ⎟ = Find ( FRx , FRy) ⎝ FRy ⎠ FR =

2

FRx + FRy

2

⎛ FRx ⎞ ⎛ 115.8 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ FRy ⎠ ⎝ 371.8 ⎠ F R = 389 N

φ = atan ⎜

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

φ = 72.7 deg

φ' = ⎡⎣φ − ( 90 deg − θ 1 )⎤⎦

φ' = 42.7 deg

Problem 2-50 Express each of the three forces acting on the column in Cartesian vector form and compute the magnitude of the resultant force. Given: F 1 = 150 lb

θ = 60 deg

F 2 = 275 lb

c = 4

F 3 = 75 lb

d = 3

Solution: Find the components of each force.

⎛

⎞ ⎟ 2 2 ⎝ c +d ⎠

F 1x = F 1 ⎜

F 1v =

d

⎛ F1x ⎞ ⎜ ⎟ ⎝ F1y ⎠

F 2x = 0 lb

F 2v =

⎛ F2x ⎞ ⎜ ⎟ ⎝ F2y ⎠

⎛

⎞ ⎟ 2 2 ⎝ c +d ⎠

F 1y = F 1 ⎜

F 1v =

−c

⎛ 90 ⎞ ⎜ ⎟ lb ⎝ −120 ⎠

F 2y = −F 2

F 2v =

⎛ 0 ⎞ ⎜ ⎟ lb ⎝ −275 ⎠ 55



F 3x = −F 3 cos ( θ )

F 3v =

Chapter 2

F 3y = −F 3 sin ( θ )

⎛ F3x ⎞ ⎜ ⎟ ⎝ F3y ⎠

F 3v =

⎛ −37.5 ⎞ ⎜ ⎟ lb ⎝ −65 ⎠

Now find the magnitude of the resultant force. F R = F 1v + F 2v + F 3v

F R = 462.9 lb

Problem 2-51 Determine the magnitude of force F so that the resultant F R of the three forces is as small as possible. What is the minimum magnitude of FR? Units Used: kN = 1000 N Given: F 1 = 5 kN F 2 = 4 kN

θ = 30 deg Solution: Scalar Notation: Suming the force components algebrically, we have + → +

↑

F Rx = ΣF x;

F Rx = F1 − F sin ( θ )

F Ry = ΣF y;

F Ry = F cos ( θ ) − F 2

The magnitude of the resultant force F R is FR =

2

2

F Rx + F Ry =

(F1 − F sin(θ ))2 + (F cos (θ ) − F2)2 56



Chapter 2

F R = F 1 + F2 + F − 2FF1 sin ( θ ) − 2F2 Fcos( θ ) 2

2FR

2

dF R dF

2

= 2F − 2F1 sin ( θ ) − 2F2 cos ( θ )

If F is a minimum, then FR =

2

⎞ ⎛ dFR = 0⎟ ⎜ ⎝ dF ⎠

F = F 1 sin ( θ ) + F2 cos ( θ )

(F1 − F sin(θ ))2 + (F cos (θ ) − F2)2

F = 5.96 kN F R = 2.3 kN

Problem 2-52 Express each of the three forces acting on the bracket in Cartesian vector form with respect to the x and y axes. Determine the magnitude and direction θ of F1 so that the resultant force is directed along the positive x' axis and has magnitude F R. Units Used: kN = 1000 N Given: F R = 600 N F 2 = 350 N F 3 = 100 N

φ = 30 deg Solution: F 2v =

⎛ F2 ⎞ ⎜ ⎟ ⎝0 ⎠

F 3v =

⎛ 0 ⎞ ⎜ ⎟ ⎝ −F 3 ⎠

F 1v =

⎛ F1 cos ( θ ) ⎞ ⎜ ⎟ ⎝ F1 sin ( θ ) ⎠

F 2v =

⎛ 350 ⎞ ⎜ ⎟N ⎝ 0 ⎠

F 3v =

⎛ 0 ⎞ ⎜ ⎟N ⎝ −100 ⎠

57



Chapter 2

F 1 = 20 N θ = 10 deg

The initial guesses: Given

⎛ F1 cos ( θ ) ⎞ ⎜ ⎟ + F2v + F3v = ( ) F sin θ 1 ⎝ ⎠ ⎛ F1 ⎞ ⎜ ⎟ = Find ( F1 , θ ) ⎝θ ⎠

⎛ FR cos ( φ ) ⎞ ⎜ ⎟ ⎝ FR sin ( φ ) ⎠

F 1 = 434.5 N

θ = 67 deg

Problem 2-53 The three concurrent forces acting on the post produce a resultant force F R = 0. If F 2 = (1/2)F 1, and F 1 is to be 90° from F 2 as shown, determine the required magnitude F 3 expressed in terms of F1 and the angle θ. Solution:

Use the primed coordiates.

ΣFRx = 0

F 3 cos ( θ − 90 deg) − F1 = 0

ΣFRy = 0

−F 3 sin ( θ − 90 deg) + F 2 = 0 tan ( θ − 90 deg) =

F2 F1

=

1 2

1 θ = 90 deg + atan ⎛⎜ ⎟⎞

⎝ 2⎠

θ = 117 deg k =

1

cos ( θ − 90 deg)

k = 1.1

F3 = k F1



Chapter 2

Problem 2-54 Three forces act on the bracket. Determine the magnitude and orientation θ of F2 so that the resultant force is directed along the positive u axis and has magnitude F R. Given: F R = 50 lb F 1 = 80 lb F 3 = 52 lb

φ = 25 deg c = 12 d = 5

Solution: Guesses F 2 = 1 lb

θ = 120 deg

Given F R cos ( φ ) = F1 + F2 cos ( φ + θ ) +

−F R sin ( φ ) = −F 2 sin ( φ + θ ) +

⎛ F2 ⎞ ⎜ ⎟ = Find ( F2 , θ ) ⎝θ ⎠

⎛ d ⎞F ⎜ 2 2⎟ 3 ⎝ c +d ⎠

c ⎛ ⎞ ⎜ 2 2 ⎟ F3 ⎝ c +d ⎠

θ = 103.3 deg

F 2 = 88.1 lb

59



Chapter 2

Problem 2-55 Determine the magnitude and orientation, measured clockwise from the positive x axis, of the resultant force of the three forces acting on the bracket. Given: F 1 = 80 lb F 2 = 150 lb F 3 = 52lb

θ = 55 deg φ = 25 deg c = 12 m d = 5m

Solution:

⎛

⎞ + F cos ( θ + φ ) 2 ⎝ c +d ⎠

F Rx = F 1 + F 3 ⎜

⎛

d

2

F Rx = 126.05 lb

2⎟

⎞ − F sin ( θ + φ ) 2 ⎟ 2 2 ⎝ c +d ⎠ c

F Ry = F 3 ⎜

FR =

2

FRx + FRy

F Ry = −99.7 lb

2

F R = 161 lb

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

β = atan ⎜

β = 38.3 deg

Problem 2-56 Three forces act on the ring. Determine the range of values for the magnitude of P so that the magnitude of the resultant force does not exceed F. Force P is always directed to the right.

60



Chapter 2

Units Used: 3

kN = 10 N Given: F = 2500 N F 1 = 1500 N F 2 = 600 N

θ 1 = 60 deg θ 2 = 45 deg Solution: Initial Guesses:

F Rx = 100 N

F Ry = 100 N

P = 100 N

Given + →

F Rx = ΣF x;

F Rx = P + F 2 cos ( θ 2 ) + F1 cos ( θ 1 + θ 2 )

+

F Ry = ΣF y;

F Ry = F2 sin ( θ 2 ) + F1 sin ( θ 1 + θ 2 )

↑

F=

2

F Rx + F Ry

⎛ FRx ⎞ ⎜ ⎟ ⎜ FRy ⎟ = Find ( FRx , FRy , P) ⎜P ⎟ ⎝ max ⎠ Initial Guesses:

2

P max = 1.6 kN

F Rx = −100 N

F Ry = 100 N

P = −2000 N

Given + →

F Rx = ΣF x;

F Rx = P + F 2 cos ( θ 2 ) + F1 cos ( θ 1 + θ 2 )

+

F Ry = ΣF y;

F Ry = F2 sin ( θ 2 ) + F1 sin ( θ 1 + θ 2 )

↑

F=

⎛ FRx ⎞ ⎜ ⎟ ⎜ FRy ⎟ = Find ( FRx , FRy , P) ⎜P ⎟ ⎝ min ⎠

2

F Rx + F Ry

2

P min = −1.7 kN

61



Since P > 0

we conclude that

Chapter 2

0 M

Yes

Problem 4-60 The lug nut on the wheel of the automobile is to be removed using the wrench and applying the vertical force F . Assume that the cheater pipe AB is slipped over the handle of the wrench and the F force can be applied at any point and in any direction on the assembly. Determine if this force is adequate, provided a torque M about the x axis is initially required to turn the nut. Given: F 1 = 30 N

M = 14 N⋅ m

a = 0.25 m

b = 0.3 m

c = 0.5 m

d = 0.1 m

255



Chapter 4

Solution: Mx = F1

a+c c

2

2

c −b

Mx = 18 N⋅ m Mx > M

Yes

Mxmax occurs when force is applied perpendicular to both the handle and the x-axis. Mxmax = F1 ( a + c) Mxmax = 22.5 N⋅ m Mxmax > M

Yes

Problem 4-61 The bevel gear is subjected to the force F which is caused from contact with another gear. Determine the moment of this force about the y axis of the gear shaft. Given: a = 30 mm b = 40 mm

⎛ 20 ⎞ ⎜ 8 ⎟N F = ⎜ ⎟ ⎝ −15 ⎠

256



Chapter 4

Solution:

⎛ −b ⎞ ⎜ ⎟ r = 0 ⎜ ⎟ ⎝a⎠

⎛0⎞ ⎜ ⎟ j = 1 ⎜ ⎟ ⎝0⎠

My = ( r × F) ⋅ j

My = 0 N⋅ m

Problem 4-62 The wooden shaft is held in a lathe. The cutting tool exerts force F on the shaft in the direction shown. Determine the moment of this force about the x axis of the shaft. Express the result as a Cartesian vector. The distance OA is a. Given: a = 25 mm

θ = 30 deg

⎛ −5 ⎞ ⎜ ⎟ F = −3 N ⎜ ⎟ ⎝8⎠ Solution:

⎡ 0 ⎤ ⎢ ⎥ r = ( a)cos ( θ ) ⎢ ⎥ ⎣ ( a)sin ( θ ) ⎦

⎛1⎞ ⎜ ⎟ i = 0 ⎜ ⎟ ⎝0⎠

Mx =

⎡⎣( r × F) ⋅ i⎤⎦ i

⎛ 0.211 ⎞ ⎜ 0 ⎟ N⋅ m Mx = ⎜ ⎟ ⎝ 0 ⎠

Problem 4-63 Determine the magnitude of the moment of the force F about the base line CA of the tripod. Given:

⎛ 50 ⎞ ⎜ ⎟ F = −20 N ⎜ ⎟ ⎝ −80 ⎠

257



Chapter 4

a = 4m b = 2.5 m c = 1m d = 0.5 m e = 2m f = 1.5 m g = 2m

Solution:

rCA

⎛ −g ⎞ = ⎜ e ⎟ ⎜ ⎟ ⎝0⎠

uCA =

rCA rCA

rCD

⎛b − g⎞ = ⎜ e ⎟ ⎜ ⎟ ⎝ a ⎠

MCA = ( rCD × F) ⋅ uCA

MCA = 226 N⋅ m

Problem 4-64 The flex-headed ratchet wrench is subjected to force P, applied perpendicular to the handle as shown. Determine the moment or torque this imparts along the vertical axis of the bolt at A. Given: P = 16 lb

a = 10 in

θ = 60 deg b = 0.75 in Solution: M = P ⎡⎣b + ( a)sin ( θ )⎤⎦

M = 150.564 lb⋅ in

258



Chapter 4

Problem 4-65 If a torque or moment M is required to loosen the bolt at A, determine the force P that must be applied perpendicular to the handle of the flex-headed ratchet wrench. Given: M = 80 lb⋅ in

θ = 60 deg a = 10 in b = 0.75 in Solution: M = P ⎡⎣b + ( a)sin ( θ )⎤⎦

P =

M

b + ( a)sin ( θ )

P = 8.50 lb

Problem 4-66 The A-frame is being hoisted into an upright position by the vertical force F . Determine the moment of this force about the y axis when the frame is in the position shown. Given: F = 80 lb a = 6 ft b = 6 ft

θ = 30 deg φ = 15 deg Solution: Using the primed coordinates we have

259



⎛ −sin ( θ ) ⎞ ⎜ ⎟ j = cos ( θ ) ⎜ ⎟ ⎝ 0 ⎠

Chapter 4

⎛0⎞ ⎜ ⎟ Fv = F 0 ⎜ ⎟ ⎝1⎠

My = ( rAC × F v ) ⋅ j

rAC

⎛ −b cos ( φ ) ⎞ ⎜ ⎟ a ⎟ = ⎜ 2 ⎜ ⎟ ⎜ b sin ( φ ) ⎟ ⎝ ⎠

My = 281.528 lb⋅ ft

Problem 4-67 Determine the moment of each force acting on the handle of the wrench about the a axis. Given:

⎛ −2 ⎞ ⎜ ⎟ F 1 = 4 lb ⎜ ⎟ ⎝ −8 ⎠

⎛3⎞ ⎜ ⎟ F 2 = 2 lb ⎜ ⎟ ⎝ −6 ⎠

b = 6 in c = 4 in d = 3.5 in

θ = 45 deg

Solution:

⎛ cos ( θ ) ⎞ ⎜ 0 ⎟ ua = ⎜ ⎟ ⎝ sin ( θ ) ⎠ ⎛ cos ( θ ) ⎞ ⎛ sin ( θ ) ⎞ ⎜ ⎟ ⎜ 0 ⎟ r1 = b 0 + ( c + d) ⎜ ⎟ ⎜ ⎟ ⎝ sin ( θ ) ⎠ ⎝ −cos ( θ ) ⎠ M1a = ( r1 × F1 ) ⋅ ua

⎛ cos ( θ ) ⎞ ⎛ sin ( θ ) ⎞ ⎜ 0 ⎟ + c⎜ 0 ⎟ r2 = b ⎜ ⎟ ⎜ ⎟ ⎝ sin ( θ ) ⎠ ⎝ −cos ( θ ) ⎠

M1a = 30 lb⋅ in

260



M2a = ( r2 × F2 ) ⋅ ua

Chapter 4

M2a = 8 lb⋅ in

Problem 4-68 Determine the moment of each force acting on the handle of the wrench about the z axis. Given:

⎛ −2 ⎞ ⎜ ⎟ F 1 = 4 lb ⎜ ⎟ ⎝ −8 ⎠

⎛3⎞ ⎜ ⎟ F 2 = 2 lb ⎜ ⎟ ⎝ −6 ⎠

b = 6 in c = 4 in d = 3.5 in

θ = 45 deg

Solution:

⎛ cos ( θ ) ⎞ ⎛ sin ( θ ) ⎞ ⎜ ⎟ ⎜ 0 ⎟ r1 = b 0 + ( c + d) ⎜ ⎟ ⎜ ⎟ ⎝ sin ( θ ) ⎠ ⎝ −cos ( θ ) ⎠

⎛ cos ( θ ) ⎞ ⎛ sin ( θ ) ⎞ ⎜ 0 ⎟ + c⎜ 0 ⎟ r2 = b ⎜ ⎟ ⎜ ⎟ ⎝ sin ( θ ) ⎠ ⎝ −cos ( θ ) ⎠

M1z = ( r1 × F 1 ) ⋅ k

M1z = 38.2 lb⋅ in

M2z = ( r2 × F 2 ) ⋅ k

M2z = 14.1 lb⋅ in

⎛0⎞ ⎜ ⎟ k = 0 ⎜ ⎟ ⎝1⎠

Problem 4-69 Determine the magnitude and sense of the couple moment. Units Used: 3

kN = 10 N 261



Chapter 4

Given: F = 5 kN

θ = 30 deg a = 0.5 m b = 4m c = 2m d = 1m Solution: MC = F cos ( θ ) ( a + c) + F sin ( θ ) ( b − d) MC = 18.325 kN⋅ m

Problem 4-70 Determine the magnitude and sense of the couple moment. Each force has a magnitude F. Given: F = 65 lb a = 2 ft b = 1.5 ft c = 4 ft d = 6 ft e = 3 ft

Solution: Mc = ΣMB;

⎡ ⎛

⎞ ( d + a)⎤ + ⎡F⎛ e ⎞ ( c + a)⎤ ⎥ ⎢ ⎜ 2 2⎟ ⎥ 2 2⎟ ⎣ ⎝ c +e ⎠ ⎦ ⎣ ⎝ c +e ⎠ ⎦

MC = ⎢F⎜

c

MC = 650 lb⋅ ft

(Counterclockwise)

Problem 4-71 Determine the magnitude and sense of the couple moment. 262



Chapter 4

Units Used: 3

kip = 10 lb Given: F = 150 lb a = 8 ft b = 6 ft c = 8 ft d = 6 ft e = 6 ft f = 8 ft Solution: MC = ΣMA;

MC = F

d 2

d + f MC = 3120 lb⋅ ft

f

( a + f) + F 2

2

d + f

( c + d) 2

MC = 3.120 kip⋅ ft

Problem 4-72 If the couple moment has magnitude M, determine the magnitude F of the couple forces. Given: M = 300 lb⋅ ft a = 6 ft b = 12 ft c = 1 ft d = 2 ft e = 12 ft f = 7 ft

263



Chapter 4

Solution:

M=F

( f − d) ( b + e) ⎤ ⎡ e( f + a) − ⎢ 2 2 2 2⎥ ( f − d) + e ⎦ ⎣ ( f − d) + e

M

F =

e( f + a) 2

−

F = 108 lb

( f − d) ( b + e)

2

2

( f − d) + e

2

( f − d) + e

Problem 4-73 A clockwise couple M is resisted by the shaft of the electric motor. Determine the magnitude of the reactive forces −R and R which act at supports A and B so that the resultant of the two couples is zero. Given: a = 150 mm

θ = 60 deg M = 5 N⋅ m

Solution: MC = −M +

2R a

tan ( θ )

=0

R =

M tan ( θ ) 2

a

R = 28.9 N

Problem 4-74 The resultant couple moment created by the two couples acting on the disk is MR. Determine the magnitude of force T.

264



Chapter 4

Units Used: 3

kip = 10 lb Given:

⎛0⎞ ⎜ ⎟ MR = 0 kip⋅ in ⎜ ⎟ ⎝ 10 ⎠ a = 4 in b = 2 in c = 3 in Solution: Initial Guess

Given

T = 1 kip

⎛ a ⎞ ⎛ 0 ⎞ ⎛ −b ⎞ ⎛ 0 ⎞ ⎛ −b − c ⎞ ⎛ 0 ⎞ ⎜ 0 ⎟ × ⎜ T ⎟ + ⎜ 0 ⎟ × ⎜ −T ⎟ + ⎜ 0 ⎟ × ⎜ −T ⎟ = M R ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠

T = Find ( T)

T = 0.909 kip

Problem 4-75 Three couple moments act on the pipe assembly. Determine the magnitude of M3 and the bend angle θ so that the resultant couple moment is zero.

Given:

θ 1 = 45 deg M1 = 900 N⋅ m M2 = 500 N⋅ m

265



Chapter 4

Solution: Initial guesses:

θ = 10 deg

M3 = 10 N⋅ m

Given + ΣMx = 0; →

M1 − M3 cos ( θ ) − M2 cos ( θ 1 ) = 0

+

M3 sin ( θ ) − M2 sin ( θ 1 ) = 0

↑ ΣMy = 0;

⎛ θ ⎞ ⎜ ⎟ = Find ( θ , M3) ⎝ M3 ⎠

θ = 32.9 deg

M3 = 651 N⋅ m

Problem 4-76 The floor causes couple moments MA and MB on the brushes of the polishing machine. Determine the magnitude of the couple forces that must be developed by the operator on the handles so that the resultant couple moment on the polisher is zero. What is the magnitude of these forces if the brush at B suddenly stops so that MB = 0? Given: a = 0.3 m MA = 40 N⋅ m MB = 30 N⋅ m Solution:

MA − MB − F 1 a = 0

F1 =

MA − F 2 a = 0

F2 =

MA − MB a MA

F 1 = 33.3 N

F 2 = 133 N

a

266



Chapter 4

Problem 4-77 The ends of the triangular plate are subjected to three couples. Determine the magnitude of the force F so that the resultant couple moment is M clockwise. Given: F 1 = 600 N F 2 = 250 N a = 1m

θ = 40 deg M = 400 N⋅ m Solution: Initial Guess

Given

F1

F = 1N

⎛ a ⎞ − F a − F ⎛ a ⎞ = −M ⎜ ⎟ 2 ⎜ ⎟ ⎝ 2 cos ( θ ) ⎠ ⎝ 2 cos ( θ ) ⎠

F = Find ( F)

F = 830 N

Problem 4-78 Two couples act on the beam. Determine the magnitude of F so that the resultant couple moment is M counterclockwise. Where on the beam does the resultant couple moment act? Given: M = 450 lb⋅ ft P = 200 lb a = 1.5 ft b = 1.25 ft c = 2 ft

θ = 30 deg

267



Chapter 4

Solution: MR = Σ M

M = F b cos ( θ ) + P a

F =

M − Pa

b cos ( θ )

F = 139 lb

The resultant couple moment is a free vector. It can act at any point on the beam.

Problem 4-79 Express the moment of the couple acting on the pipe assembly in Cartesian vector form. Solve the problem (a) using Eq. 4-13, and (b) summing the moment of each force about point O. Given:

⎛0⎞ F = ⎜0 ⎟ N ⎜ ⎟ ⎝ 25 ⎠ a = 300 mm b = 150 mm c = 400 mm d = 200 mm e = 200 mm

Solution:

( a)

( b)

rAB

⎛ −e − b ⎞ = ⎜ −c + d ⎟ ⎜ ⎟ ⎝ 0 ⎠

M = rAB × F

rOB

⎛a⎞ = ⎜d⎟ ⎜ ⎟ ⎝0⎠

⎛a + b + e⎞ ⎟ c = ⎜ ⎜ ⎟ ⎝ 0 ⎠

rOA

⎛ −5 ⎞ M = ⎜ 8.75 ⎟ N⋅ m ⎜ ⎟ ⎝ 0 ⎠

⎛ −5 ⎞ M = ⎜ 8.75 ⎟ N⋅ m ⎜ ⎟ ⎝ 0 ⎠

M = rOB × F + rOA × ( −F)

268



Chapter 4

Problem 4-80 If the couple moment acting on the pipe has magnitude M, determine the magnitude F of the vertical force applied to each wrench. Given: M = 400 N⋅ m a = 300 mm b = 150 mm c = 400 mm d = 200 mm e = 200 mm

Solution:

⎛0⎞ k = ⎜0⎟ ⎜ ⎟ ⎝1⎠

rAB

⎛ −e − b ⎞ = ⎜ −c + d ⎟ ⎜ ⎟ ⎝ 0 ⎠

Guesss

F = 1N rAB × ( Fk) = M

Given

F = Find ( F)

F = 992.278 N

Problem 4-81 Determine the resultant couple moment acting on the beam. Solve the problem two ways: (a) sum moments about point O; and (b) sum moments about point A. Units Used: 3

kN = 10 N

269



Chapter 4

Given: F 1 = 2 kN

θ 1 = 30 deg

F 2 = 8 kN

θ 2 = 45 deg

a = 0.3 m b = 1.5 m c = 1.8 m Solution: MR = ΣMO;

( a)

MRa = ( F1 cos ( θ 1 ) + F 2 cos ( θ 2 ) ) c + ( F2 cos ( θ 2 ) − F 1 sin ( θ 1 ) ) a ... + −( F2 cos ( θ 2 ) + F 1 cos ( θ 1 ) ) ( b + c) MRa = −9.69 kN⋅ m MR = ΣMA;

( b)

MRb = ( F2 sin ( θ 2 ) − F1 sin ( θ 1 ) ) a − ( F 2 cos ( θ 2 ) + F1 cos ( θ 1 ) ) b MRb = −9.69 kN⋅ m

Problem 4-82 Two couples act on the beam as shown. Determine the magnitude of F so that the resultant couple moment is M counterclockwise. Where on the beam does the resultant couple act? Given: M = 300 lb⋅ ft a = 4 ft b = 1.5 ft P = 200 lb c = 3 d = 4

270



Chapter 4

Solution: c

M=

Fa +

2

2

2

2⎛ M + P b ⎞

c +d F =

d

c +d

F = 167 lb

2

c +d

Fb − Pb 2

⎜ ⎟ ⎝ ca + d b⎠

Resultant couple can act anywhere.

Problem 4-83 Two couples act on the frame. If the resultant couple moment is to be zero, determine the distance d between the couple forces F 1. Given: F 1 = 80 lb F 2 = 50 lb a = 1 ft b = 3 ft c = 2 ft e = 3 ft f = 3 g = 4

θ = 30 deg Solution: g ⎡−F cos ( θ ) e + ⎛ ⎞ ⎤ ⎢ 2 ⎜ 2 2 ⎟ F1 d⎥ = 0 ⎣ ⎝ g +f ⎠ ⎦

d =

F2 F1

⎛ g2 + f 2 ⎞ ⎟ g ⎝ ⎠

cos ( θ ) e ⎜

d = 2.03 ft

271



Chapter 4

Problem 4-84 Two couples act on the frame. Determine the resultant couple moment. Compute the result by resolving each force into x and y components and (a) finding the moment of each couple (Eq. 4-13) and (b) summing the moments of all the force components about point A. Given: F 1 = 80 lb

c = 2 ft

g = 4

F 2 = 50 lb

d = 4 ft

θ = 30 deg

a = 1 ft

e = 3 ft

b = 3 ft

f = 3

Solution: ( a)

M = Σ ( r × F)

⎛ e ⎞ ⎡ ⎛ −sin ( θ ) ⎞⎤ ⎛ 0 ⎞ ⎡ F ⎛ −g ⎞⎤ 1 ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ − f ⎟⎥ M = 0 × F 2 −cos ( θ ) + d × ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ f + g ⎝ 0 ⎠⎦

⎛ 0 ⎞ ⎜ 0 ⎟ lb⋅ ft M= ⎜ ⎟ ⎝ 126.096 ⎠

( b)

Summing the moments of all force components aboout point A.

M1 =

g ⎛ −g ⎞ F b + ⎛ ⎞ 1 ⎜ 2 2⎟ ⎜ 2 2 ⎟ F 1 ( b + d) ⎝ f +g ⎠ ⎝ f +g ⎠

M2 = F 2 cos ( θ ) c − F 2 sin ( θ ) ( a + b + d) − F 2 cos ( θ ) ( c + e) + F 2 sin ( θ ) ( a + b + d) M = M1 + M2

M = 126.096 lb⋅ ft

Problem 4-85 Two couples act on the frame. Determine the resultant couple moment. Compute the result by resolving each force into x and y components and (a) finding the moment of each couple (Eq. 4 -13) and (b) summing the moments of all the force components about point B.

272



Chapter 4

Given: F 1 = 80 lb

d = 4ft

F 2 = 50 lb

e = 3 ft

a = 1 ft

f = 3 ft

b = 3 ft

g = 4 ft

c = 2 ft

θ = 30 deg

Solution: ( a)

M = Σ ( r × F)

⎛ e ⎞ ⎡ ⎛ −sin ( θ ) ⎞⎤ ⎛ 0 ⎞ ⎡ F ⎛ −g ⎞⎤ 1 ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ − f ⎟⎥ M = 0 × F 2 −cos ( θ ) + d × ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ f + g ⎝ 0 ⎠⎦ ( b)

⎛ 0 ⎞ ⎟ lb⋅ ft M=⎜ 0 ⎜ ⎟ ⎝ 126.096 ⎠

Summing the moments of all force components about point B. M1 =

g g ⎛ ⎛ ⎞ ⎞ ⎜ 2 2 ⎟ F1( a + d) − ⎜ 2 2 ⎟ F1 a ⎝ f +g ⎠ ⎝ f +g ⎠

M2 = F 2 cos ( θ ) c − F 2 cos ( θ ) ( c + e) M = M1 + M2

M = 126.096 lb⋅ ft

Problem 4-86 Determine the couple moment. Express the result as a Cartesian vector.

273



Chapter 4

Given:

⎛8⎞ F = ⎜ −4 ⎟ N ⎜ ⎟ ⎝ 10 ⎠ a = 5m b = 3m c = 4m d = 2m e = 3m Solution:

⎛ −b − e ⎞ r = ⎜ c+d ⎟ ⎜ ⎟ ⎝ −a ⎠

M = r× F

⎛ 40 ⎞ M = ⎜ 20 ⎟ N⋅ m ⎜ ⎟ ⎝ −24 ⎠

Problem 4-87 Determine the couple moment. Express the result as a Cartesian vector. Given: F = 80 N a = 6m b = 10 m c = 10 m d = 5m e = 4m f = 4m

274



Chapter 4

Solution: u =

⎛ b ⎞ ⎜c + d⎟ ⎟ 2 2 2⎜ a + b + ( c + d) ⎝ − a ⎠ 1

F v = Fu

⎛−f − b⎞ r = ⎜ −d − c ⎟ ⎜ ⎟ ⎝ e+a ⎠

⎛ −252.6 ⎞ M = ⎜ 67.4 ⎟ N⋅ m ⎜ ⎟ ⎝ −252.6 ⎠

M = r × Fv

Problem 4-88 If the resultant couple of the two couples acting on the fire hydrant is MR = { −15i + 30j} N ⋅ m, determine the force magnitude P. Given: a = 0.2 m b = 0.150 m

⎛ −15 ⎞ M = ⎜ 30 ⎟ N⋅ m ⎜ ⎟ ⎝ 0 ⎠ F = 75 N Solution: Initial guess

P = 1N

Given

⎛ −F a ⎞ M = ⎜ Pb ⎟ ⎜ ⎟ ⎝ 0 ⎠

P = Find ( P)

P = 200 N

Problem 4-89 If the resultant couple of the three couples acting on the triangular block is to be zero, determine the magnitude of forces F and P . 275



Chapter 4

Given: F 1 = 150 N a = 300 mm b = 400 mm d = 600 mm

Solution:

Initial guesses:

Given

F = 1N

P = 1N

⎛ d ⎞ ⎛ 0 ⎞ ⎛ d ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎛⎜ −F1 ⎟⎞ ⎛ 0 ⎞ ⎛⎜ F1 ⎞⎟ ⎜ 0 ⎟ × ⎜ 0 ⎟ + ⎜ b ⎟ × ⎜ −P ⎟ + ⎜ b ⎟ × ⎜ ⎟ + 0 × =0 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜0 ⎟ ⎝ a ⎠ ⎝ F ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎜⎝ 0 ⎟⎠ ⎝ a ⎠ ⎜⎝ 0 ⎟⎠

⎛F⎞ ⎜ ⎟ = Find ( F , P) ⎝P⎠

⎛ F ⎞ ⎛ 75 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ P ⎠ ⎝ 100 ⎠

Problem 4-90 Determine the couple moment that acts on the assembly. Express the result as a Cartesian vector. Member BA lies in the x-y plane.

276



Chapter 4

Given:

⎛ 0 ⎞ ⎜ 0 ⎟N F = ⎜ ⎟ ⎝ 100 ⎠ a = 300 mm b = 150 mm c = 200 mm d = 200 mm

θ = 60 deg

Solution:

⎡−( c + d) sin ( θ ) − b cos ( θ ) ⎤ ⎢ ⎥ r = −b sin ( θ ) + ( c + d)cos ( θ ) ⎢ ⎥ 0 ⎦ ⎣

M = r× F

⎛ 7.01 ⎞ ⎜ ⎟ M = 42.14 N⋅ m ⎜ ⎟ ⎝ 0.00 ⎠

Problem 4-91 If the magnitude of the resultant couple moment is M, determine the magnitude F of the forces applied to the wrenches. Given: M = 15 N⋅ m

c = 200 mm

a = 300 mm

d = 200 mm

b = 150 mm

θ = 60 deg

277



Chapter 4

Solution:

⎡−( c + d) sin ( θ ) − b cos ( θ ) ⎤ ⎢ ⎥ r = −b sin ( θ ) + ( c + d)cos ( θ ) ⎢ ⎥ 0 ⎦ ⎣ Guess Given

⎛0⎞ ⎜ ⎟ k = 0 ⎜ ⎟ ⎝1⎠

F = 1N r × ( Fk) = M

F = Find ( F)

F = 35.112 N

Problem 4-92 The gears are subjected to the couple moments shown. Determine the magnitude and coordinate direction angles of the resultant couple moment. Given: M1 = 40 lb⋅ ft M2 = 30 lb⋅ ft

θ 1 = 20 deg θ 2 = 15 deg θ 3 = 30 deg

Solution:

⎛ M1 cos ( θ 1) sin ( θ 2) ⎞ ⎜ ⎟ M1 = ⎜ M1 cos ( θ 1 ) cos ( θ 2 ) ⎟ ⎜ −M sin θ ( 1) ⎟⎠ 1 ⎝ MR = M1 + M2

⎛⎜ α ⎞⎟ ⎛ MR ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ MR ⎠ ⎜γ ⎟ ⎝ ⎠

⎛ −M2 sin ( θ 3) ⎞ ⎜ ⎟ M2 = ⎜ M2 cos ( θ 3 ) ⎟ ⎜ ⎟ 0 ⎝ ⎠ MR = 64.0 lb⋅ ft

⎛⎜ α ⎞⎟ ⎛ 94.7 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 13.2 ⎟ deg ⎜ γ ⎟ ⎝ 102.3 ⎠ ⎝ ⎠

278



Chapter 4

Problem 4-93 Express the moment of the couple acting on the rod in Cartesian vector form. What is the magnitude of the couple moment? Given:

⎛ 14 ⎞ ⎜ ⎟ F = −8 N ⎜ ⎟ ⎝ −6 ⎠ a = 1.5 m b = 0.5 m c = 0.5 m d = 0.8 m Solution:

⎛d⎞ ⎛0⎞ ⎜ ⎟ ⎜ ⎟ M = a × F + 0 × ( −F) ⎜ ⎟ ⎜ ⎟ ⎝ −c ⎠ ⎝b⎠

⎛ −17 ⎞ ⎜ ⎟ M = −9.2 N⋅ m ⎜ ⎟ ⎝ −27.4 ⎠

M = 33.532 N⋅ m

Problem 4-94 Express the moment of the couple acting on the pipe assembly in Cartesian vector form. Solve the problem (a) using Eq. 4-13, and (b) summing the moment of each force about point O.

279



Chapter 4

Given: a = 0.3 m b = 0.4 m c = 0.6 m

⎛ −6 ⎞ F = ⎜ 2 ⎟ N ⎜ ⎟ ⎝3⎠ Solution:

⎛0⎞ M = ⎜ c ⎟×F ⎜ ⎟ ⎝ −b ⎠

(a)

⎛ 2.6 ⎞ M = ⎜ 2.4 ⎟ N⋅ m ⎜ ⎟ ⎝ 3.6 ⎠

⎛0⎞ ⎛ 0 ⎞ ⎜ ⎟ M = 0 × ( −F ) + ⎜ c ⎟ × F ⎜ ⎟ ⎜ ⎟ ⎝ −a ⎠ ⎝ −a − b ⎠

(b)

⎛ 2.6 ⎞ M = ⎜ 2.4 ⎟ N⋅ m ⎜ ⎟ ⎝ 3.6 ⎠

Problem 4-95 A couple acts on each of the handles of the minidual valve. Determine the magnitude and coordinate direction angles of the resultant couple moment. Given: F 1 = 35 N θ = 60 deg F 2 = 25 N r1 = 175 mm r2 = 175 mm Solution:

⎛⎜ −F1 2 r1 ⎞⎟ ⎛⎜ −F2 2r2 cos ( θ ) ⎞⎟ M = ⎜ 0 ⎟ + ⎜ −F2 2 r2 sin ( θ ) ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ 0 ⎠

⎛ −16.63 ⎞ M = ⎜ −7.58 ⎟ N⋅ m ⎜ ⎟ ⎝ 0 ⎠

M = 18.3 N⋅ m

280



Chapter 4

⎛⎜ α ⎞⎟ ⎛ M ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ M ⎠ ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 155.496 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 114.504 ⎟ deg ⎜ γ ⎟ ⎝ 90 ⎠ ⎝ ⎠

Problem 4-96 Express the moment of the couple acting on the pipe in Cartesian vector form. What is the magnitude of the couple moment? Given: F = 125 N a = 150 mm b = 150 mm c = 200 mm d = 600 mm

Solution:

⎛ c ⎞ ⎛0⎞ M = ⎜a + b⎟ × ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎝F⎠

⎛ 37.5 ⎞ M = ⎜ −25 ⎟ N⋅ m ⎜ ⎟ ⎝ 0 ⎠

M = 45.1 N⋅ m

Problem 4-97 If the couple moment acting on the pipe has a magnitude M, determine the magnitude F of the forces applied to the wrenches.

281



Chapter 4

Given: M = 300 N⋅ m a = 150 mm b = 150 mm c = 200 mm d = 600 mm Solution: F = 1N

Initial guess:

Given

⎛ c ⎞ ⎛0⎞ ⎜a + b⎟ × ⎜ 0 ⎟ = M ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎝F⎠

F = Find ( F)

F = 832.1 N

Problem 4-98 Replace the force at A by an equivalent force and couple moment at point O. Given: F = 375 N a = 2m b = 4m c = 2m d = 1m

θ = 30 deg

282



Chapter 4

Solution:

⎛ sin ( θ ) ⎞ ⎜ ⎟ F v = F −cos ( θ ) ⎜ ⎟ ⎝ 0 ⎠

⎛ 187.5 ⎞ ⎜ ⎟ F v = −324.76 N ⎜ ⎟ ⎝ 0 ⎠

⎛ −a ⎞ ⎜ ⎟ MO = b × Fv ⎜ ⎟ ⎝0⎠

⎛ 0 ⎞ ⎜ ⎟ N⋅ m 0 MO = ⎜ ⎟ ⎝ −100.481 ⎠

Problem 4-99 Replace the force at A by an equivalent force and couple moment at point P . Given: F = 375 N a = 2m b = 4m c = 2m d = 1m

θ = 30 deg Solution:

⎛ sin ( θ ) ⎞ ⎜ ⎟ F v = F −cos ( θ ) ⎜ ⎟ ⎝ 0 ⎠

⎛ 187.5 ⎞ ⎜ ⎟ F v = −324.76 N ⎜ ⎟ ⎝ 0 ⎠

⎛ −a − c ⎞ ⎜ ⎟ MP = b − d × Fv ⎜ ⎟ ⎝ 0 ⎠

⎛ 0 ⎞ ⎜ 0 ⎟ N⋅ m MP = ⎜ ⎟ ⎝ 736.538 ⎠

Problem 4-100 Replace the force system by an equivalent resultant force and couple moment at point O.

283



Chapter 4

Given: F 1 = 60 lb

a = 2 ft

F 2 = 85 lb

b = 3 ft

F 3 = 25 lb

c = 6 ft

θ = 45 deg

d = 4 ft

e = 3 f = 4 Solution:

⎛⎜ 0 ⎟⎞ F = ⎜ −F1 ⎟ + ⎜ 0 ⎟ ⎝ ⎠

⎛ cos ( θ ) ⎞ ⎛ −e ⎞ ⎜ − f ⎟ + F ⎜ sin ( θ ) ⎟ 3 ⎜ ⎟ 2 2⎜ ⎟ e + f ⎝0 ⎠ ⎝ 0 ⎠ F2

⎛ −33.322 ⎞ F = ⎜ −110.322 ⎟ lb ⎜ ⎟ 0 ⎝ ⎠

F = 115.245 lb

⎛ −c ⎞ ⎛⎜ 0 ⎟⎞ ⎛ 0 ⎞ ⎡ F ⎛ −e ⎞⎤ ⎛ d ⎞ ⎡ ⎛ cos ( θ ) ⎞⎤ 2 ⎜ ⎟ ⎜ ⎟ ⎢ ⎜ − f ⎟⎥ + ⎜ b ⎟ × ⎢F ⎜ sin ( θ ) ⎟⎥ MO = 0 × ⎜ −F1 ⎟ + a × ⎜ ⎟ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ ⎢ 3⎜ ⎟⎥ ⎜ ⎟ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎣ e + f ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎛ 0 ⎞ MO = ⎜ 0 ⎟ lb⋅ ft ⎜ ⎟ ⎝ 480 ⎠

MO = 480 lb⋅ ft

Problem 4-101 Replace the force system by an equivalent resultant force and couple moment at point P.

284



Chapter 4

Given: F 1 = 60 lb

a = 2 ft

F 2 = 85 lb

b = 3 ft

F 3 = 25 lb

c = 6 ft

θ = 45 deg

d = 4 ft

e = 3

f = 4

Solution:

⎛⎜ 0 ⎟⎞ F = ⎜ −F1 ⎟ + ⎜ 0 ⎟ ⎝ ⎠

⎛ cos ( θ ) ⎞ ⎛ −e ⎞ ⎜ − f ⎟ + F ⎜ sin ( θ ) ⎟ 3 ⎜ ⎟ 2 2⎜ ⎟ e + f ⎝0 ⎠ ⎝ 0 ⎠ F2

⎛ −33.322 ⎞ ⎜ ⎟ F = −110.322 lb ⎜ ⎟ 0 ⎝ ⎠

F = 115.245 lb

⎛ −c − d ⎞ ⎛⎜ 0 ⎟⎞ ⎛ −d ⎞ ⎡ F ⎛ −e ⎞⎤ ⎛ 0 ⎞ ⎡ ⎛ cos ( θ ) ⎞⎤ 2 ⎜ ⎟ ⎜ ⎟ ⎢ ⎜ − f ⎟⎥ + ⎜ b ⎟ × ⎢F ⎜ sin ( θ ) ⎟⎥ 0 MP = × ⎜ −F 1 ⎟ + a × ⎜ ⎟ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ ⎢ 3 ⎜ ⎟⎥ ⎜ ⎟ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎣ e + f ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎛ 0 ⎞ ⎜ 0 ⎟ lb⋅ ft MP = ⎜ ⎟ ⎝ 921 ⎠

MP = 921 lb⋅ ft

Problem 4-102 Replace the force system by an equivalent force and couple moment at point O. Units Used: 3

kip = 10 lb Given: F 1 = 430 lb a = 2 ft

F 2 = 260 lb e = 5 ft 285



Chapter 4

b = 8 ft

f = 12

c = 3 ft

g = 5

d = a

θ = 60 deg

Solution:

⎛ −sin ( θ ) ⎞ ⎜ ⎟ F R = F1 −cos ( θ ) + ⎜ ⎟ ⎝ 0 ⎠

⎛g⎞ ⎜f⎟ 2 2⎜ ⎟ g + f ⎝0⎠

⎛ −272 ⎞ ⎜ 25 ⎟ lb FR = ⎜ ⎟ ⎝ 0 ⎠

F2

⎛ −d ⎞ ⎡ ⎛ −sin ( θ ) ⎞⎤ ⎛ e ⎞ ⎡ F ⎛ g ⎞⎤ 2 ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ f ⎟⎥ MO = b × F 1 −cos ( θ ) + 0 × ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ g + f ⎝ 0 ⎠⎦

F R = 274 lb

⎛ 0 ⎞ ⎜ 0 ⎟ kip⋅ ft MO = ⎜ ⎟ ⎝ 4.609 ⎠

Problem 4-103 Replace the force system by an equivalent force and couple moment at point P. Units Used: 3

kip = 10 lb Given: F 1 = 430 lb

F 2 = 260 lb

a = 2 ft

e = 5 ft

b = 8 ft

f = 12

c = 3 ft

g = 5

d = a

θ = 60 deg

Solution:

⎛ −sin ( θ ) ⎞ ⎜ ⎟ F R = F1 −cos ( θ ) + ⎜ ⎟ ⎝ 0 ⎠

⎛g⎞ ⎜f⎟ 2 2⎜ ⎟ g + f ⎝0⎠

⎛ −272 ⎞ ⎜ 25 ⎟ lb FR = ⎜ ⎟ ⎝ 0 ⎠

F2

F R = 274 lb

286



Chapter 4

⎛ 0 ⎞ ⎡ ⎛ −sin ( θ ) ⎞⎤ ⎛ d + e ⎞ ⎡ F ⎛ g ⎞⎤ 2 ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ c ⎟ × ⎢ ⎜ f ⎟⎥ MP = b + c × F 1 −cos ( θ ) + ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ g + f ⎝ 0 ⎠⎦

⎛ 0 ⎞ ⎜ 0 ⎟ kip⋅ ft MP = ⎜ ⎟ ⎝ 5.476 ⎠

Problem 4-104 Replace the loading system acting on the post by an equivalent resultant force and couple moment at point O. Given: F 1 = 30 lb

a = 1 ft

d = 3

F 2 = 40 lb

b = 3 ft

e = 4

F 3 = 60 lb

c = 2 ft

Solution:

⎛0⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ F R = F1 −1 + F2 0 + ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝0⎠ ⎛ 4 ⎞ ⎜ ⎟ F R = −78 lb ⎜ ⎟ ⎝ 0 ⎠

⎛ −d ⎞ ⎜ −e ⎟ ⎟ 2 2⎜ d +e ⎝ 0 ⎠ F3

F R = 78.1 lb

⎛ 0 ⎞ ⎡ ⎛ 0 ⎞⎤ ⎛ 0 ⎞ ⎡ ⎛ 1 ⎞⎤ ⎛ 0 ⎞ ⎡ F ⎛ −d ⎞⎤ 3 ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ −e ⎟⎥ MO = a + b + c × F1 −1 + c × F 2 0 + b + c × ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ d + e ⎝ 0 ⎠⎦ ⎛ 0 ⎞ ⎜ 0 ⎟ lb⋅ ft MO = ⎜ ⎟ ⎝ 100 ⎠

Problem 4-105 Replace the loading system acting on the post by an equivalent resultant force and couple moment at point P.

287



Chapter 4

Given: F 1 = 30 lb F 2 = 40 lb F 3 = 60 lb a = 1 ft b = 3 ft c = 2 ft d = 3 e = 4 Solution:

⎛0⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ F R = F1 −1 + F2 0 + ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝0⎠ ⎛ 4 ⎞ ⎜ ⎟ F R = −78 lb ⎜ ⎟ ⎝ 0 ⎠

⎛ −d ⎞ ⎜ −e ⎟ ⎟ 2 2⎜ d +e ⎝ 0 ⎠ F3

F R = 78.1 lb

⎡⎛ 0 ⎞ ⎤ ⎡ ⎛ 0 ⎞⎤ ⎛ 0 ⎞ ⎡ ⎛ 1 ⎞⎤ ⎛ 0 ⎞ ⎡ F ⎛ −d ⎞⎤ 3 ⎢ ⎜ ⎟ ⎥ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ −e ⎟⎥ MP = 0 ft × F1 −1 + −a − b × F2 0 + −a × ⎢⎜ ⎟ ⎥ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎣⎝ 0 ⎠ ⎦ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ d + e ⎝ 0 ⎠⎦ ⎛ 0 ⎞ ⎜ 0 ⎟ lb⋅ ft MP = ⎜ ⎟ ⎝ 124 ⎠

Problem 4-106 Replace the force and couple system by an equivalent force and couple moment at point O. Units Used: 3

kN = 10 N 288



Chapter 4

Given: M = 8 kN m

θ = 60 deg

a = 3m

f = 12

b = 3m

g = 5

c = 4m

F 1 = 6 kN

d = 4m

F 2 = 4 kN

e = 5m Solution:

FR =

⎛ −cos ( θ ) ⎞ ⎛g⎞ ⎜ f ⎟ + F ⎜ −sin ( θ ) ⎟ 2 ⎜ ⎟ 2 2⎜ ⎟ f + g ⎝0⎠ ⎝ 0 ⎠ F1

⎛ 0.308 ⎞ ⎜ ⎟ F R = 2.074 kN ⎜ ⎟ ⎝ 0 ⎠

F R = 2.097 kN

⎛ 0 ⎞ ⎛ −c ⎞ ⎡ F ⎛ g ⎞⎤ ⎛ 0 ⎞ ⎡ ⎛ −cos ( θ ) ⎞⎤ 1 ⎜ ⎟ ⎜ ⎟ ⎢ ⎜ f ⎟⎥ + ⎜ −d ⎟ × ⎢F ⎜ −sin ( θ ) ⎟⎥ MO = 0 + −e × ⎜ ⎟ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 ⎜ ⎟⎥ ⎝ M ⎠ ⎝ 0 ⎠ ⎣ f + g ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎛ 0 ⎞ ⎜ 0 ⎟ kN⋅ m MO = ⎜ ⎟ ⎝ −10.615 ⎠

Problem 4-107 Replace the force and couple system by an equivalent force and couple moment at point P. Units Used: 3

kN = 10 N

289



Chapter 4

Given: M = 8 kN⋅ m

θ = 60 deg

a = 3m

f = 12

b = 3m

g = 5

c = 4m

F 1 = 6 kN

d = 4m

F 2 = 4 kN

e = 5m Solution:

FR =

⎛ −cos ( θ ) ⎞ ⎛g⎞ ⎜ f ⎟ + F ⎜ −sin ( θ ) ⎟ 2 ⎜ ⎟ 2 2⎜ ⎟ f + g ⎝0⎠ ⎝ 0 ⎠ F1

⎛ 0.308 ⎞ F R = ⎜ 2.074 ⎟ kN ⎜ ⎟ ⎝ 0 ⎠

F R = 2.097 kN

⎛ 0 ⎞ ⎛ −c − b ⎞ ⎡ F ⎛ g ⎞⎤ ⎛ −b ⎞ ⎡ ⎛ −cos ( θ ) ⎞⎤ 1 ⎜ f ⎟⎥ + ⎜ −d ⎟ × ⎢F ⎜ −sin ( θ ) ⎟⎥ MP = ⎜ 0 ⎟ + ⎜ −e ⎟ × ⎢ ⎜ ⎟ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2⎜ ⎟⎥ ⎝ M ⎠ ⎝ 0 ⎠ ⎣ f + g ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎛ 0 ⎞ ⎟ kN⋅ m 0 MP = ⎜ ⎜ ⎟ ⎝ −16.838 ⎠

Problem 4-108 Replace the force system by a single force resultant and specify its point of application, measured along the x axis from point O. Given: F 1 = 125 lb F 2 = 350 lb F 3 = 850 lb

290



Chapter 4

a = 2 ft b = 6 ft c = 3 ft d = 4 ft

Solution: F Ry = F 3 − F 2 − F 1

F Ry = 375 lb

F Ry x = F3 ( b + c) − F 2 ( b) + F1 ( a) x =

F3 ( b + c) − F 2 ( b) + F1 a FRy

x = 15.5 ft

Problem 4-109 Replace the force system by a single force resultant and specify its point of application, measured along the x axis from point P. Given: F 1 = 125 lb

a = 2 ft

F 2 = 350 lb

b = 6 ft

F 3 = 850 lb

c = 3 ft d = 4 ft

Solution: F Ry = F 3 − F 2 − F 1

F Ry = 375 lb

F Ry x = F2 ( d + c) − F 3 ( d) + F1 ( a + b + c + d) x =

F2 d + F2 c − F3 d + F1 a + F1 b + F1 c + F1 d

x = 2.47 ft

F Ry (to the right of P)

291



Chapter 4

Problem 4-110 The forces and couple moments which are exerted on the toe and heel plates of a snow ski are F t, Mt, and F h, Mh, respectively. Replace this system by an equivalent force and couple moment acting at point O. Express the results in Cartesian vector form.

Given: a = 120 mm b = 800 mm

Solution:

⎛ −50 ⎞ F t = ⎜ 80 ⎟ N ⎜ ⎟ ⎝ −158 ⎠

⎛ −20 ⎞ F h = ⎜ 60 ⎟ N ⎜ ⎟ ⎝ −250 ⎠

⎛ −6 ⎞ Mt = ⎜ 4 ⎟ N⋅ m ⎜ ⎟ ⎝2⎠

⎛ −20 ⎞ Mh = ⎜ 8 ⎟ N⋅ m ⎜ ⎟ ⎝ 3 ⎠

FR = Ft + Fh

⎛ −70 ⎞ F R = ⎜ 140 ⎟ N ⎜ ⎟ ⎝ −408 ⎠ r0Ft

⎛a⎞ = ⎜0⎟ ⎜ ⎟ ⎝0⎠

MRP = ( r0Ft × F t) + Mt + Mh

⎛ −26 ⎞ MRP = ⎜ 31 ⎟ N⋅ m ⎜ ⎟ ⎝ 14.6 ⎠

Problem 4-111 The forces and couple moments which are exerted on the toe and heel plates of a snow ski are F t, Mt, and F h, Mh, respectively. Replace this system by an equivalent force and couple moment 292



Chapter 4

acting at point P. Express the results in Cartesian vector form. Given: a = 120 mm b = 800 mm

⎛ −50 ⎞ ⎜ 80 ⎟ N Ft = ⎜ ⎟ ⎝ −158 ⎠ ⎛ −6 ⎞ ⎜ ⎟ Mt = 4 N⋅ m ⎜ ⎟ ⎝2⎠ ⎛ −20 ⎞ ⎜ 60 ⎟ N Fh = ⎜ ⎟ ⎝ −250 ⎠ ⎛ −20 ⎞ ⎜ 8 ⎟ N⋅ m Mh = ⎜ ⎟ ⎝ 3 ⎠ Solution: FR = Ft + Fh

⎛ −70 ⎞ ⎜ ⎟ F R = 140 N ⎜ ⎟ ⎝ −408 ⎠

⎛b⎞ ⎛a + b⎞ ⎜ ⎟ ⎜ 0 ⎟×F MP = Mt + Mh + 0 × F h + ⎜ ⎟ ⎜ ⎟ t ⎝0⎠ ⎝ 0 ⎠

⎛ −26 ⎞ ⎜ ⎟ MP = 357.4 N⋅ m ⎜ ⎟ ⎝ 126.6 ⎠

Problem 4-112 Replace the three forces acting on the shaft by a single resultant force. Specify where the force acts, measured from end B.

293



Chapter 4

Given: F 1 = 500 lb F 2 = 200 lb F 3 = 260 lb a = 5 ft

e = 3

b = 3 ft

f = 4

c = 2 ft

g = 12

d = 4 ft

h = 5

Solution: FR =

⎛−f⎞ ⎛0⎞ ⎜ −e ⎟ + F ⎜ −1 ⎟ + 2 ⎜ ⎟ 2 2⎜ ⎟ e + f ⎝0 ⎠ ⎝0⎠ F1

Initial guess:

⎛h⎞ ⎜ −g ⎟ ⎟ 2 2⎜ g +h ⎝ 0 ⎠ F3

⎛ −300 ⎞ F R = ⎜ −740 ⎟ lb ⎜ ⎟ ⎝ 0 ⎠

F R = 798 lb

x = 1 ft

Given

⎛a⎞ ⎡ F ⎛ − f ⎞⎤ ⎛ a + b ⎞ ⎡ ⎛ 0 ⎞⎤ ⎛ a + b + c ⎞ ⎡ F ⎛ h ⎞⎤ ⎛ −x ⎞ 1 3 ⎜0⎟ × ⎢ ⎜ −e ⎟⎥ + ⎜ 0 ⎟ × ⎢F ⎜ −1 ⎟⎥ + ⎜ 0 ⎟ × ⎢ ⎜ −g ⎟⎥ = ⎜ 0 ⎟ × F ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ R ⎝ 0 ⎠ ⎣ e + f ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ g + h ⎝ 0 ⎠⎦ ⎝ 0 ⎠ x = Find ( x)

x = −7.432 ft

Problem 4-113 Replace the three forces acting on the shaft by a single resultant force. Specify where the force acts, measured from end B. Given: F 1 = 500 lb F 2 = 200 lb F 3 = 260 lb a = 5 ft

e = 3

294



b = 3 ft

f = 4

c = 2 ft

g = 12

d = 4 ft

h = 5

Chapter 4

Solution: FR =

⎛−f⎞ ⎛0⎞ ⎜ −e ⎟ + F ⎜ −1 ⎟ + 2 ⎜ ⎟ 2 2⎜ ⎟ e + f ⎝0 ⎠ ⎝0⎠ F1

Initial guess:

⎛h⎞ ⎜ −g ⎟ ⎟ 2 2⎜ g +h ⎝ 0 ⎠ F3

⎛ −300 ⎞ F R = ⎜ −740 ⎟ lb ⎜ ⎟ ⎝ 0 ⎠

F R = 798 lb

x = 1ft

Given

⎛ −b − c − d ⎞ ⎡ F ⎛ − f ⎞⎤ ⎛ −c − d ⎞ ⎡ ⎛ 0 ⎞⎤ ⎛ −d ⎞ ⎡ F ⎛ h ⎞⎤ ⎛ −x ⎞ 1 3 ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ −g ⎟⎥ = ⎜ 0 ⎟ × F 0 −e + × 0 × F 2 −1 + 0 × ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ 0 ⎝ ⎠ ⎣ e + f ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ g + h ⎝ 0 ⎠⎦ ⎝ 0 ⎠ x = Find ( x)

x = 6.568 ft

measured to the left of B

Problem 4-114 Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member AB, measured from A. Given: F 1 = 300 lb F 2 = 200 lb F 3 = 400 lb F 4 = 200 lb

M = 600 lb⋅ ft a = 3 ft b = 4 ft c = 2 ft d = 7 ft

Solution: F Rx = −F 4

F Rx = −200 lb

F Ry = −F 1 − F 2 − F 3

F Ry = −900 lb

295



2

F =

FRx + FRy

Chapter 4

2

F = 922 lb

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ = atan ⎜

θ = 77.5 deg

F Ry x = −F2 a − F 3 ( a + b) − F 4 c + M x = −

F2 ( a) + F 3 ( a + b) + F 4 c − M

x = 3.556 ft

F Ry

Problem 4-115 Replace the loading on the frame by a single resultant force. Specify where the force acts , measured from end A. Given: F 1 = 450 N

a = 2m

F 2 = 300 N

b = 4m

F 3 = 700 N

c = 3m

θ = 60 deg

M = 1500 N⋅ m

φ = 30 deg Solution: F Rx = F 1 cos ( θ ) − F 3 sin ( φ )

F Rx = −125 N

F Ry = −F 1 sin ( θ ) − F3 cos ( φ ) − F2

F Ry = −1.296 × 10 N

F =

2

FRx + FRy

3

3

2

F = 1.302 × 10 N

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ 1 = atan ⎜

θ 1 = 84.5 deg

F Ry( x) = −F1 sin ( θ ) a − F2 ( a + b) − F3 cos ( φ ) ( a + b + c) − M x =

−F1 sin ( θ ) a − F2 ( a + b) − F3 cos ( φ ) ( a + b + c) − M F Ry

x = 7.36 m

Problem 4-116 Replace the loading on the frame by a single resultant force. Specify where the force acts , measured from end B. 296



Chapter 4

Given: F 1 = 450 N

a = 2m

F 2 = 300 N

b = 4m

F 3 = 700 N

c = 3m

θ = 60 deg

M = 1500 N⋅ m

φ = 30deg Solution: F Rx = F 1 cos ( θ ) − F 3 sin ( φ )

F Rx = −125 N

F Ry = −F 1 sin ( θ ) − F3 cos ( φ ) − F2

F Ry = −1.296 × 10 N

F =

2

FRx + FRy

3

3

2

F = 1.302 × 10 N

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ 1 = atan ⎜

θ 1 = 84.5 deg

F Ry x = F1 sin ( θ ) b − F3 cos ( φ ) c − M x =

F1 sin ( θ ) b − F3 cos ( φ ) c − M

x = 1.36 m

F Ry

(to the right)

Problem 4-117 Replace the loading system acting on the beam by an equivalent resultant force and couple moment at point O. Given: F 1 = 200 N F 2 = 450 N M = 200 N⋅ m a = 0.2 m b = 1.5 m c = 2m d = 1.5 m

297



Chapter 4


⎛ −sin ( θ ) ⎞ ⎛0⎞ ⎜ ⎟ ⎜ ⎟ F R = F1 1 + F2 −cos ( θ ) ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝ 0 ⎠ ⎛ −225 ⎞ ⎜ ⎟ F R = −190 N ⎜ ⎟ ⎝ 0 ⎠

F R = 294 N

⎛ b + c ⎞ ⎡ ⎛ 0 ⎞⎤ ⎛ b ⎞ ⎡ ⎛ −sin ( θ ) ⎞⎤ ⎛0⎞ ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ a MO = × F 1 + a × F −cos ( θ ) + M 0 ⎜ ⎟ ⎢ 1⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 ⎜ ⎟⎥ ⎜ ⎟ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ −1 ⎠ ⎛ 0 ⎞ ⎜ 0 ⎟ N⋅ m MO = ⎜ ⎟ ⎝ −39.6 ⎠

Problem 4-118 Determine the magnitude and direction θ of force F and its placement d on the beam so that the loading system is equivalent to a resultant force FR acting vertically downward at point A and a clockwise couple moment M. Units Used: 3

kN = 10 N Given: F 1 = 5 kN

a = 3m

F 2 = 3 kN

b = 4m

F R = 12 kN

c = 6m

M = 50 kN⋅ m

e = 7

f = 24


F = 1 kN

θ = 30 deg

d = 2m

298



Given

Chapter 4

⎛ −e ⎞ F + F cos ( θ ) = 0 ⎜ 2 2⎟ 1 ⎝ e +f ⎠ ⎛ − f ⎞ F − F sin ( θ ) − F = −F 2 R ⎜ 2 2⎟ 1 ⎝ e +f ⎠ f ⎛ ⎞ ⎜ 2 2 ⎟ F1 a + F sin ( θ ) ( a + b − d) + F2 ( a + b) = M ⎝ e +f ⎠

⎛F⎞ ⎜ θ ⎟ = Find ( F , θ , d) ⎜ ⎟ ⎝d⎠

θ = 71.565 deg

F = 4.427 kN

d = 3.524 m

Problem 4-119 Determine the magnitude and direction θ of force F and its placement d on the beam so that the loading system is equivalent to a resultant force F R acting vertically downward at point A and a clockwise couple moment M. Units Used: 3


a = 3m

F 2 = 3 kN

b = 4m

F R = 10 kN

c = 6m

M = 45 kN⋅ m e = 7 f = 24 Solution: Initial guesses:

Given

F = 1 kN

θ = 30 deg

d = 1m

⎛ −e ⎞ F + F cos ( θ ) = 0 ⎜ 2 2⎟ 1 ⎝ e +f ⎠ ⎛ − f ⎞ F − F sin ( θ ) − F = −F 2 R ⎜ 2 2⎟ 1 ⎝ e +f ⎠ 299



Chapter 4

f ⎛ ⎞ ⎜ 2 2 ⎟ F1 a + F sin ( θ ) ( a + b − d) + F2 ( a + b) = M ⎝ e +f ⎠

⎛F⎞ ⎜ θ ⎟ = Find ( F , θ , d) ⎜ ⎟ ⎝d⎠

θ = 57.529 deg

F = 2.608 kN

d = 2.636 m

Problem 4-120 Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member AB, measured from A. Given: F 1 = 500 N

a = 3m b = 2m

F 2 = 300 N

c = 1m

F 3 = 250 N

d = 2m

M = 400 N⋅ m

e = 3m

θ = 60 deg

f = 3 g = 4

Solution:

⎛

⎞ − F ( cos ( θ ) ) 1 ⎟ 2 2 ⎝ g +f ⎠ g

F Rx = −F 3 ⎜

F Rx = −450 N

⎛

f

F Ry = −883.0127 N

FRx + FRy

2

F R = 991.066 N

⎞ − F sin ( θ ) 1 ⎟ 2 2 ⎝ f +g ⎠

F Ry = −F 2 − F 3 ⎜

FR =

2

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ 1 = atan ⎜

θ 1 = 62.996 deg

300



Chapter 4

−F Rx( y) = M + F 1 cos ( θ ) a + F3

g 2

g + f M + F 1 cos ( θ ) a + F3

g 2

g + f

y =

2

2

( b + a) − F2 ( d) − F 3 ⎛ ⎜

f 2

⎞ ( d + e)

2⎟

⎝ g +f ⎠

( b + a) − F2 ( d) − F 3 ⎛

f ⎞ ⎜ 2 2 ⎟ ( d + e) ⎝ g +f ⎠

−FRx y = 1.78 m

Problem 4-121 Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member CD, measured from end C. Given: F 1 = 500 N

a = 3m b = 2m

F 2 = 300 N

c = 1m

F 3 = 250 N

d = 2m

M = 400 N⋅ m

e = 3m

θ = 60 deg

f = 3 g = 4

Solution: F Rx = −F 3 ⎛ ⎜

⎞ − F ( cos ( θ ) ) 1 ⎝ g +f ⎠ g

2

F Ry = −F 2 − F 3 ⎛ ⎜

F Rx = −450 N

2⎟

⎞ − F sin ( θ ) 1 ⎝ f +g ⎠

FR =

2

f

2

FRx + FRy

F Ry = −883.0127 N

2⎟

2

F R = 991.066 N

301



Chapter 4

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ 1 = atan ⎜

θ 1 = 62.996 deg

⎛

⎞ ( c + d + e) − F ( b) cos ( θ ) − F c sin ( θ ) 1 1 ⎝ g +f ⎠

F Ry( x) = M − F2 ( d + c) − F 3 ⎜

⎛

f

2

2⎟

⎞ ( c + d + e) − F ( b) cos ( θ ) − F c sin ( θ ) 1 1 ⎝ g +f ⎠

M − F2 ( d + c) − F 3 ⎜ x =

f

2

2⎟

F Ry x = 2.64 m

Problem 4-122 Replace the force system acting on the frame by an equivalent resultant force and specify where the resultant's line of action intersects member AB, measured from point A. Given: F 1 = 35 lb a = 2 ft F 2 = 20 lb b = 4 ft F 3 = 25 lb c = 3 ft

θ = 30 deg d = 2 ft Solution: F Rx = F 1 sin ( θ ) + F3

F Rx = 42.5 lb

F Ry = −F 1 cos ( θ ) − F 2

F Ry = −50.311 lb

FR =

2

FRx + FRy

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ 1 = atan ⎜

2

F R = 65.9 lb

θ 1 = −49.8 deg

F Ry x = −F1 cos ( θ ) a − F 2 ( a + b) + F 3 ( c)

302



x =

Chapter 4

−F1 cos ( θ ) a − F 2 ( a + b) + F 3 ( c)

x = 2.099 ft

FRy

Problem 4-123 Replace the force system acting on the frame by an equivalent resultant force and specify where the resultant's line of action intersects member BC, measured from point B. Given: F 1 = 35 lb F 2 = 20 lb F 3 = 25 lb

θ = 30 deg a = 2 ft b = 4 ft c = 3 ft d = 2 ft

Solution: F Rx = F 1 sin ( θ ) + F3

F Rx = 42.5 lb

F Ry = −F 1 cos ( θ ) − F 2

F Ry = −50.311 lb

FR =

2

FRx + FRy

2

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ 1 = atan ⎜

F R = 65.9 lb

θ 1 = −49.8 deg

F Rx y = F 1 cos ( θ ) b + F3 ( c)

303



y =

F 1 cos ( θ ) b + F3 ( c) FRx

Chapter 4

y = 4.617 ft

(Below point B)

Problem 4-124 Replace the force system acting on the frame by an equivalent resultant force and couple moment acting at point A. Given: F 1 = 35 lb

a = 2 ft

F 2 = 20 lb

b = 4 ft

F 3 = 25 lb

c = 3 ft

θ = 30 deg

d = 2 ft

Solution: F Rx = F 1 sin ( θ ) + F3

F Rx = 42.5 lb

F Ry = F 1 cos ( θ ) + F 2

F Ry = 50.311 lb

FR =

2

FRx + FRy

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ 1 = atan ⎜

2

F R = 65.9 lb

θ 1 = 49.8 deg

MRA = −F1 cos ( θ ) a − F 2 ( a + b) + F 3 ( c)

MRA = −106 lb⋅ ft

Problem 4-125 Replace the force and couple-moment system by an equivalent resultant force and couple moment at point O. Express the results in Cartesian vector form.

304



Chapter 4

Units Used: 3

kN = 10 N Given:

⎛8⎞ ⎜ ⎟ F = 6 kN ⎜ ⎟ ⎝8⎠

a = 3m b = 3m

⎛ −20 ⎞ ⎜ ⎟ M = −70 kN⋅ m ⎜ ⎟ ⎝ 20 ⎠

c = 4m

e = 5m f = 6m g = 5m

d = 6m

Solution:

⎛−f⎞ ⎜ ⎟ MR = M + e × F ⎜ ⎟ ⎝g⎠

FR = F

⎛8⎞ ⎜ ⎟ F R = 6 kN ⎜ ⎟ ⎝8⎠

⎛ −10 ⎞ ⎜ ⎟ MR = 18 kN⋅ m ⎜ ⎟ ⎝ −56 ⎠

Problem 4-126 Replace the force and couple-moment system by an equivalent resultant force and couple moment at point P. Express the results in Cartesian vector form. Units Used: 3

kN = 10 N Given:

⎛8⎞ ⎜ ⎟ F = 6 kN ⎜ ⎟ ⎝8⎠ ⎛ −20 ⎞ ⎜ ⎟ M = −70 kN⋅ m ⎜ ⎟ ⎝ 20 ⎠ a = 3m b = 3m

e = 5m

c = 4m

f = 6m 305



d = 6m

Chapter 4

g = 5m

Solution:

⎛ −f ⎞ ⎜ e ⎟×F MR = M + ⎜ ⎟ ⎝d + g⎠

FR = F

⎛8⎞ ⎜ ⎟ F R = 6 kN ⎜ ⎟ ⎝8⎠

⎛ −46 ⎞ ⎜ ⎟ MR = 66 kN⋅ m ⎜ ⎟ ⎝ −56 ⎠

Problem 4-127 Replace the force and couple-moment system by an equivalent resultant force and couple moment at point Q. Express the results in Cartesian vector form. Units Used: 3

kN = 10 N Given:

⎛8⎞ ⎜ ⎟ F = 6 kN ⎜ ⎟ ⎝8⎠ ⎛ −20 ⎞ ⎜ ⎟ M = −70 kN⋅ m ⎜ ⎟ ⎝ 20 ⎠ a = 3m b = 3m

e = 5m

c = 4m

f = 6m

d = 6m

g = 5m

Solution:

FR = F

⎛0⎞ ⎜ ⎟ MR = M + e × F ⎜ ⎟ ⎝g⎠

⎛8⎞ ⎜ ⎟ F R = 6 kN ⎜ ⎟ ⎝8⎠

⎛ −10 ⎞ ⎜ ⎟ MR = −30 kN⋅ m ⎜ ⎟ ⎝ −20 ⎠

Problem 4-128 The belt passing over the pulley is subjected to forces F1 and F 2. F 1 acts in the −k direction. Replace these forces by an equivalent force and couple moment at point A. Express the result in 306



Chapter 4

p y Cartesian vector form.

q

p

p

p

Given: F 1 = 40 N

r = 80 mm

F 2 = 40 N

a = 300 mm


F 1v

F 2v

⎛0⎞ = F1 ⎜ 0 ⎟ ⎜ ⎟ ⎝ −1 ⎠

⎛ 0 ⎞ = F2 ⎜ −cos ( θ ) ⎟ ⎜ ⎟ ⎝ −sin ( θ ) ⎠

⎛ −a ⎞ r2 = ⎜ −r sin ( θ ) ⎟ ⎜ ⎟ ⎝ r cos ( θ ) ⎠

⎛ −a ⎞ r1 = ⎜ r ⎟ ⎜ ⎟ ⎝0⎠

F R = F1v + F2v

MA = r1 × F1v + r2 × F2v

⎛ 0 ⎞ F R = ⎜ −40 ⎟ N ⎜ ⎟ ⎝ −40 ⎠

⎛ 0 ⎞ MA = ⎜ −12 ⎟ N⋅ m ⎜ ⎟ ⎝ 12 ⎠

Problem 4-129 The belt passing over the pulley is subjected to forces F1 and F 2. F 1 acts in the −k direction. Replace these forces by an equivalent force and couple moment at point A. Express the result in Cartesian vector form.

307



Chapter 4

Given: F 1 = 40 N F 2 = 40 N

θ = 0 deg r = 80 mm a = 300 mm

θ = 45 deg

Solution:

F 1v

⎛0⎞ = F1 ⎜ 0 ⎟ ⎜ ⎟ ⎝ −1 ⎠

F R = F1v + F2v

⎛ 0 ⎞ F R = ⎜ −28.28 ⎟ N ⎜ ⎟ ⎝ −68.28 ⎠

F 2v

⎛ 0 ⎞ = F2 ⎜ −cos ( θ ) ⎟ ⎜ ⎟ ⎝ −sin ( θ ) ⎠

⎛ −a ⎞ r1 = ⎜ r ⎟ ⎜ ⎟ ⎝0⎠

⎛ −a ⎞ r2 = ⎜ −r sin ( θ ) ⎟ ⎜ ⎟ ⎝ r cos ( θ ) ⎠

MA = r1 × F1v + r2 × F2v

⎛ 0 ⎞ MA = ⎜ −20.49 ⎟ N⋅ m ⎜ ⎟ ⎝ 8.49 ⎠

Problem 4-130 Replace this system by an equivalent resultant force and couple moment acting at O. Express the results in Cartesian vector form. Given: F 1 = 50 N F 2 = 80 N

308



Chapter 4

F 3 = 180 N a = 1.25 m b = 0.5 m c = 0.75 m

Solution:

⎛⎜ 0 ⎞⎟ ⎛⎜ 0 ⎟⎞ ⎛⎜ 0 ⎟⎞ FR = ⎜ 0 ⎟ + ⎜ 0 ⎟ + ⎜ 0 ⎟ ⎜ F 1 ⎟ ⎜ −F2 ⎟ ⎜ −F 3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 0 ⎞ ⎜ 0 ⎟N FR = ⎜ ⎟ ⎝ −210 ⎠ ⎛ a + c ⎞ ⎛⎜ 0 ⎞⎟ ⎛ a ⎞ ⎛⎜ 0 ⎟⎞ ⎛ a ⎞ ⎛⎜ 0 ⎟⎞ ⎜ b ⎟ × 0 + ⎜b⎟ × 0 + ⎜0⎟ × 0 MO = ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ ⎟ ⎜ ⎟ ⎜ ⎝ 0 ⎠ ⎝ F1 ⎠ ⎝ 0 ⎠ ⎝ −F2 ⎠ ⎝ 0 ⎠ ⎝ −F3 ⎟⎠

⎛ −15 ⎞ ⎜ ⎟ MO = 225 N⋅ m ⎜ ⎟ ⎝ 0 ⎠

Problem 4-131 Handle forces F 1 and F 2 are applied to the electric drill. Replace this system by an equivalent resultant force and couple moment acting at point O. Express the results in Cartesian vector form. Given: a = 0.15 m b = 0.25 m c = 0.3 m

⎛ 6 ⎞ ⎜ ⎟ F 1 = −3 N ⎜ ⎟ ⎝ −10 ⎠ ⎛0⎞ ⎜ ⎟ F2 = 2 N ⎜ ⎟ ⎝ −4 ⎠ 309



Chapter 4

Solution: FR = F1 + F2

⎛ 6 ⎞ ⎜ ⎟ F R = −1 N ⎜ ⎟ ⎝ −14 ⎠

⎛a⎞ ⎛0⎞ ⎜ ⎟ ⎜ ⎟ MO = 0 × F1 + −b × F 2 ⎜ ⎟ ⎜ ⎟ ⎝c⎠ ⎝c ⎠

⎛ 1.3 ⎞ ⎜ 3.3 ⎟ N⋅ m MO = ⎜ ⎟ ⎝ −0.45 ⎠

Problem 4-132 A biomechanical model of the lumbar region of the human trunk is shown. The forces acting in the four muscle groups consist of F R for the rectus, FO for the oblique, FL for the lumbar latissimus dorsi, and F E for the erector spinae. These loadings are symmetric with respect to the y - z plane. Replace this system of parallel forces by an equivalent force and couple moment acting at the spine, point O. Express the results in Cartesian vector form. Given: F R = 35 N

a = 75 mm

F O = 45 N

b = 45 mm

F L = 23 N

c = 15 mm

F E = 32 N

d = 50 mm

e = 40 mm

f = 30 mm

Solution: F Res = ΣFi;

F Res = 2( FR + F O + F L + F E)

F Res = 270 N

MROx = ΣMOx;

MRO = −2FR a + 2F E c + 2F L b

MRO = −2.22 N⋅ m

Problem 4-133 The building slab is subjected to four parallel column loadings.Determine the equivalent resultant force and specify its location (x, y) on the slab.

310



Chapter 4

Units Used: 3

kN = 10 N Given: F 1 = 30 kN

a = 3m

F 2 = 40 kN

b = 8m

F 3 = 20 kN

c = 2m

F 4 = 50 kN

d = 6m e = 4m

Solution: +

↑ FR = ΣFx;

FR = F1 + F2 + F3 + F4 F R = 140 kN

MRx = ΣMx;

−F R( y) = −( F4 ) ( a) −

y =

⎡⎣( F1 ) ( a + b)⎤⎦ − ⎡⎣( F2 ) ( a + b + c)⎤⎦

F4 a + F1 a + F1 b + F2 a + F2 b + F2 c FR

y = 7.14 m MRy = ΣMy;

(FR)x = (F4)( e) + (F3)( d + e) + (F2)( b + c) x =

F4 e + F3 d + F3 e + F2 b + F2 c FR

x = 5.71 m

Problem 4-134 The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specify its location (x, y) on the slab.

311



Chapter 4

Units Used: 3

kN = 10 N Given: F 1 = 20 kN

a = 3m

F 2 = 50 kN

b = 8m

F 3 = 20 kN

c = 2m

F 4 = 50 kN

d = 6m e = 4m

Solution: FR = F1 + F2 + F3 + F4

F R = 140 kN

F R x = F2 e + F1 ( d + e) + F 2 ( d + e) x =

2 F2 e + F1 d + F1 e + F2 d

x = 6.43 m

FR

−F R y = −F 2 a − F3 ( a + b) − F2 ( a + b + c)

y =

2 F2 a + F3 a + F3 b + F2 b + F2 c FR

y = 7.29 m

Problem 4-135 The pipe assembly is subjected to the action of a wrench at B and a couple at A. Determine the magnitude F of the couple forces so that the system can be simplified to a wrench acting at point C. Given: a = 0.6 m

312



Chapter 4

b = 0.8 m c = 0.25 m d = 0.7 m e = 0.3 m f = 0.3 m g = 0.5 m h = 0.25 m P = 60 N Q = 40 N Solution: Initial Guess

F = 1N

MC = 1 N⋅ m

Given

⎛⎜ −MC ⎞⎟ ⎡ −P( c + h) ⎤ ⎡ 0 ⎢ ⎥+⎢ 0 ⎜ 0 ⎟=⎢ 0 ⎥ ⎢ ⎜ 0 ⎟ ⎣ 0 ⎦ ⎣ −F ( e + ⎝ ⎠ ⎛ F ⎞ ⎜ ⎟ = Find ( F , MC) ⎝ MC ⎠

⎤ ⎛ a ⎞ ⎛ −Q ⎞ ⎥ + ⎜b⎟ × ⎜ 0 ⎟ ⎥ ⎜ ⎟ ⎜ ⎟ f) ⎦ ⎝ 0 ⎠ ⎝ 0 ⎠

MC = 30 N⋅ m

F = 53.3 N

Problem 4-136 The three forces acting on the block each have a magnitude F 1 = F2 = F3. Replace this system by a wrench and specify the point where the wrench intersects the z axis, measured from point O. Given: F 1 = 10 lb

a = 6 ft

F2 = F1

b = 6 ft

F3 = F1

c = 2 ft

Solution: The vectors

313



F 1v =

⎛b⎞ ⎜ −a ⎟ ⎟ 2 2⎜ b +a ⎝ 0 ⎠ F1

Chapter 4

F 2v

⎛⎜ 0 ⎞⎟ = ⎜ −F 2 ⎟ ⎜ 0 ⎟ ⎝ ⎠

F 3v =

M = 1 lb⋅ ft

R x = 1 lb

⎛ −b ⎞ ⎜a⎟ ⎟ 2 2⎜ b +a ⎝ 0 ⎠ F3

Place the wrench in the x - z plane. x = 1ft z = 1ft

Guesses

R y = 1 lb

R z = 1 lb

⎛ Rx ⎞ ⎜ ⎟ Given ⎜ Ry ⎟ = F1v + F2v + F3v ⎜R ⎟ ⎝ z⎠ ⎛ Rx ⎞ ⎛ 0 ⎞ ⎛ x ⎞ ⎛⎜ Rx ⎟⎞ ⎛0⎞ ⎛b⎞ ⎜ ⎟ M ⎜0⎟ × R + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Ry ⎟ = ⎜ a ⎟ × F2v + ⎜ a ⎟ × F1v + ⎜ 0 ⎟ × F3v ⎜ ⎟ ⎜ y⎟ 2 2 2 Rx + Ry + Rz ⎜ R ⎟ ⎝ c ⎠ ⎝ z ⎠ ⎜⎝ Rz ⎟⎠ ⎝0⎠ ⎝c ⎠ ⎝ z⎠ ⎛x ⎞ ⎜ ⎟ ⎜ z ⎟ ⎛ Rx ⎞ ⎜ ⎟ ⎜M ⎟ M Mv = ⎜ R ⎟ = Find ( x , z , M , Rx , Ry , Rz) ⎜ Ry ⎟ 2 2 2 ⎜ x⎟ Rx + Ry + Rz ⎜ R ⎟ ⎝ z⎠ R ⎜ y⎟ ⎜ ⎟ ⎝ Rz ⎠ ⎛ Rx ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎛x⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎟ Mv = −14.142 lb⋅ ft ⎜ ⎟=⎜ ⎟ ft ⎜ Ry ⎟ = ⎜ −10 ⎟ lb ⎜ ⎟ ⎝ z ⎠ ⎝ 0.586 ⎠ ⎜R ⎟ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ z⎠ Problem 4-137 Replace the three forces acting on the plate by a wrench. Specify the magnitude of the force and couple moment for the wrench and the point P(x, y) where its line of action intersects the plate. Units Used: 3

kN = 10 N

314



Chapter 4

Given: F A = 500 N F B = 800 N F C = 300 N a = 4m b = 6m

Solution:

⎛ FA ⎞ ⎜ ⎟ FR = ⎜ FC ⎟ ⎜F ⎟ ⎝ B⎠ Guesses

F R = 0.9899 kN

x = 1m

Given

M

FR FR

y = 1m

M = 100 N⋅ m

⎛x⎞ + ⎜ y ⎟ × FR = ⎜ ⎟ ⎝0⎠

⎛M⎞ ⎜ x ⎟ = Find ( M , x , y) ⎜ ⎟ ⎝y⎠

⎛ b ⎞ ⎛⎜ 0 ⎟⎞ ⎛ 0 ⎞ ⎛⎜ 0 ⎞⎟ ⎜a⎟ × F + ⎜a⎟ × 0 ⎜ ⎟ ⎜ C⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎜⎝ 0 ⎟⎠ ⎝ 0 ⎠ ⎜⎝ FB ⎟⎠ ⎛ x ⎞ ⎛ 1.163 ⎞ ⎜ ⎟=⎜ ⎟m ⎝ y ⎠ ⎝ 2.061 ⎠

M = 3.07 kN⋅ m

Problem 4-138 Replace the three forces acting on the plate by a wrench. Specify the magnitude of the force and couple moment for the wrench and the point P(y, z) where its line of action intersects the plate. Given: F A = 80 lb

a = 12 ft

F B = 60 lb

b = 12 ft

F C = 40 lb

315



Chapter 4

Solution:

⎛ −F C ⎞ ⎜ ⎟ F R = ⎜ −F B ⎟ ⎜ −F ⎟ ⎝ A⎠

F R = 108 lb

y = 1 ft

Guesses

Given

M

FR FR

z = 1 ft

⎛0⎞ + ⎜ y ⎟ × FR = ⎜ ⎟ ⎝z⎠

⎛M⎞ ⎜ y ⎟ = Find ( M , y , z) ⎜ ⎟ ⎝z⎠

M = 1 lb⋅ ft

⎛ 0 ⎞ ⎛⎜ −FC ⎞⎟ ⎛ 0 ⎞ ⎛⎜ 0 ⎟⎞ ⎜a⎟ × ⎜ ⎟ ⎜ 0 ⎟ + a × ⎜ −FB ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎜⎝ 0 ⎟⎠ ⎝ b ⎠ ⎜⎝ 0 ⎟⎠ ⎛ y ⎞ ⎛ 0.414 ⎞ ⎜ ⎟=⎜ ⎟ ft ⎝ z ⎠ ⎝ 8.69 ⎠

M = −624 lb⋅ ft

Problem 4-139 The loading on the bookshelf is distributed as shown. Determine the magnitude of the equivalent resultant location, measured from point O. Given: w1 = 2

lb ft

w2 = 3.5

lb ft

a = 2.75 ft b = 4 ft c = 1.5 ft R = 1 lb

Solution: Guesses Given

d = 1ft

w1 b + w2 c = R w1 b⎛⎜ a −

⎝

⎛R⎞ ⎜ ⎟ = Find ( R , d) ⎝d⎠

b⎞

⎞ ⎛c ⎟ − w2 c⎜ + b − a⎟ = −d R 2⎠ ⎝2 ⎠ R = 13.25 lb

d = 0.34 ft

316



Chapter 4

Problem 4-140 Replace the loading by an equivalent resultant force and couple moment acting at point A. Units Used: 3

kN = 10 N Given: w1 = 600

N

w2 = 600

N

m

m

a = 2.5 m b = 2.5 m Solution: F R = w1 a − w2 b MRA = w1 a⎛⎜

a + b⎞

⎟ ⎝ 2 ⎠

FR = 0 N MRA = 3.75 kN⋅ m

Problem 4-141 Replace the loading by an equivalent force and couple moment acting at point O. Units Used: 3

kN = 10 N Given: w = 6

kN m

F = 15 kN M = 500 kN⋅ m a = 7.5 m b = 4.5 m

317



Chapter 4

Solution: FR =

1 2

w( a + b) + F

MR = −M −

F R = 51.0 kN

⎛ 1 w a⎞ ⎛ 2 a⎞ − ⎛ 1 w b⎞ ⎛a + ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎝ 2 ⎠⎝ 3 ⎠ ⎝ 2 ⎠⎝

b⎞

⎟ − F( a + b)

3⎠

MR = −914 kN⋅ m

Problem 4-142 Replace the loading by a single resultant force, and specify the location of the force on the beam measured from point O. Units Used: 3


kN m

F = 15 kN M = 500 kN⋅ m a = 7.5 m b = 4.5 m Solution: Initial Guesses:

F R = 1 kN

d = 1m

Given FR =

1 2

w( a + b) + F

−F R d = −M −

⎛ 1 w a⎞ ⎛ 2 a⎞ − ⎛ 1 w b⎞ ⎛ a + b ⎞ − F( a + b) ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ 3 ⎠ ⎝ 2 ⎠⎝ 3 ⎠

⎛ FR ⎞ ⎜ ⎟ = Find ( FR , d) ⎝ d ⎠

F R = 51 kN

d = 17.922 m

Problem 4-143 The column is used to support the floor which exerts a force P on the top of the column. The effect of soil pressure along its side is distributed as shown. Replace this loading by an 318



Chapter 4

p g p g y equivalent resultant force and specify where it acts along the column, measured from its base A. 3

kip = 10 lb

Units Used: Given:

P = 3000 lb w1 = 80

lb ft

w2 = 200

lb ft

h = 9 ft Solution: F Rx = w1 h +

1 (w2 − w1)h 2

F Rx = 1260 lb

F Ry = P

2

FR =

FRx + P

2

⎞ θ = atan ⎛⎜ ⎟ F ⎝ Rx ⎠ P

θ = 67.2 deg

1 h h w2 − w1 ) h + w1 h ( 2 3 2

F Rx y =

y =

F R = 3.25 kip

1 2 w2 + 2 w1 h FRx 6

y = 3.86 ft

Problem 4-144 Replace the loading by an equivalent force and couple moment at point O. Units Used: 3

kN = 10 N Given: kN m

w1 = 15

w2 = 5

kN m 319



Chapter 4

d = 9m Solution: FR =

1 (w1 + w2)d 2

MRO = w2 d

F R = 90 kN

d 1 d + ( w1 − w2 ) d 2 2 3

MRO = 338 kN⋅ m

Problem 4-145 Replace the distributed loading by an equivalent resultant force, and specify its location on the beam, measured from the pin at C. Units Used: 3

kip = 10 lb Given: w = 800

lb ft

a = 15 ft b = 15 ft

θ = 30 deg Solution: FR = w a + FR x = w a

wa x =

a 2

wb 2

F R = 18 kip

a wb⎛ b⎞ + ⎜a + ⎟ 2 ⎝ 2 3⎠ +

b⎞ ⎜a + ⎟ 2 ⎝ 3⎠

wb⎛ FR

x = 11.7 ft

Problem 4-146 The beam supports the distributed load caused by the sandbags. Determine the resultant force on the beam and specify its location measured from point A.

320



Units Used:

Chapter 4

3

kN = 10 N

Given: w1 = 1.5 w2 = 1

kN

a = 3m

m

kN

b = 3m

m

w3 = 2.5

kN

c = 1.5 m

m

Solution: F R = w1 a + w2 b + w3 c MA = w1 a

a 2

+ w2 b⎛⎜ a +

⎝

F R = 11.25 kN b⎞

c⎞ ⎛ ⎟ + w3 c⎜ a + b + ⎟ 2⎠ 2⎠ ⎝

MA = 45.563 kN⋅ m

d =

MA

d = 4.05 m

FR

Problem 4-147 Determine the length b of the triangular load and its position a on the beam such that the equivalent resultant force is zero and the resultant couple moment is M clockwise. Units Used: 3

kN = 10 N Given: w1 = 4

kN

w2 = 2.5

m

M = 8 kN⋅ m

kN m

c = 9m

Solution: Initial Guesses: Given

−1 2

a = 1m w1 b +

1 2

b = 1m

w2 c = 0

321



1 2

w1 b⎛⎜ a +

⎝

⎛a⎞ ⎜ ⎟ = Find ( a , b) ⎝b⎠

Chapter 4

2b ⎞

1

2c

⎟ − w2 c = −M 3⎠ 2 3 ⎛ a ⎞ ⎛ 1.539 ⎞ ⎜ ⎟=⎜ ⎟m ⎝ b ⎠ ⎝ 5.625 ⎠

Problem 4-148 Replace the distributed loading by an equivalent resultant force and specify its location, measured from point A. Units Used: 3

kN = 10 N Given: w1 = 800

N

w2 = 200

N

m

m

a = 2m b = 3m Solution: F R = w2 b + w1 a + x F R = w1 a

w1 a x =

a 2

+

1 2

(w1 − w2)b

F R = 3.10 kN

b b w1 − w2 ) b⎛⎜ a + ⎞⎟ + w2 b⎛⎜ a + ⎞⎟ ( 2 3 2 1

⎝

⎠

⎝

a 1 ⎛ b⎞ ⎛ b⎞ + ( w1 − w2 ) b⎜ a + ⎟ + w2 b⎜ a + ⎟ 2 2 ⎝ 3⎠ ⎝ 2⎠ FR

⎠

x = 2.06 m

Problem 4-149 The distribution of soil loading on the bottom of a building slab is shown. Replace this loading by an equivalent resultant force and specify its location, measured from point O. Units Used: 322



Chapter 4

3

kip = 10 lb Given: w1 = 50

lb ft

w2 = 300

lb

w3 = 100

lb

ft

ft

a = 12 ft b = 9 ft Solution :

(w2 − w1)a + 2 (w2 − w3)b + w3 b 2 1

F R = w1 a + F R d = w1 a

a 2

1

+

(w2 − w1)a 2 1

2

d =

2a 3

+

F R = 3.9 kip

b b w2 − w3 ) b⎛⎜ a + ⎞⎟ + w3 b⎛⎜ a + ⎞⎟ ( 2 3 2 1

2

⎝

2

⎠

3 w3 b a + 2 w3 b + w1 a + 2 a w2 + 3 b w2 a + w2 b 6FR

⎝

⎠

2

d = 11.3 ft

Problem 4-150 The beam is subjected to the distributed loading. Determine the length b of the uniform load and its position a on the beam such that the resultant force and couple moment acting on the beam are zero. Given: w1 = 40

lb

w2 = 60

lb

ft

c = 10ft d = 6 ft

ft


a = 1 ft

b = 1ft

323



Chapter 4

Given 1 2

1

w2 d − w1 b = 0

⎛a⎞ ⎜ ⎟ = Find ( a , b) ⎝b⎠

2

w2 d⎛⎜ c +

⎝

d⎞

⎛ ⎟ − w1 b⎜ a + 3⎠ ⎝

b⎞

⎟=0

2⎠

⎛ a ⎞ ⎛ 9.75 ⎞ ⎜ ⎟=⎜ ⎟ ft ⎝ b ⎠ ⎝ 4.5 ⎠

Problem 4-151 Replace the loading by an equivalent resultant force and specify its location on the beam, measured from point B. Units Used: 3


lb

w2 = 500

lb

ft

ft

a = 12 ft b = 9 ft Solution: FR =

1 2

a w1 +

1 2

(w1 − w2)b + w2 b

F R = 10.65 kip

a 1 1 b b F R x = − a w1 + ( w1 − w2 ) b + w2 b 3 2 2 3 2

x =

a 1 b b 1 − a w1 + ( w1 − w2 ) b + w2 b 3 2 3 2 2 FR

x = 0.479 ft ( to the right of B )

Problem 4-152 Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects member AB, measured from A.

324



Chapter 4

Given: w1 = 200

N m

w2 = 100

N m

w3 = 200

N m

a = 5m b = 6m Solution: F Rx = −w3 a F Ry =

F Rx = −1000 N

−1 (w1 + w2)b 2

− y FRx = w3 a

w3 a y =

a 2

F Ry = −900 N

a b 1 b − w2 b − ( w1 − w2 ) b 2 2 2 3

− w2 b

b

−

(w1 − w2)b 3 2 1

b

2 −FRx

y = 0.1 m

Problem 4-153 Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects member BC, measured from C. Units Used: 3

kN = 10 N

325



Chapter 4

Given: w1 = 200

N

w2 = 100

N

w3 = 200

N

m

m

m

a = 5m b = 6m Solution : F Rx = −w3 a F Ry =

−1 2

(w1 + w2)b

−x F Ry = −w3 a

−w3 a x =

F Rx = −1000 N

a 2

+ w2 b

F Ry = −900 N b 2

+

(w1 − w2)b 2 1

2b 3

a b 1 2b + w2 b + ( w1 − w2 ) b 2 2 2 3

x = 0.556 m

−FRy

⎛ FRx ⎞ ⎜ ⎟ = 1.345 kN ⎝ FRy ⎠ Problem 4-154 Replace the loading by an equivalent resultant force and couple moment acting at point O. Units Used: 3

kN = 10 N Given: w1 = 7.5

kN m

326



w2 = 20

Chapter 4

kN m

a = 3m b = 3m c = 4.5 m

Solution:

(w2 − w1)c + w1 c + w1 b + 2 1

FR =

1 2

w1 a

F R = 95.6 kN MRo = −

c c b w2 − w1 ) c − w1 c − w1 b⎛⎜ c + ⎞⎟ − ( 2 3 2 2 1

⎝

⎠

1 2

w1 a⎛⎜ b + c +

⎝

a⎞

⎟

3⎠

MRo = −349 kN⋅ m

Problem 4-155 Determine the equivalent resultant force and couple moment at point O. Units Used: 3

kN = 10 N Given: a = 3m wO = 3

kN m

x w ( x) = wO ⎛⎜ ⎟⎞ ⎝ a⎠

2

Solution: ⌠a F R = ⎮ w ( x) dx ⌡0

F R = 3 kN

⌠a MO = ⎮ w ( x) ( a − x) dx ⌡0

MO = 2.25 kN⋅ m

327



Chapter 4

Problem 4-156 Wind has blown sand over a platform such that the intensity of the load can be approximated by 3

⎛ x ⎞ . Simplify this distributed loading to an equivalent resultant force and ⎟ ⎝d⎠

the function w = w0 ⎜

specify the magnitude and location of the force, measured from A.

Units Used: 3

kN = 10 N Given: w0 = 500

N m

d = 10 m

⎛x⎞ ⎟ ⎝ d⎠

3

w ( x) = w0 ⎜ Solution: d

⌠ F R = ⎮ w ( x) dx ⌡0

d =

⌠d ⎮ x w ( x) dx ⌡0 FR

F R = 1.25 kN

d=8m

Problem 4-157 Determine the equivalent resultant force and its location, measured from point O.

328



Chapter 4

Solution: L

⌠ 2w0 L ⎛ πx⎞ F R = ⎮ w0 sin ⎜ ⎟ dx = ⎮ π ⎝L⎠ ⌡ 0

L

d=

⌠ ⎮ x w sin ⎛ π x ⎞ dx ⎜ ⎟ 0 ⎮ ⎝L⎠ ⌡ 0

FR

=

L 2

Problem 4-158 Determine the equivalent resultant force acting on the bottom of the wing due to air pressure and specify where it acts, measured from point A. Given: a = 3 ft k = 86

lb ft

3 2

w ( x) = k x Solution: a

⌠ F R = ⎮ w ( x) dx ⌡0

x =

⌠a ⎮ x w ( x) dx ⌡0 FR

F R = 774 lb

x = 2.25 ft

Problem 4-159 Currently eighty-five percent of all neck injuries are caused by rear-end car collisions. To 329



Chapter 4

y g y p j y alleviate this problem, an automobile seat restraint has been developed that provides additional pressure contact with the cranium. During dynamic tests the distribution of load on the cranium has been plotted and shown to be parabolic. Determine the equivalent resultant force and its location, measured from point A. Given: a = 0.5 ft w0 = 12

lb ft

lb

k = 24

ft

3

2

w ( x) = w0 + kx

Solution: ⌠a F R = ⎮ w ( x) dx ⌡0

F R = 7 lb

a

x =

⌠ ⎮ x w ( x) dx ⌡0 FR

x = 0.268 ft

Problem 4-160 Determine the equivalent resultant force of the distributed loading and its location, measured from point A. Evaluate the integrals using Simpson's rule.

Units Used: 3

kN = 10 N Given: c1 = 5 c2 = 16 a = 3

330



Chapter 4

b = 1 Solution: a+ b

⌠ FR = ⎮ ⌡0

2

c1 x +

c2 + x dx

x c1 x +

c2 + x dx

F R = 14.9

a+ b

d =

⌠ ⎮ ⌡0

2

d = 2.27

FR

Problem 4-161 Determine the coordinate direction angles of F, which is applied to the end A of the pipe assembly, so that the moment of F about O is zero. Given: F = 20 lb a = 8 in b = 6 in c = 6 in d = 10 in Solution: Require Mo = 0. This happens when force F is directed either towards or away from point O.

⎛ c ⎞ r = ⎜a + b⎟ ⎜ ⎟ ⎝ d ⎠

u =

r r

⎛ 0.329 ⎞ u = ⎜ 0.768 ⎟ ⎜ ⎟ ⎝ 0.549 ⎠

If the force points away from O, then

⎛⎜ α ⎞⎟ ⎜ β ⎟ = acos ( u) ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 70.774 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 39.794 ⎟ deg ⎜ γ ⎟ ⎝ 56.714 ⎠ ⎝ ⎠

If the force points towards O, then 331



⎛⎜ α ⎞⎟ ⎜ β ⎟ = acos ( −u) ⎜γ ⎟ ⎝ ⎠

Chapter 4

⎛⎜ α ⎞⎟ ⎛ 109.226 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 140.206 ⎟ deg ⎜ γ ⎟ ⎝ 123.286 ⎠ ⎝ ⎠

Problem 4-162 Determine the moment of the force F about point O. The force has coordinate direction angles α, β, γ. Express the result as a Cartesian vector. Given: F = 20 lb

a = 8 in

α = 60 deg

b = 6 in

β = 120 deg

c = 6 in

γ = 45 deg

d = 10 in

Solution:

⎛ c ⎞ r = ⎜a + b⎟ ⎜ ⎟ ⎝ d ⎠

⎛⎜ cos ( α ) ⎟⎞ F v = F⎜ cos ( β ) ⎟ ⎜ cos ( γ ) ⎟ ⎝ ⎠

M = r × Fv

⎛ 297.99 ⎞ M = ⎜ 15.147 ⎟ lb⋅ in ⎜ ⎟ ⎝ −200 ⎠

Problem 4-163 Replace the force at A by an equivalent resultant force and couple moment at point P. Express the results in Cartesian vector form. Units Used :

332



Chapter 4

3

kN = 10 N Given: a = 4m b = 6m c = 8m d = 4m

⎛ −300 ⎞ ⎜ ⎟ F = 200 N ⎜ ⎟ ⎝ −500 ⎠ Solution : FR = F

⎛ −300 ⎞ ⎜ ⎟ F R = 200 N ⎜ ⎟ ⎝ −500 ⎠

⎛ −a − c ⎞ ⎜ b ⎟×F MP = ⎜ ⎟ ⎝ d ⎠

⎛ −3.8 ⎞ ⎜ ⎟ MP = −7.2 kN⋅ m ⎜ ⎟ ⎝ −0.6 ⎠

Problem 4-164 Determine the moment of the force FC about the door hinge at A. Express the result as a Cartesian vector.

333



Chapter 4

Given: F = 250 N b = 1m c = 2.5 m d = 1.5 m e = 0.5 m

θ = 30 deg

Solution:

rCB

⎛ c−e ⎞ ⎜ ⎟ = b + d cos ( θ ) ⎜ ⎟ ⎝ −d sin ( θ ) ⎠

MA = rAB × F v

rAB

⎛0⎞ ⎜ ⎟ = b ⎜ ⎟ ⎝0⎠

Fv = F

rCB rCB

⎛ −59.7 ⎞ ⎜ 0.0 ⎟ N⋅ m MA = ⎜ ⎟ ⎝ −159.3 ⎠

Problem 4-165 Determine the magnitude of the moment of the force FC about the hinged axis aa of the door.

334



Chapter 4

Given: F = 250 N b = 1m c = 2.5 m d = 1.5 m e = 0.5 m

θ = 30 deg

Solution:

rCB

⎛ c−e ⎞ ⎜ ⎟ = b + d cos ( θ ) ⎜ ⎟ ⎝ −d sin ( θ ) ⎠

Maa = ( rAB × Fv) ⋅ ua

rAB

⎛0⎞ ⎜ ⎟ = b ⎜ ⎟ ⎝0⎠

Fv = F

rCB rCB

⎛1⎞ ⎜ ⎟ ua = 0 ⎜ ⎟ ⎝0⎠

Maa = −59.7 N⋅ m

Problem 4-166 A force F1 acts vertically downward on the Z-bracket. Determine the moment of this force about the bolt axis (z axis), which is directed at angle θ from the vertical.

335



Chapter 4

Given: F 1 = 80 N a = 100 mm b = 300 mm c = 200 mm


⎛ −b ⎞ ⎜ ⎟ r = a+c ⎜ ⎟ ⎝ 0 ⎠ ⎛ sin ( θ ) ⎞ ⎜ 0 ⎟ F = F1 ⎜ ⎟ ⎝ −cos ( θ ) ⎠ ⎛0⎞ ⎜ ⎟ k = 0 ⎜ ⎟ ⎝1⎠ Mz = ( r × F ) k Mz = −6.212 N⋅ m

Problem 4-167 Replace the force F having acting at point A by an equivalent force and couple moment at point C. Units Used:

3

kip = 10 lb

Given: F = 50 lb a = 10 ft b = 20 ft c = 15 ft

336



Chapter 4

d = 10 ft e = 30 ft

Solution :

rAB

⎛d⎞ = ⎜ c ⎟ ⎜ ⎟ ⎝ −e ⎠

Fv = F

rCA

rAB rAB

⎛ 0 ⎞ = ⎜a + b⎟ ⎜ ⎟ ⎝ e ⎠

FR = Fv

⎛ 14.286 ⎞ F R = ⎜ 21.429 ⎟ lb ⎜ ⎟ ⎝ −42.857 ⎠

MR = rCA × Fv

⎛ −1.929 ⎞ MR = ⎜ 0.429 ⎟ kip⋅ ft ⎜ ⎟ ⎝ −0.429 ⎠

Problem 4-168 The horizontal force F acts on the handle of the wrench. What is the magnitude of the moment of this force about the z axis? Given: F = 30 N a = 50 mm b = 200 mm c = 10 mm


⎛ sin ( θ ) ⎞ F v = F⎜ −cos ( θ ) ⎟ ⎜ ⎟ ⎝ 0 ⎠

rOA

⎛ −c ⎞ = ⎜b ⎟ ⎜ ⎟ ⎝a⎠

⎛0⎞ k = ⎜0⎟ ⎜ ⎟ ⎝1⎠ 337



Mz = ( rOA × F v ) k

Chapter 4

Mz = −4.03 N⋅ m

Problem 4-169 The horizontal force F acts on the handle of the wrench. Determine the moment of this force about point O. Specify the coordinate direction angles α, β, γ of the moment axis. Given: F = 30 N

c = 10 mm

a = 50 mm θ = 45 deg b = 200 mm Solution:

⎛ sin ( θ ) ⎞ F v = F⎜ −cos ( θ ) ⎟ ⎜ ⎟ ⎝ 0 ⎠ MO = rOA × Fv

⎛⎜ α ⎞⎟ ⎛ MO ⎞ ⎜ β ⎟ = acos ⎜ ⎟ MO ⎠ ⎝ ⎜γ ⎟ ⎝ ⎠

rOA

⎛ −c ⎞ = ⎜b ⎟ ⎜ ⎟ ⎝a⎠

⎛ 1.06 ⎞ MO = ⎜ 1.06 ⎟ N⋅ m ⎜ ⎟ ⎝ −4.03 ⎠ ⎛⎜ α ⎞⎟ ⎛ 75.7 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 75.7 ⎟ deg ⎜ γ ⎟ ⎝ 159.6 ⎠ ⎝ ⎠

Problem 4-170 If the resultant couple moment of the three couples acting on the triangular block is to be zero, determine the magnitudes of forces F and P.

338



Chapter 4

Given: F 1 = 10 lb a = 3 in b = 4 in c = 6 in d = 3 in

θ = 30 deg


F = 1 lb

P = 1 lb

Given

⎛ 0 ⎞ ⎛ 0 ⎞ ⎜ −F c ⎟ + ⎜ 0 ⎟ + ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎝ −P c ⎠

⎛0⎞ ⎜a⎟ = 0 2 2⎜ ⎟ a + b ⎝b⎠

⎛F⎞ ⎜ ⎟ = Find ( F , P) ⎝P⎠

F1 d

⎛F⎞ ⎛3⎞ ⎜ ⎟ = ⎜ ⎟ lb ⎝P⎠ ⎝4⎠

339



Chapter 5

Problem 5-1 Draw the free-body diagram of the sphere of weight W resting between the smooth inclined planes. Explain the significance of each force on the diagram.

Given: W = 10 lb

θ 1 = 105 deg θ 2 = 45 deg

Solution:

NA, NB force of plane on sphere. W force of gravity on sphere.

Problem 5-2 Draw the free-body diagram of the hand punch, which is pinned at A and bears down on the smooth surface at B. Given: F = 8 lb a = 1.5 ft b = 0.2 ft c = 2 ft

340



Chapter 5

Solution:

Problem 5-3 Draw the free-body diagram of the beam supported at A by a fixed support and at B by a roller. Explain the significance of each force on the diagram. Given: w = 40

lb ft

a = 3 ft b = 4 ft

θ = 30 deg

Solution:

A x, A y, MA effect of wall on beam. NB force of roller on beam. wa 2

resultant force of distributed load on beam.

Problem 5-4 Draw the free-body diagram of the jib crane AB, which is pin-connected at A and supported by member (link) BC.

341



Chapter 5

Units Used: 3

kN = 10 N Given: F = 8 kN a = 3m b = 4m c = 0.4 m d = 3 e = 4

Solution:

Problem 5-5 Draw the free-body diagram of the C-bracket supported at A, B, and C by rollers. Explain the significance of each forcce on the diagram. Given: a = 3 ft b = 4 ft

θ 1 = 30 deg θ 2 = 20 deg F = 200 lb 342



Chapter 5

Solution: NA , NB , NC force of rollers on beam.

Problem 5-6 Draw the free-body diagram of the smooth rod of mass M which rests inside the glass. Explain the significance of each force on the diagram. Given: M = 20 gm a = 75 mm b = 200 mm

θ = 40 deg

Solution: A x , A y , NB force of glass on rod. M(g) N force of gravity on rod.

Problem 5-7 Draw the free-body diagram of the “spanner wrench” subjected to the force F. The support at A can be considered a pin, and the surface of contact at B is smooth. Explain the significance of 343



Chapter 5

p each force on the diagram.

p

g

Given: F = 20 lb a = 1 in b = 6 in

Solution: A x, A y, NB force of cylinder on wrench.

Problem 5-8 Draw the free-body diagram of the automobile, which is being towed at constant velocity up the incline using the cable at C. The automobile has a mass M and center of mass at G. The tires are free to roll. Explain the significance of each force on the diagram. Units Used: 3

Mg = 10 kg Given: M = 5 Mg

d = 1.50 m

a = 0.3 m

e = 0.6 m

b = 0.75 m

θ 1 = 20 deg

c = 1m

θ 2 = 30 deg

g = 9.81

m 2

s

344



Chapter 5

Solution: NA, NB force of road on car. F force of cable on car. Mg force of gravity on car.

Problem 5-9 Draw the free-body diagram of the uniform bar, which has mass M and center of mass at G. The supports A, B, and C are smooth. Given: M = 100 kg a = 1.75 m b = 1.25 m c = 0.5 m d = 0.2 m g = 9.81

m 2

s Solution:



Chapter 5

Problem 5-10 Draw the free-body diagram of the beam, which is pin-connected at A and rocker-supported at B. Given: F = 500 N M = 800 N⋅ m a = 8m b = 4m c = 5m

Solution:

Problem 5-11 The sphere of weight W rests between the smooth

inclined planes. Determine the reaactions at the supports. Given: W = 10 lb

θ 1 = 105 deg θ 2 = 45 deg Solution: Initial guesses 346



NA = 1 lb

Chapter 5

NB = 1 lb

Given NB cos ( θ 1 − 90 deg) − NA cos ( θ 2 ) = 0 NA sin ( θ 2 ) − NB sin ( θ 1 − 90 deg) − W = 0

⎛ NA ⎞ ⎜ ⎟ = Find ( NA , NB) ⎝ NB ⎠

⎛ NA ⎞ ⎛ 19.3 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ NB ⎠ ⎝ 14.1 ⎠

Problem 5-12 Determine the magnitude of the resultant force acting at pin A of the handpunch. Given: F = 8 lb a = 1.5 ft b = 0.2 ft c = 2 ft

Solution:

Σ F x = 0; Σ M = 0;

Ax − F = 0

Ax = F

F a − Ay c = 0

Ay = F

Ax = 8 lb a c

Ay = 6 lb

347



FA =

Chapter 5

2

Ax + Ay

2

F A = 10 lb

Problem 5-13 The C-bracket is supported at A, B, and C by rollers. Determine the reactions at the supports. Given: a = 3 ft b = 4 ft

θ 1 = 30 deg θ 2 = 20 deg F = 200 lb Solution: Initial Guesses: NA = 1 lb NB = 1 lb NC = 1 lb Given NA a − F b = 0 NB sin ( θ 1 ) − NC sin ( θ 2 ) = 0 NB cos ( θ 1 ) + NC cos ( θ 2 ) − NA − F = 0

⎛ NA ⎞ ⎜ ⎟ ⎜ NB ⎟ = Find ( NA , NB , NC) ⎜N ⎟ ⎝ C⎠

⎛ NA ⎞ ⎛ 266.7 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ NB ⎟ = ⎜ 208.4 ⎟ lb ⎜ N ⎟ ⎝ 304.6 ⎠ ⎝ C⎠

348



Chapter 5

Problem 5-14 The smooth rod of mass M rests inside the glass. Determine the reactions on the rod. Given: M = 20 gm a = 75 mm b = 200 mm

θ = 40 deg g = 9.81

m 2

s Solution: Initial Guesses: Ax = 1 N

Ay = 1 N

NB = 1 N

Given Ax − NB sin ( θ ) = 0 Ay − M g + NB cos ( θ ) = 0 −M g

a+b cos ( θ ) + NB b = 0 2

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ = Find ( Ax , Ay , NB) ⎜N ⎟ ⎝ B⎠

⎛ Ax ⎞ ⎛ 0.066 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 0.117 ⎟ N ⎜ N ⎟ ⎝ 0.103 ⎠ ⎝ B⎠

Problem 5-15 The “spanner wrench” is subjected to the force F. The support at A can be considered a pin, and the surface of contact at B is smooth. Determine the reactions on the spanner wrench.

349



Chapter 5

Given: F = 20 lb a = 1 in b = 6 in Solution: Initial Guesses: Ax = 1 lb Ay = 1 lb NB = 1 lb Given − Ax + NB = 0

Ay − F = 0

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ = Find ( Ax , Ay , NB) ⎜N ⎟ ⎝ B⎠

−F ( a + b) + Ax a = 0

⎛ Ax ⎞ ⎛ 140 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 20 ⎟ lb ⎜ N ⎟ ⎝ 140 ⎠ ⎝ B⎠

Problem 5-16 The automobile is being towed at constant velocity up the incline using the cable at C. The automobile has a mass M and center of mass at G. The tires are free to roll. Determine the reactions on both wheels at A and B and the tension in the cable at C. Units Used: 3

Mg = 10 kg

3

kN = 10 N

350



Chapter 5

Given: M = 5 Mg

d = 1.50 m

a = 0.3 m

e = 0.6 m

b = 0.75 m

θ 1 = 20 deg

c = 1m

θ 2 = 30 deg

g = 9.81

m 2

s Solution: Guesses

F = 1 kN

NA = 1 kN

NB = 1 kN

Given NA + NB + F sin ( θ 2 ) − M g cos ( θ 1 ) = 0 −F cos ( θ 2 ) + M g sin ( θ 1 ) = 0 F cos ( θ 2 ) a − F sin ( θ 2 ) b − M g cos ( θ 1 ) c − M g sin ( θ 1 ) e + NB( c + d) = 0

⎛F ⎞ ⎜ ⎟ ⎜ NA ⎟ = Find ( F , NA , NB) ⎜ NB ⎟ ⎝ ⎠

⎛ F ⎞ ⎛ 19.37 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ NA ⎟ = ⎜ 13.05 ⎟ kN ⎜ NB ⎟ ⎝ 23.36 ⎠ ⎝ ⎠

Problem 5-17 The uniform bar has mass M and center of mass at G. The supports A, B, and C are smooth. Determine the reactions at the points of contact at A, B, and C. Given: M = 100 kg a = 1.75 m b = 1.25 m

351



Chapter 5

c = 0.5 m d = 0.2 m

θ = 30 deg g = 9.81

m 2

s Solution:

The initial guesses: NA = 20 N NB = 30 N NC = 40 N Given

ΣMA = 0; +

↑Σ Fy = 0; +

↑Σ Fy = 0;

−M g cos ( θ ) a − M g sin ( θ )

d + NB sin ( θ ) d + NC( a + b) = 0 2

NB − M g + NC cos ( θ ) = 0 NA − NC sin ( θ ) = 0

⎛ NC ⎞ ⎜ ⎟ ⎜ NB ⎟ = Find ( NC , NB , NA) ⎜N ⎟ ⎝ A⎠

⎛ NC ⎞ ⎛ 493 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ NB ⎟ = ⎜ 554 ⎟ N ⎜ N ⎟ ⎝ 247 ⎠ ⎝ A⎠

Problem 5-18 The beam is pin-connected at A and rocker-supported at B. Determine the reactions at the pin A and at the roller at B. Given: F = 500 N M = 800 N⋅ m a = 8m

352



Chapter 5

b = 4m c = 5m Solution:

⎞ ⎟ ⎝ a + b⎠

α = atan ⎛⎜

c

ΣMA = 0;

−F

a

cos ( α )

By =

− M + By a = 0

F a + M cos ( α ) cos ( α ) a

B y = 642 N + Σ F x = 0; → +

↑Σ Fy = 0;

− Ax + F sin ( α ) = 0

Ax = F sin ( α )

Ax = 192 N

− Ay − F cos ( α ) + B y = 0

Ay = −F cos ( α ) + By

Ay = 180 N

Problem 5-19 Determine the magnitude of the reactions on the beam at A and B. Neglect the thickness of the beam. Given: F 1 = 600 N F 2 = 400 N

θ = 15 deg a = 4m b = 8m c = 3 d = 4 Solution:

ΣMA = 0;

B y( a + b) − F2 cos ( θ ) ( a + b) − F1 a = 0 353



By = + Σ F x = 0; →

Chapter 5

F 2 cos ( θ ) ( a + b) + F 1 a a+b

Ax − F 2 sin ( θ ) = 0 Ax = F2 sin ( θ )

+

↑Σ Fy = 0;

B y = 586 N

Ax = 104 N

Ay − F 2 cos ( θ ) + B y − F1 = 0 Ay = F2 cos ( θ ) − B y + F1

FA =

2

Ax + Ay

2

Ay = 400 N

F A = 413 N

Problem 5-20 Determine the reactions at the supports. Given: w = 250

lb ft

a = 6 ft b = 6 ft c = 6 ft Solution: Guesses Ax = 1 lb

Ay = 1 lb

B y = 1 lb

Given Ax = 0

wa

Ay + B y − w( a + b) −

⎛ a ⎞ − w b⎛ b ⎞ − 1 w c⎛b + ⎜ ⎟ ⎜ ⎟ ⎜ ⎝ 2⎠ ⎝ 2⎠ 2 ⎝

1 2

wc = 0

c⎞

⎟ + By b = 0

3⎠

354



Chapter 5

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ = Find ( Ax , Ay , By) ⎜B ⎟ ⎝ y⎠

⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 2750 ⎟ lb ⎜ B ⎟ ⎝ 1000 ⎠ ⎝ y⎠

Problem 5-21 When holding the stone of weight W in equilibrium, the humerus H, assumed to be smooth, exerts normal forces FC and F A on the radius C and ulna A as shown. Determine these forces and the force F B that the biceps B exerts on the radius for equilibrium. The stone has a center of mass at G. Neglect the weight of the arm. Given: W = 5 lb

θ = 75 deg a = 2 in b = 0.8 in c = 14 in

Solution:

ΣMB = 0;

− W ( c − a) + F A a = 0 FA = W

⎛ c − a⎞ ⎜ ⎟ ⎝ a ⎠

F A = 30 lb +

↑Σ Fy = 0;

F B sin ( θ ) − W − FA = 0

FB =

W + FA sin ( θ )

F B = 36.2 lb + Σ F x = 0; →

F C − F B cos ( θ ) = 0 355



Chapter 5

F C = FB cos ( θ ) F C = 9.378 lb

Problem 5-22 The uniform door has a weight W and a center of gravity at G. Determine the reactions at the hinges if the hinge at A supports only a horizontal reaction on the door, whereas the hinge at B exerts both horizontal and vertical reactions. Given: W = 100 lb a = 3 ft b = 3 ft c = 0.5 ft d = 2 ft

Solution: ΣMB = 0;

W d − A x ( a + b) = 0 Ax = W

⎛ d ⎞ ⎜ ⎟ ⎝ a + b⎠

Ax = 33.3 lb

ΣF x = 0;

Bx = Ax

B x = 33.3 lb

ΣF y = 0;

By = W

B y = 100 lb

Problem 5-23 The ramp of a ship has weight W and center of gravity at G. Determine the cable force in CD needed to just start lifting the ramp, (i.e., so the reaction at B becomes zero). Also, determine the horizontal and vertical components of force at the hinge (pin) at A.

356



Chapter 5

Given: W = 200 lb

a = 4 ft

θ = 30 deg

b = 3 ft

φ = 20 deg

c = 6 ft

Solution:

ΣMA = 0; −F CD cos ( θ ) ( b + c) cos ( φ ) + FCD sin ( θ ) ( b + c) sin ( φ ) + W c cos ( φ ) = 0

F CD =

W c cos ( φ )

( b + c) ( cos ( θ ) cos ( φ ) − sin ( θ ) sin ( φ ) )

+ Σ F x = 0; →

F CD sin ( θ ) − A x = 0 Ax = FCD sin ( θ )

+

↑Σ Fy = 0;

F CD = 195 lb

Ax = 97.5 lb

Ay − W + F CD cos ( θ ) = 0 Ay = W − FCD cos ( θ )

Ay = 31.2 lb

Problem 5-24 The drainpipe of mass M is held in the tines of the fork lift. Determine the normal forces at A and B as functions of the blade angle θ and plot the results of force (ordinate) versus θ (abscissa) for 0 ≤ θ ≤ 90 deg.

357



Chapter 5

Units used: 3

Mg = 10 kg Given: M = 1.4 Mg a = 0.4 m g = 9.81

m 2

s Solution:

θ = 0 .. 90 NA ( θ ) =

M g sin ( θ deg) 3

10 NB ( θ ) =

M g cos ( θ deg) 3

10

Force in kN

15 NA( θ )10 NB( θ )

5

0

0

20

40

60

80

100

θ

Angle in Degrees

Problem 5-25 While slowly walking, a man having a total mass M places all his weight on one foot. Assuming that the normal force NC of the ground acts on his foot at C, determine the resultant vertical compressive force F B which the tibia T exerts on the astragalus B, and the vertical tension FA in the achilles tendon A at the instant shown.

358



Chapter 5

Units Used: 3

kN = 10 N Given: M = 80 kg a = 15 mm b = 5 mm c = 20 mm d = 100 mm

Solution: NC = M g NC = 785 N ΣMA = 0;

−F B c + NC ( c + d) = 0 F B = NC

⎛ c + d⎞ ⎜ ⎟ ⎝ c ⎠

F B = 4.71 kN ΣF y = 0;

F A − FB + NC = 0 F A = F B − NC F A = 3.92 kN

Problem 5-26 Determine the reactions at the roller A and pin B.

359



Chapter 5

Given: M = 800 lb ft

c = 3 ft

F = 390 lb

d = 5

a = 8 ft

e = 12

b = 4 ft

θ = 30 deg

Solution: Guesses

R A = 1 lb B x = 1 lb

B y = 1 lb

Given R A sin ( θ ) + B x −

⎛ d ⎞F = 0 ⎜ 2 2⎟ ⎝ e +d ⎠

R A cos ( θ ) + By −

e ⎛ ⎞ ⎜ 2 2 ⎟F = 0 ⎝ e +d ⎠

M − R A cos ( θ ) ( a + b) + Bx c = 0

⎛ RA ⎞ ⎜ ⎟ ⎜ Bx ⎟ = Find ( RA , Bx , By) ⎜B ⎟ ⎝ y⎠

R A = 105.1 lb

⎛ Bx ⎞ ⎛ 97.4 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ By ⎠ ⎝ 269 ⎠

Problem 5-27 The platform assembly has weight W1 and center of gravity at G1. If it is intended to support a maximum load W2 placed at point G2,,determine the smallest counterweight W that should be placed at B in order to prevent the platform from tipping over. Given: W1 = 250 lb

a = 1 ft

c = 1 ft

e = 6 ft

W2 = 400 lb

b = 6 ft

d = 8 ft

f = 2 ft

360



Chapter 5

Solution: When tipping occurs, R c = 0

ΣMD = 0;

−W2 f + W1 c + WB ( b + c) = 0

WB =

W2 f − W1 c b+c

WB = 78.6 lb

Problem 5-28 The articulated crane boom has a weight W and mass center at G. If it supports a load L, determine the force acting at the pin A and the compression in the hydraulic cylinder BC when the boom is in the position shown. Units Used: 3

kip = 10 lb Given: W = 125 lb

361



Chapter 5

L = 600 lb a = 4 ft b = 1 ft c = 1 ft d = 8 ft

θ = 40 deg

Solution: Guesses

Given

Ax = 1 lb

Ay = 1 lb

F B = 1 lb

− Ax + F B cos ( θ ) = 0

− Ay + F B sin ( θ ) − W − L = 0

F B cos ( θ ) b + F B sin ( θ ) c − W a − L( d + c) = 0

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ = Find ( Ax , Ay , FB) ⎜F ⎟ ⎝ B⎠

F B = 4.19 kip

⎛ Ax ⎞ ⎛ 3.208 ⎞ ⎜ ⎟=⎜ ⎟ kip ⎝ Ay ⎠ ⎝ 1.967 ⎠

Problem 5-29 The device is used to hold an elevator door open. If the spring has stiffness k and it is compressed a distnace δ, determine the horizontal and vertical components of reaction at the pin A and the resultant force at the wheel bearing B. Given: N m

b = 125 mm

δ = 0.2 m

c = 100 mm

a = 150 mm

θ = 30 deg

k = 40

362



Solution:

ΣMA = 0;

Chapter 5

F s = kδ −F s a + F B cos ( θ ) ( a + b) − F B sin ( θ ) c = 0 FB = Fs

a

cos ( θ ) ( a + b) − sin ( θ ) c

F B = 6.378 N + Σ F x = 0; →

Ax − F B sin ( θ ) = 0 Ax = FB sin ( θ ) Ax = 3.189 N

+

↑Σ Fy = 0;

Ay − F s + F B cos ( θ ) = 0 Ay = Fs − FB cos ( θ ) Ay = 2.477 N

Problem 5-30 Determine the reactions on the bent rod which is supported by a smooth surface at B and by a collar at A, which is fixed to the rod and is free to slide over the fixed inclined rod. Given: F = 100 lb M = 200 lb ft a = 3 ft b = 3 ft c = 2ft

363



Chapter 5

d = 3 e = 4 f = 12 g = 5 Solution: Initial Guesses: NA = 20 lb

NB = 10 lb

MA = 30 lb ft

Given f g ⎞ ⎞ ⎛ ⎛ ⎜ 2 2 ⎟ ( a + b) − NB ⎜ 2 2 ⎟ c = 0 ⎝ f +g ⎠ ⎝ f +g ⎠

ΣMA = 0;

MA − F a − M + NB

ΣF x = 0;

NA

e g ⎞ ⎞ ⎛ ⎛ ⎜ 2 2 ⎟ − NB ⎜ 2 2 ⎟ = 0 ⎝ e +d ⎠ ⎝ f +g ⎠

ΣF y = 0;

NA

f ⎞−F=0 ⎛ d ⎞+N ⎛ B⎜ ⎜ 2 2⎟ ⎟ 2 2 ⎝ e +d ⎠ ⎝ f +g ⎠

⎛ NA ⎞ ⎜ ⎟ ⎜ NB ⎟ = Find ( NA , NB , MA) ⎜M ⎟ ⎝ A⎠

⎛ NA ⎞ ⎛ 39.7 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ NB ⎠ ⎝ 82.5 ⎠

MA = 106 lb⋅ ft

Problem 5-31 The cantilevered jib crane is used to support the load F. If the trolley T can be placed anywhere in the range x1 ≤ x ≤ x2, determine the maximum magnitude of reaction at the supports A and B. Note that the supports are collars that allow the crane to rotate freely about the vertical axis. The collar at B supports a force in the vertical direction, whereas the one at A does not. Units Used: kip = 1000 lb Given: F = 780 lb

364



Chapter 5

a = 4 ft b = 8 ft x1 = 1.5 ft x2 = 7.5 ft Solution: The maximum occurs when x = x2

ΣMA = 0;

−F x2 + B x a = 0

Bx = F

x2 a 3

B x = 1.462 × 10 lb + Σ F x = 0; → +

↑Σ Fy = 0; FB =

Ax − Bx = 0 By − F = 0 2

Bx + By

Ax = Bx By = F

2

3

Ax = 1.462 × 10 lb B y = 780 lb

F B = 1.657 kip

Problem 5-32 The uniform rod AB has weight W. Determine the force in the cable when the rod is in the position shown. Given: W = 15 lb L = 5 ft

365



Chapter 5

θ 1 = 30 deg θ 2 = 10 deg

Solution:

ΣMA = 0;

NB L sin ( θ 1 + θ 2 ) − W

NB =

⎛ L ⎞ cos ( θ + θ ) = 0 ⎜ ⎟ 1 2 ⎝2⎠

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

NB = 8.938 lb ΣF x = 0;

T cos ( θ 2 ) − NB

T =

NB

cos ( θ 2 )

T = 9.08 lb

Problem 5-33 The power pole supports the three lines, each line exerting a vertical force on the pole due to its weight as shown. Determine the reactions at the fixed support D. If it is possible for wind or ice to snap the lines, determine which line(s) when removed create(s) a condition for the greatest moment reaction at D. Units Used: 3

kip = 10 lb

366



Chapter 5

Given: W1 = 800 lb W2 = 450 lb W3 = 400 lb a = 2ft b = 4 ft c = 3 ft Solution: + Σ F x = 0; → +

↑Σ Fy = 0;

Dx = 0 Dy − ( W1 + W2 + W3 ) = 0 Dy = W1 + W2 + W3

ΣMD = 0;

Dy = 1.65 kip

−W2 b − W3 c + W1 a + MD = 0 MD = W2 b + W3 c − W1 a

MD = 1.4 kip⋅ ft

Examine all cases. For these numbers we require line 1 to snap. MDmax = W2 b + W3 c

MDmax = 3 kip⋅ ft

Problem 5-34 The picnic table has a weight WT and a center of gravity at GT . If a man weighing WM has a center of gravity at GM and sits down in the centered position shown, determine the vertical reaction at each of the two legs at B.Neglect the thickness of the legs. What can you conclude from the results? Given: WT = 50 lb

367



Chapter 5

WM = 225 lb a = 6 in b = 20 in c = 20 in

Solution: ΣMA = 0;

2 NB( b + c) + WM a − WT b = 0

NB =

WT b − WM a 2 ( b + c)

NB = −4.37 lb Since NB has a negative sign, the table will tip over.

Problem 5-35 If the wheelbarrow and its contents have a mass of M and center of mass at G, determine the magnitude of the resultant force which the man must exert on each of the two handles in order to hold the wheelbarrow in equilibrium. Given: M = 60 kg a = 0.6 m b = 0.5 m c = 0.9 m d = 0.5 m g = 9.81

m 2

s

368



Chapter 5

Solution:

ΣMB = 0;

− Ay ( b + c) + M g c = 0 Ay =

Mgc b+c

Ay = 378.386 N + Σ F x = 0; →

B x = 0N

+

Ay − M g + 2 By = 0

↑Σ Fy = 0;

By =

Bx = 0

M g − Ay 2

B y = 105.107 N

Problem 5-36 The man has weight W and stands at the center of the plank. If the planes at A and B are smooth, determine the tension in the cord in terms of W and θ.

Solution: ΣMB = 0;

W

L cos ( φ ) − NA L cos ( φ ) = 0 2

NA = ΣF x = 0;

W 2

T cos ( θ ) − NB sin ( θ ) = 0

(1)

369



Chapter 5

T sin ( θ ) + NB cos ( θ ) + NA − W = 0

ΣF y = 0;

(2)

Solving Eqs. (1) and (2) yields:

T=

W sin ( θ ) 2

NB =

W cos ( θ ) 2

Problem 5-37 When no force is applied to the brake pedal of the lightweight truck, the retainer spring AB keeps the pedal in contact with the smooth brake light switch at C. If the force on the switch is F , determine the unstretched length of the spring if the stiffness of the spring is k. Given: F = 3N k = 80

N m

a = 100 mm b = 50 mm c = 40 mm d = 10 mm

θ = 30 deg Solution: ΣMD = 0; F s b − F cos ( θ ) c − F sin ( θ ) d = 0 Fs = F

cos ( θ ) c + sin ( θ ) d b

Fs = k x

x =

L0 = a − x

Fs k

F s = 2.378 N

x = 29.73 mm L0 = 70.3 mm



Chapter 5

Problem 5-38 The telephone pole of negligible thickness is subjected to the force F directed as shown. It is supported by the cable BCD and can be assumed pinned at its base A. In order to provide clearance for a sidewalk right of way, where D is located, the strut CE is attached at C, as shown by the dashed lines (cable segment CD is removed). If the tension in CD' is to be twice the tension in BCD, determine the height h for placement of the strut CE. Given: F = 80 lb

θ = 30 deg a = 30 ft b = 10 ft

Solution:

+ ΣMA = 0;

−F cos ( θ ) a +

b ⎛ ⎞ ⎜ 2 2 ⎟ TBCD a = 0 ⎝ a +b ⎠

TBCD = F cos ( θ )

2

2

a +b b

TBCD = 219.089 lb

Require TCD' = 2 TBCD + ΣMA = 0;

TCD' = 438.178 lb

TCD' d − F cos ( θ ) a = 0

d = Fa

⎛ cos ( θ ) ⎞ ⎜T ⎟ ⎝ CD' ⎠

d = 4.7434 ft

Geometry: a−h a = d b

h = a−a

⎛ d⎞ ⎜ ⎟ ⎝ b⎠

h = 15.8 ft

371



Chapter 5

Problem 5-39 The worker uses the hand truck to move material down the ramp. If the truck and its contents are held in the position shown and have weight W with center of gravity at G, determine the resultant normal force of both wheels on the ground A and the magnitude of the force required at the grip B. Given: W = 100 lb

e = 1.5 ft

a = 1 ft

f = 0.5 ft

b = 1.5 ft

θ = 60 deg

c = 2 ft

φ = 30 deg

d = 1.75 ft

Solution: ΣMB = 0; NA cos ( θ − φ ) ( b + c + d) + NA sin ( θ − φ ) ( a − f) − W cos ( θ ) ( b + c) − W sin ( θ ) ( e + a) = 0

NA =

W cos ( θ ) ( b + c) + W sin ( θ ) ( e + a)

cos ( θ − φ ) ( b + c + d) + sin ( θ − φ ) ( a − f)

NA = 81.621 lb

ΣF x = 0;

−B x + NA sin ( φ ) = 0

B x = NA sin ( φ )

B x = 40.811 lb

ΣF y = 0;

B y + NA ( cos ( φ ) − W = 0)

B y = W − NA cos ( φ )

B y = 29.314 lb

FB =

2

Bx + By

2

F B = 50.2 lb

372



Chapter 5

T sin ( θ ) + NB cos ( θ ) + NA − W = 0

ΣF y = 0;

(2)

Solving Eqs. (1) and (2) yields:

T=

W sin ( θ ) 2

NB =

W cos ( θ ) 2

Problem 5-37 When no force is applied to the brake pedal of the lightweight truck, the retainer spring AB keeps the pedal in contact with the smooth brake light switch at C. If the force on the switch is F , determine the unstretched length of the spring if the stiffness of the spring is k. Given: F = 3N k = 80

N m

a = 100 mm b = 50 mm c = 40 mm d = 10 mm

θ = 30 deg Solution: ΣMD = 0; F s b − F cos ( θ ) c − F sin ( θ ) d = 0 Fs = F

cos ( θ ) c + sin ( θ ) d b

Fs = k x

x =

L0 = a − x

Fs k

F s = 2.378 N

x = 29.73 mm L0 = 70.3 mm 373



Chapter 5

The shelf supports the electric motor which has mass m1 and mass center at Gm. The platform upon which it rests has mass m2 and mass center at Gp. Assuming that a single bolt B holds the shelf up and the bracket bears against the smooth wall at A, determine this normal force at A and the horizontal and vertical components of reaction of the bolt B on the bracket. Given: m1 = 15 kg

c = 50 mm

m2 = 4 kg

d = 200 mm

a = 60 mm

e = 150 mm

b = 40 mm g = 9.81

m 2

s Solution: ΣMA = 0;

B x a − m2 g d − m1 g( d + e) = 0

Bx = g + Σ F x = 0; →

m2 d + m1 ( d + e)

B x = 989 N

a

Ax − Bx = 0 Ax = Bx

+

↑Σ Fy = 0;

Ax = 989 N

B y − m2 g − m1 g = 0 B y = m2 g + m1 g

B y = 186 N

Problem 5-42 A cantilever beam, having an extended length L, is subjected to a vertical force F. Assuming that the wall resists this load with linearly varying distributed loads over the length a of the beam portion inside the wall, determine the intensities w1 and w2 for equilibrium.

374



Chapter 5

Units Used: 3

kN = 10 N Given: F = 500 N a = 0.15 m L = 3m Solution: The initial guesses w1 = 1

kN m

w2 = 1

kN m

Given +

↑Σ Fy = 0; ΣMA = 0;

⎛ w1 ⎞ ⎜ ⎟ = Find ( w1 , w2) ⎝ w2 ⎠

1

1 w1 a − w2 a − F = 0 2 2 −F L −

1 2

1 ⎛ 2 a⎞ ⎟ + w2 a⎜ ⎟ = 0 ⎝ 3⎠ 2 ⎝ 3 ⎠

w1 a⎛⎜

a⎞

⎛ w1 ⎞ ⎛ 413 ⎞ kN ⎜ ⎟=⎜ ⎟ ⎝ w2 ⎠ ⎝ 407 ⎠ m

Problem 5-43 The upper portion of the crane boom consists of the jib AB, which is supported by the pin at A, the guy line BC, and the backstay CD, each cable being separately attached to the mast at C. If the load F is supported by the hoist line, which passes over the pulley at B, determine the magnitude of the resultant force the pin exerts on the jib at A for equilibrium, the tension in the guy line BC, and the tension T in the hoist line. Neglect the weight of the jib. The pulley at B has a radius of r.

375



Chapter 5

Units Used: 3

kN = 10 N Given: F = 5 kN r = 0.1 m a = r b = 1.5 m c = 5m Solution: From pulley, tension in the hoist line is ΣMB = 0;

T( a) − F( r) = 0

T = F

r T = 5 kN

a

From the jib, ΣMA = 0;

b+a

−F ( c) + TBC

c=0

2

c + ( b + a) 2

c + ( b + a)

TBC = F

+

↑Σ Fy = 0;

2

2

TBC = 16.406 kN

b+a b+a ⎤−F=0 ⎢ 2 2⎥ ⎣ c + ( b + a) ⎦

− Ay + TBC ⎡

b+a ⎤−F ⎢ 2 2⎥ ⎣ c + ( b + a) ⎦

Ay = TBC⎡

+ Σ F x = 0; →

Ay = 0 kN

⎤−F=0 ⎣ c + ( b + a) ⎦

Ax − TBC ⎡ ⎢

c

2

2⎥

376



Chapter 5

c

Ax = TBC

2

c + ( b + a) FA =

2

Ax + Ay

+F 2

2

Ax = 20.6 kN

F A = 20.6 kN

Problem 5-44 The mobile crane has weight W1 and center of gravity at G1; the boom has weight W2 and center of gravity at G2. Determine the smallest angle of tilt θ of the boom, without causing the crane to overturn if the suspended load has weight W. Neglect the thickness of the tracks at A and B. Given: W1 = 120000 lb W2 = 30000 lb W = 40000 lb a = 4 ft b = 6 ft c = 3 ft d = 12 ft e = 15 ft Solution: When tipping occurs, R A = 0 ΣMB = 0;

−W2 ( d cos ( θ ) − c) − W⎡⎣( d + e)cos ( θ ) − c⎤⎦ + W1 ( b + c) = 0

⎡W2 c + W c + W1 ( b + c) ⎤ ⎥ ⎣ W2 d + W ( d + e) ⎦

θ = acos ⎢

θ = 26.4 deg

377



Chapter 5

Problem 5-45 The mobile crane has weight W1 and center of gravity at G1; the boom has weight W2 and center of gravity at G2. If the suspended load has weight W determine the normal reactions at the tracks A and B. For the calculation, neglect the thickness of the tracks . Units Used: 3

kip = 10 lb Given: W1 = 120000 lb

a = 4 ft

W2 = 30000 lb

b = 6 ft

W = 16000 lb

c = 3 ft

θ = 30 deg

d = 12 ft e = 15 ft

Solution: ΣMB = 0; −W2 ( d cos ( θ ) − c) − W⎡⎣( d + e)cos ( θ ) − c⎤⎦ − RA( a + b + c) + W1 ( b + c) = 0

RA = +

↑

−W2 ( d cos ( θ ) − c) − W⎡⎣( d + e)cos ( θ ) − c⎤⎦ + W1 ( b + c)

Σ F y = 0;

a+b+c

R A = 40.9 kip

R A + RB − W1 − W2 − W = 0 R B = −R A + W1 + W2 + W

R B = 125 kip

Problem 5-46 The man attempts to support the load of boards having a weight W and a center of gravity at G. If he is standing on a smooth floor, determine the smallest angle θ at which he can hold them up in the position shown. Neglect his weight .

378



Chapter 5

Given: a = 0.5 ft b = 3 ft c = 4 ft d = 4 ft

Solution: ΣMB = 0;

−NA( a + b) + W( b − c cos ( θ ) ) = 0

As θ becomes smaller, NA goes to 0 so that, cos ( θ ) =

b c

b θ = acos ⎛⎜ ⎟⎞

⎝c⎠

θ = 41.4 deg

Problem 5-47 The motor has a weight W. Determine the force that each of the chains exerts on the supporting hooks at A, B, and C. Neglect the size of the hooks and the thickness of the beam.

379



Chapter 5

Given: W = 850 lb a = 0.5 ft b = 1 ft c = 1.5 ft

θ 1 = 10 deg θ 2 = 30 deg θ 3 = 10 deg

Solution: Guesses F A = 1 lb

F B = 1 lb

F C = 1 lb

Given ΣMB = 0;

F A cos ( θ 3 ) b + W a − FC cos ( θ 1 ) ( a + c) = 0

ΣF x = 0;

F C sin ( θ 1 ) − F B sin ( θ 2 ) − FA sin ( θ 3 ) = 0

ΣF y = 0;

W − F A cos ( θ 3 ) − F B cos ( θ 2 ) − F C cos ( θ 1 ) = 0

⎛ FA ⎞ ⎜ ⎟ ⎜ FB ⎟ = Find ( FA , FB , FC) ⎜F ⎟ ⎝ C⎠ ⎛ FA ⎞ ⎛ 432 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FB ⎟ = ⎜ −0 ⎟ lb ⎜ F ⎟ ⎝ 432 ⎠ ⎝ C⎠

380



Chapter 5

Problem 5-48 The boom supports the two vertical loads. Neglect the size of the collars at D and B and the thickness of the boom, and compute the horizontal and vertical components of force at the pin A and the force in cable CB. Given: F 1 = 800 N F 2 = 350 N a = 1.5 m b = 1m c = 3 d = 4

θ = 30 deg Solution: ΣMA = 0; −F 1 a cos ( θ ) − F2 ( a + b) cos ( θ ) +

d 2

2

FCB ( a + b) sin ( θ ) +

c +d

F CB

d 2

c +d Ax =

2

FCB ( a + b) cos ( θ ) = 0

F CB = 0

d 2

2

F CB = 782 N

d sin ( θ ) ( a + b) + c cos ( θ ) ( a + b) Ax −

2

c +d

2 2 ⎡⎣F1 a + F2 ( a + b)⎤⎦ cos ( θ ) c + d =

+ Σ F x = 0; →

c

2

Ax = 625 N

F CB

c +d +

↑

Σ F y = 0;

Ay − F1 − F2 +

c 2

c +d Ay = F1 + F2 −

2

F CB = 0

c 2

2

FCB

Ay = 681 N

c +d

Problem 5-49 The boom is intended to support two vertical loads F1 and F 2. If the cable CB can sustain a 381



Chapter 5

maximum load Tmax before it fails, determine the critical loads if F1 = 2F2. Also, what is the magnitude of the maximum reaction at pin A? Units Used: 3

kN = 10 N Given: Tmax = 1500 N a = 1.5 m b = 1m c = 3 d = 4

θ = 30deg Solution: ΣMA = 0;

F1 = 2 F2

−2 F 2 a cos ( θ ) − F2 ( a + b) cos ( θ ) +

d 2

2

Tmax( a + b) sin ( θ ) +

c +d

F2 =

( a + b)Tmax( d sin ( θ ) + c cos ( θ ) ) 2

F1 = 2 F2 + ΣF x = 0; →

Ax −

d 2

c +d Ax =

2

2

Tmax( a + b) cos ( θ ) = 0

F 1 = 1.448 kN

Tmax = 0

d 2

2

c +d

F 2 = 724 N

c + d cos ( θ ) ( 3 a + b) 2

c

2

Ax = 1.20 kN

Tmax

c +d

+

↑Σ Fy = 0;

Ay − F2 − F1 +

c 2

c +d

2

Tmax = 0

382



Chapter 5

c

Ay = F2 + F1 −

2

2

Ay = 1.27 kN

Tmax

c +d

2

FA =

Ax + Ay

2

F A = 1.749 kN

Problem 5-50 The uniform rod of length L and weight W is supported on the smooth planes. Determine its position θ for equilibrium. Neglect the thickness of the rod.

Solution: L

cos ( θ ) + NA cos ( φ − θ ) L = 0

ΣMB = 0;

−W

ΣMA = 0;

W

ΣF x = 0;

NB sin ( ψ) − NA sin ( φ ) = 0

W cos ( θ )

2 cos ( ψ + θ )

L 2

2

NA =

cos ( θ ) − NB cos ( ψ + θ ) L = 0

sin ( ψ) −

W cos ( θ )

2 cos ( φ − θ )

NB =

W cos ( θ )

2 cos ( φ − θ ) W cos ( θ )

2 cos ( ψ + θ )

sin ( φ ) = 0

sin ( ψ) cos ( φ − θ ) − sin ( φ ) cos ( ψ + θ ) = 0 sin ( ψ) ( cos ( φ ) cos ( θ ) + sin ( φ ) sin ( θ ) ) − sin ( φ ) ( cos ( ψ) cos ( θ ) − sin ( ψ) sin ( θ ) ) = 0 2 sin ( ψ) sin ( φ ) sin ( θ ) = ( sin ( φ ) cos ( ψ) − sin ( ψ) cos ( φ ) ) cos ( θ ) tan ( θ ) =

sin ( φ ) cos ( ψ) − sin ( ψ) cos ( φ ) 2 sin ( ψ) sin ( φ )

=

cot ( ψ) − cot ( φ ) 2

⎛ cot ( ψ) − cot ( φ ) ⎞ ⎟ 2 ⎝ ⎠

θ = atan ⎜



Chapter 5

Problem 5-51 The toggle switch consists of a cocking lever that is pinned to a fixed frame at A and held in place by the spring which has unstretched length δ. Determine the magnitude of the resultant force at A and the normal force on the peg at B when the lever is in the position shown. Given:

δ = 200 mm k = 5

N m

a = 100 mm b = 300 mm c = 300 mm

θ = 30 deg

Solution: Using the law of cosines and the law of sines c + ( a + b) − 2 c( a + b) cos ( 180 deg − θ ) 2

l =

sin ( φ ) c

=

2

sin ( 180 deg − θ ) l

Fs = k s = k (l − δ )

ΣMA = 0;

−F s sin ( φ ) ( a + b) + NB a = 0

⎛ sin (180 deg − θ ) ⎞ ⎟ l ⎝ ⎠

φ = asin ⎜c

φ = 12.808 deg

F s = k( l − δ )

F s = 2.3832 N

NB = F s sin ( φ )

a+b a

NB = 2.11 N

ΣF x = 0;

Ax − F s cos ( φ ) = 0

Ax = Fs cos ( φ )

Ax = 2.3239 N

ΣF y = 0;

Ay + NB − Fs sin ( φ ) = 0

Ay = Fs sin ( φ ) − NB

Ay = −1.5850 N

FA =

2

Ax + Ay

2

F A = 2.813 N



Chapter 5

Problem 5-52 The rigid beam of negligible weight is supported horizontally by two springs and a pin. If the springs are uncompressed when the load is removed, determine the force in each spring when the load P is applied. Also, compute the vertical deflection of end C. Assume the spring stiffness k is large enough so that only small deflections occur. Hint: The beam rotates about A so the deflections in the springs can be related. Solution: ΣMA = 0; 3 F B L + FC2 L − P L = 0 2 F B + 2 FC = 1.5 P ΔC = 2 ΔB FC k

=

2 FB k

F C = 2FB 5 F B = 1.5 P F B = 0.3 P F c = 0.6 P ΔC =

0.6 P k

Problem 5-53 The rod supports a weight W and is pinned at its end A. If it is also subjected to a couple moment of M, determine the angle θ for equilibrium.The spring has an unstretched length δ and a stiffness k.

385



Chapter 5

Given: W = 200 lb M = 100 lb ft

δ = 2 ft k = 50

lb ft

a = 3 ft b = 3 ft c = 2 ft Solution: Initial Guess:

θ = 10 deg

Given k⎡⎣( a + b)sin ( θ ) + c − δ⎤⎦ ( a + b) cos ( θ ) − W a cos ( θ ) − M = 0

θ = Find ( θ )

θ = 23.2 deg

Problem 5-54 The smooth pipe rests against the wall at the points of contact A, B, and C. Determine the reactions at these points needed to support the vertical force F . Neglect the pipe's thickness in the calculation. Given: F = 45 lb

386



Chapter 5

θ = 30 deg a = 16 in b = 20 in c = 8 in

Solution: R A = 1 lb

Initial Guesses:

R B = 1 lb

R C = 1 lb

Given F cos ( θ ) ( a + b) − F sin ( θ ) c − RC b + R B c tan ( θ ) = 0

ΣMA = 0;

R C cos ( θ ) − R B cos ( θ ) − F = 0

+

↑Σ Fy = 0;

+ Σ F x = 0; →

R A + RB sin ( θ ) − RC sin ( θ ) = 0

⎛ RA ⎞ ⎜ ⎟ ⎜ RB ⎟ = Find ( RA , RB , RC) ⎜R ⎟ ⎝ C⎠

⎛ RA ⎞ ⎛ 25.981 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ RB ⎟ = ⎜ 11.945 ⎟ lb ⎜ R ⎟ ⎝ 63.907 ⎠ ⎝ C⎠

Problem 5-55 The rigid metal strip of negligible weight is used as part of an electromagnetic switch. If the stiffness of the springs at A and B is k, and the strip is originally horizontal when the springs are unstretched, determine the smallest force needed to close the contact gap at C. Units Used: mN = 10

−3

N

387



Chapter 5

Given: a = 50 mm b = 50 mm c = 10 mm k = 5

N m


F = 0.5 N

yA = 1 mm

yB = 1 mm

Given c − yA a+b

=

yB − yA a

⎛ yA ⎞ ⎜ ⎟ ⎜ yB ⎟ = Find ( yA , yB , F) ⎜F⎟ ⎝ ⎠

k yA + k yB − F = 0

⎛ yA ⎞ ⎛ −2 ⎞ ⎜ ⎟ = ⎜ ⎟ mm ⎝ yB ⎠ ⎝ 4 ⎠

k yB a − F( a + b) = 0

F = 10 mN

Problem 5-56 The rigid metal strip of negligible weight is used as part of an electromagnetic switch. Determine the maximum stiffness k of the springs at A and B so that the contact at C closes when the vertical force developed there is F . Originally the strip is horizontal as shown. Units Used: mN = 10

−3

N

Given: a = 50 mm b = 50 mm c = 10 mm F = 0.5 N

388



Chapter 5

Solution:

Initial Guesses: k = 1

N

yA = 1 mm

m

yB = 1 mm

Given c − yA a+b

=

yB − yA

k yA + k yB − F = 0

a

⎛ yA ⎞ ⎜ ⎟ ⎜ yB ⎟ = Find ( yA , yB , k) ⎜k ⎟ ⎝ ⎠

⎛ yA ⎞ ⎛ −2 ⎞ ⎜ ⎟ = ⎜ ⎟ mm ⎝ yB ⎠ ⎝ 4 ⎠

k yB a − F( a + b) = 0

k = 250

N m

Problem 5-57 Determine the distance d for placement of the load P for equilibrium of the smooth bar in the position θ as shown. Neglect the weight of the bar.

Solution: +

↑Σ Fy = 0; Σ MA = 0;

R cos ( θ ) − P = 0 −P d cos ( θ ) + R R d cos ( θ ) = R 2

d=

a

cos ( θ )

=0

a

cos ( θ )

a cos ( θ )

3

Problem 5-58 The wheelbarrow and its contents have mass m and center of mass at G. Determine the greatest 389



Chapter 5

angle of tilt θ without causing the wheelbarrow to tip over.

g

Solution: Require point G to be over the wheel axle for tipping. Thus b cos ( θ ) = a sin ( θ ) − 1⎛ b ⎞

θ = tan

⎜ ⎟ ⎝ a⎠

Problem 5-59 Determine the force P needed to pull the roller of mass M over the smooth step. Given: M = 50 kg a = 0.6 m b = 0.1 m

θ = 60 deg θ 1 = 20 deg g = 9.81

m 2

s

390



Chapter 5

Solution:

φ = acos ⎛⎜

a − b⎞

⎟ ⎝ a ⎠

φ = 33.56 deg ΣMB = 0,

M g sin ( θ 1 ) ( a − b) + M g cos ( θ 1 ) a sin ( φ ) ... = 0 + P cos ( θ ) ( a − b) − P sin ( θ ) a sin ( φ )

⎡sin ( θ 1 ) ( a − b) + cos ( θ 1 ) a sin ( φ )⎤ ⎥ ⎣ cos ( θ ) ( a − b) + sin ( θ ) a sin ( φ ) ⎦

P = M g⎢

P = 441 N

Problem 5-60 Determine the magnitude and direction θ of the minimum force P needed to pull the roller of mass M over the smooth step. Given: a = 0.6 m b = 0.1 m

θ 1 = 20 deg M = 50 kg g = 9.81

m 2

s Solution:

For Pmin, NA tends to 0

φ = acos ⎛⎜

a − b⎞

⎟ ⎝ a ⎠

φ = 33.56 deg ΣMB = 0 M g sin ( θ 1 ) ( a − b) + M g cos ( θ 1 ) a sin ( φ ) ... = 0 + ⎡⎣−P cos ( θ ) ( a − b)⎤⎦ − P sin ( θ ) a sin ( φ )

391



P=

Chapter 5

M g⎡⎣sin ( θ 1 ) ( a − b) + cos ( θ 1 ) a sin ( φ )⎤⎦ cos ( θ ) ( a − b) + a sin ( φ ) sin ( θ )

For Pmin : dP dθ

=

M g⎡⎣sin ( θ 1 ) ( a − b) + cos ( θ 1 ) a sin ( φ )⎤⎦

⎡⎣cos ( θ ) ( a − b) + a sin ( φ ) sin ( θ )⎤⎦ θ = atan ⎛⎜ sin ( φ ) ⋅

which gives,

P =

⎝

a

2

⎡⎣a sin ( φ ) cos ( θ ) − ( a − b)sin ( θ )⎤⎦ = 0

⎞ ⎟

θ = 33.6 deg

a − b⎠

M g⎡⎣sin ( θ 1 ) ( a − b) + cos ( θ 1 ) a sin ( φ )⎤⎦ cos ( θ ) ( a − b) + a sin ( φ ) sin ( θ )

P = 395 N

Problem 5-61 A uniform glass rod having a length L is placed in the smooth hemispherical bowl having a radius r. Determine the angle of inclination θ for equilibrium.

Solution: By Observation φ = θ. Equilibirium : ΣMA = 0; NB2 r cos ( θ ) − W NB =

L 2

cos ( θ ) = 0

WL 4r

ΣF x = 0; NA cos ( θ ) − W sin ( θ ) = 0

NA = W tan ( θ ) 392



Chapter 5

ΣF y = 0; W tan ( θ ) sin ( θ ) +

WL 4r

− W cos ( θ ) = 0

sin ( θ ) − cos ( θ ) = 1 − 2 cos ( θ ) = 2

2

2 cos ( θ ) − 2

cos ( θ ) =

L 4r

L+

2

−L 4r

cos ( θ )

cos ( θ ) − 1 = 0 2

⎛ L + L2 + 128 r2 ⎞ ⎟ 16 r ⎝ ⎠

2

L + 128 r

θ = acos ⎜

16 r

Problem 5-62 The disk has mass M and is supported on the smooth cylindrical surface by a spring having stiffness k and unstretched length l0. The spring remains in the horizontal position since its end A is attached to the small roller guide which has negligible weight. Determine the angle θ to the nearest degree for equilibrium of the roller. Given: M = 20 kg k = 400

N m

l0 = 1 m r = 2m g = 9.81

m 2

s a = 0.2 m Guesses

F = 10 N

Solution:

R = 10 N θ = 30 deg

Given

+ Σ F y = 0; →

R sin ( θ ) − M g = 0

+

R cos ( θ ) − F = 0

↑Σ Fx = 0;

Spring

F = k⎡⎣( r + a)cos ( θ ) − l0⎤⎦ 393



⎛F⎞ ⎜ R ⎟ = Find ( F , R , θ ) ⎜ ⎟ ⎝θ ⎠

Chapter 5

⎛ F ⎞ ⎛ 163.633 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ R ⎠ ⎝ 255.481 ⎠

θ = 50.171 deg

There is also another answer that we can find by choosing different starting guesses. Guesses

F = 200 N R = 200 N θ = 20 deg

Solution: + Σ F y = 0; → +

↑Σ Fx = 0;

Spring

Given R sin ( θ ) − M g = 0 R cos ( θ ) − F = 0 F = k⎡⎣( r + a)cos ( θ ) − l0⎤⎦

⎛F⎞ ⎜ R ⎟ = Find ( F , R , θ ) ⎜ ⎟ ⎝θ ⎠

⎛ F ⎞ ⎛ 383.372 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ R ⎠ ⎝ 430.66 ⎠

θ = 27.102 deg

Problem 5-63 Determine the x, y, z components of reaction at the fixed wall A.The force F2 is parallel to the z axis and the force F1 is parallel to the y axis. Given: a = 2m

d = 2m

b = 1m

F 1 = 200 N

c = 2.5 m

F 2 = 150 N

394



Chapter 5

Solution: ΣF x = 0;

Ax = 0

ΣF x = 0;

Ay = −F1 Ay = −200 N

ΣF x = 0;

Az = F2 Az = 150 N

ΣΜx = 0;

MAx = −F2 a + F 1 d MAx = 100 N⋅ m

ΣΜy = 0;

MAy = 0

ΣMz = 0;

MAz = F 1 c MAz = 500 N⋅ m

Problem 5-64 The wing of the jet aircraft is subjected to thrust T from its engine and the resultant lift force L. If the mass of the wing is M and the mass center is at G, determine the x, y, z components of reaction where the wing is fixed to the fuselage at A. Units Used: 3

Mg = 10 kg 3

kN = 10 N g = 9.81

m 2

s

395



Chapter 5

Given: T = 8 kN L = 45 kN M = 2.1 Mg a = 2.5 m b = 5m 395

c = 3m d = 7m Solution: ΣF x = 0;

− Ax + T = 0 Ax = T

Ax = 8 kN

ΣF y = 0;

Ay = 0

Ay = 0

ΣF z = 0;

− Az − M g + L = 0 Az = L − M g

ΣMy = 0;

Az = 24.4 kN

M y − T ( a) = 0 My = T a

ΣMx = 0;

My = 20.0 kN⋅ m

L( b + c + d) − M g b − Mx = 0 M x = L ( b + c + d) − M g b

ΣMz = 0;

Mx = 572 kN⋅ m

Mz − T( b + c) = 0 Mz = T( b + c)

Mz = 64.0 kN⋅ m

Problem 5-65 The uniform concrete slab has weight W. Determine the tension in each of the three parallel supporting cables when the slab is held in the horizontal plane as shown.

396



Chapter 5

Units Used: 3

kip = 10 lb Given: W = 5500 lb a = 6 ft b = 3 ft c = 3 ft d = 6 ft

Solution: Equations of Equilibrium : The cable tension TB can be obtained directly by summing moments about the y axis.

ΣMy = 0;

ΣMx = 0;

W

d 2

− TB d = 0

TC a + TB( a + b) − W⎛⎜

⎝

TC =

ΣF z = 0;

TB =

W 2

TB = 2.75 kip

a + b + c⎞

⎟=0 ⎠

2

1⎡ a + b + c − TB( a + b)⎥⎤ ⎢W a⎣ 2 ⎦

TA + TB + TC − W = 0

TA = − TB − TC + W

TC = 1.375 kip

TA = 1.375 kip

397



Chapter 5

Problem 5-66 The air-conditioning unit is hoisted to the roof of a building using the three cables. If the tensions in the cables are TA, TB and TC, determine the weight of the unit and the location (x, y) of its center of gravity G. Given: TA = 250 lb TB = 300 lb TC = 200 lb a = 5 ft b = 4 ft c = 3 ft d = 7 ft e = 6 ft Solution: ΣF z = 0; TA + TB + TC − W = 0 W = TA + TB + TC ΣMy = 0;

W x − T A ( c + d) − T C d = 0 x =

ΣMx = 0;

W = 750 lb

TA( c + d) + TC d W

x = 5.2 ft

TA a + TB( a + b − e) + TC( a + b) − W y = 0

y =

TA a + TB( a + b − e) + TC( a + b) W

y = 5.267 ft

Problem 5-67 The platform truck supports the three loadings shown. Determine the normal reactions on each of its three wheels. 398



Chapter 5

Given: F 1 = 380 lb F 2 = 500 lb F 3 = 800 lb a = 8 in

b = 12 in c = 10 in d = 5 in e = 12 in f = 12 in

Solution: The initail guesses are

F A = 1 lb

F B = 1 lb

F C = 1 lb

Given ΣMx = 0;

F 1 ( c + d) + F2 ( b + c + d) + F3 d − F A( a + b + c + d) = 0

ΣMy = 0;

F1 e − FB e − F2 f + FC f = 0

ΣF y = 0;

FB + FC − F2 + FA − F1 − F3 = 0

⎛ FA ⎞ ⎜ ⎟ ⎜ FB ⎟ = Find ( FA , FB , FC) ⎜F ⎟ ⎝ C⎠

⎛ FA ⎞ ⎛ 663 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FB ⎟ = ⎜ 449 ⎟ lb ⎜ F ⎟ ⎝ 569 ⎠ ⎝ C⎠

Problem 5-68 Due to an unequal distribution of fuel in the wing tanks, the centers of gravity for the airplane fuselage A and wings B and C are located as shown. If these components have weights WA, WB and WC, determine the normal reactions of the wheels D, E, and F on the ground.

399



Chapter 5

Units Used: 3

kip = 10 lb Given: WA = 45000 lb WB = 8000 lb WC = 6000 lb a = 8 ft

e = 20 ft

b = 6 ft

f = 4 ft

c = 8 ft

g = 3 ft

d = 6 ft Solution: Initial guesses: R D = 1 kip

R E = 1 kip

R F = 1 kip

Given ΣMx = 0;

WB b − RD( a + b) − WC c + R E( c + d) = 0

ΣMy = 0;

W B f + W A( g + f ) + W C f − R F( e + g + f ) = 0

ΣF z = 0;

R D + R E + RF − WA − WB − WC = 0

⎛ RD ⎞ ⎜ ⎟ ⎜ RE ⎟ = Find ( RD , RE , RF) ⎜R ⎟ ⎝ F⎠

⎛ RD ⎞ ⎛ 22.6 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ RE ⎟ = ⎜ 22.6 ⎟ kip ⎜ R ⎟ ⎝ 13.7 ⎠ ⎝ F⎠

Problem 5-69 If the cable can be subjected to a maximum tension T, determine the maximum force F which may be applied to the plate. Compute the x, y, z components of reaction at the hinge A for this loading.

400



Chapter 5

Given: a = 3 ft b = 2 ft c = 1 ft d = 3 ft e = 9 ft T = 300 lb Solution: Initial guesses: F = 10 lb

MAx = 10 lb ft

MAz = 10 lb ft

Ax = 10 lb

Ay = 10 lb

Az = 10 lb

Given

⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ + ⎜ 0 ⎟ = 0 ⎜ A ⎟ ⎝T − F⎠ ⎝ z⎠

⎛ MAx ⎞ ⎛ a ⎞ ⎛ 0 ⎞ ⎛ e ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ + ⎜ −c ⎟ × ⎜ 0 ⎟ + ⎜ −b − c ⎟ × ⎜ 0 ⎟ = 0 ⎜ MAz ⎟ ⎝ 0 ⎠ ⎝ −F ⎠ ⎝ 0 ⎠ ⎝ T ⎠ ⎝ ⎠

⎛ F ⎞ ⎜ A ⎟ ⎜ x ⎟ ⎜ Ay ⎟ ⎜ ⎟ = Find ( F , Ax , Ay , Az , MAx , MAz) Az ⎜ ⎟ ⎜ MAx ⎟ ⎟ ⎜ ⎝ MAz ⎠

⎛ MAx ⎞ ⎛ 0 ⎞ ⎜ ⎟ = ⎜ ⎟ lb ft ⎝ MAz ⎠ ⎝ 0 ⎠ ⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 0 ⎟ lb ⎜ A ⎟ ⎝ 600 ⎠ ⎝ z⎠

F = 900 lb

Problem 5-70 The boom AB is held in equilibrium by a ball-and-socket joint A and a pulley and cord system as shown. Determine the x, y, z components of reaction at A and the tension in cable DEC.

401



Chapter 5

Given:

F

⎛ 0 ⎞ = ⎜ 0 ⎟ lb ⎜ ⎟ ⎝ −1500 ⎠

a = 5 ft b = 4 ft c = b d = 5 ft e = 5 ft f = 2 ft Solution:

α = atan ⎛⎜

a

⎞ ⎟

⎝ d + e⎠

L =

2

a + ( d + e)

β = atan ⎛⎜

2

⎞ ⎟ fL ⎜L − ⎟ d+ e⎠ ⎝

Guesses

b

TBE = 1 lb TDEC = 1 lb

Ax = 1 lb Ay = 1 lb Given

Az = 1 lb

2 TDEC cos ( β ) = TBE

0 ⎞ ⎛0⎞ ⎛ 0 ⎞ ⎛⎜ ⎜ d ⎟ × F + ⎜ d + e ⎟ × −TBE cos ( α ) ⎟ = 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ ⎝0⎠ ⎝ 0 ⎠ ⎝ TBE sin ( α ) ⎟⎠

⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ + F + TBE⎜ −cos ( α ) ⎟ = 0 ⎜A ⎟ ⎝ sin ( α ) ⎠ ⎝ z⎠

402



Chapter 5

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ Az ⎟ = Find ( A , A , A , T , T x y z BE DEC) TDEC = 919 lb ⎜ ⎟ ⎜ TBE ⎟ ⎜T ⎟ ⎝ DEC ⎠

⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 1.5 × 103 ⎟ lb ⎜A ⎟ ⎜ ⎟ ⎝ z ⎠ ⎝ 750 ⎠

Problem 5-71 The cable CED can sustain a maximum tension Tmax before it fails. Determine the greatest vertical force F that can be applied to the boom. Also, what are the x, y, z components of reaction at the ball-and-socket joint A? Given: Tmax = 800 lb a = 5 ft b = 4 ft c = b d = 5 ft e = 5 ft f = 2 ft Solution: a ⎞ ⎟ ⎝ d + e⎠

α = atan ⎛⎜ L =

2

a + ( d + e)

β = atan ⎛⎜

2

⎞ fL ⎟ ⎜L − ⎟ d+ e⎠ ⎝

Guesses

b

TBE = 1 lb

Ax = 1 lb Ay = 1 lb Given

TDEC = Tmax

F = 1 lb Az = 1 lb

2 TDEC cos ( β ) = TBE

403



Chapter 5

0 ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎛⎜ ⎜ d ⎟ × ⎜ 0 ⎟ + ⎜ d + e ⎟ × −TBE cos ( α ) ⎟ = 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ ⎝ 0 ⎠ ⎝ −F ⎠ ⎝ 0 ⎠ ⎝ TBE sin ( α ) ⎟⎠

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ A ⎟ = Find ( A , A , A , T , F) x y z BE ⎜ z ⎟ ⎜ TBE ⎟ ⎜ ⎟ ⎝ F ⎠

⎛ Ax ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ + ⎜ 0 ⎟ + TBE⎜ −cos ( α ) ⎟ = 0 ⎜ A ⎟ ⎝ −F ⎠ ⎝ sin ( α ) ⎠ ⎝ z⎠

F = 1306 lb

0 ⎛ Ax ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 1.306 × 103 ⎟ lb ⎜A ⎟ ⎜ ⎟ ⎝ z ⎠ ⎝ 653.197 ⎠

Problem 5-72 The uniform table has a weight W and is supported by the framework shown. Determine the smallest vertical force P that can be applied to its surface that will cause it to tip over. Where should this force be applied? Given: W = 20 lb a = 3.5 ft b = 2.5 ft c = 3 ft e = 1.5 ft f = 1 ft

Solution: f θ = atan ⎛⎜ ⎟⎞

θ = 33.69 deg

d = e sin ( θ )

d = 0.832 ft

⎛ a −e⎞ ⎜ 2 ⎟ φ = atan ⎜ ⎟ ⎜ b ⎟ ⎝ 2 ⎠

φ = 11.31 deg

⎝ e⎠

404



2

d' =

⎛ a − e⎞ + ⎛ b ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2⎠

Chapter 5

2

d' = 1.275 ft

Tipping will occur about the g - g axis. Require P to be applied at the corner of the table for Pmin. W d = P d' sin ( 90 deg − φ + θ ) P = W

d

P = 14.1 lb

d' sin ( 90 deg − φ + θ )

Problem 5-73 The windlass is subjected to load W. Determine the horizontal force P needed to hold the handle in the position shown, and the components of reaction at the ball-and-socket joint A and the smooth journal bearing B. The bearing at B is in proper alignment and exerts only force reactions perpendicular to the shaft on the windlass. Given: W = 150 lb a = 2 ft b = 2 ft c = 1 ft d = 1 ft e = 1 ft f = 0.5 ft Solution: ΣMy = 0;

W f−P d=0 P =

W f d

ΣF y = 0;

Ay = 0 lb

ΣMx = 0;

−W a + B z( a + b) = 0

P = 75 lb Ay = 0

405



Bz = ΣF z = 0;

Chapter 5

Wa

B z = 75 lb

a+b

Az + Bz − W = 0 Az = W − Bz

ΣMz = 0;

B x( a + b) − ( a + b + c + e)P = 0 Bx =

ΣF x = 0;

Az = 75 lb

P ( a + b + c + e) a+b

B x = 112 lb

Ax − Bx + P = 0 Ax = Bx − P

Ax = 37.5 lb

Problem 5-74 A ball of mass M rests between the grooves A and B of the incline and against a vertical wall at C. If all three surfaces of contact are smooth, determine the reactions of the surfaces on the ball. Hint: Use the x, y, z axes, with origin at the center of the ball, and the z axis inclined as shown. Given: M = 2 kg

θ 1 = 10 deg θ 2 = 45 deg

Solution: ΣF x = 0; F c cos ( θ 1 ) − M g sin ( θ 1 ) = 0 F c = M g⋅ tan ( θ 1 ) F c = 0.32 kg⋅ m

406



Chapter 5

ΣF y = 0; NA cos ( θ 2 ) − NB cos ( θ 2 ) = 0 NA = NB ΣF z = 0; 2 NA sin ( θ 2 ) − M g cos ( θ 1 ) − F c sin ( θ 1 ) = 0

NA =

1 M g⋅ cos ( θ 1 ) + F c⋅ sin ( θ 1 ) ⋅ 2 sin ( θ 2 )

NA = 1.3 kg⋅ m N = NA = NB

Problem 5-75 Member AB is supported by cable BC and at A by a square rod which fits loosely through the square hole at the end joint of the member as shown. Determine the components of reaction at A and the tension in the cable needed to hold the cylinder of weight W in equilibrium. Units Used: 3

kip = 10 lb Given: W = 800 lb a = 2 ft b = 6 ft c = 3 ft Solution:

⎛

⎞=0 ⎝ c +b +a ⎠

ΣF x = 0

F BC ⎜

ΣF y = 0

Ay = 0

c

2

2

2⎟

Ay = 0lb

F BC = 0 lb

Ay = 0 lb

407



Chapter 5

ΣF z = 0

Az − W = 0

ΣMx = 0

MAx − W b = 0

ΣMy = 0 ΣMz = 0

Az = W

Az = 800 lb

MAx = W b

MAx = 4.80 kip⋅ ft

MAy = 0 lb ft

MAy = 0 lb⋅ ft

MAz = 0 lb ft

MAz = 0 lb⋅ ft

Problem 5-76 The pipe assembly supports the vertical loads shown. Determine the components of reaction at the ball-and-socket joint A and the tension in the supporting cables BC and BD. Units Used: 3

kN = 10 N Given: F 1 = 3 kN d = 2 m F 2 = 4 kN e = 1.5 m a = 1m

g = 1m

b = 1.5 m

h = 3m

c = 3m

i = 2m

f = c−e

j = 2m

Solution: The initial guesses are: TBD = 1 kN

TBC = 1 kN

Ax = 1 kN

Ay = 1 kN

Az = 1 kN

The vectors

408



⎛ 0 r1 = ⎜ a + b + ⎜ ⎝ d

rBC

⎞ f⎟ ⎟ ⎠

⎛ i ⎞ = ⎜ −a ⎟ ⎜ ⎟ ⎝h − g⎠

uBC =

Given

rBC rBC

Chapter 5

⎛ 0 ⎞ r2 = ⎜ a + b + c ⎟ ⎜ ⎟ ⎝ d ⎠

rBD

⎛ −j ⎞ = ⎜ −a ⎟ ⎜ ⎟ ⎝h − g⎠

uBD =

rBD rBD

rAB

⎛0⎞ = ⎜a⎟ ⎜ ⎟ ⎝g⎠

⎛1⎞ i = ⎜0⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ j = ⎜1⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ k = ⎜0⎟ ⎜ ⎟ ⎝1⎠

Ax i + Ay j + Az k − F1 k − F2 k + TBDuBD + TBC uBC = 0 rAB × ( TBDuBD + TBCuBC) + r1 × ( −F 1 k) + r2 × ( −F 2 k) = 0

⎛ TBD ⎞ ⎜ ⎟ ⎜ TBC ⎟ ⎜ Ax ⎟ = Find ( T , T , A , A , A ) BD BC x y z ⎜ ⎟ ⎜ Ay ⎟ ⎜ A ⎟ ⎝ z ⎠

⎛ TBD ⎞ ⎛ 17 ⎞ ⎜ ⎟ = ⎜ ⎟ kN ⎝ TBC ⎠ ⎝ 17 ⎠ ⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 11.333 ⎟ kN ⎜ A ⎟ ⎝ −15.667 ⎠ ⎝ z⎠

Problem 5-77 The hatch door has a weight W and center of gravity at G. If the force F applied to the handle at C has coordinate direction angles of α, β and γ, determine the magnitude of F needed to hold the door slightly open as shown. The hinges are in proper alignment and exert only force reactions on the door. Determine the components of these reactions if A exerts only x and z components of force and B exerts x, y, z force components. Given: W = 80 lb

α = 60 deg

409



Chapter 5

β = 45 deg γ = 60 deg a = 3 ft b = 2 ft c = 4 ft d = 3 ft Solution: Initial Guesses: Ax = 1 lb

Az = 1 lb

F = 1 lb

B x = 1 lb

B y = 1 lb

B z = 1 lb

Given

⎛ Ax ⎞ ⎛ Bx ⎞ ⎛⎜ cos ( α ) ⎟⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ + ⎜ By ⎟ + F⎜ cos ( β ) ⎟ + ⎜ 0 ⎟ = 0 ⎜ Az ⎟ ⎜ B ⎟ ⎜ cos ( γ ) ⎟ ⎝ −W ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ z⎠ ⎛ a + b ⎞ ⎢⎡ ⎛⎜ cos ( α ) ⎟⎞⎥⎤ ⎛ a ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎛⎜ Bx ⎟⎞ ⎜ 0 ⎟ × F cos ( β ) + ⎜ c ⎟ × ⎜ 0 ⎟ + ⎜ c + d ⎟ × B = 0 ⎟⎥ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ ⎟ ⎢ ⎜ ⎟ ⎜ y⎟ ⎢ ⎜ ⎟ ⎥ ⎝ 0 ⎠ ⎣ ⎝ cos ( γ ) ⎠⎦ ⎝ 0 ⎠ ⎝ −W ⎠ ⎝ 0 ⎠ ⎜⎝ Bz ⎟⎠ ⎛ Ax ⎞ ⎜ ⎟ ⎜ Az ⎟ ⎜ Bx ⎟ ⎜ ⎟ = Find ( Ax , Az , Bx , By , Bz , F) ⎜ By ⎟ ⎜ Bz ⎟ ⎜ ⎟ ⎝F ⎠

⎛ Ax ⎞ ⎛ −96.5 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ Az ⎠ ⎝ −13.7 ⎠ ⎛ Bx ⎞ ⎛ 48.5 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ By ⎟ = ⎜ −67.9 ⎟ lb ⎜ B ⎟ ⎝ 45.7 ⎠ ⎝ z⎠ F = 96 lb

410



Chapter 5

Problem 5-78 The hatch door has a weight W and center of gravity at G. If the force F applied to the handle at C has coordinate direction angles α, β, γ determine the magnitude of F needed to hold the door slightly open as shown. If the hinge at A becomes loose from its attachment and is ineffective, what are the x, y, z components of reaction at hinge B? Given: W = 80 lb

α = 60 deg β = 45 deg γ = 60 deg a = 3 ft b = 2 ft c = 4 ft d = 3 ft Solution: a

ΣMy = 0;

F = W

ΣF x = 0;

B x + F cos ( α ) = 0

cos ( γ ) ( a + b)

B x = −F cos ( α ) ΣF y = 0;

B y = −67.9 lb

B z − W + F cos ( γ ) = 0 B z = W − F cos ( γ )

ΣMx = 0;

B x = −48 lb

B y + F cos ( β ) = 0 B y = −F cos ( β )

ΣF z = 0;

F = 96 lb

B z = 32 lb

MBx + W d − F cos ( γ ) ( c + d) = 0 MBx = −W d + F cos ( γ ) ( c + d)

ΣMz = 0;

MBx = 96 lb⋅ ft

MBz + F cos ( α ) ( c + d) + F cos ( β ) ( a + b) = 0 MBz = −F cos ( α ) ( c + d) − F cos ( β ) ( a + b)

MBz = −675 lb⋅ ft

411



Chapter 5

Problem 5-79 The bent rod is supported at A, B, and C by smooth journal bearings. Compute the x, y, z components of reaction at the bearings if the rod is subjected to forces F 1 and F 2. F1 lies in the y-z plane. The bearings are in proper alignment and exert only force reactions on the rod. Given: F 1 = 300 lb

d = 3 ft

F 2 = 250 lb

e = 5 ft

a = 1 ft

α = 30 deg

b = 4 ft

β = 45 deg

c = 2 ft

θ = 45 deg

Solution: The initial guesses: Ax = 100 lb

Ay = 200 lb

B x = 300 lb

B z = 400 lb

Cy = 500 lb

Cz = 600 lb

Given Ax + B x + F2 cos ( β ) sin ( α ) = 0 Ay + Cy − F1 cos ( θ ) + F2 cos ( β ) cos ( α ) = 0 B z + Cz − F 1 sin ( θ ) − F2 sin ( β ) = 0 F 1 cos ( θ ) ( a + b) + F 1 sin ( θ ) ( c + d) − B z d − A y b = 0 Ax b + Cz e = 0 Ax( c + d) + Bx d − Cy e = 0

412



Chapter 5

⎛⎜ Ax ⎞⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎜ Bx ⎟ = Find ( A , A , B , B , C , C ) x y x z y z ⎜ Bz ⎟ ⎜ ⎟ ⎜ Cy ⎟ ⎜ Cz ⎟ ⎝ ⎠

⎛ Ax ⎞ ⎛ 632.883 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ Ay ⎠ ⎝ −141.081 ⎠ ⎛ Bx ⎞ ⎛ −721.271 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ Bz ⎠ ⎝ 895.215 ⎠ ⎛ Cy ⎞ ⎛ 200.12 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ Cz ⎠ ⎝ −506.306 ⎠

Problem 5-80 The bent rod is supported at A, B, and C by smooth journal bearings. Determine the magnitude of F2 which will cause the reaction Cy at the bearing C to be equal to zero. The bearings are in proper alignment and exert only force reactions on the rod. Given: F 1 = 300 lb

d = 3 ft

Cy = 0 lb

e = 5 ft

a = 1 ft

α = 30 deg

b = 4 ft

β = 45 deg

c = 2 ft

θ = 45 deg

Solution: The initial guesses: Ax = 100 lb

Ay = 200 lb

B x = 300 lb

B z = 400 lb

F 2 = 500 lb

Cz = 600 lb

Given Ax + B x + F2 cos ( β ) sin ( α ) = 0

413



Chapter 5

Ay + Cy − F1 cos ( θ ) + F2 cos ( β ) cos ( α ) = 0 B z + Cz − F 1 sin ( θ ) − F2 sin ( β ) = 0 F 1 cos ( θ ) ( a + b) + F 1 sin ( θ ) ( c + d) − B z d − A y b = 0 Ax b + Cz e = 0 Ax( c + d) + Bx d − Cy e = 0

⎛⎜ Ax ⎞⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎜ Bx ⎟ = Find ( A , A , B , B , C , F ) x y x z z 2 ⎜ Bz ⎟ ⎜ ⎟ ⎜ Cz ⎟ ⎜ F2 ⎟ ⎝ ⎠

F 2 = 673.704 lb

Problem 5-81 Determine the tension in cables BD and CD and the x, y, z components of reaction at the ball-and-socket joint at A. Given: F = 300 N a = 3m b = 1m c = 0.5 m d = 1.5 m Solution:

rBD

⎛ −b ⎞ ⎜ ⎟ = d ⎜ ⎟ ⎝a⎠

414



rCD

Chapter 5

⎛ −b ⎞ = ⎜ −d ⎟ ⎜ ⎟ ⎝a⎠

Initial Guesses:

TBD = 1 N

TCD = 1 N

Ax = 1 N

Ay = 1 N

Az = 1 N

Given

⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ rBD rCD + TCD +⎜ 0 ⎟ =0 ⎜ Ay ⎟ + TBD ⎜ ⎟ rBD rCD ⎜A ⎟ ⎝ −F ⎠ ⎝ z⎠ ⎛d⎞ ⎛d⎞ ⎛d − c⎞ ⎛ 0 ⎞ rBD ⎞ ⎜ ⎟ ⎛ rCD ⎞ ⎜ ⎜ −d ⎟ × ⎛ T + d × ⎜ TCD + 0 ⎟×⎜ 0 ⎟ = 0 ⎜ ⎟ ⎟ BD ⎜ ⎟ ⎝ ⎟ ⎜ ⎟ rBD ⎠ ⎜ ⎟ ⎝ rCD ⎠ ⎜ ⎝0⎠ ⎝0⎠ ⎝ 0 ⎠ ⎝ −F ⎠ ⎛ TBD ⎞ ⎜ ⎟ ⎜ TCD ⎟ ⎜ Ax ⎟ = Find ( T , T , A , A , A ) BD CD x y z ⎜ ⎟ ⎜ Ay ⎟ ⎜ A ⎟ ⎝ z ⎠

⎛ TBD ⎞ ⎛ 116.7 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ TCD ⎠ ⎝ 116.7 ⎠ ⎛ Ax ⎞ ⎛ 66.7 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 0 ⎟ N ⎜ A ⎟ ⎝ 100 ⎠ ⎝ z⎠

Problem 5-82 Determine the tensions in the cables and the components of reaction acting on the smooth collar at A necessary to hold the sign of weight W in equilibrium. The center of gravity for the sign is at G. Given: W = 50 lb

f = 2.5 ft

a = 4 ft

g = 1 ft

b = 3 ft

h = 1 ft

415



Chapter 5

c = 2 ft

i = 2 ft

d = 2 ft

j = 2 ft

e = 2.5 ft

k = 3 ft

Solution: The initial guesses are: TBC = 10 lb

Ax = 10 lb

MAx = 10 lb⋅ ft

TDE = 10 lb

Ay = 10 lb

MAy = 10 lb⋅ ft

Given TDE

( a − k)

2

2

( a − k) + i + j −i

TDE 2

( a − k) + i + j

j

TBC 2

2

2

2

TBC

2

2

2

+ Ax = 0

+ Ay = 0

−W=0 2

2

( d − b) + i + c

i 2

2

( a − k) + i + j

MAy − TDE k

2

( d − b) + i + c

+c

( a − k) + i + j

MAx + TDE j

2

( d − b) + i + c

2

TDE 2

2

−i 2

TBC

+ ( −b + d)

2

j 2

2

( a − k) + i + j

2

i

+ c TBC

2

− Wi = 0 2

2

2

2

( d − b) + i + c

d

+ TBC c

2

+ W ( k − f) = 0

( d − b) + i + c

416



Chapter 5

i

−TDE a

2

2

( a − k) + i + j

2

+ TBC b

i 2

=0 2

2

( d − b) + i + c

⎛⎜ MAx ⎞⎟ ⎜ MAy ⎟ ⎜ ⎟ ⎜ TBC ⎟ = Find ( M , M , T , T , A , A ) Ax Ay BC DE x y ⎜ TDE ⎟ ⎜ ⎟ ⎜ Ax ⎟ ⎜ Ay ⎟ ⎝ ⎠

⎛ TBC ⎞ ⎛ 42.857 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ TDE ⎠ ⎝ 32.143 ⎠ ⎛ Ax ⎞ ⎛ 3.571 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ Ay ⎠ ⎝ 50 ⎠ ⎛ MAx ⎞ ⎛ 2.698 × 10− 13 ⎞ ⎜ ⎟=⎜ ⎟ lb⋅ ft ⎝ MAy ⎠ ⎝ −17.857 ⎠

Problem 5-83 The member is supported by a pin at A and a cable BC. If the load at D is W, determine the x, y, z components of reaction at these supports. Units Used: 3

kip = 10 lb Given: W = 300 lb a = 1 ft b = 2 ft c = 6 ft d = 2 ft e = 2 ft f = 2 ft

417



Chapter 5

Solution: Initial Guesses: TBC = 1 lb

Ax = 1 lb

Ay = 1 lb

Az = 1 lb

MAy = 1 lb ft

MAz = 1 lb ft

Given

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ + ⎜A ⎟ ⎝ z⎠

⎛e + f − a⎞ ⎛ 0 ⎞ ⎜ −c ⎟ + ⎜ 0 ⎟ = 0 ⎟ ⎜ ⎟ 2 2 2⎜ b + c + ( e + f − a) ⎝ b ⎠ ⎝ −W ⎠ TBC

⎛ 0 ⎞ ⎛ −e ⎞ ⎛ 0 ⎞ ⎛ −a ⎞ ⎡ ⎛ e + f − a ⎞⎤ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎢ TBC ⎜ −c ⎟⎥ = 0 ⎜ MAy ⎟ + ⎜ c ⎟ × ⎜ 0 ⎟ + ⎜ 0 ⎟ × ⎢ ⎜ ⎟⎥ ⎜ MAz ⎟ ⎝ 0 ⎠ ⎝ −W ⎠ ⎝ b ⎠ ⎣ b2 + c2 + ( e + f − a) 2 ⎝ b ⎠⎦ ⎝ ⎠ ⎛⎜ TBC ⎞⎟ ⎜ Ax ⎟ ⎜ ⎟ ⎜ Ay ⎟ = Find ( T , A , A , A , M , M ) BC x y z Ay Az ⎜ Az ⎟ ⎜ ⎟ ⎜ MAy ⎟ ⎜ MAz ⎟ ⎝ ⎠

TBC = 1.05 kip

⎛ Ax ⎞ ⎛ −450 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 900 ⎟ lb ⎜A ⎟ ⎝ 0 ⎠ ⎝ z⎠ ⎛ MAy ⎞ ⎛ −600 ⎞ ⎜ ⎟=⎜ ⎟ lb⋅ ft ⎝ MAz ⎠ ⎝ −900 ⎠

Problem 5-84 Determine the x, y, z components of reaction at the pin A and the tension in the cable BC necessary for equilibrium of the rod.

418



Chapter 5

Given: F = 350 lb

e = 12 ft

a = 4 ft

f = 4 ft

b = 5 ft

g = 10 ft

c = 4 ft

h = 4 ft

d = 2 ft

i = 10 ft

Solution: Initial Guesses: F BC = 1 lb Ay = 1 lb MAy = 1 lb⋅ ft Ax = 1 lb Az = 1 lb MAz = 1 lb⋅ ft Given

⎛ 0 ⎞ ⎛d⎞ ⎡ ⎛ g − d ⎞⎤ ⎜ ⎟ ⎜ ⎟ ⎢ F ⎜ e − c ⎟⎥ ... = 0 ⎜ MAy ⎟ + ⎜ c ⎟ × ⎢ ⎟⎥ 2 2 2⎜ ⎜ MAz ⎟ ⎝ 0 ⎠ ⎣ ( g − d) + ( e − c) + f ⎝ − f ⎠⎦ ⎝ ⎠ ⎛ −a ⎞ ⎡ ⎛ a − h ⎞⎤ F BC ⎜ ⎟ ⎢ ⎜ −e ⎟⎥ + 0 × ⎜ ⎟ ⎢ ⎟⎥ 2 2 2⎜ ⎝ b ⎠ ⎣ ( a − h) + e + b ⎝ b ⎠⎦ ⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ + ⎜A ⎟ ⎝ z⎠

⎛g − d⎞ ⎜e− c⎟ + ⎟ 2 2 2⎜ ( g − d) + ( e − c) + f ⎝ − f ⎠ F

⎛a − h⎞ ⎜ −e ⎟ = 0 ⎟ 2 2 2⎜ ( a − h) + e + b ⎝ b ⎠ F BC

419



Chapter 5

⎛⎜ FBC ⎞⎟ ⎜ Ax ⎟ ⎜ ⎟ ⎜ Ay ⎟ = Find ( F , A , A , A , M , M ) BC x y z Ay Az ⎜ Az ⎟ ⎜ ⎟ ⎜ MAy ⎟ ⎜ MAz ⎟ ⎝ ⎠

F BC = 101 lb

⎛ Ax ⎞ ⎛ −233.3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ −140 ⎟ lb ⎜ A ⎟ ⎝ 77.8 ⎠ ⎝ z⎠ ⎛ MAy ⎞ ⎛ −388.9 ⎞ ⎜ ⎟=⎜ ⎟ lb⋅ ft ⎝ MAz ⎠ ⎝ 93.3 ⎠

Problem 5-85 Rod AB is supported by a ball-and-socket joint at A and a cable at B. Determine the x, y, z components of reaction at these supports if the rod is subjected to a vertical force F as shown. Given: F = 50 lb a = 2 ft

c = 2 ft

b = 4 ft

d = 2 ft

Solution: TB = 10 lb

Ax = 10 lb

Ay = 10 lb

Az = 10 lb

B y = 10 lb Given ΣF x = 0;

− TB + A x = 0

ΣF y = 0;

Ay + By = 0

420



ΣF z = 0;

Chapter 5

−F + A z = 0

ΣMAx = 0;

F ( c) − By( b) = 0

ΣMAy = 0;

F ( a) − TB( b) = 0

ΣMAz = 0;

B y( a) − TB( c) = 0

Solving,

⎛ TB ⎞ ⎜ ⎟ ⎜ Ax ⎟ ⎜ Ay ⎟ = Find ( T , A , A , A , B ) B x y z y ⎜ ⎟ ⎜ Az ⎟ ⎜B ⎟ ⎝ y⎠

⎛ TB ⎞ ⎛ 25 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ax ⎟ ⎜ 25 ⎟ ⎜ Ay ⎟ = ⎜ −25 ⎟ lb ⎜ ⎟ ⎜ ⎟ ⎜ Az ⎟ ⎜ 50 ⎟ ⎜ B ⎟ ⎝ 25 ⎠ ⎝ y⎠

Problem 5-86 The member is supported by a square rod which fits loosely through a smooth square hole of the attached collar at A and by a roller at B. Determine the x, y, z components of reaction at these supports when the member is subjected to the loading shown. Given: M = 50 lb⋅ ft

⎛ 20 ⎞ ⎜ ⎟ F = −40 lb ⎜ ⎟ ⎝ −30 ⎠ a = 2 ft b = 1 ft c = 2 ft Solution: Initial Guesses Ax = 1 lb

Ay = 1 lb

421



Chapter 5

MAx = 1 lb ft

MAy = 1 lb ft

MAz = 1 lb ft

B z = 1 lb

Given

⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ + ⎜ 0 ⎟ + F = 0 ⎜ 0 ⎟ ⎜ Bz ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ MAx ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎡⎛ 0 ⎞ ⎛ 0 ⎞⎤ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎢⎜ ⎟ ⎜ ⎟⎥ ⎜ MAy ⎟ + ⎜ a ⎟ × ⎜ 0 ⎟ + ⎢⎜ a + b ⎟ × F + ⎜ 0 ⎟⎥ = 0 ⎜ M ⎟ ⎝ 0 ⎠ ⎜ Bz ⎟ ⎣⎝ −c ⎠ ⎝ −M ⎠⎦ ⎝ ⎠ ⎝ Az ⎠ ⎛⎜ Ax ⎞⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎜ MAx ⎟ = Find ( A , A , M , M , M , B ) x y Ax Ay Az z ⎜ MAy ⎟ ⎜ ⎟ ⎜ MAz ⎟ ⎜ Bz ⎟ ⎝ ⎠

⎛ Ax ⎞ ⎛ −20 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ Ay ⎠ ⎝ 40 ⎠ B z = 30 lb

⎛ MAx ⎞ ⎛ 110 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ MAy ⎟ = ⎜ 40 ⎟ lb⋅ ft ⎜ M ⎟ ⎝ 110 ⎠ ⎝ Az ⎠

Problem 5-87 The platform has mass M and center of mass located at G. If it is lifted using the three cables, determine the force in each of these cables. Units Used: 3

Mg = 10 kg

3

kN = 10 N

g = 9.81

m 2

s

422



Chapter 5

Given: M = 3 Mg a = 4m b = 3m c = 3m d = 4m e = 2m Solution: The initial guesses are: F AC = 10 N

F BC = 10 N

F DE = 10 N

Given b ( F AC) 2

2

a +b

−

c ( F BC) 2

=0

2

a +c

M g e − ( F AC) a

d+e 2

2

− FBC

a +b a 2

2

a( d + e) 2

=0

2

a +c

F BC( b + c) − M g b + FDE b = 0

a +c

⎛ FAC ⎞ ⎜ ⎟ F ⎜ BC ⎟ = Find ( FAC , FBC , FDE) ⎜F ⎟ ⎝ DE ⎠

⎛ FAC ⎞ ⎜ ⎟ F ⎜ BC ⎟ = kN ⎜F ⎟ ⎝ DE ⎠

423



Chapter 5

Problem 5-88 The platform has a mass of M and center of mass located at G. If it is lifted using the three cables, determine the force in each of the cables. Solve for each force by using a single moment equation of equilibrium. Units Used: Mg = 1000 kg 3

kN = 10 N g = 9.81

m 2

s Given: M = 2 Mg

c = 3m

a = 4m

d = 4m

b = 3m

e = 2m

Solution:

rBC

⎛0⎞ = ⎜ −c ⎟ ⎜ ⎟ ⎝a⎠

rAD

⎛ −e − d ⎞ = ⎜ b ⎟ ⎜ ⎟ ⎝ 0 ⎠

rAC

⎛0⎞ = ⎜b⎟ ⎜ ⎟ ⎝a⎠

rBD

⎛ −d − e ⎞ = ⎜ −c ⎟ ⎜ ⎟ ⎝ 0 ⎠

First find FDE. ΣMy' = 0;

F DE( d + e) − M g d = 0

Next find FBC.

Guess

⎡⎛ e ⎞ ⎛ 0 Given ⎢⎜ 0 ⎟ × ⎜ 0 ⎢⎜ ⎟ ⎜ ⎣⎝ 0 ⎠ ⎝ −M

F DE =

Mgd d+e

2

F DE = 4.1 s kN

F BC = 1 kN

⎞ ⎛e + d⎞ rBC ⎟ + ⎜ c ⎟ × ⎛F ⎜ BC ⎟ ⎜ ⎟ ⎝ rBC g⎠ ⎝ 0 ⎠

⎤ ⎞⎥ ⎟⎥ rAD = 0 ⎠ ⎦

F BC = Find ( FBC )

F BC = kN Now find FAC.

Guess

F AC = 1 kN

424



⎡⎛ e ⎞ ⎛ 0 Given ⎢⎜ 0 ⎟ × ⎜ 0 ⎢⎜ ⎟ ⎜ ⎣⎝ 0 ⎠ ⎝ −M

Chapter 5

⎞ ⎛e + d⎞ rAC ⎟ + ⎜ −b ⎟ × ⎛ F ⎜ AC ⎟ ⎜ ⎟ ⎝ rAC g⎠ ⎝ 0 ⎠

⎤ ⎞⎥ ⎟⎥ rBD = 0 ⎠ ⎦

F AC = Find ( FAC )

F AC = kN

Problem 5-89 The cables exert the forces shown on the pole. Assuming the pole is supported by a ball-and-socket joint at its base, determine the components of reaction at A. The forces F 1 and F2 lie in a horizontal plane. Given: F 1 = 140 lb F 2 = 75 lb

θ = 30 deg a = 5 ft b = 10 ft c = 15 ft

Solution: The initial guesses are TBC = 100 lb

TBD = 100 lb

Ax = 100 lb

Ay = 100 lb

Az = 100 lb

Given c ⎞ − T a⎛ ⎞=0 BD ⎜ ⎟ ⎟ 2 2 2 2 2 ⎝ a +b +c ⎠ ⎝ a +c ⎠

(F1 cos (θ ) + F2)c − TBC a⎛⎜

c

⎞=0 ⎝ a +b +c ⎠

F 1 sin ( θ ) c − b TBC ⎛ ⎜

c

2

2

2⎟

425



⎛

Chapter 5

⎞ ⎟=0 2 2 2 a + b + c ⎝ ⎠

Ax + F 1 sin ( θ ) − b⎜

TBC

TBC ⎞ ⎞ + a⎛ =0 ⎜ ⎟ ⎟ 2 2 2 2 2 + c a + b + c a ⎝ ⎠ ⎝ ⎠

⎛

Ay − F 1 cos ( θ ) − F 2 + TBD⎜

a

TBC ⎛ TBD ⎞ ⎛ ⎞ − c =0 ⎟ ⎜ ⎟ 2 2 2 2 2 a + c a + b + c ⎝ ⎠ ⎝ ⎠

Az − c⎜

⎛ TBC ⎞ ⎜ ⎟ ⎜ TBD ⎟ ⎜ Ax ⎟ = Find ( T , T , A , A , A ) BC BD x y z ⎜ ⎟ ⎜ Ay ⎟ ⎜ A ⎟ ⎝ z ⎠

⎛ TBC ⎞ ⎛ 131.0 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ TBD ⎠ ⎝ 509.9 ⎠ ⎛ Ax ⎞ ⎛ −0.0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 0.0 ⎟ lb ⎜ A ⎟ ⎝ 588.7 ⎠ ⎝ z⎠

Problem 5-90 The silo has a weight W, a center of gravity at G and a radius r. Determine the vertical component of force that each of the three struts at A, B, and C exerts on the silo if it is subjected to a resultant wind loading of F which acts in the direction shown. Given: W = 3500 lb F = 250 lb

θ 1 = 30 deg θ 2 = 120 deg θ 3 = 30 deg r = 5 ft b = 12 ft c = 15 ft

426



Chapter 5


Az = 1 lb

B z = 2 lb

Cz = 31 lb

Given ΣMy = 0; B z r cos ( θ 1 ) − Cz r cos ( θ 1 ) − F sin ( θ 3 ) c = 0

[1]

ΣMx = 0; −B z r sin ( θ 1 ) − Cz r sin ( θ 1 ) + Az r − F cos ( θ 3 ) c = 0

[2]

ΣF z = 0; Az + Bz + Cz = W

[3]

Solving Eqs.[1], [2] and [3] yields:

⎛ Az ⎞ ⎜ ⎟ ⎜ Bz ⎟ = Find ( Az , Bz , Cz) ⎜C ⎟ ⎝ z⎠

⎛ Az ⎞ ⎛ 1600 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Bz ⎟ = ⎜ 1167 ⎟ lb ⎜ C ⎟ ⎝ 734 ⎠ ⎝ z⎠

Problem 5-91 The shaft assembly is supported by two smooth journal bearings A and B and a short link DC. If a couple moment is applied to the shaft as shown, determine the components of force reaction at the bearings and the force in the link. The link lies in a plane parallel to the y-z plane and the bearings are properly aligned on the shaft. Units Used: 3

kN = 10 N Given: M = 250 N⋅ m a = 400 mm b = 300 mm c = 250 mm d = 120 mm

θ = 30 deg 427



Chapter 5

φ = 20 deg Solution: Initial Guesses: Ay = 1 kN

Az = 1 kN

B z = 1 kN

F CD = 1 kN

B y = 1 kN

Given 0 ⎞ ⎛0 ⎞ ⎛0⎞ ⎛ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ + ⎜ By ⎟ + ⎜ −FCD cos ( φ ) ⎟ = 0 ⎜ Az ⎟ ⎜ Bz ⎟ ⎜ F sin ( φ ) ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ CD ⎠ 0 ⎞ ⎛ −a − b ⎞ ⎛ 0 ⎞ ⎛ −M ⎞ ⎛ d − a ⎞ ⎛⎜ ⎜ c sin ( θ ) ⎟ × −FCD cos ( φ ) ⎟ + ⎜ 0 ⎟ × ⎜ By ⎟ + ⎜ 0 ⎟ = 0 ⎟ ⎜ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ c cos ( θ ) ⎠ ⎝ FCD sin ( φ ) ⎠ ⎝ 0 ⎠ ⎜⎝ Bz ⎟⎠ ⎝ 0 ⎠

⎛ Ay ⎞ ⎜ ⎟ A z ⎜ ⎟ ⎜ By ⎟ = Find ( A , A , B , B , F ) y z y z CD ⎜ ⎟ ⎜ Bz ⎟ ⎜F ⎟ ⎝ CD ⎠

⎛ Ay ⎞ ⎛ 573 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ Az ⎠ ⎝ −208 ⎠ ⎛ By ⎞ ⎛ 382 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ Bz ⎠ ⎝ −139 ⎠ F CD = 1.015 kN

Problem 5-92 If neither the pin at A nor the roller at B can support a load no greater than F max, determine the maximum intensity of the distributed load w, so that failure of a support does not occur.

428



Chapter 5

Units Used: 3

kN = 10 N Given: F max = 6 kN a = 3m b = 3m

Solution: The greatest reaction is at A. Require ΣMB = 0;

−F max( a + b) + w a

w =

F max( a + b) a⎛⎜

a

⎝2

+ b⎞⎟ +

⎠

2

⎛ a + b⎞ + 1 w b 2 b = 0 ⎜ ⎟ 3 ⎝2 ⎠ 2 w = 2.18

b

kN m

3

Problem 5-93 If the maximum intensity of the distributed load acting on the beam is w, determine the reactions at the pin A and roller B. Units Used: 3

kN = 10 N Given: F = 6 kN a = 3m b = 3m

429



w = 4

Chapter 5

kN m


Ax = 0

ΣMA = 0;

−w a

By =

a 2 1 6

−

1 2

w b⎛⎜ a +

⎝

⎟ + By( a + b) = 0

3⎠

2

w

3a + 3 a b + b

Ay + By − w a −

ΣF y = 0;

b⎞

2

B y = 7 kN

a+b 1 2

Ay = −By + w a +

wb=0 1 2

Ay = 11 kN

wb

Problem 5-94 Determine the normal reaction at the roller A and horizontal and vertical components at pin B for equilibrium of the member. Units Used: 3

kN = 10 N Given: F 1 = 10 kN F 2 = 6 kN a = 0.6 m b = 0.6 m c = 0.8 m d = 0.4 m

θ = 60 deg

430



Chapter 5

Solution: Initial Guesses: NA = 1 kN B x = 1 kN B y = 1 kN Given B x − F2 sin ( θ ) = 0 B y + NA − F 1 − F 2 cos ( θ ) = 0 F 2 d + F1 ⎡⎣b + ( c + d)cos ( θ )⎤⎦ − NA⎡⎣a + b + ( c + d)cos ( θ )⎤⎦ = 0

⎛ NA ⎞ ⎜ ⎟ ⎜ Bx ⎟ = Find ( NA , Bx , By) ⎜B ⎟ ⎝ y⎠

⎛ NA ⎞ ⎛ 8 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Bx ⎟ = ⎜ 5.196 ⎟ kN ⎜B ⎟ ⎝ 5 ⎠ ⎝ y⎠

Problem 5-95 The symmetrical shelf is subjected to uniform pressure P. Support is provided by a bolt (or pin) located at each end A and A' and by the symmetrical brace arms, which bear against the smooth wall on both sides at B and B'. Determine the force resisted by each bolt at the wall and the normal force at B for equilibrium. Units Used: 3

kPa = 10 Pa Given: P = 4 kPa a = 0.15 m b = 0.2 m c = 1.5 m Solution: ΣMA = 0;

431



NB a − P ⎛⎜ b

Chapter 5

c⎞ b

⎟ =0 ⎝ 2⎠ 2 2

NB = P

b c

NB = 400 N

4a

ΣF x = 0; Ax = NB

Ax = 400 N

ΣF y = 0; Ay = P b

FA =

c

Ay = 600 N

2 2

Ax + Ay

2

F A = 721 N

Problem 5-96 A uniform beam having a weight W supports a vertical load F . If the ground pressure varies linearly as shown, determine the load intensities w1 and w2 measured in lb/ft, necessary for equilibrium. Given: W = 200 lb F = 800 lb a = 7 ft b = 6 ft Solution: Initial Guesses: w1 = 1

lb ft

w2 = 1

lb ft

432



Chapter 5

Given 1

w1 ( a + b) +

w1 ( a + b)

2

(w2 − w1)( a + b) − F − W = 0

a+b 2

+

(w2 − w1)( a + b) 3 ( a + b) − W 2 1

⎛ w1 ⎞ ⎜ ⎟ = Find ( w1 , w2) ⎝ w2 ⎠

2

a+b 2

− Fa = 0

⎛ w1 ⎞ ⎛ 62.7 ⎞ lb ⎜ ⎟=⎜ ⎟ ⎝ w2 ⎠ ⎝ 91.1 ⎠ ft

Problem 5-97 The uniform ladder rests along the wall of a building at A and on the roof at B. If the ladder has a weight W and the surfaces at A and B are assumed smooth, determine the angle θ for equilibrium. Given: a = 18 ft W = 25 lb

θ 1 = 40 deg Solution: Initial guesses: R A = 10 lb R B = 10 lb

θ = 10 deg Given ΣMB = 0;

−R A a sin ( θ ) + W

ΣF x = 0;

R A − RB sin ( θ 1 ) = 0

ΣF y = 0;

R B cos ( θ 1 ) − W = 0

a 2

cos ( θ ) = 0

Solving,

433



Chapter 5

⎛ RB ⎞ ⎜ ⎟ ⎜ RA ⎟ = Find ( RB , RA , θ ) ⎜ ⎟ ⎝θ ⎠

⎛ RA ⎞ ⎛ 21 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ RB ⎠ ⎝ 32.6 ⎠

θ = 30.8 deg

Problem 5-98 Determine the x, y, z components of reaction at the ball supports B and C and the ball-and-socket A (not shown) for the uniformly loaded plate. Given: P = 2

lb ft

2

a = 4 ft b = 1 ft c = 2 ft d = 2 ft

Solution: The initial guesses are

Ax = 1 lb

Ay = 1 lb

Az = 1 lb

B z = 1 lb

Cz = 1 lb

Given ΣF x = 0;

Ax = 0

ΣF y = 0;

Ay = 0

ΣF z = 0; Az + Bz + Cz − P a c = 0 c⎞ ⎟ + Cz b = 0 ⎝ 2⎠

ΣMx = 0; c B z − P a c⎛⎜

a⎞ ⎟ − Cz a = 0 ⎝ 2⎠

ΣMy = 0; −B z ( a − d) + P a c⎛⎜

434



Chapter 5

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ Az ⎟ = Find ( A , A , A , B , C ) x y z z z ⎜ ⎟ ⎜ Bz ⎟ ⎜C ⎟ ⎝ z⎠

⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 0 ⎟ lb ⎜ A ⎟ ⎝ 5.333 ⎠ ⎝ z⎠

⎛ Bz ⎞ ⎛ 5.333 ⎞ ⎜ ⎟=⎜ ⎟ lb 5.333 C ⎝ ⎠ z ⎝ ⎠

Problem 5-99 A vertical force F acts on the crankshaft. Determine the horizontal equilibrium force P that must be applied to the handle and the x, y, z components of force at the smooth journal bearing A and the thrust bearing B. The bearings are properly aligned and exert the force reactions on the shaft. Given: F = 80 lb a = 10 in b = 14 in c = 14 in d = 8 in e = 6 in f = 4 in

Solution: ΣMy = 0;

P d−F a= 0 P = F

ΣMx = 0;

P = 100 lb

B z( b + c) − F c = 0 Bz = F

ΣMz = 0;

⎛ a⎞ ⎜ ⎟ ⎝ d⎠ ⎛ c ⎞ ⎜ ⎟ ⎝ b + c⎠

B z = 40 lb

−B x( b + c) − P ( e + f) = 0

435



B x = −P ΣF x = 0;

⎛ e + f ⎞ B = −35.7 lb ⎜ ⎟ x ⎝ b + c⎠

Ax + Bx − P = 0 Ax = −Bx + P

Ax = 135.7 lb By = 0

ΣF y = 0; ΣF z = 0;

Chapter 5

Az + Bz − F = 0 Az = −B z + F

Az = 40 lb

Problem 5-100 The horizontal beam is supported by springs at its ends. If the stiffness of the spring at A is kA , determine the required stiffness of the spring at B so that if the beam is loaded with the force F , it remains in the horizontal position both before and after loading. Units Used: 3

kN = 10 N Given: kA = 5

kN m

F = 800 N

a = 1m b = 2m

Solution: Equilibrium: ΣMA = 0;

F B ( a + b) − F a = 0 FB = F

⎛ a ⎞ ⎜ ⎟ ⎝ a + b⎠

F B = 266.667 N ΣMB = 0;

F b − FA( a + b) = 0 436



Chapter 5

FA = F

⎛ b ⎞ ⎜ ⎟ ⎝ a + b⎠

F A = 533.333 N Spring force formula: xA = xB

FA kA

=

FB kB

kB =

FB FA

kA

kB = 2.5

kN m

437



Chapter 6

Problem 6-1 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: P 1 = 7 kN P 2 = 7 kN Solution:

θ = 45 deg Initial Guesses: F AB = 1 kN F DC = 1 kN

F AD = 1 kN

F DB = 1 kN

F CB = 1 kN

Given Joint A:

F AB + F AD cos ( θ ) = 0 −P 1 − F AD sin ( θ ) = 0

Joint D:

F DB cos ( θ ) − F AD cos ( θ ) + F DC cos ( θ ) = 0

(FAD + FDB − FDC)sin(θ ) − P2 = 0 Joint C:

F CB + FDC sin ( θ ) = 0

⎛⎜ FAB ⎞⎟ ⎜ FAD ⎟ ⎜ ⎟ ⎜ FDB ⎟ = Find ( FAB , FAD , FDB , FDC , FCB) ⎜F ⎟ ⎜ DC ⎟ ⎜ FCB ⎟ ⎝ ⎠

⎛⎜ FAB ⎞⎟ ⎛ 7 ⎞ ⎜ FAD ⎟ ⎜ −9.9 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FDB ⎟ = ⎜ 4.95 ⎟ kN ⎜ F ⎟ ⎜ −14.85 ⎟ ⎜ DC ⎟ ⎜ ⎟ 10.5 ⎝ ⎠ ⎜ FCB ⎟ ⎝ ⎠

Positive means Tension, Negative means Compression

438



Chapter 6


kN = 10 N Given: P 1 = 8 kN P 2 = 10 kN Solution:

θ = 45 deg Initial Guesses: F AB = 1 kN F DC = 1 kN

F AD = 1 kN

F DB = 1 kN

F CB = 1 kN

Given Joint A:


Joint D:

F DB cos ( θ ) − F AD cos ( θ ) + F DC cos ( θ ) = 0

(FAD + FDB − FDC)sin(θ ) − P2 = 0 Joint C:

F CB + FDC sin ( θ ) = 0

⎛⎜ FAB ⎞⎟ ⎜ FAD ⎟ ⎜ ⎟ F ⎜ DB ⎟ = Find ( FAB , FAD , FDB , FDC , FCB) ⎜F ⎟ ⎜ DC ⎟ ⎜ FCB ⎟ ⎝ ⎠

⎛⎜ FAB ⎞⎟ ⎛ 8 ⎞ ⎜ FAD ⎟ ⎜ −11.31 ⎟ ⎟ ⎜ ⎟ ⎜ F 7.07 = ⎟ kN ⎜ DB ⎟ ⎜ ⎜ F ⎟ ⎜ −18.38 ⎟ ⎜ DC ⎟ ⎜ ⎟ ⎜ FCB ⎟ ⎝ 13 ⎠ ⎝ ⎠


439



Chapter 6

Problem 6-3 The truss, used to support a balcony, is subjected to the loading shown. Approximate each joint as a pin and determine the force in each member. State whether the members are in tension or compression. Units Used: 3

kip = 10 lb Given: P 1 = 600 lb P 2 = 400 lb a = 4 ft

θ = 45 deg Solution: Initial Guesses F AB = 1 lb

F AD = 1 lb

F DC = 1 lb

F BC = 1 lb

F BD = 1 lb

F DE = 1 lb

Given Joint A:


Joint B:

F BC − F AB = 0 −P 2 − F BD = 0

Joint D:

(FDC − FAD)cos (θ ) + FDE = 0 (FDC + FAD)sin(θ ) + FBD = 0

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAD ⎟ ⎜F ⎟ ⎜ BC ⎟ = Find F , F , F , F , F , F ( AB AD BC BD DC DE) ⎜ FBD ⎟ ⎜ ⎟ ⎜ FDC ⎟ ⎜ ⎟ ⎝ FDE ⎠

440



⎛ FAB ⎞ ⎜ ⎟ ⎛ 600 ⎞ F AD ⎜ ⎟ ⎜ −849 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎜ BC ⎟ = 600 ⎟ lb ⎜ FBD ⎟ ⎜ −400 ⎟ ⎟ ⎜ ⎟ ⎜ 1414 ⎜ ⎟ ⎜ FDC ⎟ ⎜ ⎜ ⎟ ⎝ −1600 ⎟⎠ F ⎝ DE ⎠

Chapter 6


Problem 6-4 The truss, used to support a balcony, is subjected to the loading shown. Approximate each joint as a pin and determine the force in each member. State whether the members are in tension or compression. Units Used: 3

kip = 10 lb Given: P 1 = 800 lb P 2 = 0 lb a = 4 ft

θ = 45 deg Solution: Initial Guesses F AB = 1 lb

F AD = 1 lb

F DC = 1 lb

F BC = 1 lb

F BD = 1 lb

F DE = 1 lb

Given Joint A:


Joint B:

F BC − F AB = 0 −P 2 − F BD = 0

441



Joint D:

Chapter 6

(FDC − FAD)cos (θ ) + FDE = 0 (FDC + FAD)sin(θ ) + FBD = 0

⎛ FAB ⎞ ⎜ ⎟ F ⎜ AD ⎟ ⎜F ⎟ ⎜ BC ⎟ = Find F , F , F , F , F , F ( AB AD BC BD DC DE) ⎜ FBD ⎟ ⎜ ⎟ F ⎜ DC ⎟ ⎜ ⎟ ⎝ FDE ⎠ ⎛ FAB ⎞ ⎜ ⎟ ⎛ 800 ⎞ ⎜ FAD ⎟ ⎜ −1131 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎜ BC ⎟ = 800 ⎟ lb ⎜ FBD ⎟ ⎜ 0 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FDC ⎟ ⎜ 1131 ⎟ ⎜ ⎟ ⎜⎝ −1600 ⎟⎠ F ⎝ DE ⎠



kN = 10 N Given: P 1 = 20 kN P 2 = 10 kN a = 1.5 m e = 2m Solution:

e θ = atan ⎛⎜ ⎟⎞

⎝ a⎠

442



Chapter 6

Initial Guesses: F AB = 1 kN F AG = 1 kN F CF = 1 kN F BC = 1 kN F BG = 1 kN F DE = 1 kN F CG = 1 kN F FG = 1 kN F EF = 1 kN F CD = 1 kN F DF = 1 kN Given Joint B

F BC − F AB cos ( θ ) = 0 −F BG − F AB sin ( θ ) = 0

Joint G

F FG + F CG cos ( θ ) − F AG = 0 F CG sin ( θ ) + FBG − P1 = 0

Joint C

(

)

−F BC + F CD + FCF − F CG cos ( θ ) = 0

(

)

− FCG + FCF sin ( θ ) = 0 Joint D

−F CD + F DE cos ( θ ) = 0 −F DF − F DE sin ( θ ) = 0

Joint F

F EF − F FG − F CF cos ( θ ) = 0 F DF + F CF sin ( θ ) − P 2 = 0

Joint E

−F DE cos ( θ ) − F EF = 0

443



Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FCG ⎟ ⎜F ⎟ ⎜ CD ⎟ ⎜ FAG ⎟ ⎜ ⎟ ⎜ FBG ⎟ = Find ( FAB , FBC , FCG , FCD , FAG , FBG , FFG , FDF , FCF , FDE , FEF) ⎜F ⎟ ⎜ FG ⎟ ⎜ FDF ⎟ ⎜ ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎜F ⎟ ⎝ EF ⎠ ⎛ FAB ⎞ ⎜ ⎟ ⎛⎜ −21.88 ⎟⎞ ⎜ FBC ⎟ ⎜ −13.13 ⎟ kN ⎜ ⎟=⎜ ⎟ F 3.13 CG ⎜ ⎟ ⎜ ⎟ ⎜ F ⎟ ⎝ −9.37 ⎠ ⎝ CD ⎠

⎛ FAG ⎞ ⎜ ⎟ ⎛⎜ 13.13 ⎞⎟ ⎜ FBG ⎟ ⎜ 17.5 ⎟ kN ⎜ ⎟=⎜ ⎟ F 11.25 FG ⎜ ⎟ ⎜ ⎟ ⎜ F ⎟ ⎝ 12.5 ⎠ ⎝ DF ⎠

⎛ FCF ⎞ ⎛ −3.13 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ DE ⎟ = ⎜ −15.62 ⎟ kN ⎜ F ⎟ ⎝ 9.37 ⎠ ⎝ EF ⎠



kN = 10 N Given: P 1 = 40 kN P 2 = 20 kN a = 1.5 m e = 2m

444



Chapter 6

e θ = atan ⎛⎜ ⎞⎟

Solution:

⎝ a⎠

Initial Guesses: F AB = 1 kN F AG = 1 kN F CF = 1 kN F BC = 1 kN F BG = 1 kN F DE = 1 kN F CG = 1 kN F FG = 1 kN F EF = 1 kN F CD = 1 kN F DF = 1 kN Given Joint B

F BC − F AB cos ( θ ) = 0 −F BG − F AB sin ( θ ) = 0

Joint G

F FG + F CG cos ( θ ) − F AG = 0 F CG sin ( θ ) + FBG − P1 = 0

Joint C

(

)

−F BC + F CD + FCF − F CG cos ( θ ) = 0

(

)

− FCG + FCF sin ( θ ) = 0 Joint D

−F CD + F DE cos ( θ ) = 0 −F DF − F DE sin ( θ ) = 0

Joint F

F EF − F FG − F CF cos ( θ ) = 0 F DF + F CF sin ( θ ) − P 2 = 0

Joint E

−F DE cos ( θ ) − F EF = 0

445



Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FCG ⎟ ⎜F ⎟ ⎜ CD ⎟ ⎜ FAG ⎟ ⎜ ⎟ ⎜ FBG ⎟ = Find ( FAB , FBC , FCG , FCD , FAG , FBG , FFG , FDF , FCF , FDE , FEF) ⎜F ⎟ ⎜ FG ⎟ ⎜ FDF ⎟ ⎜ ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎜F ⎟ ⎝ EF ⎠ ⎛ FAB ⎞ ⎜ ⎟ ⎛⎜ −43.75 ⎟⎞ F ⎜ BC ⎟ ⎜ −26.25 ⎟ kN ⎜ ⎟=⎜ ⎟ ⎜ FCG ⎟ ⎜ 6.25 ⎟ ⎜ F ⎟ ⎝ −18.75 ⎠ ⎝ CD ⎠

⎛ FAG ⎞ ⎜ ⎟ ⎛⎜ 26.25 ⎞⎟ F ⎜ BG ⎟ ⎜ 35 ⎟ kN ⎜ ⎟=⎜ ⎟ ⎜ FFG ⎟ ⎜ 22.5 ⎟ ⎜ F ⎟ ⎝ 25 ⎠ ⎝ DF ⎠

⎛ FCF ⎞ ⎛ −6.25 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FDE ⎟ = ⎜ −31.25 ⎟ kN ⎜ F ⎟ ⎝ 18.75 ⎠ ⎝ EF ⎠



kN = 10 N Given: F 1 = 3 kN F 2 = 8 kN F 3 = 4 kN F 4 = 10 kN

446



Chapter 6

a = 2m b = 1.5 m b θ = atan ⎛⎜ ⎟⎞

Solution:

⎝ a⎠

Initial Guesses F BA = 1 kN

F BC = 1 kN

F AF = 1 kN

F CD = 1 kN

F DF = 1 kN

F ED = 1 kN

F AC = 1 kN F CF = 1 kN F EF = 1 kN

Given Joint B

F 1 + F BC = 0 −F 2 − F BA = 0

Joint C

F CD − F BC − F AC cos ( θ ) = 0 −F 3 − F AC sin ( θ ) − FCF = 0

Joint E

−F EF = 0

Joint D

−F CD − F DF cos ( θ ) = 0 −F 4 − F DF sin ( θ ) − FED = 0

Joint F

−F AF + F EF + F DF cos ( θ ) = 0 F CF + FDF sin ( θ ) = 0

⎛ FBA ⎞ ⎜ ⎟ ⎜ FAF ⎟ ⎜F ⎟ ⎜ DF ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FCD ⎟ = Find ( FBA , FAF , FDF , FBC , FCD , FED , FAC , FCF , FEF) ⎜F ⎟ ⎜ ED ⎟ ⎜ FAC ⎟ ⎜ ⎟ ⎜ FCF ⎟ ⎟ ⎜ ⎝ FEF ⎠

447



Chapter 6

⎛ FBA ⎞ ⎜ ⎟ ⎛ −8 ⎞ ⎜ FAF ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 4.167 ⎟ ⎜ DF ⎟ ⎜ 5.208 ⎟ ⎜ FBC ⎟ ⎜ −3 ⎟ ⎟ ⎜ ⎟ ⎜ F − 4.167 = ⎟ kN ⎜ CD ⎟ ⎜ ⎜ F ⎟ ⎜ −13.125 ⎟ ⎟ ⎜ ED ⎟ ⎜ ⎜ FAC ⎟ ⎜ −1.458 ⎟ ⎜ ⎟ ⎜ −3.125 ⎟ F ⎜ CF ⎟ ⎜ ⎟ ⎟ ⎝ 0 ⎠ ⎜ ⎝ FEF ⎠

Positive means tension, Negative means compression.

Problem 6-8 Determine the force in each member of the truss in terms of the external loading and state if the members are in tension or compression.

Solution: ΣMA = 0;

−P a + Cy2a − P a = 0 Cy = P

Joint C: ΣF x = 0;

1

ΣF y = 0;

P+

2

F BC − 1 17

4 17

F CD −

FCD = 0 1 2

F BC = 0

448



Chapter 6

F BC =

4 2P

F CD =

17 P

1

FCD +

3

3

= 1.886 P (C) = 1.374 P (T)

Joint B: ΣF x = 0;

P−

ΣFy = 0;

1

2

2

F CD +

2

2P

F AB = F BD =

1

3 5P 3

1 2

FAB = 0

F AB − F BD = 0

= 0.471P

= 1.667P

( C)

( T)

Joint D: ΣF x = 0;

F DA = F CD = 1.374P

(T)

Problem 6-9 The maximum allowable tensile force in the members of the truss is Tmax, and the maximum allowable compressive force is Cmax. Determine the maximum magnitude P of the two loads that can be applied to the truss. Given: Tmax = 1500 lb Cmax = 800 lb Solution: Set

P = 1 lb

Initial Guesses F AB = 1 lb

F AD = 1 lb

F BC = 1 lb

F CD = 1 lb

F BD = 1 lb

449



Chapter 6

Given Joint B

(FBC − FAB)

1

+P=0

2

(

−F BD − FAB + FBC Joint D

(FCD − FAD)

4

)

1

=0

2

=0

17

(

)

F BD − P − F AD + F CD Joint C

−F BC

1 2

− F CD

4

1

=0

17

=0

17

⎛⎜ FAB ⎞⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FAD ⎟ = Find ( FAB , FBC , FAD , FCD , FBD) ⎜F ⎟ ⎜ CD ⎟ ⎜ FBD ⎟ ⎝ ⎠

⎛⎜ FAB ⎞⎟ ⎛ −0.471 ⎞ ⎜ FBC ⎟ ⎜ −1.886 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FAD ⎟ = ⎜ 1.374 ⎟ lb ⎜ F ⎟ ⎜ 1.374 ⎟ ⎜ CD ⎟ ⎜ ⎟ ⎜ FBD ⎟ ⎝ 1.667 ⎠ ⎝ ⎠

Now find the critical load P1 = P

Tmax

(

max F AB , F BC , F AD , F CD , F BD

P2 = P

)

P 1 = 900 lb

)

P 2 = 424.264 lb

Cmax

(

min F AB , F BC , F AD , F CD , F BD

(

)

P = min P 1 , P 2

P = 424.3 lb

Problem 6-10 Determine the force in each member of the truss and state if the members are in tension or compression. Given: P 1 = 0 lb

450



Chapter 6

P 2 = 1000 lb a = 10 ft b = 10 ft Solution: b θ = atan ⎛⎜ ⎟⎞

⎝ a⎠

Initial Guesses: F AB = 1 lb

F AG = 1 lb

F BG = 1 lb

F BC = 1 lb

F DC = 1 lb

F DE = 1 lb

F EG = 1 lb

F EC = 1 lb

F CG = 1 lb

Given Joint B

F BC − F AB = 0 F BG − P 1 = 0

Joint G

(FCG − FAG)cos (θ ) + FEG = 0 (

)

− FCG + FAG sin ( θ ) − FBG = 0 Joint C

F DC − FBC − FCG cos ( θ ) = 0 F EC + F CG sin ( θ ) − P2 = 0

Joint E

F DE cos ( θ ) − F EG = 0 −F EC − F DE sin ( θ ) = 0

Joint D

−F DE cos ( θ ) − F DC = 0

451



Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎜ FBC ⎟ ⎜F ⎟ ⎜ EG ⎟ ⎜ FAG ⎟ ⎜ ⎟ F ⎜ DC ⎟ = Find ( FAB , FBC , FEG , FAG , FDC , FEC , FBG , FDE , FCG) ⎜F ⎟ ⎜ EC ⎟ ⎜ FBG ⎟ ⎜ ⎟ F ⎜ DE ⎟ ⎟ ⎜ ⎝ FCG ⎠ ⎛ FAB ⎞ ⎜ ⎟ ⎛ 333 ⎞ ⎜ FBC ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 333 ⎟ ⎜ EG ⎟ ⎜ −667 ⎟ ⎜ FAG ⎟ ⎜ −471 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FDC ⎟ = ⎜ 667 ⎟ lb ⎜ F ⎟ ⎜ 667 ⎟ ⎟ ⎜ EC ⎟ ⎜ ⎜ FBG ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜ −943 ⎟ ⎜ FDE ⎟ ⎜ ⎟ ⎟ ⎝ 471 ⎠ ⎜ ⎝ FCG ⎠


Problem 6-11 Determine the force in each member of the truss and state if the members are in tension or compression. Given: P 1 = 500 lb P 2 = 1500 lb

452



Chapter 6

a = 10 ft b = 10 ft Solution: b θ = atan ⎛⎜ ⎟⎞

⎝ a⎠


F AG = 1 lb

F BG = 1 lb

F BC = 1 lb

F DC = 1 lb

F DE = 1 lb

F EG = 1 lb

F EC = 1 lb

F CG = 1 lb

Given Joint B

F BC − F AB = 0 F BG − P 1 = 0

Joint G

(FCG − FAG)cos (θ ) + FEG = 0 (

)

− FCG + FAG sin ( θ ) − FBG = 0 Joint C

F DC − FBC − FCG cos ( θ ) = 0 F EC + F CG sin ( θ ) − P2 = 0

Joint E

F DE cos ( θ ) − F EG = 0 −F EC − F DE sin ( θ ) = 0

Joint D

−F DE cos ( θ ) − F DC = 0

453



Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎜ FBC ⎟ ⎜F ⎟ ⎜ EG ⎟ ⎜ FAG ⎟ ⎜ ⎟ F ⎜ DC ⎟ = Find ( FAB , FBC , FEG , FAG , FDC , FEC , FBG , FDE , FCG) ⎜F ⎟ ⎜ EC ⎟ ⎜ FBG ⎟ ⎜ ⎟ F ⎜ DE ⎟ ⎟ ⎜ ⎝ FCG ⎠ ⎛ FAB ⎞ ⎜ ⎟ ⎛ 833 ⎞ ⎜ FBC ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 833 ⎟ ⎜ EG ⎟ ⎜ −1167 ⎟ ⎜ FAG ⎟ ⎜ −1179 ⎟ ⎟ ⎜ ⎟ ⎜ F 1167 = ⎟ lb ⎜ DC ⎟ ⎜ ⎜ F ⎟ ⎜ 1167 ⎟ ⎟ ⎜ EC ⎟ ⎜ ⎜ FBG ⎟ ⎜ 500 ⎟ ⎜ ⎟ ⎜ −1650 ⎟ F ⎜ DE ⎟ ⎜ ⎟ ⎟ ⎝ 471 ⎠ ⎜ ⎝ FCG ⎠



kN = 10 N Given: P 1 = 10 kN P 2 = 15 kN

454



Chapter 6

a = 2m b = 4m c = 4m Solution:

c α = atan ⎛⎜ ⎟⎞

⎝ a⎠

c β = atan ⎛⎜ ⎟⎞

⎝ b⎠

Initial Guesses: F AB = 1 kN

F AF = 1 kN

F GB = 1 kN

F BF = 1 kN

F FC = 1 kN

F FE = 1 kN

F BC = 1 kN

F EC = 1 kN

F CD = 1 kN

F ED = 1 kN Given Joint B

−F GB + F BC − F AB cos ( α ) = 0

Joint F

−F AB sin ( α ) − FBF = 0 −F AF + F FE + F FC cos ( β ) = 0 F BF + F FC sin ( β ) − P 1 = 0

Joint C

−F BC − F FC cos ( β ) + FCD cos ( α ) = 0 −F FC sin ( β ) − F CD sin ( α ) − FEC = 0

Joint E

−F FE + F ED = 0 F EC − P 2 = 0

Joint D

−F CD cos ( α ) − F ED = 0 F CD sin ( α ) = 0

455



Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎜ FBF ⎟ ⎜F ⎟ ⎜ BC ⎟ ⎜ FED ⎟ ⎜ ⎟ ⎜ FAF ⎟ ⎜ ⎟ = Find ( FAB , FBF , FBC , FED , FAF , FFC , FEC , FGB , FFE , FCD) F FC ⎜ ⎟ ⎜F ⎟ ⎜ EC ⎟ ⎜ FGB ⎟ ⎜ ⎟ F ⎜ FE ⎟ ⎜F ⎟ ⎝ CD ⎠ ⎛⎜ FAB ⎞⎟ ⎛ −27.951 ⎞ ⎜ FBF ⎟ ⎜ 25 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FBC ⎟ = ⎜ 15 ⎟ kN ⎜F ⎟ ⎜ 0 ⎟ ⎜ ED ⎟ ⎜ ⎟ ⎜ FAF ⎟ ⎝ −15 ⎠ ⎝ ⎠

⎛⎜ FFC ⎞⎟ ⎛ −21.213 ⎞ ⎜ FEC ⎟ ⎜ 15 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FGB ⎟ = ⎜ 27.5 ⎟ kN Positive means Tension, Negative means Compression ⎜F ⎟ ⎜ 0 ⎟ FE ⎜ ⎟ ⎜ ⎟ ⎜ FCD ⎟ ⎝ 0 ⎠ ⎝ ⎠


kN = 10 N Given: P 1 = 0 kN P 2 = 20 kN a = 2m b = 4m c = 4m

456



Solution:

c α = atan ⎛⎜ ⎞⎟

⎝ a⎠

Chapter 6

c β = atan ⎛⎜ ⎞⎟

⎝ b⎠

Initial Guesses: F AB = 1 kN

F AF = 1 kN

F GB = 1 kN

F BF = 1 kN

F FC = 1 kN

F FE = 1 kN

F BC = 1 kN

F EC = 1 kN

F CD = 1 kN

F ED = 1 kN Given Joint B

−F GB + F BC − F AB cos ( α ) = 0

Joint F

−F AB sin ( α ) − FBF = 0 −F AF + F FE + F FC cos ( β ) = 0 F BF + F FC sin ( β ) − P 1 = 0

Joint C

−F BC − F FC cos ( β ) + FCD cos ( α ) = 0 −F FC sin ( β ) − F CD sin ( α ) − FEC = 0

Joint E

−F FE + F ED = 0 F EC − P 2 = 0

Joint D

−F CD cos ( α ) − F ED = 0 F CD sin ( α ) = 0

⎛ FAB ⎞ ⎜ ⎟ ⎜ FBF ⎟ ⎜F ⎟ ⎜ BC ⎟ ⎜ FED ⎟ ⎜ ⎟ ⎜ FAF ⎟ ⎜ ⎟ = Find ( FAB , FBF , FBC , FED , FAF , FFC , FEC , FGB , FFE , FCD) FFC ⎜ ⎟ ⎜F ⎟ ⎜ EC ⎟ ⎜ FGB ⎟ ⎜ ⎟ ⎜ FFE ⎟ ⎜F ⎟ ⎝ CD ⎠

457



Chapter 6

⎛⎜ FAB ⎞⎟ ⎛ −22.361 ⎞ ⎜ FBF ⎟ ⎜ 20 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FBC ⎟ = ⎜ 20 ⎟ kN ⎜F ⎟ ⎜ 0 ⎟ ⎜ ED ⎟ ⎜ ⎟ − 20 ⎝ ⎠ ⎜ FAF ⎟ ⎝ ⎠

⎛⎜ FFC ⎞⎟ ⎛ −28.284 ⎞ ⎜ FEC ⎟ ⎜ 20 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FGB ⎟ = ⎜ 30 ⎟ kN ⎜F ⎟ ⎜ 0 ⎟ ⎜ FE ⎟ ⎜ ⎟ 0 ⎝ ⎠ ⎜ FCD ⎟ ⎝ ⎠


Problem 6-14 Determine the force in each member of the truss and state if the members are in tension or compression. Given: P 1 = 100 lb P 2 = 200 lb P 3 = 300 lb a = 10 ft b = 10 ft


b φ = atan ⎛⎜ ⎟⎞

⎝ a⎠


F AF = 1 lb

F BC = 1 lb

F BF = 1 lb

F FC = 1 lb

F FE = 1 lb

F ED = 1 lb

F EC = 1 lb

F CD = 1 lb

458



Chapter 6

Given Joint B

F BC − F AB cos ( φ ) = 0 −F BF − F AB sin ( φ ) = 0

Joint F

−F AF + F FE + F FC cos ( φ ) = 0 −P 2 + F BF + F FC sin ( φ ) = 0

Joint C

−F BC + F CD cos ( φ ) − F FC cos ( φ ) = 0 −F EC − F CD sin ( φ ) − FFC sin ( φ ) = 0

Joint E

−F FE + F ED = 0 F EC − P 3 = 0

Joint D

−F ED cos ( θ ) − F CD cos ( φ + θ ) = 0

⎛ FAB ⎞ ⎜ ⎟ F AF ⎜ ⎟ ⎜F ⎟ ⎜ BC ⎟ ⎜ FBF ⎟ ⎜ ⎟ F ⎜ FC ⎟ = Find ( FAB , FAF , FBC , FBF , FFC , FFE , FED , FEC , FCD) ⎜F ⎟ ⎜ FE ⎟ ⎜ FED ⎟ ⎜ ⎟ ⎜ FEC ⎟ ⎜ ⎟ ⎝ FCD ⎠

459



Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎛ −330.0 ⎞ ⎜ FAF ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 79.4 ⎟ ⎜ BC ⎟ ⎜ −233.3 ⎟ ⎜ FBF ⎟ ⎜ 233.3 ⎟ ⎟ ⎜ ⎟ ⎜ F − 47.1 = ⎟ lb ⎜ FC ⎟ ⎜ ⎜ F ⎟ ⎜ 112.7 ⎟ ⎟ ⎜ FE ⎟ ⎜ ⎜ FED ⎟ ⎜ 112.7 ⎟ ⎜ ⎟ ⎜ 300.0 ⎟ F ⎜ EC ⎟ ⎜ ⎟ ⎟ ⎝ −377.1 ⎠ ⎜ ⎝ FCD ⎠


Problem 6-15 Determine the force in each member of the truss and state if the members are in tension or compression. Given: P 1 = 400 lb P 2 = 400 lb P 3 = 0 lb a = 10 ft b = 10 ft


b φ = atan ⎛⎜ ⎟⎞

⎝ a⎠


F AF = 1 lb

F BC = 1 lb

F BF = 1 lb

F FC = 1 lb

F FE = 1lb

F ED = 1 lb

F EC = 1 lb

F CD = 1 lb

460



Chapter 6

Given Joint B

F BC − F AB cos ( φ ) = 0 −F BF − F AB sin ( φ ) = 0

Joint F

−F AF + F FE + F FC cos ( φ ) = 0 −P 2 + F BF + F FC sin ( φ ) = 0

Joint C

−F BC + F CD cos ( φ ) − F FC cos ( φ ) = 0 −F EC − F CD sin ( φ ) − FFC sin ( φ ) = 0

Joint E

−F FE + F ED = 0 F EC − P 3 = 0

Joint D

−F ED cos ( θ ) − F CD cos ( φ + θ ) = 0

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAF ⎟ ⎜F ⎟ ⎜ BC ⎟ ⎜ FBF ⎟ ⎜ ⎟ ⎜ FFC ⎟ = Find ( FAB , FAF , FBC , FBF , FFC , FFE , FED , FEC , FCD) ⎜F ⎟ ⎜ FE ⎟ ⎜ FED ⎟ ⎜ ⎟ ⎜ FEC ⎟ ⎜ ⎟ ⎝ FCD ⎠

461



⎛ FAB ⎞ ⎜ ⎟ ⎛ −377.1 ⎞ ⎜ FAF ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 189.7 ⎟ ⎜ BC ⎟ ⎜ −266.7 ⎟ ⎜ FBF ⎟ ⎜ 266.7 ⎟ ⎟ ⎜ ⎟ ⎜ F 188.6 = ⎟ lb ⎜ FC ⎟ ⎜ ⎜ F ⎟ ⎜ 56.4 ⎟ ⎟ ⎜ FE ⎟ ⎜ ⎜ FED ⎟ ⎜ 56.4 ⎟ ⎜ ⎟ ⎜ 0.0 ⎟ F ⎜ EC ⎟ ⎜ ⎟ ⎟ ⎝ −188.6 ⎠ ⎜ ⎝ FCD ⎠

Chapter 6


Problem 6-16 Determine the force in each member of the truss in terms of the load P and state if the members are in tension or compression.

Solution: Support reactions: ΣME = 0;

3 Ax d − P d = 0 2

Ax =

2P 3

462



Chapter 6

2P

ΣF x = 0;

Ax − Ex = 0

Ex =

ΣF y = 0;

Ey − P = 0

Ey = P

3

Joint E: 2

ΣF x = 0;

F EC

ΣF y = 0;

P − FED − FEC

13

− Ex = 0

13

F EC =

3

=0

13

3

P = 1.20P (T)

F ED = 0

Joint A: 1

− F AD

1

2

=0

ΣF y = 0;

F AB

ΣF x = 0;

Ax − 2FAB

5

=0

5

F AB = F AD

F AB = F AD =

5

Joint D: ΣF x = 0;

F AD

ΣF y = 0;

2FAD

2 5

− F DC

1 5

2

=0

5

− FDB = 0

F DC = F DB =

5 6 P

5 6

P = 0.373P (C)

P = 0.373P (C)

(T)

3

Joint B: ΣF x = 0; F AB

1 5

− F BC

1 5

=0

F BC =

5 6

P = 0.373P (C)

463



Chapter 6

Problem 6-17 The maximum allowable tensile force in the members of the truss is Tmax and the maximum allowable compressive force is Cmax. Determine the maximum magnitude of the load P that can be applied to the truss. Units Used: 3

kN = 10 N Given: Tmax = 5 kN Cmax = 3 kN d = 2m Solution: P = 1 kN

Set

Initial Guesses: F AD = 1 kN

F AB = 1 kN

F BC = 1 kN

F BD = 1 kN

F CD = 1 kN

F CE = 1 kN

F DE = 1 kN

Given Joint A

Joint B

Joint D

F AD F BC

1 5 2 5

− F AB

1

− F AB

2

=0

5 =0

5

(FBC + FAB)

1

(FCD − FAD)

2

5

+ F BD = 0 =0

5

(

)

1

F DE − F BD − FAD + FCD Joint C

− FCD + FBC

(

)

(FCD − FBC )

1

2 5

5

2

− F CE

+ F CE

=0

5 =0

13 3

−P=0

13

464



Chapter 6

⎛ FAD ⎞ ⎜ ⎟ ⎜ FAB ⎟ ⎜ ⎟ FBC ⎜ ⎟ ⎜ F ⎟ = Find F , F , F , F , F , F , F ( AD AB BC BD CD CE DE) ⎜ BD ⎟ ⎜ FCD ⎟ ⎜ ⎟ ⎜ FCE ⎟ ⎜F ⎟ ⎝ DE ⎠ ⎛ FAD ⎞ ⎜ ⎟ ⎛ −0.373 ⎞ ⎟ ⎜ FAB ⎟ ⎜ ⎜ ⎟ ⎜ −0.373 ⎟ ⎜ FBC ⎟ ⎜ −0.373 ⎟ ⎜ F ⎟ = ⎜ 0.333 ⎟ kN ⎟ ⎜ BD ⎟ ⎜ ⎜ FCD ⎟ ⎜ −0.373 ⎟ ⎜ ⎟ ⎜ 1.202 ⎟ ⎟ ⎜ FCE ⎟ ⎜ 0 ⎝ ⎠ ⎜F ⎟ ⎝ DE ⎠ Now Scale the answer P1 = P

P2 = P

Tmax

(

max F AD , F AB , F BC , F BD , F CD , F CE , FDE

)

Cmax min F AD , F AB , F BC , F BD , F CD , F CE , FDE

(

(

)

P = min P 1 , P 2

)

P = 4.16 kN

Problem 6-18 Determine the force in each member of the truss and state if the members are in tension or compression. Hint: The horizontal force component at A must be zero. Why?

465



Chapter 6

Units Used: 3

kip = 10 lb Given: F 1 = 600 lb F 2 = 800 lb a = 4 ft b = 3 ft

θ = 60 deg Solution: Initial Guesses F BA = 1 lb

F BD = 1 lb

F CB = 1 lb

F CD = 1 lb

Given Joint C

Joint B

−F CB − F2 cos ( θ ) = 0 F CB + FBD

b 2

a +b

−F CD − F 2 sin ( θ ) = 0 =0

2

⎛ FBA ⎞ ⎜ ⎟ ⎜ FBD ⎟ ⎜ ⎟ = Find ( FBA , FBD , FCB , FCD) ⎜ FCB ⎟ ⎜F ⎟ ⎝ CD ⎠

−F BA − F BD

a 2

2

a +b

− F1 = 0

⎛ FBA ⎞ ⎛ 3 ⎜ ⎟ −1.133 × 10 ⎞ ⎜ ⎟ Positive means Tension ⎜ FBD ⎟ 666.667 ⎟ lb Negative means ⎜ ⎟=⎜ ⎜ ⎟ Compression F −400 ⎜ CB ⎟ ⎜ ⎜ F ⎟ ⎝ −692.82 ⎟⎠ ⎝ CD ⎠

Problem 6-19 Determine the force in each member of the truss and state if the members are in tension or compression. Hint: The resultant force at the pin E acts along member ED. Why? Units Used: 3

kN = 10 N

466



Chapter 6

Given: F 1 = 3 kN F 2 = 2 kN a = 3m b = 4m Solution: Initial Guesses: F CB = 1 kN

F CD = 1 kN

F BA = 1 kN

F BD = 1 kN

F DA = 1 kN

F DE = 1 kN

Given Joint C

−F 2 − F CD

Joint B

2a

−F CB − FCD

=0

2

( 2 a) + b b 2

2

=0 2

( 2 a) + b

−F BA + F CB = 0 −F 1 − F BD = 0

Joint D

(FCD − FDA − FDE) (

2a 2

=0 2

( 2 a) + b

F BD + FCD + FDA − FDE

)

b 2

=0 2

( 2 a) + b

467



Chapter 6

⎛ FCB ⎞ ⎜ ⎟ F CD ⎜ ⎟ ⎜F ⎟ ⎜ BA ⎟ = Find F , F , F , F , F , F ( CB CD BA BD DA DE) ⎜ FBD ⎟ ⎜ ⎟ ⎜ FDA ⎟ ⎜ ⎟ ⎝ FDE ⎠ ⎛ FCB ⎞ ⎜ ⎟ ⎛ 3 ⎞ F ⎜ CD ⎟ ⎜ −3.606 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎟ BA 3 ⎜ ⎟= ⎜ ⎟ kN ⎜ FBD ⎟ −3 ⎟ ⎜ ⎜ ⎟ ⎜ FDA ⎟ ⎜ 2.704 ⎟ ⎜ ⎟ ⎜⎝ −6.31 ⎟⎠ F ⎝ DE ⎠


Problem 6-20 Each member of the truss is uniform and has a mass density ρ. Determine the approximate force in each member due to the weight of the truss. State if the members are in tension or compression. Solve the problem by assuming the weight of each member can be represented as a vertical force, half of which is applied at each end of the member. Given:

ρ = 8

kg m

g = 9.81

m 2

s F1 = 0 N F2 = 0 N a = 3m b = 4m

468



Chapter 6

Solution: Initial Guesses: F CB = 1 N

F CD = 1 N

F BA = 1 N

F BD = 1 N

F DA = 1 N

F DE = 1 N

Given Joint C

−F CB − FCD

−F 2 − F CD

Joint B

2a 2

( 2 a) + b

=0 2

−F BA + F CB = 0

⎛ b⎞ −F 1 − F BD − ρ g⎜ a + ⎟ = 0 ⎝ 4⎠ Joint D

2 2⎤ ⎛ a⎞ + ⎛ b⎞ ⎥ = 0 ⎜ ⎟ ⎜ ⎟ ⎝ 2⎠ ⎝ 4⎠ ⎦

⎡a − ρ g⎢ + 2 2 ⎣2 ( 2 a) + b b

(

F BD + FCD + FDA − FDE

(FCD − FDA − FDE)

⎡b − ρ g⎢ + 3 2 2 ⎣4 ( 2 a) + b

)

b

2a 2

2 2⎤ ⎛ a⎞ + ⎛ b⎞ ⎥ = 0 ⎜ ⎟ ⎜ ⎟ ⎝ 2⎠ ⎝ 4⎠ ⎦

=0 2

( 2 a) + b

⎛ FCB ⎞ ⎜ ⎟ F CD ⎜ ⎟ ⎜F ⎟ ⎜ BA ⎟ = Find F , F , F , F , F , F ( CB CD BA BD DA DE) ⎜ FBD ⎟ ⎜ ⎟ ⎜ FDA ⎟ ⎜ ⎟ ⎝ FDE ⎠

469



⎛ FCB ⎞ ⎜ ⎟ ⎛ 389 ⎞ F CD ⎜ ⎟ ⎜ −467 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎜ BA ⎟ = 389 ⎟ N ⎜ FBD ⎟ ⎜ −314 ⎟ ⎟ ⎜ ⎟ ⎜ 736 ⎜ ⎟ ⎜ FDA ⎟ ⎜ ⎜ ⎟ ⎝ −1204 ⎟⎠ F ⎝ DE ⎠

Chapter 6


Problem 6-21 Determine the force in each member of the truss in terms of the external loading and state if the members are in tension or compression.

Solution: Joint B: +

↑Σ Fy = 0;

F BA sin ( 2 θ ) − P = 0 F BA = P csc ( 2 θ )

+ Σ F x = 0; →

( C)

F BAcos( 2 θ ) − FBC = 0 F BC = Pcot( 2 θ )

( C)

Joint C: + Σ F x = 0; →

P cot ( 2 θ ) + P + F CD cos ( 2 θ ) − F CA cos ( θ ) = 0

+

F CD sin ( 2 θ ) − FCA sin ( θ ) = 0

↑Σ Fy = 0;

470



F CA =

Chapter 6

cot ( 2 θ ) + 1

cos ( θ ) − sin ( θ ) cot( 2 θ )

P

F CA = ( cot ( θ ) csc ( θ ) − sin ( θ ) + 2 cos ( θ ) ) P

( T)

F CD = ( cot ( 2 θ ) + 1) P

( C)

Joint D: + Σ F x = 0; →

F DA − ⎡⎣cot( 2 θ ) + 1⎤⎦ ⎡⎣cos( 2 θ )⎤⎦ P = 0 F DA = ⎡⎣cot( 2 θ ) + 1⎤⎦ ⎡⎣cos( 2 θ )⎤⎦ P

( C)

Problem 6-22 The maximum allowable tensile force in the members of the truss is Tmax, and the maximum allowable compressive force is Cmax. Determine the maximum magnitude P of the two loads that can be applied to the truss. Units Used: 3

kN = 10 N Given: Tmax = 2 kN Cmax = 1.2 kN L = 2m

θ = 30 deg Solution: Initial guesses (assume all bars are in tension). Use a unit load for P and then scale the answer later. F BA = 1 kN

F BC = 1 kN

F CA = 1 kN

F CD = 1 kN

F DA = 1 kN

P = 1 kN

471



Chapter 6

Given Joint B +

↑ Σ Fy = 0; + Σ F x = 0; →

−F BA sin ( 2 θ ) − P = 0 −F BA cos ( 2 θ ) + F BC = 0

Joint C +

↑Σ Fy = 0;

−F CA sin ( θ ) − F CD sin ( 2 θ ) = 0

+ Σ F x = 0; →

Joint D

−F BC + P − FCD cos ( 2 θ ) − FCA cos ( θ ) = 0

+ Σ F x = 0; →

−F DA + F CD cos ( 2 θ ) = 0

⎛⎜ FBA ⎞⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FCA ⎟ = Find ( FBA , FBC , FCA , FCD , FDA) ⎜F ⎟ ⎜ CD ⎟ ⎜ FDA ⎟ ⎝ ⎠

⎛⎜ FBA ⎞⎟ ⎜ FBC ⎟ ⎜ ⎟ ans = ⎜ FCA ⎟ ⎜F ⎟ ⎜ CD ⎟ ⎜ FDA ⎟ ⎝ ⎠

⎛ −1.155 ⎞ ⎜ −0.577 ⎟ ⎜ ⎟ ans = ⎜ 2.732 ⎟ kN ⎜ −1.577 ⎟ ⎜ ⎟ ⎝ −0.789 ⎠

Now find the biggest tension and the biggest compression. T = max ( ans)

T = 2.732 kN

C = min ( ans)

C = −1.577 kN

Decide which is more important and scale the answer

⎡⎛ Tmax ⎞ ⎤ ⎢⎜ ⎟ ⎥ T ⎢ ⎜ ⎟ P⎥ P = min ⎢⎜ −Cmax ⎟ ⎥ ⎢⎜ ⎟ ⎥ ⎣⎝ C ⎠ ⎦

P = 732.051 N

Problem 6-23

472



Chapter 6

The Fink truss supports the loads shown. Determine the force in each member and state if the members are in tension or compression. Approximate each joint as a pin. Units Used: 3


a = 2.5 ft

F 2 = 1 kip

θ = 30 deg

F 3 = 1 kip Solution: Entire truss:

(

)

ΣF x = 0;

E x = F1 + F2 + F3 + F2 + F1 sin ( θ )

ΣME = 0;

− Ay 4a cos ( θ ) + F 14a + F 23a + F 32a + F 2 a = 0 Ay =

ΣF y = 0;

2 F1 + 2 F2 + F3

E x = 2000 lb

Ay = 2309.4 lb

2 cos ( θ )

E y = − Ay + 2 cos ( θ ) F1 + 2 cos ( θ ) F 2 + cos ( θ ) F3

E y = 1154.7 lb

Joint A: ΣF y = 0;

F AB =

−cos ( θ ) F 1 + Ay sin ( θ )

F AB = 3.75 kip ΣF x = 0;

(C)

F AH = −sin ( θ ) F 1 + F AB cos ( θ ) F AH = 3 kip

(T)

473



Chapter 6

Joint B: ΣF x = 0;

F BC = FAB

F BC = 3.75 kip

ΣF y = 0;

F BH = F 2

F BH = 1 kip

(C)

(C)

Joint H: Σ F y = 0; ΣF x = 0;

F HC = F 2

F HC = 1 kip

(T)

F GH = −F2 cos ( −90 deg + θ ) − FHC cos ( −90 deg + θ ) + F AH F GH = 2 kip

(T)

Joint E: ΣF y = 0;

F EF =

(

− F 1 − E x sin ( θ ) − Ey cos ( θ )

F EF = 3 kip ΣF x = 0;

sin ( θ )

)

(T)

F ED = −Ey sin ( θ ) + Ex cos ( θ ) + F EF cos ( θ ) F ED = 3.75 kip

(C)

Joint D: ΣF x = 0;

F DC = F ED F DC = 3.75 kip

ΣF y = 0;

(C)

F DF = F2 F DF = 1 kip

(C)

Joint C: ΣF x = 0;

F CF = F HC F CF = 1 kip

ΣF y = 0;

(T)

F CG = F3 + FHC cos ( 90 deg − θ ) ( 2)

F CG = 2 kip

(C)

F FG = FEF − FCF cos ( 90 deg − θ ) ( 2)

F FG = 2 kip

(T)

Joint F: ΣF x = 0;

474



Chapter 6

Problem 6-24 Determine the force in each member of the double scissors truss in terms of the load P and state if the members are in tension or compression.

Solution: ΣΜ A = 0; +

↑ ΣF y = 0;

P

L 2L +P − Dy L = 0 3 3

Ay + Dy − 2 P = 0

Joint F: +

1

↑ ΣF y = 0;

F FB

+ ΣF x = 0; →

F FD − F FE − F FB

−P=0

2 1

=0

2

475



Chapter 6

Joint E: +

↑ ΣF y = 0;

+ ΣF x = 0; →

1

F EC

−P=0

2 1

F EF − F EA + F EC

=0

2

Joint B: +

↑ ΣF y = 0;

F BA

+ ΣF x = 0; →

F BA

1 2 1 2

+ F BD

1

+ F FB

1

1

− F FB

5

=0

2

⎛ 2 ⎞=0 ⎟ ⎝ 5⎠

− F BD ⎜

2

Joint C: +

↑ ΣF y = 0;

F CA

+ ΣF x = 0; →

F CA

1 5 2 5

+ FCD

1

− FEC

1

2

2

− FEC

1

− FCD

1

=0

2 =0

2

476



Chapter 6

Joint A: + ΣF x = 0; →

F AE − F BA

1 2

− F CA

2

=0

5

Solving we find F EF = 0.667 P ( T ) F FD = 1.67 P ( T) F AB = 0.471 P ( C) F AE = 1.67 P ( T) F AC = 1.49 P ( C) F BF = 1.41 P ( T) F BD = 1.49 P ( C) F EC = 1.41 P ( T) F CD = 0.471 P ( C)

Problem 6-25 Determine the force in each member of the truss and state if the members are in tension or compression. Hint: The vertical component of force at C must equal zero. Why? Units Used: 3


477



Chapter 6

F 2 = 8 kN a = 1.5 m b = 2m c = 2m

Solution: Initial Guesses: F AB = 1 kN

F AE = 1 kN

F EB = 1 kN

F BC = 1 kN

F BD = 1 kN

F ED = 1 kN

Given Joint A

F AB

F AB

Joint E

a 2

2

a +c c 2

2

a +c

+ F AE = 0 − F1 = 0

F ED − F AE = 0 F EB − F 2 = 0

Joint B

F BC + F BD −F EB − F BD

b 2

2

b +c c 2

2

b +c

− F AB − F AB

a 2

=0 2

a +c c 2

=0 2

a +c

⎛ FAB ⎞ ⎜ ⎟ F AE ⎜ ⎟ ⎜F ⎟ ⎜ EB ⎟ = Find F , F , F , F , F , F ( AB AE EB BC BD ED) ⎜ FBC ⎟ ⎜ ⎟ ⎜ FBD ⎟ ⎜ ⎟ ⎝ FED ⎠

478



Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎛ 7.5 ⎞ F AE ⎜ ⎟ ⎜ −4.5 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎟ 8 ⎜ EB ⎟ = kN ⎜ FBC ⎟ ⎜ 18.5 ⎟ ⎟ ⎜ ⎟ ⎜ − 19.799 ⎜ ⎟ ⎜ FBD ⎟ ⎜ ⎜ ⎟ ⎝ −4.5 ⎟⎠ F ⎝ ED ⎠

Positive means Tension, Negative means Compresson.

Problem 6-26 Each member of the truss is uniform and has a mass density ρ. Remove the external loads F1 and F 2 and determine the approximate force in each member due to the weight of the truss. State if the members are in tension or compression. Solve the problem by assuming the weight of each member can be represented as a vertical force, half of which is applied at each end of the member. Given: F1 = 0 F2 = 0

ρ = 8

kg m

a = 1.5 m b = 2m c = 2m g = 9.81

m 2

s Solution:

Find the weights of each bar. 2

2

WAB = ρ g a + c

WBC = ρ g b

WBE = ρ g c

WAE = ρ g a

WBD = ρ g b + c

2

2

WDE = ρ g b

479



Guesses

Chapter 6

F AB = 1 N

F AE = 1 N

F BC = 1 N

F BE = 1 N

F BD = 1 N

F DE = 1 N

Given Joint A

a

F AE +

2

a +c

c 2

2

a +c Joint E

F AB −

F AB = 0 WAB + WAE 2

=0

F DE − F AE = 0 F BE −

Joint B

2

F BC +

WAE + WBE + WDE 2 b 2

2

b +c

−c

F BD −

F AB − F BE − 2 2 a +c

=0

a 2

2

a +c c 2

2

b +c

F AB = 0

F BD −

WAB + WBE + WBD + WBC 2

=0

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAE ⎟ ⎜F ⎟ ⎜ BC ⎟ = Find F , F , F , F , F , F ( AB AE BC BD BE DE) ⎜ FBD ⎟ ⎜ ⎟ ⎜ FBE ⎟ ⎜ ⎟ ⎝ FDE ⎠ ⎛ FAB ⎞ ⎜ ⎟ ⎛ 196 ⎞ F ⎜ AE ⎟ ⎜ −118 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎟ BC 857 ⎜ ⎟= ⎜ ⎟N ⎜ FBD ⎟ −1045 ⎜ ⎟ ⎜ ⎟ 216 ⎜ ⎟ ⎜ FBE ⎟ ⎜ ⎟ ⎜⎝ −118 ⎟⎠ FDE ⎝ ⎠

Positive means tension, Negative means Compression.

480



Chapter 6


kN = 10 N Given: P 1 = 4 kN P 2 = 0 kN a = 2m

θ = 15 deg Solution: Take advantage of the symetry. Initial Guesses: F BD = 1 kN

F CD = 1 kN

F CA = 1 kN

F BC = 1 kN

Given Joint D Joint B

−P 1 2

F AB = 1 kN

− F BD sin ( 2 θ ) − FCD sin ( 3 θ ) = 0

−P 2 cos ( 2 θ ) − F BC = 0 F BD − F AB − P 2 sin ( 2 θ ) = 0

Joint C

F CD cos ( θ ) − F CA cos ( θ ) = 0

(FCD + FCA)sin(θ ) + FBC = 0 ⎛⎜ FBD ⎞⎟ ⎜ FCD ⎟ ⎜ ⎟ F ⎜ AB ⎟ = Find ( FBD , FCD , FAB , FCA , FBC ) ⎜F ⎟ ⎜ CA ⎟ ⎜ FBC ⎟ ⎝ ⎠

⎛⎜ FFD ⎞⎟ ⎛⎜ FBD ⎞⎟ ⎜ FED ⎟ ⎜ FCD ⎟ ⎜ ⎟ ⎜ ⎟ F F = ⎜ GF ⎟ ⎜ AB ⎟ ⎜F ⎟ ⎜F ⎟ ⎜ EG ⎟ ⎜ CA ⎟ ⎜ FFE ⎟ ⎜ FBC ⎟ ⎝ ⎠ ⎝ ⎠

481



⎛⎜ FBD ⎞⎟ ⎛ −4 ⎞ ⎜ FCD ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FAB ⎟ = ⎜ −4 ⎟ kN ⎜F ⎟ ⎜ 0 ⎟ ⎜ CA ⎟ ⎜ ⎟ ⎜ FBC ⎟ ⎝ 0 ⎠ ⎝ ⎠

Chapter 6

⎛⎜ FFD ⎞⎟ ⎛ −4 ⎞ ⎜ FED ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FGF ⎟ = ⎜ −4 ⎟ kN ⎜F ⎟ ⎜ 0 ⎟ ⎜ EG ⎟ ⎜ ⎟ ⎜ FFE ⎟ ⎝ 0 ⎠ ⎝ ⎠

Positvive means Tension, Negative means Compression


kN = 10 N Given: P 1 = 2 kN P 2 = 4 kN a = 2m

θ = 15 deg Solution: Take advantage of the symmetry. Initial Guesses: F BD = 1 kN

F CD = 1 kN

F CA = 1 kN

F BC = 1 kN

Given Joint D Joint B

−P 1 2

F AB = 1 kN

− F BD sin ( 2 θ ) − FCD sin ( 3 θ ) = 0

−P 2 cos ( 2 θ ) − F BC = 0 F BD − F AB − P 2 sin ( 2θ ) = 0

Joint C

F CD cos ( θ ) − F CA cos ( θ ) = 0

(FCD + FCA)sin(θ ) + FBC = 0 482



Chapter 6

⎛⎜ FBD ⎞⎟ ⎜ FCD ⎟ ⎜ ⎟ ⎜ FAB ⎟ = Find ( FBD , FCD , FAB , FCA , FBC ) ⎜F ⎟ ⎜ CA ⎟ ⎜ FBC ⎟ ⎝ ⎠ ⎛⎜ FBD ⎞⎟ ⎛ −11.46 ⎞ ⎜ FCD ⎟ ⎜ 6.69 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FAB ⎟ = ⎜ −13.46 ⎟ kN ⎜ F ⎟ ⎜ 6.69 ⎟ ⎜ CA ⎟ ⎜ ⎟ ⎜ FBC ⎟ ⎝ −3.46 ⎠ ⎝ ⎠

⎛⎜ FFD ⎞⎟ ⎛⎜ FBD ⎞⎟ ⎜ FED ⎟ ⎜ FCD ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FGF ⎟ = ⎜ FAB ⎟ ⎜F ⎟ ⎜F ⎟ ⎜ EG ⎟ ⎜ CA ⎟ ⎜ FFE ⎟ ⎜ FBC ⎟ ⎝ ⎠ ⎝ ⎠

⎛⎜ FFD ⎞⎟ ⎛ −11.46 ⎞ ⎜ FED ⎟ ⎜ 6.69 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FGF ⎟ = ⎜ −13.46 ⎟ kN ⎜ F ⎟ ⎜ 6.69 ⎟ ⎜ EG ⎟ ⎜ ⎟ ⎜ FFE ⎟ ⎝ −3.46 ⎠ ⎝ ⎠

Positvive means Tension, Negative means Compression


kip = 10 lb Given: F 1 = 2 kip F 2 = 1.5 kip F 3 = 3 kip F 4 = 3 kip a = 4 ft b = 10 ft Solution:

a θ = atan ⎛⎜ ⎟⎞

⎝ b⎠

Initial Guesses

483



Chapter 6

F AB = 1 lb

F BC = 1 lb

F CD = 1 lb

F DE = 1 lb

F AI = 1 lb

F BI = 1 lb

F CI = 1 lb

F CG = 1 lb

F CF = 1 lb

F DF = 1 lb

F EF = 1 lb

F HI = 1 lb

F GI = 1 lb

F GH = 1 lb

F FG = 1 lb

Given Joint A

F AI cos ( θ ) + FAB = 0

Joint B

F BC − F AB = 0 F BI = 0

Joint C

(

)

F CD − F BC + FCF − F CI cos ( θ ) = 0

(

)

F CG + FCF + F CI sin ( θ ) = 0 Joint D

F DE − F CD = 0 F DF = 0

Joint I

(

)

F 2 + FGI + F CI − F AI cos ( θ ) = 0

(

)

F HI − FBI + FGI − F AI − FCI sin ( θ ) = 0 Joint H

F GH cos ( θ ) + F 1 = 0 −F GH sin ( θ ) − FHI = 0

Joint G

Joint F

(FFG − FGH − FGI )cos (θ ) = 0 −F 3 − F CG + ( FGH − FFG − FGI ) sin ( θ ) = 0 (FEF − FFG − FCF)cos (θ ) = 0 (FFG − FCF − FEF)sin(θ ) − F4 − FDF = 0

484



Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FCD ⎟ ⎜F ⎟ ⎜ DE ⎟ ⎜ FAI ⎟ ⎜ ⎟ ⎜ FBI ⎟ ⎜ ⎟ ⎜ FCI ⎟ ⎜ F ⎟ = Find F , F , F , F , F , F , F , F , F , F , F , F , F , F , F ( AB BC CD DE AI BI CI CG CF DF EF HI GI GH ⎜ CG ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDF ⎟ ⎜F ⎟ ⎜ EF ⎟ ⎜ FHI ⎟ ⎜ ⎟ ⎜ FGI ⎟ ⎜F ⎟ ⎜ GH ⎟ ⎜ FFG ⎟ ⎝ ⎠ ⎛⎜ FAB ⎞⎟ ⎛ 3.75 ⎞ ⎜ FBC ⎟ ⎜ 3.75 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FCD ⎟ = ⎜ 7.75 ⎟ kip ⎜ F ⎟ ⎜ 7.75 ⎟ ⎜ DE ⎟ ⎜ ⎟ ⎜ FAI ⎟ ⎝ −4.04 ⎠ ⎝ ⎠

⎛⎜ FBI ⎞⎟ ⎛ 0 ⎞ ⎜ FCI ⎟ ⎜ 0.27 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FCG ⎟ = ⎜ 1.4 ⎟ kip ⎜ F ⎟ ⎜ −4.04 ⎟ ⎜ CF ⎟ ⎜ ⎟ ⎜ FDF ⎟ ⎝ 0 ⎠ ⎝ ⎠

⎛⎜ FEF ⎟⎞ ⎛ −12.12 ⎞ ⎜ FHI ⎟ ⎜ 0.8 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FGI ⎟ = ⎜ −5.92 ⎟ kip ⎜ F ⎟ ⎜ −2.15 ⎟ ⎜ GH ⎟ ⎜ ⎟ ⎜ FFG ⎟ ⎝ −8.08 ⎠ ⎝ ⎠


Problem 6-30 The Howe bridge truss is subjected to the loading shown. Determine the force in members DE, EH, and HG, and state if the members are in tension or compression. Units Used: 3

kN = 10 N

485



Chapter 6

Given: F 1 = 30 kN F 2 = 20 kN F 3 = 20 kN F 4 = 40 kN a = 4m b = 4m Solution: −F 2 a − F 3( 2a) − F 4( 3a) + Gy( 4a) = 0 Gy =

F2 + 2F 3 + 3F4 4

Gy = 45 kN Guesses

F DE = 1 kN

F EH = 1 kN

F HG = 1 kN

Given −F DE − F HG = 0

Gy − F 4 − F EH = 0

F DE b + Gy a = 0

⎛ FDE ⎞ ⎜ ⎟ F ⎜ EH ⎟ = Find ( FDE , FEH , FHG) ⎜F ⎟ ⎝ HG ⎠

⎛ FDE ⎞ ⎛ −45 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ EH ⎟ = ⎜ 5 ⎟ kN ⎜ F ⎟ ⎝ 45 ⎠ ⎝ HG ⎠

Positive (T) Negative (C)

Problem 6-31 The Pratt bridge truss is subjected to the loading shown. Determine the force in members LD, LK, CD, and KD, and state if the members are in tension or compression. Units Used: 3

kN = 10 N

486



Chapter 6

Given: F 1 = 50 kN F 2 = 50 kN F 3 = 50 kN a = 4m b = 3m Solution: Ax = 0 Ay =

3F 3 + 4F2 + 5F 1 6

Guesses F LD = 1 kN

F LK = 1 kN

F CD = 1 kN

F KD = 1 kN

Given F 2 b + F 1( 2b) − Ay( 3b) − F LK a = 0 F CD a + F1 b − Ay( 2b) = 0 Ay − F1 − F2 −

a ⎞ ⎛ ⎜ 2 2 ⎟ FLD = 0 ⎝ a +b ⎠

−F 3 − F KD = 0

⎛ FLD ⎞ ⎜ ⎟ ⎜ FLK ⎟ ⎜ ⎟ = Find ( FLD , FLK , FCD , FKD) ⎜ FCD ⎟ ⎜F ⎟ ⎝ KD ⎠

⎛ FLD ⎞ ⎜ ⎟ ⎛⎜ 0 ⎟⎞ F ⎜ LK ⎟ ⎜ −112.5 ⎟ kN ⎜ ⎟=⎜ ⎟ ⎜ FCD ⎟ ⎜ 112.5 ⎟ ⎜ F ⎟ ⎝ −50 ⎠ ⎝ KD ⎠


487



Chapter 6

Problem 6-32 The Pratt bridge truss is subjected to the loading shown. Determine the force in members JI, JE, and DE, and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: F 1 = 50 kN F 2 = 50 kN F 3 = 50 kN a = 4m b = 3m Solution: Initial Guesses Gy = 1 kN F JI = 1 kN F JE = 1 kN F DE = 1 kN Given Entire Truss −F 1 b − F2( 2b) − F3( 3b) + Gy( 6b) = 0 Section −F DE − F JI = 0

F JE + Gy = 0

Gy( 2b) − F DE a = 0 488



Chapter 6

⎛ Gy ⎞ ⎜ ⎟ ⎜ FJI ⎟ ⎜ ⎟ = Find ( Gy , FJI , FJE , FDE) Gy = 50 kN F JE ⎜ ⎟ ⎜F ⎟ ⎝ DE ⎠

⎛ FJI ⎞ ⎛ −75 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ JE ⎟ = ⎜ −50 ⎟ kN ⎜ F ⎟ ⎝ 75 ⎠ ⎝ DE ⎠ Positive means Tension, Negative means Compression

Problem 6-33 The roof truss supports the vertical loading shown. Determine the force in members BC, CK, and KJ and state if these members are in tension or compression.

Units Used: 3

kN = 10 N Given: F 1 = 4 kN F 2 = 8 kN a = 2m b = 3m

489



Chapter 6

Solution: Initial Guesses Ax = 1 kN

Ay = 1 kN

F BC = 1 kN

F CK = 1 kN

F KJ = 1 kN Given Ax = 0 F 2( 3a) + F 1( 4a) − Ay( 6a) = 0

⎛ 2b ⎞ ⎟ + Ax⎜ ⎟ − Ay( 2a) = 0 ⎝3⎠ ⎝3⎠

F KJ ⎛⎜

2b ⎞

F KJ + A x +

3a ⎞F = 0 ⎛ BC ⎜ 2 ⎟ 2 ⎝ b + 9a ⎠

F CK + A y +

b ⎞F = 0 ⎛ BC ⎜ 2 2⎟ ⎝ b + 9a ⎠

⎛⎜ Ax ⎞⎟ ⎜ Ay ⎟ ⎜ ⎟ F ⎜ KJ ⎟ = Find ( Ax , Ay , FKJ , FCK , FBC ) ⎜F ⎟ ⎜ CK ⎟ ⎜ FBC ⎟ ⎝ ⎠

⎛⎜ Ax ⎞⎟ ⎛ 0 ⎞ ⎜ Ay ⎟ ⎜ 6.667 ⎟ ⎜ ⎟ ⎜ ⎟ F 13.333 = ⎜ KJ ⎟ ⎜ ⎟ kN Positive (T) Negative (C) ⎜F ⎟ ⎜ 0 ⎟ CK ⎜ ⎟ ⎜ ⎟ ⎜ FBC ⎟ ⎝ −14.907 ⎠ ⎝ ⎠

Problem 6-34 Determine the force in members CD, CJ, KJ, and DJ of the truss which serves to support the deck of a bridge. State if these members are in tension or compression. Units Used: 3

kip = 10 lb Given: F 1 = 4000 lb F 2 = 8000 lb

490



Chapter 6

F 3 = 5000 lb a = 9 ft b = 12 ft

Solution Initial Guesses:

F DJ = 1 kip

Ay = 1 kip

F CD = 1 kip

F CJ = 1 kip

F KJ = 1 kip

Given F 3 a + F 2( 4a) + F 1( 5a) − Ay( 6a) = 0 − Ay( 2a) + F 1 a + FKJ b = 0 F CD + F KJ +

a ⎛ ⎞ ⎜ 2 2 ⎟ FCJ = 0 ⎝ a +b ⎠

Ay − F1 − F2 −

b ⎛ ⎞ ⎜ 2 2 ⎟ FCJ = 0 ⎝ a +b ⎠

−F DJ = 0

⎛⎜ Ay ⎞⎟ ⎜ FKJ ⎟ ⎜ ⎟ ⎜ FCJ ⎟ = Find ( Ay , FKJ , FCJ , FDJ , FCD) ⎜F ⎟ ⎜ DJ ⎟ ⎜ FCD ⎟ ⎝ ⎠

⎛⎜ Ay ⎞⎟ ⎛ 9.5 ⎞ ⎜ FKJ ⎟ ⎜ 11.25 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FCJ ⎟ = ⎜ −3.125 ⎟ kip ⎜F ⎟ ⎜ 0 ⎟ ⎜ DJ ⎟ ⎜ ⎟ − 9.375 ⎝ ⎠ ⎜ FCD ⎟ ⎝ ⎠


491



Chapter 6

Problem 6-35 Determine the force in members EI and JI of the truss which serves to support the deck of a bridge. State if these members are in tension or compression. Units Used: 3

kip = 10 lb Given: F 1 = 4000 lb F 2 = 8000 lb F 3 = 5000 lb a = 9 ft b = 12 ft Solution: Initial Guesses: Gy = 1 kip

F EI = 1 kip

F JI = 1 kip

Given −F 1 a − F22a − F35a + Gy6a = 0 Gy2a − F 3 a − FJI b = 0 F EI − F3 + Gy = 0

⎛ Gy ⎞ ⎜ ⎟ F ⎜ JI ⎟ = Find ( Gy , FJI , FEI ) ⎜F ⎟ ⎝ EI ⎠

⎛ Gy ⎞ ⎛ 7.5 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ JI ⎟ = ⎜ 7.5 ⎟ kip ⎜ F ⎟ ⎝ −2.5 ⎠ ⎝ EI ⎠


Problem 6-36 Determine the force in members BE, EF, and CB, and state if the members are in tension or compression. Units Used: 3

kN = 10 N

492



Chapter 6

Given: F 1 = 5 kN

F 4 = 10 kN

F 2 = 10 kN

a = 4m

F 3 = 5 kN

b = 4m

a θ = atan ⎛⎜ ⎟⎞

Solution:

⎝ b⎠

Inital Guesses F CB = 1 kN F BE = 1 kN

F EF = 1 kN

Given F 1 + F 2 − F BE cos ( θ ) = 0 −F CB − FEF − FBE sin ( θ ) − F3 = 0 −F 1 a + F CB b = 0

⎛ FCB ⎞ ⎜ ⎟ F ⎜ BE ⎟ = Find ( FCB , FBE , FEF) ⎜F ⎟ ⎝ EF ⎠ ⎛ FCB ⎞ ⎛ 5 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FBE ⎟ = ⎜ 21.2 ⎟ kN ⎜ F ⎟ ⎝ −25 ⎠ ⎝ EF ⎠


Problem 6-37 Determine the force in members BF, BG, and AB, and state if the members are in tension or compression. Units Used: 3


F 4 = 10 kN

493



F 2 = 10 kN

a = 4m

F 3 = 5 kN

b = 4m

Solution:

Chapter 6

a θ = atan ⎛⎜ ⎟⎞

⎝ b⎠

Inital Guesses F AB = 1 kN F BG = 1 kN

F BF = 1 kN

Given F 1 + F 2 + F 4 + F BG cos ( θ ) = 0 −F 1 3a − F 22a − F 4 a + FAB b = 0 −F BF = 0

⎛ FAB ⎞ ⎜ ⎟ ⎜ FBG ⎟ = Find ( FAB , FBG , FBF) ⎜F ⎟ ⎝ BF ⎠ ⎛ FAB ⎞ ⎛ 45 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ BG ⎟ = ⎜ −35.4 ⎟ kN Positive (T) Negative (C) ⎜F ⎟ ⎝ 0 ⎠ BF ⎝ ⎠

Problem 6-38 Determine the force developed in members GB and GF of the bridge truss and state if these members are in tension or compression. Given: F 1 = 600 lb

494



Chapter 6

F 2 = 800 lb a = 10 ft b = 10 ft c = 4 ft Solution: Initial Guesses Ax = 1 lb

Ay = 1 lb

F GB = 1 lb

F GF = 1 lb

Given F 2 b + F1( b + 2c) − A y2( b + c) = 0 Ax = 0 Ay − F GB = 0 − Ay b − FGF a = 0

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ ⎟ = Find ( Ax , Ay , FGB , FGF) ⎜ FGB ⎟ ⎜F ⎟ ⎝ GF ⎠ ⎛ Ax ⎞ ⎞ ⎜ ⎟ ⎛⎜ 0 ⎟ ⎜ Ay ⎟ ⎜ 671.429 ⎟ lb Positive (T) ⎜ ⎟=⎜ FGB 671.429 ⎟ Negative (C) ⎜ ⎟ ⎜ ⎟ ⎜ F ⎟ ⎝ −671.429 ⎠ ⎝ GF ⎠

Problem 6-39 Determine the force members BC, FC, and FE, and state if the members are in tension or compression. Units Used: 3

kN = 10 N

495



Chapter 6

Given: F 1 = 6 kN F 2 = 6 kN a = 3m b = 3m Solution:

a θ = atan ⎛⎜ ⎟⎞

⎝ b⎠

Initial Guesses Dy = 1 kN

F BC = 1 kN

F FC = 1 kN

F FE = 1 kN

Given −F 1 b − F2( 2b) + Dy( 3b) = 0 Dy b − FFE cos ( θ ) a = 0

(

)

−F FC − FBC + FFE cos ( θ ) = 0

(

)

−F 2 + Dy + FFE + FBC sin ( θ ) = 0

⎛ Dy ⎞ ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎟ = Find ( Dy , FBC , FFC , FFE) ⎜ FFC ⎟ ⎜F ⎟ ⎝ FE ⎠ ⎛ Dy ⎞ ⎜ ⎟ ⎛⎜ 6 ⎟⎞ F ⎜ BC ⎟ ⎜ −8.49 ⎟ kN ⎜ ⎟=⎜ ⎟ ⎜ FFC ⎟ ⎜ 0 ⎟ ⎜ F ⎟ ⎝ 8.49 ⎠ ⎝ FE ⎠


496



Chapter 6

Problem 6-40 Determine the force in members IC and CG of the truss and state if these members are in tension or compression. Also, indicate all zero-force members. Units Used: 3

kN = 10 N Given: F 1 = 6 kN F 2 = 6 kN a = 1.5 m b = 2m Solution: By inspection of joints B, D, H and I. AB, BC, CD, DE, HI, and GI are all zero-force members. Guesses

Ay = 1 kN

Given

− Ay( 4a) + F 1( 2a) + F 2 a = 0 − Ay( 2a) − −a 2

2

a +b −b 2

2

a +b

F IC = 1 kN

b 2

2

a +b

FIC +

FIC −

FIC a −

a 2

a +b

2

b 2

a +b

2

⎛ Ay ⎞ ⎜ ⎟ ⎜ FIC ⎟ ⎜ ⎟ = Find ( Ay , FIC , FCG , FCJ ) ⎜ FCG ⎟ ⎜F ⎟ ⎝ CJ ⎠

F CG = 1 kN

a 2

2

a +b

F CJ = 1 kN

FIC b = 0

F CJ = 0

F CJ − FCG = 0

⎛ Ay ⎞ ⎜ ⎟ ⎛⎜ 4.5 ⎟⎞ F ⎜ IC ⎟ ⎜ −5.625 ⎟ kN ⎜ ⎟=⎜ ⎟ ⎜ FCG ⎟ ⎜ 9 ⎟ ⎜ F ⎟ ⎝ −5.625 ⎠ ⎝ CJ ⎠


Problem 6-41 Determine the force in members JE and GF of the truss and state if these members are in tension or compression. Also, indicate all zero-force members. 497



Chapter 6

Units Used: 3

kN = 10 N Given: F 1 = 6 kN F 2 = 6 kN a = 1.5 m b = 2m Solution: By inspection of joints B, D, H and I. AB, BC, CD, DE, HI, and GI are all zero-force members. E y = 1 kN

Guesses Given

F JE = 1 kN

F GF = 1 kN

−F 1( 2a) − F 2( 3a) + E y( 4a) = 0 Ey +

b 2

a +b

−a 2

2

a +b

2

F JE = 0

FJE − FGF = 0

⎛ Ey ⎞ ⎜ ⎟ F ⎜ JE ⎟ = Find ( Ey , FJE , FGF) ⎜F ⎟ ⎝ GF ⎠

⎛ Ey ⎞ ⎛ 7.5 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ JE ⎟ = ⎜ −9.375 ⎟ kN Positive (T) Negative (C) ⎜ F ⎟ ⎝ 5.625 ⎠ ⎝ GF ⎠

Problem 6-42 Determine the force in members BC, HC, and HG. After the truss is sectioned use a single equation of equilibrium for the calculation of each force. State if these members are in tension or compression. Units Used: 3

kN = 10 N

498



Chapter 6

Given: F 1 = 2 kN F 4 = 5 kN a = 5 m F 2 = 4 kN F 5 = 3 kN b = 2 m F 3 = 4 kN

c = 3m

Solution: Guesses Ax = 1 kN

Ay = 1 kN

F BC = 1 kN

F HC = 1 kN

F HG = 1 kN

d = 1m

Given c b = a+d a

− Ax = 0

(F1 − Ay)( 4a) + F2( 3a) + F3( 2a) + F4( a) = 0 (F1 − Ay)( a) + Ax( c) − FBC ( c) = 0 (F1 − Ay)( 2a) + F2( a) + ( A y − F 1) ( d) − F 2( a + d ) +

a 2

a +b

2

F HG( c) +

c 2

2

a +c

b 2

a +b

FHC ( a + d) +

⎛ Ay ⎞ ⎜ ⎟ ⎜ Ax ⎟ ⎜ ⎟ FBC ⎜ ⎟ = Find A , A , F , F , F , d ( y x BC HC HG ) ⎜F ⎟ HC ⎜ ⎟ ⎜ FHG ⎟ ⎜ ⎟ ⎝ d ⎠

2

F HG( a) = 0 a 2

2

a +c

F HC ( c) = 0

⎛⎜ Ax ⎞⎟ ⎛ 0 ⎞ = kN ⎜ Ay ⎟ ⎜⎝ 8.25 ⎟⎠ ⎝ ⎠ d = 2.5 m

⎛ FBC ⎞ ⎛ −10.417 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FHC ⎟ = ⎜ 2.235 ⎟ kN ⎜ F ⎟ ⎝ 9.155 ⎠ ⎝ HG ⎠ Positive (T) Negative (C)

Problem 6-43 Determine the force in members CD, CF, and CG and state if these members are in tension or 499



Chapter 6

compression. Units Used: 3

kN = 10 N Given: F 1 = 2 kN F 4 = 5 kN a = 5 m F 2 = 4 kN F 5 = 3 kN b = 2 m F 3 = 4 kN

c = 3m

Solution: Guesses E y = 1 kN

F CD = 1 kN

F CF = 1 kN

F CG = 1 kN

F FG = 1 kN

F GH = 1 kN

Given

(

)

−F 2( a) − F3( 2a) − F4( 3a) + E y − F5 ( 4a) = 0

(

)

(

)

F CD( c) + Ey − F 5 ( a) = 0

a

−F 4( a) − F 5 − E y ( 2a) −

a 2

2

a +b b 2

2

a +b

(F5 − Ey)

FFG −

a 2

a +b

2

2

2

a +b

FFG( b + c) = 0

F GH = 0

(FFG + FGH ) + FCG = 0 a( c − b) b

+ F4⎡⎢a +

⎣

a( c − b) ⎤ b

⎥− ⎦

c 2

2

a +c

F CF⎡⎢2 a +

⎣

a( c − b) ⎤ b

⎥=0 ⎦

500



Chapter 6

⎛ Ey ⎞ ⎜ ⎟ F CD ⎜ ⎟ ⎜F ⎟ ⎜ CF ⎟ = Find E , F , F , F , F , F ( y CD CF CG FG GH ) ⎜ FCG ⎟ ⎜ ⎟ ⎜ FFG ⎟ ⎜ ⎟ ⎝ FGH ⎠

⎛ Ey ⎞ ⎜ ⎟ ⎛ 9.75 ⎞ F CD ⎜ ⎟ ⎜ −11.25 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎜ CF ⎟ = 3.207 ⎟ kN Positive (T) ⎜ FCG ⎟ ⎜ −6.8 ⎟ Negative (C) ⎟ ⎜ ⎟ ⎜ ⎜ FFG ⎟ ⎜ 9.155 ⎟ ⎜ ⎟ ⎜⎝ 9.155 ⎟⎠ F ⎝ GH ⎠

Problem 6-44 Determine the force in members OE, LE, and LK of the Baltimore truss and state if the members are in tension or compression. Units Used: 3


a = 2m

F 2 = 2 kN

b = 2m

F 3 = 5 kN F 4 = 3 kN Solution: Ax = 0 kN Initial Guesses Ay = 1 kN

F OE = 1 kN

F DE = 1 kN F LK = 1 kN F LE = 1 kN Given F LE = 0 F 4( 3b) + F 3( 4b) + F 2( 5b) + F 1( 6b) − Ay( 8b) = 0 F LK + FDE + FOE

b 2

a +b

=0 2

501



Ay − F 1 − F 2 − F OE

Chapter 6

a 2

=0 2

a +b

−F LK ( 2a) + F2( b) + F 1( 2b) − Ay( 4b) = 0

⎛⎜ Ay ⎞⎟ ⎜ FOE ⎟ ⎜ ⎟ ⎜ FDE ⎟ = Find ( Ay , FOE , FDE , FLK , FLE) ⎜F ⎟ ⎜ LK ⎟ ⎜ FLE ⎟ ⎝ ⎠ ⎛⎜ Ay ⎞⎟ ⎛ 6.375 ⎞ = kN ⎜ FDE ⎟ ⎜⎝ 7.375 ⎟⎠ ⎝ ⎠

⎛ FOE ⎞ ⎛ 3.36 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FLE ⎟ = ⎜ 0 ⎟ kN ⎜ F ⎟ ⎝ −9.75 ⎠ ⎝ LK ⎠


Problem 6-45 Determine the force in member GJ of the truss and state if this member is in tension or compression. Units Used: 3

kip = 10 lb Given: F 1 = 1000 lb F 2 = 1000 lb F 3 = 1000 lb F 4 = 1000 lb a = 10 ft

θ = 30 deg

502


= Chapter 6


Solution: Guess

E y = 1 lb

F GJ = 1 lb

Given −F 2( a) − F3( 2a) − F4( 3a) + Ey( 4a) = 0 −F 4( a) + Ey( 2a) + F GJ sin ( θ ) ( 2a) = 0

⎛⎜ Ey ⎟⎞ = Find ( Ey , F GJ ) ⎜ FGJ ⎟ ⎝ ⎠

⎛⎜ Ey ⎟⎞ ⎛ 1.5 ⎞ = kip ⎜ FGJ ⎟ ⎜⎝ −2 ⎟⎠ ⎝ ⎠


Problem 6-46 Determine the force in member GC of the truss and state if this member is in tension or compression. Units Used: 3

kip = 10 lb Given: F 1 = 1000 lb F 2 = 1000 lb F 3 = 1000 lb F 4 = 1000 lb a = 10 ft

θ = 30 deg Solution: Guess

E y = 1 lb

F GJ = 1 lb

F HG = 1 lb

F GC = 1 lb

Given −F 2( a) − F3( 2a) − F4( 3a) + Ey( 4a) = 0 −F 4( a) + Ey( 2a) + F GJ sin ( θ ) ( 2a) = 0 503



Chapter 6

−F HG cos ( θ ) + FGJ cos ( θ ) = 0

(

)

−F 3 − F GC − F HG + FGJ sin ( θ ) = 0

⎛ Ey ⎞ ⎜ ⎟ ⎜ FGJ ⎟ ⎜ ⎟ = Find ( Ey , FGJ , FGC , FHG) F GC ⎜ ⎟ ⎜F ⎟ ⎝ HG ⎠

⎛ Ey ⎞ ⎜ ⎟ ⎛⎜ 1.5 ⎟⎞ ⎜ FGJ ⎟ ⎜ −2 ⎟ ⎜ ⎟ = ⎜ ⎟ kip F GC ⎜ ⎟ ⎜ 1 ⎟ ⎜ F ⎟ ⎝ −2 ⎠ ⎝ HG ⎠


Problem 6-47 Determine the force in members KJ, JN, and CD, and state if the members are in tension or compression. Also indicate all zero-force members. Units Used: 3

kip = 10 lb Given: F = 3 kip a = 20 ft b = 30 ft c = 20 ft Solution:

Ax = 0 2c ⎞ ⎟ ⎝ 2a + b ⎠

θ = atan ⎛⎜

2c ⎞ ⎟ ⎝b⎠

φ = atan ⎛⎜

Initial Guesses: Ay = 1 lb

F CD = 1 lb

F KJ = 1 lb

F JN = 1 lb

504



Chapter 6

Given

⎛ ⎝

F⎜a +

b⎞ ⎟ − Ay( 2a + b) = 0 2⎠

⎛ b⎞ F CD c − Ay⎜ a + ⎟ = 0 ⎝ 2⎠ F CD + F JN cos ( φ ) + F KJ cos ( θ ) = 0 Ay + F JN sin ( φ ) + FKJ sin ( θ ) = 0

⎛ Ay ⎞ ⎜ ⎟ ⎜ FCD ⎟ ⎜ ⎟ = Find ( Ay , FCD , FJN , FKJ ) ⎜ FJN ⎟ ⎜F ⎟ ⎝ KJ ⎠

Ay = 1.5 kip

⎛ FCD ⎞ ⎛ 2.625 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FJN ⎟ = ⎜ 0 ⎟ kip ⎜ F ⎟ ⎝ −3.023 ⎠ ⎝ KJ ⎠ Positive (T), Negative (C)

Problem 6-48 Determine the force in members BG, HG, and BC of the truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: F 1 = 6 kN F 2 = 7 kN F 3 = 4 kN a = 3m b = 3m c = 4.5 m

505



Chapter 6

Solution:

Initial Guesses

F BG = 1 kN

Ax = 1 kN

F HG = 1 kN

Ay = 1 kN

F BC = 1 kN

Given − Ax = 0 − Ay( a) −

a ⎡ ⎤ F ( b) = 0 HG ⎢ 2 2⎥ ⎣ ( c − b) + a ⎦

F 3( a) + F2( 2a) + F1( 3a) − A y( 4a) = 0 F BC +

a ⎡ ⎤F + ⎛ a ⎞F − A = 0 HG ⎜ BG x ⎢ 2 2⎥ 2 2⎟ ⎣ ( c − b) + a ⎦ ⎝ a +c ⎠

Ay − F1 +

c−b c ⎡ ⎤F + ⎛ ⎞F = 0 HG ⎜ BG ⎢ ⎥ ⎟ 2 2 2 2 ⎣ ( c − b) + a ⎦ ⎝ a +c ⎠ 506



Chapter 6

⎛⎜ Ax ⎞⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎜ FHG ⎟ = Find ( Ax , Ay , FHG , FBG , FBC ) ⎜F ⎟ ⎜ BG ⎟ ⎜ FBC ⎟ ⎝ ⎠

⎛⎜ Ax ⎞⎟ ⎛ 0 ⎞ ⎜ Ay ⎟ ⎜ 9 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FHG ⎟ = ⎜ −10.062 ⎟ kN ⎜ F ⎟ ⎜ 1.803 ⎟ ⎜ BG ⎟ ⎜ ⎟ 8 ⎝ ⎠ ⎜ FBC ⎟ ⎝ ⎠


Problem 6-49 The skewed truss carries the load shown. Determine the force in members CB, BE, and EF and state if these members are in tension or compression. Assume that all joints are pinned.

Solution: ΣMB = 0;

−P d − F EF d = 0

ΣME = 0;

−P d +

+ Σ F x = 0; →

2

F CB d = 0

F CB =

FCB − F BE = 0

P BE =

5 P−

1 5

F EF = −P 5 P 2 P 2

F EF = P

( C)

F CB = 1.12P

( T)

F BE = 0.5P

( T)

Problem 6-50 The skewed truss carries the load shown. Determine the force in members AB, BF, and EF and state if these members are in tension or compression. Assume that all joints are pinned. 507



Chapter 6

Solution: ΣMF = 0;

−P 2d + P d + FAB d = 0

F AB = P

F AB = P

( T)

ΣMB = 0;

−P d − F EF d = 0

F EF = −P

F EF = P

( C)

F BE = − 2 P

F BF = 1.41P

( C)

+ Σ F x = 0; →

P + FBF

1

=0

2

Problem 6-51 Determine the force developed in members BC and CH of the roof truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: F 1 = 1.5 kN F 2 = 2 kN

508



Chapter 6

a = 1.5 m b = 1m c = 2m d = 0.8 m Solution: a θ = atan ⎛⎜ ⎟⎞

⎝c⎠

a ⎞ ⎟ ⎝ c − b⎠

φ = atan ⎛⎜

Initial Guesses: E y = 1 kN

F BC = 1 kN

F CH = 1 kN

Given −F 2( d) − F1( c) + Ey( 2c) = 0 F BC sin ( θ ) ( c) + FCH sin ( φ ) ( c − b) + E y( c) = 0 −F BC sin ( θ ) − FCH sin ( φ ) − F1 + Ey = 0

⎛ Ey ⎞ ⎜ ⎟ ⎜ FBC ⎟ = Find ( Ey , FBC , FCH ) Ey = 1.15 kN ⎜F ⎟ ⎝ CH ⎠

⎛⎜ FBC ⎞⎟ ⎛ −3.25 ⎞ = kN ⎜ FCH ⎟ ⎜⎝ 1.923 ⎟⎠ ⎝ ⎠


Problem 6-52 Determine the force in members CD and GF of the truss and state if the members are in tension or compression. Also indicate all zero-force members. Units Used: 3

kN = 10 N Given: F 1 = 1.5 kN F 2 = 2 kN 509



Chapter 6

a = 1.5 m b = 1m c = 2m d = 0.8 m Solution: a θ = atan ⎛⎜ ⎟⎞

⎝c⎠

⎞ ⎟ ⎝ c − b⎠

φ = atan ⎛⎜

a

Initial Guesses: E y = 1 kN

F CD = 1 kN

F GF = 1 kN

Given −F 2( d) − F1( c) + Ey( 2c) = 0 E y( b) + F CD sin ( θ ) ( b) = 0 E y( c) − FGF( a) = 0

⎛ Ey ⎞ ⎜ ⎟ F ⎜ CD ⎟ = Find ( Ey , FCD , FGF) Ey = 1.15 kN ⎜F ⎟ ⎝ GF ⎠

⎛⎜ FCD ⎞⎟ ⎛ −1.917 ⎞ Positive (T) = kN Negative (C) ⎜ FGF ⎟ ⎜⎝ 1.533 ⎟⎠ ⎝ ⎠ DF and CF are zero force members.

Problem 6-53 Determine the force in members DE, DL, and ML of the roof truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: F 1 = 6 kN F 2 = 12 kN F 3 = 12 kN

510



Chapter 6

F 4 = 12 kN a = 4m b = 3m c = 6m Solution:

θ = atan ⎛⎜

c − b⎞

⎟ ⎝ 3a ⎠

⎡b + ⎢ φ = atan ⎢ ⎣

2 ⎤ ( c − b) ⎥ 3 a

⎥ ⎦

Initial Guesses: Ay = 1 kN

F ML = 1 kN

F DL = 1 kN

F DE = 1 kN

Given F 2( a) + F3( 2a) + F4( 3a) + F3( 4a) + F2( 5a) + F1( 6a) − A y( 6a) = 0 2 F 1( 2a) + F 2( a) − A y( 2a) + FML⎡⎢b + ( c − b)⎤⎥ = 0 ⎣ 3 ⎦ Ay − F 1 − F 2 − F 3 + F DE sin ( θ ) − FDL sin ( φ ) = 0 F ML + F DL cos ( φ ) + FDE cos ( θ ) = 0

⎛ Ay ⎞ ⎜ ⎟ ⎜ FML ⎟ ⎜ ⎟ = Find ( Ay , FML , FDE , FDL) ⎜ FDE ⎟ ⎜F ⎟ ⎝ DL ⎠

Ay = 36 kN

⎛ FML ⎞ ⎛ 38.4 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FDE ⎟ = ⎜ −37.1 ⎟ kN ⎜ F ⎟ ⎝ −3.8 ⎠ ⎝ DL ⎠ Positive (T), Negative (C)

Problem 6-54 Determine the force in members EF and EL of the roof truss and state if the members are in 511



Chapter 6

tension or compression. Units Used: 3

kN = 10 N Given: F 1 = 6 kN F 2 = 12 kN F 3 = 12 kN F 4 = 12 kN a = 4m b = 3m c = 6m Solution:

θ = atan ⎛⎜

c − b⎞

⎟ ⎝ 3a ⎠

Initial Guesses: Iy = 1 kN

F EF = 1 kN

F EL = 1 kN Given −F 2( a) − F3( 2a) − F4( 3a) − F3( 4a) − F2( 5a) − F1( 6a) + Iy( 6a) = 0 −F 3( a) − F2( 2a) − F1( 3a) + Iy( 3a) + F EF cos ( θ ) ( c) = 0 −F 4 − F EL − 2F EF sin ( θ ) = 0

⎛ Iy ⎞ ⎜ ⎟ ⎜ FEF ⎟ = Find ( Iy , FEF , FEL) ⎜F ⎟ ⎝ EL ⎠

Iy = 36 kN

⎛⎜ FEF ⎞⎟ ⎛ −37.108 ⎞ = kN ⎜ FEL ⎟ ⎜⎝ 6 ⎟⎠ ⎝ ⎠


Problem 6-55 Two space trusses are used to equally support the uniform sign of mass M. Determine the force developed in members AB, AC, and BC of truss ABCD and state if the members are in tension or compression. Horizontal short links support the truss at joints B and D and there is a ball-and512



Chapter 6

p socket joint at C.

pp

j

Given: M = 50 kg

g = 9.81

m 2

s

a = 0.25 m b = 0.5 m c = 2m Solution:

h =

2

2

b −a

⎛ −a ⎞ AB = ⎜ −c ⎟ ⎜ ⎟ ⎝h⎠

⎛a⎞ AD = ⎜ −c ⎟ ⎜ ⎟ ⎝h⎠

⎛ 2a ⎞ BD = ⎜ 0 ⎟ ⎜ ⎟ ⎝0⎠

⎛a⎞ BC = ⎜ 0 ⎟ ⎜ ⎟ ⎝ −h ⎠

⎛0⎞ AC = ⎜ −c ⎟ ⎜ ⎟ ⎝0⎠

Guesses F AB = 1 N

F AD = 1 N

F AC = 1 N

F BC = 1 N

F BD = 1 N

By = 1 N

Given

⎛⎜ 0 ⎟⎞ 0 ⎟ AB AD AC F AB + FAD + FAC +⎜ =0 ⎜ −M g ⎟ AB AD AC ⎜ 2 ⎟ ⎝ ⎠ ⎛⎜ 0 ⎟⎞ F AB + FBD + F BC + ⎜ −B y ⎟ = 0 AB BD BC ⎜ 0 ⎟ ⎝ ⎠ −AB

BD

BC

513



Chapter 6

⎛ FAB ⎞ ⎜ ⎟ F AD ⎜ ⎟ ⎜F ⎟ ⎜ AC ⎟ = Find F , F , F , F , F , B ( AB AD AC BC BD y) ⎜ FBC ⎟ ⎜ ⎟ ⎜ FBD ⎟ ⎜ ⎟ ⎝ By ⎠

⎛ By ⎞ ⎛ 566 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ AD ⎟ = ⎜ 584 ⎟ N ⎜F ⎟ ⎝ 0 ⎠ ⎝ BD ⎠ ⎛ FAB ⎞ ⎛ 584 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ AC ⎟ = ⎜ −1133 ⎟ N ⎜ F ⎟ ⎝ −142 ⎠ ⎝ BC ⎠ Positive (T), Negative (C)

Problem 6-56 Determine the force in each member of the space truss and state if the members are in tension or compression.The truss is supported by short links at B, C, and D. Given: F = 600 N a = 3m b = 1m c = 1.5 m Solution:

⎛b⎞ ⎜ ⎟ AB = −c ⎜ ⎟ ⎝ −a ⎠ ⎛0⎞ ⎜ ⎟ AC = c ⎜ ⎟ ⎝ −a ⎠ ⎛ −b ⎞ ⎜ ⎟ AD = −c ⎜ ⎟ ⎝ −a ⎠

⎛ −b ⎞ ⎜ ⎟ CD = −2c ⎜ ⎟ ⎝ 0 ⎠

514



⎛ b ⎞ CB = ⎜ −2c ⎟ ⎜ ⎟ ⎝ 0 ⎠

Chapter 6

⎛ −2b ⎞ BD = ⎜ 0 ⎟ ⎜ ⎟ ⎝ 0 ⎠

Guesses F BA = 1 N

F BC = 1 N

F CA = 1 N

F DA = 1 N

F BD = 1 N

F DC = 1 N

By = 1 N

Bz = 1 N

Cz = 1 N

Given

⎛ 0 ⎞ F BA + FCA + F DA +⎜ 0 ⎟ =0 ⎜ ⎟ AB AC AD ⎝ −F ⎠ AB

AC

AD

⎛⎜ 0 ⎟⎞ F CA + FDC + F BC +⎜0 ⎟ =0 AC CD CB ⎜ Cz ⎟ ⎝ ⎠ −AC

CD

CB

⎛0⎞ ⎜ ⎟ F BC + FBD + F BA + ⎜ By ⎟ = 0 CB BD AB ⎜B ⎟ ⎝ z⎠ −CB

BD

−AB

⎛ FBA ⎞ ⎜ ⎟ F BC ⎜ ⎟ ⎜F ⎟ ⎜ CA ⎟ ⎜ FDA ⎟ ⎜ ⎟ F ⎜ BD ⎟ = Find ( FBA , FBC , FCA , FDA , FBD , FDC , By , Bz , Cz) ⎜F ⎟ ⎜ DC ⎟ ⎜ By ⎟ ⎜ ⎟ ⎜ Bz ⎟ ⎟ ⎜ ⎝ Cz ⎠

515



⎛ By ⎞ ⎛ 1.421 × 10− 14 ⎞ ⎜ ⎟ ⎜ ⎟ B = ⎜ z⎟ ⎜ ⎟N 150 ⎜C ⎟ ⎜ ⎟ 300 ⎠ ⎝ z⎠ ⎝

Chapter 6

⎛ FBA ⎞ ⎜ ⎟ ⎛ −175 ⎞ F BC ⎜ ⎟ ⎜ 79.1 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎜ CA ⎟ = −335.4 ⎟ N ⎜ FDA ⎟ ⎜ −175 ⎟ ⎟ ⎜ ⎟ ⎜ 25 ⎜ ⎟ ⎜ FBD ⎟ ⎜ ⎜ ⎟ ⎝ 79.1 ⎟⎠ F ⎝ DC ⎠

Positive (T), Negative (C)

Problem 6-57 Determine the force in each member of the space truss and state if the members are in tension or compression.The truss is supported by short links at A, B, and C. Given: a = 4 ft b = 2 ft c = 3 ft d = 2 ft e = 8 ft

⎛ 0 ⎞ ⎜ ⎟ F = 500 lb ⎜ ⎟ ⎝ 0 ⎠ Solution:

⎛ −c ⎞ ⎜ ⎟ AD = a ⎜ ⎟ ⎝e⎠

⎛ −c ⎞ ⎜ ⎟ BD = −b ⎜ ⎟ ⎝e ⎠

⎛d⎞ ⎜ ⎟ CD = −b ⎜ ⎟ ⎝e ⎠

⎛ 0 ⎞ ⎜ ⎟ AB = a + b ⎜ ⎟ ⎝ 0 ⎠

⎛ −c − d ⎞ ⎛ −c − d ⎞ ⎜ ⎟ ⎜ 0 ⎟ AC = a + b BC = ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎝ 0 ⎠ Guesses

516



Chapter 6

F BA = 1 lb

F BC = 1 lb

F BD = 1 lb

F AD = 1 lb

F AC = 1 lb

F CD = 1 lb

Ay = 1 lb

Az = 1 lb

B x = 1 lb

B z = 1 lb

Cy = 1 lb

Cz = 1 lb

Given F + FAD

−AD AD

+ FBD

−BD BD

+ F CD

−CD CD

=0

⎛0 ⎞ ⎜ ⎟ F AD + F BA + FAC + ⎜ Ay ⎟ = 0 AD AB AC ⎜A ⎟ ⎝ z⎠ AD

AB

AC

⎛ Bx ⎞ ⎜ ⎟ −AB BC BD F BA + FBC + F BD +⎜ 0 ⎟ =0 AB BC BD ⎜B ⎟ ⎝ z⎠ ⎛0⎞ CD −AC −BC ⎜ C ⎟ F CD + F AC + F BC + ⎜ y⎟ = 0 CD AC BC ⎜C ⎟ ⎝ z⎠ ⎛⎜ FBA ⎞⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FBD ⎟ ⎜F ⎟ ⎜ AD ⎟ ⎜ FAC ⎟ ⎜ ⎟ ⎜ FCD ⎟ ⎜ ⎟ = Find ( FBA , FBC , FBD , FAD , FAC , FCD , Ay , Az , Bx , Bz , Cy , Cz) ⎜ Ay ⎟ ⎜ A ⎟ ⎜ z ⎟ ⎜ Bx ⎟ ⎜ ⎟ ⎜ Bz ⎟ ⎜ C ⎟ ⎜ y ⎟ ⎜ Cz ⎟ ⎝ ⎠

517



⎛ Ay ⎞ ⎜ ⎟ ⎛ −200 ⎞ ⎜ Az ⎟ ⎜ −667 ⎟ ⎟ ⎜B ⎟ ⎜ ⎜ ⎜ x ⎟ = 0 ⎟ lb ⎜ Bz ⎟ ⎜ 667 ⎟ ⎟ ⎜ ⎟ ⎜ − 300 ⎜ ⎟ ⎜ Cy ⎟ ⎜ ⎜ ⎟ ⎝ 0 ⎟⎠ ⎝ Cz ⎠

Chapter 6

⎛ FBA ⎞ ⎜ ⎟ ⎛ 167 ⎞ F BC ⎜ ⎟ ⎜ 250 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎜ BD ⎟ = −731 ⎟ lb ⎜ FAD ⎟ ⎜ 786 ⎟ ⎟ ⎜ ⎟ ⎜ − 391 ⎜ ⎟ ⎜ FAC ⎟ ⎜ ⎜ ⎟ ⎝ 0 ⎟⎠ F ⎝ CD ⎠


Problem 6-58 The space truss is supported by a ball-and-socket joint at D and short links at C and E. Determine the force in each member and state if the members are in tension or compression. Given:

⎛ 0 ⎞ ⎜ 0 ⎟ lb F1 = ⎜ ⎟ ⎝ −500 ⎠ ⎛ 0 ⎞ ⎜ ⎟ F 2 = 400 lb ⎜ ⎟ ⎝ 0 ⎠ a = 4 ft b = 3 ft c = 3 ft Solution: Find the external reactions Guesses E y = 1 lb

Cy = 1 lb

Cz = 1 lb

Dx = 1 lb

Dy = 1 lb

Dz = 1 lb

518



Chapter 6

Given

⎛ Dx ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Dy ⎟ + ⎜ Ey ⎟ + ⎜ Cy ⎟ + F1 + F2 = 0 ⎜D ⎟ ⎜ 0 ⎟ ⎜C ⎟ ⎝ z⎠ ⎝ ⎠ ⎝ z⎠ D ⎛0⎞ ⎛ −b ⎞ ⎛ 0 ⎞ ⎛⎜ x ⎞⎟ ⎛ −b ⎞ ⎛⎜ 0 ⎟⎞ ⎜ a ⎟ × F + ⎜ a ⎟ × F + ⎜ 0 ⎟ × D + ⎜ 0 ⎟ × Cy = 0 ⎜ ⎟ 1 ⎜ ⎟ 2 ⎜ ⎟ ⎜ y⎟ ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝0⎠ ⎝ c ⎠ ⎜⎝ Dz ⎟⎠ ⎝ c ⎠ ⎜⎝ Cz ⎟⎠

⎛ Ey ⎞ ⎜ ⎟ ⎜ Cy ⎟ ⎜C ⎟ ⎜ z ⎟ = Find E , C , C , D , D , D ( y y z x y z) ⎜ Dx ⎟ ⎜ ⎟ ⎜ Dy ⎟ ⎜ ⎟ ⎝ Dz ⎠

⎛ Ey ⎞ ⎜ ⎟ ⎛ 266.667 ⎞ ⎜ Cy ⎟ ⎜ −400 ⎟ ⎟ ⎜C ⎟ ⎜ ⎜ ⎟ 0 ⎜ z⎟ = ⎟ lb ⎜ Dx ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜ − 266.667 ⎜ ⎟ ⎜ Dy ⎟ ⎜ ⎜ ⎟ ⎝ 500 ⎟⎠ ⎝ Dz ⎠

Now find the force in each member.

⎛ −b ⎞ ⎛ −b ⎞ ⎜ ⎟ AB = 0 AC = ⎜ −a ⎟ ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝c ⎠

⎛0⎞ AD = ⎜ −a ⎟ ⎜ ⎟ ⎝c ⎠

⎛0⎞ AE = ⎜ −a ⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ ⎛b⎞ ⎜ ⎟ BC = −a BE = ⎜ −a ⎟ ⎜ ⎟ ⎜ ⎟ c ⎝ ⎠ ⎝0⎠

⎛0⎞ BF = ⎜ −a ⎟ ⎜ ⎟ ⎝0⎠

⎛b⎞ CD = ⎜ 0 ⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ ⎛0⎞ ⎜ ⎟ CF = 0 DE = ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ −c ⎠ ⎝ −c ⎠

⎛ −b ⎞ DF = ⎜ 0 ⎟ ⎜ ⎟ ⎝ −c ⎠

⎛ −b ⎞ EF = ⎜ 0 ⎟ ⎜ ⎟ ⎝0⎠

Guesses F AB = 1 lb

F AC = 1 lb

F AD = 1 lb

F AE = 1 lb

F BC = 1 lb

F BE = 1 lb

F BF = 1 lb

F CD = 1 lb

519



F CF = 1 lb

Chapter 6

F DE = 1lb

F DF = 1 lb

F EF = 1 lb

Given F 1 + F AB

AB

F 2 + F BC

BC

AB

BC

+ FAC

AC

+ FBF

BF

AC

BF

+ FAD

AD

+ FBE

BE

AD

BE

+ FAE + F AB

AE AE −AB AB

=0

=0

⎛⎜ 0 ⎟⎞ −AE −BE EF −DE + FBE + F EF + F DE =0 ⎜ Ey ⎟ + FAE AE BE EF DE ⎜0⎟ ⎝ ⎠ F BF

−BF BF

+ F CF

−CF CF

+ F DF

−DF DF

+ FEF

−EF EF

=0

⎛0⎞ ⎜ ⎟ −BC −AC CD CF + FAC + FCD + FCF =0 ⎜ Cy ⎟ + FBC BC AC CD CF ⎜C ⎟ ⎝ z⎠ ⎛⎜ FAB ⎞⎟ ⎜ FAC ⎟ ⎜ ⎟ ⎜ FAD ⎟ ⎜F ⎟ ⎜ AE ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FBE ⎟ ⎜ ⎟ = Find ( FAB , FAC , FAD , FAE , FBC , FBE , FBF , FCD , FCF , FDE , FDF , FEF) ⎜ FBF ⎟ ⎜F ⎟ ⎜ CD ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎜F ⎟ ⎜ DF ⎟ ⎜ FEF ⎟ ⎝ ⎠

520



⎛ FAB ⎞ ⎜ ⎟ ⎛ −300 ⎞ F AC ⎜ ⎟ ⎜ 583.095 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎜ AD ⎟ = 333.333 ⎟ lb ⎜ FAE ⎟ ⎜ −666.667 ⎟ ⎟ ⎜ ⎟ ⎜ 0 ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎜ ⎟ ⎝ 500 ⎟⎠ F ⎝ BE ⎠

Chapter 6

⎛ FBF ⎞ ⎜ ⎟ ⎛ 0 ⎞ F CD ⎜ ⎟ ⎜ −300 ⎟ ⎟ ⎜F ⎟ ⎜ Positive (T) ⎜ ⎜ CF ⎟ = −300 ⎟ lb Negative (C) ⎜ FDE ⎟ ⎜ 0 ⎟ ⎟ ⎜ ⎟ ⎜ 424.264 ⎜ ⎟ ⎜ FDF ⎟ ⎜ ⎜ ⎟ ⎝ −300 ⎟⎠ F ⎝ EF ⎠

Problem 6-59 The space truss is supported by a ball-and-socket joint at D and short links at C and E. Determine the force in each member and state if the members are in tension or compression. Given:

⎛ 200 ⎞ ⎜ ⎟ F 1 = 300 lb ⎜ ⎟ ⎝ −500 ⎠ ⎛ 0 ⎞ ⎜ ⎟ F 2 = 400 lb ⎜ ⎟ ⎝ 0 ⎠ a = 4 ft b = 3 ft c = 3 ft Solution: Find the external reactions Guesses E y = 1 lb

Cy = 1 lb

Cz = 1 lb

Dx = 1 lb

Dy = 1 lb

Dz = 1 lb

521



Chapter 6

Given

⎛ Dx ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Dy ⎟ + ⎜ Ey ⎟ + ⎜ Cy ⎟ + F1 + F2 = 0 ⎜D ⎟ ⎜ 0 ⎟ ⎜C ⎟ ⎝ z⎠ ⎝ ⎠ ⎝ z⎠ D ⎛0⎞ ⎛ −b ⎞ ⎛ 0 ⎞ ⎛⎜ x ⎞⎟ ⎛ −b ⎞ ⎛⎜ 0 ⎟⎞ ⎜ a ⎟ × F + ⎜ a ⎟ × F + ⎜ 0 ⎟ × D + ⎜ 0 ⎟ × Cy = 0 ⎜ ⎟ 1 ⎜ ⎟ 2 ⎜ ⎟ ⎜ y⎟ ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝0⎠ ⎝ c ⎠ ⎜⎝ Dz ⎟⎠ ⎝ c ⎠ ⎜⎝ Cz ⎟⎠

⎛ Ey ⎞ ⎜ ⎟ ⎜ Cy ⎟ ⎜C ⎟ ⎜ z ⎟ = Find E , C , C , D , D , D ( y y z x y z) ⎜ Dx ⎟ ⎜ ⎟ ⎜ Dy ⎟ ⎜ ⎟ ⎝ Dz ⎠

⎛ Ey ⎞ ⎞ ⎜ ⎟ ⎛ −33.333 ⎜ ⎟ C ⎜ y⎟ −666.667 ⎟ ⎜C ⎟ ⎜ 200 ⎜ ⎟ z ⎜ ⎟= ⎜ ⎟ lb −200 ⎜ Dx ⎟ ⎜ ⎟ ⎜ ⎟ − 13 ⎟ ⎜ −1.253 × 10 ⎜ Dy ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 300 ⎠ Dz ⎝ ⎠

Now find the force in each member.

⎛ −b ⎞ ⎛ −b ⎞ ⎜ ⎟ AB = 0 AC = ⎜ −a ⎟ ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝c ⎠

⎛0⎞ AD = ⎜ −a ⎟ ⎜ ⎟ ⎝c ⎠

⎛0⎞ AE = ⎜ −a ⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ ⎛b⎞ ⎜ ⎟ BC = −a BE = ⎜ −a ⎟ ⎜ ⎟ ⎜ ⎟ ⎝c ⎠ ⎝0⎠

⎛0⎞ BF = ⎜ −a ⎟ ⎜ ⎟ ⎝0⎠

⎛b⎞ CD = ⎜ 0 ⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ ⎛0⎞ ⎜ ⎟ CF = 0 DE = ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ −c ⎠ ⎝ −c ⎠

⎛ −b ⎞ DF = ⎜ 0 ⎟ ⎜ ⎟ ⎝ −c ⎠

⎛ −b ⎞ EF = ⎜ 0 ⎟ ⎜ ⎟ ⎝0⎠

Guesses F AB = 1 lb

F AC = 1 lb

F AD = 1 lb

F AE = 1 lb

F BC = 1 lb

F BE = 1 lb

F BF = 1 lb

F CD = 1 lb

F CF = 1 lb

F DE = 1lb

F DF = 1 lb

F EF = 1 lb

Given 522



F 1 + F AB

AB

F 2 + F BC

BC

AB

BC

Chapter 6

+ FAC

AC

+ FBF

BF

AC

BF

+ FAD

AD

+ FBE

BE

AD

BE

+ FAE + F AB

AE AE −AB AB

=0

=0

⎛⎜ 0 ⎟⎞ −AE −BE EF −DE + FBE + F EF + F DE =0 ⎜ Ey ⎟ + FAE AE BE EF DE ⎜0⎟ ⎝ ⎠ F BF

−BF BF

+ F CF

−CF CF

+ F DF

−DF DF

+ FEF

−EF EF

=0

⎛0⎞ ⎜ ⎟ −BC −AC CD CF + FAC + FCD + FCF =0 ⎜ Cy ⎟ + FBC BC AC CD CF ⎜C ⎟ ⎝ z⎠ ⎛⎜ FAB ⎞⎟ ⎜ FAC ⎟ ⎜ ⎟ ⎜ FAD ⎟ ⎜F ⎟ ⎜ AE ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FBE ⎟ ⎜ ⎟ = Find ( FAB , FAC , FAD , FAE , FBC , FBE , FBF , FCD , FCF , FDE , FDF , FEF) F BF ⎜ ⎟ ⎜F ⎟ ⎜ CD ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎜F ⎟ ⎜ DF ⎟ ⎜ FEF ⎟ ⎝ ⎠

523



Chapter 6

⎛ FAB ⎞ ⎛ FBF ⎞ −300 ⎜ ⎟ ⎛ ⎜⎞ ⎟ ⎛ 0 ⎞ ⎜ ⎟ F F AC CD ⎜ ⎟ ⎜ ⎟ ⎜ −500 ⎟ 971.825 ⎜ ⎟ ⎟ ⎜F ⎟ ⎜F ⎟ ⎜ Positive (T) − 11 ⎜ ⎟ ⎜ ⎜ AD ⎟ = 1.121 × 10 ⎜ lbCF ⎟ = −300 ⎟ lb Negative (C) ⎜ FAE ⎟ ⎜ −366.667 ⎜⎟ FDE ⎟ ⎜ 0 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜⎟ ⎟ ⎜ 424.264 ⎜ ⎟ ⎜ ⎟ 0 ⎜ FBC ⎟ ⎜ FDF ⎟ ⎜ ⎟ ⎜ ⎜ ⎟ ⎝ ⎜⎠ ⎟ ⎝ −300 ⎟⎠ 500 F F ⎝ BE ⎠ ⎝ EF ⎠

Problem 6-60 Determine the force in each member of the space truss and state if the members are in tension or compression. The truss is supported by a ball-and-socket joints at A, B, and E. Hint: The support reaction at E acts along member EC. Why? Given:

⎛ −200 ⎞ ⎜ ⎟ F = 400 N ⎜ ⎟ ⎝ 0 ⎠

a = 2m b = 1.5 m c = 5m d = 1m e = 2m

Solution:

⎛ 0 ⎞ ⎜ ⎟ AC = a + b ⎜ ⎟ ⎝ 0 ⎠

⎛d⎞ ⎜ ⎟ AD = a ⎜ ⎟ ⎝e ⎠

⎛ −c − d ⎞ ⎜ 0 ⎟ BC = ⎜ ⎟ ⎝ 0 ⎠

⎛ −c ⎞ ⎜ ⎟ BD = −b ⎜ ⎟ ⎝e ⎠

Guesses

Given

⎛d⎞ ⎜ ⎟ CD = −b ⎜ ⎟ ⎝e ⎠

F AC = 1 N

F AD = 1 N

F BC = 1 N

F CD = 1 N

F EC = 1 N

F BD = 1 N

F + FAD

−AD AD

+ FBD

−BD BD

+ F CD

−CD CD

=0

524



Chapter 6

⎛⎜ 0 ⎞⎟ F CD + F BC + FAC +⎜ 0 ⎟ =0 CD BC AC ⎜ −FEC ⎟ ⎝ ⎠ CD

−BC

−AC

⎛ FAC ⎞ ⎜ ⎟ F ⎜ AD ⎟ ⎜F ⎟ ⎜ BC ⎟ = Find F , F , F , F , F , F ( AC AD BC BD CD EC ) ⎜ FBD ⎟ ⎜ ⎟ F ⎜ CD ⎟ ⎜ ⎟ ⎝ FEC ⎠

⎛ FAC ⎞ ⎜ ⎟ ⎛ 221 ⎞ F ⎜ AD ⎟ ⎜ 343 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ BC ⎟ = ⎜ 148 ⎟ N ⎜ FBD ⎟ ⎜ 186 ⎟ ⎟ ⎜ ⎟ ⎜ − 397 ⎜ ⎟ ⎜ FCD ⎟ ⎜ ⎜ ⎟ ⎝ −295 ⎟⎠ F ⎝ EC ⎠


Problem 6-61 Determine the force in each member of the space truss and state if the members are in tension or compression. The truss is supported by ball-and-socket joints at C, D, E, and G. Units Used: 3

kN = 10 N Given: F = 3 kN a = 2m b = 1.5 m c = 2m d = 1m e = 1m Solution: Fv =

⎛0⎞ ⎜b⎟ 2 2⎜ ⎟ b + c ⎝ −c ⎠ F

525



Chapter 6

uAG =

⎛ −e ⎞ ⎜ −a ⎟ ⎟ 2 2⎜ a +e ⎝ 0 ⎠

uAE =

⎛d⎞ ⎜ −a ⎟ ⎟ 2 2⎜ a +d ⎝ 0 ⎠

1

1

⎛0⎞ ⎜ ⎟ uAB = 0 ⎜ ⎟ ⎝ −1 ⎠

uBC = uAE uBD = uAG

⎛d⎞ ⎜ −a ⎟ ⎟ 2 2 2⎜ a +c +d ⎝ c ⎠ 1

uBE =

⎛ −e ⎞ ⎜ −a ⎟ ⎟ 2 2 2⎜ a +e +c ⎝ c ⎠ 1

uBG =

Guesses F AB = 1 kN

F AE = 1 kN

F BC = 1 kN

F BD = 1 kN

F BE = 1 kN

F BG = 1 kN

F AG = 1 kN

Given −c 2

c +b

2

e 2

a

F ( a) −

2

a +e

2

2

a +d F BD( a) −

FBC ( c) −

d 2

a +d

2

a 2

2

a +e

FBD( c) = 0

F BC ( a) = 0

F v + F AEuAE + FAGuAG + F ABuAB = 0 −F AB uAB + FBGuBG + FBEuBE + F BC uBC + FBDuBD = 0

526



Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAE ⎟ ⎜ ⎟ FAG ⎜ ⎟ ⎜ F ⎟ = Find F , F , F , F , F , F , F ( AB AE AG BC BD BE BG) ⎜ BC ⎟ ⎜ FBD ⎟ ⎜ ⎟ ⎜ FBE ⎟ ⎜F ⎟ ⎝ BG ⎠

⎛ FAB ⎞ ⎜ ⎟ ⎛ −2.4 ⎞ ⎟ ⎜ FAE ⎟ ⎜ ⎜ ⎟ ⎜ 1.006 ⎟ ⎜ FAG ⎟ ⎜ 1.006 ⎟ ⎜ F ⎟ = ⎜ −1.342 ⎟ kN ⎟ ⎜ BC ⎟ ⎜ ⎜ FBD ⎟ ⎜ −1.342 ⎟ ⎜ ⎟ ⎜ 1.8 ⎟ ⎟ ⎜ FBE ⎟ ⎜ 1.8 ⎝ ⎠ ⎜F ⎟ ⎝ BG ⎠ Positive (T) Negative (C)

Problem 6-62 Determine the force in members BD, AD, and AF of the space truss and state if the members are in tension or compression. The truss is supported by short links at A, B, D, and F. Given:

⎛ 0 ⎞ ⎜ ⎟ F = 250 lb ⎜ ⎟ ⎝ −250 ⎠ a = 6 ft b = 6 ft

θ = 60 deg Solution: Find the external reactions h = b sin ( θ ) Guesses Ax = 1 lb

B y = 1 lb

B z = 1 lb

Dy = 1 lb

F y = 1 lb

F z = 1 lb

Given

⎛

⎞

⎛

⎞ 527



Chapter 6

⎛⎜ Ax ⎞⎟ ⎛⎜ 0 ⎞⎟ ⎛⎜ 0 ⎞⎟ ⎛⎜ 0 ⎞⎟ F + ⎜ 0 ⎟ + ⎜ By ⎟ + ⎜ Dy ⎟ + ⎜ F y ⎟ = 0 ⎜ 0 ⎟ ⎜B ⎟ ⎜ 0 ⎟ ⎜F ⎟ ⎝ ⎠ ⎝ z⎠ ⎝ ⎠ ⎝ z⎠ ⎛ 0.5b ⎞ ⎛ b ⎞ ⎛⎜ Ax ⎞⎟ ⎛ b ⎞ ⎛⎜ 0 ⎟⎞ ⎛ 0.5b ⎞ ⎛⎜ 0 ⎞⎟ ⎛ 0 ⎞ ⎛⎜ 0 ⎟⎞ ⎜ a ⎟ × F + ⎜a⎟ × ⎜ ⎟ B ⎜ 0 ⎟ × D + ⎜ a ⎟ × Fy = 0 ⎜ 0 ⎟ + 0 × ⎜ y⎟ + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ y⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ h ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ Bz ⎠ ⎝ h ⎠ ⎜⎝ 0 ⎟⎠ ⎝ 0 ⎠ ⎜⎝ Fz ⎟⎠ ⎛ Ax ⎞ ⎜ ⎟ ⎜ By ⎟ ⎜B ⎟ ⎜ z ⎟ = Find A , B , B , D , F , F ( x y z y y z) ⎜ Dy ⎟ ⎜ ⎟ ⎜ Fy ⎟ ⎜ ⎟ ⎝ Fz ⎠

⎛ Ax ⎞ ⎜ ⎟ ⎛ 0 ⎞ ⎜ By ⎟ ⎜ 72 ⎟ ⎟ ⎜B ⎟ ⎜ ⎜ ⎜ z ⎟ = 125 ⎟ lb ⎜ Dy ⎟ ⎜ −394 ⎟ ⎟ ⎜ ⎟ ⎜ 72 ⎜ ⎟ ⎜ Fy ⎟ ⎜ ⎜ ⎟ ⎝ 125 ⎟⎠ ⎝ Fz ⎠

Now find the forces in the members

⎛0⎞ ⎜ ⎟ AB = −a ⎜ ⎟ ⎝0⎠

⎛ −b ⎞ ⎜ ⎟ AC = −a ⎜ ⎟ ⎝0⎠

⎛ −0.5b ⎞ ⎜ −a ⎟ AD = ⎜ ⎟ ⎝ h ⎠

⎛ −0.5b ⎞ ⎜ 0 ⎟ AE = ⎜ ⎟ ⎝ h ⎠ ⎛ −0.5b ⎞ ⎜ 0 ⎟ BD = ⎜ ⎟ ⎝ h ⎠

⎛ −b ⎞ ⎜ ⎟ AF = 0 ⎜ ⎟ ⎝0⎠ ⎛ 0.5b ⎞ ⎜ 0 ⎟ CD = ⎜ ⎟ ⎝ h ⎠

⎛ −b ⎞ ⎜ ⎟ BC = 0 ⎜ ⎟ ⎝0⎠ ⎛0⎞ ⎜ ⎟ CF = a ⎜ ⎟ ⎝0⎠

⎛0⎞ ⎜ ⎟ DE = a ⎜ ⎟ ⎝0⎠

⎛ −0.5b ⎞ ⎜ a ⎟ DF = ⎜ ⎟ ⎝ −h ⎠

⎛ −0.5b ⎞ ⎜ 0 ⎟ EF = ⎜ ⎟ ⎝ −h ⎠

Guesses F AB = 1 lb

F AC = 1 lb

F AD = 1 lb

F AE = 1 lb

F AF = 1 lb

F BC = 1 lb

F BD = 1 lb

F CD = 1 lb

F CF = 1 lb

528



F DE = 1 lb

F DF = 1 lb

Chapter 6

F EF = 1 lb

Given −DE −AE EF F + FDE + F AE + FEF =0 DE AE EF F CF

CD −BC −AC + F CD + F BC + FAC =0 CF CD BC AC CF

⎛⎜ 0 ⎞⎟ F DE + FDF + F AD + F BD + FCD + ⎜ Dy ⎟ = 0 DE DF AD BD CD ⎜0 ⎟ ⎝ ⎠ ⎛0⎞ ⎜ ⎟ −AB BC BD F AB + FBC + F BD + ⎜ By ⎟ = 0 AB BC BD ⎜B ⎟ ⎝ z⎠ DE

DF

−AD

−BD

−CD

⎛⎜ Ax ⎞⎟ F AB + FAC + FAF + F AD + F AE +⎜ 0 ⎟ =0 AB AC AF AD AE ⎜0 ⎟ ⎝ ⎠ AB

AC

AF

AD

AE

⎛⎜ FAB ⎞⎟ ⎜ FAC ⎟ ⎜ ⎟ ⎜ FAD ⎟ ⎜F ⎟ ⎜ AE ⎟ ⎜ FAF ⎟ ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎟ = Find ( FAB , FAC , FAD , FAE , FAF , FBC , FBD , FCD , FCF , FDE , FDF , FEF) ⎜ FBD ⎟ ⎜F ⎟ ⎜ CD ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎜F ⎟ ⎜ DF ⎟ ⎜ FEF ⎟ ⎝ ⎠

529



⎛ FBD ⎞ ⎛ −144.3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAD ⎟ = ⎜ 204.1 ⎟ lb ⎜ F ⎟ ⎝ 72.2 ⎠ ⎝ AF ⎠

Chapter 6


Problem 6-63 Determine the force in members CF, EF, and DF of the space truss and state if the members are in tension or compression. The truss is supported by short links at A, B, D, and F. Given:

⎛ 0 ⎞ ⎜ ⎟ F = 250 lb ⎜ ⎟ ⎝ −250 ⎠ a = 6 ft b = 6 ft

θ = 60 deg Solution: Find the external reactions h = b sin ( θ ) Guesses Ax = 1 lb

B y = 1 lb

B z = 1 lb

Dy = 1 lb

F y = 1 lb

F z = 1 lb

Given

⎛⎜ Ax ⎞⎟ ⎛⎜ 0 ⎟⎞ ⎛⎜ 0 ⎞⎟ ⎛⎜ 0 ⎟⎞ F + ⎜ 0 ⎟ + ⎜ By ⎟ + ⎜ Dy ⎟ + ⎜ F y ⎟ = 0 ⎜ 0 ⎟ ⎜B ⎟ ⎜ 0 ⎟ ⎜F ⎟ ⎝ ⎠ ⎝ z⎠ ⎝ ⎠ ⎝ z⎠ ⎛ 0.5b ⎞ ⎛ b ⎞ ⎛⎜ Ax ⎞⎟ ⎛ b ⎞ ⎛⎜ 0 ⎟⎞ ⎛ 0.5b ⎞ ⎛⎜ 0 ⎞⎟ ⎛ 0 ⎞ ⎛⎜ 0 ⎟⎞ ⎜ a ⎟ × F + ⎜a⎟ × ⎜ ⎟ ⎜ 0 ⎟ × D + ⎜ a ⎟ × Fy = 0 + 0 × By + ⎜ ⎟ ⎜ ⎟ ⎜0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ y⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ h ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ Bz ⎠ ⎝ h ⎠ ⎜⎝ 0 ⎟⎠ ⎝ 0 ⎠ ⎜⎝ Fz ⎟⎠

530



Chapter 6

⎛ Ax ⎞ ⎜ ⎟ ⎜ By ⎟ ⎜B ⎟ ⎜ z ⎟ = Find A , B , B , D , F , F ( x y z y y z) ⎜ Dy ⎟ ⎜ ⎟ ⎜ Fy ⎟ ⎜ ⎟ ⎝ Fz ⎠

⎛ Ax ⎞ ⎜ ⎟ ⎛ 0 ⎞ ⎜ By ⎟ ⎜ 72 ⎟ ⎟ ⎜B ⎟ ⎜ ⎜ ⎜ z ⎟ = 125 ⎟ lb ⎜ Dy ⎟ ⎜ −394 ⎟ ⎟ ⎜ ⎟ ⎜ 72 ⎜ ⎟ ⎜ Fy ⎟ ⎜ ⎜ ⎟ ⎝ 125 ⎟⎠ ⎝ Fz ⎠

Now find the forces in the members

⎛0⎞ ⎜ ⎟ AB = −a ⎜ ⎟ ⎝0⎠

⎛ −b ⎞ ⎜ ⎟ AC = −a ⎜ ⎟ ⎝0⎠

⎛ −0.5b ⎞ ⎜ −a ⎟ AD = ⎜ ⎟ ⎝ h ⎠

⎛ −0.5b ⎞ ⎜ 0 ⎟ AE = ⎜ ⎟ ⎝ h ⎠ ⎛ −0.5b ⎞ ⎜ 0 ⎟ BD = ⎜ ⎟ ⎝ h ⎠

⎛ −b ⎞ ⎜ ⎟ AF = 0 ⎜ ⎟ ⎝0⎠ ⎛ 0.5b ⎞ ⎜ 0 ⎟ CD = ⎜ ⎟ ⎝ h ⎠

⎛ −b ⎞ ⎜ ⎟ BC = 0 ⎜ ⎟ ⎝0⎠ ⎛0⎞ ⎜ ⎟ CF = a ⎜ ⎟ ⎝0⎠

⎛0⎞ ⎜ ⎟ DE = a ⎜ ⎟ ⎝0⎠

⎛ −0.5b ⎞ ⎜ a ⎟ DF = ⎜ ⎟ ⎝ −h ⎠

⎛ −0.5b ⎞ ⎜ 0 ⎟ EF = ⎜ ⎟ ⎝ −h ⎠

Guesses F AB = 1 lb

F AC = 1 lb

F AD = 1 lb

F AE = 1 lb

F AF = 1 lb

F BC = 1 lb

F BD = 1 lb

F CD = 1 lb

F CF = 1 lb

F DE = 1 lb

F DF = 1 lb

F EF = 1 lb

Given F + FDE

F CF

CF CF

−DE DE

+ F AE

+ F CD

−AE

CD CD

AE

+ FEF

+ F BC

EF EF

−BC BC

=0

+ FAC

−AC AC

=0

531



Chapter 6

⎛⎜ 0 ⎞⎟ F DE + FDF + F AD + F BD + FCD + ⎜ Dy ⎟ = 0 DE DF AD BD CD ⎜0 ⎟ ⎝ ⎠ DE

DF

−AD

−BD

−CD

⎛0⎞ ⎜ ⎟ F AB + FBC + F BD + ⎜ By ⎟ = 0 AB BC BD ⎜B ⎟ ⎝ z⎠ −AB

BC

BD

⎛⎜ Ax ⎞⎟ F AB + FAC + FAF + F AD + F AE +⎜ 0 ⎟ =0 AB AC AF AD AE ⎜0 ⎟ ⎝ ⎠ AB

AC

AF

AD

AE

⎛⎜ FAB ⎞⎟ ⎜ FAC ⎟ ⎜ ⎟ ⎜ FAD ⎟ ⎜F ⎟ ⎜ AE ⎟ ⎜ FAF ⎟ ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎟ = Find ( FAB , FAC , FAD , FAE , FAF , FBC , FBD , FCD , FCF , FDE , FDF , FEF) ⎜ FBD ⎟ ⎜F ⎟ ⎜ CD ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎜F ⎟ ⎜ DF ⎟ ⎜ FEF ⎟ ⎝ ⎠

⎛ FCF ⎞ ⎛ 72.2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FEF ⎟ = ⎜ −144.3 ⎟ lb ⎜F ⎟ ⎝ 0 ⎠ ⎝ DF ⎠


532



Chapter 6

Problem 6-64 Determine the force developed in each member of the space truss and state if the members are in tension or compression. The crate has weight W. Given: W = 150 lb a = 6 ft b = 6 ft c = 6 ft

Solution: h =

2

Unit Vectors

c −

⎛ a⎞ ⎜ ⎟ ⎝ 2⎠

2

uAD =

⎛⎜ −a ⎟⎞ 1 ⎜ 2 ⎟ 2⎜ 0 ⎟ 2 ⎛ a⎞ ⎜ ⎟ h +⎜ ⎟ ⎝ 2⎠ ⎝ h ⎠

uBD =

⎛⎜ a ⎟⎞ 1 ⎜2⎟ 2⎜ 0 ⎟ a 2 h + ⎛⎜ ⎟⎞ ⎜ h ⎟ ⎝ 2⎠ ⎝ ⎠

uAC =

⎛⎜ − a ⎟⎞ 1 ⎜ 2⎟ 2⎜ b ⎟ a 2 2 h + b + ⎛⎜ ⎟⎞ ⎜ h ⎟ ⎝ 2⎠ ⎝ ⎠

uBC =

⎛⎜ a ⎟⎞ 1 ⎜2⎟ 2⎜ b ⎟ a 2 2 h + b + ⎛⎜ ⎟⎞ ⎜ h ⎟ ⎝ 2⎠ ⎝ ⎠

Guesses F AB = 1 lb

B y = 1 lb

Ax = 1 lb

F AC = 1 lb

Ay = 1 lb

F AD = 1 lb

F BC = 1 lb

F BD = 1 lb

F CD = 1 lb

533



Chapter 6

Given

⎛⎜ 0 ⎟⎞ ⎜ −FCD ⎟ − FAC uAC − FBC uBC = 0 ⎜ −W ⎟ ⎝ ⎠ ⎛ FAB ⎞ ⎜ ⎟ B ⎜ y ⎟ + FBC uBC + FBDuBD = 0 ⎜ ⎟ ⎝ 0 ⎠ ⎛ Ax − FAB ⎞ ⎜ ⎟ A ⎜ ⎟ + FAC uAC + FADuAD = 0 y ⎜ ⎟ 0 ⎝ ⎠ ⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ B ⎟ ⎜ y ⎟ ⎜ FAB ⎟ ⎜ ⎟ ⎜ FAC ⎟ = Find ( Ax , Ay , By , FAB , FAC , FAD , FBC , FBD , FCD) ⎜F ⎟ ⎜ AD ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FBD ⎟ ⎟ ⎛ FAB ⎞ ⎜ ⎜ ⎟ ⎛ 0.0 ⎞ ⎝ FCD ⎠ F ⎜ AC ⎟ ⎜ −122.5 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎟ AD 86.6 Positive (T) ⎜ ⎟= lb ⎜ ⎟ Negative (C) ⎜ FBC ⎟ −122.5 ⎜ ⎟ ⎜ ⎟ ⎜ FBD ⎟ ⎜ 86.6 ⎟ ⎜ ⎟ ⎜⎝ 173.2 ⎟⎠ FCD ⎝ ⎠

Problem 6-65 The space truss is used to support vertical forces at joints B, C, and D. Determine the force in each member and state if the members are in tension or compression.

534



Chapter 6

Units Used: 3


a = 0.75 m

F 2 = 8 kN

b = 1.00 m

F 3 = 9 kN

c = 1.5 m

Solution: Assume that the connections at A, E, and F are rollers Guesses F BC = 1 kN

F CF = 1 kN

F CD = 1 kN

F AD = 1 kN

F DF = 1 kN

F DE = 1 kN

F BD = 1 kN

F BA = 1 kN

F EF = 1 kN

F AE = 1 kN

F AF = 1 kN Given Joint C F BC = 0

F CD = 0

−F 2 − F CF = 0 Joint D a 2

2

a +b

FBD +

a 2

2

2

a +b +c

FAD = 0

535



b

−F CD −

2

−b

+ 2

a +b

2

2

2

2

2

b +c

−c 2

F AD −

c

−F 3 − F DE − +

FBD ...

2

a +b +c

2

a +b +c

Chapter 6

=0 b 2

2

b +c

F DF

F DF ... = 0

F AD

Joint B a

−F BC −

2

2

a +b

FBD = 0

b 2

2

a +b

FBD = 0

−F 1 − F BA = 0

Joint E a 2

2

a +b

FAE = 0

−F EF −

b 2

2

a +b

FAE = 0

⎛ FBC ⎞ ⎜ ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FCD ⎟ ⎜F ⎟ ⎜ AD ⎟ ⎜ FDF ⎟ ⎜ ⎟ ⎜ FDE ⎟ = Find ( FBC , FCF , FCD , FAD , FDF , FDE , FBD , FBA , FEF , FAE , FAF) ⎜F ⎟ ⎜ BD ⎟ ⎜ FBA ⎟ ⎜ ⎟ ⎜ FEF ⎟ ⎜ ⎟ ⎜ FAE ⎟ ⎜F ⎟ ⎝ AF ⎠

536



⎛ FBC ⎞ ⎜ ⎟ ⎜ FCF ⎟ ⎛⎜ 0.00 ⎟⎞ ⎜ ⎟ ⎜ −8.00 ⎟ ⎜ FCD ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 0.00 ⎟ AD ⎜ ⎟ ⎜ 0.00 ⎟ ⎜ FDF ⎟ ⎜ 0.00 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FDE ⎟ = ⎜ −9.00 ⎟ kN ⎜ F ⎟ ⎜ 0.00 ⎟ ⎜ BD ⎟ ⎜ ⎟ ⎜ FBA ⎟ ⎜ −6.00 ⎟ ⎜ ⎟ ⎜ 0.00 ⎟ ⎜ FEF ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0.00 ⎟ F AE ⎜ ⎟ ⎝ 0.00 ⎠ ⎜F ⎟ ⎝ AF ⎠

Chapter 6


Problem 6-66 A force P is applied to the handles of the pliers. Determine the force developed on the smooth bolt B and the reaction that pin A exerts on its attached members. Given: P = 8 lb a = 1.25 in b = 5 in c = 1.5 in Solution: ΣMA = 0;

−R B c + P b = 0 RB = P

b c

R B = 26.7 lb ΣF x = 0;

Ax = 0

ΣF y = 0;

Ay − P − RB = 0 537



Chapter 6

Ay = P + RB Ay = 34.7 lb

Problem 6-67 The eye hook has a positive locking latch when it supports the load because its two parts are pin-connected at A and they bear against one another along the smooth surface at B. Determine the resultant force at the pin and the normal force at B when the eye hook supports load F . Given: F = 800 lb a = 0.25 in b = 3 in c = 2 in

θ = 30 deg

Solution: Σ MA = 0;

−F B cos ( 90 deg − θ ) ( b) − F B sin ( 90 deg − θ ) ( c) + F a = 0 FB = F

+

↑Σ Fy = 0;

+ Σ F x = 0; →

a

cos ( 90 deg − θ ) b + sin ( 90 deg − θ ) c

F B = 61.9 lb

−F − FB sin ( 90 deg − θ ) + A y = 0 Ay = F + F B sin ( 90 deg − θ )

Ay = 854 lb

Ax − F B cos ( 90 deg − θ ) = 0 Ax = FB cos ( 90 deg − θ ) FA =

2

2

Ax + Ay

Ax = 30.9 lb F A = 854 lb

Problem 6-68 Determine the force P needed to hold the block of mass F in equilibrium. 538



Chapter 6

Given: F = 20 lb Solution: Pulley B:

ΣF y = 0;

2P − T = 0

Pulley A:

ΣF y = 0;

2T − F = 0

T =

1 F 2

2P = T

T = 10 lb P =

1 T 2

P = 5 lb

Problem 6-69 The link is used to hold the rod in place. Determine the required axial force on the screw at E if the largest force to be exerted on the rod at B, C or D is to be F max. Also, find the magnitude of the force reaction at pin A. Assume all surfaces of contact are smooth. Given: F max = 100 lb a = 100 mm b = 80 mm c = 50 mm

θ = 45 deg

539



Chapter 6

Solution: Assign an initial value for R E. This will be scaled at the end of the problem. Guesses

Given

Ax = 1 lb

Ay = 1 lb

R B = 1 lb

R C = 1 lb

R D = 1 lb

R E = 1 lb

− Ax + R E − R B cos ( θ ) = 0

Ay − R B sin ( θ ) = 0

R E a − RB cos ( θ ) ( a + b + c cos ( θ ) ) − RB sin ( θ ) c sin ( θ ) = 0 R B cos ( θ ) − R D = 0

R B sin ( θ ) − RC = 0

⎛⎜ Ax ⎟⎞ ⎜ Ay ⎟ ⎜ ⎟ ⎜ RB ⎟ = Find ( Ax , Ay , RB , RC , RD) ⎜R ⎟ ⎜ C⎟ ⎜ RD ⎟ ⎝ ⎠

⎛⎜ Ax ⎟⎞ ⎛ 0.601 ⎞ ⎜ Ay ⎟ ⎜ 0.399 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ RB ⎟ = ⎜ 0.564 ⎟ lb ⎜ R ⎟ ⎜ 0.399 ⎟ ⎜ C⎟ ⎜ ⎟ 0.399 ⎝ ⎠ ⎜ RD ⎟ ⎝ ⎠

Now find the critical load and scale the problem

⎛ RB ⎞ ⎜ ⎟ ans = ⎜ R C ⎟ ⎜R ⎟ ⎝ D⎠

F scale =

Fmax

R E = Fscale RE

max ( ans )

F A = Fscale

2

2

Ax + Ay

R E = 177.3 lb

F A = 127.9 lb

Problem 6-70 The man of weight W1 attempts to lift himself and the seat of weight W2 using the rope and pulley system shown. Determine the force at A needed to do so, and also find his reaction on the seat.

540



Chapter 6

Given: W1 = 150 lb W2 = 10 lb Solution: Pulley C: ΣF y = 0;

3T − R = 0

Pulley B: ΣF y = 0;

3R − P = 0

Thus,

P = 9T

Man and seat: ΣF y = 0;

T + P − W1 − W2 = 0 10T = W1 + W2 T =

W1 + W2 10

P = 9T

T = 16 lb P = 144 lb

Seat: ΣF y = 0;

P − N − W2 = 0 N = P − W2

N = 134 lb

Problem 6-71 Determine the horizontal and vertical components of force that pins A and C exert on the frame. Given: F = 500 N a = 0.8 m

d = 0.4 m

541



Chapter 6

b = 0.9 m

e = 1.2 m

c = 0.5 m

θ = 45deg

Solution: BC is a two-force member Member AB : ΣMA = 0;

e

−F c + FBC

2

2

a +e

b + F BC

2

a 2

( c + d) = 0 2

a +e

2

a +e F BC = F c eb+a c+a d

F BC = 200.3 N

Thus, Cx = F BC

Cy = F BC

ΣF x = 0;

ΣF y = 0;

Ax − F BC

e 2

Cx = 167 N

2

a +e a 2

Cy = 111 N

2

a +e e 2

=0

Ax = FBC

2

a +e

Ay − F + FBC

a 2

=0 2

a +e

e 2

Ax = 167 N

2

a +e

Ay = F − F BC

a 2

2

a +e

Ay = 389 N

Problem 6-72 Determine the horizontal and vertical components of force that pins A and C exert on the frame. Units Used: 3


542



Chapter 6

F 2 = 500 N

θ = 45 deg a = 0.2 m b = 0.2 m c = 0.4 m d = 0.4 m Solution: Guesses Ax = 1 N

Ay = 1 N

Cx = 1 N

Cy = 1 N

Given Ax − Cx = 0

Ay + Cy − F1 − F2 = 0

F1 a − Ay 2 a + Ax d = 0

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ ⎟ = Find ( Ax , Ay , Cx , Cy) ⎜ Cx ⎟ ⎜C ⎟ ⎝ y⎠

−F 2 b + Cy( b + c) − Cx d = 0

⎛ Ax ⎞ ⎜ ⎟ ⎛⎜ 500 ⎞⎟ ⎜ Ay ⎟ ⎜ 1000 ⎟ N ⎜ ⎟=⎜ ⎟ C 500 x ⎜ ⎟ ⎜ ⎟ ⎜ C ⎟ ⎝ 500 ⎠ ⎝ y⎠

Problem 6-73 The truck exerts the three forces shown on the girders of the bridge. Determine the reactions at the supports when the truck is in the position shown. The girders are connected together by a short vertical link DC. Units Used: 3

kip = 10 lb

543



Chapter 6

Given: a = 55 ft

f = 12 ft

b = 10 ft

F 1 = 5 kip

c = 48 ft

F 2 = 4 kip

d = 5 ft

F 3 = 2 kip

e = 2 ft Solution:

Member CE: ΣMC = 0;

−F 3 e + E y( e + c) = 0

e Ey = F3 e+c

E y = 80 lb

ΣF y = 0;

Cy − F3 + Ey = 0

Cy = F 3 − E y

Cy = 1920 lb

Member ABD: ΣMA = 0;

−F 1 a − F2( d + a) − Cy( a + d + b) + By( a + d) = 0 By =

ΣF y = 0;

F 1 a + F2( d + a) + Cy( a + d + b) d+a

B y = 10.8 kip

Ay − F 1 + B y − F2 − Cy = 0 Ay = Cy + F 1 − B y + F2

Ay = 96.7 lb

Problem 6-74 Determine the greatest force P that can be applied to the frame if the largest force resultant acting at A can have a magnitude F max. Units Used: 3

kN = 10 N

544



Chapter 6

Given: F max = 2 kN a = 0.75 m b = 0.75 m c = 0.5 m d = 0.1 m Solution: Σ MA = 0;

T( c + d) − P( a + b) = 0

+ Σ F x = 0; →

Ax − T = 0

+

↑Σ Fy = 0;

Ay − P = 0

Thus,

T=

a+b c+d

P

Ay = P

Ax =

a+b c+d

P

Require, F max = P =

2

2

Ax + Ay F max 2

⎛ a + b⎞ + 1 ⎜ ⎟ ⎝c + d⎠ P = 743 N

Problem 6-75 The compound beam is pin supported at B and supported by rockers at A and C. There is a hinge (pin) at D. Determine the reactions at the supports. Units Used: 3

kN = 10 N

545



Chapter 6

Given: F 1 = 7 kN

a = 4m

F 2 = 6 kN

b = 2m

F 3 = 16 kN

c = 3m

θ = 60 deg

d = 4m

Solution: Member DC : ΣMD = 0;

−F 1 sin ( θ ) ( a − c) + Cy a = 0 Cy = F 1 sin ( θ )

ΣF y = 0;

a−c

Cy = 1.52 kN

a

Dy − F 1 sin ( θ ) + Cy = 0 Dy = F1 sin ( θ ) − Cy

ΣF x = 0;

Dy = 4.55 kN

Dx − F 1 cos ( θ ) = 0 Dx = F1 cos ( θ )

Dx = 3.5 kN

Member ABD : ΣMA = 0;

−F 3 a − F2( 2 a + b) − Dy( 3 a + b) + By2 a = 0 By =

ΣF y = 0;

F 3 a + F2( 2a + b) + Dy ( 3a + b) 2a

Ay − F 3 + B y − F2 − Dy = 0 Ay = Dy + F3 − By + F 2

ΣF x = 0;

B y = 23.5 kN

Ay = 3.09 kN

B x − F1 cos ( θ ) = 0 B x = F 1 cos ( θ )

B x = 3.5 kN

Problem 6-76 The compound beam is fixed supported at A and supported by rockers at B and C. If there are hinges (pins) at D and E, determine the reactions at the supports A, B, and C.

546



Chapter 6

Units Used: 3

kN = 10 N Given: a = 2 m M = 48 kN⋅ m b = 4m

kN w1 = 8 m

c = 2m

kN w2 = 6 d = 6m m e = 3m

Solution: Guesses Ax = 1 N

Ay = 1 N

MA = 1 N m

Dx = 1 N

Dy = 1 N

By = 1 N

Ey = 1 N

Ex = 1 N

Cy = 1 N

Given Ay − w2 a − Dy = 0 a MA − w2 a − Dy a = 0 2 −w1

( b + c) 2

E y − w1 −w1⎛⎜

2

d+e 2

+ B y b − E y( b + c) = 0 + Cy = 0

− Ax − Dx = 0 Dy − w1( b + c) + By − E y = 0 Dx + E x = 0 −E x = 0

d + e⎞⎛ d + e⎞

⎟⎜ ⎟ + Cy d − M = 0 ⎝ 2 ⎠⎝ 3 ⎠

547



Chapter 6

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜M ⎟ ⎜ A⎟ ⎜ Dx ⎟ ⎜ ⎟ D ⎜ y ⎟ = Find ( Ax , Ay , MA , Dx , Dy , By , Ey , Ex , Cy) ⎜B ⎟ ⎜ y⎟ ⎜ Ey ⎟ ⎜ ⎟ E ⎜ x⎟ ⎟ ⎜ ⎝ Cy ⎠

⎛⎜ Ax ⎞⎟ ⎛ 0 ⎞ = kN ⎜ Ay ⎟ ⎜⎝ 19 ⎟⎠ ⎝ ⎠ MA = 26 kN m B y = 51 kN Cy = 26 kN

Problem 6-77 Determine the reactions at supports A and B. Units Used: 3

kip = 10 lb Given: lb w1 = 500 ft lb w2 = 700 ft a = 6 ft b = 8 ft c = 9 ft d = 6 ft Solution: Guesses Ax = 1 lb

Ay = 1 lb

B x = 1 lb

B y = 1 lb

F CD = 1 lb MA = 1 lb⋅ ft

548



Chapter 6

Given Ay − w1 a −

MA − w1 a a −

d 2

b +d

2

F CD = 0

d 2

d +b

2

Ax −

b 2

b +d

b

F CD2 a = 0

2

2

b +d

d

2

FCD = 0

FCD − Bx = 0

c c B y c − w2 =0 2 3

c FCD − w2 + B y = 0 2 2 2 b +d

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ B ⎟ ⎜ x ⎟ = Find A , A , B , B , F , M ( x y x y CD A) ⎜ By ⎟ ⎜ ⎟ ⎜ FCD ⎟ ⎜ ⎟ ⎝ MA ⎠

⎛⎜ Ax ⎞⎟ ⎛ 2.8 ⎞ = kip ⎜ Ay ⎟ ⎜⎝ 5.1 ⎟⎠ ⎝ ⎠ MA = 43.2 kip⋅ ft

⎛⎜ Bx ⎟⎞ ⎛ 2.8 ⎞ = kip ⎜ By ⎟ ⎜⎝ 1.05 ⎟⎠ ⎝ ⎠

Problem 6-78 Determine the horizontal and vertical components of force at C which member ABC exerts on member CEF. Given: F = 300 lb a = 4 ft b = 6 ft c = 3 ft r = 1 ft Solution: Guesses Ax = 1 lb

Ay = 1 lb

F y = 1 lb

549



Cx = 1 lb

Chapter 6

Cy = 1 lb

Given b b Ax a − A y − Cx a − Cy = 0 2 2 Cx a − F r = 0 Ax = 0 Ay + Fy − F = 0 F y b − F ( b + c + r) = 0

⎛⎜ Ax ⎞⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎜ Fy ⎟ = Find ( Ax , Ay , Fy , Cx , Cy) ⎜C ⎟ ⎜ x⎟ ⎜ Cy ⎟ ⎝ ⎠

⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ −200 ⎟ lb ⎜ F ⎟ ⎝ 500 ⎠ ⎝ y⎠

⎛⎜ Cx ⎟⎞ ⎛ 75 ⎞ = lb ⎜ Cy ⎟ ⎜⎝ 100 ⎟⎠ ⎝ ⎠

Problem 6-79 Determine the horizontal and vertical components of force that the pins at A, B, and C exert on their connecting members. Units Used: 3

kN = 10 N Given: F = 800 N a = 1m r = 50 mm b = 0.2 m Solution: −F ( a + r) + A x b = 0 550



Ax = F

a+r b

Chapter 6

Ax = 4.2 kN

− Ax + Bx = 0 Bx = Ax

B x = 4.2 kN

−F r − Ay b + A x b = 0 Ay =

−F r + A x b b

Ay = 4 kN

Ay − By − F = 0 By = Ay − F

B y = 3.2 kN

− Ax + F + Cx = 0 Cx = Ax − F

Cx = 3.4 kN

Ay − Cy = 0 Cy = Ay

Cy = 4 kN

Problem 6-80 Operation of exhaust and intake valves in an automobile engine consists of the cam C, push rod DE, rocker arm EFG which is pinned at F, and a spring and valve, V. If the spring is compressed a distance δ when the valve is open as shown, determine the normal force acting on the cam lobe at C. Assume the cam and bearings at H, I, and J are smooth.The spring has a stiffness k. Given: a = 25 mm b = 40 mm

δ = 20 mm k = 300

N m

551



Chapter 6

Solution: F s = kδ

Fs = 6 N

ΣF y = 0; −F G + Fs = 0 FG = Fs

FG = 6 N

ΣMF = 0; FG b + T a = 0 b T = FG a

T = 9.60 N

552



Chapter 6

Problem 6-81 Determine the force P on the cord, and the angle θ that the pulley-supporting link AB makes with the vertical. Neglect the mass of the pulleys and the link. The block has weight W and the cord is attached to the pin at B. The pulleys have radii of r1 and r2. Given: W = 200 lb r1 = 2 in r2 = 1 in

φ = 45 deg Solution: The initial guesses are

θ = 30 deg +

↑Σ Fy = 0;

F AB = 30 lb 2P − W = 0 P =

1 2

W

P = 100 lb

Given + Σ F x = 0; →

P cos ( φ ) − FAB sin ( θ ) = 0

+

F AB cos ( θ ) − P − P − P sin ( φ ) = 0

↑Σ Fy = 0;

⎛ θ ⎞ ⎜ F ⎟ = Find ( θ , FAB) ⎝ AB ⎠

θ = 14.6 deg

F AB = 280 lb

Problem 6-82 The nail cutter consists of the handle and the two cutting blades. Assuming the blades are pin connected at B and the surface at D is smooth, determine the normal force on the fingernail when a force F is applied to the handles as shown.The pin AC slides through a smooth hole at A and is attached to the bottom member at C.

553



Chapter 6

Given: F = 1 lb a = 0.25 in b = 1.5 in Solution: Handle : ΣMD = 0;

FA a − F b = 0 FA = F

ΣF y = 0;

⎛ b⎞ ⎜ ⎟ ⎝ a⎠

F A = 6 lb

ND − FA − F = 0 ND = F A + F

ND = 7 lb

Top blade : ΣMB = 0;

ND b − F N( 2 a + b) = 0 b ⎞ F N = ND ⎛⎜ ⎟ ⎝ 2 a + b⎠

F N = 5.25 lb

Problem 6-83 The wall crane supports load F. Determine the horizontal and vertical components of reaction at the pins A and D. Also, what is the force in the cable at the winch W? Units Used:

3

kip = 10 lb

Given: F = 700 lb

554



Chapter 6

a = 4 ft b = 4 ft c = 4 ft

θ = 60 deg Solution: Pulley E: +

↑Σ Fy = 0; T =

1 2

2T − F = 0 T = 350 lb

F

This is the force in the cable at the winch W a φ = atan ⎛⎜ ⎟⎞

Member ABC:

⎝ b⎠

Σ MA = 0;

(

)

−F ( b + c) + TBD sin ( φ ) − T sin ( θ ) b = 0

⎛ b + c ⎞ + T sin ( θ ) ⎟ ⎝ b ⎠

F⎜ TBD =

sin ( φ ) 3

TBD = 2.409 × 10 lb +

↑Σ Fy = 0; − Ay + TBD sin ( φ ) − T sin ( θ ) − F = 0 Ay = TBD sin ( φ ) − T sin ( θ ) − F

Ay = 700 lb

+ Σ F x = 0; →

Ax − TBD cos ( φ ) − T cos ( θ ) = 0 Ax = TBD cos ( φ ) + T cos ( θ )

Ax = 1.878 kip

At D: 555



Chapter 6

Dx = TBD cos ( φ )

Dx = 1.703 kip

Dy = TBD sin ( φ )

Dy = 1.703 kip

Problem 6-84 Determine the force that the smooth roller C exerts on beam AB. Also, what are the horizontal and vertical components of reaction at pin A? Neglect the weight of the frame and roller. Given: M = 60 lb⋅ ft a = 3 ft b = 4 ft c = 0.5 ft Solution: Σ MA = 0;

−M + Dx c = 0 M Dx = c Dx = 120 lb

+ Σ F x = 0; →

+

↑Σ Fy = 0; Σ MB = 0;

Ax − Dx = 0 Ax = Dx

Ax = 120 lb

Ay = 0

Ay = 0 lb

−Nc b + Dx c = 0 c Nc = Dx b

Nc = 15.0 lb

Problem 6-85 Determine the horizontal and vertical components of force which the pins exert on member ABC.

556



Chapter 6

Given: W = 80 lb a = 6 ft b = 9 ft c = 3 ft r = 0.5 ft Solution: + Σ F x = 0; →

− Ax + W = 0 Ax = W

+

↑Σ Fy = 0;

Ax = 80 lb

Ay − W = 0 Ay = W

Σ MC = 0;

Ay = 80 lb

Ay( a + b) − B y b = 0 By = Ay

a+b b

B y = 133 lb Σ MD = 0;

−W( c − r) + By b − Bx c = 0 Bx =

+ Σ F x = 0; →

B y b − W ( c − r)

B x = 333 lb

c

Ax + B x − Cx = 0 Cx = Ax + B x

+

↑Σ Fy = 0;

Cx = 413 lb

− Ay + B y − Cy = 0 Cy = B y − A y

Cy = 53.3 lb

Problem 6-86 The floor beams AB and BC are stiffened using the two tie rods CD and AD. Determine the force along each rod when the floor beams are subjected to a uniform load w. Assume the three contacting members at B are smooth and the joints at A, C, and D are pins. Hint: 557



Chapter 6

g j Members AD, CD, and BD are two-force members

p

3

kip = 10 lb

Units Used: Given: w = 80

lb ft

b = 5 ft a = 12 ft Solution: Due to summetry: Cy =

w( 2a) 2

Cy = 960 lb

Member BC : ΣMB = 0;

a Cy( a) − w a⎛⎜ ⎟⎞ − T⎛ ⎜ 2

⎝ ⎠

a T = ⎛⎜ Cy − w ⎟⎞ 2⎠ ⎝

⎞a = 0 2 2⎟ ⎝ a +b ⎠

2

b

2

a +b

T = 1.248 kip

b

Problem 6-87 Determine the horizontal and vertical components of force at pins B and C. Given: F = 50 lb

c = 6 ft

a = 4 ft

d = 1.5 ft

558



b = 4 ft Solution:

Chapter 6

r = 0.5 ft Guesses

Cx = 1 lb

Cy = 1 lb

B x = 1 lb

B y = 1 lb

Given F ( a − r) + Cx c − Cy( a + b) = 0 −F ( a − r) − F( d + r) + Cy( a + b) = 0 −B x + F + Cx = 0 B y − F + Cy = 0

⎛ Bx ⎞ ⎜ ⎟ ⎜ By ⎟ ⎜ ⎟ = Find ( Bx , By , Cx , Cy) ⎜ Cx ⎟ ⎜C ⎟ ⎝ y⎠ ⎛ Bx ⎞ ⎜ ⎟ ⎛⎜ 66.667 ⎞⎟ ⎜ By ⎟ ⎜ 15.625 ⎟ lb ⎜ ⎟=⎜ ⎟ ⎜ Cx ⎟ ⎜ 16.667 ⎟ ⎜ C ⎟ ⎝ 34.375 ⎠ ⎝ y⎠

559



Chapter 6

Problem 6-88 The skid steer loader has a mass M1, and in the position shown the center of mass is at G1. If there is a stone of mass M2 in the bucket, with center of mass at G2 determine the reactions of each pair of wheels A and B on the ground and the force in the hydraulic cylinder CD and at the pin E. There is a similar linkage on each side of the loader. Units Used: 3

Mg = 10 kg 3

kN = 10 N Given: M1 = 1.18 Mg M2 = 300 kg a = 1.25 m

d = 0.15 m

b = 1.5 m

e = 0.5 m

c = 0.75 m

θ = 30 deg

Solution:

Entire System:

ΣMA = 0;

M2 g b − M1 g( c − d) + NB c = 0 NB =

ΣF y = 0;

M1 g( c − d) − M2 g b c

NB = 3.37 kN

(Both wheels)

NA = 11.1 kN

(Both wheels)

NB − M2 g − M1 g + NA = 0 NA = −NB + M2 g + M1 g

Upper member: ΣME = 0;

M2 g( a + b) − 2 F CD sin ( θ ) a = 0

560



F CD = ΣF x = 0; ΣF y = 0;

Chapter 6

M 2 g( a + b)

F CD = 6.5 kN

2 sin ( θ ) a

E x = F CD( cos ( θ ) ) M2 g Ey − + FCD sin ( θ ) = 0 2 Ey = FR =

M2 g 2

E x = 5607 N

− F CD sin ( θ )

2

E y = −1766 N

2

Ex + Ey

F R = 5.879 kN

Problem 6-89 Determine the horizontal and vertical components of force at each pin. The suspended cylinder has a weight W. Given: W = 80 lb

d = 6 ft

a = 3 ft

e = 2 ft

b = 4 ft

r = 1 ft

c = 4 ft Solution: Guesses Ax = 1 lb

B x = 1 lb

B y = 1 lb

F CD = 1 lb

E x = 1 lb

E y = 1 lb

Given Ex − Bx + W = 0 −E y + By = 0

561



Chapter 6

−W r + Ey a − Ex d = 0 −B y +

−B y a +

c 2

FCD − W = 0

2

c +d c 2

c +d

− Ax + Bx +

2

F CD d − W( d + e − r) = 0

d 2

2

c +d

FCD − W = 0

⎛ Ax ⎞ ⎜ ⎟ B x ⎜ ⎟ ⎜ B ⎟ ⎜ y ⎟ = Find A , B , B , F , E , E ( x x y CD x y) ⎜ FCD ⎟ ⎜ ⎟ ⎜ Ex ⎟ ⎜ ⎟ ⎝ Ey ⎠ Cx = F CD

d 2

c +d

2

Cy = F CD

c 2

c +d

2

Dx = −Cx

Dy = −Cy

⎛ Ax ⎞ ⎜ ⎟ ⎛ 160 ⎞ ⎜ Bx ⎟ ⎜ ⎟ 80 ⎜B ⎟ ⎜ ⎟ 26.667 ⎜ y⎟ ⎜ ⎟ ⎜ Cx ⎟ ⎜ ⎟ 160 ⎜ ⎟ ⎜ ⎟ 106.667 C = ⎜ y⎟ ⎜ ⎟ lb ⎜D ⎟ ⎜ ⎟ −160 ⎜ x⎟ ⎜ ⎟ ⎜ Dy ⎟ ⎜ −106.667 ⎟ ⎜ ⎟ ⎜ −8.694 × 10− 13 ⎟ ⎜ Ex ⎟ ⎜ ⎟ 26.667 ⎠ ⎜ ⎟ ⎝ Ey ⎝ ⎠

562



Chapter 6

Problem 6-90 The two-member frame is pin connected at C, D, and E. The cable is attached to A, passes over the smooth peg at B, and is attached to a load W. Determine the horizontal and vertical reactions at each pin. Given: a = 2 ft b = 1 ft c = 0.75 ft W = 100 lb Solution: d =

c

( a + 2 b) b Initial guesses: Cx = 1 lb

Cy = 1 lb

Dx = 1 lb

Dy = 1 lb

E x = 1 lb

E y = 1 lb

Given −Dx + Cx − W = 0 −Dy + Cy − W = 0 −Cx c + Cy b + W d − W( a + 2 b) = 0 W − Cx + Ex = 0 E y − Cy = 0 −W d + Cx c + Cy b = 0

562

563



Chapter 6

⎛ Cx ⎞ ⎜ ⎟ ⎜ Cy ⎟ ⎜D ⎟ ⎜ x ⎟ = Find C , C , D , D , E , E ( x y x y x y) ⎜ Dy ⎟ ⎜ ⎟ ⎜ Ex ⎟ ⎜ ⎟ ⎝ Ey ⎠

⎛ Cx ⎞ ⎜ ⎟ ⎛ 133 ⎞ ⎜ Cy ⎟ ⎜ 200 ⎟ ⎟ ⎜D ⎟ ⎜ ⎜ ⎜ x ⎟ = 33 ⎟ lb ⎜ Dy ⎟ ⎜ 100 ⎟ ⎟ ⎜ ⎟ ⎜ 33 ⎜ ⎟ ⎜ Ex ⎟ ⎜ ⎜ ⎟ ⎝ 200 ⎟⎠ ⎝ Ey ⎠

Problem 6-91 Determine the horizontal and vertical components of force which the pins at A, B, and C exert on member ABC of the frame. Given: F 1 = 400 N F 2 = 300 N F 3 = 300 N a = 1.5 m b = 2m c = 1.5 m d = 2.5 m f = 1.5 m g = 2m e = a+b+c−d Solution: Guesses Ay = 1 N F BD = 1 N

Cx = 1 N

Cy = 1 N

F BE = 1 N

564



Chapter 6

Given F 1 g + F2( a + b) + F3 a − Ay( f + g) = 0 F 1 g − Cy( f + g) = 0 Cx e = 0 −Cx −

f+g 2

e + ( f + g)

Ay − Cy −

2

FBD +

e 2

e + ( f + g)

2

f+g 2

d + ( f + g)

FBD −

2

F BE = 0

d 2

d + ( f + g)

2

F BE = 0

⎛⎜ Ay ⎞⎟ ⎜ Cx ⎟ ⎜ ⎟ C ⎜ y ⎟ = Find ( Ay , Cx , Cy , FBD , FBE) ⎜F ⎟ ⎜ BD ⎟ ⎜ FBE ⎟ ⎝ ⎠ Bx = −

By =

f+g 2

e + ( f + g) e 2

e + ( f + g)

2

2

F BD +

F BD +

f+g 2

d + ( f + g) d 2

d + ( f + g)

2

2

FBE

FBE

Ay = 657 N

⎛⎜ Bx ⎟⎞ ⎛ 0 ⎞ = N ⎜ By ⎟ ⎜⎝ 429 ⎟⎠ ⎝ ⎠ ⎛⎜ Cx ⎟⎞ ⎛ 0 ⎞ = N ⎜ Cy ⎟ ⎜⎝ 229 ⎟⎠ ⎝ ⎠

565



Chapter 6

Problem 6-92 The derrick is pin-connected to the pivot at A. Determine the largest mass that can be supported by the derrick if the maximum force that can be sustained by the pin at A is Fmax. Units Used: 3

kN = 10 N m

g = 9.81

2

s 3

Mg = 10 kg Given: F max = 18 kN L = 5m

θ = 60 deg Solution: AB is a two-force member. Require F AB = Fmax +

↑

Σ F y = 0;

F AB sin ( θ ) − M = 2

Mg

⎛ FAB ⎞ ⎜ ⎟ ⎝ g ⎠

2

sin ( θ ) − W = 0

⎛ sin ( θ ) ⎞ ⎜ ⎟ ⎝ sin ( θ ) + 2 ⎠

M = 5.439

1 2

Mg

s

Problem 6-93 Determine the required mass of the suspended cylinder if the tension in the chain wrapped around the freely turning gear is T. Also, what is the magnitude of the resultant force on pin A? Units Used: 3

kN = 10 N g = 9.8

m 2

s

566



Chapter 6

Given: T = 2 kN L = 2 ft

θ = 30 deg φ = 45 deg

Solution: Σ MA = 0;

−2 T L cos ( θ ) + M g cos ( φ ) L cos ( θ ) + M g sin ( φ ) L sin ( θ ) = 0 M =

2 T cos ( θ )

(cos ( φ ) cos ( θ ) + sin (φ ) sin ( θ )) g

M = 1793 +

→ Σ Fx = 0;

1 2

kg

s

2 T − M g cos ( φ ) − A x = 0 Ax = 2 T − M g cos ( φ )

+

↑Σ Fy = 0;

M g sin ( φ ) − Ay = 0 Ay = M g sin ( φ ) FA =

2

2

Ax + Ay

F A = 2.928 kN

567



Chapter 6

Problem 6-94 The tongs consist of two jaws pinned to links at A, B, C, and D. Determine the horizontal and vertical components of force exerted on the stone of weight W at F and G in order to lift it. Given: a = 1 ft b = 2 ft c = 1.5 ft d = 1 ft W = 500 lb Solution: Guesses F x = 1 lb F y = 1 lb F AD = 1 lb F BE = 1 lb Given

2 Fy − W = 0 F AD − F x −

F AD b − Fx( b + c) = 0 a ⎞ ⎛ ⎜ 2 2 ⎟ FBE = 0 ⎝ a +d ⎠

⎛ Fx ⎞ ⎜ ⎟ ⎜ Fy ⎟ ⎜ ⎟ = Find ( Fx , Fy , FAD , FBE) ⎜ FAD ⎟ ⎜F ⎟ ⎝ BE ⎠

−F y +

d ⎞ ⎛ ⎜ 2 2 ⎟ FBE = 0 ⎝ a +d ⎠

⎛⎜ Gx ⎞⎟ ⎛⎜ Fx ⎟⎞ = ⎜ Gy ⎟ ⎜ Fy ⎟ ⎝ ⎠ ⎝ ⎠

⎛ Fx ⎞ ⎜ ⎟ ⎛⎜ 333 ⎟⎞ ⎜ Fy ⎟ ⎜ 250 ⎟ lb ⎜ ⎟=⎜ ⎟ ⎜ Gx ⎟ ⎜ 333 ⎟ ⎜ G ⎟ ⎝ 250 ⎠ ⎝ y⎠

Problem 6-95 Determine the force P on the cable if the spring is compressed a distance δ when the mechanism is in the position shown. The spring has a stiffness k.

568



Chapter 6

Given:

δ = 0.5 in

c = 6 in

lb

d = 6 in

k = 800

ft

a = 24 in

e = 4 in

b = 6 in

θ = 30 deg

Solution: F E = kδ

F E = 33.333 lb

The initial guesses are P = 20 lb

B x = 11 lb

B y = 34 lb

F CD = 34 lb

Given ΣMA = 0;

B x b + B y c − F E ( a + b) = 0

ΣMD = 0;

By d − P e = 0

+ Σ F x = 0; →

ΣMB = 0;

−B x + FCD cos ( θ ) = 0 F CD sin ( θ ) d − P ( d + e) = 0

⎛⎜ FCD ⎞⎟ ⎜ Bx ⎟ ⎜ ⎟ = Find ( FCD , Bx , By , P) B ⎜ y ⎟ ⎜ P ⎟ ⎝ ⎠ B x = 135.398 lb

B y = 31.269 lb

F CD = 156.344 lb

P = 46.903 lb

569



Chapter 6

Problem 6-96 The scale consists of five pin-connected members. Determine the load W on the pan EG if a weight F is suspended from the hook at A. Given: F = 3 lb

b = 3 in

a = 5 in

c = 4 in

d = 6 in

f = 2 in

e = 8 in Solution: Guesses

TC = 10 lb

TD = 10 lb

TG = 10 lb

W = 10 lb

Given Member ABCD:

ΣMB = 0;

F a − TC b − TD( b + e − d) = 0 Member EG: ΣMG = 0;

− TC e + W c = 0

ΣF y = 0;

TG − W + TC = 0

Member FH: ΣMH = 0;

−TD( d + f) + TG f = 0

⎛⎜ TC ⎟⎞ ⎜ TD ⎟ ⎜ ⎟ = Find ( TC , TD , TG , W) ⎜ TG ⎟ ⎜W⎟ ⎝ ⎠ W = 7.06 lb

Problem 6-97 The machine shown is used for forming metal plates. It consists of two toggles ABC and DEF, 570



Chapter 6

g p gg which are operated by the hydraulic cylinder H. The toggles push the movable bar G forward, pressing the plate p into the cavity. If the force which the plate exerts on the head is P, determine the force F in the hydraulic cylinder for the given angle θ. Units Used: 3

kN = 10 N Given: P = 12 kN a = 200 mm

θ = 30 deg Solution: Member EF: ΣME = 0;

−F y a cos ( θ ) + Fy =

ΣF x = 0;

Ex − Ex =

ΣF y = 0;

2

2

2

a sin ( θ ) = 0

tan ( θ )

P

P

P

F y = 3.464 kN

=0

P

E x = 6 kN

2

Ey − Fy = 0 Ey = Fy

E y = 3.464 kN

Joint E: ΣF x = 0;

−F DE cos ( θ ) − E x = 0 F DE =

−E x

cos ( θ )

F DE = −6.928 kN

F − Ey + F DE sin ( θ ) = 0

ΣF y = 0;

F = E y − FDE sin ( θ )

F = 6.93 kN

Problem 6-98 Determine the horizontal and vertical components of force at pins A and C of the two-member frame.

571



Chapter 6

Given: N w1 = 500 m N w2 = 400 m N w3 = 600 m a = 3m b = 3m


Ay = 1 N Cx = 1 N

Cy = 1 N

Given 1 Ay + Cy − w1 a − w2 a = 0 2

Ax a −

1 − Ax + Cx + w3 b = 0 2

− Ay a +

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ ⎟ = Find ( Ax , Ay , Cx , Cy) ⎜ Cx ⎟ ⎜C ⎟ ⎝ y⎠

1 2a 1 b a w1 a − w3 b − w2 a = 0 2 3 2 3 2 1 a w1 a = 0 2 3

⎛ Ax ⎞ ⎜ ⎟ ⎛⎜ 1400 ⎞⎟ ⎜ Ay ⎟ ⎜ 250 ⎟ N ⎜ ⎟=⎜ ⎟ C 500 x ⎜ ⎟ ⎜ ⎟ ⎜ C ⎟ ⎝ 1700 ⎠ ⎝ y⎠

572



Chapter 6

Problem 6-99 The truck rests on the scale, which consists of a series of compound levers. If a mass M1 is placed on the pan P and it is required that the weight is located at a distance x to balance the “beam” ABC, determine the mass of the truck. There are pins at all lettered points. Is it necessary for the truck to be symmetrically placed on the scale? Explain. Units Used: 3

Mg = 10 kg

g = 9.81

m 2

s Given: M1 = 15 kg

FD = 3 m

x = 0.480 m

EF = 0.2 m

a = 0.2 m HI = 0.1 m

GH = 2.5 m

KJ = HI

KG = GH

Solution: Member ABC : ΣMB = 0;

−M1 g x + F AD a = 0

x F AD = M1 g a Member EFD : ΣME = 0;

2

F AD = 72 s N

−F y EF + F AD( FD + EF) = 0

⎛ FD + EF ⎞ ⎟ ⎝ EF ⎠

F y = F AD⎜

2

F y = 1152 s N

573



Chapter 6

Member GHI : ΣMI = 0;

Hy HI − GY( GH + HI ) = 0

Member JKG : ΣMJ = 0;

(Fy − Gy)( KJ + GH ) − Ky( KJ ) = 0 Ky + Hy = Fy

KJ + KG HI

Scale Platform : ΣF y = 0;

Ky + Hy = W W = Fy⎛⎜

⎝

M =

W g

KJ + KG ⎞ HI

⎟ ⎠

M = 14.98 Mg

Because KJ = HI and KG = GH it doesn't matter where the truck is on the scale.

Problem 6-100 By squeezing on the hand brake of the bicycle, the rider subjects the brake cable to a tension T If the caliper mechanism is pin-connected to the bicycle frame at B, determine the normal force each brake pad exerts on the rim of the wheel. Is this the force that stops the wheel from turning? Explain. Given: T = 50 lb a = 2.5 in b = 3 in

574



Chapter 6

Solution: Σ MB = 0;

−N b + T a = 0

N = T

a b

N = 41.7 lb

This normal force does not stop the wheel from turning. A frictional force (see Chapter 8), which acts along the wheel's rim stops the wheel.

Problem 6-101 If a force of magnitude P is applied perpendicular to the handle of the mechanism, determine the magnitude of force F for equilibrium. The members are pin-connected at A, B, C, and D. Given: P = 6 lb a = 25 in b = 4 in c = 5 in d = 4 in e = 5 in f = 5 in g = 30 in Solution: Σ MA = 0;

F BC b − P a = 0 F BC =

Pa b

575



Chapter 6

F BC = 37.5 lb + Σ F x = 0; →

− Ax + P = 0 Ax = P

+

↑Σ Fy = 0;

Ax = 6 lb

− Ay + F BC = 0 Ay = FBC

Σ MD = 0;

Ay = 37.5 lb

−e Ax − Ay( b + c) + ( g + b + c)F = 0 F =

e Ax + Ay( b + c)

F = 9.423 lb

g+b+c

Problem 6-102 The pillar crane is subjected to the load having a mass M. Determine the force developed in the tie rod AB and the horizontal and vertical reactions at the pin support C when the boom is tied in the position shown. Units Used: 3

kN = 10 N Given: M = 500 kg a = 1.8 m b = 2.4 m

θ 1 = 10 deg θ 2 = 20 deg g = 9.81

m 2

s Solution:

initial guesses: F CB = 10 kN

F AB = 10 kN

576



Chapter 6

Given −M

( )

( )

( )

( )

g cos θ 1 − F AB cos θ 2 + FCB 2

−M

g sin θ 1 − FAB sin θ 2 + FCB 2

⎛ FAB ⎞

⎜ ⎟ = Find F , F ( AB CB) ⎜ FCB ⎟ ⎝ ⎠

⎛⎜ Cx ⎟⎞ ⎜ Cy ⎟ ⎝ ⎠

b

=0

2

2

a +b a 2

− Mg = 0 2

a +b

=

⎛b⎞ ⎜ ⎟ 2 2 a a +b ⎝ ⎠ FCB

⎛ FAB ⎞ ⎛ 9.7 ⎞ ⎜ ⎟ ⎜ ⎟ C ⎜ x ⎟ = ⎜ 11.53 ⎟ kN ⎜ C ⎟ ⎝ 8.65 ⎠ ⎝ y ⎠

Problem 6-103 The tower truss has a weight W and a center of gravity at G. The rope system is used to hoist it into the vertical position. If rope CB is attached to the top of the shear leg AC and a second rope CD is attached to the truss, determine the required tension in BC to hold the truss in the position shown. The base of the truss and the shear leg bears against the stake at A, which can be considered as a pin. Also, compute the compressive force acting along the shear leg. Given: W = 575 lb

θ = 40 deg a = 5 ft b = 3 ft c = 10 ft d = 4 ft e = 8 ft Solution: Entire system:

ΣMA = 0;

TBC cos ( θ ) ( d + e) − TBC sin ( θ ) a − W( a + b) = 0 TBC =

W( a + b)

cos ( θ ) ( d + e) − sin ( θ ) a

TBC = 769 lb 577



Chapter 6

CA is a two-force member.

At C:

e ⎞ ⎟ ⎝ b + c⎠

φ = atan ⎛⎜ initial guesses:

F CA = 500 lb

TCD = 300 lb

Given ΣF x = 0;

ΣF y = 0;

F CA

F CA

a 2

a + ( d + e)

2

d+e 2

a + ( d + e)

⎛⎜ FCA ⎟⎞ = Find ( FCA , TCD) ⎜ TCD ⎟ ⎝ ⎠

2

+ TCD cos ( φ ) − TBC cos ( θ ) = 0

− TCD sin ( φ ) − TBC sin ( θ ) = 0

TCD = 358 lb

F CA = 739 lb

Problem 6-104 The constant moment M is applied to the crank shaft. Determine the compressive force P that is exerted on the piston for equilibrium as a function of θ. Plot the results of P (ordinate) versus θ (abscissa) for 0 deg ≤ θ ≤ 90 deg. Given: a = 0.2 m b = 0.45 m M = 50 N⋅ m Solution: a cos ( θ ) = b sin ( φ )

φ = asin ⎛⎜ cos ( θ )⎟⎞ a ⎝b

⎠

−M + F BC cos ( θ − φ ) a = 0

578



F BC =

Chapter 6

M

a cos ( θ − φ )

P = FBC cos ( φ ) =

M cos ( φ )

a cos ( θ − φ )

This function goes to infinity at θ = 90 deg, so we will only plot it to θ = 80 deg.

θ = 0 , 0.1 .. 80 φ ( θ ) = asin ⎛⎜ cos ( θ deg)⎟⎞

P (θ) =

a

⎝b

⎠

M cos ( φ ( θ ) )

a cos ( θ deg − φ ( θ ) )

Newtons

1500

P( θ )

1000 500 0

0

20

40

60

80

θ

Degrees

Problem 6-105 Five coins are stacked in the smooth plastic container shown. If each coin has weight W, determine the normal reactions of the bottom coin on the container at points A and B. Given: W = 0.0235 lb a = 3 b = 4

579



Chapter 6

Solution: All coins : ΣF y = 0;

NB = 5W NB = 0.1175 lb

Bottom coin : ΣF y = 0;

⎛

⎞=0 ⎝ a +b ⎠ b

NB − W − N⎜

(

N = NB − W

2⎟

2

⎛ a2 + b2 ⎞ )⎜⎝ b ⎟⎠

N = 0.1175 lb ΣF x = 0;

⎛

NA = N⎜

a 2

⎞

2⎟

⎝ a +b ⎠

NA = 0.0705 lb

Problem 6-106 Determine the horizontal and vertical components of force at pin B and the normal force the pin at C exerts on the smooth slot. Also, determine the moment and horizontal and vertical reactions of force at A. There is a pulley at E. Given: F = 50 lb a = 4 ft b = 3 ft Solution: Guesses B x = 1 lb B y = 1 lb NC = 1 lb 580



Ax = 1 lb

Ay = 1 lb

Chapter 6

MA = 1 lb⋅ ft

Given Bx +

a ⎛ ⎞ ⎜ 2 2 ⎟ NC − F = 0 ⎝ a +b ⎠

By −

b ⎛ ⎞ ⎜ 2 2 ⎟ NC − F = 0 ⎝ a +b ⎠

(F + Bx)a − (F + By)b = 0 F−

a ⎛ ⎞ ⎜ 2 2 ⎟ NC − Ax = 0 ⎝ a +b ⎠

b ⎛ ⎞ ⎜ 2 2 ⎟ NC − Ay = 0 ⎝ a +b ⎠ −F 2 a +

a ⎛ ⎞ ⎜ 2 2 ⎟ NC a + MA = 0 ⎝ a +b ⎠

⎛ Bx ⎞ ⎜ ⎟ B y ⎜ ⎟ ⎜N ⎟ ⎜ C ⎟ = Find B , B , N , A , A , M ( x y C x y A) ⎜ Ax ⎟ ⎜ ⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎝ MA ⎠

NC = 20 lb

⎛⎜ Bx ⎟⎞ ⎛ 34 ⎞ = lb ⎜ By ⎟ ⎜⎝ 62 ⎟⎠ ⎝ ⎠ ⎛⎜ Ax ⎞⎟ ⎛ 34 ⎞ = lb ⎜ Ay ⎟ ⎜⎝ 12 ⎟⎠ ⎝ ⎠ MA = 336 lb⋅ ft

Problem 6-107 A force F is applied to the handles of the vise grip. Determine the compressive force developed on the smooth bolt shank A at the jaws. 581



Chapter 6

Given: F = 5 lb

b = 1 in

a = 1.5 in

c = 3 in

d = 0.75 in

e = 1 in

θ = 20 deg Solution: From FBD (a) ΣME = 0;

⎡

⎤b = 0 ⎣ c + ( d + e) ⎦

F ( b + c) − F CD⎢

d+e

2

2⎥

⎡ c2 + ( d + e) 2⎤ ⎥ ⎣ b( d + e) ⎦

F CD = F( b + c) ⎢ ΣF x = 0;

⎡

F CD = 39.693 lb

⎤ ⎥ 2 2 ⎣ c + ( d + e) ⎦

E x = F CD⎢

c

E x = 34.286 lb From FBD (b) ΣMB = 0;

NA sin ( θ ) d + NA cos ( θ ) a − Ex( d + e) = 0

NA = Ex

d+e ⎛ ⎜ ( ) ⎝ sin θ d + cos ( θ )

⎞ ⎟

a⎠

NA = 36.0 lb

Problem 6-108 If a force of magnitude P is applied to the grip of the clamp, determine the compressive force F that the wood block exerts on the clamp.

582



Chapter 6

Given: P = 10 lb a = 2 in b = 2 in c = 0.5 in d = 0.75 in e = 1.5 in

Solution: Define

b φ = atan ⎛⎜ ⎟⎞

φ = 69.444 deg

⎝ d⎠

From FBD (a), ΣMB = 0;

F CD cos ( φ ) c − P ( a + b + c) = 0 F CD =

+

↑Σ Fy = 0;

P( a + b + c)

F CD = 256.32 lb

cos ( φ ) ( c)

F CD sin ( φ ) − By = 0 B y = F CD sin ( φ )

B y = 240 lb

From FBD (b), ΣMA = 0;

By d − F e = 0 F =

By d

F = 120 lb

e

583



Chapter 6

Problem 6-109 The hoist supports the engine of mass M. Determine the force in member DB and in the hydraulic cylinder H of member FB. Units Used: 3

kN = 10 N Given: M = 125 kg

d = 1m

a = 1m

e = 1m

b = 2m

f = 2m

c = 2m

g = 9.81

m 2

s Solution: Member GFE: ΣME = 0; −F FB⎡ ⎢

⎤ b + M g ( a + b) = 0 2 2⎥ ⎣ ( c + d) + ( b − e) ⎦

F FB = M g

c+d

⎡ a + b ⎤ ( c + d) 2 + ( b − e) 2 ⎢ ⎥ ⎣ b( c + d) ⎦

F FB = 1.94 kN ΣF x = 0;

⎡

⎤=0 ⎣ ( c + d) + ( b − e) ⎦ b−e

E x − FFB⎢

2⎥

2

⎡

⎤

b−e

E x = F FB⎢

2⎥

2

⎣ ( c + d) + ( b − e) ⎦

Member EDC: ΣΜc = 0;

⎛

⎞d = 0 ⎟ 2 2 ⎝ e +d ⎠

E x( c + d) − F DB⎜

F DB = Ex

⎛ c + d⎞ ⎜ ⎟ ⎝ ed ⎠

e

2

2

e +d

F DB = 2.601 kN

584



Chapter 6

Problem 6-110 The flat-bed trailer has weight W1 and center of gravity at GT . It is pin-connected to the cab at D. The cab has a weight W2 and center of gravity at GC. Determine the range of values x for the position of the load L of weight W3 so that no axle is subjected to a force greater than FMax. The load has a center of gravity at GL . Given: W1 = 7000 lb

a = 4 ft

W2 = 6000 lb

b = 6 ft

W3 = 2000 lb

c = 3 ft

F max = 5500 lb d = 10 ft e = 12 ft Solution: Case 1:

Assume

Guesses

Ay = Fmax x = 1 ft

Given

Ay = Fmax B y = F max

Cy = F max

Dy = Fmax

Ay + B y − W2 − Dy = 0 −W2 a − Dy( a + b) + B y( a + b + c) = 0 Dy − W1 − W3 + Cy = 0 W3 x + W1 e − Dy( c + d + e) = 0

⎛⎜ By ⎞⎟ ⎜ Cy ⎟ ⎜ ⎟ = Find ( By , Cy , Dy , x) ⎜ Dy ⎟ ⎜ x ⎟ ⎝ ⎠ ⎛ Ay ⎞ ⎛⎜ 5.5 × 103 ⎟⎞ ⎜ ⎟ x1 = 30.917 ft ⎜ By ⎟ = ⎜ 6.333 × 103 ⎟ lb x1 = x ⎟ ⎜C ⎟ ⎜ Since By > Fmax then this solution is no good. ⎝ y ⎠ ⎜⎝ 3.167 × 103 ⎟⎠ Case 2:

Assume

Guesses

Ay = Fmax

B y = F max B y = F max

Cy = F max 585



x = 1 ft Given

Chapter 6

Dy = Fmax


⎛⎜ Ay ⎞⎟ ⎜ Cy ⎟ ⎜ ⎟ = Find ( Ay , Cy , Dy , x) ⎜ Dy ⎟ ⎜ x ⎟ ⎝ ⎠

⎛ Ay ⎞ ⎛⎜ 5.25 × 103 ⎟⎞ ⎜ ⎟ ⎜ By ⎟ = ⎜ 5.5 × 103 ⎟ lb x2 = x ⎟ ⎜C ⎟ ⎜ 3 ⎜ ⎝ y ⎠ ⎝ 4.25 × 10 ⎟⎠

x2 = 17.375 ft

Since Ay < Fmax and Cy < F max then this solution is good. Cy = F max

Case 3:

Assume

Guesses

Ay = Fmax

B y = F max

x = 1 ft

Dy = Fmax

Given

Cy = F max


⎛⎜ Ay ⎞⎟ ⎜ By ⎟ ⎜ ⎟ = Find ( Ay , By , Dy , x) ⎜ Dy ⎟ ⎜ x ⎟ ⎝ ⎠

⎛ Ay ⎞ ⎛⎜ 4.962 × 103 ⎟⎞ ⎜ ⎟ ⎜ By ⎟ = ⎜ 4.538 × 103 ⎟ lb x3 = x ⎟ ⎜C ⎟ ⎜ 3 ⎜ ⎝ y ⎠ ⎝ 5.5 × 10 ⎟⎠

x3 = 1.75 ft

Since Ay < Fmax and B y < F max then this solution is good. We conclude that x3 = 1.75 ft < x < x2 = 17.375 ft

586



Chapter 6

Problem 6-111 Determine the force created in the hydraulic cylinders EF and AD in order to hold the shovel in equilibrium. The shovel load has a mass W and a center of gravity at G. All joints are pin connected. Units Used: 3

Mg = 10 kg 3

kN = 10 N Given: a = 0.25 m θ 1 = 30 deg b = 0.25 m θ 2 = 10 deg c = 1.5 m

θ 3 = 60 deg

d = 2m

W = 1.25 Mg

e = 0.5 m Solution: Assembly FHG :

(

( )) = 0

ΣMH = 0; −[ W g( e) ] + FEF c sin θ 1 F EF = W g

⎛ e ⎞ F = 8.175 kN (T) ⎜ c sin θ ⎟ EF ( ) 1 ⎝ ⎠

Assembly CEFHG:

(

)

( )

ΣMC = 0; F AD cos θ 1 + θ 2 b − W g⎡( a + b + c)cos θ 2 + e⎤ = 0 ⎣ ⎦ F AD = W g

⎛ cos ( θ 2) a + cos ( θ 2) b + cos ( θ 2) c + e ⎞ ⎜ ⎟ cos ( θ 1 + θ 2) b ⎝ ⎠

F AD = 158 kN (C)

587



Chapter 6

Problem 6-112 The aircraft-hangar door opens and closes slowly by means of a motor which draws in the cable AB. If the door is made in two sections (bifold) and each section has a uniform weight W and length L, determine the force in the cable as a function of the door's position θ. The sections are pin-connected at C and D and the bottom is attached to a roller that travels along the vertical track.

587

Solution: L

⎛ θ ⎞ − 2L sin ⎛ θ ⎞ N = 0 ⎟ ⎜ ⎟ A ⎝2⎠ ⎝2⎠

cos ⎜

ΣMD = 0;

2W

ΣΜC = 0;

T L cos ⎜

2

⎛ θ ⎞ − N L sin ⎛ θ ⎞ − W L cos ⎛ θ ⎞ = 0 ⎟ ⎜ ⎟ ⎜ ⎟ A 2 ⎝2⎠ ⎝2⎠ ⎝2⎠

W ⎛θ⎞ NA = cot ⎜ ⎟ 2 ⎝2⎠ T=W

Problem 6-113 A man having weight W attempts to lift himself using one of the two methods shown. Determine the total force he must exert on bar AB in each case and the normal reaction he exerts on the platform at C. Neglect the weight of the platform. Given: W = 175 lb

588



Chapter 6

Solution:

(a) Bar: +

↑Σ Fy = 0;

2

⎛ F⎞ − 2 ⎛ W⎞ = 0 ⎜ ⎟ ⎜ ⎟ ⎝ 2⎠ ⎝2⎠

F = W

F = 175 lb

Man: +

↑Σ Fy = 0;

NC − W − 2

⎛ F⎞ = 0 ⎜ ⎟ ⎝ 2⎠

NC = W + F

NC = 350 lb

( b) Bar: +

↑

Σ F y = 0;

2

⎛ W⎞ − 2 F = 0 ⎜ ⎟ 2 ⎝4⎠

F =

W 2

Man: +

↑Σ Fy = 0;

NC − W + 2

F = 87.5 lb

⎛ F⎞ = 0 ⎜ ⎟ ⎝ 2⎠

NC = W − F

NC = 87.5 lb

589



Chapter 6

Problem 6-114 A man having weight W1 attempts to lift himself using one of the two methods shown. Determine the total force he must exert on bar AB in each case and the normal reaction he exerts on the platform at C. The platform has weight W2.

Given: W1 = 175 lb W2 = 30 lb Solution: (a) Bar: +

↑Σ Fy = 0;

2

F 2

(

)

− W1 + W2 = 0

F = W1 + W2 F = 205 lb Man: +

↑

Σ F y = 0;

F NC − W1 − 2 =0 2 NC = F + W1 NC = 380 lb

( b) Bar: +

↑

Σ F y = 0;

−2

⎛ F ⎞ + 2 ⎛⎜ W1 + W2 ⎞⎟ = 0 ⎜ ⎟ ⎝ 2⎠ ⎝ 4 ⎠

590



F =

Chapter 6

W1 + W2 2

F = 102 lb Man: +

↑

Σ F y = 0;

NC − W1 + 2

⎛F⎞ = 0 ⎜ ⎟ ⎝ 2⎠

NC = W1 − F NC = 72.5 lb

Problem 6-115 The piston C moves vertically between the two smooth walls. If the spring has stiffness k and is unstretched when θ = 0, determine the couple M that must be applied to AB to hold the mechanism in equilibrium. Given: k = 15

lb in

θ = 30 deg a = 8 in b = 12 in

Solution: Geometry: b sin ( ψ) = a sin ( θ )

ψ = asin ⎛⎜ sin ( θ )

⎝

a⎞ ⎟ b⎠

φ = 180 deg − ψ − θ

ψ = 19.471 deg φ = 130.529 deg

591



rAC

sin ( φ )

=

b

Chapter 6

rAC = b

sin ( θ )

⎛ sin ( φ ) ⎞ ⎜ ⎟ ⎝ sin ( θ ) ⎠

rAC = 18.242 in

Free Body Diagram: The solution for this problem will be simplified if one realizes that member CB is a two force member. Since the spring stretches x = ( a + b) − rAC

x = 1.758 in F sp = k x

the spring force is

F sp = 26.371 lb

Equations of Equilibrium: Using the method of joints +

↑

F CB cos ( ψ) − Fsp = 0

Σ F y = 0;

F CB =

Fsp

cos ( ψ)

F CB = 27.971 lb

From FBD of bar AB F CB sin ( φ ) a − M = 0

+ ΣMA = 0;

M = FCB sin ( φ ) a

M = 14.2 lb⋅ ft

Problem 6-116 The compound shears are used to cut metal parts. Determine the vertical cutting force exerted on the rod R if a force F is applied at the grip G. The lobe CDE is in smooth contact with the head of the shear blade at E. Given: F = 20 lb

e = 0.5 ft

a = 1.4 ft

f = 0.5 ft

b = 0.2 ft

g = 0.5 ft

c = 2 ft

h = 2.5 ft

d = 0.75 ft

θ = 60 deg

Solution: Member AG: ΣMA = 0;

F ( a + b) − FBC b sin ( θ ) = 0

592



F BC = F

⎛ a+b ⎞ ⎜ ⎟ ⎝ b sin ( θ ) ⎠

Chapter 6

F BC = 184.75 lb

Lobe: ΣMD = 0;

F BC f − NE e = 0 f NE = FBC e

NE = 184.75 lb

Head: ΣMF = 0;

−NE( h + g) + h NR = 0 NR = NE

⎛ h + g⎞ ⎜ ⎟ ⎝ h ⎠

NR = 222 lb

Problem 6-117 The handle of the sector press is fixed to gear G, which in turn is in mesh with the sector gear C. Note that AB is pinned at its ends to gear C and the underside of the table EF, which is allowed to move vertically due to the smooth guides at E and F. If the gears exert tangential forces between them, determine the compressive force developed on the cylinder S when a vertical force F is applied to the handle of the press. Given: F = 40 N a = 0.5 m b = 0.2 m c = 1.2 m d = 0.35 m e = 0.65 m Solution: Member GD: ΣMG = 0;

−F a + F CG b = 0

593



F CG = F

Chapter 6

a

F CG = 100 N

b

Sector gear :

⎞d = 0 2 2⎟ ⎝ c +d ⎠

ΣMH = 0; F CG( d + e) − FAB⎛ ⎜

c

⎛ c2 + d2 ⎞ ⎟ F = 297.62 N F AB = FCG ( d + e) ⎜ ⎝ c d ⎠ AB Table: ΣF y = 0;

c ⎞ ⎜ 2 2 ⎟ − Fs = 0 ⎝ c +d ⎠

F AB⎛

F s = FAB

c ⎞ ⎛ ⎜ 2 2⎟ ⎝ c +d ⎠

F s = 286 N

Problem 6-118 The mechanism is used to hide kitchen appliances under a cabinet by allowing the shelf to rotate downward. If the mixer has weight W, is centered on the shelf, and has a mass center at G, determine the stretch in the spring necessary to hold the shelf in the equilibrium position shown. There is a similar mechanism on each side of the shelf, so that each mechanism supports half of the load W. The springs each have stiffness k. Given: W = 10 lb k = 4

a = 2 in

lb

b = 4 in

in

φ = 30 deg

c = 15 in

θ = 30 deg

d = 6 in

Solution: ΣMF = 0;

W 2

b − a F ED cos ( φ ) = 0

594



F ED =

Wb

2 a cos ( φ )

+ Σ F x = 0; →

↑Σ Fy = 0;

Fy =

W 2

F ED = 11.547 lb

−F x + FED cos ( φ ) = 0

F x = F ED cos ( φ ) +

Chapter 6

−W 2

F x = 10 lb + F y − FED sin ( φ ) = 0

+ F ED⋅ sin ( φ )

F y = 10.774 lb

Member FBA: ΣMA = 0;

F s = k s;

F y( c + d) cos ( φ ) − F x( c + d) sin ( φ ) − Fs sin ( θ + φ ) d = 0 Fy( c + d) cos ( φ ) − Fx( c + d) sin ( φ ) Fs = F s = 17.5 lb d sin ( θ + φ ) Fs = k x

x =

Fs

x = 4.375 in

k

Problem 6-119 If each of the three links of the mechanism has a weight W, determine the angle θ for equilibrium.The spring, which always remains horizontal, is unstretched when θ = 0°.

595



Chapter 6

Given: W = 25 lb k = 60

lb ft

a = 4 ft b = 4 ft Solution: Guesses

θ = 30 deg

B x = 10 lb

B y = 10 lb

Cx = 10 lb

Cy = 10 lb

Given

⎛ a ⎞ sin ( θ ) − C a sin ( θ ) + C a cos ( θ ) = 0 ⎟ y x ⎝ 2⎠

−W⎜

⎛ b⎞ + C b = 0 ⎟ y ⎝ 2⎠

−W⎜

B y + Cy − W = 0 −B x + Cx = 0

⎛ a⎞ ⎛ a⎞ ⎛ a⎞ −B x a cos ( θ ) − B y a sin ( θ ) − W⎜ ⎟ sin ( θ ) + k⎜ ⎟ sin ( θ ) ⎜ ⎟ cos ( θ ) = 0 ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ ⎛ Bx ⎞ ⎜ ⎟ ⎜ By ⎟ ⎜ C ⎟ = Find B , B , C , C , θ (x y x y ) ⎜ x⎟ ⎜ Cy ⎟ ⎜ ⎟ ⎝θ ⎠

⎛ Bx ⎞ ⎜ ⎟ ⎛⎜ 16.583 ⎞⎟ ⎜ By ⎟ ⎜ 12.5 ⎟ lb ⎜ ⎟=⎜ ⎟ ⎜ Cx ⎟ ⎜ 16.583 ⎟ ⎜ C ⎟ ⎝ 12.5 ⎠ ⎝ y⎠

θ = 33.6 deg

Problem 6-120 Determine the required force P that must be applied at the blade of the pruning shears so that the blade exerts a normal force F on the twig at E.

596



Chapter 6

Given: F = 20 lb a = 0.5 in b = 4 in c = 0.75 in d = 0.75 in e = 1 in Solution: initial guesses: Ax = 1 lb

Ay = 1 lb

Dy = 10 lb P = 20 lb

Dx = 10 lb F CB = 20 lb

Given − P ( b + c + d) − A x a + F e = 0 Dy − P − A y − F = 0 Dx − Ax = 0 − Ay( d) − A x( a) + ( b + c)P = 0

⎞=0 2 2⎟ ⎝ c +a ⎠

Ax − F CB⎛ ⎜

c

⎞=0 ⎟ 2 2 ⎝ a +c ⎠

Ay + P − FCB⎛ ⎜

a

597



Chapter 6

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ ⎟ Dx ⎜ ⎟ = Find A , A , D , D , P , F ( x y x y CB) ⎜ D ⎟ y ⎜ ⎟ ⎜ P ⎟ ⎜F ⎟ ⎝ CB ⎠

⎛⎜ Ax ⎟⎞ ⎛ 13.333 ⎞ ⎜ Ay ⎟ ⎜ 6.465 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Dx ⎟ = ⎜ 13.333 ⎟ lb ⎜ D ⎟ ⎜ 28.889 ⎟ ⎜ y ⎟ ⎜ ⎟ 16.025 ⎝ ⎠ ⎜ FCB ⎟ ⎝ ⎠

P = 2.424 lb

Problem 6-121 The three power lines exert the forces shown on the truss joints, which in turn are pin-connected to the poles AH and EG. Determine the force in the guy cable AI and the pin reaction at the support H. Units Used: 3


d = 125 ft

F 2 = 800 lb

e = 50 ft

a = 40 ft

f = 30 ft

b = 20 ft

g = 30 ft

c = 20 ft Solution: AH is a two-force member. c θ = atan ⎛⎜ ⎟⎞

φ = atan ⎛⎜

d β = atan ⎛⎜ ⎟⎞

e γ = atan ⎛⎜ ⎟⎞

⎝ b⎠

⎝ f⎠

⎞ ⎟ ⎝ a + b⎠ c

⎝ d⎠

α = 90 deg − β + γ

Guesses F AB = 1 lb

F BC = 1 lb

F CA = 1 lb

F AI = 1 lb

F H = 1 lb

598



Chapter 6

Given F BC − F AB cos ( θ ) = 0 F AB sin ( θ ) − F1 = 0 2 F CA sin ( φ ) − F 2 = 0 −F AI sin ( α ) + F AB sin ( β − θ ) + F CA sin ( β − φ ) = 0 −F AI cos ( α ) − FAB cos ( β − θ ) − F CA cos ( β − φ ) + F H = 0

⎛⎜ FAB ⎞⎟ ⎜ FCA ⎟ ⎜ ⎟ F ⎜ BC ⎟ = Find ( FAB , FCA , FBC , FAI , FH) ⎜F ⎟ ⎜ AI ⎟ ⎜ FH ⎟ ⎝ ⎠ ⎛ FAB ⎞ ⎛ 1.131 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FCA ⎟ = ⎜ 1.265 ⎟ kip ⎜ F ⎟ ⎝ 0.8 ⎠ ⎝ BC ⎠

⎛⎜ FAI ⎞⎟ ⎛ 2.881 ⎞ = kip ⎜ FH ⎟ ⎜⎝ 3.985 ⎟⎠ ⎝ ⎠

Problem 6-122 The hydraulic crane is used to lift the load of weight W. Determine the force in the hydraulic cylinder AB and the force in links AC and AD when the load is held in the position shown. Units Used: 3

kip = 10 lb Given:

W = 1400 lb a = 8 ft

c = 1 ft

b = 7 ft

γ = 70 deg

599



Chapter 6

Solution: ΣMD = 0; F CA sin ( 60 deg)c − W a = 0 F CA =

Wa sin ( 60 deg) c

+

↑Σ Fy = 0;

F CA sin ( 60 deg) − FAB sin ( γ ) = 0

F AB = FCA + Σ F x = 0; →

F CA = 12.9 kip

sin ( 60 deg) sin ( γ )

F AB = 11.9 kip

−F AB cos ( γ ) + FCA cos ( 60 deg) − FAD = 0

F AD = −FAB cos ( γ ) + F CA cos ( 60 deg)

F AD = 2.39 kip

Problem 6-123 The kinetic sculpture requires that each of the three pinned beams be in perfect balance at all times during its slow motion. If each member has a uniform weight density γ and length L, determine the necessary counterweights W1, W2 and W3 which must be added to the ends of each member to keep the system in balance for any position. Neglect the size of the counterweights. Given:

γ = 2

lb ft

L = 3 ft a = 1 ft

Solution: ΣMA = 0;

W1 a cos ( θ ) − γ L cos ( θ ) ⎛⎜

L

⎝2

− a⎟⎞ = 0

⎠

600



Chapter 6

γ L⎛⎜

⎝2

W1 = +

↑Σ Fy = 0;

L

− a⎞⎟

⎠

a

W1 = 3 lb

R A − W1 − γ L = 0 R A = W1 + γ L

R A = 9 lb

ΣMB = 0;

⎛ L − a⎞ − R ( L − a) cos ( φ ) = 0 ⎟ A ⎝2 ⎠

W2 a cos ( φ ) − γ L cos ( φ ) ⎜

γ L⎛⎜ W2 =

L

⎝2

+

↑Σ Fy = 0;

− a⎟⎞ + R A( L − a)

⎠

a

W2 = 21 lb

R B − W2 − RA − γ L = 0 R B = W2 + R A + γ L

R B = 36 lb

L ΣMC = 0; R B( L − a) cos ( φ ) + γ L⎛⎜ − a⎟⎞ cos ( φ ) − W3 a cos ( φ ) = 0 2

⎝ ⎠ L R B( L − a) + γL⎛⎜ − a⎟⎞ ⎝2 ⎠ W3 = a

W3 = 75 lb

Problem 6-124 The three-member frame is connected at its ends using ball-and-socket joints. Determine the x, y, z components of reaction at B and the tension in member ED. The force acting at D is F. Given:

⎛ 135 ⎞ ⎜ ⎟ F = 200 lb ⎜ ⎟ ⎝ −180 ⎠ a = 6 ft

e = 3 ft

b = 4 ft

f = 1 ft

601



d = 6 ft

Chapter 6

g = 2 ft

c = g+ f Solution: AC and DE are two-force members. Define some vectors

⎛ −e ⎞ ⎜ ⎟ rDE = −b − g ⎜ ⎟ ⎝ a ⎠

uDE =

⎛ −d − e ⎞ ⎜ −b ⎟ rAC = ⎜ ⎟ ⎝ 0 ⎠

uAC =

⎛e⎞ ⎜ ⎟ rBD = − f ⎜ ⎟ ⎝0⎠

⎛e + d⎞ ⎜ −c ⎟ rBA = ⎜ ⎟ ⎝ 0 ⎠

rDE rDE

rAC rAC

Guesses B x = 1 lb

B y = 1 lb

B z = 1 lb

F DE = 1 lb

F AC = 1 lb

Given

⎛ Bx ⎞ ⎜ ⎟ ⎜ By ⎟ + FDE uDE + FAC uAC + F = ⎜B ⎟ ⎝ z⎠

0

⎛⎜ Bx ⎞⎟ ⎜ By ⎟ ⎜ ⎟ ⎜ Bz ⎟ = Find ( Bx , By , Bz , FDE , FAC ) ⎜F ⎟ ⎜ DE ⎟ ⎜ FAC ⎟ ⎝ ⎠

(

)

(

)

rBD × FDEuDE + F + rBA × FAC uAC = 0

⎛ Bx ⎞ ⎛ −30 ⎞ F ⎜ ⎟ ⎜ ⎟ ⎛⎜ DE ⎞⎟ ⎛ 270 ⎞ = lb ⎜ By ⎟ = ⎜ −13.333 ⎟ lb⎜ ⎟ ⎜ 16.415 ⎟⎠ F ⎜B ⎟ ⎜ − 12 ⎟ ⎝ AC ⎠ ⎝ ⎝ z ⎠ ⎝ 3.039 × 10 ⎠

602



Chapter 6

Problem 6-125 The four-member "A" frame is supported at A and E by smooth collars and at G by a pin. All the other joints are ball-and-sockets. If the pin at G will fail when the resultant force there is F max, determine the largest vertical force P that can be supported by the frame. Also, what are the x, y, z force components which member BD exerts on members EDC and ABC? The collars at A and E and the pon at G only exert force components on the frame. Given: F max = 800 N a = 300 mm b = 600 mm c = 600 mm Solution: ΣMx = 0; b

−P 2 c +

2

2

b +c

Fmax c = 0

F max b

P = 2

2

P = 282.843 N 2

b +c

c

B z + Dz − Fmax

2

b +c

F max c

Bz = 2

2

=0 2

Dz = B z

2

b +c

Dz = Bz

B z = 283 N Dz = 283 N

b

B y + Dy − Fmax

2

b +c

F max b

By = 2

2

=0 2

2

b +c

Dy = By

Dy = B y

B y = 283 N

Dy = 283 N

B x = Dx = 0

Problem 6-126 The structure is subjected to the loading shown. Member AD is supported by a cable AB and a roller at C and fits through a smooth circular hole at D. Member ED is supported by a roller at 603



Chapter 6

g pp y D and a pole that fits in a smooth snug circular hole at E. Determine the x, y, z components of reaction at E and the tension in cable AB. Units Used: 3

kN = 10 N Given:

⎛ 0 ⎞ F = ⎜ 0 ⎟ kN ⎜ ⎟ ⎝ −2.5 ⎠ a = 0.5 m

d = 0.3 m

b = 0.4 m

e = 0.8 m

c = 0.3 m Solution: Guesses

⎛ −c − d ⎞ AB = ⎜ 0 ⎟ ⎜ ⎟ ⎝ e ⎠

F AB = 1 kN

Dx = 1 kN

Dz = 1 kN

Dz2 = 1 kN

E x = 1 kN

E y = 1 kN

MDx = 1 kN⋅ m

MDz = 1 kN⋅ m

Cx = 1 kN

MEx = 1 kN⋅ m

MEy = 1⋅ kN m

Given

⎛⎜ Cx ⎟⎞ ⎛⎜ Dx ⎞⎟ F + FAB +⎜ 0 ⎟+⎜ 0 ⎟ =0 AB ⎜ 0 ⎟ ⎜D ⎟ ⎝ ⎠ ⎝ z⎠ AB

⎛ MDx ⎞ ⎛ 0 ⎞ ⎛ Cx ⎞ ⎛ d ⎞ ⎛c + d⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ AB ⎞ ⎜ ⎟ ⎛ ⎜ 0 ⎟ + ⎜ b ⎟ × ⎜ 0 ⎟ + ⎜ b ⎟ × F + ⎜ b ⎟ × ⎜ FAB ⎟=0 AB ⎠ ⎝ ⎜M ⎟ ⎝0⎠ ⎜ 0 ⎟ ⎝0⎠ ⎝ 0 ⎠ ⎝ ⎠ ⎝ Dz ⎠

⎛ −Dx ⎞ ⎛ Ex ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ + ⎜ Ey ⎟ = 0 ⎜D − D ⎟ ⎜ ⎟ z⎠ ⎝ 0 ⎠ ⎝ z2

⎛ −MDx ⎞ ⎛ MEx ⎞ ⎛ 0 ⎞ ⎛ Dx ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ + ⎜ MEy ⎟ + ⎜ a ⎟ × ⎜ 0 ⎟ = 0 ⎜ −M ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ Dz ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ Dz − Dz2 ⎠ 604



Chapter 6

⎛ Cx ⎞ ⎜ ⎟ ⎜ Dx ⎟ ⎜ ⎟ D z ⎜ ⎟ ⎜D ⎟ ⎜ z2 ⎟ ⎜ Ex ⎟ ⎜ ⎟ E ⎜ y ⎟ = Find ( Cx , Dx , Dz , Dz2 , Ex , Ey , FAB , MDx , MDz , MEx , MEy) ⎜F ⎟ ⎜ AB ⎟ ⎜ MDx ⎟ ⎜ ⎟ ⎜ MDz ⎟ ⎜ ⎟ ⎜ MEx ⎟ ⎜M ⎟ ⎝ Ey ⎠ ⎛ Cx ⎞ ⎜ ⎟ ⎛⎜ 0.937 ⎞⎟ D ⎜ x⎟ ⎜ 0 ⎟ kN ⎜ ⎟=⎜ ⎟ ⎜ Dz ⎟ ⎜ 1.25 ⎟ ⎜ D ⎟ ⎝ 1.25 ⎠ ⎝ z2 ⎠ ⎛⎜ Ex ⎟⎞ ⎛ 0 ⎞ = kN ⎜ Ey ⎟ ⎜⎝ 0 ⎟⎠ ⎝ ⎠

⎛⎜ MDx ⎞⎟ ⎛ 0.5 ⎞ = kN⋅ m ⎜ MDz ⎟ ⎜⎝ 0 ⎟⎠ ⎝ ⎠

⎛⎜ MEx ⎟⎞ ⎛ 0.5 ⎞ = kN⋅ m ⎜ MEy ⎟ ⎜⎝ 0 ⎟⎠ ⎝ ⎠

F AB = 1.562 kN

Problem 6-127 The structure is subjected to the loadings shown.Member AB is supported by a ball-and-socket at A and smooth collar at B. Member CD is supported by a pin at C. Determine the x, y, z components of reaction at A and C.

605



Chapter 6

Given: a = 2m

M = 800 N⋅ m

b = 1.5 m

F = 250 N

c = 3m

θ 1 = 60 deg

d = 4m

θ 2 = 45 deg

Solution:

θ 3 = 60 deg

Guesses Bx = 1 N

By = 1 N

Ax = 1 N

Ay = 1 N

Az = 1 N

Cx = 1 N

Cy = 1 N

Cz = 1 N

MBx = 1 N⋅ m

MBy = 1 N⋅ m

MCy = 1 N⋅ m

MCz = 1 N⋅ m

Given

⎛ Ax ⎞ ⎛ Bx ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ + ⎜ By ⎟ = 0 ⎜A ⎟ ⎜ ⎟ ⎝ z⎠ ⎝ 0 ⎠ ⎛ c ⎞ ⎛⎜ Bx ⎟⎞ ⎛ M ⎞ ⎛⎜ −MBx ⎞⎟ ⎜ a ⎟ × B + ⎜ 0 ⎟ + −M =0 ⎜ ⎟ ⎜ y ⎟ ⎜ ⎟ ⎜ By ⎟ ⎝ 0 ⎠ ⎜⎝ 0 ⎟⎠ ⎝ 0 ⎠ ⎜⎝ 0 ⎟⎠ ⎛ cos ( θ 1) ⎞ ⎛ −Bx ⎞ ⎛ Cx ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ F ⎜ cos ( θ 2) ⎟ + ⎜ −B y ⎟ + ⎜ Cy ⎟ = 0 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ cos ( θ 3) ⎠ ⎝ 0 ⎠ ⎝ Cz ⎠ ⎡ ⎛ cos θ ⎞⎤ ⎛ 0 ⎞ ⎢ ⎜ ( 1) ⎟⎥ ⎛ 0 ⎞ ⎛⎜ −Bx ⎞⎟ ⎛⎜ MBx ⎟⎞ ⎛⎜ 0 ⎞⎟ ⎜ 0 ⎟ × F cos ( θ ) + ⎜ 0 ⎟ × −B + M + MCy ⎟ = 0 2 ⎟⎥ ⎜ ⎟ ⎢ ⎜ ⎜ ⎟ ⎜ y ⎟ ⎜ By ⎟ ⎜ ⎝ b + d ⎠ ⎢ ⎜ cos ( θ 3) ⎟⎥ ⎝ b ⎠ ⎜⎝ 0 ⎟⎠ ⎜⎝ 0 ⎟⎠ ⎜⎝ MCz ⎟⎠ ⎣ ⎝ ⎠⎦

606



Chapter 6

⎛⎜ Ax ⎞⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎜ Az ⎟ ⎜ C ⎟ ⎜ x ⎟ ⎜ Cy ⎟ ⎜ ⎟ ⎜ Cz ⎟ ⎜ ⎟ = Find ( Ax , Ay , Az , Cx , Cy , Cz , Bx , By , MBx , MBy , MCy , MCz ) ⎜ Bx ⎟ ⎜ B ⎟ ⎜ y ⎟ ⎜ MBx ⎟ ⎜ ⎟ ⎜ MBy ⎟ ⎜M ⎟ ⎜ Cy ⎟ ⎜ MCz ⎟ ⎝ ⎠ ⎛ Ax ⎞ ⎜ ⎟ ⎛ −172.3 ⎞ ⎜ Ay ⎟ ⎜ −114.8 ⎟ ⎟ ⎜A ⎟ ⎜ ⎜ ⎟ z 0 ⎜ ⎟= ⎜ ⎟N ⎜ Cx ⎟ 47.3 ⎜ ⎟ ⎜ ⎟ − 61.9 ⎜ ⎟ ⎜ Cy ⎟ ⎜ ⎜ ⎟ ⎝ −125 ⎟⎠ ⎝ Cz ⎠

⎛⎜ MCy ⎞⎟ ⎛ −429 ⎞ = N⋅ m ⎜ MCz ⎟ ⎜⎝ 0 ⎟⎠ ⎝ ⎠

Problem 6-128 Determine the resultant forces at pins B and C on member ABC of the four-member frame. Given: w = 150

lb ft

a = 5 ft b = 2 ft

607



Chapter 6

c = 2 ft e = 4 ft d = a+b−c

Solution: The initial guesses are F CD = 20 lb

F BE = 40 lb

Given F CD( c + d) − FBE

⎛ a + b⎞ + ⎟ ⎝ 2 ⎠

−w( a + b) ⎜

ec

=0 2

2

( d − b) + e F BE e 2

2

( d − b) + e

⎛⎜ FCD ⎞⎟ = Find ( F CD , F BE) ⎜ FBE ⎟ ⎝ ⎠

a − FCD( a + b) = 0

⎛⎜ FCD ⎞⎟ ⎛ 350 ⎞ = lb ⎜ FBE ⎟ ⎜⎝ 1531 ⎟⎠ ⎝ ⎠

Problem 6-129 The mechanism consists of identical meshed gears A and B and arms which are fixed to the gears. The spring attached to the ends of the arms has an unstretched length δ and a stiffness k. If a torque M is applied to gear A, determine the angle θ through which each arm rotates. The gears are each pinned to fixed supports at their centers.

608



Chapter 6

Given:

δ = 100 mm k = 250

N m

M = 6 N⋅ m r =

δ 2

a = 150 mm Solution: ΣMA = 0;

−F r − P a cos ( θ ) + M = 0

ΣMB = 0;

P a cos ( θ ) − F r = 0 2P a cos ( θ ) = M 2k( 2a) sin ( θ ) a cos ( θ ) = M 2k a sin ( 2θ ) = M 2

θ =

1 2

⎞ ⎟ 2 ⎝ 2ka ⎠

asin ⎛⎜

M

θ = 16.1 deg

Problem 6-130 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: kN = 1000 N

609



Chapter 6

Given: F 1 = 20 kN F 2 = 10 kN a = 1.5 m b = 2m Solution: b θ = atan ⎛⎜ ⎟⎞

⎝ a⎠

Guesses

F AG = 1 kN

F BG = 1 kN

F GC = 1 kN

F GF = 1 kN

F AB = 1 kN

F BC = 1 kN

F CD = 1 kN

F CF = 1 kN

F DF = 1 kN

F DE = 1 kN

F EF = 1 kN

Given −F AB cos ( θ ) + F BC = 0 −F AB sin ( θ ) − FBG = 0 F GC cos ( θ ) + FGF − FAG = 0 F GC sin ( θ ) + F BG − F 1 = 0 −F BC + F CD − F GC cos ( θ ) + FCF cos ( θ ) = 0 −F GC sin ( θ ) − F CF sin ( θ ) = 0 −F CD + F DE cos ( θ ) = 0 −F DF − F DE sin ( θ ) = 0 −F GF − F CF cos ( θ ) + FEF = 0

610



Chapter 6

F DF + F CF sin ( θ ) − F 2 = 0 −F DE cos ( θ ) − F EF = 0

611



Chapter 6

⎛ FAG ⎞ ⎜ ⎟ ⎜ FBG ⎟ ⎜ ⎟ ⎜ FGC ⎟ ⎜F ⎟ ⎜ GF ⎟ ⎜ FAB ⎟ ⎜ ⎟ ⎜ FBC ⎟ = Find ( FAG , FBG , FGC , FGF , FAB , FBC , FCD , FCF , FDF , FDE , FEF) ⎜F ⎟ ⎜ CD ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDF ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎜F ⎟ ⎝ EF ⎠ ⎛ FAG ⎞ ⎜ ⎟ ⎜ FBG ⎟ ⎛⎜ 13.13 ⎟⎞ ⎜ ⎟ ⎜ 17.50 ⎟ F GC ⎜ ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 3.13 ⎟ ⎜ GF ⎟ ⎜ 11.25 ⎟ ⎜ FAB ⎟ ⎜ −21.88 ⎟ ⎜ ⎟ ⎜ ⎟ F ⎜ BC ⎟ = ⎜ −13.13 ⎟ kN ⎜ F ⎟ ⎜ −9.37 ⎟ ⎜ CD ⎟ ⎜ ⎟ ⎜ FCF ⎟ ⎜ −3.13 ⎟ ⎜ ⎟ ⎜ 12.50 ⎟ F ⎜ DF ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ −15.62 ⎟ ⎜ FDE ⎟ ⎝ 9.37 ⎠ ⎜F ⎟ ⎝ EF ⎠


Problem 6-131 The spring has an unstretched length δ. Determine the angle θ for equilibrium if the uniform links each have a mass mlink.

612



Chapter 6

Given: mlink = 5 kg

δ = 0.3 m N

k = 400

m

a = 0.1 m b = 0.6 m g = 9.81

m 2

s Solution: Guesses

θ = 10 deg F BD = 1 N Ex = 1 N Given mlink g

a+b

cos ( θ ) − FBD b cos ( θ ) + E x b sin ( θ ) = 0

2

−2 mlink g

a+b 2

cos ( θ ) + Ex 2 b sin ( θ ) = 0

F BD = k( 2 b sin ( θ ) − δ )

⎛ FBD ⎞ ⎜ ⎟ ⎜ Ex ⎟ = Find ( FBD , Ex , θ ) ⎜ ⎟ ⎝ θ ⎠

θ = 21.7 deg

Problem 6-132 The spring has an unstretched length δ. Determine the mass mlink of each uniform link if the angle for equilibrium is θ. Given:

δ = 0.3 m

613



Chapter 6

θ = 20 deg k = 400

N m

a = 0.1 m b = 0.6 m g = 9.81

m 2

s Solution: Guesses

Ey = 1 N

mlink = 1 kg

Fs = 1 N Given F s = ( 2 b sin ( θ ) − δ ) k mlink g

a+b cos ( θ ) − Fs b cos ( θ ) + E y b sin ( θ ) = 0 2

−2mlink g

a+b cos ( θ ) + E y2b sin ( θ ) = 0 2

⎛ mlink ⎞ ⎜ ⎟ ⎜ Fs ⎟ = Find ( mlink , Fs , Ey) ⎜ E ⎟ ⎝ y ⎠ mlink = 3.859 kg y = sin ( θ ) 2( b) y = 2 b sin ( θ ) F s = ( y − δ ) ( k) F s = 44.17 N

614



ΣMA = 0;

Chapter 6

a+b

( cos (θ ) ) = 0 ⎡⎣Ey ( 2) ( a + b) sin ( θ ) − 2( M) ( g)⎤⎦ 2 ⎡E ( 2) sin ( θ ) − 2 ( m) ⎢ y ⎣

g⎤

⎥ ( cos ( θ ) = 0

2⎦

⇒ E y ( 2 sin ( θ ) ) = m ( g) ( cos ( θ ) ) Ey = ΣMC = 0;

m( g) ( cos ( θ ) ) 2 sin ( θ )

m( g) ( cos ( θ ) ) 2 sin ( θ )

( a + b) sin ( θ ) + m( g) ⎛⎜

a + b⎞

⎟ cos ( θ ) − Fs( b cos ( θ ) ) = 0 ⎝ 2 ⎠

b m = Fs g ( a + b) m = 3.859 kg

Problem 6-133 Determine the horizontal and vertical components of force that the pins A and B exert on the two-member frame. Given: w = 400

N m

a = 1.5 m b = 1m c = 1m F = 0N

θ = 60 deg Solution: Guesses Ax = 1 N

Ay = 1 N

Bx = 1 N

615



By = 1 N

Cx = 1 N

Chapter 6

Cy = 1 N

Given −w a

a + Cx a sin ( θ ) − Cy a cos ( θ ) = 0 2

F c − Cx c − Cy b = 0

Ay − Cy − w a cos ( θ ) = 0

− Ax − Cx + w a sin ( θ ) = 0

Cx − Bx − F = 0 Cy + By = 0

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜B ⎟ ⎜ x ⎟ = Find A , A , B , B , C , C ( x y x y x y) ⎜ By ⎟ ⎜ ⎟ ⎜ Cx ⎟ ⎜ ⎟ ⎝ Cy ⎠

⎛⎜ Ax ⎞⎟ ⎛ 300.0 ⎞ = N ⎜ Ay ⎟ ⎜⎝ 80.4 ⎟⎠ ⎝ ⎠

⎛⎜ Bx ⎟⎞ ⎛ 220 ⎞ = N ⎜ By ⎟ ⎜⎝ 220 ⎟⎠ ⎝ ⎠

Problem 6-134 Determine the horizontal and vertical components of force that the pins A and B exert on the two-member frame. Given: w = 400

N m

a = 1.5 m b = 1m c = 1m F = 500 N

θ = 60 deg

616



Chapter 6


Ay = 1 N

Bx = 1 N

By = 1 N

Cx = 1 N

Cy = 1 N

Given −w a

a 2

+ Cx a sin ( θ ) − Cy a cos ( θ ) = 0

Ay − Cy − w a cos ( θ ) = 0 Cx + Bx − F = 0

F c − Cx c − Cy b = 0 − Ax − Cx + w a sin ( θ ) = 0

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜B ⎟ ⎜ x ⎟ = Find A , A , B , B , C , C ( x y x y x y) ⎜ By ⎟ ⎜ ⎟ ⎜ Cx ⎟ ⎜ ⎟ ⎝ Cy ⎠

Cy − By = 0

⎛⎜ Ax ⎞⎟ ⎛ 117.0 ⎞ = N ⎜ Ay ⎟ ⎜⎝ 397.4 ⎟⎠ ⎝ ⎠

⎛⎜ Bx ⎟⎞ ⎛ 97.4 ⎞ = N ⎜ By ⎟ ⎜⎝ 97.4 ⎟⎠ ⎝ ⎠

Problem 6-135 Determine the force in each member of the truss and indicate whether the members are in tension or compression. Units Used: kip = 1000 lb Given: F 1 = 1000 lb

b = 8 ft

F 2 = 500 lb

c = 4 ft

a = 4 ft

Solution: Joint B: 617



Chapter 6

Initial Guesses: Given F BC = 100 lb

F BA = 150 lb

ΣF x = 0;

b c F 1 − F BC cos ⎛⎜ atan ⎛⎜ ⎟⎞ ⎞⎟ − FBA cos ⎜⎛ atan ⎛⎜ ⎟⎞ ⎞⎟ = 0 a a

ΣF y = 0;

−F BC sin ⎜⎛ atan ⎛⎜

⎝

⎝ ⎠⎠

⎝

⎝ ⎠⎠

⎛ ⎛ c ⎞⎞ ⎟ ⎟ + FBA sin ⎜atan ⎜ ⎟ ⎟ − F2 = 0 ⎝ a ⎠⎠ ⎝ ⎝ a ⎠⎠

⎝

b ⎞⎞

⎛⎜ FBC ⎞⎟ = Find ( F BC , F BA) ⎜ FBA ⎟ ⎝ ⎠ F BC = 373 lb(C) F BA = 1178.51 lb F BA = 1.179 kip (C) Joint A: ΣF y = 0;

c F AC − FBA sin ⎜⎛ atan ⎛⎜ ⎟⎞ ⎞⎟ = 0 a

(

)

⎝

c

F AC = FBA

⎝ ⎠⎠

2

a

2

a +c 2

a F AC = 833 lb (T)


kip = 10 lb Given: F = 1000 lb a = 10 ft b = 10 ft

618



Solution:

Chapter 6

b θ = atan ⎛⎜ ⎞⎟

⎝ a⎠

Guesses F AB = 1 lb

F AG = 1 lb

F BC = 1 lb

F BG = 1 lb

F CD = 1 lb

F CE = 1 lb

F CG = 1 lb

F DE = 1 lb

F EG = 1 lb Given F AB + F AG cos ( θ ) = 0 −F AB + F BC = 0 F BG = 0 F CD − F BC − F CG cos ( θ ) = 0 F CE − F + F CG sin ( θ ) = 0 −F CD − F DE cos ( θ ) = 0 F DE cos ( θ ) − F EG = 0 −F CE − FDE sin ( θ ) = 0 F EG − F AG cos ( θ ) + F CG cos ( θ ) = 0 −F AG sin ( θ ) − FBG − FCG sin ( θ ) = 0

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAG ⎟ ⎜F ⎟ ⎜ BC ⎟ ⎜ FBG ⎟ ⎜ ⎟ ⎜ FCD ⎟ = Find ( FAB , FAG , FBC , FBG , FCD , FCE , FCG , FDE , FEG) ⎜F ⎟ ⎜ CE ⎟ ⎜ FCG ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎟ ⎜ ⎝ FEG ⎠

619



Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎛ 333 ⎞ ⎜ FAG ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ −471 ⎟ ⎜ BC ⎟ ⎜ 333 ⎟ ⎜ FBG ⎟ ⎜ 0 ⎟ ⎟ ⎜ ⎟ ⎜ F 667 = ⎟ lb ⎜ CD ⎟ ⎜ ⎜ F ⎟ ⎜ 667 ⎟ ⎟ ⎜ CE ⎟ ⎜ ⎜ FCG ⎟ ⎜ 471 ⎟ ⎜ ⎟ ⎜ −943 ⎟ F ⎜ DE ⎟ ⎜ ⎟ ⎟ ⎝ −667 ⎠ ⎜ ⎝ FEG ⎠

Positive (T), Negative (C)

Problem 6-137 Determine the force in members AB, AD, and AC of the space truss and state if the members are in tension or compression. The force F is vertical. Units Used: 3

kip = 10 lb Given: F = 600 lb a = 1.5 ft b = 2 ft c = 8 ft Solution:

⎛a⎞ AB = ⎜ −c ⎟ ⎜ ⎟ ⎝0⎠

⎛ −a ⎞ AC = ⎜ −c ⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ AD = ⎜ −c ⎟ ⎜ ⎟ ⎝b⎠ Guesses

F AB = 1 lb

F AC = 1 lb

F AD = 1 lb 620



Given

Chapter 6

⎛ 0 ⎞ F AB + FAC + FAD +⎜ 0 ⎟ =0 ⎜ ⎟ AB AC AD ⎝ −F ⎠ AB

AC

AD

⎛ FAB ⎞ ⎜ ⎟ F ⎜ AC ⎟ = Find ( FAB , FAC , FAD) ⎜F ⎟ ⎝ AD ⎠ ⎛ FAB ⎞ ⎛ −1.221 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ AC ⎟ = ⎜ −1.221 ⎟ kip ⎜ F ⎟ ⎝ 2.474 ⎠ ⎝ AD ⎠


621



Chapter 7

Problem 7-1 The column is fixed to the floor and is subjected to the loads shown. Determine the internal normal force, shear force, and moment at points A and B. Units Used: 3

kN = 10 N Given: F 1 = 6 kN F 2 = 6 kN F 3 = 8 kN a = 150 mm b = 150 mm c = 150 mm Solution: Free body Diagram: The support reaction need not be computed in this case. Internal Forces: Applying equations of equillibrium to the top segment sectioned through point A, we have + Σ F x = 0; →

VA = 0

+

NA − F1 − F2 = 0

NA = F 1 + F 2

NA = 12.0 kN

F 1 a − F2 b − MA = 0

MA = F1 a − F 2 b

MA = 0 kN⋅ m

↑Σ Fy = 0; ΣMA = 0;

VA = 0

Applying equations of equillibrium to the top segment sectioned through point B, we have + Σ F x = 0; →

VB = 0

+

NB − F1 − F2 − F3 = 0

NB = F 1 + F 2 + F 3

NB = 20.0 kN

F 1 a − F 2 b − F3 c + MB = 0

MB = −F1 a + F 2 b + F3 c

MB = 1.20 kN⋅ m

↑Σ Fy = 0;

+ΣMB = 0;

VB = 0

Problem 7-2 The axial forces act on the shaft as shown. Determine the internal normal forces at points A and B. 622



Chapter 7

Given: F 1 = 20 lb F 2 = 50 lb F 3 = 10 lb

Solution: Section A: ΣF z = 0;

F 2 − 2 F1 − NA = 0 NA = F 2 − 2 F 1 NA = 10.00 lb

Section B: ΣF z = 0;

F 2 − 2 F1 − NA + NB = 0 NB = −F 2 + 2 F1 + NA NB = 0.00 lb

Problem 7-3 The shaft is supported by smooth bearings at A and B and subjected to the torques shown. Determine the internal torque at points C, D, and E. Given: M1 = 400 N⋅ m M2 = 150 N⋅ m M3 = 550 N⋅ m

623



Chapter 7

Solution: Section C: ΣMx = 0;

TC = 0

Section D: ΣMx = 0;

TD − M 1 = 0 TD = M 1 TD = 400.00 N⋅ m

Section E: ΣMx = 0;

M 1 + M 2 − TE = 0 TE = M 1 + M 2 TE = 550.00 N⋅ m

Problem 7-4 Three torques act on the shaft. Determine the internal torque at points A, B, C, and D. Given: M1 = 300 N⋅ m M2 = 400 N⋅ m M3 = 200 N⋅ m Solution: Section A: ΣΜx = 0;

− TA + M 1 − M 2 + M 3 = 0 TA = M 1 − M 2 + M 3 TA = 100.00 N⋅ m

Section B: ΣMx = 0;

TB + M 3 − M 2 = 0

624



Chapter 7

TB = − M 3 + M 2 TB = 200.00 N⋅ m Section C: ΣΜx = 0;

− TC + M 3 = 0 TC = M 3 TC = 200.00 N⋅ m

Section D: ΣΜx = 0;

TD = 0

Problem 7-5 The shaft is supported by a journal bearing at A and a thrust bearing at B. Determine the normal force, shear force, and moment at a section passing through (a) point C, which is just to the right of the bearing at A, and (b) point D, which is just to the left of the force F 2. Units Used: 3

kip = 10 lb Given: F 1 = 2.5 kip F 2 = 3 kip w = 75

lb ft

a = 6 ft

b = 12 ft c = 2 ft


625



Chapter 7

b − Ay( b + c) + F1 ( a + b + c) + w b⎛⎜ c + ⎞⎟ + F2 c = 0 ⎝ 2⎠ F1 ( a + b + c) + w b⎛⎜ c +

⎝

Ay =

b⎞

⎟ + F2 c

2⎠

b+c

Ay = 4514 lb + Σ F x = 0; → +

↑Σ Fy = 0;

B x = 0 lb Ay − F1 − w b − F2 + By = 0 By = − Ay + F1 + w b + F2

Σ MC = 0; + Σ F x = 0; → +

↑

Σ F y = 0; Σ MD = 0;

+ → +

↑

Σ F x = 0; Σ F y = 0;

F 1 a + Mc = 0

B y = 1886 lb

MC = −F 1 a

NC = 0 lb

MC = −15.0 kip⋅ ft NC = 0.00 lb

−F 1 + Ay − V C = 0 −MD + By c = 0

VC = Ay − F1

V C = 2.01 kip

MD = B y c

MD = 3.77 kip⋅ ft

ND = 0 lb

ND = 0.00 lb

VD − F2 + By = 0

VD = F2 − By

V D = 1.11 kip

Problem 7-6 Determine the internal normal force, shear force, and moment at point C. Given: M = 400 lb⋅ ft a = 4 ft b = 12 ft

626



Chapter 7

Solution: Beam: ΣMB = 0;

M − Ay( a + b) = 0 Ay =

M a+b

Ay = 25.00 lb

Segment AC: ΣF x = 0;

NC = 0

ΣF y = 0;

Ay − VC = 0 VC = Ay V C = 25.00 lb

ΣMC = 0;

− Ay a + MC = 0 MC = Ay a MC = 100.00 lb⋅ ft

Problem 7-7 Determine the internal normal force, shear force, and moment at point C. Units Used: kN = 103 N Given: F 1 = 30 kN F 2 = 50 kN F 3 = 25 kN a = 1.5 m b = 3m

θ = 30 deg Solution: ΣF x = 0;

−NC + F 3 cos ( θ ) = 0 627



Chapter 7

NC = F3 cos ( θ ) NC = 21.7 kN V C − F 2 − F 3 sin ( θ ) = 0

ΣF y = 0;

V C = F2 + F3 sin ( θ ) V C = 62.50 kN ΣMC = 0; −MC − F2 b − F 3 sin ( θ ) 2 b = 0 MC = −F 2 b − F3 sin ( θ ) 2b MC = −225.00 kN⋅ m

Problem 7-8 Determine the normal force, shear force, and moment at a section passing through point C. Assume the support at A can be approximated by a pin and B as a roller. Units used: 3

kip = 10 lb Given: F 1 = 10 kip

a = 6 ft

F 2 = 8 kip

b = 12 ft

kip ft

c = 12 ft

w = 0.8

d = 6 ft

Solution: ΣMA = 0;

⎛ b + c ⎞ − F ( b + c + d) + B ( b + c) + F a = ⎟ 2 y 1 ⎝ 2 ⎠

−w( b + c) ⎜

w

( b + c)

By = + Σ F x = 0; →

2

0

2

+ F 2 ( b + c + d) − F 1 a B y = 17.1 kip

b+c NC = 0 628



+

↑Σ Fy = 0;

Chapter 7

VC − w c + By − F2 = 0 VC = w c − By + F2

V C = 0.5 kip

⎛ c ⎞ + B c − F ( c + d) = ⎟ y 2 ⎝ 2⎠

−MC − w c⎜

ΣMC = 0;

⎛ c2 ⎞ ⎟ + B y c − F 2 ( c + d) ⎝2⎠

MC = −w⎜

0

MC = 3.6 kip⋅ ft

Problem 7-9 The beam AB will fail if the maximum internal moment at D reaches Mmax or the normal force in member BC becomes Pmax. Determine the largest load w it can support. Given: Mmax = 800 N⋅ m P max = 1500 N a = 4m b = 4m c = 4m d = 3m Solution:

⎛ a + b ⎞ − A ( a + b) = ⎟ y ⎝ 2 ⎠

w ( a + b) ⎜ Ay =

w( a + b) 2

⎛ a⎞ − A a + M = ⎟ y D ⎝2⎠

w a⎜

MD = w

0

⎛ a b⎞ ⎜2 ⎟ ⎝ ⎠

Ay − w( a + b) +

T=

0

w( a + b)

⎛ d ⎞T = ⎜ 2 2⎟ ⎝ c +d ⎠ 2

c +d

0

2

2d

Assume the maximum moment has been reached 629



Chapter 7

MD = Mmax

2 MD

w1 =

ab

w1 = 100

N m

w2 = 225

N m

Assume that the maximum normal force in BC has been reached T = Pmax

T2 d

w2 =

2

( a + b) c + d

2

w = min ( w1 , w2 )

Now choose the critical load

w = 100

N m

Problem 7-10 Determine the shear force and moment acting at a section passing through point C in the beam. Units Used: 3


kip ft

a = 6 ft b = 18 ft Solution: ΣMB = 0;

− Ay b + Ay =

ΣMC = 0;

1 6

− Ay a +

1 2

wb

⎛ b⎞ = ⎜3⎟ ⎝ ⎠

0

Ay = 9 kip

wb

a⎞ ⎛ a⎞ w ⎟ a ⎜ ⎟ + MC = ⎜ 2 ⎝ b⎠ ⎝ 3 ⎠ 1⎛

0

3

wa MC = Ay a − 6b MC = 48 kip⋅ ft +

↑

Σ F y = 0;

Ay −

1 2

⎛w a ⎞ a − V = ⎜ ⎟ C ⎝ b⎠

0

2

wa VC = Ay − 2b

630



Chapter 7

V C = 6 kip

Problem 7-11 Determine the internal normal force, shear force, and moment at points E and D of the compound beam. Given: M = 200 N⋅ m

c = 4m

F = 800 N

d = 2m

a = 2m

e = 2m

b = 2m Solution: Segment BC : M d+e

−M + Cy( d + e) = 0

Cy =

−B y + Cy = 0

B y = Cy

Segment EC : −NE = 0

NE = 0 N

V E + Cy = 0

NE = 0.00

V E = −Cy

−ME − M + Cy e = 0

V E = −50.00 N

ME = Cy e − M

ME = −100.00 N⋅ m

Segment DB : −ND = 0

ND = 0 N

ND = 0.00

VD − F + By = 0

VD = F − By

V D = 750.00 N

−MD − F b + By( b + c) = 0 MD = −F b + B y( b + c)

MD = −1300 N⋅ m

631



Chapter 7

Problem 7-12 The boom DF of the jib crane and the column DE have a uniform weight density γ. If the hoist and load have weight W, determine the normal force, shear force, and moment in the crane at sections passing through points A, B, and C. Treat the boom tip, beyond the hoist, as weightless. Given: W = 300 lb

γ = 50

lb ft

a = 7 ft b = 5 ft c = 2 ft d = 8 ft e = 3 ft Solution: + ΣF x = 0; →

−NA = 0

+

VA − W − γ e = 0

↑ ΣFy = 0;

NA = 0 lb

VA = W + γ e ΣΜΑ = 0;

+ → +

ΣF x = 0;

↑ ΣFy = 0; ΣΜΒ = 0;

NA = 0.00 lb

V A = 450 lb

⎛e⎞ − We = ⎟ ⎝ 2⎠ ⎛ e2 ⎞ MA = γ ⎜ ⎟ + W e ⎝2⎠ MA − γ e⎜

0

−NB = 0

MA = 1125.00 lb⋅ ft NB = 0 lb

V B − γ ( d + e) − W = 0

V B = γ ( d + e) + W

⎛ d + e ⎞ − W( d + e) = ⎟ ⎝ 2 ⎠

MB − γ ( d + e) ⎜ MB =

1 2

NB = 0.00 lb

2

γ ( d + e) + W( d + e)

V B = 850 lb

0

MB = 6325.00 lb⋅ ft

632



Chapter 7

+ →

ΣF x = 0;

VC = 0

+

ΣF y = 0;

NC − ( c + d + e) γ − W − γ ( b) = 0

↑

V C = 0 lb

V C = 0.00 lb

NC = γ ( c + d + e + b) + W ΣΜC = 0;

NC = 1200.00 lb

MC − ( c + d + e) γ

⎛ c + d + e ⎞ − W( c + d + e) = ⎜ 2 ⎟ ⎝ ⎠

MC = ( c + d + e) γ

⎛ c + d + e ⎞ + W( c + d + e) ⎜ 2 ⎟ ⎝ ⎠

0

MC = 8125.00 lb⋅ ft

Problem 7-13 Determine the internal normal force, shear force, and moment at point C. Units Used: 3

kip = 10 lb Given: a = 0.5 ft

d = 8 ft

b = 2 ft

e = 4 ft

c = 3 ft

w = 150

lb ft

Solution: Entire beam: ΣMA = 0;

⎛ d + e ⎞ + T ( a + b) = ⎟ ⎝ 2 ⎠

−w( d + e) ⎜

w ( d + e) T = 2 ( a + b) ΣF x = 0;

2

T = 4.32 kip

Ax − T = 0 Ax = T

ΣF y = 0;

0

Ax = 4.32 kip

Ay − w( d + e) = 0 Ay = w( d + e)

Ay = 1.80 kip

Segment AC: 633



Chapter 7

ΣF x = 0;

Ax + NC = 0

NC = − A x

NC = −4.32 kip

ΣF y = 0;

Ay − w c − VC = 0

VC = Ay − w c

V C = 1.35 kip

ΣMC = 0;

− Ay c + w c⎜

⎛c⎞ + M = ⎟ C ⎝2⎠

MC = Ay c − w

0

⎛ c2 ⎞ ⎜ ⎟ ⎝2⎠

MC = 4.72 kip⋅ ft

Problem 7-14 Determine the normal force, shear force, and moment at a section passing through point D of the two-member frame. Units Used: 3

kN = 10 N Given: w = 400

N m

a = 2.5 m b = 3m c = 6m

Solution: ΣMA = 0;

−1 2

⎛ 2 c⎞ + F ⎛ a ⎞ c = ⎜3 ⎟ BC ⎜ ⎝ ⎠ 2 2⎟ ⎝ a +c ⎠

wc

F BC = +

→ Σ Fx = 0;

↑Σ Fy = 0;

3

wc

⎛ a2 + c2 ⎞ ⎜ ⎟ ⎝ a ⎠

⎛ c ⎞F − A = x ⎜ 2 2 ⎟ BC ⎝ a +c ⎠ Ax =

+

1

Ay −

⎛ c ⎞F ⎜ 2 2 ⎟ BC ⎝ a +c ⎠ 1 2

wc +

0

F BC = 2080 N

0

Ax = 1920 N

⎛ a ⎞F = ⎜ 2 2 ⎟ BC ⎝ a +c ⎠

0

634



Ay = + Σ F x = 0; → +

↑Σ Fy = 0;

Chapter 7

1 2

wc −

⎛ a ⎞F ⎜ 2 2 ⎟ BC ⎝ a +c ⎠

ND − Ax = 0 Ay −

Ay = 400 N

ND = A x

b⎞ ⎜w ⎟ b − VD = c⎠

1⎛ 2⎝

0

2⎞

1 ⎛b V D = A y − w⎜ 2 ⎝ c

ΣF y = 0;

− Ay b +

⎟ ⎠

V D = 100 N

b b w ⎞⎟ b⎛⎜ ⎟⎞ + MD = ⎜ 2⎝ c⎠ ⎝3⎠ 1⎛

MD = Ay b −

1 6

ND = 1.920 kN

0

⎛ b3 ⎞ w⎜ ⎟ ⎝c⎠

MD = 900 N m

Problem 7-15 The beam has weight density γ. Determine the internal normal force, shear force, and moment at point C. Units Used: kip = 103 lb Given:

γ = 280

lb ft

a = 3 ft b = 7 ft c = 8 ft d = 6 ft

635



Chapter 7

Solution: c θ = atan ⎛⎜ ⎟⎞

W = γ ( a + b)

⎝ d⎠

Guesses Ax = 1 lb

Ay = 1 lb

B x = 1 lb

NC = 1 lb

V C = 1 lb

MC = 1 lb ft

Given Entire beam: Ax − Bx = 0

Ay − W = 0

Bx c − W

⎛ d⎞ = ⎜ 2⎟ ⎝ ⎠

0

Bottom Section Ax − NC cos ( θ ) + V C sin ( θ ) = 0 Ay − W

⎛ a ⎞ − N sin ( θ ) − V cos ( θ ) = ⎜ ⎟ C C ⎝ a + b⎠

MC − VC a − W

⎛ a ⎞ ⎜ ⎟ ⎝ a + b⎠

⎛ a ⎞ cos ( θ ) = ⎜ 2⎟ ⎝ ⎠

⎛⎜ Ax ⎟⎞ ⎜ Ay ⎟ ⎜ ⎟ ⎜ Bx ⎟ = Find ( A , A , B , N , V , M ) x y x C C C ⎜ NC ⎟ ⎜ ⎟ ⎜ VC ⎟ ⎜ MC ⎟ ⎝ ⎠

0

0

⎛ Ax ⎞ ⎛ 1.05 ⎞ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 2.80 ⎟ kip ⎜ B ⎟ ⎜⎝ 1.05 ⎟⎠ ⎝ x⎠

⎛ NC ⎞ ⎛ 2.20 ⎞ ⎜ ⎟=⎜ ⎟ kip ⎝ VC ⎠ ⎝ 0.34 ⎠ MC = 1.76 kip⋅ ft

Problem 7-16 Determine the internal normal force, shear force, and moment at points C and D of the beam.

636



Chapter 7

Units Used: 3


lb ft

a = 12 ft b = 15 ft

lb w2 = 40 ft

c = 10 ft d = 5 ft

F = 690 lb e = 12

f = 5 e θ = atan ⎛⎜ ⎟⎞

Solution:

⎝ f⎠

Guesses B y = 1 lb

NC = 1 lb

V C = 1 lb

MC = 1 lb⋅ ft

ND = 1 lb

V D = 1 lb

MD = 1 lb⋅ ft Given

⎛ b ⎞ ... = ⎟ ⎝ 2⎠

B y b − F sin ( θ ) ( b + c) − w2 b⎜ +

−1 2

(w1 − w2)b⎛⎜ 3 ⎟⎞

0

b

⎝ ⎠

−NC − F cos ( θ ) = 0 b − a⎞ ⎟ ( b − a) ... = 0 ⎝ b ⎠ + B y − w2 ( b − a) − F sin ( θ ) VC −

(w1 − w2) ⎛⎜ 2 1

−ND − F cos ( θ ) = 0 V D − F sin ( θ ) = 0 −MD − F sin ( θ ) d = 0

⎛ b − a⎞ − ⎟ ⎝ 2 ⎠

−MC − w2 ( b − a) ⎜

b − a⎞ ⎛ b − a⎞ ⎟ ( b − a) ⎜ 3 ⎟ ... = 0 ⎝ ⎠ ⎝ b ⎠

⎛ w1 − w2 ) ⎜ ( 2 1

+ B y ( b − a) − F sin ( θ ) ( c + b − a)

637



Chapter 7

⎛ By ⎞ ⎜ ⎟ ⎜ NC ⎟ ⎜ VC ⎟ ⎜ ⎟ M ⎜ C ⎟ = Find ( By , NC , VC , MC , ND , VD , MD) ⎜N ⎟ ⎜ D⎟ ⎜ VD ⎟ ⎜ ⎟ ⎝ MD ⎠

⎛ NC ⎞ ⎛ −265 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ VC ⎟ = ⎜ −649 ⎟ lb ⎜ ND ⎟ ⎜ −265 ⎟ ⎜ ⎟ ⎜ 637 ⎟ ⎠ ⎝ VD ⎠ ⎝ ⎛ MC ⎞ ⎛ −4.23 ⎞ ⎜ ⎟=⎜ ⎟ kip⋅ ft ⎝ MD ⎠ ⎝ −3.18 ⎠

Problem 7-17 Determine the normal force, shear force, and moment acting at a section passing through point C. 3

kip = 10 lb

Units Used: Given:

F 1 = 800 lb F 2 = 700 lb F 3 = 600 lb

θ = 30 deg a = 1.5 ft b = 1.5 ft c = 3 ft d = 2 ft e = 1 ft f = a+b+c−d−e Solution: Guesses

B y = 1 lb

Ax = 1 lb

Ay = 1 lb

NC = 1 lb

V C = 1 lb

MC = 1 lb⋅ ft

638



Chapter 7

Given − Ax + V C sin ( θ ) + NC cos ( θ ) = 0 Ay − V C cos ( θ ) + NC sin ( θ ) = 0 MC − A x ( a) sin ( θ ) − Ay ( a) cos ( θ ) = 0 − Ax + F 1 sin ( θ ) − F3 sin ( θ ) = 0 Ay + B y − F2 − F1 cos ( θ ) − F3 cos ( θ ) = 0 −F 1 ( a + b) − F 2 ( a + b + c) cos ( θ ) − F 3 cos ( θ ) ( a + b + c + d + e) cos ( θ ) ... = 0 + F 3 sin ( θ ) f sin ( θ ) + B y2 ( a + b + c) cos ( θ )

⎛⎜ Ax ⎟⎞ ⎜ Ay ⎟ ⎜ ⎟ ⎜ By ⎟ = Find ( A , A , B , N , V , M ) x y y C C C ⎜ NC ⎟ ⎜ ⎟ ⎜ VC ⎟ ⎜ MC ⎟ ⎝ ⎠

⎛ Ax ⎞ ⎛ 100 ⎞ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 985 ⎟ lb ⎜ B ⎟ ⎜⎝ 927 ⎟⎠ ⎝ y⎠ ⎛ NC ⎞ ⎛ −406 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ VC ⎠ ⎝ 903 ⎠ MC = 1.355 kip⋅ ft

Problem 7-18 Determine the normal force, shear force, and moment acting at a section passing through point D. Units Used:

kip = 103 lb

639



Chapter 7

Given: F 1 = 800 lb F 2 = 700 lb F 3 = 600 lb

θ = 30 deg a = 1.5 ft b = 1.5 ft c = 3 ft d = 2 ft e = 1 ft f = a+b+c−d−e Solution: Guesses

B y = 1 lb

Ax = 1 lb

Ay = 1 lb

ND = 1 lb

V D = 1 lb

MD = 1 lb⋅ ft

Given V D sin ( θ ) − ND cos ( θ ) − F3 sin ( θ ) = 0 B y + VD cos ( θ ) + ND sin ( θ ) − F 3 cos ( θ ) = 0 −MD − F3 e + By( e + f) cos ( θ ) = 0 Ay + B y − F2 − F1 cos ( θ ) − F3 cos ( θ ) = 0 − Ax + F 1 sin ( θ ) − F3 sin ( θ ) = 0 −F 1 ( a + b) − F 2 ( a + b + c) cos ( θ ) − F 3 cos ( θ ) ( a + b + c + d + e) cos ( θ ) ... = 0 + F 3 sin ( θ ) f sin ( θ ) + B y2 ( a + b + c) cos ( θ )

⎛⎜ Ax ⎟⎞ ⎜ Ay ⎟ ⎜ ⎟ ⎜ By ⎟ = Find ( A , A , B , N , V , M ) x y y D D D ⎜ ND ⎟ ⎜ ⎟ ⎜ VD ⎟ ⎜ MD ⎟ ⎝ ⎠

⎛ Ax ⎞ ⎛ 100 ⎞ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 985 ⎟ lb ⎜ B ⎟ ⎜⎝ 927 ⎟⎠ ⎝ y⎠ ⎛ ND ⎞ ⎛ −464 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ VD ⎠ ⎝ −203 ⎠ MD = 2.61 kip⋅ ft

640



Chapter 7

Problem 7-19 Determine the normal force, shear force, and moment at a section passing through point C. Units Used: 3

kN = 10 N Given: P = 8 kN

c = 0.75 m

a = 0.75m

d = 0.5 m

b = 0.75m

r = 0.1 m

Solution: ΣMA = 0;

−T ( d + r) + P( a + b + c) = 0 T = P

⎛ a + b + c⎞ T = ⎜ ⎟ ⎝ d+r ⎠

30 kN

+ Σ F x = 0; →

Ax = T

Ax = 30 kN

+

Ay = P

Ay = 8 kN

↑Σ Fy = 0;

+ Σ F x = 0; →

−NC − T = 0 NC = −T

+

↑Σ Fy = 0;

VC + P = 0 V C = −P

ΣMC = 0;

NC = −30 kN

V C = −8 kN

−MC + P c = 0 MC = P c

MC = 6 kN⋅ m

Problem 7-20 The cable will fail when subjected to a tension Tmax. Determine the largest vertical load P the frame will support and calculate the internal normal force, shear force, and moment at a section passing through point C for this loading. Units Used: 3

kN = 10 N 641



Chapter 7

Given: Tmax = 2 kN a = 0.75 m b = 0.75 m c = 0.75 m d = 0.5 m r = 0.1 m Solution: ΣMA = 0;

−Tmax( r + d) + P ( a + b + c) = 0

⎛ d+r ⎞ ⎟ ⎝ a + b + c⎠

P = Tmax⎜

P = 0.533 kN

+ Σ F x = 0; →

Tmax − Ax = 0

Ax = Tmax

Ax = 2 kN

+

Ay − P = 0

Ay = P

Ay = 0.533 kN

+ Σ F x = 0; →

−NC − Ax = 0

NC = − A x

NC = −2 kN

VC = Ay

V C = 0.533 kN

MC = Ay c

MC = 0.400 kN⋅ m

↑Σ Fy = 0;

+

↑Σ Fy = 0; ΣMC = 0;

−V C + Ay = 0 −MC + A y c = 0

Problem 7-21 Determine the internal shear force and moment acting at point C of the beam. Units Used: 3


kip ft

a = 9 ft

642



Chapter 7


NC = 0

ΣF y = 0;

wa

ΣMC = 0;

MC −

2

NC = 0

wa

−

− VC = 0

2

⎛ w a⎞a + ⎛ w a ⎞ ⎜ 2 ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠

VC = 0

⎛ a⎞ = ⎜3⎟ ⎝ ⎠

0

2

MC =

wa

MC = 54.00 kip⋅ ft

3

Problem 7-22 Determine the internal shear force and moment acting at point D of the beam. Units Used: kip = 103 lb Given: w = 2

kip ft

a = 6 ft b = 9 ft Solution: ΣF x = 0; ΣF y = 0;

ND = 0 wb 2

− w⎛⎜

VD =

a⎞

⎛ a⎞ ⎟ ⎜ 2 ⎟ − VD = ⎝ b⎠ ⎝ ⎠

wb 2

− w⎛⎜

0

a⎞

⎛ a⎞ ⎟ ⎜ 2⎟ ⎝ b⎠ ⎝ ⎠

V D = 5.00 kip ΣMD = 0;

⎛ w b ⎞a + ⎛ w a ⎞⎛ a ⎞ ⎜ 2 ⎟ ⎜ ⎟⎜ 2 ⎟ ⎝ ⎠ ⎝ b ⎠⎝ ⎠ ⎛ w a3 ⎞ wb⎞ ⎟ MD = ⎛⎜ a−⎜ ⎟ ⎝ 2 ⎠ ⎝ 6b ⎠ MD −

⎛ a⎞ = ⎜ 3⎟ ⎝ ⎠

0

MD = 46.00 kip⋅ ft

643



Chapter 7

Problem 7-23 The shaft is supported by a journal bearing at A and a thrust bearing at B. Determine the internal normal force, shear force, and moment at (a) point C, which is just to the right of the bearing at A, and (b) point D, which is just to the left of the F2 force. Units Used: kip = 103 lb Given: F 1 = 2500 lb

a = 6 ft

F 2 = 3000 lb

b = 12 ft

w = 75

lb ft

c = 2 ft

Solution: ΣMB = 0;

− Ay( b + c) + F1 ( a + b + c) + w b

Ay =

(2

2

b+c

B x = 0 lb

ΣF y = 0;

Ay − F1 − w b − F2 + By = 0 By = − Ay + F1 + w b + F2

ΣMC = 0;

ΣF x = 0; ΣF y = 0;

)

1 2 F 1 ( a + b + c) + w b + 2 c b + 2 F 2 c

ΣF x = 0;

Segment AC :

⎛ b + c⎞ + F c = ⎜2 ⎟ 2 ⎝ ⎠

0

Ay = 4514 lb

B y = 1886 lb

F 1 a + MC = 0 MC = −F 1 a

MC = −15 kip⋅ ft

NC = 0

NC = 0

−F 1 + Ay − V C = 0 VC = Ay − F1

V C = 2.01 kip

Segment BD: ΣMD = 0;

−MD + By c = 0 MD = B y c

MD = 3.77 kip⋅ ft

ΣF x = 0;

ND = 0

ND = 0

ΣF y = 0;

VD − F2 + By = 0 VD = F2 − By

V D = 1.11 kip

644



Chapter 7

Problem 7-24 The jack AB is used to straighten the bent beam DE using the arrangement shown. If the axial compressive force in the jack is P, determine the internal moment developed at point C of the top beam. Neglect the weight of the beams. Units Used: kip = 103 lb Given: P = 5000 lb a = 2 ft b = 10 ft Solution: Segment: ΣMC = 0; MC +

⎛ P ⎞b = ⎜2⎟ ⎝ ⎠

MC = −

P 2

0

b

MC = −25.00 kip⋅ ft

Problem 7-25 The jack AB is used to straighten the bent beam DE using the arrangement shown. If the axial compressive force in the jack is P, determine the internal moment developed at point C of the top beam. Assume that each beam has a uniform weight density γ. Units Used: kip = 103 lb Given: P = 5000 lb 645



γ = 150

Chapter 7

lb ft

a = 2 ft b = 10 ft Solution: Beam: +

↑Σ Fy = 0;

P − 2 γ ( a + b) − 2 R = 0 R =

P 2

− γ ( a + b)

R = 700 lb Segment: ΣMC = 0;

M C + R b + γ ( a + b) MC = −R b − γ

⎛ a + b⎞ = ⎜ 2 ⎟ ⎝ ⎠

( a + b)

0

2

2

MC = −17.8 kip⋅ ft

Problem 7-26 Determine the normal force, shear force, and moment in the beam at sections passing through points D and E. Point E is just to the right of the F load. Units Used: kip = 103 lb Given: w = 1.5

a = 6 ft kip ft

F = 3 kip

b = 6 ft c = 4 ft d = 4 ft

646



Chapter 7


1 2

⎛ a + b ⎞ − A ( a + b) = ⎟ y ⎝ 3 ⎠

w( a + b) ⎜ 1 2

Ay =

w( a + b)

0

⎛ a + b⎞ ⎜ 3 ⎟ ⎝ ⎠

a+b

Ay = 3 kip + Σ F x = 0; →

Bx = 0

+

By + Ay −

↑

Σ F y = 0;

1

w( a + b) = 0

2

1

By = − Ay +

2

w( a + b)

B y = 6 kip + Σ F x = 0; →

ND = 0

+

Ay −

↑

Σ F y = 0;

⎞a − V = ⎜ ⎟ D 2 ⎝ a + b⎠ 1⎛ aw

VD = Ay −

Σ MD = 0;

MD +

1⎛ aw

⎜

⎞a ⎟

V D = 0.75 kip

2 ⎝ a + b⎠

⎞a ⎛ a ⎞ − A a = ⎜ ⎟ ⎜ ⎟ y 2 ⎝ a + b⎠ ⎝ 3⎠ 1⎛ aw

MD =

−1 ⎛ a w ⎞ ⎛ a ⎞ ⎜ ⎟ a ⎜ ⎟ + Ay a 2 ⎝ a + b⎠ ⎝ 3 ⎠

+ Σ F x = 0; →

NE = 0

+

−V E − F − B y = 0

↑Σ Fy = 0;

0

V E = −F − By ΣME = 0;

0

MD = 13.5 kip⋅ ft

V E = −9 kip

ME + B y c = 0 ME = −By c

ME = −24.0 kip⋅ ft

647



Chapter 7

Problem 7-27 Determine the normal force, shear force, and moment at a section passing through point D of the two-member frame. Units Used: kN = 103 N Given: w1 = 200

N

w2 = 400

N

m

m

a = 2.5 m b = 3m c = 6m Solution: Σ MA = 0;

F BC + Σ F x = 0; → +

↑Σ Fy = 0;

−w1 c⎛⎜

c⎞

→ Σ Fx = 0; +

↑

Σ F y = 0;

2

(w2 − w1)c ⎛⎜ 3 ⎟⎞ + 2c

Ax =

2

1 2

2

2

(FBC c) = 0

a +c

2

a +c

F BC = 2600 N

ac

⎛ c ⎞F ⎜ 2 2 ⎟ BC ⎝ a +c ⎠

Ay − w1 c −

a

⎝ ⎠

2 ⎡ c2 c⎤ ⎢ = w1 + ( w2 − w1 ) ⎥ 3⎦ ⎣ 2

Ay = w1 c + +

1

⎟− ⎝ 2⎠

Ax = 2400 N

(w2 − w1)c + ⎛⎜

⎞F = BC ⎟ 2 2 + c a ⎝ ⎠

(w2 − w1)c − ⎛⎜ 2 1

a

⎞F

a

2⎟

2

⎝ a +c ⎠

− Ax + ND = 0

0

BC

ND = A x

ND = 2.40 kN

1 b Ay − w1 b − ( w2 − w1 ) ⎛⎜ ⎟⎞ b − V D = 0 2 ⎝c⎠ 2⎞

⎛b 1 V D = A y − w1 b − ( w2 − w1 ) ⎜ 2 ⎝c Σ MD = 0;

Ay = 800 N

− Ay b + w1 b⎛⎜

b⎞

⎟+ ⎝2⎠

⎟ ⎠

V D = 50 N

(w2 − w1)⎛⎜ c ⎟⎞ b ⎛⎜ 3 ⎟⎞ + MD = 0 2 1

⎛ b2 ⎞ MD = Ay( b) − w1 ⎜ ⎟ − ⎝2⎠

b

b

⎝ ⎠

⎝ ⎠

⎛ b3 ⎞ (w2 − w1)⎜ 3c ⎟ 2 ⎝ ⎠ 1

648

MD = 1.35 kN⋅ m



Chapter 7

Problem 7-28 Determine the normal force, shear force, and moment at sections passing through points E and F. Member BC is pinned at B and there is a smooth slot in it at C. The pin at C is fixed to member CD. Units Used: 3

kip = 10 lb Given: M = 350 lb⋅ ft w = 80

lb ft

c = 2 ft

F = 500 lb

d = 3 ft

θ = 60 deg

e = 2 ft

a = 2 ft

f = 4 ft

b = 1 ft

g = 2 ft

Solution: Σ MB = 0; −1 2

wd

⎛ 2d ⎞ − F sin ( θ ) d + C ( d + e) = ⎜3⎟ y ⎝ ⎠

⎛ w d2 ⎞ ⎜ ⎟ + F sin ( θ ) d 3 ⎠ ⎝ Cy =

Cy = 307.8 lb

d+e

+ Σ F x = 0; →

B x − F cos ( θ ) = 0

+

By −

↑

Σ F y = 0;

By = + Σ F x = 0; →

0

1 2

B x = F cos ( θ )

B x = 250 lb

w d − F sin ( θ ) + Cy = 0

1 2

w d + F sin ( θ ) − Cy

−NE − Bx = 0

NE = −B x

B y = 245.2 lb NE = −250 lb

649



+

↑Σ Fy = 0;

Chapter 7

VE − By = 0

VE = By

V E = 245 lb

−ME − B y c = 0

ME = −By c

ME = −490 lb⋅ ft

+ Σ F x = 0; →

NF = 0

NF = 0 lb

NF = 0.00 lb

+

−Cy − VF = 0

V F = −Cy

V F = −308 lb

Cy( f) + MF = 0

MF = − f Cy

MF = −1.23 kip⋅ ft

Σ ME = 0

↑Σ Fy = 0; Σ MF = 0;

Problem 7-29 The bolt shank is subjected to a tension F. Determine the internal normal force, shear force, and moment at point C. Given: F = 80 lb a = 6 in


NC + F = 0 NC = −F NC = −80.00 lb

ΣF y = 0;

VC = 0

ΣMC = 0;

MC + F a = 0 MC = −F a MC = −480.00 lb⋅ in

Problem 7-30 Determine the normal force, shear force, and moment acting at sections passing through points B and C on the curved rod.

650



Chapter 7

Units Used: kip = 103 lb Given: F 1 = 300 lb F 2 = 400 lb

θ = 30 deg

r = 2 ft

φ = 45 deg

Solution: Σ F x = 0;

F 2 sin ( θ ) − F1 cos ( θ ) + NB = 0 NB = −F 2 sin ( θ ) + F1 cos ( θ ) NB = 59.8 lb

Σ F y = 0;

V B + F2 cos ( θ ) + F1 sin ( θ ) = 0 V B = −F 2 cos ( θ ) − F 1 sin ( θ ) V B = −496 lb

Σ MB = 0;

MB + F 2 r sin ( θ ) + F 1 ( r − r cos ( θ ) ) = 0 MB = −F2 r sin ( θ ) − F1 r( 1 − cos ( θ ) ) MB = −480 lb⋅ ft

+ Σ F x = 0; →

F2 − Ax = 0

Ax = F2

Ax = 400 lb

+

Ay − F1 = 0

Ay = F1

Ay = 300 lb

↑Σ Fy = 0; Σ MA = 0;

−MA + F 1 2 r = 0

MA = 2 F 1 r

Σ F x = 0;

NC + Ax sin ( φ ) + A y cos ( φ ) = 0

MA = 1200 lb⋅ ft

NC = − A x sin ( φ ) − Ay cos ( φ ) Σ F y = 0;

V C − Ax cos ( φ ) + Ay sin ( φ ) = 0 V C = A x cos ( φ ) − A y sin ( φ )

Σ MC = 0;

NC = −495 lb

V C = 70.7 lb

−MC − MA − A x r sin ( φ ) + A y( r − r cos ( φ ) ) = 0

651



Chapter 7

MC = −MA − Ax r sin ( φ ) + Ay r( 1 − cos ( φ ) )

MC = −1.59 kip⋅ ft

Problem 7-31 The cantilevered rack is used to support each end of a smooth pipe that has total weight W. Determine the normal force, shear force, and moment that act in the arm at its fixed support A along a vertical section. Units Used: 3

kip = 10 lb Given: W = 300 lb r = 6 in

θ = 30 deg

Solution: Pipe: +

↑

Σ F y = 0;

NB cos ( θ ) −

NB =

1 2

W 2

=0

⎛ W ⎞ ⎜ ( )⎟ ⎝ cos θ ⎠

NB = 173.205 lb

Rack: + Σ F x = 0; →

−NA + NB sin ( θ ) = 0 NA = NB sin ( θ )

+

↑Σ Fy = 0;

NA = 86.6 lb

V A − NB cos ( θ ) = 0 V A = NB cos ( θ )

Σ MA = 0;

MA − NB

⎛ r + r sin ( θ ) ⎞ = ⎜ ⎟ ⎝ cos ( θ ) ⎠

MA = NB

⎛ r + r sin ( θ ) ⎞ ⎜ ⎟ ⎝ cos ( θ ) ⎠

V A = 150 lb 0

MA = 1.800 kip⋅ in

652



Chapter 7

Problem 7-32 Determine the normal force, shear force, and moment at a section passing through point D of the two-member frame. Units Used: 3

kN = 10 N

Given: kN m

w = 0.75 F = 4 kN a = 1.5 m

d = 1.5 m

b = 1.5 m

e = 3

c = 2.5 m

f = 4

Solution: Σ MC = 0; −B x( c + d) +

f ⎛ ⎞ ⎜ 2 2 ⎟F d = ⎝ e +f ⎠ fdF

Bx =

2

2

0

B x = 1.2 kN

e + f ( c + d) Σ MA = 0;

⎛ c + d ⎞ + B ( a + b) + B ( c + d) = ⎟ y x ⎝ 2 ⎠

−w( c + d) ⎜

⎡ ( c + d) 2⎤ ⎥ − B x ( c + d) w⎢ 2 ⎣ ⎦ By = a+b

0

B y = 0.40 kN

+ Σ F x = 0; →

−ND − B x = 0

ND = −Bx

ND = −1.2 kN

+

VD + By = 0

V D = −By

V D = −0.4 kN

↑Σ Fy = 0;

653



Σ MD = 0;

−MD + By b = 0

Chapter 7

MD = B y b

MD = 0.6 kN⋅ m

Problem 7-33 Determine the internal normal force, shear force, and moment acting at point A of the smooth hook. Given:

θ = 45 deg a = 2 in F = 20 lb Solution: ΣF x = 0;

NA − F cos ( θ ) = 0 NA = F cos ( θ )

ΣF y = 0;

V A − F sin ( θ ) = 0 V A = F sin ( θ )

ΣMB = 0;

NA = 14.1 lb

V A = 14.1 lb

MA − NA a = 0 MA = NA a

MA = 28.3 lb⋅ in

Problem 7-34 Determine the internal normal force, shear force, and moment acting at points B and C on the curved rod.

654



Chapter 7

Units Used: kip = 103 lb Given:

θ 2 = 30 deg

F = 500 lb r = 2 ft

a = 3

θ 1 = 45 deg

b = 4

Solution: ΣF N = 0;

⎛ F b ⎞ sin ( θ ) − ⎛ F a ⎞ cos ( θ ) + N = 2 2 B ⎜ 2 2⎟ ⎜ 2 2⎟ ⎝ a +b ⎠ ⎝ a +b ⎠ ⎡( a)cos ( θ 2 ) − b sin ( θ 2 )⎥⎤ 2 2 ⎢ ⎥ a +b ⎣ ⎦

NB = F ⎢

ΣF V = 0;

VB +

ΣMB = 0;

NB = 59.8 lb

⎛ F b ⎞ cos ( θ ) + ⎛ F a ⎞ sin ( θ ) = 2 2 ⎜ 2 2⎟ ⎜ 2 2⎟ + b + b a a ⎝ ⎠ ⎝ ⎠

V B = −F

MB +

0

⎡⎢ b cos ( θ 2) + ( a)sin ( θ 2 )⎥⎤ 2 2 ⎢ ⎥ a +b ⎣ ⎦

0

V B = −496 lb

⎛ F b ⎞ r sin ( θ ) + F⎛ a ⎞ ( r − r cos ( θ ) ) = 2 2 ⎜ 2 2⎟ ⎜ 2 2⎟ + b + b a a ⎝ ⎠ ⎝ ⎠

MB = F r

⎡⎢−b sin ( θ 2) − a + ( a)cos ( θ 2 )⎥⎤ 2 2 ⎢ ⎥ a +b ⎣ ⎦

0

MB = −480 lb⋅ ft

Also, ΣF x = 0;

Fb

− Ax +

=0

2

a +b Ax = F

ΣF y = 0;

Ay −

2

⎛ b ⎞ ⎜ 2 2⎟ ⎝ a +b ⎠ Fa 2

a +b

Ax = 400.00 lb

=0 2

655



Ay = F

ΣMA = 0;

−MA +

MA =

Chapter 7

⎛ a ⎞ ⎜ 2 2⎟ ⎝ a +b ⎠

⎛ F a ⎞ 2r = ⎜ 2 2⎟ ⎝ a +b ⎠ 2F r a 2

2

Ay = 300.00 lb

0

MA = 1200 lb⋅ ft

a +b ΣF x = 0;

NC + Ax sin ( θ 1 ) + Ay cos ( θ 1 ) = 0 NC = − A x sin ( θ 1 ) − A y cos ( θ 1 )

ΣF y = 0;

NC = −495 lb

V C − Ax cos ( θ 1 ) + A y sin ( θ 1 ) = 0 V C = A x cos ( θ 1 ) − Ay sin ( θ 1 )

ΣMC = 0;

V C = 70.7 lb

−MC − MA + A y( r − r cos ( θ 1 ) ) − Ax r sin ( θ 1 ) = 0 MC = −MA + Ay( r − r cos ( θ 1 ) ) − A x r sin ( θ 1 )

MC = −1.59 kip⋅ ft

Problem 7-35 Determine the ratio a/b for which the shear force will be zero at the midpoint C of the beam.

Solution: Find Ay:

ΣMB = 0;

656



Chapter 7

( 2 a + b) w ⎡⎢ ( b − a)⎤⎥ − Ay b = 0

1

1

⎣3

2

Ay =

⎦

w ( 2 a + b) ( b − a) 6b

This problem requires

V C = 0.

Summing forces vertically for the section, we have +

↑Σ Fy = 0; w 1 ( 2a + b) ( b − a) − 2 6b

⎛a + ⎜ ⎝

b⎞ w

⎟

2⎠ 2

=0

w w ( 2a + b) ( b − a) = ( 2a + b) 8 6b 4 ( b − a) = 3 b

b = 4a a 1 = 4 b

Problem 7-36 The semicircular arch is subjected to a uniform distributed load along its axis of w0 per unit length. Determine the internal normal force, shear force, and moment in the arch at angle θ. Given:

θ = 45 deg

Solution: Resultants of distributed load:

657



Chapter 7

θ ⌠ F Rx = ⎮ w0 r dθ sin ( θ ) = r w0 ( 1 − cos ( θ ) ) ⌡0

F Rx = r w0 ( 1 − cos ( θ ) ) θ ⌠ F Ry = ⎮ w0 r dθ cos ( θ ) = r w0 sin ( θ ) ⌡0

F Rx = r w0 ( sin ( θ ) ) θ ⌠ 2 MRo = ⎮ w0 r dθ r = r w0 θ ⌡0

Σ F x = 0;

−V + FRx cos ( θ ) − F Ry sin ( θ ) = 0

V = ⎡⎣r w0 ( 1 − cos ( θ ) )⎤⎦ cos ( θ ) − ⎡⎣r w0 ( sin ( θ ) )⎤⎦ sin ( θ ) V = w0 r( cos ( θ ) − 1 ) a = cos ( θ ) − 1 a = −0.293 Σ F y = 0;

V = a r w0

N + FRy cos ( θ ) + F Rx sin ( θ ) = 0 N = −⎡⎣r w0 ( 1 − cos ( θ ) )⎤⎦ sin ( θ ) − ⎡⎣r w0 ( sin ( θ ) )⎤⎦ cos ( θ ) N = −w0 r sin ( θ ) b = −sin ( θ ) b = −0.707

ΣMo = 0;

N = w0 r b

−M + r w0 ( θ ) + b r w0 r = 0 2

M = w0 r ( θ + b) 2

c = θ+b c = 0.0783

2

M = c r w0

658



Chapter 7

Problem 7-37 The semicircular arch is subjected to a uniform distributed load along its axis of w0 per unit length. Determine the internal normal force, shear force, and moment in the arch at angle θ. Given:

θ = 120 deg Solution: Resultants of distributed load: θ ⌠ F Rx = ⎮ w0 r dθ sin ( θ ) = r w0 ( 1 − cos ( θ ) ) ⌡0

F Rx = r w0 ( 1 − cos ( θ ) ) θ ⌠ F Ry = ⎮ w0 r dθ cos ( θ ) = r w0 sin ( θ ) ⌡0

F Rx = r w0 ( sin ( θ ) ) θ ⌠ 2 MRo = ⎮ w0 r dθ r = r w0 θ ⌡0

Σ F x = 0;

−V + FRx cos ( θ ) − F Ry sin ( θ ) = 0 V = ⎡⎣r w0 ( 1 − cos ( θ ) )⎤⎦ cos ( θ ) − ⎡⎣r w0 ( sin ( θ ) )⎤⎦ sin ( θ ) V = w0 r( cos ( θ ) − 1 ) a = cos ( θ ) − 1 a = −1.500

Σ F y = 0;

V = a r w0

N + FRy cos ( θ ) + F Rx sin ( θ ) = 0 N = −⎡⎣r w0 ( 1 − cos ( θ ) )⎤⎦ sin ( θ ) − ⎡⎣r w0 ( sin ( θ ) )⎤⎦ cos ( θ ) N = −w0 r sin ( θ ) b = −sin ( θ ) 659



Chapter 7

b = −0.866 Σ Mo = 0;

N = w0 ⋅ r⋅ b

−M + r w0 ( θ ) + b r w0 r = 0 2

M = w0 r ( θ + b) 2

c = θ+b c = 1.2284

2

M = c r w0

Problem 7-38 Determine the x, y, z components of internal loading at a section passing through point C in the pipe assembly. Neglect the weight of the pipe. Units Used: kip = 103 lb Given:

⎛ 0 ⎞ F 1 = ⎜ 350 ⎟ lb ⎜ ⎟ ⎝ −400 ⎠ ⎛ 150 ⎞ F 2 = ⎜ 0 ⎟ lb ⎜ ⎟ ⎝ −300 ⎠ a = 1.5 ft

b = 2 ft

c = 3 ft

Solution:

⎛c⎞ r1 = ⎜ b ⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ r2 = ⎜ b ⎟ ⎜ ⎟ ⎝0⎠

F C = −F1 − F2

⎛ −150.00 ⎞ F C = ⎜ −350.00 ⎟ lb ⎜ ⎟ ⎝ 700.00 ⎠

660



MC = −r1 × F1 − r2 × F2

Chapter 7

⎛ 1400.00 ⎞ MC = ⎜ −1200.00 ⎟ lb⋅ ft ⎜ ⎟ ⎝ −750.00 ⎠

Problem 7-39 Determine the x, y, z components of internal loading at a section passing through point C in the pipe assembly. Neglect the weight of the pipe. Units Used: kip = 103 lb Given:

⎛ −80 ⎞ F 1 = ⎜ 200 ⎟ lb ⎜ ⎟ ⎝ −300 ⎠ ⎛ 250 ⎞ F 2 = ⎜ −150 ⎟ lb ⎜ ⎟ ⎝ −200 ⎠ a = 1.5 ft

b = 2 ft

c = 3 ft

Solution:

⎛c⎞ r1 = ⎜ b ⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ r2 = ⎜ b ⎟ ⎜ ⎟ ⎝0⎠

F C = −F1 − F2

⎛ −170.00 ⎞ F C = ⎜ −50.00 ⎟ lb ⎜ ⎟ ⎝ 500.00 ⎠

MC = −r1 × F1 − r2 × F2

⎛ 1000.00 ⎞ MC = ⎜ −900.00 ⎟ lb⋅ ft ⎜ ⎟ ⎝ −260.00 ⎠

Problem 7-40 Determine the x, y, z components of internal loading in the rod at point D.

661



Chapter 7

Units Used: kN = 103 N Given: M = 3 kN⋅ m

⎛ 7 ⎞ F = ⎜ −12 ⎟ kN ⎜ ⎟ ⎝ −5 ⎠ a = 0.75 m b = 0.2 m c = 0.2 m d = 0.6 m e = 1m Solution: Guesses Cx = 1 N

Cy = 1 N

Bx = 1 N

Bz = 1 N

Ay = 1 N

Az = 1 N

Given

⎛⎜ 0 ⎞⎟ ⎛⎜ Bx ⎟⎞ ⎛⎜ Cx ⎟⎞ ⎜ Ay ⎟ + ⎜ 0 ⎟ + ⎜ Cy ⎟ + F = 0 ⎜ Az ⎟ ⎜ Bz ⎟ ⎜ 0 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ −e ⎞ ⎛⎜ 0 ⎞⎟ ⎛ 0 ⎞ ⎛⎜ Bx ⎟⎞ ⎛ 0 ⎞ ⎛⎜ Cx ⎟⎞ ⎛ 0 ⎛ 0 ⎞ ⎞ ⎜ b + c + d ⎟ × ⎜ Ay ⎟ + ⎜ b + c ⎟ × ⎜ 0 ⎟ + ⎜ 0 ⎟ × ⎜ C ⎟ + ⎜ b + c + d ⎟ × F + ⎜ 0 ⎟ = 0 y ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ ⎟ ⎜ 0 a ⎠ ⎝ Bz ⎠ ⎝ ⎠ ⎝ 0 ⎟⎠ ⎝ 0 ⎠ ⎝ −M ⎠ ⎝ 0 ⎠ ⎝ Az ⎠ ⎝

662



⎛⎜ Ay ⎞⎟ ⎜ Az ⎟ ⎜ ⎟ ⎜ Bx ⎟ = Find ( A , A , B , B , C , C ) y z x z x y ⎜ Bz ⎟ ⎜ ⎟ ⎜ Cx ⎟ ⎜ Cy ⎟ ⎝ ⎠

Chapter 7

⎛⎜ Ay ⎞⎟ ⎛ −53.60 ⎞ ⎜ Az ⎟ ⎜ 87.00 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Bx ⎟ = ⎜ 109.00 ⎟ kN ⎜ Bz ⎟ ⎜ −82.00 ⎟ ⎜ ⎟ ⎜ −116.00 ⎟ ⎜ Cx ⎟ ⎜ ⎟ 65.60 ⎝ ⎠ ⎜ Cy ⎟ ⎝ ⎠

Guesses V Dx = 1 N

NDy = 1 N

V Dz = 1 N

MDx = 1 N⋅ m

MDy = 1 N⋅ m

MDz = 1 N⋅ m

Given

⎛⎜ Cx ⎟⎞ ⎛⎜ VDx ⎟⎞ ⎜ Cy ⎟ + ⎜ NDy ⎟ = 0 ⎜ 0 ⎟ ⎜V ⎟ ⎝ ⎠ ⎝ Dz ⎠ ⎛ 0 ⎞ ⎛⎜ Cx ⎟⎞ ⎛⎜ MDx ⎟⎞ ⎛ 0 ⎞ ⎜ −b ⎟ × ⎜ C ⎟ + ⎜ MDy ⎟ + ⎜ 0 ⎟ = 0 ⎜ ⎟ ⎜ ⎟ ⎜ y⎟ ⎜ ⎝ a ⎠ ⎝ 0 ⎠ ⎝ MDz ⎟⎠ ⎝ −M ⎠ ⎛⎜ VDx ⎟⎞ ⎜ NDy ⎟ ⎜ ⎟ ⎜ VDz ⎟ = Find ( V , N , V , M , M , M ) Dx Dy Dz Dx Dy Dz ⎜ MDx ⎟ ⎜ ⎟ ⎜ MDy ⎟ ⎜ MDz ⎟ ⎝ ⎠

⎛ VDx ⎞ ⎛ 116.00 ⎞ ⎜ ⎟ ⎜ NDy ⎟ = ⎜ −65.60 ⎟ kN ⎜ V ⎟ ⎜⎝ 0.00 ⎟⎠ ⎝ Dz ⎠ ⎛ MDx ⎞ ⎛ 49.20 ⎞ ⎜ ⎟ ⎜ MDy ⎟ = ⎜ 87.00 ⎟ kN⋅ m ⎜ M ⎟ ⎜⎝ 26.20 ⎟⎠ ⎝ Dz ⎠

663



Chapter 7

Problem 7-41 Determine the x, y, z components of internal loading in the rod at point E. Units Used: 3

kN = 10 N Given: M = 3 kN⋅ m

⎛ 7 ⎞ F = ⎜ −12 ⎟ kN ⎜ ⎟ ⎝ −5 ⎠ a = 0.75 m b = 0.4 m c = 0.6 m d = 0.5 m e = 0.5 m Solution: Guesses Cx = 1 N

Cy = 1 N

Bx = 1 N

Bz = 1 N

Ay = 1 N

Az = 1 N

Given

⎛⎜ 0 ⎞⎟ ⎛⎜ Bx ⎟⎞ ⎛⎜ Cx ⎟⎞ ⎜ Ay ⎟ + ⎜ 0 ⎟ + ⎜ Cy ⎟ + F = 0 ⎜ Az ⎟ ⎜ Bz ⎟ ⎜ 0 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ −d − e ⎞ ⎛⎜ 0 ⎞⎟ ⎛ 0 ⎞ ⎛⎜ Bx ⎟⎞ ⎛ 0 ⎞ ⎛⎜ Cx ⎟⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎜ b + c ⎟ × ⎜ Ay ⎟ + ⎜ b ⎟ × ⎜ 0 ⎟ + ⎜ 0 ⎟ × ⎜ C ⎟ + ⎜ b + c ⎟ × F + ⎜ 0 ⎟ = 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ y⎟ ⎜ 0 ⎝ −M ⎠ ⎝ ⎠ ⎝ Az ⎠ ⎝ 0 ⎠ ⎝ Bz ⎠ ⎝ a ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠

664



⎛⎜ Ay ⎞⎟ ⎜ Az ⎟ ⎜ ⎟ ⎜ Bx ⎟ = Find ( A , A , B , B , C , C ) y z x z x y ⎜ Bz ⎟ ⎜ ⎟ ⎜ Cx ⎟ ⎜ Cy ⎟ ⎝ ⎠

Chapter 7

⎛⎜ Ay ⎞⎟ ⎛ −53.60 ⎞ ⎜ Az ⎟ ⎜ 87.00 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Bx ⎟ = ⎜ 109.00 ⎟ kN ⎜ Bz ⎟ ⎜ −82.00 ⎟ ⎜ ⎟ ⎜ −116.00 ⎟ ⎜ Cx ⎟ ⎜ ⎟ 65.60 ⎝ ⎠ ⎜ Cy ⎟ ⎝ ⎠

Guesses NEx = 1 N

V Ey = 1 N

V Ez = 1 N

MEx = 1 N⋅ m

MEy = 1 N⋅ m

MEz = 1 N⋅ m

Given

⎛⎜ 0 ⎞⎟ ⎛⎜ NEx ⎟⎞ ⎜ Ay ⎟ + ⎜ VEy ⎟ = 0 ⎜ Az ⎟ ⎜ V ⎟ ⎝ ⎠ ⎝ Ez ⎠ ⎛ −e ⎞ ⎛⎜ 0 ⎞⎟ ⎛⎜ MEx ⎞⎟ ⎜ 0 ⎟ × Ay + M =0 ⎜ ⎟ ⎜ ⎟ ⎜ Ey ⎟ ⎝ 0 ⎠ ⎜⎝ Az ⎟⎠ ⎜⎝ MEz ⎟⎠ ⎛⎜ NEx ⎞⎟ ⎜ VEy ⎟ ⎜ ⎟ ⎜ VEz ⎟ = Find ( N , V , V , M , M , M ) Ex Ey Ez Ex Ey Ez ⎜ MEx ⎟ ⎜ ⎟ ⎜ MEy ⎟ ⎜ MEz ⎟ ⎝ ⎠

⎛ NEx ⎞ ⎛ 0.00 ⎞ ⎜ ⎟ ⎜ VEy ⎟ = ⎜ 53.60 ⎟ kN ⎜ V ⎟ ⎜⎝ −87.00 ⎟⎠ ⎝ Ez ⎠ ⎛ MEx ⎞ ⎛ 0.00 ⎞ ⎜ ⎟ ⎜ MEy ⎟ = ⎜ −43.50 ⎟ kN⋅ m ⎜ M ⎟ ⎜⎝ −26.80 ⎟⎠ ⎝ Ez ⎠

Problem 7-42 Draw the shear and moment diagrams for the shaft in terms of the parameters shown; There is 665



Chapter 7

g a thrust bearing at A and a journal bearing at B.

p

Units Used: 3

kN = 10 N Given: P = 9 kN a = 2m L = 6m Solution: P ( L − a) − Ay L = 0

Ay = P

L−a L

x1 = 0 , 0.01a .. a Ay − V 1 ( x) = 0

V 1 ( x) =

M1 ( x) − Ay x = 0

M1 ( x) =

Ay kN

Ay x kN⋅ m

x2 = a , 1.01a .. L Ay − P − V2 ( x) = 0

V 2 ( x) =

Ay − P kN

M2 ( x) − Ay x + P ( x − a) = 0

M2 ( x) =

A y x − P ( x − a) kN⋅ m

666



Chapter 7

Force (kN)

10

V1( x1) 5 V2( x2) 0

5

0

1

2

3

4

5

6

x1 , x2

Distance (m)

Moment (kN-m)

15 10

M1( x1) M2( x2)

5 0 5

0

1

2

3

4

5

6

x1 , x2

Distance (m)

Problem 7-43 Draw the shear and moment diagrams for the beam in terms of the parameters shown. Given: P = 800 lb

a = 5 ft

L = 12 ft

Solution:

667



x1 = 0 , 0.01a .. a

Chapter 7

x2 = a , 1.01a .. L − a

x3 = L − a , 1.01( L − a) .. L

V 2 ( x) = 0

V 3 ( x) =

P

V 1 ( x) =

lb Px

M1 ( x) =

M2 ( x) =

lb⋅ ft

Pa

M3 ( x) =

lb⋅ ft

−P lb P( L − x) lb⋅ ft

Force (lb)

1000

V1( x1)

500

V2( x2)

0

V3( x3)

500

1000

0

2

4

6

8

10

8

10

12

x1 x2 x3 , , ft ft ft

Distance (ft)

Moment (lb-ft)

4000

M1( x1) M2( x2)

M3( x3)2000

0

0

2

4

6

12


Distance (ft)

Problem 7-44 Draw the shear and moment diagrams for the beam (a) in terms of the parameters 668



Chapter 7

g shown; (b) set M0 and L as given.

( )

p

Given: M0 = 500 N⋅ m L = 8m Solution: For

0 ≤ x≤

L 3

+

↑Σ Fy = 0;

V1 = 0

ΣMx = 0;

For

L 3

≤x≤

M1 = 0

2L 3

+

↑Σ Fy = 0; ΣMx = 0;

For

2L 3

M2 = M0

≤x≤L

+

↑Σ Fy = 0; ΣMx = 0;

( b)

V2 = 0

V3 = 0 M3 = 0

x1 = 0 , 0.01L ..

L 3

x2 =

L L 2L , 1.01 .. 3

3

3

x3 =

2L 2L 3

,

3

V 1 ( x1 ) = 0

V 2 ( x2 ) = 0

V 3 ( x3 ) = 0

M1 ( x1 ) = 0

M2 ( x2 ) = M0

M3 ( x3 ) = 0

1.01 .. L

669



Chapter 7

1

Shear force in N

0.5

V1( x1) V2( x2)

0

V3( x3)

0.5

1

0

1

2

3

4

5

6

7

8

9

x1 , x2 , x3

Distance in m

600

Moment in N - m

400

M1( x1) M2( x2)

M3( x3)200

0

0

1

2

3

4

5

6

7

8

9

x1 , x2 , x3

Distance in m

Problem 7-45 The beam will fail when the maximum shear force is V max or the maximum bending moment is Mmax. Determine the magnitude M0 of the largest couple moments it will support.

670



Chapter 7

Units Used: 3

kN = 10 N Given: L = 9m V max = 5 kN Mmax = 2 kN⋅ m

Solution: The shear force is zero everywhere in the beam. The moment is zero in the first third and the last third of the bam. In the middle section of the beam the moment is

M = M0

M0 = Mmax

Thus the beam will fail when

M0 = 2.00 kN⋅ m

Problem 7-46 The shaft is supported by a thrust bearing at A and a journal bearing at B. Draw the shear and moment diagrams for the shaft in terms of the parameters shown. Given: w = 500

lb ft

L = 10 ft Solution:

For

ΣF y = 0;

wL

ΣM = 0;

2

0 ≤ x< L

− wx − V = 0

−w L 2

⎛x⎞ + M = ⎟ ⎝ 2⎠

x + w x⎜

V ( x) =

0

w 2

( L − 2 x)

M ( x) =

1

lb

1 2 ( Lx − x ) 2 lb⋅ ft

w

671



Chapter 7

Force (lb)

5000

V( x )

0

5000

0

2

4

6

8

10

x ft

Distance (ft) 1 .10

Moment (lb-ft)

4

M( x)

0

1 .10

4

0

2

4

6

8

10

x ft

Distance (ft)

Problem 7-47 The shaft is supported by a thrust bearing at A and a journal bearing at B. The shaft will fail when the maximum moment is Mmax. Determine the largest uniformly distributed load w the shaft will support. Units Used: 3

kip = 10 lb Given: L = 10 ft Mmax = 5 kip⋅ ft

672



Chapter 7

Solution: wL 2

−

− wx − V = 0

wL 2

V = −w x +

⎛x⎞ x + wx ⎜ ⎟ + M = ⎝2⎠

wL 2

2 wL⎞ wx ⎛ M=⎜ ⎟x − 2 ⎝ 2 ⎠

0

From the Moment Diagram, wL

Mmax =

2

w =

8

8 Mmax

L

2

w = 400.00

lb ft

Problem 7-48 Draw the shear and moment diagrams for the beam. 3

kN = 10 N


kN m

L = 5m MB = 5 kN⋅ m Solution: ΣF y = 0; −V ( x) + wL − wx = 0

ΣM = 0;

V ( x) = ( w L − w x)

⎛ L ⎞ + M − w L x + w x⎛ x ⎞ = ⎟ ⎜ 2⎟ B ⎝2⎠ ⎝ ⎠

M ( x) + w L⎜

⎡ ⎣

1

kN

0

⎛ x ⎞ − w L⎛ L ⎞ − M ⎤ 1 ⎟ ⎜2⎟ B⎥ ⎝2⎠ ⎝ ⎠ ⎦ kN⋅ m

M ( x) = ⎢w L x − w x⎜

673



Chapter 7

Force (kN)

10

V( x ) 5

0

0

1

2

3

4

5

4

5

x

Distance (m)

Moment (kN-m)

0

M( x) 20

40

0

1

2

3

x

Distance (m)


kN = 10 N


kN m

F = 10 kN

L = 6m Solution: V ( x) − w( L − x) − F = 0 V ( x) = [ w( L − x) + F ]

1

kN

⎛ L − x ⎞ − F( L − x) = ⎟ ⎝ 2 ⎠

−M ( x) − w( L − x) ⎜

0

674



⎡

( L − x)

⎣

2

M ( x) = ⎢−w

2

Chapter 7

⎤

− F ( L − x)⎥

1

⎦ kN⋅ m

Force (kN)

40

V( x) 20

0

0

1

2

3

4

5

6

x

Distance (m) Moment (kN-m)

0 50

M( x) 100 150

0

1

2

3

4

5

6

x

Distance (m)

Problem 7-50 Draw the shear and moment diagrams for the beam. Units Used: 3

kN = 10 N Given: a = 2m

b = 4m

w = 1.5

kN m

Solution:

⎛ b − a⎞ − A b = ⎟ y ⎝ 2 ⎠

w( b − a) ⎜

0

Ay =

w ( b − a) 2b

2

Ay = 0.75 kN

x1 = 0 , 0.01a .. a Ay − V 1 ( x) = 0

V 1 ( x) = Ay

1

kN 675



Chapter 7

− Ay x + M1 ( x) = 0

M1 ( x) = A y x

1

kN⋅ m

x2 = b − a , 1.01( b − a) .. b 1

Ay − w( x − a) − V 2 ( x) = 0

V 2 ( x) = ⎡⎣A y − w( x − a)⎤⎦

⎛ x − a ⎞ + M ( x) = − Ay x + w( x − a) ⎜ ⎟ 2 ⎝ 2 ⎠

2 ⎡ ( x − a) ⎤ ⎢ ⎥ 1 M2 ( x) = Ay x − w 2 ⎣ ⎦ kN⋅ m

0

kN

Force (kN)

2

V1( x1) 0 V2( x2)

2

4

0

0.5

1

1.5

2

2.5

3

2.5

3

3.5

4

x1 , x2

Distance (m)

Moment (kN-m)

2

M1( x1) 1 M2( x2) 0

1

0

0.5

1

1.5

2

3.5

4

x1 , x2

Distance (m)

676



Chapter 7

Problem 7-51 Draw the shear and moment diagrams for the beam. Given: L = 20 ft

w = 250

lb ft

M1 = 150 lb⋅ ft

Solution:

⎛ L⎞ − M − A l = ⎟ 2 y ⎝2⎠

M 1 + w L⎜

V ( x) = ( Ay − w x)

M2 = 150 lb⋅ ft

M1 − M2 + w Ay =

0

⎛ L2 ⎞ ⎜ ⎟ ⎝2⎠

L

Ay = 2500 lb

⎡ ⎛ x2 ⎞ ⎤ 1 ⎢ M ( x) = A y x − w⎜ ⎟ − M1⎥ ⎣ ⎝2⎠ ⎦ lb⋅ ft

1

lb

Force in lb

2000

V( x )

0

2000 0

5

10

15

20

x ft

Distance in ft

677



Chapter 7

1.5 .10

Moment in lb ft

4

1 .10

4

M( x) 5000

0 0

5

10

15

20

x ft

Distance in ft


kN = 10 N Given: w = 40

kN m

F = 20 kN

M = 150 kN⋅ m

Solution: − Ay a + w a

⎛ a⎞ − F b − M = ⎜ 2⎟ ⎝ ⎠

0

x1 = 0 , 0.01a .. a 1

kN

⎛x

M1 ( x) = ⎢Ay x − w⎜

⎣

b = 3m

⎡ ⎛ a2 ⎞ ⎤ ⎟ − F b − M⎥ ⎛⎜ 1 ⎟⎞ ⎣ ⎝2⎠ ⎦⎝ a ⎠

Ay = ⎢w⎜

Ay = 133.75 kN

x2 = a , 1.01a .. a + b

V 1 ( x) = ( A y − w x)

⎡

a = 8m

2 ⎞⎤

⎟⎥ 1 ⎝ 2 ⎠⎦ kN⋅ m

V 2 ( x) = F

1

kN

M2 ( x) = [ −F( a + b − x) − M]

1

kN⋅ m

678



Chapter 7

Force in kN

200

V1( x1) V2( x2)

0

200

0

2

4

6

8

10

x1 , x2

Distance in m

Moment (kN-m)

400

M1( x1)

200

M2( x2)

0 200 400

0

2

4

6

8

10

x1 , x2

Distance (m)

Problem 7-53 Draw the shear and moment diagrams for the beam.

679



Chapter 7

Solution: 0 ≤ x< a ΣF y = 0; −V − w x = 0

V = −w x ΣM = 0;

M + w x⎛⎜

x⎞

M = −w

x

⎟= ⎝2⎠

0

2

2

a < x ≤ 2a ΣF y = 0; −V + 2 w a − w x = 0

V = w( 2 a − x) ΣM = 0;

M + w x⎛⎜

x⎞

⎟ − 2 w a( x − a) = ⎝2⎠

0

2

M = 2 w a( x − a) −

wx 2

Problem 7-54 Draw the shear and bending-moment diagrams for beam ABC. Note that there is a pin at B.

680



Chapter 7

Solution: Support Reactions: From FBD (b), wL ⎛ L⎞

⎛ L⎞ ⎜ 4 ⎟ − B y⎜ 2 ⎟ = ⎝ ⎠ ⎝ ⎠

2

wL

By =

4

Ay −

From FBD (a),

wL 2

−

B y

=0

3w L

Ay =

−B y

0

4

L 2

− w⎛⎜

MA = w

⎛ L⎞ + M = ⎟ ⎜ ⎟ A ⎝2⎠ ⎝4⎠ L⎞

0

⎛ L2 ⎞ ⎜ ⎟ ⎝4⎠

Shear and Moment Functions: From FBD (c) For 0 ≤ x≤ L

V=

w 4

Ay − w x − V = 0

( 3 L − 4 x)

x MA − Ay x + w x ⎛⎜ ⎟⎞ + M = 0 ⎝ 2⎠

M=

w 4

(3L x − 2x2 − L2)

681



Chapter 7

Problem 7-55 The beam has depth a and is subjected to a uniform distributed loading w which acts at an angle θ from the vertical as shown. Determine the internal normal force, shear force, and moment in the beam as a function of x. Hint:The moment loading is to be determined from a point along the centerline of the beam (x axis). Given: a = 2 ft L = 10 ft

θ = 30 deg w = 50

lb ft

Solution: 0 ≤ x≤ L

N + w sin ( θ ) x = 0

ΣF x = 0;

N ( x) = −w sin ( θ ) x −V − w cos ( θ ) x = 0

ΣF y = 0;

V = −w cos ( θ ) x

⎛ x ⎞ − w sin ( θ ) x⎛ a ⎞ + M = ⎟ ⎜ 2⎟ ⎝2⎠ ⎝ ⎠

w cos ( θ ) x⎜

ΣM = 0;

M ( x) = −w cos ( θ )

0

⎛ x2 ⎞ ⎜ ⎟ + w sin ( θ ) ⎛⎜ x a ⎞⎟ ⎝2⎠ ⎝2⎠

Problem 7-56 Draw the shear and moment diagrams for the beam. Given: w = 250

lb ft

L = 12 ft

682



Chapter 7

Solution: 1

2

wx =0 L

ΣF y = 0;

−V −

ΣM = 0;

x ⎛ w x⎞ 1 M + ⎜ ⎟ ⎛⎜ x⎟⎞ = 0 2 ⎝ L ⎠⎝ 3 ⎠

2

x

V ( x) = −

wx 1 2 L lb 3

−w x M ( x) = 6L

1

lb⋅ ft

Force (lb)

0

V( x) 1000

2000

0

2

4

6

8

10

12

x ft

Distance (ft)

Moment (lb-ft)

0

M( x)

5000

1 .10

4

0

2

4

6

8

10

12

x ft

Distance (ft)

Problem 7-57 The beam will fail when the maximum shear force is V max or the maximum moment is Mmax. Determine the largest intensity w of the distributed loading it will support.

683



Chapter 7

Given: L = 18 ft V max = 800 lb Mmax = 1200 lb⋅ ft Solution: For 0 ≤ x ≤ L 2

V=

M=

2L

w1 = 2

6L

2

⎛ Vmax ⎞ ⎜ ⎟ ⎝ L ⎠

Mmax =

−w x

wL

V max =

w2 = 6

3

−w x

w2 L

w1 = 88.9

lb ft

2

6

⎛ Mmax ⎞ ⎜ 2 ⎟ ⎝ L ⎠

w2 = 22.2

Now choose the critical case

lb ft

w = min ( w1 , w2 )

w = 22.22

lb ft

Problem 7-58 The beam will fail when the maximum internal moment is Mmax. Determine the position x of the concentrated force P and its smallest magnitude that will cause failure.

684



Chapter 7

Solution: For ξ < x,

M1 =

For ξ > x,

M2 =

Pξ ( L − x) L P x( L − ξ ) L

Note that M1 = M2 when x = ξ Mmax = M1 = M2 =

d dx

P x( L − x) L

(L x − x2) = L − 2x

Thus,

Mmax =

x=

=

P L

(L x − x2)

L 2

P ⎛ L ⎞⎛

L⎞ P ⎛ L⎞ ⎜ 2 ⎟ ⎜L − 2 ⎟ = 2 ⎜ 2 ⎟ L ⎝ ⎠⎝ ⎠ ⎝ ⎠

P=

4 Mmax

L

Problem 7-59 Draw the shear and moment diagrams for the beam. Given: w = 30

lb

MC = 180 lb⋅ ft

ft

a = 9 ft

b = 4.5 ft

Solution: − Ay a +

1 2

wa

⎛ a⎞ − M = ⎜ 3⎟ C ⎝ ⎠

0

685



Ay =

wa 6

Ay + By −

By =

wa 2

MC

− 1 2

Chapter 7

a wa = 0

− Ay

x1 = 0 , 0.01a .. a Ay −

1 2

⎛x⎞ w ⎜ ⎟ x − V1 ( x) = ⎝ a⎠

2 ⎛ wx ⎞ 1 ⎜ ⎟ V 1 ( x) = A y − 2 a ⎠ lb ⎝

0

⎛x⎞ ⎛x⎞ − Ay x + w⎜ ⎟ x⎜ ⎟ + M1 ( x) = 2 ⎝ a⎠ ⎝ 3 ⎠ 1

3 ⎛ wx ⎞ 1 ⎜ ⎟ M1 ( x) = Ay x − 6 a ⎠ lb⋅ ft ⎝

0

x2 = a , 1.01a .. a + b Ay −

1 2

w a + By − V 2 ( x) = 0

− Ay x +

1 2

⎛ ⎝

w a ⎜x −

V 2 ( x) =

2a ⎞ 3

⎟ − By( x − a) + M2 ( x) = ⎠

⎛A + B − wa⎞ 1 ⎜ y y 2 ⎟ ⎝ ⎠ lb

0

⎡ ⎣

M2 ( x) = ⎢Ay x + B y( x − a) −

wa 2

⎛ ⎝

⋅ ⎜x −

2 a ⎞⎤ 3

1

⎟⎥ ⎠⎦ lb⋅ ft

Force (lb)

100

V1( x1) V2( x2)

0

100

200

0

2

4

6

8

10

12

x1 x2 , ft ft

Distance (ft)

686



Chapter 7

Moment (lb-ft)

100

M1( x1) M2( x2)

0

100

200

0

2

4

6

8

10

12

x1 x2 , ft ft

Diastance (ft)

Problem 7-60 The cantilevered beam is made of material having a specific weight γ. Determine the shear and moment in the beam as a function of x. Solution: By similar triangles y h = x d

y=

h x d

W = γV = γ

⎛⎜ 1 y x t⎟⎞ = γ ⎡⎢ 1 ⎛⎜ h x⎟⎞ x t⎤⎥ = ⎛⎜ γ h t ⎟⎞ x2 ⎝2 ⎠ ⎣2 ⎝ d ⎠ ⎦ ⎝ 2d ⎠

ΣF y = 0;

V−

⎛ γ h t ⎞ x2 = ⎜ ⎟ ⎝ 2d ⎠

V=

ΣM = 0;

−M −

0

⎛ γ h t ⎞ x2 ⎜ ⎟ ⎝ 2d ⎠ ⎛ γ h t ⎞ x2 ⎛ x ⎞ = ⎜ ⎟ ⎜3⎟ ⎝ 2d ⎠ ⎝ ⎠

0

687



M=−

Chapter 7

γh t 3 6d

x


kip = 10 lb

Given:

w1 = 30

lb ft

w2 = 120

lb ft

L = 12 ft Solution:

w1 L

⎛ L⎞ + ⎜2⎟ ⎝ ⎠

(w2 − w1)L ⎛⎜ 3 ⎞⎟ − Ay L = 0 2 L

1

⎝ ⎠

⎛ L ⎞ + ⎛⎜ w2 − w1 ⎞⎟ ⎛ L ⎞ ⎟ ⎜ ⎟ ⎝2⎠ ⎝ 2 ⎠ ⎝3⎠

Ay = w1 ⎜

Ay − w1 x −

(w2 − w1)⎛⎜ L ⎞⎟ x − V ( x) 2 x

1

⎝ ⎠

⎡ V ( x) = ⎢Ay − w1 x − ⎣ ⎛x⎞ + ⎟ ⎝ 2⎠

− Ay x + w1 x⎜

=0

⎛ x2 ⎞⎤ 1 (w2 − w1)⎜ L ⎟⎥ lb 2 ⎝ ⎠⎦ 1

(w2 − w1)⎛⎜ L ⎞⎟ x ⎛⎜ 3 ⎟⎞ + M ( x) 2 1

x

x

⎝ ⎠

⎝ ⎠

=0

⎡ ⎛ x2 ⎞ ⎛ x3 ⎞⎤ 1 ⎜ ⎟ M ( x) = ⎢A y x − w1 − ( w2 − w1 ) ⎜ ⎟⎥ ⎝2⎠ ⎣ ⎝ 6 L ⎠⎦ kip⋅ ft

688



Chapter 7

Force (lb)

500

0

V( x )

500 0

2

4

6

8

10

12

x ft

Distance (ft)

Moment (kip-ft)

2

M( x) 1

0

0

2

4

6

8

10

12

x ft

Distance (ft)


kip = 10 lb Given: F 1 = 20 kip

F 2 = 20 kip

w = 4

kip ft

a = 15 ft

b = 30 ft

c = 15 ft

Solution:

⎛ b⎞ − F c − A b = ⎟ 2 y ⎝ 2⎠

F 1 ( a + b) + w b⎜

⎛ a + b⎞ + wb − F ⎛ c ⎞ ⎟ 2⎜ ⎟ ⎝ b ⎠ 2 ⎝ b⎠

Ay = F1 ⎜

0

Ay + By − F1 − F2 − w b = 0

By = F1 + F2 + w b − Ay 689



Chapter 7

x1 = 0 , 0.01a .. a 1

−F 1 − V 1 ( x) = 0

V 1 ( x) = −F 1

F 1 x + M1 ( x) = 0

M1 ( x) = −F1 x

kip 1

kip⋅ ft

x2 = a , 1.01a .. a + b 1

V 2 ( x) = ⎡⎣−F1 − w( x − a) + A y⎤⎦

−F 1 − w( x − a) + Ay − V 2 ( x) = 0

⎛ x − a ⎞ + M ( x) = ⎟ 2 ⎝ 2 ⎠

F 1 x − Ay( x − a) + w( x − a) ⎜

kip

0

⎡

( x − a)

⎣

2

M2 ( x) = ⎢−F 1 x + Ay( x − a) − w

2⎤

⎥ 1 ⎦ kip⋅ ft

x3 = a + b , 1.01( a + b) .. a + b + c 1

V 3 ( x) − F2 = 0

V 3 ( x) = F 2

−M3 ( x) − F 2 ( a + b + c − x) = 0

M3 ( x) = −F2 ( a + b + c − x)

kip 1

kip⋅ ft

Force (kip)

100

V1( x1) V2( x2) V3( x3)

50

0

50

100

0

10

20

30

40

50

60


Distance (ft)

690



Chapter 7

200

Moment (kip-ft)

M1( x1)

0

M2( x2) M3( x3)

200

400

0

10

20

30

40

50

60


Distance (ft)

Problem 7-63 Express the x, y, z components of internal loading in the rod at the specific value for y, where 0 < y< a Units Used: 3

kip = 10 lb Given:

y = 2.5 ft

w = 800

lb F = 1500 lb ft

a = 4 ft

b = 2 ft

Solution:

In general we have

⎡ F ⎤ ⎥ V = ⎢ 0 ⎢ ⎥ ⎣w( a − y) ⎦

691


.


⎡⎢ w( a − y) ⎛ a − ⎜ 2 ⎝ ⎢ M = ⎢ −F b ⎢ −F( a − y) ⎣

Chapter 7

y⎞⎤

⎟⎥ ⎠⎥ ⎥ ⎥ ⎦

For these values we have

⎛ 1500.00 ⎞ V = ⎜ 0.00 ⎟ lb ⎜ ⎟ ⎝ 1200.00 ⎠

⎛ 900.00 ⎞ M = ⎜ −3000.00 ⎟ lb⋅ ft ⎜ ⎟ ⎝ −2250.00 ⎠

Problem 7-64 Determine the normal force, shear force, and moment in the curved rod as a function of θ. Given: c = 3 d = 4

Solution: For

0 ≤ θ ≤ π

ΣF x = 0;

ΣF y = 0;

N−

⎛ d ⎞ P cos ( θ ) − ⎛ c ⎞ P sin ( θ ) = ⎜ 2 2⎟ ⎜ 2 2⎟ ⎝ c +d ⎠ ⎝ c +d ⎠

N=

⎛ P ⎞ ( d cos ( θ ) + c sin ( θ ) ) ⎜ 2 2⎟ ⎝ c +d ⎠

V−

⎛ d ⎞ P sin ( θ ) + ⎛ c ⎞ P cos ( θ ) = ⎜ 2 2⎟ ⎜ 2 2⎟ ⎝ c +d ⎠ ⎝ c +d ⎠

V=

⎛ P ⎞ ( d sin ( θ ) − c cos ( θ ) ) ⎜ 2 2⎟ ⎝ c +d ⎠

0

0

692



ΣM = 0;

Chapter 7

⎛ −d ⎞ P( r − r cos ( θ ) ) + ⎛ c ⎞ P r sin ( θ ) + M = ⎜ 2 2⎟ ⎜ 2 2⎟ ⎝ c +d ⎠ ⎝ c +d ⎠ M=

0

⎛ P r ⎞ ( d − d cos ( θ ) − c sin ( θ ) ) ⎜ 2 2⎟ ⎝ c +d ⎠

Problem 7-65 The quarter circular rod lies in the horizontal plane and supports a vertical force P at its end. Determine the magnitudes of the components of the internal shear force, moment, and torque acting in the rod as a function of the angle θ.

Solution: ΣF z = 0;

V= P

ΣMx = 0;

M + P r cos ( θ ) = 0 M = −P r cos ( θ ) M = P r cos ( θ )

ΣMy = 0;

T + P r( l − sin ( θ ) ) = 0 T = −P r ( l − sin ( θ ) T = P r ( 1 − sin ( θ )

693



Chapter 7

Problem 7-66 Draw the shear and moment diagrams for the beam. Given: MB = 800 lb⋅ ft a = 5 ft F = 100 lb

b = 5 ft

Solution: V 1 ( x) = F

x2 = a , 1.01a .. a + b

V 2 ( x) = F

Force (lb)

x1 = 0 , 0.01a .. a

1

M1 ( x) = ⎡⎣−F ( a + b − x) − MB⎤⎦

lb 1

M2 ( x) = −F( a + b − x)

lb

1

lb⋅ ft

1

lb⋅ ft

V1( x1) 100 V2( x2)

99.9

99.8

0

2

4

6

8

10

x1 x2 , ft ft

Distance (ft)

Moment (lb-ft)

0

M1( x1) M2( x2)

1000

2000

0

2

4

6

8

10

x1 x2 , ft ft

Distane (ft)

694



Chapter 7

Problem 7-67 Draw the shear and moment diagrams for the beam. kN = 103 N


kN

F = 10 kN

m

L = 6m Solution: V ( x) = [ w( L − x) + F ]

⎡

( L − x)

⎣

2

M ( x) = ⎢−w

2

1

kN

⎤

− F ( L − x)⎥

1

⎦ kN⋅ m

Force (kN)

40

V( x) 20

0

0

1

2

3

4

5

6

x

Distance (m)

Moment (kN-m)

0 50

M( x) 100 150

0

1

2

3

4

5

6

x

Distance (m)

695



Chapter 7


kN = 10 N

Units Used: Given:

F = 7 kN

Solution: Given

M = 12 kN⋅ m A = 1N

Guesses A+B−F=0

a = 2m

b = 2m

c = 4m

B = 1N

⎛A⎞ ⎜ ⎟ = Find ( A , B) ⎝B ⎠

−F a − M + B ( a + b + c) = 0

x1 = 0 , 0.01a .. a

x2 = a , 1.01a .. a + b

x3 = a + b , 1.01( a + b) .. a + b + c

V 1 ( x1 ) = A

V 2 ( x2 ) = ( A − F)

V 3 ( x3 ) = −B

M1 ( x1 ) =

1

kN A x1

kN⋅ m

M2 ( x2 ) =

1

kN

A x2 − F( x2 − a) kN⋅ m

1

kN

M3 ( x3 ) = B ( a + b + c − x3 )

1

kN⋅ m

Force (kN)

5

V1( x1) V2( x2) V3( x3)

0

5

0

1

2

3

4

5

6

7

x1 , x2 , x3

Distance (m)

696


8


Chapter 7

Moment (kN-m)

15

M1( x1)

10

M2( x2)

5

M3( x3)

0

5

0

1

2

3

4

5

6

7

8

x1 , x2 , x3

Distance (m)


kN = 10 N Given: a = 2m

b = 4m

w = 1.5

kN m

Solution:

⎛ b − a⎞ − A b = ⎟ y ⎝ 2 ⎠

w( b − a) ⎜

x1 = 0 , 0.01a .. a

V 1 ( x) = Ay

x2 = b − a , 1.01( b − a) .. b

Ay =

0 1

kN

w ( b − a) 2b

2

Ay = 0.75 kN

M1 ( x) = A y x

V 2 ( x) = ⎡⎣A y − w( x − a)⎤⎦

1

kN⋅ m

2 ⎡ ( x − a) ⎤ 1 ⎢ ⎥ kN M2 ( x) = Ay x − w 2 ⎣ ⎦ kN⋅ m

1

697



Chapter 7

Force (kN)

2

V1( x1) 0 V2( x2)

2

4

0

0.5

1

1.5

2

2.5

3

2.5

3

3.5

4

x1 , x2

Distance (m)

Moment (kN-m)

2

M1( x1) 1 M2( x2) 0

1

0

0.5

1

1.5

2

3.5

4

x1 , x2

Distance (m)

Problem 7-70 Draw the shear and moment diagrams for the beam. Given: lb ft

w = 30

MC = 180 lb⋅ ft

a = 9 ft

b = 4.5 ft

Solution: − Ay a +

Ay =

1 2

wa 6

⎛ a⎞ − M = ⎟ C ⎝ 3⎠

w a⎜

−

0

MC a 698



Ay + By − By =

wa 2

1

Chapter 7

wa = 0

2

− Ay

x1 = 0 , 0.01a .. a 2 ⎛ wx ⎞ 1 ⎜ Ay − ⎟ 2 a ⎠ lb ⎝

V 1 ( x) =

M1 ( x) =

3 ⎛ wx ⎞ 1 ⎜ Ay x − ⎟ 6 a ⎠ lb⋅ ft ⎝

x2 = a , 1.01a .. a + b

⎛A + B − wa⎞ 1 ⎜ y y 2 ⎟ ⎝ ⎠ lb

V 2 ( x) =

⎡ ⎣

M2 ( x) = ⎢Ay x + B y( x − a) −

wa⎛ 2

⎜x − ⎝

2 a ⎞⎤ 1 3

⎟⎥ ⎠⎦ lb⋅ ft

Force (lb)

100

V1( x1) V2( x2)

0

100

200

0

2

4

6

8

10

12

x1 x2 , ft ft

Distance (ft)

699



Chapter 7

Moment (lb-ft)

100

M1( x1) M2( x2)

0

100

200

0

2

4

6

8

10

12

x1 x2 , ft ft

Diastance (ft)


kip = 10 lb

Given:

w1 = 30

lb ft

w2 = 120

lb ft

L = 12 ft Solution:

⎛ L⎞ + ⎟ ⎝2⎠

w1 L⎜

(w2 − w1)L⎛⎜ 3 ⎞⎟ − Ay L = 0 2 L

1

⎝ ⎠

⎛ L ⎞ + ⎛⎜ w2 − w1 ⎞⎟ ⎛ L ⎞ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠⎝ 3 ⎠

Ay = w1 ⎜

⎡ ⎣

⎦

⎡

x

⎣

2

M ( x) = ⎢A y x − w1

2⎤

x 1 w2 − w1 ) ⎥ ( 2 L lb 1

V ( x) = ⎢Ay − w1 x −

2

− ( w2 − w1 )

x ⎤ 1 ⎥ 6 L⎦ kip⋅ ft 3

700



Chapter 7

Force (lb)

500

0

V( x )

500 0

2

4

6

8

10

12

x ft

Distance (ft)

Moment (kip-ft)

2

M( x) 1

0

0

2

4

6

8

10

12

x ft

Distance (ft)

Problem 7-72 Draw the shear and moment diagrams for the shaft. The support at A is a journal bearing and at B it is a thrust bearing. Given: F 1 = 400 lb w = 100

Solution:

lb in

F 2 = 800 lb a = 4 in

b = 12 in

c = 4 in

⎛ b ⎞ − F b + B( b + c) = ⎟ 2 ⎝2⎠

F 1 a − w b⎜

⎛ b2 ⎞ ⎟ + F2 b − F1 a ⎝2⎠ B=

w⎜ 0

B =

b+c

950.00 lb

701



x1 = 0 , 0.01a .. a

Chapter 7

V 1 ( x) = −F 1

x2 = a , 1.01a .. a + b M2 ( x2 )

1

M1 ( x) = −F1 x

lb

V 2 ( x2 ) = ⎡⎣−B + F2 + w( a + b − x2 )⎤⎦

1

lb⋅ in

1

lb

2 ⎡ a + b − x2 ) ⎤ 1 ( ⎥ = ⎢B( a + b + c − x2 ) − F2 ( a + b − x2 ) − w⋅ 2 ⎣ ⎦ lb⋅ in

V 3 ( x3 ) =

x3 = a + b , 1.01( a + b) .. a + b + c

−B lb

M3 ( x3 ) = B ( a + b + c − x3 )

1

lb⋅ in

Force (lb)

2000

V1( x1) V2( x2)

1000

V3( x3)

0

1000

0

5

10

15

20

x1 x2 x3 , , in in in

Distance (in)

Moment (lb-in)

4000

M1( x1) M2( x2) M3( x3)

2000

0

2000

0

5

10

15

x1 x2 x3 , , in in in

Distance (ini)

702


20


Chapter 7


kN = 10 N Given: F 1 = 10 kN

F 2 = 10 kN

M0 = 12 kN⋅ m

a = 2m

b = 2m

c = 2m

d = 2m

Solution: F 1 ( b + c + d) − M0 + F2 d − Ay( a + b + c + d) = 0

Ay =

F 1 ( b + c + d) − M 0 + F 2 d

Ay = 8.50 kN

a+b+c+d

Ay + By − F1 − F2 = 0

By = F1 + F2 − Ay

B y = 11.50 kN

x1 = 0 , 0.01a .. a V 1 ( x) = Ay

1

M1 ( x) = A y x

kN

1

kN⋅ m

x2 = a , 1.01a .. a + b V 2 ( x) = ( A y − F1 )

1

kN

M2 ( x) = ⎡⎣Ay x − F 1 ( x − a)⎤⎦

1

kN⋅ m

x3 = a + b , 1.01( a + b) .. a + b + c V 3 ( x) = ( A y − F1 )

1

kN

M3 ( x) = ⎡⎣Ay x − F 1 ( x − a) + M0⎤⎦

1

kN⋅ m

x4 = a + b + c , 1.01( a + b + c) .. a + b + c + d V 4 ( x) = −B y

1

kN

M4 ( x) = By( a + b + c + d − x)

1

kN⋅ m

703



Chapter 7

10

Force (kN)

V1( x1)

5

V2( x2) 0 V3( x3) 5

V4( x4)

10 15

0

1

2

3

4

5

6

7

5

6

7

8

x1 , x2 , x3 , x4

Distance (m)

Moment (kN-m)

30

M1( x1) M2( x2)

20

M3( x3) M4( x4)10

0

0

1

2

3

4

8

x1 , x2 , x3 , x4

Distance (m)

Problem 7-74 Draw the shear and moment diagrams for the shaft. The support at A is a journal bearing and at B it is a thrust bearing.

704



Chapter 7

Given: F = 200 lb

w = 100

lb

M = 300 lb⋅ ft

ft

Solution: F ( a + b) − A b + w b⎛⎜

F( a + b) +

b⎞

⎟−M= ⎝ 2⎠

x1 = 0 , 0.01a .. a

a = 1 ft

A =

0

V 1 ( x1 ) = −F

⎛ w b2 ⎞ ⎜ ⎟−M ⎝ 2 ⎠

M1 ( x1 ) = −F x1

1

lb

M2 ( x2 )

c = 1 ft

A = 375.00 lb

b

V 2 ( x2 ) = ⎡⎣−F + A − w( x2 − a)⎤⎦

x2 = a , 1.01a .. a + b

b = 4 ft

1

lb⋅ ft

1

lb

⎡ (x2 − a)2⎥⎤ 1 ⎢ = −F x2 + A ( x2 − a) − w 2 ⎣ ⎦ lb⋅ ft V 3 ( x3 ) = 0

x3 = a + b , 1.01( a + b) .. a + b + c

1

lb

M3 ( x3 ) = −M

1

lb⋅ ft

Force (lb)

200

V1( x1) V2( x2) V3( x3)

0

200

400

0

1

2

3

4

5


Distance (ft)

705


6


Chapter 7

0

Moment (lb-ft)

M1( x1)

100

M2( x2) M3( x3)

200

300

0

1

2

3

4

5


Distance (ft)


kN = 10 N Given:

F = 8 kN c = 2m

M = 20 kN⋅ m

w = 15

kN m

a = 2m

b = 1m

d = 3m

Solution:

⎛ d⎞ = ⎟ ⎝2⎠

− A( a + b + c) − M + F c − w d⎜

A =

0

x1 = 0 , 0.01a .. a

V 1 ( x1 ) = A

x2 = a , 1.01a .. a + b

V 2 ( x2 ) = A

x3 = a + b , 1.01( a + b) .. a + b + c

⎛ d2 ⎞ ⎟ ⎝2⎠

F c − M − w⎜

1

kN 1

kN

A = −14.30 kN

a+b+c M1 ( x1 ) = A x1

1

kN⋅ m

M2 ( x2 ) = ( A x2 + M)

V 3 ( x3 ) = ( A − F)

1

kN⋅ m

1

kN

M3 ( x3 ) = ⎡⎣A x3 + M − F ( x3 − a − b)⎤⎦

1

kN⋅ m

706


6


Chapter 7

x4 = a + b + c , 1.01( a + b + c) .. a + b + c + d V 4 ( x4 ) = w( a + b + c + d − x4 )

M4 ( x4 ) = −w

1

kN

(a + b + c + d − x4)2

1

2

kN⋅ m

60

Force (kN)

V1( x1)

40

V2( x2) 20 V3( x3) 0

V4( x4)

20 40

0

1

2

3

4

5

6

7

8

5

6

7

8

x1 , x2 , x3 , x4

Distance (m)

Moment (kN-m)

0

M1( x1) 20 M2( x2) M3( x3) 40 M4( x4) 60

80

0

1

2

3

4

x1 , x2 , x3 , x4

Distance (m)

707



Chapter 7

Problem 7-76 Draw the shear and moment diagrams for the shaft. The support at A is a thrust bearing and at B it is a journal bearing. Units Used: 3

kN = 10 N Given:

w = 2

kN m

F = 4 kN

⎛ a⎞ = ⎟ ⎝2⎠

B ( a + b ) − F a − w a⎜

Solution:

a = 0.8 m

x2 = a , 1.01a .. a + b

B =

0

a+b

A = wa + F − B

V 1 ( x1 ) = ( A − w x1 ) V 2 ( x2 ) = −B

⎛ a2 ⎞ ⎟ ⎝2⎠

F a + w⎜

A + B − wa − F = 0 x1 = 0 , 0.01a .. a

b = 0.2 m

M1 ( x1 )

1

kN

B = 3.84 kN A = 1.76 kN

⎡ ⎛ x1 2 ⎞⎤ 1 ⎢ ⎟⎥ = A x1 − w⎜ ⎣ ⎝ 2 ⎠⎦ kN⋅ m

M2 ( x2 ) = B ( a + b − x2 )

1

kN

1

kN⋅ m

Force (kN)

5

V1( x1) V2( x2)

0

5

0

0.2

0.4

0.6

0.8

x1 , x2

Distance (m)

Moment (kN-m)

1

M1( x1) 0.5 M2( x2)

0

0.5

0

0.2

0.4

0.6

0.8

x1 , x2

Distance (m)



Chapter 7

Problem 7-77 Draw the shear and moment diagrams for the beam. Given:

w = 20

lb ft

MB = 160 lb⋅ ft

a = 20 ft

b = 20 ft

Solution:

⎛ a⎞ −w a⎜ ⎟ − MB + Cy( a + b) = ⎝ 2⎠

0

Ay − w a + Cy = 0

⎛ a2 ⎞ 1 ⎞ Cy = ⎜ w + MB⎟ ⎛⎜ ⎟ ⎝ 2 ⎠⎝ a + b ⎠

Cy = 104.00 lb

Ay = w a − Cy

Ay = 296.00 lb

x1 = 0 , 0.01a .. a V 1 ( x) = ( A y − w x)

2 ⎛ x ⎞ 1 ⎜ M1 ( x) = Ay x − w ⎟ 2 ⎠ lb⋅ ft ⎝

1

lb

x2 = a , 1.01a .. a + b V 2 ( x) = −Cy

1

M2 ( x) = Cy( a + b − x)

lb

1

lb⋅ ft

Force (lb)

400

V1( x1) 200 V2( x2)

0

200

0

5

10

15

20

25

30

35

40

x1 x2 , ft ft

Distance (ft)

709



Chapter 7

Moment (lb-ft)

3000

M1( x1)2000 M2( x2) 1000

0

0

10

20

30

40

x1 x2 , ft ft

Distance (ft)

Problem 7-78 The beam will fail when the maximum moment is Mmax or the maximum shear is Vmax. Determine the largest distributed load w the beam will support. 3

kip = 10 lb

Units used: Given:

Mmax = 30 kip⋅ ft

Solution:

−A b + w

kip ft

w = 1

Set

V max = 8 kip

b ⎞ + b⎟ + w b = ⎜ 2⎝3 2 ⎠ a 2

b = 6 ft

and then scale the answer at the end

a⎛a

A + B − wb − w

a = 6 ft

w A =

0

b ⎞ + b⎟ + w ⎜ 2⎝3 2 ⎠ b

⎛ ⎝

B = w⎜ b +

=0

2

a⎛a

a⎞

⎟−A

2⎠

A = 7.00 kip

B = 2.00 kip

Shear limit - check critical points to the left and right of A and at B

⎛ ⎝

V big = max ⎜ B , w

wshear =

⎛ Vmax ⎞ ⎜ ⎟w ⎝ Vbig ⎠

a 2

, w

a 2

−A

⎞ ⎟ ⎠

wshear = 2.00

V big = 4.00 kip

kip ft

710



Chapter 7

Moment limit - check critical points at A and betwen A and B B ⎛ a⎞⎛ a ⎞ MA = −w⎜ ⎟ ⎜ ⎟ x = w ⎝ 2 ⎠⎝ 3 ⎠ Mbig = max ( MA , MAB wmoment =

MAB

)

⎛ x2 ⎞ = B x − w⎜ ⎟ ⎝2⎠

⎛ MA ⎞ ⎛ −6.00 ⎞ ⎜ ⎟=⎜ ⎟ kip⋅ ft ⎝ MAB ⎠ ⎝ 2.00 ⎠

Mbig = 6.00 kip⋅ ft

⎛ Mmax ⎞ ⎜ ⎟w ⎝ Mbig ⎠

wmoment = 5.00

kip ft

wans = min ( wshear , wmoment)

Choose the critical case

wans = 2.00

kip ft

Problem 7-79 The beam consists of two segments pin connected at B. Draw the shear and moment diagrams for the beam.

Given: F = 700 lb

w = 150

lb ft

Solution:

⎛ c ⎞ − M + Cc = ⎟ ⎝2⎠

−w c⎜

B + C − wc = 0 x1 = 0 , 0.01a .. a x2 = a , 1.01a .. a + b

M = 800 lb ft

⎛ c2 ⎞ ⎟+M ⎝2⎠

a = 8 ft

b = 4 ft

c = 6 ft

w⎜ 0

C =

C = 583.33 lb

c

B = wc − C V 1 ( x1 ) = ( B + F ) V 2 ( x2 ) = B

x3 = a + b , 1.01( a + b) .. a + b + c M3 ( x3 )

B = 316.67 lb M1 ( x1 ) = ⎡⎣−F( a − x1 ) − B ( a + b − x1 )⎤⎦

1

lb

M2 ( x2 ) = −B ( a + b − x2 )

1

lb

V 3 ( x3 ) = ⎡⎣−C + w( a + b + c − x3 )⎤⎦

1

lb⋅ ft 1

lb

2 ⎡ ⎤ 1 a + b + c − x3 ) ( = ⎢C( a + b + c − x3 ) − w − M⎥ 2 ⎣ ⎦ lb⋅ ft

711


1

lb⋅ ft


Chapter 7

Force (lb)

1500

V1( x1)

1000

V2( x2)

500

V3( x3)

0 500 1000

0

2

4

6

8

10

12

14

16

18

20


Distance (ft)

Moment (lb-ft)

5000

M1( x1)

0

M2( x2) M3( x3)

5000

1 .10

4

0

2

4

6

8

10

12

14

16

18


Distance (ft)

Problem 7-80 The beam consists of three segments pin connected at B and E. Draw the shear and moment diagrams for the beam. 3

Units Used:

kN = 10 N

Given:

MA = 8 kN⋅ m

F = 15 kN

w = 3

c = 2m

d = 2m

e = 2m

kN m

a = 3m

b = 2m

f = 4m

712


20


Guesses

Chapter 7

Ay = 1 N

By = 1 N

Cy = 1 N

Dy = 1 N

Ey = 1 N

Fy = 1 N

Given Ay + Cy + Dy + Fy − F − w f = 0

F b − MA − Ay( a + b) = 0

−w f⎛⎜

M A + F a + B y ( a + b) = 0

f⎞

⎟ + Fy f = ⎝2⎠

0

B y + Cy + Dy + E y = 0

−B y c + Dy d + E y( d + e) = 0

⎛⎜ Ay ⎞⎟ ⎜ By ⎟ ⎜ ⎟ ⎜ Cy ⎟ = Find ( A , B , C , D , E , F ) y y y y y y ⎜ Dy ⎟ ⎜ ⎟ ⎜ Ey ⎟ ⎜ Fy ⎟ ⎝ ⎠

⎛⎜ Ay ⎞⎟ ⎛ 4.40 ⎞ ⎜ By ⎟ ⎜ −10.60 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ Cy ⎟ = ⎜ 15.20 ⎟ kN ⎜ Dy ⎟ ⎜ 1.40 ⎟ ⎜ ⎟ ⎜ −6.00 ⎟ ⎟ ⎜ Ey ⎟ ⎜ ⎜ Fy ⎟ ⎝ 6.00 ⎠ ⎝ ⎠

x1 = 0 , 0.01a .. a M1 ( x) = ( Ay x + MA)

1

V 1 ( x) = Ay

kN x2 = a , 1.01a .. a + b V 2 ( x) = ( A y − F)

1

kN

1

kN⋅ m

M2 ( x) = ⎡⎣Ay x + MA − F ( x − a)⎤⎦

1

kN⋅ m

x3 = a + b , 1.01( a + b) .. a + b + c V 3 ( x) = B y

1

M3 ( x) = By( x − a − b)

kN

1

kN⋅ m

x4 = a + b + c , 1.01( a + b + c) .. a + b + c + d V 4 ( x) = ( By + Cy)

1

kN

M4 ( x) = ⎡⎣B y( x − a − b) + Cy( x − a − b − c)⎤⎦

1

kN⋅ m

x5 = a + b + c + d , 1.01( a + b + c + d) .. a + b + c + d + e V 5 ( x) = −E y

1

kN

M5 ( x) = Ey( a + b + c + d + e − x)

1

kN⋅ m

x6 = a + b + c + d + e , 1.01( a + b + c + d + e) .. a + b + c + d + e + f 713



Chapter 7

V 6 ( x) = ⎡⎣−Fy + w( a + b + c + d + e + f − x)⎤⎦

1

kN

2 ⎡ ( a + b + c + d + e + f − x) ⎤ 1 ⎢ ⎥ M6 ( x) = F y( a + b + c + d + e + f − x) − w 2 ⎣ ⎦ kN⋅ m

10

V1( x1) 5

Force (kN)

V2( x2) V3( x3) 0 V4( x4) V5( x5) 5 V6( x6) 10

15

0

2

4

6

8

10

12

14

x1 , x2 , x3 , x4 , x5 , x6

Distance (m)

714



Chapter 7

30

M1( x1) 20

Moment (kN-m)

M2( x2) M3( x3)

10

M4( x4) 0 M5( x5) M6( x6)

10

20

30

0

2

4

6

8

10

12

14

x1 , x2 , x3 , x4 , x5 , x6

Distance (m)


Solutions: Support Reactions: ΣMx = 0;

⎛ L ⎞ − w0 L ⎛ 4 L ⎞ = ⎟ ⎜ ⎟ 2 ⎝ 3 ⎠ ⎝2⎠

B y L − w0 L⎜

0

By =

7 w0 L 6

715



+

↑

Σ F y = 0;

Ay +

Chapter 7

⎛ 7w0 L ⎞ ⎛ w0 L ⎞ ⎜ ⎟ − w0 L − ⎜ ⎟= ⎝ 6 ⎠ ⎝ 2 ⎠

0

Ay =

w0 L 3

716



Chapter 7


kip = 10 lb Given: F = 2000 lb

a = 9 ft

lb ft

b = 9 ft

w = 500 Solution:

⎛ 2b ⎞ ⎛ a⎞ w b⎜ ⎟ + F b + w a⎜ b + ⎟ − A y ( a + b ) = 2 ⎝3⎠ ⎝ 2⎠ 1

Ay + By − F − w a − Ay = 5.13 kip

1 2

0

wb = 0

⎛ w b2 ⎞ ⎜ ⎟ + F b + w a⎛⎜b + 3 ⎠ ⎝ ⎝ Ay =

a⎞

a+b

By = w a +

1 2

⎟

2⎠

w b − Ay + F

B y = 3.63 kip

x1 = 0 , 0.01a .. a

717



V 1 ( x) = ( A y − w x)

Chapter 7

x M1 ( x) = ⎡⎢Ay x − w x⎛⎜ ⎞⎟⎥⎤ 2

1

⎣

kip

1

⎝ ⎠⎦ kip⋅ ft

x2 = a , 1.01a .. a + b V 2 ( x) = ⎡⎢−By +

⎣

1 2

w⎛⎜

⎝

a + b − x⎞ b

M2 ( x) = ⎡⎢B y( a + b − x) −

⎣

⎤ 1 ⎟ ( a + b − x)⎥ ⎠ ⎦ kip

1 2

w⎛⎜

⎝

a + b − x⎞ b

⎛ a + b − x ⎞⎤ 1 ⎟ ( a + b − x) ⎜ 3 ⎟⎥ ⎝ ⎠⎦ kip⋅ ft ⎠

Force (kip)

10

V1( x1) 5 V2( x2) 0

5

0

5

10

15

x1 x2 , ft ft

Distance (ft)

Moment (kip-ft)

30

M1( x1)

20

M2( x2)

10 0 10

0

5

10

15

x1 x2 , ft ft

Distance (ft)

718



Chapter 7



kN m

a = 3m

b = 3m

Solution:

wb

− Ay( a + b) + w

Ay + By −

2

⎛ w b ⎞ ⎛ 2b ⎞ ⎜ ⎝

⎟⎜ ⎠⎝

2

3

⎛ w a ⎞⎛

⎟+⎜ ⎠ ⎝

2

a⎞

⎟ ⎜b + 3 ⎟ = ⎠⎝ ⎠

0

( a + b) = 0

+

3

Ay = By =

2

⎛ w a ⎞ ⎛b + a ⎞ ⎜ 2 ⎟⎜ 3 ⎟ ⎝ ⎠⎝ ⎠ a+b

w 2

( a + b) − A y

x1 = 0 , 0.01a .. a

⎡ ⎣

V 1 ( x) = ⎢A y −

1 2

⎛ x ⎞ x⎤ 1 ⎟⎥ ⎝ a ⎠ ⎦ kN

⎡ ⎣

w⎜

M1 ( x) = ⎢Ay x −

1 2

⎛ x ⎞ x⎛ x ⎞⎤ 1 ⎟ ⎜ ⎟⎥ ⎝ a ⎠ ⎝ 3 ⎠⎦ kN⋅ m

w⎜

x2 = a , 1.01a .. a + b

⎡ ⎣

V 2 ( x) = ⎢−By +

1 2

⎛ a + b − x ⎞ ( a + b − x)⎤ 1 ⎟ ⎥ ⎝ b ⎠ ⎦ kN

w⎜

⎡ ⎣

M2 ( x) = ⎢B y( a + b − x) −

1 2

⎛ a + b − x ⎞ ( a + b − x) ⎛ a + b − x ⎞⎤ 1 ⎟ ⎜ 3 ⎟⎥ ⎝ ⎠⎦ kN⋅ m ⎝ b ⎠

w⎜

Force (kN)

5

V1( x1) V2( x2)

0

5

0

1

2

3

4

5

6

x1 , x2

Distance (m)

719



Chapter 7

Moment (kN-m)

10

M1( x1) 5 M2( x2) 0

5

0

1

2

3

4

5

6

x1 , x2

Distance (m)


kip = 10 lb Given: w = 100 Guesses: Given

lb ft

M0 = 9 kip⋅ ft

Ay = 1 lb

1 2

⎛ Ay ⎞ ⎜ ⎟ = Find ( Ay , By) ⎝ By ⎠

b = 6 ft

c = 4 ft

B y = 1 lb

−M0 + By( a + b) − Ay + By −

a = 6 ft

1 2

⎛ 2 a ⎞ − 1 w b⎛ a + ⎟ ⎜ ⎝3⎠ 2 ⎝

w a⎜

b⎞

⎟=

0

M1 ( x) = ⎢Ay x −

1

3⎠

w( a + b) = 0

⎛ Ay ⎞ ⎛ −0.45 ⎞ ⎜ ⎟=⎜ ⎟ kip ⎝ By ⎠ ⎝ 1.05 ⎠

x1 = 0 , 0.01a .. a

⎡ ⎣

V 1 ( x) = ⎢A y −

1 2

⎛ x ⎞ x⎤ 1 ⎟⎥ ⎝ a ⎠ ⎦ lb

w⎜

⎡ ⎣

2

⎛ x ⎞ x⎛ x ⎞⎤ 1 ⎟ ⎜ ⎟⎥ ⎝ a ⎠ ⎝ 3 ⎠⎦ kip⋅ ft

w⎜

x2 = a , 1.01a .. a + b

720



V 2 ( x) = ⎡⎢−By + w⎛⎜ 2 ⎝ ⎣ 1

Chapter 7

a + b − x⎞ b

⎤1 ⎟ ( a + b − x)⎥ ⎠ ⎦ lb

M2 ( x) = ⎡⎢B y( a + b − x) − M0 −

⎣

1 2

w⎛⎜

⎝

a + b − x⎞

⎛ a + b − x ⎞⎤ 1 ⎟ ( a + b − x) ⎜ 3 ⎟⎥ ⎝ ⎠⎦ kip⋅ ft ⎠

b

x3 = a + b , 1.01( a + b) .. a + b + c V 3 ( x) = 0

M3 ( x) = −M0

1

kip⋅ ft

V1( x1) V2( x2)

500

V3( x3) 1000

1500

0

2

4

6

8

10

12

14

16


Distance (ft) 0

Moment (kip-ft)

Force (lb)

0

M1( x1) M2( x2) M3( x3)

5

10

0

5

10

15


Distance (ft)

721



Chapter 7


Solution:

722



Chapter 7


kN = 10 N Given:

kN m

w = 2

a = 3m

b = 3m

Solution: x1 = 0 , 0.01a .. a V 1 ( x) =

⎡ 1 w b + 1 w⎛ a − x ⎞ ( a − x)⎤ 1 ⎢2 ⎜ ⎟ ⎥ 2 ⎝ a ⎠ ⎣ ⎦ kN

M1 ( x) =

⎡−1 w b⎛ 2 b + a − x⎞ − 1 w⎛ a − x ⎞ ( a − x) ⎛ a − x ⎞⎤ 1 ⎢2 ⎜3 ⎟ 2 ⎜ ⎟ ⎜ 3 ⎟⎥ ⎝ ⎠ ⎝ ⎠⎦ kN⋅ m ⎣ ⎝ a ⎠

x2 = a , 1.01a .. a + b V 2 ( x) =

⎡ 1 w b − 1 w⎛ x − a ⎞ ( x − a)⎤ 1 ⎢2 ⎜ ⎟ ⎥ 2 ⎝ a ⎠ ⎣ ⎦ kN

M2 ( x) =

⎡−1 w b⎛a + 2b − x⎞ − 1 w⎛ x − a ⎞ ( x − a) ⎛ x − a ⎞⎤ 1 ⎢2 ⎜ ⎟ 2 ⎜ ⎟ ⎜ 3 ⎟⎥ 3 ⎝ ⎠ ⎝ ⎠⎦ kN⋅ m ⎣ ⎝ b ⎠

Force (kN)

10

V1( x1) 5 V2( x2) 0

5

0

1

2

3

4

5

6

x1 , x2

Distance (m)

723



Chapter 7

Moment (kN-m)

0

M1( x1) M2( x2)

10

20

0

1

2

3

4

5

6

x1 , x2

Distance (m)


kip = 10 lb

Given: w = 5

kip ft

M1 = 15 kip⋅ ft

M2 = 15 kip⋅ ft

a = 6 ft

b = 10 ft

c = 6 ft

Solution:

⎛ a ⎞ ⎛b + ⎟⎜ ⎝ 2 ⎠⎝

M1 − A b − M2 + w⎜

a⎞

⎛ b⎞ ⎛ c ⎞⎛ c ⎞ + w b⎜ ⎟ − w⎜ ⎟ ⎜ ⎟ = ⎟ 3⎠ ⎝2⎠ ⎝ 2 ⎠⎝ 3 ⎠

0

⎛ a + c⎞ = ⎟ ⎝ 2 ⎠

A + B − w b − w⎜

2 2 ⎛ a ⎞ ⎛b + a ⎞ + w⎛⎜ b ⎟⎞ − w⎛⎜ c ⎞⎟ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ 3 ⎠ ⎝2⎠ ⎝6⎠

M1 − M2 + w⎜ A =

b

⎛ ⎝

B = w⎜ b +

a + c⎞ 2

x1 = 0 , 0.01a .. a

⎟−A ⎠

⎛ A ⎞ ⎛ 40.00 ⎞ ⎜ ⎟=⎜ ⎟ kip ⎝ B ⎠ ⎝ 40.00 ⎠ ⎛ ⎝

V 1 ( x) = ⎜ −w

M1p ( x) =

x⎞ x 1 ⎟ a ⎠ 2 kip

⎡⎛ −w x ⎞ x x − M ⎤ 1 ⎢⎜ ⎟ 1⎥ ⎣⎝ a ⎠ 2 3 ⎦ kip⋅ ft 724


0


Chapter 7

x2 = a , 1.01a .. a + b

⎡ ⎣

V 2 ( x) = ⎢A − w

⎡ ⎣

a 2

⎤ 1 ⎦ kip

− w( x − a)⎥

M2p ( x) = ⎢−M1 − w

a⎛

⎜x − 2⎝

2a ⎞ 3

⎛ x − a ⎞⎤ 1 ⎟ + A( x − a) − w( x − a) ⎜ 2 ⎟⎥ ⎠ ⎝ ⎠⎦ kip⋅ ft

x3 = a + b , 1.01( a + b) .. a + b + c

⎛ a + b + c − x⎞⎛ a + b + c − x⎞ 1 ⎟⎜ ⎟ 2 c ⎠ kip ⎝ ⎠⎝

V 3 ( x) = w⎜

⎡ ⎛ a + b + c − x⎞⎛ a + b + c − x⎞⎛ a + b + c − x⎞ − M ⎤ 1 ⎟⎜ ⎟⎜ ⎟ 2⎥ 2 3 c ⎠⎝ ⎠ ⎣ ⎝ ⎠⎝ ⎦ kip⋅ ft

M3p ( x) = ⎢−w⎜

Force (lb)

40

V1( x1) V2( x2) V3( x3)

20

0

20

40

0

5

10

15

20


Distance (ft)

725



Chapter 7

20

Moment (kip-ft)

M1p( x1) M2p( x2) M3p( x3)

0

20

40

60

0

5

10

15

20


Distance (ft)


kip = 10 lb

Given: w1 = 2

kip ft

w2 = 1

kip ft

a = 15 ft

Solution: x = 0 , 0.01a .. a

⎡ ⎣

⎛ ⎝

V ( x) = ⎢w2 x − ⎜ w1

x ⎞ x⎤ 1 ⎟ ⎥ a ⎠ 2⎦ kip

⎡ ⎣

M ( x) = ⎢w2 x

x 2

⎛ ⎝

− ⎜ w1

x ⎞ x x⎤ 1 ⎟ ⎥ a ⎠ 2 3⎦ kip⋅ ft

726



Chapter 7

Force (kip)

4

V( x ) 2

0

0

2

4

6

8

10

12

14

x ft

Distance (ft) Moment (kip-ft)

40

M( x) 20

0

0

2

4

6

8

10

12

14

x ft

Distance (ft)

Problem 7-89 Determine the force P needed to hold the cable in the position shown, i.e., so segment BC remains horizontal. Also, compute the sag yB and the maximum tension in the cable. Units Used: 3

kN = 10 N

727



Chapter 7

Given: a = 4m

F 1 = 4 kN

b = 6m

F 2 = 6 kN

c = 3m d = 2m e = 3m Solution: Initial guesses: yB = 1 m P = 1 kN

TAB = 1 kN TBC = 1 kN TCD = 1 kN TDE = 1 kN

Given

⎛ −a ⎞ T + T = AB BC ⎜ 2 2⎟ + y a B ⎠ ⎝ yB ⎛⎜ ⎞⎟ TAB − F1 = ⎜ a2 + yB2 ⎟ ⎝ ⎠ −TBC +

0

0

c ⎡ ⎤T = CD ⎢ 2 2⎥ + y − e c (B )⎦ ⎣

yB − e ⎡⎢ ⎥⎤ TCD − P = ⎢ c2 + ( yB − e) 2⎥ ⎣ ⎦

0

0

−c ⎡ ⎤T + ⎛ d ⎞T = CD ⎜ DE ⎢ 2 ⎥ 2 2 2⎟ + y − e + e c d (B )⎦ ⎝ ⎠ ⎣

0

⎡⎢ −( yB − e) ⎥⎤ e ⎞T − F = TCD + ⎛ ⎜ ⎟ DE 2 2 2 2 2 ⎢ c + ( yB − e) ⎥ + d e ⎝ ⎠ ⎣ ⎦

0

⎛ yB ⎞ ⎜ ⎟ ⎜ P ⎟ ⎜ TAB ⎟ ⎜ ⎟ = Find ( yB , P , TAB , TBC , TCD , TDE) TBC ⎜ ⎟ ⎜ TCD ⎟ ⎜ ⎟ ⎝ TDE ⎠ 728



Chapter 7

Tmax = max ( TAB , TBC , TCD , TDE)

yB = 3.53 m P = 800.00 N Tmax = 8.17 kN

Problem 7-90 Cable ABCD supports the lamp of mass M1 and the lamp of mass M2. Determine the maximum tension in the cable and the sag of point B. Given: M1 = 10 kg M2 = 15 kg a = 1m b = 3m c = 0.5 m d = 2m Solution: Guesses

Given

yB = 1 m

TAB = 1 N

TBC = 1 N

b ⎛ −a ⎞ T + ⎡ ⎤T = AB BC ⎜ 2 ⎢ 2 2⎟ 2⎥ + y + y − d a b ( ) B B ⎝ ⎠ ⎣ ⎦

TCD = 1 N 0

yB yB − d ⎛⎜ ⎡ ⎞⎟ ⎥⎤ TBC − M1 g = TAB + ⎢ ⎜ a2 + yB2 ⎟ ⎢ b2 + ( yB − d) 2⎥ ⎝ ⎠ ⎣ ⎦ −b ⎡ ⎤T + ⎛ c ⎞T = BC ⎜ CD ⎢ 2 2⎥ 2 2⎟ + y − d + d b c ( ) ⎝ ⎠ B ⎣ ⎦

0

0

⎡⎢ −( yB − d) ⎥⎤ d ⎞T − M g = TBC + ⎛ CD 2 ⎜ ⎟ 2 2 ⎢ b2 + ( yB − d) 2⎥ + d c ⎝ ⎠ ⎣ ⎦

0

729



Chapter 7

⎛ yB ⎞ ⎜ ⎟ ⎜ TAB ⎟ = Find ( y , T , T , T ) B AB BC CD ⎜ TBC ⎟ ⎜ ⎟ ⎝ TCD ⎠ Tmax = max ( TAB , TBC , TCD)

⎛ TAB ⎞ ⎛ 100.163 ⎞ ⎜ ⎟ ⎜ TBC ⎟ = ⎜ 38.524 ⎟ N ⎜ T ⎟ ⎜⎝ 157.243 ⎟⎠ ⎝ CD ⎠ Tmax = 157.2 N

yB = 2.43 m

Problem 7-91 The cable supports the three loads shown. Determine the sags yB and yD of points B and D. Given: a = 4 ft

e = 12 ft

b = 12 ft

f = 14 ft

c = 20 ft

P 1 = 400 lb

d = 15 ft

P 2 = 250 lb

Solution: Guesses

Given

yB = 1 ft

yD = 1 ft

TAB = 1 lb

TBC = 1 lb

TCD = 1 lb

TDE = 1 lb

c ⎛ −b ⎞ T + ⎡ ⎤T = AB ⎢ BC ⎜ 2 ⎟ 2 2 2⎥ + y + f − y b c ( ) B ⎠ B ⎦ ⎝ ⎣

0

yB f − yB ⎛⎜ ⎡ ⎞⎟ ⎤⎥ TAB − ⎢ TBC − P 2 = ⎜ b2 + yB2 ⎟ ⎢ c2 + ( f − yB) 2⎥ ⎝ ⎠ ⎣ ⎦

0

−c d ⎡ ⎤T + ⎡ ⎤T = BC ⎢ CD ⎢ 2 ⎥ 2 2 2⎥ + f − y + f − y c d ( ) ( ) B D ⎣ ⎦ ⎣ ⎦

0

730



Chapter 7

f − yB f − yD ⎡⎢ ⎥⎤ TBC + ⎡⎢ ⎥⎤ TCD − P1 = ⎢ c2 + ( f − yB) 2⎥ ⎢ d2 + ( f − yD) 2⎥ ⎣ ⎦ ⎣ ⎦ −d e ⎡ ⎤T + ⎡ ⎤T = CD ⎢ DE ⎢ 2 ⎥ ⎥ 2 2 2 + f − y + a + y d e ( D) ⎦ ( D) ⎦ ⎣ ⎣

0

a + yD ⎡⎢ −( f − yD) ⎥⎤ ⎡ ⎥⎤ TDE − P2 = TCD + ⎢ ⎢ d2 + ( f − yD) 2⎥ ⎢ e2 + ( a + yD) 2⎥ ⎣ ⎦ ⎣ ⎦

⎛⎜ TAB ⎞⎟ ⎜ TBC ⎟ ⎜ ⎟ ⎜ TCD ⎟ = Find ( T , T , T , T , y , y ) AB BC CD DE B D ⎜ TDE ⎟ ⎜ ⎟ ⎜ yB ⎟ ⎜ yD ⎟ ⎝ ⎠

0

0

⎛ TAB ⎞ ⎛ 675.89 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ TBC ⎟ = ⎜ 566.90 ⎟ lb ⎜ TCD ⎟ ⎜ 603.86 ⎟ ⎜ ⎟ ⎜ 744.44 ⎟ ⎠ ⎝ TDE ⎠ ⎝ ⎛ yB ⎞ ⎛ 8.67 ⎞ ⎜ ⎟ =⎜ ⎟ ft ⎝ yD ⎠ ⎝ 7.04 ⎠

Problem 7-92 The cable supports the three loads shown. Determine the magnitude of P 1 and find the sag yD for the given data. Given: P 2 = 300 lb

c = 20 ft

yB = 8 ft

d = 15 ft

a = 4 ft

e = 12 ft

b = 12 ft

f = 14 ft

Solution: Guesses

P 1 = 1 lb

TAB = 1 lb

TBC = 1 lb

TCD = 1 lb

731



TDE = 1 lb

Chapter 7

yD = 1 ft

Given c ⎛ −b ⎞ T + ⎡ ⎤T = AB ⎢ BC ⎜ 2 ⎟ 2 2 2⎥ + y + f − y b c ( B) ⎦ B ⎠ ⎝ ⎣

0

yB f − yB ⎛⎜ ⎡ ⎞⎟ ⎤⎥ TAB − ⎢ TBC − P 2 = ⎜ b2 + yB2 ⎟ ⎢ c2 + ( f − yB) 2⎥ ⎝ ⎠ ⎣ ⎦

0

−c d ⎡ ⎤T + ⎡ ⎤T = BC CD ⎢ 2 ⎢ 2 2⎥ 2⎥ + f − y + f − y c d ( ) ( ) B D ⎣ ⎦ ⎣ ⎦

0

f − yB f − yD ⎡⎢ ⎥⎤ TBC + ⎡⎢ ⎥⎤ TCD − P1 = ⎢ c2 + ( f − yB) 2⎥ ⎢ d2 + ( f − yD) 2⎥ ⎣ ⎦ ⎣ ⎦ −d e ⎡ ⎤T + ⎡ ⎤T = CD DE ⎢ 2 ⎢ 2 2⎥ 2⎥ + f − y + a + y d e ( ) ( ) D D ⎣ ⎦ ⎣ ⎦

0

a + yD ⎡⎢ −( f − yD) ⎥⎤ ⎡ ⎥⎤ TDE − P2 = TCD + ⎢ ⎢ d2 + ( f − yD) 2⎥ ⎢ e2 + ( a + yD) 2⎥ ⎣ ⎦ ⎣ ⎦

⎛⎜ TAB ⎞⎟ ⎜ TBC ⎟ ⎜ ⎟ ⎜ TCD ⎟ = Find ( T , T , T , T , P , y ) AB BC CD DE 1 D ⎜ TDE ⎟ ⎜ ⎟ ⎜ P1 ⎟ ⎜ yD ⎟ ⎝ ⎠

0

0

⎛ TAB ⎞ ⎛ 983.33 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ TBC ⎟ = ⎜ 854.21 ⎟ lb ⎜ TCD ⎟ ⎜ 916.11 ⎟ ⎜ ⎟ ⎜ 1084.68 ⎟ ⎠ ⎝ TDE ⎠ ⎝ P 1 = 658 lb yD = 6.44 ft

Problem 7-93 The cable supports the loading shown. Determine the distance xB the force at point B acts from A.

732



Chapter 7

Given: P = 40 lb

c = 2 ft

F = 30 lb

d = 3 ft

a = 5 ft

e = 3

b = 8 ft

f = 4

Solution: The initial guesses: TAB = 10 lb

TCD = 30 lb

TBC = 20 lb

xB = 5 ft

Given xB − d ⎛⎜ −xB ⎟⎞ ⎡ ⎤⎥ TAB − ⎢ TBC + P = ⎜ xB2 + a2 ⎟ ⎢ ( xB − d) 2 + b2⎥ ⎝ ⎠ ⎣ ⎦ a b ⎛ ⎞T − ⎡ ⎤T = AB BC ⎜ ⎢ 2 2⎟ 2 2⎥ + a − d + b x x ( ) B B ⎝ ⎠ ⎣ ⎦

0

0

xB − d ⎡⎢ d ⎤⎥ f ⎞T + ⎛ ⎞F = TBC − ⎛ CD ⎜ ⎜ ⎟ ⎟ 2 2 2 2 ⎢ ( xB − d) 2 + b2⎥ ⎝ c +d ⎠ ⎝ e +f ⎠ ⎣ ⎦

0

b ⎡ ⎤T − ⎛ c ⎞T − ⎛ e ⎞F = BC ⎜ CD ⎜ ⎢ 2 2⎥ 2 2⎟ 2 2⎟ − d + b + d + f x c e ( ) ⎝ ⎠ ⎝ ⎠ B ⎣ ⎦

0

⎛ TAB ⎞ ⎜ ⎟ ⎜ TCD ⎟ = Find ( T , T , T , x ) AB CD BC B ⎜ TBC ⎟ ⎜ ⎟ ⎝ xB ⎠

⎛ TAB ⎞ ⎛ 50.90 ⎞ ⎜ ⎟ ⎜ TCD ⎟ = ⎜ 36.70 ⎟ lb ⎜ T ⎟ ⎜⎝ 38.91 ⎟⎠ ⎝ BC ⎠

xB = 4.36 ft

Problem 7-94 The cable supports the loading shown. Determine the magnitude of the horizontal force P.

733



Chapter 7

Given: F = 30 lb

c = 2 ft

xB = 6 ft

d = 3 ft

a = 5 ft

e = 3

b = 8 ft

f = 4

Solution: The initial guesses: TAB = 10 lb

TCD = 30 lb

TBC = 20 lb

P = 10 lb

Given xB − d ⎛⎜ −xB ⎟⎞ ⎡ ⎤⎥ TAB − ⎢ TBC + P = ⎜ xB2 + a2 ⎟ ⎢ ( xB − d) 2 + b2⎥ ⎝ ⎠ ⎣ ⎦ a b ⎛ ⎞T − ⎡ ⎤T = AB BC ⎜ 2 ⎢ 2 2⎟ 2⎥ + x + x − d a b ( ) B B ⎝ ⎠ ⎣ ⎦

0

0

xB − d ⎤⎥ f ⎛ −d ⎞ T + ⎡⎢ ⎞F = TBC + ⎛ CD ⎜ 2 2⎟ ⎜ ⎟ 2 2 ⎢ b2 + ( xB − d) 2⎥ ⎝ c +d ⎠ ⎝ e +f ⎠ ⎣ ⎦

0

b ⎛ −c ⎞ T + ⎡ ⎤T − ⎛ e ⎞F = BC ⎜ ⎜ 2 2 ⎟ CD ⎢ 2 2⎥ 2 2⎟ + x − d + f b e ( ) ⎝ c +d ⎠ ⎝ ⎠ B ⎣ ⎦

0

⎛ TAB ⎞ ⎜ ⎟ T BC ⎜ ⎟ = Find ( T , T , T , P) AB BC CD ⎜ TCD ⎟ ⎜ ⎟ ⎝ P ⎠

⎛ TAB ⎞ ⎛ 70.81 ⎞ ⎜ ⎟ ⎜ TBC ⎟ = ⎜ 48.42 ⎟ lb ⎜ T ⎟ ⎜⎝ 49.28 ⎟⎠ ⎝ CD ⎠

P = 71.40 lb

734



Chapter 7

Problem 7-95 Determine the forces P1 and P2 needed to hold the cable in the position shown, i.e., so segment CD remains horizontal. Also, compute the maximum tension in the cable. Given:

3

kN = 10 N

F = 5 kN

d = 4m

a = 1.5 m

e = 5m

b = 1m

f = 4m

c = 2m Solution: Guesses F AB = 1 kN

F BC = 1 kN

F CD = 1 kN

F DE = 1 kN

P 1 = 1 kN

P 2 = 1 kN

Given

⎛ −c ⎞ F + ⎛ d ⎞ F = ⎜ 2 2 ⎟ AB ⎜ 2 2 ⎟ BC ⎝ a +c ⎠ ⎝ b +d ⎠

0

735



Chapter 7

⎛ a ⎞F − ⎛ b ⎞F − F = ⎜ 2 2 ⎟ AB ⎜ 2 2 ⎟ BC ⎝ a +c ⎠ ⎝ b +d ⎠ ⎛ −d ⎞ F + F = ⎜ 2 2 ⎟ BC CD ⎝ b +d ⎠ ⎛ b ⎞F − P = ⎜ 2 2 ⎟ BC 1 ⎝ b +d ⎠ −F CD +

0

0

0

f ⎡ ⎤F = DE ⎢ 2 2⎥ ⎣ f + ( a + b) ⎦

a+b ⎡ ⎤F − P = DE 2 ⎢ 2 2⎥ + ( a + b ) f ⎣ ⎦

0

0

⎛⎜ FAB ⎟⎞ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FCD ⎟ = Find ( F , F , F , F , P , P ) AB BC CD DE 1 2 ⎜ FDE ⎟ ⎜ ⎟ ⎜ P1 ⎟ ⎜ P2 ⎟ ⎝ ⎠ ⎛ FAB ⎞ ⎛ 12.50 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FBC ⎟ = ⎜ 10.31 ⎟ kN ⎜ FCD ⎟ ⎜ 10.00 ⎟ ⎜ ⎟ ⎜ 11.79 ⎟ ⎠ ⎝ FDE ⎠ ⎝

⎛ P1 ⎞ ⎛ 2.50 ⎞ ⎜ ⎟=⎜ ⎟ kN ⎝ P2 ⎠ ⎝ 6.25 ⎠

Tmax = max ( FAB , F BC , F CD , FDE) F max = max ( F AB , FBC , FCD , F DE)

Tmax = 12.50 kN F max = 12.50 kN

736



Chapter 7

Problem 7-96 The cable supports the loading shown. Determine the distance xB from the wall to point B. Given: W1 = 8 lb W2 = 15 lb a = 5 ft b = 8 ft c = 2 ft d = 3 ft Solution: Guesses TAB = 1 lb

TBC = 1 lb

TCD = 1 lb

xB = 1 ft

Given xB − d ⎛⎜ −xB ⎟⎞ ⎡ ⎤⎥ TAB − ⎢ TBC + W2 = ⎜ a2 + xB2 ⎟ ⎢ b2 + ( xB − d) 2⎥ ⎝ ⎠ ⎣ ⎦ a b ⎛ ⎞T − ⎡ ⎤T = AB ⎢ BC ⎜ 2 ⎟ 2 2 2⎥ + x + x − d a b ( ) B ⎠ B ⎝ ⎣ ⎦ xB − d ⎡⎢ ⎤⎥ ⎛ d ⎞T = TBC − ⎜ CD 2 2⎟ ⎢ b2 + ( xB − d) 2⎥ + d c ⎝ ⎠ ⎣ ⎦

0

0

b ⎡ ⎤T − ⎛ c ⎞T − W = BC ⎜ CD 1 ⎢ 2 2⎥ 2 2⎟ + x − d + d b c ( ) ⎝ ⎠ B ⎣ ⎦

⎛ TAB ⎞ ⎜ ⎟ TBC ⎜ ⎟ = Find ( T , T , T , x ) AB BC CD B ⎜ TCD ⎟ ⎜ ⎟ ⎝ xB ⎠

0

0

⎛ TAB ⎞ ⎛ 15.49 ⎞ ⎜ ⎟ ⎜ TBC ⎟ = ⎜ 10.82 ⎟ lb ⎜ T ⎟ ⎜⎝ 4.09 ⎟⎠ ⎝ CD ⎠

xB = 5.65 ft

Problem 7-97 Determine the maximum uniform loading w, measured in lb/ft, that the cable can support if it is 737



Chapter 7

g capable of sustaining a maximum tension Tmax before it will break.

pp

Given: Tmax = 3000 lb a = 50 ft b = 6 ft Solution: 1 ⌠ ⌠

2

y=

wx ⎮ ⎮ w dx dx = FH ⌡ ⌡ 2 FH

y=

⎛ w ⎞ x2 ⎜ 2F ⎟ ⎝ H⎠

x=

2

a

y=b

2

wa 8b

FH =

⎛ dy ⎞ = tan ( θ ) = w ⎛ a ⎞ = 4b ⎜ ⎟ ⎜ ⎟ max a FH ⎝ 2 ⎠ ⎝ dx ⎠ Tmax =

FH

cos ( θ max)

θ max = atan ⎛⎜

⎟ ⎝a⎠

2

=

wa

(

8 b cos θ max

)

4b ⎞

w =

Tmax8 b cos ( θ max) a

2

θ max = 25.64 deg

w = 51.93

lb ft

Problem 7-98 The cable is subjected to a uniform loading w. Determine the maximum and minimum tension in the cable. Units Used: 3


lb ft

a = 50 ft b = 6 ft

738



Chapter 7

Solution: 2

y=

⎛ a⎞ b= ⎜ ⎟ 2FH ⎝ 2 ⎠

wx

w

2FH

tan ( θ max) =

Tmax =

d dx

y ⎛⎜ x =

⎝

2

FH =

w ⎛ a⎞ = ⎟ ⎜ ⎟ 2⎠ FH ⎝ 2 ⎠

a⎞

FH

wa

2

F H = 13021 lb

8b

⎞ θ max = atan ⎛⎜ ⎟ 2 F ⎝ H⎠ wa

θ max = 25.64 deg

Tmax = 14.44 kip

cos ( θ max)

The minimum tension occurs at

θ = 0 deg

Tmin = FH

Tmin = 13.0 kip

Problem 7-99 The cable is subjected to the triangular loading. If the slope of the cable at A is zero, determine the equation of the curve y = f(x) which defines the cable shape AB, and the maximum tension developed in the cable. Units Used: kip = 103 lb Given: w = 250

lb ft

a = 20 ft b = 30 ft Solution: ⌠ ⌠

1 ⎮ ⎮ y= FH ⎮ ⎮

⌡ ⌡

y=

wx b

dx dx

⎞ ⎛ w x3 ⎜ + c1 x + c2⎟ FH ⎝ 6 b ⎠ 1

739



Chapter 7

Apply boundary conditions y = x = 0 and

d

y = 0,x = 0

3

C1 = C2 = 0

3

wx y= 6FH b

set

y=a

x=b

2

wb FH = 6a Tmax =

Thus

dx

F H = 1.875 kip FH

cos ( θ max)

θ max

wb a= 6FH b

⎛ w b2 ⎞ = atan ⎜ ⎟ ⎝ 2 FH b ⎠

θ max = 63.43 deg

Tmax = 4.19 kip

Problem 7-100 The cable supports a girder which has weight density γ. Determine the tension in the cable at points A, B, and C. Units used: 3

kip = 10 lb Given:

γ = 850

lb ft

a = 40 ft b = 100 ft c = 20 ft

Solution:

y=

1 ⌠ ⌠

⎮ ⎮ γ dx dx FH ⌡ ⌡ 2

y=

γx

2FH

γx d y = FH dx

x1 = 1 ft

Guesses

F H = 1 lb

740



c=

Given

tan ( θ A) =

γ FH

tan ( θ C) =

TA =

γ x1

Chapter 7

2

2 FH

a=

γ

θ A = atan ⎢

⎛ γ x⎞ 1⎟ ⎝ FH ⎠

θ C = atan ⎜

x1

FH

cos ( θ A)

2FH

⎛ x1 ⎞ ⎜ ⎟ = Find ( x1 , FH) ⎝ FH ⎠

(b − x1)2

⎡ γ x − b⎤ ( 1 )⎥ ⎣FH ⎦

(x1 − b)

FH

γ

TB = F H

TC =

F H = 36.46 kip

θ A = −53.79 deg

θ C = 44.00 deg

FH

cos ( θ C)

⎛ TA ⎞ ⎛ 61.71 ⎞ ⎜ ⎟ ⎜ TB ⎟ = ⎜ 36.46 ⎟ kip ⎜ T ⎟ ⎜⎝ 50.68 ⎟⎠ ⎝ C⎠

Problem 7-101 The cable is subjected to the triangular loading. If the slope of the cable at point O is zero, determine the equation of the curve y = f(x) which defines the cable shape OB, and the maximum tension developed in the cable. Units used: kip = 103 lb Given: w = 500

lb ft

b = 8 ft

a = 15 ft

741



Chapter 7

Solution:

⎛⌠ ⎜⎮ y= FH ⎜⎮ ⎝⌡

⎞

⌠ ⎮ ⎮ ⌡

1

wx dx dx⎟ a ⎟

⎠

⎤ ⎡w ⎛ x3 ⎞ ⎢ ⎜ ⎟ + C1 x + C2⎥ y= FH ⎣ a ⎝ 6 ⎠ ⎦ 1

2⎞

1 ⎛ wx d ⎜ y = FH ⎝ 2a dx

At x = 0,

d dx

⎛ C1 ⎞ ⎟+⎜ ⎟ ⎠ ⎝ FH ⎠

y = 0, C1 = 0

At x = 0, y = 0, C2 = 0 3

y=

2

wx d y = 2a FH dx

wx 6a FH

3

At x = a, y = b

b=

wa 6a FH

FH =

1 6

⎛ a2 ⎞ ⎟ ⎝b⎠

w⎜

F H = 2343.75 lb

742



Chapter 7

2

wa d y = tan ( θ max) = 2a FH dx Tmax =

(θ max)

FH

⎛ w a2 ⎞ = atan ⎜ ⎟ ⎝ 2a FH ⎠

(θ max) = 57.99 deg

Tmax = 4.42 kip

cos ( θ max)

Problem 7-102 The cable is subjected to the parabolic loading w = w0(1− (2x/a)2). Determine the equation y = f(x) which defines the cable shape AB and the maximum tension in the cable. Units Used: 3

kip = 10 lb Given:

⎡

w = w0 ⎢1 −

⎣

a = 100 ft

2 ⎛ 2x ⎞ ⎥⎤ ⎜ ⎟ ⎝a⎠⎦

w0 = 150

lb ft

b = 20 ft Solution: y=

1 ⌠ ⌠

⎮ ⎮ w ( x) dx dx FH ⌡ ⌡ ⌠

y=

⎛⎜ 4 x ⎞⎟ + C1 dx ⎮ w0 x − 2⎟ FH ⎜ ⎮ ⎝ 3a ⎠ ⌡ 1 ⎮

3

⎞ ⎛ w0 x2 x4w0 ⎜ ⎟ y= − + C1 x + C2 ⎜ ⎟ 2 2 FH 3a ⎝ ⎠ 1

⎞ ⎛ 4 w0 x dy 1 ⎜ ⎟ = w0 x − + C1 ⎟ 2 dx FH ⎜ 3a ⎝ ⎠ 3

743



dy

x=0

At

dx

At x = 0

=0

y=0

Chapter 7

C1 = 0 C2 = 0

Thus

⎛ w0 x2 x4w0 ⎞ ⎜ ⎟ y= − ⎜ ⎟ 2 2 FH 3a ⎠ ⎝

3 ⎛ 4 w0 x ⎞ ⎜ ⎟ = w0 x − ⎜ ⎟ 2 FH dx 3a ⎠ ⎝

dy

1

x=

At

a

we have

2

⎡⎢w0 ⎛ a ⎞ 2 w0 ⎛ a ⎞ 4⎥⎤ b= ⎜ ⎟ − 2 ⎜2⎟ = FH ⎢ 2 ⎝ 2 ⎠ ⎥ 3a ⎝ ⎠ ⎦ ⎣ 1

tan ( θ max)

⎛⎜ w0 a2 ⎟⎞ ⎟ 48 ⎜ ⎝ FH ⎠ 5

FH =

5 w0 a

2

F H = 7812.50 lb

48b

3 w0 a a⎞ 1 ⎡ a ⎞ 4 w0 ⎛ a ⎞ ⎥⎤ ⎛ ⎛ ⎢ = y⎜ ⎟ = w0 ⎜ ⎟ − = ⎜ ⎟ FH ⎢ ⎝ 2 ⎠ 3 a2 ⎝ 2 ⎠ ⎥ 3 FH dx ⎝ 2 ⎠ ⎣ ⎦

d

⎛ w0 a ⎞ ⎟ ⎝ 3 FH ⎠

θ max = atan ⎜

Tmax =

1

FH

cos ( θ max)

θ max = 32.62 deg

Tmax = 9.28 kip

Problem 7-103 The cable will break when the maximum tension reaches Tmax. Determine the minimum sag h if it supports the uniform distributed load w. Given:

kN = 103 N

Tmax = 10 kN w = 600

N m

a = 25 m Solution: The equation of the cable: y=

1 ⌠ ⌠

⎮ ⎮ w dx dx FH ⌡ ⌡

y=

⎛ w x2 ⎞ ⎜ + C1 x + C2⎟ FH ⎝ 2 ⎠ 1

dy dx

=

1

FH

(w x + C1)

744



Chapter 7

Boundary Conditions: y = 0 at x = 0, then from Eq.[1]

0 =

d

y = 0 at x = 0, then from Eq.[2]

0 =

w ⎞ 2 y = ⎛⎜ ⎟x ⎝ 2 FH ⎠

w

dx Thus,

tan ( θ max) =

Tmax =

dy dx

2FH

cos ( θ max)

Guess

h = 1m

Given

Tmax =

FH

FH 1

FH

(C2)

C2 = 0

(C1)

C1 = 0

⎛ a⎞ h= ⎜ ⎟ 2FH ⎝ 2 ⎠ w

x

2

4 F H + ( w a)

=

wa 2

2

FH +

( w a) 4

2

=

a

wa 2

2

⎛ a⎞ FH = ⎜ ⎟ 2h ⎝ 2 ⎠ w

2

2 FH

cos ( θ max) =

wa

FH

=

1

2

2 2

+1

16h

2

a

16h

2

+1

h = Find ( h)

h = 7.09 m

Problem 7-104 A fiber optic cable is suspended over the poles so that the angle at the supports is θ. Determine the minimum tension in the cable and the sag. The cable has a mass density ρ and the supports are at the same elevation. Given:

θ = 22 deg ρ = 0.9

kg m

a = 30 m g = 9.81

m 2

s Solution:

θ max = θ 745



Chapter 7

w0 = ρ g a ⎞ ⎛ ⎜ 2 ⎟ d y = tan ( θ ) = sinh ⎜ w0 ⎟ dx ⎝ FH ⎠

w0 ⎛⎜ FH =

a⎞

⎟ ⎝2⎠

F H = 336 N

asinh ( tan ( θ ) )

Tmax =

FH

Tmax = 363 N

cos ( θ )

⎛ ⎛w a ⎞ ⎞ ⎜ ⎜ 02⎟ ⎟ FH h = ⎜ cosh ⎜ ⎟ − 1⎟ w0 ⎝ ⎝ FH ⎠ ⎠

h = 2.99 m

Problem 7-105 A cable has a weight density γ and is supported at points that are a distance d apart and at the same elevation. If it has a length L, determine the sag. Given:

γ = 3

lb

d = 500 ft

ft

L = 600 ft

Solution: Guess

Given

h =

F H = 100 lb L 2

−

⎡FH ⎡ γ ⎛ d ⎞⎤⎤ ⎢ sinh ⎢ ⎜ 2 ⎟⎥⎥ = ⎣γ ⎣FH ⎝ ⎠⎦⎦

F H = Find ( FH)

0

⎛1 γ ⎞ ⎞ ⎜ cosh ⎜ 2 F d⎟ − 1⎟ γ ⎝ ⎝ H ⎠ ⎠

FH ⎛

F H = 704.3 lb

h = 146 ft

Problem 7-106 Show that the deflection curve of the cable discussed in Example 7.15 reduces to Eq. (4) in Example 7.14 when the hyperbolic cosine function is expanded in terms of a series and only the 746



Chapter 7

p yp f p y first two terms are retained. (The answer indicates that the catenary may be replaced by a parabola in the analysis of problems in which the sag is small. In this case, the cable weight is assumed to be uniformly distributed along the horizontal.) Solution: 2

cosh ( x) = 1 +

x

2!

+ ..

Substituting into 2 2 ⎞ w0 x2 ⎛ w0 ⎞ ⎞ FH ⎛⎜ w0 x ⎟ y= 1+ + .. − 1 = ⎜cosh ⎜ x⎟ − 1⎟ = 2 w0 ⎝ ⎟ 2 FH ⎝ FH ⎠ ⎠ w0 ⎜⎝ 2FH ⎠

FH ⎛

Using the boundary conditions y = h at

h=

⎛ L⎞ ⎜ ⎟ 2FH ⎝ 2 ⎠ w0

2

FH =

w0 L 8h

x=

L 2

2

We get

y=

4h 2

L

2

x

Problem 7-107 A uniform cord is suspended between two points having the same elevation. Determine the sag-to-span ratio so that the maximum tension in the cord equals the cord's total weight. Solution: s=

y=

FH w0

⎛ w0 ⎞ x⎟ ⎝ FH ⎠

sinh ⎜

FH ⎛

⎛ w0 ⎞ ⎞ ⎜cosh ⎜ x⎟ − 1⎟ w0 ⎝ ⎝ FH ⎠ ⎠

At x =

L 2

⎛ w0 L ⎞ d y = tan ( θ max) = sinh ⎜ ⎟ dx max ⎝ 2FH ⎠ cos ( θ max) =

1

⎛ w0 L ⎞ ⎟ ⎝ 2 FH ⎠

cosh ⎜

747



⎛ w0 L ⎞ ⎟ ⎝ 2 FH ⎠

FH

Tmax =

Chapter 7

w0 2 s = F H cosh ⎜

cos ( θ max)

⎛ w0 L ⎞ ⎛ w0 L ⎞ ⎟ = FH cosh ⎜ ⎟ ⎝ 2 FH ⎠ ⎝ 2FH ⎠

⎛ w0 L ⎞ ⎟= ⎝ 2 FH ⎠

2 F H sinh ⎜

k1 = atanh ( 0.5)

when x =

L

⎛ FH ⎞ ⎟ ⎝ w0 ⎠

h=

⎛ 2 Fh ⎞ ⎟ ⎝ w0 ⎠

h = k2 ⎜

w0 L

k1 = 0.55

y=h

2

tanh ⎜

L = k1 ⎜

W0

(cosh (k1) − 1)

ratio =

2

= k1

2 FH

FH

1

k2 h = 2 k1 L

k2 = cosh ( k1 ) − 1

ratio =

k2 = 0.15

k2 2 k1

ratio = 0.14

Problem 7-108 A cable has a weight denisty γ. If it can span a distance L and has a sag h determine the length of the cable. The ends of the cable are supported from the same elevation. Given:

γ = 2

lb ft

L = 100 ft

h = 12 ft

Solution: From Eq. (5) of Example 7-15 :

⎡⎢⎛ γ L ⎞ 2⎤⎥ ⎜ ⎟ FH ⎢⎝ 2 FH ⎠ ⎥ h= ⎢ 2 ⎥ γ ⎣ ⎦

FH =

1 8

⎛ L2 ⎞ ⎟ ⎝h⎠

γ⎜

F H = 208.33 lb

From Eq. (3) of Example 7-15: l 2

=

⎛ FH ⎞ ⎡ γ ⎛ L ⎞⎤ ⎜ ⎟ sinh ⎢ ⎜ ⎟⎥ ⎝ γ ⎠ ⎣FH ⎝ 2 ⎠⎦

⎛ FH ⎞ ⎛ 1 L ⎞ ⎟ sinh ⎜ γ ⎟ ⎝ γ ⎠ ⎝ 2 FH ⎠

l = 2⎜

l = 104 ft

748



Chapter 7

Problem 7-109 The transmission cable having a weight density γ is strung across the river as shown. Determine the required force that must be applied to the cable at its points of attachment to the towers at B and C. Units Used: 3

kip = 10 lb Given:

γ = 20

lb ft

b = 75 ft

a = 50 ft

c = 10 ft

Solution: From Example 7-15,

y=

FH ⎡ ⎛ γ ⎞ ⎤ ⎢cosh⎜ F x⎟ − 1⎥ γ ⎣ ⎝ H ⎠ ⎦

Guess

Given

dy ⎛ γx ⎞ = sinh ⎜ ⎟ dx ⎝ FH ⎠

F H = 1000 lb c=

At B:

⎛ γa ⎞ ⎞ ⎜ cosh ⎜ − F ⎟ − 1⎟ γ ⎝ ⎝ H⎠ ⎠

FH ⎛

⎛ γa ⎞ ⎟ ⎝ FH ⎠

θ B = atan ⎜ sinh ⎜

⎛ γb ⎞ ⎟ ⎝ FH ⎠

θ C = atan ⎜ sinh ⎜

tan ( θ B) = sinh ⎜ −

tan ( θ C) = sinh ⎜

TB =

FH

cos ( θ B)

TC =

FH

cos ( θ C)

F H = Find ( FH)

F H = 2.53 kip

⎛ ⎝

⎛ −γ a ⎞ ⎞ ⎟⎟ ⎝ FH ⎠⎠

θ B = −22.06 deg

⎛ ⎝

⎛ γ b ⎞⎞ ⎟⎟ ⎝ FH ⎠⎠

θ C = 32.11 deg

⎛ TB ⎞ ⎛ 2.73 ⎞ ⎜ ⎟=⎜ ⎟ kip ⎝ TC ⎠ ⎝ 2.99 ⎠

Problem 7-110 Determine the maximum tension developed in the cable if it is subjected to a uniform load w.

749



Chapter 7

Units Used: MN = 106 N Given: N

w = 600

m

a = 100 m b = 20 m

θ = 10 deg Solution: The Equation of the Cable:

⎛ w x2 ⎞ ⎜ y= ⎮ ⎮ w( x) dx dx = + C1 x + C2⎟ FH ⌡ ⌡ FH ⎝ 2 ⎠ 1 ⌠ ⌠

dy dx

=

1

FH

1

(w x + C1)

Initial Guesses: Given

C1 = 1 N

FH = 1 N

Boundary Conditions: 0 =

x=0

1

FH

tan ( θ ) =

C2

b=

y = b at x = a

tan ( θ max) =

1

FH FH

(w a + C1)

cos ( θ max)

1

FH

(C1)

⎛ w ⎞ a2 + ⎛ C1 ⎞ a ⎜ ⎟ ⎜ 2F ⎟ ⎝ H⎠ ⎝ FH ⎠

⎛ C1 ⎞ ⎜ ⎟ ⎜ C2 ⎟ = Find ( C1 , C2 , FH) ⎜F ⎟ ⎝ H⎠

Tmax =

C2 = 1 N⋅ m

C1 = 0.22 MN

C2 = 0.00 N⋅ m

⎛ w a + C1 ⎞ ⎟ ⎝ FH ⎠

θ max = atan ⎜

F H = 1.27 MN

θ max = 12.61 deg

Tmax = 1.30 MN

750



Chapter 7

Problem 7-111 A chain of length L has a total mass M and is suspended between two points a distance d apart. Determine the maximum tension and the sag in the chain. Given: L = 40 m

M = 100 kg

d = 10 m

g = 9.81

m 2

s Solution: w = M

s=

g L

⎛ FH ⎞ ⎛ w ⎞ ⎜ ⎟ sinh ⎜ x⎟ ⎝ w ⎠ ⎝ FH ⎠ F H = 10 N

Guesses

Given

L

=

2

y=

θ max = atan ⎛⎜ sinh ⎛⎜

F H = 37.57 N

d dx

y = sinh ⎛⎜

w

⎝ FH

x⎞⎟

⎠

h = 10 m

⎛ FH ⎞ ⎛ w ⎜ ⎟ sinh ⎜ ⎝ w ⎠ ⎝ FH ⎝

⎛ FH ⎞ ⎛ w ⎜ ⎟ ⎜cosh ⎛⎜ x⎟⎞ − 1⎞⎟ ⎝ w ⎠⎝ ⎝ FH ⎠ ⎠

d⎞ 2⎟ ⎠

h=

FH ⎛

w cosh ⎛⎜ ⎜ w ⎝ ⎝ FH

w d ⎞⎞

Tmax =

⎟⎟ ⎝ FH 2 ⎠⎠

h = 18.53 m

d⎞

− 1⎞⎟ 2⎟ ⎠ ⎠

⎛ FH ⎞ ⎜ ⎟ = Find ( FH , h) ⎝ h ⎠

FH

cos ( θ max)

Tmax = 492 N

Problem 7-112 The cable has a mass density ρ and has length L. Determine the vertical and horizontal components of force it exerts on the top of the tower. Given:

ρ = 0.5

kg m

L = 25 m

θ = 30 deg d = 15 m g = 9.81

m 2

s

751



Chapter 7

Solution: ⌠ ⎮ x=⎮ ⎮ ⎮ ⎮ ⎮ ⌡

1 1+

⎞ ⎛⌠ ⎜ ⎮ ρ g ds⎟ 2⎜ ⎟ FH ⎝⌡ ⎠

ds 2

1

Performing the integration yields:

⎞ ⎛ ρ g s + C1 ⎞ ⎜ asinh ⎜ ⎟ + C2⎟ ρg ⎝ ⎝ FH ⎠ ⎠

FH ⎛

x=

dy dx

=

1 ⌠

⎮ ρ g ds = (ρ g s + C1) FH ⌡ FH dy

At s = 0; dy dx

=

1

dx

ρg s FH

= tan( θ )

C1 = F H tan ( θ )

Hence

+ tan ( θ )

Applying boundary conditions at x = 0; s = 0 to Eq.[1] and using the result C1 = F H tan ( θ ) yields C2 = −asinh ( tan ( θ ) ). Hence Guess

FH = 1 N

Given

d=

⎛ FH ⎞ ⎡ 1 ⎜ ⎟ ⎢asinh ⎡⎢⎛⎜ ⎞⎟ ( ρ g L + FH tan ( θ ) )⎥⎤ − ( asinh ( tan ( θ ) ) )⎤⎥ ⎝ ρg ⎠⎣ ⎣⎝ FH ⎠ ⎦ ⎦

F H = Find ( FH)

At A

FA =

F H = 73.94 N

tan ( θ A) =

ρg L

FH

F Ax = F A cos ( θ A)

cos ( θ A)

FH

+ tan ( θ )

⎛ ρ g L + tan ( θ )⎞ ⎟ ⎝ FH ⎠

θ A = atan ⎜

F Ay = F A sin ( θ A)

θ A = 65.90 deg

⎛ FAx ⎞ ⎛ 73.94 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ FAy ⎠ ⎝ 165.31 ⎠



Chapter 7

Problem 7-113 A cable of length L is suspended between two points a distance d apart and at the same elevation. If the minimum tension in the cable is Tmin, determine the total weight of the cable and the maximum tension developed in the cable. Units Used:

kip = 103 lb

Given:

L = 50 ft

d = 15 ft

Tmin = 200 lb

Solution:

Tmin = F H

F H = Tmin

F H = 200 lb

s=

From Example 7-15:

w0 = 1

Guess

L

Given

2

=

⎛ FH ⎞ ⎛ w0 x ⎞ ⎜ ⎟ sinh ⎜ ⎟ ⎝ w0 ⎠ ⎝ FH ⎠

lb ft

⎛ FH ⎞ ⎛ w0 d ⎞ ⎜ ⎟ sinh ⎜ ⎟ ⎝ w0 ⎠ ⎝ FH 2 ⎠

w0 = Find ( w0 )

Totalweight = w0 L

Totalweight = 4.00 kip

w0 L

⎡w ⎛ L ⎞⎤ ⎢ 0⎜⎝ 2 ⎟⎠⎥ = atan ⎢ ⎥ ⎣ FH ⎦

tan ( θ max) =

FH 2

θ max

w0 = 79.93

lb ft

θ max = 84.28 deg

Then, Tmax =

FH

cos ( θ max)

Tmax = 2.01 kip

753



Chapter 7

Problem 7-114 The chain of length L is fixed at its ends and hoisted at its midpoint B using a crane. If the chain has a weight density w, determine the minimum height h of the hook in order to lift the chain completely off the ground. What is the horizontal force at pin A or C when the chain is in this position? Hint: When h is a minimum, the slope at A and C is zero.

Given: L = 80 ft d = 60 ft w = 0.5

lb ft

Solution: F H = 10 lb

Guesses

Given

h=

FH ⎛ w

h = 1 ft

⎛w ⎝ FH

⎜cosh ⎜ ⎝

⎛ h ⎞ ⎜ ⎟ = Find ( h , FH) ⎝ FH ⎠

d⎞ 2⎟ ⎠

⎞ ⎠

L

− 1⎟

FA = FH

2

FC = FH

=

⎛ FH ⎞ ⎛ w ⎜ ⎟ sinh ⎜ ⎝ w ⎠ ⎝ FH

d⎞ 2⎟ ⎠

⎛ FA ⎞ ⎛ 11.1 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ FC ⎠ ⎝ 11.1 ⎠ h = 23.5 ft

Problem 7-115 A steel tape used for measurement in surveying has a length L and a total weight W. How much horizontal tension must be applied to the tape so that the distance marked on the ground is a? In practice the calculation should also include the effects of elastic stretching and temperature changes on the tape’s length. Given: L = 100 ft W = 2 lb a = 99.90 ft

754



Chapter 7

Solution: w0 =

W L

w0 = 0.02

lb ft

F H = 10 lb

Guess

L

Given

2

−

⎛ w0 a ⎞⎞ ⎛ FH ⎜ sinh ⎜ ⎟⎟ = ⎝ w0 ⎝ FH 2 ⎠⎠

F H = Find ( FH)

0

F H = 12.9 lb

Problem 7-116 A cable of weight W is attached between two points that are a distance d apart, having equal elevations. If the maximum tension developed in the cable is Tmax determine the length L of the cable and the sag h. W = 100 lb

Given:

d = 50 ft

Tmax = 75 lb

Solution: F H = 20 lb

Guesses

L = 20 ft

θ max = 20 deg

h = 2 ft

Given h=

FH L ⎛

W cosh ⎛⎜ ⎜ W ⎝ ⎝ FH L

Tmax =

d⎞

− 1⎟⎞ 2⎟ ⎠ ⎠

FH

cos ( θ max)

⎛ FH ⎞ ⎜ ⎟ ⎜ L ⎟ = Find ( F , L , θ , h) H max ⎜ θ max ⎟ ⎜ ⎟ ⎝ h ⎠

tan ( θ max) = sinh ⎛⎜

W

d⎞

⎟ ⎝ FH L 2 ⎠

L 2

=

⎛ FH L ⎞ ⎛ W ⎜ ⎟ sinh ⎜ ⎝ W ⎠ ⎝ FH L

F H = 55.90 lb

d⎞ 2⎟ ⎠

⎛ L ⎞ ⎛ 55.57 ⎞ ⎜ ⎟=⎜ ⎟ ft ⎝ h ⎠ ⎝ 10.61 ⎠

θ max = 41.81 deg

Problem 7-117 Determine the distance a between the supports in terms of the beam's length L so that the moment in the symmetric beam is zero at the beam's center.

755



Chapter 7

Solution: Support Reactions: ΣMD = 0; w 2

( L + a) ⎛⎜

By =

a⎞

⎟ − B y ( a) = ⎝ 2⎠

w 4

0

( L + a)

Internal Forces: ΣMC = 0;

1 L − a ⎞ ⎛ 2a + L ⎞ w a + w⎛⎜ − ( L + a) ⎛⎜ ⎟⎞ = ⎟ ⎜ ⎟ ⎟ ⎜ ⎟ ⎝ 2 ⎠⎝ 4 ⎠ 2 ⎝ 2 ⎠⎝ 6 ⎠ 4 ⎝ 2⎠

w⎛⎜

a ⎞⎛ a ⎞

2

2

2a + 2a L − L = 0

b =

−2 + 4

12

0

b = 0.366

a = bL

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws 756 as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


Chapter 7

Problem 7-118 Determine the internal normal force, shear force, and moment at point D.

Given: w = 150

N m

a = 4m b = 4m c = 3m Solution: Guesses

Ax = 1 N

Ay = 1 N

F BC = 1 N

Given Ax −

⎛ b ⎞F = ⎜ 2 2 ⎟ BC ⎝ b +c ⎠

Ay − w( 2 a) +

0

⎛ c ⎞F = ⎜ 2 2 ⎟ BC ⎝ b +c ⎠

0

w2 a a − Ay( 2 a) = 0

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ = Find ( Ax , Ay , FBC) ⎜F ⎟ ⎝ BC ⎠ Guesses Given

ND = 1 N Ax + ND = 0

⎛ Ax ⎞ ⎛ 800 ⎞ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 600 ⎟ N ⎜ F ⎟ ⎜⎝ 1000 ⎟⎠ ⎝ BC ⎠

VD = 1 N

MD = 1 N⋅ m

Ay − w a − VD = 0

− Ay a + w a⎛⎜

a⎞

⎟ + MD = ⎝2⎠

0

757



Chapter 7

⎛ ND ⎞ ⎜ ⎟ ⎜ VD ⎟ = Find ( ND , VD , MD) ⎜M ⎟ ⎝ D⎠

⎛ ND ⎞ ⎛ −800.00 ⎞ ⎜ ⎟=⎜ ⎟N 0.00 V ⎝ ⎠ D ⎝ ⎠

MD = 1200 N⋅ m

Problem 7-119 The beam is supported by a pin at C and a rod AB. Determine the internal normal force, shear force, and moment at point D. Units Used: kN = 103 N Given: F = 4 kN a = 6m b = 5m c = 3m d = 6m

θ = 60 deg Solution: Guesses F AB = 1 N Given

ND = 1 N

VD = 1 N

−F sin ( θ ) ( b + c + d) +

−ND −

VD +

MD = 1 N⋅ m

a ⎡ ⎤ F ( b + c) = AB ⎢ 2 2⎥ + ( b + c ) a ⎣ ⎦

b+c ⎡ ⎤ F + F cos ( θ ) = AB ⎢ 2 2⎥ + ( b + c ) a ⎣ ⎦

a ⎡ ⎤ F − F sin ( θ ) = AB ⎢ 2 2⎥ + ( b + c ) a ⎣ ⎦

0

0

0

a ⎡ ⎤ F c − F sin ( θ ) ( c + d) − M = AB D ⎢ 2 2⎥ ⎣ a + ( b + c) ⎦

0

758



Chapter 7

⎛ FAB ⎞ ⎜ ⎟ ⎜ ND ⎟ = Find ( F , N , V , M ) AB D D D ⎜ VD ⎟ ⎜ ⎟ ⎝ MD ⎠

⎛ ND ⎞ ⎛ −6.08 ⎞ ⎜ ⎟=⎜ ⎟ kN ⎝ VD ⎠ ⎝ −2.6 ⎠

MD = −12.99 kN⋅ m

Problem 7-120 Express the shear and moment acting in the pipe as a function of y, where 0 ≤ y ≤ b ft.

Given: w = 4

lb ft

a = 2 ft b = 4 ft Solution: ΣF y = 0;

wb − w y − V = 0 V ( y) = w( b − y) V ( y) = 4

ΣM = 0;

M + w y⎛⎜

lb ft

( 4 ft − y)

b a + w b⎛⎜ ⎟⎞ + w a⎛⎜ ⎟⎞ − w b y = ⎟ ⎝ 2⎠ ⎝2⎠ ⎝ 2⎠ y⎞

M ( y) = w b y −

1 2

M ( y) = 16 lb y − 2

2

w y − lb ft

1 2

2

wb −

1 2

0

wa

2

2

y − 40 lb⋅ ft

Problem 7-121 Determine the normal force, shear force, and moment at points B and C of the beam. Given:

kN = 103 N

759



a = 5m

Chapter 7

F 2 = 6 kN

b = 5m

w1 = 2

kN m

w2 = 1

kN m

c = 1m d = 3m

F 1 = 7.5 kN M = 40 kN⋅ m Solution: Guesses NC = 1 N

VC = 1 N

MC = 1 N⋅ m

Given −NC = 0

V C − w2 d − F2 = 0

⎛ d⎞ − F d − M = ⎟ 2 ⎝ 2⎠

−MC − w2 d⎜

0

⎛ NC ⎞ ⎜ ⎟ V ⎜ C ⎟ = Find ( NC , VC , MC) ⎜M ⎟ ⎝ C⎠ ⎛ NC ⎞ ⎛ 0.00 ⎞ ⎜ ⎟=⎜ ⎟ kN ⎝ VC ⎠ ⎝ 9.00 ⎠ Guesses

NB = 1 N

MC = −62.50 kN⋅ m

VB = 1 N

MB = 1 N⋅ m

Given −NB = 0

V B − w1 b − w2 ( c + d) − F1 − F2 = 0

⎛ b ⎞ − F b − w ( c + d) ⎛ b + c + d ⎞ − F ( b + c + d) − M = ⎟ ⎜ ⎟ 1 2 2 2 ⎠ ⎝2⎠ ⎝

−MB − w1 b⎜

⎛ NB ⎞ ⎜ ⎟ ⎜ VB ⎟ = Find ( NB , VB , MB) ⎜M ⎟ ⎝ B⎠

⎛ NB ⎞ ⎛ 0.00 ⎞ ⎜ ⎟=⎜ ⎟ kN ⎝ VB ⎠ ⎝ 27.50 ⎠

0

MB = −184.50 kN⋅ m

Problem 7-122 The chain is suspended between points A and B. If it has a weight weight density w and the 760



Chapter 7

p p sag is h, determine the maximum tension in the chain.

g

g

y

Given: lb ft

w = 0.5

L = 60 ft h = 3 ft Solution: Form Example 7-15 y=

FH ⎛

⎛ w x ⎞ − 1⎞ cosh ⎜ ⎜ ⎟ ⎟ w ⎝ ⎝ FH ⎠ ⎠

d ⎛ wx ⎞ y = sinh ⎜ ⎟ dx ⎝ FH ⎠ Guess

Given

F H = 1 lb h=

FH ⎛

⎛ w L ⎞ − 1⎞ cosh ⎜ ⎜ ⎟ ⎟ w ⎝ ⎝ FH 2 ⎠ ⎠

θ max = atan ⎜⎛ sinh ⎛⎜ F

w L ⎞⎞

⎝

⎝

H

⎟ 2⎟ ⎠⎠

F H = Find ( FH)

Tmax =

FH

cos ( θ max)

F H = 75.2 lb

Tmax = 76.7 lb


kN = 10 N Given: w = 2 Solution:

kN m

a = 5m

Guesses

A = 1N C = 1N

b = 5m

M = 50 kN⋅ m

761



Given

⎛ ⎝

w a⎜ b +

Chapter 7

a⎞

⎟ − A( a + b) − M = 2⎠

x1 = 0 , 0.01a .. a x2 = a , 1.01a .. a + b

A + C − wa = 0

0

V 1 ( x1 ) = ( A − w x1 ) V 2 ( x2 ) = −C

1

kN

M1 ( x1 ) =

⎛A⎞ ⎜ ⎟ = Find ( A , C) ⎝C⎠

x1 ⎞ 1 ⎛ ⎜ A x1 − w x1 ⎟ 2 ⎠ kN⋅ m ⎝

M2 ( x2 ) = ⎡⎣−M + C( a + b − x2 )⎤⎦

1

kN

1

kN⋅ m

Force (kN)

5

V1( x1) 0 V2( x2)

5

10

0

2

4

6

8

10

6

8

10

x1 , x2

Distance (m)

Moment (kN-m)

20

M1( x1)

0

M2( x2)

20 40 60

0

2

4

x1 , x2

Distance (m)

762



Chapter 8

Problem 8-1 The horizontal force is P. Determine the normal and frictional forces acting on the crate of weight W. The friction coefficients are μk and μs. Given: W = 300 lb P = 80 lb

μ s = 0.3 μ k = 0.2 θ = 20 deg Solution: Assume no slipping: ΣF x = 0;

P cos ( θ ) − W sin ( θ ) + F c = 0 F c = −P cos ( θ ) + W sin ( θ )

ΣF y = 0;

Check

F c = 27.4 lb

Nc − W cos ( θ ) − P sin ( θ ) = 0 Nc = W cos ( θ ) + P⋅ sin ( θ )

Nc = 309 lb

F cmax = μ s Nc

F cmax = 92.8 lb F cmax > Fc

Problem 8-2 Determine the magnitude of force P needed to start towing the crate of mass M. Also determine the location of the resultant normal force acting on the crate, measured from point A. Given: M = 40 kg

c = 200 mm

μ s = 0.3

d = 3

a = 400 mm

e = 4

b = 800 mm

763



Chapter 8


NC = 200 N

P = 50 N

Given ΣF x = 0;

⎛ d ⎞P − μ N = 0 s C ⎜ 2 2⎟ + e d ⎝ ⎠

ΣF y = 0;

NC − M g +

eP 2

=0 2

d +e

⎛ NC ⎞ ⎜ ⎟ = Find ( NC , P) ⎝P ⎠ NC = 280.2 N ΣMO = 0;

P = 140 N

⎛ e P ⎞⎛ b ⎞ = 0 ⎟ − N1 x + ⎜ ⎜ ⎟ 2 2 ⎟⎝ 2 ⎠ ⎝ 2⎠ + e d ⎝ ⎠

−μ s NC⎛⎜

x =

a⎞

−1 μ s NC a 2

Thus, the distance from A is

NC

2

2

d +e −ePb 2

x = 123.51 mm

2

d +e

A = x+

b

A = 523.51 mm

2

Problem 8-3 Determine the friction force on the crate of mass M, and the resultant normal force and its position x, measured from point A, if the force is P. Given: M = 40 kg

μ s = 0.5

a = 400 mm

μ k = 0.2

b = 800 mm

d = 3

c = 200 mm

e = 4

P = 300 N Solution: Initial guesses:

F C = 25 N

NC = 100 N

764



Chapter 8

Given ΣF x = 0;

ΣF y = 0;

P⎛

d ⎞ ⎜ 2 2 ⎟ − FC = 0 ⎝ d +e ⎠

NC − Mg + P⎛

e ⎞ ⎜ 2 2⎟ = 0 ⎝ d +e ⎠

⎛ FC ⎞ ⎜ ⎟ = Find ( FC , NC) ⎝ NC ⎠

F Cmax = μ s NC

Since FC = 180.00 N > F Cmax = 76.13 N then the crate slips

ΣMO = 0;

F C = μ k NC

⎛ FC ⎞ ⎛ 30.5 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ NC ⎠ ⎝ 152.3 ⎠

−NC x + P⎛

e ⎛ d ⎞ ⎞ ⎜ 2 2 ⎟ a − P⎜ 2 2 ⎟ c = 0 ⎝ d +e ⎠ ⎝ d +e ⎠

x = −P⎛

−e a + d c

⎞ ⎜ 2 2⎟ ⎝ NC d + e ⎠

b = 0.40 m 2 Then the block does not tip. Since x = 0.39 m

P Reqd = 39.81 N

then the break will hold the wheel

Problem 8-14 The block brake consists of a pin-connected lever and friction block at B. The coefficient of static friction between the wheel and the lever is μs and a torque M is applied to the wheel. Determine if the brake can hold the wheel stationary when the force applied to the lever is (a) P1 (b) P2. Assume that the torque M is applied counter-clockwise.

774



Chapter 8

Given:

μ s = 0.3 M = 5 N⋅ m a = 50 mm b = 200 mm c = 400 mm r = 150 mm P 1 = 30 N P 2 = 70 N

Solution:

To hold lever:

ΣMO = 0;

FB r − M = 0 FB =

Require

NB =

M r

F B = 33.333 N

FB

NB = 111.1 N

μs

Lever, ΣMA = 0;

P Reqd ( b + c) − NB b + FB a = 0

P Reqd =

NB b − F B a b+c

P Reqd = 34.3 N

(a) If P 1 = 30.00 N

> PReqd = 34.26 N


(b) If P2 = 70.00 N

> P Reqd = 34.26 N


775



Chapter 8

Problem 8-15 The doorstop of negligible weight is pin connected at A and the coefficient of static friction at B is μs. Determine the required distance s from A to the floor so that the stop will resist opening of the door for any force P applied to the handle.

Given:

μ s = 0.3 a = 1.5 in

Solution: ΣF y = 0; NB −

s ⎛ ⎞ ⎜ 2 2 ⎟ FA = 0 ⎝ s +a ⎠

ΣF x = 0; μ s NB −

⎛ a ⎞F =0 ⎜ 2 2⎟ A ⎝ s +a ⎠

⎛⎜ μ s s ⎟⎞ ⎛ a ⎞F = 0 FA − ⎜ A 2 2⎟ ⎜ s2 + a2 ⎟ ⎝ ⎠ ⎝ s +a ⎠ μs s = a

s =

a

μs

s = 5.00 in

Problem 8-16 The chair has a weight W and center of gravity at G. It is propped against the door as shown. If the coefficient of static friction at A is μA, determine the smallest force P that must be applied to the handle to open the door.

776



Chapter 8

Given:

μ A = 0.3 a = 1.20 ft b = 0.75 ft c = 3ft

θ = 30 deg W = 10 lb

Solution: B y = 1 lb

Guesses

NA = 1 lb

P = 1 lb

Given ΣF x = 0;

−P + μ A NA = 0

ΣF y = 0;

NA − W − By = 0

ΣΜΒ = 0; μ A NA c cos ( θ ) − NA c sin ( θ ) + W ⎡⎣( c − a)sin ( θ ) + b cos ( θ )⎤⎦ = 0

⎛ By ⎞ ⎜ ⎟ ⎜ NA ⎟ = Find ( By , NA , P) ⎜P ⎟ ⎝ ⎠ B y = 11.5 lb NA = 21.5 lb P = 6.45 lb

Problem 8-17 The uniform hoop of weight W is suspended from the peg at A and a horizontal force P is slowly applied at B. If the hoop begins to slip at A when the angle is θ , determine the coefficient of static friction between the hoop and the peg. 777



Chapter 8

Given:

θ = 30 deg


μ NA cos ( θ ) + P − NA sin ( θ ) = 0 P = ( μ cos ( θ ) − sin ( θ ) ) NA

ΣF y = 0;

μ NA sin ( θ ) − W + NA cos ( θ ) = 0 W = ( μ sin ( θ ) + cos ( θ ) ) NA

ΣΜΑ = 0; −W r sin ( θ ) + P ( r + r cos ( θ ) ) = 0 W sin ( θ ) = P( 1 + cos ( θ ) )

(μ

sin ( θ ) + cos ( θ ) ) sin ( θ ) = ( sin ( θ ) − μ cos ( θ ) ) ( 1 + cos ( θ ) )

μ =

sin ( θ )

1 + cos ( θ )

μ = 0.27

Problem 8-18 The uniform hoop of weight W is suspended from the peg at A and a horizontal force P is slowly applied at B. If the coefficient of static friction between the hoop and peg is μs, determine if it is possible for the hoop to reach an angle θ before the hoop begins to slip.

778



Chapter 8

Given:

μ s = 0.2 θ = 30 deg


μ NA cos ( θ ) + P − NA sin ( θ ) = 0 P = ( μ cos ( θ ) − sin ( θ ) ) NA

ΣF y = 0;

μ NA sin ( θ ) − W + NA cos ( θ ) = 0 W = ( μ sin ( θ ) + cos ( θ ) ) NA

ΣΜΑ = 0; −W r sin ( θ ) + P ( r + r cos ( θ ) ) = 0 W sin ( θ ) = P ( 1 + cos ( θ ) )

(μ

sin ( θ ) + cos ( θ ) ) sin ( θ ) = ( sin ( θ ) − μ cos ( θ ) ) ( 1 + cos ( θ ) )

μ =

sin ( θ )

1 + cos ( θ )

μ = 0.27

If μ s = 0.20 < μ = 0.27 then it is not possible to reach θ = 30.00 deg.

Problem 8-19 The coefficient of static friction between the shoes at A and B of the tongs and the pallet is μs1 and between the pallet and the floor μs2. If a horizontal towing force P is applied to the tongs, determine the largest mass that can be towed.

779



Chapter 8

Given:

μ s1 = 0.5

a = 75 mm

μ s2 = 0.4

b = 20 mm

P = 300 N

c = 30 mm

m

θ = 60 deg

g = 9.81

2

s Solution:

Assume that we are on the verge of slipping at every surface. Guesses T = 1N

NA = 1N

F = 1N

Nground = 1N

F A = 1N

mass = 1kg

Given 2 T sin ( θ ) − P = 0 −T sin ( θ ) ( b + c) − T cos ( θ ) a − FA b + NA a = 0 F A = μ s1 NA 2 FA − F = 0 Nground − mass g = 0 F = μ s2 Nground

⎛ T ⎞ ⎜ ⎟ ⎜ NA ⎟ ⎜ FA ⎟ ⎜ ⎟ = Find ( T , NA , FA , F , Nground , mass) F ⎜ ⎟ ⎜ Nground ⎟ ⎜ ⎟ ⎝ mass ⎠

780



Chapter 8

⎛ T ⎞ ⎛ 173.21 ⎞ ⎜ ⎟ ⎜ ⎟ N A ⎜ ⎟ ⎜ 215.31 ⎟ ⎜ FA ⎟ = ⎜ 107.66 ⎟ N ⎜ ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 215.31 ⎟ ⎜ Nground ⎟ ⎝ 538.28 ⎠ ⎝ ⎠

mass = 54.9 kg

Problem *8-20 The pipe is hoisted using the tongs. If the coefficient of static friction at A and B is μs, determine the smallest dimension b so that any pipe of inner diameter d can be lifted. Solution: W − 2 FB = 0

⎛ W⎞ b − N h − F ⎛ d⎞ = 0 ⎜ ⎟ B B⎜ ⎟ ⎝2⎠ ⎝ 2⎠ Thus FB =

NB =

W 2 W ( 2 b − d) 4h

Require F B ≤ μ s NB

W 2

≤

μ s W ( 2b − d)

2 h ≤ μ s ( 2b − d)

4h

b>

h

μs

+

d 2

Problem 8-21 A very thin bookmark having a width a. is in the middle of a dictionary of weight W. If the pages are b by c, determine the force P needed to start to pull the bookmark out.The coefficient of static friction between the bookmark and the paper is μs. Assume the pressure on each page and the bookmark is uniform.

781



Chapter 8

Given: a = 1 in W = 10 lb b = 8 in c = 10 in

μ s = 0.7

Solution: Pressure on book mark : P =

1 W 2 bc

P = 0.06

in

Normal force on bookmark: F = μs N ΣF x = 0;

lb 2

N = Pca

F = 0.44 lb P − 2F = 0

P = 2F

P = 0.88 lb

Problem 8-22 The uniform dresser has weight W and rests on a tile floor for which the coefficient of friction is μs. If the man pushes on it in the direction θ, determine the smallest magnitude of force F needed to move the dresser. Also, if the man has a weight Wman,, determine the smallest coefficient of static friction between his shoes and the floor so that he does not slip. Given: W = 90 lb

μ s = 0.25

782



Chapter 8

Wman = 150 lb

θ = 0 deg

Solution: Dresser:

ND = 1lb

Guesses

F = 1lb

Given +

↑Σ Fy = 0;

ND − W − F sin ( θ ) = 0

+ Σ F x = 0; →

F cos ( θ ) − μ s ND = 0

⎛ ND ⎞ ⎜ ⎟ = Find ( ND , F) ⎝F ⎠ Man:

F = 22.50 lb

Nm = 1lb

Guesses

μ m = 0.2

Given +

↑Σ Fy = 0;

+ Σ F x = 0; →

Nm − Wman + F sin ( θ ) = 0 −F cos ( θ ) + μ m Nm = 0

⎛ Nm ⎞ ⎜ ⎟ = Find ( Nm , μ m) ⎝ μm ⎠

μ m = 0.15

Problem 8-23 The uniform dresser has weight W and rests on a tile floor for which the coefficient of friction is μs. If the man pushes on it in the direction θ, determine the smallest magnitude of force F needed to move the dresser. Also, if the man has a weight Wman, determine the smallest coefficient of static friction between his shoes and the floor so that he does not slip. Given: W = 90 lb 783



Chapter 8

μ s = 0.25 Wman = 150 lb

θ = 30 deg

Solution: Dresser:

ND = 1lb

Guesses

F = 1lb

Given +

↑Σ Fy = 0;

ND − W − F sin ( θ ) = 0

+ Σ F x = 0; →

F cos ( θ ) − μ s ND = 0

⎛ ND ⎞ ⎜ ⎟ = Find ( ND , F) ⎝F ⎠ Man:

F = 30.36 lb

Nm = 1lb

Guesses

μ m = 0.2

Given +

↑Σ Fy = 0;

+ Σ F x = 0; →

Nm − Wman + F sin ( θ ) = 0 −F cos ( θ ) + μ m Nm = 0

⎛ Nm ⎞ ⎜ ⎟ = Find ( Nm , μ m) ⎝ μm ⎠

μ m = 0.195

Problem 8-24 The cam is subjected to a couple moment of M. Determine the minimum force P that should be applied to the follower in order to hold the cam in the position shown.The coefficient of static friction between the cam and the follower is μs. The guide at A is smooth.

784



Chapter 8

Given: a = 10 mm b = 60 mm M = 5 N⋅ m

μ s = 0.4 Solution: ΣM0 = 0;

M − μ s NB b − a NB = 0 NB =

M

μs b + a

NB = 147.06 N Follower: ΣF y = 0;

NB − P = 0 P = NB P = 147 N

Problem 8-25 The board can be adjusted vertically by tilting it up and sliding the smooth pin A along the vertical guide G. When placed horizontally, the bottom C then bears along the edge of the guide, where the coefficient of friction is μs. Determine the largest dimension d which will support any applied force F without causing the board to slip downward. Given:

μ s = 0.4 a = 0.75 in b = 6 in

785



Chapter 8

Solution: +

↑Σ Fy = 0;

μ s NC − F = 0 ΣMA = 0;

−F b + d NC − μ s NC a = 0

Solving we find

−μ s b + d − μ s a = 0

d = μ s ( a + b)

d = 2.70 in

Problem 8-26 The homogeneous semicylinder has a mass m and mass center at G. Determine the largest angle θ of the inclined plane upon which it rests so that it does not slip down the plane. The coefficient of static friction between the plane and the cylinder is μs. Also, what is the angle φ for this case? Given:

μ s = 0.3 Solution: The semicylinder is a two-force member: Since

F=μ N tan ( θ ) =

μs N N

= μS

θ = atan ( μ s) 786



Chapter 8

θ = 16.7 deg Law of sines 4r

r

sin ( 180 deg − φ )

=

3π

sin ( θ )

⎛ 3π sin ( θ )⎞ ⎟ ⎝4 ⎠

φ = asin ⎜

φ = 42.6 deg

Problem 8-27 A chain having a length L and weight W rests on a street for which the coefficient of static friction is μs. If a crane is used to hoist the chain, determine the force P it applies to the chain if the length of chain remaining on the ground begins to slip when the horizontal component is P x. What length of chain remains on the ground? Given: L = 20 ft W = 8

lb ft

μ s = 0.2 P x = 10 lb Solution: ΣF x = 0;

−P x + μ s Nc = 0

Nc =

Px

μs

Nc = 50.00 lb ΣF y = 0;

P y − W L + Nc = 0

787



Chapter 8

P y = W L − Nc P y = 110.00 lb 2

P =

Px + Py

2

P = 110 lb The length on the ground is supported by L =

Nc = 50.00 lbthus

Nc W

L = 6.25 ft

Problem 8-28 The fork lift has a weight W1 and center of gravity at G. If the rear wheels are powered, whereas the front wheels are free to roll, determine the maximum number of crates, each of weight W2 that the fork lift can push forward. The coefficient of static friction between the wheels and the ground is μs and between each crate and the ground is μ's. Given: W1 = 2400 lb W2 = 300 lb

μ s = 0.4 μ's = 0.35 a = 2.5 ft b = 1.25 ft c = 3.50 ft Solution: Fork lift: ΣMB = 0;

W1 c − NA ( b + c) = 0 NA = W1

ΣF x = 0;

⎛ c ⎞ ⎜ ⎟ ⎝ b + c⎠

NA = 1768.4 lb

μ s NA − P = 0 788



P = μ s NA

Chapter 8

P = 707.37 lb

Crate: Nc − W2 = 0

ΣF y = 0;

Nc = W2 ΣF x = 0;

Nc = 300.00 lb

P' − μ's Nc = 0 P' = μ's Nc

Thus

n =

P P'

P' = 105.00 lb

n = 6.74

n = floor ( n)

n = 6.00

Problem 8-29 The brake is to be designed to be self locking, that is, it will not rotate when no load P is applied to it when the disk is subjected to a clockwise couple moment MO. Determine the distance d of the lever that will allow this to happen. The coefficient of static friction at B is μs.

Given: a = 1.5 ft b = 1 ft

μ s = 0.5

Solution: ΣM0 = 0;

M0 − μ s NB b = 0

NB =

ΣMA = 0;

M0

μs b

P 2 a − NB a + μ s NB d = 0 789



Chapter 8

P=0 d =

a

μs

d = 3.00 ft

Problem 8-30 The concrete pipe of weight W is being lowered from the truck bed when it is in the position shown. If the coefficient of static friction at the points of support A and B is μs determine where it begins to slip first: at A or B, or both at A and B. Given: W = 800 lb

a = 30 in

μ s = 0.4

b = 18 in

θ = 30 deg

c = 5 in r = 15 in

Solution: initial guesses are

NA = 10 lb Given

NB = 10 lb

F A = 10 lb

Assume slipping at A:

ΣF x = 0;

NA + FB − W sin ( θ ) = 0

ΣF y = 0;

F A + NB − W cos ( θ ) = 0

ΣM0= 0;

F B = 10 lb

FB r − FA r = 0 F A = μ s NA

⎛ NA ⎞ ⎜ ⎟ ⎜ NB ⎟ = Find ( N , N , F , F ) A B A B ⎜ FA ⎟ ⎜ ⎟ ⎝ FB ⎠

⎛ NA ⎞ ⎛ 285.71 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ NB ⎟ = ⎜ 578.53 ⎟ lb ⎜ FA ⎟ ⎜ 114.29 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ FB ⎠ ⎝ 114.29 ⎠ 790



Chapter 8

At B, F Bmax = μ s NB Since F B = 114.29 lb < F Bmax = 231.41 lb then we conclude that slipping begins at A.

Problem 8-31 A wedge of mass M is placed in the grooved slot of an inclined plane. Determine the maximum angle θ for the incline without causing the wedge to slip. The coefficient of static friction between the wedge and the surfaces of contact is μs. Given: M = 5 kg

μ s = 0.2 φ = 60 deg g = 9.81

m 2

s Solution:

Initial guesses: NW = 10 N

θ = 10 deg

Given ΣF x = 0;

M g sin ( θ ) − 2 μ s NW = 0

ΣF z = 0;

2 NW sin ⎜

⎛ φ ⎞ − M g cos ( θ ) = 0 ⎟ ⎝2⎠

Solving,

⎛ NW ⎞ ⎜ ⎟ = Find ( NW , θ ) ⎝ θ ⎠ NW = 45.5 N

θ = 21.8 deg

791



Chapter 8

Problem 8-32 A roll of paper has a uniform weight W and is suspended from the wire hanger so that it rests against the wall. If the hanger has a negligible weight and the bearing at O can be considered frictionless, determine the force P needed to start turning the roll. The coefficient of static friction between the wall and the paper is μs. Given: W = 0.75 lb

θ = 30 deg φ = 30 deg μ s = 0.25 a = 3 in Solution: Initial guesses: R = 100 lb

NA = 100 lb

P = 100 lb

Given ΣF x = 0;

NA − R sin ( φ ) + P sin ( θ ) = 0

ΣF y = 0;

R cos ( φ ) − W − P cos ( θ ) − μ s NA = 0

ΣM0 = 0;

μ s NA a − P a = 0

Solving for P,

⎛⎜ R ⎞⎟ ⎜ NA ⎟ = Find ( R , NA , P) ⎜P ⎟ ⎝ ⎠ R = 1.14 lb

NA = 0.51 lb

P = 0.13 lb

Problem 8-33 A roll of paper has a uniform weight W and is suspended from the wire hanger so that it rests against the wall. If the hanger has a negligible weight and the bearing at O can be considered frictionless, determine the minimum force P and the associated angle θ needed to start turning the roll. The coefficient of static friction between the wall and the paper is μs. 792



Chapter 8

Given: W = 0.75 lb

φ = 30 deg μ s = 0.25 r = 3 in Solution: ΣF x = 0;

NA − R sin ( φ ) + P sin ( θ ) = 0

ΣF y = 0;

R cos ( φ ) − W − P cos ( θ ) − μ s NA = 0

ΣM0 = 0;

μ s NA r − P r = 0

Solving for P, P=

μ s W sin ( φ )

cos ( φ ) + μ sin ( θ − φ ) − μ sin ( φ )

For minimum P we must have dP dθ

−μ s W sin ( φ ) cos ( θ − φ ) 2

=

(cos (φ ) + μ s sin (θ − φ ) − μ s sin (φ ))

One answer is

P =

2

θ = φ + 90 deg

=0

Implies

cos ( θ − φ ) = 0

θ = 120.00 deg

μ s W sin ( φ )

P = 0.0946 lb

cos ( φ ) + μ s sin ( θ − φ ) − μ s sin ( φ )

Problem 8-34 The door brace AB is to be designed to prevent opening the door. If the brace forms a pin connection under the doorknob and the coefficient of static friction with the floor is μs determine the largest length L the brace can have to prevent the door from being opened. Neglect the weight of the brace. Given:

μ s = 0.5 793



Chapter 8

a = 3 ft Solution: The brace is a two-force member.

μs N N

2

L −a a

=

2

μs a =

L −a

L = a

1 + μs

2

2

2

L = 3.35 ft

Problem 8-35 The man has a weight W, and the coefficient of static friction between his shoes and the floor is

μs. Determine where he should position his center of gravity G at d in order to exert the maximum horizontal force on the door. What is this force? Given: W = 200 lb

μ s = 0.5 h = 3 ft Solution: N−W = 0

N = W

F max = μ s N + Σ F x = 0; →

F max = 100 lb P − Fmax = 0 P = F max

ΣMO = 0;

N = 200.00 lb

P = 100 lb

W d−P h= 0 d = P

h W

d = 1.50 ft

794



Chapter 8

Problem 8-36 In an effort to move the two crates, each of weight W, which are stacked on top of one another, the man pushes horizontally on them at the bottom of crate A as shown. Determine the smallest force P that must be applied in order to cause impending motion. Explain what happens. The coefficient of static friction between the crates is μs and between the bottom crate and the floor is μs'. Given: W = 100 lb

μ s = 0.8 μ's = 0.3 a = 2 ft b = 3 ft

Solution: Assume crate A slips: ΣF y = 0;

NA − W = 0

NA = W

NA = 100.00 lb

ΣF x = 0;

P − μ s NA = 0

P 1 = μ s NA

P 1 = 80.00 lb

Assume crate B slips: ΣF y = 0;

NB − 2 W = 0

NB = 2 W

NB = 200.00 lb

ΣF x = 0;

P − μ's NB = 0

P 2 = μ's NB

P 2 = 60.00 lb

Assume both crates A and B tip: ΣM = 0;

2W

⎛ a⎞ − P b = 0 ⎜ ⎟ ⎝ 2⎠

P = min ( P 1 , P 2 , P 3 )

P3 = W

⎛ a⎞ ⎜ ⎟ ⎝ b⎠

P 3 = 66.7 lb

P = 60.00 lb

Problem 8-37 The man having a weight of W1 pushes horizontally on the bottom of crate A, which is stacked on top of crate B. Each crate has a weight W2. If the coefficient of static friction between each crate is μs and between the bottom crate, his shoes, and the floor is μ's, determine if he can cause impending motion. 795



Chapter 8

Given: W1 = 150 lb W2 = 100 lb a = 2 ft b = 3 ft

μ s = 0.8 μ's = 0.3

Assume crate A slips: ΣF y = 0;

NA − W2 = 0

NA = W2

NA = 100.00 lb

ΣF x = 0;

P − μ s NA = 0

P 1 = μ s NA

P 1 = 80.00 lb

Assume crate B slips: ΣF y = 0;

NB − 2 W2 = 0

NB = 2 W2

NB = 200.00 lb

ΣF x = 0;

P − μ's NB = 0

P 2 = μ's NB

P 2 = 60.00 lb

Assume both crates A and B tip: ΣM = 0;

2 W2

⎛ a⎞ − P b = 0 ⎜ ⎟ ⎝ 2⎠

P min = min ( P 1 , P 2 , P 3 )

P 3 = W2

a b

P 3 = 66.7 lb

P min = 60.00 lb

Now check to see if he can create this force ΣF y = 0;

Nm − W1 = 0

Nm = W1

ΣF x = 0;

F m − P min = 0

F m = Pmin

F mmax = μ's Nm Since Fm = 60.00 lb > F mmax = 45.00 lb then the man cannot create the motion.

796



Chapter 8

Problem 8-38 The crate has a weight W and a center of gravity at G. Determine the horizontal force P required to tow it.Also, determine the location of the resultant normal force measured from A. Given: a = 3.5 ft b = 3 ft c = 2 ft W = 200 lb h = 4 ft

μ s = 0.4 Solution: ΣF x = 0;

P = FO

ΣF y = 0;

NO = W NO = 200.00 lb

ΣMo = 0;

−P h + W x = 0 F O = μ s NO F O = 80.00 lb P = FO P = 80.00 lb x = P

h W

x = 1.60 ft The distance of NO from A is c − x = 0.40 ft

797



Chapter 8

Problem 8-39 The crate has a weight W and a center of gravity at G. Determine the height h of the tow rope so that the crate slips and tips at the same time. What horizontal force P is required to do this? Given: a = 3.5 ft b = 3 ft

c = 2 ft W = 200 lb h = 4 ft

μ s = 0.4

Solution: ΣF y = 0;

NA = W NA = 200.00 lb

ΣF x = 0;

P = FA

F s = μsN;

FA = μ s W F A = 80.00 lb

ΣMA = 0;

P = 80 lb

−P h + W c = 0 h = W

c P

h = 5.00 ft

Problem 8-40 Determine the smallest force the man must exert on the rope in order to move the crate of mass M. Also, what is the angle θ at this moment? The coefficient of static friction between the crate and the floor is μs. Given: M = 80 kg 798



Chapter 8

μ s = 0.3 α = 30 deg β = 45 deg m

g = 9.81

2

s

Solution :

The initial guesses are

T1 = 1 N

NC = 1 N

θ = 30 deg

T = 1N

Given NC − M g + T1 cos ( θ ) = 0

μ s NC − T1 sin ( θ ) = 0 T cos ( β ) − T cos ( α ) + T1 sin ( θ ) = 0 T sin ( β ) + T sin ( α ) − T1 cos ( θ ) = 0

⎛ T ⎞ ⎜T ⎟ ⎜ 1 ⎟ = Find ( T , T , N , θ ) 1 C ⎜ NC ⎟ ⎜ ⎟ ⎝ θ ⎠

T = 451.86 N

θ = 7.50 deg

Problem 8-41 The symmetrical crab hook is used to lift packages by means of friction developed between the shoes Aand B and a package. Determine the smallest coefficient of static friction at the shoes so that the package of weight W can be lifted. Given: a = 1 ft b = 2 ft c = 0.8 ft 799



Chapter 8

d = 1 ft

θ = 45 deg Solution: From FBD (a) ΣF y = 0;

W = 2 F sin ( θ ) F=

W

2 sin ( θ )

From FBD (b) ΣMD = 0; μ NB d + NB a − F sin ( θ ) c − F cos ( θ ) b = 0

μ NB d + NB a − NB =

W

2 ( μ d + a)

W

2 sin ( θ )

sin ( θ ) c −

W

2 sin ( θ )

cos ( θ ) b = 0

(c + cot ( θ ) b)

From FBD (c) ΣF y = 0; 2 μ NB − W = 0 2μ

μ =

⎡ W ( c + cot (θ ) b)⎥⎤ − W = 0 ⎢ ( ⎣ 2 μ d + a) ⎦ a

c + b cot ( θ ) − d

μ = 0.56

Problem 8-42 The friction hook is made from a fixed frame which is shown colored and a cylinder of negligible weight. A piece of paper is placed between the smooth wall and the cylinder. D etermine the smallest coefficient of static friction μ at all points of contact so that any weight W of paper p 800



Chapter 8

μ

p

y

g

p p

p

can be held. Given:

θ = 20 deg Solution: Paper: +

↑

Σ F y = 0; F+F−W=0

F= μ N

F=

W 2

N=

W 2μ

Cylinder: F' r − N' −

⎛ W⎞ r = 0 ⎜ ⎟ ⎝2⎠

⎛ W ⎞ sin ( θ ) − W cos ( θ ) = 0 ⎜ ⎟ 2μ ⎝2⎠

F' = μ N'

W

2

=μ

F' =

W 2

N' =

W 2

⎛ sin ( θ ) + 1 cos ( θ )⎞ ⎜ ⎟ μ ⎝ ⎠

⎛ W ⎞ ⎛ sin ( θ ) + 1 cos ( θ )⎞ ⎜ ⎟⎜ ⎟ μ ⎝ 2 ⎠⎝ ⎠

1 = μ sin ( θ ) + cos ( θ )

μ =

1 − cos ( θ ) sin ( θ )

μ = 0.176

Problem 8-43 The crate has a weight W1 and a center of gravity at G. If the coefficient of static friction between the crate and the floor is μs, determine if the man of weight W2 can push the crate to the left. The coefficient of static friction between his shoes and the floor is μ's. Assume the man exerts only a horizontal force on the crate.

801



Chapter 8

Given: W1 = 300 lb W2 = 200 lb

μ s = 0.2 μ's = 0.35 a = 4.5 ft

c = 3 ft

b = 3.5 ft

d = 4.5 ft


NC − W1 = 0

NC = W1

ΣF x = 0;

μ s NC − P = 0

P = μ s NC

ΣΜO = 0;

−W1 x + P d = 0

x =

Pd W1

Since x = 0.90 ft < a = 4.50 ft there will not be any tipping. ΣF y = 0;

Nm − W2 = 0

Nm = W2

Nm = 200.00 lb

ΣF x = 0;

P − Fm = 0

Fm = P

F m = 60.00 lb

F mmax = μ's Nm

F mmax = 70.00 lb

Since F m = 60.00 lb < F mmax = 70.00 lb then the man can push the crate.

Problem 8-44 The crate has a weight W1 and a center of gravity at G. If the coefficient of static friction between the crate and the floor is μs, determine the smallest weight of the man so that he can push the crate to the left. The coefficient of static friction between his shoes and the floor is μ's. Assume the man exerts only a horizontal force on the crate.

802



Chapter 8

Given: W1 = 300 lb W2 = 200 lb

μ s = 0.2 μ's = 0.35 a = 4.5 ft

c = 3 ft

b = 3.5 ft

d = 4.5 ft


NC − W1 = 0

NC = W1

ΣF x = 0;

μ s NC − P = 0

ΣΜO = 0;

−W1 x + P d = 0

P = μ s NC Pd x = W1

Since x = 0.90 ft < a = 4.50 ft there will not be any tipping. ΣF x = 0;

ΣF y = 0;

P − Fm = 0

Fm = P

F m = μ's Nm

Nm =

Nm − W2 = 0

W2 = Nm

F m = 60.00 lb

Fm

Nm = 171.4 lb

μ's

W2 = 171.4 lb

Problem 8-45 The wheel has weight WA and rests on a surface for which the coefficient of friction is μB. A cord wrapped around the wheel is attached to the top of the homogeneous block of weight WC. 803



Chapter 8

If the coefficient of static friction at D is μD determine the smallest vertical force that can be applied tangentially to the wheel which will cause motion to impend. Given: WA = 20 lb

μ B = 0.2 WC = 30 lb

μ D = 0.3 h = 3 ft b = 1.5 ft Solution:

Assume that slipping occurs at B, but that the block does not move.

Guesses

P = 1 lb

NB = 1 lb

F B = 1 lb

T = 1 lb

ND = 1 lb

F D = 1 lb

x = 1 ft Given

NB − WA − P = 0

T − FB = 0

F B = μ B NB

−T + F D = 0

( P − T − F B) 2 h

=0

ND − WC = 0

T h − ND x = 0

⎛P ⎞ ⎜ ⎟ ⎜ NB ⎟ ⎜ FB ⎟ ⎜ ⎟ ⎜ T ⎟ = Find ( P , NB , FB , T , ND , FD , x) ⎜ ND ⎟ ⎜ ⎟ ⎜ FD ⎟ ⎜ x ⎟ ⎝ ⎠ Now checke the assumptions

⎛ P ⎞ ⎛ 13.33 ⎞ ⎜N ⎟ ⎜ ⎟ ⎜ B ⎟ ⎜ 33.33 ⎟ ⎜ FB ⎟ ⎜ 6.67 ⎟ ⎜ ⎟=⎜ ⎟ lb 6.67 T ⎜ ⎟ ⎜ ⎟ ⎜ ND ⎟ ⎜ 30.00 ⎟ ⎜ ⎟ ⎜ 6.67 ⎟ ⎠ ⎝ FD ⎠ ⎝

x = 0.67 ft

F Dmax = μ D ND

Since F D = 6.67 lb < FDmax = 9.00 lb then the block does not slip

804



Since x = 0.67 ft
θ p = 2.19 deg , screw is self locking.

Problem 8-73 The vise is used to grip the pipe. If a horizontal force F 1 is applied perpendicular to the end of the handle of length l, determine the compressive force F developed in the pipe. The square threads have a mean diameter d and a lead a. How much force must be applied perpendicular to the handle to loosen the vise? Given: F 1 = 25 lb d = 1.5 in

μ s = 0.3 L = 10 in

834



Chapter 8

a = 0.2 in Solution: r =

d 2

⎞ ⎟ ⎝ 2π r ⎠

θ = atan ⎛⎜

a

φ = atan ( μ s)

θ = 2.43 deg φ = 16.70 deg

F 1 L = F r tan ( θ + φ ) F = F1

L ⎞ ⎛ ⎜ ⎟ ( ) r tan θ + φ ⎝ ⎠

F = 961 lb

To loosen screw, P L = F r tan ( φ − θ ) P = Fr

tan ( φ − θ ) L

P = 18.3 lb

Problem 8-74 Determine the couple forces F that must be applied to the handle of the machinist’s vise in order to create a compressive force F A in the block. Neglect friction at the bearing A. The guide at B is smooth so that the axial force on the screw is F A. The single square-threaded screw has a mean radius b and a lead c, and the coefficient of static friction is μs. Given: a = 125 mm F A = 400 N b = 6 mm c = 8 mm

μ s = 0.27

835



Chapter 8

Solution:

φ = atan ( μ s) θ = atan ⎛⎜

c

φ = 15.11 deg

⎞ ⎟

⎝ 2 π b⎠

θ = 11.98 deg

F 2 a = FA b tan ( θ + φ ) F = FA

⎛ b ⎞ tan ( θ + φ ) ⎜ ⎟ ⎝ 2a ⎠

F = 4.91 N

Problem 8-75 If couple forces F are applied to the handle of the machinist’s vise, determine the compressive force developed in the block. Neglect friction at the bearing A. The guide at B is smooth. The single square-threaded screw has a mean radius of r1 and a lead of r2, and the coefficient of static friction is μs. Units Used: 3

kN = 10 N Given: F = 35 N a = 125 mm r1 = 6 mm r2 = 8 mm

μ s = 0.27 Solution:

φ = atan ( μ s)

⎛ r2 ⎞ ⎟ ⎝ 2 π r1 ⎠

θ = atan ⎜

φ = 15.11 deg θ = 11.98 deg

F 2 a = P r1 tan ( θ + φ ) P = 2F

a ⎞ ⎛ ⎜ r tan ( θ + φ ) ⎟ ⎝ 1 ⎠

P = 2.85 kN

836



Chapter 8

Problem 8-76 The machine part is held in place using the double-end clamp.The bolt at B has square threads with a mean radius r and a lead r1, and the coefficient of static friction with the nut is μs. If a torque M is applied to the nut to tighten it, determine the normal force of the clamp at the smooth contacts A and C. Given: a = 260 mm b = 90 mm r = 4 mm rl = 2 mm

μ s = 0.5 M = 0.4 N⋅ m Solution:

φ = atan ( μ s) φ = 26.57 deg

⎛ rl ⎞ ⎟ ⎝ 2π r ⎠

θ = atan ⎜

θ = 4.55 deg M = W r tan ( θ + φ ) W =

M

r tan ( θ + φ )

W = 165.67 N ΣMA = 0;

NC( a + b) − W a = 0 NC = W

ΣF y = 0;

a a+b

NC = 123 N

NA − W + NC = 0 NA = W − NC

NA = 42.6 N

837



Chapter 8

Problem 8-77 Determine the clamping force on the board A if the screw of the “C” clamp is tightened with a twist M. The single square-threaded screw has a mean radius r, a lead h, and the coefficient of static friction is μs. Units Used: 3

kN = 10 N Given: M = 8 N⋅ m r = 10 mm h = 3 mm

μ s = 0.35 Solution:

φ s = atan ( μ s)

φ s = 19.29 deg

⎞ ⎟ ⎝ 2 π r⎠

θ p = atan ⎛⎜

1

h

M = P r tan ( φ s + θ p )

θ p = 2.734 deg P =

M

r tan ( φ s + θ p )

P = 1.98 kN

Problem 8-78 If the required clamping force at the board A is to be P, determine the torque M that must be applied to the handle of the “C” clamp to tighten it down. The single square-threaded screw has a mean radius r, a lead h, and the coefficient of static friction is μs. Given: P = 50 N r = 10 mm h = 3 mm

μ s = 0.35

838



Chapter 8

Solution:

φ s = atan ( μ s) θ P = atan ⎛⎜

1

φ s = 19.29 deg h

⎞ ⎟

⎝ 2 π r⎠

M = P r tan ( φ s + θ P)

θ P = 2.73 deg M = 0.202 N⋅ m

Problem 8-79 Determine the clamping force on the board at A if the screw of the hold-down clamp is tightened with a twist M. The single square-threaded screw has a mean radius of r and a lead of rl, and the coefficient of static friction is μs. Given: M = 0.2 N m r = 8 mm rl = 2 mm

μ s = 0.38

Solution:

φ = atan ( μ s)

⎛ rl ⎞ ⎟ ⎝ 2π r ⎠

θ = atan ⎜

φ = 20.81 deg

θ = 2.28 deg

M = F r tan ( θ + φ ) F =

M

r tan ( θ + φ )

F = 58.7 N

839



Chapter 8

Problem 8-80 If the required clamping force at the board A is to be F, determine the torque M that must be applied to the handle of the hold-down clamp to tighten it down.The single square-threaded screw has a mean radius r and a lead r1, and the coefficient of static friction is μs. Given: F = 70 N r = 8 mm rl = 2 mm

μ s = 0.38

Solution:

φ = atan ( μ s)

φ = 20.81 deg

⎛ rl ⎞ ⎟ ⎝ 2π r ⎠

θ = atan ⎜

θ = 2.2785 deg

M = F r tan ( θ + φ )

M = 0.24 N⋅ m

Problem 8-81 The fixture clamp consist of a square-threaded screw having a coefficient of static friction μs mean diameter d, and a lead h. The five points indicated are pin connections. Determine the clamping force at the smooth blocks D and E when a torque M is applied to the handle of the screw. Given:

μ s = 0.3 d = 3 mm h = 1 mm M = 0.08 N⋅ m

a = 30 mm b = 40 mm c = 40 mm

β = 45 deg

840



Chapter 8

Frictional Forces on Screw: Here

θ = atan ⎡⎢

h

⎤ ⎥

d ⎢2 π ⎛⎜ ⎟⎞ ⎥ ⎣ ⎝ 2 ⎠⎦

φ s = atan ( μ s)

θ = 6.06 deg

φ s = 16.70 deg

Applying Eq.8-3, we have M = P ⎛⎜

d⎞

⎟ tan ( θ + φ s) ⎝ 2⎠

P = 2

M ⎞ ⎛ ⎜ d tan θ + φ ⎟ ( s) ⎠ ⎝

P = 127.15 N

Note since φ s = 16.70 deg > θ = 6.06 deg , the screw is self-locking. It will not unscrew even if the moment M is removed. Equations of Equilibrium and Friction: ΣMc = 0; c ⎛ ⎞ ⎜ 2 2 ⎟ P b − FE cos ( β ) b − FE sin ( β ) a = 0 ⎝ b +c ⎠

841



Chapter 8

P cb

FE =

F E = 72.7 N

b + c ( cos ( β ) b + sin ( β ) a) 2

2

The equilibrium of clamped block requires that FD = FE

F D = 72.7 N

Problem 8-82 The clamp provides pressure from several directions on the edges of the board. If the square-threaded screw has a lead h, radius r, and the coefficient of static friction is μs, determine the horizontal force developed on the board at A and the vertical forces developed at B and C if a torque M is applied to the handle to tighten it further. The blocks at B and C are pin-connected to the board. Given: h = 3 mm r = 10 mm

μ s = 0.4 M = 1.5 N⋅ m

β = 45 deg Solution:

φ s = atan ( μ s) θ = atan ⎛⎜

h

φ s = 21.801 deg

⎞ ⎟

⎝ 2π r ⎠

M = Ax r tan ( φ s + θ )

θ = 2.734 deg M

Ax =

r tan ( φ s + θ )

+ Σ F x = 0; →

⎛ Ax ⎞ ⎜ ⎟ 2 ⎝ cos ( β ) ⎠

Ax − 2T cos ( β ) = 0

T =

Cy = T sin ( β )

B y = Cy

1

Ax = 329 N

T = 232.36 N

⎛ By ⎞ ⎛ 164.3 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ Cy ⎠ ⎝ 164.3 ⎠

842



Chapter 8

Problem 8-83 The two blocks under the double wedge are brought together using a left and right square-threaded screw. If the mean diameter is d, the lead is rl, and the coefficient of static friction is μs, determine the torque needed to draw the blocks together. The coefficient of static friction between each block and its surfaces of contact is μ's. Units Used: 3

kN = 10 N Given: F = 5 kN

θ = 20 deg d = 20 mm rl = 5 mm

μ s = 0.4 μ's = 0.4 Solution: Top block: −F + 2N1 cos ( θ ) − 2μ's N1 sin ( θ ) = 0 N1 =

F

2( cos ( θ ) − μ's sin ( θ ) )

N1 = 3.1138 kN Bottom block: N' − N1 cos ( θ ) + μ's N1 sin ( θ ) = 0 N' = N1 cos ( θ ) − μ's N1 sin ( θ ) N' = 2.50 kN −N1 sin ( θ ) − μ's N1 cos ( θ ) + T − μ's N' = 0 T = N1 sin ( θ ) + μ's N1 cos ( θ ) + μ's N' T = 3.2354 kN

843



φ = atan ( μ s)

Chapter 8

φ = 21.80 deg

⎛ rl ⎞ ⎟ ⎝ πd ⎠

θ = atan ⎜

θ = 4.55 deg

Since there are two blocks, M = 2T

d 2

tan ( θ + φ )

M = 32 N⋅ m

Problem 8-84 The two blocks under the double wedge are brought together using a left and right square-threaded screw. If the mean diameter is d, the lead is rl, and the coefficient of static friction is μs, determine the torque needed to spread the blocks apart. The coefficient of static friction between each block and its surfaces of contact is μ's. Units Used: 3

kN = 10 N Given: F = 5 kN

θ = 20 deg d = 20 mm rl = 5 mm

μ s = 0.4 μ's = 0.4 Solution: Top block: −F + 2N1 cos ( θ ) + 2μ's N1 sin ( θ ) = 0 N1 =

F

2( cos ( θ ) + μ's sin ( θ ) )

N1 = 2.3223 kN

844



Chapter 8

Bottom block : N' − N1 cos ( θ ) − μ's N1 sin ( θ ) = 0 N' = N1 cos ( θ ) + μ's N1 sin ( θ ) N' = 2.50 kN −N1 sin ( θ ) + μ's N1 cos ( θ ) − T + μ's N' = 0 T = −N1 sin ( θ ) + μ's N1 cos ( θ ) + μ's N' T = 1.0786 kN

φ = atan ( μ s)

φ = 21.80 deg

⎛ rl ⎞ ⎟ ⎝ πd⎠

θ = atan ⎜

θ = 4.55 deg

Since there are two blocks, M = 2T

d 2

tan ( φ − θ )

M = 6.7 N⋅ m

Problem 8-85 The cord supporting the cylinder of mass M passes around three pegs, A, B, C, where the coefficient of friction is μs. Determine the range of values for the magnitude of the horizontal force P for which the cylinder will not move up or down. Given: M = 6 kg

θ = 45 deg μ s = 0.2 g = 9.81

m 2

s Solution: Total angle

β =

5 2

π − 4θ

β = 270.00 deg

845



Forces

Chapter 8

− μs β

μ sβ P max = M g e

P min = M g e

P min = 15.9 N < P < P max = 217.4 N

Answer

Problem 8-86 The truck, which has mass mt , is to be lowered down the slope by a rope that is wrapped around a tree. If the wheels are free to roll and the man at A can resist a pull P, determine the minimum number of turns the rope should be wrapped around the tree to lower the truck at a constant speed. The coefficient of kinetic friction between the tree and rope is μk. Units Used: Mg = 1000 kg Given: mt = 3.4 Mg P = 300 N

θ = 20 deg μ k = 0.3 g = 9.81

m 2

s Solution: ΣF x = 0;

T2 − mt g sin ( θ ) = 0

μ kβ T2 = P e

T2 = mt g sin ( θ )

T2 = 11407.74 N

⎛ T2 ⎞ ⎟ ⎝P⎠

ln ⎜

β =

β = 694.86 deg

μk

⎛ β ⎞ = 2.00 turns ⎟ ⎝ 360 deg ⎠

Use ceil ⎜

846



Chapter 8

Problem 8-87 The wheel is subjected to a torque M. If the coefficient of kinetic friction between the band brake and the rim of the wheel is μk, determine the smallest horizontal force P that must be applied to the lever to stop the wheel. Given: a = 400 mm d = 25 mm b = 100 mm r = 150 mm c = 50 mm

M = 50 N⋅ m

μ k = 0.3 Solution: Initial guesses:

T1 = 5 N

T2 = 10 N

Given

⎛ 3π ⎞ μ k⎜ ⎟ 2 T2 = T1 e ⎝ ⎠

Wheel: ΣM0 = 0;

− T2 r + T1 r + M = 0

⎛ T1 ⎞ ⎜ ⎟ = Find ( T1 , T2 ) ⎝ T2 ⎠

T1 = 54.66 N

T1 c − F d = 0

F = T1 ⎛⎜

−P a + F b = 0

P = F ⎛⎜

Link: ΣMB = 0;

c⎞

⎟ ⎝ d⎠

F = 109.32 N

Lever: ΣMA = 0;

b⎞

⎟ ⎝ a⎠

P = 27.3 N

Problem 8-88 A cylinder A has a mass M. Determine the smallest force P applied to the handle of the lever required for equilibrium. The coefficient of static friction between the belt and the wheel is μs. The drum is pin connected at its center, B.

847



Chapter 8

Given: M = 75 kg a = 700 mm b = 25 mm c = 300 mm d = 200 mm e1 = 60 mm

μ s = 0.3 e = 2.718 Solution: Initial guesses: T1 = 1 N

T2 = 1 N

P = 1N

Given Drum:

⎛ 3π ⎞ μ s⎜ ⎟ 2 T2 = T1 e ⎝ ⎠ − T2 c + T1 c + M g d = 0 Lever: −T1 e1 + T2 b − P a = 0

⎛ T1 ⎞ ⎜ ⎟ ⎜ T2 ⎟ = Find ( T1 , T2 , P) ⎜P⎟ ⎝ ⎠

⎛ T1 ⎞ ⎛ 157.7 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ T2 ⎠ ⎝ 648.2 ⎠

P = 9.63 N

Problem 8-89 Determine the largest mass of cylinder A that can be supported from the drum if a force P is applied to the handle of the lever. The coefficient of static friction between the belt and the wheel is μs. The drum is pin supported at its center, B. Given: P = 20 N 848



Chapter 8

a = 700 mm b = 25 mm c = 300 mm d = 200 mm e1 = 60 mm

μ s = 0.3 e = 2.718

Solution: Initial guesses: T1 = 1 N

T2 = 1 N

M = 1 kg

Given Drum:

⎛ 3π ⎞ μ s⎜ ⎟ 2 T2 = T1 e ⎝ ⎠ − T2 c + T1 c + M g d = 0 Lever: −T1 e1 + T2 b − P a = 0

⎛ T1 ⎞ ⎜ ⎟ ⎜ T2 ⎟ = Find ( T1 , T2 , M) ⎜M ⎟ ⎝ ⎠

⎛ T1 ⎞ ⎛ 327.4 ⎞ N ⎜ ⎟=⎜ 3⎟ ⎝ T2 ⎠ ⎝ 1.3 × 10 ⎠

M = 155.7 kg

Problem 8-90 The uniform bar AB is supported by a rope that passes over a frictionless pulley at C and a fixed peg at D. If the coefficient of static friction between the rope and the peg is μD, determine the smallest distance x from the end of the bar at which a force F may be placed and not cause the bar to move. Given: F = 20 N

a = 1m 849



Chapter 8

μ D = 0.3 Solution: Initial guesses: TA = 5 N

TB = 10 N x = 10 m

Given ΣMA = 0;

− F x + TB a = 0

ΣF y = 0;

TA + TB − F = 0

⎛π⎞ μ D⎜ ⎟ 2 TA = TB e ⎝ ⎠

⎛ TA ⎞ ⎜ ⎟ ⎜ TB ⎟ = Find ( TA , TB , x) ⎜ x ⎟ ⎝ ⎠ x = 0.38 m

Problem 8-91 Determine the smallest lever force P needed to prevent the wheel from rotating if it is subjected to a torque M. The coefficient of static friction between the belt and the wheel is μs. The wheel is pin-connected at its center, B. Given: M = 250 N m

μ s = 0.3 r = 400 mm a = 200 mm b = 750 mm Solution: ΣMA = 0;

− F a + P ( a + b) = 0

850



F=P

β =

Chapter 8

⎛ a + b⎞ ⎜ ⎟ ⎝ a ⎠

3π 2

μ sβ F' = F e

ΣMB = 0;

−P

⎛ a + b ⎞ eμ sβ r + M + P ⎜ ⎟ ⎝ a ⎠

P =

Ma μ sβ −1 ( a + b)r e

(

⎛ a + b⎞r = 0 ⎜ ⎟ ⎝ a ⎠ P = 42.3 N

)

Problem 8-92 Determine the torque M that can be resisted by the band brake if a force P is applied to the handle of the lever. The coefficient of static friction between the belt and the wheel is μs. The wheel is pin-connected at its center, B. Given: P = 30 N

μ s = 0.3 r = 400 mm a = 200 mm b = 750 mm Solution:

ΣMA = 0;

− F a + P ( a + b) = 0

⎛ a + b⎞ ⎟ ⎝ a ⎠

F = P⎜

F = 142.5 N

3π μs 2 F' = F e

F' = 585.8 N

851



ΣMB = 0;

Chapter 8

−F' r + F r + M = 0 M = F' r − F r

M = 177 N⋅ m

Problem 8-93 Blocks A and B weigh WA and WB and respectively. Using the coefficients of static friction indicated, determine the greatest weight of block D without causing motion. Given: WA = 50 lb WB = 30 lb

μ = 0.5 μ BA = 0.6 μ AC = 0.4 θ = 20 deg Assume that B slips on A, but A does not move. Guesses NB = 1 lb

Given

WD = 1 lb

TB = 1 lb

NC = 1 lb

F C = 1 lb

π μ 2 W D = TB e

− TB + F C = 0 NC − WA − WB = 0 −TB + μ BA NB cos ( θ ) − NB sin ( θ ) = 0 NB cos ( θ ) + μ BA NB sin ( θ ) − WB = 0

852



Chapter 8

⎛ WD ⎞ ⎜ ⎟ ⎜ TB ⎟ ⎜ NB ⎟ = Find ( W , T , N , N , F ) D B B C C ⎜ ⎟ ⎜ NC ⎟ ⎜F ⎟ ⎝ C⎠

⎛ WD ⎞ ⎛ 12.75 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ TB ⎟ ⎜ 5.81 ⎟ ⎜ NB ⎟ = ⎜ 26.20 ⎟ lb ⎜ ⎟ ⎜ ⎟ N ⎜ C ⎟ ⎜ 80.00 ⎟ ⎜ F ⎟ ⎝ 5.81 ⎠ ⎝ C⎠

Now check the assumption that A does not move F Cmax = μ AC NC

F Cmax = 32.00 lb

Since F C = 5.81 lb < F Cmax = 32.00 lb then our assumption is good.

WD = 12.75 lb

Problem 8-94 Blocks A and B have weight W, and D weighs WD. Using the coefficients of static friction indicated, determine the frictional force between blocks A and B and between block A and the floor C. Given: W = 75 lb

μ BA = 0.6

WD = 30 lb

μ AC = 0.4

μ = 0.5

θ = 20 deg

Solution: π μ 2 W D = TB e

WD

TB =

1

e

Check

2

μπ

F C = TB

F C = 13.68 lb

NC − 2W = 0

NC = 2W

F Cmax = μ AC NC

F Cmax = 60.00 lb

TB = 13.679 lb

NC = 150.00 lb

Since F C = 13.68 lb < F Cmax = 60.00 lb then the system does not slip at C. 853



Chapter 8

For block B: The initial guessess:

NB = 1 lb

F B = 1 lb

Given +

↑Σ Fy = 0; + Σ F x = 0; → ⎛ NB ⎞ ⎜ ⎟ = Find ( NB , FB) ⎝ FB ⎠

NB cos ( θ ) + FB sin ( θ ) − W = 0 F B cos ( θ ) − NB sin ( θ ) − TB = 0 NB = 65.80 lb

F B = 38.51 lb

Check F Bmax = μ BA NB

F Bmax = 39.48 lb

Since F B = 38.51 lb < FBmax = 39.48 lb then no slipping occurs between the blocks

Problem 8-95 Show that the frictional relationship between the belt tensions, the coefficient of friction μ, and the angular contacts α and β for the V-belt is T2=T1eμβ/sin(α/2) when the belt is on the verge of slipping.

Solution: FBD of a section of the belt is shown. Proceeding in the general manner:

ΣF x = 0;

⎛ dθ ⎞ + T cos ⎛ dθ ⎞ + 2μ dN = 0 ⎟ ⎜ ⎟ ⎝2⎠ ⎝2⎠

−( T + dT) cos ⎜

854



Chapter 8

⎛ dθ ⎞ − T sin ⎛ dθ ⎞ + 2dN sin ⎛ α ⎞ = 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝2⎠ ⎝2⎠ ⎝2⎠

−( T + dT) sin ⎜

ΣF y = 0;

Since dθ, dN, and dT are small, these become

⎛α⎞ ⎟ ⎝2⎠

Tdθ = 2dN sin ⎜

dT = 2μ dN dT =μ T

Combine

dθ

⎛α⎞ ⎟ ⎝2⎠

sin⎜

θ = 0 , T = T1

Integrate from

to

θ = β , T = T2

μβ

We get, T2 = T1 e

⎛α⎞ ⎟ ⎝2⎠

sin⎜

Q.E.D

Problem 8-96 A V-fan-belt (V-angle θ) of an automobile engine passes around the hub H of a generator G and over the housing F to a fan. If the generator locks, and the maximum tension the belt can sustain is Tmax, determine the maximum possible torque M resisted by the axle as the belt slips over the hub. Assume that slipping of the belt occurs only at H and that the coefficient of kinetic friction for the hub is μs. Given:

θ = 60 deg

b = 2 in

a = 2 in

c = 1.25 ft

Tmax = 175 lb

μ s = 0.25

Solution: −T1 a + Tmax a − M = 0

⎞ ⎟ 1 ⎜ sin⎛⎜ θ ⎞⎟ ⎟ ⎝ ⎝2 ⎠⎠ ⎛

− μ s⎜

T1 = Tmax e

π

T1 = 36.4 lb

855



Chapter 8

M = −T1 a + Tmax a

M = 23.1 lb⋅ ft

Problem 8-97 A cable is attached to the plate B of mass MB, passes over a fixed peg at C, and is attached to the block at A. Using the coefficients of static friction shown, determine the smallest mass of block A so that it will prevent sliding motion of B down the plane. Given: MB = 20 kg

μ A = 0.2

θ = 30 deg

μ B = 0.3

m

μ C = 0.3

g = 9.81

2

s Solution: Iniitial guesses:

T1 = 1 N

T2 = 1 N NA = 1 N

NB = 1 N

MA = 1 kg

Given Block A: ΣF x = 0;

T1 − μ A NA − MA g sin ( θ ) = 0

ΣF y = 0;

NA − MA g cos ( θ ) = 0

Plate B: ΣF x = 0; ΣF y = 0; Peg C:

T2 − MB g sin ( θ ) + μ B NB + μ A NA = 0 NB − NA − MB g cos ( θ ) = 0 μ Cπ T2 = T1 e 856



Chapter 8

⎛ T1 ⎞ ⎜ ⎟ ⎜ T2 ⎟ ⎜ NA ⎟ = Find ( T , T , N , N , M ) 1 2 A B A ⎜ ⎟ ⎜ NB ⎟ ⎜M ⎟ ⎝ A⎠ MA = 2.22 kg

Problem 8-98 The simple band brake is constructed so that the ends of the friction strap are connected to the pin at A and the lever arm at B. If the wheel is subjected to a torque M, determine the smallest force P applied to the lever that is required to hold the wheel stationary. The coefficient of static friction between the strap and wheel is μs. Given: M = 80 lb⋅ ft

β = 45 deg

μ s = 0.5

r = 1.25 ft

α = 20 deg

a = 1.5 ft

b = 3 ft Solution: The initial guesses: T1 = 10 lb

T2 = 20 lb

P = 30 lb

Given T1 r + M − T2 r = 0 μ s( π + α + β ) T2 = T1 e

T2 sin ( β ) a − ( a + b)P = 0

857



Chapter 8

⎛ T1 ⎞ ⎜ ⎟ ⎜ T2 ⎟ = Find ( T1 , T2 , P) ⎜P⎟ ⎝ ⎠ ⎛ T1 ⎞ ⎛ 8.56 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ T2 ⎠ ⎝ 72.56 ⎠

P = 17.10 lb

Problem 8-99 The uniform beam of weight W1 is supported by the rope which is attached to the end of the beam, wraps over the rough peg, and is then connected to the block of weight W2. If the coefficient of static friction between the beam and the block, and between the rope and the peg, is μs, determine the maximum distance that the block can be placed from A and still remain in equilibrium. Assume the block will not tip. Given: W1 = 50 lb W2 = 100 lb

μ s = 0.4 a = 1 ft b = 10 ft Solution: Block: ΣF y = 0; N − W2 = 0 N = W2

N = 100.00 lb

ΣF x = 0; T1 − μ s N = 0 T1 = μ s N

T1 = 40.00 lb

⎛π⎞ μ s⎜ ⎟ 2 T2 = T1 e ⎝ ⎠

T2 = 74.97 lb

System: b⎞ ⎟ + T2 b = 0 ⎝ 2⎠

ΣΜA = 0; −W2 d − T1 a − W1 ⎛⎜

858



Chapter 8

T2 b − T1 a − W1 ⎛⎜

b⎞

⎟ ⎝ 2⎠

d =

W2

d = 4.6 ft

Problem 8-100 The uniform concrete pipe has weight W and is unloaded slowly from the truck bed using the rope and skids shown. If the coefficient of kinetic friction between the rope and pipe is μk ,determine the force the worker must exert on the rope to lower the pipe at constant speed. There is a pulley at B, and the pipe does not slip on the skids. The lower portion of the rope is parallel to the skids. Given: W = 800 lb

μ k = 0.3 α = 15 deg β = 30 deg Solution: −W r sin ( β ) + T2 cos ( α ) ( r cos ( α ) + r cos ( β ) ) + T2 sin ( α ) ( r sin ( α ) + r sin ( β ) ) = 0

T2 =

W sin ( β )

T2 = 203.47 lb

1 + cos ( α ) cos ( β ) + sin ( α ) sin ( β ) − μ k( π + β − α )

T1 = T2 e

T1 = 73.3 lb

Problem 8-101 A cord having a weight density γ and a total length L is suspended over a peg P as shown. If the coefficient of static friction between the peg and cord is μs ,determine the longest length h which one side of the suspended cord can have without causing motion. Neglect the size of the peg and the length of cord draped over it.

859



Chapter 8

Given:

γ = 0.5

lb ft

L = 10 ft

μ s = 0.5 Solution: μβ T2 = T1 e μ sπ γ h = γ ( L − h) e

T 1 = γ ( L − h)

T2 = γ h

⎛ eμ sπ ⎞ ⎟ h = L⎜ ⎜ μ sπ ⎟ ⎝1 + e ⎠

h = 8.28 ft

Problem 8-102 Granular material, having a density ρ is transported on a conveyor belt that slides over the fixed surface, having a coefficient of kinetic friction of μk. Operation of the belt is provided by a motor that supplies a torque M to wheel A.The wheel at B is free to turn, and the coefficient of static friction between the wheel at A and the belt is μΑ. If the belt is subjected to a pretension T when no load is on the belt, determine the greatest volume V of material that is permitted on the belt at any time without allowing the belt to stop. What is the torque M required to drive the belt when it is subjected to this maximum load? Units used: 6

Mg = 10 g Given: r = 100 mm

μ A = 0.4 μ k = 0.3 kg

ρ = 1500

m

3

T = 300 N g = 9.81

m 2

s

860



Solution:

Chapter 8

(μ A)π

T2 = T e

T2 = 1053.9 N

− M − T r + T2 r = 0

M = − T r + T2 r

T2 − μ k m1 g − T = 0

m1 =

Wheel A : ΣMA = 0; Belt ΣF x = 0;

V =

m1

T2 − T

μk g

M = 75.4 N⋅ m

m1 = 256.2 kg

V = 0.17 m

ρ

3

Problem 8-103 Blocks A and B have a mass MA and MB, respectively. If the coefficient of static friction between A and B and between B and C is μs and between the ropes and the pegs D and E μ's, determine the smallest force F needed to cause motion of block B . Units Used: 3

kN = 10 N Given:

θ = 45 deg

μ s = 0.25

MA = 100 kg

μ's = 0.5

MB = 150 kg

P = 30 N

g = 9.81

m 2

s Solution:

Assume no slipping between A & B. Guesses F = 1N NBC = 1 N

NAB = 1 N F BE = 1 N

F AB = 1 N F AD = 1 N

Given

⎛π ⎞ μ 's⎜ + θ ⎟ 2 ⎠ F = FBE e ⎝

π μ 's 2 F AD = P e 861



Chapter 8

F BE cos ( θ ) − μ s NBC − FAB = 0 F BE sin ( θ ) − NAB − MB g + NBC = 0 −F AD + F AB = 0 NAB − MA g = 0

⎛ F ⎞ ⎜N ⎟ ⎜ AB ⎟ ⎜ FAB ⎟ ⎜ ⎟ = Find ( F , NAB , FAB , NBC , FBE , FAD) NBC ⎜ ⎟ ⎜ FBE ⎟ ⎟ ⎜ ⎝ FAD ⎠ F ABmax = μ s NAB

Now check assumption Since F AB = 65.8 N

< FABmax = 245.3 N

⎛ F ⎞ ⎛ 2.49 ⎞ ⎜N ⎟ ⎜ ⎟ ⎜ AB ⎟ ⎜ 0.98 ⎟ ⎜ FAB ⎟ ⎜ 0.07 ⎟ ⎜ ⎟=⎜ ⎟ kN ⎜ NBC ⎟ ⎜ 1.91 ⎟ ⎜ FBE ⎟ ⎜ 0.77 ⎟ ⎟ ⎜ 0.07 ⎟ ⎜ ⎠ ⎝ FAD ⎠ ⎝ F ABmax = 245.25 N

then our assumption is correct

F = 2.49 kN

Problem 8-104 Blocks A and B weigh W1 and W2, respectively. Using the coefficients of static friction indicated, determine the greatest weight W of block E without causing motion. Given: W1 = 50 lb W2 = 30 lb

d = 12

μ A = 0.3

a = 1.5 ft

μ B = 0.5

b = 2 ft

μ C = 0.2

c = 5

μ D = 0.3

862



Chapter 8

Solution:

Assume that the wedge slips on the ground, but the block does not slip on the wedge and the block does not tip.

Guesses

W = 1 lb

TD = 1 lb TC = 1 lb

NA = 1 lb

F A = 1 lb NB = 1 lb

x = 1 ft

F Amax = 1 lb

Given π μC W 2 = TC e 2

⎛

π μD W 2 = TD e 2

c ⎞+N ⎛ ⎞+T =0 B D ⎟ ⎜ ⎟ 2 2 2 2 ⎝ c +d ⎠ ⎝ c +d ⎠

F A − μ B NB⎜

⎛

d

c ⎞+μ N ⎛ ⎞−W −N =0 B B 2 A ⎟ ⎜ 2 2 2 2⎟ ⎝ c +d ⎠ ⎝ c +d ⎠

NB⎜

d

TC − F A = 0

NA − W1 = 0

−TC b + NA x = 0

⎛ W ⎞ ⎜ T ⎟ ⎜ D ⎟ ⎜ TC ⎟ ⎜ ⎟ NA ⎜ ⎟ = Find ( W , T , T , N , F , N , x , F D C A A B Amax ) ⎜ FA ⎟ ⎜ ⎟ ⎜ NB ⎟ ⎜ x ⎟ ⎜ ⎟ F Amax ⎠ ⎝

F Amax = μ A NA

W = 8.15 lb

863



Chapter 8

Check assumptions Since FA = 2.97 lb < FAmax = 15.00 lb then the block does not slip relative to the wedge.

Since x = 0.12 ft
0, then the vertical position is stable.

Problem 11-38 If each of the three links of the mechanism has a weight W, determine the angle θ for equilibrium. The spring, which always remains vertical, is unstretched when θ = 0°. Solution: V=

1 2 k ( a sin ( θ ) ) − 2W a sin ( θ ) − W( 2a) sin ( θ ) 2

V=

ka 2 sin ( θ ) − 4W a sin ( θ ) 2

2

d dθ

V = k a sin ( θ ) cos ( θ ) − 4W a cos ( θ ) = 0 2

cos ( θ ) = 0

θ = 90 deg

sin ( θ ) =

θ = asin ⎛⎜

4W ka

4W ⎞ ⎟ ⎝ ka ⎠

Problem 11-39 The small postal scale consists of a counterweight W1 connected to the members having negligible weight. Determine the weight W2 that is on the pan in terms of the angles θ and φ and the dimensions shown. All members are pin connected. 1107



Chapter 11

Solution:

φ = −θ + constant V = W2 a sin ( φ ) − W1 b cos ( θ ) V' =

d dθ

V = −W2 a cos ( φ ) + W1 b sin ( θ ) = 0

⎛ b sin ( θ ) ⎞ ⎟ ⎝ a cos ( φ ) ⎠

W2 = W1 ⎜

Problem 11-40 The uniform right circular cone having a mass m is suspended from the cord as shown. Determine the angle θ at which it hangs from the wall for equilibrium. Is the cone in stable equilibrium? Solution:

⎛ 3a cos ( θ ) + a sin ( θ )⎞ m g ⎟ 4 ⎝2 ⎠

V = −⎜

⎛ −3a sin ( θ ) + a cos ( θ )⎞ m g ⎟ 4 ⎝ 2 ⎠

V' =

dV

V'' =

d V

= −⎜

dθ 2

dθ

2

⎛ −3a cos ( θ ) − a sin ( θ )⎞ m g ⎟ 4 ⎝ 2 ⎠

= −⎜

Equilibrium V' = 0

3 1 sin ( θ ) = cos ( θ ) 2 4

tan ( θ ) =

1 6

1 θ = atan ⎛⎜ ⎟⎞

⎝ 6⎠

θ = 9.462 deg

1108



V'' = −⎛⎜

−3

⎝2

cos ( θ ) −

Chapter 11

1 4

sin ( θ )⎞⎟ a m g

V'' = 1.521 amg

⎠

Stable

Problem 11-41 The homogeneous cylinder has a conical cavity cut into its base as shown. Determine the depth d of the cavity so that the cylinder balances on the pivot and remains in neutral equilibrium. Given: a = 50 mm b = 150 mm Solution:

yc =

b 2 d⎛1 2 ⎞ a π b − ⎜ π a d⎟ 2 4⎝3 ⎠ 2

πa b −

1 2 πa d 3

V = ( yc − d) cos ( θ ) W

d dθ

V = −W sin ( θ ) ( yc − d)

θ = 0 deg d2 dθ

2

(equilibrium position)

V = −W cos ( θ ) ( yc − d) = 0

d = yc Guess

Given

d = 10 mm

d=

d⎛1 2 ⎞ b 2 a π b − ⎜ π a d⎟ 4⎝3 2 ⎠ 2

πa b −

1 2 πa d 3

d = Find ( d)

d = 87.868 mm

1109



Chapter 11

Problem 11-42 The conical manhole cap is made of concrete and has the dimensions shown. Determine the critical location h = hcr of the pick-up connectors at A and B so that when hoisted with constant velocity the cap is in neutral equilibrium. Explain what would happen if the connectors were placed at a point h > hcr. Given: a = 2 ft b = 2.5 ft c = 3 ft d = 5 ft e = d − ( b − a) Solution: V = W( yc − h) cos ( θ ) d dθ

V = W( h − yc) sin ( θ ) = 0

Equilibrium at sin ( θ ) = 0

θ = 0deg

For neutral equilibrium require d

2

dθ

2

V = W( h − yc) cos ( θ ) = 0

Thus

yc = h

Thus, A and B must be at the elevation of the center of gravity of the cap. c1 =

cd d−b

c2 =

2 2 2 2 ⎛ d ⎞ ⎛⎜ c1 ⎟⎞ ⎛⎜ c1 ⎟⎞ − ⎛ b ⎞ ⎛⎜ c1 − c ⎟⎞ ⎛⎜ c1 + 3c ⎞⎟ − ⎛ e ⎞ ⎛⎜ c2 ⎟⎞ ⎛⎜ c2 ⎟⎞ + ⎛ a ⎞ ⎛⎜ c2 − c ⎟⎞ ⎛⎜ c2 + 3c ⎞⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ 3 ⎠⎝ 4 ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠⎝ 4 ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠⎝ 4 ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠⎝ 4 ⎠ yc = 2 2 2 2 ⎛ d ⎞ ⎛⎜ c1 ⎟⎞ − ⎛ b ⎞ ⎛⎜ c1 − c ⎟⎞ − ⎛ e ⎞ ⎛⎜ c2 ⎟⎞ + ⎛ a ⎞ ⎛⎜ c2 − c ⎟⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 2⎠ ⎝ 3 ⎠ ⎝ 2⎠ ⎝ 3 ⎠ ⎝ 2⎠ ⎝ 3 ⎠ ⎝ 2⎠ ⎝ 3 ⎠

hcr = yc

hcr = 1.32 ft

If h > hcr then stable.

1110


ce e−a


Chapter 11

Problem 11-43 Each bar has a mass per length of m0. Determine the angles θ and φ at which they are suspended in equilibrium. The contact at A is smooth, and both are pin con-nected at B. Solution: 1 θ + φ = atan ⎛⎜ ⎟⎞

⎝ 2⎠

V=−

d dθ

l 3l l ⎞ ⎛ 3l ⎞ ⎛l⎞ ⎛ m0 ⎜ ⎟ cos ( θ ) − l m0 ⎜ ⎟ cos ( φ ) − m0 ⎜ l cos ( φ ) + sin ( φ )⎟ 4 2 2 ⎝ ⎝4⎠ ⎝ 2⎠ ⎠

V =

Guess

9m0 l 8

2

sin ( θ ) − m0 l sin ( φ ) +

θ = 10 deg

Given

2

8

2

cos ( φ ) = 0

φ = 10 deg

1 θ + φ = atan ⎛⎜ ⎟⎞

⎛θ⎞ ⎜ ⎟ = Find ( θ , φ ) ⎝φ⎠

m0 l

⎝ 2⎠

9 1 sin ( θ ) − sin ( φ ) + cos ( φ ) = 0 8 8

⎛ θ ⎞ ⎛ 9.18 ⎞ ⎜ ⎟=⎜ ⎟ deg ⎝ φ ⎠ ⎝ 17.38 ⎠

Problem 11-44 The triangular block of weight W rests on the smooth corners which are a distance a apart. If the block has three equal sides of length d, determine the angle θ for equilibrium.

1111



Chapter 11

a 60(+

60(-

c b

60(

Solution: a b = sin ( 60 deg) sin ( 60 deg − θ )

b=a

sin ( 60 deg − θ ) sin ( 60 deg)

⎞ ⎛2 V = W⎜ d sin ( 60 deg) cos ( θ ) − b cos ( 30 deg − θ )⎟ 3 ⎝ ⎠ V=

( 2d cos (θ ) − 2a cos (2θ ) − a)

W 2 3

d dθ

V =

W

(−2d sin (θ ) + 8a sin ( θ ) cos ( θ )) = 0

2 3

θ 1 = asin ( 0)

θ 1 = 0 deg d⎞ ⎟ ⎝ 4a ⎠

θ 2 = acos ⎛⎜

Problem 11-45 A homogeneous cone rests on top of the cylindrical surface. Derive a relationship between the radius r of the cylinder and the height h of the cone for neutral equilibrium. Hint: Establish the potential function for a small angle θ of tilt of the cone, i.e., approximate sin θ ≈ 0 and cos θ ≈ 1−θ 2/2.

1112



Chapter 11

Solution: V=

V app

d dθ

h⎞ ⎤ ⎟ cos ( θ ) + rθ sin ( θ )⎥ W 4⎠ ⎦

⎡⎛r + ⎢⎜ ⎣⎝

⎡⎛ h ⎞ ⎛ θ 2 ⎞ ⎤ ⎟ + rθ 2⎥ W = ⎢⎜ r + ⎟ ⎜ 1 − 2⎠ ⎣⎝ 4 ⎠ ⎝ ⎦ ⎡⎛ ⎣⎝

V app = ⎢−⎜ r +

dVapp dθ d

⎛ ⎝

= ⎜r −

h⎞ ⎤ ⎟ θ + 2rθ⎥ W = 0 4⎠ ⎦

h⎞ ⎟ θW = 0 4⎠

2

h

V = r− = 0 2 app 4 dθ Equilibrium

θ = 0 deg For neutral equilibrium: r=

h 4

Problem 11-46 The door has a uniform weight W1. It is hinged at A and is held open by the weight W2 and the pulley. Determine the angle θ for equilibrium. Given: W1 = 50 lb W2 = 30 lb

1113



Chapter 11

a = 6 ft b = 6 ft

Solution: V = W1 ⎛⎜

b⎞

2 2 ⎟ sin ( θ ) + W2 a + b − 2a b sin ( θ )

⎝ 2⎠

d dθ

a b cos ( θ ) ⎞=0 ⎛ ⎟ cos ( θ ) − W2 ⎜ ⎟ 2 2 ⎝ 2⎠ ⎝ a + b − 2a b sin ( θ ) ⎠

V = W1 ⎛⎜

b⎞

Guess

θ = 10 deg

Given

W1 ⎛⎜

a b cos ( θ ) ⎞=0 ⎛ ⎟ cos ( θ ) − W2 ⎜ ⎟ 2 2 2 ⎝ ⎠ ⎝ a + b − 2a b sin ( θ ) ⎠ b⎞

θ = Find ( θ ) θ = 16.26 deg

Problem 11-47 The hemisphere of weight W supports a cylinder having a specific weight γ. If the radii of the cylinder and hemisphere are both a., determine the height h of the cylinder which will produce neutral equilibrium in the position shown. Given: W = 60 lb a = 5 in

γ = 311

lb ft

3

Solution: 2 ⎛ h⎞ ⎟ cos ( θ ) + γ π a h⎜ ⎟ cos ( θ ) ⎝8⎠ ⎝ 2⎠

V = −W⎛⎜

3a ⎞

1114



V=

Chapter 11

⎛ γ π a2 h2 W3a ⎞ ⎜ ⎟ cos ( θ ) − 8 ⎠ ⎝ 2

⎛ γ π a2 h2 W3a ⎞ ⎟ sin ( θ ) V = −⎜ − 8 ⎠ ⎝ 2 dθ d

⎛ γ π a2 h2 W3a ⎞ ⎟ cos ( θ ) V = −⎜ − 2 8 ⎠ ⎝ 2 dθ d

2

For neutral equilibrium we must have 2 2

γ πa h 2

−

W3a =0 8

h =

W3 4π γ a

h = 3.99 in

Problem 11-48 Compute the force developed in the spring required to keep the rod of mass Mrod in equilibrium at θ. The spring remains horizontal due to the roller guide. Given: k = 200

N m

M = 40 N⋅ m a = 0.5 m

θ = 30 deg Mrod = 6 kg

Solution:

⎛ a ⎞ sin ( θ ) + 1 k ( a cos ( θ ) − δ ) 2 ⎟ 2 ⎝ 2⎠

V = Mθ + Mrod g⎜ d dθ

⎛ a ⎞ cos ( θ ) − k( a cos ( θ ) − δ ) a sin ( θ ) = 0 ⎟ ⎝ 2⎠

V = M + Mrod g⎜

1115



Guess

Chapter 11

δ = 100 mm

⎛ a ⎞ cos ( θ ) − k( a cos ( θ ) − δ ) a sin ( θ ) = 0 ⎟ ⎝ 2⎠

Given M + Mrod g⎜

δ = Find ( δ )

F = k( a cos ( θ ) − δ )

δ = −0.622 m

F = 211.0 N

Problem 11-49 Determine the force P acting on the cord which is required to maintain equilibrium of the horizontal bar CB of mass M. Hint: First show that the coordinates sA and sB are related to the constant vertical length l of the cord by the equation 5sB − sA = L. Given: M = 20 kg Solution: L = 4sB + ( sB − sA) L = 5sB − sA

Δ L = 5Δ sB − Δ sA = 0 Δ sA = 5ΔsB V = −M g sB + P sA

Δ V = −M gΔ sB + P ΔsA = ( −M g + 5P) Δ sB = 0 P =

Mg 5

P = 39.2 N

1116



Chapter 11

Problem 11-50 The uniform bar AB has weight W. If the attached spring is unstretched when θ = 90 deg, use the method of virtual work and determine the angle θ for equilibrium. Note that the spring always remains in the vertical position due to the roller guide. Given: W = 10 lb k = 5

lb ft

a = 4 ft Solution: y = a sin ( θ )

δy = a cos ( θ ) δθ

δU = ( −W + Fs) δ y = ⎣⎡k( a − a sin ( θ ) ) − W⎦⎤ a cos ( θ ) δθ = 0 cos ( θ 1 ) = 0 sin ( θ 2 ) = 1 −

θ 1 = acos ( 0) W ka

θ 2 = asin ⎛⎜ 1 −

⎝

θ 1 = 90 deg W⎞ ⎟ ka⎠

θ 2 = 30 deg

Problem 11-51 The uniform bar AB has weight W. If the attached spring is unstretched when θ = 90 deg, use the principle of potential energy and determine the angle θ for equilibrium. Investigate the stability of the equilibrium positions. Note that the spring always remains in the vertical position due to the roller guide. Given: W = 10 lb k = 5

lb ft

a = 4 ft

1117



Chapter 11

Solution: V = W a sin ( θ ) +

1 1 2 2 2 k ( a − a sin ( θ ) ) = W a sin ( θ ) + k a ( 1 − sin ( θ ) ) 2 2

Equilibrium d

V = W a cos ( θ ) − k a ( 1 − sin ( θ ) ) cos ( θ ) = 0 2

dθ

cos ( θ 1 ) = 0 sin ( θ 2 ) = 1 −

θ 1 = acos ( 0)

Check Stability 2

dθ

⎝

W⎞ ⎟ ka⎠

θ 2 = 30 deg

If V'' > 0 the equilibrium point is stable. If V'' < 0, then unstable

= −W a sin ( θ ) + k a sin ( θ ) + k a cos ( 2θ )

d V

V'' =

θ 2 = asin ⎛⎜ 1 −

W ka

θ 1 = 90 deg

2

2

2

V'' 1 = −W a sin ( θ 1 ) + k a sin ( θ 1 ) + k a cos ( 2θ 1 )

V'' 1 = −40 lb⋅ ft

V'' 2 = −W a sin ( θ 2 ) + k a sin ( θ 2 ) + k a cos ( 2θ 2 )

V'' 2 = 60 lb⋅ ft

2 2

2 2

Problem 11-52 The punch press consists of the ram R, connecting rod AB, and a flywheel. If a torque M is applied to the flywheel, determine the force F applied at the ram to hold the rod in the position θ = θ0. Given: M = 50 N⋅ m

θ 0 = 60 deg r = 0.1 m a = 0.4 m Solution:

θ = θ0

Free Body Diagram: The system has only one degree of freedom defined by the independent coordinate θ. When θ undergoes a positive displacement δθ, only force F and Moment M do work. a = x + r − 2x r cos ( θ ) 2

2

2

⎛ x r sin ( θ ) ⎞ δθ ⎟ ⎝ r cos ( θ ) − x ⎠

0 = 2xδx − 2r cos ( θ ) δx + 2x r sin ( θ ) δθ

δx = ⎜

⎡ ⎛ x r sin ( θ ) ⎞ − M⎤ δθ = 0 ⎟ ⎥ ⎣ ⎝ r cos ( θ ) − x ⎠ ⎦

δU = −F δx − Mδθ = ⎢−F ⎜

1118



Guesses Given

Chapter 11

F = 1N

x = 0.1 m

a = x + r − 2x r cos ( θ ) 2

⎛x⎞ ⎜ ⎟ = Find ( x , F) ⎝F⎠

2

2

x = 0.441 m

⎛ x r sin ( θ ) ⎞ − M = 0 ⎟ ⎝ r cos ( θ ) − x ⎠

−F ⎜

F = 512 N

1119


Engineering Mechanics Statics Eleventh Edition- R - Instructor's Solutions Manual

Engineering Mechanics - Statics (10th Edition) SOLUTION MANUAL

Engineering Mechanics: Statics (12th Edition)