Mechanics Of Materials 9th Edition Hibbeler Solutions Manual

Mechanics©Of Materials 9th Edition Hibbeler Solutions Manual 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Full Download:exist. https://testbanklive.com/download/mechanics-of-materials-9th-edition-hibbeler-solutions-manual/ 3–1. A tension test was performed on a steel specimen having an original diameter of 0.503 in. and gauge length of 2.00 in. The data is listed in the table. Plot the stress–strain diagram and determine approximately the modulus of elasticity, the yield stress, the ultimate stress, and the rupture stress. Use a scale of 1 in. = 20 ksi and 1 in. = 0.05 in.> in. Redraw the elastic region, using the same stress scale but a strain scale of 1 in. = 0.001 in.> in.

A =

Load (kip) Elongation (in.)

0 1.50 4.60 8.00 11.00 11.80 11.80 12.00 16.60 20.00 21.50 19.50 18.50

1 p(0.503)2 = 0.1987 in2 4

L = 2.00 in. s(ksi)

0 0.0005 0.0015 0.0025 0.0035 0.0050 0.0080 0.0200 0.0400 0.1000 0.2800 0.4000 0.4600

e(in.>in.)

0

0

7.55

0.00025

23.15

0.00075

40.26

0.00125

55.36

0.00175

59.38

0.0025

59.38

0.0040

60.39

0.010

83.54

0.020

100.65

0.050

108.20

0.140

98.13

0.200

93.10 Eapprox

0.230 48 = = 32.0(103) ksi 0.0015

Ans.

Ans: (sult)approx = 110 ksi, (sR)approx = 93.1 ksi, (sY)approx = 55 ksi, Eapprox = 32.0(103) ksi 139

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3–2. Data taken from a stress–strain test for a ceramic are given in the table. The curve is linear between the origin and the first point. Plot the diagram, and determine the modulus of elasticity and the modulus of resilience.

Modulus of Elasticity: From the stress–strain diagram E =

33.2 - 0 = 55.3 A 103 B ksi 0.0006 - 0

S (ksi)

P (in./in.)

0 33.2 45.5 49.4 51.5 53.4

0 0.0006 0.0010 0.0014 0.0018 0.0022

Ans.

Modulus of Resilience: The modulus of resilience is equal to the area under the linear portion of the stress–strain diagram (shown shaded). ur =

1 lb in. in # lb (33.2) A 103 B ¢ 2 ≤ ¢ 0.0006 ≤ = 9.96 2 in. in in3

Ans.

Ans: in # lb E = 55.3 A 103 B ksi, ur = 9.96 in3 140

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3–3. Data taken from a stress–strain test for a ceramic are given in the table. The curve is linear between the origin and the first point. Plot the diagram, and determine approximately the modulus of toughness. The rupture stress is sr = 53.4 ksi.

Modulus of Toughness: The modulus of toughness is equal to the area under the stress-strain diagram (shown shaded). (ut)approx =

S (ksi)

P (in./in.)

0 33.2 45.5 49.4 51.5 53.4

0 0.0006 0.0010 0.0014 0.0018 0.0022

1 lb in.

(33.2) A 103 B ¢ 2 ≤ (0.0004 + 0.0010) ¢ ≤ 2 in. in + 45.5 A 103 B ¢ +

1 lb in.

(7.90) A 103 B ¢ 2 ≤ (0.0012) ¢ ≤ 2 in. in +

= 85.0

lb in. ≤ (0.0012) ¢ ≤ in. in2

lb in.

1 (12.3) A 103 B ¢ 2 ≤ (0.0004) ¢ ≤ 2 in. in

in # lb in3

Ans.

Ans: in # lb (ut)approx = 85.0 in3 141

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*3–4. A tension test was performed on a steel specimen having an original diameter of 0.503 in. and a gauge length of 2.00 in. The data is listed in the table. Plot the stress–strain diagram and determine approximately the modulus of elasticity, the ultimate stress, and the rupture stress. Use a scale of 1 in. = 15 ksi and 1 in. = 0.05 in.> in.

Redraw the linear-elastic region, using the same stress scale but a strain scale of 1 in. = 0.001 in.

A =

1 p(0.503)2 = 0.19871 in2 4

L = 2.00 in. s =

P A (ksi)

P =

¢L L (in.>in.)

0

0

12.58

0.00045

32.71

0.00125

42.78

0.0020

46.30

0.00325

49.32

0.0049

60.39

0.02

70.45

0.06

72.97

0.125

70.45

0.175

66.43

0.235

Eapprox =

32.71 = 26.2(103) ksi 0.00125

Ans.

142

Load (kip)

Elongation (in.)

0 2.50 6.50 8.50 9.20 9.80 12.0 14.0 14.5 14.0 13.2

0 0.0009 0.0025 0.0040 0.0065 0.0098 0.0400 0.1200 0.2500 0.3500 0.4700

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3–5. A tension test was performed on a steel specimen having an original diameter of 0.503 in. and gauge length of 2.00 in. Using the data listed in the table, plot the stress–strain diagram and determine approximately the modulus of toughness.

Modulus of toughness (approx) ut = total area under the curve (1)

= 87 (7.5) (0.025) = 16.3

in. # kip

Load (kip)

Elongation (in.)

0 2.50 6.50 8.50 9.20 9.80 12.0 14.0 14.5 14.0 13.2

0 0.0009 0.0025 0.0040 0.0065 0.0098 0.0400 0.1200 0.2500 0.3500 0.4700

Ans.

in3

In Eq.(1), 87 is the number of squares under the curve. s =

P A (ksi)

P =

¢L L (in.>in.)

0

0

12.58

0.00045

32.71

0.00125

42.78

0.0020

46.30

0.00325

49.32

0.0049

60.39

0.02

70.45

0.06

72.97

0.125

70.45

0.175

66.43

0.235

Ans: in. # kip ut = 16.3 in3 143

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3–6. A specimen is originally 1 ft long, has a diameter of 0.5 in., and is subjected to a force of 500 lb. When the force is increased from 500 lb to 1800 lb, the specimen elongates 0.009 in. Determine the modulus of elasticity for the material if it remains linear elastic.

Normal Stress and Strain: Applying s =

s1 =

s2 =

¢P =

0.500 p 2 4 (0.5 )

1.80 p 2 4 (0.5 )

dL P and e = .

A L

= 2.546 ksi

= 9.167 ksi

0.009 = 0.000750 in.>in.

12

Modulus of Elasticity: E =

¢s 9.167 - 2.546 = = 8.83 A 103 B ksi ¢P 0.000750

Ans.

Ans:

E = 8.83 A 103 B ksi

144

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3–7. A structural member in a nuclear reactor is made of a zirconium alloy. If an axial load of 4 kip is to be supported by the member, determine its required cross-sectional area.

Use a factor of safety of 3 relative to yielding. What is the load on the member if it is 3 ft long and its elongation is 0.02 in.? Ezr = 14(103) ksi, sY = 57.5 ksi. The material has elastic behavior.

Allowable Normal Stress: F.S. =

3 =

sy sallow 57.5 sallow

sallow = 19.17 ksi sallow =

P A

19.17 =

4 A

A = 0.2087 in2 = 0.209 in2

Ans.

Stress–Strain Relationship: Applying Hooke’s law with P =

0.02 d = = 0.000555 in.>in.

L 3 (12) s = EP = 14 A 103 B (0.000555) = 7.778 ksi

Normal Force: Applying equation s =

P .

A

P = sA = 7.778 (0.2087) = 1.62 kip

Ans.

Ans: A = 0.209 in2, P = 1.62 kip 145

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*3–8. The strut is supported by a pin at C and an A-36 steel guy wire AB. If the wire has a diameter of 0.2 in., determine how much it stretches when the distributed load acts on the strut.

A

60⬚ 200 lb/ft

a + ©MC = 0;

1 FAB cos 60°(9) - (200)(9)(3) = 0 2

9 ft

FAB = 600 lb

The normal stress the wire is sAB =

FAB = AAB

p 4

600 = 19.10(103) psi = 19.10 ksi (0.22)

Since sAB 6 sy = 36 ksi, Hooke’s Law can be applied to determine the strain in wire. sAB = EPAB;

19.10 = 29.0(103)PAB PAB = 0.6586(10 - 3) in>in

9(12) = 124.71 in. Thus, the wire The unstretched length of the wire is LAB = sin 60° stretches dAB = PAB LAB = 0.6586(10 - 3)(124.71) = 0.0821 in.

Ans.

146

B

C

Here, we are only interested in determining the force in wire AB.

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s (psi)

3–9. The s-P diagram for elastic fibers that make up human skin and muscle is shown. Determine the modulus of elasticity of the fibers and estimate their modulus of toughness and modulus of resilience.

55

11 1

E =

11 = 5.5 psi 2

Ans.

ut =

1 1 (2)(11) + (55 + 11)(2.25 - 2) = 19.25 psi 2 2

Ans.

ur =

1 (2)(11) = 11 psi 2

Ans.

2 2.25

P (in./in.)

Ans: E = 5.5 psi, ut = 19.25 psi, ur = 11 psi 147

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s (ksi)

3–10. The stress–strain diagram for a metal alloy having an original diameter of 0.5 in. and a gauge length of 2 in. is given in the figure. Determine approximately the modulus of elasticity for the material, the load on the specimen that causes yielding, and the ultimate load the specimen will support.

105 90 75 60 45 30 15 0

0 0

0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.001 0.002 0.003 0.004 0.005 0.006 0.007

P (in./in.)

From the stress–strain diagram, Fig. a, 60 ksi - 0 E = ; 1 0.002 - 0 sy = 60 ksi

E = 30.0(103) ksi

Ans.

sult = 100 ksi

Thus, PY = sYA = 60 C p4 (0.52) D = 11.78 kip = 11.8 kip

Ans.

Pult = sult A = 100 C p4 (0.52) D = 19.63 kip = 19.6 kip

Ans.

Ans: E = 30.0(103) ksi, PY = 11.8 kip, Pult = 19.6 kip 148

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s (ksi)

3–11. The stress–strain diagram for a steel alloy having an original diameter of 0.5 in. and a gauge length of 2 in. is given in the figure. If the specimen is loaded until it is stressed to 90 ksi, determine the approximate amount of elastic recovery and the increase in the gauge length after it is unloaded.

105 90 75 60 45 30 15 0

0 0

0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.001 0.002 0.003 0.004 0.005 0.006 0.007

P (in./in.)

From the stress–strain diagram Fig. a, the modulus of elasticity for the steel alloy is E 60 ksi - 0 = ; 1 0.002 - 0

E = 30.0(103) ksi

when the specimen is unloaded, its normal strain recovers along line AB, Fig. a, which has a slope of E. Thus Elastic Recovery =

90 90 ksi = 0.003 in>in. = E 30.0(103) ksi

Ans.

Thus, the permanent set is PP = 0.05 - 0.003 = 0.047 in>in.

Then, the increase in gauge length is ¢L = PPL = 0.047(2) = 0.094 in.

Ans.

Ans: Elastic Recovery = 0.003 in.>in., ¢L = 0.094 in.

149

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s (ksi)

*3–12. The stress–strain diagram for a steel alloy having an original diameter of 0.5 in. and a gauge length of 2 in. is given in the figure. Determine approximately the modulus of resilience and the modulus of toughness for the material.

105 90 75 60 45 30 15 0

The Modulus of resilience is equal to the area under the stress–strain diagram up to the proportional limit. sPL = 60 ksi

PPL = 0.002 in.>in.

Thus, (ui)r =

1 1 in. # lb sPLPPL = C 60(103) D (0.002) = 60.0 2 2 in3

Ans.

The modulus of toughness is equal to the area under the entire stress–strain diagram. This area can be approximated by counting the number of squares. The total number is 38. Thus,

C (ui)t D approx = 38 c15(103)

lb in. in. # lb d a 0.05 b = 28.5(103) 2 in. in in3

150

Ans.

0 0

0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.001 0.002 0.003 0.004 0.005 0.006 0.007

P (in./in.)

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3–13. A bar having a length of 5 in. and cross-sectional area of 0.7 in.2 is subjected to an axial force of 8000 lb. If the bar stretches 0.002 in., determine the modulus of elasticity of the material. The material has linear-elastic behavior.

8000 lb

8000 lb 5 in.

Normal Stress and Strain:

s =

8.00 P = = 11.43 ksi A 0.7

P =

d 0.002 = = 0.000400 in.>in.

L 5

Modulus of Elasticity: E =

s 11.43 = = 28.6(103) ksi P 0.000400

Ans.

Ans: E = 28.6(103) ksi 151

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3–14. The rigid pipe is supported by a pin at A and an A-36 steel guy wire BD. If the wire has a diameter of 0.25 in., determine how much it stretches when a load of P = 600 lb acts on the pipe.

B

4 ft

P

A

D C 3 ft

3 ft

Here, we are only interested in determining the force in wire BD. Referring to the FBD in Fig. a a + ©MA = 0;

FBD A 45 B (3) - 600(6) = 0

FBD = 1500 lb

The normal stress developed in the wire is sBD =

FBD = ABD

p 4

1500 = 30.56(103) psi = 30.56 ksi (0.252)

Since sBD 6 sy = 36 ksi, Hooke’s Law can be applied to determine the strain in the wire. sBD = EPBD;

30.56 = 29.0(103)PBD PBD = 1.054(10 - 3) in.>in.

The unstretched length of the wire is LBD = 232 + 42 = 5 ft = 60 in. Thus, the wire stretches dBD = PBD LBD = 1.054(10 - 3)(60) = 0.0632 in.

Ans.

Ans: dBD = 0.0632 in.

152

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3–15. The rigid pipe is supported by a pin at A and an A-36 guy wire BD. If the wire has a diameter of 0.25 in., determine the load P if the end C is displaced 0.15 in. downward.

B

4 ft

P

A

D C 3 ft

3 ft

Here, we are only interested in determining the force in wire BD. Referring to the FBD in Fig. a FBD A 45 B (3) - P(6) = 0

a + ©MA = 0;

FBD = 2.50 P

The unstretched length for wire BD is LBD = 232 + 42 = 5 ft = 60 in. From the geometry shown in Fig. b, the stretched length of wire BD is LBD¿ = 2602 + 0.0752 - 2(60)(0.075) cos 143.13° = 60.060017 Thus, the normal strain is PBD =

LBD¿ - LBD 60.060017 - 60 = = 1.0003(10 - 3) in.>in.

LBD 60

Then, the normal stress can be obtain by applying Hooke’s Law. sBD = EPBD = 29(103) C 1.0003(10 - 3) D = 29.01 ksi Since sBD 6 sy = 36 ksi, the result is valid. sBD =

FBD ; ABD

29.01(103) =

2.50 P (0.252)

p 4

P = 569.57 lb = 570 lb

Ans.

Ans: P = 570 lb 153

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*3–16. The wire has a diameter of 5 mm and is made from A-36 steel. If a 80-kg man is sitting on seat C, determine the elongation of wire DE.

E W 600 mm D A

B 800 mm

Equations of Equilibrium: The force developed in wire DE can be determined by writing the moment equation of equilibrium about A with reference to the freebody diagram shown in Fig. a, a + ©MA = 0;

3 FDE a b(0.8) - 80(9.81)(1.4) = 0 5 FDE = 2289 N

Normal Stress and Strain: sDE =

FDE 2289 = = 116.58 MPa p ADE (0.0052) 4

Since sDE < sY , Hooke’s Law can be applied sDE = EPDE 116.58(106) = 200(109)PDE PDE = 0.5829(10-3) mm>mm The unstretched length of wire DE is LDE = 26002 + 8002 = 1000 mm. Thus, the elongation of this wire is given by dDE = PDELDE = 0.5829(10-3)(1000) = 0.583 mm

Ans.

154

C 600 mm

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s (ksi)

3–17. A tension test was performed on a magnesium alloy specimen having a diameter 0.5 in. and gauge length 2 in.

The resulting stress–strain diagram is shown in the figure.

Determine the approximate modulus of elasticity and the yield strength of the alloy using the 0.2% strain offset method.

40 35 30 25 20 15 10 5 0

0.002

0.004

0.006

0.008

0.010

P (in./in.)

Modulus of Elasticity: From the stress–strain diagram, when P = 0.002 in.>in., its corresponding stress is s = 13.0 ksi. Thus, Eapprox =

13.0 - 0 = 6.50(103) ksi 0.002 - 0

Ans.

Yield Strength: The intersection point between the stress–strain diagram and the straight line drawn parallel to the initial straight portion of the stress–strain diagram from the offset strain of P = 0.002 in.>in. is the yield strength of the alloy. From the stress–strain diagram, sYS = 25.9 ksi

Ans.

Ans: Eapprox = 6.50(103) ksi, sYS = 25.9 ksi 155

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s (ksi)

3–18. A tension test was performed on a magnesium alloy specimen having a diameter 0.5 in. and gauge length of 2 in.

The resulting stress–strain diagram is shown in the figure. If the specimen is stressed to 30 ksi and unloaded, determine the permanent elongation of the specimen.

40 35 30 25 20 15 10 5 0

0.002

0.004

0.006

0.008

0.010

P (in./in.)

Permanent Elongation: From the stress–strain diagram, the strain recovered is along the straight line BC which is parallel to the straight line OA. Since 13.0 - 0 = 6.50(103) ksi, then the permanent set for the specimen is Eapprox = 0.002 - 0 30(103) PP = 0.0078 = 0.00318 in.>in.

6.5(106) Thus, dP = PPL = 0.00318(2) = 0.00637 in.

Ans.

Ans: dP = 0.00637 in.

156

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3–19. The stress–strain diagram for a bone is shown, and can be described by the equation P = 0.45110-62 s ⫹ 0.36110-122 s3, where s is in kPa. Determine the yield strength assuming a 0.3% offset.

P

s

P ⫽ 0.45(10⫺6)s + 0.36(10⫺12)s3 P

P

P = 0.45(10-6)s + 0.36(10-12)s3, dP = A 0.45(10-6) + 1.08(10-12) s2 B ds

E =

ds 1 2 = = 2.22(106) kPa = 2.22 GPa dP 0.45(10 - 6) s=0

The equation for the recovery line is s = 2.22(106)(P - 0.003).

This line intersects the stress–strain curve at sYS = 2027 kPa = 2.03 MPa

Ans.

Ans: sYS = 2.03 MPa 157

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*3–20. The stress–strain diagram for a bone is shown and can be described by the equation P = 0.45110-62 s ⫹ 0.36110-122 s3, where s is in kPa. Determine the modulus of toughness and the amount of elongation of a 200-mm-long region just before it fractures if failure occurs at P = 0.12 mm>mm.

P

s

P ⫽ 0.45(10⫺6)s + 0.36(10⫺12)s3 P

When P = 0.12

120(10-3) = 0.45 s + 0.36(10-6)s3 Solving for the real root: s = 6873.52 kPa 6873.52

ut =

LA

dA =

L0

(0.12 - P)ds

6873.52

ut =

L0

(0.12 - 0.45(10-6)s - 0.36(10-12)s3)ds 6873.52

= 0.12 s - 0.225(10-6)s2 - 0.09(10-12)s4|0 = 613 kJ>m3

Ans.

d = PL = 0.12(200) = 24 mm

Ans.

158

P

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3–21. The two bars are made of polystyrene, which has the stress–strain diagram shown. If the cross-sectional area of bar AB is 1.5 in2 and BC is 4 in2, determine the largest force P that can be supported before any member ruptures.

Assume that buckling does not occur.

P 4 ft C

B

3 ft

A s (ksi) 25

+ c gFy = 0; + ; ©Fx = 0;

3 F - P = 0; 5 AB FBC

4 - (1.6667P) = 0; 5

FAB = 1.6667 P

20

(1)

15

FBC = 1.333 P

(2)

0

From the stress–strain diagram (sR)t = 5 ksi FBC ; ABC

5 =

tension

5

Assuming failure of bar BC:

s =

compression

10

FBC ; 4

0

0.20

0.40

0.60

0.80

P (in./in.)

FBC = 20.0 kip

From Eq. (2), P = 15.0 kip Assuming failure of bar AB: From stress–strain diagram (sR)c = 25.0 ksi s =

FAB ; AAB

25.0 =

FAB ; 1.5

FAB = 37.5 kip

From Eq. (1), P ⫽ 22.5 kip Choose the smallest value P = 15.0 kip

Ans.

Ans: P = 15.0 kip 159

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3–22. The two bars are made of polystyrene, which has the stress–strain diagram shown. Determine the cross-sectional area of each bar so that the bars rupture simultaneously when the load P = 3 kip. Assume that buckling does not occur.

P 4 ft C

B

3 ft

A s (ksi) 25

+ c ©Fy = 0; + : ©Fx = 0;

3 FBA a b - 3 = 0; 5 -FBC

4 + 5a b = 0; 5

20

FBA = 5 kip

15

FBC = 4 kip

compression

10 tension

5

For member BC:

0

(smax)t =

4 kip FBC ; ABC = = 0.8 in2 ABC 5 ksi

(smax)c =

FBA ; ABA

0

0.20

0.40

0.60

0.80

P (in./in.)

Ans.

For member BA:

ABA =

5 kip = 0.2 in2 25 ksi

Ans.

Ans: ABC = 0.8 in2, ABA = 0.2 in2 160

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3–23. The stress–strain diagram for many metal alloys can be described analytically using the Ramberg-Osgood three parameter equation P = s>E + ksn, where E, k, and n are determined from measurements taken from the diagram.

Using the stress–strain diagram shown in the figure, take E = 30(103) ksi and determine the other two parameters k and n and thereby obtain an analytical expression for the curve.

s (ksi) 80 60 40 20 0.1 0.2 0.3 0.4 0.5

P (10 – 6 )

Choose, s = 40 ksi, e = 0.1 s = 60 ksi, e = 0.3 0.1 =

40 + k(40)n 30(103)

0.3 =

60 + k(60)n 30(103)

0.098667 = k(40)n 0.29800 = k(60)n 0.3310962 = (0.6667)n ln (0.3310962) = n ln (0.6667) n = 2.73

Ans.

k = 4.23(10 - 6)

Ans.

Ans: n = 2.73, k = 4.23(10 - 6) 161

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3–24. The wires AB and BC have original lengths of 2 ft 3 and 3 ft, and diameters of 81 in. and 16 in., respectively. If these wires are made of a material that has the approximate stress–strain diagram shown, determine the elongations of the wires after the 1500-lb load is placed on the platform.

C

Equations of Equilibrium: The forces developed in wires AB and BC can be determined by analyzing the equilibrium of joint B, Fig. a, + : ©Fx = 0; + c ©Fy = 0;

FBC sin 30° - FAB sin 45° = 0

(1)

FBC cos 30° + FAB cos 45° = 1500

(2)

A

3 ft 45⬚

30⬚

2 ft B

Solving Eqs. (1) and (2), FAB = 776.46 lb

FBC = 1098.08 lb

Normal Stress and Strain: sAB =

FAB 776.46 = = 63.27 ksi p AAB (1>8)2 4 s (ksi)

sBC =

FBC 1098.08 = = 39.77 ksi p ABC 2 (3>16) 4

80 58

The corresponding normal strain can be determined from the stress–strain diagram, Fig. b.

39.77 58 ; = PBC 0.002

PBC = 0.001371 in.>in.

63.27 - 58 80 - 58 = ; PAB - 0.002 0.01 - 0.002

PAB = 0.003917 in.>in.

0.002

Thus, the elongations of wires AB and BC are dAB = PABLAB = 0.003917(24) = 0.0940

Ans.

dBC = PBCLBC = 0.001371(36) = 0.0494

Ans.

162

0.01

P (in./in.)

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3–25. The acrylic plastic rod is 200 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it, determine the change in its length and the change in its diameter.

Ep = 2.70 GPa, np = 0.4.

s =

P = A

Plong =

300 p 2 4 (0.015)

300 N

300 N 200 mm

= 1.678 MPa

1.678(106) s = 0.0006288 = E 2.70(109)

d = Plong L = 0.0006288 (200) = 0.126 mm

Ans.

Plat = -nPlong = -0.4(0.0006288) = -0.0002515 ¢d = Platd = -0.0002515 (15) = -0.00377 mm

Ans.

Ans: d = 0.126 mm, ¢d = -0.00377 mm 163

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3–26. The thin-walled tube is subjected to an axial force of 40 kN. If the tube elongates 3 mm and its circumference decreases 0.09 mm, determine the modulus of elasticity, Poisson’s ratio, and the shear modulus of the tube’s material. The material behaves elastically.

40 kN 900 mm

10 mm

40 kN 12.5 mm

Normal Stress and Strain: s =

40(103) P = 226.35 MPa = A p(0.01252 - 0.012)

Pa =

3 d = = 3.3333 (10-3) mm>mm L 900

Applying Hooke’s law, s = EPa;

226.35(106) = E [3.3333(10-3)] E = 67.91(106) Pa = 67.9 GPa

Ans.

Poisson’s Ratio: The circumference of the loaded tube is 2p(12.5) - 0.09 = 78.4498 mm. Thus, the outer radius of the tube is r =

78.4498 = 12.4857 mm 2p

The lateral strain is Plat =

r - r0 12.4857 - 12.5 = = -1.1459(10-3) mm>mm r0 12.5

n = -

-1.1459(10-3) Plat d = 0.3438 = 0.344 = -c Pa 3.3333(10-3)

G =

Ans.

67.91(109) E = = 25.27(109) Pa = 25.3 GPa 2(1 + n) 2(1 + 0.3438)

Ans.

Ans: E = 67.9 GPa, v = 0.344, G = 25.3 GPa 164

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3–27. When the two forces are placed on the beam, the diameter of the A-36 steel rod BC decreases from 40 mm to 39.99 mm. Determine the magnitude of each force P.

C P 1m

A

P 1m

1m

1m

B 0.75 m

Equations of Equilibrium: The force developed in rod BC can be determined by writing the moment equation of equilibrium about A with reference to the free-body diagram of the beam shown in Fig. a.

4 FBC a b (3) - P(2) - P(1) = 0 5

a + ©MA = 0;

FBC = 1.25P

Normal Stress and Strain: The lateral strain of rod BC is Plat =

d - d0 39.99 - 40 = = -0.25(10 - 3) mm>mm d0 40

Plat = -nPa;

-0.25(10-3) = -(0.32)Pa Pa = 0.78125(10-3) mm>mm

Assuming that Hooke’s Law applies, sBC = EPa;

sBC = 200(109)(0.78125)(10-3) = 156.25 MPa

Since s 6 sY, the assumption is correct. sBC =

FBC ; ABC

156.25(106) =

1.25P p A 0.042 B 4

P = 157.08(103)N = 157 kN

Ans.

Ans: P = 157 kN 165

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*3–28. If P = 150 kN, determine the elastic elongation of rod BC and the decrease in its diameter. Rod BC is made of A-36 steel and has a diameter of 40 mm.

C P 1m

A

P 1m

1m

1m

B 0.75 m

Equations of Equilibrium: The force developed in rod BC can be determined by writing the moment equation of equilibrium about A with reference to the freebody diagram of the beam shown in Fig. a. a + ©MA = 0;

4 FBC a b (3) - 150(2) - 150(1) = 0 5

FBC = 187.5 kN

Normal Stress and Strain: The lateral strain of rod BC is sBC =

187.5(103) FBC = = 149.21 MPa p ABC A 0.042 B 4

Since s 6 sY, Hooke’s Law can be applied. Thus, sBC = EPBC;

149.21(106) = 200(109)PBC PBC = 0.7460(10-3) mm>mm

The unstretched length of rod BC is LBC = 27502 + 10002 = 1250 mm. Thus the elongation of this rod is given by dBC = PBCLBC = 0.7460(10-3)(1250) = 0.933 mm

Ans.

We obtain, Plat = -nPa ;

Plat = -(0.32)(0.7460)(10-3) = -0.2387(10-3) mm>mm

Thus, dd = Plat dBC = -0.2387(10-3)(40) = -9.55(10-3) mm

166

Ans.

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3–29. The friction pad A is used to support the member, which is subjected to an axial force of P = 2 kN. The pad is made from a material having a modulus of elasticity of E = 4 MPa and Poisson’s ratio n = 0.4. If slipping does not occur, determine the normal and shear strains in the pad.

The width is 50 mm. Assume that the material is linearly elastic. Also, neglect the effect of the moment acting on the pad.

P

60⬚

25 mm

A

100 mm

Internal Loading: The normal force and shear force acting on the friction pad can be determined by considering the equilibrium of the pin shown in Fig. a. + : ©Fx = 0;

V - 2 cos 60° = 0

V = 1 kN

+ c ©Fy = 0;

N - 2 sin 60° = 0

N = 1.732 kN

Normal and Shear Stress: t =

1(103) V = = 200 kPa A 0.1(0.05)

s =

1.732(103) N = = 346.41 kPa A 0.1(0.05)

Normal and Shear Strain: The shear modulus of the friction pad is G =

4 E = = 1.429 MPa 2(1 + n) 2(1 + 0.4)

Applying Hooke’s Law, s = EP;

346.41(103) = 4(106)P

P = 0.08660 mm>mm

Ans.

t = Gg;

200(103) = 1.429(106)g

g = 0.140 rad

Ans.

Ans: P = 0.08660 mm>mm, g = 0.140 rad 167

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3–30. The lap joint is connected together using a 1.25 in. diameter bolt. If the bolt is made from a material having a shear stress–strain diagram that is approximated as shown, determine the shear strain developed in the shear plane of the bolt when P = 75 kip.

P 2

P 2

P

t (ksi) 75 50

Internal Loadings: The shear force developed in the shear planes of the bolt can be determined by considering the equilibrium of the free-body diagram shown in Fig. a. + : ©Fx = 0;

75 - 2V = 0

0.005

0.05

g (rad)

V = 37.5 kip

Shear Stress and Strain: t =

V 37.5 = = 30.56 ksi p A A 1.252 B 4

Using this result, the corresponding shear strain can be obtained from the shear stress–strain diagram, Fig. b.

30.56 50 = ; g 0.005

g = 3.06(10-3) rad

Ans.

Ans: g = 3.06(10-3) rad 168

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3–31. The lap joint is connected together using a 1.25 in. diameter bolt. If the bolt is made from a material having a shear stress–strain diagram that is approximated as shown, determine the permanent shear strain in the shear plane of the bolt when the applied force P = 150 kip is removed.

P 2

P 2

P

t (ksi) 75 50

Internal Loadings: The shear force developed in the shear planes of the bolt can be determined by considering the equilibrium of the free-body diagram shown in Fig. a. + : ©Fx = 0;

150 - 2V = 0

V = 75 kip 0.005

Shear Stress and Strain: t =

0.05

g (rad)

V 75 = = 61.12 ksi p A A 1.252 B 4

Using this result, the corresponding shear strain can be obtained from the shear stress–strain diagram, Fig. b.

61.12 - 50 75 - 50 = ; g - 0.005 0.05 - 0.005

g = 0.02501 rad

When force P is removed, the shear strain recovers linearly along line BC, Fig. b, with a slope that is the same as line OA. This slope represents the shear modulus.

G =

50 = 10(103) ksi 0.005

Thus, the elastic recovery of shear strain is t = Ggr;

61.12 = (10)(103)gr

gr = 6.112(10-3) rad

And the permanent shear strain is gP = g - gr = 0.02501 - 6.112(10-3) = 0.0189 rad

Ans.

Ans: gP = 0.0189 rad 169

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*3–32. A shear spring is made by bonding the rubber annulus to a rigid fixed ring and a plug. When an axial load P is placed on the plug, show that the slope at point y in the rubber is dy>dr = -tan g = -tan1P>12phGr22. For small angles we can write dy>dr = -P>12phGr2. Integrate this expression and evaluate the constant of integration using the condition that y = 0 at r = ro. From the result compute the deflection y = d of the plug.

P

h

ro

y

d

ri r y

Shear Stress–Strain Relationship: Applying Hooke’s law with tA

g =

P = .

2p r h

tA P = G 2p h G r

dy P = -tan g = -tan a b dr 2p h G r

(Q.E.D)

If g is small, then tan g = g. Therefore, dy P = dr 2p h G r

At r = ro,

y = -

dr P 2p h G L r

y = -

P ln r + C 2p h G

0 = -

P ln ro + C 2p h G

y = 0

C =

Then, y =

ro P ln r 2p h G

At r = ri,

y = d d =

P ln ro 2p h G

ro P ln ri 2p h G

Ans.

170

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3–33. The aluminum block has a rectangular cross section and is subjected to an axial compressive force of 8 kip. If the 1.5-in. side changed its length to 1.500132 in., determine Poisson’s ratio and the new length of the 2-in. side.

Eal = 10(103) ksi.

s =

2 in.

8 kip

8 kip 3 in.

P 8 = = 2.667 ksi A (2)(1.5)

Plong =

Plat =

n =

1.5 in.

s -2.667 = -0.0002667 = E 10(103)

1.500132 - 1.5 = 0.0000880 1.5

-0.0000880 = 0.330 -0.0002667

Ans.

h¿ = 2 + 0.0000880(2) = 2.000176 in.

Ans.

Ans: n = 0.330, h¿ = 2.000176 in.

171

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3–34. A shear spring is made from two blocks of rubber, each having a height h, width b, and thickness a. The blocks are bonded to three plates as shown. If the plates are rigid and the shear modulus of the rubber is G, determine the displacement of plate A if a vertical load P is applied to this plate. Assume that the displacement is small so that d = a tan g L ag.

P

d A

h

Average Shear Stress: The rubber block is subjected to a shear force of V =

P .

2

a

a

P

t =

V P 2 = = A bh 2bh

Shear Strain: Applying Hooke’s law for shear P

g =

t P 2bh = = G G 2bhG

Thus, d = ag = =

Pa 2bhG

Ans.

Ans: d =

172

Pa 2bhG

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s (ksi)

3–35. The elastic portion of the tension stress–strain diagram for an aluminum alloy is shown in the figure. The specimen used for the test has a gauge length of 2 in. and a diameter of 0.5 in. When the applied load is 9 kip, the new diameter of the specimen is 0.49935 in. Compute the shear modulus Gal for the aluminum.

70

0.00614

From the stress–strain diagram,

P (in./in.)

s 70 = = 11400.65 ksi P 0.00614

Eal =

When specimen is loaded with a 9 - kip load, s =

P = A

0.49935 - 0.5 d¿ - d = = - 0.0013 in.>in. d 0.5

V = -

Gal =

9 = 45.84 ksi (0.5)2

45.84 s = = 0.0040205 in.>in.

E 11400.65

Plong =

Plat =

p 4

Plat -0.0013 = 0.32334 = Plong 0.0040205

11.4(103) Eat = = 4.31(103) ksi 2(1 + v) 2(1 + 0.32334)

Ans.

Ans: Gal = 4.31(103) ksi 173

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s (ksi)

*3–36. The elastic portion of the tension stress–strain diagram for an aluminum alloy is shown in the figure. The specimen used for the test has a gauge length of 2 in. and a diameter of 0.5 in. If the applied load is 10 kip, determine the new diameter of the specimen. The shear modulus is Gal = 3.811032 ksi.

P s = = A

70

10 = 50.9296 ksi p 2 (0.5) 4

0.00614

From the stress–strain diagram E =

70 = 11400.65 ksi 0.00614

Plong =

G =

50.9296 s = = 0.0044673 in.>in.

E 11400.65

E ; 2(1 + v)

3.8(103) =

11400.65 ; 2(1 + v)

v = 0.500

Plat = - vPlong = - 0.500(0.0044673) = - 0.002234 in.>in. ¢d = Plat d = - 0.002234(0.5) = - 0.001117 in. d¿ = d + ¢d = 0.5 - 0.001117 = 0.4989 in.

Ans.

174

P (in./in.)

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3–37. The rigid beam rests in the horizontal position on two 2014-T6 aluminum cylinders having the unloaded lengths shown. If each cylinder has a diameter of 30 mm. determine the placement x of the applied 80-kN load so that the beam remains horizontal. What is the new diameter of cylinder A after the load is applied? nal = 0.35.

80 kN x

A

B

220 mm

210 mm

3m

a +©MA = 0;

FB(3) - 80(x) = 0;

a +©MB = 0;

-FA(3) + 80(3 - x) = 0;

FB =

80x 3 FA =

(1) 80(3 - x) 3

(2)

Since the beam is held horizontally, dA = dB s =

P

P ; A

s A = E E

P =

d = PL = a

P A

E

bL =

80(3 - x) 3

dA = dB;

PL AE (220)

AE

=

80x 3

(210)

AE

80(3 - x)(220) = 80x(210) x = 1.53 m

Ans.

From Eq. (2), FA = 39.07 kN sA =

39.07(103) FA = = 55.27 MPa p A (0.032) 4

Plong =

sA E

55.27(106) = -

73.1(109)

= -0.000756

Plat = -nPlong = -0.35(-0.000756) = 0.0002646 dA¿ = dA + d Plat = 30 + 30(0.0002646) = 30.008 mm

Ans.

Ans: x = 1.53 m, dA ¿ = 30.008 mm 175

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3–38. The wires each have a diameter of 12 in., length of 2 ft, and are made from 304 stainless steel. If P = 6 kip, determine the angle of tilt of the rigid beam AB.

D

C

2 ft

P 2 ft A

1 ft B

Equations of Equilibrium: Referring to the free-body diagram of beam AB shown in Fig. a, a +©MA = 0;

FBC(3) - 6(2) = 0

FBC = 4 kip

+ c ©MB = 0;

6(1) - FAD(3) = 0

FAD = 2 kip

Normal Stress and Strain: sBC =

sAD =

4(103) FBC = = 20.37 ksi ABC p 1 2 a b 4 2 2(103) FAD = = 10.19 ksi AAD p 1 2 a b 4 2

Since sBC 6 sY and sA 6 sY, Hooke’s Law can be applied. sBC = EPBC;

20.37 = 28.0(103)PBC

PBC = 0.7276(10-3) in.>in.

sAD = EPAD;

10.19 = 28.0(103)PAD

PAD = 0.3638(10-3) in.>in.

Thus, the elongation of cables BC and AD are given by dBC = PBCLBC = 0.7276(10-3)(24) = 0.017462 in. dAD = PADLAD = 0.3638(10-3)(24) = 0.008731 in.

Referring to the geometry shown in Fig. b and using small angle analysis, u =

dBC - dAD 0.017462 - 0.008731 180° = = 0.2425(10-3) rada b = 0.0139° 36 36 prad

Ans.

Ans: u = 0.0139° 176

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3–39. The wires each have a diameter of 12 in., length of 2 ft, and are made from 304 stainless steel. Determine the magnitude of force P so that the rigid beam tilts 0.015°.

D

C

2 ft

P 2 ft A

1 ft B

Equations of Equilibrium: Referring to the free-body diagram of beam AB shown in Fig. a, a +©MA = 0;

FBC(3) - P(2) = 0

FBC = 0.6667P

+ c ©MB = 0;

P(1) - FAD(3) = 0

FAD = 0.3333P

Normal Stress and Strain: sBC =

sAD =

FBC 0.6667P = = 3.3953P ABC p 1 2 a b 4 2 FAD 0.3333P = = 1.6977P AAD p 1 2 a b 4 2

Assuming that sBC 6 sY and sAD 6 sY and applying Hooke’s Law, sBC = EPBC;

3.3953P = 28.0(106)PBC

PBC = 0.12126(10-6)P

sAD = EPAD;

1.6977P = 28.0(106)PAD

PAD = 60.6305(10-9)P

Thus, the elongation of cables BC and AD are given by dBC = PBCLBC = 0.12126(10-6)P(24) = 2.9103(10-6)P dAD = PADLAD = 60.6305(10-9)P(24) = 1.4551(10-6)P Here, the angle of the tile is u = 0.015°a

prad b = 0.2618(10-3) rad. Using small 180°

angle analysis, u =

dBC - dAD ; 36

0.2618(10-3) =

2.9103(10-6)P - 1.4551(10-6)P 36

P = 6476.93 lb = 6.48 kip

Ans.

Since sBC = 3.3953(6476.93) = 21.99 ksi 6 sY and sAD = 1.6977(6476.93) = 11.00 ksi 6 sY, the assumption is correct.

Ans: P = 6.48 kip 177

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*3–40. The head H is connected to the cylinder of a compressor using six steel bolts. If the clamping force in each bolt is 800 lb, determine the normal strain in the 3 bolts. Each bolt has a diameter of 16 in. If sY = 40 ksi and 3 Est = 29110 2 ksi, what is the strain in each bolt when the nut is unscrewed so that the clamping force is released?

C L

H

Normal Stress: s =

P = A

800

A B

p 3 2 4 16

= 28.97 ksi 6 sg = 40 ksi

Normal Strain: Since s 6 sg, Hooke’s law is still valid.

P =

s 28.97 = 0.000999 in.>in. = E 29(103)

Ans.

If the nut is unscrewed, the load is zero. Therefore, the strain P = 0

178

Ans.

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s (ksi)

3–41. The stress–strain diagram for polyethylene, which is used to sheath coaxial cables, is determined from testing a specimen that has a gauge length of 10 in. If a load P on the specimen develops a strain of P = 0.024 in.>in., determine the approximate length of the specimen, measured between the gauge points, when the load is removed. Assume the specimen recovers elastically.

P 5 4 3 2 1 0

P 0

0.008

0.016

0.024

0.032

0.040

0.048

P (in./in.)

Modulus of Elasticity: From the stress–strain diagram, s = 2 ksi when P = 0.004 in.>in.

E =

2 - 0 = 0.500(103) ksi 0.004 - 0

Elastic Recovery: From the stress–strain diagram, s = 3.70 ksi when P = 0.024 in.>in.

Elastic recovery =

s 3.70 = 0.00740 in.>in. = E 0.500(103)

Permanent Set: Permanent set = 0.024 - 0.00740 = 0.0166 in.>in.

Thus, Permanent elongation = 0.0166(10) = 0.166 in.

L = L0 + permanent elongation = 10 + 0.166 = 10.17 in.

Ans.

Ans: L = 10.17 in.

179

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3–42. The pipe with two rigid caps attached to its ends is subjected to an axial force P. If the pipe is made from a material having a modulus of elasticity E and Poisson’s ratio n, determine the change in volume of the material.

ri ro L P

a

Section a – a

a P

Normal Stress: The rod is subjected to uniaxial loading. Thus, slong =

P and slat = 0.

A

dV = AdL + 2prLdr = APlong L + 2prLPlatr Using Poisson’s ratio and noting that AL = pr2L = V, dV = PlongV - 2nPlongV = Plong (1 - 2n)V slong =

E

(1 - 2n)V

Since slong = P>A, dV =

=

P (1 - 2n)AL AE PL (1 - 2n) E

Ans.

Ans: dV =

180

PL (1 - 2n) E

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3–43. The 8-mm-diameter bolt is made of an aluminum alloy. It fits through a magnesium sleeve that has an inner diameter of 12 mm and an outer diameter of 20 mm. If the original lengths of the bolt and sleeve are 80 mm and 50 mm, respectively, determine the strains in the sleeve and the bolt if the nut on the bolt is tightened so that the tension in the bolt is 8 kN. Assume the material at A is rigid.

Eal = 70 GPa, Emg = 45 GPa.

50 mm A

30 mm

Normal Stress: 8(103)

sb =

P = Ab

p 2 4 (0.008 )

ss =

P = As

p 2 4 (0.02

= 159.15 MPa

8(103) - 0.0122)

= 39.79 MPa

Normal Strain: Applying Hooke’s Law Pb =

159.15(106) sb = 0.00227 mm>mm = Eal 70(109)

Ans.

Ps =

39.79(106) ss = 0.000884 mm>mm = Emg 45(109)

Ans.

Ans: Pb = 0.00227 mm>mm, Ps = 0.000884 mm>mm 181

Mechanics Of Materials 9th Edition Hibbeler Solutions Manual © 2014 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.

Full Download: https://testbanklive.com/download/mechanics-of-materials-9th-edition-hibbeler-solutions-manual/ *3–44. An acetal polymer block is fixed to the rigid plates at its top and bottom surfaces. If the top plate displaces 2 mm horizontally when it is subjected to a horizontal force P = 2 kN, determine the shear modulus of the polymer.

The width of the block is 100 mm. Assume that the polymer is linearly elastic and use small angle analysis.

400 mm P ⫽ 2 kN

200 mm

Normal and Shear Stress: t =

2(103) V = = 50 kPa A 0.4(0.1)

Referring to the geometry of the undeformed and deformed shape of the block shown in Fig. a, g =

2 = 0.01 rad 200

Applying Hooke’s Law, t = Gg;

50(103) = G(0.01) G = 5 MPa

Ans.

182

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