Ch01-03 stress & strain & properties

FM_TOC 46060

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Page iii

CONTENTS To the Instructor

iv

1 Stress

1

2 Strain

73

3 Mechanical Properties of Materials

92

4 Axial Load

122

5 Torsion

214

6 Bending

329

7 Transverse Shear

472

8 Combined Loadings

532

9 Stress Transformation

619

10 Strain Transformation

738

11 Design of Beams and Shafts

830

12 Deflection of Beams and Shafts

883

13 Buckling of Columns

1038

14 Energy Methods

1159

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© 2010 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.

1–1. Determine the resultant internal normal force acting on the cross section through point A in each column. In (a), segment BC weighs 180 lb>ft and segment CD weighs 250 lb>ft. In (b), the column has a mass of 200 kg>m. (a) + c ©Fy = 0;

5 kip

8 kN

B

200 mm

200 mm

6 kN

6 kN

FA - 1.0 - 3 - 3 - 1.8 - 5 = 0 10 ft

FA = 13.8 kip

8 in.

Ans.

3m

8 in. 200 mm

(b) + c ©Fy = 0;

3 kip

FA - 4.5 - 4.5 - 5.89 - 6 - 6 - 8 = 0 FA = 34.9 kN

3 kip

200 mm

4.5 kN

4.5 kN

C

Ans. 4 ft

A

A

1m

4 ft D (a)

1–2. Determine the resultant internal torque acting on the cross sections through points C and D. The support bearings at A and B allow free turning of the shaft. ©Mx = 0;

A 250 Nm 300 mm

TC - 250 = 0 TC = 250 N # m

©Mx = 0;

(b)

400 Nm

200 mm

Ans.

TD = 0

150 Nm

C

150 mm

Ans.

200 mm

B

D

250 mm 150 mm

1–3. Determine the resultant internal torque acting on the cross sections through points B and C. A

©Mx = 0;

TB = 150 lb # ft ©Mx = 0;

600 lbft B 350 lbft

TB + 350 - 500 = 0 Ans.

TC - 500 = 0 TC =

3 ft

C 500 lbft

1 ft

500 lb # ft

Ans.

2 ft 2 ft

1

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*1–4. A force of 80 N is supported by the bracket as shown. Determine the resultant internal loadings acting on the section through point A.

0.3 m A 30 0.1 m

80 N

Equations of Equilibrium: +

Q©Fx¿ = 0;

NA - 80 cos 15° = 0 NA = 77.3 N

a+ ©Fy¿ = 0;

Ans.

VA - 80 sin 15° = 0 VA = 20.7 N

a+

©MA = 0;

Ans.

MA + 80 cos 45°(0.3 cos 30°) - 80 sin 45°(0.1 + 0.3 sin 30°) = 0 MA = - 0.555 N # m

Ans.

or a+

©MA = 0;

MA + 80 sin 15°(0.3 + 0.1 sin 30°) -80 cos 15°(0.1 cos 30°) = 0 MA = - 0.555 N # m

Ans.

Negative sign indicates that MA acts in the opposite direction to that shown on FBD.

2

45

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•1–5.

Determine the resultant internal loadings in the beam at cross sections through points D and E. Point E is just to the right of the 3-kip load.

3 kip 1.5 kip/ ft

A D 6 ft

Support Reactions: For member AB a + ©MB = 0; + : ©Fx = 0; + c ©Fy = 0;

9.00(4) - A y(12) = 0

A y = 3.00 kip

Bx = 0 By + 3.00 - 9.00 = 0

By = 6.00 kip

Equations of Equilibrium: For point D + : ©Fx = 0; + c ©Fy = 0;

ND = 0

Ans.

3.00 - 2.25 - VD = 0 VD = 0.750 kip

a + ©MD = 0;

Ans.

MD + 2.25(2) - 3.00(6) = 0 MD = 13.5 kip # ft

Ans.

Equations of Equilibrium: For point E + : ©Fx = 0; + c ©Fy = 0;

NE = 0

Ans.

- 6.00 - 3 - VE = 0 VE = - 9.00 kip

a + ©ME = 0;

Ans.

ME + 6.00(4) = 0 ME = - 24.0 kip # ft

Ans.

Negative signs indicate that ME and VE act in the opposite direction to that shown on FBD.

3

E

B 6 ft

4 ft

4 ft

C

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1–6. Determine the normal force, shear force, and moment at a section through point C. Take P = 8 kN.

B

0.1 m

Support Reactions: a + ©MA = 0;

8(2.25) - T(0.6) = 0

0.5 m C

T = 30.0 kN

+ : ©Fx = 0;

30.0 - A x = 0

A x = 30.0 kN

+ c ©Fy = 0;

Ay - 8 = 0

A y = 8.00 kN

0.75 m

0.75 m

A 0.75 m

P

Equations of Equilibrium: For point C + : ©Fx = 0;

- NC - 30.0 = 0 NC = - 30.0 kN

+ c ©Fy = 0;

Ans.

VC + 8.00 = 0 VC = - 8.00 kN

a + ©MC = 0;

Ans.

8.00(0.75) - MC = 0 MC = 6.00 kN # m

Ans.

Negative signs indicate that NC and VC act in the opposite direction to that shown on FBD.

1–7. The cable will fail when subjected to a tension of 2 kN. Determine the largest vertical load P the frame will support and calculate the internal normal force, shear force, and moment at the cross section through point C for this loading.

B

0.1 m

0.5 m C


0.75 m

P(2.25) - 2(0.6) = 0 P

P = 0.5333 kN = 0.533 kN + : ©Fx = 0;

2 - Ax = 0

+ c ©Fy = 0;

A y - 0.5333 = 0

Ans.

A x = 2.00 kN A y = 0.5333 kN

Equations of Equilibrium: For point C + : ©Fx = 0;

- NC - 2.00 = 0 NC = - 2.00 kN

+ c ©Fy = 0;

Ans.

VC + 0.5333 = 0 VC = - 0.533 kN

a + ©MC = 0;

Ans.

0.5333(0.75) - MC = 0 MC = 0.400 kN # m

Ans.

Negative signs indicate that NC and VC act in the opposite direction to that shown on FBD.

4

0.75 m

A 0.75 m

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*1–8. Determine the resultant internal loadings on the cross section through point C. Assume the reactions at the supports A and B are vertical.

6 kN 3 kN/m

Referring to the FBD of the entire beam, Fig. a, a + ©MB = 0;

1 - A y(4) + 6(3.5) + (3)(3)(2) = 0 2

+ c ©Fy = 0; a + ©MC = 0;

C

A y = 7.50 kN

NC = 0

D 1.5 m

0.5 m 0.5 m

Referring to the FBD of this segment, Fig. b, + : ©Fx = 0;

B

A

1.5 m

Ans.

7.50 - 6 - VC = 0

VC = 1.50 kN

MC + 6(0.5) - 7.5(1) = 0

Ans.

MC = 4.50 kN # m

Ans.

•1–9.

Determine the resultant internal loadings on the cross section through point D. Assume the reactions at the supports A and B are vertical.

6 kN 3 kN/m

Referring to the FBD of the entire beam, Fig. a,

B

A

a + ©MA = 0;

By(4) - 6(0.5) -

1 (3)(3)(2) = 0 2

By = 3.00 kN 0.5 m 0.5 m

Referring to the FBD of this segment, Fig. b, + : ©Fx = 0; + c ©Fy = 0;

ND = 0 VD -

1 (1.5)(1.5) + 3.00 = 0 2

a + ©MD = 0; 3.00(1.5) -

C

Ans. VD = - 1.875 kN

Ans.

1 (1.5)(1.5)(0.5) - MD = 0 MD = 3.9375 kN # m 2 = 3.94 kN # m

5

Ans.

D 1.5 m

1.5 m

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1–10. The boom DF of the jib crane and the column DE have a uniform weight of 50 lb/ft. If the hoist and load weigh 300 lb, determine the resultant internal loadings in the crane on cross sections through points A, B, and C.

D 2 ft

F

A

B 8 ft

3 ft

5 ft C 300 lb 7 ft

E

Equations of Equilibrium: For point A + ; © Fx = 0; + c © Fy = 0;

NA = 0

Ans.

VA - 150 - 300 = 0 VA = 450 lb

a + ©MA = 0;

Ans.

- MA - 150(1.5) - 300(3) = 0 MA = - 1125 lb # ft = - 1.125 kip # ft

Ans.

Negative sign indicates that MA acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point B + ; © Fx = 0;

NB = 0

+ c © Fy = 0;

VB - 550 - 300 = 0

Ans.

VB = 850 lb a + © MB = 0;

Ans.

- MB - 550(5.5) - 300(11) = 0 MB = - 6325 lb # ft = - 6.325 kip # ft

Ans.

Negative sign indicates that MB acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point C + ; © Fx = 0; + c © Fy = 0;

VC = 0

Ans.

- NC - 250 - 650 - 300 = 0 NC = - 1200 lb = - 1.20 kip

a + ©MC = 0;

Ans.

- MC - 650(6.5) - 300(13) = 0 MC = - 8125 lb # ft = - 8.125 kip # ft

Ans.

Negative signs indicate that NC and MC act in the opposite direction to that shown on FBD.

6

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1–11. The force F = 80 lb acts on the gear tooth. Determine the resultant internal loadings on the root of the tooth, i.e., at the centroid point A of section a–a.

a

F 80 lb 30

Equations of Equilibrium: For section a–a +

Q©Fx¿ = 0;

VA - 80 cos 15° = 0

0.23 in.

VA = 77.3 lb a+ ©Fy¿ = 0;

Ans. A

NA - 80 sin 15° = 0

0.16 in.

NA = 20.7 lb a + ©MA = 0;

Ans.

- MA - 80 sin 15°(0.16) + 80 cos 15°(0.23) = 0 MA = 14.5 lb # in.

45

Ans.

*1–12. The sky hook is used to support the cable of a scaffold over the side of a building. If it consists of a smooth rod that contacts the parapet of a wall at points A, B, and C, determine the normal force, shear force, and moment on the cross section at points D and E.

a

0.2 m

0.2 m B

0.2 m

0.2 m D

E

Support Reactions: + c ©Fy = 0;

NB - 18 = 0

d+ ©MC = 0;

18(0.7) - 18.0(0.2) - NA(0.1) = 0

0.2 m

0.3 m

NB = 18.0 kN

A

NA = 90.0 kN + : ©Fx = 0;

NC - 90.0 = 0

0.3 m

NC = 90.0 kN

Equations of Equilibrium: For point D + : © Fx = 0;

18 kN

VD - 90.0 = 0 VD = 90.0 kN

+ c © Fy = 0;

Ans.

ND - 18 = 0 ND = 18.0 kN

d+ © MD = 0;

Ans.

MD + 18(0.3) - 90.0(0.3) = 0 MD = 21.6 kN # m

Ans.

Equations of Equilibrium: For point E + : © Fx = 0; 90.0 - VE = 0 VE = 90.0 kN + c © Fy = 0; d + © ME = 0;

Ans.

NE = 0

Ans.

90.0(0.2) - ME = 0 ME = 18.0 kN # m

Ans.

7

C

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•1–13.

The 800-lb load is being hoisted at a constant speed using the motor M, which has a weight of 90 lb. Determine the resultant internal loadings acting on the cross section through point B in the beam. The beam has a weight of 40 lb>ft and is fixed to the wall at A.

M 1.5 ft A D 4 ft

+ : ©Fx = 0;

3 ft

3 ft

4 ft

0.25 ft

Ans.

VB - 0.8 - 0.16 = 0 VB = 0.960 kip

a + ©MB = 0;

B

- NB - 0.4 = 0 NB = - 0.4 kip

+ c ©Fy = 0;

4 ft

C

Ans.

- MB - 0.16(2) - 0.8(4.25) + 0.4(1.5) = 0 MB = - 3.12 kip # ft

Ans.

1–14. Determine the resultant internal loadings acting on the cross section through points C and D of the beam in Prob. 1–13.

M 1.5 ft A D 4 ft

4 ft

C

B

3 ft

3 ft

4 ft

For point C: + ; ©Fx = 0; + c ©Fy = 0; a + ©MC = 0;

NC + 0.4 = 0;

NC = - 0.4kip

VC - 0.8 - 0.04 (7) = 0;

Ans.

VC = 1.08 kip

Ans.

- MC - 0.8(7.25) - 0.04(7)(3.5) + 0.4(1.5) = 0 MC = - 6.18 kip # ft

Ans.

ND = 0

Ans.

For point D: + ; ©Fx = 0; + c ©Fy = 0; a + ©MD = 0;

VD - 0.09 - 0.04(14) - 0.8 = 0;

VD = 1.45 kip

Ans.

- MD - 0.09(4) - 0.04(14)(7) - 0.8(14.25) = 0 MD = - 15.7 kip # ft

Ans.

8

0.25 ft

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1–15. Determine the resultant internal loading on the cross section through point C of the pliers. There is a pin at A, and the jaws at B are smooth.

20 N 40 mm

120 mm

15 mm C

+ c ©Fy = 0; + : ©Fx = 0; +d ©MC = 0;

- VC + 60 = 0;

VC = 60 N

Ans.

NC = 0 - MC + 60(0.015) = 0;

MC = 0.9 N.m

D

Ans. 80 mm 20 N

*1–16. Determine the resultant internal loading on the cross section through point D of the pliers.

B

A

Ans.

30

20 N 40 mm

120 mm

15 mm

R+ ©Fy = 0;

VD - 20 cos 30° = 0;

VD = 17.3 N

Ans.

+b©Fx = 0;

ND - 20 sin 30° = 0;

ND = 10 N

Ans.

+d ©MD = 0;

MD - 20(0.08) = 0;

MD = 1.60 N.m

Ans.

C A D 80 mm 20 N

9

30

B

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•1–17.

Determine resultant internal loadings acting on section a–a and section b–b. Each section passes through the centerline at point C.

5 kN

B

b a

Referring to the FBD of the entire beam, Fig. a, a + ©MA = 0;

NB sin 45°(6) - 5(4.5) = 0

1.5 m

NB = 5.303 kN

C

Referring to the FBD of this segment (section a–a), Fig. b, b

+b©Fx¿ = 0;

Na - a + 5.303 cos 45° = 0

Na - a = - 3.75 kN Va - a = 1.25 kN

Ans.

+a ©Fy¿ = 0;

Va - a + 5.303 sin 45° - 5 = 0

a + ©MC = 0;

5.303 sin 45°(3) - 5(1.5) - Ma - a = 0 Ma - a = 3.75 kN # m Ans.

Ans.

Referring to the FBD (section b–b) in Fig. c, + ; ©Fx = 0;

Nb - b - 5 cos 45° + 5.303 = 0 Nb - b = - 1.768 kN = - 1.77 kN

+ c ©Fy = 0; a + ©MC = 0;

Vb - b - 5 sin 45° = 0

Vb - b = 3.536 kN = 3.54 kN

Ans. Ans.

5.303 sin 45° (3) - 5(1.5) - Mb - b = 0 Mb - b = 3.75 kN # m

10

Ans.

A

45

3m

45

1.5 m a

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1–18. The bolt shank is subjected to a tension of 80 lb. Determine the resultant internal loadings acting on the cross section at point C.

C 6 in. 90

A

B

Segment AC: + : ©Fx = 0;

NC + 80 = 0;

+ c ©Fy = 0;

VC = 0

NC = - 80 lb

Ans. Ans.

MC = - 480 lb # in.

Ans.

1–19. Determine the resultant internal loadings acting on the cross section through point C. Assume the reactions at the supports A and B are vertical.

6 kip/ft

a + ©MC = 0;

MC + 80(6) = 0;

6 kip/ft

A

C 3 ft

Referring to the FBD of the entire beam, Fig. a, a + ©MB = 0;

1 1 (6)(6)(2) + (6)(6)(10) - A y(12) = 0 A y = 18.0 kip 2 2

Referring to the FBD of this segment, Fig. b, + : ©Fx = 0;

NC = 0

+ c ©Fy = 0;

18.0 -

a + ©MC = 0;

Ans. 1 (3)(3) - (3)(3) - VC = 0 2

MC + (3)(3)(1.5) +

VC = 4.50 kip

Ans.

1 (3)(3)(2) - 18.0(3) = 0 2

MC = 31.5 kip # ft

Ans.

11

B

D 3 ft

6 ft

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*1–20. Determine the resultant internal loadings acting on the cross section through point D. Assume the reactions at the supports A and B are vertical.

6 kip/ft

Referring to the FBD of the entire beam, Fig. a,

A

a + ©MB = 0;

+ c ©Fy = 0; a + ©MA = 0;

C

1 1 (6)(6)(2) + (6)(6)(10) - A y(12) = 0 A y = 18.0 kip 2 2

Referring to the FBD of this segment, Fig. b, + : ©Fx = 0; ND = 0 18.0 -

1 (6)(6) - VD = 0 2

MD - 18.0 (2) = 0

6 kip/ft

3 ft

B

D 3 ft

6 ft

Ans. VD = 0

Ans.

MD = 36.0 kip # ft

Ans.

The forged steel clamp exerts a force of F = 900 N on the wooden block. Determine the resultant internal loadings acting on section a–a passing through point A.

•1–21.

200 mm F 900 N

Internal Loadings: Referring to the free-body diagram of the section of the clamp shown in Fig. a, ©Fy¿ = 0;

900 cos 30° - Na - a = 0

Na - a = 779 N

Ans.

©Fx¿ = 0;

Va - a - 900 sin 30° = 0

Va - a = 450 N

Ans.

a + ©MA = 0;

900(0.2) - Ma - a = 0

Ma - a = 180 N # m

12

Ans.

a

30 a

A

F 900 N

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1–22. The floor crane is used to lift a 600-kg concrete pipe. Determine the resultant internal loadings acting on the cross section at G.

0.2 m 0.2 m

0.4 m

0.6 m

G E B

F 0.3 m

H

C 0.5 m 75

A

Support Reactions: We will only need to compute FEF by writing the moment equation of equilibrium about D with reference to the free-body diagram of the hook, Fig. a. a + ©MD = 0;

FEF(0.3) - 600(9.81)(0.5) = 0

FEF = 9810 N

Internal Loadings: Using the result for FEF, section FG of member EF will be considered. Referring to the free-body diagram, Fig. b, + : ©Fx = 0; + c ©Fy = 0; a + ©MG = 0;

9810 - NG = 0

NG = 9810 N = 9.81 kN

VG = 0

Ans. Ans.

MG = 0

Ans.

13

D

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1–23. The floor crane is used to lift a 600-kg concrete pipe. Determine the resultant internal loadings acting on the cross section at H.

0.2 m 0.2 m

0.4 m

0.6 m

G E B

F 0.3 m

H

C 0.5 m 75

A

Support Reactions: Referring to the free-body diagram of the hook, Fig. a. a + ©MF = 0;

Dx(0.3) - 600(9.81)(0.5) = 0

Dx = 9810 N

+ c ©Fy = 0;

Dy - 600(9.81) = 0

Dy = 5886 N

Subsequently, referring to the free-body diagram of member BCD, Fig. b, a + ©MB = 0;

FAC sin 75°(0.4) - 5886(1.8) = 0

+ : ©Fx = 0;

Bx + 27 421.36 cos 75° - 9810 = 0 Bx = 2712.83 N

+ c ©Fy = 0;

27 421.36 sin 75° - 5886 - By = 0

FAC = 27 421.36 N

By = 20 601 N

Internal Loadings: Using the results of Bx and By, section BH of member BCD will be considered. Referring to the free-body diagram of this part shown in Fig. c, + : ©Fx = 0;

NH + 2712.83 = 0

NH = - 2712.83 N = - 2.71 kN

Ans.

+ c ©Fy = 0;

- VH - 2060 = 0

VH = - 20601 N = - 20.6 kN

Ans.

a + ©MD = 0;

MH + 20601(0.2) = 0

MH = - 4120.2 N # m = - 4.12 kN # m Ans.

The negative signs indicates that NH, VH, and MH act in the opposite sense to that shown on the free-body diagram.

14

D

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*1–24. The machine is moving with a constant velocity. It has a total mass of 20 Mg, and its center of mass is located at G, excluding the front roller. If the front roller has a mass of 5 Mg, determine the resultant internal loadings acting on point C of each of the two side members that support the roller. Neglect the mass of the side members. The front roller is free to roll.

2m

G

C

B

A

1.5 m

Support Reactions: We will only need to compute NA by writing the moment equation of equilibrium about B with reference to the free-body diagram of the steamroller, Fig. a. a + ©MB = 0; NA (5.5) - 20(103)(9.81)(1.5) = 0

NA = 53.51(103) N

Internal Loadings: Using the result for NA, the free-body diagram of the front roller shown in Fig. b will be considered. + ; ©Fx = 0; 2NC = 0

NC = 0

+ c ©Fy = 0; 2VC + 53.51(103) - 5(103)(9.81) = 0

VC = - 2229.55 N = - 2.23 kN

Ans.

Ans.

a + ©MC = 0; 53.51(103)(2) - 5(103)(9.81)(2) - 2MC = 0 MC = 4459.10 N # m = 4.46 kN # m Ans.

15

4m

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z

•1–25.

Determine the resultant internal loadings acting on the cross section through point B of the signpost. The post is fixed to the ground and a uniform pressure of 7 lb>ft2 acts perpendicular to the face of the sign.

3 ft 2 ft

©Fx = 0;

(VB)x - 105 = 0;

(VB)x = 105 lb

©Fy = 0;

(VB)y = 0

Ans.

©Fz = 0;

(NB)z = 0

Ans.

Ans. 3 ft 7 lb/ft2

Ans.

©Mx = 0;

(MB)x = 0

©My = 0;

(MB)y - 105(7.5) = 0;

©Mz = 0;

(TB)z - 105(0.5) = 0;

(MB)y = 788 lb # ft

6 ft

Ans.

(TB)z = 52.5 lb # ft

B

Ans. A

4 ft

y

x

1–26. The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the pulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section located at point C. The 300-N forces act in the z direction and the 500-N forces act in the x direction. The journal bearings at A and B exert only x and z components of force on the shaft.

z

A

400 mm 150 mm 200 mm C 250 mm

x

300 N

300 N B

500 N 500 N y

©Fx = 0;

(VC)x + 1000 - 750 = 0;

©Fy = 0;

(NC)y = 0

©Fz = 0;

(VC)z + 240 = 0;

(VC)x = - 250 N

Ans. Ans. Ans.

(VC)z = - 240 N (MC)x = - 108 N # m

©Mx = 0;

(MC)x + 240(0.45) = 0;

©My = 0;

(TC)y = 0

©Mz = 0;

(MC)z - 1000(0.2) + 750(0.45) = 0;

Ans. Ans.

(MC)z = - 138 N # m Ans.

16

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1–27. The pipe has a mass of 12 kg>m. If it is fixed to the wall at A, determine the resultant internal loadings acting on the cross section at B. Neglect the weight of the wrench CD.

z

A

300 mm

200 mm B 60 N

x

D 400 mm

60 N 150 mm C

©Fx = 0;

(NB)x = 0

Ans.

©Fy = 0;

(VB)y = 0

Ans.

©Fz = 0;

(VB)z - 60 + 60 - (0.2)(12)(9.81) - (0.4)(12)(9.81) = 0 Ans.

(VB)z = 70.6 N ©Mx = 0;

(TB)x + 60(0.4) - 60(0.4) - (0.4)(12)(9.81)(0.2) = 0 (TB)x = 9.42 N # m

©My = 0;

Ans.

(MB)y + (0.2)(12)(9.81)(0.1) + (0.4)(12)(9.81)(0.2) - 60(0.3) = 0 (MB)y = 6.23 N # m

©Mz = 0;

Ans. Ans.

(MB)z = 0

17

150 mm

y

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z

*1–28. The brace and drill bit is used to drill a hole at O. If the drill bit jams when the brace is subjected to the forces shown, determine the resultant internal loadings acting on the cross section of the drill bit at A.

O

x

Internal Loading: Referring to the free-body diagram of the section of the drill and brace shown in Fig. a, ©Fx = 0; ©Fy = 0; ©Fz = 0; ©Mx = 0; ©My = 0; ©Mz = 0;

A VA B x - 30 = 0

A NA B y - 50 = 0 A VA B z - 10 = 0

A MA B x - 10(2.25) = 0

A TA B y - 30(0.75) = 0

A MA B z + 30(1.25) = 0

A VA B x = 30 lb

Ans.

A NA B y = 50 lb

Ans.

A VA B z = 10 lb

Ans.

A MA B x = 22.5 lb # ft A TA B y = 22.5 lb # ft

A MA B z = - 37.5 lb # ft

Ans. Ans. Ans.

The negative sign indicates that (MA)Z acts in the opposite sense to that shown on the free-body diagram.

18

3 in. 9 in.

Fx 30 lb

A

Fz 10 lb 9 in.

6 in.

6 in.

6 in.

Fy 50 lb y

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•1–29. The curved rod has a radius r and is fixed to the wall at B. Determine the resultant internal loadings acting on the cross section through A which is located at an angle u from the horizontal.

B

A r

U

Equations of Equilibrium: For point A R+ ©Fx = 0;

P

P cos u - NA = 0 NA = P cos u

Q+ ©Fy = 0;

Ans.

VA - P sin u = 0 VA = P sin u

d+ ©MA = 0;

Ans.

MA - P[r(1 - cos u)] = 0 MA = Pr(1 - cos u)

Ans.

19

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1–30. A differential element taken from a curved bar is shown in the figure. Show that dN>du = V, dV>du = - N, dM>du = - T, and dT>du = M.

M dM V dV

©Fx = 0; (1)

N

©Fy = 0; N sin

du du du du - V cos + (N + dN) sin + (V + dV) cos = 0 2 2 2 2

(2)

©Mx = 0; T cos

du du du du + M sin - (T + dT) cos + (M + dM) sin = 0 2 2 2 2

(3)

©My = 0; du du du du - M cos + (T + dT) sin + (M + dM) cos = 0 2 2 2 2 du du du du Since is can add, then sin , cos = = 1 2 2 2 2 T sin

Eq. (1) becomes Vdu - dN +

dVdu = 0 2

Neglecting the second order term, Vdu - dN = 0 dN = V du Eq. (2) becomes Ndu + dV +

QED dNdu = 0 2

Neglecting the second order term, Ndu + dV = 0 dV = -N du Eq. (3) becomes Mdu - dT +

QED dMdu = 0 2

Neglecting the second order term, Mdu - dT = 0 dT = M du Eq. (4) becomes Tdu + dM +

QED dTdu = 0 2

Neglecting the second order term, Tdu + dM = 0 dM = -T du

N dN

M V

du du du du + V sin - (N + dN) cos + (V + dV) sin = 0 N cos 2 2 2 2

QED

20

(4)

T

T dT

du

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1–31. The column is subjected to an axial force of 8 kN, which is applied through the centroid of the cross-sectional area. Determine the average normal stress acting at section a–a. Show this distribution of stress acting over the area’s cross section.

8 kN 75 mm 75 mm

10 mm

10 mm

70 mm

10 mm

70 mm a a

A = (2)(150)(10) + (140)(10) = 4400 mm2 = 4.4 (10-3) m2 s =

8 (103) P = 1.82 MPa = A 4.4 (10 - 3)

Ans.

*1–32. The lever is held to the fixed shaft using a tapered pin AB, which has a mean diameter of 6 mm. If a couple is applied to the lever, determine the average shear stress in the pin between the pin and lever.

B 12 mm A 250 mm 20 N

a + ©MO = 0; tavg =

- F(12) + 20(500) = 0;

F = 833.33 N

V 833.33 = p 6 = 29.5 MPa 2 A 4 (1000 )

Ans.

21

250 mm 20 N

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•1–33.

The bar has a cross-sectional area A and is subjected to the axial load P. Determine the average normal and average shear stresses acting over the shaded section, which is oriented at u from the horizontal. Plot the variation of these stresses as a function of u 10 … u … 90°2.

P

P u A

Equations of Equilibrium: R+ ©Fx = 0;

V - P cos u = 0

V = P cos u

Q+ ©Fy = 0;

N - P sin u = 0

N = P sin u

Average Normal Stress and Shear Stress: Area at u plane, A¿ =

s =

A . sin u

P sin u N P = = sin2 u A A¿ A sin u

tavg =

Ans.

V P cos u = A A¿ sin u =

P P sin u cos u = sin 2u A 2A

Ans.

1–34. The built-up shaft consists of a pipe AB and solid rod BC. The pipe has an inner diameter of 20 mm and outer diameter of 28 mm. The rod has a diameter of 12 mm. Determine the average normal stress at points D and E and represent the stress on a volume element located at each of these points.

4 kN

D

At D: sD =

4(103)

P = A

p 2 4 (0.028

P = A

p 4

- 0.02 2)

= 13.3 MPa (C)

Ans.

At E: sE =

8(103) (0.012 2)

B

A

= 70.7 MPa (T)

Ans.

22

6 kN 6 kN E

C

8 kN

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1–35. The bars of the truss each have a cross-sectional area of 1.25 in2. Determine the average normal stress in each member due to the loading P = 8 kip. State whether the stress is tensile or compressive.

B

3 ft

A 4 ft

P

Joint A: sAB =

FAB 13.33 = = 10.7 ksi A AB 1.25

(T)

Ans.

sAE =

FAE 10.67 = = 8.53 ksi A AE 1.25

(C)

Ans.

(C)

Ans.

Joint E: sED =

FED 10.67 = = 8.53 ksi A ED 1.25

sEB =

FEB 6.0 = = 4.80 ksi A EB 1.25

C

Ans.

(T)

Joint B: sBC =

FBC 29.33 = 23.5 ksi = A BC 1.25

(T)

Ans.

sBD =

FBD 23.33 = = 18.7 ksi A BD 1.25

(C)

Ans.

23

E 0.75 P

4 ft

D

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*1–36. The bars of the truss each have a cross-sectional area of 1.25 in2. If the maximum average normal stress in any bar is not to exceed 20 ksi, determine the maximum magnitude P of the loads that can be applied to the truss.

B

3 ft

A 4 ft

P

Joint A: + c ©Fy = 0;

3 - P + a b FAB = 0 5

FAB = (1.667)P + : ©Fx = 0;

4 - FAE + (1.667)Pa b = 0 5 FAE = (1.333)P

Joint E: + c ©Fy = 0;

FEB - (0.75)P = 0 FEB = (0.75)P

+ : ©Fx = 0;

(1.333)P - FED = 0 FED = (1.333)P

Joint B: + c ©Fy = 0;

3 3 a b FBD - (0.75)P - (1.667)Pa b = 0 5 5 FBD = (2.9167)P

+ : ©Fx = 0;

4 4 FBC - (2.9167)Pa b - (1.667)P a b = 0 5 5

FBC = (3.67)P The highest stressed member is BC:

sBC =

C

(3.67)P = 20 1.25

P = 6.82 kip

Ans.

24

E 0.75 P

4 ft

D

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•1–37.

The plate has a width of 0.5 m. If the stress distribution at the support varies as shown, determine the force P applied to the plate and the distance d to where it is applied.

4m P d x

s (15x1/2) MPa

The resultant force dF of the bearing pressure acting on the plate of area dA = b dx = 0.5 dx, Fig. a, 1

1

dF = sb dA = (15x2)(106)(0.5dx) = 7.5(106)x2 dx + c ©Fy = 0;

L

dF - P = 0 4m 1

L0

7.5(106)x2 dx - P = 0

P = 40(106) N = 40 MN

Ans.

Equilibrium requires a + ©MO = 0;

L

xdF - Pd = 0

4m 1

L0

x[7.5(106)x2 dx] - 40(106) d = 0 d = 2.40 m

Ans.

25

30 MPa

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1–38. The two members used in the construction of an aircraft fuselage are joined together using a 30° fish-mouth weld. Determine the average normal and average shear stress on the plane of each weld. Assume each inclined plane supports a horizontal force of 400 lb. N - 400 sin 30° = 0;

N = 200 lb

400 cos 30° - V = 0;

V = 346.41 lb

A¿ =

1.5 in.

30

1 in. 1 in.

800 lb

800 lb 30

1.5(1) = 3 in2 sin 30°

s =

N 200 = = 66.7 psi A¿ 3

Ans.

t =

V 346.41 = = 115 psi A¿ 3

Ans.

1–39. If the block is subjected to the centrally applied force of 600 kN, determine the average normal stress in the material. Show the stress acting on a differential volume element of the material.

150 mm 600 kN

150 mm 150 mm 150 mm

The cross-sectional area of the block is A = 0.6(0.3) - 0.3(0.2) = 0.12 m2. savg =

600(103) P = = 5(106) Pa = 5 MPa A 0.12

Ans.

The average normal stress distribution over the cross-section of the block and the state of stress of a point in the block represented by a differential volume element are shown in Fig. a

26

50 mm 100 mm 100 mm 50 mm

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*1–40. The pins on the frame at B and C each have a diameter of 0.25 in. If these pins are subjected to double shear, determine the average shear stress in each pin.

3 ft 500 lb

3 ft A 3 ft

Support Reactions: FBD(a) a + ©Mg = 0;

500(6) + 300(3) - Dy (6) = 0 Dy = 650 lb

+ ; ©Fx = 0;

500 - Ex = 0

Ex = 500 lb

+ c ©Fy = 0;

650 - 300 - Ey = 0

Ey = 350 lb

a + ©MB = 0;

Cy (3) - 300(1.5) = 0

Cy = 150 lb

+ c ©Fy = 0;

By + 150 - 300 = 0

By = 150 lb

1.5 ft

From FBD (b) 150(1.5) + Bx(3) - 650(3) = 0 Bx = 575 lb From FBD (c), + : ©Fx = 0;

Cx - 575 = 0

Cx = 575 lb

Hence, FB = FC = 2 5752 + 1502 = 594.24 lb Average shear stress: Pins B and C are subjected to double shear as shown on FBD (d) (tB)avg = (tC)avg =

1.5 ft

300 lb D

From FBD (c),

a + ©MA = 0;

C

B

V 297.12 = p 2 A 4 (0.25 ) = 6053 psi = 6.05 ksi

Ans.

27

3 ft E

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•1–41.

Solve Prob. 1–40 assuming that pins B and C are subjected to single shear.

3 ft 500 lb

3 ft A 3 ft C

B

Support Reactions: FBD(a) a + ©Mg = 0;

1.5 ft

500(6) + 300(3) - Dy (6) = 0

300 lb D

Dy = 650 lb + ; ©Fx = 0;

500 - Ex = 0

Ex = 500 lb

+ c ©Fy = 0;

650 - 300 - Ey = 0

Ey = 350 lb

From FBD (c), a + ©MB = 0; + c ©Fy = 0;

Cy (3) - 300(1.5) = 0 By + 150 - 300 = 0

Cy = 150 lb By = 150 lb

From FBD (b) d+ ©MA = 0;

150(1.5) + Bx(3) - 650(3) = 0 Bx = 575 lb

From FBD (c), + : ©Fx = 0;

Cx - 575 = 0

Cx = 575 lb

Hence, FB = FC = 2 5752 + 1502 = 594.24 lb Average shear stress: Pins B and C are subjected to single shear as shown on FBD (d) (tB)avg = (tC)avg =

1.5 ft

594.24 V = p 2 A 4 (0.25 ) = 12106 psi = 12.1 ksi

Ans.

28

3 ft E

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1–42. The pins on the frame at D and E each have a diameter of 0.25 in. If these pins are subjected to double shear, determine the average shear stress in each pin.

3 ft 500 lb

3 ft A 3 ft C

B 1.5 ft

1.5 ft

Support Reactions: FBD(a) a + ©ME = 0;

300 lb

500(6) + 300(3) - Dy(6) = 0

D

Dy = 650 lb + ; ©Fx = 0;

500 - Ex = 0

Ex = 500 lb

+ c ©Fy = 0;

650 - 300 - Ey = 0

Ey = 350 lb

Average shear stress: Pins D and E are subjected to double shear as shown on FBD (b) and (c). For Pin D, FD = Dy = 650 lb then VD = (pD)avg =

VD = AD

FD z

= 325 lb

325 p 2 4 (0.25)

Ans.

= 6621 psi = 6.62 ksi For Pin E, FE = 2 5002 + 3502 = 610.32 lb then VE = (tE)avg =

Fg z

= 305.16 lb

VE 305.16 = p 2 AE 4 (0.25 ) = 6217 psi = 6.22 ksi

Ans.

29

3 ft E

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1–43. Solve Prob. 1–42 assuming that pins D and E are subjected to single shear.

3 ft 500 lb

3 ft A

Support Reactions: FBD(a) a + ©ME = 0;

3 ft

500(6) + 300(3) - Dy(6) = 0

+ ; ©Fx = 0;

500 - Ex = 0 650 - 300 - Ey = 0

+ c ©Fy = 0;

C

B

Dy = 650 lb

1.5 ft

Ex = 500 lb Ey = 350 lb

1.5 ft

300 lb D

Average shear stress: Pins D and E are subjected to single shear as shown on FBD (b) and (c). For Pin D, VD = FD = Dy = 650 lb (tD)avg =

VD = AD

650 p 2 4 (0.25 )

Ans.

= 13242 psi = 13.2 ksi For Pin E, VE = FE = 2 5002 + 3502 = 610.32 lb (tE)avg =

VE 610.32 = p 2 AE 4 (0.25 ) = 12433 psi = 12.4 ksi

Ans.

*1–44. A 175-lb woman stands on a vinyl floor wearing stiletto high-heel shoes. If the heel has the dimensions shown, determine the average normal stress she exerts on the floor and compare it with the average normal stress developed when a man having the same weight is wearing flat-heeled shoes. Assume the load is applied slowly, so that dynamic effects can be ignored. Also, assume the entire weight is supported only by the heel of one shoe. Stiletto shoes: A =

1.2 in.

1 (p)(0.3)2 + (0.6)(0.1) = 0.2014 in2 2

0.3 in. 0.1 in. 0.5 in.

P 175 lb = s = = 869 psi A 0.2014 in2

Ans.

Flat-heeled shoes: A =

1 (p)(1.2)2 + 2.4(0.5) = 3.462 in2 2

s =

P 175 lb = = 50.5 psi A 3.462 in2

Ans.

30

3 ft E

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•1–45.

The truss is made from three pin-connected members having the cross-sectional areas shown in the figure. Determine the average normal stress developed in each member when the truss is subjected to the load shown. State whether the stress is tensile or compressive.

500 lb

C

sBC =

FBC 375 = = 469 psi A BC 0.8

(T)

Ans.

(T)

Ans.

4 ft

Joint A: FAC 500 = = 833 psi A AC 0.6

.2 in

A

1–46. Determine the average normal stress developed in links AB and CD of the smooth two-tine grapple that supports the log having a mass of 3 Mg. The cross-sectional area of each link is 400 mm2. + c ©Fy = 0;

1.5

Ans.

(C)

AB

FAB 625 = = 417 psi A AB 1.5

A

sAB =

AAC 0.6 in.2

ABC 0.8 in.2

Joint B:

œ sAC =

3 ft

A

C

20 B

E

D

2(F sin 30°) - 29.43 = 0 0.2 m

F = 29.43 kN a + ©ME = 0; P cos 20°(0.2) - (29.43 cos 30°)(1.2) + (29.43 sin 30°)(0.4 cos 30°)

1.2 m

= 0 P = 135.61 kN

30

30 0.4 m

s =

135.61(103) P = 339 MPa = A 400(10 - 6)

Ans.

31

B

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1–47. Determine the average shear stress developed in pins A and B of the smooth two-tine grapple that supports the log having a mass of 3 Mg. Each pin has a diameter of 25 mm and is subjected to double shear. + c ©Fy = 0;

A

C

20 E

B

D

2(F sin 30°) - 29.43 = 0 0.2 m

F = 29.43 kN 1.2 m

a + ©ME = 0; P cos 20°(0.2) - (29.43 cos 30°)(1.2) + (29.43 sin 30°)(0.4 cos 30°) = 0 P = 135.61 kN tA = tB =

V = A

135.61(103) 2 p 2 (0.025) 4

30

30 0.4 m

= 138 MPa

Ans.

*1–48. The beam is supported by a pin at A and a short link BC. If P = 15 kN, determine the average shear stress developed in the pins at A, B, and C. All pins are in double shear as shown, and each has a diameter of 18 mm.

P

4P

4P

2P

0.5m

0.5 m 1m

1.5 m

1.5 m

C 30 B

For pins B and C: 82.5 (103) V tB = tC = = p 18 2 = 324 MPa A 4 (1000 )

A

Ans.

For pin A: FA = 2 (82.5)2 + (142.9)2 = 165 kN tA =

82.5 (103) V = p 18 2 = 324 MPa A 4 (1000 )

Ans.

32

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•1–49. The beam is supported by a pin at A and a short link BC. Determine the maximum magnitude P of the loads the beam will support if the average shear stress in each pin is not to exceed 80 MPa. All pins are in double shear as shown, and each has a diameter of 18 mm.

P

4P

4P

2P

0.5m

0.5 m 1m

1.5 m

1.5 m

C 30 B

a + ©MA = 0;

A

2P(0.5) + 4P(2) + 4P(3.5) + P(4.5) - (TCB sin 30°)(5) = 0

TCB = 11P + : ©Fx = 0;

A x - 11P cos 30° = 0

A x = 9.5263P + c ©Fy = 0;

A y - 11P + 11P sin 30° = 0

A y = 5.5P FA = 2 (9.5263P)2 + (5.5P)2 = 11P Require; t =

V ; A

80(106) =

11P>2 p 2 4 (0.018)

P = 3.70 kN

Ans.

33

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1–50. The block is subjected to a compressive force of 2 kN. Determine the average normal and average shear stress developed in the wood fibers that are oriented along section a–a at 30° with the axis of the block.

50 mm a

150 mm

2 kN

2 kN 30

a

Force equilibrium equations written perpendicular and parallel to section a–a gives +Q©Fx¿ = 0;

Va - a - 2 cos 30° = 0

Va - a = 1.732 kN

+a©Fy¿ = 0;

2 sin 30° - Na - a = 0

Na - a = 1.00 kN

The cross sectional area of section a–a is A = a

0.15 b (0.05) = 0.015 m2. Thus sin 30°

(sa - a)avg =

1.00(103) Na - a = = 66.67(103)Pa = 66.7 kPa A 0.015

Ans.

(ta - a)avg =

1.732(103) Va - a = = 115.47(103)Pa = 115 kPa A 0.015

Ans.

34

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1–51. During the tension test, the wooden specimen is subjected to an average normal stress of 2 ksi. Determine the axial force P applied to the specimen. Also, find the average shear stress developed along section a–a of the specimen.

P

a

4 in.

a 2 in.

1 in.

Internal Loading: The normal force developed on the cross section of the middle portion of the specimen can be obtained by considering the free-body diagram shown in Fig. a. + c ©Fy = 0;

P P + - N = 0 2 2

4 in.

N = P

Referring to the free-body diagram shown in fig. b, the shear force developed in the shear plane a–a is + c ©Fy = 0;

P - Va - a = 0 2

Va - a =

P 2

Average Normal Stress and Shear Stress: The cross-sectional area of the specimen is A = 1(2) = 2 in2. We have savg =

N ; A

2(103) =

P 2

P = 4(103)lb = 4 kip

Ans.

4(103) P = = 2(103) lb. The area of the shear plane is 2 2 = 2(4) = 8 in2. We obtain

Using the result of P, Va - a = Aa - a

A ta - a B avg =

2(103) Va - a = = 250 psi Aa - a 8

Ans.

35

P

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*1–52. If the joint is subjected to an axial force of P = 9 kN, determine the average shear stress developed in each of the 6-mm diameter bolts between the plates and the members and along each of the four shaded shear planes.

P

P 100 mm

Internal Loadings: The shear force developed on each shear plane of the bolt and the member can be determined by writing the force equation of equilibrium along the member’s axis with reference to the free-body diagrams shown in Figs. a. and b, respectively. ©Fy = 0; 4Vb - 9 = 0

Vb = 2.25 kN

©Fy = 0; 4Vp - 9 = 0

Vp = 2.25 kN

Average Shear Stress: The areas of each shear plane of the bolt and the member p are A b = (0.0062) = 28.274(10 - 6)m2 and A p = 0.1(0.1) = 0.01 m2, respectively. 4 We obtain

A tavg B b =

2.25(103) Vb = 79.6 MPa = Ab 28.274(10 - 6)

A tavg B p =

Vp Ap

=

Ans.

2.25(103) = 225 kPa 0.01

Ans.

36

100 mm

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•1–53.

The average shear stress in each of the 6-mm diameter bolts and along each of the four shaded shear planes is not allowed to exceed 80 MPa and 500 kPa, respectively. Determine the maximum axial force P that can be applied to the joint.

P

P 100 mm 100 mm

Internal Loadings: The shear force developed on each shear plane of the bolt and the member can be determined by writing the force equation of equilibrium along the member’s axis with reference to the free-body diagrams shown in Figs. a. and b, respectively. ©Fy = 0;

4Vb - P = 0

Vb = P>4

©Fy = 0;

4Vp - P = 0

Vp = P>4

Average Shear Stress: The areas of each shear plane of the bolts and the members p are A b = (0.0062) = 28.274(10 - 6)m2 and A p = 0.1(0.1) = 0.01m2, respectively. 4 We obtain

A tallow B b =

Vb ; Ab

80(106) =

P>4 28.274(10 - 6)

P = 9047 N = 9.05 kN (controls)

A tallow B p =

Vp Ap

;

500(103) =

Ans.

P>4 0.01

P = 20 000 N = 20 kN

37

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Page 38


1–54. The shaft is subjected to the axial force of 40 kN. Determine the average bearing stress acting on the collar C and the normal stress in the shaft.

40 kN

30 mm

C

40 mm

Referring to the FBDs in Fig. a, + c ©Fy = 0;

Ns - 40 = 0

Ns = 40 kN

+ c ©Fy = 0;

Nb - 40 = 0

Nb = 40 kN

Here, the cross-sectional area of the shaft and the bearing area of the collar are p p A s = (0.032) = 0.225(10 - 3)p m2 and A b = (0.04 2) = 0.4(10 - 3)p m2. Thus, 4 4

A savg B s =

40(103) Ns = 56.59(106) Pa = 56.6 MPa = As 0.225(10 - 3)p

Ans.

A savg B b =

40(103) Nb = 31.83(106)Pa = 31.8 MPa = Ab 0.4(10 - 3)p

Ans.

38

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Page 39


1–55. Rods AB and BC each have a diameter of 5 mm. If the load of P = 2 kN is applied to the ring, determine the average normal stress in each rod if u = 60°.

A u P B

Consider the equilibrium of joint B, Fig. a, + : ©Fx = 0;

2 - FAB sin 60° = 0

+ c ©Fy = 0;

2.309 cos 60° - FBC = 0

FAB = 2.309 kN

C

FBC = 1.155 kN

The cross-sectional area of wires AB and BC are A AB = A BC =

p (0.0052) 4

= 6.25(10 - 6)p m2. Thus,

A savg B AB =

2.309(103) FAB = 117.62(106) Pa = 118 MPa = A AB 6.25(10 - 6)p

Ans.

A savg B BC =

1.155(103) FBC = 58.81(106) Pa = 58.8 MPa = A BC 6.25(10 - 6)p

Ans.

39

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*1–56. Rods AB and BC each have a diameter of 5 mm. Determine the angle u of rod BC so that the average normal stress in rod AB is 1.5 times that in rod BC. What is the load P that will cause this to happen if the average normal stress in each rod is not allowed to exceed 100 MPa?

A u P B

Consider the equilibrium of joint B, Fig. a, FAB cos u - FBC = 0

+ c ©Fy = 0; + : ©Fx = 0;

(1)

P - FAB sin u = 0

(2)

p (0.0052) 4 = 6.25(10 - 6)p m2. Since the average normal stress in rod AB is required to be

The cross-sectional area of rods AB and BC are A AB = A BC =

1.5 times to that of rod BC, then

A savg B AB = 1.5 A savg B BC FAB FBC = 1.5 a b A AB A BC FAB 6.25(10 - 6)p

= 1.5 c

FBC 6.25(10 - 6)p

d

FAB = 1.5 FBC

(3)

Solving Eqs (1) and (3), u = 48.19° = 48.2°

Ans.

Since wire AB will achieve the average normal stress of 100 MPa first when P increases, then FAB = sallow A AB = C 100(106) D C 6.25(10 - 6)p D = 1963.50 N Substitute the result of FAB and u into Eq (2), P = 1.46 kN

Ans.

40

C

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•1–57.

The specimen failed in a tension test at an angle of 52° when the axial load was 19.80 kip. If the diameter of the specimen is 0.5 in., determine the average normal and average shear stress acting on the area of the inclined failure plane. Also, what is the average normal stress acting on the cross section when failure occurs? +

b © Fx = 0;

52 0.5 in.

V - 19.80 cos 52° = 0 V = 12.19 kip

+a © Fy = 0;

N - 19.80 sin 52° = 0 N = 15.603 kip

Inclined plane: s¿ =

P ; A

œ tavg =

s¿ =

V ; A

15.603 p(0.25)2 sin 52°

œ tavg =

Ans.

= 62.6 ksi

12.19 p(0.25)2 sin 52°

Ans.

= 48.9 ksi

Cross section: s =

P ; A

tavg =

s =

V ; A

19.80 = 101 ksi p(0.25)2

Ans.

tavg = 0

Ans.

1–58. The anchor bolt was pulled out of the concrete wall and the failure surface formed part of a frustum and cylinder. This indicates a shear failure occurred along the cylinder BC and tension failure along the frustum AB. If the shear and normal stresses along these surfaces have the magnitudes shown, determine the force P that must have been applied to the bolt.

P

A 45

45 50 mm

Average Normal Stress: For the frustum, A = 2pxL = 2p(0.025 + 0.025) A 2 0.05 + 0.05 2

2

B

3 MPa

3 MPa B

= 0.02221 m2 s =

P ; A

3 A 106 B =

4.5 MPa

F1 0.02221

C 25 mm 25 mm

F1 = 66.64 kN Average Shear Stress: For the cylinder, A = p(0.05)(0.03) = 0.004712 m2 F2 V ; 4.5 A 106 B = tavg = A 0.004712 F2 = 21.21 kN Equation of Equilibrium: + c ©Fy = 0;

P - 21.21 - 66.64 sin 45° = 0 P = 68.3 kN

Ans.

41

30 mm

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1–59. The open square butt joint is used to transmit a force of 50 kip from one plate to the other. Determine the average normal and average shear stress components that this loading creates on the face of the weld, section AB.

50 kip 30 30 2 in.

Equations of Equilibrium: a+ © Fy = 0; +

Q© Fx = 0;

N - 50 cos 30° = 0 - V + 50 sin 30° = 0

B

N = 43.30 kip

A 6 in.

V = 25.0 kip

50 kip

Average Normal and Shear Stress: A¿ = a s = tavg

2 b (6) = 13.86 in2 sin 60°

N 43.30 = = 3.125 ksi A¿ 13.86 V 25.0 = = = 1.80 ksi A¿ 13.86

Ans. Ans.

*1–60. If P = 20 kN, determine the average shear stress developed in the pins at A and C. The pins are subjected to double shear as shown, and each has a diameter of 18 mm.

C

Referring to the FBD of member AB, Fig. a a + ©MA = 0;

30

FBC sin 30° (6) - 20(2) - 20(4) = 0

FBC = 40 kN A

+ : ©Fx = 0;

A x - 40 cos 30° = 0

+ c ©Fy = 0;

A y - 20 - 20 + 40 sin 30°

A x = 34.64 kN A y = 20 kN

2m

P

Thus, the force acting on pin A is FA = 2 A x 2 + A y 2 = 2 34.64 2 + 202 = 40 kN Pins A and C are subjected to double shear. Referring to their FBDs in Figs. b and c, FA FBC 40 40 VA = VC = = = 20 kN = = 20 kN 2 2 2 2 p The cross-sectional area of Pins A and C are A A = A C = (0.0182) 4 = 81(10 - 6)p m2. Thus tA =

20(103) VA = 78.59(106) Pa = 78.6 MPa = AA 81(10 - 6)p

Ans.

tC =

20(103) VC = 78.59(106) Pa = 78.6 MPa = AC 81(10 - 6)p

Ans.

42

B 2m

2m P

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•1–61. Determine the maximum magnitude P of the load the beam will support if the average shear stress in each pin is not to allowed to exceed 60 MPa. All pins are subjected to double shear as shown, and each has a diameter of 18 mm.

C 30

Referring to the FBD of member AB, Fig. a, a + ©MA = 0; + : ©Fx = 0;

FBC sin 30°(6) - P(2) - P(4) = 0 A x - 2P cos 30° = 0

A x = 1.732P

A y - P - P + 2P sin 30° = 0

+ c ©Fy = 0;

A

FBC = 2P

FA = 2 A x 2 + A y 2 = 2 (1.732P)2 + P2 = 2P All pins are subjected to same force and double shear. Referring to the FBD of the pin, Fig. b, 2P F = = P 2 2

The cross-sectional area of the pin is A = tallow =

V ; A

60(106) =

2m

P

Ay = P

Thus, the force acting on pin A is

V =

B 2m

p (0.0182) = 81.0(10 - 6)p m2. Thus, 4

P 81.0(10 - 6)p

P = 15 268 N = 15.3 kN

Ans.

43

2m P

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1–62. The crimping tool is used to crimp the end of the wire E. If a force of 20 lb is applied to the handles, determine the average shear stress in the pin at A. The pin is subjected to double shear and has a diameter of 0.2 in. Only a vertical force is exerted on the wire.

20 lb

C

E A

Support Reactions:

5 in.

From FBD(a) a + ©MD = 0;

B D

1.5 in. 2 in. 1 in.

20(5) - By (1) = 0

+ : ©Fx = 0;

20 lb

By = 100 lb

Bx = 0

From FBD(b) + : ©Fx = 0; a + ©ME = 0;

Ax = 0 A y (1.5) - 100(3.5) = 0 A y = 233.33 lb

Average Shear Stress: Pin A is subjected to double shear. Hence, Ay FA VA = = = 116.67 lb 2 2 (tA)avg =

VA 116.67 = p 2 AA 4 (0.2 )

= 3714 psi = 3.71 ksi

Ans.

1–63. Solve Prob. 1–62 for pin B. The pin is subjected to double shear and has a diameter of 0.2 in.

20 lb

Support Reactions:

a + ©MD = 0; + : ©Fx = 0;

C

E

From FBD(a)

A

20(5) - By (1) = 0

By = 100 lb

5 in.

Bx = 0

1.5 in. 2 in. 1 in.

Average Shear Stress: Pin B is subjected to double shear. Hence, By FB VB = = = 50.0 lb 2 2 (tB)avg =

B D

VB = AB

p 4

50.0 (0.2 2)

= 1592 psi = 1.59 ksi

Ans.

44

20 lb

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*1–64. The triangular blocks are glued along each side of the joint. A C-clamp placed between two of the blocks is used to draw the joint tight. If the glue can withstand a maximum average shear stress of 800 kPa, determine the maximum allowable clamping force F.

50 mm

F glue

45

25 mm

Internal Loadings: The shear force developed on the glued shear plane can be obtained by writing the force equation of equilibrium along the x axis with reference to the free-body diagram of the triangular block, Fig. a. + : ©Fx = 0;

F cos 45° - V = 0

V =

F

2 2 F 2

Average Normal and Shear Stress: The area of the glued shear plane is A = 0.05(0.025) = 1.25(10 - 3)m2. We obtain

tavg =

V ; A

2 2 F 2 800(103) = 1.25(10 - 3) F = 1414 N = 1.41 kN

Ans.

•1–65.

The triangular blocks are glued along each side of the joint. A C-clamp placed between two of the blocks is used to draw the joint tight. If the clamping force is F = 900 N, determine the average shear stress developed in the glued shear plane.

50 mm 45

F glue 25 mm

Internal Loadings: The shear force developed on the glued shear plane can be obtained by writing the force equation of equilibrium along the x axis with reference to the free-body diagram of the triangular block, Fig. a. + : ©Fx = 0;

900 cos 45° - V = 0

V = 636.40 N

Average Normal and Shear Stress: The area of the glued shear plane is A = 0.05(0.025) = 1.25(10 - 3)m2. We obtain tavg =

V 636.40 = 509 kPa = A 1.25(10 - 3)

Ans.

45

F

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1–66. Determine the largest load P that can be a applied to the frame without causing either the average normal stress or the average shear stress at section a–a to exceed s = 150 MPa and t = 60 MPa, respectively. Member CB has a square cross section of 25 mm on each side.

B

Analyse the equilibrium of joint C using the FBD Shown in Fig. a, + c ©Fy = 0;

4 FBC a b - P = 0 5

FBC = 1.25P

2m

a

Referring to the FBD of the cut segment of member BC Fig. b. + : ©Fx = 0;

+ c ©Fy = 0; The

3 Na - a - 1.25Pa b = 0 5 4 1.25Pa b - Va - a = 0 5

cross-sectional

area

of

section

Na - a = 0.75P a A

Va - a = P a–a

is

A a - a = (0.025) a

0.025 b 3>5

= 1.0417(10 - 3)m2. For Normal stress, sallow =

Na - a ; Aa - a

150(106) =

0.75P 1.0417(10 - 3)

P = 208.33(103) N = 208.33 kN For Shear Stress Va - a ; tallow = Aa - a

60(106) =

P 1.0417(10 - 3)

P = 62.5(103) N = 62.5 kN (Controls!)

Ans.

46

C 1.5 m P

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1–67. The prismatic bar has a cross-sectional area A. If it is subjected to a distributed axial loading that increases linearly from w = 0 at x = 0 to w = w0 at x = a, and then decreases linearly to w = 0 at x = 2a, determine the average normal stress in the bar as a function of x for 0 … x 6 a.

w0

x a

a

Equation of Equilibrium: + : ©Fx = 0;

1 w0 1 a x + w0 b(a - x) + w0a = 0 2 a 2

-N +

N =

w0 A 2a2 - x2 B 2a

Average Normal Stress: N s = = A

w0 2a

(2a2 - x2) =

A

w0 A 2a2 - x2 B 2aA

Ans.

*1–68. The prismatic bar has a cross-sectional area A. If it is subjected to a distributed axial loading that increases linearly from w = 0 at x = 0 to w = w0 at x = a, and then decreases linearly to w = 0 at x = 2a, determine the average normal stress in the bar as a function of x for a 6 x … 2a.

w0

x a

a

Equation of Equilibrium: + : ©Fx = 0;

-N +

1 w0 c (2a - x) d(2a - x) = 0 2 a N =

w0 (2a - x)2 2a

Average Normal Stress: s =

N = A

w0 2a

(2a - x)2 =

A

w0 (2a - x)2 2aA

Ans.

The tapered rod has a radius of r = (2 - x>6) in. and is subjected to the distributed loading of w = (60 + 40x) lb>in. Determine the average normal stress at the center of the rod, B.

•1–69.

A = pa 2 -

r w (60 40x) lb/ in. x r = (2 — ) in. 6

3 2 b = 7.069 in2 6

x B

6

© Fx = 0;

N -

L3

(60 + 40x) dx = 0;

3 in.

N = 720 lb

720 s = = 102 psi 7.069

Ans.

47

3 in.

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1–70. The pedestal supports a load P at its center. If the material has a mass density r, determine the radial dimension r as a function of z so that the average normal stress in the pedestal remains constant. The cross section is circular.

P r1

z r

Require: P + W1 P + W1 + dW s = = A A + dA P dA + W1 dA = A dW P + W1 dW = = s dA A

(1)

dA = p(r + dr)2 - pr2 = 2p r dr dW = pr2(rg) dt From Eq. (1) pr2(rg) dz = s 2p r dr r rg dz = s 2 dr z r rg dr dz = 2s L0 Lr1 r

rg z r = ln ; r1 2s

p

r = r1 e(2a)z

However, s =

P p r21

r = r1 e

(p r12rg )z

Ans.

2P

48

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1–71. Determine the average normal stress at section a–a and the average shear stress at section b–b in member AB. The cross section is square, 0.5 in. on each side.

150 lb/ft

Consider the FBD of member BC, Fig. a, a + ©MC = 0;

B

4 ft

C

60 a

FAB sin 60°(4) - 150(4)(2) = 0

FAB = 346.41 lb a

Referring to the FBD in Fig. b, +

b©Fx¿ = 0;

Na - a + 346.41 = 0

b

Na - a = - 346.41 lb

Referring to the FBD in Fig. c. + c ©Fy = 0;

Vb - b - 346.41 sin 60° = 0

b

Vb - b = 300 lb

The cross-sectional areas of section a–a and b–b are A a - a = 0.5(0.5) = 0.25 in2 and 0.5 b = 0.5 in2. Thus A b - b = 0.5 a cos 60° Na - a 346.41 Ans. = = 1385.64 psi = 1.39 ksi sa - a = Aa - a 0.25 tb - b =

Vb - b 300 = = 600 psi Ab - b 0.5

Ans.

49

A

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•1–73. Member B is subjected to a compressive force of 800 lb. If A and B are both made of wood and are 38 in. thick, determine to the nearest 14 in. the smallest dimension h of the horizontal segment so that it does not fail in shear. The average shear stress for the segment is tallow = 300 psi.

tallow = 300 =

B

307.7

13

(32)

12

h h

3 in. 4

Ans.

1–74. The lever is attached to the shaft A using a key that has a width d and length of 25 mm. If the shaft is fixed and a vertical force of 200 N is applied perpendicular to the handle, determine the dimension d if the allowable shear stress for the key is tallow = 35 MPa. a + ©MA = 0;

5

A

h = 2.74 in. Use h = 2

800 lb

a A

d a 20 mm

500 mm 200 N

Fa - a (20) - 200(500) = 0 Fa - a = 5000 N

tallow =

Fa - a ; Aa - a

35(106) =

5000 d(0.025)

d = 0.00571 m = 5.71 mm

Ans.

1–75. The joint is fastened together using two bolts. Determine the required diameter of the bolts if the failure shear stress for the bolts is tfail = 350 MPa. Use a factor of safety for shear of F.S. = 2.5.

30 mm

30 mm

350(106) = 140(105) 2.5

40 kN 3

tallow = 140(106) =

20(10 ) p 4

80 kN

40 kN

d2

d = 0.0135 m = 13.5 mm

Ans.

50

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*1–76. The lapbelt assembly is to be subjected to a force of 800 N. Determine (a) the required thickness t of the belt if the allowable tensile stress for the material is (st)allow = 10 MPa, (b) the required lap length dl if the glue can sustain an allowable shear stress of (tallow)g = 0.75 MPa, and (c) the required diameter dr of the pin if the allowable shear stress for the pin is (tallow)p = 30 MPa.

800 N 45 mm

t dl

dr 800 N

Allowable Normal Stress: Design of belt thickness. 10 A 106 B =

P ; A

(st)allow =

800 (0.045)t

t = 0.001778 m = 1.78 mm

Ans.

Allowable Shear Stress: Design of lap length. VA 400 ; 0.750 A 106 B = (tallow)g = A (0.045) dt dt = 0.01185 m = 11.9 mm

Ans.

Allowable Shear Stress: Design of pin size. VB 400 ; 30 A 106 B = p 2 (tallow)P = A 4 dr dr = 0.004120 m = 4.12 mm

Ans.

•1–77.

The wood specimen is subjected to the pull of 10 kN in a tension testing machine. If the allowable normal stress for the wood is (st)allow = 12 MPa and the allowable shear stress is tallow = 1.2 MPa, determine the required dimensions b and t so that the specimen reaches these stresses simultaneously. The specimen has a width of 25 mm.

10 kN

Allowable Shear Stress: Shear limitation tallow =

V ; A

1.2 A 106 B =

t

A

b

3

5.00(10 ) (0.025) t

t = 0.1667 m = 167 mm

Ans.

Allowable Normal Stress: Tension limitation sallow =

P ; A

12.0 A 106 B =

10 kN

10(103) (0.025) b

b = 0.03333 m = 33.3 mm

Ans.

51

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1–78. Member B is subjected to a compressive force of 600 lb. If A and B are both made of wood and are 1.5 in. thick, determine to the nearest 1>8 in. the smallest dimension a of the support so that the average shear stress along the blue line does not exceed tallow = 50 psi. Neglect friction.

600 lb

3

5 4

B A

Consider the equilibrium of the FBD of member B, Fig. a, + : ©Fx = 0;

4 600 a b - Fh = 0 5

Fh = 480 lb

Referring to the FBD of the wood segment sectioned through glue line, Fig. b + : ©Fx = 0;

480 - V = 0

V = 480 lb

The area of shear plane is A = 1.5(a). Thus, tallow =

V ; A

50 =

480 1.5a

a = 6.40 in Use a = 612 in.

Ans.

52

a

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1–79. The joint is used to transmit a torque of T = 3 kN # m. Determine the required minimum diameter of the shear pin A if it is made from a material having a shear failure stress of tfail = 150 MPa. Apply a factor of safety of 3 against failure.

100 mm T

A

T

Internal Loadings: The shear force developed on the shear plane of pin A can be determined by writing the moment equation of equilibrium along the y axis with reference to the free-body diagram of the shaft, Fig. a. ©My = 0; V(0.1) - 3(103) = 0

V = 30(103)N

Allowable Shear Stress: tfail 150 tallow = = = 50 MPa F.S. 3 Using this result, tallow =

V ; A

50(106) =

30(103) p d 2 4 A

dA = 0.02764 m = 27.6 mm

Ans.

*1–80. Determine the maximum allowable torque T that can be transmitted by the joint. The shear pin A has a diameter of 25 mm, and it is made from a material having a failure shear stress of tfail = 150 MPa. Apply a factor of safety of 3 against failure.

100 mm T

Internal Loadings: The shear force developed on the shear plane of pin A can be determined by writing the moment equation of equilibrium along the y axis with reference to the free-body diagram of the shaft, Fig. a. ©My = 0; V(0.1) - T = 0

T

V = 10T

Allowable Shear Stress: tfail 150 tallow = = = 50 MPa F.S. 3 The area of the shear plane for pin A is A A =

p (0.0252) = 0.4909(10 - 3)m2. Using 4

these results, tallow =

V ; AA

50(106) =

10T 0.4909(10 - 3)

T = 2454.37 N # m = 2.45 kN # m

Ans.

53

A

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•1–81.

The tension member is fastened together using two bolts, one on each side of the member as shown. Each bolt has a diameter of 0.3 in. Determine the maximum load P that can be applied to the member if the allowable shear stress for the bolts is tallow = 12 ksi and the allowable average normal stress is sallow = 20 ksi.

60 P

P

N - P sin 60° = 0

a+ ©Fy = 0;

P = 1.1547 N

(1)

V - P cos 60° = 0

b+ ©Fx = 0;

P = 2V

(2)

Assume failure due to shear: tallow = 12 =

V (2) (0.3)2 p 4

V = 1.696 kip From Eq. (2), P = 3.39 kip Assume failure due to normal force: sallow = 20 =

N (2) p4 (0.3)2

N = 2.827 kip From Eq. (1), P = 3.26 kip

Ans.

(controls)

54

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1–82. The three steel wires are used to support the load. If the wires have an allowable tensile stress of sallow = 165 MPa, determine the required diameter of each wire if the applied load is P = 6 kN.

A C

The force in wire BD is equal to the applied load; ie, FBD = P = 6 kN. Analysing the equilibrium of joint B by referring to its FBD, Fig. a, + : ©Fx = 0;

FBC cos 30° - FAB cos 45° = 0

(1)

+ c ©Fy = 0;

FBC sin 30° + FAB sin 45° - 6 = 0

(2)

Solving Eqs. (1) and (2), FAB = 5.379 kN For wire BD, FBD ; sallow = A BD

FBC = 4.392 kN

165(106) =

6(103) p 2 4 dBD

dBD = 0.006804 m = 6.804 mm Use dBD = 7.00 mm For wire AB, FAB ; sallow = A AB

165(106) =

Ans.

5.379(103) p 4

dAB 2

dAB = 0.006443 m = 6.443 mm Use dAB = 6.50 mm For wire BC, FBC ; sallow = A BC

165(106) =

Ans.

4.392(103) p 4

dBC 2

dBC = 0.005822 m = 5.822 mm dBC = 6.00 mm

Ans.

55

45

B

30

D P

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1–83. The three steel wires are used to support the load. If the wires have an allowable tensile stress of sallow = 165 MPa, and wire AB has a diameter of 6 mm, BC has a diameter of 5 mm, and BD has a diameter of 7 mm, determine the greatest force P that can be applied before one of the wires fails.

A C

45

The force in wire BD is equal to the applied load; ie, FBD = P. Analysing the equilibrium of joint B by referring to its FBD, Fig. a, + : ©Fx = 0;

FBC cos 30° - FAB cos 45° = 0

(1)

+ c ©Fy = 0;

FBC sin 30° + FAB sin 45° - P = 0

(2)

Solving Eqs. (1) and (2), FAB = 0.8966 P For wire BD, FBD sallow = ; A BD

FBC = 0.7321 P

165(106) =

p 4

P (0.0072)

P = 6349.94 N = 6.350 kN For wire AB, FAB ; sallow = A AB

165(106) =

0.8966 P (0.0062)

p 4

P = 5203.42 N = 5.203 kN For wire BC, FBC sallow = ; A BC

165(106) =

0.7321 P (0.0052)

p 4

P = 4425.60 N = 4.43 kN (Controls!)

Ans.

56

B

30

D P

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*1–84. The assembly consists of three disks A, B, and C that are used to support the load of 140 kN. Determine the smallest diameter d1 of the top disk, the diameter d2 within the support space, and the diameter d3 of the hole in the bottom disk. The allowable bearing stress for the material is 1sallow2b = 350 MPa and allowable shear stress is tallow = 125 MPa.

140 kN d1

20 mm A

10 mm

B C d3 d2

Solution Allowable Bearing Stress: Assume bearing failure for disk B. (sb)allow =

P ; A

350 A 106 B =

140(103) p 4

d21

d1 = 0.02257 m = 22.6 mm Allowable Shear Stress: Assume shear failure for disk C. tallow =

125 A 106 B =

V ; A

140(103) pd2 (0.01)

d2 = 0.03565 m = 35.7 mm

Ans.

Allowable Bearing Stress: Assume bearing failure for disk C. 140(103) P ; 350 A 106 B = (sb)allow = p A A 0.035652 - d23 B 4

d3 = 0.02760 m = 27.6 mm

Ans.

Since d3 = 27.6 mm 7 d1 = 22.6 mm, disk B might fail due to shear. t =

140(103) V = = 98.7 MPa 6 tallow = 125 MPa (O. K !) A p(0.02257)(0.02) d1 = 22.6 mm

Therefore,

Ans.

•1–85.

The boom is supported by the winch cable that has a diameter of 0.25 in. and an allowable normal stress of sallow = 24 ksi. Determine the greatest load that can be supported without causing the cable to fail when u = 30° and f = 45°. Neglect the size of the winch.

B

u 20 ft

s =

P ; A

24(103) =

p 4

T ; (0.25)2

A

T = 1178.10 lb d

+ : ©Fx = 0;

- 1178.10 cos 30° + FAB sin 45° = 0

+ c ©Fy = 0;

- W + FAB cos 45° - 1178.10 sin 30° = 0 W = 431 lb

Ans.

FAB = 1442.9 lb

57

f

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1–86. The boom is supported by the winch cable that has an allowable normal stress of sallow = 24 ksi. If it is required that it be able to slowly lift 5000 lb, from u = 20° to u = 50°, determine the smallest diameter of the cable to 1 the nearest 16 in. The boom AB has a length of 20 ft. Neglect the size of the winch. Set d = 12 ft.

B

u 20 ft f

A

Maximum tension in cable occurs when u = 20°. sin c sin 20° = 20 12

d

c = 11.842° + : © Fx = 0;

- T cos 20° + FAB cos 31.842° = 0

+ c ©Fy = 0;

FAB sin 31.842° - T sin 20° - 5000 = 0 T = 20 698.3 lb FAB = 22 896 lb

P ; s = A

Use

20 698.3 p 2 4 (d) d = 1.048 in.

24(103) =

d = 1

1 in. 16

Ans.

1–87. The 60 mm * 60 mm oak post is supported on the pine block. If the allowable bearing stresses for these materials are soak = 43 MPa and spine = 25 MPa, determine the greatest load P that can be supported. If a rigid bearing plate is used between these materials, determine its required area so that the maximum load P can be supported. What is this load?

P

For failure of pine block: s =

P ; A

25(106) =

P (0.06)(0.06)

P = 90 kN

Ans.

For failure of oak post: s =

P ; A

43(106) =

P (0.06)(0.06)

P = 154.8 kN Area of plate based on strength of pine block: 154.8(10)3 P s = ; 25(106) = A A A = 6.19(10 - 3)m2

Ans.

Pmax = 155 kN

Ans.

58

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*1–88. The frame is subjected to the load of 4 kN which acts on member ABD at D. Determine the required diameter of the pins at D and C if the allowable shear stress for the material is tallow = 40 MPa. Pin C is subjected to double shear, whereas pin D is subjected to single shear.

4 kN 1m E

1.5 m C

45 D 1.5 m

Referring to the FBD of member DCE, Fig. a, a + ©ME = 0;

Dy(2.5) - FBC sin 45° (1) = 0

(1)

+ : ©Fx = 0

FBC cos 45° - Dx = 0

(2)

B 1.5 m

Referring to the FBD of member ABD, Fig. b, a + ©MA = 0;

4 cos 45° (3) + FBC sin 45° (1.5) - Dx (3) = 0

(3)

Solving Eqs (2) and (3), FBC = 8.00 kN

Dx = 5.657 kN

Substitute the result of FBC into (1) Dy = 2.263 kN Thus, the force acting on pin D is FD = 2 Dx 2 + Dy 2 = 2 5.6572 + 2.2632 = 6.093 kN Pin C is subjected to double shear white pin D is subjected to single shear. Referring to the FBDs of pins C, and D in Fig c and d, respectively, FBC 8.00 = = 4.00 kN VC = VD = FD = 6.093 kN 2 2 For pin C, tallow =

VC ; AC

40(106) =

4.00(103) p 4

dC 2

dC = 0.01128 m = 11.28 mm Use dC = 12 mm For pin D, VD ; tallow = AD

40(106) =

Ans.

6.093(103) p 4

dD 2

dD = 0.01393 m = 13.93 mm Use

dD = 14 mm

Ans.

59

A

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•1–89. The eye bolt is used to support the load of 5 kip. Determine its diameter d to the nearest 18 in. and the required thickness h to the nearest 18 in. of the support so that the washer will not penetrate or shear through it. The allowable normal stress for the bolt is sallow = 21 ksi and the allowable shear stress for the supporting material is tallow = 5 ksi.

1 in. h

d

5 kip

Allowable Normal Stress: Design of bolt size 5(103) P sallow = ; 21.0(103) = p 2 Ab 4 d d = 0.5506 in. Use d =

5 in. 8

Ans.

Allowable Shear Stress: Design of support thickness 5(103) V ; 5(103) = tallow = A p(1)(h) Use h =

3 in. 8

Ans.

60

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1–90. The soft-ride suspension system of the mountain bike is pinned at C and supported by the shock absorber BD. If it is designed to support a load P = 1500 N, determine the required minimum diameter of pins B and C. Use a factor of safety of 2 against failure. The pins are made of material having a failure shear stress of tfail = 150 MPa, and each pin is subjected to double shear.

P A

Internal Loadings: The forces acting on pins B and C can be determined by considering the equilibrium of the free-body diagram of the soft-ride suspension system shown in Fig. a. a + ©MC = 0;

FBD = 5905.36 N + : ©Fx = 0;

Cx - 5905.36 cos 60° = 0

+ c ©Fy = 0;

5905.36 sin 60° - 1500 - Cy = 0 Cy = 3614.20 N

Cx = 2952.68 N

Thus, FC = 2 Cx 2 + Cy 2 = 2 2952.682 + 3614.202 = 4666.98 N Since both pins are in double shear, FB 5905.36 VB = = = 2952.68 N 2 2

VC =

FC 4666.98 = = 2333.49 N 2 2

Allowable Shear Stress: tfail 150 tallow = = = 75 MPa F.S. 2 Using this result, VB tallow = ; AB

75(106) =

2952.68 p 2 d 4 B

dB = 0.007080 m = 7.08 mm tallow =

VC ; AC

75(106) =

Ans.

2333.49 p 2 d 4 C

dC = 0.006294 m = 6.29 mm

Ans.

61

30 mm

B 60

1500(0.4) - FBD sin 60°(0.1) - FBD cos 60°(0.03) = 0

FB = FBD = 5905.36 N

100 mm 300 mm

C

D

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1–91. The soft-ride suspension system of the mountain bike is pinned at C and supported by the shock absorber BD. If it is designed to support a load of P = 1500 N, determine the factor of safety of pins B and C against failure if they are made of a material having a shear failure stress of tfail = 150 MPa. Pin B has a diameter of 7.5 mm, and pin C has a diameter of 6.5 mm. Both pins are subjected to double shear.

P A

Internal Loadings: The forces acting on pins B and C can be determined by considerning the equilibrium of the free-body diagram of the soft-ride suspension system shown in Fig. a. + ©MC = 0;

1500(0.4) - FBD sin 60°(0.1) - FBD cos 60°(0.03) = 0 FBD = 5905.36 N

+ : ©Fx = 0;

Cx - 5905.36 cos 60° = 0

+ c ©Fy = 0;

5905.36 sin 60° - 1500 - Cy = 0 Cy = 3614.20 N

Cx = 2952.68 N

Thus, FB = FBD = 5905.36 N

FC = 2 Cx 2 + Cy 2 = 2 2952.682 + 3614.202 = 4666.98 N

Since both pins are in double shear, VB =

FB 5905.36 = = 2952.68N 2 2

VC =

FC 4666.98 = = 2333.49 N 2 2

Allowable Shear Stress: The areas of the shear plane for pins B and C are p p A B = (0.00752) = 44.179(10 - 6)m2 and A C = (0.00652) = 33.183(10 - 6)m2. 4 4 We obtain

A tavg B B =

VB 2952.68 = 66.84 MPa = AB 44.179(10 - 6)

A tavg B C =

VC 2333.49 = 70.32 MPa = AC 33.183(10 - 6)

Using these results, tfail = (F.S.)B = A tavg B B tfail (F.S.)C = = A tavg B C

100 mm 300 mm

150 = 2.24 66.84

Ans.

150 = 2.13 70.32

Ans.

62

30 mm

B 60

C

D

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*1–92. The compound wooden beam is connected together by a bolt at B. Assuming that the connections at A, B, C, and D exert only vertical forces on the beam, determine the required diameter of the bolt at B and the required outer diameter of its washers if the allowable tensile stress for the bolt is 1st2allow = 150 MPa and the allowable bearing stress for the wood is 1sb2allow = 28 MPa. Assume that the hole in the washers has the same diameter as the bolt.

2 kN 1.5 kN 1.5 m 1.5 m 1.5 m

3 kN 2m

2m

1.5 m C

D

A B

From FBD (a): a + ©MD = 0;

FB(4.5) + 1.5(3) + 2(1.5) - FC(6) = 0 4.5 FB - 6 FC = - 7.5

(1)

From FBD (b): a + ©MD = 0;

FB(5.5) - FC(4) - 3(2) = 0 5.5 FB - 4 FC = 6

(2)

Solving Eqs. (1) and (2) yields FB = 4.40 kN;

FC = 4.55 kN

For bolt: sallow = 150(106) =

4.40(103) p 2 4 (dB)

dB = 0.00611 m = 6.11 mm

Ans.

For washer: sallow = 28 (104) =

4.40(103) p 2 4 (d w

- 0.006112)

dw = 0.0154 m = 15.4 mm

Ans.

•1–93.

The assembly is used to support the distributed loading of w = 500 lb>ft. Determine the factor of safety with respect to yielding for the steel rod BC and the pins at B and C if the yield stress for the steel in tension is sy = 36 ksi and in shear ty = 18 ksi. The rod has a diameter of 0.40 in., and the pins each have a diameter of 0.30 in.

C

4 ft

For rod BC: s =

A

1.667 P = 13.26 ksi = p 2 A 4 (0.4 )

F. S. =

sy s

=

36 = 2.71 13.26

Ans. 3 ft

For pins B and C:

w

0.8333 V = 11.79 ksi = p t = 2 A 4 (0.3 ) F. S. =

ty t

=

B

1 ft

18 = 1.53 11.79

Ans.

63

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1–94. If the allowable shear stress for each of the 0.30in.-diameter steel pins at A, B, and C is tallow = 12.5 ksi, and the allowable normal stress for the 0.40-in.-diameter rod is sallow = 22 ksi, determine the largest intensity w of the uniform distributed load that can be suspended from the beam.

C

4 ft

Assume failure of pins B and C: tallow = 12.5 =

1.667w

A

p 2 4 (0.3 )

w = 0.530 kip>ft

Ans.

(controls)

B

Assume failure of pins A: 3 ft

FA = 2 (2w)2 + (1.333w)2 = 2.404 w tallow

w

1.202w = 12.5 = p 2 4 (0.3 )

1 ft

w = 0.735 kip>ft Assume failure of rod BC: sallow = 22 =

3.333w p 2 4 (0.4 )

w = 0.829 kip>ft

1–95. If the allowable bearing stress for the material under the supports at A and B is 1sb2allow = 1.5 MPa, determine the size of square bearing plates A¿ and B¿ required to support the load. Dimension the plates to the nearest mm. The reactions at the supports are vertical. Take P = 100 kN.

40 kN/m

Referring to the FBD of the bean, Fig. a

1.5 m

A

a + ©MA = 0;

NB(3) + 40(1.5)(0.75) - 100(4.5) = 0

NB = 135 kN

a + ©MB = 0;

40(1.5)(3.75) - 100(1.5) - NA(3) = 0

NA = 25.0 kN

For plate A¿ , NA (sb)allow = ; A A¿

1.5(106) =

25.0(103) a2A¿

aA¿ = 0.1291 m = 130 mm

Ans.

For plate B¿ , sallow =

NB ; A B¿

1.5(106) =

135(103) a2B¿

aB¿ = 0.300 m = 300 mm

Ans.

64

P

A¿

B¿ 3m

B

1.5 m

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40 kN/m

*1–96. If the allowable bearing stress for the material under the supports at A and B is 1sb2allow = 1.5 MPa, determine the maximum load P that can be applied to the beam. The bearing plates A¿ and B¿ have square cross sections of 150 mm * 150 mm and 250 mm * 250 mm, respectively.

A

P

A¿

B¿ 3m

1.5 m

B

1.5 m

Referring to the FBD of the beam, Fig. a, a + ©MA = 0;

NB(3) + 40(1.5)(0.75) - P(4.5) = 0

NB = 1.5P - 15

a + ©MB = 0;

40(1.5)(3.75) - P(1.5) - NA(3) = 0

NA = 75 - 0.5P

For plate A¿ , NA (sb)allow = ; A A¿

1.5(106) =

(75 - 0.5P)(103) 0.15(0.15)

P = 82.5 kN For plate B¿ , NB ; (sb)allow = A B¿

1.5(106) =

(1.5P - 15)(103) 0.25(0.25)

P = 72.5 kN (Controls!)

Ans.

•1–97.

The rods AB and CD are made of steel having a failure tensile stress of sfail = 510 MPa. Using a factor of safety of F.S. = 1.75 for tension, determine their smallest diameter so that they can support the load shown. The beam is assumed to be pin connected at A and C.

B

D 6 kN 5 kN

4 kN


FCD(10) - 5(7) - 6(4) - 4(2) = 0 A

FCD = 6.70 kN a + ©MC = 0;

4(8) + 6(6) + 5(3) - FAB(10) = 0 FAB = 8.30 kN

Allowable Normal Stress: Design of rod sizes For rod AB sallow =

sfail FAB = ; F.S A AB

510(106) 8.30(103) = p 2 1.75 4 d AB dAB = 0.006022 m = 6.02 mm

Ans.

For rod CD sallow =

FCD sfail = ; F.S A CD

C 2m

510(106) 6.70(103) = p 2 1.75 4 d CD

dCD = 0.005410 m = 5.41 mm

Ans.

65

2m

3m

3m

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1–98. The aluminum bracket A is used to support the centrally applied load of 8 kip. If it has a constant thickness of 0.5 in., determine the smallest height h in order to prevent a shear failure. The failure shear stress is tfail = 23 ksi. Use a factor of safety for shear of F.S. = 2.5.

A

Equation of Equilibrium: + c ©Fy = 0;

V - 8 = 0

V = 8.00 kip

Allowable Shear Stress: Design of the support size tallow =

tfail V = ; F.S A

23(103) 8.00(103) = 2.5 h(0.5)

8 kip

h = 1.74 in.

Ans.

1–99. The hanger is supported using the rectangular pin. Determine the magnitude of the allowable suspended load P if the allowable bearing stress is (sb)allow = 220 MPa, the allowable tensile stress is (st)allow = 150 MPa, and the allowable shear stress is tallow = 130 MPa. Take t = 6 mm, a = 5 mm, and b = 25 mm.

20 mm 75 mm 10 mm a

a

b

Allowable Normal Stress: For the hanger (st)allow =

P ; A

150 A 106 B =

P (0.075)(0.006)

P

P = 67.5 kN

37.5 mm

Allowable Shear Stress: The pin is subjected to double shear. Therefore, V = tallow =

130 A 106 B =

V ; A

P 2

P>2 (0.01)(0.025)

P = 65.0 kN Allowable Bearing Stress: For the bearing area (sb)allow =

P ; A

220 A 106 B =

37.5 mm

t

P>2 (0.005)(0.025)

P = 55.0 kN (Controls!)

Ans.

66

h

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*1–100. The hanger is supported using the rectangular pin. Determine the required thickness t of the hanger, and dimensions a and b if the suspended load is P = 60 kN. The allowable tensile stress is (st)allow = 150 MPa, the allowable bearing stress is (sb)allow = 290 MPa, and the allowable shear stress is tallow = 125 MPa.

20 mm 75 mm 10 mm a

a

37.5 mm

t P 37.5 mm

Allowable Normal Stress: For the hanger (st)allow =

P ; A

150 A 106 B =

60(103) (0.075)t

t = 0.005333 m = 5.33 mm

Ans.

Allowable Shear Stress: For the pin tallow =

125 A 106 B =

V ; A

30(103) (0.01)b

b = 0.0240 m = 24.0 mm

Ans.

Allowable Bearing Stress: For the bearing area (sb)allow =

P ; A

290 A 106 B =

30(103) (0.0240) a

a = 0.00431 m = 4.31 mm

Ans.

67

b

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•1–101. The 200-mm-diameter aluminum cylinder supports a compressive load of 300 kN. Determine the average normal and shear stress acting on section a–a. Show the results on a differential element located on the section.

300 kN

a

Referring to the FBD of the upper segment of the cylinder sectional through a–a shown in Fig. a, +Q©Fx¿ = 0;

Na - a - 300 cos 30° = 0

+a©Fy¿ = 0;

Va - a - 300 sin 30° = 0

30

Na - a = 259.81 kN

a

Va - a = 150 kN

0.1 Section a–a of the cylinder is an ellipse with a = 0.1 m and b = m. Thus, cos 30° 0.1 b = 0.03628 m2. A a - a = pab = p(0.1)a cos 30°

A sa - a B avg =

259.81(103) Na - a = = 7.162(106) Pa = 7.16 MPa Aa - a 0.03628

Ans.

A ta - a B avg =

150(103) Va - a = = 4.135(106) Pa = 4.13 MPa Aa - a 0.03628

Ans.

d

The differential element representing the state of stress of a point on section a–a is shown in Fig. b

1–102. The long bolt passes through the 30-mm-thick plate. If the force in the bolt shank is 8 kN, determine the average normal stress in the shank, the average shear stress along the cylindrical area of the plate defined by the section lines a–a, and the average shear stress in the bolt head along the cylindrical area defined by the section lines b–b.

8 mm

a 7 mm

b

8 kN

18 mm b a

P = ss = A

8 (103)

= 208 MPa

Ans.

(tavg)a =

8 (103) V = = 4.72 MPa A p (0.018)(0.030)

Ans.

(tavg)b =

8 (103) V = = 45.5 MPa A p (0.007)(0.008)

Ans.

p 4

(0.007)2

68

30 mm

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1–103. Determine the required thickness of member BC and the diameter of the pins at A and B if the allowable normal stress for member BC is sallow = 29 ksi and the allowable shear stress for the pins is tallow = 10 ksi.

C

1.5 in.

Referring to the FBD of member AB, Fig. a, a + ©MA = 0;

2(8)(4) - FBC sin 60° (8) = 0

+ : ©Fx = 0;

9.238 cos 60° - A x = 0

+ c ©Fy = 0;

9.238 sin 60° - 2(8) + A y = 0

FBC = 9.238 kip

60 B

A x = 4.619 kip

8 ft

A y = 8.00 kip 2 kip/ft

Thus, the force acting on pin A is FA = 2 A 2x + A 2y = 2 4.6192 + 8.002 = 9.238 kip Pin A is subjected to single shear, Fig. c, while pin B is subjected to double shear, Fig. b. FBC 9.238 VA = FA = 9.238 kip VB = = = 4.619 kip 2 2 For member BC FBC ; sallow = A BC

29 =

9.238 1.5(t)

t = 0.2124 in.

Use t = For pin A, VA ; tallow = AA

10 =

9.238 p 2 4 dA

1 in. 4

Ans.

dA = 1.085 in. 1 Use dA = 1 in 8

For pin B, VB ; tallow = AB

10 =

4.619 p 2 4 dB

Ans.

dB = 0.7669 in Use dB =

13 in 16

Ans.

69

A

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*1–104. Determine the resultant internal loadings acting on the cross sections located through points D and E of the frame.

150 lb/ft

Segment AD:

1.5 ft

+

: ©Fx = 0;

ND - 1.2 = 0;

+ T ©Fy = 0;

VD + 0.225 + 0.4 = 0;

a + ©MD = 0;

ND = 1.20 kip

Ans.

VD = - 0.625 kip

B

E

Ans. 2.5 ft

MD + 0.225(0.75) + 0.4(1.5) = 0 MD =

A 4 ft

D

C

- 0.769 kip # ft

Ans.

3 ft

5 ft

Segment CE: Q+ ©Fx = 0;

NE + 2.0 = 0;

R+ ©Fy = 0;

VE = 0

a + ©ME = 0;

NE = - 2.00 kip

Ans. Ans.

ME = 0

Ans.

•1–105.

The pulley is held fixed to the 20-mm-diameter shaft using a key that fits within a groove cut into the pulley and shaft. If the suspended load has a mass of 50 kg, determine the average shear stress in the key along section a–a. The key is 5 mm by 5 mm square and 12 mm long.

a + ©MO = 0;

a

75 mm

F (10) - 490.5 (75) = 0

F = 3678.75 N tavg =

a

3678.75 V = = 61.3 MPa A (0.005)(0.012)

Ans.

70

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1–106. The bearing pad consists of a 150 mm by 150 mm block of aluminum that supports a compressive load of 6 kN. Determine the average normal and shear stress acting on the plane through section a–a. Show the results on a differential volume element located on the plane.

6 kN

a

Equation of Equilibrium: +Q©Fx = 0;

Va - a - 6 cos 60° = 0

Va - a = 3.00 kN

a+ ©Fy = 0;

Na - a - 6 sin 60° = 0

Na - a = 5.196 kN

30 a

150 mm

Averge Normal Stress And Shear Stress: The cross sectional Area at section a–a is 0.15 b (0.15) = 0.02598 m2. A = a sin 60° sa - a =

5.196(103) Na - a = = 200 kPa A 0.02598

Ans.

ta - a =

3.00(103) Va - a = = 115 kPa A 0.02598

Ans.

1–107. The yoke-and-rod connection is subjected to a tensile force of 5 kN. Determine the average normal stress in each rod and the average shear stress in the pin A between the members.

5 kN

40 mm

For the 40 – mm – dia rod: s40

30 mm

5 (103) P = p = = 3.98 MPa 2 A 4 (0.04)

Ans.

For the 30 – mm – dia rod:

5 kN

3

s30 =

5 (10 ) V = p = 7.07 MPa 2 A 4 (0.03)

Ans.

Average shear stress for pin A: tavg =

A 25 mm

2.5 (103) P = p = 5.09 MPa 2 A 4 (0.025)

Ans.

71

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*1–108. The cable has a specific weight g (weight>volume) and cross-sectional area A. If the sag s is small, so that its length is approximately L and its weight can be distributed uniformly along the horizontal axis, determine the average normal stress in the cable at its lowest point C.

A

s C L/2

Equation of Equilibrium: a + ©MA = 0;

Ts -

gAL L a b = 0 2 4 T =

B

gAL2 8s

Average Normal Stress: gAL2

gL2 T 8s s = = = A A 8s

Ans.

72

L/2

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2–1. An air-filled rubber ball has a diameter of 6 in. If the air pressure within it is increased until the ball’s diameter becomes 7 in., determine the average normal strain in the rubber. d0 = 6 in. d = 7 in. e =

pd - pd0 7 - 6 = = 0.167 in./in. pd0 6

Ans.

2–2. A thin strip of rubber has an unstretched length of 15 in. If it is stretched around a pipe having an outer diameter of 5 in., determine the average normal strain in the strip. L0 = 15 in. L = p(5 in.) e =

L - L0 5p - 15 = = 0.0472 in.>in. L0 15

Ans.

2–3. The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam causes the end C to be displaced 10 mm downward, determine the normal strain developed in wires CE and BD.

D

E

4m

P

¢LBD ¢LCE = 3 7

A

3 (10) = 4.286 mm 7 ¢LCE 10 = = = 0.00250 mm>mm L 4000

3m

¢LBD = eCE

eBD =

B

Ans.

¢LBD 4.286 = = 0.00107 mm>mm L 4000

Ans.

1

C

2m

2m

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*2–4. The two wires are connected together at A. If the force P causes point A to be displaced horizontally 2 mm, determine the normal strain developed in each wire.

C 300

œ = 23002 + 22 - 2(300)(2) cos 150° = 301.734 mm LAC

eAC = eAB

œ - LAC LAC 301.734 - 300 = = = 0.00578 mm>mm LAC 300

mm

30⬚

Ans.

30⬚

300

A

P

mm

B

•2–5. The rigid beam is supported by a pin at A and wires BD and CE. If the distributed load causes the end C to be displaced 10 mm downward, determine the normal strain developed in wires CE and BD.

E D 2m 1.5 m 3m

2m A

B

C

w

Since the vertical displacement of end C is small compared to the length of member AC, the vertical displacement dB of point B, can be approximated by referring to the similar triangle shown in Fig. a dB 10 = ; dB = 4 mm 2 5 The unstretched lengths of wires BD and CE are LBD = 1500 mm and LCE = 2000 mm. dB 4 Ans. = = 0.00267 mm>mm A eavg B BD = LBD 1500

A eavg B CE =

dC 10 = = 0.005 mm>mm LCE 2000

Ans.

2

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2–6. Nylon strips are fused to glass plates. When moderately heated the nylon will become soft while the glass stays approximately rigid. Determine the average shear strain in the nylon due to the load P when the assembly deforms as indicated.

y 2 mm P

3 mm 5 mm 3 mm 5 mm 3 mm

g = tan - 1 a

2 b = 11.31° = 0.197 rad 10

x

Ans.

2–7. If the unstretched length of the bowstring is 35.5 in., determine the average normal strain in the string when it is stretched to the position shown. 18 in.

6 in. 18 in.

Geometry: Referring to Fig. a, the stretched length of the string is L = 2L¿ = 2 2182 + 62 = 37.947 in. Average Normal Strain: eavg =

L - L0 37.947 - 35.5 = = 0.0689 in.>in. L0 35.5

Ans.

3

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u

*2–8. Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If a force is applied to the end D of the member and causes it to rotate by u = 0.3°, determine the normal strain in the cable. Originally the cable is unstretched.

D

P 300 mm

B

AB = 24002 + 3002 = 500 mm

300 mm

AB¿ = 2400 + 300 - 2(400)(300) cos 90.3° 2

2

A

C

= 501.255 mm eAB =

AB¿ - AB 501.255 - 500 = AB 500

400 mm

= 0.00251 mm>mm

Ans.

•2–9.

Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If a force is applied to the end D of the member and causes a normal strain in the cable of 0.0035 mm>mm, determine the displacement of point D. Originally the cable is unstretched.

u D

300 mm

B

AB = 23002 + 4002 = 500 mm

300 mm

AB¿ = AB + eABAB A

= 500 + 0.0035(500) = 501.75 mm

C

501.752 = 3002 + 4002 - 2(300)(400) cos a a = 90.4185°

400 mm

p (0.4185) rad u = 90.4185° - 90° = 0.4185° = 180° ¢ D = 600(u) = 600(

P

p )(0.4185) = 4.38 mm 180°

Ans.

4

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2–10. The corners B and D of the square plate are given the displacements indicated. Determine the shear strains at A and B.

y A

16 mm D

B

3 mm 3 mm 16 mm

16 mm

Applying trigonometry to Fig. a f = tan - 1 a

13 p rad b = 39.09° a b = 0.6823 rad 16 180°

a = tan - 1 a

16 p rad b = 50.91° a b = 0.8885 rad 13 180°

By the definition of shear strain,

A gxy B A =

p p - 2f = - 2(0.6823) = 0.206 rad 2 2

Ans.

A gxy B B =

p p - 2a = - 2(0.8885) = - 0.206 rad 2 2

Ans.

5

C

16 mm

x

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2–11. The corners B and D of the square plate are given the displacements indicated. Determine the average normal strains along side AB and diagonal DB.

y A

16 mm D

B

3 mm 3 mm 16 mm

16 mm

Referring to Fig. a, LAB = 2162 + 162 = 2512 mm LAB¿ = 2162 + 132 = 2425 mm LBD = 16 + 16 = 32 mm LB¿D¿ = 13 + 13 = 26 mm Thus,

A eavg B AB =

LAB¿ - LAB 2425 - 2512 = = - 0.0889 mm>mm LAB 2512

Ans.

A eavg B BD =

LB¿D¿ - LBD 26 - 32 = = - 0.1875 mm>mm LBD 32

Ans.

6

C

16 mm

x

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*2–12. The piece of rubber is originally rectangular. Determine the average shear strain gxy at A if the corners B and D are subjected to the displacements that cause the rubber to distort as shown by the dashed lines.

y 3 mm

C D

2 = 0.006667 rad 300 3 u2 = tan u2 = = 0.0075 rad 400 u1 = tan u1 =

400 mm

gxy = u1 + u2

A

= 0.006667 + 0.0075 = 0.0142 rad

Ans.

•2–13.

The piece of rubber is originally rectangular and subjected to the deformation shown by the dashed lines. Determine the average normal strain along the diagonal DB and side AD.

f = tan

B 2 mm

3 mm

C D

400 mm

3 b = 0.42971° a 400

AB¿ = 2(300)2 + (2)2 = 300.00667 w = tan - 1 a

x

y

AD¿ = 2(400)2 + (3)2 = 400.01125 mm -1

300 mm

A

2 b = 0.381966° 300

a = 90° - 0.42971° - 0.381966° = 89.18832° D¿B¿ = 2(400.01125)2 + (300.00667)2 - 2(400.01125)(300.00667) cos (89.18832°) D¿B¿ = 496.6014 mm DB = 2(300)2 + (400)2 = 500 mm 496.6014 - 500 = - 0.00680 mm>mm 500 400.01125 - 400 = = 0.0281(10 - 3) mm>mm 400

eDB =

Ans.

eAD

Ans.

7

300 mm

B 2 mm

x

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2–14. Two bars are used to support a load. When unloaded, AB is 5 in. long, AC is 8 in. long, and the ring at A has coordinates (0, 0). If a load P acts on the ring at A, the normal strain in AB becomes PAB = 0.02 in.>in., and the normal strain in AC becomes PAC = 0.035 in.>in. Determine the coordinate position of the ring due to the load.

y

B

C

60⬚

5 in.

8 in.

A

x

P

Average Normal Strain: œ = LAB + eAB LAB = 5 + (0.02)(5) = 5.10 in. LAB œ = LAC + eACLAC = 8 + (0.035)(8) = 8.28 in. LAC

Geometry: a = 282 - 4.33012 = 6.7268 in. 5.102 = 9.22682 + 8.282 - 2(9.2268)(8.28) cos u u = 33.317° x¿ = 8.28 cos 33.317° = 6.9191 in. y¿ = 8.28 sin 33.317° = 4.5480 in. x = - (x¿ - a) = - (6.9191 - 6.7268) = - 0.192 in.

Ans.

y = - (y¿ - 4.3301) = - (4.5480 - 4.3301) = - 0.218 in.

Ans.

8

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2–15. Two bars are used to support a load P. When unloaded, AB is 5 in. long, AC is 8 in. long, and the ring at A has coordinates (0, 0). If a load is applied to the ring at A, so that it moves it to the coordinate position (0.25 in., -0.73 in.), determine the normal strain in each bar.

y

B

C

60⬚

5 in.

8 in.

A

x

P

Geometry: a = 282 - 4.33012 = 6.7268 in. LA¿B = 2(2.5 + 0.25)2 + (4.3301 + 0.73)2 = 5.7591 in. LA¿C = 2(6.7268 - 0.25)2 + (4.3301 + 0.73)2 = 8.2191 in. Average Normal Strain: eAB =

=

eAC =

=

LA¿B - LAB LAB 5.7591 - 5 = 0.152 in.>in. 5

Ans.

LA¿C - LAC LAC 8.2191 - 8 = 0.0274 in.>in. 8

Ans.

9

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*2–16. The square deforms into the position shown by the dashed lines. Determine the average normal strain along each diagonal, AB and CD. Side D¿B¿ remains horizontal.

y 3 mm D¿

B¿

B

D

Geometry: AB = CD = 2502 + 502 = 70.7107 mm 53 mm

C¿D¿ = 2532 + 582 - 2(53)(58) cos 91.5°

50 mm 91.5⬚

= 79.5860 mm C

B¿D¿ = 50 + 53 sin 1.5° - 3 = 48.3874 mm

A

x

C¿

AB¿ = 2532 + 48.38742 - 2(53)(48.3874) cos 88.5°

50 mm 8 mm

= 70.8243 mm Average Normal Strain:

eAB =

=

eCD =

=

AB¿ - AB AB 70.8243 - 70.7107 = 1.61 A 10 - 3 B mm>mm 70.7107

Ans.

C¿D¿ - CD CD 79.5860 - 70.7107 = 126 A 10 - 3 B mm>mm 70.7107

Ans.

•2–17. The three cords are attached to the ring at B. When a force is applied to the ring it moves it to point B¿ , such that the normal strain in AB is PAB and the normal strain in CB is PCB. Provided these strains are small, determine the normal strain in DB. Note that AB and CB remain horizontal and vertical, respectively, due to the roller guides at A and C.

A¿

B¿

A

B

L

Coordinates of B (L cos u, L sin u)

u

Coordinates of B¿ (L cos u + eAB L cos u, L sin u + eCB L sin u)

C¿ D

LDB¿ = 2(L cos u + eAB L cos u) + (L sin u + eCB L sin u) 2

2

LDB¿ = L2cos2 u(1 + 2eAB + e2AB) + sin2 u(1 + 2eCB + e2CB) Since eAB and eCB are small, LDB¿ = L21 + (2 eAB cos2 u + 2eCB sin2 u) Use the binomial theorem, LDB¿ = L ( 1 +

1 (2 eAB cos2 u + 2eCB sin2 u)) 2

= L ( 1 + eAB cos2 u + eCB sin2 u) Thus, eDB =

L( 1 + eAB cos2 u + eCB sin2 u) - L L

eDB = eAB cos2 u + eCB sin2 u

Ans.

10

C

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2–18. The piece of plastic is originally rectangular. Determine the shear strain gxy at corners A and B if the plastic distorts as shown by the dashed lines.

y 5 mm 2 mm 2 mm

B

4 mm

C 300 mm

Geometry: For small angles, 2 mm D

2 a = c = = 0.00662252 rad 302 b = u =

A

x

400 mm 3 mm

2 = 0.00496278 rad 403

Shear Strain: (gB)xy = a + b = 0.0116 rad = 11.6 A 10 - 3 B rad

Ans.

(gA)xy = - (u + c) = - 0.0116 rad = - 11.6 A 10 - 3 B rad

Ans.

2–19. The piece of plastic is originally rectangular. Determine the shear strain gxy at corners D and C if the plastic distorts as shown by the dashed lines.

y 5 mm 2 mm 2 mm

B

4 mm

C 300 mm 2 mm D

A 400 mm 3 mm

Geometry: For small angles, 2 = 0.00496278 rad 403 2 = 0.00662252 rad b = u = 302 Shear Strain: a = c =

(gC)xy = - (a + b) = - 0.0116 rad = - 11.6 A 10 - 3 B rad

Ans.

(gD)xy = u + c = 0.0116 rad = 11.6 A 10 - 3 B rad

Ans.

11

x

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*2–20. The piece of plastic is originally rectangular. Determine the average normal strain that occurs along the diagonals AC and DB.

y 5 mm 2 mm 2 mm

Geometry:

B

4 mm

C

AC = DB = 24002 + 3002 = 500 mm

300 mm

DB¿ = 24052 + 3042 = 506.4 mm

2 mm D

A¿C¿ = 2401 + 300 = 500.8 mm 2

2

x

A 400 mm 3 mm

Average Normal Strain: eAC =

A¿C¿ - AC 500.8 - 500 = AC 500

= 0.00160 mm>mm = 1.60 A 10 - 3 B mm>mm eDB =

Ans.

DB¿ - DB 506.4 - 500 = DB 500

= 0.0128 mm>mm = 12.8 A 10 - 3 B mm>mm

Ans.

•2–21.

The force applied to the handle of the rigid lever arm causes the arm to rotate clockwise through an angle of 3° about pin A. Determine the average normal strain developed in the wire. Originally, the wire is unstretched.

D

600 mm

Geometry: Referring to Fig. a, the stretched length of LB¿D can be determined using the consine law, A

LB¿D = 2(0.6 cos 45°)2 + (0.6 sin 45°)2 - 2(0.6 cos 45°)(0.6 sin 45°) cos 93°

B

= 0.6155 m Average Normal Strain: The unstretched length of wire BD is LBD = 0.6 m. We obtain eavg =

C

45⬚

LB¿D - LBD 0.6155 - 0.6 = 0.0258 m>m = LBD 0.6

Ans.

12

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2–22. A square piece of material is deformed into the dashed position. Determine the shear strain gxy at A.

y 15.18 mm B

Shear Strain: (gA)xy =

89.7° p - ¢ ≤p 2 180°

C

15.24 mm

15 mm

= 5.24 A 10 - 3 B rad

Ans.

89.7⬚ A

2–23. A square piece of material is deformed into the dashed parallelogram. Determine the average normal strain that occurs along the diagonals AC and BD.

15 mm 15.18 mm

x

D

y 15.18 mm B

C

15.24 mm

15 mm 89.7⬚ A

Geometry: AC = BD = 2152 + 152 = 21.2132 mm AC¿ = 215.182 + 15.242 - 2(15.18)(15.24) cos 90.3° = 21.5665 mm B¿D¿ = 215.182 + 15.242 - 2(15.18)(15.24) cos 89.7° = 21.4538 mm Average Normal Strain: eAC =

eBD

AC¿ - AC 21.5665 - 21.2132 = AC 21.2132

= 0.01665 mm>mm = 16.7 A 10 - 3 B mm>mm

Ans.

= 0.01134 mm>mm = 11.3 A 10 - 3 B mm>mm

Ans.

B¿D¿ - BD 21.4538 - 21.2132 = = BD 21.2132

13

15 mm 15.18 mm

D

x

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*2–24. A square piece of material is deformed into the dashed position. Determine the shear strain gxy at C.

y 15.18 mm B

C

15.24 mm

15 mm 89.7⬚ A

(gC)xy =

15 mm 15.18 mm

x

D

p 89.7° - ¢ ≤p 2 180° = 5.24 A 10 - 3 B rad

Ans.

u ⫽ 2⬚

•2–25.

The guy wire AB of a building frame is originally unstretched. Due to an earthquake, the two columns of the frame tilt u = 2°. Determine the approximate normal strain in the wire when the frame is in this position. Assume the columns are rigid and rotate about their lower supports.

u ⫽ 2⬚

B

Geometry: The vertical displacement is negligible 3m

xA

2° = (1) ¢ ≤ p = 0.03491 m 180° A

2° xB = (4) ¢ ≤ p = 0.13963 m 180°

1m

x = 4 + xB - xA = 4.10472 m A¿B¿ = 232 + 4.104722 = 5.08416 m AB = 232 + 42 = 5.00 m Average Normal Strain: eAB =

=

A¿B¿ - AB AB 5.08416 - 5 = 16.8 A 10 - 3 B m>m 5

Ans.

14

4m

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2–26. The material distorts into the dashed position shown. Determine (a) the average normal strains along sides AC and CD and the shear strain gxy at F, and (b) the average normal strain along line BE.

y 15 mm C

25 mm D

10 mm B E

75 mm

90 mm

A

Referring to Fig. a, LBE = 2(90 - 75)2 + 802 = 26625 mm LAC¿ = 21002 + 152 = 210225 mm LC¿D¿ = 80 - 15 + 25 = 90 mm f = tan-1 ¢

25 p rad ≤ = 14.04° ¢ ≤ = 0.2450 rad. 100 180°

When the plate deforms, the vertical position of point B and E do not change. LBB¿ 15 = ; LBB¿ = 13.5 mm 90 100 LEE¿ 25 = ; 75 100

LEE¿ = 18.75 mm

LB¿E¿ = 2(90 - 75)2 + (80 - 13.5 + 18.75)2 = 27492.5625 mm Thus,

A eavg B AC =

LAC¿ - LAC 210225 - 100 = = 0.0112 mm>mm LAC 100

Ans.

A eavg B CD =

LC¿D¿ - LCD 90 - 80 = = 0.125 mm>mm LCD 80

Ans.

A eavg B BE =

LB¿E¿ - LBE 27492.5625 - 26625 = = 0.0635 mm>mm LBE 26625

Ans.

Referring to Fig. a, the angle at corner F becomes larger than 90° after the plate deforms. Thus, the shear strain is negative. 0.245 rad

Ans.

15

80 mm

F

x

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2–27. The material distorts into the dashed position shown. Determine the average normal strain that occurs along the diagonals AD and CF.

y 15 mm

25 mm D

C 10 mm

The undeformed length of diagonals AD and CF are

B E

LAD = LCF = 280 + 100 = 216400 mm 2

2

The deformed length of diagonals AD and CF are

75 mm

90 mm

LAD¿ = 2(80 + 25) + 100 = 221025 mm 2

2

LC¿F = 2(80 - 15)2 + 1002 = 214225 mm A

Thus,

A eavg B AD =

LAD¿ - LAD 221025 - 216400 = = 0.132 mm>mm LAD 216400

Ans.

A eavg B CF =

LC¿F - LCF 214225 - 216400 = = - 0.0687 mm>mm LCF 216400

Ans.

*2–28. The wire is subjected to a normal strain that is 2 defined by P = xe - x , where x is in millimeters. If the wire has an initial length L, determine the increase in its length.

80 mm

P ⫽ xe⫺x

L

2

dL = e dx = x e-x dx L 2

L0

x e-x dx

L 1 1 1 2 2 = - c e-x d 冷 = - c e-L - d 2 2 2 0

=

x

2

x x

¢L =

F

1 2 [1 - e-L ] 2

Ans.

16

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•2–29. The curved pipe has an original radius of 2 ft. If it is heated nonuniformly, so that the normal strain along its length is P = 0.05 cos u, determine the increase in length of the pipe.

e = 0.05 cos u ¢L =

L

2 ft

e dL

=

A

u

90°

(0.05 cos u)(2 du)

L0

90°

= 0.1

90°

cos u du = [0.1[sin u] 0冷 ] = 0.100 ft

L0

Ans.

Solve Prob. 2–29 if P = 0.08 sin u.

2–30.

dL = 2 due = 0.08 sin u ¢L =

e dL

L

90°

=

2 ft

(0.08 sin u)(2 du)

L0

= 0.16

L0

90°

sin u du = 0.16[- cos u] 0冷 = 0.16 ft

Ans.

2–31. The rubber band AB has an unstretched length of 1 ft. If it is fixed at B and attached to the surface at point A¿, determine the average normal strain in the band. The surface is defined by the function y = (x2) ft, where x is in feet.

y y ⫽ x2

A¿

Geometry: 1 ft

L =

L0

A

1 + a

dy 2 b dx dx

However y = x2 then

1 ft

dy = 2x dx

B

1 ft

L =

=

L0

A

u 90°

21 + 4 x2 dx

1 1 ft C 2x 21 + 4 x2 + ln A 2x + 21 + 4x2 B D 冷0 4

= 1.47894 ft Average Normal Strain: L - L0 1.47894 - 1 = = 0.479 ft>ft eavg = L0 1

Ans.

17

A 1 ft

x

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*2–32. The bar is originally 300 mm long when it is flat. If it is subjected to a shear strain defined by gxy = 0.02x, where x is in meters, determine the displacement ¢y at the end of its bottom edge. It is distorted into the shape shown, where no elongation of the bar occurs in the x direction.

y

⌬y x 300 mm

Shear Strain: dy = tan gxy ; dx

dy = tan (0.02 x) dx 300 mm

¢y

dy =

L0

L0

tan (0.02 x)dx

mm ¢y = - 50[ln cos (0.02x)]|300 0

= 2.03 mm

Ans.

•2–33. The fiber AB has a length L and orientation u. If its ends A and B undergo very small displacements uA and vB , respectively, determine the normal strain in the fiber when it is in position A¿B¿.

y B¿ vB B L

Geometry: LA¿B¿ = 2(L cos u - uA) + (L sin u + yB) 2

2

A

= 2L3 + u2A + y2B + 2L(yB sin u - uA cos u) Average Normal Strain: LA¿B¿ - L eAB = L =

A

1 +

2(yB sin u - uA cos u) u2A + y2B + - 1 L L2

Neglecting higher terms u2A and y2B 1

eAB

2(yB sin u - uA cos u) 2 = B1 + R - 1 L

Using the binomial theorem: eAB = 1 +

=

2uA cos u 1 2yB sin u ¢ ≤ + ... - 1 2 L L

yB sin u uA cos u L L

Ans.

18

uA A¿

u x

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2–34. If the normal strain is defined in reference to the final length, that is, Pnœ = lim a p : p¿

¢s¿ - ¢s b ¢s¿

instead of in reference to the original length, Eq. 2–2, show that the difference in these strains is represented as a second-order term, namely, Pn - Pnœ = PnPnœ .

eB =

¢S¿ - ¢S ¢S

œ = eB - eA

¢S¿ - ¢S ¢S¿ - ¢S ¢S ¢S¿

¢S¿ 2 - ¢S¢S¿ - ¢S¿¢S + ¢S2 ¢S¢S¿ 2 2 ¢S¿ + ¢S - 2¢S¿¢S = ¢S¢S¿ =

=

(¢S¿ - ¢S)2 ¢S¿ - ¢S ¢S¿ - ¢S = ¢ ≤¢ ≤ ¢S¢S¿ ¢S ¢S¿

= eA eBœ (Q.E.D)

19

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•3–1.

A concrete cylinder having a diameter of 6.00 in. and gauge length of 12 in. is tested in compression. The results of the test are reported in the table as load versus contraction. Draw the stress–strain diagram using scales of 1 in. = 0.5 ksi and 1 in. = 0.2110-32 in.>in. From the diagram, determine approximately the modulus of elasticity.

Stress and Strain: s =

P (ksi) A

e =

dL (in./in.) L

0

0

0.177

0.00005

0.336

0.00010

0.584

0.000167

0.725

0.000217

0.902

0.000283

1.061

0.000333

1.220

0.000375

1.362

0.000417

1.645

0.000517

1.768

0.000583

1.874

0.000625

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

1.31 - 0 = 3.275 A 103 B ksi 0.0004 - 0

Ans.

1

Load (kip)

Contraction (in.)

0 5.0 9.5 16.5 20.5 25.5 30.0 34.5 38.5 46.5 50.0 53.0

0 0.0006 0.0012 0.0020 0.0026 0.0034 0.0040 0.0045 0.0050 0.0062 0.0070 0.0075

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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

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). ut =

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

Ans.

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 =

lb in. 1 (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

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

in # lb in3

Ans.

2

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*3–4. A tension test was performed on a specimen having an original diameter of 12.5 mm and a gauge length of 50 mm. The data are listed in the table. Plot the stress–strain diagram, and determine approximately the modulus of elasticity, the ultimate stress, and the fracture stress. Use a scale of 20 mm = 50 MPa and 20 mm = 0.05 mm>mm. Redraw the linear-elastic region, using the same stress scale but a strain scale of 20 mm = 0.001 mm>mm. Stress and Strain: s =

dL P (MPa) e = (mm/mm) A L 0

0

90.45

0.00035

259.9

0.00120

308.0

0.00204

333.3

0.00330

355.3

0.00498

435.1

0.02032

507.7

0.06096

525.6

0.12700

507.7

0.17780

479.1

0.23876

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

228.75(106) - 0 = 229 GPa 0.001 - 0

Ans.

Ultimate and Fracture Stress: From the stress–strain diagram (sm)approx = 528 MPa

Ans.

(sf)approx = 479 MPa

Ans.

3

Load (kN)

Elongation (mm)

0 11.1 31.9 37.8 40.9 43.6 53.4 62.3 64.5 62.3 58.8

0 0.0175 0.0600 0.1020 0.1650 0.2490 1.0160 3.0480 6.3500 8.8900 11.9380

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3–5. A tension test was performed on a steel specimen having an original diameter of 12.5 mm and gauge length of 50 mm. Using the data listed in the table, plot the stress–strain diagram, and determine approximately the modulus of toughness. Use a scale of 20 mm = 50 MPa and 20 mm = 0.05 mm>mm. Stress and Strain: s =

P dL (MPa) e = (mm/mm) A L 0

0

90.45

0.00035

259.9

0.00120

308.0

0.00204

333.3

0.00330

355.3

0.00498

435.1

0.02032

507.7

0.06096

525.6

0.12700

507.7

0.17780

479.1

0.23876

Modulus of Toughness: The modulus of toughness is equal to the total area under the stress–strain diagram and can be approximated by counting the number of squares. The total number of squares is 187. (ut)approx = 187(25) A 106 B ¢

N m ≤ a 0.025 b = 117 MJ>m3 m m2

Ans.

4

Load (kN)

Elongation (mm)

0 11.1 31.9 37.8 40.9 43.6 53.4 62.3 64.5 62.3 58.8

0 0.0175 0.0600 0.1020 0.1650 0.2490 1.0160 3.0480 6.3500 8.8900 11.9380

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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 =

¢e =

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 ¢e 0.000750

Ans.

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 e =

0.02 d = = 0.000555 in.>in. L 3 (12) s = Ee = 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.

5

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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.

6

B

C

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

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The s –P diagram for a collagen fiber bundle from which a human tendon is composed is shown. If a segment of the Achilles tendon at A has a length of 6.5 in. and an approximate cross-sectional area of 0.229 in2, determine its elongation if the foot supports a load of 125 lb, which causes a tension in the tendon of 343.75 lb.

•3–9.

s =

s (ksi) 4.50 A

3.75 3.00 2.25 1.50

P 343.75 = = 1.50 ksi A 0.229

125 lb

0.75 0.05

From the graph e = 0.035 in.>in. d = eL = 0.035(6.5) = 0.228 in.

0.10

P (in./in.)

Ans.

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

From the stress–strain diagram, Fig. a,

45

60 ksi - 0 E = ; 1 0.002 - 0 sy = 60 ksi

E = 30.0(103) ksi

Ans.

30 15

su>t = 100 ksi

0

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

Ans.

Pu>t = su>t A = 100 C p4 (0.52) D = 19.63 kip = 19.6 kip

Ans.

7

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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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

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 recovered along line AB, Fig. a, which has a gradient 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.

8

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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*3–12. 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. 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 c 15(103)

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

s (ksi) 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.)

9

Ans.

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•3–13.

A bar having a length of 5 in. and cross-sectional area of 0.7 in2 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: 8.00 P = = 11.43 ksi A 0.7

s =

e =

dL 0.002 = = 0.000400 in.>in. L 5

Modulus of Elasticity: E =

s 11.43 = = 28.6(103) ksi e 0.000400

Ans.

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

Here, we are only interested in determining the force in wire BD. Referring 4 ft to the FBD in Fig. a a + ©MA = 0;

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

FBD = 1500 lb

A

sBD

3 ft

1500 = 30.56(103) psi = 30.56 ksi p 2 (0.25 ) 4

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

D C

The normal stress developed in the wire is FBD = = ABD

P

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

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

Ans.

10

3 ft

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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 0.075 in. downward.

B

4 ft

P

A

D C 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.

11

3 ft

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s (MPa)

*3–16. Determine the elongation of the square hollow bar when it is subjected to the axial force P = 100 kN. If this axial force is increased to P = 360 kN and released, find the permanent elongation of the bar. The bar is made of a metal alloy having a stress–strain diagram which can be approximated as shown.

500 600 mm P

250

50 mm 5 mm

0.00125

Normal Stress and Strain: The cross-sectional area of the hollow bar is A = 0.052 - 0.042 = 0.9(10 - 3)m2. When P = 100 kN, s1 =

100(103) P = 111.11 MPa = A 0.9(10 - 3)

From the stress–strain diagram shown in Fig. a, the slope of the straight line OA which represents the modulus of elasticity of the metal alloy is E =

250(106) - 0 = 200 GPa 0.00125 - 0

Since s1 6 250 MPa, Hooke’s Law can be applied. Thus s1 = Ee1; 111.11(106) = 200(109)e1 e1 = 0.5556(10 - 3) mm>mm Thus, the elongation of the bar is d1 = e1L = 0.5556(10 - 3)(600) = 0.333 mm

Ans.

When P = 360 kN, s2 =

360(103) P = 400 MPa = A 0.9(10 - 3)

From the geometry of the stress–strain diagram, Fig. a, e2 - 0.00125 0.05 - 0.00125 = 400 - 250 500 - 250

e2 = 0.0305 mm>mm

When P = 360 kN is removed, the strain recovers linearly along line BC, Fig. a, parallel to OA. Thus, the elastic recovery of strain is given by s2 = Eer;

400(106) = 200(109)er er = 0.002 mm>mm

The permanent set is eP = e2 - er = 0.0305 - 0.002 = 0.0285 mm>mm Thus, the permanent elongation of the bar is dP = ePL = 0.0285(600) = 17.1 mm

Ans.

12

0.05

P (mm/mm) 50 mm

P 5 mm

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3–16. Continued

13

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s (ksi)

3–17. A tension test was performed on an aluminum 2014-T6 alloy specimen. The resulting stress–strain diagram is shown in the figure. Estimate (a) the proportional limit, (b) the modulus of elasticity, and (c) the yield strength based on a 0.2% strain offset method.

70 60 50 40 30 20 10 0

0.02 0.002

Proportional Limit and Yield Strength: From the stress–strain diagram, Fig. a, spl = 44 ksi

Ans.

sY = 60 ksi

Ans.

Modulus of Elasticity: From the stress–strain diagram, the corresponding strain for sPL = 44 ksi is epl = 0.004 in.>in. Thus, E =

44 - 0 = 11.0(103) ksi 0.004 - 0

Ans.

Modulus of Resilience: The modulus of resilience is equal to the area under the

14

0.04 0.004

0.06 0.006

0.08 0.008

0.10 0.010

P (in./in.)

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s (ksi)

3–18. A tension test was performed on an aluminum 2014-T6 alloy specimen. The resulting stress–strain diagram is shown in the figure. Estimate (a) the modulus of resilience; and (b) modulus of toughness.

70 60 50 40 30 20 10 0

0.02 0.002

0.04 0.004

0.06 0.006

0.08 0.008

0.10 0.010

P (in./in.)

stress–strain diagram up to the proportional limit. From the stress–strain diagram, spl = 44 ksi

epl = 0.004 in.>in.

Thus,

A Ui B r = splepl = (44)(103)(0.004) = 88 1 2

1 2

in # lb in3

Ans.

Modulus of Toughness: 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 of squares is 65. Thus,

C A Ui B t D approx = 65 B 10(103)

lb in. in # lb c 0.01 d = 6.50(103) 2R in. in in3

Ans.

The stress–strain diagram for a bone is shown, and can be described by the equation

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

e = 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 MPa dP 0.45(10 - 6)

Ans.

s=0

15

P

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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-mmlong 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 e = 0.12

120(103) = 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 - e)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 = eL = 0.12(200) = 24 mm

Ans.

16

P

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•3–21.

The stress–strain diagram for a polyester resin is given in the figure. If the rigid beam is supported by a strut AB and post CD, both made from this material, and subjected to a load of P = 80 kN, determine the angle of tilt of the beam when the load is applied. The diameter of the strut is 40 mm and the diameter of the post is 80 mm.

B

2m P

A

C 0.75 m 0.75 m

D

0.5 m

From the stress–strain diagram, E =

32.2(10)6 = 3.22(109) Pa 0.01

s (MPa) 100 95

Thus,

70 60

40(10 ) FAB = p = 31.83 MPa 2 AAB 4 (0.04)

sAB =

eAB

50

31.83(106) sAB = 0.009885 mm>mm = = E 3.22(109)

20 0

7.958(106) sCD = 0.002471 mm>mm = E 3.22(109)

dAB = eABLAB = 0.009885(2000) = 19.771 mm dCD = eCDLCD = 0.002471(500) = 1.236 mm Angle of tilt a: tan a =

18.535 ; 1500

tension

40 32.2

40(103) FCD = p = 7.958 MPa 2 ACD 4 (0.08)

sCD =

eCD =

compression

80 3

a = 0.708°

Ans.

17

0

0.01 0.02 0.03 0.04

P (mm/mm)

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3–22. The stress–strain diagram for a polyester resin is given in the figure. If the rigid beam is supported by a strut AB and post CD made from this material, determine the largest load P that can be applied to the beam before it ruptures. The diameter of the strut is 12 mm and the diameter of the post is 40 mm.

B

2m P

Rupture of strut AB: sR =

FAB ; AAB

50(106) =

P>2

A

; p 2 4 (0.012)

0.75 m 0.75 m

P = 11.3 kN (controls)

D

0.5 m

Ans. s (MPa)

Rupture of post CD: FCD sR = ; ACD

C

95(10 ) =

100 95

P>2

6

p 2 4 (0.04)

compression

80 70 60

P = 239 kN

50 tension

40 32.2 20 0

0

0.01 0.02 0.03 0.04

P (mm/mm)

s (ksi)

3–23. By adding plasticizers to polyvinyl chloride, it is possible to reduce its stiffness. The stress–strain diagrams for three types of this material showing this effect are given below. Specify the type that should be used in the manufacture of a rod having a length of 5 in. and a diameter of 2 in., that is required to support at least an axial load of 20 kip and also be able to stretch at most 14 in.

15 P unplasticized 10

copolymer

flexible

5

(plasticized)

Normal Stress:

P

P s = = A

20 p 2 = 6.366 ksi (2 ) 4

0

0

Normal Strain: e =

0.25 = 0.0500 in.>in. 5

From the stress–strain diagram, the copolymer will satisfy both stress and strain requirements. Ans.

18

0.10

0.20

0.30

P (in./in.)

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*3–24. 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 = 3011032 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.

•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

elong =

300 p 2 4 (0.015)

300 N

300 N 200 mm

= 1.697 MPa

1.697(106) s = 0.0006288 = E 2.70(109)

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

Ans.

elat = - Velong = - 0.4(0.0006288) = - 0.0002515 ¢d = elatd = - 0.0002515 (15) = - 0.00377 mm

Ans.

19

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3–26. The short cylindrical block of 2014-T6 aluminum, having an original diameter of 0.5 in. and a length of 1.5 in., is placed in the smooth jaws of a vise and squeezed until the axial load applied is 800 lb. Determine (a) the decrease in its length and (b) its new diameter.

800 lb

800 lb

a) s =

P = A

elong =

p 4

800 = 4074.37 psi (0.5)2

s - 4074.37 = - 0.0003844 = E 10.6(106)

d = elong L = - 0.0003844 (1.5) = - 0.577 (10 - 3) in.

Ans.

b) V =

-elat = 0.35 elong

elat = - 0.35 ( -0.0003844) = 0.00013453 ¢d = elat d = 0.00013453 (0.5) = 0.00006727 d¿ = d + ¢d = 0.5000673 in.

Ans.

s(MPa)

3–27. The elastic portion of the stress–strain diagram for a steel alloy is shown in the figure. The specimen from which it was obtained had an original diameter of 13 mm and a gauge length of 50 mm. When the applied load on the specimen is 50 kN, the diameter is 12.99265 mm. Determine Poisson’s ratio for the material.

400

Normal Stress: s =

P = A

50(103) p 4

(0.0132)

= 376.70 Mpa 0.002

Normal Strain: From the stress–strain diagram, the modulus of elasticity 400(106) = 200 GPa. Applying Hooke’s law E = 0.002 elong =

elat =

376.70(106) s = 1.8835 A 10 - 3 B mm>mm = E 200(104)

d - d0 12.99265 - 13 = = - 0.56538 A 10 - 3 B mm>mm d0 13

Poisson’s Ratio: The lateral and longitudinal strain can be related using Poisson’s ratio. V = -

- 0.56538(10 - 3) elat = 0.300 = elong 1.8835(10 - 3)

Ans.

20

P(mm/mm)

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s(MPa)

*3–28. The elastic portion of the stress–strain diagram for a steel alloy is shown in the figure. The specimen from which it was obtained had an original diameter of 13 mm and a gauge length of 50 mm. If a load of P = 20 kN is applied to the specimen, determine its diameter and gauge length. Take n = 0.4.

400

Normal Stress: s =

P = A

20(103) p 4

(0.0132)

= 150.68Mpa 0.002

P(mm/mm)

Normal Strain: From the Stress–Strain diagram, the modulus of elasticity 400(106) E = = 200 GPa. Applying Hooke’s Law 0.002 elong =

150.68(106) s = 0.7534 A 10 - 3 B mm>mm = E 200(109)

Thus, dL = elong L0 = 0.7534 A 10 - 3 B (50) = 0.03767 mm L = L0 + dL = 50 + 0.03767 = 50.0377 mm

Ans.

Poisson’s Ratio: The lateral and longitudinal can be related using poisson’s ratio. elat = - velong = - 0.4(0.7534) A 10 - 3 B

= - 0.3014 A 10 - 3 B mm>mm

dd = elat d = - 0.3014 A 10 - 3 B (13) = - 0.003918 mm d = d0 + dd = 13 + ( - 0.003918) = 12.99608 mm

Ans.

•3–29.

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 =

elat =

2 in.

8 kip

8 kip 3 in.

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

elong =

v =

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.

21

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3–30. The block is made of titanium Ti-6A1-4V and is subjected to a compression of 0.06 in. along the y axis, and its shape is given a tilt of u = 89.7°. Determine Px, Py, and gxy.

y

Normal Strain: ey =

dLy Ly

=

4 in. u

-0.06 = - 0.0150 in.>in. 4

Ans.

Poisson’s Ratio: The lateral and longitudinal strain can be related using Poisson’s ratio.

x

5 in.

ex = - vey = - 0.36( -0.0150) = 0.00540 in. >in.

Ans.

Shear Strain: b = 180° - 89.7° = 90.3° = 1.576032 rad gxy =

p p - b = - 1.576032 = - 0.00524 rad 2 2

Ans.

3–31. The shear stress–strain diagram for a steel alloy is shown in the figure. If a bolt having a diameter of 0.75 in. is made of this material and used in the double lap joint, determine the modulus of elasticity E and the force P required to cause the material to yield. Take n = 0.3.

P/2 P/2

P

t(ksi) 60

The shear force developed on the shear planes of the bolt can be determined by considering the equilibrium of the FBD shown in Fig. a + ©F = 0; : x

V + V - P = 0

V = =

P

0.00545

2

From the shear stress–strain diagram, the yield stress is ty = 60 ksi. Thus, ty =

Vy A

;

60 =

P>2

p 4

A 0.752 B

P = 53.01 kip = 53.0 kip

Ans.

From the shear stress–strain diagram, the shear modulus is G =

60 ksi = 11.01(103) ksi 0.00545

Thus, the modulus of elasticity is G =

E ; 2(1 + y)

11.01(103) =

E 2(1 + 0.3)

E = 28.6(103) ksi

Ans.

22

g(rad)

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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.

23

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•3–33.

The support consists of three rigid plates, which are connected together using two symmetrically placed rubber pads. If a vertical force of 5 N is applied to plate A, determine the approximate vertical displacement of this plate due to shear strains in the rubber. Each pad has cross-sectional dimensions of 30 mm and 20 mm. Gr = 0.20 MPa.

C

B

40 mm

40 mm

A

tavg =

g =

V 2.5 = = 4166.7 Pa A (0.03)(0.02)

5N

4166.7 t = 0.02083 rad = G 0.2(106)

d = 40(0.02083) = 0.833 mm

Ans.

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

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.

24

a

a

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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.)

70 s = = 11400.65 ksi e 0.00614

Eal =

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

P = A

p 4

9 = 45.84 ksi (0.5)2

s 45.84 = = 0.0040208 in.>in. E 11400.65

elong =

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

elat =

V = -

Gal =

elat - 0.0013 = 0.32332 = elong 0.0040208

11.4(103) Eat = = 4.31(103) ksi 2(1 + v) 2(1 + 0.32332)

Ans.

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

0.00614

10 = 50.9296 ksi p 2 (0.5) 4

From the stress–strain diagram E =

70 = 11400.65 ksi 0.00614

elong =

G =

s 50.9296 = = 0.0044673 in.>in. E 11400.65

E ; 2(1 + v)

3.8(103) =

11400.65 ; 2(1 + v)

v = 0.500

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

Ans.

25

P (in./in.)

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s(psi)

3–37. 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.

ut =

1 (2)(11) = 11 psi 2

Ans.

3–38. A short cylindrical block of 6061-T6 aluminum, having an original diameter of 20 mm and a length of 75 mm, is placed in a compression machine and squeezed until the axial load applied is 5 kN. Determine (a) the decrease in its length and (b) its new diameter.

a)

s =

- 5(103) P = p = - 15.915 MPa 2 A 4 (0.02)

s = E elong ;

- 15.915(106) = 68.9(109) elong elong = - 0.0002310 mm>mm

d = elong L = - 0.0002310(75) = - 0.0173 mm b)

v = -

elat ; elong

0.35 = -

Ans.

elat - 0.0002310

elat = 0.00008085 mm>mm ¢d = elat d = 0.00008085(20) = 0.0016 mm d¿ = d + ¢d = 20 + 0.0016 = 20.0016 mm

Ans.

26

2 2.25

P(in./in.)

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3–39. 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. a + ©MA = 0;

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

a + ©MB = 0;

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

FB =

80 kN x

A

80x 3 FA =

B

210 mm

220 mm

(1) 80(3 - x) 3

3m

(2)

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

P ; A

d = eL = a

P

e = P A

E

dA = dB ;

s A = E E

bL =

PL AE

80(3 - x) (220) 3

80x 3 (210)

=

AE

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 2 A 4 (0.03 )

elong =

55.27(106) sA = - 0.000756 = E 73.1(109)

elat = - velong = - 0.35( - 0.000756) = 0.0002646 œ dA = dA + d elat = 30 + 30(0.0002646) = 30.008 mm

Ans.

*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. e =

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

Ans.

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

27

Ans.

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•3–41. The stone has a mass of 800 kg and center of gravity at G. It rests on a pad at A and a roller at B. The pad is fixed to the ground and has a compressed height of 30 mm, a width of 140 mm, and a length of 150 mm. If the coefficient of static friction between the pad and the stone is ms = 0.8, determine the approximate horizontal displacement of the stone, caused by the shear strains in the pad, before the stone begins to slip. Assume the normal force at A acts 1.5 m from G as shown. The pad is made from a material having E = 4 MPa and n = 0.35.

0.4 m

B

Equations of Equilibrium: a + ©MB = 0; + ©F = 0; : x

FA(2.75) - 7848(1.25) - P(0.3) = 0

[1]

P - F = 0

[2]

Note: The normal force at A does not act exactly at A. It has to shift due to friction. Friction Equation: F = ms FA = 0.8 FA

[3]

Solving Eqs. [1], [2] and [3] yields: FA = 3908.37 N

F = P = 3126.69 N

Average Shear Stress: The pad is subjected to a shear force of V = F = 3126.69 N. t =

V 3126.69 = = 148.89 kPa A (0.14)(0.15)

Modulus of Rigidity: G =

E 4 = = 1.481 MPa 2(1 + v) 2(1 + 0.35)

Shear Strain: Applying Hooke’s law for shear g =

148.89(103) t = 0.1005 rad = G 1.481(106)

Thus, dh = hg = 30(0.1005) = 3.02 mm

Ans.

28

P

G

1.25 m

0.3 m 1.5 m

A

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3–42. The bar DA is rigid and is originally held in the horizontal position when the weight W is supported from C. If the weight causes B to be displaced downward 0.025 in., determine the strain in wires DE and BC. Also, if the wires are made of A-36 steel and have a cross-sectional area of 0.002 in2, determine the weight W.

E 3 ft 2 ft D

3 ft B

5 3 = 0.025 d

A

4 ft

d = 0.0417 in eDE =

C

0.0417 d = = 0.00116 in.>in. L 3(12)

Ans. W

3

sDE = EeDE = 29(10 )(0.00116) = 33.56 ksi FDE = sDEADE = 33.56 (0.002) = 0.0672 kip a + ©MA = 0;

- (0.0672) (5) + 3(W) = 0

W = 0.112 kip = 112 lb

Ans.

sBC =

W 0.112 = = 55.94 ksi ABC 0.002

eBC =

sBC 55.94 = 0.00193 in.>in. = E 29 (103)

Ans.

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 eb =

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

Ans.

es =

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

Ans.

29

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*3–44. The A-36 steel wire AB has a cross-sectional area of 10 mm2 and is unstretched when u = 45.0°. Determine the applied load P needed to cause u = 44.9°.

A

400 mm

u 400 m

m

B P

¿ LAB 400 = sin 90.2° sin 44.9° ¿ = 566.67 mm LAB

LAB =

e =

400 = 565.69 sin 45°

¿ - LAB LAB 566.67 - 565.69 = = 0.001744 LAB 565.69

s = Ee = 200(109) (0.001744) = 348.76 MPa a + ©MA = 0 P(400 cos 0.2°) - FAB sin 44.9° (400) = 0

(1)

However, FAB = sA = 348.76(106)(10)(10 - 6) = 3.488 kN From Eq. (1), P = 2.46 kN

Ans.

30

Ch01-03 stress & strain & properties

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