Solution Manual for Mechanics of Materials 9th Edition by Hibbeler PDF

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Summary

We have assumed so far that, in an axially loaded member, the nor- mal stresses are uniformly distributed in any section perpendicular to the axis of the member. As we saw in Sec. 1.5, such an assumption may be quite in error in the immediate vicinity of the points of application of the loads. Howev...

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full file at http://testbankinstant.com 1–1. The shaft is supported by a smooth thrust bearing at B and a journal bearing at C. Determine the resultant internal loadings acting on the cross section at E.

B

A

4 ft

C

E

4 ft

4 ft

D

4 ft

400 lb 800 lb

Support Reactions: We will only need to compute Cy by writing the moment equation of equilibrium about B with reference to the free-body diagram of the entire shaft, Fig. a. a + ©MB = 0;

Cy(8) + 400(4) - 800(12) = 0

Cy = 1000 lb

Internal Loadings: Using the result for Cy, section DE of the shaft will be considered. Referring to the free-body diagram, Fig. b, + : ©Fx = 0;

NE = 0

+ c ©Fy = 0;

VE + 1000 - 800 = 0

Ans. VE = -200 lb

Ans.

a + ©ME = 0; 1000(4) - 800(8) - ME = 0 ME = - 2400 lb # ft = - 2.40 kip # ft Ans. The negative signs indicates that VE and ME act in the opposite sense to that shown on the free-body diagram.

Ans: NE = 0 , VE = -200 lb , ME = - 2.40 kip # ft

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full file at http://testbankinstant com 1–2. Determine the resultant internal normal and shear force in the member at (a) section a–a and (b) section b–b, each of which passes through point A. The 500-lb load is applied along the centroidal axis of the member.

a

b

30⬚

500 lb

500 lb

A b

a

(a) + : ©Fx = 0;

Na - 500 = 0 Na = 500 lb

Ans.

+ T©Fy = 0;

Va = 0

Ans.

R+ ©Fx = 0;

Nb - 500 cos 30° = 0

(b)

Nb = 433 lb +Q©F = 0; y

Ans.

Vb - 500 sin 30° = 0 Vb = 250 lb

Ans.

Ans: Na = 500 lb, Va = 0, Nb = 433 lb, Vb = 250 lb

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full file at http://testbankinstant com 1–3. The beam AB is fixed to the wall and has a uniform weight of 80 lb>ft. If the trolley supports a load of 1500 lb, determine the resultant internal loadings acting on the cross sections through points C and D.

5 ft

20 ft

10 ft

3 ft

A

B C

D

1500 lb

Segment BC: + ; ©Fx = 0;

NC = 0

+ c ©Fy = 0;

VC - 2.0 - 1.5 = 0

Ans.

VC = 3.50 kip a + ©MC = 0;

Ans.

-MC - 2(12.5) - 1.5 (15) = 0 MC = -47.5 kip # ft

Ans.

Segment BD: + ; ©Fx = 0;

ND = 0

Ans.

+ c ©Fy = 0;

VD - 0.24 = 0 VD = 0.240 kip

a + ©MD = 0;

Ans.

-MD - 0.24 (1.5) = 0 MD = -0.360 kip # ft

Ans.

Ans: NC = 0, VC = 3.50 kip , MC = -47.5 kip # ft,

ND = 0, VD = 0.240 kip, MD = -0.360 kip # ft

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full file at http://testbankinstant com *1–4. The shaft is supported by a smooth thrust bearing at A and a smooth journal bearing at B. Determine the resultant internal loadings acting on the cross section at C.

600 N/m

A

B

D

C

1m

1m

1m

1.5 m

1.5 m

900 N

Support Reactions: We will only need to compute By by writing the moment equation of equilibrium about A with reference to the free-body diagram of the entire shaft, Fig. a. a + ©MA = 0;

By(4.5) - 600(2)(2) - 900(6) = 0

By = 1733.33 N

Internal Loadings: Using the result of By, section CD of the shaft will be considered. Referring to the free-body diagram of this part, Fig. b, + Ans. ; ©Fx = 0; NC = 0 + c ©Fy = 0;

VC - 600(1) + 1733.33 - 900 = 0

a + ©MC = 0;

1733.33(2.5) - 600(1)(0.5) - 900(4) - MC = 0

VC = -233 N MC = 433 N # m

Ans.

Ans.

The negative sign indicates that VC act in the opposite sense to that shown on the free-body diagram.

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full file at http://testbankinstant com 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

B 6 ft

C

E 4 ft

4 ft

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

9.00(4) - Ay(12) = 0

Ay = 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.

Ans: ND = 0, VD = 0.750 kip , MD = 13.5 kip # ft, NE = 0 , VE = -9.00 kip , ME = -24.0 kip # ft

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full file at http://testbankinstant com 1–6. Determine the normal force, shear force, and moment at a section through point C. Take P = 8 kN.

B

0.1 m

0.5 m A

C 0.75 m

0.75 m

0.75 m

P

Support Reactions: a + ©MA = 0;

8(2.25) - T(0.6) = 0

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

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.

Ans: NC = -30.0 kN, VC = -8.00 kN, MC = 6.00 kN # m

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full file at http://testbankinstant com 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

0.75 m

A 0.75 m

P

Support Reactions: a + ©MA = 0;

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

Ans: P = 0.533 kN , NC = -2.00 kN , VC = -0.533 kN , MC = 0.400 kN # m

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full file at http://testbankinstant com *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

a + ©MB = 0;

-Ay(4) + 6(3.5) +

B

A

Referring to the FBD of the entire beam, Fig. a, 1 (3)(3)(2) = 0 2

Ay = 7.50 kN

D

C 0.5 m 0.5 m

1.5 m

1.5 m

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

NC = 0 7.50 - 6 - VC = 0

Ans. VC = 1.50 kN

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

Ans.

MC = 4.50 kN # m

Ans.

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full file at http://testbankinstant com 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

B

A

C 0.5 m 0.5 m

D 1.5 m

1.5 m

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

By(4) - 6(0.5) -

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

By = 3.00 kN

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

ND = 0

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

+ c ©F y = 0;

VD -

a + ©MD = 0;

3.00(1.5) -

Ans.

VD = -1.875 kN

1 (1.5)(1.5)(0.5) - MD = 0 2

Ans.

MD = 3.9375 kN # m = 3.94 kN # m

Ans.

Ans: ND = 0, VD = -1.875 kN, MD = 3.94 kN # m

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full file at http://testbankinstant com 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

B

2 ft

F

A

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; + c © Fy = 0;

NB = 0

Ans.

VB - 550 - 300 = 0 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.

Ans:

NA = 0 ,VA = 450 lb , MA = -1.125 kip # ft, NB = 0, VB = 850 lb, MB = -6.325 kip # ft, VC = 0, NC = -1.20 kip, MC = -8.125 kip # ft

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full file at http://testbankinstant com 1–11. The forearm and biceps support the 2-kg load at A.If C can be assumed as a pin support, determine the resultant internal loadings acting on the cross section of the bone of the forearm at E.The biceps pulls on the bone along BD.

D

75⬚

A C E B

Support Reactions: In this case, all the support reactions will be completed. Referring to the free-body diagram of the forearm, Fig. a, a + ©MC = 0;

FBD sin 75°(0.07) - 2(9.81)(0.3) = 0

+ : ©Fx = 0;

Cx - 87.05 cos 75° = 0

Cx = 22.53 N

+ c ©Fy = 0;

87.05 sin 75° - 2(9.81) - Cy = 0

Cy = 64.47 N

230 mm 35 mm 35 mm

FBD = 87.05 N

Internal Loadings: Using the results of Cx and Cy, section CE of the forearm will be considered. Referring to the free-body diagram of this part shown in Fig. b, + : ©Fx = 0;

NE + 22.53 = 0

NE = -22.5 N

Ans.

+ c ©Fy = 0;

-VE - 64.47 = 0

VE = - 64.5 N

Ans.

a + ©ME = 0;

ME + 64.47(0.035) = 0

ME = - 2.26 N # m

Ans.

The negative signs indicate that NE, VE and ME act in the opposite sense to that shown on the free-body diagram.

Ans: NE = -22.5 N , VE = - 64.5 N, ME = - 2.26 N # m

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full file at http://testbankinstant com *1–12. The serving tray T used on an airplane is supported on each side by an arm.The tray is pin connected to the arm at A, and at B there is a smooth pin. (The pin can move within the slot in the arms to permit folding the tray against the front passenger seat when not in use.) Determine the resultant internal loadings acting on the cross section of the arm through point C when the tray arm supports the loads shown.

12 N

9N 15 mm B 60⬚

100 mm A

150 mm

T

500 mm

VC

C

MC NC

b+ ©Fx = 0;

NC + 9 cos 30° + 12 cos 30° = 0;

NC = - 18.2 N

Ans.

a+ ©Fy = 0;

VC - 9 sin 30° - 12 sin 30° = 0;

VC = 10.5 N

Ans.

a + ©MC = 0; - MC - 9(0.5 cos 60° + 0.115) - 12(0.5 cos 60° + 0.265) = 0 MC = - 9.46 N

#

m

Ans.

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full file at http://testbankinstant com 1–13. The blade of the hacksaw is subjected to a pretension force of F = 100 N. Determine the resultant internal loadings acting on section a–a that passes through point D.

a

225 mm 30⬚ b B

A D

150 mm

b F

a

F C

Internal Loadings: Referring to the free-body diagram of the section of the hacksaw shown in Fig. a, + ; ©Fx = 0;

Na - a + 100 = 0

+ c ©Fy = 0;

Va - a = 0

a + ©MD = 0;

- Ma - a - 100(0.15) = 0

Na - a = -100 N

Ans. Ans.

Ma - a = -15 N

#

m

Ans.

The negative sign indicates that Na–a and Ma–a act in the opposite sense to that shown on the free-body diagram.

Ans: Na - a = -100 N, Va - a = 0, Ma - a = -15 N # m

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full file at http://testbankinstant com 1–14. The blade of the hacksaw is subjected to a pretension force of F = 100 N. Determine the resultant internal loadings acting on section b–b that passes through point D.

a

225 mm 30⬚ b B

A D

150 mm

b F

a

F C

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

Nb - b + 100 cos 30° = 0

Nb - b = -86.6 N

Ans.

©Fy¿ = 0;

Vb - b - 100 sin 30° = 0

Vb - b = 50 N

Ans.

a + ©MD = 0;

- Mb - b - 100(0.15) = 0

Mb - b...