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Mechanics of Materials - Instructor Solutions Manual 9th Edition, Exercises of Mechanics of Materials

Solution Manual for Mechanics of Materials Hibbeler 9th Edition

Typology: Exercises

2020/2021

Uploaded on 05/27/2021

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© 2014 Pearson Education, Inc.,Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
Support Reactions: We will only need to compute Cyby writing the moment
equation of equilibrium about Bwith reference to the free-body diagram of the
entire shaft, Fig.a.
a
Internal Loadings: Using the result for Cy, section DE of the shaft will be
considered. Referring to the free-body diagram, Fig. b,
Ans.
Ans.
a
Ans.
The negative signs indicates that V
Eand MEact in the opposite sense to that shown
on the free-body diagram.
ME =-2400 lb #ft =-2.40 kip #ft
1000(4) -800(8) -ME = 0ME = 0;
VE=-200 lbVE+1000 - 800 =0+c©Fy = 0;
NE=0
:
+©Fx=0;
Cy=1000 lbCy(8) +400(4) -800(12) =0MB = 0;
A
ED
BC
4 ft
400 lb
800 lb
4 ft 4 ft 4 ft
Ans:
,,ME =-2.40 kip #ft VE=-200 lbNE=0
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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.

Support Reactions: We will only need to compute C y by writing the moment equation of equilibrium about B with reference to the free-body diagram of the entire shaft, Fig. a.

a

Internal Loadings: Using the result for C y , section DE of the shaft will be considered. Referring to the free-body diagram, Fig. b ,

Ans.

Ans.

a

Ans.

The negative signs indicates that V E and M E act in the opposite sense to that shown on the free-body diagram.

M E = - 2400 lb #^ ft = - 2.40 kip #^ ft

+ © ME = 0;1000(4) - 800(8) - ME = 0

  • c © Fy = 0; VE + 1000 - 800 = 0 V (^) E = -200 lb

:^ + ©F (^) x = 0; N (^) E = 0

  • © MB = 0; C y(8) + 400(4) - 800(12) = 0 Cy = 1000 lb

A B^ E C D

4 ft 400 lb 800 lb

4 ft 4 ft 4 ft

Ans:

NE = 0 , VE = -200 lb, M E = - 2.40 kip #^ ft

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)

Ans.

Ans.

(b)

Ans.

V b = 250 lb Ans.

+Q© Fy = 0; Vb - 500 sin 30° = 0

Nb = 433 lb

R+ © F x = 0; N b - 500 cos 30° = 0

  • Fy = 0; V (^) a = 0

Na = 500 lb

:^ + © Fx = 0; N (^) a - 500 = 0

30 

A

a b

b (^) a

500 lb 500 lb

Ans: , , Nb = 433 lb, V b = 250 lb

N (^) a = 500 lb Va = 0

*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.

Support Reactions: We will only need to compute B y by writing the moment equation of equilibrium about A with reference to the free-body diagram of the entire shaft, Fig. a.

a

Internal Loadings: Using the result of B y , section CD of the shaft will be considered. Referring to the free-body diagram of this part, Fig. b ,

Ans.

Ans.

a

Ans.

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

M C = 433 N #^ m

+ ©M C = 0; 1733.33(2.5) - 600(1)(0.5) - 900(4) - MC = 0

  • c©Fy = 0; VC - 600(1) + 1733.33 - 900 = 0 VC = -233 N

; NC = 0

©F (^) x = 0;

  • ©MA = 0; By(4.5) - 600(2)(2) - 900(6) = 0 By = 1733.33 N

A (^) B D

C

900 N

1.5 m

600 N/m

1 m 1 m 1 m 1.5 m

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.

Support Reactions: For member AB

a

Equations of Equilibrium: For point D

Ans.

Ans.

a

Ans.

Equations of Equilibrium: For point E

Ans.

Ans.

a

Ans.

Negative signs indicate that M (^) E and VE act in the opposite direction to that shown on FBD.

ME = -24.0 kip #^ ft

+ ©ME = 0; ME + 6.00(4) = 0

VE = -9.00 kip

  • c ©Fy = 0; - 6.00 - 3 - VE = 0

:^ + ©Fx = 0; NE = 0

MD = 13.5 kip #^ ft

+ ©MD = 0; MD + 2.25(2) - 3.00(6) = 0

VD = 0.750 kip

  • c ©Fy = 0; 3.00 - 2.25 - VD = 0

:^ +^ ©Fx = 0; ND = 0

  • c ©Fy = 0; By + 3.00 - 9.00 = 0 By = 6.00 kip

:^ + ©Fx = 0; Bx = 0

  • ©MB = 0; 9.00(4) - Ay(12) = 0 Ay = 3.00 kip

6 ft 4 ft

A

4 ft

D B E C

6 ft

3 kip 1.5 kip/ ft

Ans: , ,

NE = 0 , VE = -9.00 kip, ME = -24.0 kip #^ ft

N D = 0 VD = 0.750 kip MD = 13.5 kip #^ ft,

Support Reactions:

a

Ans.

Equations of Equilibrium: For point C

Ans.

Ans.

a

Ans.

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

MC = 0.400 kN #^ m

+ ©MC = 0; 0.5333(0.75) - MC = 0

VC = -0.533 kN

  • c ©Fy = 0; VC + 0.5333 = 0

NC = -2.00 kN

:^ + ©Fx = 0; - NC - 2.00 = 0

  • c ©Fy = 0; A (^) y - 0.5333 = 0 A (^) y = 0.5333 kN

:^ + ©Fx = 0; 2 - A (^) x = 0 A (^) x = 2.00 kN

P = 0.5333 kN = 0.533 kN

+ ©MA = 0; P(2.25) - 2(0.6) = 0

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.

0.75 m

C

P

A

B

0.1 m 0.5 m

0.75 m 0.75 m

Ans: , , ,

MC = 0.400 kN #^ m

P = 0.533 kNNC = -2.00 kN VC = -0.533 kN

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

a

Referring to the FBD of this segment, Fig. b ,

Ans.

Ans.

a + ©MC = 0; MC + 6(0.5) - 7.5(1) = 0 MC = 4.50 kN #^ m Ans.

  • c ©Fy = 0; 7.50 - 6 - VC = 0 VC = 1.50 kN

:^ + ©Fx = 0; NC = 0

  • ©MB = 0; - Ay(4) + 6(3.5) +

(3)(3)(2) = 0 Ay = 7.50 kN

*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.

0.5 m 0.5 m 1.5 m 1.5 m

C

A B

3 kN/m

6 kN

D

Equations of Equilibrium: For point A

Ans.

Ans.

a

Ans.

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

Equations of Equilibrium: For point B

Ans.

Ans.

a

Ans.

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

Equations of Equilibrium: For point C

Ans.

Ans.

a

Ans.

Negative signs indicate that N (^) C and MC act in the opposite direction to that shown on FBD.

MC = -8125 lb #^ ft = -8.125 kip #^ ft

+ ©MC = 0; - MC - 650(6.5) - 300(13) = 0

NC = -1200 lb = -1.20 kip

  • c © Fy = 0; - NC - 250 - 650 - 300 = 0

;^ + © Fx = 0; VC = 0

MB = -6325 lb #^ ft = -6.325 kip #^ ft

+ © MB = 0; - MB - 550(5.5) - 300(11) = 0

VB = 850 lb

  • c © Fy = 0; VB - 550 - 300 = 0

;^ + © Fx = 0; NB = 0

MA = -1125 lb #^ ft = -1.125 kip #^ ft

+ ©MA = 0; - MA - 150(1.5) - 300(3) = 0

VA = 450 lb

  • c © Fy = 0; VA - 150 - 300 = 0

;^ + © Fx = 0; NA = 0

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.

5 ft

7 ft

C

D (^) F

E

B A

300 lb

2 ft 8 ft 3 ft

Ans: , , , ,

VC = 0 , NC = -1.20 kip,MC = -8.125 kip #^ ft

NB = 0 VB = 850 lb MB = -6.325 kip #^ ft,

NA = 0 VA = 450 lb MA = -1.125 kip #^ ft,

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.

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

a

Internal Loadings: Using the results of C x and C y , section CE of the forearm will be considered. Referring to the free-body diagram of this part shown in Fig. b ,

Ans.

Ans.

a Ans.

The negative signs indicate that N E , V E and M E act in the opposite sense to that shown on the free-body diagram.

+ ©ME = 0; ME + 64.47(0.035) = 0 ME = - 2.26 N #^ m

  • c©Fy = 0; -VE - 64.47 = 0 VE = - 64.5 N

: NE + 22.53 = 0 N E = -22.5 N

©Fx = 0;

  • c ©Fy = 0; 87.05 sin 75° - 2(9.81) - Cy = 0 C (^) y = 64.47 N

: C (^) x - 87.05 cos 75° = 0 Cx = 22.53 N

©Fx = 0;

  • ©MC = 0; FBD sin 75°(0.07) - 2(9.81)(0.3) = 0 FBD = 87.05 N

75 

230 mm 35 mm35 mm

C (^) E B

D

A

Ans:

NE = -22.5 N , VE = - 64.5 N, M E = - 2.26 N #^ m

Internal Loadings: Referring to the free-body diagram of the section of the hacksaw shown in Fig. a ,

Ans.

Ans.

a Ans.

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

+ ©MD = 0; - M a - a - 100(0.15) = 0 Ma - a = -15 N #^ m

  • c ©Fy = 0; Va - a = 0

; Na - a + 100 = 0 N (^) a - a = -100 N

©Fx = 0;

1–13. The blade of the hacksaw is subjected to a pretension force of Determine the resultant internal loadings acting on section a–a that passes through point D.

F = 100 N.

A B

C

D

F F

a

b

b a

30 

225 mm

150 mm

Ans:

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

1–14. The blade of the hacksaw is subjected to a pretension force of. Determine the resultant internal loadings acting on section b–b that passes through point D.

F = 100 N

Internal Loadings: Referring to the free-body diagram of the section of the hacksaw shown in Fig. a ,

Ans.

Ans.

a Ans.

The negative sign indicates that N b–b and M b–b act in the opposite sense to that shown on the free-body diagram.

+ ©M D = 0; - Mb - b - 100(0.15) = 0 Mb - b = -15 N #^ m

©Fy¿ = 0; Vb - b - 100 sin 30° = 0 Vb - b = 50 N

©Fx¿ = 0; Nb - b + 100 cos 30° = 0 Nb - b = - 86.6 N

A B

C

D

F F

a

b

b a

30 

225 mm

150 mm

Ans: , ,

Mb - b = -15 N #^ m

N (^) b - b = - 86.6 N Vb - b = 50 N

*1–16. A 150-lb bucket is suspended from a cable on the wooden frame. Determine the resultant internal loadings acting on the cross section at E.

Support Reactions: We will only need to compute A x , A y , and F BI. Referring to the free-body diagram of the frame, Fig. a ,

a

Internal Loadings: Using the results of A x and A y , section AE of member AB will be considered. Referring to the free-body diagram of this part shown in Fig. b ,

Ans.

Ans.

a Ans.

The negative sign indicates that N E acts in the opposite sense to that shown on the free-body diagram.

+ ©MD = 0; 100(3) - ME = 0 ME = 300 lb # ft

  • c ©Fy = 0; 100 - VE = 0 VE = 100 lb

: NE + 323.21 = 0 NE = - 323 lb

©Fx = 0;

  • c ©Fy = 0; A (^) y - 200 cos 30° - 150 = 0 A (^) y = 323.21 lb

:^ + ©Fx = 0; A^ x -^ 200 sin 30°^ =^0 A^ x =^ 100 lb

  • ©MA = 0; F (^) BI sin 30°(6) - 150(4) = 0 FBI = 200 lb

2 ft

2 ft

3 ft

1 ft 1 ft

1 ft E

D (^) C

B

I A

30  G

H

45 

1.5 m

1.5 m

3 m

45 

A

C

B

b a

a

b

5 kN

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

a

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

Ans.

Ans.

a Ans.

Referring to the FBD (section b–b ) in Fig. c ,

Ans.

Ans.

a

Mb - b = 3.75 kN #^ m Ans.

  • ©MC = 0; 5.303 sin 45° (3) - 5(1.5) - Mb - b = 0

  • c ©Fy = 0; Vb - b - 5 sin 45° = 0 Vb - b = 3.536 kN = 3.54 kN

= -1.77 kN

;^ + ©Fx = 0; Nb - b - 5 cos 45° + 5.303 = 0 Nb - b = -1.768 kN

+ ©MC = 0; 5.303 sin 45°(3) - 5(1.5) - Ma - a = 0 Ma - a = 3.75 kN #^ m

+a ©Fy¿ = 0; Va - a + 5.303 sin 45° - 5 = 0 Va - a = 1.25 kN

+b©Fx¿ = 0; Na - a + 5.303 cos 45° = 0 Na - a = -3.75 kN

  • ©MA = 0; NB sin 45°(6) - 5(4.5) = 0 NB = 5.303 kN

1–17. Determine resultant internal loadings acting on section a–a and section b–b. Each section passes through the centerline at point C.

Ans:

V b - b = 3.54 kN #^ m, Mb - b = 3.75 kN #^ m

Ma - a = 3.75 kN #^ m, Nb - b = -1.77 kN,

Na - a = -3.75 kN, V (^) a - a = 1.25 kN,

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

a

Referring to the FBD of this segment, Fig. b ,

Ans.

Ans.

a

MC = 31.5 kip #^ ft Ans.

+ ©MC = 0; MC + (3)(3)(1.5) +

  • c ©Fy = 0; 18.0 -

(3)(3) - (3)(3) - VC = 0 VC = 4.50 kip

:^ + ©Fx = 0; NC = 0

+ ©MB = 0;

(6)(6)(10) - A (^) y(12) = 0 Ay = 18.0 kip

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.

3 ft 3 ft

C D A B

6 ft

6 kip/ft 6 kip/ft

Ans:

NC = 0 , V C = 4.50 kip, MC = 31.5 kip #^ ft

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

a

Referring to the FBD of this segment, Fig. b ,

Ans.

Ans.

a + ©MA = 0; MD - 18.0 (2) = 0 MD = 36.0 kip #^ ft Ans.

  • c ©Fy = 0; 18.0 -

(6)(6) - VD = 0 VD = 0

:^ + ©Fx = 0; ND = 0

+ ©MB = 0;

(6)(6)(10) - A (^) y(12) = 0 Ay = 18.0 kip

*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.

3 ft 3 ft

C D A B

6 ft

6 kip/ft 6 kip/ft