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Solution Manual for Mechanics of Materials Hibbeler 9th Edition
Typology: Exercises
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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.
:^ + ©F (^) x = 0; N (^) E = 0
A B^ E C D
4 ft 400 lb 800 lb
4 ft 4 ft 4 ft
Ans:
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
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.
©F (^) x = 0;
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.
VE = -9.00 kip
:^ + ©Fx = 0; NE = 0
VD = 0.750 kip
:^ +^ ©Fx = 0; ND = 0
:^ + ©Fx = 0; Bx = 0
6 ft 4 ft
A
4 ft
D B E C
6 ft
3 kip 1.5 kip/ ft
Ans: , ,
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.
VC = -0.533 kN
NC = -2.00 kN
:^ + ©Fx = 0; - NC - 2.00 = 0
:^ + ©Fx = 0; 2 - A (^) x = 0 A (^) x = 2.00 kN
P = 0.5333 kN = 0.533 kN
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: , , ,
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.
:^ + ©Fx = 0; NC = 0
(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.
NC = -1200 lb = -1.20 kip
;^ + © Fx = 0; VC = 0
VB = 850 lb
;^ + © Fx = 0; NB = 0
VA = 450 lb
;^ + © 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: , , , ,
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.
©Fx = 0;
: C (^) x - 87.05 cos 75° = 0 Cx = 22.53 N
©Fx = 0;
75
230 mm 35 mm35 mm
C (^) E B
D
A
Ans:
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.
; 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.
A B
C
D
F F
a
b
b a
30
225 mm
150 mm
Ans:
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.
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.
©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: , ,
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.
: NE + 323.21 = 0 NE = - 323 lb
©Fx = 0;
:^ + ©Fx = 0; A^ x -^ 200 sin 30°^ =^0 A^ x =^ 100 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
©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
+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
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:
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
(3)(3) - (3)(3) - VC = 0 VC = 4.50 kip
:^ + ©Fx = 0; NC = 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:
Referring to the FBD of the entire beam, Fig. a ,
a
Referring to the FBD of this segment, Fig. b ,
Ans.
Ans.
:^ + ©Fx = 0; ND = 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