Solution Manual for Mechanics of Materials SI, 9th Edition

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1
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 Cy 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 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 VE and ME act 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 = 0+ ©ME = 0;
VE = -200 lbVE + 1000 - 800 = 0+ c ©Fy = 0;
NE = 0:
+ ©F x = 0;
Cy = 1000 lbCy(8) + 400(4) - 800(12) = 0+ ©MB = 0;
A E DB C
4 ft
400 lb
800 lb
4 ft 4 ft 4 ft
Ans:
, , ME = - 2.40 kip # ftVE = -200 lbNE = 0

2
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.
Ans.Vb = 250 lb
Vb - 500 sin 30° = 0+Q©Fy = 0;
Nb = 433 lb
Nb - 500 cos 30° = 0R+ ©Fx = 0;
Va = 0+ T©Fy = 0;
Na = 500 lb
Na - 500 = 0:
+ ©Fx = 0;
30
A
ba
b a
500 lb500 lb
Ans:
, ,
Vb = 250 lbNb = 433 lb,
Va = 0Na = 500 lb

3
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.
>
Segment BC:
Ans.
Ans.
a
Ans.
Segment BD:
Ans.
Ans.
a
Ans.MD = -0.360 kip # ft
-M D - 0.24 (1.5) = 0+ ©M D = 0;
VD = 0.240 kip
VD - 0.24 = 0+ c ©Fy = 0;
ND = 0;
+ ©Fx = 0;
M C = -47.5 kip # ft
-MC - 2(12.5) - 1.5 (15) = 0+ ©MC = 0;
VC = 3.50 kip
VC - 2.0 - 1.5 = 0+ c©Fy = 0;
N C = 0;
+ ©Fx = 0;
D
5 ft 20 ft 3 ft
10 ft
C
BA
1500 lb
Ans:
, ,
, , M D = -0.360 kip # ftVD = 0.240 kipND = 0
MC = -47.5 kip # ft,VC = 3.50 kipNC = 0

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4
*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 By 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 By, 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 VC act in the opposite sense to that shown on the
free-body diagram.
MC = 433 N # m
1733.33(2.5) - 600(1)(0.5) - 900(4) - MC = 0+ ©M C = 0;
VC = -233 NVC - 600(1) + 1733.33 - 900 = 0+ c©Fy = 0;
NC = 0;
+ ©F x = 0;
By = 1733.33 NBy(4.5) - 600(2)(2) - 900(6) = 0+ ©MA = 0;
A DB
C
900 N
1.5 m
600 N/m
1.5 m1 m1 m1 m

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5
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 ME 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
B CD E
6 ft
3 kip
1.5 kip/ ft
Ans:
, ,
, , ME = -24.0 kip # ftVE = -9.00 kipNE = 0
MD = 13.5 kip # ft,VD = 0.750 kipN D = 0

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6
1–6. Determine the normal force, shear force, and
moment at a section through point C. Take P = 8 kN.
Support Reactions:
a
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 = 6.00 kN # m
+ ©MC = 0; 8.00(0.75) - MC = 0
VC = -8.00 kN
+ c ©Fy = 0; VC + 8.00 = 0
NC = -30.0 kN
:
+ ©Fx = 0; -NC - 30.0 = 0
+ c ©Fy = 0; A y - 8 = 0 A y = 8.00 kN
:
+ ©Fx = 0; 30.0 - A x = 0 A x = 30.0 kN
+ ©MA = 0; 8(2.25) - T(0.6) = 0 T = 30.0 kN
0.75 m
C
P
A
B
0.5 m
0.1 m
0.75 m 0.75 m
Ans:
MC = 6.00 kN # m
V C = -8.00 kN,NC = -30.0 kN,

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7
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
0.5 m
0.1 m
0.75 m 0.75 m
Ans:
, , ,
MC = 0.400 kN # m
VC = -0.533 kNNC = -2.00 kNP = 0.533 kN

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8
Referring to the FBD of the entire beam, Fig. a,
a
Referring to the FBD of this segment, Fig. b,
Ans.
Ans.
a Ans.+ ©MC = 0; MC + 6(0.5) - 7.5(1) = 0 MC = 4.50 kN # m
+ c ©Fy = 0; 7.50 - 6 - VC = 0 VC = 1.50 kN
:
+ ©Fx = 0; NC = 0
+ ©MB = 0; -Ay(4) + 6(3.5) + 1
2 (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 m1.5 m
C
A B
3 kN/m
6 kN
D

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9
Referring to the FBD of the entire beam, Fig. a,
a
Referring to the FBD of this segment, Fig. b,
Ans.
Ans.
a
Ans.= 3.94 kN # m
+ ©MD = 0; 3.00(1.5) - 1
2 (1.5)(1.5)(0.5) - MD = 0 MD = 3.9375 kN # m
+ c ©Fy = 0; VD - 1
2 (1.5)(1.5) + 3.00 = 0 VD = -1.875 kN
:
+ ©Fx = 0; ND = 0
+ ©MA = 0; By(4) - 6(0.5) - 1
2 (3)(3)(2) = 0 By = 3.00 kN
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.
Ans:
MD = 3.94 kN # m
ND = 0, VD = -1.875 kN,
0.5 m 0.5 m 1.5 m1.5 m
C
A B
3 kN/m
6 kN
D

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10
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 NC 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:
, ,
, ,
, MC = -8.125 kip # ftNC = -1.20 kip,VC = 0
MB = -6.325 kip # ft,VB = 850 lbNB = 0
MA = -1.125 kip # ft,VA = 450 lbNA = 0

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11
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 Cx and Cy, 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 NE, VE and ME act in the opposite sense to that
shown on the free-body diagram.
ME = - 2.26 N # mME + 64.47(0.035) = 0+ ©ME = 0;
VE = - 64.5 N-VE - 64.47 = 0+ c©Fy = 0;
N E = -22.5 NNE + 22.53 = 0:
+ ©Fx = 0;
Cy = 64.47 N87.05 sin 75° - 2(9.81) - Cy = 0+ c ©Fy = 0;
Cx = 22.53 NCx - 87.05 cos 75° = 0:
+ ©Fx = 0;
FBD = 87.05 NFBD sin 75°(0.07) - 2(9.81)(0.3) = 0+ ©MC = 0;
75
230 mm
35 mm35 mm
C E B
D
A
Ans:
, , ME = - 2.26 N # mVE = - 64.5 NNE = -22.5 N

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12
*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.
Ans.
Ans.
a
Ans.MC = - 9.46 N # m
+ ©MC = 0; - MC - 9(0.5 cos 60° + 0.115) - 12(0.5 cos 60° + 0.265) = 0
VC = 10.5 NVC - 9 sin 30° - 12 sin 30° = 0;a+ ©Fy = 0;
NC = - 18.2 NNC + 9 cos 30° + 12 cos 30° = 0;b+ ©Fx = 0;
9 N
500 mm
12 N
15 mm 150 mm
60
AB
C
T
VC
MC
NC
100 mm

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13
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 Na–a and Ma–a act in the opposite sense to that
shown on the free-body diagram.
Ma - a = -15 N # m- M a - a - 100(0.15) = 0+ ©MD = 0;
Va - a = 0+ c ©Fy = 0;
N a - a = -100 NNa - a + 100 = 0;
+ ©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:
, , Ma - a = -15 N # mVa - a = 0Na - a = -100 N

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14
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 Nb–b and Mb–b act in the opposite sense to that
shown on the free-body diagram.
Mb - b = -15 N # m- Mb - b - 100(0.15) = 0+ ©MD = 0;
Vb - b = 50 N©Fy¿ = 0; Vb - b - 100 sin 30° = 0
Nb - b = - 86.6 N©Fx¿ = 0; Nb - b + 100 cos 30° = 0
A B
C
D
F F
a
b
b a
30
225 mm
150 mm
Ans:
, ,
Mb - b = -15 N # m
Vb - b = 50 NN b - b = - 86.6 N

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15
1–15. A 150-lb bucket is suspended from a cable on the
wooden frame. Determine the resultant internal loadings
on the cross section at D.
Support Reactions: We will only need to compute Bx, By, and FGH . Referring to the
free-body diagram of member BC, Fig. a,
a
Internal Loadings: Using the results of Bx and By , section BD of member BC will be
considered. Referring to the free-body diagram of this part shown in Fig. b,
Ans.
Ans.
a Ans.
The negative signs indicates that VD and MD act in the opposite sense to that shown
on the free-body diagram.
MD = -150 lb # ft150(1) + MD = 0+ ©MD = 0;
VD = - 150 lb- VD - 150 = 0+ c ©Fy = 0;
ND = 300 lbND - 300 = 0:
+ ©Fx = 0;
By = 150 lb424.26 sin 45° - 150 - By = 0+ c©Fy = 0;
B x = 300 lb424.26 cos 45° - B x = 0:
+ ©Fx = 0;
F GH = 424.26 lbFGH sin 45°(2) - 150(4) = 0+ ©MB = 0:
2 ft
2 ft
3 ft
1 ft1 ft
1 ft
E
D C
B
AI
30 G
H
Ans:
, , MD = -150 lb # ftVD = - 150 lbND = 300 lb

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16
*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 Ax, Ay, and FBI. Referring to the
free-body diagram of the frame, Fig. a,
a
Internal Loadings: Using the results of Ax and Ay, 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 NE acts in the opposite sense to that shown on the
free-body diagram.
ME = 300 lb # ft100(3) - ME = 0+ ©MD = 0;
VE = 100 lb100 - VE = 0+ c ©Fy = 0;
NE = - 323 lbNE + 323.21 = 0:
+ ©Fx = 0;
A y = 323.21 lbA y - 200 cos 30° - 150 = 0+ c ©Fy = 0;
A x = 100 lbAx - 200 sin 30° = 0:
+ ©Fx = 0;
FBI = 200 lbF BI sin 30°(6) - 150(4) = 0+ ©MA = 0;
2 ft
2 ft
3 ft
1 ft1 ft
1 ft
E
D C
B
AI
30 G
H

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17
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
Ans.Mb - b = 3.75 kN # m
+ ©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; N B 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:
Mb - b = 3.75 kN # mV b - b = 3.54 kN # m,
Nb - b = -1.77 kN,Ma - a = 3.75 kN # m,
V a - a = 1.25 kN,Na - a = -3.75 kN,

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19
Referring to the FBD of the entire beam, Fig. a,
a
Referring to the FBD of this segment, Fig. b,
Ans.
Ans.
a
Ans.MC = 31.5 kip # ft
+ ©MC = 0; MC + (3)(3)(1.5) + 1
2 (3)(3)(2) - 18.0(3) = 0
+ c ©Fy = 0; 18.0 - 1
2 (3)(3) - (3)(3) - VC = 0 VC = 4.50 kip
:
+ ©Fx = 0; NC = 0
+ ©MB = 0; 1
2 (6)(6)(2) + 1
2 (6)(6)(10) - Ay(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
DCA B
6 ft
6 kip/ft6 kip/ft
Ans:
, , MC = 31.5 kip # ftV C = 4.50 kipN C = 0

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20
Referring to the FBD of the entire beam, Fig. a,
a
Referring to the FBD of this segment, Fig. b,
Ans.
Ans.
a Ans.+ ©MA = 0; MD - 18.0 (2) = 0 MD = 36.0 kip # ft
+ c ©Fy = 0; 18.0 - 1
2 (6)(6) - VD = 0 VD = 0
:
+ ©Fx = 0; ND = 0
+ ©MB = 0; 1
2 (6)(6)(2) + 1
2 (6)(6)(10) - Ay(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
DCA B
6 ft
6 kip/ft6 kip/ft

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21
Internal Loadings: Referring to the free-body diagram of the section of the clamp
shown in Fig. a,
Ans.
Ans.
a Ans.+ ©MA = 0; 900(0.2) - Ma - a = 0 Ma - a = 180 N # m
©Fx¿ = 0; Va - a - 900 sin 30° = 0 Va - a = 450 N
©Fy¿ = 0; 900 cos 30° - Na - a = 0 Na - a = 779 N
1–21. The forged steel clamp exerts a force of N
on the wooden block. Determine the resultant internal
loadings acting on section a–a passing through point A.
F = 900 200 mm
a
a F 900 N
F 900 N
30 A
Ans:
Ma - a = 180 N # m:V a - a = 450 N,Na - a = 779 N,

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22
1–22. The metal stud punch is subjected to a force of 120 N
on the handle. Determine the magnitude of the reactive force
at the pin A and in the short link BC. Also, determine the
internal resultant loadings acting on the cross section passing
through the handle arm at D.
Member:
a
Ans.
Ans.
Segment:
Ans.
Ans.
a
Ans.MD = 36.0 N # m
MD - 120(0.3) = 0+ ©MD = 0;
V D = 0+Q©F y¿ = 0;
ND = 120 N
ND - 120 = 0a+ ©Fx¿ = 0;
= 1.49 kN= 1491 N
F A = 21489.562 + 602
Ax = 60 NA x - 120 sin 30° = 0;;
+ ©Fx = 0;
A y = 1489.56 N
Ay - 1385.6 - 120 cos 30° = 0+ c ©Fy = 0;
FBC = 1385.6 N = 1.39 kN
FBC cos 30°(50) - 120(500) = 0+©MA = 0;
60 50 mm
100 mm
200 mm
300 mm
B
C
D
120 N
50 mm 100 mm
E
30
A
Ans:
, ,
MD = 36.0 N # mVD = 0,
ND = 120 N,FA = 1.49 kNFBC = 1.39 kN

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23
1–23. Solve Prob. 1–22 for the resultant internal loadings
acting on the cross section passing through the handle arm
at E and at a cross section of the short link BC.
Member:
a
Segment:
Ans.
Ans.
a Ans.
Short link:
Ans.
Ans.
a Ans.M = 0+ ©MH = 0;
N = 1.39 kN1.3856 - N = 0;+ c ©Fy = 0;
V = 0;
+ ©Fx = 0;
ME = 48.0 N # mM E - 120(0.4) = 0;+ ©ME = 0;
VE = 120 NVE - 120 = 0;a+ ©Fy¿ = 0;
NE = 0+b©Fx¿ = 0;
FBC = 1385.6 N = 1.3856 kN
FBC cos 30°(50) - 120(500) = 0+ ©MA = 0;
60 50 mm
100 mm
200 mm
300 mm
B
C
D
120 N
50 mm 100 mm
E
30
A
Ans:
, ,
Short link: M = 0N = 1.39 kN,V = 0,
ME = 48.0 N # m,V E = 120 NN E = 0

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24
*1–24. Determine the resultant internal loadings acting
on the cross section of the semicircular arch at C.
a
By
Ans.
Ans.
a
Ans.MC = 0
w0 r(r) - MC + (-w0 r)(r) = 0+ ©M0 = 0;
V2 = 0w0 r + VC - w0 r(-cos u) L
p
2
0
= 0;
w0 r + VC - w0 r L
p
2
0
sin u du = 0+ c ©Fy = 0;
NC = -w0 r sin u L
p
2
0
= -w0 r
-NC - w0 r L
p
2
0
cos u du = 0:
+ ©Fx = 0;
= w0 r
By (2r) - w0 r2 (-cos u)]0
p = 0
By (2r) - w0 r2
L
p
0
sin u du = 0
- L
p
0
(w0 r du)(sin u)r(1 - cos u) = 0
+ ©MA = 0; By (2r) - L
p
0
(w0 r du)(cos u)(r sin u)
C
A B
w0
u
r

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25
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.©Mz = 0; (TB)z - 105(0.5) = 0; (TB)z = 52.5 lb # ft
©My = 0; (MB)y - 105(7.5) = 0; (MB)y = 788 lb # ft
©Mx = 0; (MB)x = 0
©Fz = 0; (NB)z = 0
©Fy = 0; (VB)y = 0
©Fx = 0; (VB)x - 105 = 0; (VB)x = 105 lb
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 > acts
perpendicular to the face of the sign.
ft2
lb
4 ft
z
y
6 ft
x
B
A
3 ft
2 ft
3 ft
7 lb/ft2
Ans:
, ,
, ,
(TB)z = 52.5 lb # ft
(MB)y = 788 lb # ft(MB)x = 0
(NB)z = 0(VB)y = 0(VB)x = 105 lb,

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26
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.
+
-
y
B
C
400 mm
150 mm
200 mm
250 mm
A
x
z
300 N 300 N
500 N
500 N
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.©Mz = 0; (MC)z - 1000(0.2) + 750(0.45) = 0; (MC)z = -138 N # m
©My = 0; (TC)y = 0
©Mx = 0; (MC)x + 240(0.45) = 0; (MC)x = -108 N # m
©Fz = 0; (VC)z + 240 = 0; (VC)z = -240 N
©Fy = 0; (NC)y = 0
©Fx = 0; (VC)x + 1000 - 750 = 0; (VC)x = -250 N
Ans:
, , ,
,
(MC)z = -138 N # m
(TC)y = 0(MC)x = -108 N # m,
(VC)z = -240 N(N C)y = 0(VC)x = -250 N

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27
1–27. The pipe assembly is subjected to a force of 600 N
at B. Determine the resultant internal loadings acting on
the cross section at C.
Internal Loading: Referring to the free-body diagram of the section of the pipe
shown in Fig. a,
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
The negative signs indicate that and act in the opposite
sense to that shown on the free-body diagram.
(MC)z(VC)y, (VC)z, (TC)x,
(MC)z = -99.0 N # m
(MC)z + 600 cos 60° cos 30°(0.15) + 600 cos 60° sin 30°(0.4) = 0©M z = 0;
(MC)y = 153 N # m
(MC)y - 600 sin 60° (0.15) - 600 cos 60° sin 30°(0.5) = 0©My = 0;
(TC)x = -77.9 N # m
(TC)x + 600 sin 60°(0.4) - 600 cos 60° cos 30°(0.5) = 0©Mx = 0;
(VC)z = -520 N(VC)z + 600 sin 60° = 0©Fz = 0;
(V C)y = -260 N(VC)y + 600 cos 60° cos 30° = 0©Fy = 0;
(NC)x = 150 N(N C)x - 600 cos 60° sin 30° = 0©Fx = 0;
A
C
B
yx
z
400 mm
150 mm
500 mm
600 N
150 mm
30
60
Ans:Ans:
,, ,,
,,
(MC)z = -99.0 N # m(MC)y = 153 N # m,
(TC)x = -77.9 N # m,(VC)z = -520 N
(VC)y = -260 N(NC)x = 150 N

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28
Internal Loading: Referring to the free-body diagram of the section of the drill and
brace shown in Fig. a,
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
The negative sign indicates that (MA)z acts in the opposite sense to that shown on
the free-body diagram.
©Mz = 0; AMA Bz + 30(1.25) = 0 AMA Bz = -37.5 lb # ft
©My = 0; ATA By - 30(0.75) = 0 ATA By = 22.5 lb # ft
©Mx = 0; AMA Bx - 10(2.25) = 0 AMA Bx = 22.5 lb # ft
©Fz = 0; AVA Bz - 10 = 0 AVA Bz = 10 lb
©Fy = 0; ANA By - 50 = 0 ANA By = 50 lb
©Fx = 0; AVA Bx - 30 = 0 AVA Bx = 30 lb
*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.
z
x
y
AO
9 in.6 in. 6 in. 6 in.
9 in.3 in.
Fx 30 lb
Fy 50 lb
Fz 10 lb

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29
1–29. The curved rod AD of radius r has a weight per
length of w. If it lies in the vertical plane, determine the
resultant internal loadings acting on the cross section
through point B. Hint: The distance from the centroid C of
segment AB to point O is OC = [2r sin (u>2)]>u.
Ans.
Ans.
a
Ans.MB = wr2
(u cos u - sin u)
MB = -NBr - wr2 2 sin (u/2) cos (u/2)
wruacos u
2 b a 2r sin (u/2)
u b + (NB)r + MB = 0+ ©M0 = 0;
VB = -wru sin u
- VB - wru sin u = 0+Q©Fy = 0;
NB = -wru cos u
NB + wru cos u = 0R+ ©Fx = 0;
O
r
C
B
D
A
u
u
2
Ans:Ans:
, ,
MB = wr2
(u cos u - sin u)
VB = -wru sin uNB = -wru cos u

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30
(1)
(2)
(3)
(4)
Since is can add, then ,
Eq. (1) becomes
Neglecting the second order term,
QED
Eq. (2) becomes
Neglecting the second order term,
QED
Eq. (3) becomes
Neglecting the second order term,
QED
Eq. (4) becomes
Neglecting the second order term,
QED
dM
du = -T
Tdu + dM = 0
Tdu + dM + dTdu
2 = 0
dT
du = M
Mdu - dT = 0
Mdu - dT + dMdu
2 = 0
dV
du = -N
Ndu + dV = 0
Ndu + dV + dNdu
2 = 0
dN
du = V
Vdu - dN = 0
Vdu - dN + dVdu
2 = 0
cos du
2 = 1sin du
2 = du
2
du
2
T sin du
2 - M cos du
2 + (T + dT) sin du
2 + (M + dM) cos du
2 = 0
©My = 0;
T cos du
2 + M sin du
2 - (T + dT) cos du
2 + (M + dM) sin du
2 = 0
©Mx = 0;
N sin du
2 - V cos du
2 + (N + dN) sin du
2 + (V + dV) cos du
2 = 0
©Fy = 0;
N cos du
2 + V sin du
2 - (N + dN) cos du
2 + (V + dV) sin du
2 = 0
©Fx = 0;
1–30. A differential element taken from a curved bar is
shown in the figure. Show that
and dT>du = M.dM>du = -T,
dV>du = -N,dN>du = V,
M V
N
d
u
M dM T dT
N dN
V dV
T

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Mechanical Engineering

Solution Manual for Mechanics of Materials SI, 9th Edition

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