Mechanics of Materials, 10th Edition Solution Manual

Name: Mechanics of Materials, 10th Edition Solution Manual
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Solution
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.
A E DB C
4 ft
400 lb
800 lb
4 ft 4 ft 4 ft
Solution
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+ ΣMB = 0; C y(8) + 400(4) - 800(12) = 0 C y = 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,
S+ ΣFx = 0; NE = 0 Ans.
+ c ΣFy = 0; VE + 1000 - 800 = 0 VE = -200 lb Ans.
a+ ΣME = 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

Solution
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 the centroid A. The 500-lb load is
applied along the centroidal axis of the member.
30°
A
ba
b a
500 lb500 lb
Solution
(a)
+S ΣFx = 0; Na - 500 = 0
Na = 500 lb Ans.
+ T ΣFy = 0; Va = 0 Ans.
(b)
R+ ΣFx = 0; Nb - 500 cos 30° = 0
Nb = 433 lb Ans.
+Q ΣFy = 0; Vb - 500 sin 30° = 0
Vb = 250 lb Ans.
Ans:
(a) Na = 500 lb, Va = 0,
(b) Nb = 433 lb, Vb = 250 lb

Solution
3
1–3.
Determine the resultant internal loadings acting on section
b–b through the centroid C on the beam.
30° 6 ft
3 ft60°
bb900 lb/ft
C
A
B
Solution
Support Reaction:
a+ ΣMA = 0; NB(9 sin 30°) - 1
2 (900)(9)(3) = 0
NB = 2700 lb
Equations of Equilibrium: For section b–b
S+ ΣFx = 0; Vb - b + 1
2 (300)(3) sin 30° - 2700 = 0
Vb - b = 2475 lb = 2.475 kip Ans.
+ c ΣFy = 0; Nb - b - 1
2 (300)(3) cos 30° = 0
Nb - b = 389.7 lb = 0.390 kip Ans.
a+ ΣMC = 0; 2700(3 sin 30°)
- 1
2 (300)(3)(1) - Mb - b = 0
Mb - b = 3600 lb # ft = 3.60 kip # ft Ans.
Ans:
Vb - b = 2.475 kip,
Nb - b = 0.390 kip,
Mb - b = 3.60 kip # ft

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. A DB
C
900 N
1.5 m
600 N/m
1.5 m1 m1 m1 m
Solution
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+ ΣMA = 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,
S+ ΣFx = 0; NC = 0 Ans.
+ cΣFy = 0; VC - 600(1) + 1733.33 - 900 = 0 VC = -233 N Ans.
a+ ΣMC = 0; 1733.33(2.5) - 600(1)(0.5) - 900(4) - MC = 0
MC = 433 N # m Ans.
The negative sign indicates that VC acts in the opposite sense to that shown on the
free-body diagram.
*
Ans:
NC = 0,
VC = -233 N,
MC = 433 N # m

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6
1–6.
Determine the resultant internal loadings on the cross
section at point D.
2 m
D
1.25 kN/m
1.5 m
1 m
E B
C
A
F
0.5 m 0.5 m 0.5 m
Solution
Support Reactions: Member BC is the two force member.
a+ ΣMA = 0; 4
5FBC (1.5) - 1.875(0.75) = 0
FBC = 1.1719 kN
+ c ΣFy = 0; Ay + 4
5 (1.1719) - 1.875 = 0
Ay = 0.9375 kN
S+ ΣFx = 0; 3
5 (1.1719) - Ax = 0
Ax = 0.7031 kN
Equations of Equilibrium: For point D
S+ ΣFx = 0; ND - 0.7031 = 0
ND = 0.703 kN Ans.
+ c ΣFy = 0; 0.9375 - 0.625 - VD = 0
VD = 0.3125 kN Ans.
a+ ΣMD = 0; MD + 0.625(0.25) - 0.9375(0.5) = 0
MD = 0.3125 kN # m Ans.
Ans:
ND = 0.703 kN,
VD = 0.3125 kN,
MD = 0.3125 kN # m

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7
1–7.
Determine the resultant internal loadings at cross sections
at points E and F on the assembly.
2 m
D
0.5 m 0.5 m
1.25 kN/m
0.5 m 1.5 m
1 m
E B
C
A
F
Solution
Support Reactions: Member BC is the two-force member.
a+ ΣMA = 0; 4
5 FBC (1.5) - 1.875(0.75) = 0
FBC = 1.1719 kN
+ c ΣFy = 0; Ay + 4
5 (1.1719) - 1.875 = 0
Ay = 0.9375 kN
S+ ΣFx = 0; 3
5 (1.1719) - Ax = 0
Ax = 0.7031 kN
Equations of Equilibrium: For point F
+ bΣFx′ = 0; NF - 1.1719 = 0
NF = 1.17 kN Ans.
a + ΣFy′ = 0; VF = 0 Ans.
a+ ΣMF = 0; MF = 0 Ans.
Equations of Equilibrium: For point E
+d ΣFx = 0; NE - 3
5 (1.1719) = 0
NE = 0.703 kN Ans.
+ c ΣFy = 0; VE - 0.625 + 4
5 (1.1719) = 0
VE = -0.3125 kN Ans.
a+ ΣME = 0; -ME - 0.625(0.25) + 4
5 (1.1719)(0.5) = 0
ME = 0.3125 kN # m Ans.
Negative sign indicates that VE acts in the opposite direction to that shown on FBD.
Ans:
NF = 1.17 kN,
VF = 0,
MF = 0,
NE = 0.703 kN,
VE = -0.3125 kN,
ME = 0.3125 kN # m

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8
*1–8.
The beam supports the distributed load shown. Determine
the resultant internal loadings acting on the cross section at
point C. Assume the reactions at the supports A and B are
vertical.
1.5 m 3 m
DC
A B
4 kN/m
1.5 m
Solution
Support Reactions: Referring to the FBD of the entire beam, Fig. a,
a+ ΣMA = 0; By(6) - 1
2 (4)(6)(2) = 0 By = 4.00 kN
Internal Loadings: Referring to the FBD of the right segment of the beam sectioned
through C, Fig. b,
S+ ΣFx = 0; NC = 0 Ans.
+ c ΣFy = 0; VC + 4.00 - 1
2 (3)(4.5) = 0 VC = 2.75 kN Ans.
a+ ΣMC = 0; 4.00(4.5) - 1
2 (3)(4.5)(1.5) - MC = 0
MC = 7.875 kN # m Ans.
Ans:
NC = 0,
VC = 2.75 kN,
MC = 7.875 kN # m

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9
1–9.
The beam supports the distributed load shown. Determine
the resultant internal loadings acting on the cross section at
point D. Assume the reactions at the supports A and B are
vertical.
1.5 m 3 m
DC
A B
4 kN/m
1.5 m
Solution
Support Reactions: Referring to the FBD of the entire beam, Fig. a,
a+ ΣMA = 0; By(6) - 1
2 (4)(6)(2) = 0 By = 4.00 kN
Internal Loadings: Referring to the FBD of the right segment of the beam sectioned
through D, Fig. b,
S+ ΣFx = 0; ND = 0 Ans.
+ c ΣFy = 0; VD + 4.00 - 1
2 (1.00)(1.5) = 0 VD = -3.25 kN Ans.
a+ ΣMD = 0; 4.00(1.5) - 1
2 (1.00)(1.5)(0.5) - MD = 0
MD = 5.625 kN # m Ans.
The negative sign indicates that VD acts in the sense opposite to that shown on
the FBD.
Ans:
ND = 0,
VD = -3.25 kN,
MD = 5.625 kN # m

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10
1–10.
The boom DF of the jib crane and the column DE have a
uniform weight of 50 lb/ft. If the supported load is 300 lb,
determine the resultant internal loadings in the crane on
cross sections at points A, B, and C. 5 ft
7 ft
C
D F
E
B A
300 lb
2 ft 8 ft 3 ft
Solution
Equations of Equilibrium: For point A
d+ ΣFx = 0; NA = 0 Ans.
+ c ΣFy = 0; VA - 150 - 300 = 0
VA = 450 lb Ans.
a+ ΣMA = 0; -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
d+ ΣFx = 0; NB = 0 Ans.
+ c ΣFy = 0; VB - 550 - 300 = 0
VB = 850 lb Ans.
a+ ΣMB = 0; -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
d+ ΣFx = 0; VC = 0 Ans.
+ c ΣFy = 0; -NC - 250 - 650 - 300 = 0
NC =

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11
1–11.
Determine the resultant internal loadings acting on the
cross sections at points D and E of the frame.
4 ft
A
D2 ft
C
B
75 lb/ft
1 ft
150 lb
1 ft
E
30°
2 ft 1 ft
G
1 ft
F
Solution
Member AG:
a+ ΣMA = 0; 4
5 FBC (3) - 75(4)(5) - 150 cos 30°(7) = 0; FBC = 1003.89 lb
a+ ΣMB = 0; Ay (3) - 75(4)(2) - 150 cos 30°(4) = 0; Ay = 373.20 lb
S+ ΣFx = 0; Ax - 3
5 (1003.89) + 150 sin 30° = 0; Ax = 527.33 lb
For point D:
S+ ΣFx = 0; ND + 527.33 = 0
ND = -527 lb Ans.
+ c ΣFy = 0; -373.20 - VD = 0
VD = -373 lb Ans.
a+ ΣMD = 0; MD + 373.20(1) = 0
MD = -373 lb # ft Ans.
For point E:
S+ ΣFx = 0; 150 sin 30° - NE = 0
NE = 75.0 lb Ans.
+ c ΣFy = 0; VE - 75(3) - 150 cos 30° = 0
VE = 355 lb Ans.
a+ ΣM

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12
1–12.
Determine the resultant internal loadings acting on the
cross sections at points F and G of the frame.
4 ft
A
D2 ft
C
B
75 lb/ft
1 ft
150 lb
1 ft
E
30°
2 ft 1 ft
G
1 ft
F
Solution
Member AG:
a+ ΣMA = 0; 4
5 FBF (3) - 300(5) - 150 cos 30°(7) = 0
FBF = 1003.9 lb
For point F:
+Q ΣFx′ = 0; VF = 0 Ans.
a +ΣFy′ = 0; NF - 1003.9 = 0
NF = 1004 lb Ans.
a+ ΣMF = 0; MF = 0 Ans.
For point G:
d+ ΣFx = 0; NG - 150 sin 30° = 0
NG = 75.0 lb Ans.
+ c ΣFy = 0; VG - 75(1) - 150 cos 30° = 0
VG = 205 lb Ans.
a+ ΣMG = 0; -MG - 75(1)(0.5) - 150 cos 30°(1) = 0
MG = -167 lb # ft Ans.
*
Ans:
VF = 0,
NF = 1004 lb,
MF = 0 ,
NG = 75.0 lb,
VG

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13
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 B
C
D
F F
a
b
b a
30°
225 mm
150 mm
E
Solution
Internal Loadings: Referring to the free-body diagram of the section of the hacksaw
shown in Fig. a,
d+ ΣFx = 0; Na - a + 100 = 0 Na - a = -100 N Ans.
+ c ΣFy = 0; Va - a = 0 Ans.
a+ ΣMD = 0; - Ma - a - 100(0.15) = 0 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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14
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 B
C
D
F F
a
b
b a
30°
225 mm
150 mm
E
Solution
Internal Loadings: Referring to the free-body diagram of the section of the hacksaw
shown in Fig. a,
ΣFx′ = 0; Nb - b + 100 cos 30° = 0 Nb - b = - 86.6 N Ans.
ΣFy′ = 0; Vb - b - 100 sin 30° = 0 Vb - b = 50 N Ans.
a+ ΣMD = 0; - Mb - b - 100(0.15) = 0 Mb - b = -15 N # m Ans.
The negative sign indicates that Nb–b and Mb–b act in the opposite sense to that
shown on the free-body diagram.
Ans:
Nb - b = - 86.6 N, Vb - b = 50 N,
Mb - b = -15 N # m

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15
1–15.
The beam supports the triangular distributed load shown.
Determine the resultant internal loadings on the cross
section at point C. Assume the reactions at the supports A
and B are vertical.
6 ft 6 ft
C
A B
4.5 ft
800 lb/ft
6 ft 4.5 ft
ED
Solution
Support Reactions: Referring to the FBD of the entire beam, Fig. a,
a+ ΣMB = 0; 1
2 (0.8)(18)(6) - 1
2 (0.8)(9)(3) - Ay(18) = 0 Ay = 1.80 kip
Internal Loadings: Referring to the FBD of the left beam segment sectioned through
point C, Fig. b,
S+ ΣFx = 0; NC = 0 Ans.
+ c ΣFy = 0; 1.80 - 1
2 (0.5333)(12) - VC = 0 VC = -1.40 kip Ans.
a+ ΣMC = 0; MC + 1
2 (0.5333)(12)(4) - 1.80(12) = 0
MC = 8.80 kip # ft Ans.
The negative sign indicates that VC acts in the sense opposite to that shown on
the FBD.
Ans:
NC = 0,
VC = -1.40 kip,
MC = 8.80 kip # ft

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Mechanics of Materials, 10th Edition Solution Manual

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