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Solution Manual Mechanics of Materials By Hibbeler
11th Edition
All Chapters |Q&A Verified |Grade A+ Assured
M01_HIBB5613_11_SE_C01.indd 1 21/01/22 4:34 PM
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1–1
A B E C
B and
a journal bearing at C. Determine the resultant
internal loadings acting on the cross section at E.
4 ft 4 ft 4 ft 4 ft
400 lb
800 lb
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; 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,
+SΣFx = 0; NE = 0 Ans.
+ cΣFy = 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:
Cy 1000 lb, NE 0, VE 200lb, ME 2.40kip ft
1
M01_HIBB5613_11_SE_C01.indd 2 21/01/22 4:34 PM
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1–2.
1–2.
Determine the resultant internal normal and shear force in the a b
member on (a) section a–a and (b) section b–b, each of which 30
passes through the centroid A. The 500-lb load is applied along
the centroidal axis of the member.
500 lb 500 lb
A
b a
SOLUTION
(a)
+ ΣFx = 0;
➞ Na - 500 = 0
Na = 500 lb Ans.
+T ΣFy = 0; Va = 0 Ans.
(b)
\+ ΣFx = 0; Nb - 500 cos 30° = 0
Nb = 433 lb Ans.
+fΣFy = 0; Vb - 500 sin 30° = 0
Vb = 250 lb Ans.
Ans:
(a) Na = 500 lb, Va = 0,
(b) Nb = 433 lb, Vb = 250 lb
2
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1–3.
Determine the resultant internal torque acting on
the cross sections through points B and C.
A 600 lb ft
B
350 lb ft
C
500 lb ft
SOLUTION
\Mx = 0; TB + 350 - 500 = 0
TB = 150 lb . ft Ans.
\Mx = 0; TC - 500 = 0
TC = 500 lb . ft Ans.
Ans:
TB 150 lb ft
TC 500 lb ft
3
M01_HIBB5613_11_SE_C01.indd 4 21/01/22 4:34 PM
Solution Manual Mechanics of Materials By Hibbeler
11th Edition
All Chapters |Q&A Verified |Grade A+ Assured
M01_HIBB5613_11_SE_C01.indd 1 21/01/22 4:34 PM
, THIS DOCUMENT BELONGS TO TESTBANKSPROF
1–1
A B E C
B and
a journal bearing at C. Determine the resultant
internal loadings acting on the cross section at E.
4 ft 4 ft 4 ft 4 ft
400 lb
800 lb
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; 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,
+SΣFx = 0; NE = 0 Ans.
+ cΣFy = 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:
Cy 1000 lb, NE 0, VE 200lb, ME 2.40kip ft
1
M01_HIBB5613_11_SE_C01.indd 2 21/01/22 4:34 PM
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1–2.
1–2.
Determine the resultant internal normal and shear force in the a b
member on (a) section a–a and (b) section b–b, each of which 30
passes through the centroid A. The 500-lb load is applied along
the centroidal axis of the member.
500 lb 500 lb
A
b a
SOLUTION
(a)
+ ΣFx = 0;
➞ Na - 500 = 0
Na = 500 lb Ans.
+T ΣFy = 0; Va = 0 Ans.
(b)
\+ ΣFx = 0; Nb - 500 cos 30° = 0
Nb = 433 lb Ans.
+fΣFy = 0; Vb - 500 sin 30° = 0
Vb = 250 lb Ans.
Ans:
(a) Na = 500 lb, Va = 0,
(b) Nb = 433 lb, Vb = 250 lb
2
M01_HIBB5613_11_SE_C01.indd 3 21/01/22 4:34 PM
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1–3.
Determine the resultant internal torque acting on
the cross sections through points B and C.
A 600 lb ft
B
350 lb ft
C
500 lb ft
SOLUTION
\Mx = 0; TB + 350 - 500 = 0
TB = 150 lb . ft Ans.
\Mx = 0; TC - 500 = 0
TC = 500 lb . ft Ans.
Ans:
TB 150 lb ft
TC 500 lb ft
3
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