ch01-03 stress & strain & properties

01 Solutions 46060 5/6/10 2:43 PM Page 7
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1–11. The force F = 80 lb acts on the gear tooth.
Determine the resultant internal loadings on the root of the
tooth, i.e., at the centroid point A of section a–a.
Equations of Equilibrium: For section a–a
30
F 80 lb
a
+ Q©F x¿ = 0; V A - 80 cos 15° = 0
V A = 77.3 lb
a + ©F y¿ = 0; N A - 80 sin 15° = 0
N A = 20.7 lb
Ans.
Ans.
0.23 in.
A
0.16 in.
a +©M A = 0; -M A - 80 sin 15°(0.16) + 80 cos 15°(0.23) = 0
M A = 14.5 lb # in.
Ans.
a
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*1–12. The sky hook is used to support the cable of a
scaffold over the side of a building. If it consists of a smooth
rod that contacts the parapet of a wall at points A, B, and C,
determine the normal force, shear force, and moment on
the cross section at points D and E.
Support Reactions:
+ c ©F y = 0; N B - 18 = 0 N B = 18.0 kN
d+©M C = 0; 18(0.7) - 18.0(0.2) - N A (0.1) = 0
0.2 m
0.3 m
0.2 m 0.2 m
B
D
E
A
C
0.2 m
0.2 m
N A = 90.0 kN
0.3 m
: + ©F x = 0; N C - 90.0 = 0 N C = 90.0 kN
Equations of Equilibrium: For point D
18 kN
: + © F x = 0; V D - 90.0 = 0
+ c © F y = 0; N D - 18 = 0
V D = 90.0 kN
N D = 18.0 kN
Ans.
Ans.
d+© M D = 0; M D + 18(0.3) - 90.0(0.3) = 0
M D = 21.6 kN # m
Equations of Equilibrium: For point E
: + © F x = 0; 90.0 - V E = 0
V E = 90.0 kN
+ c © F y = 0; N E = 0
d+© M E = 0; 90.0(0.2) - M E = 0
M E = 18.0 kN # m
Ans.
Ans.
Ans.
Ans.
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