ch01-03 stress & strain & properties

01 Solutions 46060 5/6/10 2:43 PM Page 1
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1–1. Determine the resultant internal normal force acting
on the cross section through point A in each column. In
(a), segment BC weighs 180 lb>ft and segment CD weighs
250 lb>ft. In (b), the column has a mass of 200 kg>m.
5 kip
B
200 mm
8 kN
200 mm
(a)
(b)
+ c ©F y = 0; F A - 1.0 - 3 - 3 - 1.8 - 5 = 0
F A = 13.8 kip
+ c ©F y = 0; F A - 4.5 - 4.5 - 5.89 - 6 - 6 - 8 = 0
F A = 34.9 kN
Ans.
Ans.
10 ft
8 in.
3 kip
4 ft
8 in.
3 kip
C
6 kN
200 mm
4.5 kN
6 kN
3 m
200 mm
4.5 kN
4 ft
A
D
A
1 m
(a)
(b)
1–2. Determine the resultant internal torque acting on the
cross sections through points C and D.The support bearings
at A and B allow free turning of the shaft.
A
250 Nm
©M x = 0; T C - 250 = 0
©M x = 0; T D = 0
T C = 250 N # m
Ans.
Ans.
C
300 mm
200 mm
150 mm
200 mm
150 Nm
250 mm
400 Nm
D
B
150 mm
1–3. Determine the resultant internal torque acting on the
cross sections through points B and C.
©M x = 0; T B + 350 - 500 = 0
T B = 150 lb # ft
©M x = 0; T C - 500 = 0
T C = 500 lb # ft
Ans.
Ans.
A
3 ft
1 ft
600 lbft
B
350 lbft
2 ft
C
500 lbft
2 ft
1