sm9 69

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PROBLEM (9.69) The sign of Fig. P9.68 is supported by a steel pipe of an outer-diameter D and
inner-diameter d. Determine the maximum shearing stress in the pipe.
Given: D = 100 mm, d = 75 mm, P = 800 N, W = 500 N
SOLUTION
Refer to solution of Prob.9.68:
We now have
44 64
(100 75 ) 6.71 10
32
Jmm
π
=−=× 6 10355.32 ×== JI
22 3
(100 75 ) 3.43 10
4
A
π
=−=×
Point A:
6
1.2(800)(0.05) 7.15
6.71 10
t
M
Pa τ
−
=− =−
×
36
500 3(800)(0.05) 35.92
3.43 10 3.355 10
x
M
Pa σ
−−
=− − =−
××
22
max 35.92
( ) ( 7.15) 19.33
2
M
Pa τ
−
=+−=
Point B:
For semi-circle: 23
42
23 3
rr
QAy r
π
π
== ⋅=
Thus, for half thin-tube:
33 33
2(50 37.5 ) 48.18 10
3
Qmm=−=×
6
6
800(48.18 10 ) 0.46
3.355 10 (0.025)
dVQ
M
Pa
Ib
τ
−
−
×
=− =− =−
×
MPa61.746.015.7 −=−−=
τ
6
1.2(500)(0.05)
0.146 9.09
3.355 10
x
M
Pa σ
−
=− − =
×
22
max 9.09
( ) ( 7.61) 8.86
2
M
Pa τ =− +− =
B
y
x
M
z
W
A
I
σ =+
td
τ
ττ
=
+
A
z
x
My
W
A
I
σ =+
Tc
J
τ =
x
x