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PROBLEM (9.72) The tube shown in Fig. P9.71 is of t = 20-mm uniform thickness. What is the
minimum permissible distance d?
Given: P = 250 kN, ( t
σ
)all = 90 MPa, ( c
σ
)all = 130 MPa
SOLUTION
See solution of Prob.9.71:
mkNPMz
⋅
=
== 5.12)250(05.005.0
)07.0(10250)07.0( 3ddPM y−×=−=
d
33 10250105.17 ×−×=
Maximum tensile stress is at D:
yz
Dyz
Mz
M
yP
AI I
σ
=− + +
333
6366
250 10 (17.5 250 )10 (0.07) 12.5 10 (0.05)
90 10 8 10 17.87 10 9.87 10
d
−−−
−× − ×
×= + +
×××
or
32.6329.97955.6825.3190
+
−+−= d, mmd 8.10
=
Maximum compressive stress is at B:
32.6329.97955.6825.31130
−
+−−=− d, mmd 8.33
=
Thus,
mmd 8.33
min =
1
/
1
100%