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PROBLEM (8.102) A thick-walled steel cylindrical tank of internal radius a is subject to an internal
pressure pi (Fig. P8.101). The ultimate strength of the material in tension and compression is equal to
all
σ
. factor of safety is to be
s
n . Calculate the wall thickness.
Given: a = 3 ft, p = 2 ksi , all
σ
= 40 ksi,
s
n = 1.5
SOLUTION
The maximum normal stress is given by Eq. (8.30a).
Thus, with 3 12 36 .:
ai=× = n
22
max 22
()
uti
s
ab
p
nb
σσ
a
+
==
−
Substituting the given numerical values
322
322
40(10 ) 36
(2 10 )
1.5 36
b
b
+
=× −38.81 .bin,
=
Hence
38.81 36 2.81 .
tba in=−= − =
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