Sol'n: If the maximum stress of the steel is 50 ksi and
the Factor of Safety is 2.0, the allowable (maximum working) stress in
the pressure vessel is:
sw
=
|
sf
F.S.
|
=
|
50 ksi
2
|
= 25 ksi
|
In a cylindrical pressure vessel, the hoop stress is always twice the longitudinal stress. Therefore in this problem we want to limit the hoop stress to 25 ksi
sH
=
|
pR
t
|
pw =
|
swt
R
|
pw =
|
(25 ksi)(0.10 in.)
10 in.
|
= 0.25 ksi = 250 psi
|
The change in radius of the pressure vessel is given by:
DR =
|
pR2
Et
|
|
1 -
|
n
2
|
|
DR =
|
(250 psi)(10 in)2
(30 Msi)(0.10 in)
|
|
1 -
|
0.3
2
|
|
= 0.0071 in
|
|