_______________________________________________________________________
PROBLEM (9.17) Rework Prob. 9.16 for the case in which the vessel is subjected to an axial
compression load P = -10
π
kN.
SOLUTION
Refer to solution of Prob. 9.16. We now have
6
1.5625(10 ) 25 50
xy
p
p
σσ
=− + =
Equations (8.4):
6
'25(10 ) (25 50) (25 50)cos300
22 o
xpp
σ
==++−
66
1.5625(10 ) 1.5625(10 ) cos300
22o
−−
or 838
p
kPa=
6
6
'' 1.5625(10 )
10 10 (25 50)sin300 sin300
22
oo
xy p
τ
=− × =− − +
Solving,
866
p
kPa=
So 836
all
p
kPa=
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