Question 7.S-P.6: A cylindrical storage tank used to transport gas under press...
A cylindrical storage tank used to transport gas under pressure has an inner diameter of 24 in. and a wall thickness of \frac{3}{4} in. Strain gages attached to the surface of the tank in transverse and longitudinal directions indicate strains of 255 × 10^{-6} and 60 × 10^{-6} in./in. respectively. Knowing that a torsion test has shown that the modulus of rigidity of the material used in the tank is G = 11.2 × 10^6 psi, determine (a) the gage pressure inside the tank, (b) the principal stresses and the maximum shearing stress in the wall of the tank.
a. Gage Pressure Inside Tank. We note that the given strains are the principal strains at the surface of the tank. Plotting the corresponding points A and B, we draw Mohr’s circle for strain. The maximum in-plane shearing strain is equal to the diameter of the circle.
g _{\max (\text { in plane })}=\epsilon_{1}-\epsilon_{2}=255 \times 10^{-6}-60 \times 10^{-6}=195 \times 10^{-6} rad
From Hooke’s law for shearing stress and strain, we have
t _{\max (\text { in plane })}=G g _{\max (\text { in plane })}=\left(11.2 \times 10^{6} psi \right)\left(195 \times 10^{-6} rad \right)
= 2184 psi = 2.184 ksi
Substituting this value and the given data in Eq. (7.33), we write
t _{\max (\text { in plane })}=\frac{1}{2} s _{2}=\frac{p r^{\circ}}{4 t} (7.33)
t _{\max (\text { in plane })}=\frac{p r}{4 t} 2184 \text { psi }=\frac{p(12 \text { in. })}{4(0.75 \text { in. })}
Solving for the gage pressure p, we have
p = 546 psi
b. Principal Stresses and Maximum Shearing Stress. Recalling that, for a thin-walled cylindrical pressure vessel, s _{1}=2 s _{2}, we draw Mohr’s circle for stress and obtain
s _{2}=2 t _{\max (\text { in plane })}=2(2.184 ksi )=4.368 ksi s_{2}=4.37 ksi
s _{1}=2 s _{2}=2(4.368 ksi ) s_{1}=8.74 ksi
The maximum shearing stress is equal to the radius of the circle of diameter OA and corresponds to a rotation of 45° about a longitudinal axis.
t _{\max }=\frac{1}{2} s _{1}= s _{2}=4.368 ksi t _{\max }=4.37 ksi
