A cylindrical storage tank used to transport gas under pressure has an inner diameter of 24 in. and a wall thickness of ¾ 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

Question

A cylindrical storage tank used to transport gas under pressure has an inner diameter of 24 in. and a wall thickness of [latex]\frac{3}{4}[/latex] in. Strain gages attached to the surface of the tank in transverse and longitudinal directions indicate strains of 255 × 10[latex]^{-6}[/latex] and 60 × 10[latex]^{-6}[/latex] 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[latex]^6[/latex] psi, determine (a) the gage pressure inside the tank, (b) the principal stresses and the maximum shearing stress in the wall of the tank.

Accepted Answer

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.

[latex]g _{\max ( ext { in plane })}=\epsilon_{1}-\epsilon_{2}=255 imes 10^{-6}-60 imes 10^{-6}=195 imes 10^{-6} rad[/latex]

From Hooke’s law for shearing stress and strain, we have

[latex]t _{\max ( ext { in plane })}=G g _{\max ( ext { in plane })}[/latex]

[latex]=\left(11.2 imes 10^{6} psi ight)\left(195 imes 10^{-6} rad ight)[/latex]

= 2184 psi = 2.184 ksi

Substituting this value and the given data in Eq. (7.33), we write

[latex]t _{\max ( ext { in plane })}=\frac{1}{2} s _{2}=\frac{p r^{\circ}}{4 t}[/latex] (7.33)

[latex]t _{\max ( ext { in plane })}=\frac{p r}{4 t}[/latex] [latex]2184 ext { psi }=\frac{p(12 ext { in. })}{4(0.75 ext { in. })}[/latex]

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, [latex]s _{1}=2 s _{2}[/latex], we draw Mohr’s circle for stress and obtain

[latex]s _{2}=2 t _{\max ( ext { in plane })}=2(2.184 ksi )=4.368 ksi[/latex] [latex]s_{2}=4.37 ksi[/latex]

[latex]s _{1}=2 s _{2}=2(4.368 ksi )[/latex] [latex]s_{1}=8.74 ksi[/latex]

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.

[latex]t _{\max }=\frac{1}{2} s _{1}= s _{2}=4.368 ksi[/latex] [latex]t _{\max }=4.37 ksi[/latex]

Question 7.S-P.6: A cylindrical storage tank used to transport gas under press...

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