Question 3.14: An aluminum-alloy pressure vessel is made of tubing having a...

(a) Here d_{i} = 8 − 2(0.25) = 7.5 in, r_{i}= 7.5/2 = 3.75 in, and r_{o}= 8/2 = 4 in. Then t/r_{i}= 0.25/3.75 = 0.067. Since this ratio is greater than \frac {1}{20} , the theory for thin-walled vessels may not yield safe results.

We first solve Eq. (3–53) to obtain the allowable pressure. This gives

(σ_{t} )_{max} =\frac {p(d_{i} + t)}{2t}              (3–53)

p =\frac {2t (σ_{t} )_{max}}{d_{i} + t} =\frac {2(0.25)(12)(10)^{3}}{7.5 + 0.25 }= 774  psi

Then, from Eq. (3–54), we find the average longitudinal stress to be

σ_{l} =\frac {pd_{i}}{4t}             (3–54)

σ_{l} =\frac {pd_{i}}{4t}=\frac {774(7.5)}{4(0.25)} = 5810  psi

(b) The maximum tangential stress will occur at the inside radius, and so we use r = r_{i} in the first equation of Eq. (3–50). This gives

σ_{t} =\frac {r^{2}_{i}p_{i}}{r^{2}_{o}− r^{2}_{i}} (1 +\frac {r^{2}_{o}}{r^{2}})

σ_{r} =\frac {r^{2}_{i}p_{i}}{r^{2}_{o}− r^{2}_{i}} (1 -\frac {r^{2}_{o}}{r^{2}})             (3.50)

(σ_{t} )_{max} =\frac {r^{2}_{i} p_{i}}{r^{2}_{o} − r^{2}_{i}}(1 +\frac {r^{2}_{o}}{r^{2}_{i}})= p_{i} \frac {r^{2}_{o} + r^{2}_{i}}{r^{2}_{o}− r^{2}_{i}}= 774 \frac {4^{2} + 3.75^{2}} {4^{2} − 3.75^{2}} = 12 000   psi

Similarly, the maximum radial stress is found, from the second equation of Eq. (3–50) to be

σ_{r} = −p_{i} = −774  psi

Equation (3–51) gives the longitudinal stress as

σ_{l} =\frac {p_{i}r^{2}_{i}}{r^{2}_{o} − r^{2}_{i}}          (3-51)

σ_{l} =\frac {p_{i}r^{2}_{i}}{r^{2}_{o} − r^{2}_{i}} =\frac {774(3.75)^{2}}{4^{2} − 3.75^{2}} = 5620  psi

These three stresses, σ_{t} , σ_{r} ,  and  σ_{l} , are principal stresses, since there is no shear on these surfaces. Note that there is no significant difference in the tangential stresses in parts (a) and (b), and so the thin-wall theory can be considered satisfactory.