Question 8.4: The length of a cylindrical shell is 15 ft, outside diameter...

(a) Assume

Then,

From Figure 8.10, factor A=0.00018. From Figure 8.11, the modulus of elasticity at room temperature is 29,000,000 psi. Hence, from Eq. (8.6),

P=\frac{2 A E_{0}}{3\left(D_{ o } / t\right)}                           (8.6)

P=\frac{(2)(0.00018)\left(29 \times 10^{6}\right)}{(3)(320)}=10.9 psi    acceptable

A check is needed to ascertain that the buckling is in the elastic rather than the inelastic region. From Figure 8.11 with A=0.0018, a value of B=2600 psi is obtained in the elastic region of the curve. Hence, the aforementioned solution of 10.9 psi is adequate.

Use  t=\frac{3}{8} in

(b) For a DF 2.0, assume

t = 0.3125 in.

Then,

From Figure 8.10, the factor A=0.00014, and from Eq. (8.6),

= \frac{2(0.00014) \times 29 \times 10^{6}}{3(384)} \times \frac{3.0}{2.0}

= 10.6 psi acceptable

Returning to Figure 8.11 with A=0.00014, it is seen that the value from the chart indicates an elastic behavior.

Use  t=\frac{5}{16} \operatorname{in}