Question 5.15: A thin cylindrical vessel is heated by a jacket from the out...
A thin cylindrical vessel is heated by a jacket from the outside such that the temperature distribution is as shown in Figure 5.26. If E=27 \times 10^{6} psi, \alpha=9.5 \times 10^{-6} in./in.- °F, and \mu=0.28, determine the maximum thermal stress (a) using Eq. (5.40) and (b) using Eq. (5.41).
\sigma_{\theta}=\sigma_{z}=\left\{\begin{array}{l} \frac{-E \alpha T_{i}}{1-\mu}\left[\frac{2 r_{ o }+r_{ i }}{3\left(r_{ o }+r_{ i }\right)}\right] \text { for inside surface } \\ \frac{E \alpha T_{ i }}{1-\mu}\left[\frac{r_{ o }+2 r_{ i }}{3\left(r_{ o }+r_{ i }\right)}\right] \text { for outside surface } \end{array}\right. (5.40)
\sigma_{\theta}=\sigma_{z}=\left\{\begin{array}{ll} \frac{-E \alpha T_{ i }}{2(1-\mu)} & \text { for inside surface } \\ \frac{E \alpha T_{ i }}{2(1-\mu)} & \text { for outside surface } \end{array}\right. (5.41)
(a): T_{ i } =400–700= −300 °F. Hence, at the inside surface,
\sigma=\frac{-\left(27 \times 10^{6}\right)\left(9.5 \times 10^{-6}\right)(-300)}{(1-0.28)} \times\left(\frac{2(13)+10}{3(13+10)}\right)= 55,800 psi
and at the outer surface,
\sigma=\frac{\left(27 \times 10^{6}\right)\left(9.5 \times 10^{-6}\right)(-300)}{(1-0.28)} \times\left(\frac{13+2 \times 10}{3(13+10)}\right)= −51,000 psi
(b): At the inner surface,
\sigma=\frac{\left(-27 \times 10^{6}\right)\left(9.5 \times 10^{-6}\right)(-300)}{2(1-0.28)}= 53,400 psi,
and at the outer surface, \sigma = −53,400 psi.