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## Q. 5.14

A vessel that operates at 800°F is supported by an insulated skirt. The thermal distribution in the skirt is shown in Fig. 5.24a. If the top and bottom of the skirt are assumed fixed with respect to rotation, what is the maximum stress due to temperature gradient? α = 7 × 10$^{-6}$ in./in.°F , μ = 0.3, E = 30 × 10$^6$ psi.

## Verified Solution

The equation for linear temperature gradient is

$T_{y}=T_{t}+\frac{T_{b}-T_{t}}{l} x$

The temperature change can be expressed as

$T=\frac{T_{b}-T_{t}}{l} x$

and the circumferential stress due to ring action obtained from Eq. 5.30 a is

$\sigma_{\theta}=\frac{-P r}{t}=-E \alpha T_{x}$               (5.30a)

$\frac{d^{4} w}{d x^{4}}+4 \beta^{4} w=\frac{E t \alpha T_{x}}{r D}$                       (5.30b)

$\sigma_{\theta}=-E \alpha\left(\frac{T_{b}-T_{t}}{l}\right) x$                                 (1)

Equation 5.30 b gives

$\frac{d^{4} w}{d x^{4}}+4 \beta^{4} w=\frac{E t \alpha}{r D}\left(\frac{T_{b}-T_{t}}{l}\right) x$

A particular solution takes the form

$w=c_{1} \frac{E t \alpha}{r D}\left(\frac{T_{b}-T_{t}}{l}\right) x+c_{2}$

which upon substituting into the differential equation gives

$c_{1}=\frac{1}{4 \beta^{4}}$                     and                $c_{2}=0$

and w reduces to

$w=r \alpha\left(\frac{T_{b}-T_{t}}{l}\right) x$

From Eq. 5.19

$N_{\theta}=\frac{-E t w}{r}$                         (5.19)

$\sigma_{\theta}=\frac{N_{\theta}}{t}=E \alpha\left(\frac{T_{b}-T_{t}}{l}\right) x$                          (2)

Adding Eqs. 1 and 2 results in

$\sigma_{\theta}=0$

which means that for a linear distribution the thermal stress along the skirt is zero.

The slope due to axial gradient is given by

$\theta=\frac{d w}{d x}=r \alpha\left(\frac{T_{b}-T_{t}}{l}\right)$

Because the ends are fixed against rotation, a moment must be applied at the ends to reduce θ to zero. From Eq. 5.24

\begin{aligned}w &=\frac{1}{2 \beta^{3} D}\left(\beta M_{0} B_{\beta x}+Q_{0} C_{\beta x}\right) \\\frac{d w}{d x} &=\frac{-1}{2 \beta^{2} d}\left(2 \beta M_{0} C_{\beta x}+Q_{0} A_{\beta x}\right) \\\frac{d^{2} w}{d x^{2}} &=\frac{1}{2 \beta D}\left(2 \beta M_{0} A_{\beta x}+2 Q_{0} D_{\beta x}\right) \\\frac{d^{3} w}{d x^{3}} &=\frac{-1}{D}\left(2 \beta M_{0} D_{\beta x}-Q_{0} B_{\beta x}\right)\end{aligned}                          (5.24)

$r \alpha\left(\frac{T_{b}-T_{t}}{l}\right)=\frac{-M_{0}}{\beta D}$

or

$M_{0}=-\beta D\left(\frac{\alpha r}{l}\right)\left(T_{b}-T_{t}\right)$

Since $\beta=0.2142, D=2.747 \times 10^{6}$,

$M_{0}=742 \text { in.- lb / in . }$

and

σ = 4450 psi