Question 5.14: A vessel that operates at 800 °F is supported by an insulate...

The equation for a linear temperature gradient is

The temperature change can be expressed as

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

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

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

Eq. (5.30b) gives

\frac{ d ^{4} w}{ d x^{4}}+3 \beta^{4} w=\frac{E t \alpha T_{x}}{r D}                                    (5.30b)

A particular solution takes the form

which upon substituting into the differential equation gives

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

and w reduces to

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

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

The slope due to axial gradient is given by

Because the ends are fixed against rotation, a moment must be applied at the ends to reduce \theta to zero as shown in Figure 5.24b. From Eq. (5.24),

\frac{ d ^{3} w}{ d x^{3}}=\frac{-1}{D}\left(2 \beta M_{0} D_{\beta x}-Q_{0} B_{\beta x}\right)                              (5.24)

or

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

M_{0} = 742in.-lb∕in.

and

\sigma = 4450 psi.