Question 5.12: A pipe at 10 °F is partly filled with liquid at 40 °F and ga...

A solution can be obtained by taking a free-body diagram at the gas–liquid boundary as shown in Figure 5.22b. Compatibility at the interface requires that the deflection in (1) equals the deflection in (2). Hence, 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)

= (\alpha)\left(\Delta T_{2}\right)(r)+\frac{H_{0}}{2 \beta^{3} D}+\frac{M_{0}}{2 \beta^{2} D}

from which

= \left(6.5 \times 10^{-6}\right)(240-30)(6)(1.4843)^{3}(5366)

= 144 lb /in.

M_{0} can be obtained from the second compatibility equation, whereby the slope in (1) at the interface is equal to the slope in (2):

and M_{0} =0

The circumferential force in the pipe due to H_{0} is obtained from Eqs. (5.19) and (5.24):

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

The maximum value of C_{\beta x} is obtained from Table 5.1 as 1.0. Hence,

N_{\theta} = 2565 lb/ in. at interface.

Also,

And

The maximum bending moment due to H_{0} was derived in Example 5.4 as

M_{x}=\frac{0.322 H_{0}}{\beta}     at     \beta x=\frac{\pi}{4}

The stress due to the bending moment at \beta x=\pi / 4 is

The deflection due to H_{0} at \beta x=\pi / 4 is obtained from Eqs. (5.23) and (5.24) as

D_{\beta x}= e ^{-\beta x} \sin \beta x                            (5.23)

Hence,

N_{\theta} = 827lb/in.,

and the circumferential stress is

\sigma_{\theta}=\frac{827}{0.125} 0.3 × 12,000 = 10,200 psi.

Thus, the maximum stress occurs at the interface with magnitude of 20,500 psi.