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## Q. 5.12

A pipe at 10°F is partly filled with liquid at 40°F and gas at 250°F as shown in Fig. 5.22 a. What is the maximum thermal stress if α = 6.5 × 10$^{-6}$ in./in.°F , E = 30 × 10$^6$ psi, and μ = 0.3 ?

## Verified Solution

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

$(\alpha)\left(\Delta T_{1}\right)(r)-\frac{H_{0}}{2 \beta^{3} D}+\frac{M_{0}}{2 \beta^{2} D}=(\alpha)\left(\Delta T_{2}\right)(r)+\frac{H_{0}}{2 \beta^{3} D}+\frac{M_{0}}{2 \beta^{2} D}$

from which

\begin{aligned}H_{0} &=(\alpha)\left(\Delta T_{1}-\Delta T_{2}\right)(r)\left(\beta^{3}\right)(D) \\&=\left(6.5 \times 10^{-6}\right)(240-30)(6)(1.4843)^{3}(5366) \\&=144 \text { lb./in. }\end{aligned}

$M_{0}$ can be obtained from the second compatibility equation whereby the slope in Fig. 5.22 a at the interface is equal to the slope in (b). Or

$\frac{M_{0}}{\beta D}-\frac{H_{0}}{2 \beta^{2} D}=\frac{M_{0}}{\beta D}+\frac{H_{0}}{2 \beta^{2} D}$

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)

$N_{\theta}=\frac{E t}{r} \cdot \frac{H_{0}}{2 \beta^{3} D} C_{\beta x}$

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

 Table 5.1 Values of Functions $A_{\beta x}, B_{\beta x}, C_{\beta x}, D_{\beta x}$ $\beta _{x}$ $A_{\beta x}$ $B_{\beta x}$ $C_{\beta x}$ $D_{\beta x}$ 0 1.0000 1.0000 1.0000 1.0000 0.05 0.9976 0.9025 0.9500 0.0475 0.10 0.9907 0.8100 0.9003 0.0903 0.15 0.9797 0.7224 0.8510 0.1286 0.20 0.9651 0.6398 0.8024 0.1627 0.30 0.9267 0.4888 0.7077 0.2189 0.40 0.8784 0.3564 0.6174 0.2610 0.50 0.8231 0.2415 0.5323 0.2908 0.55 0.7934 0.1903 0.4919 0.3016 0.60 0.7628 0.1431 0.4530 0.3099 0.80 0.6354 -0.0093 0.3131 0.3223 1.00 0.5083 -0.1108 0.1988 0.3096 1.20 0.3899 -0.1716 0.1091 0.2807 1.40 0.2849 -0.2011 0.0419 0.2430 1.60 0.1959 -0.2077 -0.0059 0.2018 1.80 0.1234 -0.1985 -0.0376 0.1610 2.00 0.0667 -0.1794 -0.0563 0.1231 2.50 -0.0166 -0.1149 -0.0658 0.0491 3.00 -0.0423 -0.0563 -0.0493 0.0070 3.5 -0.0389 -0.0177 -0.0283 -0.0106 4.0 -0.0258 0.0019 -0.0120 -0.0139 5.0 -0.0045 0.0084 0.0019 -0.0065 6.0 0.0017 0.0031 0.0024 -0.0007 7.0 0.0013 0.0001 0.0007 0.0006

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

Also

$M_{x}=0$

and

$\operatorname{Max} \sigma=\frac{2565}{0.125}=20,500 psi$

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

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

Stress due to bending moment at $\beta x=\pi / 4$ is

$\sigma_{x}=\frac{6 M}{t^{2}}=\frac{6}{t^{2}}\left(\frac{0.322 H_{0}}{\beta}\right)=12,000 psi$

Deflection due to $H_{0}$ at $\beta x=\pi / 4$ is obtained from Eqs. 5.23 and 5.24 as

\begin{aligned}&A_{\beta x}=e^{-\beta x}(\cos \beta x+\sin \beta x) \\&B_{\beta x}=e^{-\beta x}(\cos \beta x-\sin \beta x) \\&C_{\beta x}=e^{-\beta x} \cos \beta x \\&D_{\beta x}=e^{-\beta x} \sin \beta x\end{aligned}                            (5.23)

$w=\frac{0.322 H_{0}}{2 \beta^{3} D}$

Hence,

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

and circumferential stress is

$\sigma_{\theta}=\frac{827}{0.125}+0.3 \times 12,000=10,200 \text { psi }$

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