Question 5.16: A pressure vessel operating at 300 °F is subjected to a shor...

A pressure vessel operating at 300 °F is subjected to a short excursion temperature of 600 °F. At a given time, the temperature distribution in the wall is shown in Figure 5.27. Find the maximum thermal stress at that instance. Let \mu=0.3, E=30 \times 10^{6} psi, and \alpha=6.0 \times 10^{-6} in./in.- °F.

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This problem can be visualized as a biaxial case where the inside surface heats quickly while the rest of the wall remains at 300 °F. Using Eq. (5.29b) results in

\sigma_{x}=\sigma_{y}=-\frac{\alpha \Delta T E}{1-\mu}  for a biaxial case.                          (5.29b)

\sigma=\frac{\left(6 \times 10^{-6}\right)(600-300)\left(30 \times 10^{6}\right)}{1-0.3}

= − 77,100 psi,

which is extremely high and is based on very limiting assumptions. A more realistic approach is that based on Eq. (5.46). The mean temperature is obtained from Figure 5.27 and tabulated as follows:

\sigma_{\theta}=\frac{E \alpha}{1-\mu}\left(T_{ m }-T\right)                            (5.46)

Locations as ratios of thickness

Temperature (°F)

Area

0

600  
0.1 460

53.0

0.2

400 43.0
0.3 370

38.5

0.4

340 35.5
0.5 320

33.0

0.6

310 31.5
0.7 305

30.8

0.8

300 30.3
0.9 300

30.0

1.0

300 30.0
   

\sum 355.6

And T_{ m } \approx 356^{\circ} F

From Eq. (5.46), at the inside surface,

\sigma_{\theta}=\frac{E \alpha}{1-\mu}\left(T_{ m }-T\right)                  (5.46)

\sigma=\frac{\left(30 \times 10^{6}\right)\left(6.0 \times 10^{-6}\right)}{1-0.3}(356-600)

= −62,700 psi,

and at the outside surface

\sigma=\frac{\left(30 \times 10^{6}\right)\left(6.0 \times 10^{-6}\right)}{1-0.3}(356-300)

= 14,400 psi.

It is of interest to note that the high stress occurs at the surface only.Thus at one-tenth of the thickness inside the surface, the stress is

\sigma=\frac{\left(30 \times 10^{6}\right)\left(6 \times 10^{-6}\right)}{1-0.3}(356-460)

= −26,700 psi.

The high stress at the inside surface indicates that local yielding will occur.

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