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Chapter 5

Q. 5.16

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 Fig. 5.27. Find the maximum thermal stress at that instance. Let μ = 0.3, E = 30 × 10^6 psi, and α = 6.0 × 10^{-6} in./in.°F

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 Fig. 5.27. Find the maximum thermal stress at that instance. Let μ = 0.3, E = 30 × 10^6 psi, and α = 6.0 × 10^-6 in./in.°F

Step-by-Step

Verified Solution

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

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

\begin{aligned}\sigma &=\frac{\left(6 \times 10^{-6}\right)(600-300)\left(30 \times 10^{6}\right)}{1-0.3} \\&=-77,100  psi\end{aligned}

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 Fig. 5.27 and tabulated as follows:

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

where      T_{m} = mean value of temperature distribution through the wall

T =  temperature at desired location

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
∑355.6

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

From Eq. 5.46, at inner surface,

\begin{aligned}\sigma &=\frac{\left(30 \times 10^{6}\right)\left(6.0 \times 10^{-6}\right)}{1-0.3}(356-600) \\&=-62,700  psi\end{aligned}

and at outer surface

\begin{aligned}\sigma &=\frac{\left(30 \times 10^{6}\right)\left(6.0 \times 10^{-6}\right)}{1-0.3}(356-300) \\&=14,400  psi\end{aligned}

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

\begin{aligned}\sigma &=\frac{\left(30 \times 10^{6}\right)\left(6 \times 10^{-6}\right)}{1-0.3}(356-460) \\&=-26,700  psi\end{aligned}

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