Steel spherical vessel: circumferential stress and diameter change

Data: radius $r = 1000$ mm, so $d = 2000$ mm; $t = 10$ mm, $p = 1\ \text{MPa}$, $E = 200\ \text{GPa} = 2\times10^5\ \text{MPa}$, $\nu = 0.25$.

Stress

In a thin sphere the stress in the wall is the same in every direction (tangential stress). Balancing the pressure force on a diametral section, $p\frac{\pi}{4}d^2$, with the wall force $\sigma\pi d t$:

$$\sigma = \frac{pd}{4t} = \frac{1(2000)}{4(10)} = 50.0\ \text{MPa}$$

Change in diameter

$$\varepsilon = \frac{\sigma}{E} - \nu\frac{\sigma}{E} = \frac{\sigma}{E}(1-\nu) = \frac{50.0(0.75)}{2\times10^5} = 1.875\times10^{-4}$$ $$\delta d = \varepsilon\,d = 1.875\times10^{-4}\times2000 = 0.3750\ \text{mm}$$

Answer: circumferential stress $= 50.0$ MPa; change in diameter $= 0.375$ mm (increase).