Thin shell: stresses and dimension changes

Data: $d = 1000$ mm, $L = 5000$ mm, $t = 20$ mm, $p = 5\ \text{N/mm}^2$, $E = 200\ \text{GPa} = 2\times10^5\ \text{N/mm}^2$, $\nu = 0.3$.

Stresses

$$\sigma_c = \frac{pd}{2t} = \frac{5(1000)}{2(20)} = 125.0\ \text{N/mm}^2, \qquad \sigma_l = \frac{pd}{4t} = 62.5\ \text{N/mm}^2$$ $$\tau_{max} = \frac{\sigma_c-\sigma_l}{2} = 31.25\ \text{N/mm}^2$$

(maximum shear stress in the plane of the wall; it equals half the longitudinal stress).

Strains

$$\varepsilon_c = \frac{125.0-0.3(62.5)}{2\times10^5} = 5.3125\times10^{-4}, \qquad \varepsilon_l = \frac{62.5-0.3(125.0)}{2\times10^5} = 1.2500\times10^{-4}$$ $$\varepsilon_v = 2\varepsilon_c + \varepsilon_l = 1.1875\times10^{-3}$$

Changes in dimensions

$$\delta d = \varepsilon_c d = 0.5312\ \text{mm}, \qquad \delta L = \varepsilon_l L = 0.6250\ \text{mm}$$ $$V = \frac{\pi}{4}(1000)^2(5000) = 3.9270\times10^{9}\ \text{mm}^3, \qquad \delta V = \varepsilon_v V = 4.6633\times10^{6}\ \text{mm}^3$$

Answer: $\sigma_c = 125.0$, $\sigma_l = 62.5$, $\tau_{max} = 31.25$ N/mm²; $\delta d = 0.531$ mm, $\delta L = 0.625$ mm, $\delta V = 4.663\times10^6$ mm³ ($\approx 4663$ cm³).