Thin cylindrical shell: max shear, dimension change

Data: $d = 1000$ mm, $L = 4000$ mm, $t = 12$ mm, $p = 2\ \text{N/mm}^2$, $E = 2\times10^5\ \text{N/mm}^2$, $\nu = 0.3$.

Stresses

$$\sigma_c = \frac{pd}{2t} = \frac{2(1000)}{2(12)} = 83.33\ \text{N/mm}^2, \qquad \sigma_l = \frac{pd}{4t} = 41.67\ \text{N/mm}^2$$

Maximum shear stress

The principal stresses are $\sigma_c$ and $\sigma_l$ (the radial stress is neglected), so

$$\tau_{max} = \frac{\sigma_c - \sigma_l}{2} = \frac{83.33 - 41.67}{2} = 20.83\ \text{N/mm}^2$$

(this is the maximum in the plane of the wall; it equals $\sigma_l/2$ and $\sigma_c/4$).

Change in dimensions

$$\varepsilon_c = \frac{\sigma_c - \nu\sigma_l}{E} = \frac{83.33 - 0.3(41.67)}{2\times10^5} = 3.542\times10^{-4}$$ $$\varepsilon_l = \frac{\sigma_l - \nu\sigma_c}{E} = \frac{41.67 - 0.3(83.33)}{2\times10^5} = 8.333\times10^{-5}$$ $$\delta d = \varepsilon_c\,d = 0.3542\ \text{mm}, \qquad \delta L = \varepsilon_l\,L = 0.3333\ \text{mm}$$

Volumetric strain $\varepsilon_v = 2\varepsilon_c + \varepsilon_l = 7.917\times10^{-4}$, with $V = \frac{\pi}{4}d^2L = 3.1416\times10^{9}\ \text{mm}^3$:

$$\delta V = \varepsilon_v V = 2.4871\times10^{6}\ \text{mm}^3 \;(2487.1\ \text{cm}^3)$$

Answer: $\tau_{max} = 20.83$ N/mm²; $\delta d = 0.354$ mm (increase), $\delta L = 0.333$ mm (increase), $\delta V = 2487$ cm³.