Thin cylindrical shell: change in length and diameter

Data (as given): $L = 4000$ mm, $t = 15$ mm, $d = 15$ mm, $p = 25\ \text{N/mm}^2$, $E = 2.05\times10^5\ \text{N/mm}^2$, $\nu = 0.3$.

Stresses

$$\sigma_c = \frac{pd}{2t} = \frac{25(15)}{2(15)} = 12.50\ \text{N/mm}^2, \qquad \sigma_l = \frac{pd}{4t} = 6.25\ \text{N/mm}^2$$

Strains

$$\varepsilon_c = \frac{\sigma_c-\nu\sigma_l}{E} = \frac{12.50-0.3(6.25)}{2.05\times10^5} = 5.1829\times10^{-5}$$ $$\varepsilon_l = \frac{\sigma_l-\nu\sigma_c}{E} = \frac{6.25-0.3(12.50)}{2.05\times10^5} = 1.2195\times10^{-5}$$

Changes

$$\delta d = \varepsilon_c d = 0.00078\ \text{mm}, \qquad \delta L = \varepsilon_l L = 0.0488\ \text{mm}$$

Answer (data as given): $\delta L \approx 0.049\ \text{mm}$ and $\delta d \approx 0.0008\ \text{mm}$ (both increases).

Note: with $t = d$ the shell is not really thin, so the thin-shell formulas are only approximate here. If the internal diameter was meant to be 150 mm, the same method gives $\sigma_c = 125.0$, $\sigma_l = 62.5\ \text{N/mm}^2$, $\delta d = 0.078$ mm and $\delta L = 0.488$ mm.