Cylindrical shell with spherical ends: volume change

Assumption: the 1 m length is the cylindrical portion; the two spherical (hemispherical) ends have the same diameter, 200 mm, and thickness, 6 mm. Total change in volume = change in the cylindrical part + change in the two ends (which together make one full sphere).

Data: $d = 200$ mm, $t = 6$ mm, $L = 1000$ mm, $p = 1.5\ \text{N/mm}^2$, $E = 2\times10^5\ \text{N/mm}^2$, $\nu = 0.25$.

Cylindrical part

$$\sigma_c = \frac{pd}{2t} = 25.00\ \text{N/mm}^2, \qquad \sigma_l = \frac{pd}{4t} = 12.50\ \text{N/mm}^2$$ $$\varepsilon_c = \frac{25.00 - 0.25(12.50)}{2\times10^5} = 1.0938\times10^{-4}, \qquad \varepsilon_l = \frac{12.50 - 0.25(25.00)}{2\times10^5} = 3.1250\times10^{-5}$$ $$\varepsilon_v = 2\varepsilon_c + \varepsilon_l = 2.5000\times10^{-4}, \quad V_{cyl} = \frac{\pi}{4}(200)^2(1000) = 3.1416\times10^{7}\ \text{mm}^3$$ $$\delta V_{cyl} = 2.5000\times10^{-4}\times3.1416\times10^{7} = 7854.0\ \text{mm}^3$$

Spherical ends (one complete sphere)

$$\sigma = \frac{pd}{4t} = 12.50\ \text{N/mm}^2, \qquad \varepsilon = \frac{\sigma}{E}(1-\nu) = 4.6875\times10^{-5}$$ $$V_s = \frac{\pi d^3}{6} = 4.1888\times10^{6}\ \text{mm}^3, \qquad \delta V_s = 3\varepsilon V_s = 589.0\ \text{mm}^3$$

Total

$$\delta V = 7854.0 + 589.0 = 8443.0\ \text{mm}^3 = 8.44\ \text{cm}^3$$

Answer: increase in volume $\approx 8.44\ \text{cm}^3$.