Cylindrical shell with joint efficiencies: change in dimensions

Data: external diameter 260 mm, $t = 5$ mm, so internal diameter $d = 260 - 2(5) = 250$ mm; $L = 2500$ mm, $p = 1.6\ \text{MPa}$, $E = 2\times10^5\ \text{MPa}$, $\nu = 0.3$, longitudinal joint efficiency $\eta_l = 0.85$, circumferential joint efficiency $\eta_c = 0.65$.

Stresses

In the solid plate (away from the joints):

$$\sigma_c = \frac{pd}{2t} = \frac{1.6(250)}{2(5)} = 40.0\ \text{MPa}, \qquad \sigma_l = \frac{pd}{4t} = 20.0\ \text{MPa}$$

At the joints (for strength, the efficiency reduces the effective resisting area):

$$\sigma_{c,joint} = \frac{pd}{2t\eta_l} = \frac{40.0}{0.85} = 47.06\ \text{MPa}, \qquad \sigma_{l,joint} = \frac{pd}{4t\eta_c} = \frac{20.0}{0.65} = 30.77\ \text{MPa}$$

Changes in dimensions

The joint efficiencies affect only the local stress at the riveted or welded seams. The overall deformation of the shell is governed by the stresses in the solid plate, so the efficiencies are not used in the strain calculation.

$$\varepsilon_c = \frac{40.0-0.3(20.0)}{2\times10^5} = 1.700\times10^{-4}, \qquad \varepsilon_l = \frac{20.0-0.3(40.0)}{2\times10^5} = 4.000\times10^{-5}$$ $$\delta d = \varepsilon_c d = 0.04250\ \text{mm}, \qquad \delta L = \varepsilon_l L = 0.10000\ \text{mm}$$ $$\varepsilon_v = 2\varepsilon_c + \varepsilon_l = 3.800\times10^{-4}, \qquad V = \frac{\pi}{4}(250)^2(2500) = 1.2272\times10^{8}\ \text{mm}^3$$ $$\delta V = \varepsilon_v V = 46633.0\ \text{mm}^3 = 46.63\ \text{cm}^3$$

Answer: $\delta d = 0.0425$ mm, $\delta L = 0.1000$ mm, $\delta V = 46.63$ cm³ (all increases).