Volumetric strain of thin cylindrical vessel

To show: $\varepsilon_v = 2\varepsilon_c + \varepsilon_l$, where $\varepsilon_c$ is the circumferential (hoop) strain and $\varepsilon_l$ the longitudinal strain.

Consider a thin cylinder of internal diameter $d$ and length $L$. Its volume is

$$V = \frac{\pi}{4}d^2L$$

Derivation

Under internal pressure the diameter changes to $d + \delta d$ and the length to $L + \delta L$. The strains are

$$\varepsilon_c = \frac{\pi(d+\delta d) - \pi d}{\pi d} = \frac{\delta d}{d}, \qquad \varepsilon_l = \frac{\delta L}{L}$$

(circumferential strain equals diametral strain, since circumference $= \pi d$). The new volume is

$$V' = \frac{\pi}{4}(d+\delta d)^2(L+\delta L) = V\left(1+\varepsilon_c\right)^2\left(1+\varepsilon_l\right)$$

Expand and neglect products of small strains:

$$V' = V\left(1 + 2\varepsilon_c + \varepsilon_l + \varepsilon_c^2 + 2\varepsilon_c\varepsilon_l + \varepsilon_c^2\varepsilon_l\right) \approx V\left(1 + 2\varepsilon_c + \varepsilon_l\right)$$ $$\varepsilon_v = \frac{V'-V}{V} = 2\varepsilon_c + \varepsilon_l$$

Alternative (differentiation): from $V = \frac{\pi}{4}d^2L$,

$$\frac{dV}{V} = 2\frac{dd}{d} + \frac{dL}{L} = 2\varepsilon_c + \varepsilon_l$$

In terms of pressure

With $\sigma_c = \dfrac{pd}{2t}$ and $\sigma_l = \dfrac{pd}{4t}$:

$$\varepsilon_c = \frac{pd}{2tE}\left(1-\frac{\nu}{2}\right), \qquad \varepsilon_l = \frac{pd}{2tE}\left(\frac{1}{2}-\nu\right)$$ $$\varepsilon_v = 2\varepsilon_c + \varepsilon_l = \frac{pd}{2tE}\left(\frac{5}{2} - 2\nu\right) = \frac{pd}{4tE}\left(5-4\nu\right)$$