Max shear stress and volumetric strain in thin cylinder

Maximum shear stress is half of the longitudinal stress

In a thin cylinder of diameter $d$, thickness $t$ and pressure $p$, the two principal stresses are the hoop stress and the longitudinal stress (the radial stress is neglected):

$$\sigma_1 = \sigma_c = \frac{pd}{2t}, \qquad \sigma_2 = \sigma_l = \frac{pd}{4t}$$

The maximum shear stress (in the plane of the wall) is half the difference of the principal stresses:

$$\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \frac{1}{2}\left(\frac{pd}{2t} - \frac{pd}{4t}\right) = \frac{pd}{8t}$$

Since $\sigma_l = \dfrac{pd}{4t}$:

$$\tau_{max} = \frac{pd}{8t} = \frac{1}{2}\,\sigma_l$$

Hence the maximum shear stress is half of the longitudinal stress (and one quarter of the hoop stress). It acts on planes at $45^\circ$ to the axis.

Volumetric strain

Volume $V = \dfrac{\pi}{4}d^2L$. Taking the differential,

$$\frac{\delta V}{V} = 2\frac{\delta d}{d} + \frac{\delta L}{L} = 2\varepsilon_c + \varepsilon_l$$

The strains (including Poisson's effect) are

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

so the change in volume is $\delta V = \dfrac{pd}{4tE}(5-4\nu)\,V$.