= Solution
Assume a thin cylindrical wall, uniform internal pressure and closed ends carrying the pressure load. The radial <stress> is small compared with the membrane stresses when $t\ll r$.
For an end cap, the outward pressure <force> is $P\pi r^2$. The wall's axial tensile <force> is $\sigma_z(2\pi rt)$. Equilibrium therefore gives
$$
\boxed{\sigma_z=\frac{Pr}{2t}.}
$$
For a longitudinally split half-cylinder of length $L$, pressure has resultant $P(2rL)$ across the projected rectangular area. Two cut wall sections each carry $\sigma_\theta tL$, so
$$
\boxed{\sigma_\theta=\frac{Pr}{t}=2\sigma_z.}
$$
Thus \b[<hoop stress> is twice the closed-end axial <stress>] in this model. An open-ended tube without an external end load does not acquire the same pressure-induced <axial membrane stress>.
\Image[/past-exam-of-the-materials-science-course-of-the-university-of-cambridge/2018/paper-1-pipe-fracture.png]
{title=Original thin-pipe force balances and schematic crack-tip stress regularization by plastic deformation; the plastic-zone curve is qualitative}
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