Membrane Stress in a Thin-Walled Sphere (σ = pr/2t)

Cut a pressurised sphere in half through any great circle and the argument writes itself. The pressure pushes the two halves apart with a force $p\pi r^2$, the area of the flat disc it acts on. The wall resists across a thin ring of circumference $2\pi r$ and thickness $t$, so $\sigma \cdot 2\pi r t = p \pi r^2$ and $\sigma = pr/2t$. The cut could have been made anywhere, in any orientation, which is the whole point: a sphere has no worse direction. A 3 m diameter sphere at 1.5 MPa with a 12 mm wall carries $1.5 \times 10^6 \times 1.5/(2 \times 0.012) = 93.75$ MPa, the same in every direction in the wall.

Set that beside a cylinder of the same radius and wall. The cylinder's longitudinal stress is also $pr/2t$ — but its hoop stress is $pr/t$, twice as much, and it is the hoop stress that governs. So a sphere of the same material and pressure needs half the wall a cylinder needs, and holds about twice the pressure at the same thickness. That is why LPG and cryogenic storage past a certain size goes spherical, and why the ends of a cylindrical vessel are dished rather than flat.

The membrane solution assumes the stress is uniform through the thickness and that there is no bending. Both assumptions expire together once $t$ is more than about a tenth of $r$, at which point the real bore stress is meaningfully higher and the thick-wall solution is needed. Bending also reappears wherever the membrane cannot follow its own shape — at nozzles, at supports, and at the seam where a dished head meets a cylindrical shell. Those discontinuity stresses are local, they are often several times the membrane value, and no version of $pr/2t$ will find them.