Thin & Thick Cylinders Notes

Last reviewed 16 Sept 2026 · 6 min read

Thin and thick cylinders

Boilers, water mains, penstocks, gas cylinders and storage tanks are pressure vessels. A cylinder is treated as thin when its wall thickness $t$ is small compared with its diameter — commonly $t < d /20$ (some books use $d / t > 20$ ). The stresses are then taken as uniform through the thickness and radial stress is neglected. Otherwise it is a thick cylinder, analysed by Lamé's theory.

Stresses in a thin cylinder

Internal pressure $p$ , internal diameter $d$ , thickness $t$ .

Hoop (circumferential) stress

Cut the cylinder along its length. Bursting force on a length $L$ $= p d L$ ; resisting force of the two walls $= 2 σ_{c} t L$ .

σ_{c} = \frac{p d}{2 t}

Longitudinal stress

Cut across the cylinder. Bursting force on the end $= p π d^{2} /4$ ; resisting area $= π d t$ .

σ_{l} = \frac{p d}{4 t}

FormulaThin cylinder

σ_{c} = \frac{p d}{2 t}, σ_{l} = \frac{p d}{4 t} = \frac{σ _{c}}{2}, τ_{ma x (in - pl an e)} = \frac{σ _{c} - σ _{l}}{2} = \frac{p d}{8 t}

Hoop stress is twice the longitudinal stress, so a thin cylinder bursts along a longitudinal seam. Considering the third (radial) principal stress ≈ 0, the absolute maximum shear stress is $σ_{c} /2 = p d /4 t$ .

Joint efficiency

Riveted or welded seams are weaker than the plate. With longitudinal joint efficiency $η_{l}$ and circumferential joint efficiency $η_{c}$ :

σ_{c} = \frac{p d}{2 t η _{l}}, σ_{l} = \frac{p d}{4 t η _{c}}

Note the pairing: the longitudinal joint resists the hoop stress; the circumferential joint resists the longitudinal stress.

Thin sphere

By symmetry the stress is the same in every direction in the wall:

σ = \frac{p d}{4 t}

For the same diameter, pressure and allowable stress, a sphere needs half the thickness of a cylinder — spheres are used for high-pressure gas storage.

Changes in dimensions of a thin cylinder

With $E$ and Poisson's ratio $μ$ :

ε_{c} = \frac{σ _{c} - μ σ _{l}}{E} = \frac{p d}{4 tE} (2 - μ), ε_{l} = \frac{σ _{l} - μ σ _{c}}{E} = \frac{p d}{4 tE} (1 - 2 μ)

FormulaChanges in size

Change in diameter: $δ d = ε_{c} d = \frac{p d ^{2}}{4 tE} (2 - μ)$
Change in length: $δ L = ε_{l} L = \frac{p d L}{4 tE} (1 - 2 μ)$
Volumetric strain (cylinder): $ε_{v} = 2 ε_{c} + ε_{l} = \frac{p d}{4 tE} (5 - 4 μ)$
Volumetric strain (sphere): $ε_{v} = 3 ε = \frac{3 p d}{4 tE} (1 - μ)$

The extra volume of fluid that can be pumped in at pressure $p$ = increase in vessel volume + compression of the fluid ( $p V / K$ ).

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Thin & Thick Cylinders