Spherical Vessel Stress⚠ unverified
Mechanical / Pressure Vessels · Compute the membrane stress in a thin-walled spherical vessel
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| P | P | Pa | 1.0 | Internal pressure |
| r | r | m | 1.0 | Sphere radius |
| t | t | m | 1.0 | Wall thickness |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| result | σ | Pa | Membrane stress, in pascals (Pa). Returns 0.0 when ``t`` is not positive |
The science & history
Understanding the Parameters
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Internal pressure $P$ and radius $r$ — as for a cylinder, the stress scales linearly with both; larger spheres need proportionally thicker walls.
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Uniform biaxial stress — unlike a cylinder (hoop $\ne$ longitudinal), a sphere has the same membrane stress $Pr/2t$ in every direction by symmetry. There is no "hoop vs longitudinal" distinction.
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Efficiency — the sphere's stress equals the cylinder's longitudinal stress and is half its hoop stress, so a sphere needs about half the wall thickness of a cylinder of equal radius and pressure. Its drawback is fabrication cost and poor volumetric packing.
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Same form as $\sigma_l$ — numerically identical to Thin-Wall Longitudinal Stress ($Pr/2t$), but the physics differs: here it is the equal biaxial stress of a sphere, not the axial stress of a cylinder.
Derivation (Approaching a Proof)
Cut the sphere through any great circle (a plane through its centre), isolating a hemisphere. The internal pressure acts on the projected circular area $\pi r^2$, giving a force pushing the two halves apart:
$$F_P = P\,(\pi r^2).$$
This is resisted by the membrane stress acting on the circular wall cross-section exposed by the cut, area $\approx 2\pi r\,t$:
$$F_\sigma = \sigma\,(2\pi r t).$$
Force balance $P\pi r^2 = \sigma\,2\pi r t$, cancelling $\pi r$:
$$\sigma = \frac{P r}{2 t}.$$
By symmetry any great-circle cut gives the same result, confirming the stress is equal in all directions — the hallmark of the sphere's efficiency.
Dimensional check. $\sigma = \dfrac{P r}{2 t} = \dfrac{\text{Pa}\cdot\text{m}}{\text{m}} = \text{Pa}$ — a stress, as required.
History and Development
The sphere's efficiency as a pressure container has been exploited since early gas storage and is enshrined in the ASME code's separate (thinner) rules for spherical shells and hemispherical heads (Head Thickness Hemispherical). Spherical LNG and LPG storage tanks, gas-cylinder domed ends, and deep-sea/space pressure hulls all exploit the uniform biaxial stress state. The trade-off — spheres are harder and costlier to fabricate than cylinders — is why most vessels are cylinders with spherical or ellipsoidal heads.
Related Concepts: Thin-Wall Hoop Stress, Thin-Wall Longitudinal Stress, Head Thickness Hemispherical, Pressure Vessel Design, Burst Pressure
Notes: Uniform in all directions (symmetry). Half the cylinder hoop stress → ~half the wall thickness = most material-efficient shape. Numerically equals Thin-Wall Longitudinal Stress ($Pr/2t$) but different physics. Thin-wall ($t/r \lesssim 0.1$).