How this instrument works
Cut a pressurized cylinder in half lengthwise, the way you would slice a hot dog, and look at what holds the two halves together against the pressure trying to force them apart. The internal pressure pushes outward across the cylinder's full projected width, 2r, over whatever length L you consider, for a total force of P·2r·L. Only the wall resists that force, split across two cut strips of area t·L apiece, so 2·σ·t·L = P·2r·L. The length terms cancel, the 2s cancel each other, and what remains is σ_hoop = Pr ⁄ t — the stress stretching the wall circumferentially, at right angles to the cylinder's axis.
This is strictly a thin-wall result, valid once the wall is slender enough that loading can be treated as uniform through its thickness — a rule of thumb used across pressure vessel codes caps wall thickness at roughly a tenth of the inner radius. Thicker walls, like a gun barrel or a deep submersible's hull, need the Lamé equations instead, which let the value vary from a peak at the bore down to a lower one at the outer surface. Cutting the same cylinder the other way, straight across rather than along its length, gives the longitudinal stress carried by a closed end: σ_long = Pr ⁄ 2t, exactly half the hoop value — which is why an overpressurized pipe or tank tends to split lengthwise rather than pop off an end cap.
An engineer sizing the wall of a propane tank, a scuba cylinder, or a run of high-pressure pipeline reaches for this formula first, because the hoop figure is the largest load the wall sees in normal service and so is what governs minimum thickness. Codes such as the ASME Boiler and Pressure Vessel Code build on it directly, adding a joint efficiency factor for welded seams and a corrosion allowance on top of whatever bare thickness this formula returns. The recurring mistake is reading a nameplate diameter as the radius, or an outer diameter as the inner one — since the result scales linearly with r, a slip like that silently misstates the answer by a factor of two.
- Enter the operating pressure into Internal pressure — gauge pressure above atmospheric, in kPa, MPa, or psi.
- Enter Inner radius, the cylinder's bore measured to the inside surface of the wall, not the outer diameter.
- Enter Wall thickness, the radial thickness of material actually resisting the pressure, in mm or cm.
- Read Hoop stress, then compare it against your material's allowable limit with whatever safety factor your design code requires.
Worked example — a 5 MPa vessel with a 200 mm bore
Take a steel pressure vessel holding 5 MPa of internal gauge pressure — about 725 psi, a fairly typical air-receiver or small process-vessel duty — with an inner radius of 200 mm and a wall 10 mm thick. Feed 5 MPa into Internal pressure, 200 mm into Inner radius, and 10 mm into Wall thickness. The instrument converts everything to consistent SI units first: P = 5,000,000 Pa, r = 0.2 m, t = 0.01 m. Hoop stress then comes out as σ_hoop = (5,000,000 × 0.2) ⁄ 0.01 = 100,000,000 Pa, a clean 100 MPa.
That figure is worth sitting with. Against a mild steel yield strength near 275 MPa, it leaves a factor of safety of about 2.75, ordinary territory for a static vessel. The same tank's longitudinal stress, carried by its dished ends, is exactly half that: 50 MPa. Double the radius alone, to a 400 mm bore at the same pressure and thickness, and hoop stress doubles again to 200 MPa — the reason large-diameter pipelines need proportionally thicker walls to hold the same working pressure as a narrow one.
Questions
Why does the formula use radius instead of diameter?
Because the derivation balances forces across a lengthwise cut, where pressure acts over the full width 2r — the radius appears naturally and the 2 cancels out of the result. Written with diameter instead it becomes σ_hoop = PD ⁄ 2t, an identical answer; just don't mix the two forms, or you will accidentally halve or double what you get.
How is hoop stress different from the longitudinal component in the same vessel?
Hoop stress acts circumferentially, around the cylinder; the longitudinal component acts along its axis, carried by the end caps of a closed vessel, and is exactly half that value: σ_long = Pr ⁄ 2t. Because the circumferential figure is always the larger of the two, it usually governs wall thickness, and it is why an overpressurized pipe typically splits lengthwise rather than blowing off an end cap.
When does the thin-wall assumption behind this formula stop applying?
Once wall thickness exceeds roughly a tenth of the inner radius, the loading starts varying meaningfully between the bore and the outer surface, and this uniform-stress formula understates the true peak at the inner wall. Thick-walled cases — gun barrels, high-pressure hydraulic cylinders, deep-submergence hulls — need the Lamé equations instead, which describe that internal variation directly.
Does this formula already include a safety margin?
No — it returns the raw figure the material physically carries, with no factor of safety, corrosion allowance, or weld joint efficiency folded in. Design codes such as the ASME Boiler and Pressure Vessel Code apply those separately: they divide material strength by a safety factor to get an allowable limit, then require wall thickness that keeps the calculated hoop stress below it.
Can I use this for a spherical tank instead of a cylindrical one?
No — a sphere carries the same load in every direction across its wall, expressed as σ = Pr ⁄ 2t, exactly half of a cylinder's hoop stress at the same pressure, radius, and thickness. That is one reason spherical pressure vessels, though harder to fabricate, use less material than cylindrical ones for the same volume and pressure rating.
What happens to hoop stress if the pressure doubles?
It doubles exactly, since hoop stress is directly proportional to pressure with radius and thickness held fixed — 5 MPa producing 100 MPa becomes 200 MPa at 10 MPa. That linearity is also why pressure-relief valves and burst-disc ratings matter so much: a modest overpressure event translates one-for-one into wall stress, with no cushioning from the formula itself.