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Engineering · Mechanics of Materials

Hoop Stress Calculator

Find the hoop and longitudinal stress in a thin-walled pressure vessel from pressure, radius, and wall thickness.

Enter the bore as
MPa
Gauge pressure inside the vessel.
mm
Bore radius of the vessel.
mm
Thickness of the vessel wall.
Try a scenario
Hoop (circumferential) stress (σ_hoop)
100MPaThin wall

Thin-wall assumption holds: r/t = 50 (≳ 10). Longitudinal stress is half the hoop stress.

50
Longitudinal stress σ_long (MPa)
50
r ÷ t ratio (≳ 10 for thin wall)

Hoop (circumferential) stress in a thin-walled pressure vessel is σ_hoop = p·r ÷ t, and longitudinal stress is σ_long = p·r ÷ 2t. For p = 2 MPa, r = 500 mm, t = 10 mm, σ_hoop = 2×500 ÷ 10 = 100 MPa and σ_long = 50 MPa. Hoop stress is always twice the longitudinal stress.

Why hoop stress governs pressure-vessel design

Internal pressure pushes outward on every part of a cylinder’s wall. That outward push creates two distinct membrane stresses. The hoop stress (σ_hoop), also called circumferential stress, acts around the circumference and tries to burst the cylinder open along its length. The longitudinal stress (σ_long) acts along the axis and tries to pull the end caps off. For a cylinder the hoop stress is exactly twice the longitudinal stress, so it is the hoop direction that sets the limit on how much pressure the vessel can hold.

This is why a sausage splits along its length before its ends blow off, and why a burst water pipe tears open in a long axial seam rather than snapping in two. The wall is working twice as hard in the hoop direction, so failure follows the line of highest stress — a crack running along the pipe, perpendicular to the hoop stress that drives it.

σ_hoop = p·r ÷ t  ·  σ_long = p·r ÷ 2t

p = internal gauge pressure, r = inside radius (d = 2r), t = wall thickness. With p in MPa and r, t in mm, both stresses come out in MPa. Valid for thin walls where r/t ≳ 10.

Worked example

A cylindrical steel vessel holds gas at 2 MPa gauge pressure. Its inside radius is 500 mm and the wall is 10 mm thick. Find the hoop and longitudinal stress, and confirm the thin-wall assumption is valid.

  1. 1
    Gather p, r, and t in consistent units. Use p = 2 MPa, r = 500 mm, t = 10 mm. Keeping pressure in MPa and lengths in mm makes the mm units cancel, so both stresses come out directly in MPa.
  2. 2
    If you have the diameter, halve it. The radius is what enters the formula: r = d ÷ 2. A 1000 mm inside diameter gives r = 500 mm.
  3. 3
    Compute the hoop stress. σ_hoop = p·r ÷ t = 2 × 500 ÷ 10 = 100 MPa. This is the circumferential stress that tries to split the cylinder along its length.
  4. 4
    Compute the longitudinal stress. σ_long = p·r ÷ 2t = 2 × 500 ÷ (2 × 10) = 50 MPa — exactly half the hoop stress.
  5. 5
    Check the thin-wall ratio. r ÷ t = 500 ÷ 10 = 50, which is comfortably above 10, so the thin-wall (membrane) formulas apply and the stress is essentially uniform through the wall.

Hoop vs longitudinal stress

The two membrane stresses in a thin-walled cylinder, plus the ratio that tells you when these formulas are valid.

QuantityFormulaNotes
Hoop (circumferential) stressσ_hoop = p·r ÷ t = p·d ÷ 2tLargest stress; acts around the circumference and drives axial (lengthwise) cracks
Longitudinal (axial) stressσ_long = p·r ÷ 2tExactly half the hoop stress; acts along the axis
Stress ratioσ_hoop ÷ σ_long = 2Hoop stress is always twice the longitudinal stress in a cylinder
Thin-wall validityr ÷ t ≳ 10Membrane formulas assume a thin wall; below ~10 use thick-wall (Lamé) equations
Spherical vesselσ = p·r ÷ 2tA sphere carries equal stress in all directions — half a cylinder’s hoop stress

When the thin-wall assumption breaks down

These formulas treat the wall as a thin membrane in which the stress is uniform from the inside surface to the outside surface. That approximation is accurate while the radius-to-thickness ratio r/t is roughly 10 or more. Below that the wall is “thick”: the stress varies significantly across it, peaking on the inside surface, and the simple membrane values understate the true maximum. For thick-walled cylinders you need the Lamé equations, which give the radial and hoop stress as functions of position through the wall.

A few other assumptions are worth remembering. The pressure must be gauge pressure — the internal pressure above the surrounding atmosphere, since that difference is what the wall resists. The formulas describe the membrane stress well away from discontinuities; near end caps, nozzles, welds, and supports, local stress concentrations can push the real stress well above these values, which is exactly where real vessels tend to crack. Always compare the hoop stress against the material strength with an appropriate factor of safety before trusting a design.

Why is hoop stress twice the longitudinal stress?
It comes from the geometry of how pressure is resisted. Splitting a cylinder lengthwise, the hoop stress carries pressure over a rectangle of area 2r×L against a wall area of 2t×L. Splitting it across, the longitudinal stress carries pressure over the circular end (πr²) against a wall ring of area 2πr×t. Working the force balances through leaves σ_hoop = p·r ÷ t and σ_long = p·r ÷ 2t, so the hoop value is exactly double.
Why do pipes and pressure vessels split along their length?
Because the hoop stress is twice the longitudinal stress, the wall is most highly stressed in the circumferential direction. Cracks open perpendicular to the largest tensile stress, so failure runs along the axis — a long lengthwise seam — rather than cutting the pipe in two across its diameter.
When is the thin-wall assumption valid?
When the inside radius is at least about ten times the wall thickness, i.e. r/t ≳ 10. Then the stress is nearly uniform through the wall and the membrane formulas are accurate. Below that the wall is “thick”, the stress varies across it and peaks on the inner surface, and you should switch to the thick-wall Lamé equations.
Should I use gauge or absolute pressure?
Use gauge pressure — the internal pressure above the surrounding atmosphere. The wall only resists the difference between inside and outside, so the pressure driving the stress is p_internal minus p_atmospheric. That difference is exactly what gauge pressure measures.
Can I enter the diameter instead of the radius?
Yes. Toggle the input to diameter and the calculator halves it, since r = d ÷ 2. Written in terms of diameter the hoop stress is σ_hoop = p·d ÷ 2t, which is the same formula as p·r ÷ t. Just be sure it is the inside diameter.
How is a spherical vessel different from a cylinder?
A sphere is symmetric in every direction, so it carries the same membrane stress all around: σ = p·r ÷ 2t. That is half the hoop stress of a cylinder of the same radius and thickness, which is why spheres are the most material-efficient shape for high-pressure storage.
What do I do with the hoop stress once I have it?
Compare it with the material’s strength using a factor of safety — the hoop stress, being the larger of the two, is the value that must stay safely below the allowable stress. If it is too high, increase the wall thickness or reduce the radius or pressure, since σ_hoop scales with p·r ÷ t.