Hoop Stress in a Thin-Walled Cylinder

Slice a pressurised pipe lengthwise and the pressure acting on the projected area d × L must be held by two wall strips of area t × L: p·d·L = 2·σ·t·L, so σ = pd/2t. Edme Mariotte worked this out in the 1680s while sorting out burst pipes in the waterworks feeding the fountains at Versailles. A 1 m diameter vessel at 2 MPa with a 10 mm wall carries σ = 2 × 10⁶ × 1 ÷ (2 × 0.01) = 100 MPa. In field units, 150 psi in a 24-inch line needs t = 150 × 24 ÷ (2 × 15 000) = 0.12 inch of wall before any corrosion or mill-tolerance allowance.

Hoop stress is twice the longitudinal stress, which is why an overpressured pipe splits along its length rather than snapping in two, and why a boiled sausage always bursts lengthwise. The 1954 de Havilland Comet crashes made the point at altitude: repeated pressurisation cycles drove fatigue cracks from the corners of the fuselage cutouts, where the hoop stress concentrated far above the nominal value — the investigation rewrote how the world thinks about fatigue and stress concentration. Two limits on the formula: it is a thin-wall result, valid when d/t is greater than about 20, and it uses gauge pressure, the difference across the wall.