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.