Fuselage Hoop Stress⚠ unverified

Understanding the Parameters

• Internal gauge pressure $p$ — the difference between cabin and ambient pressure, not the absolute cabin pressure. At cruise the cabin holds roughly an $8{,}000$-ft altitude ($\sim 75$ kPa) while outside is near $24{,}000$ ft ($\sim 37$ kPa), giving a differential of about $0.5$–$0.6\,\text{bar}$ ($\sim 55$ kPa). Stress scales linearly with $p$.

• Radius $r$ — the hoop stress grows with radius: a wide-body fuselage carries far more skin tension than a narrow-body at the same pressure and thickness. This is why large fuselages need thicker skins or more frames, and why pressure vessels are made as slender as mission allows.

• Skin thickness $t$ — in the denominator, so doubling the skin halves the stress. The thin-wall assumption ($r/t \gtrsim 10$; fuselages run $r/t \sim 100$–$500$) is what makes the simple formula valid — the stress is essentially uniform through the thin skin.

• The output $\sigma_{hoop}$ — the circumferential tension trying to split the fuselage along a longitudinal seam (which is why longitudinal lap joints are the fatigue-critical detail). It is compared to the material allowable via a margin of safety and, crucially, drives fatigue and damage-tolerance analysis of the pressurised shell.

Derivation (Approaching a Proof)

Cut the cylinder with a longitudinal plane through its axis and consider the free body of one half of a length $L$ of shell. The internal pressure acts on the projected rectangular area $2rL$ (diameter $\times$ length), producing an outward force

$$F_p = p\,(2r\,L).$$

This is resisted by the hoop tension in the two cut edges of the skin, each of area $t\,L$, carrying stress $\sigma_{hoop}$:

$$F_\sigma = 2\,(\sigma_{hoop}\,t\,L).$$

Equilibrium $F_p = F_\sigma$ gives

$$p\,(2rL) = 2\,\sigma_{hoop}\,t\,L \quad\Longrightarrow\quad \sigma_{hoop} = \frac{p\,r}{t}. \qquad\blacksquare$$

The length $L$ cancels — hoop stress is independent of how long a slice you take, as it must be. The longitudinal stress comes from a transverse cut and works out to half this value, $\sigma_{long} = pr/(2t)$, because the pressure there acts on a circle $\pi r^2$ resisted by a hoop-shaped ring $2\pi r t$. Hence $\sigma_{hoop} = 2\,\sigma_{long}$ — the fundamental 2:1 ratio for a closed cylinder.

Dimensional check. $$\left[\frac{p\,r}{t}\right] = \frac{(\text{Pa})(\text{m})}{\text{m}} = \text{Pa}.\ \checkmark$$

History and Development

• Thin-walled pressure-vessel theory. The hoop/longitudinal formulas are classical results of shell theory, standard since the 19th century and identical to those used for boilers and tanks (Thin-Wall Hoop Stress, Pressure Vessel Design). The 2:1 ratio explains why cylindrical vessels split lengthwise, not around.

• The Comet lesson. The de Havilland Comet disasters of 1954 — fuselages failing from fatigue cracks at stress concentrations near pressurised window/hatch corners — were a watershed. They established that hoop-stress fatigue and fracture mechanics, not just static strength, govern pressurised fuselage design, and gave rise to the damage-tolerance philosophy now central to aircraft certification.

• Fail-safe design. Modern fuselages use crack-stopping frames ("tear straps"), bonded/riveted lap joints designed against the hoop tension, and inspection intervals set by crack-growth analysis under the cyclic hoop stress of each pressurisation. The humble $pr/t$ is the starting point for all of it.

Related Concepts: Fuselage Longitudinal Stress, Thin-Wall Hoop Stress, Thin-Wall Longitudinal Stress, Spherical Vessel Stress, Burst Pressure, Pressure Vessel Design, Margin Of Safety

Notes: Registry calculator fuselage-hoop-stress (unverified). Thin-wall hoop stress $pr/t$ — correct as shipped; duplicates the Mechanical Thin-Wall Hoop Stress. $p$ is the cabin–ambient differential. $\sigma_{hoop} = 2\,\sigma_{long}$. Fatigue (not static strength) usually governs — the Comet legacy. Returns $0$ if $t \le 0$. All defaults $1.0$.