Fuselage Longitudinal Stress⚠ unverified

Understanding the Parameters

• Internal gauge pressure $p$ — the cabin–ambient differential, the same driver as for hoop stress. The end pressure acting over the fuselage cross-section produces the axial pull.

• Radius $r$ — as with hoop stress, bigger fuselages carry more axial tension for the same pressure and thickness.

• Skin thickness $t$ — the factor of 2 in the denominator is the whole story: the axial stress is only half the hoop stress, because the geometry that resists it is more favourable. So a skin sized by the hoop stress automatically has generous margin against the longitudinal stress in pure pressurisation.

• The output $\sigma_{long}$ — the lengthwise membrane tension. In a real fuselage it superposes with the much larger axial stresses from fuselage bending (the whole airframe acting as a beam under tail loads, landing, and gusts), so the total axial stress in a skin is pressure plus bending — the pressure part being this $pr/2t$. Circumferential (butt) joints and frames carry it.

Derivation (Approaching a Proof)

Cut the cylinder with a transverse plane perpendicular to its axis and take the free body of the capped end. Internal pressure acts on the full circular cross-section of area $\pi r^2$, producing an axial force

$$F_p = p\,(\pi r^2).$$

This is resisted by the longitudinal tension in the skin, which forms a thin ring of circumference $2\pi r$ and thickness $t$, i.e. area $2\pi r\,t$, carrying stress $\sigma_{long}$:

$$F_\sigma = \sigma_{long}\,(2\pi r\,t).$$

Equilibrium $F_p = F_\sigma$ gives

$$p\,\pi r^2 = \sigma_{long}\,2\pi r\,t \quad\Longrightarrow\quad \sigma_{long} = \frac{p\,r}{2 t}. \qquad\blacksquare$$

Comparing with the hoop result $\sigma_{hoop} = pr/t$ shows the 2:1 ratio: the circular end cap ($\pi r^2$) is resisted by a full ring ($2\pi r t$), whereas the longitudinal split ($2rL$) is resisted by only two edges ($2tL$) — the geometry makes the hoop direction the more highly stressed. This is the universal reason cylindrical vessels rupture along a longitudinal seam.

Dimensional check. $$\left[\frac{p\,r}{2 t}\right] = \frac{(\text{Pa})(\text{m})}{\text{m}} = \text{Pa}.\ \checkmark$$

History and Development

• Membrane theory of shells. The hoop and longitudinal stresses are the two principal membrane stresses of a thin cylindrical shell under internal pressure, a classical result underpinning boiler, tank, and airframe design (Thin-Wall Longitudinal Stress, Pressure Vessel Design). Bending stresses appear only near discontinuities (bulkheads, cutouts), where local shell-bending analysis is added.

• Superposition in a real fuselage. A fuselage is simultaneously a pressure vessel and a beam. The longitudinal skin stress is the pressure term $pr/2t$ superposed on the axial stress from global fuselage bending — the latter often dominant near the tail and over the wing. Stress reports carry both.

• Joint orientation. Because hoop stress is twice longitudinal, longitudinal lap joints (which carry hoop load) are the fatigue-critical ones, while circumferential joints (carrying the lighter longitudinal load) are comparatively benign — a direct design consequence of the 2:1 ratio.

Related Concepts: Fuselage Hoop Stress, Thin-Wall Longitudinal Stress, Thin-Wall Hoop Stress, Spherical Vessel Stress, Burst Pressure, Pressure Vessel Design, Wing Bending Stress

Notes: Registry calculator fuselage-longitudinal-stress (unverified). Thin-wall axial stress $pr/2t$ — correct as shipped; duplicates the Mechanical Thin-Wall Longitudinal Stress. Exactly half the hoop stress ($\sigma_{hoop} = 2\sigma_{long}$). Superposes with global fuselage-bending axial stress in service. Returns $0$ if $t \le 0$. All defaults $1.0$.