Thin-walled pressure vessels

Why it matters

A pressurised fuselage is a thin-walled pressure vessel cycled once every flight, and so are rocket propellant tanks, oxygen bottles, hydraulic accumulators and engine casings. Thin-wall theory gives the membrane stresses in a few lines, sets the skin thickness for pressurisation, and explains why longitudinal joints and cracks are the most dangerous.

Key ideas

When is a vessel thin? When the wall thickness t is small compared with the radius, usually t/r below about 1/10 (some texts use d/t above 20). Then the stress can be taken as uniform through the wall and the radial stress (between p on the inside and 0 on the outside) is negligible compared with the membrane stresses. Thicker vessels need Lamé's thick-cylinder theory.

Hoop (circumferential) stress in a cylinder. Cut the cylinder along a diametral plane over length L. The pressure force on the projected area, p·d·L, is resisted by two wall strips of area t·L: σh = p·d/(2t) = p·r/t.

Longitudinal (axial) stress in a closed cylinder. Cut across the axis. The pressure on the end, p·πd²/4, is resisted by the ring of wall π·d·t: σl = p·d/(4t) = p·r/(2t). The hoop stress is twice the longitudinal stress. That is why a pressurised tube splits along its length, and why longitudinal seams need the higher joint efficiency.

Sphere. Any diametral cut gives p·πd²/4 = σ·π·d·t, so σ = p·d/(4t) = p·r/(2t) in every tangential direction. For the same p, r and t a sphere has half the cylinder's hoop stress, so it needs half the thickness. Spheres are the lightest pressure vessels, which is why many spacecraft tanks are spherical.

State of stress. The wall element is in biaxial tension (σh, σl) with no shear on those planes, so these are the principal stresses. Radial stress σ₃ ≈ 0 at the outer surface (−p at the inner surface). The maximum in-plane shear is (σh − σl)/2 = p·r/(4t), but the absolute maximum shear is σh/2 = p·r/(2t), acting out of plane.

Joint efficiency. Riveted or welded seams are weaker than the plate. With efficiency η, the hoop stress at a longitudinal seam is p·r/(t·η_l) and the longitudinal stress at a circumferential seam is p·r/(2t·η_c).

Strains and volume change. With Hooke's law for biaxial stress, εh = (σh − ν·σl)/E and εl = (σl − ν·σh)/E. The change in volume of a cylinder is ΔV/V = 2εh + εl; for a sphere it is 3εh.

Limits. Membrane theory ignores bending near end closures, cut-outs, frames and joints, where local stresses can be much higher, and it does not cover buckling under external pressure.

Formulas

σh = p·d / (2·t) = p·r / t Hoop stress (Pa) in a thin cylinder; p = internal gauge pressure (Pa), d = internal (or mean) diameter (m), r = radius (m), t = thickness (m).

σl = p·d / (4·t) = p·r / (2·t) Longitudinal stress in a closed thin cylinder (Pa).

σ = p·d / (4·t) = p·r / (2·t) Membrane stress in a thin sphere, all tangential directions (Pa).

εh = (σh − ν·σl) / E, εl = (σl − ν·σh) / E Hoop and longitudinal strains; E in Pa, ν dimensionless.

ΔV / V = 2·εh + εl = p·d·(5 − 4·ν) / (4·t·E) Volumetric strain of a thin closed cylinder.

ΔV / V = 3·εh = 3·p·d·(1 − ν) / (4·t·E) Volumetric strain of a thin sphere.

τmax,abs = p·r / (2·t) Absolute maximum shear stress in a thin cylinder wall (outer surface).

Worked examples

Example 1 (standard). A transport fuselage of radius 1.8 m has a skin 1.2 mm thick and a cabin pressure differential of 60 kPa. Find the hoop and longitudinal stresses and the maximum shear stress (membrane theory).

1. σh = p·r/t = 60 000 × 1.8/0.0012 = 90 × 10⁶ Pa = 90 MPa.
2. σl = p·r/(2t) = 45 MPa.
3. In-plane maximum shear = (90 − 45)/2 = 22.5 MPa.
4. Radial stress ≈ 0, so the absolute maximum shear = 90/2 = 45 MPa.

Answer: σh = 90 MPa, σl = 45 MPa, τmax (absolute) = 45 MPa. In a real fuselage, frames and stringers share the load and pressure cycling makes fatigue of longitudinal skin joints critical.

Example 2 (GATE level). A closed thin steel cylinder of internal diameter 1 m, length 3 m and wall 10 mm holds a gas at 2 MPa. E = 200 GPa, ν = 0.3. Find the stresses, the strains and the increase in volume.

1. σh = p·d/(2t) = 2 × 1000/(2 × 10) = 100 MPa; σl = p·d/(4t) = 50 MPa.
2. εh = (σh − ν·σl)/E = (100 − 15)/200 000 = 4.25 × 10⁻⁴.
3. εl = (σl − ν·σh)/E = (50 − 30)/200 000 = 1.0 × 10⁻⁴.
4. ΔV/V = 2εh + εl = 8.5 × 10⁻⁴ + 1.0 × 10⁻⁴ = 9.5 × 10⁻⁴.
5. Check: p·d·(5 − 4ν)/(4tE) = 2 × 1000 × 3.8/(4 × 10 × 200 000) = 9.5 × 10⁻⁴.
6. V = π/4 × 1² × 3 = 2.356 m³, so ΔV = 9.5 × 10⁻⁴ × 2.356 = 2.24 × 10⁻³ m³.

Answer: σh = 100 MPa, σl = 50 MPa; εh = 4.25 × 10⁻⁴, εl = 1.0 × 10⁻⁴; ΔV ≈ 2.24 litres. Diameter grows by εh·d = 0.425 mm and length by εl·L = 0.30 mm.

Common mistakes

• Swapping the formulas: the hoop stress has 2t in the denominator with d (or t with r), the longitudinal stress has 4t with d (or 2t with r).
• Mixing radius and diameter in the same formula.
• Using the sphere formula pr/t; for a sphere it is pr/(2t).
• Forgetting the Poisson term in the strains, or taking ΔV/V = εh + εl instead of 2εh + εl.
• Quoting the in-plane maximum shear as the absolute maximum.
• Applying thin-wall formulas to thick vessels or near end closures.

For GATE AE

Expect numericals on hoop and longitudinal stress, thickness required for a given allowable stress or joint efficiency, comparison of cylinder and sphere, change in diameter, length and volume, and fuselage-style pressurisation problems. Conceptual questions test why longitudinal cracks are critical and the absolute maximum shear stress. Practise keeping p in MPa and dimensions in mm so stresses come out directly in MPa.

Quick check

1. What is the ratio of hoop to longitudinal stress in a thin closed cylinder?
2. For the same p, r and t, how does the sphere's stress compare with the cylinder's hoop stress?
3. A cylinder has p = 1 MPa, d = 500 mm, t = 5 mm. What is σh?
4. Write ΔV/V for a thin cylinder in terms of εh and εl.

Answers: 1. 2 : 1. 2. It is half. 3. 50 MPa. 4. 2εh + εl.