Thin cylinders and spheres under internal pressure

Why it matters

Air-brake reservoirs, LPG and CNG cylinders, fuel tanks, hydraulic accumulators, tyres and radiator tanks are all thin-walled shells holding pressure. Their wall thickness is set by two simple stress formulas, and the same formulas explain why cylinders split lengthwise, why spheres need only half the wall thickness and how much extra fluid a vessel takes up when it swells under pressure.

Key ideas

What "thin" means. A shell is thin when its wall thickness t is small compared with its diameter, commonly t < d/20 (equivalently t < r/10). Then the stress through the thickness can be taken as uniform, and the radial stress (equal to −p at the inside, zero outside) is small compared with the membrane stresses and is neglected.

Hoop (circumferential) stress. Cut the cylinder along its length through a diameter. The pressure force on the projected area, p·d·L, is resisted by two wall strips of area t·L: σ_h = p·d/(2t). This stress tends to split the cylinder along a longitudinal line, which is why burst pipes and frozen tubes crack lengthwise.

Longitudinal stress. Cut across the cylinder. The pressure on the end, p·(π/4)·d², is resisted by the ring of wall π·d·t: σ_l = p·d/(4t), half the hoop stress. It acts on circumferential joints.

Joint efficiency. Riveted or welded seams are weaker than the plate. With efficiency η, the effective stresses become σ_h = p·d/(2t·η_l) (longitudinal seam) and σ_l = p·d/(4t·η_c) (circumferential seam). Allowable stresses, joint efficiencies and corrosion allowances come from the applicable pressure-vessel code.

Sphere. By symmetry every diametral cut is the same: σ = p·d/(4t) in all directions in the wall. For the same pressure, diameter and allowable stress, a sphere needs half the wall thickness of a cylinder, which is why high-pressure gas storage favours spheres (though they are harder to make and package).

Strains and volume change. The wall is in biaxial stress, so use the generalised Hooke's law:

• Cylinder: ε_h = (σ_h − ν·σ_l)/E = p·d·(2 − ν)/(4tE); ε_l = p·d·(1 − 2ν)/(4tE); volumetric strain ε_v = 2ε_h + ε_l = p·d·(5 − 4ν)/(4tE).
• Sphere: ε = p·d·(1 − ν)/(4tE) in every direction; ε_v = 3ε. The volume change matters for how much extra fluid must be pumped in to reach a test pressure (together with the fluid's own compressibility).

Cylinder with hemispherical ends. If both parts have the same thickness, the hemisphere expands less than the cylinder, causing bending at the junction. Equal hoop strains require t_sphere/t_cylinder = (1 − ν)/(2 − ν), but then check that the thinner head's stress stays within the allowable.

Maximum shear stress. In a cylinder wall, in-plane τ_max = (σ_h − σ_l)/2 = p·d/(8t). Including the near-zero radial stress, the absolute maximum is σ_h/2 = p·d/(4t), which is what Tresca uses.

Limits. These formulas ignore end-closure bending, openings (nozzles need reinforcement) and thick-wall effects; for t > d/20 use Lamé's thick-cylinder equations.

Formulas

σ_h = p·d / (2·t), σ_l = p·d / (4·t)

• σ_h, σ_l: hoop and longitudinal stresses (Pa); p: internal gauge pressure (Pa); d: internal diameter (m); t: wall thickness (m). Thin cylinder, t < d/20.

σ_h = p·d / (2·t·η_l), σ_l = p·d / (4·t·η_c)

• η_l, η_c: efficiencies of longitudinal and circumferential joints.

σ = p·d / (4·t)

• Thin sphere, all directions.

ε_h = p·d·(2 − ν) / (4·t·E), ε_l = p·d·(1 − 2ν) / (4·t·E), δV/V = p·d·(5 − 4ν) / (4·t·E)

• Cylinder strains; ν: Poisson's ratio; E: Young's modulus.

δV/V = 3·p·d·(1 − ν) / (4·t·E)

• Sphere volumetric strain.

t_sphere / t_cylinder = (1 − ν) / (2 − ν)

• Hemispherical ends with no differential hoop strain.

τ_max,in-plane = p·d / (8·t), τ_abs,max = p·d / (4·t) (cylinder)

Worked examples

Example 1 (standard): air-brake reservoir. Given: a cylindrical air reservoir, d = 400 mm, length 1.2 m, t = 4 mm, p = 1.0 MPa, E = 200 GPa, ν = 0.3. Longitudinal seam efficiency 0.8, circumferential seam efficiency 0.5. Find the stresses at the seams, and the changes in diameter, length and volume (use plate stresses for strains).

1. Plate stresses: σ_h = p·d/(2t) = 1 × 400/8 = 50 MPa; σ_l = p·d/(4t) = 25 MPa.
2. At seams: σ_h = 50/0.8 = 62.5 MPa; σ_l = 25/0.5 = 50 MPa.
3. ε_h = (50 − 0.3 × 25)/200 000 = 2.125 × 10⁻⁴, so Δd = 2.125 × 10⁻⁴ × 400 = 0.085 mm.
4. ε_l = (25 − 0.3 × 50)/200 000 = 5.0 × 10⁻⁵, so ΔL = 5.0 × 10⁻⁵ × 1200 = 0.060 mm.
5. ε_v = 2ε_h + ε_l = 4.75 × 10⁻⁴. V = π/4 × 0.4² × 1.2 = 0.1508 m³, so ΔV = 7.16 × 10⁻⁵ m³ = 71.6 cm³.
6. Check with the formula: p·d·(5 − 4ν)/(4tE) = 1 × 400 × 3.8/(16 × 200 000) = 4.75 × 10⁻⁴. Answer: 62.5 MPa and 50 MPa at the seams; Δd = 0.085 mm, ΔL = 0.060 mm, ΔV ≈ 71.6 cm³.

Example 2 (GATE level): cylinder, sphere and end design. Given: a vessel of internal diameter 1.5 m holds 1.5 MPa. Allowable tensile stress 120 MPa, ν = 0.3. Find (a) the cylinder thickness, (b) the thickness of a sphere of the same diameter, (c) the thickness of hemispherical ends for equal hoop strain, and check its stress, (d) the absolute maximum shear in the cylinder wall.

1. (a) t = p·d/(2σ) = 1.5 × 1500/240 = 9.375 mm.
2. (b) t = p·d/(4σ) = 1.5 × 1500/480 = 4.69 mm.
3. (c) t_s = t_c·(1 − ν)/(2 − ν) = 9.375 × 0.7/1.7 = 3.86 mm. Its stress would be p·d/(4t_s) = 2250/15.44 = 145.7 MPa, above 120 MPa, so the heads must be at least 4.69 mm thick; equal strain cannot be achieved here without overstressing the heads.
4. (d) With t = 9.375 mm: σ_h = 120 MPa, σ_l = 60 MPa, σ_r ≈ 0. τ_abs,max = (120 − 0)/2 = 60 MPa (in-plane value only 30 MPa). Answer: (a) 9.38 mm, (b) 4.69 mm, (c) 3.86 mm for equal strain but 4.69 mm needed for strength, (d) 60 MPa.

Common mistakes

• Using the radius instead of the diameter in p·d/(2t), which halves the stress.
• Swapping the joint efficiencies: the longitudinal joint carries the hoop stress.
• Ignoring Poisson's ratio in strain and volume-change calculations.
• Taking volumetric strain as 3ε for a cylinder (that is only for a sphere).
• Using thin-shell formulas for thick walls.
• Quoting in-plane maximum shear as the absolute maximum.

For GATE ME

Expect hoop and longitudinal stresses, ratio of stresses or strains, change in diameter or volume, sphere versus cylinder thickness, required thickness for a given allowable stress, and maximum shear stress in the wall. Practise the volumetric strain formulas and quick ratio checks such as σ_h/σ_l = 2.

Quick check

1. What is σ_h/σ_l for a thin cylinder?
2. Sphere of diameter 2 m, t = 15 mm, p = 3 MPa. Find the wall stress.
3. If p doubles, what happens to the hoop stress?
4. Which seam, longitudinal or circumferential, carries the hoop stress?

Answers: 1. 2. 2. 3 × 2000/(4 × 15) = 100 MPa. 3. It doubles. 4. The longitudinal seam.