Thin and thick cylinders

Why it matters

Pneumatic and hydraulic systems are everywhere in automation: air receivers, actuator barrels, accumulators, hoses, manifolds and pressure-sensor housings all contain fluid under pressure. Sizing their walls correctly is a safety issue, since a burst vessel releases stored energy violently. Thin-cylinder theory sizes low-pressure air equipment quickly; thick-cylinder (Lamé) theory is needed for high-pressure hydraulic barrels, where the stress varies strongly through the wall.

Key ideas

Thin or thick? A cylinder is treated as thin when its wall thickness t is small compared with its diameter, commonly t < d/20 (some texts use d/t > 20 or t < r/10). Then the hoop stress is nearly uniform through the wall and the radial stress (equal to −p at the inside, zero at the outside) is small compared with the hoop stress and is neglected. Otherwise use thick-cylinder theory.

Thin cylinder stresses (closed ends, internal pressure p).

• Hoop (circumferential) stress σ_h = p·d/(2t), from equilibrium of half the cylinder cut along its length: pressure force p·d·L is resisted by 2·σ_h·t·L.
• Longitudinal stress σ_l = p·d/(4t), from equilibrium across a transverse cut: p·πd²/4 resisted by σ_l·π·d·t.
• So σ_h = 2·σ_l. The longitudinal seam (which carries hoop stress) is the critical joint, and a cylinder bursts by splitting lengthwise.
• Maximum in-plane shear (σ_h − σ_l)/2 = p·d/(8t); absolute maximum shear, using radial stress ≈ 0, is σ_h/2 = p·d/(4t).

Joint efficiency. Riveted or welded seams are weaker than the plate. With efficiency η_l for the longitudinal seam and η_c for the circumferential seam: σ_h = p·d/(2t·η_l), σ_l = p·d/(4t·η_c). Values of η come from the relevant pressure-vessel code.

Thin sphere. By symmetry the stress is the same in every direction: σ = p·d/(4t). A sphere needs only half the wall thickness of a cylinder of the same diameter and material, which is why gas tanks are often spherical or have hemispherical ends.

Strains and volume change. The wall is in biaxial stress, so use generalised Hooke's law: ε_h = (σ_h − ν·σ_l)/E and ε_l = (σ_l − ν·σ_h)/E. The volumetric strain of a cylinder is ε_v = 2ε_h + ε_l = (p·d/(4t·E))·(5 − 4ν); for a sphere ε_v = 3ε_h = (3p·d/(4t·E))·(1 − ν). This gives the extra fluid needed to pressurise a vessel and affects the stiffness of hydraulic systems.

Thick cylinders (Lamé's theory). Assumptions: homogeneous, isotropic, linear-elastic material; plane transverse sections remain plane (longitudinal strain uniform). Radial equilibrium of a ring element plus compatibility give σ_r = A − B/r² and σ_h = A + B/r², where A and B are constants from the boundary conditions σ_r = −p_i at r = a (inner radius) and σ_r = −p_o at r = b (outer radius). Note that σ_h − σ_r = 2B/r² and σ_h + σ_r = 2A, constant through the wall.

Internal pressure only. A = p·a²/(b² − a²) and B = p·a²·b²/(b² − a²). The hoop stress is greatest at the inner surface, σ_h,max = p·(b² + a²)/(b² − a²), and falls towards the outside; the radial stress is −p at the inside and zero outside. The longitudinal stress for closed ends is A, uniform. Increasing the wall thickness has diminishing returns: even an infinitely thick wall has σ_h,max = p at the bore. That is why high-pressure cylinders are compounded (shrink-fitted) or autofrettaged to put the bore into residual compression.

Formulas

σ_h = p·d / (2t), σ_l = p·d / (4t)

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

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

• η_l, η_c: efficiencies of longitudinal and circumferential joints (dimensionless, from the applicable code).

σ = p·d / (4t)

• Thin sphere, all directions.

ε_h = (σ_h − ν·σ_l)/E, ε_l = (σ_l − ν·σ_h)/E, ε_v = 2ε_h + ε_l

• E: Young's modulus (Pa); ν: Poisson's ratio. ΔV = ε_v·V.

σ_r = A − B/r², σ_h = A + B/r²

• Lamé equations; r: radius (m), a ≤ r ≤ b; A (Pa) and B (N) from boundary conditions.

σ_h,max = p·(b² + a²) / (b² − a²) at r = a

• Thick cylinder under internal pressure p only; a, b: inner and outer radii (m).

Worked examples

Example 1 (standard): compressed-air receiver. Given: a closed cylindrical air receiver, internal diameter 600 mm, length 1.5 m, wall 6 mm, working pressure 1.2 MPa; E = 200 GPa, ν = 0.3. Find the stresses and the increase in volume (treat as thin; d/t = 100).

1. σ_h = p·d/(2t) = 1.2 × 600/12 = 60 MPa.
2. σ_l = p·d/(4t) = 30 MPa.
3. ε_h = (60 − 0.3 × 30)/200 000 = 2.55 × 10⁻⁴.
4. ε_l = (30 − 0.3 × 60)/200 000 = 6.0 × 10⁻⁵.
5. ε_v = 2 × 2.55 × 10⁻⁴ + 6.0 × 10⁻⁵ = 5.70 × 10⁻⁴.
6. V = π/4 × 0.6² × 1.5 = 0.4241 m³, so ΔV = 5.70 × 10⁻⁴ × 0.4241 = 2.42 × 10⁻⁴ m³. Answer: σ_h = 60 MPa, σ_l = 30 MPa, volume increase ≈ 0.24 litre. Compare with the allowable stress for the plate, reduced by the joint efficiency from your pressure-vessel code.

Example 2 (GATE level): hydraulic cylinder barrel. Given: a hydraulic barrel with bore 80 mm and outside diameter 120 mm works at 20 MPa, closed ends. Find the hoop and radial stresses at the inner and outer surfaces, the longitudinal stress and the maximum shear stress, and compare with thin-cylinder theory.

1. a = 40 mm, b = 60 mm, b² − a² = 3600 − 1600 = 2000 mm².
2. A = p·a²/(b² − a²) = 20 × 1600/2000 = 16 MPa. B = A·b² = 16 × 3600 = 57 600 N.
3. Inner surface: σ_h = 16 + 57 600/1600 = 52 MPa; σ_r = 16 − 36 = −20 MPa (equals −p, a check).
4. Outer surface: σ_h = 16 + 57 600/3600 = 32 MPa; σ_r = 0 (a check).
5. Longitudinal stress = A = 16 MPa.
6. Maximum shear, at the bore: (σ_h − σ_r)/2 = (52 + 20)/2 = 36 MPa.
7. Thin theory with the bore diameter: 20 × 80/(2 × 20) = 40 MPa, which underestimates the true 52 MPa by 23 %. With the mean diameter, 20 × 100/40 = 50 MPa, closer but still low. Answer: σ_h = 52 MPa (bore) to 32 MPa (outside); σ_r = −20 MPa to 0; σ_l = 16 MPa; τ_max = 36 MPa at the bore. For d/t = 4, thin theory is unsafe.

Common mistakes

• Swapping the hoop and longitudinal formulas (hoop has 2t, longitudinal 4t).
• Using radius in place of diameter in p·d/(2t), which halves the stress.
• Using thin theory for thick walls; check d/t first.
• Forgetting joint efficiency, or applying η_l to the longitudinal stress (η_l belongs with hoop stress).
• Computing volumetric strain as ε_h + ε_l, or ignoring the Poisson terms.
• Using the radii in mm and pressures in MPa, then giving B in the wrong units; keep a consistent set.
• Forgetting that radial stress is compressive (negative) under internal pressure.

For GATE ME

Expect hoop and longitudinal stress in thin cylinders and spheres, ratios of stresses and strains, change in diameter, length and volume, joint efficiency, and Lamé stresses at the inner and outer surfaces of thick cylinders (including the maximum hoop stress formula). Practise volumetric-strain derivations and the boundary-condition step for A and B.

Quick check

1. p = 2 MPa, d = 1 m, t = 10 mm. Find the hoop stress.
2. What is the ratio of hoop to longitudinal stress in a thin closed cylinder?
3. Thin sphere, p = 1.5 MPa, d = 0.8 m, t = 8 mm. Find the stress.
4. Where is the hoop stress maximum in a thick cylinder under internal pressure?
5. What is the radial stress at the outer surface of a thick cylinder with no external pressure?

Answers: 1. 100 MPa. 2. 2:1. 3. 1.5 × 800/32 = 37.5 MPa. 4. At the inner surface. 5. Zero.