Thin cylinders under internal pressure

Why it matters

Boilers, air receivers, gas cylinders, pipelines, hydraulic cylinders and tanks all hold fluid under pressure. Their walls are thin compared with their diameter, which allows a simple and accurate analysis. A wrong wall thickness either wastes steel or causes a burst, so these formulas, together with joint efficiency and code allowances, are the first step in pressure-vessel design.

Key ideas

Thin cylinder. A cylindrical shell is treated as thin when the wall thickness t is small compared with the diameter, commonly t ≤ d/20 (some texts use d/10). The stresses can then be taken as uniform through the thickness, and the radial stress (from p at the inner face to zero at the outer face) is small compared with the other two and is neglected. Thicker walls need Lamé's thick-cylinder equations.

Hoop (circumferential) stress. Cut the cylinder along its length by a diametral plane. The pressure acting on the projected area d·L is resisted by the two wall strips 2·t·L: p d L = σh · 2 t L, so σh = p d / (2t). This stress tends to split the cylinder along its length, so it acts on the longitudinal joints.

Longitudinal stress. Cut across the cylinder. The pressure on the end area πd²/4 is resisted by the ring of wall πd·t: σl = p d / (4t). It acts on the circumferential joints.

So σh = 2 σl: hoop stress governs, and a cylinder bursts lengthwise. These are principal stresses (no shear on these planes). The maximum in-plane shear is (σh − σl)/2 = p d/(8t); taking the radial stress as zero at the outer surface, the absolute maximum shear is σh/2 = p d/(4t).

Joint efficiency. Riveted or welded joints are weaker than the plate. With efficiency η, the stress at a joint is the plate stress divided by η. Use the longitudinal-joint efficiency ηl with σh and the circumferential-joint efficiency ηc with σl. Values come from the pressure-vessel code.

Thin sphere. By symmetry every direction in the wall is like the longitudinal direction of a cylinder: σ = p d/(4t). For the same p, d and allowable stress, a sphere needs half the thickness of a cylinder.

Strains and volume change. With biaxial stress and Poisson's effect:

• hoop strain εh = (σh − ν σl)/E (equals the strain in diameter);
• longitudinal strain εl = (σl − ν σh)/E;
• volumetric strain εv = 2εh + εl, giving the extra fluid that must be pumped in to reach pressure (plus fluid compressibility, if asked).

Design. Choose t so that the larger stress, divided by the right joint efficiency, does not exceed the allowable stress; add a corrosion allowance as the code specifies.

Formulas

σh = p d / (2t) (hoop stress, cylinder) σl = p d / (4t) (longitudinal stress, cylinder)

• p = internal gauge pressure (N/mm² = MPa), d = internal diameter (mm), t = wall thickness (mm).

σh = p d / (2 t ηl), σl = p d / (4 t ηc) (with joint efficiencies)

τmax,in-plane = p d / (8t); τabs,max ≈ p d / (4t) (radial stress taken as zero)

σ = p d / (4t) (thin sphere, all directions)

εh = (σh − ν σl)/E, εl = (σl − ν σh)/E

εv = 2εh + εl = (p d / (4 t E)) (5 − 4ν) (cylinder); εv = 3 (p d / (4 t E)) (1 − ν) (sphere)

δd = εh d, δL = εl L, δV = εv V

Worked examples

Example 1 (standard). A boiler shell of 1.5 m internal diameter and 15 mm thick carries steam at 1.2 MPa. Find the hoop and longitudinal stresses in the plate, the absolute maximum shear stress, and the stresses at the joints if ηl = 0.8 and ηc = 0.5.

1. Check thin: t/d = 15/1500 = 1/100 ✓.
2. σh = 1.2 × 1500 / (2 × 15) = 60 MPa.
3. σl = 1.2 × 1500 / (4 × 15) = 30 MPa.
4. τabs,max ≈ σh/2 = 30 MPa (in-plane maximum (60 − 30)/2 = 15 MPa).
5. At the longitudinal joint: 60 / 0.8 = 75 MPa; at the circumferential joint: 30 / 0.5 = 60 MPa.

Answer: σh = 60 MPa, σl = 30 MPa, τmax ≈ 30 MPa; 75 MPa at the longitudinal joint and 60 MPa at the circumferential joint.

Example 2 (GATE level, volume change). A closed steel cylinder 1 m internal diameter, 3 m long and 10 mm thick is pressurised to 1.5 MPa. E = 200 GPa, ν = 0.3. Find the changes in diameter, length and volume.

1. σh = 1.5 × 1000 / 20 = 75 MPa; σl = 37.5 MPa.
2. εh = (75 − 0.3 × 37.5)/200 000 = 63.75/200 000 = 3.1875 × 10⁻⁴.
3. εl = (37.5 − 0.3 × 75)/200 000 = 15/200 000 = 7.5 × 10⁻⁵.
4. δd = 3.1875 × 10⁻⁴ × 1000 = 0.319 mm; δL = 7.5 × 10⁻⁵ × 3000 = 0.225 mm.
5. εv = 2 × 3.1875 × 10⁻⁴ + 7.5 × 10⁻⁵ = 7.125 × 10⁻⁴. Check: (1.5 × 1000 / (4 × 10 × 200 000)) × (5 − 1.2) = 1.875 × 10⁻⁴ × 3.8 = 7.125 × 10⁻⁴ ✓.
6. V = π/4 × 1² × 3 = 2.356 m³; δV = 7.125 × 10⁻⁴ × 2.356 = 1.679 × 10⁻³ m³ = 1.68 litres.

Answer: δd ≈ 0.319 mm, δL ≈ 0.225 mm, δV ≈ 1.68 L.

Common mistakes

• Swapping the formulas: hoop is pd/2t (larger), longitudinal pd/4t.
• Using the outer diameter or radius instead of the internal diameter.
• Applying the efficiency of the longitudinal joint to the longitudinal stress; it goes with the hoop stress.
• Forgetting Poisson's effect in strain and volume calculations.
• Using thin-cylinder formulas when t is more than about d/20.
• Taking the in-plane shear (pd/8t) as the maximum shear for Tresca design.

For GATE PI

Expect hoop and longitudinal stresses, the ratio 2:1, wall thickness for a given allowable stress and joint efficiency, thin spheres, changes in diameter and volume, and maximum shear stress. Practise the volumetric strain formula (pd/4tE)(5 − 4ν) and the sphere version.

Quick check

1. p = 2 MPa, d = 1 m, t = 10 mm: hoop stress?
2. Ratio of hoop to longitudinal stress in a thin cylinder?
3. Stress in a thin sphere with p = 2 MPa, d = 1 m, t = 10 mm?
4. Which joint does the hoop stress act on?

Answers: 1. 100 MPa. 2. 2. 3. 50 MPa. 4. The longitudinal joint.