Thin-walled pressure vessels under internal pressure

Membrane (hoop and longitudinal) stresses in thin cylinders and spheres, the code thickness formulas that follow from them, joint efficiency, strains and volume change, and the limits of thin-wall theory.

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Why it matters

Most process vessels — reactors, drums, columns, storage bullets — are thin-walled cylinders or spheres, and their shell thickness comes straight from the two membrane-stress equations in this topic. Every code thickness formula is a dressed-up version of them, so understanding where they come from tells you which stress governs, which weld seam is critical and why spheres are used for high-pressure gas storage.

Key ideas

Thin-wall assumption. When the wall is thin compared with the radius (commonly t/D below about 1/20, i.e. t below about r/10), the stress can be taken as uniform through the thickness and bending is negligible. The wall then carries the pressure as a stretched membrane. The radial stress varies only from −P at the inner surface to zero at the outer surface, which is small compared with the hoop stress (of order P·r/t), so it is neglected.

Hoop (circumferential) stress. Cut a cylinder of length L along a diametral plane. The pressure force on the projected area, P·D·L, is resisted by two wall strips, 2·σ_h·t·L. So σ_h = P·D/(2t) = P·r/t.

Longitudinal (axial) stress. Cut across the cylinder. The pressure on the end, P·πD²/4, is resisted by the ring of wall, σ_l·πD·t. So σ_l = P·D/(4t) = P·r/(2t), exactly half the hoop stress. That is why hoop stress governs the thickness of a plain cylinder, and why the longitudinal weld seam (which carries hoop stress) is the critical one; a cylinder tends to split along its length.

Sphere. By symmetry every direction is the same, and cutting any great circle gives σ = P·r/(2t). A sphere therefore needs only half the thickness of a cylinder of the same diameter and allowable stress — the reason Horton spheres store LPG and other high-pressure gases. The price is harder fabrication (pressed petals) and poor use of plot area.

Which radius? Using the mean diameter (Di + t) instead of Di in σ_h = P·D/(2t) and rearranging gives the familiar code form t = P·Di/(2fJ − P). The "− P" is a small correction for the difference between inside and mean diameter; it is not an empirical fudge.

Joint efficiency in the formula. A weld is assumed weaker than plate, so the allowable stress is multiplied by J. For the hoop-stress check use J of the longitudinal seams; for the longitudinal-stress check use J of the circumferential seams.

Strain and volume change. Using Hooke's law in two dimensions (radial stress neglected): the hoop strain ε_h = (σ_h − ν·σ_l)/E and the longitudinal strain ε_l = (σ_l − ν·σ_h)/E. The diameter grows by ε_h·D, the length by ε_l·L, and the volume by (ε_l + 2ε_h)·V. These matter for estimating the extra water needed in a hydrotest, and for designing attachments that must move with the shell.

Maximum shear. In the plane of the wall the maximum shear stress is (σ_h − σ_l)/2 = P·r/(4t); considering the radial direction (σ_r ≈ 0), the absolute maximum is σ_h/2 = P·r/(2t).

Limits. When t/r is larger, the stress varies across the wall and Lamé's thick-cylinder equations are used (hoop stress maximum at the bore). Discontinuities (shell-to-head junctions, nozzles, supports) create local bending stresses not captured here; codes handle them with head shapes, reinforcement rules and stress analysis.

Formulas

σ_h = P·D / (2·t) = P·r / t (cylinder, hoop) σ_l = P·D / (4·t) = P·r / (2·t) (cylinder, longitudinal) σ = P·D / (4·t) (sphere, any direction)

  • σ = membrane stress (MPa); P = internal gauge pressure (MPa); D, r = diameter, radius (mm, use mean values for accuracy); t = wall thickness (mm). Valid for t/D below about 1/20, away from discontinuities.

t = P·Di / (2·f·J − P) (cylinder design thickness, excluding corrosion allowance) t = P·Di / (4·f·J − P) (sphere design thickness)

  • f = allowable stress at design temperature (MPa, from code book); J = weld joint efficiency (–); Di = inside diameter (mm).

ε_h = (σ_h − ν·σ_l) / E, ε_l = (σ_l − ν·σ_h) / E ΔV/V = ε_l + 2·ε_h = (P·r / (2·t·E))·(5 − 4ν) (cylinder) ΔV/V = 3·(P·r / (2·t·E))·(1 − ν) (sphere)

  • E = Young's modulus (MPa, about 2 × 10⁵ MPa for steel); ν = Poisson's ratio (about 0.3 for steel).

Worked examples

Example 1 (standard): cylinder versus sphere Given: P = 1.5 MPa, Di = 2000 mm, f = 120 MPa, J = 0.85, CA = 2 mm, plates in even millimetres.

  1. Cylinder: t = P·Di / (2·f·J − P) = 1.5 × 2000 / (204 − 1.5) = 3000 / 202.5 = 14.81 mm; with CA 16.81 mm → 18 mm.
  2. Sphere: t = P·Di / (4·f·J − P) = 3000 / (408 − 1.5) = 3000 / 406.5 = 7.38 mm; with CA 9.38 mm → 10 mm.
  3. The sphere needs about half the pressure thickness, as predicted.

Example 2 (GATE level): volume increase of a pressurised cylinder Given: closed thin cylinder, inside diameter 1.0 m (r = 500 mm), length 3.0 m, t = 10 mm, P = 2 MPa, E = 200 GPa = 2 × 10⁵ MPa, ν = 0.3. Find the increase in volume.

  1. Stresses: σ_h = P·r/t = 2 × 500 / 10 = 100 MPa; σ_l = 50 MPa.
  2. Hoop strain: ε_h = (100 − 0.3 × 50) / (2 × 10⁵) = 85 / (2 × 10⁵) = 4.25 × 10⁻⁴.
  3. Longitudinal strain: ε_l = (50 − 0.3 × 100) / (2 × 10⁵) = 1.0 × 10⁻⁴.
  4. Volumetric strain: ε_v = ε_l + 2·ε_h = 1.0 × 10⁻⁴ + 8.5 × 10⁻⁴ = 9.5 × 10⁻⁴.
  5. Volume: V = π/4 × 1.0² × 3.0 = 2.356 m³.
  6. ΔV = 9.5 × 10⁻⁴ × 2.356 = 2.24 × 10⁻³ m³ = 2.24 litres. Check with the closed form: (P·r/(2·t·E))·(5 − 4ν) = (1000 / (4 × 10⁶)) × 3.8 = 9.5 × 10⁻⁴ ✓.

Common mistakes

  • Writing σ_h = P·D/t (forgetting the 2) or confusing radius and diameter.
  • Thinking longitudinal stress governs because the vessel is long: length does not appear in either membrane stress.
  • Using the sphere formula for a cylinder's hemispherical head without noting it needs only about half the shell thickness.
  • Applying thin-wall formulas when t/r is large; switch to Lamé's equations.
  • Using ε_h = σ_h/E, ignoring the Poisson effect of the other stress.
  • Using absolute rather than gauge pressure.

For GATE CH

Expect direct numericals for hoop and longitudinal stress or required thickness, ratio questions (hoop/longitudinal = 2; cylinder/sphere thickness = 2), and occasionally strain or volume-change calculations using E and ν. Practise the free-body derivations so you can handle a variation, such as a vessel with a different end condition or a sphere.

Quick check

  1. A cylinder with r = 0.4 m and t = 8 mm is at 2 MPa. Find σ_h.
  2. Which weld seam of a cylinder carries the hoop stress?
  3. For the same D, P and f, how does the sphere's thickness compare with the cylinder's?
  4. Why is radial stress neglected in thin-wall theory?

Answers: 1. 2 × 400 / 8 = 100 MPa. 2. The longitudinal seam. 3. About half. 4. It is at most P, far smaller than P·r/t when t ≪ r.

Interview questions

All Process Equipment Design interview questions

Try answering each one aloud before you open it.

  1. 1.What is a thin-walled pressure vessel?Concept

    A vessel whose wall thickness is small compared with its radius, commonly t below about r/10 (t/D below about 1/20). The wall then behaves as a membrane: hoop and longitudinal stresses can be taken as uniform through the thickness, bending is negligible and the radial stress (at most P) is ignored. Most process drums, columns and reactors fall in this range; thick vessels need Lamé's equations.

  2. 2.Explain the significance of hoop stress and longitudinal stress in thin-walled pressure vessels.Concept

    Hoop stress, σ_h = P·r/t, acts around the circumference and tends to split the cylinder along its length; longitudinal stress, σ_l = P·r/(2t), acts along the axis and tends to pull the ends off. Because hoop stress is twice the longitudinal stress, it governs the shell thickness, and the longitudinal weld seams (which carry hoop stress) are the critical joints. In a sphere both stresses equal P·r/(2t).

  3. 3.Why are thin-walled pressure vessels often used in industrial applications?Application

    For the pressures of most process plants, a thin membrane shell carries the load efficiently in pure tension with little material, so it is light and cheap to roll and weld from standard plate. Its stresses follow from simple equilibrium, which makes design by code formula straightforward and reliable. Thick-walled construction is reserved for very high pressures, where the extra cost and analysis are unavoidable.

  4. 4.What happens if the wall thickness of a pressure vessel exceeds the thin-wall assumption?Application

    If the wall thickness exceeds the thin-wall assumption, the stress distribution through the thickness becomes non-uniform. This requires a more complex analysis using thick-walled pressure vessel theory, which considers radial stress and the variation of hoop and longitudinal stresses across the wall thickness.

  5. 5.How does internal pressure affect the stresses in a thin-walled pressure vessel?Application

    Internal pressure creates tensile stresses in the walls of a thin-walled pressure vessel. The hoop stress is the most significant and is calculated as σ_hoop = (P·r) / t, where P is the internal pressure, r is the internal radius, and t is the wall thickness. Longitudinal stress is half of the hoop stress and is calculated as σ_longitudinal = (P·r) / (2·t).

  6. 6.Why is it important to consider the material properties when designing a thin-walled pressure vessel?Application

    Material properties such as yield strength, tensile strength, and ductility are crucial in determining the vessel's ability to withstand internal pressure without failure. The material must be able to handle the calculated stresses without yielding or fracturing, ensuring safety and reliability in operation.

  7. 7.A cylindrical thin-walled pressure vessel has an internal radius of 0.5 m and a wall thickness of 0.01 m. If the internal pressure is 2 MPa, calculate the hoop stress.Numerical

    To calculate the hoop stress, use the formula σ_hoop = (P·r) / t. Here, P = 2 MPa = 2 × 10^6 Pa, r = 0.5 m, and t = 0.01 m. Thus, σ_hoop = (2 × 10^6 Pa × 0.5 m) / 0.01 m = 100 × 10^6 Pa = 100 MPa.

  8. 8.A spherical thin-walled pressure vessel has an internal radius of 1 m and a wall thickness of 0.02 m. If the internal pressure is 1.5 MPa, calculate the hoop stress.Numerical

    For a spherical vessel, the hoop stress is calculated using the formula σ_hoop = (P·r) / (2·t). Here, P = 1.5 MPa = 1.5 × 10^6 Pa, r = 1 m, and t = 0.02 m. Thus, σ_hoop = (1.5 × 10^6 Pa × 1 m) / (2 × 0.02 m) = 37.5 × 10^6 Pa = 37.5 MPa.

  9. 9.Explain why spherical pressure vessels are preferred over cylindrical ones for high-pressure applications.Application

    Spherical pressure vessels are preferred for high-pressure applications because they have a uniform stress distribution, with hoop stress being the same in all directions. This makes them more efficient in handling pressure, as the material is used more effectively, reducing the risk of failure. Additionally, spherical vessels require less material for the same internal volume compared to cylindrical vessels, making them lighter and potentially more cost-effective.

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PreviousDesign codes: IS 2825 and ASME Section VIIINext Vessels under external pressure and stiffening rings