Langford Analytic · Knowledge Base

Thin-Walled Cylinder Hoop Stress — Worked Example

A complete thin-walled pressure vessel calculation: hoop stress, longitudinal stress, combined stress check and margin of safety for an aluminium cylinder.

Article 12.01Pressure & Structural Loading6 min read
pressurehoop stressthin wallcylindervesselaluminiumworked example

1. Problem

A thin-walled cylindrical pressure vessel made from 6061-T6 aluminium has an internal diameter of 300 mm and a wall thickness of 3 mm. The internal pressure is 2.0 MPa. Determine the hoop stress, longitudinal stress, von Mises equivalent stress and margin of safety.

2. Given

ParameterValueUnits
Material6061-T6 aluminium—
Fty276MPa
Ftu310MPa
Internal diameter d300mm
Wall thickness t3mm
Internal pressure p2.0MPa
Radius r (mean)151.5mm (d/2 + t/2)

3. Required

  • Hoop (circumferential) stress
  • Longitudinal stress
  • Von Mises equivalent stress
  • Margin of safety

4. Assumptions

  • Thin-wall assumption valid: r/t = 151.5/3 = 50.5 >> 10
  • Uniform internal pressure
  • Closed-ended cylinder (longitudinal stress exists)
  • No end-cap discontinuity stresses (away from the ends)
  • No external pressure
  • Linear elastic material behaviour

5. Governing Equations

Hoop stress:       σ_h  =  p × r / t
Longitudinal stress: σ_l  =  p × r / (2 × t)

Principal stresses (plane stress in shell):
  σ₁  =  σ_h   (hoop)
  σ₂  =  σ_l   (longitudinal)
  σ₃  =  0     (radial, negligible for thin wall)
  (or σ₃ ≈ −p/2 at inner surface)

Von Mises (2D, σ₃ = 0):
  σ_vm  =  √[σ₁² − σ₁·σ₂ + σ₂²]

Margin of safety:
  MS  =  (Fty / σ_vm) − 1

6. Calculation

σ_h  =  2.0 × 151.5 / 3  =  101.0 MPa
σ_l  =  2.0 × 151.5 / (2 × 3)  =  50.5 MPa

σ_vm  =  √[101² − 101 × 50.5 + 50.5²]
      =  √[10,201 − 5,100.5 + 2,550.25]
      =  √7,650.75
      =  87.5 MPa

MS  =  (276 / 87.5) − 1  =  3.15 − 1  =  +2.15

7. Result

Hoop stress = 101 MPa. Longitudinal stress = 50.5 MPa. Von Mises = 87.5 MPa. MS = +2.15 — the vessel meets requirements with a large margin.

8. Check

  • r/t = 50.5 >> 10 — thin-wall assumption is valid
  • σ_h = 2 × σ_l — the 2:1 ratio is a fundamental thin-wall relationship, confirmed
  • σ_h = 101 MPa < Fty = 276 MPa — no yielding
  • Dimensional check: MPa × mm / mm = MPa — consistent
  • The vessel could be thinned to 1.5 mm: σ_h = 202 MPa, σ_vm = 175 MPa, MS = +0.58 — still positive but with less reserve

9. Interpretation

The vessel has a large safety margin (+2.15). The hoop stress is twice the longitudinal stress — this is why pressure vessels fail by longitudinal cracking (splitting along the length). The vessel could be significantly thinned for weight reduction, but the margin would decrease rapidly (stress scales as 1/t). At t = 1.5 mm, MS = +0.58; at t = 1.2 mm, MS = +0.15.

For a production vessel, also check: (1) stress concentrations at nozzle openings and junctions; (2) buckling under external pressure (if applicable); (3) fatigue from pressure cycling; (4) creep at elevated temperature; (5) end-cap discontinuity stresses. The simple thin-wall formula does not capture any of these.

10. Limitations

  • No stress concentration at nozzles, openings or junctions — these require detailed FEA
  • No end-cap discontinuity stresses (radial bending at the cylinder-to-end junction)
  • No external pressure buckling check
  • No fatigue from pressure cycling
  • No creep or temperature-dependent properties
  • No weld efficiency factor (reduces allowable for welded vessels)
  • No corrosion allowance

11. Related Resources