Technical guide · method and practical tips

Pressure pipe stresses and wall thickness

Membrane formulas provide an excellent magnitude check, but real piping must also satisfy a code, tolerances, corrosion and mechanical loads.

Reviewed

End viewpσθσθcircumferential stressLongitudinal viewσzσzplongitudinal stressσθ = pDm/(2t) ≈ 2σz
The end view isolates circumferential stress σθ; the longitudinal view isolates longitudinal stress σz. For this closed thin-wall model, σθ ≈ 2σz.

1. Use gauge pressure

Stresses depend on the pressure difference between inside and outside. Absolute pressure should not be used directly when the outside is also at atmospheric pressure.

Practical tip: always write p = pin − pout explicitly in the calculation note.

2. Obtain both membrane stresses

σθ = pDm/(2t) ; σz = pDm/(4t)

For a closed pipe, pressure creates hoop stress σθ and longitudinal stress σz. In the thin-wall model, σθ is twice σz.

Radial stress is neglected in this approximation, which is no longer valid for a thick wall.

3. Check the thin-wall assumption

The Dm/t ratio is a quick check. There is no universal threshold independent of required accuracy, but a high ratio better supports the model.

If the wall becomes thick, use Lamé equations or a method compliant with the applicable code.

4. Move from net to real thickness

A thickness from a simple formula should not be ordered directly. Add corrosion, erosion or machining allowance, account for negative product tolerance and select an actually available wall.

Nominal thickness may also be governed by stiffness, handling, welding or code minimum rules.

5. What the pre-calculation does not cover

Industrial piping is a system, not only a straight pipe.

  • External pressure and buckling.
  • Weight, thermal expansion, support reactions and flexibility.
  • Branches, bends, reducers, welds and stress intensification.
  • Flanges, bolts, gaskets and leakage.
  • Fatigue, water hammer, vibration and occasional loads.
  • Allowable stresses depending on material and temperature.

Numerical application

Example: 114.3 mm OD pipe at 10 bar

Wall thickness is 4 mm and selected allowable stress is 120 MPa.

  1. Dm = 114.3 − 4 = 110.3 mm.
  2. σθ = 10 bar × 110.3/(2×4) = 13.7875 MPa.
  3. σz = 6.89375 MPa.
  4. σVM = 11.940 MPa.
  5. Utilization = 9.9503%.

Result: The magnitude is low compared with 120 MPa, but this does not validate the piping code or other loads.

Method linked to MIT thin-wall vessel notes and the scope of ASME B31.3. Open exact source ↗

For checking and further study

Direct references

Each link points to the exact course, standard or publication page used, rather than a generic homepage.

  1. Mechanics & Materials I — lecture notesMIT OpenCourseWareLecture notes covering multiaxial stress, failure criteria, thin-walled pressure vessels and torsion.
  2. B31.3 — Process PipingAmerican Society of Mechanical EngineersOfficial code page covering materials, design, fabrication, inspection and testing of process piping.
  3. The International System of Units, 9th editionInternational Bureau of Weights and MeasuresOfficial reference for units and symbols.