Longitudinal Stress in a Thin-Walled Cylinder

σl=pd4t\sigma_{l} = \frac{p d}{4 t}

Worked example: 1 m vessel, 10 mm wall, 2 MPa → 50 MPa longitudinal — press Try an example to run it live, then adjust anything.

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Longitudinal Stress in a Thin-Walled Cylinder explained

σlpdt

Cap the ends of a pressurised cylinder and the pressure pushes them apart with a force p × πd²/4, resisted by the full ring of wall material, area πdt. Equate the two and σ = pd/4t — precisely half the hoop stress in the same vessel. The same 1 m vessel at 2 MPa with a 10 mm wall carries 100 MPa around the circumference but only 50 MPa along its axis.

That 2:1 split explains a lot of everyday behaviour: a pipe under excessive pressure splits along a longitudinal seam, a sausage skin splits the long way, and the girth (circumferential) welds joining pipe spools see only half the stress that a longitudinal seam weld does — which is why pipe manufacturing standards obsess over the long seam and why spiral-welded pipe places its seam on a helix to keep it off the worst direction. The catch: only a closed vessel develops this stress. A pipe with an expansion joint, a slip coupling or an open end passes the end thrust into anchors and thrust blocks instead — the buried-main equivalent, where an unrestrained bend will walk out of the ground if the thrust block is undersized.

Longitudinal Stress in a Thin-Walled Cylinder formula

σl=pd4t\sigma_{l} = \frac{p d}{4 t}
Where
  • σl\sigma_{l}= Longitudinal stress (kPa)
  • pp= Internal gauge pressure (kPa)
  • dd= Internal diameter (mm)
  • tt= Wall thickness (mm)