Longitudinal Stress in a Thin-Walled Cylinder

σl=pd4t\sigma_{l} = \frac{p d}{4 t}

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

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