Longitudinal Stress in a Thin-Walled Cylinder
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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The vessel, two ways →
UniversityMechanics of Materials
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Longitudinal Stress in a Thin-Walled Cylinder explained
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
- = Longitudinal stress (kPa)
- = Internal gauge pressure (kPa)
- = Internal diameter (mm)
- = Wall thickness (mm)
Missing one of these? Work it out first, then come back
- Longitudinal stress — Normal (Axial) Stress, Young's Modulus (E = σ/ε)
- Internal gauge pressure — Hoop Stress in a Thin-Walled Cylinder, Membrane Stress in a Thin-Walled Sphere (σ = pr/2t)
- Internal diameter — Hoop Stress in a Thin-Walled Cylinder, Area Moment of Inertia — Solid Round Bar
- Wall thickness — Hoop Stress in a Thin-Walled Cylinder, Membrane Stress in a Thin-Walled Sphere (σ = pr/2t)