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Lamé Hoop Stress in a Thick-Walled Cylinder

Also known as Lame equation hoop stress · thick walled cylinder hoop stress · thick cylinder circumferential stress · gun barrel stress · hydraulic cylinder wall stress · autofrettage bore stress · Lame thick wall

σθ=piri2ro2−ri2(1+ro2r2)\sigma_{\theta} = \frac{p_i r_i^{2}}{r_o^{2} - r_i^{2}} \left(1 + \frac{r_o^{2}}{r^{2}}\right)

Worked example: 100/200 mm cylinder at 60 MPa, bore → 100 MPa hoop — press Try an example to run it live, then adjust anything.

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See also
Lamé Radial Stress in a Thick-Walled Cylinder — the radial companion — the two together are the complete Lamé solutionHoop Stress in a Thin-Walled Cylinder — what this collapses to once the wall is thin enough to ignore the gradientVon Mises Equivalent Stress (Plane Stress) — the bore is a biaxial stress state, and this is how you turn it into one number to compare with yield

Learning zone

The thin-wall formula pretends the stress is the same all the way through the wall. In a thick cylinder it plainly is not: the bore is being pushed on directly and the outside surface is not being pushed on at all. Lamé's solution, from 1833, gives the whole profile. For internal pressure only, σθ=piri2ro2−ri2(1+ro2r2)\sigma_\theta = \frac{p_i r_i^2}{r_o^2 - r_i^2}\left(1 + \frac{r_o^2}{r^2}\right) — largest at the bore and falling as 1/r21/r^2 toward the outside.

Take a cylinder bored to 100 mm radius, 200 mm outside, at 60 MPa internal. At the bore the hoop stress is 60(0.050.03)=10060\left(\frac{0.05}{0.03}\right) = 100 MPa; at the outside surface it has dropped to 2×60×0.01/0.03=402 \times 60 \times 0.01/0.03 = 40 MPa. The difference between those two, 60 MPa, is exactly the applied pressure — and that is not a coincidence of these numbers but an identity of the solution, which makes it a free check on any pair of answers you compute.

The engineering consequence is the one that catches people out. Doubling the wall thickness does not halve the bore stress. Push ror_o to infinity in the expression and the bore hoop stress tends to pip_i, not to zero: a solid block of steel with a hole in it still sees a hoop stress equal to the pressure. So for very high pressures there is a wall thickness past which adding metal buys almost nothing, and the answer has to come from somewhere else — a compound cylinder shrunk together, or autofrettage, both of which put the bore into residual compression before it ever sees service. Those residual fields are invisible to this equation, which knows only about a homogeneous elastic wall with nothing locked into it.

Lamé Hoop Stress in a Thick-Walled Cylinder
σθ=piri2ro2−ri2(1+ro2r2)\sigma_{\theta} = \frac{p_i r_i^{2}}{r_o^{2} - r_i^{2}} \left(1 + \frac{r_o^{2}}{r^{2}}\right)
Where
  • σθ\sigma_{\theta}= Hoop stress at radius r (kPa)
  • pip_i= Internal gauge pressure (kPa)
  • rir_i= Inside radius (mm)
  • ror_o= Outside radius (mm)
  • rr= Radius at which the stress is wanted (mm)
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