*Originally posted by twhitehead*

**Yes. I took the assumption of an idealised thin walled vessel a bit too far. Perhaps an oval shape would be more accurate. My point, however, is that the tension being calculated is not dependant on the exact shape of the vessel - hence pi isn't involved. I believe a square pipe whose side was the same as the diameter of the round pipe should have the same tension as the round pipe at the centre of its sides.**

Your statement depends on the symmetries present. In the case with circular cross-section we have symmetry along both the axial and circumferential (hoop) directions. If we alter the profile so that we lose circular symmetry then we'd expect the stress component σ_θθ to pick up a dependence on θ. What your statement hangs on is an assumption that the ratio of the length of a line passing through the centre to a point on the exterior surface and the shorter distance to the interior surface doesn't depend on θ. Imagine filing one side of a circular pipe down so it's half the thickness along a line parallel to the axial direction we start increasing the load, intuitively we expect the pipe to fail along that line which demonstrates that the stress must be highest at the weakness. However for a pipe with elliptical cross-section and the two ellipses with the same eccentricity the diagonal components of the stress don't change.

In the above calculation for a pipe of square cross-section my L is doing the same job as 2R, they come out the same all along the faces, not just at the centre points. So you are right about that. What varies is the shear stress, which is completely absent in the cylindrical case, it's zero at the centre of faces and a maximum at the corners, as we'd expect.

For an oval shape the pressure is trying to turn our oval cross-section into a circular one, so there'll still be shear components. The area of an ellipse is πab where a and b are the radii of the major and minor axes respectively. Suppose R1 and R2 are the inner major and minor radii and λR1 and λR2 the outer ones and let λ = 1 + s and choose s=t/R1 where t is the thickness along the major axis. The area enclosed by the pipe is A = πR1R2, and the area bounded by the outer surface of the pipe is πλR1λR2 = λ²A, so the area bounded by the inner and outer ellipses is πR1R2(λ² - 1) = λR1R2(s² + 2s). We can neglect the term in s² to obtain:

σ_zz = pπR1R2 / 2sπR1R2 = p / 2s

recalling our definition of s = t/R1, we get

σ_zz = pa/2t

Which is the same as for the circular case, up to a correction term of order s².

The formula for the distance to the edge of an ellipse from the centre point is:

r(θ; a, b) = ab/√[(b cos( θ ))² + (a sin( θ ))²]

Assuming the circumferential stress doesn't vary in the z and r directions we we can try to calculate it. Using the formula for an ellipse:

r(θ; a, b) = ab/√[(b cos( θ ))² + (a sin( θ ))²]

we get a handy expression for r(θ; λR1, λR2):

r(θ; λR1, λR2) = λ²R1R2/√[( λR2 cos( θ ))² + (λR2 sin( θ ))²] = λ r(θ, R1, R2)

so we can write:

σ_θθ( θ ) = p r(θ; R1, R2) / [r(θ; λR1, λR2) - r(θ; R1, R2)] = p / ([r(θ; λR1, λR2)/r(θ; R1, R2) - 1] = p/(λ - 1) = p/s = pa/t

which is identical to the result for the circular case.

By axial symmetry we have that:

σ_zr = σ_rz = σ_θz = σ_zθ = 0

Which leaves σ_θr = σ_rθ and the calculation for this is painful.