Lamé Radial Stress in a Thick-Walled Cylinder

The radial stress is the half of Lamé's solution that people forget, and it is the half that tells you the boundary conditions are being obeyed. At the bore the wall is being pushed on by the fluid, so $\sigma_r = -p_i$ exactly — the pressure, in compression. At the outside surface there is nothing pushing, so $\sigma_r = 0$ exactly. In between it climbs from one to the other along $\sigma_r = \frac{p_i r_i^2}{r_o^2 - r_i^2}\left(1 - \frac{r_o^2}{r^2}\right)$, and it is compressive the entire way.

On the 100/200 mm cylinder at 60 MPa, midwall at $r = 150$ mm gives $\sigma_r = -15.56$ MPa against a hoop stress of $+55.56$ MPa. Add them: 40 MPa, which is the same sum you get at the bore ($100 - 60$) and at the outside ($40 + 0$). The sum $\sigma_\theta + \sigma_r$ is constant through the entire wall thickness. Their difference is not — $\sigma_\theta - \sigma_r$ is largest at the bore, and since that difference is what drives yielding under both Tresca and von Mises, the bore is where the cylinder gives up first.

Which is the practical reason to bother with the radial stress at all. A yield check that uses only the hoop stress ignores a compression of magnitude $p_i$ acting at right angles to it, and at high pressures that is not a small correction. Take both to a von Mises or Tresca comparison and the answer changes materially. The sign convention matters too: this page returns $\sigma_r$ negative, because it is compression, and dropping the minus sign on the way into a combined-stress calculation is a quiet and expensive mistake.