Kirsch equations for a hole in a plate

Example

Consider an infinite thin plate of homogeneous isotropic linear elastic material in plane stress, with a circular hole of radius $a > 0$ centred at the origin. The edge of the hole is free of traction, and far from the hole the plate is under uniform tension $s$ in the $x$ direction. In polar coordinates $(r, \theta)$, $r \ge a$, the Kirsch solution for the stress is $$\sigma_{rr} = \frac{s}{2} \left( 1 - \frac{a^2}{r^2} \right) + \frac{s}{2} \left( 1 - \frac{4a^2}{r^2} + \frac{3a^4}{r^4} \right) \cos 2\theta,$$ $$\sigma_{\theta\theta} = \frac{s}{2} \left( 1 + \frac{a^2}{r^2} \right) - \frac{s}{2} \left( 1 + \frac{3a^4}{r^4} \right) \cos 2\theta,$$ $$\sigma_{r\theta} = -\frac{s}{2} \left( 1 + \frac{2a^2}{r^2} - \frac{3a^4}{r^4} \right) \sin 2\theta.$$

Check of the boundary conditions: at $r = a$, $\sigma_{rr} = \frac{s}{2} \cdot 0 + \frac{s}{2}(1 - 4 + 3) \cos 2\theta = 0$ and $\sigma_{r\theta} = -\frac{s}{2}(1 + 2 - 3) \sin 2\theta = 0$, so the hole is traction-free. As $r \to \infty$, $\sigma_{rr} \to \frac{s}{2}(1 + \cos 2\theta) = s \cos^2 \theta$, $\sigma_{\theta\theta} \to s \sin^2 \theta$ and $\sigma_{r\theta} \to -s \sin \theta \cos \theta$, which are the polar components of the uniform tension $\sigma_{xx} = s$. The fields come from an Airy stress function that is biharmonic, so equilibrium and compatibility hold.

At the edge of the hole, $\sigma_{\theta\theta}(a, \theta) = s (1 - 2 \cos 2\theta)$. Its maximum is $3s$, at $\theta = \pm \pi/2$, the points of the hole on the axis across the load. At $\theta = 0$ and $\theta = \pi$ it equals $-s$, a compression. So the stress concentration factor of a circular hole is $3$, whatever its radius.