 Torsion stress function for a simply connected bar

For uniform twist $\Omega$ and displacement $(-\Omega yz,\Omega xz,w(x,y))$, linear elasticity gives ${\tau }_{xz}=\mu (w_x-\Omega y)$ and ${\tau }_{yz}=\mu (w_y+\Omega x)$. Their planar divergence vanishes by axial equilibrium. On a simply connected section there is therefore a stress function $\varphi$ with ${\tau }_{xz}={\varphi }_y$ and ${\tau }_{yz}=-{\varphi }_x$.

The mixed derivatives of $w$ imply $\Delta \varphi =-2\mu \Omega$. The lateral axial traction is the tangential derivative of $\varphi$, so a traction-free connected boundary makes $\varphi$ constant; choose it to be zero. Finally the torque is $M={\int }_D(x{\tau }_{yz}-y{\tau }_{xz})dA=-{\int }_D(x{\varphi }_x+y{\varphi }_y)dA=2{\int }_D\varphi dA$, since its boundary term vanishes. Thus one scalar Poisson equation determines both the shear field and torsional stiffness. Multiply connected sections require additional constants and compatibility conditions on the separate boundary components.