Torsion stress function for a simply connected bar ($\Delta \varphi =-2\mu \Omega ,\varphi {|}_{\partial D}=0,M=2{\int }_D\varphi dA$)

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.

In uniform torsion about the $z$-axis, the axial displacement $w(x,y)$ describes departure of a cross-section from a plane. It is determined up to an additive axial translation by $w_x=\Omega y+{\varphi }_y/\mu$, $w_y=-\Omega x-{\varphi }_x/\mu$, where $\varphi$ is the torsion stress function for a simply connected bar. For a circular section these derivatives vanish. For an ellipse they give $w=\Omega (b^2-a^2)xy/(a^2+b^2)+\text{constant}$, so a noncircular section generally warps. The stress-function Poisson equation is exactly the mixed-derivative compatibility condition for this axial displacement.

On $x^2/a^2+y^2/b^2<1$, the torsion stress function for a simply connected bar is $\varphi =\mu \Omega a^2{b}^2(1-x^2/a^2-y^2/b^2)/(a^2+b^2)$. Its Laplacian is $-2\mu \Omega$ and its boundary value is zero, proving the formula by uniqueness of the Poisson equation with Dirichlet data. Integration gives $M=\pi \mu \Omega a^3{b}^3/(a^2+b^2)$, hence the displayed stiffness $R=M/\Omega$.

The three principal stresses are $0,\pm |\nabla \varphi |$. With ${\tau }_Y$ defined as the shear yield stress, the Tresca yield criterion is $|\nabla \varphi |\le {\tau }_Y$. For $a>b$ its maximum is $2|C|/b$ at $(0,\pm b)$, where $C=\mu \Omega a^2{b}^2/(a^2+b^2)$. Equating that maximum to ${\tau }_Y$ gives the displayed first-yield twist. The ends of the minor axis yield first; geometric distance from the bar axis alone does not determine stress in a noncircular section.