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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.

Use the torsion stress function for a simply connected bar. The only nonzero strain components are $e_{xz}=(w_x-\Omega y)/2$ and $e_{yz}=(w_y+\Omega x)/2$. ThusThe axial equilibrium equation is ${\partial }_x{\tau }_{xz}+{\partial }_y{\tau }_{yz}=0$. Since the section is simply connected, the planar divergence-free stress field has a stream function:Equality of the mixed warping derivatives gives $\boxed{\Delta \varphi =-2\mu \Omega }$. On the lateral boundary the axial traction is ${\varphi }_y{n}_x-{\varphi }_x{n}_y$, the tangential derivative of $\varphi$. Traction-free walls therefore make $\varphi$ constant on the connected boundary, and its arbitrary additive constant can be chosen so $\boxed{\varphi =0\text{ on }\partial D}$.

The axial torque isThe divergence theorem removes the boundary term because $\varphi =0$.

Tresca's yield criterion requires the maximum shear stress, half the largest principal-stress difference, not to exceed the material's shear yield stress ${\tau }_Y$. The stress eigenvalues here are $0,\pm \sqrt{{\tau }_{xz}^2+{\tau }_{yz}^2}$, so the maximum shear is $|\nabla \varphi |$ andIf yield is instead quoted as a uniaxial tensile stress, that parameter is $2{\tau }_Y$; the present convention is the source's shear parameter.

For the ellipse, the proposed quadratic vanishes on its boundary. Its Laplacian fixesThe corresponding warping is $w=\Omega (b^2-a^2)xy/(a^2+b^2)+\text{constant}$. Elliptic polar coordinates $x=ar\cos \vartheta$, $y=br\sin \vartheta$ give ${\int }_D\varphi =C\pi ab/2$, henceTo locate first yield, maximize $|\nabla \varphi |=2|C|\sqrt{x^2/a^4+y^2/b^4}$. It increases radially to the boundary. For $a>b$, the maximum there occurs at $(0,\pm b)$ and equals $2|C|/b$. ThereforeThe positive-twist value is the requested expression. The ends of the minor axis yield first, not the ends of the longer axis. This is elliptical-bar torsion and first yield.

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.