 Elliptical-bar torsion and first yield

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.