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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.

Take the trace of the constitutive relation at zero stress. It gives $0=(2\mu +3\lambda )e_{kk}-3c\alpha (T-T_0)$. With the observed stress-free volume strain this requiresThis is the bulk modulus. A stable solid resists uniform compression, so $\boxed{K>0}$; equivalently the elastic energy of a nonzero pure dilation is positive. The thermal strain is isotropic, $e_{ij}^{\mathrm{th}}=\alpha (T-T_0){\delta }_{ij}/3$, because $\alpha$ here is the volumetric, not linear, thermal expansion coefficient.

Let $\beta =K\alpha$. Using $e_{ij}=(u_{i,j}+u_{j,i})/2$, stress divergence isZero body force and equilibrium therefore giveFor an applied body force per unit volume $b$, equilibrium is $\nabla \cdot \tau +b=0$, so its right-hand side becomes $\beta \nabla T-b$. Consequentlycancels the effective thermal forcing in the displacement equation. Per unit mass the force is $\beta \nabla T/\rho$. This sign follows from the divergence of the thermal stress $-\beta (T-T_0)I$, which is $-\beta \nabla T$. Cancelling that divergence is distinct from removing the stress tensor itself: boundary tractions and thermal compatibility still matter.

For spatially uniform heating with the entire boundary fixed, $u=0$ is the unique equilibrium displacement in a stable elastic solid, andThe stress is compressive hydrostatic stress for a temperature rise. The fixed boundary supplies the tractions that prevent thermal expansion. If $T$ varies spatially and no compensating body force is imposed, fixing the boundary alone does not generally force the interior displacement to vanish: one must solve the displayed forced Navier-Cauchy equation with $u=0$ on the boundary, then recover the stress. The simple uniform hydrostatic expression should not be asserted for arbitrary temperature fields. This is uniform thermal stress in a fully constrained solid.