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.

The in-plane equilibrium equations are ${\tau }_{xx,x}+{\tau }_{xy,y}=0$ and ${\tau }_{xy,x}+{\tau }_{yy,y}=0$. On a simply connected patch, each divergence-free row has a streamfunction: write ${\tau }_{xx}={\varphi }_y$, ${\tau }_{xy}=-{\varphi }_x=-{\psi }_y$, ${\tau }_{yy}={\psi }_x$. Symmetry gives ${\varphi }_x={\psi }_y$, so another potential exists with $A_y=\varphi$, $A_x=\psi$. HenceIf two such Airy stress functions give the same stress, their difference has zero Hessian and is therefore an arbitrary affine function $a+bx+cy$ on a connected domain. This is the full nonuniqueness for a fixed stress. The existence argument is local, or global on a simply connected domain; domains that are multiply connected regions require the associated force and moment period conditions for a single-valued global potential. Equilibrium alone should not silently erase that topological qualification.

For plane strain put $s=e_{xx}+e_{yy}$. The constitutive equation gives ${\tau }_{xx}+{\tau }_{yy}=2(\lambda +\mu )s$. ConsequentlyThe displacement Saint-Venant compatibility equations $e_{xx,yy}+e_{yy,xx}-2e_{xy,xy}=0$ then becomesFor stable elastic moduli the coefficient is nonzero, so $\boxed{{\Delta }^{2}A=0}$. If $D$ is bounded and smooth and $A={\partial }_{n}A=0$ on its boundary, integrating twice by parts gives $0={\int }_{D}A{\Delta }^{2}AdV={\int }_D(\Delta A{)}^{2}dV$. Thus $A$ is harmonic with zero boundary value and $\boxed{A=0}$ by uniqueness of the Dirichlet problem. The normal-derivative condition is what removes the extra fourth-order boundary terms.

For the pressurized annulus, substituting the radial constitutive equations into equilibrium gives $u^{\prime \prime }+u^{\prime }/r-u/r^2=0$, so $u=Cr+D/r$. The stresses areUsing gauge stresses relative to ambient, the boundaries are ${\tau }_{rr}(a)=-P$, ${\tau }_{rr}(b)=0$. ThereforeandThe maximum occurs at the bore. Once the chosen yield criterion is reached, Hooke's law cannot describe the total strain alone: write total strain as elastic plus plastic strain and supplement the elastic stress law with a yield condition, a plastic flow rule and any required work hardening law. A plastic region starts near the bore and must be matched to the remaining elastic region. The exact conversion between a principal-stress difference and the quoted yield stress depends on whether the material criterion is Tresca or von Mises. This is the elastic stress concentration in a pressurized thick cylinder.