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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.

Thermal stress is stress generated when thermal expansion or contraction is constrained or incompatible. A uniformly heated isotropic body whose entire boundary is fixed has $e=0$ and $\tau =-K{\alpha }_v(T-T_0)I$. For a nonuniform temperature field, interior displacement generally remains nonzero even with a fixed outer boundary; equilibrium and compatibility must be solved together. Cancelling the divergence of a thermal stress term with a body force is not the same as making that stress vanish.

For isotropic linear elasticity with volumetric expansion coefficient $\alpha$, the stress-free thermal strain is $e^{\mathrm{th}}=\alpha (T-T_0)I/3$. Thus the thermal stress term is $-K\alpha (T-T_0)I$, where $K=\lambda +2\mu /3$ is the bulk modulus. A uniformly heated body with its full boundary fixed has zero displacement and the displayed compressive stress. For spatially varying temperature and zero body force, the displacement instead solves $\mu \Delta u+(\lambda +\mu )\nabla \mathrm{div}u=K\alpha \nabla T$ with its boundary values; the uniform formula is not automatic. A body force per volume $+K\alpha \nabla T$ cancels this thermal forcing in the equilibrium equation, but does not by itself cancel the stress or its boundary tractions.