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Write the Cauchy stress tensor in the local fibre frame as $\sigma =se\otimes e+h(e\otimes n+n\otimes e)+tn\otimes n$. The compatible geometry has $\nabla \theta =\kappa e$, $\nabla \cdot e=0$ and $\nabla \cdot n=-\kappa$. With zero body force, force balance becomes $e\cdot \nabla s=2\kappa h-n\cdot \nabla h$ and $n\cdot \nabla t-\kappa t=-e\cdot \nabla h-\kappa s$. For specified shear stress $h$, the first equation determines $s$ along fibres, then the second determines $t$ along transverse curves. Pressure and fibre tension supply these two normal constraint reactions. This is a local stress construction; boundary tractions must be chosen consistently with its integration data.

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.