Thermoelasticity

Thermoelasticity couples elasticity and temperature through a specific Helmholtz free energy $\psi (F,\theta )$. Without constraints its reversible laws are $P_{Ii}={\rho }_0{\psi }_{,F_{iI}}$ and $\eta =-{\psi }_{,\theta }$. Mixed derivatives give the same coupling tensor in the thermal stress response and the strain dependence of specific entropy, with opposite signs.

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.

Thermal strain is the strain that heating would produce without mechanical stress. In isotropic linear thermoelasticity it is a pure dilation: a volumetric expansion coefficient ${\alpha }_v$ corresponds to a linear expansion coefficient ${\alpha }_v/3$. The constitutive law acts on the mechanical strain $e-e^{\mathrm{th}}$. Spatially varying thermal strain need not be globally compatible with a stress-free displacement.

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.

Let $fdx$ be reversible work supplied to stretch a band. Its Helmholtz free energy obeys $dF=-SdT+fdx$. Equality of mixed derivatives gives the displayed relation. An entropy decrease on stretching therefore corresponds to increasing tensile force on heating at fixed extension. The sign follows the force/work convention; a restoring force exerted by the band has the opposite sign.

The admissible rates obey ${\varphi }_{,F}:\dot{F}=h^{\prime }(\theta )\dot{\theta }$. A reaction multiplier $q$ in the reversible nominal stress tensor therefore has an accompanying specific entropy contribution:Their reaction contributions cancel in the Clausius-Duhem inequality on admissible rates. Omitting the entropy term would generally violate the constraint's power balance.

Linearize $\det F=h(\theta )$ at $F=I$, $\theta ={\theta }_0$, $h({\theta }_0)=1$, and zero reaction. With displacement gradient $H$ and temperature increment $\vartheta$, one has $\mathrm{tr}H=h^{\prime }({\theta }_0)\vartheta$ andwhere $C_{jilk}={\rho }_0{\psi }_{,F_{ij}{F}_{kl}}$, ${\beta }_{ji}={\rho }_0{\psi }_{,F_{ij}\theta }$ and $C_e=-{\rho }_0{\psi }_{,\theta \theta }$ at the reference state. The temperature factor converts this entropy coefficient to a volumetric heat capacity: ${\theta }_0{C}_e$. Strictly constant volume is the case $h^{\prime }=0$.