Strong and weak formulation for classical linear thermoelasticity

We refer the reader to this page of the wiki to understand the signification of the various symbols.

Hypotheses

We assume the material is linear isotropic with respect to mechanical as well as thermal properties. That is to say that:

$$\underline{\underline{\sigma}} = \lambda \; \textrm{tr}(\underline{\underline{\varepsilon}}) \; \underline{\underline{I}} + 2\mu \; \underline{\underline{\varepsilon}}$$ $$\underline{\underline{k}} = k \; \underline{\underline{I}}$$

and

$$\underline{\underline{\beta}} = \beta \; \underline{\underline{I}}$$

The first of these equations stipulates that only 2 parameters are required to describe the materials' elasticity tensor $\mathbb{C}$, but also serves as the constitutive equation for the mechanical part.

Strong form of the coupled governing equations

The coupled equations in classical thermoelasticity consist in the equation of motion and the energy balance equation, but before obtaining the strong (local) form of the equations, we have to introduce a few equations:

• the kinematic relation relating the linearized strain tensor to the displacement field:

$$\underline{\underline{\varepsilon}} = \frac{1}{2} \; \left( \underline{\underline{\nabla}} \underline{u} + \underline{\underline{\nabla}} \underline{u}^T \right)$$

• the Cauchy stress tensor is symmetric (balance of momentum equation)

$$\underline{\underline{\sigma}} = \underline{\underline{\sigma}}^T$$

• the thermal constitutive equation which here is Fourier's law

$$\underline{q} = -k \; \underline{\nabla} \theta$$

• the entropy relation

$$\rho \; s = \frac{\rho \; c}{T_0} \; \theta + \beta \; \textrm{tr}(\underline{\underline{\varepsilon}})$$

Now, the equation of motion is written

$$\rho \; \underline{\ddot{u}} = \underline{f}_v + \underline{\textrm{div}}(\underline{\underline{\sigma}}) + \beta \; \underline{\nabla} \theta$$

and, using the entropy relation, the energy balance equation is written

$$\rho \; c \; \dot{\theta} - \beta \; T_0 \; \textrm{div}(\underline{\dot{u}}) = R - \textrm{div}(\underline{q})$$

Injecting the constitutive equations in those governing equation, we obtain the final version of the strong formulation:

$$\rho \; \underline{\ddot{u}} - \beta \; \underline{\nabla} \theta = \underline{f}_v + \underline{\textrm{div}}(\mathbb{C} : \underline{\underline{\varepsilon}})$$ $$\rho \; c \; \dot{\theta} - \beta \; T_0 \; \textrm{div}(\underline{\dot{u}}) = R + k \; \Delta \theta$$

Weak form of the coupled governing equations

Here, the standard procedure to obtain the weak form from the strong form is applied:

• Scalar product with an appropriate test function
• Integration over the whole domain $\Omega$
• Integration by parts on the terms with second-order space derivatives to reduce the derivation order and make Neumann boundary conditions appear

The Neumann boundary condition in the mechanical part is assumed to be defined for part $\Gamma_p$ of the domain's boundary. The Neumann boundary condition in the thermal part is assumed to be defined for part $\Gamma_q$ of the domain's boundary.

Using this procedure, we obtain the weak form of the coupled governing equations, ready to be discretized:

$$\int_{\Omega} \rho \; \underline{\ddot{u}} . \underline{u}^* \; dV + \int_{\Omega} (\mathbb{C} : \underline{\underline{\varepsilon}}(\underline{u})) : \underline{\underline{\varepsilon}}(\underline{u}^*) \; dV - \int_{\Omega} \beta \; \underline{\nabla} \theta . \underline{u}^* \; dV = \int_{\Omega} \underline{f}_v . \underline{u}^* \; dV + \int_{\Gamma_p} (\underline{\underline{\sigma}} . \underline{n}) . \underline{u}^* \; dS$$ $$\int_{\Omega} \rho \; c \; \dot{\theta} . \theta^* \; dV + \int_{\Omega} k \; \underline{\nabla} \theta . \underline{\nabla} \theta^* \; dV - \int_{\Omega} \beta \; T_0 \; \textrm{div}(\underline{\dot{u}}) . \theta^* \; dV = \int_{\Omega} R . \theta^* \; dV + \int_{\Gamma_q} (\underline{q} . \underline{n}) . \theta^* \; dS$$