I am trying to model 2D thermal expansion of a square area inside another square using FEATool. I have simulated plane strain by incorporating forces pointing along the $[1 \,\,\, -1]^T$ direction inside the smaller square . However, what I really want to simulate is this "shrinking" of the material as a thermal expansion process. What kind of PDE should I use to model the subtraction of heat from that small square?

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1 Answer 1


In the most general case you want to solve elasticity and heat conduction equations simultaneously, what is normally termed as a strongly coupled model.

More commonly people use a weakly coupled model, where you first compute the temperature increase ($\Delta T$) for each point in your body and then you solve the following set of equations, where I am using index notation:

  • Equilibrium equations

$$\sigma_{ij, j} + b_i = 0\, ,$$

$\sigma$ is the stress tensor, $b$ is the body forces vector.

  • Constitutive equations

$$ \sigma_{ij} = \frac{E}{1 + \nu}\left[\epsilon_{ij} + \frac{\nu}{1 - 2\nu}\epsilon_{kk} \delta_{ij}\right] - \frac{E \alpha \Delta T}{1 - 2\nu}\delta_{ij}\, ,$$

$E$ is the Young modulus, $\nu$ the Poisson coefficient, $\epsilon$ the strain tensor, $\Delta T$ is the temperature increase, $\alpha$ the linear coefficient of thermal expansion, and $\delta_{ij}$ is the Kronecker delta.

  • Infinitesimal strain definition

$$\epsilon_{ij} = \frac{1}{2}\left(u_{i,j} + u_{j, i}\right)\, ,$$

$u$ is the displacement vector.

You can write the equations in terms of displacements, and this is the most common formulation in FEM software. In that case you can think that the change in temperature add a body force to your differential equation.

As a final comment, the simplest case is to consider $\Delta T$ to be uniform over your whole domain.

  • $\begingroup$ I wanted to calculate the extra body force due to temperature from the constitutive. If I'm not mistaken it should be the divergence of the last term in the expression for $\sigma_{ij}$. However, every term in that expression is constant in space (for uniform $\Delta T$), which means it will have no divergence and $\sigma_{ij,j}$ will reduce to the usual term. Am I getting something wrong? $\endgroup$
    – Zegpi
    Commented May 30 at 12:09
  • 1
    $\begingroup$ No, I think that's how it is $\endgroup$
    – nicoguaro
    Commented May 31 at 11:48

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