I want to solve the thermal expansion of solid by using FEM approach. When I developed the model based on the the principle the minimum potential energy, the solutions for thermal expansion are not correct, although the results of static bending problems under external mechanical loads shows good results. The basic equations I employed as follows

$\delta U = \int\limits_V {\delta {{\bf{\varepsilon }}^T}\left( {{\bf{C\varepsilon }} + {\bf{\bar \sigma }}} \right)dV} = \int\limits_V {\left( {\delta {{\bf{\varepsilon }}^T}{\bf{C\varepsilon }} + \delta {{\bf{\varepsilon }}^T}{\bf{\bar \sigma }}} \right)dV} =0 $

where C is the constituve matrix,

${\bf{\sigma }} = \left\{ \begin{array}{l} {\sigma _{xx}}\\ {\sigma _{yy}}\\ {\sigma _{zz}}\\ {\sigma _{yz}}\\ {\sigma _{xz}}\\ {\sigma _{xy}} \end{array} \right\}$

${\bf{\varepsilon }} = \left\{ \begin{array}{l} {\varepsilon _{xx}}\\ {\varepsilon _{yy}}\\ {\varepsilon _{zz}}\\ {\gamma _{yz}}\\ {\gamma _{xz}}\\ {\gamma _{xy}} \end{array} \right\}$

${\bf{\bar \sigma }} = - \frac{{E\alpha \Delta T}}{{1 - 2\nu }}\left\{ \begin{array}{l} 1\\ 1\\ 1\\ 0\\ 0\\ 0 \end{array} \right\}$

I saw that most of the references dealing with thermal bending problems are developed based on equilibrium equations $\frac{{\partial {\sigma _{xx}}}}{{\partial x}} + \frac{{\partial {\sigma _{xy}}}}{{\partial y}} + \frac{{\partial {\sigma _{xz}}}}{{\partial z}} = 0$

$\frac{{\partial {\sigma _{xy}}}}{{\partial y}} + \frac{{\partial {\sigma _{yy}}}}{{\partial y}} + \frac{{\partial {\sigma _{yz}}}}{{\partial z}} = 0$

$\frac{{\partial {\sigma _{xz}}}}{{\partial x}} + \frac{{\partial {\sigma _{yz}}}}{{\partial y}} + \frac{{\partial {\sigma _{zz}}}}{{\partial z}} = 0$

I'm not sure that I might have missed something in the energy equation or is there a condition that the problem must be dealt with basis of equilibrium equation. Could someone help me out?


1 Answer 1


The virtual strain energy should be

\begin{equation} \delta U = \int\limits_V {\delta {{\bf{\varepsilon }}^T}{\bf{C}\left(\varepsilon - {\bf{\bar \varepsilon }} \right)dV}} \end{equation}


\begin{equation} {\bf{\bar \varepsilon }} = \alpha \Delta T\left\{ \begin{array}{l} 1\\ 1\\ 1\\ 0\\ 0\\ 0 \end{array} \right\} \end{equation}

I like to think about this as follows. $\varepsilon$ is the strain calculated from the total displacements of the body. Some of this strain is due to the forces acting on the body and some is due to thermal expansion. The stress in the body and also the strain energy is due only due the applied forces. So we subtract off the thermal strain from the total strain when calculating the stress. If, for example, all the strain is due to thermal expansion, the stress and strain energy are zero.

  • 2
    $\begingroup$ I would add that the "equilibrium equations" the OP is referring to are obtained as the Euler-Lagrange equations of the functional. So, they are (in some sense) equivalent. $\endgroup$
    – nicoguaro
    Sep 27, 2021 at 20:59

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