How one can get the strain or stress from the strain energy function ? And if one cannot do it, what is the use of that function ?

  • $\begingroup$ You can't. You want to get 6 components of the stress tensor, plus 6 of the strain tensor, from one scalar value. Unless you know a lot more about the stress and strain in the material, you are trying to solve one equation for 12 unknowns! $\endgroup$
    – alephzero
    Oct 12, 2018 at 12:43
  • $\begingroup$ Ok. So what is the usefulness of that energy ? $\endgroup$
    – J.A
    Oct 12, 2018 at 12:56
  • $\begingroup$ A system is in mechanical equilibrium when the strain energy is minimized subject to the given constraints. This feature is useful because many engineering and scientific problems can be reduced to finding a system's equilibrium state. $\endgroup$ Oct 12, 2018 at 18:42

2 Answers 2


You can determine the 3D stress-strain equation for a material from its strain energy function. To get a certain component of the stress, you take the partial derivative of the strain energy function with respect to the corresponding component of strain. And, to get a certain component of the strain, you take the partial derivative of the strain energy function with respect to the corresponding component of stress. The following reference gives the strain energy function for Hooke's law in Eqn. 8.2.19:


Try taking the partial derivative of this function with respect to any of the strain components and see what you get.


The Strain Energy Deformation Function (SEDF) is a scalar $\psi(\varepsilon_{ij},\xi_k)$ defined in term of some strain tensor $\varepsilon_{ij}$ and possibly some internal variables $\xi_k$ (representing non-reversible or dissipative phenomena). This functions is a thermodynamic potential, that can be interpreted as Gibbs free energy per volum unit. The stress tensor components are related to SEDF by means of:

$$\sigma_{ij} = \frac{\partial \psi}{\partial \varepsilon_{ij}}$$

There are several possibilities for defining it, in terms of what representation is taken for stress and strain. For example, a possible choice is to use the Green–Saint-Venant strain tensor and the second Piola–Kirchhoff stress tensor, or, more commonly, the right CauchyGreen deformation tensor and the second Piola–Kirchhoff stress tensor:

$$S_{ij} = 2\frac{\partial \psi}{\partial C_{ij}}, \tag{1}$$

An important property in the case of materials that can undergo irreversible deformation is that their entropy must increase, which is equivalent to the positiveness of the internal dissipation power:

$$\mathcal{D}_{int} = \sum_{i,j} \left(S_{ij}-2\frac{\partial \psi}{\partial C_{ij}}\right)\frac{\partial{C}_{ij}}{\partial t} + \sum_{k} \frac{\partial \psi}{\partial \xi_k}\frac{\partial \xi_k}{\partial t} \ge 0$$

because of $(1)$ the first term is zero.


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