2
$\begingroup$

I am interested in calculating the total potential energy stored in a finite element mesh given its nodal point displacements alone. The forces that created the displacements are irrelevant because the objective is to calculate the strain energy stored in the mesh post-deformation.

From what I understand the total potential energy can be computed by integrating the strain energy density $W(x, \epsilon(x))$

$$ \Pi = \int_\Omega W(x, \epsilon(x)) d\Omega $$

In a two-dimensional discrete setting

$$ \Pi = \sum_e W_e(x, \epsilon(x)) \cdot A_e $$

where $A_e$ is the area of the element $e$. Is this correct?

If it matters the material is hyperelastic.

$\endgroup$
16
  • 2
    $\begingroup$ I'm missing something. I'm not sure what your question is - it seems you already know the answer. $\endgroup$
    – Nachiket
    Jan 11 at 9:56
  • $\begingroup$ @Nachiket I'm not sure if my thinking is right. For example in three-dimensions would I be multiplying $W_e$ by the volume $V_e$ of an element? $\endgroup$
    – Olumide
    Jan 11 at 11:31
  • 1
    $\begingroup$ I cannot understand the link you referred to. Why don't you take a look at equations (3.9.1)-(3.9.9) from TJR Hughes's "The finite element method: linear static and dynamic finite element analysis" $\endgroup$
    – Nachiket
    Jan 15 at 9:58
  • 1
    $\begingroup$ That's from Continuum Mechanics Volume 1 by C.S. Jog -- Equations (2.5b) and (2.6b). The point, as I understand it, is that $u(x)$ is over the deformed configuration, not the reference configuration. (Happy to continue this conversation by email if you like.) $\endgroup$
    – Olumide
    Jan 16 at 3:03
  • 1
    $\begingroup$ @Nachiket I did a few days ago. Please check your spam bin (gok*****@gmail.com) $\endgroup$
    – Olumide
    Jan 19 at 8:18
3
$\begingroup$

The elastic energy stored in your solid is computed as

$$\Pi = \int_\Omega \sigma : \epsilon\, \mathrm{d}\Omega\, ,$$

where $\sigma$ is the stress tensor, $\epsilon$ is the strain tensor, and $:$ is the double contraction over the tensors.

When you discretize the solid, the strain energy is

$$\Pi_h = \mathbf{F}^T \mathbf{U}\, ,$$

where $h$ represents the discretization, $\mathbf{F}$ is the vector of nodal forces, and $\mathbf{U}$ is the vector of nodal displacements. Furthermore, we know that

$$\mathbf{F} = [\mathbf{K}] \mathbf{U}\, ,$$

thus, we can also compute the energy as

$$\Pi_h = \mathbf{U}^T [\mathbf{K}] \mathbf{U}\, .$$

$\endgroup$
2
  • $\begingroup$ Perhaps I don't fully appreciate your answer but nodal forces are not given -- just the nodal displacements and the strain energy density function. $\endgroup$
    – Olumide
    Jan 12 at 0:22
  • 1
    $\begingroup$ @Olumide once the nodal displacements and material properties are known you can calculate $\mathbf{F}=[\mathbf{K}]\mathbf{U}$ which is commony referred to as the vector of nodal forces. In the general hyperelastic case, you won't be able to get such a nice linear relation. You would need to evaluate the integral I mentioned in the previous comment. $\endgroup$
    – Nachiket
    Jan 12 at 3:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.