I hope someone could help me.

Let $W\left( \lambda_1,\lambda_2,\lambda_3 \right) = \sum_{p=1}^N \frac{\mu_p}{\alpha_p}\left( \lambda_1^{\alpha_p} + \lambda_2^{\alpha_p} + \lambda_3^{\alpha_p} -3 \right)$ be a strain energy density function. How can I find the non-diagonal elements of the Cauchy tensor?

Edit 1: Basically I want to derive the equations motion for hyperelastic material. In order to achieve that I first need the Cauchy Tensor.

Edit 2: As far as I know, the $ \lambda_k \frac{\partial W(\lambda_1,\lambda_2,\lambda_3)}{\partial \lambda_k} + p = \sigma_{kk} = \sigma_{k}$ (no summing convention) for incompressible and isotropic hyperelastic materials[Ogden 2001]. But what abount the $\sigma_{kj}$ | $k \neq j$? How can I calculate them?

Edit 3: I think I have found the answer. When calculating $\sigma_k$, using the description above, I am implicitly using a frame of reference which the tensor $\sigma$ is diagonal. This frame of reference concedes with the principal directions of the deformation.

[Ogden 2001] Nonlinear Elasticity: Theory and Applications (Ogden)


  • $\begingroup$ "A strain energy density function is used to define a hyperelastic material by postulating that the stress in the material can be obtained by taking the derivative of $W$ with respect to the strain." source. I honestly think that using wiki before asking would be nice. $\endgroup$
    – Sanya
    May 6 '17 at 10:47
  • $\begingroup$ Good that you think that, because I honestly used the wiki before asking. The partial derivative of W in relation to "lambda i" gives (i,i) element of the Cauchy tensor right? But how can I calculate the element (i,j) for i different of j? $\endgroup$
    – Caslu
    May 6 '17 at 15:03
  • 1
    $\begingroup$ If you want to avoid annoying comments of that sort, just include all relevant information in your question the next time ;) it will make your life on this website a lot easier. $\endgroup$
    – Sanya
    May 8 '17 at 20:54
  • $\begingroup$ Thank you for the tip, I am a beginner :-) Can you tell me if it is correct my answer? $\endgroup$
    – Caslu
    May 9 '17 at 7:58

Dvorkin and Goldschmit ([1], p.120) define a hyperelastic material as one whose stress power is an exact differential and for which thus the strain $ \boldsymbol{\sigma} $ is given by $$ \boldsymbol{\sigma} = \frac{\partial U}{\partial \boldsymbol{E}} $$ with $\boldsymbol{E}$ the strain and $U$ the elastic energy function per volume.
In general, this translates into $$\sigma_{ij} = \frac{\partial U}{\partial E^{ij}} $$ Thus, your concern about off-diagonal components is quite right in general.
Usually, the variables $\lambda_1,\lambda_2,\lambda_3$ are used for the principal strains and there is a coordinate system in which $\boldsymbol{E}=\sum_j \lambda_j \phantom{a} \vec{e}_j \otimes \vec{e}_j$. In this coordinate system, there should be no off-diagonal components of $\boldsymbol{\sigma}$ either because of its definition via the partial derivatives.
So yeah, I think your edits are correct.

[1] Eduardo N. Dvorkin, Marcela B. Goldschmit: Nonlinear Continua; Springer-Verlag Berlin Heidelberg 2005


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.