# Cauchy tensor tensor components from Energy function

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)

Thanks

• "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. May 6 '17 at 10:47
• 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? May 6 '17 at 15:03
• 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. May 8 '17 at 20:54
• Thank you for the tip, I am a beginner :-) Can you tell me if it is correct my answer? 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.