I need to find a relation between the tangent stiffness $L_1$ of the first Piola-Kirchhoff stress tensor with the tangents stiffness $L_2$ of the second Piola-kirchoff stress tensor. They are defined as

\begin{align} &L_1 = \frac{dP}{dF}\, ,\\ &L_2 = 2\frac{dP}{dC}\, , \end{align}

with $F$ being the deformation gradient and $C$ being the right Cauchy-Green tensor.

If I got $L_1$, there is a way to compute $L_2$?

  • 1
    $\begingroup$ I have formatted your equations. Can you define $P$? $\endgroup$ – nicoguaro Feb 2 at 17:37

Ignoring subtleties regarding differentiation in two separate tangent spaces, and also avoiding complications arising from non-Euclidean coordinates, we can proceed as follows.

The derivative of $\boldsymbol{P}$ with respect to $\boldsymbol{C}$ is $$ \frac{\partial\boldsymbol{P}}{\partial\boldsymbol{C}} = \frac{\partial\boldsymbol{P}}{\partial\boldsymbol{F}} : \frac{\partial\boldsymbol{F}}{\partial\boldsymbol{C}} $$ where, for two fourth-order tensors $\mathbb{A}$ and $\mathbb{B}$, the contraction operaion is defined as (summation implied on repeated indices) $$ \mathbb{A} : \mathbb{B} = A_{ijk\ell} B_{k\ell mn} $$ We need to find $\partial\boldsymbol{F}/\partial\boldsymbol{C}$. Let us start with the definition of $\boldsymbol{C}$: $$ \boldsymbol{C} = \tfrac{1}{2} (\boldsymbol{F}^T \cdot \boldsymbol{F} - \boldsymbol{I}) $$ where the partial inner product of two second order tensors $\boldsymbol{A}$ and $\boldsymbol{B}$ is defined as $$ \boldsymbol{A}\cdot\boldsymbol{B} = A_{ij} B_{jk} $$ Then, $$ \frac{\partial\boldsymbol{C}}{\partial\boldsymbol{F}} = \tfrac{1}{2}\left[ \frac{\partial\boldsymbol{F}^T}{\partial\boldsymbol{F}}\cdot\boldsymbol{F} + \boldsymbol{F}^T \cdot\frac{\partial\boldsymbol{F}}{\partial\boldsymbol{F}}\right] $$ where the contraction between a fourth order tensor $\mathbb{A}$ and a second order tensor $\boldsymbol{B}$ is defined as $$ \mathbb{A} \cdot \boldsymbol{B} = A_{ijkm} B_{m\ell} \quad \text{and} \quad \boldsymbol{B} \cdot \mathbb{A} = B_{im} A_{mjk\ell} $$ Expressions for derivatives can be found at https://en.wikipedia.org/wiki/Tensor_derivative_(continuum_mechanics)#Derivative_of_a_second-order_tensor_with_respect_to_itself. Using these, we have $$ \frac{\partial\boldsymbol{F}^T}{\partial\boldsymbol{F}}\cdot\boldsymbol{F} = \delta_{im}\delta_{jk} F_{m\ell} = F_{i\ell} \delta_{jk} $$ and $$ \boldsymbol{F}^T \cdot\frac{\partial\boldsymbol{F}}{\partial\boldsymbol{F}} = F_{mi} \delta_{mk} \delta_{j\ell} = F_{ki} \delta_{j\ell} $$ which leads to $$ \left[\frac{\partial\boldsymbol{C}}{\partial\boldsymbol{F}}\right]_{ijk\ell} = \tfrac{1}{2}(F_{i\ell} \delta_{jk} + F_{ki} \delta_{j\ell}) $$ Finally, $$ \mathbb{L}_2 = 2\frac{\partial\boldsymbol{P}}{\partial\boldsymbol{F}} :\left[\frac{\partial\boldsymbol{C}}{\partial\boldsymbol{F}}\right]^{-1} $$ where the inverse of the fourth-order tensor can be computed using the approach given in https://math.stackexchange.com/questions/1624955/the-inverse-of-a-fourth-order-tensor.

It may actually be simpler to start with the definition of $\boldsymbol{P}$ and $\boldsymbol{S}$, e.g., $$ \boldsymbol{P} = \det(\boldsymbol{F}) \,\boldsymbol{F}^{-1}\cdot\boldsymbol{\sigma} ~,~~ \boldsymbol{S} = \det(\boldsymbol{F})~\boldsymbol{F}^{-1}\cdot\boldsymbol{\sigma}\cdot\boldsymbol{F}^{-T} $$ and compute the derivative directly using $$ \frac{\partial}{\partial\boldsymbol{A}}\det(\boldsymbol{A}) = \det(\boldsymbol{A})~\left[\boldsymbol{A}^{-1}\right]^\textsf{T} $$ and $$ \frac{\partial }{\partial \boldsymbol{A}} \left(\boldsymbol{A}^{-1}\right) : \boldsymbol{T} = - \boldsymbol{A}^{-1}\cdot\boldsymbol{T}\cdot\boldsymbol{A}^{-1} \implies \frac{\partial A^{-1}_{ij}}{\partial A_{kl}} = - A^{-1}_{ik}~A^{-1}_{lj} $$ That's left as an exercise for the reader :)

  • $\begingroup$ See section 6.1.2 of Odgen's book "Non-linear elastic deformations" for an alternative approach. $\endgroup$ – Biswajit Banerjee Feb 3 at 16:10

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.