# relation between different tangent stiffness

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 $$L_2$$?

• I have formatted your equations. Can you define $P$? Feb 2, 2021 at 17:37

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 :)