In the finite-deformation theory, the elastic Cauchy-Green strain $\mathbf{E}_e$ is defined as

$\mathbf{E}_e=\frac{1}{2}(\mathbf{F}_e^T \mathbf{F}_e-\mathbf{I})$,

where the superscript $T$ denotes the transpose, $\mathbf{F}_e$ is the elastic component of the deformation gradient tensor $\mathbf{F}$ and $\mathbf{I}$ is the identity tensor.

With reference to the multiplicative decomposition of the deformation gradient $\mathbf{F}$, i.e. $\mathbf{F}=\mathbf{F}_e\mathbf{F}_p$, where $\mathbf{F}_p$ is the plastic part of the deformation gradient, $\mathbf{F}_e$ is calculated as


One can assume the formulation of $\mathbf{F}_p$ as following

$\mathbf{F}_p=\exp((1-p_0)(p \log \mathbf{U}_I+(1-p)\log \mathbf{U}_J))$,

where $p_0$ and $p$ are the plastic variables and $\mathbf{U}_I$ and $\mathbf{U}_J$ are the plastic stretch tensors, which are tensors possessing real numbers. Now, the question is how the differentiation of $\mathbf{E}_e$ with respect to $p_0$, i.e. $\displaystyle \frac{\partial \mathbf{E}_e}{\partial p_0}$, can be calculated.


For ease of typing, denote the operation $\frac{\partial}{\partial p_0}$ by $d$ $$\eqalign{ A &= p\log U_I + (1-p)\log U_J \cr F_p &= e^{(1-p_0)A} \cr dF_p &= -A F_p \cr\cr F_e &= FF_p^{-1} \cr dF_e &= F\,dF_p^{-1} \cr &= -FF_p^{-1}\,dF_p\,F_p^{-1} \cr &= FF_p^{-1}\,A F_p\,F_p^{-1} \cr &= F_eA \cr\cr E_e &= \tfrac{1}{2}(F_e^TF_e-I) \cr dE_e &= \tfrac{1}{2}(F_e^T\,dF_e+dF_e^T\,F_e) \cr &= \tfrac{1}{2}(F_e^TF_eA+A^TF_e^TF_e) \cr &= \tfrac{1}{2}((2E_e+I)A+A^T(2E_e+I)) \cr &= (E_eA + A^TE_e) + \tfrac{1}{2}(A+A^T) \cr }$$

  • $\begingroup$ nice way to do it! Thanks! $\endgroup$ – KratosMath Apr 23 at 16:32

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