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For this criterion it is assumed that damage is initiated by increases in the von Mises (or effective) stress, $\sigma_{Y}$ . Thus, $$\begin{equation} \Xi\left(\mathbf{U}\right)=\sigma_{Y}\left(\mathbf{U}\right)=\sqrt{\frac{3}{2}\dev\boldsymbol{\sigma}:\dev\boldsymbol{\sigma}}\,,\label{eq:dmg-model-VMS} \end{equation}$$ where $\dev\boldsymbol{\sigma}$ is the deviatoric part of $\boldsymbol{\sigma}$ . To evaluate the damage surface normal in this case, we must use the chain rule, $$\begin{equation} \mathbf{N}=\frac{\partial\Xi}{\partial\boldsymbol{\sigma}}:\frac{\partial\boldsymbol{\sigma}}{\partial\mathbf{F}}:\frac{\partial\mathbf{F}}{\partial\mathbf{U}}\,.\label{eq:dmg-normal-chain-rule} \end{equation}$$ From the hyperelasticity relation in Eq.(5.6.2-5) it can be shown that $$\begin{equation} \frac{\partial\boldsymbol{\sigma}}{\partial\mathbf{F}}=\left(\boldsymbol{\mathcal{C}}+\mathbf{I}\,\underline{\otimes}\,\boldsymbol{\sigma}+\boldsymbol{\sigma}\,\overline{\otimes}\mathbf{I}-\boldsymbol{\sigma}\otimes\mathbf{I}\right)\cdot\mathbf{F}^{-T}\,,\label{eq:dmg-stress-tangent} \end{equation}$$ where $\boldsymbol{\mathcal{C}}$ is the fourth-order spatial elasticity tensor associated with the strain energy density $\Psi_{r}$ . Then, it can be shown that $$\begin{equation} \mathbf{N}=\frac{1}{2}\mathbf{R}^{T}\cdot\mathbf{M}\cdot\mathbf{R}\cdot\mathbf{U}^{-1}+\frac{1}{2}\mathbf{U}^{-1}\cdot\mathbf{R}^{T}\cdot\mathbf{M}^{T}\cdot\mathbf{R}\,,\label{eq:dmg-N-redux} \end{equation}$$ where $$\begin{equation} \mathbf{M}=\frac{\partial\Xi}{\partial\boldsymbol{\sigma}}:\boldsymbol{\mathcal{C}}+2\frac{\partial\Xi}{\partial\boldsymbol{\sigma}}\cdot\boldsymbol{\sigma}-\left(\frac{\partial\Xi}{\partial\boldsymbol{\sigma}}:\boldsymbol{\sigma}\right)\mathbf{I}\,.\label{eq:dmg-N-M} \end{equation}$$ From the von Mises criterion in Eq.(5.6.5.4-1), it can be shown that $$\begin{equation} \frac{\partial\Xi}{\partial\boldsymbol{\sigma}}=\frac{3}{2\sigma_{Y}}\dev\boldsymbol{\sigma}\,.\label{eq:dmg-VMS-tangent} \end{equation}$$ Substituting Eqs.(5.6.5.4-5)-(5.6.5.4-6) into Eq.(5.6.5.4-4) provides the surface normal $\mathbf{N}$ .