$\newcommand{\Dev}{\operatorname{Dev}}$ $\newcommand{\dev}{\operatorname{dev}}$ $\newcommand{\grad}{\operatorname{grad}}$ $\newcommand{\Grad}{\operatorname{Grad}}$ $\newcommand{\divg}{\operatorname{div}}$ $\newcommand{\Ei}{\operatorname{Ei}}$ $\newcommand{\tr}{\operatorname{tr}}$ $\newcommand{\Divg}{\operatorname{Div}}$ $\newcommand{\cay}{\operatorname{cay}}$ $\newcommand{\rot}{\operatorname{rot}}$

In this case, the active stress is isotropic in a plane transverse to the direction $\mathbf{a}_{r}$ , $$\begin{equation} \mathbf{S}^{a}=T^{a}\left(\mathbf{I}-\mathbf{a}_{r}\otimes\mathbf{a}_{r}\right)\,,\label{eq521} \end{equation}$$ and the corresponding Cauchy stress is $$\begin{equation} \boldsymbol{\sigma}^{a}=J^{-1}T^{a}\left(\mathbf{B}-\mathbf{a}\otimes\mathbf{a}\right)\,,\label{eq522} \end{equation}$$ where $\mathbf{B}=\mathbf{F}\cdot\mathbf{F}^{T}$ is the left Cauchy-Green tensor. The material and spatial tangents are zero.