no slip boundary condition for stress tensor
Deviatoric solid stress components, \( {\bf S}_s\), are set to zero in all tangential directions:
\begin{align} {\bf t_1} \cdot {\bf S_{s}} \cdot {\bf t_1} = 0 \\ {\bf t_2} \cdot {\bf S_{s}} \cdot {\bf t_2} = 0\end{align} \(\bf n\) is the boundary normal, \(\bf t_1\) and \(\bf t_2\) are the associated tangentials, i.e. \begin{align} {\bf t_1} \cdot {\bf n} = 0 \\ {\bf t_2} \cdot {\bf n} = 0\end{align}with
\begin{align} \left\Vert{\bf t_1}\right\Vert = \left\Vert{\bf t_2}\right\Vert = \left\Vert{\bf n}\right\Vert = 1\end{align}