In case of sliding conditions the shear stress is zero: t_{nt}=0. The stress vector now amounts to:
\begin{bmatrix} t_{nn} & 0 \\0 & t_{tt} \end{bmatrix} \cdot \begin{bmatrix} -1 \\0 \end{bmatrix} = \begin{bmatrix} -t_{nn} \\0 \end{bmatrix} | (628) |
or
\boldsymbol{t}= -t_{nn} \boldsymbol{e}_n = t_{nn} \boldsymbol{n} , | (629) |
where
t_{nn} = 2 \mu^T v_{n,n} - \frac{2}{3} (\mu^T v_{k,k} + \rho k). | (630) |
Consequently, \boldsymbol{t} can be approximated by:
\boldsymbol{t} \approx 2 \mu^T \frac{\boldsymbol{v}_P \cdot \boldsymbol{e}_n }{(\boldsymbol{r}_S - \boldsymbol{r}_P) \cdot \boldsymbol{n}} \boldsymbol{n} - \frac{2}{3} \left( \mu^T v_{k,k} + \rho k \right ) \boldsymbol{n}. | (631) |
This finally amounts to:
\boldsymbol{t} \approx -2 \mu^{T(m-1)} \frac{\boldsymbol{v}_P^{(m-1)} \cdot \boldsymbol{n} }{(\boldsymbol{r}_S - \boldsymbol{r}_P) \cdot \boldsymbol{n}} \boldsymbol{n} - \frac{2}{3} \left( \mu^{T(m-1)} )(\nabla \cdot \boldsymbol{v})^{(m-1)} + \rho^{(m-1)} k^{(m-1)} \right ) \boldsymbol{n}. | (632) |