Sliding conditions

In case of sliding conditions the shear stress is zero: $ t_{nt}=0$. The stress vector now amounts to:

$\displaystyle \begin{bmatrix}t_{nn} & 0 \\ 0 & t_{tt} \end{bmatrix} \cdot \begin{bmatrix}-1 \\ 0 \end{bmatrix} = \begin{bmatrix}-t_{nn} \\ 0 \end{bmatrix}$ (628)

or

$\displaystyle \boldsymbol{t}= -t_{nn} \boldsymbol{e}_n = t_{nn} \boldsymbol{n} ,$ (629)

where

$\displaystyle t_{nn} = 2 \mu^T v_{n,n} - \frac{2}{3} (\mu^T v_{k,k} + \rho k).$ (630)

Consequently, $ \boldsymbol{t}$ can be approximated by:

$\displaystyle \boldsymbol{t} \approx 2 \mu^T \frac{\boldsymbol{v}_P \cdot \bold...
...dsymbol{n} - \frac{2}{3} \left( \mu^T v_{k,k} + \rho k \right ) \boldsymbol{n}.$ (631)

This finally amounts to:

$\displaystyle \boldsymbol{t} \approx -2 \mu^{T(m-1)} \frac{\boldsymbol{v}_P^{(m...
...\cdot \boldsymbol{v})^{(m-1)} + \rho^{(m-1)} k^{(m-1)} \right ) \boldsymbol{n}.$ (632)