Traction on the boundary - how to compute from velocity field

Question

can you help me? I have a fluid simulation in 2D, where I compute by finite differences on the grid the velocity field and pressure. So this is my output from the program. Now I have to show how the vertical component of traction on the boundary looks like. But I am not sure what the word traction means, what I have to compute. I can compute the stress from the velocity field as

$$ \sigma_{xy}=\eta (\frac{dv_x}{dy}+\frac{dv_y}{dx}) $$

$$ \sigma_{xx}=2\eta \frac{dv_x}{dx} $$

$$ \sigma_{yy}=2\eta \frac{dv_y}{dy} $$

So this I would know how to compute from my results - by finite differences. What does the vertical traction mathematically means, what I have to compute?
What I about know is, that the traction in vertical (y-axis) could be: $$ \vec{t_y} = \sigma_{xy}\vec{e_x} + \sigma_{yy}\vec{e_y} $$ Am I right? Many thanks

Thsi is about traction:
"Note a convention that we have implicitly established: the sense of the force per unit area $t$ across the oblique surface is that the fluid penetrated by the unit normal $n$ acts on the fluid on the other side of the surface. This force per unit area is called the traction across the surface. Note further that $t_x$ indicates the traction across the y − z plane, not the x component of a traction. Representation of components requires a second subscript: $T_{xy}$ is the y component of the traction $t_x$ across the y − z face of the tetrahedron, whereas $T_{xy}$ is the x component of the traction ty across the x − z face. Thus, the first subscript represents the vector component of the force and the second subscript indicates the face on which the force is acting. " But how to compute it?

Answer 1

First, the stress in fluid dynamics must also include the pressure. In fact, it is $$ \sigma = 2 \eta \varepsilon(\mathbf v) + p I $$ where $\varepsilon(\mathbf v)$ is the symmetric gradient of the velocity $\mathbf v$. Then, the traction is the normal component of the stress at the boundary: $$ \mathbf t = \sigma \mathbf n $$ If you want you can further decompose the traction into the normal force $f = \mathbf n \cdot \mathbf t = \mathbf n \cdot \sigma \mathbf n=2\eta \mathbf n \cdot \varepsilon(\mathbf v) \mathbf n + p$ and the tangential friction force $\mathbf t - f \mathbf n$ for which it is easy to show that it doesn't depend on the pressure any more.