Why traction vector depends on surface(section) orientration? Need help with stress tensors. Every book says that traction vector at a point P depends on orientation of surface cutting this point. But as far as I know traction is defined in this way: Traction is force over area it is acting on, so $\vec{T}$ equals $\vec{F}/| \vec{n}|$. In this case traction is a physical vector (not a coordinate vector) and should not depend on anything at all!
Why the hell physical vector depends on orientation of the surface? I guess my problem lies in my miss understanding what a traction vector is, maybe it's a resulatant of all forces acting on a cut sufface.
Please, explain in excruciating details since I've tried like 20 or 30 sources (intoduction to solid dynamics and stuff) and everywhere I've looked authors just say that "traction depends..." and no details why physical vector all of a sudden depends on something..
Here is a picture in my head:

In it we see a traction vector acting on a point P. Let's make a cut SurfaceH and a cut SurfaceV. So... Traction vector stays the same just the projections on to the different cuts change but not the vector itself.
 A: Before starting to outline my understanding, let me link two related questions on Physics SE here and here. Further, let me give my main sources for learning continuum mechanics which my answer will mainly be inspired by:


*

*Haupt, Continuum Mechanics and Theory of Materials, Berlin Heidelberg: Springer,
2000

*Liu, Continuum Mechanics, Berlin, Heidelberg: Springer, 2002


Let $\mathcal{P}$ be a part of the material body with surface $\partial\mathcal{P}$. We now assume that there are two types of forces that can act on this body part. On the one hand, there are forces that act on the bulk of the material ("on each of the overcountable small particles the body is made up of") and we can characterise them by a body force density. On the other hand, there are forces that are actually transmitted through the material body as contact forces and thus for the body part $\mathcal{P}$ they only act on its surface $\partial\mathcal{P}$. A typical force like this is pressure throughout a fluid. These force contributions on the surface are the surface traction $\vec{\mathbf{t}}$. Cauchy's theorem states that there is a tensor field, the Cauchy stress tensor $\mathbf{T}$, which for a surface with surface normal $\vec{\mathbf{n}}$ gives the traction on that surface at that point as $\mathbf{T}\vec{\mathbf{n}}$. The important point here is that the traction vector depends on the chosen surface by definition because it represents the force contribution onto a body part which is enclosed by this chosen surface. If we choose a different surface, we also get a physically different force because it is the force on another body part.
A: I will tell you what would happen if traction vector did not depend on orientation of surface element. Let us a take simple case: a body of stationary water in the absence of gravity. Since the water is stationary, there are no shear forces acting on or inside it. This means that traction vector must be normal to any given surface element. This alone shows a case where traction vector depends on orientation of surface element. 
But let us go a step further, and consider a infinitesimal cubical volume of water, which must be in equilibrium. If traction vectors were all to point in the same direction irrespective of orientation of surface element then forces on all six faces of the cube would add up to give a resultant force on the fluid element, and so it could not be in equilibrium. This argument can be generalized to any material in equilibrium. Therefore traction vector must in general depend on orientation of surface area element.
A: It is a good question. One example for an intuitive idea is an uniaxial tensile loading of a rod in the $x$ direction. The stress tensor has only one non zero component: $\sigma_{xx}$.
If we multiply the tensor matrix by the direction (1,0,0) the result is exactly the tensile stress $\sigma_{xx}$.
If the surface has any other orientation:
$n = (sin(\theta)cos(\phi),sin(\theta)sin(\phi),cos(\theta))$, the result is $sin(\theta)cos(\phi)\sigma_{xx} < \sigma_{xx}$.
The meaning is that if we design a rod with a weaker plane for some orientation, to induce a premature and controlled rupture, the minimum necessary load corresponds to a transverse section. Any other orientation has a bigger area, and results in a smaller traction for the same load, not enough to break the rod.
