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: enter image description here

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.

  • $\begingroup$ Are you talking about classical Cauchy stress? $\endgroup$ – Sanya Aug 22 '16 at 9:47
  • $\begingroup$ mmm. I don't know if there is any other. In every book I've read the talk was about Cauchy stress tensor $\sigma$ (where matrix form of $\sigma$ contains information about 3 tractions across 3 perpendicular planes) $\endgroup$ – coobit Aug 22 '16 at 9:52
  • $\begingroup$ good, then we are talking about the same stress/traction :) I will write an answer, but it might be a bit before I come around to doing so $\endgroup$ – Sanya Aug 22 '16 at 12:56
  • $\begingroup$ Hey coobit, it has been a bit but I've finally come around to writing an answer. I hope this is not too late for you and I hope it adresses your issue. Please ask for any clarification you might want and adress any incomprehensible parts of what I wrote in the comment to my answer. $\endgroup$ – Sanya Sep 4 '16 at 11:06

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.

  • $\begingroup$ "The important point here is that the traction vector depends on the chosen surface by definition" Now, that's a problem! I just don't get it. Welll, while waiting for your answer I started to grasp the reason for WHY the DEFINITION is the way it is. Let' load body B. This causes external forces to be transmitted to internal elements of B. Now let's cut B in arbitrarry orientation. Immediatly after the cut a chasm will appear and will start to widen or shrink. Now, you need to apply forces to keep this chasm from appearing. But still my understanding is shaky here. $\endgroup$ – coobit Sep 4 '16 at 15:13
  • $\begingroup$ Just to be sure - I am talking about mathematical surfaces, not physical surfaces (inside a continous material body, there are no surfaces). I have no clue about fracture mechanics. But taking your example: if we load body B from one side, the external forces get transmitted with a direction throughout the body. For small loads, there is probably no force inside the body normal to the applied external force. So the traction on surfaces parallel to the direction of the app. ex. force is probably zero, while it is nonzero on surfaces which have the external load direction as normals. $\endgroup$ – Sanya Sep 4 '16 at 15:18
  • $\begingroup$ You've missed the point. If I make a cut in a body under load the cut will expirience strain i.e. the straight line of a cut will deform in some way. If the load was "pulling body B apart" the cut will open up (a cut would not be felt in a body BEFORE any loading, but after a loading a cut will widen). Now of corse this all is imaginary and has nothing to do with fracture mechanics. Look her for the example: maps.unomaha.edu/maher/GEOL3300/week4/stress.html $\endgroup$ – coobit Sep 4 '16 at 17:43
  • $\begingroup$ now I understand the concept of the imaginary cut and I actually like the picture. But let's say we apply a load in the $x$-direction and cut the material along the $x$ axis. The cut will not widen because the traction in the $y$/$z$-direction is zero. If we cut along $y$/$z$ it will widen because there is traction in the $x$ direction and exactly in that way is the traction vector dependent on the plane/cut. $\endgroup$ – Sanya Sep 4 '16 at 19:05
  • $\begingroup$ Who is answering whose question? :) by the way it looks to me that under this "chasm" setup, traction(resultant traction) can NEVER be parralel to the cut! Since cut is not widening when forses act along it (though it should not be the case for real cut because it has non zero width as a result of the knife's blade thickness) $\endgroup$ – coobit Sep 4 '16 at 21:00

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.

  • $\begingroup$ definition: traction is a force divided by a surface area. As far as I know traction is a one part of a pair of forces acting on a surface in opposite directions, so for each surface(orientation) there are 2 opposite and equal forces acting on it. And how can tractions add up to a force? I thought tractions can add up to a traction only! Now if we consider only tractions (not forces) and add them upp across every possible orientation (surface) obtaining sum non zero resultant traction vector then we will must have the opposite traction for this sum so the body remains at rest. $\endgroup$ – coobit Sep 28 '16 at 7:40
  • $\begingroup$ @coobit When you draw a free body diagram for a portion of a body in equilibrium, the sum of forces acting on that portion must be zero. You don't say there are a pair of opposing forces at every surface of the part and so it must be in equilibrium, because equality of force pairs is a necessary due to 3rd law, irrespective of whether or not there is equilibrium. Thus when I consider cubical volume of water, at each surface, I must consider one of the pair, and that which is being exerted by the outside fluid on the cubical volume of water. $\endgroup$ – Deep Sep 28 '16 at 8:39
  • $\begingroup$ @coobit Traction as you have defined it, when multiplied by area gives your usual force, and you may add them up; if all those areas are equal you may simply sum the traction vectors. $\endgroup$ – Deep Sep 28 '16 at 8:39

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.


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