Why traction vector is zero? I'm reading a book about continuum mechanics and I can't understand why traction vector depends on orientation. Here is an extract from the book: 




As far as I understand no matter what orientation you make a cut the forces acting on the system do not change. So the F in invariant. Now the author says that if you do a cut perpendicular to a truss then you will see that a traction vector $t(i)$ (not a coordinate vector) acting along x axis (that is along i cut normal). But if you make a cut in the same point perpendicular to x axis then traction dissapears suddenly!!! This is absurd! the traction will be tangential to the new cut. 
Help me. I see similar claims by many other authors and I don't understand how this is possible.
 A: 
As far as I understand no matter what orientation you make a cut the
  forces acting on the system do not change.

I think this is your main point of confusion. The force does depend on where you make the cut. For example, imagine you are holding two ropes, one in each hand, and people pull equally on the ropes in opposite directions. Now what is the force on you? It is zero. But what happens to the force on you if we cut one of the ropes? The force does not stay the same. Instead you will find that if you cut the rope on your right, then you will be pulled to the left. Meanwhile if you cut the rope on the right, you will be pulled to the left. So we see that where you make the cut does affect the forces on the system.
Now let's consider a slightly more complicated example where you additionally have ropes pulling you forward and backward. Now if you just cut the rope pulling you forward you will be pulled backward, and if you just cut the rope pulling you backward, you will be pulled forward.
The difference between cutting the rope to your left and cutting the rope in front of you corresponds to a difference in the orientation of the cut. Thus the orientation of the cut does affect the direction of the resulting force.
If we translate the books example into this language, we would imagine a grid of people who have ropes attached to them in all four directions, but only the left and right ropes have tension; the forward and backward are just loose. Now if we go in and make a big cut through this network cutting along the "$y$" direction as shown in the original picture, then we are cutting ropes that have tension in them, so the people at the surface where the cut was made feel an uncompensated tension away from the surface, and they will be pulled away. Thus there is a non-zero traction force. Now if instead we cut in the "$x$" direction, we are cutting ropes that don't have tension in them, so the force on each person remains zero and so there is no traction force.
A: Suppose you have two identical bars with rectangular sections, and a identical forces applied to the ends of each bar. There are no "traction forces" at the surface along the length of the bars. 
Now stick the two bars together, side by side.  Nothing changes. The stress in the bars is the same as before
Now cut them apart again. It would be "absurd" if sticking them together and cutting them apart magically created a traction force from nowhere.
Of course there is a traction force on cut a made  in any other direction, except the axial direction.
And for a general (triaxial) stress field in the bar there will be a non-zero traction force on a cut made in every possible direction.
The extract from the book that you quoted doesn't seem to mention that the constant uniaxial stress field in the example is a special (and simple) situation, not the general case.
