Uniaxial stress question Let's have a rectangular profiled bar. Let us introduce force $\vec{F}$ which pull the bar apart. In the picture below let us make a virtual horizontal cut $A$.

Well, everything is in the picture. Nothing fancy. But the part I'm stuck with is this: 
Let's instead of cut $A$ make a cut $B$ which will be perpendicular to $A$'s normal. That is, $B$'s normal is perpendicular to $\vec{F}$. From my point of view, the force $\vec{F}$ will now be shearing plane $B$. But, of cource, every textbook say that there will be NO stress (neither normal nor tangental) on the plane $B$. 
And that's where I'm stuck: My intuition says that $\vec{F}$ will shear $B$, but theory says -- it will not. 
I guess my problem lies in the fact that I don't understand why Tractrions(Forces) on cuts with different normals can't add up. But nowhere I've seen any thorough explanation about this inability of comparing tractions on different cuts.
Please, help.
 A: There will be no shear across plane B. To see why, imagine a very small 'needle' embedded in the solid block, with the needle pointing from left to right in your picture. Also, with the center of the needle residing in the center of the block. I'm imagining the needle penetrating plane B. If there were shear, the needle would have to rotate. But it doesn't rotate, it's just displaced.
Something like this (excuse the poor drawing):

I'm representing plane B as the thin black line, and the needle as the thin red line intersecting it. You can imagine tensioning the block on the top and bottom points of plane B. You can see that the needle doesn't rotate.
Now if you apply force asymmetrically, for example on the top-left corner and the bottom-right corner, then the needle would indeed rotate a bit.
A: I think the force is assumed to be uniformly distributed over the cross section.If uniformly distributed, ther will be no shearing effect.But if applied at a point then the analysis becomes complicated.Refer to the image attached.
A: To supplement the answers by Al Nejati and Mohan, a few comments on your question:


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*"My intuition says that $\vec{F}$  will shear $B$, but theory says -- it will not." Because of the symmetry of the object, we can immediately rule out vertical shear at the centerline because if you performed a mirror reflection, the shear would be acting in an incompatible direction.

*"I guess my problem lies in the fact that I don't understand why [tractions (forces)] on cuts with different normals can't add up." It's required that forces applied to any particular element must add up to zero in every direction (otherwise the element would accelerate away). But there's no requirement that one can add up various traction forces across and within an object and obtain some constant value.

*"But nowhere [have I] seen any thorough explanation about this inability of comparing tractions on different cuts." Well, this is what Mohr's circle is all about, among other graphical interpretations of the stress transformation equations. Shear requires not one surface load nor two but four to prevent rotation. These four forces can't all be applied in the 0° or 90° orientation for the geometry you describe because there's nothing pulling or pushing laterally (ignoring Poisson effects). However, you can obtain a state of nonzero shear in the 45° orientation (see the photo of stretched rubber here).
A: The figure below shows a dashed line representing the cut made by vertical plane B.  The fraction of the rod on this side of the cut is $\alpha$, so the tensile force on each of the ends of the truncated section is $\alpha F$.  The shear force on the cut (caused by the portion of the rod on the other side of the cut) is $Q_B$.
The force balance in the vertical direction on this portion of the rod is:  $$\alpha F+Q_B-\alpha F=0$$Therefore, it follows that $$Q_B=0$$

