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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 $A$.

enter image description here

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.

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  • $\begingroup$ Let's see your free body diagrams when the cut is made at B. I contend that these diagrams will show that there is no shear on plane B. $\endgroup$
    – Chet Miller
    Oct 10, 2018 at 14:59
  • $\begingroup$ Those pics are from Wiki. And I know that there will be no shear stress, I just don't understand why. $\endgroup$
    – coobit
    Oct 10, 2018 at 19:58
  • $\begingroup$ Do you not have PowerPoint? $\endgroup$
    – Chet Miller
    Oct 10, 2018 at 22:04

4 Answers 4

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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):

enter image description here

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.

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  • $\begingroup$ Yes, I see your point. Does it mean that Tractions are not .... true vectors and you can't add them up? Or you can't add tractions from different planes? $\endgroup$
    – coobit
    Oct 8, 2018 at 21:14
  • $\begingroup$ The stresses are identical on the 'left' and 'right' of the plane. There is no change in direction of stress as you move through the plane, thus there is no shear. If you want to think about it in terms of traction, at any point on plane B, the traction on the left side is the same as the traction on the right side. $\endgroup$
    – Al Nejati
    Oct 8, 2018 at 21:37
  • $\begingroup$ Sorry, I don't understand this 'left' and 'right'. Can you elaborate? Tnx. $\endgroup$
    – coobit
    Oct 9, 2018 at 5:34
  • $\begingroup$ Imagine instead of one plane B, you had two planes parallel to each other, a small distance apart. You would observe that the stresses and strains on both planes look identical. This would not be the case if there were shear. $\endgroup$
    – Al Nejati
    Oct 9, 2018 at 5:56
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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 $$Q_B=0$$

Shear Force on Surface B

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  • $\begingroup$ Thanks for the picture. I'm starting to understand all of you. $\endgroup$
    – coobit
    Oct 11, 2018 at 7:37
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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.enter image description here

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  • $\begingroup$ Although, the picture is just great, I still do not feel that I undestand the crux of my misunderstanding. Why there will be no force on $B$? Is it because you can't add tractions from different cuts or is it because any shear stress must rotate while normal stress $\vec{F}$ on $A$ does not rotate material, thus it can't rotate from the point of $B$? $\endgroup$
    – coobit
    Oct 9, 2018 at 8:59
  • $\begingroup$ When a fluid flows,viscous shear occurs due to relative motion between layers.Similarly for shear between sections , there should be tendency of relative motion between several vertical section(like B).If there is no relative motion, then there wont be any traction between vertical planes.The points on the adjacent vertical sections move together.So no shear stress along the vertical planes.But this is a vague way of imagining it(I think this is wrong). $\endgroup$
    – Mohan
    Oct 9, 2018 at 12:08
  • $\begingroup$ You can get a clear idea if you study the state of stress at a point(for uniaxial loading) and Mohrs circle.I think you will get better answers in engineering stack exchange(I think there were a few people experienced in structural engineering and machine design). $\endgroup$
    – Mohan
    Oct 9, 2018 at 12:08
  • $\begingroup$ So, basiclly, you tell me that one can't add pure pressure vector in the direction of $\vec{F}$ and pure shear vector in the same direction but on the other cut-plane. Right? Pressure and shear are so different in their effect, so there is no meaning in adding them up even if they point in the same direction? $\endgroup$
    – coobit
    Oct 9, 2018 at 13:26
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To supplement the answers by Al Nejati and Mohan, a few comments on your question:

  • "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).

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