# How to understand shear stress

1. Sep 9, 2014

### Will Flannery

This is not a text book problem, instead it is how to understand a common diagram. The diagram occurs in the wiki entry for shear stress

http://en.wikipedia.org/wiki/Shear_stress

It shows a 2D square, anchored on the bottom side, and with positive (rightward) shear stress applied to the top side.

The only stress shown explicitly is the top side shear stress. My question is, what other stresses are applied to the block that keep it in equilibrium, i.e. not moving and not rotating.

The only other side where a stress can be applied is the bottom. So there must be a stress to counter the top stress to keep the square from moving rightward, and it will be negative and pointing leftward. But there must be other stresses to keep the block from rotating, as both these stresses apply counter-clockwise torques to the square.

It seems to me that there must be an upward normal stress on the bottom right side and a downward normal stress on the left side of the bottom, so that the stress is not constant on the bottom.

This is all new to me and I'm not confident of my understanding of the matter.

2. Sep 9, 2014

### SteamKing

Staff Emeritus
The diagram shown in the Wiki article is perhaps not the best illustration of shear stress.

This diagram may be a better illustration of the nature of shear:

Examine the diagram titled 'Single Shear'. Without getting too complicated in the description (and ignoring other forces at work here), the rivet shank (the circular bit connecting the two heads, top and bottom) is said to be undergoing pure shear at the plane where the two plates lap together. The upper portion of the rivet wants to move to the left, while the lower portion of the rivet wants to move to the right, under the action of the forces P as shown.

Unlike axial tensile stresses, which are constant over the cross section of the member, the shear stress in the rivet shank is distributed in roughly a parabolic fashion across the circular cross section.

These concepts of axial loading, shear loading, and bending loading, are usually presented initially in a highly idealized fashion for instruction purposes. In real structural situations, all three types of loading can be present simultaneously to some degree.

3. Sep 9, 2014

### Will Flannery

@SteamKing,
that's interesting, but I'd like to understand the diagram in the wiki article. In particular, they show a shear stress tau and the caption says that tau produces the deformation shown, but, there must other stresses acting through the base of the cube, and I'd like to understand exactly and explicitly what they are.

4. Sep 9, 2014

### AlephZero

The wiki diagram is poor and (as you seem to have realized) incomplete.

A constant shear stress in the rectangle will produce equal and opposite shear forces on opposite sides, so add an arrow point left at the bottom of the diagram.

But that is only half the story, because those two forces would form a couple and make the element rotate. The components of shear stress always occur in pairs, so if the rectangle is in pure shear (i.e. there are no direct stresses) there will be another pair of equal and opposite shear forces acting on the sloping edges, so the two couples cancel each other out.

This is a complete diagram: http://engineering-references.sbainvent.com/strength_of_materials/stress-element.php

5. Sep 9, 2014

### Will Flannery

Well, that's what I was thinking, however, correct me if I'm wrong but there cannot be shear forces on the sides because nothing is touching the sides, there is nothing to apply the force. So, the stress that opposes the rotation has to come from the base it seems to me.

Not only that confuses me, but shear stress is symmetric, right, so there does have to be a shear stress on the sides as you say, but there is nothing to apply it besides the base. So, I think a diagram showing all the forces/stresses acting in the diagram on wiki would be interesting.

I also see that if you define shear stress as du/dy + dv/dx then it's symmetric and there is a shear stress on the side even though nothing is touching it? Weird !

On the other hand, I think if you actually did the experiment, there would be a downward force on the right side of the block on the table, but .... maybe not, maybe there would be no vertical forces on the table at all?

Last edited: Sep 9, 2014
6. Sep 9, 2014

### nasu

I think you are mixing the stress with forces.
The forces may be applied only on the top and bottom faces of the cube.
However the stress is distributed throughout the cube. You can have stress along the oblique faces as well.

Edit. In this case however it seems that the shear stress is close to zero near the oblique sides. This is what a quick simulation with finite elements shows me.

Last edited: Sep 9, 2014
7. Sep 9, 2014

### Will Flannery

"I think you are mixing the stress with forces."

I think that is part of it. In particular I think of 2 definitions of stress (?), as force/area which seems to me to be an applied stress, and stress calculated by Hooke's law, i.e. as a multiple of strain, which is calculated from the deformation function, which seems to me to be 'experienced' stress or 'resultant' stress rather than applied stress.

So, it seems to me the Hooke's law or resultant stress given by G(du/dy + dv/dx), that is the decrease in angle between the i and j vectors in the stressed block times the shear modulus, should be constant throughtout the block. But, on the top surface the stress is caused by an applied stress or traction, and on the sides it is purely a resultant stress.

So, it seems a little wierd to me that a resultant stress can cancel an applied stress, I'm looking for the force (traction) that produces the resultant stress, and it's the same force (traction) that produced the applied stress), so it's cancelling itself ????

So, maybe you say (even thought it's not what wiki says) it's the couple of tractions, the ones on the top and bottom surfaces, that are producing the shear stresses, and that seems right. And intuitively I suppose these tractions are not really producing any torques on the block, so it's all OK.* (I'm not sure of this, what if the block is tall and thin instead of square).

I would be suspicious of the sim results.

Last edited: Sep 9, 2014
8. Sep 9, 2014

### AlephZero

That is the basic problem with the wiki diagram. If shows one unbalanced force acting on the object, so it is not in equilibrium.

You could apply more forces to to keep it in equilibrium in lots of different ways. For example you could apply horizontal and vertical forces at one bottom corner, and a vertical force at the other corner to balance out the moments. But the only way for the stress in the object to be pure shear, with no direct stresses, is to have shear forces acting along all four sides.

9. Sep 9, 2014

### AlephZero

What you learn in a first mechanics course, that "stress = force / area", is usually only applied to a special situation where it "works". That is a direct stress, for example a wire in tension, and where the direction of the stress is normal to the "area", for example at the end of the wire.

If you think about it, it should be clear that the stress is the same everywhere in the material of the wire, but there is no "force / area" at the cylindrical surface along the length of the wire! But in a first course, you probably didn't stop to think about that, because it was "obvious" how to use the "force/area" formula correctly to get the right answers to problems.

For the general 2-D situation, you have four components of the stress field: two direct stresses and two shear stresses (which are always numerically equal, but generate forces that act in different directions). If you imagine cutting out a small part of the material - small enough so you can assume the stresses are constant over it - the stresses are balanced by forces normal to all the edges (i.e. what you would probably call "pressure", corresponding to "force / area"), and also shear forces acting along all the edges, as shown in the two links in my previous post.

The wiki diagram is an attempt to explain where one of those forces comes from, while ignoring the others, but in reality you can't have just one of the forces acting on its own.

In SteamKing's diagrams, there are other stresses acting as well as shear stress, but if you want to decide what size the rivet should be, the shear stress is the critical factor so you ignore everything else. But the other forces and stresses are not zero, just because you decided to ignore them!

Of course the "art of Engineering" is often about understanding what you can and can't ignore in a given situation.

10. Sep 9, 2014

### Will Flannery

I assume you simulated the base as fixed, given that, what did the sim say about stresses along the base? Any normal stresses? I'll guess yes.

Also, what if you don't fix the base but specify shear stresses on both the top and bottom boundaries, and no other stresses, a rightward stress on the top as in the wiki diagram, and a leftward stress along the bottom. Does the sim work, i.e. produce a steady state non-rotating solution, or does it blow up? I'll guess blow up.

11. Sep 9, 2014

### Staff: Mentor

Actually, there are shear stresses on the sides. What you're missing is that you don't need a solid surface to apply a shear stress. The shear stress on the sides is being applied by the adjacent fluid. This adjacent fluid is what is touching the sides.

Chet

12. Sep 14, 2014

### Will Flannery

With a little help from AlephZero in another thread I got my stress/strain sim to run, and ran it on the wiki diagram example. The results are near the bottom on

www.berkeleyscience.com/stress.htm [Broken]

According to the sim the shear stress is constant over the square and there are no normal stresses at all. So there are shear stresses on the sides even without any applied tractions on those sides.

Last edited by a moderator: May 6, 2017
13. Sep 14, 2014

### Staff: Mentor

To be precise about terminology, the term "traction" refers to the vector force per unit area, including both normal and tangential components. So the tractions on the sides are equal (in this case) to the shear stresses. The normal components of the tractions are zero.

Chet