# Shearing Stress at the Boundaries of two infinite Planes

This is a worked example on a book I am reading (Fundamentals of fluid Mechanics).

"Can you explain the change in sign? [Hint: Sketch a sheared fluid particle adjacent to the top plate.]"

They calculate the shearing stress in the top layer, bottom layer and middle layer.

The calculate $$\tau > 0$$ (in the fluid direction, they explain) for bottom layer, $$\tau = 0$$ for middle layer and $$\tau < 0$$ for top layer.

I do not understand how the shear can go be in opposite directions at the top and bottom, especially given the symmetry of the problem and the flow being in the same direction.

Hope you can help me out.

If I get your point correctly, It is only a sign convention.

say in a pipe with the positive flow gradient we consider a plane C parallel to the pipe axis.

The shear stress above the Yc plane is considered positive and the shear stress below that same plane is considered negative. both of these stresses are due to $$\tau \$$ which is positive. and the sum of these two stressed is zero because plane C is moving with the flow. But on the top wall, it is negative because the pipe wall is stationary. for the same reason, the bottom wall shear stress is positive. look at the sign convention on this figure.

.

• It is more fundamental than being "only a sign convention". The 3D stress in the fluid is a second-order tensor (like all stress fields!) and to find the force on a surface, you take the scalar product of the tensor and the normal to the surface. The normal has opposite directions on the top and bottom surface, and the shear stress terms du/dy also have opposite signs, so the shear force on the top and bottom is in the same direction. This is no different from the fact that if there is a constant axial stress in a rod, the forces on the ends of the rod are equal and opposite. Jan 10 at 3:35
• Following the reasoning around the surface normal vector, I do not understand why the author talks about direction of the stress, when talks about their values. As such, doesn't a negative stress $\tau < 0$ imply the stress goes in the opposite direction? Or he made a bad explanation, and stress sign convention depends solely on the normal of a surface.
– RSM
Jan 10 at 14:48

If I understand your question correctly your qualm is about $$\tau$$ being negative.

I will use the same image as karman because it is a very good one:

you will notice that shear stress is defined as:

$$\tau = \mu\frac{d u}{dy}$$

that for increasing y, the velocity increases up to the middle of the pipe. (That is because the fluid in the center transfers momentum to the fluid that is closer to the wall, although this is incosequential to your question).

The important thing is the gradient of the velocity $$\frac{d u}{dy}$$. As you can see for increasing y, from the lower wall up to the middle of the cross-section the gradient is positive (ie. velocity increases).

Right at the middle, the velocity gradient is zero (local extreme/local maximum), therefore the shear stress is zero.

Above the middle of the section, velocity drops, therefore the shear stress is negative.

This is all a mater of convention. If you had y, increasing downwards then the upper half would be positive, and the lower half negative.

UPDATE 1 (The way I perceive it) shear stress is a measure of the forces between adjacent layers of the fluid (in laminar flow specifically). High shear stress, means that between each layers there are high forces and therefore there will be a greater change in velocity. However, as you can see in order to determine the change in velocity you need to consider a direction orthogonal to the velocity (hence the gradient). So, (as in other cases of shear stress definitions), shear stress is positive when:

• for increasing y ($$\Uparrow$$), the velocity increases($$\Uparrow$$).
• for decreasing y($$\Downarrow$$), the velocity decreases($$\Downarrow$$).

while it is negative when:

• for increasing y($$\Uparrow$$), the velocity decreases($$\Downarrow$$).
• for decreasing y($$\Downarrow$$), the velocity increases($$\Uparrow$$).

UPDATE 2:

if you notice the graph with the definition of the shear stress, there is a very interesting detail. (see bollow for detail). You can see that $$\tau$$ is presented as half arrow.

The half arrow (among other things) make sense because $$\tau$$ is applied on a surface/plane. So the top half arrow is applied on the top of the fluid surface "layer". Also notice that immediately below there is another half arrow pointing the opposite direction. It is useful to note that this $$\tau$$ is also positive. You might ask: "How can it be?"

The convention is that the sign of the shear stress (Actually all stresses but that;s another matter) is defined by two things:

1. the direction of the shear stress.
2. the direction of the normal to the surface compared to the positive direction (in that respective axis).

if both directions are positive or negative then the shear stress is positive. If one direction is positive and the other is negative the shear stress is negative. (I tend to think about it as multiplication +1*+1 = 1, (-1)(-1) =1, and (1)(-1) =-1)

So for example: the top half arrow:

• $$\tau$$ points to the positive x
• the normal of the surface that $$\tau$$ (the top half) is applied is pointing upwards (the normal to a surface is always points to the side that the stress is applied so top). Upwards is positive y.

So the top half of $$\tau$$ is pointing to positive x and the direction of the normal to the surface point to positive y. Since both are positive then the $$\tau$$ is positive

Also for the bottom half $$\tau$$ arrow:

• $$\tau$$ points to the negative x
• the normal of the surface (that $$\tau$$ (the top half) is applied) is pointing downwards. Downwards is negative y.

So the bottom half of $$\tau$$ is pointing to negative x (-1) and the direction of the normal to the surface point to negative y(-1). Therefore, the bottom half is also positive.

This so far applies for the bottom half of your original post. For the top half, the direction of the shear stresses would be reversed, and as such the sign of the shear stresses with this convention that is solely based on the directions would be negative (try it out for your self).

Following this convention for shear stress, uses the positive direction on the horizontal axis (where velocity is positive).

• I can see it, as a matter of convention. The sign is supposed to indicate the direction of the flow. In the example I provided, the y-axis is centered, by inspection I can see the slope decreasing in the +y direction or in the -y direction. Once, I have a positive or negative shear, how do i interpret its sing. The author said, positive shear was shear in the direction of the flow of the fluid, but that would tell me that the shear at the top goes in opposite direction, which does not seem to be right due to the symmetry of the problem.
– RSM
Jan 10 at 14:47
• could you please update in your post the exact wording of the author? Jan 10 at 15:03
• I've updated the answer, in the meantime, while waiting for the exact author wording. Jan 10 at 15:40
• I have updated the question, thank you for all the help.
– RSM
Jan 10 at 18:01
• After the previous comment I looked for a text that might explain it better. I found this. See bullet point 3 in page 2. Jan 10 at 19:47