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I encountered the following elementary problem a couple of days back:

A long pipe of radius $h$ carries a fluid through it. The co-ordinate system is setup such that the origin lies on the axis of the pipe, with the x-axis being the axis of the pipe. The velocity gradient inside the pipe is given as $$u=\frac{3v}{2}[1-(\frac{y}{h}) ^2]$$ where $v$ is a constant. We are supposed to find the shear stress on the top and bottom walls of the pipe. I recreated the diagram given as follows: diagram for problem

Attempt at the solution:

$\tau=\mu\frac{du}{dy}$, where $\mu$ is the dynamic viscosity coefficient.

$\frac{du}{dy}=\frac{-3vy}{h^2}$

Hence, at the bottom wall, $\tau=\frac{3v}{h}$ and at the top wall, $\tau=-\frac{3v}{h}$

Now, while the magnitudes look sane to me, I really can't interpret what the negative sign is all about. Isn't the stress in the same direction for both the walls, or is it about something else?

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    $\begingroup$ To state Chester Miller's answer in a different way, top wall and bottom wall have oppositely directed normals pointing into the fluid. This gives rise to the sign difference because stress tensor is a linear function of the area vector. $\endgroup$
    – Deep
    Jul 22, 2017 at 12:43
  • $\begingroup$ @Deep can you post this an answer so that I can cite you? Actually, I had thought of the exact same thing, but someone (much senior to me) told me that we consider only magnitude of the area in this case, which doesn't really make sense to me. $\endgroup$
    – pulsejet
    Jul 22, 2017 at 16:29
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    $\begingroup$ You cite answers from this site? Alright, I will post this as an answer although Chester's answer is equivalent. $\endgroup$
    – Deep
    Jul 23, 2017 at 4:35
  • $\begingroup$ If you consider the stress to be the frictional force between layers of fluid it should be obvious that, like force, the stress has magnitude and direction. As the gradient in the $y+$-direction is negative (as can be deduced from your diagram), the stress is negative indicating it is pointing from the top wall into the fluid. The reverse is the case for the bottom wall. $\endgroup$
    – nluigi
    Jul 25, 2017 at 10:15

2 Answers 2

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To state Chester Miller's answer in a different way, top wall and bottom wall have oppositely directed normals pointing into the fluid. This gives rise to the sign difference because stress tensor is a linear function of the area vector. That is if $\mathbf{\tau}$ is the stress tensor and $\mathbf{n}$ is the area normal then $\mathbf{\tau}(-\mathbf{n})=-\mathbf{\tau}(\mathbf{n})$.

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The way you are using it here, $\tau$ is the shear stress in the x-direction on a plane of constant y, exerted by the material at higher y on the material at lower y. So, at the upper boundary, it is the shear stress exerted by the wall on the fluid, and, at the lower boundary, it is the shear stress exerted by the fluid on the wall. The latter is positive in the positive x direction. However, by Newton's 3 rd law, the shear stress exerted by the wall on the fluid at the lower boundary is in the opposite (negative x) direction.

I agree with @Deep's comment above, although it may be a little beyond the experience base of the OP.

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  • $\begingroup$ +1 I understood your explanation, though I am even more confused about the definitions now :P. As for my experience, I am majoring in Mechanical Engineering, and I need to cite from here, so I am fine if the answer get slightly more complex. $\endgroup$
    – pulsejet
    Jul 22, 2017 at 16:34
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    $\begingroup$ OK. If you apply the Cauchy stress relationship at the lower boundary by dotting the stress tensor with a unit vector in the negative y-direction (an outward directed normal from the fluid), you obtain the stress vector (traction vector) exerted by lower solid boundary on the fluid. This comes out to be $-\tau_{xy}\mathbf{i}_x-\sigma_{yy}\mathbf{i}_{y}$. In this expression, $\tau_{xy}$ is what you are calling the shear stress $\tau$ and $\sigma_{yy}$ is the y-y component of the stress tensor (in the y-direction). This is what we call the normal stress. $\endgroup$ Jul 22, 2017 at 22:56

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