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This is the canonical example, where the velocity $v(x_2)$ is parabolic. Assuming the plates have a distance of $h$, where the bottom one is at $x_2=0$ and the upper plate is at $x_2=h$. Let the viscosity be $\mu$ and let the pressure gradient $\frac{\partial \pi}{\partial x_1} = -\delta$, so that $$v(x_2)=\frac{\delta x_2}{2 \mu} (h - x_2)$$

Therefore, the non-zero components of the Cauchy stress tensor are $T_{11}=T_{22}=T_{33}= -\pi$ and $T_{12}=T_{21}=\delta (\frac{h}{2} - x_2 )$ since this is

$$[T]=- \pi \begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix} + \delta (\frac{h}{2} - x_2 )\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$

So far so good.

Question: my professor wrote in the notes: shearing forces have opposite direction, but I am not able to prove this.


My attempt:

  • Let me be at the upper plate. To compute the force, per unit area, acting on the plate by the fluid, I need to consider as normal $-e_2$, and hence I obtain $$T(-e_2)=(-\delta(\frac{h}{2} - h),+\pi,0)$$ and the first component is positive. (I evaluated at $x_2=h$)

  • Let's consider now the lower plate: There we need to consider as normal $e_2$, and so $$T e_2=(\delta(\frac{h}{2} - 0),-\pi,0)$$ but as you can see the first component is still positive!!!

What am I missing?

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You are not missing anything. The shear force exerted by the fluid on the plate is in the positive e1 direction at both the upper plate and the lower plate. I have no idea what your professor meant by his statement, except possibly that the shearing force exerted by the fluid on a plate is in the opposite direction of the shearing force exerted by the plate on the fluid.

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  • $\begingroup$ Thanks a lot Chet. However, it seems that the other answer to my post is saying that I'm wrong and the two stresses from the fluid to the wall have opposite direction, right? @ChetMiller $\endgroup$ May 20, 2021 at 12:43
  • $\begingroup$ I stand by what I said. I have tons of experience with such flows. $\endgroup$ May 20, 2021 at 13:30
  • $\begingroup$ I didn't mean to object to your kind answer, with whom I agree. I was only asking if what Chemomechanics means in his answer is the same thing you mean. I didn't want to sound rude, I'm sorry $\endgroup$ May 20, 2021 at 13:39
  • $\begingroup$ I mean, if I look at his first image, it seems the two stresses have opposite direction, which Is wrong in this case... $\endgroup$ May 20, 2021 at 13:40
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    $\begingroup$ I think you did. $\endgroup$ May 21, 2021 at 12:29
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Question: my professor wrote in the notes: shearing forces have opposite direction, but I am not able to prove this.

It's possible that your professor simply meant that a state of (positive or negative) shear stress requires four forces in different directions to maintain equilibrium; note that any pair of forces acting in the same axis point in opposite directions:

(images from my website)

Elimination of any of these forces would result in rotational and translation acceleration:

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  • $\begingroup$ I see your point, but as you can see from my computations I obtain the same direction. So the question is: why is that? What am I doing wrong? Thanks $\endgroup$ May 20, 2021 at 6:42
  • $\begingroup$ You’re comparing shear forces at the upper and lower plates; I’m comparing shear forces on either side of a differential element anywhere shear exists. The squares above do not show your channel or tube! $\endgroup$ May 20, 2021 at 14:29
  • $\begingroup$ Okay, so you agree with the fact that the shearing forces applied by the fluid on the plates are both in the positive $e_1$ direction? @Chemomechanics $\endgroup$ May 20, 2021 at 16:15
  • $\begingroup$ Of course; symmetry requires this as well. $\endgroup$ May 20, 2021 at 16:52

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