Shear stress direction and the velocity gradient

In summary, the conversation discusses the direction of stress in a flowing fluid and how it affects different elements within the fluid. It mentions the Cauchy stress relationship and the stress tensor as tools to determine stress on a surface of arbitrary orientation. The conversation also touches on the concept of a stress vector and its components in different directions.
  • #1
Rahulx084
99
1
1)In a flowing fluid in laminar fashion we know that it flows in planes which slides over each other, Let's take a fluid element (cylindrical) in a pipe(Radius=R) the resistive force is (stress)(cross section area of cylinder with radius 'r') acting in backward direction, now if I take Flow between two parallel fixed plates and I take a element(cubical with thickness dy and depth (inside screen) to be unity we see that the direction of stress on the top and bottom face of element is opposite. How we are determining the direction of the stress?
2)lets take a fluid slab in a pipe (flowing fluid in it), the difference in velocity of top and bottom plane of slab is du and thickness between them is dy , then we know stress= (viscocity) (du/dy), what does this stress mean and where this stress is acting , on top of slab or bottom of slab , in forward or backward direction?
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  • #2
Are you asking about a radial shell of fluid situated between r and ##r+\delta r##, and are you asking about the shear stresses on the fluid within the shell at these inner and outer surfaces?
 
  • #3
Umm no sir this is not the case , but I think understanding what you are telling can make me understand what I am asking.
 
  • #4
Well, at ##r+\delta r##, the shear stress exerted by the fluid at larger radii on the fluid in the shell is given by:
$$\mu\frac{dv}{dr}\mathbf{i_z}$$where v is the axial velocity and ##\mathbf{i_z}## is a unit vector in the positive axial direction (flow direction). Since the fluid at radial locations greater than ##r+\delta r## is traveling slower than at radial location ##r+\delta r##, dv/dr is negative. This means that shear stress at this surface is in the negative axial direction. The fluid at greater radii are acting to slow down the fluid in the shell.

At radial location r (the inner surface of the shell), the shear stress exerted by the fluid at lesser radii on the fluid in the shell is given by:
$$\mu\frac{dv}{dr}(-\mathbf{i_z})$$. Since dv/dr is negative, this means that the shear stress at this surface is in the positive axial direction. The fluid at lesser radii are acting to speed up the fluid in the shell.
 
  • #5
why the unit vector in second expression is negative?And the stress at r+dr, why we are only taking the effects of fluid which are greater than r+dr , can't the fluid inside also going to apply stress on it?
 
  • #6
Rahulx084 said:
why the unit vector in second expression is negative?And the stress at r+dr, why we are only taking the effects of fluid which are greater than r+dr , can't the fluid inside also going to apply stress on it?
I'll answer your 2nd question first. We are doing a force balance on the fluid within the shell (like we did on rigid bodies in freshman physics). So we only include externally applied forces on the free body.

Your first question is a little more difficult to answer. It follows from the so-called Cauchy stress relationship which enables one to use the stress tensor to determine the stress on a surface of arbitrary orientation, exerted by the material on a specified side of the surface. Please google Cauchy stress relationship (aka Cauchy's Law), and see if you can follow what they are saying. I'll try to help if you have problems.
 
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  • #7
I tried to understand this relation , what I'm able to grasp is , It states that you don't have to find stress in all orientation(which can be in large number) , all you have to do is find the stress in the direction of the coordinate system and multiply it with unit normal vector (In matrix form), but sir how we going to apply this concept in our problem
 
  • #8
Have you learned about the stress tensor and how to represent a 2nd order tensor using dyadic notation? If not, please see Transport Phenomena by Bird, Stewart, and Lightfoot, Appendix A, section A.3 Tensor Operations in Terms of Components. After you are comfortable with that, I will show you how to determine the stress vector on a surface of arbitrary orientation, and then how to apply this to the specific problem at hand.
 
  • #9
##\tau_{xy}##= This means the normal vector is in x direction and stress component is in y direction(If this is a cubical element face in x direction there will be ##\tau_{xx}## and ##\tau_{xz}## as well ). Is this what you asked? But still I will refer to the refrence .
 
  • #10
Rahulx084 said:
##\tau_{xy}##= This means the normal vector is in x direction and stress component is in y direction(If this is a cubical element face in x direction there will be ##\tau_{xx}## and ##\tau_{xz}## as well ). Is this what you asked? But still I will refer to the refrence .
It's more general than that. Please be patient. What I'm going to show you is worth waiting for.
 
  • #11
okay sir
 
  • #13
This is really a great post you shared , It will take me a bit time to digest(I think I understood like 70%) . I need few hours to digest it completely , I will get back to you as soon as I get it . Thanks again
 
  • #14
I have a doubt, sum of the traction vector=net force acting on the element?
 
  • #15
Rahulx084 said:
I have a doubt, sum of the traction vector=net force acting on the element?
I don't quite understand your question.
 
  • #16
I actually meant resultant of traction vector=net force acting on element?
 
  • #17
Rahulx084 said:
I actually meant resultant of traction vector=net force acting on element?
I still don't understand. Can you state your doubt in the form of an equation?
 
  • #18
like what I understood is ,you get stress vector of a plane(traction vector) by dotting the stress tensor with the normal vector . so I was thinking if we take resultant (like triangle law or parallelogram law) of all the traction vectors is this going to yield the net force per unit area on the element? .I also got that traction vector have 3 components in which two are shear stresses and one is normal stress .
 
  • #19
Let's try this for your problem. The stress tensor for your laminar flow in a pipe system is given by: $$\boldsymbol{\sigma}=-p(\mathbf{i_r}\mathbf{i_r}+\mathbf{i_{\theta}}\mathbf{i_{\theta}}+\mathbf{i_z}\mathbf{i_z})+\tau_{rz}(\mathbf{i_r}\mathbf{i_z}+\mathbf{i_z}\mathbf{i_r})$$
OK so far?
 
  • #20
Actually no , I don't get how this expression came. I can understand the terms that we took pressure in r,theta and z direction (normal and pressure are counter parallel to each other) , next term says shear stress in axial direction when normal is in radial and shear stress in radial direction when normal is in axial and the term inside bracket are dyadic they aren't operated as vectors multiplication(dot or cross). This is all I can get for now . Sorry this is literally very new for me ,maybe I'm going to take more time than you expect from me.
 
  • #21
No need to apologize. Let's continue assuming it is correct. What do you get when you dot this with ##\mathbf{i_r}## to determine the traction exerted on our shell at ##r+\delta r## by the fluid at radial locations ##>(r+\delta r)##?
 
  • #22
##-pi_{r}## + ##\tau_{rz}i_{z}##
 
  • #23
Rahulx084 said:
##-pi_{r}## + ##\tau_{rz}i_{z}##
Excellent. So this is the traction on our shell at ##r+\delta r## by the fluid external to the shell at larger radii. Note that since ##\tau_{rz}=\mu\frac{dv}{dr}## and since dv/dr is negative, the z-component of the traction vector is in the negative z direction.

Next, we would like to get the traction vector on the inner surface of the shell at r, exerted by the fluid at lesser radii. To get this traction vector, what would be the unit normal vector you use? What do you get when you dot the stress tensor at r with this unit normal vector?
 
  • #24
I guess it will be ##-i_{r}## and this will give ##pi_{r}+\tau_{rz}(-i_{z})##
 
  • #25
Rahulx084 said:
I guess it will be ##-i_{r}## and this will give ##pi_{r}+\tau_{rz}(-i_{z})##
Excellent again. So now I'm going to write the equilibrium force balance on your shell which is situated between r and ##r+\delta r## in the radial direction and between z and ##z+\delta z## in the axial direction:

$$[2\pi r \tau_{rz}]_{r+\delta r}\delta z\mathbf{i_z}+[2\pi r \tau_{rz}]_{r}\delta z(\mathbf{-i_z})+[(\pi(r+\delta r)^2-\pi r^2)p]_z\delta z\mathbf{i_z}+[(\pi(r+\delta r)^2-\pi r^2)p]_{z+\delta z}(-\mathbf{i_z})=\mathbf{0}$$

OK so far? Does this make sense to you?
 
  • #26
yes, understood , this is for steady flow right?
 
  • #27
Rahulx084 said:
yes, understood , this is for steady flow right?
Yes. So now what do you get if you divide the equation by ##2\pi r \delta r \delta z## and then take the limit as ##\delta r## and ##\delta z## approach zero?
 
  • #28
Sir , I think there should not be ##\delta_{z}## in the third term of force balance in which we have pressure
 
  • #29
##\frac 1 r## ##\frac {dr\tau_{rz}} {dr}## + ##\frac {dp} {dz}## , this is what I'm getting sir .
 
  • #30
Rahulx084 said:
##\frac 1 r## ##\frac {dr\tau_{rz}} {dr}## + ##\frac {dp} {dz}## , this is what I'm getting sir .
Excellent. So now you have some actual experience at applying the Cauchy Stress Relationship to a problem. Does that help answering your questions?
 
  • #31
sir but how to get the term in #19 ? and sir one more doubt If we have a cubical element how we are going to write the term #19 ? Is it going to remain the same??
 
  • #32
Rahulx084 said:
sir but how to get the term in #19 ? and sir one more doubt If we have a cubical element how we are going to write the term #19 ? Is it going to remain the same??
The term in #19 follows from the equations for the stress tensor components for a viscous Newtonian fluid. Are you familiar with these equations?

When you have a cubical element (or rectangular parallelepiped), you use the stress tensor in component form for Cartesian coordinates.
 
  • #33
I think I'm not familiar with that. Actually sir this isn't in my course but I'm learning it because it seems so interesting plus your great explanations . Can you provide me with only the results of cubical element or if there is any source where I can find it so that I can look after , or maybe you tell me sir.
 
  • #34
Rahulx084 said:
I think I'm not familiar with that. Actually sir this isn't in my course but I'm learning it because it seems so interesting plus your great explanations . Can you provide me with only the results of cubical element or if there is any source where I can find it so that I can look after , or maybe you tell me sir.
See page 29 of http://web.mit.edu/2.25/www/pdf/viscous_flow_eqn.pdf
 
  • #35
can you just give me the result for stress tensor of cubical and parallelopiped one ? It would be so nice of you . Thanks
 
<h2>1. What is shear stress direction?</h2><p>Shear stress direction refers to the direction in which a force is applied to an object, causing it to deform or change shape. In the context of velocity gradient, it is the direction in which the fluid is moving relative to the surface of an object.</p><h2>2. How is shear stress direction related to velocity gradient?</h2><p>Shear stress direction and velocity gradient are closely related because the velocity gradient is the change in velocity over a given distance, and the direction of this change determines the direction of the shear stress. In other words, the velocity gradient determines the direction of the shear stress.</p><h2>3. What is the significance of shear stress direction in fluid mechanics?</h2><p>Shear stress direction is important in fluid mechanics because it affects the flow behavior of fluids. The direction of shear stress can determine the type of flow (e.g. laminar or turbulent) and can also impact the efficiency of fluid flow in pipes and channels.</p><h2>4. How is shear stress direction measured?</h2><p>Shear stress direction can be measured using various instruments such as a strain gauge, a viscometer, or a rheometer. These instruments measure the force or strain on a surface and can determine the direction of the shear stress.</p><h2>5. Can the shear stress direction be changed?</h2><p>Yes, the shear stress direction can be changed by altering the velocity gradient. For example, if the velocity gradient is increased in a certain direction, the shear stress will also increase in that direction. This can be achieved by changing the shape of the object or by changing the properties of the fluid, such as its viscosity.</p>

1. What is shear stress direction?

Shear stress direction refers to the direction in which a force is applied to an object, causing it to deform or change shape. In the context of velocity gradient, it is the direction in which the fluid is moving relative to the surface of an object.

2. How is shear stress direction related to velocity gradient?

Shear stress direction and velocity gradient are closely related because the velocity gradient is the change in velocity over a given distance, and the direction of this change determines the direction of the shear stress. In other words, the velocity gradient determines the direction of the shear stress.

3. What is the significance of shear stress direction in fluid mechanics?

Shear stress direction is important in fluid mechanics because it affects the flow behavior of fluids. The direction of shear stress can determine the type of flow (e.g. laminar or turbulent) and can also impact the efficiency of fluid flow in pipes and channels.

4. How is shear stress direction measured?

Shear stress direction can be measured using various instruments such as a strain gauge, a viscometer, or a rheometer. These instruments measure the force or strain on a surface and can determine the direction of the shear stress.

5. Can the shear stress direction be changed?

Yes, the shear stress direction can be changed by altering the velocity gradient. For example, if the velocity gradient is increased in a certain direction, the shear stress will also increase in that direction. This can be achieved by changing the shape of the object or by changing the properties of the fluid, such as its viscosity.

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