Shear stress at boundary of wall = 0

In summary: So there is no need to cancel.In summary, the homework statement is saying that when you are calculating the moment of inertia about the neutral axis (Ixx), you dont have to put a negative sign for ybar even though the area is below the neutral axis.
  • #1
chetzread
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Homework Statement



in the formula of shear stress τ = (V)(Q) / It ,
Q=Ay = first moment of inertia of area, the area can be located above(or bottom) at the point of interest)
when the chosen point is at the wall(boundary) , why shear stress = 0?

Homework Equations

The Attempt at a Solution


When i choose area above the point of interest, there's nothing above A, so i understand that Q=0, shear stress = 0
But, when I choose the area below the point of interest, it's the whole coloured part, right? how can Q= 0 ?
 

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  • #2
At the top of the section, Q is 0 regardless of whether you use the area above or below that section of interest in its calculation. Please show how you calculate Q. What is the value of y-bar when using the lower area?
 
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  • #3
PhanthomJay said:
At the top of the section, Q is 0 regardless of whether you use the area above or below that section of interest in its calculation. Please show how you calculate Q. What is the value of y-bar when using the lower area?
Q = (102.1x 10)(120.2 +5) + (240.4 x 6.4)(0 )+ (102.1x 10)(120.2 +5) , which is not 0
 
  • #4
It is simpler for this special case to look at the entire area and recognize that since it's centroid lies along the neutral axis, then y-bar of the entire area is 0. But in general, your method is fine, except that first moments of area have positive and negative signs associated with them. So if you consider the first moment of the top flange area as plus, then the first moment of the bottom flange area has the same numerical value, but it is minus.
 
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  • #5
PhanthomJay said:
It is simpler for this special case to look at the entire area and recognize that since it's centroid lies along the neutral axis, then y-bar of the entire area is 0. But in general, your method is fine, except that first moments of area have positive and negative signs associated with them. So if you consider the first moment of the top flange area as plus, then the first moment of the bottom flange area has the same numerical value, but it is minus.
Why the moment at upper and below parts have different sign?
 
  • #6
Because ybar is above the neutral axis in one case, and below it in the other. The area moments thus cancel.
 
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  • #7
PhanthomJay said:
Because ybar is above the neutral axis in one case, and below it in the other. The area moments thus cancel.
why we don't have to cancel off them when we are calculating the first area moment of inertia in normal case(not this case) ?
Why it is necessary to cancel off them in this case?
 
  • #8
When the neutral axis is below the section of interest where you want to calculate Q, and you use the area above the section of interest in your calculation for Q, then all ybar values of the area parts are of the same sign, and there is nothing to cancel. Although you have a choice to use the area above or below the section of interest, use the easier of these choices. There is less chance of error this way.
 
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  • #9
PhanthomJay said:
When the neutral axis is below the section of interest where you want to calculate Q, and you use the area above the section of interest in your calculation for Q, then all ybar values of the area parts are of the same sign, and there is nothing to cancel. Although you have a choice to use the area above or below the section of interest, use the easier of these choices. There is less chance of error this way.
What do you mean ?
If i consider the area above the point of interest , the area above point A ( in the notes) will be 0 ...Q = 0
why you said that same sign ? and cancel ? there's nothing to be canceled off , right ?
 
  • #10
Oh I thought you meant in general . At the top of the beam, yes, Q is 0.
 
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  • #11
PhanthomJay said:
Oh I thought you meant in general . At the top of the beam, yes, Q is 0.
how if we considering the area below point A? why there's a need to put negative value for ybar?
I have an example here...When we are calcualting the moment of inertia about the neutral axis (Ixx), we dont have to put negative sign for ybar although it's the area below the neutral axis( x - axis)
refer to the example below
 

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  • #12
First moment of area Q is A times y. It can be plus or minus. Second moment of area I is A times y squared. Square a negative and you get a positive, right?
 
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  • #13
PhanthomJay said:
First moment of area Q is A times y. It can be plus or minus. Second moment of area I is A times y squared. Square a negative and you get a positive, right?
Sorry, refer to this example, there
PhanthomJay said:
First moment of area Q is A times y. It can be plus or minus. Second moment of area I is A times y squared. Square a negative and you get a positive, right?
sorry, i posted the wrong example...
anyway, here's an example, the author doesn't consider ybar of the area below the neutral axis as negative...Everything is positive...
 

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  • #14
I think you posted the wrong example. Anyway, you keep fighting the minus sign. Whether using moment areas to calculate deflections, or Q, or centroids, you must consider the signage or you will continue to get the wrong answers.
 
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  • #15
PhanthomJay said:
I think you posted the wrong example. Anyway, you keep fighting the minus sign. Whether using moment areas to calculate deflections, or Q, or centroids, you must consider the signage or you will continue to get the wrong answers.
I am very sorry for that.
Here's the example, can you refer to it again...There's no negative sign involve for ybar in this question
 

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  • #16
In determining the centroid of the section, the author is summing moment areas about the bottom, so ybar is always positive for all areas. There's a couple of half holes that complicate the problem because the area of those holes subtract and are considered negative. I urge you again to start with the simpler cases before tackling the complex.
 
  • #17
PhanthomJay said:
In determining the centroid of the section, the author is summing moment areas about the bottom, so ybar is always positive for all areas. There's a couple of half holes that complicate the problem because the area of those holes subtract and are considered negative. I urge you again to start with the simpler cases before tackling the complex.
ok, i noticed the difference between the 2 cases already...
 

1. What is shear stress at the boundary of a wall?

Shear stress at the boundary of a wall refers to the force that acts parallel to the surface of the wall, at the point where the wall meets another fluid or solid. It is a measure of the frictional force between the wall and the fluid or solid.

2. What does it mean when shear stress at the boundary of a wall is equal to 0?

When shear stress at the boundary of a wall is equal to 0, it means that there is no frictional force between the wall and the fluid or solid. This can occur when the fluid is moving at a constant velocity or when the wall is smooth and there is no roughness to create friction.

3. How is shear stress at the boundary of a wall measured?

Shear stress at the boundary of a wall can be measured using a device called a viscometer, which measures the force required to move a fluid at a certain velocity through a tube. The shear stress can also be calculated using mathematical equations based on the fluid's viscosity and the velocity gradient near the wall.

4. What factors can affect shear stress at the boundary of a wall?

Shear stress at the boundary of a wall can be affected by factors such as the viscosity of the fluid, the velocity of the fluid, the roughness of the wall, and the temperature of the system. Increasing any of these factors can lead to an increase in shear stress.

5. Why is shear stress at the boundary of a wall important in fluid mechanics?

Shear stress at the boundary of a wall is important in fluid mechanics because it plays a role in determining the flow behavior of fluids. It can affect the drag force on objects in a fluid, the resistance of pipes and channels, and the stability of flow. Understanding and controlling shear stress is crucial in many engineering and scientific applications.

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