# Homework Help: Shear stress at boundary of wall = 0

1. Aug 11, 2016

1. The problem statement, all variables and given/known data

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?
2. Relevant equations

3. 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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Last edited: Aug 11, 2016
2. Aug 11, 2016

### PhanthomJay

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?

3. Aug 11, 2016

Q = (102.1x 10)(120.2 +5) + (240.4 x 6.4)(0 )+ (102.1x 10)(120.2 +5) , which is not 0

4. Aug 12, 2016

### PhanthomJay

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.

5. Aug 12, 2016

Why the moment at upper and below parts have different sign?

6. Aug 12, 2016

### PhanthomJay

Because ybar is above the neutral axis in one case, and below it in the other. The area moments thus cancel.

7. Aug 12, 2016

why we dont 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. Aug 12, 2016

### PhanthomJay

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.

9. Aug 13, 2016

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 cancelled off , right ?

10. Aug 13, 2016

### PhanthomJay

Oh I thought you meant in general . At the top of the beam, yes, Q is 0.

11. Aug 13, 2016

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. Aug 13, 2016

### PhanthomJay

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?

13. Aug 13, 2016

Sorry, refer to this example, there
sorry, i posted the wrong example....
anyway, here's an example, the author doesnt consider ybar of the area below the neutral axis as negative...Everything is positive...

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14. Aug 13, 2016

### PhanthomJay

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.

15. Aug 14, 2016

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. Aug 14, 2016

### PhanthomJay

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. Aug 14, 2016