Calculating Shear Stress: Understanding the Role of Y in Shear Force Diagrams

[chetzread]

Homework Statement

Refer to the second photo , Q= Ay
Does the author mean point P is at the center of the region (62.5mm x 100mm) ?The figure is not labelled well i am confused

Homework Equations

The Attempt at a Solution

In the notes , I was told that y is the location of the shear stress is measured from the neutral axis

 

[chetzread]

why the value of y is 62.5 / 2 ?
I have no idea...

 

[SteamKing]

why the value of y is 62.5 / 2 ?
I have no idea...

If the area of interest is a rectangle measuring 100 mm wide by 62.5 mm deep, where is the vertical centroid located, relative to the neutral axis?

 

[chetzread]

If the area of interest is a rectangle measuring 100 mm wide by 62.5 mm deep, where is the vertical centroid located, relative to the neutral axis?

do you mean y is the distance from the centorid of specific area to neutral axis ?

 

[chetzread]

If the area of interest is a rectangle measuring 100 mm wide by 62.5 mm deep, where is the vertical centroid located, relative to the neutral axis?

why it's 62.5 / 2 in B ? and it's 12.5 +(50/2) in part A ?

 

[chetzread]

I'm confused where is the location of point P now ...

 

[SteamKing]

why it's 62.5 / 2 in B ? and it's 12.5 +(50/2) in part A ?

Because one calculation is for the shear stress for part a) of the problem and the other is for the shear stress in part b).

Part a) wants to know the shear stress in the beam for a point which is located 50 mm below the top of the beam. Remember, Q must be calculated using the neutral axis as the reference point, which is why the y-bar is 12.5 mm + 50 mm / 2.

Part b) wants to know the shear stress in the beam, which the box in the illustration indicates occurs at the neutral axis. The y-bar here is 62.5 mm / 2.

 

[chetzread]

Because one calculation is for the shear stress for part a) of the problem and the other is for the shear stress in part b).

Part a) wants to know the shear stress in the beam for a point which is located 50 mm below the top of the beam. Remember, Q must be calculated using the neutral axis as the reference point, which is why the y-bar is 12.5 mm + 50 mm / 2.

Part b) wants to know the shear stress in the beam, which the box in the illustration indicates occurs at the neutral axis. The y-bar here is 62.5 mm / 2.

then, for part a , why the value of y isn't 12.5mm? Since it's 12.5mm away from neutral axis.
neutral axis is located 62.5mm way from the top of beam, why value of y is not 62.5 mm?

 

[SteamKing]

then, for part a , why the value of y isn't 12.5mm? Since it's 12.5mm away from neutral axis.

Because you are supposed to calculate the first moment Q = A * y-bar for the area of the beam which lies above the line which is 12.5 mm above the neutral axis.

And it's not y you are dealing with, but y-bar, which is the location of the centroid of the area A.

Haven't you understood anything about the calculation of the shear stress using the formula $τ = \frac{V ⋅ Q}{I ⋅ t}$ ?

neutral axis is located 62.5mm way from the top of beam, why value of y is not 62.5 mm?

Again, it's not y you are dealing with, but y-bar, which is the location of the centroid of the area A above the neutral axis.

 

[chetzread]

Because you are supposed to calculate the first moment Q = A * y-bar for the area of the beam which lies above the line which is 12.5 mm above the neutral axis.

And it's not y you are dealing with, but y-bar, which is the location of the centroid of the area A.

Haven't you understood anything about the calculation of the shear stress using the formula $τ = \frac{V ⋅ Q}{I ⋅ t}$ ?
Again, it's not y you are dealing with, but y-bar, which is the location of the centroid of the area A above the neutral axis.

[/QUOTE]

the maximum shear stress occur at the neutral axis of the beam , am i right? why we need to consider the centroid of area which is above the neutral axis?

 

[SteamKing]

the maximum shear stress occur at the neutral axis of the beam , am i right? why we need to consider the centroid of area which is above the neutral axis?[/QUOTE]
This is getting really circular. I don't know why you can't understand the shear stress formula, when it is laid out and explained.

The formula for calculating the shear stress is a beam is $τ = \frac{V ⋅ Q}{I ⋅ t}$

τ - shear stress
V - shear force
Q - first moment of area above the location where the shear stress is calculated.
I - second moment of area for the entire beam about the N.A.
t - width of the beam where the shear stress is calculated.

Now V, I, and t are all pretty easy to find for a given beam.

The first moment Q can be calculated using Q = A * y-bar, since that is the definition of the first moment of area.

If the beam cross section is a nice square or rectangle, then y-bar can be determined very easily.

 

[chetzread]

deleted

 

[chetzread]

the maximum shear stress occur at the neutral axis of the beam , am i right? why we need to consider the centroid of area which is above the neutral axis?

This is getting really circular. I don't know why you can't understand the shear stress formula, when it is laid out and explained.

The formula for calculating the shear stress is a beam is $τ = \frac{V ⋅ Q}{I ⋅ t}$

τ - shear stress
V - shear force
Q - first moment of area above the location where the shear stress is calculated.
I - second moment of area for the entire beam about the N.A.
t - width of the beam where the shear stress is calculated.

Now V, I, and t are all pretty easy to find for a given beam.

The first moment Q can be calculated using Q = A * y-bar, since that is the definition of the first moment of area.

If the beam cross section is a nice square or rectangle, then y-bar can be determined very easily.[/QUOTE]
do you mean the shear stress is to be measured at neutral axis , so the first moment of area is taken to be the area above the neutral axis (100mm)(62.5mm) ?

 

[SteamKing]

This is getting really circular. I don't know why you can't understand the shear stress formula, when it is laid out and explained.

The formula for calculating the shear stress is a beam is $τ = \frac{V ⋅ Q}{I ⋅ t}$

τ - shear stress
V - shear force
Q - first moment of area above the location where the shear stress is calculated.
I - second moment of area for the entire beam about the N.A.
t - width of the beam where the shear stress is calculated.

Now V, I, and t are all pretty easy to find for a given beam.

The first moment Q can be calculated using Q = A * y-bar, since that is the definition of the first moment of area.

If the beam cross section is a nice square or rectangle, then y-bar can be determined very easily.

do you mean the shear stress is to be measured at neutral axis , so the first moment of area is taken to be the area above the neutral axis (100mm)(62.5mm) ?[/QUOTE]
Yes, for part b) of the problem. The shear stress for part a) is measured at a different location, and its area is slightly different.

 

[chetzread]

do you mean the shear stress is to be measured at neutral axis , so the first moment of area is taken to be the area above the neutral axis (100mm)(62.5mm) ?

Yes, for part b) of the problem. The shear stress for part a) is measured at a different location, and its area is slightly different.[/QUOTE]
for part b , could it be area below the neutral axis?

 

[SteamKing]

Yes, for part b) of the problem. The shear stress for part a) is measured at a different location, and its area is slightly different.

for part b , could it be area below the neutral axis?[/QUOTE]
The neutral axis of the section splits the section into two parts, such that the first moment of area of the part above the N.A. is the same as the first moment of area of the part below the N.A. This is why the shear stress is a maximum at the N.A. of the cross section.

 

[chetzread]

for part b , could it be area below the neutral axis?

The neutral axis of the section splits the section into two parts, such that the first moment of area of the part above the N.A. is the same as the first moment of area of the part below the N.A. This is why the shear stress is a maximum at the N.A. of the cross section.[/QUOTE]
so,taking moment about an area below or above the neutral axis is acceptable?

How did the same area below and above neutral axis related to the maximum shear stress?

 

[SteamKing]

so,taking moment about an area below or above the neutral axis is acceptable?

It doesn't matter if the area is above or below the neutral axis. Q for either area is the same.

How did the same area below and above neutral axis related to the maximum shear stress?

Because Q is the same for the area above or below the neutral axis, that's as large as Q can get for that cross section, hence, the shear stress is a maximum at the N.A.

 

[chetzread]

It doesn't matter if the area is above or below the neutral axis. Q for either area is the same.

Because Q is the same for the area above or below the neutral axis, that's as large as Q can get for that cross section, hence, the shear stress is a maximum at the N.A.

What do you mean Q is the same for the area above or below the neutral axis, that's as large as Q can get for that cross sectio

 

[SteamKing]

What do you mean Q is the same for the area above or below the neutral axis, that's as large as Q can get for that cross sectio

If you don't believe me, work out Q above and below the N.A. for the wooden beam in the problem described in the OP.

The beam is 125 mm deep and 100 mm wide.

The N.A. is located 62.5 mm from the top or bottom of the beam.

A = 62.5 * 100 = 6250 mm²

y-bar = 62.5 / 2 = 31.25 mm

Q = A * y-bar = 6250 * 31.25 = 195,312.50 mm³

The only way Q gets bigger than 195,312.5 mm³ is if the N.A. moves up or down, but this is impossible, since it won't be the N.A. anymore.

 

[chetzread]

If you don't believe me, work out Q above and below the N.A. for the wooden beam in the problem described in the OP.

The beam is 125 mm deep and 100 mm wide.

The N.A. is located 62.5 mm from the top or bottom of the beam.

A = 62.5 * 100 = 6250 mm²

y-bar = 62.5 / 2 = 31.25 mm

Q = A * y-bar = 6250 * 31.25 = 195,312.50 mm³

The only way Q gets bigger than 195,312.5 mm³ is if the N.A. moves up or down, but this is impossible, since it won't be the N.A. anymore.

do you mean shear stress is maximum at neutral axis because the value of y-bar is smallest , thus the 'surface area' is smallest , stress is highest ?

 

[chetzread]

If you don't believe me, work out Q above and below the N.A. for the wooden beam in the problem described in the OP.

The beam is 125 mm deep and 100 mm wide.

The N.A. is located 62.5 mm from the top or bottom of the beam.

A = 62.5 * 100 = 6250 mm²

y-bar = 62.5 / 2 = 31.25 mm

Q = A * y-bar = 6250 * 31.25 = 195,312.50 mm³

The only way Q gets bigger than 195,312.5 mm³ is if the N.A. moves up or down, but this is impossible, since it won't be the N.A. anymore.

do you mean shear stress is maximum at neutral axis because the value of y-bar is smallest , thus the 'surface area' is smallest , stress is highest ?

 

[SteamKing]

do you mean shear stress is maximum at neutral axis because the value of y-bar is smallest , thus the 'surface area' is smallest , stress is highest ?

What you're saying doesn't make any sense.

The shear stress is a maximum at the N.A. because the product A * y-bar = Q is a maximum there.

After all, $τ = \frac {V ⋅ Q}{I ⋅ t}$, and the only quanty which changes for the wooden beam section is Q.

 

[chetzread]

deleted

 

[chetzread]

What you're saying doesn't make any sense.

The shear stress is a maximum at the N.A. because the product A * y-bar = Q is a maximum there.

After all, $τ = \frac {V ⋅ Q}{I ⋅ t}$, and the only quanty which changes for the wooden beam section is Q.

the shear stress at a point where lower than the neutral axis is even higher , isn't it ? why it is max at neutral axis ?

 

[SteamKing]

the shear stress at a point where lower than the neutral axis is even higher , isn't it ? why it is max at neutral axis ?

No, it's not. The first moment Q is always calculated about the neutral axis as the reference point.

 

[chetzread]

No, it's not. The first moment Q is always calculated about the neutral axis as the reference point.

then , how to prove that shear stress is maximum at neutral axis ? Since you said that Q can be only calculated at neutral axis ?

 

[SteamKing]

then , how to prove that shear stress is maximum at neutral axis ? Since you said that Q can be only calculated at neutral axis ?

No, I said the neutral axis is used as the reference axis for the calculation of Q. There's a difference.

You can set up a calculation for different values of Q as the location of the shear stress moves from the outer fiber to the neutral axis.

​

When the line B-B is located at the outer fiber of the beam, A = 0 and Q = 0, which also implies that τ = 0.

By bringing the axis B-B closer to the N.A., the area A above B-B is increasing, while y-bar is getting smaller. What happens to the product A * y-bar?

The N.A.is located at d/2 from the top or bottom of the section.

If the axis B-B is y units from the N.A.,then A = (d/2 - y) * width and y-bar = (d/2 + y) / 2.

Q = A * y-bar = [(d/2 - y) * width ] * [(d/4 + y/2)]

Q = [d²/8 + d*y/4 - d*y/4 - y²/2] * width

Q = [d² / 8 - y² / 2] * width

When the axis B-B coincides with the N.A., then y = 0 and Q = (d² / 8) * width.

For any other value of y, Q will be less than this amount. Hence, Q is a maximum when B-B is at the N.A., and so is the shear stress τ.

 

[chetzread]

No, I said the neutral axis is used as the reference axis for the calculation of Q. There's a difference.

You can set up a calculation for different values of Q as the location of the shear stress moves from the outer fiber to the neutral axis.

​

When the line B-B is located at the outer fiber of the beam, A = 0 and Q = 0, which also implies that τ = 0.

By bringing the axis B-B closer to the N.A., the area A above B-B is increasing, while y-bar is getting smaller. What happens to the product A * y-bar?

The N.A.is located at d/2 from the top or bottom of the section.

If the axis B-B is y units from the N.A.,then A = (d/2 - y) * width and y-bar = (d/2 + y) / 2.

Q = A * y-bar = [(d/2 - y) * width ] * [(d/4 + y/2)]

Q = [d²/8 + d*y/4 - d*y/4 - y²/2] * width

Q = [d² / 8 - y² / 2] * width

When the axis B-B coincides with the N.A., then y = 0 and Q = (d² / 8) * width.

For any other value of y, Q will be less than this amount. Hence, Q is a maximum when B-B is at the N.A., and so is the shear stress τ.

Thanks , everything is clear now

 

[chetzread]

No, I said the neutral axis is used as the reference axis for the calculation of Q. There's a difference.

You can set up a calculation for different values of Q as the location of the shear stress moves from the outer fiber to the neutral axis.

​

When the line B-B is located at the outer fiber of the beam, A = 0 and Q = 0, which also implies that τ = 0.

By bringing the axis B-B closer to the N.A., the area A above B-B is increasing, while y-bar is getting smaller. What happens to the product A * y-bar?

The N.A.is located at d/2 from the top or bottom of the section.

If the axis B-B is y units from the N.A.,then A = (d/2 - y) * width and y-bar = (d/2 + y) / 2.

Q = A * y-bar = [(d/2 - y) * width ] * [(d/4 + y/2)]

Q = [d²/8 + d*y/4 - d*y/4 - y²/2] * width

Q = [d² / 8 - y² / 2] * width

When the axis B-B coincides with the N.A., then y = 0 and Q = (d² / 8) * width.

For any other value of y, Q will be less than this amount. Hence, Q is a maximum when B-B is at the N.A., and so is the shear stress τ.

ybar should be y , right ? it's measured from the neutral axis to the centroid of B-B , right ? as in the figure...