Why use Ixx instead of Iyy in the shear stress formula τ=VQ/It

[chetzread]

Homework Statement

i know that to calculate the shear stress, formula is τ= (V)(Q)/ (I)(t)
So, why the author choose to use Ixx, but not Iyy for shear stress at P?

Homework Equations

The Attempt at a Solution

IMO, we should done the question in 2 ways, which are by using Ixx and Iyy respectively...Am i right?

 

[David Lewis]

Yes, but the reason is to see which one is greater.

 

[chetzread]

Yes, but the reason is to see which one is greater.

but, the problem ask detremine shear stress at P , not maximum shear stress at P...

 

[chetzread]

Yes, but the reason is to see which one is greater.

why we need to choose the greater ?

 

[SteamKing]

Homework Statement

i know that to calculate the shear stress, formula is τ= (V)(Q)/ (I)(t)
So, why the author choose to use Ixx, but not Iyy for shear stress at P?

Homework Equations

The Attempt at a Solution

IMO, we should done the question in 2 ways, which are by using Ixx and Iyy respectively...Am i right?

The last part of the first line of the attachment is unclear. Does it say "... vertical shear force V = 3 kN" ?

 

[chetzread]

The last part of the first line of the attachment is unclear. Does it say "... vertical shear force V = 3 kN" ?

Ya , so , what are you trying to say ?

 

[SteamKing]

Ya , so , what are you trying to say ?

I'm asking for clarification of the problem statement is all. Part of image is too blurry for me to read.

 

[chetzread]

I'm asking for clarification of the problem statement is all. Part of image is too blurry for me to read.

Yes, it is...

 

[chetzread]

Yes, it is...

Do u know why Ixx is used here? Why not Iyy ?

 

[SteamKing]

Do u know why Ixx is used here? Why not Iyy ?

If you would answer my questions in Posts 5 and 7, I could probably tell you.

Why are you being so evasive in your replies?

 

[chetzread]

If you would answer my questions in Posts 5 and 7, I could probably tell you.

Why are you being so evasive in your replies?

Yes, it's vertical shear force of 3kN...

 

[David Lewis]

Because if a structural member can withstand the greater stress, it can also withstand lesser ones.

 

[SteamKing]

Yes, it's vertical shear force of 3kN...

Thanks.

Since the shear force is applied in the vertical direction, the shear stress depends on Ixx rather than Iyy. From the diagram of the cross section, the neutral axis runs parallel with the x-axis.

Therefore, $I_{xx} = \int y^2 \, dA$

which just happens to be greater than Iyy due to the orientation of the cross section.

The proper value of I is chosen not because it is greater, but because of how the shear is applied. If the shear were applied horizontally, then the shear stress at P would depend on Iyy instead of Ixx.

The shear stress at P is $\tau = \frac {V ⋅ Q_x}{I_{xx} ⋅ t}$, where V is the shear force, Q_x is the first moment of the area between P and the top of the cross section, calculated w.r.t. the neutral axis, I_xx is as discussed above, and the thickness of the section t is equal to the width of 100 mm.