Calculating Shear Stress: Choosing Between Ixx and Iyy for Point P

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SUMMARY

The discussion focuses on calculating shear stress at point P using the formula τ = (V)(Q)/(I)(t). The participants clarify that Ixx is used instead of Iyy due to the vertical shear force of 3 kN applied to the structural member, which aligns with the orientation of the neutral axis parallel to the x-axis. The choice of Ixx is justified not by its greater value but by the direction of the applied shear force. The correct formula for shear stress at P is τ = (V)(Qx)/(Ixx)(t), where Qx is the first moment of the area above point P.

PREREQUISITES
  • Understanding of shear stress calculations in structural engineering
  • Familiarity with the concepts of moment of inertia (Ixx and Iyy)
  • Knowledge of vertical shear forces and their effects on structural members
  • Ability to interpret cross-sectional diagrams of structural elements
NEXT STEPS
  • Study the derivation of the shear stress formula τ = (V)(Q)/(I)(t)
  • Learn about the significance of the neutral axis in beam theory
  • Explore the differences between Ixx and Iyy in various loading conditions
  • Investigate the effects of shear force direction on stress distribution in beams
USEFUL FOR

Structural engineers, civil engineering students, and professionals involved in analyzing shear stress in beams and other structural elements will benefit from this discussion.

chetzread
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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?
 

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Yes, but the reason is to see which one is greater.
 
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David Lewis said:
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...
 
David Lewis said:
Yes, but the reason is to see which one is greater.
why we need to choose the greater ?
 
chetzread said:

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" ?
 
SteamKing said:
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 ?
 
chetzread said:
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.
 
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SteamKing said:
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 said:
Yes, it is...
Do u know why Ixx is used here? Why not Iyy ?
 
  • #10
chetzread said:
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?
 
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  • #11
SteamKing said:
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...
 
  • #12
Because if a structural member can withstand the greater stress, it can also withstand lesser ones.
 
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  • #13
chetzread said:
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, Qx is the first moment of the area between P and the top of the cross section, calculated w.r.t. the neutral axis, Ixx is as discussed above, and the thickness of the section t is equal to the width of 100 mm.
 

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