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Shear stress

  1. Jul 31, 2016 #1
    1. The problem statement, all variables and given/known data
    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?
    2. Relevant equations


    3. 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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  3. Jul 31, 2016 #2

    David Lewis

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    Yes, but the reason is to see which one is greater.
     
  4. Jul 31, 2016 #3
    but, the problem ask detremine shear stress at P , not maximum shear stress at P....
     
  5. Aug 26, 2016 #4
    why we need to choose the greater ?
     
  6. Aug 26, 2016 #5

    SteamKing

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    The last part of the first line of the attachment is unclear. Does it say "... vertical shear force V = 3 kN" ?
     
  7. Aug 26, 2016 #6
    Ya , so , what are you trying to say ?
     
  8. Aug 26, 2016 #7

    SteamKing

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    I'm asking for clarification of the problem statement is all. Part of image is too blurry for me to read.
     
  9. Aug 26, 2016 #8
    Yes, it is....
     
  10. Aug 26, 2016 #9
    Do u know why Ixx is used here? Why not Iyy ?
     
  11. Aug 26, 2016 #10

    SteamKing

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    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?
     
  12. Aug 26, 2016 #11
    Yes, it's vertical shear force of 3kN....
     
  13. Aug 26, 2016 #12

    David Lewis

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    Because if a structural member can withstand the greater stress, it can also withstand lesser ones.
     
  14. Aug 26, 2016 #13

    SteamKing

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    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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