First moment and transversal shear

  • Thread starter etotheix
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  • #1
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Homework Statement



[PLAIN]http://img823.imageshack.us/img823/6601/unledpicturex.png [Broken]

Homework Equations



[itex]q = \tau/b[/itex]
[itex]\tau = (VQ)/(Ib)[/itex]

I would like to calculate the shear flux in one row of nail. By taking the first moment for the upper board, calculating q and dividing by 2 I get the correct value.

My question is the following, why can't I take for example the left board? I would get Q=0, but what is the physical meaning of this? Why can't I use the left or right board to calculate the flux in the nails?

Thank you.
 
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Answers and Replies

  • #2
PhanthomJay
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I am not sure how you are getting Q = 0 for the web, but regardless, you are looking for the longitudinal shear flow along the axis of the beam across the nails, where the flanges and webs are nailed together, which requires determining Q using the flange area, the piece that the nails must hold together without shearing. For perhaps a better insight, you might want to consider a similar box beam of the same overall dimensions, but with the flanges nailed from the top (and bottom) into the webs (flange width = 310 mm, vertical board lengths = 90 mm). The results will not be the same, but the visualization of why you use the flange Q and not the web Q might be easier.
 
  • #3
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A further way of getting insight in this is to return to the derivation from first principles of the formula τ=(VQ)/(Ib).
 

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