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Shear stress?

  1. Aug 16, 2011 #1
    1. The problem statement, all variables and given/known data

    I just want to know the difference between certain equations for shear stress. I'm trying to find the factor of safety for pins that are in double shear and have a circular cross section.

    2. Relevant equations

    So far I've used stress = 2F/pi*d^2 for double shear. Following what I found on this website: http://www.roymech.co.uk/Useful_Tables/Screws/Bolted_Joint.html. This comes from the general equation stress = F/A. Then I used n = 0.6*UTS/stress for safety factor.

    3. The attempt at a solution

    I've also seen the equation stress = 4F/3A defined as 'max' shear stress when dealing with a circular cross section. So then is F/A just 'average' shear stress? The website I used seems to use F/A as a max stress though. How is 4F/3A derived, and should it be used for safety factor calculations since its a maximum? Also, since it is double shear would it be written differently?
     
    Last edited: Aug 16, 2011
  2. jcsd
  3. Aug 16, 2011 #2

    PhanthomJay

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    this is correct
    This is largely a matter of Code requirements..I usually use 0.4 * (Yield Stress) as a safe shear stress
    This is for a circular cross section subject to bending stresses and shear. The shears are maximum at the neutral axis and zero at the top and bottom points for this case.
    For a bolt in direct shear, F/A (or F/2A in double shear) is considered as uniform throughout the cross section (average can be considered as max , or simply 'shear stress'.
    the shear stress formula VQ/It is not used for bolts subject to shear forces only, with allowable max shear stresses (.4*Yield stress).
     
  4. Aug 16, 2011 #3
    PhanthomJay gave a very good reply already, but I would clarify one detail:

    The quantity [itex].6*f_y[/itex] has nothing to due with safety factors.
    It comes from the assumption of a Von Mises yield surface.
    In LRFD, the "resistance factor" is taken as .75 for direct shear. This is to be applied to the quantity [itex].6*f_y[/itex].


    Extra credit:
    If you're interested, one way of describing your total stress demand would be in matrix form (maybe you've seen this before?). For direct shear of magnitude [itex]\tau[/itex]:

    stress [itex]\sigma = [/itex] deviatoric stress [itex] \sigma ' = \begin{bmatrix}0 & \tau & 0\\ \tau & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}[/itex]

    The Von Mises formula is then:

    [itex]\sqrt{\frac{1}{2} \sigma_{ij}' \sigma_{ij}'}=\frac{f_y}{\sqrt{3}}[/itex]

    If you like, you can prove that the LHS of the above equation becomes [itex]\tau[/itex], and the RHS of the equation can be rounded to [itex].6*f_y[/itex]. This is the equation you are using, where you are taking [itex]\tau[/itex] equal to [itex]\frac{F}{A}[/itex]
    :)

    note: You would still need to apply some sort of safety factor.. perhaps a resistance factor, [itex]\phi[/itex], of .75 if you are using LRFD.
     
  5. Aug 17, 2011 #4

    nvn

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    gomerpyle: In response to post 1, as mentioned by PhanthomJay, for pins (bolts), you usually use average shear stress, tau_av = F/A, not peak shear stress, tau_pk = 1.3333*F/A, where A = 0.50*pi*d^2 for double shear.
    This is correct, if you ensure n = 2.4, provided you do not have threads in the shear plane, and assuming F = unfactored, applied shear load.
     
    Last edited: Aug 17, 2011
  6. Aug 17, 2011 #5
    Thank you all for the responses, I understand it now.
     
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