1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Science Advisor
    Homework Helper

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook