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Torsion Shear of Shaft with Transverse Hole

  1. Aug 9, 2011 #1
    Looking to figure out the allowable stress (psi) of a shaft which has two pin holes in it on the transverse axis of the shaft.

    tau = allowable stress, psi
    T = torque, in-lb
    c = distance from center of gravity to extreme fiber
    J = polar moment of inertia
    D = diameter of the shaft
    d = diameter of the pin hole

    tau = Tc / J = (T*(D/2)) / (((pi*D^4)/32) - (d*D(d^2 + D^2))/12)

    I think this is right. but i don't know why it is right. Can someone please help me prove the backend of the equation? ie: (d*D(d^2 + D^2))/12)
     
  2. jcsd
  3. Apr 24, 2012 #2
    jmart157: I am not yet an expert but whould give some input.

    The tau = Tc / J equation is indeed correct

    Tc = T*(D/2), which I think you understand

    So the J = (((pi*D^4)/32) - (d*D(d^2 + D^2))/12) part I think is where you are struggling...

    When looking at the shaft: Jcircle = (pi*D^4)/32) which is the first part
    and the hole the pin creates has a square cross sectional area:
    Jsquare = (d*D(d^2 + D^2))/12
    So the total polar moment of the cross sectional area: Jtotal=Jcircle-Jsquare
    Hope this helps, check the attackment that I've put together to help you
     

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