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!

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

    Attached Files:

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