# Torsion Shear of Shaft with Transverse Hole

1. Aug 9, 2011

### jmart157

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. Apr 24, 2012

### WillemBouwer

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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