- #1
jmart157
- 20
- 0
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)
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)