The formatting of my previous post went a bit haywire due to my post being written in MS Word and then posted across. For clarity;
Ip, or the polar moment of inertia, is a basically a mathematical quantity that describes the distribution of the cross sections area about the cross section axis. For a solid shaft;
Ip = [pi*(r^4)] / 2
= [pi*(d^4)] / 32
r = radius (mm)
d = diameter (mm)
For a hollow shaft;
Ip = [pi/2]*[ro^4 – ri^4]
= [pi/32]*[do^4 – di^4]
ri = inner radius (mm)
ro = outer radius (mm)
di= inner diameter(mm)
do = outer diameter(mm)
It is true that the inner elements of a solid shaft carry very little shear stress. In fact, for a solid shaft, the shear stress varies in a linear manner from zero in the centre to maximum at the outer radius.
A hollow shaft is much more structually efficient in terms of amount of torque carried for the weight of material in the shaft, but the outer diameter will be greater. Removing the core of a solid shaft "shifts" the shear stress distribution, thus, to get the same maximum shear stress on the outside, a hollow shaft will need to be of greater radius.
The shear stress distribution in a hollow shaft is still linear, but it now starts at some nonzero value on the inner radius, and increases (linearly) to the maximum value on the outer radius.
Last edited by JetMech; 5th Nov 2006 at 04:24.
