1
$\begingroup$

I have a shape that I'd like to calculate the polar moment of inertia for:

enter image description here

the diameter is 600mm; the hexagon in the middle sides are 70mm each; the holes around the hexagon are 130mm from the center of the disk and they are 16mm.

$$ \frac{\pi D^4}{32} - \frac{5\sqrt3}{8}s^4$$

How can I remove the 5 holes from the calculation? Is that even something that's done?

[EDIT] The disk is 30mm thickness. The mass of the disk is 66.5Kg

The mass of the small holes is total 0.3Kg

The hexagon mass is 3Kg

weight without holes is 66Kg

mass with the circular cutouts is 66.5-0.3 = 66.2kg mass with the hexagon cutout is 66.5-3 = 63.5Kg mass with all the cutouts is 63.3Kg

Which percent change in weight would help calculate the polar moment of inertia for this shape?

$\endgroup$
  • $\begingroup$ How thick is the disc? What is the change in mass with and without the holes - how much difference will this make in % terms? $\endgroup$ – Solar Mike Aug 24 '17 at 16:03
3
$\begingroup$

Without looking up all the formula's for you, the approach to this problem is rather simple. Find the polar moment of a solid cylinder, and subtract off the polar moment of the holes.

For the off center holes you need to use the parallel axis therem, so you will have

$$I_\text{assembly} = I_\text{solid cylinder} - I_\text{hexagon} - 6 (I_\text{small cylinder} + m r^2 )$$

the expressions you need are all on Wikipedia.

| improve this answer | |
$\endgroup$
  • $\begingroup$ thank you, the parallel axis theorem is exactly what I was looking for! $\endgroup$ – user1610950 Aug 26 '17 at 4:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.