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This is a pretty simple problem, but I'm just not sure how to go about it. I want to find the moment of inertia of a discontinuous hollow circle. But since the shape is so irregular, I'm not sure how to go about solving for it. I've posted the shape in mind, so hopefully that'll clear up any confusion.

discontinuous hollow circle

Would an equivalence be needed to calculate the moment of inertia? For example, would I assume that the "new" diameter to calculate with for the outer circle is (0.625 - 0.375)?

Note: All these numbers have inches as units.

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  • $\begingroup$ Do you want to calculate the MOI about the horizontal axis? $\endgroup$ – Andrew Mar 16 '18 at 5:36
  • $\begingroup$ Yes, the horizontal axis. $\endgroup$ – tempestwing0101 Mar 16 '18 at 6:03
  • $\begingroup$ I think it's quite a tough one to calculate by hand. What program did you use for that drawing? AutoCAD for example offers a tool to calculate MOI for objects you've drawn. $\endgroup$ – Andrew Mar 16 '18 at 6:48
  • $\begingroup$ What's holding it together? If nothing, then calculate for the single shape & "double it" $\endgroup$ – Carl Witthoft Mar 16 '18 at 14:18
  • $\begingroup$ Take a element using polar coordinate and integrate it. $\endgroup$ – Boris Mar 16 '18 at 16:02
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Edit: There seems to be a much easier way I overlooked, which I'll explain. My first answer is kept below for reference.

Your assembly consists of a small sector subtracted from a larger sector as shown below:

enter image description here

You calculate the moment of inertia of the sector about the horizontal axis as follows:

$$I=\frac{R^4}{24}(3\phi-3sin(\phi)-2sin(\phi)sin^2(\frac{\phi}{2}))$$

So you will have:

$$I_{xx}=2(I_{out}-I_{in})$$


Old Answer:

It will take some effort, but it can be done analytically:

enter image description here

You first calculate the moment of inertia of the complete annulus with:

$$I_{ann}=\frac{\pi}{4}(R^4-r^4)$$

Then you subract twice the moment of inertia, $I_{rec}$, of the lxb rectangle calculated with $\frac{lb^3}{12}$ and the parallel axis theorem.

Now you have subtracted the red hatched part which shouldn't be, but not yet the green part, which should be, so lets fix this quickly by adding the red moment twice, and subtracting the green moment twice.

The image below explains how to calculate the moment of inertia of the sectors about the centreline:

enter image description here

$$I_{sect}=\frac{r^4}{8}(\phi_i-sin(\phi_i)+2sin(\phi_i)sin^2(\frac{\phi_i}{2}))$$

$\phi_i$ is measured in radians.

So in the end you have:

$$I_{xx}=I_{ann}-2(I_{rec}+I_{sect-green}-I_{sect-red})$$

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very close to 1E-3 in^4

More precisely: 997E-6 in^4

I wouldn't (didn't) use either of the other algebraic proposals. It's relatively easy to write an equation for the width of cut face at a defined offset from the centre:

R.o = outer radius = 0.3125"

R.i = inner radius = 0.25"

G = gap = 0.375"

width across the outer circle (whether it's actually there or not) at y from centre = 2*sqrt(R.o^2 - y^2) by pythagoras

width across the inner circle = 2*sqrt(R.i^2 - y^2)

cut width = max(0, width across outer-max(G, width across inner)).

That's trivial (5 mins) on excel with a row calcuating the width of each of (say) 500 slices from -0.3125" to 0.3125".

I actually did it in a numerical computer algebra system to get 997E-6, when I do it on Excel and 500 slices I get 996E-6.

Hand check - by eye, it's similar to two rectangles each 0.0625" wide and 0.5" deep, which would have I 1.3E-3, so I probably trust my answer.

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Assuming you want the moment of inertia, "$Ixx$", about the horizontal axis, you do as follows:

You find the "I" of a complete ring about xx axis first and that is done by calculating the "I" of the big circle: R = 0.625 subtracted by the "I" of small inner disk: R = 0.5.

Now you subtract the "I" of two missing parts from the top and bottom. These can be broken into segments and small triangles on their sides. one may need the help of Matlab, or Mathematica or such tools.

But generally for each part you find its "I" about its own CG then add that to its area times the square of its CG height from xx axis. Wolfram has the equation for these "$ y^-$"s, of a segment as well as its area. You can search for CG and area of a circle segment.

Similar routine can be applied to find:$Iyy$.

Edit:

As mentioned in other answers another way of calculating Ixx is two subtract the I of smaller segment from that of bigger segment.

$$Iyy = R^4/24 ( 3\theta -3sin\theta -2sin\theta sin^2(\theta/2) $$

Based on this equation there is an online calculator which you can plug in first the big segment and find its Iy then the small one and subtract the results to find the I of your section. Finaly you need to multiplay it by 2, Because you have 2 sides. Hear is the link to online calculator, the blue bottom on top of the page turns the calculator on/off. circle segment geometric properties

You may want to bookmark this page, It has many calculators and shapes!

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  • $\begingroup$ Matlab/Mathematica? this problem has an analytic solution. $\endgroup$ – Carl Witthoft Mar 16 '18 at 14:20
  • $\begingroup$ @CarlWitthoft, did you read my answer? I did not recommend using Matlab for a numerical solution; I recommended they may have the formulas. However, I modified my answer and added both an equation for Ixx and an online calculator which utilizes that same equation. $\endgroup$ – kamran Mar 16 '18 at 16:40

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