Second moment of area (moment of inertia) for annulus segment

Question

I am a Mechanical Engineering student and I'm working on some calculations for an assignment at a company. As a part of this assignment I need to calculate the moment of inertia (Ixx) of a segment of an annulus (so a circle with inside and outside radius, but not fully closed).

The start (top) of the shape is always perpendicular to the y axis. The shape then forms an arc counterclockwise (shouldn't matter for the value I need) until an angle α is reached. So the following variables are needed in the formula:

• Inner radius/diameter
• Outer radius/diameter
• Angle α

The angle α always lies somewhere between 270 and 360 degrees.

I've been looking everywhere but I can't seem to gather all the data I need. I thought of two options to do this:

1: I can get the moment of inertia for a circle and subtract the inner circle from that, but then I have a closed annulus. I want an open annulus. So if I do that I need to remove a segment from this shape, of which I can't find any formulas for calculating this.

2: I can also remove a wedge- shape from both circles and then subtract them from each other. I could find formula's for the wedge shapes (circle sectors), but I could only find a formula for a situation where the x-axis goes through the centroid of the shape and the origin of the radius. To rotate the axis I need the moment of inertia for Ixx, Iyy and Ixy, but I can only find the one for Ixx.

3: Two wedges (circle sectors) with the same angle and different radii.

These are the formula's I found (they're all from Wikipedia, but I searched through a lot of websites and none of them had additional information):

TL;DR: I require the formula for the moment of inertia over the x axis for a circle segment rotated with one side straight up, where the radius and the angle are variable. I have searched everywhere but can't find it.

If anyone could help me in the right direction that would be very much appreciated!

Answer 1

Moment of Inertia of Circular Sector (Origin of Axes at Center of Circle):

Rotate axes so y axis bisects the sector as shown.

$Ix = \dfrac{r_i^4 (\alpha + sin\alpha cos\alpha)}{4}$

$Iy = \dfrac{r_i^4 (\alpha - sin\alpha cos\alpha)}{4}$

$Ixy = 0$

Now the result can be obtained simply by subtracting the void areas from the full circle.

Answer 2

I have put together a comprehensive solution to this here: https://www.adamlebaigue.com/engineering

The method involves working out the second moment of area of the large and smaller circular sectors and subtracting one from the other. This is done when finding the geometric centroid and when applying the parallel axis theorem.

Additionally this is only done for the local axis of the annular sector. The reason for this is that it greatly simplifies the problem due to symmetry being applied.

To get the second moment of area when the annular sector is rotated you can then apply a transformation to the second moments of area you've previously calculated.

So overall it's a two step process

For step one the local second moment of areas of the annular sector can be found as follows:

Once you have found the second moment of area about these local axis you can then apply a global transformation to find the second moment of area when the annulus is rotated i.e.: