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

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:

• 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!

• I will point you in the right direction. This problem can be solved quite easily if you calculate Ixx by using the polar form of the second moment of area formula. If you try to use the cartesian form of the formula, it can get quite nasty. In polar form, the distance from the x axis is r * sin(theta) and dA = r * dr * dtheta Jul 24 at 16:05
• I'm just gently wondering what you intend to do with this number when you've worked it out. The behavior of slitted tubes if used as beams is NOT a case of plugging the MoI into the beam formula. Jul 24 at 22:10
• @RocketScience Oh thanks! I've been looking all over for a standard formula, that's the only way we learned to calculate Ix in school. Didn't think about integrating the formula myself using the polar form. I'll have to look into how to use the double integral method for calculating Ix. Thank you!
– Lars
Jul 25 at 7:23
• @GregLocock This is part of a bent plate (kind of like a hinge in cross-section). So this Ix gets added to the Ix of a plate to use in the beam formula to estimate the deflection of the plate like this for many different lenghts, widths and material thicknesses. Also, like a hinge, it includes a pin through the hole in the middle. Does it make a big difference using a slit pipe moment of inertia vs reality? I only need an estimate, but if it is wrong by a lot than I might need to do some things differently.
– Lars
Jul 25 at 7:27
• I guess the issue is that it might fail in bending by local buckling near the slit. I don't remember how to calculate that, I think the aero boys call it crippling. Jul 25 at 7:35

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.

• Thank you very much! I need to add this Ix to another Ix (where the X axis is horizontal in the drawing). Do I need to use Ix = (Ix+Iy)/2 + ((Ix-Iy)/2)*cos(-angle)-Ixy*sin(-angle) to rotate the moment of intertia values back to the regular carthesian system? Or is the formula you gave already a rotated version of the tilted values?
– Lars
Jul 25 at 7:34
• Yes, you need to rotate the axes accordingly.
– r13
Jul 25 at 12:55