# How to calculate principal moment of inertia of an unequal angle?

Is there a formula to calculate the moment of inertia of an unequal angle? I am not talking about x or y direction but the maximum and minimum moment of inertia about its rotated axes (usually labelled u and v) through its centroid.

I can get this using Massprop function in AutoCAD, but I would like to have a formula so that I can use it in excel.

For any shape, if you know the two moments, $I_{xx}$ and $I_{yy}$, there is a fascinating amount of study to arrive at the principal moments. You'd need to know $I_{xy}$ as well, which is harder to find tables on. Fortunately, an angle simply can be broken into two rectangles. Rectangles by definition have 0 for their product of inertia. So, the product of intertia for the angle is simply the area of the rectangles times the distance between the centroid of each rectangle in both x and y directions: $$I_{xy} = A_1d_{x_1}d_{y_1} + A_2d_{x_2}d_{y_2}$$

Note that dimensions are, unlike we're used to, relative vectors - as such for the first rectangle, $d_{y_1}$ is positive, but $d_{x_1}$ is negative. For rectangle 2, the reverse is true.

With this, then the principal moments are simply:

$$\frac{I_{xx}+I_{yy} \pm \sqrt{I_{xx}^2+I_{yy}^2+4I_{xy}^2-2I_{xx}I_{yy}}}{2}$$

The angle can be found as

$$\theta = \frac{1}{2}\arctan{\frac{2I_{xy}}{I_{yy}-I_{xx}}}$$

• Please ignore my edit – Nathan Apr 22 '19 at 8:33

In principle, it involves figuring out the orientation at which the product second moment of area is equal to zero. This orientation gives the principal second moments of area.