Thin Wall Moment of Area

Question

Hi there, how do you find the moment of area about the x and y axes (at centroid)?

Dimensions are given as in the picture and thickness "c" is constant throughout.

Answer 1

I'm here going to assume that by moment of area you mean the second moment of area (aka moment of inertia). That being said, the argument I'll be making is valid for the first moment as well.

The planar second moment of area around an axis (say, the x-axis) is calculated by:

$$I_x = \iint y^2 \text{d}x\text{d}y$$

You'll notice this means the moment of inertia around the x-axis is a measure of the square of the distance from the origin along the y-axis. So, what's the impact of the shape being angled? Well, none. The bottom point is to the left of the centroid, but the moment of inertia around the x-axis doesn't care.

So, when calculating the moment of inertia around the x-axis, you can pretend you're dealing with a rectangular beam, where the classic equation $I_x = bh^3/12$ applies. The same applies when calculating the moment of inertia around the y-axis.

Now, there is a nuance: the beam's thickness $c$ is also inclined. So the thickness to be used when calculating $I_x$ and $I_y$ is not $c$, but the horizontal and vertical projections of $c$ according to the beam's inclination. (This projection fails if the beam were horizontal, but is otherwise accurate)

By calculating the beam as rectangular, you are effectively cutting off tiny triangles from the ends of the beam, but those are certainly negligible.