# Does a cross really have a smaller area moment of inertia than a square of the same area?

I need to calculate the area moment of inertia of a cross for a homework assignment. The cross is a symmetric plus sign:

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where the length/width of each arm is 1" (so the total area is 5 square in.).

My understanding is that I can just separate the cross into three sections (left, middle, right), so the area moment of inertia would be

$$I = \sum \frac{bh^3}{12} = \frac{1\times 1^3 + 3\times 1^3 + 1\times 1^3}{12} = \frac{5}{12}$$

This is surprising to me, because a 2.2" x 2.2" square would have the same area but a much larger moment of inertia:

$$I = \frac{bh^3}{12} = \frac{(\sqrt{5})^4}{12} = 2.08$$

Since the arms of the cross span 3 inches, I would expect there to be more material further from the center, whereas the square only has a small amount of area in the corners that the cross doesn't have.

Are these formulas correct or am I doing something wrong? Thanks in advance.

• Don't you mean $1 \times 3^3$? – Peter Shor Apr 15 '13 at 3:44
• No, I was determining the area moment of inertia about the y-axis. The problem lies with the two square sections at the left and right of the cross. Because their centroids are a unit away from the centroid of the cross, the parallel axis theorem applies. – mgiuffrida Apr 15 '13 at 5:31

$$I = \sum \frac{bh^3}{12} = 2[\frac{1\times 1^3}{12}+(1\times1^2)] + \frac{3\times 1^3}{12} = \frac{29}{12}$$