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

   | |
|==   ==|
   | |

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.

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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 figured out what was wrong with my calculation. The area moment of inertia of this cross is indeed larger than the square example.

If I were to separate the cross into three sections and sum up the area moments of inertia of these component sections, because two of the sections would not be centered about the centroid of the cross, the parallel axis theorem must be applied. The equation becomes:

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

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I am a bit rusty at moments of inertia, but I believe both your calculations are off. Calculating the moment of the cross would be easier if you thought of it as two rectangles and then subtracted off the moment of the intersection, I.e. the center square. I believe your value for the moment of the square is off by a factor of two.

I compliment you on your initiative

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The cross and the square have the same centroid. But they have different area the cross have less area than the square. The square have large moment because the its larger area. You do this using the parallel axis theorem. Consider a reference axis such that the objects in the lies in positive plane. Both of them have same moment on the centroidal axis but when you transfer it to the reference axis square have higher value

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