Moment of Inertia / Section Modulus for a cruciform (rectangular cross)

[rubadello]

Hi,

I'm trying to find a formula to help me find either the Moment of Inertia for a cruciform (rectangular cross) or better yet, its Section Modulus (S).

Please see attached image.

 

[PhanthomJay]

Per forum rules, you must show some attempt at a solution before we can help.

 

[GodBlessTexas]

Jay,

Wouldn't this require parallel axis theorem?

Or without getting into that could you just say [(b*h^2)/6] + [(h*b^2)/6] <-- (wait, isn't this the same as perpendicular axis theorem?)

where b is the length from end to end of the cruciform, and h is the width of the segment?

(for section modulus)

 

[PhanthomJay]

Jay,

Wouldn't this require parallel axis theorem?

Or without getting into that could you just say [(b*h^2)/6] + [(h*b^2)/6] <-- (wait, isn't this the same as perpendicular axis theorem?)

where b is the length from end to end of the cruciform, and h is the width of the segment?

(for section modulus)

You don't really need the parallel axis theorem because the cross section is symmetric, so you can just take the moment of inertia about the centroid, which is [(b*h^3)/12] + [(h*b^3)/12] (actually, a bit less, because you have to take out that small square where the rectangles cross). But to get the section modulus, your calculation is not correct, you have to take the total I and divide it by b/2. Note that if b is much greater than h, that first term is insignificant, and essentially the moment of inertia approximates hb^3/12, and the section modulus approximates hb^2/6.