The discussion focuses on calculating the Moment of Inertia and Section Modulus for a cruciform (rectangular cross) section. The formula for the Moment of Inertia is established as [(b*h^3)/12] + [(h*b^3)/12], where b represents the length and h the width of the segments. The Section Modulus is derived by dividing the total Moment of Inertia by b/2. It is noted that if b is significantly larger than h, the Moment of Inertia simplifies to hb^3/12 and the Section Modulus to hb^2/6.
PREREQUISITESStructural engineers, mechanical engineers, and students studying mechanics of materials who are involved in calculating properties of complex cross-sections.
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.GodBlessTexas said: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)