# Homework Help: Second Moment of Area

1. Dec 29, 2008

### v_pino

1. The problem statement, all variables and given/known data
I have to find the second moment of area, Iy.
The answer is 351.88 x 10^6

2. Relevant equations

3. The attempt at a solution
I divided the shape into 2 sections and found the second moment of area for each of them then added the two. I have attached my work but is rather poor quality. I'll explain my working more if it is unclear.

Thanks :D

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2. Dec 29, 2008

### PhanthomJay

It looks like using calculus you are attempting to calculate the moment of inertia about the right edge of the left wide flange; if so, your second part integration limits seem like they ought to be from 0 to .3 rather than .5 to .35. But in any case, unless the problem specifically states, you are trying to find the second moment of area about the centroid of the overall section. So you must first calculate the centroid using the moment area method, then calculate the momemt of inertia about the centroid using the parallel axis theorem. Are you familiar with both? You don't need calculus if you know that the moment fo inertia of a rectangle about its own centroid is bh^3/12. I try to steer away from using calculus as much as possible..it's a great learning tool, but it can throw you off if not used properly.

3. Dec 29, 2008

### v_pino

I am a little confused with the difference between second moment of area and moment of inertia.

Isn't the second moment of area = integral of y^2 dA

and the moment of inertia = integral of y^2 dm

So in my solution for my problem, didn't I calculate the moment of area instead of inertia?

thanks

4. Dec 30, 2008

### nvn

v_pino: Second moment of area, integral[(y^2)dA], is sometimes called area moment of inertia. Mass moment of inertia is integral[(y^2)dm]. You got the correct answer for second moment of area about the left-hand edge of your cross section. Nice work. But as PhanthomJay mentioned, the question is probably asking you to compute the second moment of area about a vertical axis passing through the centroid of the cross section. However, when you do this, the answer is not 351.88e6 mm^4.