Calculating Q (first moment of inertia) above and below the neutral axis

1. Jul 26, 2010

tn3003

1. The problem statement, all variables and given/known data

This is the cross sectional are of the shape: http://img38.imageshack.us/i/shapep.jpg/" [Broken]
It's made of 2 10"x1" plates. (Picture is not to scale)

Q (first moment of inertia) above and below the neutral axis should be the same. For some reasons, my calculated Q above is not equal to Q below the NA.

3. The attempt at a solution

Taking the datum at the bottom

y bar = (10*0.5 + 10*6) / (20) = 3.25"

Q above = 7.75 * 2.75 = 21.31

Q below = 2.25 * (2.25/2) + 10 * 2.75 = 30.03

21.31 =/= 30.03

Someone please point out my mistake.

Last edited by a moderator: May 4, 2017
2. Jul 27, 2010

PhanthomJay

Your Q below is corect, but in calculating thte Q above, the distance from the centroid to the N.A. is not 2.75. I think you may have just written down the wrong number.

3. Jul 27, 2010

tn3003

OK I see it now. The distance for Q above should be 7.75"/2 = 3.875"
I was using the centroid of the whole shape, which gave me 7.75" - 5" = 2.75" and that's wrong.

Thank you PhanthomJay for not giving out the answer and let me figure it out myself. I really appreciate it.

Last edited: Jul 27, 2010