Moment of inertia for T shaped object

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Dell
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h have a T shaped object (2-d) for which i need to find the moment of enertia,
i used a parallel axis theorem, then to check myself i found i site which had a direct equation for it
http://www.efunda.com/designstandards/beams/SquareTbeam.cfm

what i did

b1=100mm
h1=12mm
b2=12mm
h2=75mm
q=12.643mm(the distance from the joint of the 2 shapes to the centroid of the T)

>> ((b1*h1^3)/12)+((b1*h1)*(q+h1/2)^2)+((b2*h2^3)/12)+((b2*h2)*(h2/2-q)^2)

1.4094e-006 m4

using their equation

t=12mm
y=75-12.643=62.357mm
b=100mm
s=12mm
d=87mm

>> (t*y^3+b*(d-y)^3-(b-t)*(d-y-s)^3)/3

7.6247e-006m4


where have i gone wrong here?
 
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I don't know what you may have done wrong, but did you calculate the centroid correctly? Once you have calculated the centroid of the T-shape using moment areas, the I of the T-shape, about the xx axis thru its centroid, is the sum of the I + Ad^2 of each rectangle, where d is the distance from the centroid of that rectangle to the centroid of the T shape. It's all number or letter variable crunching from there. I'd avoid the cookbook solution, as you lose track of how the equation is derived.
 
the centroid is correct, using their"recipe" i got the same centroid, only i took the centroid from the joint of the 2 shapes and they took from the base of the T
 
Dell said:
the centroid is correct, using their"recipe" i got the same centroid, only i took the centroid from the joint of the 2 shapes and they took from the base of the T
Ok, if you are sure of the centroid, calculate I, using I_T = sum of I + Ad^2. Don't forget that d is the distance from the centroid of each rectangle to the centroid of the T. Please show your calculation using numbers, and define h1, h2, etc., so we can check your math. Your equation appears correct, although i haven't checked the numbers yet.
 
I checked your numbers and your result looks good. I didn't check the recipe nor the numbers you may have entered into it. Your way is better anyway. Toss out the cookbook!:mad: