Second moment of area of a cross section?

Question

Having a bit of trouble finding Ixx for the following cross section:

http://img261.imageshack.us/img261/7579/27453636wy6.png

The line dividing the two triangles is of the length w.

Any help would be appreciated.

Answer 1

Make a table as follows:

Item_Area_y____Ay____Ay^2__Ixx

1.___30w_50_1500w_75000w_6000w

2.___15w_20__300w__6000w__750w

SUM_45w___1800w_81000w__6750w

Yn = Ay/At = 1800w/45w = 40

Ixx = Sum Ay^2 + Sum Ixx - At(Yn)^2

Ixx(total) = 81000w + 6750w - 45w(40)^2 = 15250w

The first column is the item number where the upper triangle is taken as the first item.

The second column is the area ot the item.

The third column is the distance of the centroid of the area of the item to the bottom of the entire figure.

The fourth column is the product of the entries in column 2 and 3.

The fifth column is the product of the entries in column 4 and 3.

The sixt column is the moment of inertia of the item at its neutral axis. For a triangle this is equal to (bd^3)/36.

The first row is for values for the upper triangle

The second row is for the values for the lower triangle

The third row is the sum of the entries in columns 2, 4,5 and 6

The location of the neutral axis of the entire figure (Yn), therefore is obtained by dividing the sum of Ay by the sum of A

The Moment of inertia of the entire figure is obtained by:

Ixx(total) = Sum Ay^2  Sum Ixx - Sum A(Yn)^2