Answer
$I_y=91.3(10^6)mm^4$
Work Step by Step
We can find the required moment of inertia as follows:
$I_y=\Sigma (I+Ad^2_x)$
$\implies I_y=\Sigma (\frac{bh^3}{12}+Ad^2_x )$
We plug in the known values to obtain:
$I_y=\frac{(30)(70)^3}{12}+30(70)(65)^2+\frac{200(30)^3}{12}+200(30)(15)^2+\frac{(30)(170)^3}{12}+30(170)(115)^2$
This simplifies to:
$I_y=91.3(10^6)mm^4$
