Answer
$I_x=154(10)^6mm^4$
Work Step by Step
The required moment of inertia can be determined as follows:
$I_{x}=\Sigma (I+Ad^2_y)$
$\implies I_x=\Sigma (\frac{bh^3}{12}+Ad^2y)$
We plug in the known values to otbain:
$I_x=\frac{70(30)^3}{12}+70(30)(185)^2+\frac{30(200)^2}{12}+30(200)(100)^2+\frac{170(30)^3}{12}+170(30)(15)^2$
This simplifies to:
$I_x=154(10)^6mm^4$
