Answer
$I_x^{\prime}=520(10^6)mm^4$
Work Step by Step
We can find the required moment of inertia as follows:
$I_x^{\prime}=\Sigma (I+Ad^2_y)$
We plug in the known values to obtain:
$I_x^{\prime}=\frac{(100+100)(2\cdot 25+2\cdot 200 sin 45^{\circ})^3}{36}+4(\frac{200 cos 45^{\circ}(200 sin 45^{\circ})^3}{36})+200 cos 45^{\circ}200 sin 45^{\circ}\cdot \frac{200 sin45 }{3^2}-2(\frac{(200)^4(\frac{\pi}{4}-\frac{sin 45}{2})}{4})$
This simplifies to:
$I_x^{\prime}=520(10^6)mm^4$
