## Engineering Mechanics: Statics & Dynamics (14th Edition)

$1971in^4$
We can find the required moment of inertia as follows: We know that $I_{triangle}=\frac{bh^3}{36}$ and $I_{rectangle}=\frac{bh^3}{12}$ We find the area for each part in the given figure $A_1=\frac{hb}{2}$ $A_1=\frac{(3)(6)}{2}=9in^2$ $A_2=hb$ $\implies A_2=(6)(6)=36in^2$ and $A_3=\frac{hb}{2}=\frac{(9)(6)}{2}=27in^2$ Now $I_{sum}=\Sigma (I+Ad^2_x)$ We plug in the known values to obtain: $I_{sum}=\frac{bh^3}{36}+9(7)^2+(\frac{bh^3}{12}+36(3)^2)+(\frac{bh^3}{36}+27(6)^2)$ This simplifies to: $I_{sum}=1971in^4$