Question A.04: For the rectangular area of Fig. A.16, determine (a) the mom...

(a) Moment of Inertia I_{x}. We select as an element of area a horizontal strip of length b and thickness d y (Fig. A.17). Since all the points within the strip are at the same distance y from the x axis, the moment of inertia of the strip with respect to that axis is

d I_{x}=y^{2} d A=y^{2}(b d y)

Integrating from y=-h / 2 to y=+h / 2, we write

\begin{aligned} I_{x}=\int_{A} y^{2} d A & =\int_{-h / 2}^{+h / 2} y^{2}(b d y)=\frac{1}{3} b\left[y^{3}\right]_{-h / 2}^{+h / 2} \\ & =\frac{1}{3} b\left(\frac{h^{3}}{8}+\frac{h^{3}}{8}\right) \end{aligned}

or

I_{x}=\frac{1}{12} b h^{3}

(b) Radius of Gyration r_{x}. From Eq. (A.10), we have

I_{x}=r_{x}^{2} A \quad \frac{1}{12} b h^{3}=r_{x}^{2}(b h)

and, solving for r_{x},

r_{x}=h / \sqrt{12}