Question 12.3: Determine the moments of inertia Ix and Iy for the parabolic......

To determine the moments of inertia by integration, we will use Eqs. (12-9a) and (12-9b). The differential element of area dA is selected as a vertical strip of width dx and height v, as shown in Fig. 12-12. The area of this element is

$I_x=\int{y^2dA}\quad \quad I_y=\int{x^2dA}$                      (12-9a,b)

$dA=y\ dx=h\left(1\ -\ \frac{x^2}{b^2}\right)dx$                           (f)

Since every point in this element is at the same distance from the y axis, the moment of inertia of the element with respect to the y axis is x²dA. Therefore, the moment of inertia of the entire area with respect to the y axis is obtained as follows:

$I_y=\int{x^2dA}=\int_{0}^{b}{x^2h\left(1\ -\ \frac{x^2}{b^2}\right)dx}=\frac{2hb^3}{15}$                   (g)

To obtain the moment of inertia with respect to the x axis, we note that the differential element of area dA has a moment of inertia $dI_x$ with respect to the x axis equal to

$dI_x=\frac{1}{3}(dx)y^3=\frac{y^3}{3}dx$

as obtained from Eq. (c). Hence, the moment of inertia of the entire area with respect to the x axis is

$I_x=\int_{0}^{b}{\frac{y^3}{3}dx}=\int_{0}^{b}{\left(1\ -\ \frac{x^2}{b^2}\right)^3dx}=\frac{16bh^3}{105}$                    (h)

These same results for $I_x$ and $I_y$ can be obtained by using an element in the form of a horizontal strip of area dA = x dy or by using a rectangular element of area dA = dx dy and performing a double integration. Also, note that the preceding formulas for $I_x$ and $I_y$ agree with those given in Case 17 of Appendix D.