Question 12.3: Determine the moments of inertia Ix and Iy for the parabolic......
Determine the moments of inertia I_x and I_y for the parabolic semisegment OAB shown in Fig. 12-12. The equation of the parabolic boundary is
y=f(x)=h\left(1\ -\ \frac{x^2}{b^2}\right) (e)
(This same area was considered previously in Example 12-1.)
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.