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) I_{x}=\int y^{2} d A and (12-9b)I_{y}=\int x^{2} d A. The differential element of area dA is selected as a vertical strip of width dx and height y, as shown in Fig. 12-12. The area of this element is

d A=y d x=h\left(1-\frac{x^{2}}{b^{2}}\right) d x   (b)

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 x^{2} dA. Therefore, the moment of inertia of the entire area with respect to the y axis is obtained as follows:

I_{y}=\int x^{2} d A=\int_{0}^{b} x^{2} h\left(1-\frac{x^{2}}{b^{2}}\right) d x=\frac{2 h b^{3}}{15}  (c)

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

as obtained from Eq. (12-12) I_{B B}=\int y^{2} d A=\int_{0}^{h} y^{2} b d y=\frac{b h^{3}}{3}. 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} d x=\int_{0}^{b} \frac{h^{3}}{3}\left(1-\frac{x^{2}}{b^{2}}\right)^{3} d x=\frac{16 b h^{3}}{105}    (d)

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 = xdy or by using a rectangular element of area dA = dxdy 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.

Appendix D ( case 17)

Parabolic semisegment (Origin of axes at corner) \begin{aligned}&y=f(x)=h\left(1-\frac{x^{2}}{b^{2}}\right) \\&A=\frac{2 b h}{3} \quad \bar{x}=\frac{3 b}{8} \quad \bar{y}=\frac{2 h}{5} \\&I_{x}=\frac{16 b h^{3}}{105} \quad I_{y}=\frac{2 h b^{3}}{15} \quad I_{x y}=\frac{b^{2} h^{2}}{12}\end{aligned}