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 y, as shown in Fig. 12-12. The area of this element is

I_{x}=\int{y^{2}dA}          I_{y}=\int{x^{2}dA}          (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^{2}dA}=\int_{0}^{b}{x^{2}h\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

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

I_{BB}=\int{y^{2}dA}=\int_{0}^{h}{y^{2}b  dy=\frac{bh^{3}}{3}}                  (c)

I_{x}=\int_{0}^{b}{\frac{y^{3}}{3}dx}=\int_{0}^{b}{\frac{h^{3}}{3}\left(1-\frac{x^{2}}{b^{2}}\right)^{3}}dx=\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.

Parabolic semisegment   (Origin of axes at corner) y=f(x)=h\left(1-\frac{x^{2}}{b^{2}}\right)   A=\frac{2bh}{3}      \overline{x}=\frac{3b}{8}         \overline{y}=\frac{2h}{5}   I_{x}=\frac{16bh^{3}}{105}    I_{y}=\frac{2hb^{3}}{15}    I_{xy}=\frac{b^{2}h^{2}}{12}