Question D.4: The parabolic semisegment OAB shown in Fig. D-15 has base b ...
The parabolic semisegment OAB shown in Fig. D-15 has base b and height h. Using the parallel-axis theorem, determine the moments of inertia I_{xc} and I_{yc} with respect to the centroidal axes x_{c} and y_{c}.
Use the parallel-axis theorem (rather than integration) to find the centroidal moments of inertia because the area A, the centroidal coordinates \bar{x} and \bar{y}, and the moments of inertia I_{x} and I_{y} with respect to the x and y axes are already known. These quantities were obtained earlier in Examples D-1 and D-3 (they also are listed in Case 17 of Appendix E) and are repeated here:
A=\frac{2 b h}{3} \quad \bar{x}=\frac{3 b}{8} \quad \bar{y}=\frac{2 h}{5} \quad I_{x}=\frac{16 b h^{3}}{105} \quad I_{y}=\frac{2 h b^{3}}{15}To obtain the moment of inertia with respect to the x_{c} axis, use Eq. (D-17) and write the parallel-axis theorem as
I_{x c}=I_{1}-A d_{1}^{2} (D-17)
I_{x c}=I_{x}-A \bar{y}^{2}=\frac{16 b h^{3}}{105}-\frac{2 b h}{3}\left(\frac{2 h}{5}\right)^{2}=\frac{8 b h^{3}}{175} (D-19a)
In a similar manner, obtain the moment of inertia with respect to the y_{c} axis:
I_{y c}=I_{y}-A \bar{x}^{2}=\frac{2 h b^{3}}{15}-\frac{2 b h}{3}\left(\frac{3 b}{8}\right)^{2}=\frac{19 h b^{3}}{480} (D-19b)