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:
\quad\quad\quad\quad A = \frac{2bh}{3} \quad \overline{x}=\frac{3b}{8}\quad \overline{y}= \frac{2h}{5}\quad I_x = \frac{16bh^3}{105}\quad I_y = \frac{2hb^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
\quad\quad\quad\quad I_{x c}=I_{x}-A\overline{{{y}}}^{2}\,=\,\frac{16b h^{3}}{105}\,-\,\frac{2b h}{3}\!{\Bigg\lgroup}\frac{2h}{5}{\Bigg\rgroup}^{2}\,=\,\frac{8b h^{3}}{175}\quad\quad (D-19a)
In a similar manner, obtain the moment of inertia with respect to the y_c axis:
\quad\quad\quad\quad I_{y c} = I_{y}-A\overline{{{x}}}^{2}=\frac{2h b^{3}}{15}-\frac{2 b{h}}{3}{\Bigg\lgroup}\frac{3b}{8}{\Bigg\rgroup}^{2}=\frac{19 h b^{3}}{480}\quad\quad(D_19b)