Question 12.4: The parabolic semisegment OAB shown in Fig. 12-15 has base b...
The parabolic semisegment OAB shown in Fig. 12-15 has base b and height h. Using the parallel-axis theorem, determine the moments of inertia I_{x_{c}} and I_{y_{c}} with respect to the centroidal axes x_{c} and y_{c}.
We can use the parallel-axis theorem (rather than integration) to find the centroidal moments of inertia because we already know the area A, the centroidal coordinates \overline{x} and \overline{y}, and the moments of inertia I_{x} and I_{y} with respect to the x and y axes. These quantities were obtained earlier in Examples 12-1 and 12-3. They also are listed in Case 17 of Appendix D and are repeated here:
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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} |
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}
To obtain the moment of inertia with respect to the x_{c} axis, we use Eq. (b) and write the parallel-axis theorem as follows:
I_{x_{c}}=I_{1}-Ad^{2}_{1} (b)
I_{x_{c}}=I_{x}-A\overline{y}^{2}=\frac{16bh^{3}}{105}-\frac{2bh}{3}\left(\frac{2h}{5}\right)^{2}=\frac{8bh^{3}}{175} (12-13a)
In a similar manner, we obtain the moment of inertia with respect to the y_{c} axis:
I_{y_{c}}=I_{y}-A\overline{x}^{2}=\frac{2hb^{3}}{15}-\frac{2bh}{3}\left(\frac{3b}{8}\right)^{2}=\frac{19hb^{3}}{480} (12-13b)
Thus, we have found the centroidal moments of inertia of the semisegment.