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 \bar{x} \text { and } \bar{y}_{e} , 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:
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}Appendix D (case 17):
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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} |
To obtain the moment of inertia with respect to the x_{c} axis, we use Eq. (12-17) I_{x_{c}}=I_{1}-A d_{1}^{2} and write the parallel-axis theorem as follows:
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} (12-19a)
In a similar manner, we obtain the moment of inertia with respect to the y_{c} axis:
I_{y_{c}}=I_{c}-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} (12-19b)
Thus, we have found the centroidal moments of inertia of the semisegment.