Chapter 9, Problem 9.65P

Compute the principal centroidal moments of inertia for the plane area.

To determine

Compute the principal centroidal moments of inertia for the plane area.

Answer to Problem 9.65P

The principal centroidal moments of inertia:

  $I_1=1405i{n}^4$

  $I_2=458i{n}^4$

Explanation of Solution

Given information:

The plane area shown in figure P9.65.

Calculations:

  $\begin{array}{l}\text{For the figure shown above, finding the centroid of the plane area:}\\ A_1=48i{n}^2\\ {\overline{x}}_1=4in.\\ {\overline{y}}_1=-3in.\\ A_2=\frac{\pi \left(8^2\right)}{4}\text{ =5027 }i{n}^2\\ {\overline{x}}_2={\overline{y}}_2=\frac{4\left(8\right)}{3\pi }=3.395in.\\ A={{\displaystyle \sum A}}_i=48+50.27=98.27i{n}^2\\ \overline{x}=\frac{{\displaystyle \sum A_i}{\overline{x}}_i}{A}=\frac{48\left(4\right)+50.27\left(3.395\right)}{98.27}=3.691in.\\ \overline{y}=\frac{{\displaystyle \sum A_i}{\overline{y}}_i}{A}=\frac{48\left(-3\right)+50.27\left(3.395\right)}{98.27}=0.2714in.\end{array}$

  $\begin{array}{l}\text{The moments of inertia with respect to }x-\text{ and }y-\text{axes:}\\ I_x=\frac{8\left(6^3\right)}{3}+\frac{\pi \left(8^4\right)}{16}=1380.2i{n}^4\\ I_y=\frac{6\left(8^3\right)}{3}+\frac{\pi \left(8^4\right)}{16}=1828.2i{n}^4\\ \text{And, the product of inertia:}\\ I_{xy}=48\left(4\right)\left(-3\right)+\frac{8^4}{8}=64.0i{n}^4\end{array}$

  $\begin{array}{l}\text{Now, defining the moments and product of inertia with respect to centroidal }x-\text{ and }y-\text{axes:}\\ {\overline{I}}_x=I_x-A{\overline{y}}^2=1380.2-98.27{\left(0.2714\right)}^2=1373.0i{n}^4\\ {\overline{I}}_y=I_y-A{\overline{x}}^2=1828.2-98.27{\left(3.691\right)}^2=489.4i{n}^4\\ {\overline{I}}_{xy}=I_{xy}-A\overline{x}\overline{y}=-64.0-98.27{\left(3.691\right)}^2\left(0.2714\right)=-169.64i{n}^4\end{array}$

  $\begin{array}{l}\text{Using equation 9}\text{.22:}\\ R=\sqrt{{\left(\frac{{\overline{I}}_x-{\overline{I}}_y}{2}\right)}^2+{\overline{I}}^2{}_{xy}}\\ R=\sqrt{{\left(\frac{1373.0-489.4}{2}\right)}^2+{\left(-169.64\right)}^2}\\ R=473.25i{n}^4\\ \text{Finally using equation 9}\text{.23:}\\ I_1=\frac{{\overline{I}}_x+{\overline{I}}_y}{2}+R=\frac{1373.0+489.4}{2}+473.25\\ I_1=1405i{n}^4\\ and,\\ I_2=\frac{{\overline{I}}_x-{\overline{I}}_y}{2}-R=\frac{1373.0-489.4}{2}-473.25\\ I_2=458i{n}^4\end{array}$

Conclusion:

The principal centroidal moments of inertia for the plane area shown in figure are:

$I_1=1405i{n}^4$ and $I_2=458i{n}^4$.