Chapter 9, Problem 9.47P

To determine

  ${\overline{I}}_{xy}$ for the region shown.

Answer to Problem 9.47P

  ${\overline{I}}_{xy}=1175i{n}^4$

Explanation of Solution

Given information:

The region:

Calculations:

Find the centroid first. From Table 8.1:

  $\begin{array}{l}{A}_2=-\frac{2}{3}{\left(12\right)}^2=-96i{n}^2\\ {\overline{x}}_2=\frac{3}{8}\left(12\right)=4.5in.\\ {\overline{y}}_2=6+\frac{3}{5}\left(12\right)=13.2in.\end{array}$

$Part$	$A\left(i{n}^2\right)$	$\overline{x}\left(in.\right)$	$\overline{y}\left(in.\right)$	$A\overline{x}\left(i{n}^3\right)$	$A\overline{y}\left(i{n}^3\right)$
1	216	6.0	9.0	1296	1944.0
2	-96	4.5	13.2	-432	-1267.2
Sum	120			864	676.8

  $\begin{array}{l}\overline{X}=\frac{{\displaystyle \sum A\overline{x}}}{{\displaystyle \sum A}}=\frac{864}{120}=7.200in.\\ \overline{Y}=\frac{{\displaystyle \sum A\overline{y}}}{{\displaystyle \sum A}}=\frac{676.8}{120}=5.640in.\end{array}$

The product of inertias with respect to x- and y-axis:

  $\begin{array}{l}Part1:{\left(I_{xy}\right)}_1=\frac{b^2{h}^2}{4}=\frac{{\left(12\right)}^2{\left(18\right)}^2}{4}=11664i{n}^4\\ Part2:{\left(I_{xy}\right)}_2={\left({\overline{I}}_{xy}\right)}_2+A_2{\overline{x}}_2{\overline{y}}_2=\frac{b^2{h}^2}{60}+\frac{2bh}{3}\left(\frac{3h}{8}\right)\left(d+\frac{3h}{5}\right)\\ =\frac{{\left(12\right)}^2{\left(12\right)}^2}{60}+\frac{2\left(12\right)\left(12\right)}{3}\left[\frac{3\left(12\right)}{8}\right]\left[6+\frac{3\left(12\right)}{5}\right]=6048i{n}^4\end{array}$

Hence, for assembly:

  $\begin{array}{l}{I}_{xy}={\left(I_{xy}\right)}_1-{\left(I_{xy}\right)}_2=11664-6048=5620i{n}^4\\ \text{Hence, the product of inertia with respect to centroidal axis is}\\ {\overline{I}}_{xy}=I_{xy}-\overline{x}\overline{y}A=6048-\left(7.200\right)\left(5.640\right)\left(120\right)\\ \Rightarrow {\overline{I}}_{xy}=1175i{n}^4\end{array}$

Conclusion:

For the region shown, ${\overline{I}}_{xy}=1175i{n}^4$.