Chapter 9, Problem 9.17P

A $\text{W}8\times 67$ section is joined to a $\text{C}10\times 20$ section to form a structural member that has the cross section shown. Calculate ${\bar{I}}_x$ and ${\bar{I}}_y$ for this cross section. (See Probs. 9.13 and 9.16 for the properties of the structural sections.)

To determine

The values of ${\overline{I}}_{x}and{\overline{I}}_y$ for the given cross section shown.

Answer to Problem 9.17P

  ${\overline{I}}_x=358i{n}^4$

  ${\overline{I}}_y=167.5i{n}^4$

Explanation of Solution

Given information:

The cross-sectional properties of the structural shape $W8\times 67$

  $A=19.7in{.}^2,{\stackrel{-}{I}}_x=272in{.}^4,{\stackrel{-}{I}}_y=88.6in{.}^4$

Cross sectional properties of structural steel section $C10\times 20$ :

  $A=5.88in{.}^2,{\stackrel{-}{I}}_x=78.9in{.}^4,{\stackrel{-}{I}}_y=2.81in{.}^4$

Calculations:

For the composite area: (representing the data in tabular form)

Part	${\overline{I}}_x\left(i{n}^4\right)$	$A\left(i{n}^2\right)$	$\overline{y}\left(in\right)$	$A\overline{y}\left(i{n}^3\right)$	${\overline{I}}_x+A{\overline{y}}^2$	$I_y\left(i{n}^4\right)$
1	2.81	5.88	8.773	51.59	455.4	78.9
2	272	19.70	4.50	88.65	670.9	88.6
sum		25.58		140.24	1126.3	167.5

For the assembly:

  $\begin{array}{l}Thecentroid:\\ \overline{y}=\frac{{\displaystyle \sum A\overline{y}}}{{\displaystyle \sum A}}=\frac{140.24}{25.58}=5.482in\\ Momentofinertiaaboutthex-axis:\\ I_x={\displaystyle \sum \left({\overline{I}}_x-A{\overline{y}}^2\right)=1126.3i{n}^4}\\ Momentofinertiaaboutthey-axis:\\ I_y={\displaystyle \sum I_y}=167.5i{n}^4\\ Hence,\\ Momentofinertiaaboutthecentroidalx-andy-axis:\\ {\overline{I}}_x=I_x-A{\overline{y}}^2=1126.3-\left(25.58\right)\left({5.482}^2\right)\\ \Rightarrow {\overline{I}}_x=358i{n}^4\\ and,\\ {\overline{I}}_y=I_y\left(\text{since section is symmetric about }y-\text{axis}\right)\\ \Rightarrow {\overline{I}}_y=167.5i{n}^4\end{array}$

Conclusion:

For the composite cross-section, ${\overline{I}}_x=358i{n}^4$ and ${\overline{I}}_y=167.5i{n}^4$.