Chapter 9.2, Problem 9.43P

9.41 through 9.44 Determine the moments of inertia $\overline{I_x}$ and $\overline{I_y}$ of the area shown with respect to centroidal axes respectively parallel and perpendicular to side AB.

Fig. P9.43

To determine

Find the moment of inertia about x and y axis of the area with respect to centroid axes.

Answer to Problem 9.43P

The moment of inertia about x axis is $\underset{\_}{191.3{\text{in}}^{\text{4}}}$

The moment of inertia about y axis is $\underset{\_}{75.2{\text{in}}^{\text{4}}}$

Explanation of Solution

Calculation:

Sketch the cross section as shown in Figure 1.

Refer to Figure 1.

Find the area $\left(A_1\right)$ of section ABCD as shown in below:

$A_1=bh$ (1)

Substitute $5\text{in}\text{.}$ for b and $8\text{in}\text{.}$ for h in Equation (1).

$\begin{array}{c}{A}_1=5\times 8\\ =40{\text{in}}^{\text{2}}\end{array}$

Find the area $\left(A_2\right)$ of section abcd as shown in below:

$A_2=bh$ (2)

Substitute $-2\text{in}\text{.}$ for b and $5\text{in}\text{.}$ for h in Equation (2).

$\begin{array}{c}{A}_2=\left(-2\times 5\right)\\ =-10{\text{in}}^{\text{2}}\end{array}$

Find the total area (A) using the relation as follows:

$A=A_1+A_2$ (3)

Here, $A_1$ is area of section ABCD and $A_2$ is area of section abcd.

Substitute $40{\text{in}}^{\text{2}}$ for $A_1$, and $-10{\text{in}}^{\text{2}}$ for $A_2$ in Equation (3).

$\begin{array}{c}A=40-10\\ =30{\text{in}}^{\text{2}}\end{array}$

Refer to Figure 1.

Find the centroid $\left(x_1\right)$ section ABCD as shown below:

$\begin{array}{c}{x}_1=\frac{5}{2}\\ =2.5\text{in}\text{.}\end{array}$

Find the centroid $\left(x_2\right)$ section abcd as shown below:

$\begin{array}{c}{x}_2=\frac{2}{2}+0.9\\ =1.9\text{in}\text{.}\end{array}$

Find the centroid $\left(y_1\right)$ section ABCD as shown below:

$\begin{array}{c}{y}_1=\frac{8}{2}\\ =4\text{in}\text{.}\end{array}$

Find the centroid $\left(y_2\right)$ section abcd as shown below:

$\begin{array}{c}{y}_2=1.8+2.5\\ =4.3\text{in}\text{.}\end{array}$

Find the centroid $\left(\overline{x}\right)$ using the relation as follows:

$\overline{x}=\frac{A_1{x}_1+A_2{x}_2}{A_1+A_2}$ (4)

Substitute $40{\text{in}}^{\text{2}}$ for $A_1$, $-10{\text{in}}^{\text{2}}$ for $A_2$, $2.5\text{in}\text{.}$ for $x_1$, and $1.9\text{in}\text{.}$ for $x_2$ in Equation (4).

$\begin{array}{c}\overline{x}=\frac{40\times 2.5+\left(-10\times 1.9\right)}{40-10}\\ =\frac{100-19}{30}\\ =2.70\text{in}\text{.}\end{array}$

Find the centroid $\left(\overline{y}\right)$ using the relation as follows:

$\overline{y}=\frac{A_1{y}_1+A_2{y}_2}{A_1+A_2}$ (5)

Substitute $40{\text{in}}^{\text{2}}$ for $A_1$, $-10{\text{in}}^{\text{2}}$ for $A_2$, $4\text{in}\text{.}$ for $y_1$, and $4.3\text{in}\text{.}$ for $y_2$ in Equation (5).

$\begin{array}{c}\overline{y}=\frac{40\times 4+\left(-10\times 4.3\right)}{40-10}\\ =\frac{160-43}{30}\\ =3.9\text{in}\text{.}\end{array}$

Find the moment of inertia ${\left(I_x\right)}_1$ for section ABCD as shown below:

$\begin{array}{c}{\left(I_x\right)}_1=\frac{1}{12}b{h}^3+A{\left(\overline{h}\right)}^2\\ =\frac{1}{12}b{h}^3+A{\left(y-\overline{y}\right)}^2\end{array}$ (6)

Substitute $5\text{in}\text{.}$ for b, $8\text{in}\text{.}$ for h, $40{\text{in}}^{\text{2}}$ for A, $4\text{in}\text{.}$ for y, and $3.9\text{in}\text{.}$ for $\overline{y}$ in Equation (6).

$\begin{array}{c}{\left(I_x\right)}_1=\frac{1}{12}\left(5\right)\times {\left(8\right)}^3+\left(40\right){\left(4-3.9\right)}^2\\ =213.3333+0.4\\ =213.733{\text{in}}^{\text{4}}\end{array}$

Find the moment of inertia ${\left(I_x\right)}_2$ for section abcd as shown below:

$\begin{array}{c}{\left(I_x\right)}_2=\frac{1}{12}b{h}^3+A{\left(\overline{h}\right)}^2\\ =\frac{1}{12}b{h}^3+A{\left(y-\overline{y}\right)}^2\end{array}$ (7)

Substitute $2\text{in}\text{.}$ for b, $5\text{in}\text{.}$ for h in, $10{\text{in}}^{\text{2}}$ for A, $4.3\text{in}\text{.}$ for y, and $3.9\text{in}\text{.}$ for $\overline{y}$ in Equation (7).

$\begin{array}{c}{\left(I_x\right)}_2=\frac{1}{12}\left(2\right)\times {\left(5\right)}^3+\left(10\right){\left(4.3-3.9\right)}^2\\ =20.83333+1.6\\ =22.433{\text{in}}^{\text{4}}\end{array}$

Find the total moment of inertia $\left({\overline{I}}_x\right)$ using the relation as shown below:

$\left({\overline{I}}_x\right)={\left(I_x\right)}_1-{\left(I_x\right)}_2$ (8)

Substitute $213.733{\text{in}}^{\text{4}}$ for ${\left(I_x\right)}_1$ and $22.433{\text{in}}^{\text{4}}$ for ${\left(I_x\right)}_2$ in Equation (8).

$\begin{array}{c}\left({\overline{I}}_x\right)=213.7333-22.4333\\ =191.3{\text{in}}^{\text{4}}\end{array}$

Thus, the moment of inertia $\left({\overline{I}}_x\right)$ about x axis is $\underset{\_}{191.3{\text{in}}^{\text{4}}}$

Find the moment of inertia ${\left(I_y\right)}_1$ for section ABCD as shown below:

$\begin{array}{c}{\left(I_y\right)}_1=\frac{1}{12}{b}^{3}h+A{\left(\overline{h}\right)}^2\\ =\frac{1}{12}{b}^{3}h+A{\left(\overline{x}-x\right)}^2\end{array}$ (9)

Substitute $5\text{in}\text{.}$ for b, $8\text{in}\text{.}$ for h in, $40{\text{in}}^{\text{2}}$ for A, $2.5\text{in}\text{.}$ for x, and $2.7\text{in}\text{.}$ for $\overline{x}$ in Equation (9).

$\begin{array}{c}{\left(I_y\right)}_1=\frac{1}{12}\left(5^3\right)\times \left(8\right)+\left(40\right){\left(2.7-2.5\right)}^2\\ =83.3333+1.6\\ =84.93{\text{in}}^{\text{4}}\end{array}$

Find the moment of inertia ${\left(I_y\right)}_2$ for section abcd as shown below:

$\begin{array}{c}{\left(I_y\right)}_2=\frac{1}{12}{b}^{3}h+A{\left(\overline{h}\right)}^2\\ =\frac{1}{12}{b}^{3}h+A{\left(\overline{x}-x\right)}^2\end{array}$ (10)

Substitute $2\text{in}\text{.}$ for b, $5\text{in}\text{.}$ for h in, $10{\text{in}}^{\text{2}}$ for A, $1.9\text{in}\text{.}$ for x, and $2.7\text{in}\text{.}$ for $\overline{x}$ in Equation (10).

$\begin{array}{c}{\left(I_y\right)}_2=\frac{1}{12}{\left(2\right)}^3\times \left(5\right)+\left(10\right){\left(2.7-1.9\right)}^2\\ =3.3333+6.4\\ =9.733{\text{in}}^{\text{4}}\end{array}$

Find the total moment of inertia $\left({\overline{I}}_y\right)$ using the relation as shown below:

$\left({\overline{I}}_y\right)={\left(I_y\right)}_1-{\left(I_y\right)}_2$ (11)

Substitute $84.93{\text{in}}^{\text{4}}$ for ${\left(I_y\right)}_1$ and $9.733{\text{in}}^{\text{4}}$ for ${\left(I_y\right)}_2$ in Equation (11).

$\begin{array}{c}\left({\overline{I}}_y\right)=84.93-9.733\\ =75.1967\\ =75.2{\text{in}}^{\text{4}}\end{array}$

Thus, the moment of inertia $\left({\overline{I}}_y\right)$ about y axis is $\underset{\_}{191.3{\text{in}}^{\text{4}}}$.