Chapter 9.2, Problem 9.44P

To determine

Find the moment of inertia $\left({\overline{I}}_x\right)$ and $\left({\overline{I}}_y\right)$ of the area with respect to centroidal axes.

Answer to Problem 9.44P

The moment of inertia $\left({\overline{I}}_x\right)$ and $\left({\overline{I}}_y\right)$ of the area are $\underset{\_}{18.13\text{in}{\text{.}}^{\text{4}}}$ and $\underset{\_}{4.51\text{in}{\text{.}}^{\text{4}}}$ respectively.

Explanation of Solution

Calculation:

Show the area with parts of the section as Figure 1.

Calculate the centroid of the area as in Table 1.

Section	Area, A $\left(\text{in}{\text{.}}^{\text{2}}\right)$	$\left(\overline{x}\right)$ (in.)	$\left(\overline{y}\right)$ (in.)	$\left(A\overline{x}\right)$ $\left(\text{in}{\text{.}}^3\right)$	$\left(A\overline{y}\right)$ $\left(\text{in}{\text{.}}^3\right)$
1	$3.6\times 0.5=1.8$	$\frac{3.6}{2}=1.8$	$\frac{0.5}{2}=0.25$	3.24	0.45
2	$3.8\times 0.5=1.9$	$\frac{0.5}{2}=0.25$	$0.5+\frac{3.8}{2}=2.4$	0.475	4.56
3	$1.3\times 1.0=1.3$	$\frac{1.3}{2}=0.65$	$0.5+3.8+\frac{1}{2}=4.8$	0.845	6.24
Total $(\sum )$	5.0			4.560	11.25

Calculate the centroid of the area with respect to x axis $\left(\overline{X}\right)$ using the formula:

$\overline{X}=\frac{{\displaystyle \sum A\overline{x}}}{{\displaystyle \sum A}}$

Substitute $4.56\text{in}{\text{.}}^{\text{3}}$ for ${\displaystyle \sum A\overline{x}}$ and $5\text{in}{\text{.}}^{\text{2}}$ for ${\displaystyle \sum A}$.

$\begin{array}{c}\overline{X}=\frac{4.56}{5}\\ =0.912\text{in}\text{.}\end{array}$

Calculate the centroid of the area with respect to y axis $\left(\overline{Y}\right)$ using the formula:

$\overline{Y}=\frac{{\displaystyle \sum A\overline{y}}}{{\displaystyle \sum A}}$

Substitute $11.25\text{in}{\text{.}}^{\text{3}}$ for ${\displaystyle \sum A\overline{y}}$ and $5\text{in}{\text{.}}^{\text{2}}$ for ${\displaystyle \sum A}$.

$\begin{array}{c}\overline{Y}=\frac{11.25}{5}\\ =2.25\text{in}\text{.}\end{array}$

Consider the x axis.

Consider the section 1.

Calculate the moment of inertia of section 1 $\left({\overline{I}}_1\right)$ using the formula:

$\begin{array}{c}{\overline{I}}_1=\frac{3.6\times {0.5}^3}{12}\\ =0.0375\text{in}{\text{.}}^{\text{4}}\end{array}$

Calculate the centroid of section 1 $\left(d_1\right)$ from the centroid of entire section using the relation:

$d_1=\overline{Y}-{\overline{y}}_1$

Substitute 2.25 in. for $\overline{Y}$ and 0.25 in. for ${\overline{y}}_1$.

$\begin{array}{c}{d}_1=2.25-0.25\\ =2\text{in}\text{.}\end{array}$

Calculate the moment of inertia of section 1 ${\left(I_x\right)}_1$ with respect to x axis using the formula:

${\left(I_x\right)}_1={\overline{I}}_1+A_1{d}_1^2$

Substitute $0.0375\text{in}{\text{.}}^{\text{4}}$ for ${\overline{I}}_1$, $1.8\text{in}{\text{.}}^{\text{2}}$ for $A_1$ and 2 in. for $d_1$.

$\begin{array}{c}{\left(I_x\right)}_1=0.0375\text{in}{\text{.}}^{\text{4}}+\left[1.8\times {\left(2\right)}^2\right]\\ =7.2375\text{in}{\text{.}}^{\text{4}}\end{array}$

Consider the section 2.

Calculate the moment of inertia of section 2 $\left({\overline{I}}_2\right)$ using the formula:

$\begin{array}{c}{\overline{I}}_2=\frac{0.5\times {3.8}^3}{12}\\ =2.2863\text{in}{\text{.}}^{\text{4}}\end{array}$

Calculate the centroid of section 2 $\left(d_2\right)$ from the centroid of entire section using the relation:

$d_2={\overline{y}}_2-\overline{Y}$

Substitute 2.25 in. for $\overline{Y}$ and 2.4 in. for ${\overline{y}}_2$.

$\begin{array}{c}{d}_2=2.4-2.25\\ =0.15\text{in}\text{.}\end{array}$

Calculate the moment of inertia of section 2 ${\left(I_x\right)}_2$ with respect to x axis using the formula:

${\left(I_x\right)}_2={\overline{I}}_2+A_2{d}_2^2$

Substitute $2.2863\text{in}{\text{.}}^{\text{4}}$ for ${\overline{I}}_2$, $1.9\text{in}{\text{.}}^{\text{2}}$ for $A_2$ and 0.15 in. for $d_2$.

$\begin{array}{c}{\left(I_x\right)}_2=2.2863\text{in}{\text{.}}^{\text{4}}+\left[1.9\times {\left(0.15\right)}^2\right]\\ =2.3291\text{in}{\text{.}}^{\text{4}}\end{array}$

Consider the section 3.

Calculate the moment of inertia of section 3 $\left({\overline{I}}_3\right)$ using the formula:

$\begin{array}{c}{\overline{I}}_3=\frac{1.3\times {1.0}^3}{12}\\ =0.1083\text{in}{\text{.}}^{\text{4}}\end{array}$

Calculate the centroid of section 3 $\left(d_3\right)$ from the centroid of entire section using the relation:

$d_3={\overline{y}}_3-\overline{Y}$

Substitute 2.25 in. for $\overline{Y}$ and 4.8 in. for ${\overline{y}}_3$.

$\begin{array}{c}{d}_3=4.8-2.25\\ =2.55\text{in}\text{.}\end{array}$

Calculate the moment of inertia of section 3 ${\left(I_x\right)}_3$ with respect to x axis using the formula:

${\left(I_x\right)}_3={\overline{I}}_3+A_3{d}_3^2$

Substitute $0.1083\text{in}{\text{.}}^{\text{4}}$ for ${\overline{I}}_3$, $1.3\text{in}{\text{.}}^{\text{2}}$ for $A_3$ and 2.55 in. for $d_3$.

$\begin{array}{c}{\left(I_x\right)}_3=0.1083\text{in}{\text{.}}^{\text{4}}+\left[1.3\times {\left(2.55\right)}^2\right]\\ =8.5616\text{in}{\text{.}}^{\text{4}}\end{array}$

Calculate the moment of inertia of entire section $\left({\overline{I}}_x\right)$ with respect to x axis using the relation:

${\overline{I}}_x={\left(I_x\right)}_1+{\left(I_x\right)}_2+{\left(I_x\right)}_3$

Substitute $7.2375\text{in}{\text{.}}^{\text{4}}$ for ${\left(I_x\right)}_1$, $2.3291\text{in}{\text{.}}^{\text{4}}$ for ${\left(I_x\right)}_2$ and $8.5616\text{in}{\text{.}}^{\text{4}}$ for ${\left(I_x\right)}_3$.

$\begin{array}{c}{\overline{I}}_x=7.2375\text{in}{\text{.}}^{\text{4}}+2.3291\text{in}{\text{.}}^{\text{4}}+8.5616\text{in}{\text{.}}^{\text{4}}\\ =18.13\text{in}{\text{.}}^{\text{4}}\end{array}$

Consider the y axis.

Consider the section 1.

Calculate the moment of inertia of section 1 $\left({\overline{I}}_1\right)$ using the formula:

$\begin{array}{c}{\overline{I}}_1=\frac{0.5\times {3.6}^3}{12}\\ =1.944\text{in}{\text{.}}^{\text{4}}\end{array}$

Calculate the centroid of section 1 $\left(d_1\right)$ from the centroid of entire section using the relation:

$d_1={\overline{x}}_1-\overline{X}$

Substitute 0.912 in. for $\overline{X}$ and 1.8 in. for ${\overline{x}}_1$.

$\begin{array}{c}{d}_1=1.8-0.912\\ =0.888\text{in}\text{.}\end{array}$

Calculate the moment of inertia of section 1 ${\left(I_y\right)}_1$ with respect to y axis using the formula:

${\left(I_y\right)}_1={\overline{I}}_1+A_1{d}_1^2$

Substitute $1.944\text{in}{\text{.}}^{\text{4}}$ for ${\overline{I}}_1$, $1.8\text{in}{\text{.}}^{\text{2}}$ for $A_1$ and 0.888 in. for $d_1$.

$\begin{array}{c}{\left(I_y\right)}_1=1.944\text{in}{\text{.}}^{\text{4}}+\left[1.8\times {\left(0.888\right)}^2\right]\\ =3.3634\text{in}{\text{.}}^{\text{4}}\end{array}$

Consider the section 2.

Calculate the moment of inertia of section 2 $\left({\overline{I}}_2\right)$ using the formula:

$\begin{array}{c}{\overline{I}}_2=\frac{3.8\times {0.5}^3}{12}\\ =0.0396\text{in}{\text{.}}^{\text{4}}\end{array}$

Calculate the centroid of section 2 $\left(d_2\right)$ from the centroid of entire section using the relation:

$d_2=\overline{X}-{\overline{x}}_2$

Substitute 0.912 in. for $\overline{X}$ and 0.25 in. for ${\overline{x}}_2$.

$\begin{array}{c}{d}_2=0.912-0.25\\ =0.662\text{in}\text{.}\end{array}$

Calculate the moment of inertia of section 2 ${\left(I_y\right)}_2$ with respect to y axis using the formula:

${\left(I_y\right)}_2={\overline{I}}_2+A_2{d}_2^2$

Substitute $0.0396\text{in}{\text{.}}^{\text{4}}$ for ${\overline{I}}_2$, $1.9\text{in}{\text{.}}^{\text{2}}$ for $A_2$ and 0.662 in. for $d_2$.

$\begin{array}{c}{\left(I_y\right)}_2=0.0396\text{in}{\text{.}}^{\text{4}}+\left[1.9\times {\left(0.662\right)}^2\right]\\ =0.8723\text{in}{\text{.}}^{\text{4}}\end{array}$

Consider the section 3.

Calculate the moment of inertia of section 3 $\left({\overline{I}}_3\right)$ using the formula:

$\begin{array}{c}{\overline{I}}_3=\frac{1.0\times {1.3}^3}{12}\\ =0.1831\text{in}{\text{.}}^{\text{4}}\end{array}$

Calculate the centroid of section 3 $\left(d_3\right)$ from the centroid of entire section using the relation:

$d_3=\overline{X}-{\overline{x}}_3$

Substitute 0.912 in. for $\overline{X}$ and 0.65 in. for ${\overline{x}}_3$.

$\begin{array}{c}{d}_3=0.912-0.65\\ =0.262\text{in}\text{.}\end{array}$

Calculate the moment of inertia of section 3 ${\left(I_y\right)}_3$ with respect to y axis using the formula:

${\left(I_y\right)}_3={\overline{I}}_3+A_3{d}_3^2$

Substitute $0.1831\text{in}{\text{.}}^{\text{4}}$ for ${\overline{I}}_3$, $1.3\text{in}{\text{.}}^{\text{2}}$ for $A_3$ and 0.262 in. for $d_3$.

$\begin{array}{c}{\left(I_y\right)}_3=0.1831\text{in}{\text{.}}^{\text{4}}+\left[1.3\times {\left(0.262\right)}^2\right]\\ =0.2723\text{in}{\text{.}}^{\text{4}}\end{array}$

Calculate the moment of inertia of entire section $\left({\overline{I}}_y\right)$ with respect to y axis using the relation:

${\overline{I}}_y={\left(I_y\right)}_1+{\left(I_y\right)}_2+{\left(I_y\right)}_3$

Substitute $3.3634\text{in}{\text{.}}^{\text{4}}$ for ${\left(I_y\right)}_1$, $0.8723\text{in}{\text{.}}^{\text{4}}$ for ${\left(I_y\right)}_2$ and $0.2723\text{in}{\text{.}}^{\text{4}}$ for ${\left(I_y\right)}_3$.

$\begin{array}{c}{\overline{I}}_y=3.3634\text{in}{\text{.}}^{\text{4}}+0.8723\text{in}{\text{.}}^{\text{4}}+0.2723\text{in}{\text{.}}^{\text{4}}\\ =4.51\text{in}{\text{.}}^{\text{4}}\end{array}$

Therefore, the moment of inertia $\left({\overline{I}}_x\right)$ and $\left({\overline{I}}_y\right)$ of the area are $\underset{\_}{18.13\text{in}{\text{.}}^{\text{4}}}$ and $\underset{\_}{4.51\text{in}{\text{.}}^{\text{4}}}$ respectively.