Chapter 9.2, Problem 9.36P

To determine

Find the moment of inertia of the shaded area with respect to x axis $\left(I_x\right)$ and y axis $\left(I_y\right)$.

Answer to Problem 9.36P

The moment of inertia of the shaded area with respect to x axis $\left(I_x\right)$ and y axis $\left(I_y\right)$ are $\underset{\_}{69.9a^4}$ and $\underset{\_}{72a^4}$ respectively.

Explanation of Solution

Show the centroidal location of the given section as Figure 1.

Consider x axis.

Consider the section 1.

Calculate the moment of inertia of the section 1 ${\left({\overline{I}}_{x_1}\right)}_1$ using the formula:

$\begin{array}{c}{\left({\overline{I}}_{x_1}\right)}_1=\frac{4a\times {\left(4a\right)}^3}{12}\\ =\frac{256a^4}{12}\end{array}$

The expression for the area of the section 1 (A) as follows;

$\begin{array}{c}A=4a\times 4a\\ =16a^2\end{array}$

The expression for the centroid of the section 1 (d) as follows;

$\begin{array}{c}d=\frac{4a}{2}\\ =2a\end{array}$

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

${\left(I_x\right)}_1={\left({\overline{I}}_{x_1}\right)}_1+A{d}^2$

Substitute $\frac{256a^4}{12}$ for ${\left({\overline{I}}_{x_1}\right)}_1$ and $16a^2$ for A and 2a for d.

$\begin{array}{c}{\left(I_x\right)}_1=\frac{256a^4}{12}+16a^2{\left(2a\right)}^2\\ =\frac{256a^4}{12}+64a^4\\ =a^4\left(\frac{256}{12}+64\right)\\ =\frac{256a^4}{3}\end{array}$

The expression for the moment of inertia of the section AA ${\left(I_{AA}\right)}_2$ and section BB ${\left(I_{BB}\right)}_3$ as follows;

The section AA and BB compressed of semi-circle.

${\left(I_{AA}\right)}_2={\left(I_{BB}\right)}_3=\frac{\pi }{8}{a}^4$

The expression for the area of the semi-circle (A) as follows;

$A=\frac{\pi }{2}{a}^2$

The expression for the centroid of the semi-circle (d) as follows;

$d=\frac{4a}{3\pi }$

Calculate the moment of inertia of the section 2 ${\left({\overline{I}}_{x_2}\right)}_2$ and section 3 ${\left({\overline{I}}_{x_3}\right)}_3$ using the formula:

Since both sections (2) and (3) are semi-circle which has same moment of inertia.

$\begin{array}{c}{\left(I_{AA}\right)}_2={\left({\overline{I}}_{x_2}\right)}_2+A{d}^2\\ {\left({\overline{I}}_{x_3}\right)}_3={\left({\overline{I}}_{x_2}\right)}_2={\left(I_{AA}\right)}_2-A{d}^2\end{array}$

Substitute $\frac{\pi }{8}{a}^4$ for ${\left(I_{AA}\right)}_2$, $\frac{\pi }{2}{a}^2$ for A and $\frac{4a}{3\pi }$ for d.

$\begin{array}{c}{\left({\overline{I}}_{x_3}\right)}_3={\left({\overline{I}}_{x_2}\right)}_2=\frac{\pi }{8}{a}^4-\frac{\pi }{2}{a}^2{\left(\frac{4a}{3\pi }\right)}^2\\ =\frac{\pi }{8}{a}^4-\frac{\pi a^2}{2}\left(\frac{16a^2}{9{\pi }^2}\right)\\ =\frac{\pi }{8}{a}^4-\frac{8a^4}{9\pi }\\ =a^4\left(\frac{\pi }{8}-\frac{8}{9\pi }\right)\end{array}$

Consider the section 2.

The expression for the centroid of the section 2 (d) from neutral axis as follows;

$d=\frac{5}{2}a+\frac{4a}{3\pi }$

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

${\left(I_x\right)}_2={\left({\overline{I}}_{x_2}\right)}_2+A{d}^2$

Substitute $a^4\left(\frac{\pi }{8}-\frac{8}{9\pi }\right)$ for ${\left({\overline{I}}_{x_2}\right)}_2$, $\frac{\pi }{2}{a}^2$ for A and $\left(\frac{5}{2}a+\frac{4a}{3\pi }\right)$ for d.

$\begin{array}{c}{\left(I_x\right)}_2=a^4\left(\frac{\pi }{8}-\frac{8}{9\pi }\right)+\left[\frac{\pi }{2}{a}^2{\left(\frac{5}{2}a+\frac{4a}{3\pi }\right)}^2\right]\\ =a^4\left(\frac{\pi }{8}-\frac{8}{9\pi }\right)+\frac{\pi }{2}{a}^2\left({\left(\frac{5}{2}a\right)}^2+{\left(\frac{4a}{3\pi }\right)}^2+2\left(\frac{5a}{2}\right)\frac{4a}{3\pi }\right)\\ =a^4\left(\frac{10}{3}+\frac{13\pi }{4}\right)\end{array}$

Consider the section 3.

The expression for the centroid of the section 3 (d) from neutral axis as follows;

$d=\frac{3}{2}a-\frac{4a}{3\pi }$

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

${\left(I_x\right)}_3={\left({\overline{I}}_{x_3}\right)}_3+A{d}^2$

Substitute $a^4\left(\frac{\pi }{8}-\frac{8}{9\pi }\right)$ for ${\left({\overline{I}}_{x_3}\right)}_3$, $\frac{\pi }{2}{a}^2$ for A and $\left(\frac{3}{2}a-\frac{4a}{3\pi }\right)$ for d.

$\begin{array}{c}{\left(I_x\right)}_3=a^4\left(\frac{\pi }{8}-\frac{8}{9\pi }\right)+\left[\frac{\pi }{2}{a}^2{\left(\frac{3}{2}a-\frac{4a}{3\pi }\right)}^2\right]\\ =a^4\left(\frac{\pi }{8}-\frac{8}{9\pi }\right)+\frac{\pi }{2}{a}^2\left({\left(\frac{3}{2}a\right)}^2+{\left(\frac{4a}{3\pi }\right)}^2-2\left(\frac{3}{2}a\right)\frac{4a}{3\pi }\right)\\ =a^4\left(-2+\frac{5\pi }{4}\right)\end{array}$

Calculate the total moment of inertia of the entire section $\left(I_x\right)$ about x axis using the relation:

$I_x={\left(I_x\right)}_1-{\left(I_x\right)}_2-{\left(I_x\right)}_3$

Substitute $\frac{256}{3}{a}^4$ for ${\left(I_x\right)}_1$, $a^4\left(\frac{10}{3}+\frac{13\pi }{4}\right)$ for ${\left(I_x\right)}_2$ and $a^4\left(-2+\frac{5\pi }{4}\right)$ for ${\left(I_x\right)}_3$.

$\begin{array}{c}{I}_x=\frac{256}{3}{a}^4-a^4\left(\frac{10}{3}+\frac{13\pi }{4}\right)-a^4\left(-2+\frac{5\pi }{4}\right)\\ =\frac{256}{3}{a}^4-\frac{10a^4}{3}-\frac{13a^4\pi }{4}+2a^4-\frac{5a^4\pi }{4}\\ =\frac{168a^4-9\pi a^4}{2}\\ =69.9a^4\end{array}$

Consider y axis.

Consider the section 1.

Calculate the moment of inertia of the section 1 ${\left({\overline{I}}_{y_1}\right)}_1$ about y axis using the formula:

$\begin{array}{c}{\left({\overline{I}}_{y_1}\right)}_1=\frac{4a\times {\left(4a\right)}^3}{12}\\ =\frac{256a^4}{12}\end{array}$

The expression for the area of the section 1 (A) as follows;

$\begin{array}{c}A=4a\times 4a\\ =16a^2\end{array}$

The expression for the centroid of the section 1 (d) as follows;

$\begin{array}{c}d=\frac{4a}{2}\\ =2a\end{array}$

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

${\left(I_y\right)}_1={\left({\overline{I}}_{y_1}\right)}_1+A{d}^2$

Substitute $\frac{256a^4}{12}$ for ${\left({\overline{I}}_{y_1}\right)}_1$ and $16a^2$ for A and 2a for d.

$\begin{array}{c}{\left(I_y\right)}_1=\frac{256a^4}{12}+16a^2{\left(2a\right)}^2\\ =\frac{256a^4}{12}+64a^4\\ =a^4\left(\frac{256}{12}+64\right)\\ =\frac{256a^4}{3}\end{array}$

Consider the section 2.

Calculate the moment of inertia of the section 2 ${\left({\overline{I}}_{y_2}\right)}_2$ about y axis using the relation:

${\left({\overline{I}}_{y_2}\right)}_2={\left(I_{AA}\right)}_2$

Substitute $\frac{\pi }{8}{a}^4$ for ${\left(I_{AA}\right)}_2$.

${\left({\overline{I}}_{y_2}\right)}_2=\frac{\pi }{8}{a}^4$

The expression for the area of the semi-circle (A) as follows;

$A=\frac{\pi }{2}{a}^2$

The expression for the centroid of the semi-circle (d) as follows;

$\begin{array}{c}d=\frac{4a}{2}\\ =2a\end{array}$

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

${\left(I_y\right)}_2={\left({\overline{I}}_{y_2}\right)}_2+A{d}^2$

Substitute $\frac{\pi }{8}{a}^4$ for ${\left({\overline{I}}_{y_2}\right)}_2$ and $\frac{\pi }{2}{a}^2$ for A and 2a for d.

$\begin{array}{c}{\left(I_y\right)}_2=\frac{\pi }{8}{a}^4+\frac{\pi }{2}{a}^2{\left(2a\right)}^2\\ =\frac{\pi }{8}{a}^4+2\pi a^4\\ =\frac{17\pi a^4}{8}\end{array}$

Since the section 2 and section 3 are same. Hence, ${\left(I_y\right)}_3={\left(I_y\right)}_2=\frac{17\pi a^4}{8}$.

Calculate the total moment of inertia of the entire section $\left(I_y\right)$ about y axis using the relation:

$I_y={\left(I_y\right)}_1-{\left(I_y\right)}_2-{\left(I_y\right)}_3$

Substitute $\frac{256a^4}{3}$ for ${\left(I_y\right)}_1$ and $\frac{17\pi a^4}{8}$ for ${\left(I_y\right)}_2$ and $\frac{17\pi a^4}{8}$ for ${\left(I_y\right)}_3$.

$\begin{array}{c}{I}_y=\frac{256a^4}{3}-\frac{17\pi a^4}{8}-\frac{17\pi a^4}{8}\\ =\frac{256a^4}{3}-2\left(\frac{17\pi a^4}{8}\right)\\ =a^4\left(\frac{256}{3}-\frac{17\pi }{4}\right)\\ =72a^4\end{array}$

Therefore, the moment of inertia of the shaded area with respect to x axis $\left(I_x\right)$ and y axis $\left(I_y\right)$ are $\underset{\_}{69.9a^4}$ and $\underset{\_}{72a^4}$ respectively.