Chapter 9.4, Problem 9.99P

To determine

Find the orientation of the principal axes at the origin and the maximum minimum value of moment of inertia.

Answer to Problem 9.99P

The orientation of the principal axes at the origin is $\underset{\_}{33.4°\text{counter}\text{clockwise}}$.

The maximum moment of inertia is $\underset{\_}{22.1\times {10}^3{\text{in}}^{\text{4}}}$.

The minimum moment of inertia is $\underset{\_}{2,490{\text{in}}^{\text{4}}}$

Explanation of Solution

Sketch the cross section as shown in Figure 1.

Express the product of inertia as shown below:

${\overline{I}}_{xy}={\left(I_{xy}\right)}_1+{\left({\overline{I}}_{xy}\right)}_2+{\left(I_{xy}\right)}_3$

Here, $\left[{\left(I_{xy}\right)}_1,{\left(I_{xy}\right)}_2,{\left(I_{xy}\right)}_3\right]$ is product of inertia section 1, 2, and 3.

Applying parallel axis theorem for each triangle,

$I_{xy}=I_{x^{\prime }{y}^{\prime }}+\overline{x}\overline{y}A$

When the x and y axis is symmetry.

${\left(I_{x^{\prime }{y}^{\prime }}\right)}_1=0$

Refer to sample problem 9.6 in the textbook.

${\overline{I}}_{x^{\prime }{y}^{\prime }}=-\frac{1}{72}{b}^2{h}^2$

The sign of ${\overline{I}}_{x^{\prime }{y}^{\prime }}$ is unchanged because the angles of rotation are $0°$ and $180°$ for triangle 2,3 respectively.

Refer to Figure 1.

Find the area of rectangular section 2 as shown below:

$A_2=b_2{h}_2$ (1)

Here, $b_2$ is width of the rectangular section and $h_2$ is height of the rectangular section.

Substitute $9\text{in}\text{.}$ for $b_2$ and $15\text{in}\text{.}$ for $h_2$ in Equation (1).

$\begin{array}{c}{A}_2=9\times 15\\ =67.5{\text{in}}^{\text{2}}\end{array}$

Find the area of rectangle section 3 as shown below:

$A_3=b_3{h}_3$ (2)

Here, $b_3$ is width of the rectangular section and $h_3$ is height of the rectangular section.

Substitute $9\text{in}\text{.}$ for $b_3$ and $15\text{in}\text{.}$ for $h_3$ in Equation (2).

$\begin{array}{c}{A}_3=9\times 15\\ =67.5{\text{in}}^{\text{2}}\end{array}$

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

$\left(x_2\right)=-9\text{in}\text{.}$

Find the centroid $\left(x_3\right)$ as shown below:

$\left(x_3\right)=9\text{in}\text{.}$

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

$\left(y_2\right)=7\text{in}\text{.}$

Find the centroid $\left(y_3\right)$ as shown below:

$\left(y_3\right)=-7\text{in}\text{.}$

Find the product of inertia of the area with respect to x and y axes by using parallel axis theorem as shown below:

$\begin{array}{c}{\overline{I}}_{xy}={\left(\overline{x}\overline{y}A\right)}_1+{\left(\overline{x}\overline{y}A\right)}_2\\ =\left(x_1{y}_{1}A\right)+\left({\overline{x}}_2{\overline{y}}_{2}A\right)\end{array}$ (3)

Substitute $-9\text{in}\text{.}$ for $x_2$, $9\text{in}\text{.}$ for $x_3$, $7\text{in}\text{.}$ for $\left(y_2\right)$, $-7\text{in}\text{.}$ for $y_3$, $67.5{\text{in}}^{\text{2}}$ for $A_2$, and $67.5{\text{in}}^{\text{2}}$ for $A_3$ in Equation (3).

$\begin{array}{c}{\overline{I}}_{xy}=\left(-9\times \left(7\right)\times 67.5\right)+\left(9\times \left(-7\right)\times 67.5\right)\\ =-4,252.5-4,252.5\\ =-8,505{\text{in}}^{\text{4}}\end{array}$

Find the product of inertia as follows:

$I_{xy}=2\left(-\frac{1}{72}{b}^2{h}^2\right)-{\overline{I}}_{xy}$ (4)

Substitute $9\text{in}\text{.}$ for b, $15\text{in}\text{.}$ for h, and $-8,505{\text{in}}^{\text{4}}$ for ${\overline{I}}_{xy}$ in Equation (4).

$\begin{array}{c}{I}_{xy}=2\left(-\frac{1}{72}{\left(9\right)}^2{\left(15\right)}^2\right)-8,505\\ =-506.25-8,505\\ =-9011.25{\text{in}}^{\text{4}}\\ \simeq -9010{\text{in}}^{\text{4}}\end{array}$

Find the moment of inertia for section 1 about x axis as shown below:

${\left({\overline{I}}_x\right)}_1=\frac{1}{12}\left(b{h}^3\right)$ (5)

Here, b is the width of the section 1 about x axis and h is the height of the section 1about x axis.

Substitute $24\text{in}\text{.}$ for b and $4\text{in}\text{.}$ for h in Equation (5).

$\begin{array}{c}{\left({\overline{I}}_x\right)}_1=\frac{1}{12}\left(24\times 4^3\right)\\ =\frac{1}{12}\left(1,536\right)\\ =128{\text{in}}^{\text{4}}\end{array}$

Find the moment of inertia for section 2 and 3 about x axis as shown below:

${\left({\overline{I}}_x\right)}_3={\left({\overline{I}}_x\right)}_2=\frac{1}{36}\left(b{h}^3\right)+\left[\frac{1}{2}bh\right]{\overline{y}}^2$ (6)

Here, b is the width of the section 2 and h is the height of the section 2.

Substitute $9\text{in}\text{.}$ for b, $7\text{in}\text{.}$ for $\overline{y}$, and $15\text{in}\text{.}$ for h in Equation (6).

$\begin{array}{c}{\left({\overline{I}}_x\right)}_3={\left({\overline{I}}_x\right)}_2=\frac{1}{36}\left(9\times {15}^3\right)+\left[\frac{1}{2}\left(9\times 15\right)\right]7^2\\ =843.75+3,307.5\\ =4,151.25{\text{in}}^{\text{4}}\end{array}$

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

${\overline{I}}_x={\left({\overline{I}}_x\right)}_1+{\left({\overline{I}}_x\right)}_2+{\left({\overline{I}}_x\right)}_3$ (7)

Substitute $128{\text{in}}^{\text{4}}$ for ${\left({\overline{I}}_x\right)}_1$, $4,151.25{\text{in}}^{\text{4}}$ for ${\left({\overline{I}}_x\right)}_2$, and $4,151.25{\text{in}}^{\text{4}}$ for ${\left({\overline{I}}_x\right)}_2$ in Equation (7).

$\begin{array}{c}{\overline{I}}_x=128+4,151.25+4,151.25\\ =8,430.5{\text{in}}^{\text{4}}\end{array}$

Find the moment of inertia for section 1 about y axis as shown below:

${\left({\overline{I}}_y\right)}_1=\frac{1}{12}\left(b^{3}h\right)$ (8)

Here, b is the width of the section 1 about y axis and h is the height of the section 1 about y axis.

Substitute $24\text{in}\text{.}$ for b and $4\text{in}\text{.}$ for h in Equation (8).

$\begin{array}{c}{\left({\overline{I}}_x\right)}_1=\frac{1}{12}\left({24}^3\times 4\right)\\ =\frac{1}{12}\left(55,296\right)\\ =4,608{\text{in}}^{\text{4}}\end{array}$

Find the moment of inertia for section 2 and 3 about y axis as shown below:

${\left({\overline{I}}_y\right)}_3={\left({\overline{I}}_y\right)}_2=\frac{1}{36}\left(b{h}^3\right)+\left[\frac{1}{2}bh\right]{\overline{y}}^2$ (9)

Substitute $9\text{in}\text{.}$ for b, $9\text{in}\text{.}$ for $\overline{y}$, and $15\text{in}\text{.}$ for h in Equation (9).

$\begin{array}{c}{\left({\overline{I}}_y\right)}_3={\left({\overline{I}}_y\right)}_2=\frac{1}{36}\left(9^3\times 15\right)+\left[\frac{1}{2}\left(9\times 15\right)\right]9^2\\ =303.75+5,467.5\\ =5,771.25{\text{in}}^{\text{4}}\end{array}$

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

${\overline{I}}_y={\left({\overline{I}}_y\right)}_1+{\left({\overline{I}}_y\right)}_2+{\left({\overline{I}}_y\right)}_3$ (10)

Substitute $4,608{\text{in}}^{\text{4}}$ for ${\left({\overline{I}}_y\right)}_1$, $5,771.25{\text{in}}^{\text{4}}$ for ${\left({\overline{I}}_y\right)}_2$, and $5,771.25{\text{in}}^{\text{4}}$ for ${\left({\overline{I}}_y\right)}_2$ in Equation (10).

$\begin{array}{c}{\overline{I}}_y=4,608+5,771.25+5,771.25\\ =16,150.5{\text{in}}^{\text{4}}\end{array}$

Consider the point X with co-ordinates $\left(I_x,I_{xy}\right)$ and point Y with co-ordinates $\left(I_y,-I_{xy}\right)$.

The diameter of the Mohr circle is defined by XY distance. It is expressed as follows:

$X\left(8,403.5,-9,011.25\right)$ and $Y\left(16,150.5,9,011.25\right)$.

Consider that the average moment of inertia $\left(I_{ave}\right)$ is represents the distance from origin O to the center of the circle C.

Find the average moment of inertia $\left(I_{ave}\right)$ using the relation as shown below:

$I_{ave}=\frac{1}{2}\left({\overline{I}}_x+{\overline{I}}_y\right)$ (11)

Here, ${\overline{I}}_x$ is moment of inertia about x axis and ${\overline{I}}_y$ is moment of inertia about y axis.

Substitute $8,430.5{\text{in}}^{\text{4}}$ for ${\overline{I}}_x$ and $16,150.5{\text{in}}^{\text{4}}$ for ${\overline{I}}_y$ in Equation (11).

$\begin{array}{c}{I}_{\text{ave}}=\frac{1}{2}\left(8,430.5+16,150.5\right)\\ =\frac{1}{2}\left(24,581\right)\\ =12,290.5{\text{in}}^{\text{4}}\end{array}$

Find the radius (R) using the relation as shown below:

$R=\sqrt{{\left(\frac{{\overline{I}}_x-{\overline{I}}_y}{2}\right)}^2+{\overline{I}}_{xy}^2}$ (12)

Here, R is radius and ${\overline{I}}_{xy}$ is product of inertia.

Substitute $8,430.5{\text{in}}^{\text{4}}$ for ${\overline{I}}_x$, $-9,011.25{\text{in}}^{\text{4}}$ for ${\overline{I}}_{xy}$, and $16,150.5{\text{in}}^{\text{4}}$ for ${\overline{I}}_y$ in Equation (12).

$\begin{array}{c}R=\sqrt{{\left(\frac{1}{2}\left(8,430.5-16,150.5\right)\right)}^2+{\left(-9,011.25\right)}^2}\\ =\sqrt{14,899,600+81,202,626.56}\\ =9,803.17{\text{in}}^{\text{4}}\end{array}$

Draw the Mohr circle using the procedure as follows:

• Plot the point X with coordinate $\left(8,403.5,-9,011.25\right)$ and plot the point Y with co-ordinate $\left(16,150.5,9,011.25\right)$.
• Joint the points X and Y with straight line, which is defined as center of the circle (C).
• Plot the circle using the center point and diameter (XY).

Sketch the Mohr circle as shown in Figure 2.

Refer to Figure 2.

$\tan 2{\theta }_m=-\frac{2I_{xy}}{I_x-I_y}$ (13)

Substitute $8,430.5{\text{in}}^{\text{4}}$ for ${\overline{I}}_x$, $-9,011.25{\text{in}}^{\text{4}}$ for ${\overline{I}}_{xy}$, and $16,150.5{\text{in}}^{\text{4}}$ for ${\overline{I}}_y$ in Equation (13).

$\begin{array}{c}\tan 2{\theta }_m=-\frac{2\left(-9,011.25\right)}{8,430.5-16,150.5}\\ =-\frac{-18,022.5}{-7,720}\\ \tan 2{\theta }_m=-2.33452\\ 2{\theta }_m={\tan }^{-1}\left(-2.33452\right)\\ 2{\theta }_m=-66.812°\\ {\theta }_m=-33.4°\end{array}$

Thus, the orientation of the principal axes at the origin is $\underset{\_}{33.4°\text{counter}\text{clockwise}}$.about C.

Sketch the orientation of the principal axes as shown in Figure 3.

Sketch the cross section as shown in Figure 3.

Find the maximum moment $I_{\max }$ of inertia using the relation:

$I_{\max }=I_{\text{ave}}+R$ (14)

Substitute $12,290.5{\text{in}}^{\text{4}}$ for $I_{\text{ave}}$ and $9,803.17{\text{in}}^{\text{4}}$ for R in Equation (14).

$\begin{array}{c}{I}_{\max }=12,290.5+9,803.17\\ =22.1\times {10}^3{\text{in}}^{\text{4}}\end{array}$

Thus, the maximum moment of inertia is $\underset{\_}{22.1\times {10}^3{\text{in}}^{\text{4}}}$.

Find the minimum moment $I_{\min }$ of inertia using the relation:

$I_{\min }=I_{\text{ave}}-R$ (15)

Substitute $12,290.5{\text{in}}^{\text{4}}$ for $I_{\text{ave}}$ and $9,803.17{\text{in}}^{\text{4}}$ for R in Equation (15).

$\begin{array}{c}{I}_{\min }=12,290.5-9,803.17\\ =2,490{\text{in}}^{\text{4}}\end{array}$

Thus, the minimum moment of inertia is $\underset{\_}{2,490{\text{in}}^{\text{4}}}$

From the Mohr’s circle, the axis a represents ${\overline{I}}_{\min }$ and b axis represents ${\overline{I}}_{\max }$.