Chapter 9.4, Problem 9.102P

9.98 through 9.102 Using Mohr’s circle, determine for the area indicated the orientation of the principal centroidal axes and the corresponding values of the moments of inertia.

9.102 Area of Prob. 9.77

9.75 through 9.78 Using the parallel-axis theorem, determine the product of inertia of the area shown with respect to the centroidal x and y axes.

Fig. P9.77

To determine

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

Answer to Problem 9.102P

The orientation of the principal axes at the origin is $\underset{\_}{15.99°\text{counter}\text{clockwise}}$.

The maximum moment of inertia is $\underset{\_}{19.35{\text{in}}^{\text{4}}}$.

The minimum moment of inertia is $\underset{\_}{3.29{\text{in}}^{\text{4}}}$

Explanation of Solution

Calculation:

Sketch the cross section shown in Figure 1.

Refer to Figure 1.

Find the area $\left(A_1\right)$ of rectangle section 1 as shown below:

$A_1=b_1{h}_1$ (1)

Here, $b_1$ is width of the rectangular section and $h_1$ is height of the rectangular section.

Substitute $3.6\text{in}\text{.}$ for $b_1$ and $0.5\text{in}.$ for $h_1$ in Equation (1).

$\begin{array}{c}{A}_1=36\times 0.5\\ =18{\text{in}}^{\text{2}}\end{array}$

Find the area $\left(A_2\right)$ of rectangle section 2 as shown below:

$A_2=b_2{h}_2$ (2)

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

Substitute $0.5\text{in}\text{.}$ for $b_2$ and $3.8\text{in}.$ for $h_2$ in Equation (2).

$\begin{array}{c}{A}_2=0.5\times 3.8\\ =1.9{\text{in}}^{\text{2}}\end{array}$

Find the area of rectangle section 3 as shown below:

$A_3=b_3{h}_3$ (3)

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

Substitute $1.3\text{in}\text{.}$ for $b_3$ and $1.0\text{in}\text{.}$ for $h_3$ in Equation (3).

$\begin{array}{c}{A}_3=1.3\times 1.0\\ =1.3{\text{in}}^{\text{2}}\end{array}$

Refer to Figure 1.

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

$\begin{array}{c}\left(x_1\right)=\frac{3.6}{2}\\ =1.8\text{in}\text{.}\end{array}$

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

$\begin{array}{c}\left(x_2\right)=\frac{0.5}{2}\\ =0.25\text{in}\text{.}\end{array}$

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

$\begin{array}{c}\left(x_3\right)=\frac{1.3}{2}\\ =0.65\text{in}\text{.}\end{array}$

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

$\begin{array}{c}\left(y_1\right)=\frac{0.5}{2}\\ =0.25\text{in}\text{.}\end{array}$

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

$\begin{array}{c}\left(y_2\right)=0.5+1.9\\ =2.4\text{in}\text{.}\end{array}$

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

$\begin{array}{c}\left(y_3\right)=\frac{1.3}{2}\\ =0.65\text{in}\text{.}\end{array}$

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

$\overline{x}=\frac{A_1{x}_1+A_2{x}_2+A_3{x}_3}{A_1+A_2+A_3}$ (6)

Substitute $18{\text{in}}^{\text{2}}$ for $A_1$, $1.9{\text{in}}^{\text{2}}$ for $A_2$, $1.3{\text{in}}^{\text{2}}$ for $A_1$, $1.8\text{in}\text{.}$ for $x_1$, $0.25\text{in}\text{.}$ for $x_2$, and $0.65\text{in}\text{.}$ for $x_3$ in Equation (6).

$\begin{array}{c}\overline{x}=\frac{\left(1.8\times 1.8\right)+\left(1.9\times 0.25\right)+\left(1.3\times 0.65\right)}{1.8+1.9+1.3}\\ =\frac{4.56}{5}\\ =0.912\text{in}\text{.}\end{array}$

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

$\overline{y}=\frac{A_1{y}_1+A_2{y}_2+A_3{y}_3}{A_1+A_2+A_3}$ (7)

Substitute $18{\text{in}}^{\text{2}}$ for $A_1$, $1.9{\text{in}}^{\text{2}}$ for $A_2$, $1.3{\text{in}}^{\text{2}}$ for $A_1$, $0.25\text{in}\text{.}$ for $y_1$, $2.4\text{in}\text{.}$ for $y_2$, and $4.8\text{in}\text{.}$ for $y_3$ in Equation (7).

$\begin{array}{c}\overline{y}=\frac{\left(1.8\times 0.25\right)+\left(1.9\times 2.4\right)+\left(1.3\times 4.8\right)}{1.8+1.9+1.3}\\ =\frac{11.25}{5}\\ =2.25\text{in}\text{.}\end{array}$

Find the moment of inertia ${\left(I_x\right)}_1$ about x axis using the relation as shown below:

$\begin{array}{l}{\left(I_x\right)}_1=\frac{1}{12}{b}_1{h}_1^3+\left(b_1{h}_1\right){\left(\overline{h}\right)}^2\\ {\left(I_x\right)}_1=\frac{1}{12}{b}_1{h}_1^3+\left(b_1{h}_1\right){\left(\overline{y}-y_1\right)}^2\end{array}$ (8)

Here, $\overline{h}$ is vertical distance from the centroid of the segment to the neutral axis.

Substitute $3.6\text{in}\text{.}$ for $b_1$,$2.25\text{in}\text{.}$ for $\overline{y}$, $0.25\text{in}\text{.}$ for $y_1$ and $0.5\text{in}.$ for $h_1$ in Equation (8).

$\begin{array}{c}{\left(I_x\right)}_1=\frac{1}{12}\left(3.6\right){\left(0.5\right)}^3+\left(3.6\times 0.5\right){\left(2.25-0.25\right)}^2\\ =0.0375+7.2\\ =7.2375{\text{in}}^{\text{4}}\end{array}$

Find the moment of inertia ${\left(I_x\right)}_2$ about x axis using the relation as shown below:

$\begin{array}{l}{\left(I_x\right)}_2=\frac{1}{12}{b}_2{h}_2^3+\left(b_2{h}_2\right){\left(\overline{h}\right)}^2\\ {\left(I_x\right)}_2=\frac{1}{12}{b}_2{h}_2^3+\left(b_2{h}_2\right){\left(\overline{y}-y_2\right)}^2\end{array}$ (9)

Here, $\overline{h}$ is vertical distance from the centroid of the segment to the neutral axis.

Substitute $0.5\text{in}\text{.}$ for $b_2$,$2.25\text{in}\text{.}$ for $\overline{y}$, $2.4\text{in}\text{.}$ for $y_2$ and $3.8\text{in}.$ for $h_2$ in Equation (9).

$\begin{array}{c}{\left(I_x\right)}_2=\frac{1}{12}\left(0.5\right){\left(3.8\right)}^3+\left(1.9\right){\left(2.25-2.4\right)}^2\\ =2.28633+0.04275\\ =2.329{\text{in}}^{\text{4}}\end{array}$

Find the moment of inertia ${\left(I_x\right)}_3$ about x axis using the relation as shown below:

$\begin{array}{l}{\left(I_x\right)}_3=\frac{1}{12}{b}_3{h}_3^3+\left(b_3{h}_3\right){\left(\overline{h}\right)}^2\\ {\left(I_x\right)}_3=\frac{1}{12}{b}_3{h}_3^3+\left(b_3{h}_3\right){\left(\overline{y}-y_3\right)}^2\end{array}$ (10)

Here, $\overline{h}$ is vertical distance from the centroid of the segment to the neutral axis.

Substitute $1.3\text{in}\text{.}$ for $b_3$,$2.25\text{in}\text{.}$ for $\overline{y}$, $4.8\text{in}\text{.}$ for $y_3$ and $1\text{in}.$ for $h_3$ in Equation (10).

$\begin{array}{c}{\left(I_x\right)}_3=\frac{1}{12}\left(1.3\right){\left(1\right)}^3+\left(1.3\right){\left(4.8-2.25\right)}^2\\ =0.10833+8.45325\\ =8.5616{\text{in}}^{\text{4}}\end{array}$

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

${\overline{I}}_x={\left(I_x\right)}_1+{\left(I_x\right)}_2+{\left(I_x\right)}_3$ (11)

Substitute $7.2375{\text{in}}^{\text{4}}$ for ${\left(I_x\right)}_1$, $2.329{\text{in}}^{\text{4}}$ for ${\left(I_x\right)}_2$, and $8.5616{\text{in}}^{\text{4}}$ for ${\left(I_x\right)}_3$ in Equation (11).

$\begin{array}{c}{\overline{I}}_x=7.2375+2.329+8.5616\\ =18.128{\text{in}}^{\text{4}}\end{array}$

Find the moment of inertia ${\left(I_y\right)}_1$ about y axis using the relation as shown below:

${\left(I_y\right)}_1=\frac{1}{12}{b}_1^3{h}_1+\left(b_1{h}_1\right){\left(\overline{x}-x_1\right)}^2$ (12)

Substitute $3.6\text{in}\text{.}$ for $b_1$,$0.912\text{in}\text{.}$ for $\overline{x}$, $0.25\text{in}\text{.}$ for $x_1$ and $0.5\text{in}.$ for $h_1$ in Equation (12).

$\begin{array}{c}{\left(I_y\right)}_1=\frac{1}{12}{\left(3.6\right)}^3\left(0.5\right)+\left(3.6\times 0.5\right){\left(1.8-0.912\right)}^2\\ =1.944+1.419\\ =3.363{\text{in}}^{\text{4}}\end{array}$

Find the moment of inertia ${\left(I_y\right)}_2$ about y axis using the relation as shown below:

${\left(I_y\right)}_2=\frac{1}{12}{b}_2^3{h}^3+\left(b_2{h}_2\right){\left(\overline{x}-x_2\right)}^2$ (13)

Substitute $0.5\text{in}\text{.}$ for $b_2$,$0.912\text{in}\text{.}$ for $\overline{x}$, $\text{0}\text{.25}\text{in}\text{.}$ for $x_2$ and $3.8\text{in}.$ for $h_2$ in Equation (13).

$\begin{array}{c}{\left(I_y\right)}_2=\frac{1}{12}{\left(0.5\right)}^3\left(3.8\right)+\left(1.9\right){\left(0.912-0.25\right)}^2\\ =0.03958+0.83266\\ =0.8723{\text{in}}^{\text{4}}\end{array}$

Find the moment of inertia ${\left(I_y\right)}_3$ about y axis using the relation as shown below:

${\left(I_y\right)}_3=\frac{1}{12}{b}_3^3{h}^3+\left(b_3{h}_3\right){\left(\overline{x}-x_3\right)}^2$ (14)

Substitute $1.3\text{in}\text{.}$ for $b_3$,$0.912\text{in}\text{.}$ for $\overline{x}$, $0.65\text{in}\text{.}$ for $x_3$ and $1\text{in}.$ for $h_3$ in Equation (14).

$\begin{array}{c}{\left(I_y\right)}_3=\frac{1}{12}\left(1\right){\left(1.3\right)}^3+{\left(1.3\right)}^2{\left(0.912-0.65\right)}^2\\ =0.1831+0.0892\\ =0.2723{\text{in}}^{\text{4}}\end{array}$

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

${\overline{I}}_y={\left(I_y\right)}_1+{\left(I_y\right)}_2+{\left(I_y\right)}_3$ (15)

Substitute $3.3634{\text{in}}^{\text{4}}$ for ${\left(I_y\right)}_1$, $0.8723{\text{in}}^{\text{4}}$ for ${\left(I_y\right)}_2$, and $0.2723{\text{in}}^{\text{4}}$ for ${\left(I_y\right)}_3$ in Equation (15).

$\begin{array}{c}{\overline{I}}_y=\left(3.3634+0.8723+0.2723\right)\\ =\text{4}\text{.5080}{\text{in}}^{\text{4}}\end{array}$

Refer to problem 9.77.

The product of inertia $\left({\overline{I}}_{xy}\right)$ about xy axis is $-4.25320{\text{in}}^{\text{4}}$

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:

$X\left(18.1282,-4.25320\right)$ and $Y\left(4.5080,4.25320\right)$.

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 $\left({\overline{I}}_{ave}\right)$ using the relation as shown below:

${\overline{I}}_{ave}=\frac{1}{2}\left({\overline{I}}_x+{\overline{I}}_y\right)$ (16)

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

Substitute $18.128{\text{in}}^{\text{4}}$ for ${\overline{I}}_x$ and $\text{4}\text{.5080}{\text{in}}^{\text{4}}$ for ${\overline{I}}_y$ in Equation (16).

$\begin{array}{c}{I}_{\text{ave}}=\frac{1}{2}\left(18.1282+4.5080\right)\\ =\frac{1}{2}\left(22.6362\right)\\ =11.3181{\text{in}}^{\text{4}}\end{array}$

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

$R=\sqrt{{\left(\frac{{\overline{I}}_x-{\overline{I}}_y}{2}\right)}^2+{\overline{I}}_{xy}^2}$ (17)

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

Substitute $18.128{\text{in}}^{\text{4}}$ for ${\overline{I}}_x$, $-4.25320{\text{in}}^{\text{4}}$ for ${\overline{I}}_{xy}$,  and $\text{4}\text{.5080}{\text{in}}^{\text{4}}$ for ${\overline{I}}_y$ in Equation (17).

$\begin{array}{c}R=\sqrt{{\left(\frac{1}{2}\left(18.1282-4.5080\right)\right)}^2+{\left(-4.25320\right)}^2}\\ =\sqrt{46.37774+18.0897}\\ =8.02915{\text{in}}^{\text{4}}\end{array}$

Draw the Mohr circle using the procedure as follows:

• Plot the point X with coordinate $\left(18.1282,-4.25320\right)$ and plot the point Y with co-ordinate $\left(4.5080,4.25320\right)$.
• 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).
• Obtain the line $XY$ by rotating the angle $2{\theta }_m$ and also obtain by joining the points X and Y.

Sketch the Mohr circle as shown in Figure 2.

Refer to Figure 2.

$\tan 2{\theta }_m=-\frac{2I_{xy}}{I_x-I_y}$ (18)

Substitute $18.128{\text{in}}^{\text{4}}$ for ${\overline{I}}_x$, $-4.25320{\text{in}}^{\text{4}}$ for ${\overline{I}}_{xy}$,  and $\text{4}\text{.5080}{\text{in}}^{\text{4}}$ for ${\overline{I}}_y$ in Equation (18).

$\begin{array}{c}\tan 2{\theta }_m=-\frac{2\left(-4.25320\right)}{18.1282-4.5080}\\ =-\frac{8.5064}{13.6202}\\ \tan 2{\theta }_m=0.62454\\ 2{\theta }_m={\tan }^{-1}\left(0.62454\right)\\ 2{\theta }_m=31.986°\\ {\theta }_m=15.99°\end{array}$

Thus, the orientation of the principal axes at the origin is $\underset{\_}{15.99°\text{counter}\text{clockwise}}$.

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

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

$I_{\max }=I_{\text{ave}}+R$ (19)

Substitute $11.3181{\text{in}}^{\text{4}}$ for $I_{\text{ave}}$ and $8.02915{\text{in}}^{\text{4}}$ for R in Equation (19).

$\begin{array}{c}{I}_{\max }=11.3181+8.02915\\ =19.35{\text{in}}^{\text{4}}\end{array}$

Thus, the maximum moment of inertia is $\underset{\_}{19.35{\text{in}}^{\text{4}}}$.

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

$I_{\min }=I_{\text{ave}}-R$ (20)

Substitute $11.3181{\text{in}}^{\text{4}}$ for $I_{\text{ave}}$ and $8.02915{\text{in}}^{\text{4}}$ for R in Equation (20).

$\begin{array}{c}{I}_{\min }=11.3181-8.02915\\ =3.29{\text{in}}^{\text{4}}\end{array}$

Thus, the minimum moment of inertia is $\underset{\_}{3.29{\text{in}}^{\text{4}}}$

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