Chapter 9.3, Problem 9.82P

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. P9.75

9.82 Determine the moments of inertia and the product of inertia of the area of Prob. 9.75 with respect to new centroidal axes obtained by rotating the x and y axes 45° clockwise.

To determine

Find the moment of inertia and product of inertia with respect new centroid axes obtained x and y axes $45°$ clockwise.

Answer to Problem 9.82P

The moment of inertia with respect new centroid axes obtained x axes $45°$ clockwise is $\underset{\_}{1.474\times {10}^6{\text{mm}}^{\text{4}}}$.

The moment of inertia with respect new centroid axes obtained y axes $45°$ clockwise is $\underset{\_}{0.532\times {10}^6{\text{mm}}^{\text{4}}}$.

The product of inertia with respect new centroid axes obtained x axes $45°$ clockwise is $\underset{\_}{0.750\times {10}^6{\text{mm}}^{\text{4}}}$

Explanation of Solution

Calculation:

Refer to problem 9.75.

The product of inertia of the area with respect to x and y axes by using direct parallel axis theorem is $471,040{\text{in}}^{\text{4}}$.

Sketch the rectangular section as shown in Figure 1.

Find the moment of inertia ${\left({\overline{I}}_x\right)}_1$ using the relation as shown below:

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

Here, $b_1$ is width of the section 1 and $h_1$ is height of the section 1.

Substitute $100\text{mm}$ for $b_1$ and $8\text{mm}$ for $h_1$ in Equation (1).

$\begin{array}{c}{\left({\overline{I}}_x\right)}_1=\frac{1}{12}\left(100\right)\times 8^3\\ =4,266.67{\text{mm}}^{\text{4}}\end{array}$

Find the moment of inertia ${\left({\overline{I}}_x\right)}_2$ and ${\left({\overline{I}}_x\right)}_3$ using the relation as shown below:

${\left({\overline{I}}_x\right)}_2={\left({\overline{I}}_x\right)}_3=\frac{1}{12}{b}_2{h}_2^3+\left(bd\right){\overline{y}}^2$ (2)

Here, $b_2$ is width of the section 2 and $h_2$ is height of the section 2.

Substitute $8\text{mm}$ for $b_2$,$20\text{mm}$ for ${\overline{y}}^2$, and $32\text{mm}$ for $h_2$ in Equation (2).

$\begin{array}{c}{\left({\overline{I}}_x\right)}_2={\left({\overline{I}}_x\right)}_3=\frac{1}{12}\left(8\right){\left(32\right)}^3+\left(8\times 32\right){\left(20\right)}^2\\ =21,845.333+102,400\\ =124,245.333{\text{mm}}^{\text{4}}\end{array}$

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

$\left({\overline{I}}_x\right)={\left({\overline{I}}_x\right)}_1+{\left({\overline{I}}_x\right)}_2+{\left({\overline{I}}_x\right)}_3$ (3)

Substitute $4,266.67{\text{mm}}^{\text{4}}$ for ${\left({\overline{I}}_x\right)}_1$, $124,245.333{\text{mm}}^{\text{4}}$ for ${\left({\overline{I}}_x\right)}_2$, and $124,245.333{\text{mm}}^{\text{4}}$ for ${\left({\overline{I}}_x\right)}_3$ in Equation (3).

$\begin{array}{c}\left({\overline{I}}_x\right)=4,266.67+124,245.333+124,245.333\\ =252,757{\text{mm}}^{\text{2}}\end{array}$

Find the moment of inertia ${\left({\overline{I}}_y\right)}_1$ using the relation as shown below:

${\left({\overline{I}}_y\right)}_1=\frac{1}{12}{b}_1^3{h}_1$ (4)

Substitute $100\text{mm}$ for $b_1$ and $8\text{mm}$ for $h_1$ in Equation (4).

$\begin{array}{c}{\left({\overline{I}}_y\right)}_1=\frac{1}{12}\left({100}^3\right)\times 8\\ =666,666.667{\text{mm}}^{\text{4}}\end{array}$

Find the moment of inertia ${\left({\overline{I}}_y\right)}_2$ and ${\left({\overline{I}}_y\right)}_3$ using the relation as shown below:

${\left({\overline{I}}_y\right)}_2={\left({\overline{I}}_y\right)}_3=\frac{1}{12}{b}_2^3{h}_2+\left(bd\right){\overline{y}}^2$ (5)

Substitute $8\text{mm}$ for $b_2$,$46\text{mm}$ for ${\overline{y}}^2$, and $32\text{mm}$ for $h_2$ in Equation (5).

$\begin{array}{c}{\left({\overline{I}}_y\right)}_2={\left({\overline{I}}_y\right)}_3=\frac{1}{12}{\left(8\right)}^3\left(32\right)+\left(8\times 32\right){\left(46\right)}^2\\ =1,365.333+541,696\\ =543,061.3{\text{mm}}^{\text{4}}\end{array}$

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

$\left({\overline{I}}_y\right)={\left({\overline{I}}_y\right)}_1+{\left({\overline{I}}_y\right)}_2+{\left({\overline{I}}_y\right)}_3$ (6)

Substitute $666,666.667{\text{mm}}^{\text{4}}$ for ${\left({\overline{I}}_y\right)}_1$, $543,061.3{\text{mm}}^{\text{4}}$ for ${\left({\overline{I}}_y\right)}_2$, and $543,061.3{\text{mm}}^{\text{4}}$ for ${\left({\overline{I}}_y\right)}_3$ in Equation (6).

$\begin{array}{c}\left({\overline{I}}_y\right)=666,666.667+543,061.3+543,061.3\\ =1,752,789{\text{mm}}^{\text{4}}\end{array}$

Find the value of $\frac{1}{2}\left({\overline{I}}_x+{\overline{I}}_y\right)$:

$\begin{array}{c}\frac{1}{2}\left({\overline{I}}_x+{\overline{I}}_y\right)=\frac{1}{2}\left(252,757+1,752,789\right)\\ =1,002,773{\text{mm}}^{\text{4}}\end{array}$

Find the value of $\frac{1}{2}\left({\overline{I}}_x-{\overline{I}}_y\right)$:

$\begin{array}{c}\frac{1}{2}\left({\overline{I}}_x-{\overline{I}}_y\right)=\frac{1}{2}\left(252,757-1,752,789\right)\\ =-750,016{\text{mm}}^{\text{4}}\end{array}$

Find the moment of inertia with respect new centroid axes obtained x axes $45°$ clockwise using the Equation.

Refer to Equation 9.18 in section 9.3B in the textbook.

$I_{x^{\prime }}=\frac{1}{2}\left({\overline{I}}_x+{\overline{I}}_y\right)+\frac{1}{2}\left({\overline{I}}_x-{\overline{I}}_y\right)\cos 2\theta -{\overline{I}}_{xy}\sin 2\theta$ (7)

Substitute $1,002,773{\text{mm}}^{\text{4}}$ for $\frac{1}{2}\left(I_x+I_y\right)$, $-750,016{\text{mm}}^{\text{4}}$ for $\frac{1}{2}\left(I_x-I_y\right)$, $471,040{\text{mm}}^{\text{4}}$ for ${\overline{I}}_{xy}$, and $-45°$ for $\theta$ in Equation (7).

$\begin{array}{c}{I}_{x^{\prime }}=1,002,773-750,016\cos 2\left(-45°\right)-471,040\sin 2\left(-45°\right)\\ =1,002,773-0+471,040\\ =1.474\times {10}^6{\text{mm}}^{\text{4}}\end{array}$

Thus, the moment of inertia with respect new centroid axes obtained x axes $45°$ clockwise is $\underset{\_}{1.474\times {10}^6{\text{mm}}^{\text{4}}}$.

Find the moment of inertia with respect new centroid axes obtained y axes $45°$ clockwise using the Equation.

Refer to Equation 9.19 in section 9.3B in the textbook.

$I_{y^{\prime }}=\frac{1}{2}\left({\overline{I}}_x+{\overline{I}}_y\right)-\frac{1}{2}\left({\overline{I}}_x-{\overline{I}}_y\right)\cos 2\theta +{\overline{I}}_{xy}\sin 2\theta$ (8)

Substitute $1,002,773{\text{mm}}^{\text{4}}$ for $\frac{1}{2}\left(I_x+I_y\right)$, $-750,016{\text{mm}}^{\text{4}}$ for $\frac{1}{2}\left(I_x-I_y\right)$, $471,040{\text{mm}}^{\text{4}}$ for ${\overline{I}}_{xy}$, and $45°$ for $\theta$ in Equation (8).

$\begin{array}{c}{I}_y=1,002,773-\left(-750,016\cos 2\left(-45°\right)\right)-471,040\sin 2\left(-45°\right)\\ =1,002,773-0-471,040\\ =0.532\times {10}^6{\text{mm}}^{\text{4}}\end{array}$

Thus, the moment of inertia with respect new centroid axes obtained y axes $45°$ clockwise is $\underset{\_}{0.532\times {10}^6{\text{mm}}^{\text{4}}}$.

Find the product of inertia with respect new centroid axes obtained x and y axes $45°$ clockwise using the Equation.

${\overline{I}}_{x^{\prime }{y}^{\prime }}=\frac{1}{2}\left({\overline{I}}_x-{\overline{I}}_y\right)\sin 2\theta +{\overline{I}}_{xy}\cos 2\theta$ (9)

Substitute $-750,016{\text{mm}}^{\text{4}}$ for $\frac{1}{2}\left(I_x-I_y\right)$, $471,040{\text{mm}}^{\text{4}}$ for ${\overline{I}}_{xy}$, and $45°$ for $\theta$ in Equation (9).

$\begin{array}{c}{\overline{I}}_{x^{\prime }{y}^{\prime }}=-750,016\sin 2\left(45°\right)+\left(471,040\right)\cos 2\left(45°\right)\\ =-750,016\sin \left(90°\right)+\left(471,040\right)\cos \left(90°\right)\\ =0.750\times {10}^6{\text{mm}}^{\text{4}}\end{array}$

Thus, the product of inertia with respect new centroid axes obtained x axes $45°$ clockwise is $\underset{\_}{0.750\times {10}^6{\text{mm}}^{\text{4}}}$