Chapter 9.5, Problem 9.135P

To determine

Find the mass moment of inertia of the component with respect to $x,y,z$ coordinate axes.

Answer to Problem 9.135P

The mass moment of inertia of the component with respect to $x$ coordinate axes is $\underset{\_}{7.11\times {10}^{-3}\text{kg}\cdot {\text{m}}^2}$

The mass moment of inertia of the component with respect to $y$ coordinate axes is $\underset{\_}{16.96\times {10}^{-3}\text{kg}\cdot {\text{m}}^2}$

The mass moment of inertia of the component with respect to $z$ coordinate axes is $\underset{\_}{15.27\times {10}^{-3}\text{kg}\cdot {\text{m}}^2}$.

Explanation of Solution

Given information:

The thickness (t) of sheet steel is $2\text{mm}$.

The density $\rho$ of steel is $7,850\text{kg}/{\text{m}}^{\text{3}}$.

Calculation:

Sketch the sheet steel as shown in Figure 1.

Find the mass of rectangular section 1 as shown in below:

$m_1=\rho V_1$ (1)

Here, $V_1$ is volume of rectangular section 1.

Express the volume of rectangular section 1 as follows:

$V_1=lwt$

Substitute $120\text{mm}$ for l, $200\text{mm}$ for w, $7,850\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, and $2\text{mm}$ for t in Equation (1).

$\begin{array}{c}{m}_1=7,850\times \left[120\text{mm}\left(\frac{1\text{m}}{{10}^3\text{m}}\right)\times 200\text{mm}\left(\frac{1\text{m}}{{10}^3\text{m}}\right)\times 2\text{mm}\left(\frac{1\text{m}}{{10}^3\text{m}}\right)\right]\\ =7,850\left[0.120\times 0.200\times 0.002\right]\\ =7,850\left[0.000048\right]\\ =0.3768\text{kg}\end{array}$

Find the mass of rectangular section 2 as shown in below:

$m_2=\rho V_2$ (2)

Here, $V_2$ is volume of rectangular section 2.

Express the volume of rectangular section 2 as follows:

$V_2=lwt$

Substitute $120\text{mm}$ for l, $100\text{mm}$ for w, $7,850\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, and $2\text{mm}$ for t in Equation (2).

$\begin{array}{c}{m}_2=7,850\times \left[120\text{mm}\left(\frac{1\text{m}}{{10}^3\text{m}}\right)\times 100\text{mm}\left(\frac{1\text{m}}{{10}^3\text{m}}\right)\times 2\text{mm}\left(\frac{1\text{m}}{{10}^3\text{m}}\right)\right]\\ =7,850\left[0.120\times 0.100\times 0.002\right]\\ =7,850\left[0.000024\right]\\ =0.1884\text{kg}\end{array}$

Panel 1:

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

$\begin{array}{c}{I}_x={\overline{I}}_x+m{d}^2\\ =\frac{1}{12}\left(m_1{h}^2\right)+m_1\left(d^2\right)\end{array}$ (3)

Substitute $0.3768\text{kg}$ for $m_1$, $120\text{mm}$ for h, and $\left(10\text{+60}\right)\text{mm}$ for d in Equation (3).

$\begin{array}{c}{I}_x=\frac{1}{12}\left(0.3768\right){\left(120\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}\right)}^2+\left(0.3768\right)\left({\left(100\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}\right)}^2+{\left(60\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\right)\\ =\frac{1}{12}\left(0.3768\right){\left(0.12\right)}^2+\left(0.3768\right)\left({0.1}^2+{0.06}^2\right)\\ =0.00045216+0.00512448\\ =5.5766\times {10}^{-3}\end{array}$

$=5.577\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$

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

$\begin{array}{c}{I}_y={\overline{I}}_y+m{d}^2\\ =\frac{1}{12}\left(m_1{h}^2\right)+m_1\left(d^2\right)\end{array}$ (4)

Substitute $0.3768\text{kg}$ for $m_1$, $200\text{mm}$ for h, and $\left(100+100\right)\text{mm}$ for d in Equation (4).

$\begin{array}{c}{I}_y=\frac{1}{12}\left(0.3768\right){\left(200\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}\right)}^2+\left(0.3768\right)\left(100\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}+100\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}\right)\\ =\frac{1}{12}\left(0.3768\right){\left(0.2\right)}^2+\left(0.3768\right)\left({0.1}^2+{0.1}^2\right)\\ =0.001256+0.007536\\ =8.792\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Find the moment of inertia about z axis for panel 1 as shown below:

$I_z={\overline{I}}_z+m{d}^2$ (5)

Substitute $0.3768\text{kg}$ for $m_1$, $\frac{1}{12}\left(0.3768\left[{0.2}^2+{0.12}^2\right]\right)$ for ${\overline{I}}_z$, and $\left(100+60\right)\text{mm}$ for d in Equation (5).

$\begin{array}{c}{I}_z=\frac{1}{12}\left(0.3768\left[{0.2}^2+{0.12}^2\right]\right)+0.3768\left[\left(100\text{mm}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2+60\text{mm}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\right)\right]\\ =0.00170816+0.005124\\ =6.833\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Panel 2:

Find the moment of inertia about x axis for panel 2 as shown below:

$I_x={\overline{I}}_x+m{d}^2$ (6)

Substitute $0.1884\text{kg}$ for $m_2$, $\frac{1}{12}\left(0.1884\right)\left[{0.12}^2+{0.1}^2\right]$ for ${\overline{I}}_x$, and $\left(50+60\right)\text{mm}$ for d in Equation (6).

$\begin{array}{c}{I}_x=\frac{1}{12}\left(0.1884\right)\left[{0.12}^2+{0.1}^2\right]+\left(0.1884\right)\left({\left(50\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}\right)}^2+{\left(60\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\right)\\ =\frac{1}{12}\left(0.1884\right)\left[{0.12}^2+{0.1}^2\right]+\left(0.1884\right)\left({0.05}^2+{0.06}^2\right)\\ =3.8308\times {10}^{-4}+0.00114924\\ =1.532\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Find the moment of inertia about y axis for panel 2 as shown below:

$\begin{array}{c}{I}_y={\overline{I}}_y+m{d}^2\\ =\frac{1}{12}\left(m_2{h}^2\right)+m_2\left(d^2\right)\end{array}$ (7)

Substitute $0.1884\text{kg}$ for $m_2$, $100\text{mm}$ for h, and $\left(5+200\right)\text{mm}$ for d in Equation (7).

$\begin{array}{c}{I}_y=\frac{1}{12}\left(0.1884\right){\left(100\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}\right)}^2+\left(0.1884\right)5\left(50\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}+200\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}\right)\\ =\frac{1}{12}\left(0.1884\right){\left(0.1\right)}^2+\left(0.1884\right)\left({0.05}^2+{0.2}^2\right)\\ =0.000157+0.00800\\ =8.164\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Find the moment of inertia about z axis for panel 2 as shown below:

$\begin{array}{c}{I}_z={\overline{I}}_z+m{d}^2\\ =\frac{1}{12}\left(m_2{h}^2\right)+m_2\left(d^2\right)\end{array}$ (8)

Substitute $0.1884\text{kg}$ for $m_2$, $120\text{mm}$ for h, and $\left(6+200\right)\text{mm}$ for d in Equation (8).

$\begin{array}{c}{I}_z=\frac{1}{12}\left(0.1884\right){\left(120\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}\right)}^2+\left(0.1884\right)5\left(60\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}+200\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}\right)\\ =\frac{1}{12}\left(0.1884\right){\left(0.12\right)}^2+\left(0.1884\right)\left({0.06}^2+{0.2}^2\right)\\ =0.00022608+0.008214\\ =8.44\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Find the total mass of inertia $\left(I_x\right)$ about x axis as shown below:

$I_x={\left(I_x\right)}_1+{\left(I_x\right)}_2$ (9)

Here, ${\left(I_x\right)}_1$ is moment of inertia about x panel 1 and ${\left(I_x\right)}_2$ is moment of inertia about x panel 2.

Substitute $5.577\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for ${\left(I_x\right)}_1$ and $1.532\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for ${\left(I_x\right)}_2$ in Equation (9).

$\begin{array}{c}{I}_x=5.577\times {10}^{-3}+1.532\times {10}^{-3}\\ =7.11\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Thus, the mass moment of inertia of the component with respect to $x$ coordinate axes is $\underset{\_}{7.11\times {10}^{-3}\text{kg}\cdot {\text{m}}^2}$

Find the total mass of inertia $\left(I_y\right)$ about y axis as shown below:

$I_y={\left(I_y\right)}_1+{\left(I_y\right)}_2$ (10)

Here, ${\left(I_y\right)}_1$ is moment of inertia about x panel 1 and ${\left(I_y\right)}_2$ is moment of inertia about x panel 2.

Substitute $8.792\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for  ${\left(I_y\right)}_1$ and $8.164\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for ${\left(I_y\right)}_2$ in Equation (10).

$\begin{array}{c}{I}_y=8.792\times {10}^{-3}+8.164\times {10}^{-3}\\ =16.96\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Thus, the mass moment of inertia of the component with respect to $y$ coordinate axes is $\underset{\_}{16.96\times {10}^{-3}\text{kg}\cdot {\text{m}}^2}$

Find the total mass of inertia $\left(I_z\right)$ about z axis as shown below:

$I_z={\left(I_z\right)}_1+{\left(I_z\right)}_2$ (11)

Here, ${\left(I_z\right)}_1$ is moment of inertia about x panel 2 and ${\left(I_z\right)}_2$ is moment of inertia about x panel 2.

Substitute $6.833\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for ${\left(I_z\right)}_1$ and $8.440\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for ${\left(I_z\right)}_2$ in Equation (11).

$\begin{array}{c}{I}_y=6.833\times {10}^{-3}+8.440\times {10}^{-3}\\ =15.27\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Thus, the mass moment of inertia of the component with respect to $z$ coordinate axes is $\underset{\_}{15.27\times {10}^{-3}\text{kg}\cdot {\text{m}}^2}$.