Chapter 9.5, Problem 9.135P

9.135 and 9.136 A 2-mm thick piece of sheet steel is cut and bent into the machine component shown. Knowing that the density of steel is 7850 kg/m³, determine the mass moment of inertia of the component with respect to each of the coordinate axes.

Fig. P9.135

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 $\underset{\_}{15.27\times {10}^{-3}\text{kg}\cdot {\text{m}}^2}$.