Chapter 9.5, Problem 9.137P

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. P9.137

To determine

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

Answer to Problem 9.137P

The mass moment of inertia with respect to $x$ coordinate axes is $\underset{\_}{5.14\times {10}^{-3}\text{kg}\cdot {\text{m}}^2}$.

The mass moment of inertia with respect to $y$ coordinate axes is $\underset{\_}{7.54\times {10}^{-3}\text{kg}\cdot {\text{m}}^2}$.

The mass moment of inertia with respect to $z$ coordinate axes is $\underset{\_}{3.47\times {10}^{-3}\text{kg}\cdot {\text{m}}^2}$.

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:

Divided the section into three geometric portions as shown below:

• Upper Flange
• Lower Flange
• Horizontal base

Find the mass of the upper flange component using the relation as shown below:

$\begin{array}{c}m=\rho V\\ =\rho tA\end{array}$ (1)

Here, V is volume of component of upper flange and A is the area of the section.

Find the area A of upper flange as shown below:

$A=bh$ (2)

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

Substitute $80\text{mm}$ for b and $100\text{mm}$ for h in Equation (2).

$\begin{array}{c}A=80\times 100\\ =8,000{\text{mm}}^{\text{2}}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\\ =0.008{\text{m}}^{\text{2}}\end{array}$

Substitute $2\text{mm}$ for t, $7,850\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, and $0.008{\text{m}}^{\text{2}}$ for A in Equation (1).

$\begin{array}{c}m=7,850\times 0.008\times 2\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\\ =7,850\times 0.008\times 0.002\\ =0.1256\text{kg}\end{array}$

Find the mass of the lower flange component using the relation as shown below:

$\begin{array}{c}m=\rho V\\ =\rho tA\end{array}$ (3)

Here, V is volume of component of lower flange.

Find the area A of lower flange as shown below:

$A=bh$ (4)

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

Substitute $80\text{mm}$ for b and $100\text{mm}$ for h in Equation (4).

$\begin{array}{c}A=80\times 100\\ =8,000{\text{mm}}^{\text{2}}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\\ =0.008{\text{m}}^{\text{2}}\end{array}$

Substitute $2\text{mm}$ for t, $7,850\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, and $0.008{\text{m}}^{\text{2}}$ for A in Equation (3).

$\begin{array}{c}m=7,850\times 0.008\times 2\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\\ =7,850\times 0.008\times 0.002\\ =0.1256\text{kg}\end{array}$

Find the moment of inertia about x axis of upper lower flange section as shown below:

$\begin{array}{c}{I}_x=I_x+m{d}^2\\ =\frac{1}{12}\times m{d}^2+m\left[b^2+h^2\right]\end{array}$ (5)

Substitute $0.1256\text{kg}$ for m, $0.1\text{m}$ for b, $\text{0}\text{.08}\text{m}$ for d, and $0.04\text{m}$ for h in Equation (5).

$\begin{array}{c}{I}_x=\frac{1}{12}\times 0.12560\times {0.08}^2+0.12560\left[{0.1}^2+{0.04}^2\right]\\ =6.6986\times {10}^{-5}+0.001456\\ =1.52395\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Find the moment of inertia about y axis of upper lower flange section as shown below:

$\begin{array}{c}{I}_y=I_y+m{d}^2\\ =\frac{1}{12}\times m{d}^2+m\left[b^2+h^2\right]\end{array}$ (6)

Substitute $0.1256\text{kg}$ for m, $0.1\text{m}$ for d, $0.05\text{m}$ for b , and $0.1\text{m}$ for h in Equation (6).

$\begin{array}{c}{I}_y=\frac{1}{12}\times 0.12560\times {0.1}^2+0.12560\left[{0.05}^2+{0.1}^2\right]\\ =1.04666\times {10}^{-4}+0.00157\\ =1.67466\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Find the moment of inertia about z axis of upper lower flange section as shown below:

$\begin{array}{c}{I}_z=I_z+m{d}^2\\ =\frac{1}{12}\times m\left[b^2+h^2\right]+m\left[b_1^2+h_1^2\right]\end{array}$ (7)

Substitute $0.1256\text{kg}$ for m, $100\text{mm}$ for b, $80\text{mm}$ for h , $40\text{mm}$ for $h_1$, and $50\text{mm}$ for $b_1$ in Equation (7).

$\begin{array}{c}{I}_z=\left[\begin{array}{l}\frac{1}{12}\times 0.12560\times \left[100\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)+80\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right]\\ +0.12560\left[50\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)+40\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right]\end{array}\right]\\ =\frac{1}{12}\times 0.12560\times \left[{0.1}^2+{0.08}^2\right]+0.12560\left[{0.05}^2+{0.04}^2\right]\\ =1.716533\times {10}^{-4}+5.1496\times {10}^{-4}\\ =0.68661\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Find the mass of the horizontal base using the relation as shown below:

$\begin{array}{c}m=\rho V\\ =\rho tA\end{array}$ (8)

Find the area A of horizontal base as shown below:

$A=bh$ (9)

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

Substitute $200\text{mm}$ for b and $200\text{mm}$ for h in Equation (9).

$\begin{array}{c}A=200\times 200\\ =40,000{\text{mm}}^{\text{2}}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\\ =0.04{\text{m}}^{\text{2}}\end{array}$

Substitute $2\text{mm}$ for t, $7,850\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, and $0.04{\text{m}}^{\text{2}}$ for A in Equation (8).

$\begin{array}{c}m=7,850\times 0.008\times 2\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\\ =7,850\times 0.04\times 0.002\\ =0.628\text{kg}\end{array}$

Find the moment of inertia about x axis of horizontal base as shown in below:

$I_x=\frac{1}{12}m{a}^2$ (10)

Substitute $0.6280\text{kg}$ for m and $200\text{mm}$ for a in Equation (10).

$\begin{array}{c}{I}_x=\frac{1}{12}\times 0.1256\times 200\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\\ =\frac{1}{12}\times 0.1256\times 0.2\\ =2.0933\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Find the moment of inertia about y axis of horizontal base as shown in below:

$I_y=\frac{1}{12}m\left(a^2+b^2\right)$ (11)

Substitute $0.6280\text{kg}$ for m, $200\text{mm}$ for a, and $200\text{mm}$ for b, in Equation (11).

$\begin{array}{c}{I}_y=\frac{1}{12}\times 0.6280\times \left[200\text{mm}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2+\times 200\text{mm}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\right]\\ =\frac{1}{12}\times 0.1256\times \left[{0.2}^2+{0.2}^2\right]\\ =4.1867\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Find the moment of inertia about z axis of horizontal base as shown in below:

$I_z=\frac{1}{12}m{b}^2$ (12)

Substitute $0.6280\text{kg}$ for m and $200\text{mm}$ for b in Equation (12).

$\begin{array}{c}{I}_z=\frac{1}{12}\times 0.1256\times 200\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\\ =\frac{1}{12}\times 0.1256\times 0.2\\ =2.0933\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Find the total moment of inertia $I_x$ of composite shape as shown below:

$I_x=2{\left(I_x\right)}_{f\text{lange}}+{\left(I_x\right)}_{\text{base}}$ (13)

Here, ${\left(I_x\right)}_{f\text{lange}}$ is moment of inertia of flange section about x axis and ${\left(I_x\right)}_{\text{base}}$ is moment of inertia of base section x axis.

Substitute $1.52395\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for ${\left(I_x\right)}_{f\text{lange}}$ and $2.0933\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for ${\left(I_x\right)}_{\text{base}}$ in Equation (13).

$\begin{array}{c}{I}_x=2\left(1.52395\times {10}^{-3}\right)+2.0933\times {10}^{-3}\\ =3.0479\times {10}^{-3}+2.0933\times {10}^{-3}\\ =5.14\times {10}^{-3}\end{array}$

Thus, the mass moment of inertia with respect to $x$ coordinate axes is $\underset{\_}{5.14\times {10}^{-3}\text{kg}\cdot {\text{m}}^2}$.

Find the total moment of inertia $I_y$ of composite shape as shown below:

$I_y=2{\left(I_y\right)}_{f\text{lange}}+{\left(I_y\right)}_{\text{base}}$ (14)

Here, ${\left(I_x\right)}_{f\text{lange}}$ is moment of inertia of flange section about y axis and ${\left(I_x\right)}_{\text{base}}$ is moment of inertia of base about y axis section.

Substitute $1.67467\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for ${\left(I_x\right)}_{f\text{lange}}$ and $4.1867\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for ${\left(I_x\right)}_{\text{base}}$ in Equation (14).

$\begin{array}{c}{I}_y=2\left(1.67467\times {10}^{-3}\right)+\left(4.1867\times {10}^{-3}\right)\\ =3.34934\times {10}^{-3}+4.1867\times {10}^{-3}\\ =7.536\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Thus, the mass moment of inertia with respect to $y$ coordinate axes is $\underset{\_}{7.54\times {10}^{-3}\text{kg}\cdot {\text{m}}^2}$.

Find the total moment of inertia $I_z$ of composite shape as shown below:

$I_z=2{\left(I_z\right)}_{f\text{lange}}+{\left(I_z\right)}_{\text{base}}$ (15)

Here, ${\left(I_z\right)}_{f\text{lange}}$ is moment of inertia of flange section about z axis and ${\left(I_z\right)}_{\text{base}}$ is moment of inertia of base about z axis section.

Substitute $0.68661\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for ${\left(I_x\right)}_{f\text{lange}}$ and $2.0933\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for ${\left(I_x\right)}_{\text{base}}$ in Equation (15).

$\begin{array}{c}{I}_y=2\left(0.68661\times {10}^{-3}\right)+\left(2.0933\times {10}^{-3}\right)\\ =1.373\times {10}^{-3}+2.0933\times {10}^{-3}\\ =3.466\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Thus, the mass moment of inertia with respect to $z$ coordinate axes is $\underset{\_}{3.47\times {10}^{-3}\text{kg}\cdot {\text{m}}^2}$.