Chapter 9.5, Problem 9.145P

Determine the mass moment of inertia of the steel fixture shown with respect to (a) the x axis, (b) the y axis, (c) the z axis. (The density of steel is 7850 kg/m³.)

Fig. P9.145

To determine

Find the mass moment of inertia with respect to x axis.

Answer to Problem 9.145P

The mass moment of inertia with respect to x axis is $\underset{\_}{26.4\times {10}^{-3}\text{kg}\cdot {\text{m}}^2}$.

Explanation of Solution

Given information:

The density $\left({\rho }_{\text{ST}}\right)$ of steel is $7,850\text{kg}/{\text{m}}^{\text{3}}$.

Calculation:

Sketch the section of steel fixture as shown in Figure 1.

Find the mass $\left(m_1\right)$ of component 1 as shown below:

$\left(m_1\right)=\rho V$ (1)

Here, V is volume of rectangular section 1.

Modify Equation (1).

$\left(m_1\right)=\rho \left(wlt\right)$ (2)

Here, w is the width of the section 1, l is the length of the rectangular section 1, and t is the thickness of the section 1.

Substitute $80\text{mm}$ for w, $50\text{mm}$ for l, and $16\text{mm}$ for t in Equation (2).

$\begin{array}{c}\left(m_1\right)=7,850\times \left(80\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\times 50\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\times \text{16}\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)\\ =7,850\times 0.08\times 0.05\times 0.16\\ =5.021\text{kg}\end{array}$

Find the mass $\left(m_2\right)$ of component 2 as shown below:

$\left(m_2\right)=\rho V$ (3)

Here, V is volume of rectangular section 2.

Modify Equation (3).

$\left(m_2\right)=\rho \left(wlt\right)$ (4)

Here, w is the width of the section 2, l is the length of the rectangular section 2, and t is the thickness of the section 2.

Substitute $80\text{mm}$ for w, $38\text{mm}$ for l, and $70\text{mm}$ for t in Equation (4).

$\begin{array}{c}\left(m_2\right)=7,850\times \left[80\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\times 38\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\times \text{70}\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right]\\ =7,850\times \left(0.08\times 0.038\times 0.07\right)\\ =1.67048\text{kg}\end{array}$

Find the mass $\left(m_3\right)$ of component 3 as shown below:

$\begin{array}{c}\left(m_3\right)=\rho V\\ =\rho \left(\frac{\pi }{2}\times r^{2}h\right)\end{array}$ (5)

Substitute $24\text{mm}$ for r and $40\text{mm}$ for h in Equation (5).

$\begin{array}{c}\left(m_3\right)=7,850\times \left(\frac{\pi }{2}\times 24\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\times 40\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)\\ =7,850\times \left(\frac{\pi }{2}\times {0.024}^2\times 0.04\right)\\ =7,850\times 3.619\times {10}^{-5}\\ =0.28410\text{kg}\end{array}$

Refer to Figure 9.28, “Mass moment of inertia for common geometric shapes” in the text book.

Find the mass moment of inertia with respect to x axis as shown below:

$\begin{array}{c}{I}_x={\left(I_x\right)}_1-{\left(I_x\right)}_2-{\left(I_x\right)}_3\\ =\left[\begin{array}{l}\frac{1}{12}\times m_1\left[b_1^2+h_1^2\right]+m_1\left[{\left(\frac{b}{2}\right)}^2+{\left(\frac{h}{2}\right)}^2\right]-\\ \frac{1}{12}\left(m_2\left[b_2^2+h_2^2\right]\right)+m_2\left[{\left(b_1^2-\frac{b_2}{2}\right)}^2+\left({\left(b_1^2+\frac{h_2}{2}\right)}^2\right)\right]\\ -m_3\left[\left(\frac{1}{4}-\frac{16}{9{\pi }^2}\right)r^2+\frac{1}{12}{h}_3^2+m_3\left[b_1-\frac{4r}{3\pi }\right]+{\left(h_1-\frac{h_3}{2}\right)}^2\right]\end{array}\right]\end{array}$ (6)

Convert the dimension in mm to m.

$\begin{array}{c}{b}_1=50\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\\ \text{=0}\text{.05}\text{m}\end{array}$

Similarly calculate the remaining values.

Substitute $0.28410\text{kg}$ for $m_3$, $1.67048\text{kg}$ for $m_2$, $5.021\text{kg}$ for $m_1$, $\text{0}\text{.05}\text{m}$ for $b_1$, $\text{0}\text{.16}\text{m}$ for $h_1$, $\text{0}\text{.038}\text{m}$ for $b_2$, $\text{0}\text{.07}\text{m}$ for $h_2$, $0.024\text{m}$ for r, $0.04\text{m}$ for $h_3$ in Equation (6).

$\begin{array}{c}{I}_x=\left[\begin{array}{l}\frac{1}{12}\times 5.02400\times \left({0.05}^2\right)+{\left(0.16\right)}^2+5.02400\left[{\left(\frac{0.05}{2}\right)}^2+{\left(\frac{0.16}{2}\right)}^2\right]\\ -\frac{1}{12}\times 1.67048\left[{0.038}^2+{0.07}^2\right]+1.67048\left[\begin{array}{l}{\left(0.05-\frac{0.038}{2}\right)}^2+\\ {\left(0.05-\frac{0.07}{2}\right)}^2\end{array}\right]\\ -0.28410\left[\left(\frac{1}{4}-\frac{16}{9{\pi }^2}\right){0.024}^2+\frac{1}{12}\left({0.04}^2\right)\right]+0.28410\left[\begin{array}{l}{\left(0.05-\frac{4\times 0.024}{3\pi }\right)}^2+\\ {\left(0.16-\frac{0.04}{2}\right)}^2\end{array}\right]\end{array}\right]\\ =\left[11.76456+35.2936-0.8831+13.6745-0.0493+6.0187\right]\\ =26.4325\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Thus, the mass moment of inertia with respect to x axis is $\underset{\_}{26.4\times {10}^{-3}\text{kg}\cdot {\text{m}}^2}$.

To determine

Find the mass moment of inertia with respect to y axis.

Answer to Problem 9.145P

The mass moment of inertia with respect to y axis is $\underset{\_}{31.2\times {10}^{-3}\text{kg}\cdot {\text{m}}^2}$.

Explanation of Solution

Calculation:

Find the mass moment of inertia with respect to y axis as shown below:

$\begin{array}{c}{I}_y={\left(I_y\right)}_1-{\left(I_y\right)}_2-{\left(I_y\right)}_3\\ =\left[\begin{array}{l}\frac{1}{12}\times m_1\left[b_1^2+h_1^2\right]+m_1\left[{\left(\frac{b_1}{2}\right)}^2+{\left(\frac{h_1}{2}\right)}^2\right]-\\ \frac{1}{12}\left(m_2\left[b_2^2+h_2^2\right]\right)+m_2\left[{\left(\frac{b_2}{2}\right)}^2+\left({\left(b+\frac{h_2}{2}\right)}^2\right)\right]\\ -\frac{1}{12}{m}_3\left[3r^3+h_3^2\right]+m_3\left[{\left(\frac{b_3}{2}\right)}^2+{\left(h_1+\frac{h_3}{2}\right)}^2\right]\end{array}\right]\end{array}$ (7)

Convert the dimensions in mm to m.

$\begin{array}{c}{b}_1=50\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\\ \text{=0}\text{.05}\text{m}\end{array}$

Substitute $0.28410\text{kg}$ for $m_3$, $1.67048\text{kg}$ for $m_2$, $5.021\text{kg}$ for $m_1$, $\text{0}\text{.08m}$ for $b_1$, $\text{0}\text{.16}\text{m}$ for $h_1$, $\text{0}\text{.08}\text{m}$ for $b_2$, $\text{0}\text{.07}\text{m}$ for $h_2$, $0.024\text{m}$ for r, $0.084\text{m}$ for $b_3$, $\text{0}\text{.05}\text{m}$ for b, and $0.04\text{m}$ for $h_3$ in Equation (7).

$\begin{array}{c}{I}_y=\left[\begin{array}{l}\frac{1}{12}\times 5.02400\left[{0.08}^2+{0.16}^2\right]+5.02400\times \left[{\left(\frac{0.08}{2}\right)}^2+{\left(\frac{0.16}{2}\right)}^2\right]-\\ \frac{1}{12}\times 1.67048\left[{0.08}^2+{0.07}^2\right]+1.67048\times \left[{\left(\frac{0.08}{2}\right)}^2+{\left(0.05+\frac{0.07}{2}\right)}^2\right]\\ -\frac{1}{12}\times 0.28410\left[3\left({0.024}^2+{0.04}^2\right)+0.28410\left[{\left(\frac{0.08}{2}\right)}^2+{\left(0.16+\frac{0.04}{2}\right)}^2\right]\right]\end{array}\right]\\ =\left[13.3973+40.1920-1.5730+14.7420-0.0788+6.0229\right]\\ =31.1726\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

To determine

Find the mass moment of inertia with respect to z axis.

Answer to Problem 9.145P

The mass moment of inertia with respect to z axis is $\underset{\_}{8.58\times {10}^{-3}\text{kg}\cdot {\text{m}}^2}$.

Explanation of Solution

Calculation:

Find the mass moment of inertia with respect to z axis as shown below:

$\begin{array}{c}{I}_z={\left(I_z\right)}_1-{\left(I_z\right)}_2-{\left(I_z\right)}_3\\ =\left[\begin{array}{l}\frac{1}{12}\times m_1\left[b_1^2+h_1^2\right]+m_1\left[{\left(\frac{b_1}{2}\right)}^2+{\left(\frac{h_1}{2}\right)}^2\right]-\\ \frac{1}{12}\left(m_2\left[b_2^2+h_2^2\right]\right)+m_2\left[{\left(\frac{b_2}{2}\right)}^2+\left({\left(b-\frac{h_2}{2}\right)}^2\right)\right]\\ -m_3\left(\frac{1}{2}-\frac{16}{9{\pi }^2}\right)r^2+m_3\left[{\left(\frac{b_3}{2}\right)}^2+{\left(h_1-\frac{4r}{3\pi }\right)}^2\right]\end{array}\right]\end{array}$ (8)

Convert the dimensions in mm to m.

$\begin{array}{c}{b}_1=80\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\\ \text{=0}\text{.08}\text{m}\end{array}$

Substitute $0.28410\text{kg}$ for $m_3$, $1.67048\text{kg}$ for $m_2$, $5.021\text{kg}$ for $m_1$, $\text{0}\text{.08m}$ for $b_1$, $\text{0}\text{.05}\text{m}$ for $h_1$, $\text{0}\text{.08}\text{m}$ for $b_2$, $\text{0}\text{.38}\text{m}$ for $h_2$, $0.024\text{m}$ for r, $0.08\text{m}$ for $b_3$, and $\text{0}\text{.05}\text{m}$ for b, in Equation (8).

$\begin{array}{c}{I}_z=\left[\begin{array}{l}\frac{1}{12}\times 5.0240\left[{0.08}^2+{0.05}^2\right]+5.02400\left[{\left(\frac{0.08}{2}\right)}^2+{\left(\frac{0.05}{2}\right)}^2\right]-\\ \frac{1}{12}\times 1.67048\times \left[{0.08}^2+{0.38}^2\right]+1.67048\left[{\left(\frac{0.08}{2}\right)}^2+{\left(0.05-\frac{0.038}{2}\right)}^2\right]-\\ \left(-0.28410\left(\frac{1}{2}-\frac{16}{9{\pi }^2}\right)\times {0.024}^2+0.28410\left[{\left(\frac{0.08}{2}\right)}^2+{\left(0.05-\frac{4\times 0.024}{3\pi }\right)}^2\right]\right)\end{array}\right]\\ =3.7261+11.1784-1.0919+4.2781-\left(0.0523+0.9049\right)\\ =8.5773\times {10}^{-3}\\ =8.58\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$