Chapter 9.5, Problem 9.139P

To determine

Find the mass moment of inertia of the anchor with respect to each of the coordinate axes.

Answer to Problem 9.139P

The mass moment of inertia $\left(I_x\right)$ of the anchor with respect to x axis is $\underset{\_}{344\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^2}$.

The mass moment of inertia $\left(I_y\right)$ of the anchor with respect to y axis is $\underset{\_}{132.1\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^2}$.

The mass moment of inertia $\left(I_z\right)$ of the anchor with respect to z axis is $\underset{\_}{453\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^2}$.

Explanation of Solution

Given information:

The thickness (t) of the anchor is $0.05\text{in}\text{.}$

The specific weight $\left(\gamma \right)$ of the galvanized steel is $470{\text{lb/ft}}^3$.

Assume the acceleration due to gravity (g) as $32.2{\text{ft/s}}^{\text{2}}$

Calculation:

Show the section 1, section 2 and section 3 of the framing anchor as in Figure 1.

Calculate the density $\left(\rho \right)$ of the galvanized steel using the relation:

$\rho =\frac{\gamma }{g}$

Substitute $470{\text{lb/ft}}^3$ for $\gamma$ and $32.2{\text{ft/s}}^{\text{2}}$ for g.

$\begin{array}{c}\rho =\frac{470}{32.2}\\ =14.596\text{lb}\cdot {\text{s}}^2/{\text{ft}}^4\end{array}$

Calculate the volume of the section 1 $\left(V_1\right)$ using the relation:

$\begin{array}{c}{V}_1=A_{1}t\\ =3.5\times 2.25\times t\end{array}$

Substitute $0.05\text{in}\text{.}$ for t.

$\begin{array}{c}{V}_1=3.5\times 2.25\times 0.05\\ =0.39375\text{in}{\text{.}}^3\times {\left(\frac{1\text{ft}}{12\text{in}\text{.}}\right)}^3\\ =2.2786\times {10}^{-4}{\text{ft}}^3\end{array}$

Calculate the volume of the section 2 $\left(V_2\right)$ as below:

$\begin{array}{c}{V}_2=A_{2}t\\ =1\times 2.25\times t\end{array}$

Substitute $0.05\text{in}\text{.}$ for t.

$\begin{array}{c}{V}_2=1\times 2.25\times 0.05\\ =0.1125\text{in}{\text{.}}^3\times {\left(\frac{1\text{ft}}{12\text{in}\text{.}}\right)}^3\\ =6.5104\times {10}^{-5}{\text{ft}}^3\end{array}$

Calculate the volume $\left(V_3\right)$ of the section 3 as below:

$\begin{array}{c}{V}_3=A_{3}t\\ =\frac{1}{2}\left(3.5+1.25\right)\times 2\times t\end{array}$

Substitute $0.05\text{in}\text{.}$ for t.

$\begin{array}{c}{V}_3=\frac{1}{2}\times \left(3.5+1.25\right)\times 2\times 0.05\\ =0.2375\text{in}{\text{.}}^3\times {\left(\frac{1\text{ft}}{12\text{in}\text{.}}\right)}^3\\ =1.3744\times {10}^{-4}{\text{ft}}^3\end{array}$

Calculate the mass of section 1 $\left(m_1\right)$ using the relation:

$m_1=\rho V_1$

Substitute $2.2786\times {10}^{-4}{\text{ft}}^3$ for $V_1$ and $14.596\text{lb}\cdot {\text{s}}^2/{\text{ft}}^4$ for $\rho$.

$\begin{array}{c}{m}_1=14.596\times 2.2786\times {10}^{-4}\\ =3325.845\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}\end{array}$

Calculate the mass of section 2 $\left(m_2\right)$ using the relation:

$m_2=\rho V_2$

Substitute $6.5104\times {10}^{-5}{\text{ft}}^3$ for $V_2$ and $14.596\text{lb}\cdot {\text{s}}^2/{\text{ft}}^4$ for $\rho$.

$\begin{array}{c}{m}_2=14.596\times 6.5104\times {10}^{-5}\\ =950.26\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}\end{array}$

Calculate the mass of section 3 $\left(m_3\right)$ using the relation:

$m_3=\rho V_3$

Substitute $1.3744\times {10}^{-4}{\text{ft}}^3$ for $V_3$ and $14.596\text{lb}\cdot {\text{s}}^2/{\text{ft}}^4$ for $\rho$.

$\begin{array}{c}{m}_3=14.596\times 1.3744\times {10}^{-4}\\ =2006.14\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}\end{array}$

Calculate the mass moment of inertia with respect to x axis for section 1 ${\left(I_x\right)}_1$ as below:

${\left(I_x\right)}_1=\frac{1}{12}\left(m_1\right){\left(3.5\right)}^2+\left(m_1\times {\left(\frac{3.5}{2}\right)}^2\right)$

Substitute $3325.845\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_1$.

$\begin{array}{c}{\left(I_x\right)}_1=\frac{1}{12}\left(3325.845\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}\right){\left(3.5\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2+\left(3325.845\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}\times {\left(\frac{\left(3.5\text{in}\text{.}\right)\times \frac{1\text{ft}}{12\text{in}\text{.}}}{2}\right)}^2\right)\\ =2.3577\times {10}^{-5}+7.073\times {10}^{-5}\\ =9.4307\times {10}^{-5}\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}\end{array}$

Calculate the mass moment of inertia with respect to x axis for  section 2 ${\left(I_x\right)}_2$ as below:

${\left(I_x\right)}_2=\frac{1}{12}\left(m_2\right){\left(1\right)}^2+\left(m_2\times \left({3.5}^2+{\left(\frac{1}{2}\right)}^2\right)\right)$

Substitute $950.26\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_2$.

$\begin{array}{c}{\left(I_x\right)}_2=\frac{1}{12}\left(950.26\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}\right){\left(1\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2+\left(\begin{array}{l}950.26\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}\times \\ \left({\left(3.5\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2+{\left(\frac{\left(1\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}{2}\right)}^2\right)\end{array}\right)\\ =5.4992\times {10}^{-7}+8.2488\times {10}^{-5}\\ =83.037\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}\end{array}$

Calculate the mass moment of inertia with respect to x axis for section 3 ${\left(I_x\right)}_3$ as below:

${\left(I_x\right)}_3=\frac{1}{18}\left(m_3\right)\left({\left(3.5+1.25\right)}^2+2^2\right)+\left(m_3\times \left({\left(\frac{2}{3}\left(3.5+1.25\right)\right)}^2+{\left(\frac{1}{3}\times 2\right)}^2\right)\right)$

Substitute $2006.14\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_3$.

$\begin{array}{c}{\left(I_x\right)}_3=\frac{1}{18}\left(2006.14\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}\right)\left({\left(\left(3.5+1.25\right)\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2+{\left(2\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2\right)+\\ +\left(2006.14\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}\times \left({\left(\frac{2}{3}\left(\left(3.5+1.25\right)\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\right)}^2+{\left(\frac{1}{3}\times 2\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2\right)\right)\\ =2.0558\times {10}^{-5}+1.4589\times {10}^{-4}\\ =166.452\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}\end{array}$

Calculate the mass moment of inertia with respect to x axis for anchor $\left(I_x\right)$ using the parallel axis theorem as below:

$I_x={\left(I_x\right)}_1+{\left(I_x\right)}_2+{\left(I_x\right)}_3$

Substitute $9.4307\times {10}^{-5}\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}$ for ${\left(I_x\right)}_1$, $83.037\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}$ for ${\left(I_x\right)}_2$ and $166.452\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}$ for ${\left(I_x\right)}_3$.

$\begin{array}{c}{I}_x=9.4307\times {10}^{-5}+83.037\times {10}^{-6}+166.452\times {10}^{-6}\\ =3.437\times {10}^{-4}\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}\\ \approx 344\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}\end{array}$

Calculate the mass moment of inertia with respect to y axis for section 1 ${\left(I_y\right)}_1$ as below:

${\left(I_y\right)}_1=\frac{1}{12}{m}_1{\left(2.25\right)}^2+m_1{\left(\frac{2.25}{2}\right)}^2$

Substitute $3325.845\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_1$.

$\begin{array}{c}{\left(I_y\right)}_1=\left\{\begin{array}{c}\frac{1}{12}\times 3325.845\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}\times {\left(2.25\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2\\ +3325.845\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}\times {\left(\frac{\left(2.25\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}{2}\right)}^2\end{array}\right\}\\ =9.7436\times {10}^{-6}+2.923\times {10}^{-5}\\ =38.975\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}\end{array}$

Calculate the mass moment of inertia with respect to y axis for section 2 ${\left(I_y\right)}_2$ as below:

${\left(I_y\right)}_2=\frac{1}{12}{m}_2\left({\left(2.25\right)}^2+{\left(1\right)}^2\right)+m_2\left[{\left(\frac{2.25}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2\right]$

Substitute $950.26\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_2$.

$\begin{array}{c}{\left(I_y\right)}_2=\left\{\begin{array}{l}\frac{1}{12}\left(950.26\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}\right)\left({\left(2.25\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2+{\left(1\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2\right)\\ +950.26\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}\left[{\left(\frac{\left(2.25\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}{2}\right)}^2+{\left(\frac{1\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}}{2}\right)}^2\right]\end{array}\right\}\\ =3.334\times {10}^{-6}+1.00017\times {10}^{-5}\\ =13.336\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}\end{array}$

Calculate the mass moment of inertia with respect to y axis for section 3 ${\left(I_y\right)}_3$ as below:

${\left(I_y\right)}_3=\frac{1}{18}{m}_3{\left(2\right)}^2+m_3\left[{2.25}^2+{\left(\frac{1}{3}2\right)}^2\right]$

Substitute $2006.14\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_3$.

$\begin{array}{c}{\left(I_y\right)}_3=\left\{\begin{array}{l}\frac{1}{18}\times 2006.14\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}{\left(2\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2\\ +2006.14\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}\left[{\left(2.25\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2+{\left(\frac{1}{3}\times \left(2\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\right)}^2\right]\end{array}\right\}\\ =3.096\times {10}^{-6}+7.672\times {10}^{-5}\\ =79.816\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}\end{array}$

Calculate the mass moment of inertia with respect to y axis for anchor $\left(I_y\right)$ using the parallel axis theorem as below:

$I_y={\left(I_y\right)}_1+{\left(I_y\right)}_2+{\left(I_y\right)}_3$

Substitute $38.975\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}$ for ${\left(I_y\right)}_1$ $13.336\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}$ for ${\left(I_y\right)}_2$ and $79.816\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}$ for ${\left(I_y\right)}_3$.

$\begin{array}{c}{I}_y=38.975\times {10}^{-6}+13.336\times {10}^{-6}+79.816\times {10}^{-6}\\ =132.127\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}\\ \approx 132.1\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}\end{array}$

Calculate the mass moment of inertia with respect to z axis for section 1 ${\left(I_z\right)}_1$ as below:

${\left(I_z\right)}_1=\frac{1}{12}{m}_1\left[{\left(2.25\right)}^2+{\left(3.5\right)}^2\right]+m_1\left[{\left(\frac{2.25}{2}\right)}^2+{\left(\frac{3.5}{2}\right)}^2\right]$

Substitute $3325.845\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_1$.

$\begin{array}{c}{\left(I_z\right)}_1=\left\{\begin{array}{l}\frac{1}{12}3325.845\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}\left[{\left(2.25\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2+{\left(3.5\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2\right]\\ +3325.845\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}\left[{\left(\frac{2.25\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}}{2}\right)}^2+{\left(\frac{3.5\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}}{2}\right)}^2\right]\end{array}\right\}\\ =3.332\times {10}^{-5}+9.996\times {10}^{-5}\\ =133.283\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}\end{array}$

Calculate the mass moment of inertia with respect to z axis for section 2 ${\left(I_x\right)}_2$ as below:

${\left(I_z\right)}_2=\frac{1}{12}{m}_2{\left(2.25\right)}^2+m_2\left[{\left(\frac{2.25}{2}\right)}^2+{\left(3.5\right)}^2\right]$

Substitute $950.26\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_2$.

$\begin{array}{c}{\left(I_z\right)}_2=\left\{\begin{array}{l}\frac{1}{12}\times 950.26\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}\times {\left(2.25\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2\\ +950.26\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}\times \left[{\left(\frac{2.25\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}}{2}\right)}^2+{\left(3.5\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2\right]\end{array}\right\}\\ =2.784\times {10}^{-6}+8.918\times {10}^{-5}\\ =91.974\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}\end{array}$

Calculate the mass moment of inertia with respect to z axis for section 3 ${\left(I_y\right)}_3$ as below:

${\left(I_z\right)}_3=\frac{1}{18}{m}_3{\left(3.5+1.25\right)}^2+m_3\left[{\left(2.25\right)}^2+{\left(\frac{2}{3}\left(3.5+1.25\right)\right)}^2\right]$

Substitute $2006.14\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_3$.

$\begin{array}{c}{\left(I_z\right)}_3=\left\{\begin{array}{l}\frac{1}{18}\times 2006.14\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}{\left(\left(3.5+1.25\right)\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2\\ +2006.14\times {10}^{-6}\text{lb}\cdot {\text{s}}^2/\text{ft}\times \left[{\left(2.25\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2+{\left(\frac{2}{3}\left(\left(3.5+1.25\right)\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\right)}^2\right]\end{array}\right\}\\ =1.746\times {10}^{-6}+2.102\times {10}^{-4}\\ =227.694\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}\end{array}$

Calculate the mass moment of inertia with respect to z axis for anchor $\left(I_x\right)$ using the parallel axis theorem as below:

$I_z={\left(I_z\right)}_1+{\left(I_z\right)}_2+{\left(I_z\right)}_3$

Substitute $133.283\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}$ for ${\left(I_z\right)}_1$ $91.974\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}$ for ${\left(I_z\right)}_2$ and $227.694\times {10}^{-5}\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}$ for ${\left(I_z\right)}_3$.

$\begin{array}{c}{I}_z=133.283\times {10}^{-6}+91.974\times {10}^{-6}+\mathrm{2.27.694}\times {10}^{-6}\\ =452.951\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}\\ \approx 453\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}\end{array}$

Therefore, the mass moment of inertia $\left(I_x\right)$ of the anchor with respect to x axis is $\underset{\_}{344\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^2}$.

Therefore, the mass moment of inertia $\left(I_y\right)$ of the anchor with respect to y axis is $\underset{\_}{132.1\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^2}$.

Therefore, the mass moment of inertia $\left(I_z\right)$ of the anchor with respect to z axis is $\underset{\_}{453\times {10}^{-6}\text{lb}\cdot \text{ft}\cdot {\text{s}}^2}$.