Chapter 19.5, Problem 19.143P

To determine

(a)

The combined spring constant.

Answer to Problem 19.143P

The combined spring constant is $147.11\text{kip}/\text{ft}$.

Explanation of Solution

Given information:

The weight of two counter rotating eccentric mass exciters is $14\text{oz}$, the radius of rotating mass is $6\text{in}$, the rotational speed of rotating mass is $1200\text{rpm}$ and the amplitude of the motion is $0.6\text{in}$ and the total weight of the system is $300\text{lb}$.

Write the expression of mass of imbalanced weight.

$m=\frac{w}{g}$ ..... (I)

Here, the imbalanced mass is the acceleration due to gravity is $g$ and the imbalanced weight is $w$.

Write the expression of mass of the system.

$M=\frac{W}{g}$ ...... (II)

Here, the mass of the system is $M$ and the total weight of the system is $W$.

Write the expression of the frequency of periodic force.

${\omega }_{\text{f}}=\frac{2\pi N}{60}$ ...... (III)

Here, the rotational speed is $N$ and the frequency of periodic force is ${\omega }_{\text{f}}$.

Write the expression of phase difference between the amplitude of vibration and periodic force.

$\varphi ={\tan }^{-1}\left(\frac{c{\omega }_{\text{f}}}{k-M{\omega }_{\text{f}}^2}\right)$ ...... (IV)

Here, the damping coefficient is $c$, the spring constant is $k$ and the phase difference is $\varphi$.

Calculation:

Substitute $14\text{oz}$ for $w$, $32.185\text{ft}/{\text{s}}^2$ for $g$ in Equation (I):

$\begin{array}{c}m=\frac{14\text{oz}\left(\frac{1\text{lb}}{16\text{oz}}\right)}{32.185\text{ft}/{\text{s}}^2}\\ =0.02717\text{lb}\cdot {\text{s}}^2/\text{ft}\end{array}$

Substitute $300\text{lb}$ for $W$ and $32.185\text{ft}/{\text{s}}^2$ for $g$ in Equation (II):

$\begin{array}{c}M=\frac{300\text{lb}}{32.185\text{ft}/{\text{s}}^2}\\ =9.3167\text{lb}\cdot {\text{s}}^2/\text{ft}\end{array}$

Substitute $1200\text{rpm}$ for ${\omega }_{\text{f}}$ in Equation (III):

$\begin{array}{c}{\omega }_{\text{f}}=\frac{2\pi \left(1200\text{rpm}\right)}{60}\\ =0.1047\left(1200\text{rpm}\right)\\ =125.66\text{rad}/\text{s}\end{array}$

Since, the phase difference between the amplitude of vibration and periodic force is $\frac{\pi }{2}$.

Hence, $k-M{\omega }_{\text{f}}^2=0$ ...... (V)

Substitute $9.3167\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $M$, $125.66\text{rad}/\text{s}$ for ${\omega }_{\text{f}}$ in Equation (V):

$\begin{array}{l}k-\left(9.3167\text{lb}\cdot {\text{s}}^2/\text{ft}\right){\left(125.66\text{rad}/\text{s}\right)}^2=0\\ k=147114.7514\text{lb}/\text{ft}\\ k=147114.7514\text{lb}/\text{ft}\left(\frac{1\text{kip}/\text{ft}}{{10}^3\text{lb}/\text{ft}}\right)\\ k=147.11\text{kip}/\text{ft}\end{array}$

Conclusion:

The combined spring constant is $147.11\text{kip}/\text{ft}$.

To determine

(b)

The damping factor.

Answer to Problem 19.143P

The damping factor is $0.029$.

Explanation of Solution

Given information:

Write the expression of amplitude of steady state response of the system.

$X_{\text{m}}=\frac{P_{\text{m}}}{\sqrt{{\left[k-{\left(M{\omega }_{\text{f}}^2\right)}^2\right]}^2+{\left(c{\omega }_{\text{f}}\right)}^2}}$ ...... (VI)

Here, the frequency of periodic force is ${\omega }_{\text{f}}$, the damping coefficient is $c$, the amplitude of applied force is $P_{\text{m}}$, the spring constant is $k$, the amplitude is $X_{\text{m}}$.

Write the expression for the amplitude of applied force.

$P_{\text{m}}=2mr{\omega }_{\text{f}}^2$ ...... (VII)

Here, the radius of rotating mass is $r$.

Write the expression of critical damping coefficient.

$c_{\text{c}}=2\sqrt{kM}$ ....... (VIII)

Here, the critical damping coefficient is $c_{\text{c}}$.

Write the expression of damping factor.

$\zeta =\frac{c}{c_{\text{c}}}$ ...... (IX)

Here, the damping factor is $\zeta$.

Calculation:

Substitute $0.02717\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m$, $6\text{in}$ for $r$, and $125.66\text{rad}/\text{s}$ for ${\omega }_{\text{f}}$ in Equation (VII):

$\begin{array}{c}{P}_{\text{m}}=2\left(0.02717\text{lb}\cdot {\text{s}}^2/\text{ft}\right)\left(6\text{in}\right){\left(125.66\text{rad}/\text{s}\right)}^2\\ =0.05434\text{lb}\cdot {\text{s}}^2/\text{ft}\left(\left(6\text{in}\right)\left(\frac{1\text{ft}}{12\text{in}}\right)\right)\left(15790.4356\right)\\ =429.026\text{lb}\end{array}$

Substitute $429.026\text{lb}$ for $P_{\text{m}}$, $0.6\text{in}$ for $X_{\text{m}}$, $0$ for $k-M{\omega }_{\text{f}}^2$, $125.66\text{rad}/\text{s}$ for ${\omega }_{\text{f}}$ in Equation (VI):

$\begin{array}{l}0.6\text{in}=\frac{429.026\text{lb}}{\sqrt{{\left[0\right]}^2+{\left(c\left(125.66\text{rad}/\text{s}\right)\right)}^2}}\\ 0.6\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)=\frac{429.026\text{lb}}{\sqrt{{\left[0\right]}^2+{\left(c\left(125.66\text{rad}/\text{s}\right)\right)}^2}}\\ 0.6\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)=\frac{429.026\text{lb}}{125.66\text{rad}/\text{s}\left(c\right)}\\ 0.05\text{ft}=\frac{429.026\text{lb}}{125.66\text{rad}/\text{s}\left(c\right)}\end{array}$

$\begin{array}{l}\left(c\right)=\frac{429.026\text{lb}}{\left(0.05\text{ft}\right)125.66\text{rad}/\text{s}}\\ c=68.2836\text{lb}\cdot \text{s}/\text{ft}\end{array}$

Substitute $147.11\text{kip}/\text{ft}$ for $k$, $9.3167\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $M$ in Equation (VIII):

$\begin{array}{c}{c}_{\text{c}}=2\sqrt{\left(147.11\text{kip}/\text{ft}\right)\left(9.3167\text{lb}\cdot {\text{s}}^2/\text{ft}\right)}\\ =2\sqrt{\left(\left(147.11\text{kip}/\text{ft}\right)\left(\frac{{10}^3\text{ft}}{1\text{kip}}\right)\right)\left(9.3167\text{lb}\cdot {\text{s}}^2/\text{ft}\right)}\\ =2\sqrt{1370579.737{\text{lb}}^2\cdot {\text{s}}^2/{\text{ft}}^2}\\ =2341.43\text{lb}\cdot \text{s}/\text{ft}\end{array}$

Substitute $2341.43\text{lb}\cdot \text{s}/\text{ft}$ for $c_{\text{c}}$ and $68.2836\text{lb}\cdot \text{s}/\text{ft}$ for $c$ in Equation (IX):

$\begin{array}{c}\zeta =\frac{68.2836\text{lb}\cdot \text{s}/\text{ft}}{2341.43\text{lb}\cdot \text{s}/\text{ft}}\\ =0.029\end{array}$

Conclusion:

The damping factor is $0.029$.