Chapter 19, Problem 19.166RP

To determine

(a)

The amplitude of the fluctuating force transmitted to the foundation.

Answer to Problem 19.166RP

The amplitude of the fluctuating force transmitted to the foundation is $7.09\times 10^{-6}\text{m}$.

Explanation of Solution

Given information:

The mass of the motor is $400\text{kg}$, the constant of the spring is $150\text{kN}/\text{m}$, the damping coefficient of the dashpot is $6500\text{N}\cdot \text{s}/\text{m}$, mass of the unbalance of the rotor is $23\text{g}$, the distance of the mass is $100\text{mm}$, and the speed of the rotor is $800\text{rpm}$.

Expression for unbalanced force.

$P=mr{\omega }^2$ ...... (I)

Here, the mass of the unbalance of the rotor is $m$, distance of this mass is $r$, and the speed of the rotor is $\omega$.

Expression for the maximum amplitude of the vertical motion.

$x_{\text{m}}=\frac{P}{\sqrt{\left(4k-M{\omega }^2\right)+{\left(c\omega \right)}^2}}$ ...... (II)

Calculation:

Substitute $23\text{g}$ for $m$, $100\text{mm}$ for $r$, and $800\text{rpm}$ for $\omega$ in Equation (I).

$\begin{array}{c}P=23\text{g}\times 100\text{mm}\times {\left(800\text{rpm}\right)}^2\\ =23\text{g}\times \frac{1\text{kg}}{1000\text{g}}\times 100\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\times {\left(80\text{rpm}\times \frac{0.105\text{rad}/\text{s}}{1\text{rpm}}\right)}^2\\ =16.23\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\times \frac{1\text{N}}{\text{1}\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\\ =16.23\text{N}\end{array}$

Substitute $16.23\text{N}$, for $P$, $150\text{kN}/\text{m}$ for $k$, $400\text{kg}$ for $M$, $6500\text{N}\cdot \text{s}/\text{m}$ for $c$, and $800\text{rpm}$ for $\omega$ in Equation (II).

$\begin{array}{c}{x}_{\text{m}}=\frac{16.23\text{N}}{\sqrt{\left(4\times 150\text{kN}/\text{m}-400\text{kg}\times {\left(800\text{rpm}\right)}^2\right)+{\left(6500\text{N}\cdot \text{s}/\text{m}\times 800\text{rpm}\right)}^2}}\\ =\frac{16.23\text{N}\times \frac{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{1\text{N}}}{\sqrt{\begin{array}{l}{\left[4\times 150\text{kN}/\text{m}\times \frac{1000\text{kg}/{\text{s}}^{\text{2}}}{\text{1}\text{kN}/\text{m}}-400\text{kg}\times {\left(800\text{rpm}\times \frac{0.105\text{rad}/\text{s}}{1\text{rpm}}\right)}^2\right]}^2+\\ {\left(6500\text{N}\cdot \text{s}/\text{m}\times \frac{\text{1}\text{kg}/\text{s}}{\text{1}\text{N}\cdot \text{s}/\text{m}}\times 800\text{rpm}\times \frac{0.105\text{rad}/\text{s}}{1\text{rpm}}\right)}^2\end{array}}}\\ =\frac{16.23\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{\sqrt{{\left(-2222400\text{kg}/{\text{s}}^{\text{2}}\right)}^2+{\left(546000\text{kg}/{\text{s}}^2\right)}^2}}\\ =7.09\times 10^{-6}\text{m}\end{array}$

Expression for the amplitude of the fluctuating force transmitted to the foundation.

$F_{\text{m}}=\sqrt{{\left(4k{x}_{\text{m}}\right)}^2+{\left(c\omega x_{\text{m}}\right)}^2}$ ...... (III)

Substitute $150\text{kN}/\text{m}$ for $k$, $6500\text{N}\cdot \text{s}/\text{m}$ for $c$, $7.09\times 10^{-6}\text{m}$ for $x_{\text{m}}$ and $800\text{rpm}$ for $\omega$ in Equation (III).

$\begin{array}{c}{F}_{\text{m}}=\sqrt{{\left(4\times 150\text{kN}/\text{m}\times 7.09\times 10^{-6}\text{m}\right)}^2+{\left(6500\text{N}\cdot \text{s}/\text{m}\times 800\text{rpm}\times 7.09\times 10^{-6}\text{m}\right)}^2}\\ =\sqrt{\begin{array}{l}{\left(4\times 150\text{kN}/\text{m}\times \frac{1000\text{N}/\text{m}}{\text{1}\text{kN}/\text{m}}\times 7.09\times 10^{-6}\text{m}\right)}^2+\\ {\left(6500\text{N}\cdot \text{s}/\text{m}\times 800\text{rpm}\times \frac{0.105\text{rad}/\text{s}}{1\text{rpm}}\times 7.09\times 10^{-6}\text{m}\right)}^2\end{array}}\\ =\sqrt{{4.25}^2{\text{N}}^2+{3.87}^2{\text{N}}^2}\\ =5.75\text{N}\end{array}$

Conclusion:

The amplitude of the fluctuating force transmitted to the foundation is $5.75\text{N}$.

To determine

(b)

The amplitude of the vertical motion of the rotor.

Answer to Problem 19.166RP

The amplitude of the vertical motion of the rotor is $7.09\times 10^{-6}\text{m}$.

Explanation of Solution

Given information:

Given information:

The mass of the motor is $400\text{kg}$, the constant of the spring is $150\text{kN}/\text{m}$, the damping coefficient of the dashpot is $6500\text{N}\cdot \text{s}/\text{m}$, mass of the unbalance of the rotor is $23\text{g}$, the distance of the mass is $100\text{mm}$, and the speed of the rotor is $800\text{rpm}$.

Expression for unbalanced force.

$P=mr{\omega }^2$ ...... (I)

Here, the mass of the unbalance of the rotor is $m$, distance of this mass is $r$, and the speed of the rotor is $\omega$.

Expression for the amplitude of the fluctuating force.

$x_{\text{m}}=\frac{P}{\sqrt{\left(4k-M{\omega }^2\right)+{\left(c\omega \right)}^2}}$

Calculation:

Substitute $23\text{g}$ for $m$, $100\text{mm}$ for $r$, and $800\text{rpm}$ for $\omega$ in Equation (I).

$\begin{array}{c}P=23\text{g}\times 100\text{mm}\times {\left(800\text{rpm}\right)}^2\\ =23\text{g}\times \frac{1\text{kg}}{1000\text{g}}\times 100\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\times {\left(80\text{rpm}\times \frac{0.105\text{rad}/\text{s}}{1\text{rpm}}\right)}^2\\ =16.23\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\times \frac{1\text{N}}{\text{1}\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\\ =16.23\text{N}\end{array}$

Substitute $16.23\text{N}$, for $P$, $150\text{kN}/\text{m}$ for $k$, $400\text{kg}$ for $M$, $6500\text{N}\cdot \text{s}/\text{m}$ for $c$, and $800\text{rpm}$ for $\omega$ in Equation (II).

$\begin{array}{c}{x}_{\text{m}}=\frac{16.23\text{N}}{\sqrt{\left(4\times 150\text{kN}/\text{m}-400\text{kg}\times {\left(800\text{rpm}\right)}^2\right)+{\left(6500\text{N}\cdot \text{s}/\text{m}\times 800\text{rpm}\right)}^2}}\\ =\frac{16.23\text{N}\times \frac{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{1\text{N}}}{\sqrt{\begin{array}{l}{\left[4\times 150\text{kN}/\text{m}\times \frac{1000\text{kg}/{\text{s}}^{\text{2}}}{\text{1}\text{kN}/\text{m}}-400\text{kg}\times {\left(800\text{rpm}\times \frac{0.105\text{rad}/\text{s}}{1\text{rpm}}\right)}^2\right]}^2+\\ {\left(6500\text{N}\cdot \text{s}/\text{m}\times \frac{\text{1}\text{kg}/\text{s}}{\text{1}\text{N}\cdot \text{s}/\text{m}}\times 800\text{rpm}\times \frac{0.105\text{rad}/\text{s}}{1\text{rpm}}\right)}^2\end{array}}}\\ =\frac{16.23\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{\sqrt{{\left(-2222400\text{kg}/{\text{s}}^{\text{2}}\right)}^2+{\left(546000\text{kg}/{\text{s}}^2\right)}^2}}\\ =7.09\times 10^{-6}\text{m}\end{array}$

Conclusion:

The amplitude of the vertical motion of the rotor is $7.09\times 10^{-6}\text{m}$.