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A machine element weighing 800 lb is supported by two springs, each having a constant of 200 lb/in. A periodic force of maximum value 30 lb is applied to the element with a frequency of 2.5 cycles per second. Knowing that the coefficient of damping is 8 lb-s/n. determine the amplitude of the steady- state vibration of the element.

Short Answer

Expert verified
The amplitude of the steady-state vibration is approximately 0.235 inches.

Step by step solution

01

Identify Given Values

First, list all the given values from the problem statement.- Weight of the machine element, \( W = 800 \text{ lb} \)- Spring constant for each spring, \( k = 200 \text{ lb/in} \)- Maximum periodic force applied, \( F_0 = 30 \text{ lb} \)- Frequency of force, \( f = 2.5 \text{ cycles/s} \)- Damping coefficient, \( c = 8 \text{ lb-s/in} \)
02

Calculate Total Spring Constant

Since the machine element is supported by two springs in parallel, the total spring constant \( K \) is the sum of the individual spring constants.\[ K = 2 \times k = 2 \times 200 = 400 \text{ lb/in} \]
03

Find Natural Frequency

Calculate the natural frequency of the system using the formula for the natural frequency \( \omega_n \) of a spring-mass system.\[ \omega_n = \sqrt{\frac{K}{m}} \] Note that we need to convert the weight to mass using \( g = 386.4 \text{ in/s}^2 \).\[ m = \frac{800}{386.4} \approx 2.07 \text{ slug} \]Substitute into the natural frequency formula:\[ \omega_n = \sqrt{\frac{400}{2.07}} \approx 13.91 \text{ rad/s} \]
04

Find Forced Frequency

Convert the frequency from cycles per second to radians per second for the forced response by using \( \omega = 2\pi f \).\[ \omega = 2\pi \times 2.5 = 15.71 \text{ rad/s} \]
05

Calculate Damping Ratio

Use the formula for the damping ratio \( \zeta \).\[ \zeta = \frac{c}{2 \sqrt{Km}} \]Substitute the known values:\[ \zeta = \frac{8}{2 \sqrt{400 \times 2.07}} = \frac{8}{2 \times 28.76} \approx 0.139 \]
06

Calculate Steady-State Amplitude

Use the formula for steady-state amplitude \( X \), given by\[ X = \frac{F_0}{\sqrt{(K - m\omega^2)^2 + (c\omega)^2}} \].Substitute the values into the formula:\[ X = \frac{30}{\sqrt{(400 - 2.07 \times 15.71^2)^2 + (8 \times 15.71)^2}} \approx \frac{30}{\sqrt{17.75^2 + 125.68^2}} \approx \frac{30}{127.54} \approx 0.235 \text{ in} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Damping Ratio
The damping ratio, denoted as \( \zeta \), plays a crucial role in understanding how oscillations of a system decay over time. It helps us figure out how quickly or slowly an oscillating system comes to a stop after it's been disturbed. Specifically, it measures the amount of damping in a system relative to critical damping. When the damping is exactly at the critical level, the system returns to equilibrium as quickly as possible without oscillating.

In our exercise, the damping ratio is calculated using the formula:\[ \zeta = \frac{c}{2 \sqrt{Km}} \]where \( c \) is the damping coefficient, \( K \) is the total spring constant, and \( m \) is the mass. For our problem, it was found to be approximately 0.139. This indicates underdamping since the ratio is less than 1, meaning the system will oscillate but the amplitude of oscillation will gradually decrease over time.

The damping ratio is important because:
  • A ratio of 0 means the system oscillates indefinitely.
  • A ratio less than 1 (like ours) means underdamping, resulting in oscillations that decrease over time.
  • A ratio of 1 indicates critical damping, the system returns to rest without oscillating.
  • A ratio greater than 1 results in overdamping, where the system returns to rest slower but without oscillating.
Natural Frequency
Natural frequency \( \omega_n \) is the rate at which a system vibrates when it is not subjected to an external force or damping. Imagine plucking a guitar string — the sound you hear is tied to its natural frequency. It's determined by the system's mass and stiffness, showcasing how they resist acceleration and displacement respectively.

The formula for natural frequency is given by:\[ \omega_n = \sqrt{\frac{K}{m}} \]where \( K \) is again the total spring constant, and \( m \) is the mass of the system. In our calculation, the natural frequency was approximately \( 13.91 \text{ rad/s} \).

Understanding the natural frequency helps engineers and designers to:
  • Predict how systems respond to vibrations.
  • Avoid resonant frequencies that can lead to failure.
  • Ensure comfort and stability in mechanical designs.
Natural frequency is key in so many applications, from buildings in earthquake zones to the tuning of musical instruments.
Spring Constant
The spring constant \( k \) is a measure of a spring's stiffness. It's a crucial parameter that tells us how much force is required to stretch or compress a spring by a certain distance. This fundamentality is at the heart of Hooke's Law, where the force exerted by the spring is proportional to the change in length:

\[ F = k \cdot x \]

In the exercise, because there are two springs in parallel, the total spring constant is calculated as \( K = 2 \times k = 400 \text{ lb/in} \). The stiffer the spring, the larger the force needed to deform it.

The significance of the spring constant includes:
  • Designing suspension systems for vehicles.
  • Creating stable support structures.
  • Protecting sensitive components from excessive force.
Having the correct spring constant ensures that your system can handle expected loads without excessive movement, providing both functionality and safety.
Forced Vibration
Forced vibration occurs when a system is continuously driven by an external force, rather than freely vibrating on its own. Such external forces often have a periodic nature, like a repeating engine cycle or rhythmic seismic waves. Understanding forced vibrations is essential in engineering to mitigate potential damage from unwanted external disturbances.

In the exercise, a periodic force of maximum value \( F_0 = 30 \text{ lb} \) is applied with a frequency of \( 2.5 \text{ cycles/s} \). It's crucial to convert this frequency into radians per second for calculation by using \( \omega = 2\pi f \) to get \( 15.71 \text{ rad/s} \).

Forced vibrations are important because they help us to:
  • Analyze how systems respond to constant forcing.
  • Design structures to reduce unwanted vibrations.
  • Ensure systems are safe under operative conditions.
The study of forced vibrations ensures that even with ongoing external forces, systems can remain stable and reliable.

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Most popular questions from this chapter

A uniform rod \(A B\) can rotate in a vertical plane about a horizontal axis at \(C\) located at a distance \(c\) above the mass center \(G\) of the rod. For small oscillations determine the value of \(c\) for which the frequency of the motion will be maximum.

For a steady-state vibration with damping under a harmonic force, show that the mechanical energy dissipated per cycle by the dashpot is \(E=\pi c x_{m}^{2} \omega_{f},\) where \(c\) is the coefficient of damping, \(x_{m}\) is the amplitude of the motion, and \(\omega_{f}\) is the circular frequency of the harmonic force.

A simple pendulum of length \(l\) is suspended from collar \(C\) that is forced to move horizontally according to the relation \(x_{C}=\delta_{m} \sin \omega_{f} t\) Determine the range of values of \(\omega_{f}\) for which the amplitude of the motion of the bob is less than \(\delta_{m}\) (Assume that \(\delta_{m}\) is small compared with the length \(l\) of the pendulum.)

Denoting by \(\delta_{\mathrm{st}}\) the static deflection of a beam under a given load, show that the frequency of vibration of the load is $$ f=\frac{1}{2 \pi} \sqrt{\frac{g}{\delta_{\mathrm{st}}}} $$ Neglect the mass of the beam, and assume that the load remains in contact with the beam.

A small trailer and its load have a total mass of \(250 \mathrm{kg}\). The trailer is supported by two springs, each of constant \(10 \mathrm{kN} / \mathrm{m},\) and is pulled over a road, the surface of which can be approximated by a sine curve with an amplitude of \(40 \mathrm{mm}\) and a wavelength of \(5 \mathrm{m}\) (i.e., the distance between successive crests is \(5 \mathrm{m}\) and the vertical distance from crest to trough is \(80 \mathrm{mm}\) ). Determine (a) the speed at which resonance will occur, (b) the amplitude of the vibration of the trailer at a speed of \(50 \mathrm{km} / \mathrm{h}\).

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