EN4: Dynamics and Vibrations

 


Division of Engineering
Brown University

 

6.9 Forced vibration of damped, single degree of freedom, linear spring mass systems.

Finally, we solve the most important vibration problems of all. In engineering practice, we are almost invariably interested in predicting the response of a structure or mechanical system to external forcing. For example, we may need to predict the response of a bridge or tall building to wind loading, earthquakes, or ground vibrations due to traffic. Another typical problem you are likely to encounter is to isolate a sensitive system from vibrations. For example, the suspension of your car is designed to isolate a sensitive system (you) from bumps in the road. Electron microscopes are another example of sensitive instruments that must be isolated from vibrations. Electron microscopes are designed to resolve features a few nanometers in size. If the specimen vibrates with amplitude of only a few nanometers, it will be impossible to see! Great care is taken to isolate this kind of instrument from vibrations. That is one reason they are almost always in the basement of a building: the basement vibrates much less than the floors above.

We will again use a spring-mass system as a model of a real engineering system. As before, the spring-mass system can be thought of as representing a single mode of vibration in a real system, whose natural frequency and damping coefficient coincide with that of our spring-mass system.

We will consider three types of forcing applied to the spring-mass system, as shown below:

 

 

External Forcing models the behavior of a system which has a time varying force acting on it. An example might be an offshore structure subjected to wave loading.

Base Excitation models the behavior of a vibration isolation system. The base of the spring is given a prescribed motion, causing the mass to vibrate. This system can be used to model a vehicle suspension system, or the earthquake response of a structure.

Rotor Excitation models the effect of a rotating machine mounted on a flexible floor. The crank with small mass rotates at constant angular velocity, causing the mass m to vibrate.

 

 

 

Notice that in each case, we will restrict our analysis to harmonic excitation. For example, the external force applied to the first system is given by

The force varies harmonically, with amplitude and frequency . Similarly, the base motion for the second system is

and the distance between the small mass and the large mass m for the third system has the same form.

We assume that at time t=0, the initial position and velocity of each system is

In each case, we wish to calculate the displacement of the mass x from its static equilibrium configuration, as a function of time t. It is of particular interest to determine the influence of forcing amplitude and frequency on the motion of the mass.

We follow the same approach to analyze each system: we set up, and solve the equation of motion.

 

6.9.1 Equations of Motion for Forced Spring Mass Systems

Equation of Motion for External Forcing

We have no problem setting up and solving equations of motion by now. First draw a free body diagram for the system (neglecting gravity and reactions for simplicity)

Newton II states that

Rearrange and susbstitute for F(t)

Check out our list of solutions to standard ODEs. We find that if we set

,

our equation can be reduced to the form

which is on the list.

The (horrible) solution to this equation is given in the list of solutions. We will discuss the solution later, after we have analyzed the other two systems.

Equation of Motion for Base Excitation

Exactly the same approach works for this system. The free body diagram is shown below, again with gravity and reactions neglected

The force in the spring is now k(x-y) because the length of the spring is d+x-y. Similarly, the rate of change of length of the dashpot is d(x-y)/dt.

Newton’s second law then tells us that

Make the following substitutions

and the equation reduces to the standard form

Given the initial conditions

and the base motion

we can look up the solution in our handy list of solutions to ODEs.

Equation of motion for Rotor Excitation

Finally, we will derive the equation of motion for the third case. Free body diagrams are shown below for both the rotor and the mass

Note that the horizontal acceleration of the mass is

Hence, applying Newton’s second law in the horizontal direction for both masses:

Add these two equations to eliminate H and rearrange

To arrange this into standard form, make the following substitutions

whereupon the equation of motion reduces to

Finally, look at the picture below to convince yourself that if the crank rotates with angular velocity , then

where is the length of the crank.

The solution can once again be found in the list of solutions to ODEs.

6.9.2 Definition of Transient and Steady State Response.

If you have looked at the list of solutions to the equations of motion we derived in the preceding section, you will have discovered that they look horrible. Unless you have a great deal of experience with visualizing equations, it is extremely difficult to work out what the equations are telling us.

This applet should help. The applet simply calculates the solution to the equations of motion using the formulae given in the list of solutions, and plots graphs showing features of the motion. You can use the sliders to set various parameters in the system, including the type of forcing, its amplitude and frequency; spring constant, damping coefficient and mass; as well as the position and velocity of the mass at time t=0. Note that you can control the properties of the spring-mass system in two ways: you can either set values for k, m and using the sliders, or you can set , K and instead.

We will use the applet to demonstrate a number of important features of forced vibrations, including the following:

The steady state response of a forced, damped, spring mass system is independent of the initial conditions.

To convince yourself of this, run the applet (click on `start’ and let the system run for a while). Now, press `stop’; change the initial position of the mass, and press `start’ again.

You will see that, after a while, the solution with the new initial conditions is exactly the same as it was before. Change the type of forcing, and repeat this test. You can change the initial velocity too, if you wish.

We call the behavior of the system as time gets very large the `steady state’ response; and as you see, it is independent of the initial position and velocity of the mass.

The behavior of the system while it is approaching the steady state is called the `transient’ response. The transient response depends on everything…

Now, reduce the damping coefficient and repeat the test. You will find that the system takes longer to reach steady state. Thus, the length of time to reach steady state depends on the properties of the system (and also the initial conditions).

The observation that the system always settles to a steady state has two important consequences. Firstly, we rarely know the initial conditions for a real engineering system (who knows what the position and velocity of a bridge is at time t=0?) . Now we know this doesn’t matter – the response is not sensitive to the initial conditions. Secondly, if we aren’t interested in the transient response, it turns out we can greatly simplify the horrible solutions to our equations of motion.

When analyzing forced vibrations, we (almost) always neglect the transient response of the system, and calculate only the steady state behavior.

If you look at the solutions to the equations of motion we calculated in the preceding sections, you will see that each solution has the form

The term accounts for the transient response, and is always zero for large time. The second term gives the steady state response of the system.

Following standard convention, we will list only the steady state solutions below. You should bear in mind, however, that the steady state is only part of the solution, and is only valid if the time is large enough that the transient term can be neglected.

 

6.9.3 Summary of Steady-State Response of Forced Spring Mass Systems.

This section should be all you need to solve problems involving forced vibrations. The same information is also summarized on a handout (pdf format), which may be referred to in examinations.

Solution for External Forcing

Equation of Motion

with

Steady State Solution:

The expressions for and are graphed below, as a function of

Steady state vibration amplitude of a forced spring—mass system.

 

Steady state phase of a forced spring—mass system

Solution for Base Excitation

 

Equation of Motion

with

Steady State solution

The expressions for and are graphed below, as a function of

Steady state vibration amplitude for a base excited spring—mass system

 

 

Steady state phase for a base excited spring—mass system

 

 

Solution for Rotor Excitation

Equation of Motion

with

Steady state solution

 

The expressions for and are graphed below, as a function of

Steady state vibration amplitude for a rotor excited spring—mass system

 

Steady state phase lead for a rotor excited spring—mass system

 

6.9.4 Features of the Steady State Response of Spring Mass Systems to Forced Vibrations.

Now, we will discuss the implications of the results in the preceding section.

 The steady state response is always harmonic, and has the same frequency as that of the forcing.

 The amplitude of vibration is strongly dependent on the frequency of excitation, and on the properties of the spring—mass system.

 If the forcing frequency is close to the natural frequency of the system, and the system is lightly damped, huge vibration amplitudes may occur. This phenomenon is known as resonance.

There is a phase lag between the forcing and the system response, which depends on the frequency of excitation and the properties of the spring-mass system.

6.9.5. Using Forced Vibration Response to Measure Properties of a System.

We often measure the natural frequency and damping coefficient for a mode of vibration in a structure or component, by measuring the forced vibration response of the system.

Here is how this is done. We find some way to apply a harmonic excitation to the system (base excitation might work; or you can apply a force using some kind of actuator, or you could deliberately mount an unbalanced rotor on the system).

Then, we mount accelerometers on our system, and use them to measure the displacement of the structure, at the point where it is being excited, as a function of frequency.

We then plot a graph, which usually looks something like the picture below.

Then, read off the maximum response , and draw a horizontal line at amplitude . Measure the frequencies , and as shown in the picture.

We define the bandwidth of the response as

Like the logarithmic decrement, the bandwidth of the forced harmonic response is a measure of the damping in a system.

It turns out that we can estimate the natural frequency of the system and its damping coefficient using the following formulae

The formulae are accurate for small - say .

To understand the origin of these formulae, recall that the amplitude of vibration due to external forcing is given by

We can find the frequency at which the amplitude is a maximum by differentiating with respect to , setting the derivative equal to zero and solving the resulting equation for frequency. It turns out that the maximum amplitude occurs at a frequency

For small , we see that

 

 

Next, to get an expression relating the bandwidth to , we first calculate the frequencies and . Note that the maximum amplitude of vibration can be calculated by setting , which gives

Now, at the two frequencies of interest, we know , so that and must be solutions of the equation

Rearrange this equation to see that

This is a quadratic equation for and has solutions

Expand both expressions in a Taylor series about to see that

so, finally, we confirm that

 

 

6.9.6 Example Problems in Forced Vibrations

Example 1: A structure is idealized as a damped spring—mass system with stiffness 10 kN/m; mass 2Mg; and dashpot coefficient 2 kNs/m. It is subjected to a harmonic force of amplitude 500N at frequency 0.5Hz. Calculate the steady state amplitude of vibration.

Start by calculating the properties of the system:

Now, the list of solutions to forced vibration problems gives

For the present problem:

Substituting numbers into the expression for the vibration amplitude shows that

 

Example 2: A car and its suspension system are idealized as a damped spring—mass system, with natural frequency 0.5Hz and damping coefficient 0.2. Suppose the car drives at speed V over a road with sinusoidal roughness. Assume the roughness wavelength is 10m, and its amplitude is 20cm. At what speed does the maximum amplitude of vibration occur, and what is the corresponding vibration amplitude?

Let s denote the distance traveled by the car, and let L denote the wavelength of the roughness and H the roughness amplitude. Then, the height of the wheel above the mean road height may be expressed as

Noting that , we have that

i.e., the wheel oscillates vertically with harmonic motion, at frequency .

Now, the suspension has been idealized as a spring—mass system subjected to base excitation. The steady state vibration is

For light damping, the maximum amplitude of vibration occurs at around the natural frequency. Therefore, the critical speed follows from

Note that K=1 for base excitation, so that the amplitude of vibration at is approximately

Note that at this speed, the suspension system is making the vibration worse. The amplitude of the car’s vibration is greater than the roughness of the road. Suspensions work best if they are excited at frequencies well above their resonant frequencies.

Example 3: The suspension system discussed in the preceding problem has the following specifications. For the roadway described in the preceding section, the amplitude of vibration may not exceed 35cm at any speed. At 55 miles per hour, the amplitude of vibration must be less than 10cm. The car weighs 3000lb. Select values for the spring stiffness and the dashpot coefficient.

The specification requires that

at resonance. Examine the graphs of shown with the solutions to the equations of motion. Recall that K=1 for a base excited spring—mass system. Observe that, with , the amplification factor never exceeds 1.75.

Now, the frequency of excitation at 55mph is

We must choose system parameters so that, at this excitation frequency,

Examine the graph showing the response of a base excited spring—mass system again.

We observe that, for , for . Therefore, we pick

.

Finally, we can compute properties of the system. We have that

Similarly