baxter - understanding amplification factor - part i

8/10/2019 Baxter - Understanding Amplification Factor - Part i

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PART I-UNDERSTANDING THE DERIVATION OF THE AMPLIFICATION FACTOR AND ITSIMPORTANCE IN FREQUENCY RESPONSE RELATED EQUATIONS

Nelson Baxter & Brad Barton

ABM Technical Services

[email protected]

Abstract: The response of structures depends upon the amount of the force applied divided by thedynamic stiffness characteristics of the structure. In the world of dynamics, the response of the structureis dependent not just on the magnitude of the force, but also on the frequency of the force. This paperstarts with a physical visualization of a mechanical system then progresses into a graphical interpretationof the force response characteristics that then eventually results in the formula for the amplification factor.This paper highly references work by Dr. Ron Eshleman(Forced Harmonic Vibrations) and the graphicalinterpretations of W. T. Thomson ( Mechanical Vibrations, Second Edition, Prentice-Hall, INC., EnglewoodCliffs, NJ, 1953). A second paper which shows the use of the equations developed in this paper to solve

field problems will also be presented in this conference.

Keywords : Amplification Factor, Force, Response, Eccentricity, Transmissibility, Isolation, HarmonicMotion, Damping

In Mechanical Engineering, we were taught you only need to know two things. These things were F=maand you cannot push on a rope. For this paper, Hooks law of F=kX will be added. From the basic lawsof Newton and Hook, and with extensive help from Dr. Ron Eshleman and W.T. Thomson, a simplifiedexplanation of the concept of the frequency dependence of the resistance to motion will be presented.The you cant push on a rope bit of knowledge will not be of much use for this discussion.

It is helpful to start any learning process with a visualization of the interaction of the elements that makeup a system. Albert Einstein was riding a bus one day when he viewed a clock as he was moving awayfrom it. He began to ponder time and its relationship to speed and out of that thought process came forthhis theory of relativity. Making mental images of things can therefore be a powerful aid to understandinghow things work. Ultimately, mathematics must be used to describe a process, but it takes observationand our innate ability to have a feel for how things work to truly understand the world around us. Whenthat observation based model is combined with the associated mathematics, then there is a solid basisfrom which we can work to achieve better understanding.

STEP 1- A MENTAL OBSERVATION BASED MODEL

THE CONCEPT OF MECHANICAL LAG

The lower carrier moves at a constant velocity then stops,

but the mass continues to move deflecting the support.

In the figure to the left, a mass is on avertical leaf spring with stiffness k and ismoving at constant velocity when it

decelerates and comes to a stop. It isnot hard to imagine that the masscontinues to move, thereby deflectingthe spring. If the lower carrier thenaccelerates in the opposite direction, theupper mass follows behind. This delayillustrates mechanical lag, which canarise under acceleration or deceleration


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The preceding illustration is intuitive. If there is a situation where there is a spring and a mass and we tryto accelerate the mass by applying a force (kX ) from the spring, then the motion of the mass followsforce applied by the spring. It is also intuitive that if the spring is weak or the rate at which we attempt toaccelerate the mass is fast that the amount of lag will increase. From a slightly more technical approachto looking at things, we know that the resistance to acceleration is a function of the frequency squared.We also know that the force generated by the spring is kX. For a given spring force, it therefore becomesapparent that there is a frequency at which this kX force cannot accelerate the mass through a givendisplacement. The kX force just cannot overcome the (Ma )resistance force at higher frequencies and themotion of the mass thereby lags the force. At high frequencies (high meaning well above the naturalfrequency) the mass will lag the input force 180 degrees and the amount of the transmitted motion will bereduced. This is an intuitive interpretation of the concept of isolation, which will be discussed later on.

THE CONCEPT OF AMPLIFICATION

MAKING A MODEL THAT SHOWS HOW THE RESISTIVE FORCES ARE EQUAL TO THE DRIVINGFORCE

Mental models are very useful in describing how a system responds and allow us to get an intuitive feelfor the important parameters. In the mental exercise above, for an un-damped system the importantparameters turned out to be the amount of mass (M), the stiffness (k), the Force that was applied (F) andthe frequency of the force, w hich we will refer to as (). Mental models unfortunately do not allow us toadequately design a system or predict the response to a known force. Our understanding musteventually be expressed in mathematics if we want to make any practical use of it.

For this discussion, rather than jumping directly into the mathematics, a graphical representation of theforces will be utilized. This graphical interpretation will serve as a bridge to aid in the understanding of themathematical formulae that describe simple dynamic systems. W.T. Thomson who is referenced in theabstract, pioneered this graphical approach. It shows the derivation of all of the important amplificationfactor related equations. The intuitive model will be useful in summing up the force vectors which will inturn lead to the formulation of a mathematical representation of the amplification factor. The diagramsand vector drawings were obtained from the referenced work by Dr. Ron Eshleman.

K

If a static force F is applied to the systemto the right, then the amount of deflectionX will be F/K. This is Hooks law an d is

simple to comprehend. This law,however; only works for static deflectionor when the frequency of the appliedforce is well below the systems naturalfrequency. Intuitively we know that ifwe start pushing on the mass faster andfaster that there will be a point where themass moves with very little force. Theamount of motion per unit of force at acertain frequency gets very high. This isreferred to as resonance amplification.

FX As force is applied at

the natural frequency,the amount of motion

per unit of force getsvery high. Hooks lawcan therefore by itselfno longer be used todescribe the responseof the mass to theforce, when the force isdynamic in nature.


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STEP 3-A GRAPHICAL REPRESENTATION OF THE DYNAMICS OF A MECHANICAL SPRING MASSDAMPER SYSTEM

The vector diagrams below that will eventually be used to derive the amplification factor equation showthe relationships of the force vectors. For the first case, the forcing frequency is below the naturalfrequency. The second case is at the natural frequency. This special case of when the forcing frequency

is exactly at the natural frequency illustrates how the well known can be derived . The thirdcase shows the vector relationship when the forcing frequency is above the natural frequency.Important Graphical Vector Relationships

1: The spring restoration force is always 180 degrees out of phase from the mass resistance force.

2: Since velocity peaks at 90 degrees from when the displacement peaks, its vector will be 90 degreesout of phase from the displacement vector. The damping force vector, which is velocity dependent, istherefore always 90 degrees out from the stiffness and mass force vectors.

3: The lag angle is less than 90 degrees when the excitation frequency is below the natural frequency, 90

degrees at the natural frequency and between 90 and 180 degrees above the natural frequency.

Vector drawings are courtesy of Dr. Eshlemen s referenced paper-Forced Harmonic Vibrations

VECTORS SHOWING FORCE RELATIONSHIPS BELOW RESONANCE

Note that from the above plot, which represents the forcing frequency being below resonance that the

spring force( kX) is greater than the mass force ( 2

mX )and the Lag Angle ( ) is less than 90

degrees.

Driving Forceleadsdisplacement

Displacement

DampingForce

Mass resistance

force=

SpringForce


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VECTORS SHOWING FORCE RELATIONSHIPS AT RESONANCE

The figure below shows what happens at resonance

1: The spring and mass forces are equal and opposite in direction and cancel each other out.

2: The displacement lags the applied force by 90 degrees.3: Since the spring and mass forces cancel each other out, the only force opposing motion is thedamping.

When the spring and mass forces are equal and cancel each other out then and if thisrelationship is solved for then under these conditions = which is the well known formulafor the natural frequency. For this unique case is therefore named .

VECTORS SHOWING FORCE RELATIONSHIPS ABOVE RESONANCE

For above resonance the lag angleis greater than 90 degrees.

For above resonance, the mass term

m X is now greater than the kXstiffness term.

Angle betweenforce and responseis 90 degrees

At resonance the kXspring force and

mass forcecancel each otherout.


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The next step in deriving the Amplification Factor is to sum the stiffness and mass terms.

Since the kX and xm 2 are opposite in direction those terms can be combined as follows:

Using the Pythagorean Theorem, the following equation can be derived

X m X Derivation by W.T. Thompson (1) For a quick review lets look at w hat has been accomplished to this point:

1: By looking at a basic model, the forces have been identified and summed

2: A vector representation of these forces has been formed

3: By representing the forces present in the spring-mass-damper model as vectors, we have derived auseful mathematical expression that defines the relationship of all the forces with each other.

The next step is to put these mathematical relationships into a form that can be used for some design andanalytical purposes, which we get to later. This is accomplished by algebraic manipulation andsubstitution of equal terms.

STEP 4- SOLVING THE FORCE VECTOR EQUATIONS FOR THE AMPLIFICATION FACTOR

It is necessary to rearrange the terms of the solution for the summation of the force vectors.Remember that the formula below is nothing more than the Pythagorean Theorem solution for thesummation of the force vectors shown above.

X m X Divide both sides by [

- ] and take square root of both sides to get

X m (2)Divide numerator and denominator by k to get the equation (3)on the following page

This equation comes directly from theforce summation triangle shown above.

Summation of stiffness and mass

forces to form one combined term.


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X (3) Since which is then substituted in the above equation.

This above substitution makes the first term under the radical (4)

The Damping c in the term is very difficult to determine, so the concept of the damping ratio

term is introduced. The damping ratio can easily be measured in the field, so this concept isvery useful in the derivation of a solution that can be solved using field supplied data. Later on in thispaper four methods of measuring the damping ratio in the field are presented.

and Substitute k with this expression

The term then becomes or so =

The second term under the radical therefore becomes (5)

Replace the terms under the radical of equation (3)with expressions (4) & (5) above and finally theequation that we have been looking for, the Amplification Factor has been derived.

Since X X (6)

The essence of the Amplification Factor comes down to knowing where the forcing frequency isrelative to the natural frequency and the damping ratio. This is an incredibly useful equation thatis the basis for understanding dynamic systems.

To get the response for a known force and stiffness another substitution is made.

Multiply through by and substitute back in for and the result is the force response equationX (7)

What we now have is basically Hooks Law times the Amplification Factor. This equation describes howa spring mass system will respond to both a static force where the frequency equals zero or from adynamic force of any frequency.

Hooks Law times the Amplification Factor

Amplification Factor

Hooks Law times the Amplification Factor


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A COMPARISON OF THE RESPONSE CHARACTERISTICS OF SOME OF THE IMPORTANTEQUATIONS THAT CONTAIN THE AMPLIFICATION FACTOR

FORCED RESPONSE EQUATION

X

(7)

Equation(7) is used when an analyst needs to know how much a structure will respond to a known forcegiven the stiffness of the structure(k), the amount of force , where the machine operates at relative to its natural frequency ( and the percent of critical damping( .

Example: Lets take a spring mass system and solve for the amount of response at 4 differentfrequencies. Assume that the spring stiffness is 10000 lb/in, the damping ratio is .05 and the force is 400lb.

Frequency Response of Structure to an applied force

0.0 X natural frequency .040 This is the solution for Hooks Law =

.5 X natural frequency .053 There is some amplification present at a frequency = to .5

1.0 X natural frequency .400 There is an amplification of 10:1 In this case AF= = 1/(2 *.05)

2.0 X natural frequency .013 The response is lower than & is thus in the isolation region

This curve shows the solution forall frequency ratios to a force of

400 lb for a system with astiffness of 10000 lb/in and adamping ratio of .05 . Thecircles show the solution for thefour frequency ratios listedabove.


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COMPARISON BETWEEN AMPLIFICATION FACTOR AND FORCED RESPONSE CURVES

Amplification Factor Curve for .05 damping ratio Forced Response Curve for .05 damping ratio

As can be seen from the above, the Amplification Factor curve and the Forced Response curves look

identical for a system with the same damping. This is as would be expected since the Force Responseequation is just the Amplification Factor times the static response predicted by Hooks Law.

RESPONSE TO ECCENTRICITY AND UNBALANCE

In vibration work, there are some commonly used variations of the force response equation. One is theresponse to an unbalance force and the other is eccentricity response, which is a close cousin to theunbalance response case. The main difference between the unbalance and eccentricity responses, ascompared to the standard force response solution, is that the forces from unbalance and eccentricityare not constant, but instead vary with the speed. The response to a force at a given frequency is stillgoverned by the amplification factor, but since the force for unbalance varies as a function of the

frequency squared, the response curves look different than they do with the standard forced responsefunction.

Remember that F=ma. So for a rotating unbalance the force is F=mr and the response X is:

X (8)The curve to the left shows the response to anunbalance force. Instead of a constant force, theforce varies as the square of the frequency. Notethat the response starts at zero because therotational force is zero at a 0 frequency. Unlikethe general force response curve, at higherfrequencies, the unbalance response does nottaper off towards zero. This is because both theunbalance force and the masss resistance to

acceleration increase with the square of thefrequency. These offsetting effects yield a flatcurve in this region.


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COMPARISON BETWEEN GENERAL FORCE RESPONSE AND UNBALANCE RESPONSE CURVES

AMPLIFICATION FACTOR AND FORCE RESPONSE CURVE

UNBALANCE RESPONSE AND ECCENTRICITY RESPONSE CURVE

Starts at 1.0ratio

Tapers downtowards 0 due tothe isolationeffect.

Starts at 0 because there is nounbalance force at 0 frequency

Ends at a fixed value becauseunbalance force and massresistance are bothfrequency squared functions.

Important Point- Remember fromearlier in the paper that as thefrequency is increased that therebecomes a frequency where the kXforce from the spring cannotovercome the ma resistance force.That is why isolation occurs and theresponse ratios drop at the higherfrequencies and thus approachzero.

As noted previously, there is noresponse due to unbalance at thebeginning of the curve. This is becauseat zero speed, there is no forcegenerated by the unbalance.

At the higher frequencies, the forcegenerated by the unbalance increases as the square of the speedand so does the resistance to

motion(ma), so they equal one anotherand the response reaches anequilibrium value rather than droppingto zero.


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Force Transmissibility /Base Motion

Another important equation that is used in the field of vibration analysis and vibration control is theForce Transmissibility & Base Motion equation. It is used when an analyst or designer wants to calculateeither the amount of force transmitted from a vibrating mass to a structure or alternatively the amountof motion transmitted from a floor to an isolated mass. That equation is shown below. As can be seen,

it also contains the amplification factor. It is times the amplification factor.The derivation of the transmissibility equation is not covered in this paper, however since the formulacontains the Amplification Factor and is used extensively in the field of vibration analysis, it is beingincluded in this compilation of Amplification Factor related equations.

Force Transmissibility / Base Motion Equation

(10)

This equation can be used to calculate the transmissibility of motion or force across a set of isolationsprings. The efficiency of the isolators is simply 1-minus the transmissibility. At low damping values, theresponse curve of the above equation is similar to the amplification factor. At higher damping values,

the upper term that contains the damping ratio starts to have an effect on the results.Comparison of Amplification Factor and Transmissibility at .05 damping ratio

When the frequencyratio is 1.414, forcetransmission is 1.0.Above a frequencyratio of 1.414 isolationbegins.


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At low damping ratios, the Amplification Factor and the Force Transmission/Base Motion response lookvery similar. As the amount of damping gets higher, the curves start to show differences in theirshape. The next three sets of plots show what happens as damping is increased. As would beexpected, one of the effects of adding damping, is to reduce the amplification at the natural frequency.As noted in the plots below, another effect of the additional damping is that more force and motion is

transmitted across the springs. Consequently another side effect of increased damping is the reducedeffectiveness of the isolators at frequency ratios above 1.414 of .

Comparison of Amplification Factor and Force Transmission at .3 damping ratio

Comparison of Amplification Factor and Force Transmission at 1.0 damping ratio

At the 1.414frequency ratiothe value is 1.0

Added dampingreduces responseat naturalfrequency

As damping increases,

the force & motiontransmitted across theisolators increases.


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Comparison of Amplification Factor and Force Transmission at 2.0 damping ratio

As noted in the graphs above, the transmissibility is always 1 at the 1.414 ratio.

The above equations include the amplification factor, forced response given a fixed force, responseto unbalance, eccentricity response and force and base transmissibility. These are very importantconcepts in the field of vibration analysis. Understanding them is vital in working with theresponse to dynamic forces. All these equations contain the Amplification Factor. Its DNA can beseen in each of them. It is what allows us to determine the dynamic stiffness of a structure. TheAmplification Factor equation looks complicated, but in simple terms comes down to just knowingthe following: Where does the machine operate at ( ) relative to its natural frequency ( ) andwhat is the damping ratio( ). We usually know the speed ( ) and can often obtain the naturalfrequency( ) using an impact test or a structural model. That leaves the need to determine thedamping ratio ( ). In order to provide a full set of basic tools for the analyst, four well known

methods of measuring the damping ratio in the field are outlined below.

FOUR METHODS OF THE FIELD DETERMINATION OF THE DAMPING RATIO

POWER METHOD OF DETERMINING DAMPING RATIO

The power approach to computing the damping uses the following equation. This data can beobtained by measuring the response due to an impact or from coast down data.

AMPLIFICATION FACTOR Q N

N N C

2 1

(11)

N 2

N C = NATURAL FREQUENCY

N 2 = FREQUENCY ABOVE Nc AMPLITUDE = .707 Nc AMPLITUDE

N 1 = FREQUENCY BELOW Nc AMPLITUDE = .707 Nc AMPLITUDE

At the 1.414frequency ratiothe value is still

1.0

Greater damping meansmore force transmission


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TRANSFER FUNCTION PHASE SLOPE APPROACH TO DETERMINING DAMPING RATIO

Using the rate of change of phase shift in the transfer function phase plot to calculate the dampingratio. This data can be obtained with a dual channel analyzer by viewing the phase displaygenerated by the transfer function calculation.

AMPLIFICATION FACTOR f

f Q *

360

* (14)

WHERE f n = NATURAL FREQUENCY

=CHANGE IN PHASE

f =CHANGE IN FREQUENCY

PHASE SHIFT PLOT USED TO CALCULATE DAMPING RATIO

Example from above phase plot Q * * ..

266360

82 619

101 12 * Q

12 101

00495*

.

Summary:

Mechanical spring mass systems respond differently to dynamic forces than they do to static loads.These systems have a frequency dependent resistance to motion. This paper, through the use ofintuition, the mathematical summation of forces, the vector representation of these forces and somealgebraic substitutions has shown the derivation of the Amplification Factor.

The Amplification Factor forms the basis for forced response calculations, unbalance response, responseto eccentricity, isolation calculations and the determination of force and motion transmissibility. Byknowing just the rotational speed, the natural frequency and the damping ratio, it is possible to predicthow a simple mechanical system will react to a given dynamic force. The analyst usually knows thefrequency of a force. The natural frequency can often be obtained with an impact test and as shown inthis paper, there are several ways to determine the damping ratio. This means that the analyst orengineer has some relatively powerful tools to work with in the design and analysis of mechanicalsystems.

Damping Ratio

Slope of phase shiftdetermines damping ratio.


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As a quick review, the following are the response equations derived and or discussed in this paper. Theyare all based upon the Amplification Factor.

AMPLIFICATION FACTOR- The basis for the response of mechanical systems over a range offrequencies. The Amplification Factor is contained in all the following equations.

X X FORCED RESPONSE-Calculates response to a given force at various frequencies

X RESPONSE TO UNBALANCE- Calculates response to mass(m) unbalance at a radius r

X m RESPONSE TO ECCENTRICITY- Calculates motion due to a given eccentricity (e)

X FORCE & BASE MOTION TRANSMISSIBILITY- Calculates force & motion through a spring dampersystem at various frequencies. This equation can be used for isolation calculations

X

Understanding the frequency dependent resistance to motion as described by the Amplification Factor isvital in the field of vibration analysis. It is hoped that the tools presented in this paper will enable theanalyst to understand the effects of the Amplification Factor, and its value in solving real world problems.That will be the intended subject of the paper that follows. A thank you again to Dr. Ron Eshleman &W.T. Thomson for their work in this area.

baxter - understanding amplification factor - part i

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