1

MEEG 5113MEEG 5113

Set 3

2

3

4

5

Vibration AnalysisIt is quite common for engineering structures to be exposed to excitation which is the result of the operation of one or more mechanical systems. Many times this excitation will be periodic or at least can be treated as if it is periodic. Non-periodic excitation (e.g. shock, earthquake, impulse, and random excitation) also occurs in specific situations. These forms of excitation require a more complex analysis and will not be dealt with here.

Periodic excitation can result from unbalanced rotating or reciprocating components of machinery or equipment, wind or current effects, or a shaking foundation. Usually it is required to keep the amplitude of any vibration low so as to avoid substantial dynamic stresses, noise, fatigue, and other effects. Periodic excitation can be either harmonic or non-harmonic. Non-harmonic forms are handled by using Fourier analysis techniques to replace the original wave form with a series of sine and cosine terms which when added together reproduce the original excitation. In this manner, a complex wave is replaced with a

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Vibration Analysisset of harmonic functions and the analysis can be easily performed.

Shown below is a damped single degree-of-freedom system which is being acted on by a time-varying harmonic force, F sin t where is the excitation or driving frequency of the external force. A free body diagram of the system now contains three external forces. Summing forces and gathering all system terms together gives an equation of motion of the form

sinmx cx kx F t The solution to this linear, constant coefficient, second order differential equation is called the particular or forced or steady-state solution which gives x(t) due to the continuous application of the excitation.

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Vibration AnalysisBy steady-state it is meant that whatever transient vibration occurs due to the initial application of the excitation has decayed away and only the response due to the excitation remains. If the magnitude of either F or changes, the transient response will return and remain active for some amount of time which depends on the amount of damping present. This is shown in the figure below which displays the free, forced, and total responses.

( ) ( ) ( )Total Free Forced

x t x t x t As can be seen in the figure, each case is solved independently and the results combined to produce the total response.

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Vibration AnalysisReturning to the forced response solution, since the excitation is a sine wave, the solution will be as well except that it will have a different phase angle than the excitation due to the presence of the damper. This means that the solution will have the form

12

2 2

( ) sin( )

( ) cos( ) sin( )

( ) sin( ) sin( )

AND

AND

x t X t

x t X t X t

x t X t X t

The equation of motion can now be written as

2 12

sin( ) sin( ) sin( ) sinmX t cX t kX t F t

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Vibration AnalysisA vector diagram containing each of these forces is shown below. The phase angle is negative because response cannot occur until the excitation is applied, i.e. the response lags the excitation. Because the inertia and spring forces differ by radians, the previous equation may be rewritten as

2 12

( ) sin( ) sin( ) sink m X t cX t F t

The force polygon can now be seen to form a right triangle and following equation relates F and X:

2 2 2 2( ) ( )kX mX cX F

This may be solved for X and to get

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Vibration Analysis

1

22 22

tanANDcFX

k mk m c

This yields the steady-state solution of

2 22

( ) sin( )Fx t tk m c

And a complete solution of

2 22

( ) sin( ) sin( )t

Free dFx t X e t t

k m c

2 2 22 2 22 2n n

F FXk c mmm m

The steady-state solution can be non-dimensionalized as follows:

11

Vibration Analysis

where XS = Static response (deflection) of the system = Dynamic amplification (magnification) factor r = Frequency ratio (/) = Damping ratio (c/cc)

A non-dimensional plot is obtained when the magnitude of the

2 2 222 22

2

2 22

1

1

221

2

nn

n n

SS

OR

FkFX

r rm

XX Xr r

Continuing the process yields

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amplitude ratio (X/XS) is plotted as a function of the frequency ratio (r = /) for various values of damping ratio () as shown below. Notice that if the system is either undamped or very lightly damped and the frequency ratio is close to one, the amplitude ratio becomes extremely large meaning that it is very likely that the system will be damaged or destroyed.

Vibration Analysis

Also note that this plot is simply an edge view of a three dimensional surface which is a function of r and . Each line represents a slice through this surface at a given value of . When viewed in perspective and from an elevation, the surface can be easily seen.

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Vibration Analysis

16

1116

2126

3136

4146

51

S1

S7 S1

3 S1

9

0

1

2

3

4

5

X/XS

Frequency Ratio Damping Ratio

Dynamic Amplification Factor

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Vibration AnalysisAlso note that as increases, the maximum amplitude is reduced and occurs closer to the origin. To determine value of r at which the maximum value occurs, the equation for m is differentiated with respect to r and set equal to zero. The resulting equation allows one to determine the value of r at the maximum for a given value of .

0.52 22

0.52 22

2

2 2

2

1 2 0

0.5 1 2 *

2 1 2 2 2 2 0

1 2 0

1 2 0 0.707

d r rdr

r r

r r r

r

r

For > 0.707, the static response is the largest system response.

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Vibration AnalysisThe phase angle () is also a function of the frequency ratio (r = /) and damping ratio (). Shown below is the variation in phase angle with frequency for different amounts of damping.

This graph can be divided into three areas of interest. These areas are easier to understand for the undamped case where and are equal to

1 1

2 22 2

0 01 1 tan tan11 k m rk m r

When r is less than 1, is positive or k > m2 and the mass moves in the same direction as F. This means the phase angle between F and x is zero as the stiffness dominates the response.

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Vibration Analysis

For the damped case and are given by

When r > 1, is negative or m2 > k and the mass moves in the opposite direction of F. This means that the phase angle between F and x is 180o and this results in the excitation force acting as a “brake” to help bring the mass to rest at its extreme displacement value.

When r = 1, is undefined which requires x(t) to be redefined as a cosine function in order to solve this problem. Redefining x(t) in this manner makes = 90o.

2 22

2 22

1 1

2 2

1

( ) ( )1

1 2

2tan tan1

k m c

r r

rck m r

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Vibration AnalysisNotice that the phase angle now continuously changes with r due to presence of the damping term.

The three cases r < 1, r = 1, and r > 1 are illustrated in the three figures below, (a), (b), and (c) respectively. As r, and hence , increases from zero, both the damping and inertia terms also increase but the inertia term increases faster than the damping term which causes to increase at an increasingly non-linear rate between 0o and90o.

When r = 1, the inertia and spring terms have equal magnitudes so = 90o.

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Vibration Analysis

1 6 11 16 21 26 31 36 41 46 51S1

S12

S230

45

90

135

180

Frequency Ratio

Damping Ratio

Phase Angle

When r > 1, the inertia term increases faster than the damping term which causes to increase at a decreasingly non-linear rate.

The phase equation also represents a surface as shown below. By rotating this plot it is possible to view the non-linear behavior of .

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Vibration AnalysisAs mentioned earlier for the undamped case, when r = 1, is undefined which means that X is infinite. However, the question becomes “How quickly does X grow to these extremely large values?”

To answer this question requires x(t) to be redefined as a cosine function and the equation of motion to be solved for this new definition of x(t). Therefore, let x(t) be given by

2

( ) cos( ) ( ) cos( ) sin( )( ) 2 sin( ) cos( )

x t Et t x t E t Et tx t E t Et t

Substituting into the undamped equation of motion yields2( 2 sin( ) cos( )) ( cos( )) sin( )m E t Et t k Et t F t

2 20

22

kmEt kEtm

FmE F Em

Equating like terms yields

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Vibration Analysis

Undamped Resonance Response

-30

-20

-10

0

10

20

30

0 1 2 3 4 5

Time

X

Et

-Et

Et cos(wt)

For m = 1, = 4, and F = 40, the resonance response is given in the figure below.

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Vibration AnalysisFor the damped case, is determined from

2 22

sin( )( ) ( cos( ) sin( ))(1 ) (2 )

t STotal d d

X tx t e A t B tr r

The amplitude becomes

When r = 1, the amplitude of vibration reduces to

After the transient response decays out

2 22

sin( )( )(1 ) (2 )

STotal

X tx tr r

2 22(1 ) (2 )S

S

XX Xr r

12 2 2S

S

FkX XX

X

This means that the amplitude is bounded and the presence of the mass causes the amplitude to build up to the peak value over time.

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Vibration Analysis

Damped Resonance Response

-10

-8

-6

-4

-2

0

2

4

6

8

10

0 1 2 3 4 5 6 7 8 9 10

Time

X

For k = 16, c = 1, m = 1, = 4, and F = 40, the complete resonance response (r =1, = , and = /2) is given in the figure below.

2

sin(0) (0) 0 ( ) sin2 1

tdtFx x x t e t

k

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Vibration AnalysisFor rotating unbalance, F(t) = mee2 sin(t) and the response becomes

2

2

2 2 2 2 2 2

( )sin( )sin( )( )( ) ( ) (1 ) (2 )

e

e

m er tm e t mx tk m c r r

This response plots as the mirror image of F(t) = F sin(t) which means that when r = 0, x(t) = 0 and when r is large, x(t) approaches a constant.

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Vibration Analysis