Vibes 2

8/11/2019 Vibes 2

1/52

8/11/2019 Vibes 2

2/52

8/11/2019 Vibes 2

3/52

Viscous Damping

Schematically, a SDOF mass-spring-dampersystem can be represented as

k

m

)(tx

The damper has neither mass nor elasticity

Forces acting on mass m inx direction

Newtons second law gives

c

m

kF

cF

ck FFxm =&&

0=++ kxxcxm &&& 44

8/11/2019 Vibes 2

4/52

Equations of Motion for

Rectilinear and RotationalSystems

Similarly, torsional SDOF inertia-spring-damper systems can be represented by theequation of motion

0=++ TT kcJ &&&

Rectilinear System Rotational System

Spring Force

Damping Force

Inertia Force

sp acement

Equation of Motion

0=++ TT kcJ &&&

xm &&

0=++ kxxcxm &&&

xc&

kx

x

&Tc

Tk

&&

Spring Torque

Damping Torque

Inertia Torque

Rotation

Equation of Motion

45

8/11/2019 Vibes 2

5/52

Units for Rectilinear and

Rotational Systems

Rectilinear System Rotational System

Stiffness

Damping

Displacement

c

k

x

Tc

Tk

Stiffness

Damping

Rotation

N.m/rad

N.m.s/rad

rad

N/m

N.s/m

m

Force Torque

.

N.mN

46

8/11/2019 Vibes 2

6/52

Viscous Damping

Consider the SDOF mass-spring-dampersystem which is represented as

k

m

)(tx

Let the natural frequency of the undampedmass-spring system be

Note: natural frequency has been shown as

up to now.

c

0)()()( =++ tkxtxctxm &&&

00)( xtx

t =

=

00)( vtx

t =

=&

m

kn =

47

8/11/2019 Vibes 2

7/52

Beware

4difference between Inman and Thomson

for natural frequency symbol4always check definition of variables

4notation even changes throughout thesesubject notes

Define a damping ratio

ccr= critical damping coefficient

km

c

m

c

c

c

ncr 22===

n,

Then

can be written as

Assume solution of the form

Recall derivatives of exponential functions

0)()()( =++ tkxtxctxm &&&

0)()(2)( 2 =++ txtxtx nn &&&

tCetx =)(

[ ] teCtxdt

txd == )()(

&

[ ] teCtxdt

txd 2)()( == &&

& 48

8/11/2019 Vibes 2

8/52

Hence

becomes

Solve

using the quadratic formula

02 22 =++ nn

0)()(2)( 2 =++ txtxtx nn &&&

02 22 =++ tnn Ce

002 22 =++ tnn Ce

For positive mass, damping and stiffnesscoefficients there are three cases

1. Underdamped motion

2. Critically damped motion

3. Overdamped motion

122,1 = nn

10 49

8/11/2019 Vibes 2

9/52

Underdamped Motion

use relationship

where

10

8/11/2019 Vibes 2

10/52

Specifying the damped natural frequency

the solution can be reduced to (Solution 1)

Using the Euler relations

the solution can be expressed as (Solution 2)

sincos je j += sincos je j =

2

1 = nd

( )tBtBetx ddtn

sincos)( 21 +=

tjtjt ddn eCeCetx += 21)(

Defining the constants

the solution can also be expressed as(Solution 3)

22

21 BBA +=

=

2

11tanB

B

( ) += tAetx dtn sin)(

( )

+=

++=

00

012

0022

0 tanxv

xxvxA

n

d

d

nd

51

8/11/2019 Vibes 2

11/52

The damping ratio determines the rate ofdecay of the systems motion.

The damping ratio also determines the shiftfrom the undamped natural frequency to thedamped natural frequency

Underdamped motion is the most commontype of motion for mechanical systems

21 = nd

tne

52

8/11/2019 Vibes 2

12/52

Underdamped Motion

10

8/11/2019 Vibes 2

13/52

8/11/2019 Vibes 2

14/52

Critically Damped Motion

1=

0.5

0.6

1=)(tx

0,4.0 >= vx

( ) tnetCCtx += 21)(

-0.1

0

0.1

0.2

0.3

0.4

t

0,4.0 00

8/11/2019 Vibes 2

15/52

Overdamped Motion

two distinct real roots

1>

12

2,1 = nn

012

>

121 = nn

122 += nn

The solution becomes

which represents a non-oscillatory response

t

Cetx

=)(

+= + ttt nnn eCeCetx 121

1

22

)(

12

1

2

02

01

++=

n

nxvC

12

1

2

02

02

++=

n

nxvC

56

8/11/2019 Vibes 2

16/52

Overdamped Motion

1>

0.1

0.2

0.3

0.4

0.5

)(t

0,4.0 00 == vx

1,0 00 == vx

1>

+=

+ ttt nnneCeCetx

1

2

1

1

22

)(

12

1

2

02

01

++=

n

nxvC

12

1

2

02

0

2

++=

n

nxv

C

-0.5

-0.4

-0.3

-0.2

-0.1

0,4.0 00 == vx

57

8/11/2019 Vibes 2

17/52

Example

Recall earlier Example

4A small spring 30 mm long is welded to the

underside of a table so that it is fixed at thepoint of contact, with a 12 mm bolt weldedto the free end. The bolt has a mass of 50grams and the spring stiffness is measuredto be 800 N/m. The initial displacementfrom the static equilibrium position is 10mm.

Solution was given by

4natural frequency

Now the spring damping coefficient ismeasured as c = 0.11 kg/s

4Calculate the damping ratio and determineif the free motion of the spring-bolt systemis overdamped, underdamped, or criticallydamped

5.1261050

8003=

==

m

kn rad/s

58

8/11/2019 Vibes 2

18/52

59

8/11/2019 Vibes 2

19/52

Evaluating from

measurements

For underdamped oscillatory motion

4solution given by

1

8/11/2019 Vibes 2

20/52

Hence

Now

Define logarithmic decrement as

21

22

==

nd

dT

)(2

1

1

1

dn

n

Tt

t

Aex

Aex

+

=

=

dnTex

x =2

1

=

+=

2

1ln)(

)(ln

x

x

Ttx

tx

2 === Tdn

4 rearranging

Measuring displacement of any two successivepeaks can be used to produce a measuredvalue of damping ratio

4with known m and kcan determine c

2

1 n

nn

21

2

=

224

+=

kmcc cr 2==

61

8/11/2019 Vibes 2

21/52

Evaluating from

measurements (revisited)

For underdamped oscillatory motion

4solution given by

1

8/11/2019 Vibes 2

22/52

For starters

Can express

ex

x

x

x

x

x

x

x

n

n

===== +14

3

3

2

2

1L

( )nn

n

n

ex

x

x

x

x

x

x

x

x

x =

=

++ 14

3

3

2

2

1

1

1L

n

n

ex

x=

+1

1

nx

x=

1ln

In general

Check forn = 1

,...2,1ln1

1

1 =

=

+

nx

x

n n

= 2

1

ln x

x

,...2,1ln1 =

=

+

nxx

n ni

i

63

8/11/2019 Vibes 2

23/52

Alternative measures of

damping

The loss factor is defined as the ratio of theaverage energy dissipated per radian to theenergy in the system

potE

2=

= damping energy per cycle

potE = maximum potential energy

or v scous y ampe systems

4off resonance

4at resonance (r = 1), and for small

Quality Factor, the Q of a system

2

1

2

2

2

11Q

drk

c=

rdr

22 ==

=

drr=

64

8/11/2019 Vibes 2

24/52

Measuring Q

From a plot of displacement ratio for thesystem

4determine value at peak of the systemresponse and its frequency,

4determine the two frequencies to the leftand right of the peak where the level hasdropped to times the peak value1

4 represents the difference between thetwo frequencies (called the half powerpoints)

0 0.5 1 1.5 2 2.50

2

4

6

DisplacementRatio

Frequency Ratio

stX

dr

=Q

2A

65

8/11/2019 Vibes 2

25/52

Energy is proportional to the square of theamplitude

4hence half power points

If system response is displayed in dB, then halfpower points are 3 dB down from the peak

22

22

AAE =

dBA

A3

2

1log20

1

2log20 1010 =

=

66

8/11/2019 Vibes 2

26/52

Usually measure m by weighing the object

Sometimes it is not possible to measure m and

kdirectly4measure natural frequency of the system

and of a modified system

Example

Evaluating m from

measurements

k k

m

m

0m

m

k=1

02

mm

k

+=

( )022

21 mmmk +==

=

12

2

21

0

mm

67

8/11/2019 Vibes 2

27/52

Examine static deflection

Evaluating k from

measurements

k

m

m

1

2

0kk

In linear range

linear

non-linear

= kFk

68

8/11/2019 Vibes 2

28/52

Examples

4 longitudinal vibration along a slender bar

Calculating k from geometry

and material properties

l

Ak=

m

E= Youngs modulus [N/m2]l= length of bar [m]

A = cross-sectional area of bar [m2]

4torsional vibration of a slender bar

G = shear modulus [N/m2]

l= length of bar [m]

Jp = polar moment of inertia of rod [m4]

J= polar mass moment of inertia of disk [kg/m2]

)(t

l

GJk

pT=pJ

For a solid cylinder

32

4dJp

=

d= diameter of bar [m]

)1(2 +=G

= Poissons ratio69

8/11/2019 Vibes 2

29/52

Examples (continued)

4cantilever beam

Calculating k from geometry

and material

33lIk=m

E= Youngs modulus [N/m2]l= length of beam [m]

I= second moment of area of beam [m4]

4helical spring

G = shear modulus [N/m2]

d= diameter of spring material [m]

2R = diameter of turns [m]

n = number of turns

3

4

64nR

Gdk=

2

70

8/11/2019 Vibes 2

30/52

Spring Combinations

In parallel

1k 2k

m

21 kkkeq +=

In series

1k

2k

m

21

111kkkeq

+=

71

8/11/2019 Vibes 2

31/52

SAMPLE SPRING CONSTANTS

Table from [1].

72

8/11/2019 Vibes 2

32/52

TABLE OF SPRING STIFFNESS

Table from [2].

73

8/11/2019 Vibes 2

33/52

Harmonic Excitation of

Undamped Systems

So far we have considered the response ofsystems to free vibration

4if a system, after an initial disturbance, isleft to vibrate on its own, the ensuingvibration is known as free vibration. Norepeated external force acts on the system.

Now consider the case of forced vibration

4

the resulting vibration is known as forcedvibration

4 the external force is often a repeating force

e.g. IC engines, aircraft propellers

4 in fact, many external forcing functions canbe represented as an infinite sum of

harmonic terms

74

8/11/2019 Vibes 2

34/52


Undamped Systems

For the mass-spring system, we assume thatthe driving forceF(t) has the form of a sine orcosine function of a single frequency

k )(txk)cos()( 0 tFtF dr=

F0 = constant amplitude

dris also known as the input frequency or theforcing frequency

rearrange as

4Solution

xh = homogeneous solutionxp = particular solution

)(tx

)(t )(tdr= driving frequency

)()()( tkxttxm =&&

)()()( ttkxtxm =+&&

m

Ff

m

ktftxtx dr

000

2 ,),cos()()( ===+ &&

ph xxx +=

75

8/11/2019 Vibes 2

35/52

The homogeneous solution has the harmonicform used earlier

Assume particular solution has same form asthe driving force

A0 = amplitude of the forced response Substitutingxp into the differential equation

gives

)sin()cos()( 21 tBtBtxh +=

)cos()( 0 tAtx drp =

= drdr

fA ,

220

0

B1 andB2 found from initial conditions

m

kvB

fxB

dr

==

=

,, 02220

01

)cos()sin()cos()(22

021 t

ftBtBtx dr

dr

++=

220

10)0(dr

fBxx

+==

20)0( Bvv ==

)()()( txtxtx ph +=

76

8/11/2019 Vibes 2

36/52

Consider the case where the driving frequencyis very close to the systems natural frequency

For initial conditionsx0 =0 v0 =0, the solution to

is given by

0 dr

)cos(

)sin()cos()(

220

022

00

tf

tv

tf

xtx

dr

dr

dr

+

+

=

[ ])cos()cos()(22

0 ttf

tx dr

=

Using trig identities this can be written as

4two periods of oscillation

The resulting motion is a rapid oscillation withslowly varying amplitude a beat

r

)2

sin()2

sin(2

)(22

0 ttf

tx drdr

dr

+

=

drdr

TT

+=

=

4,

4

77

8/11/2019 Vibes 2

37/52

Beats

An example of a beat

dr

T

+=

4)(tx

0 dr

This phenomenon can be used to matchfrequencies

4e.g. tuning of a piano

t

dr

T

=

4

78

8/11/2019 Vibes 2

38/52

Consider the case where the driving frequencyequals the systems natural frequency

For this case the particular solution of the formis not valid because it is

also a solution of the homogeneous solution

It can be shown that in this case a particular

solution is of the form

Substituting into the equation of motion yields

0= dr

)cos()( 0 tAtx drp =

)sin(2

)( 0 ttf

tx drp

=

)sin()( 0 tAttx drp =

and

Solving for initial conditions

So for a mass-spring system driven at itsnatural frequency,x(t) grows without boundwith time, t. This phenomenon is called aresonance.

)sin(2

)sin()cos()( 021 ttf

tBtBtx dr

++=

)sin(

2

)sin()cos()( 000 ttf

tv

txtx

++=

79

8/11/2019 Vibes 2

39/52

Resonance

An example of a resonance

0= dr

)(tx

tf

20

dr=

Amplitude of vibration becomes unbounded at

Eventually spring would fail and break

m

kdr ==

tf

20

80

8/11/2019 Vibes 2

40/52


Damped Systems

For the mass-spring-damper system, weassume that the driving forceF(t) has the formof a sine or cosine function of a singlefrequency

)cos()( 0 tFtF dr=

F0 = constant amplitudek xkc xc &

Applying Newtons second law on m

rearrange as

dr= driving frequency

)()()()( txctkxttxm &&& =

)()()()( ttkxtxctxm =++ &&&

)cos()()(2)( 02 tftxtxtx dr =++ &&&

m )(tx

)(t

m

)(t

m

c

m

Ff

m

k

2,, 00 ===

81

8/11/2019 Vibes 2

41/52

8/11/2019 Vibes 2

42/52

For light damping the transient response maylast a significant amount of time and should not

be ignored

Transient response also important inapplications where the amplitude is large

e.g. earthquakes

or where the positioning accuracy is important

e.g. satellite analysis, Hubble spacetelescope

Devices are usually designed and analysedbased on the steady-state response

a ways c ec t at t s reasona e to gnore

the transient response

)cos()sin()( 0 ++= tAtAetx drd

t

Transient ResponseSteady-state Response

( ) ( )22220

0

2 drdr

fA

+=

= 22

1 2tandr

dr

83

8/11/2019 Vibes 2

43/52

After some manipulation the expressions forthe magnitude and phase of the response can

be written as

( ) ( )2220

20

0

0

21

1

rrf

A

F

kA

+==

=

21 1

2tan

r

r

r= frequency ratio = dr/

0k

is known as the amplitude ratio, theamplification factor, or the magnification factor

4 represents the ratio of the dynamicamplitude of motion to the static amplitudeof motion

Static deflection of a mass by a forceFo is

0F

k0=

84

8/11/2019 Vibes 2

44/52

0 0.5 1 1.5 2 2.50

1

2

3

4

5

6

7

8

9

10

NormalisedAmplitude

Frequency Ratio

0

20

f

A

dr

05.0=

1.0=

5.0=

1=

05.0=

1.0=

1=

5.0=

0 0.5 1 1.5 2 2.50

Phase

Frequency Ratio

2

dr

85

8/11/2019 Vibes 2

45/52

Points to note from graphs

As the driving frequency approaches theundamped natural frequency (r1)

4 the magnitude approaches a maximumvalue for light damping

4 the phase shift crosses through 90

this defines resonance for the damped

case As the driving frequency approaches zero

(r0)

4 the amplitudek

fA 0

20

0 =

As the driving frequency becomes large (r)

4 the amplitude approaches zero

As the damping ratio is increased

4 the peak in the magnitude curve decreasesand eventually disappears

As the damping ratio is decreased

4 the peak value increases and becomesnarrower

4as (0) the peak value

86

8/11/2019 Vibes 2

46/52

Example

Consider the mass-spring system consisting ofa spring and a bolt from previous examples

Calculate the amplitude of the steady stateresponse iffo = 10 N/kg and the drivingfrequency is 126.5 rad/s, and the amplitude ifthe driving frequency is 120 rad/s

4Calculated previously 0087.0

=

87

8/11/2019 Vibes 2

47/52

To find r at the maximum value of

Set

4

4

Define the peak frequency

0

0

F

k

00

0

=

dr

FkAd

210

2

121 2

>=

8/11/2019 Vibes 2

48/52

Rotating Unbalance

Many machines and devices have rotatingcomponents and small irregularities in thedistribution of the rotating masses can causesubstantial vibration

4called a rotating unbalance, e.g.

uneven load in a washing machine

small imperfections in machining, ,

blades etc.

Machine mount

2

kc

2

k

tre

0m

Friction free guide

Machine of mass mMass of the unbalance

89

8/11/2019 Vibes 2

49/52

Model equivalent system

k c

tre

0m

Machine of mass m

)(tx

r= frequency of rotation (constant)

m = mass of the unbalance

x component of motion of m0

Reaction forcex component

Reaction forcey component cancelled by the

guides

m = total mass of the machine including m o

Fr = reaction force generated by the unbalance

x = displacement of the non-rotating mass (m-m0)

tex rr sin=

( )tdt

demxmF rrr sin2

2

00 == &&

tem rr sin2

0=

90

8/11/2019 Vibes 2

50/52

Displacement of m0 from the static equilibriumposition is

Summing forces in thex direction yields

This form of the equation of motion has beenanalysed before (Forced Vibration of DampedSystems)

The steady state solution is given by

tex rsin+

( ) xckxtexdt

dmxmm r &&& =++ sin)( 2

2

00

temkxxcxm rr sin2

0=++ &&&

where

Rearranging gives the dimensionlessdisplacement magnitude

( ) =

tXtx rp sin)(

( ) ( )2222

0

rr

r

cmk

emX

+=

=

2

1tanr

r

mk

c

( ) ( )2222

0 21 rr

r

em

Xm

+=

=

2

1

1

2tan

r

r

r= frequency ratio = r/91

8/11/2019 Vibes 2

51/52

0 0.5 1 1.5 2 2.50

1

2

3

4

5

6

NormalisedAmplitude

Frequency Ratio

em

m

0

r

1.0=

25.0=

5.0=

1=

0 0.5 1 1.5 2 2.50

Phase

Frequency Ratio

1.0=

25.0=

1=

5.0=2

r

92

8/11/2019 Vibes 2

52/52

Points to note from graphs

For (>1) the maximum deflection is less than

or equal to 14 indicates that the increase in amplification

of the displacement caused by theunbalance can be eliminated by increasingthe system damping

4 large damping not always practical $$

As the rotational frequency becomes large(r)

4 the amplitude ratio 1

if rotational frequency is such that r>>1the effect of the unbalance is limited

4

approach 1 choice of damping coefficient not

important for large r

As the rotational frequency approaches theundamped natural frequency (r1)

4 the magnitude approaches a maximum

value for light damping4 the phase shift crosses through 90

As the rotational frequency approaches zero(r0)

4 the amplitude 0

Documents

Vibes 2