Vibration analysis lectures

8/7/2019 Vibration analysis lectures

http://slidepdf.com/reader/full/vibration-analysis-lectures 1/283

Indian Institute of Technology Kanpur

Lecture 2

Basic Vibratory Phenomena




Simple Mechanical System

Physical System Mathematical Model

Can be a Modal Model




WHAT IS A DOF ?

• The ability to move in any one direction is a DOF.

• The number of co-ordinates required to specify the

motion of a system uniquely determines the order of that system.

6 DOFs1 DOF




NATURAL FREQUENCY AND TIME RESPONSE

frequency. natural theis where

)sin(sincos :Solution

0 :motion of Equation

m

k

t C t Bt Ax(t)

kxxm

n

nnn

=

+=+=

=+

ω

φ ω ω ω

&&

xm &&




2

max

2

max

0

0

00

sin

cos

)sinsincos

:case general In the

tstcos

:shown that becanit algebra, someAfter

0 and0thatassume uslet instance,For

.conditionsinitialon the depend),(or ),( constants The

nnn

nnn

nnn

nn

n

C ωx)t ( ωC ω(t)x

C ωx )t ( ωC ω(t)xt ( ωC t Bt Ax(t)

in

x

xx(t)

x)( x x)x(

C BA

=→+−=

=→+=+=+=

+=

==

&&&&

&&

&

&&

φ

φ

φ ω ω

ω ω ω

φ

DISPLACEMENT, VELOCITY & ACCELERATION - I




)sin

)cos

)sin

2

φ

φ

φ

+−=

+=

+=

t ( ωC ω(t)x

ON ACCELERATI

t ( ωC ω(t)x

VELOCITY

t ( ωC x(t)

NT DISPLACEME

nn

nn

n

&&

&

DISPLACEMENT, VELOCITY & ACCELERATION - II




SPRINGS - I

DISPLACEMENT x

FORCE F F=kx

F=kx 0

x0

Strain energy=1/2

kx 02




• What provides the spring (or restoring )

force?

• A spring

• A float mechanism

• Beam flexibility

SPRINGS - II

• Gravity




STIFFNESS FORMULAE






POSITION OF EQUILIBRIUM - II

kxxmk

mg xk mg kymg xm

F xm

k

mg xxy

x

k

mg

stiffnes

Weight

−=

+−=−=

=

+=+=

==

∑

&&

&&

&&

)(

:law second sNewton'

:be willdeflection total, is amplitude vibrationtheIf

:deflection Static

δ

δ




TRANSLATIONAL, TORSIONAL & ROTATIONAL SDOF SYSTEMS




PENDULUM WITH SPRING

0mgL)(

small assuming and arrangingRe

2 =++

−

θ θ

θ

kaI &&

L mgsin)sin(

: arranging-Re

L mgsindistanceweight xEffective

:moment restoringGravity)sin(distancent xdisplacemexStiffness

:moment restoring Spring

:apply willWe

θ θ θ

θ

θ

θ

−−=

−=

−=

=∑

aak I

aak

Moment I

&&

&&

I

mgL)( 2

n

+=

kaω

Note: I=mL2

L

m

a

θ

mg




TORSIONAL SYSTEM

srad I

k

rad NmLGJ k

r

J L

G

J

T

T Ang dispTorquek

T n

T

T

/ 4.4016.0

261

/261/ :Combining

2 where Also

stiffness Torsional

4

===

====

===

ω

π θ

θ

G= 80 GPa

D=10 mm

L= 300 mm

k T θ

0=+ θ θ T k I &&

I=0.16 kgm2




KINETIC ENERGY vs STRAIN ENERGY

2

2

2

12

1

kxSE

mvKE

=

=

x

v




ENERGY METHOD TO DETERMINE ωn

MAX KINETIC ENERGY = MAX STRAIN ENERGY

Max displacement,

zero velocity

Max velocity,

zero displacement

nnn

n

ωxMaxt xt x

xMaxt xt x

00

00

cos)( :Velocity

sin)( :ntDisplaceme

==

==

ω ω

ω

&

m

k kxωxm nn == ω 2

0)0 :Hence2

1(

2

1 2

FOR THE SDOF MASS-SPRING SYSTEM

0 0 AND xxωxx MAX nMAX →→&




MORE COMPLEX SYSTEMS

i

iiii

QxD

xV

xT

xT

dt d =∂∂+∂∂+∂∂−

∂∂ &&

GENERAL LAGRANGE EQUATION




VERY COMPLEX SYSTEMS Finite element model of a car




Simple Vibratory Phenomena

External Force – Time dependent usually periodicOne harmonic – rotational or multiples is sufficient

Mass times Acceleration

opposite to acceleration

Damping coefficient times

Velocity – dissipates energy

Mean Equilibrium position

under self weight

Static Deflection d = mg/k

Linear System

Inertia Force proportional to accln

Damping force proportional to vel

Stiffness force proportional to displ

System that can be described by one coordinate, say, x




Torsional Vibration

Angular vibrations of any drive train

• A serious problem in reciprocating machines limiting

the speeds, multi cylinder engines are better to even out

the highly pulsating torque

• Usually not a problem in rotating machinery as the drive

torque is fairly uniform

• Torsional vibrations can be very severe under suddenly

applied loads, e.g., rolling mills, electrical short circuitconditions … Under these conditions, couplings, gear

boxes … are susceptible for failures

• Choose proper coupling to make it work well under

normal conditions and act as a fuse under severe loads

and protect the machinery

• Torsional (Angular) stiffness Nm/rad

• Mass moment of inertia Kg-m2

rad/s

02

2

I

k p

k dt

d I

=

=+ θ θ




Bending Vibration

Bending (Flexural) vibrations of a drive train

• Most common problem in all rotating and

reciprocating machinery

• All heavy duty machinery operate above first

critical speed

• A small unbalance (residual balance or

imbalance) can cause serious problems atcritical speeds

• Bending vibrations effected by

misalignment, loosely mounted parts, bearing

stiffness, gears, asymmetry, instabilities due tooil film, etc… They cause most machinery

problems




FREE MOTION OF A DAMPED SDOF SYSTEM

.ratio damping critical viscoustheis2

frequency, natural undamped theis where

02 :by Dividing

0 :motion of Equation

0

2

kmc

cc

mk

xxxm

kxxcxm

n

nn

==

=

=++=++

ζ

ω

ω ζω &&&

&&&




SOLUTION OF THE EQUATION OF MOTION

)( :form theof isSolution

02 :EOM

21

2

t Be

t Aet x

xxx nn

α α

ω ζω

+=

=++ &&&

n

nnt Ae

αt Aex(t)

ω ζ ζ α

ω α ζω α α

α α

)1(

0)2(

EOM. into insert , & find To

2

2 ,1

22

2 1

−±−=→

=++

=

A & B are two constants depending on initial

conditions.




frequency. natural damped theis1 where

)sin()1sin(

)]1sin()1cos([)(

:becomessolution general The)1(

)1(:complex are rootsBoth

2

2

22

2

2

2

1

nd

d n

nn

nn

n

n

n

t C t

et C t

e

t Bt A

t

et x

i

i

ω ζ ω

φ ω ζω

φ ω ζ ζω

ω ζ ω ζ

ζω

ω ζ ζ α

ω ζ ζ α

−=

+−=+−−=−+−

−=

−−−=

−+−=< motion yoscillator withcase dUnderdampe 1.ζ 1. Case

POSSIBILITIES, DEPENDING ON THE VALUE OF ζ




TIME HISTORY FOR OSCILLATORY MOTION

Exponential term

t e nζω −

)sin( φ ω +t C d

Oscillatory term

t




CASE 2

t Be

t Aet x

n

n

n

n

ω ζ ζ

ω ζ ζ

ω ζ ζ α

ω ζ ζ α

)1(

)1()(

:becomessolution The

)1(

)1(:real are rootsBoth

.

2

2

22

2

1

−−−+

−+−=

−−−=

−+−=

> motion yoscillator no withcase Overdamped 1.ζ




EFFECT OF OVERDAMPING




CASE 3

t eBt At x n

n

ω

ω α α

−+=

−==→

=

)()(

:becomessolution The

:root Double 21

decay of rate Max

motion damped-Critically 1.ζ




0)4(

0)5.2( ,0)1(0)0(

:conditions Initial

=

===

x

xxx&

EXAMPLE: SYSTEM IDENTIFICATION FROM TIME RESPONSE

t

4000 kg

Find k and c.

1.0 s 2.5 s 4.0s

x(t)

T=2π/ωd=3

s

ωd= 2π/3

I di I tit t f T h l K




LOGARITHMIC DECREMENT - I

n

n

d

nn

mmn

AMP

md mn

md mn

eeT

e

t t e

t C t

e

t C t e

ω ζ

π ζω

ω π ζω ζω

ζω

φ ω ζω

φ ω ζω

21

2

2

)()x(t

)x(t

)sin(

)sin(

)x(t

)x(t

:cycles successive obetween tw ratio amplitude The

1

1m

m

111m

m

−===

−−=

+−+−

=

+

+

+++





LOGARITHMIC DECREMENT - II

)x(t

)x(tln

1

:cycles 1)-(Nby separated are amplitudes twotheIf

221

2

)x(t

)x(tln

:sidesboth of logarithms Taking

2

1

2

)x(t)x(t

Nm

m

1m

m

1m

m

+

+

+

=

≈−

==

−=

N

e

δ

πζ

ζ

πζ δ

ζ

π ζ





Viscous Damping Principle

• Damping force is proportional to velocity and

= Damping Coefficient C times Velocity dx/dt

– dissipates energy• Dashpots can de designed as in shock

absorbers or the equivalent effect of energy

dissipating capacity determined from tests to

find the value of this coefficient c

s/m-N

)(

δ

µ π

δ µ π

µ τ

Dt c

xcv

Dt F

dz

dv

=

==

=

&





DAMPING IS NOT ALWAYS VISCOUS !

• Viscous damping ratio: ζ = c/c0

0 =++ kxxcxm &&&

0 )1( =++ xik xm η &&

•Aerodynamic damping: δ (e. g. gas pressure on a blade)

• Friction damping (energy dissipation via contact mechanism)

π δ η ζ 2/2/ ==

• Material damping (what the material can dissipate in onecycle)

AT RESONANCE

Dashpot





EXAMPLE OF DAMPED MOTION - PARACHUTE

Find the maximum compression in thespring if m=20 kg, k=10kN/m, c=540Ns/m

and v=8 m/s.

Compression = static compression + dynamic compression

mx

x

k

mg

stiffness

weight d 0196.0

100010

81.9201 ====

2d





USE INITIAL CONDITIONS TO FIND A & B IN EOM:

x

xo=-d1=-0.0196 m & v0= 8m/s

434.0 gives A,for onSubstituti

/8)0(

]sin)(cos)[(

)sinsin()sincos()(

)sincos()( :atingdifferentibyocityObtain vel

, =

=−=→

+−−−=

+−−++−−=

+−=

Band

smABx

t BAt ABt

e

t Bt At

et Bt At

et x

t Bt At

et x

d n

nd

d nd d nd n

d d n

d d d n

n

d d n

ζ ω ω

ζ ω ω

ω ζ ω ω ω ζ ω ω ζω

ω ω ζω

ω ω ω ζω

ζω

ω ω ζω

&

&

mxAx

t Bt At et x d d n

0196.0)0(

)sincos()(

0 −===+−= ω ω ζω

d1 Datum for motion

t=0





i

iiiiQx

D

x

V

x

T

x

T

dt

d

=∂∂

+∂∂

+∂∂

−

∂∂

&&

0D :damping No

0Q :vibration-Free

systems)SDOFfor 1(inumber ordinate-Co :i

force External :Q

functionndissipatio Damping :D

energy spring & Potential :energy Kinetic :

=

=

=

V T

0

x is some general arbitrary co-ordinate.

0=∂∂

+

∂∂

x

V

x

T

dt

d

&

SDOF

LAGRANGE’S EQUATION OF MOTION





EXAMPLE 1 – SDOF SYSTEM

FIND THE EQUATION OF MOTION

00

2

1

)(2

1

2

2

=+→=∂

∂+

∂

∂=∂

∂

→=

==

∂∂

→=∂∂

→=

kxxmx

V

x

T

dt

d

kxx

V

kxV

xmxmdt

d

x

T

dt

d xmx

T xmT

&&

&

&&&

&

&

&

&





EX 2 - PENDULUM WITH SPRING

energy potential:PE energy, Spring:

Change: where

0

2

1 2

SE

PE SE V

D

I T

∆∆+∆=

=

= θ &

)cos1()(2

1

)cos1(

y where

2

1SE

2

2

θ θ

θ

θ

−+=

−=

==

mgLak V

mgLPE

aky

L

m

a

θ

mg

L

mgLcosθ





OBTAIN DERIVATIVES

θ θ θ θ

θ θ

θ θ

θ

θ

θ

θ

)(sin

)cos1()(

2

1

)(

2

1

22

2

2

mgLkamgLkaV

mgLak V

I I

dt

d T

dt

d I

T

I T

+≈+=∂∂

→

−+=

==

∂

∂→=

∂

∂→

=

&&&

&

&

&

&





INSERT INTO LAGRANGE EQUATION

0)(

0

2 =++

=∂

∂

+

∂

∂

θ θ

θ θ

mgLkaI

V T

dt

d

ii

&&

&

I

mgL)(kaωn

+=

2




gy p

FORCED VIBRATION

PI CF +=x(t):Solution

Steady-state

x(t)=xCF(t)+ xPI(t)

xCF(t)Transient

xPI(t)

Periodic solution

RHS)withODEorder -(2nd

)( :EOM t F kxxcxm =++ &&&




gy p

PI. a guess toneed now We

sin)(

:only excitation of typeoneconsider willWe

F(t). RHS, theknowmust wePI, obtain the To

)sincos()(

:bygivenisCF the

0 t F t F kxxcxm

t Bt At

et x d d n

ω

ω ω ζω

==++

+−=

<

&&&

,1ζ onlyconsider we If

)( OF SOLUTION t F kxxcxm =++ &&&




gy p

PROPERTIES OF THE PI

When a linear system is subjected to a harmonic

excitation of the formFsinωt

,• It will respond harmonically at the same frequency.

• There will be a phase lag between the force and the

response.

)sin()( :0 sin)( :

0

0

φ ω ω ω

−=∞<<=

t xt xOutput t F t F Input

PI




PHASE LAG

t F t F ω sin)( 0=

)sin()( 0 φ ω −= t xt x

-100

-50

0

50

100

0 0.1 0.2 0.3-100

-50

0

50

100φ

Tim

e




SOLUTION FOR THE STEADY-STATE VIBRATION

22220

0

2222

00

00

1

2tan& )2()1(

1

2 and Let

tan&

)()(

sin EOM theinto )sin()( PI the

insertingby found becan vibrationstatesteady for thesolution The

r

r

r r k F

x

km

cr

mk

c

cmk

F x

t F kxxcxmt xt x

n

PI

−=

+−=

==

−=

+−

=

=++−=

−

ζ φ ζ

ζ ω

ω

ω

ω φ

ω ω

ω φ ω &&&




FREQUENCY RESPONSE FUNCTION (FRF)

)&properties (systemFunction)&c m, (k,FunctionInput

Output

:form theof isIt

)()(1

)2()1(1

:expression heConsider t

2222220

0

ω ω

ω ω ζ

==

+−=+−==cmk r r k F

xH

Such a function is called Transfer Function in general

It is called Frequency Response Function (FRF) in

vibration analysis.




FRF PLOT

I

RESONANCE

φ=900 at resonance




EFFECTS OF DAMPING

Reduces

response at

resonance.

Has little effectelsewhere.

Has relatively

little effect on

resonantfrequency.




Another look at RESONANCE




Another look at RESONANCE……contd…
















Q-FACTOR

ζ

ζ

ζ

2

1

)2()1(

1

& )2()1(

:have We

1r

0

1

222

1r

0

0

1r

0

0

222

00

=

=→

+−=

=

=

=+−

=

=

==

=

d

xQ

r r k

F

x

d

xQ

k

F d

r r k

F x

r

1r

deflection Static

amplitudeResonant 0

=

==d

xQ

Inverse of Damping Ratio x 2




HALF-POWER POINTS

Half-power pointsX0 /d

ω

Q

Q/1.414

r =1

1.0

points.power half called are

2

amplitude

toingcorrespond sFrequencie

−

=Q




DAMPING ESTIMATION FROM AN FRF

2)(such that Determine

/at 1

)()(

1222

0

0

res

n

n

res

H

H

mk c

H

cmk F

xH

=

===

+−==

ω ω

ω ω ω

ω ω

Q

H H H t

ccmk

n

res

nn

1 2kmc shown that becanIt

2)()( such tha & sfrequencie 2 are There

) (inequation quadratic a Yields

2

1

)()(

1

12

212 1

2

222

==−

==

→=

+−

→

≈ ζ ω ω ω

ω ω ω ω

ω

ω

ω ω ω

ω1

ω2

ωn

resH H

2)()( 21

resH H H == ω ω

ω




PHASOR DIAGRAM FOR F=F 0sinωt

π π ω += 2n t

sin0 t F ω 0F t ω

π ω 2n =t

-50

50

0 0.04 0.08 0.12

π π ω 22n +=t

F0 is rotating with speed ω




RELATIVE PHASE BETWEEN DISP, VEL & ACCEL

. by and ,2by leads ie

)sin( )sin(2

by leads ie

)2sin()cos(

)sin(

0

22

0

2

00

0

π

π

π φ ω ω ω φ ω ω

π

π

φ ω ω φ ω ω

φ ω

xxx

t xxt xx

xx

t xt xx

t xx

&&&

&&

&

&

+−=−=−−=

+−=−=

−=

x&&

x&x

ωt-φ




3 CASES TO CONSIDER

m 0

2

xω

m 02xω

0xcω 0kx

0F

ω> ωn

-> Kx 0< mω2x

0

Inertia control

m 0

2xω

0xcω

0kx0F

ω= ωn

-> Kx 0= mω2x

0

Damping control

0xcω 0kx

0F ω < ω

n-> Kx

0> mω2x

0

Stiffness control




A O SO A O



VIBRATION ISOLATION


CAR ENGINE



CAR ENGINE

• Large forces over a wide

frequency range.• They arise from the crank-

connecting-rod-piston system

and combustion process.

• If transmitted to the car body,

severe noise and vibration in

the passenger compartment.

• The engine is therefore

mounted on rubber blocks.


TYPICAL ISOLATION MOUNTS



TYPICAL ISOLATION MOUNTS

UNDAMPED

SPRING MOUNT

DAMPED SPRING

MOUNT

PNEUMATIC

RUBBER MOUNT


HIGH SPEED PUNCH PRESS ON RUBBER MOUNTS



HIGH-SPEED PUNCH PRESS ON RUBBER MOUNTS

MOUNTS


BASIC THEORY



BASIC THEORY

Source of

vibration

0Force Excitation

edtransmittForce :ratio theis know want toWhat we

:damper and spring thetodue is ground toedtransmittforce The

F

F xckxF

T

T

=+=

&

FT sin(ωt+θ)




m 0

2xω

0

xcω

0kxωt-φ

φ

0F

F T

km

c

r

r r r

cmk ck

F F T

n

T

2 and where

) 2()1() 2(1

)()()(

:bygiven is T, ibility,transmissSo,

222

2

222

22

0

==

+− +=+− +==

ζ ω

ω

ζ ζ

ω ω ω

)()(

:thatknow we4, Lecture From)(F

:diagramphasor theFrom

222

0

0

22

0T

ω ω

ω

cmk

F

x

ck x

+−=

+=


HOW TO OBTAIN LOW TRANSMISSIBILITY ?



HOW TO OBTAIN LOW TRANSMISSIBILITY ?

222

22

0 )()(

)(

ω ω

ω

cmk

ck

F

F T T

+−+

==

• WE WANT T TO BE AS LOW AS POSSIBLE.

• T IS SMALL IF ω>>ωn.

• WE WANT TO LOWER ωn

• WE WANT LOW STIFFNESS AND/OR HIGH MASS.




VARIATION OF F0 WITH ω m 0

2xω m 0

2xω

0xcω

0kx

0F

F T 0xcω

0

kx

0F F T

m 0

2xω

0xcω

0kx0F F T

m 0

2xω

0xcω

0

kx0F

F T

ω < ωn

ω = ω n

ω > ω n

ω >> ω n


TRANSMISSIBILITY CURVES



TRANSMISSIBILITY CURVES

TRAN

SMISS

IBILTY

FT/F0

AMPLIFICATION T> 1 ISOLATION T< 1


EFFECT OF DAMPING IN THE ISOLATION REGION



EFFECT OF DAMPING IN THE ISOLATION REGION

TRA

NSMISS

IBILTY FT/F0

LESS ISOLATION WITH INCREASING DAMPING

DECREASING

T


EXAMPLE: ISOLATION OF RAIL NOISE



EXAMPLE: ISOLATION OF RAIL NOISE

RUBBERPADS

Before isolation

After isolation




Lecture –3

Multi-Degree of Freedom Systems

+ Modal Analysis




EXAMPLES OF SDOF AND 2-DOF SYSTEMS



EXAMPLES OF SDOF AND 2-DOF SYSTEMS

SDOF

SYSTEM

2-DOF SYSTEMS


Multi Degree of Freedom Systems



Multi Degree of Freedom Systems

• Real life systems are complex, they can bend, twist and elongate in

axial direction, the mass is distributed, not discrete as assumed inthe simple models, similarly, elasticity is distributed, there are no

perfect springs without mass …

• In reality we have infinite degrees of freedom in a system, for

convenience, we can model them as finite degrees of freedomsystems.

• The methods of modeling have been refined over the years

depending on the computational facilities available at respective

times.

• We will illustrate some methods that allowed us to understand the

way real life practical systems behave and derive (rather study)

some properties of significance to practical vibration engineers and

diagnostics.


DETAILED ANALYSIS OF A 2-DOF SYSTEM



DETAILED ANALYSIS OF A 2 DOF SYSTEM

ELONGATION OF k 1: x1

STEP 1: SPRING DEFLECTIONS

COMPRESSION OF k 2: (x1-

x2) DUE TO DISPLACEMENT

OF BOTH ENDS

REFERENCE

DEFLECTED

k1 force

k2 forcek2 force




STEP 2: FREE-BODY DIAGRAMS

m1 m2

k 2(x1-x2)k 1x1 k 2(x1-x2)

x

1

x2




STEP 3: OBTAIN THE EQUATIONS OF MOTION

0 0)(

)( )(

:2 Mass :1 Mass

:law 2nd sNewton'

2212222212111

212222121111

=+−=−+

−=−−−=

=

+

∑

xk xk xmxk xk k xm

xxk xmxxk xk xm

F xm

&&&&

&&&&

&&

m1 m2

k 2(x1-x2)k 1x1 k 2(x1-x2)

x1 x2




STEP 4: ASSUME SHM WRITE THE MATRIX EOM

=

−

−+

−

=+−

=−+−

=+−

=−+

+

+

−=−=

+

0

0

0

0

0

0

thatgRememberin

0

0)( :have We

2

1

22

221

2

1

2

12

221222

2

22121121

22

212

1

221222

2212111

x

x

k k

k k k

x

x

m

mω

xk xk xω-m

k )xk (k xωm

xωx&xωx

xk xk xm

xk xk k xm

x

&&&&

&&

&&




STEP 4: CONTINUED

0 ])[ ([K] 2 =− xM ω

2

1

0

0

m

m

2

1

x

x

−

−+

22

221

k k

k k k

STIFFNESSMATRIX

MASSMATRIX

MODE SHAPEVECTOR

EIGENVALUE = (NATURAL FREQUENCY)2

2-DOF system-> 2 modes -> 2 natural frequencies & 2 mode shapes




STEP 5: CHECK SYMMETRY & POSITIVE MAIN DIAGONAL

THE MASS & STIFFNESS MATRICES MUST BE SYMMETRIC.

THE MAIN DIAGONAL ELEMENTS MUST BE POSITIVE.

−

−+

22

221

k k

k k k

2

1

0

0

m

m




STEP 6: OBTAIN THE NATURAL FREQUENCIES

FOR LARGE N, THERE ARE MANY NUMERICAL SOLUTION TECHNIQUES.

USE DET=0 FOR SMALL SYSTEMS.

n2)n1( n2.n1

2

2

222

212

21

2

2

22

212

212-

2-

2-

& :sfrequencie natural 2 in Quadratic

0))((

0 0 ])[ ([K]det

:solution trivialnon aFor

0or 0 ])[ ([K]det

0 ])[ ([K]

ω ω ω ω

ω ω ω

ω ω

ω

ω

≤→

=−−→

=

−−

−→=

−

==→=

−+

−+

ω

k mk mk k mk k

k mk k M

xM

xM




STEP 7: OBTAIN THE MODE SHAPES

Insert ωn1 into ([K]-ω 2[M])x=0

By definition, det([K]- ω n1

2

[M])=0x1 & x2 are linearly dependent, but we can obtain x1/x2

2

2

2121

1

2

2

2

1121

1

2

2212112

1

2

1

:mode 2nd for the Similarly,

:

0 :result previous theUsing

k

mk k

x

x

k

mk k

x

xHence

xk )xk (k xω

m

n

n

n

n

ω

ω

ω ω

ω ω

−=

−=

=−+−

+

=

+

=

+


MODE SHAPE INTERPRETATION



1 & 1

:gives k, m,for valuesinsertingthat,Assume

211

2

1

2

−== == nnx

x

x

x

ω ω ω ω

ω

MODE 1 MODE 2

The masses move in

phase. X1 and X2 move by+1 unit each.

The masses move out of

phase. X1 moves by +1 unit,X2 moves by –1 unit.






3 DOF SYSTEM

FREQUENCYTIME




MODAL SUPERPOSITIONMODE 1

MODE 2 MODE 3

= + +

= + +

TIME DOMAIN

FREQUENCY DOMAIN


Torsional System



One of the earliest fatigue failures experienced

is that of the propeller shaft of a steam engine

driven naval ship during the I world war. It was

reported that the propeller shaft (which has the

lowest torsional stiffness in the system because

of its length) upon its failure was stiffened by

increasing its diameter, however, it failed

earlier. Then it was identified that the excitationbecame closer to the new natural frequency

causing fatigue failure in lesser time. From then

onwards, torsional analysis became mandatory

for all reciprocating installations.

• Briefly, we will talk about simple modeling adopted for torsional analysis of a reciprocating diesel engine

driving a generator.

• All the 8 cylinders are considered as discs, whose rotational mass moments of inertia can be determined

and connected by equivalent torsional stiffnesses of the crank shaft.

• The damper connected to cylinder 8 is divided into two separate disks.

• The coupling stiffness is usually the lowest when compared to the stiffness of any of the shaft sections in

the train.

• The generator is modeled as one rotor

• A model thus derived (the details to arrive these values is out of current scope) is given in the next slide.


Reciprocating Engine Installation



The system given here has 13

inertias connected by 12angular stiffnesses, therefore,

we have a system with 13

degrees of freedom.

• We will attempt to understand

the behavior of such a systemand study some important

vibrational terms that are

regularly used in routine

testing and analysis.

• It is suggested that themathematical intricacies in this

process may be ignored by an

engineer in the field – we will

emphasize the physical

concepts that are of concern tofield vibration engineers and

just brush aside the

mathematical stuff (unless, of

course you are otherwise

interested)




• These are governing differential

equations of motion written from

equilibrium conditions, 13 for 13 inertias

• These equations are written in a compactmatrix form

• Mass Matrix

• Stiffness Matrix

[ ] [ ]

[ ]

[ ]

− −−+−

−−

−

=

=

=+

=+−

=−++−

=−+

1212

12...3322

211

11

13

3

2

1

0

0

131312121313

...

0332)21(1122

0211111

k k k k k k k

k k k

k k

M

I

I

I I

M

K M

K k I

K k k k I

k k I

θ θ

θ θ θ

θ θ θ θ

θ θ θ

&&

&&

&&

&&


Free Vibration - Mode Shapes



Let us assume that the system vibrates at a natural

frequency p and when it does, each disk has a

specified amplitude, capital theta with a subscript

denoting the disk number. This assumption gives

what is called an eigen value problem, given below.

• On expansion the above gives, a thirteenth degree

polynomial equation and therefore, thirteen natural

frequencies p1, p2, .. p13. This shows, a n degree of

freedom has n natural frequencies.

• Each frequency gives a specific pattern for the thirteen

amplitudes, with any one amplitude arbitrarily fixed,for example one unit for the first disk

Arranging each shape in corresponding columns, we get the modal matrix.

[ ] [ ][ ]

[ ] [ ][ ] 0

013,...2,1cos

2

2

=−

=−==

M pK

M pK ipt ii

θ θ θ

[ ]

=

NN N N N

N

N

N

θ θ θ θ

θ θ θ θ

θ θ θ θ

θ θ θ θ

θ

321

3333231

2232221

1131211

1313

3

2

1

1313

213

3

2

1

22

113

3

2

1

11

...

1

...

1

...

1

=

==

=

==

=

==

θ

θ

θ θ

θ

θ

θ

θ

θ

θ

θ

θ

θ θ

θ

pp

pp

pp


Bending Vibration



• The conclusions in the previous slide are not restricted to

torsional vibrations alone, they are true for all kinds of

vibration, bending, torsion, axial, combined bending and

torsion etc.

• Here, in this slide we show how the same equations can be

derived for bending, by using influence coefficient approach.• In all vibration problems, the first thing is to set up a

workable

mathematical model, write the eigen value formulation,

determine the natural frequencies and mode shapes.

• Modern FE codes, ANSYS, NASTRAN …use finite

elements, make a CAD model, mesh and ask for the natural

frequencies and mode shapes.

[ ][ ] [ ]

[ ] [ ] [ ]

[ ] [ ] 0

0

0

0......

0...0...

1

1222111

2222221121

1122121111

=+

=+

=+

=++++

=++++

=++++

−

xK xM

xI xM

xI xM

xxmxmxm

xxmxmxmxxmxmxm

nnnnnn

nnn

nnn

&&

&&

&&

&&&&&&

&&&&&&

&&&&&&

α

α

α α α

α α α α α α


FORCED RESPONSE



THE RESPONSE OF THE SYSTEM TO SOME GIVEN

HARMONIC EXCITATION CAN BE FOUND USING A

TRANSFER FUNCTION APPROACH:

][])[ ([K]

:have weresponse, forcedFor orcex)

INPUTxFUNCTIONSYSTEMOUTPUT

12-

2

F H F M X

F[M]) Xω([K]F

ω

roperties,H(system pX -

==→

=

=

=

−ω

WE WANT THE NORMALIZED RESPONSE TO A SINGLE

EXCITATION, APPLIED TO EACH CO-ORDINATE IN

TURN SO THAT WE CAN OBTAIN THE TOTAL

RESPONSE BY SUMMATION.


Orthogonality Conditions



• The mode shapes of a system have specialand useful properties that explain significant

physical observations. We will just state

them.

• Let us take two different modes, say rth and

sth modes. When they are multiplied withmass matrix or stiffness matrix, the result is

zero. This multiplication is orthogonalization

as written in a compact matrix form here.

• Generalized Mass – However, if we choose

the mode shapes r and s to be the same, say,r, we get rth mode generalized mass, similarly,

sth mode generalized stiffness.

• Remember, the mode shapes are

proportional, therefore the generalized mass

and stiffness depend upon the original shapesthat you choose – a unique way of choosing

the shapes is such that the generalized mass is

one unit and the stiffness is p2. They are then

called orthogonolized mode shapes.

[ ] [ ]

[ ] [ ]

[ ]

r

T

r

r

r T r

r T s

r T s

r r T s

r T s

um

u

uM u

sr k uM u

sr uK u

sr muM u

sr uM u

1~

1~~

0

0

=

=

==

≠=

==

≠=




Proportional Damping



• Damping matrix does not obey the

orthogonality properties stated earlier, hence, the

concept of a proportional damping is evolved,wherein, the damping matrix is taken to be the

sum of a mass matrix and stiffness matrix with

appropriate proportionality constants, a and b.

• If we know a and b we can find the

viscous damping ratio in the correspondingmode and thus use experimental value to

write damped modal equations.

• General finite element codes such as

ANSYS, NASTRAN … adopt betadamping in place of viscous damping..

[ ] [ ] [ ]

[ ] [ ] [ ] [ ]( )[ ] [ ]

[ ] [ ] [ ]( ) [ ]

K&&&

&&&

&&&

,2,1 02

22

2

0

0~~

2

2

22

2

==++

+=

+=

=+++

=+++

+=

r pp

p

p

pp

ppI I

pU K M U I

K M C

r r r r r r

r

r

r

r r r

T

η η ξ η

β α ξ

β α ξ

η η β α η

η η β α η

β α


Modal Analysis




Modal Analysis




Modal Analysis




Modal Analysis - example




Modal Analysis – example.. contd…
























VIBRATION ABSORBER








LECTURE 4

Continuous Systems Approach




CONTINUOUS SYSTEMS WAVE EQUATIONcontd….



Axial vibration of Bars


CONTINUOUS SYSTEMS WAVE EQUATIONcontd….



Torsional vibration of RODS


Solution of WAVE EQUATION




Solution of WAVE EQUATION contd….






Solution of WAVE EQUATION contd….

F Vib ti f B



Free Vibration of Bars




BEAMS Bending Vibrations




BEAMS Bending Vibrations contd….




























Example………..RAYLEIGH’s Method




Example………..RAYLEIGH’s Method …. contd…




Example………..RAYLEIGH’s Method …. contd…










RAYLEIGH – Ritz Method contd….




Example………..RAYLEIGH – RITZ Method




Example………..RAYLEIGH – RITZ Method …. contd…






GALERKIN Method






GALERKIN Method …. contd…








Lecture 5

Stability Considerations




STABILITY Considerations Phase-Plane






STABILITY Considerations Phase-Plane ..contd..






STABILITY Considerations Phase-Plane ..contd..








Lecture 6

FINITE ELEMENT FORMULATION

&

COMPONENT MODE SYNTHESIS


Finite Element Formulation – Axial Vibration of Beams









Reference Systems




Example






Guyan Reduction Scheme






Component Mode Synthesis







Example







Stiffness and Mass Matrices for substructures 1 and 2



Reduced Equation




Lecture 7

Rotor Dynamics


Rotor Dynamics is different from Structural

Rotor Dynamics



Dynamics, as we deal with a rotatingstructure. Basically, all the vibration

phenomena will be valid, however, there are

several differences and we have to set up

procedures on handling the rotors and their

vibratory phenomena.

• Rankine is attributed to have mentioned theexistence of a critical speed of a rotor in

1869. He defined this as a limit of speed for

centrifugal whirling.

• There were many doubts whether a rotor can

cross such a critical speed? It was presumed

that it will be unstable after crossing thecritical speed. This is somewhat similar to

Speed of sound and whether one can cross

this barrier in flying.

• We have to wait for nearly 50 years to have

a clear understanding on this topic.

William John Macquorn Rankine (1820-1872)




Jeffcott’s fundamental contribution



• Jeffcott in 1919 treated the problem asforced vibration and identified the basic

principles of rotor dynamics. He showed

that the shaft did not rotate about rest

position but about its own center line, which

is spinning. The spinning rotor whirled

about the mean bearing center line.

• He identified the unbalance in the system

to be the driving force setting the whirl in

forced vibration. He derived simple

differential equations of the system and

solved them.• Jeffcott through this model has shown that

one can cross the critical speed without any

instability.

• In English speaking countries, a rotor

such as the one shown is named after Jeffcott, though in Germany, Scandinavia,

Holland, they prefer to call it as Laval rotor,

because, it was Laval who derived the

expression in the previous page.

• In this model, the shaft is assumed mass less and

the disk to be rigid. The total mass M of the rotor is

put as disk and the stiffness K is represented asshaft. The eccentricity is denoted by EG = a. The

rotor spins about its own axis with an angular velocity

w and whirls with angular velocity n. Jeffcott

considered synchronous whirl, i.e., n=w.






Translatory and Conical Whirl



•Rigid rotor on flexible bearings is same as

a flexible rotor on rigid bearings - however it provides for translatory and conical whirl

modes. The first flexure of the rotor comes

after these whirls.


Bowed Rotor - Bow r 0 at

)(2

2

2

0KreeMaKr dt

dr C

dt

r d M

aaxit i +=++ +ω ω

0α



Rotors get statically bent or bowed or warped

due to sudden thermal loads, or leaving the rotor unattended for long periods without barring.

Sometimes, the rotor is balanced in a tunnel and

left for long periods in a crate without adequate

support to avoid gravity sag. In all these cases,

the rotor comes to a halt at the heavy spot, with

the rotor sag and eccentricity in one line.•In the above, OO’=R is whirl radius, O’E=r0 is

the bow, EG=a is the eccentricity, a0 is the bow

location with the eccentricity.

•If the rotor is dropped from a height, the bow

location angle is 180 deg, which is in a directionopposite to the mass center.

( )( )

( )( )

bow of angle phase

unbalance of angle phase1

2tan

factor bow

2121

0b

2

1-

00

)(

222

0)(

222

2 0

a

a

r R

eeR

e

a

RR

bt iia

t i

+=

Ω−

Ω=

=

Ω+Ω−+

Ω+Ω−

Ω=

=

−−

φ φ

ξ φ

ξ ξ

φ ω φ ω

Response due to bow unbalance.

•Response due to conventional unbalance.


Response - Self Balancing Speed

( )[ ]+= +−− t iaii eBeAeR ω φ φ 0 0

0 0

a =



•A is response due to conventional unbalance and

B due to the bow unbalance.

•Usually, the bow gives a0 = 0, then these two

responses get added, i.e., the unbalance increases.•For a dropped rotor the bow usually is a0 = 180,

then these two responses oppose.

•The above condition leads to a self balancing

speed, Ws, speed at which the response becomes

zero.

[ ] ( )

( ) ( )

( ) ( )

++=

Ω+Ω−=

Ω+Ω−Ω=

+=

−

−−

b

b

t iia

BABA

RB

A

eBeA

φ φ φ φ ψ

ξ

ξ

ψ ω

coscossinsintan

21

21

1

222

0

222

2

0

( ) ( )

( ) ( )

0

222

0

2

0

0

222

0

2

21

180

21

R

R

a

RR

a

R

a

RR

s=Ω

Ω+Ω−

−Ω==

=

Ω+Ω−

+Ω==

ξ

ξ




Rotor with bow phase 1800



Phase relationship of a bowed rotor Whirl Amplitude of a bowed rotor


Rigid Rotors in Flexible Anisotropic Bearings




Rigid Rotors in Flexible Anisotropic Bearings ..contd..












Rigid Rotors in Flexible Anisotropic Bearings

with Cross-Coupling & Damping




Rigid Rotors in Flexible Anisotropic Bearings .contd..





Rigid Rotors in Flexible Anisotropic Bearings .contd..





.contd..




UNEQUAL MOMENTS OF INERTIA




UNEQUAL MOMENTS OF INERTIA …. contd…














Instability in Torsional Systems




Instability in Torsional Systems …. contd…




Hill’s Equation; Mathieu‘s Equation








GRAVITATIONAL Effect …. contd…






GRAVITATIONAL Effect …. contd…






OIL WHIRL




ROTOR STABILITY IN FLUID FILM BEARINGS contd..




ROUTH-HURWITZ CRITERION




ROUTH-HURWITZ CRITERION contd..




Example ROUTH-HURWITZ CRITERION




Example ROUTH-HURWITZ CRITERION …. contd…




Holzer’s Method for Torsional Systems




Holzer’s Method for Torsional Systems …. contd…




Holzer’s Method for Torsional Systems …. contd…




Example …… Holzer’s Method




Myklestad-Prohl Method for Beams






Myklestad-Prohl Method for Beams ..contd…




Myklestad-Prohl Method for Beams ..contd…




GYROSCOPIC EFFECTS




GYROSCOPIC EFFECTS Freely Spinning Disc




GYROSCOPIC EFFECTS Freely Spinning Disc ..contd..




GYROSCOPIC EFFECTS Disc on Shaft



Additional term in place of 0






Dynamic Unbalance




Single Plane Balancing

•Balancing of single discs is a fairly simple task

as

all the unbalance can be considered as confinedto be in one plane.



p

•If you run the rotor with the residual unbalance

as it existed and which is to be corrected, a

vibration pickup on the bearing will sense a

response due to this unbalance. In order to relatethe vibration signal to the rotor, one can have a

phasor placed on the shaft, relative to which the

vibration signal is measured, that gives the

magnitude and phase angle of the response with

the residual unbalance. Let that be a vector Ov as

shown. Note the 0, 90, 180 and 270o positionstaken for the given rotation with 0 as the phasor

location.

•Though we know a phasor location marked on

the shaft, we have no idea where the unbalance

is, now we place a known mass at a known

location as per the chart and measure theresponse which is O+T. The effect of trial mass is

now given by (O+T) – Ov = Tv.

•The correction mass is therefore to be placed at

35o in the direction of rotation from the location of

trial mass, the magnitude is decided by thelengths of the vectors Tv and Ov.




Rigid Rotor – Two Plane Balancing

•Let Fi be the unbalance force in ith plane. Introduce

two equal and opposite forces Fi in plane L. Sincethe rotor is rigid these two equal and opposite forces

ff f



have no effect on the equilibrium of the system. We

have three forces now, two of them form a couple

ML as shown.

•Now split the couple into two equal and opposite

forces in the planes L and R separated by distance a.•In plane L we have now two forces, combine them

to form into one and let this be Fi L and let the force

in the right plane be denoted as Fi R.

•Repeat this to all the n disks and form a set of n

concurrent forces in each of the planes L and R.Find the resultants of these concurrent forces and

denote them FL and FR.

•Thus we reduced the original distributed unbalance

in the system to two unbalance forces in any two

convenient planes – remember this is valid only for

rigid rotors and a process of removing these twounbalance forces is called Two Plane Balancing or

Dynamic Balancing, in place of simple statics used

in Single Plane Balancing or Static Balancing.


Influence Coefficients

•As in single plane balancing fix the

angular locations and use a phasor to



angular locations and use a phasor to

measure the phase. Choose the

measurement planes to be, say left

bearing and right bearing. Let the

responses without any trial mass be asshown.

•Now introduce a trial mass in right plane

and let the responses be as denoted here.

•Similarly, let a trial mass in left plane

give the responses as shown here.

•Using these six responses, we derive the

influence coefficients, measures that tell us

how the responses will be arising out of known forces. The example in the

following pages illustrates how to achieve

the balancing of a rigid rotor.


Example

iL 144.9 90at microns 9.144:L PlaneLeft

MassTrialNo :1Run

1o =



iR

iL

iT

iR

R

274.89906.0 99at microns 6.35 :R PlaneRight

311.2521.4 27at microns 5.08

:L PlaneLeft

608.2294.6

PlaneRightin22.5at 6.8g Mass Trial :2Run

188.7188.7 45at microns 10.16

:R PlaneRight

2o

2o

o

1o

+−=

+=

+=

+=


Example – contd.

T

RRa

RbR

12 −

=



•Influence coefficient denoting the

response in plane b due to a unit force in

plane R

•Influence coefficient denoting the

response in plane a due to a unit force in

plane R

( ) ( )

( ) ( )

i

i

ii

T

LLa

i

i

ii

R

aR

181.1229.0

608.2294.6

144.9311.2521.4

336.0160.1

608.2294.6

188.7188.7274.69906.0

12

+=

+

−+=

−=

+−=

+

+−+−=


Example – contd.

iT L 997.35.5

36at 6.8g :3 RunFinalo

+=



•Influence coefficient denoting the response in

plane b due to a unit force in plane L

•Influence coefficient denoting the response in

plane a due to a unit force in plane L

( ) ( )

i

i

ia

T

LLa

i

iiia

T

RRa

iR

iL

aL

L

aL

bL

L

bL

9.1327.0

997.35.5

144.94.9

76.3558.0

997.35.5188.7188.71.30775.4

1.30775.490at microns 30.5:R eRight Plan

04.9

0at microns 9.4:LLeft Plane

13

13

3

o

3

o

−=

+

−=

−=

+−=

+ +−+−=

−=

+−=

+=


Example – contd.

•We need to eliminate the original response vectors amamR bLLbRR1 +=



g p

by using correction masses in planes L and R, to

estimate these masses, we form the set of

equations given here using the influence

coefficients determined earlier.

•Solving the above two equations, we get

the expressions for correction masses.

•When these correction masses are added, (we may

have to do this in two masses each by placing them

in the nearest locations) the response theoreticallyshould be zero. This does not happen as there are

several assumptions made in this analysis, the first

correction should bring the responses to be low, one

or two additional balancing runs may be needed to

achieve the desired grade quality.

imim

aaaa

aLaRm

aaaa

aRaLm

amamL

amamR

L

R

bLaRaLbR

bRaRL

bLaRaLbR

aLbLR

aLLaRR

bLLbRR

9.181.2

56.194.10

11

11

1

1

−−=+=

−−=

−

−=

+=−

+=−


Shop Balancing set up






Influence Coefficients




Balancing Masses

vv 112 Planein MassTrial



[ ] vaU

T

vva

T

vva

T

vva

k iijk

ij

iij

ij

iii

1

1

th

11

1

21

2

Tests Speedk

...

j Planein MassTrial

−=

−=

−=

−=


Classification of Rotors

•Class 1 – Rigid Rotors: Rotors that can be corrected in any two arbitrary planes and after

correction, its unbalance does not significantly change at any speed upto the maximumoperating speed and when running under conditions which approximate closely to the final

supporting system Rotors which do not satisfy this condition are classified as flexible rotors



supporting system. Rotors which do not satisfy this condition are classified as flexible rotors.

•Class 2 – Quasi Flexible Rotors: Rotors that cannot be considered rigid but can be balanced

adequately in a low speed balancing machine. These are rotors, (1) where the axial distribution

of unbalance is known – e.g., 2A – shaft with a grinding wheel; 2B – Shaft with a grinding wheeland pulley; 2C – Jet engine compressor rotor; 2D – Printing roller; 2E – rotors with a long rigid

mass supported by a flexible shaft, whose unbalance can be neglected, such as computer

memory drum; (2) where axial distribution is not known – e.g., 2F – Symmetrical rotors with two

end correction planes, whose maximum speed does not significantly approach II critical speed

and whose service speed does not contain I critical speed and with controlled initial unbalance;

2G – same as 2F but with an additional central correction plane and that it may have its service

speed in I critical speed range; 2H – Same as 2F rotors but unsymmetrical.

•Class 3 – Flexible Rotors: Rotors that cannot be balanced in a low speed balancing machine

and that require some special flexible rotor balancing technique – e.g., Generator rotors•

Class 4 – Flexible Attachment Rotors: Rotors that could fall in categories 1 or 2 but have in

addition one or two more components that are themselves flexible or flexibly attached – e.g., a

rotor with centrifugal switch

•Class 5 – Single Speed Flexible Rotors: Rotors that could fall into category 3, but for some

reason, e.g., economy, are balanced only for one speed of operation – e.g., high speed motor






L-8



Vibration Measurement

Pickups, Analyzers, Modulation,

Cepstrum Analysis,

Digital Measurement


Why Vibration Measurements?

• We studied basic principles of structural vibrations, and special applications to rotors and their

behavior under dynamic conditions.

• While all this information and much more is basic to the understanding of machine behavior, thecurrent course is concerned primarily with the health of a machi ne. Every machine deteriorates in

condition however well it has been designed/ The rotating machinery is very expensive and they



condition, however well it has been designed/ The rotating machinery is very expensive and they

should be available for their operation for long uninterrupted intervals (years) and without any failures

and unexpected shut downs.

• Asset management is an important aspect to any heavy industry with rotating machinery. We find

several indications reflecting the condition of a machine in its life time; e.g., we know that an

automobile engine needs a change of lubricating oil as it becomes contaminated with dirt over a

period of time, so if we can find a measure of the state of the lubricating oil, an appropriate action can

be planned – or if one finds the bearing temperatures are going up, there is some rubbing and

inadequacy of lubricating oil, so on and so forth. Some of these measures, however, do not tell us the

problem in a machine sufficiently well in time for a proper asset management.• Over years of asset management by maintenance people, we know that the earliest indications of any

fault in a rotating machine are detected by an increase in vibration and sound levels, that is why every

asset management practice adopts vibration measurement as the first and foremost step that may be

further assisted by other measurements such as lub oil particle counting, bearing pressure and

temperature measurement, process parameter variations … Basically it is the vibration level, its

signature in time and frequency domains, orbits, and trends over short and long durations that help usin understanding the health of the machine and predicting any impending problem so that timely

action is taken. Therefore vibration measurement and its analysis is important first step in asset

management of rotating machinery.


Vibration Units of Measure

AMPLITUDE



Displacement in microns, mils, below 10 Hz

Is a measure of the distance the object moves

Velocity in mm/sec, inch/sec, between 10-1000 Hz

Is a measure of how fast it moves - (Speed) - most destructive

energy

Acceleration in mm/sec², Spike Energy - g above all

The force imparted on the vibration object as it changes its velocity.

TIME (FREQUENCY) Cycles / Sec, Hertz, RPM

PHASE Radian, Degree


Displacement, Velocity and Acceleration

• Displacement, velocity



and acceleration for a

given frequency are all

related through• V = Xp

• A = Vp = Xp2

• Usually velocity is taken

as a standard as itrepresents the energy

associated in the system;

for a given velocity and

frequency, we can findthe associated amplitudes

of displacement and

acceleration.





Type of Measurement:

1) Contact (Seismic)

Transducers Selection



1) Contact (Seismic)

2) Non Non-Contact (Relative) Contact (Relative)

•Displacement: Eddy Current Proximity Pickup

•Velocity (Seismic): Electro Dynamic Transducer

•Acceleration: Piezoelectric Pickup

Direction & Location of Measurement: Direction & Location of Measurement:

• Horizontal, Vertical and Axial

• Bearing Pedestal, Shaft, Journal housing, Structure


• Measure Relative Distance Between Two Surfaces

• Accurate Low Frequency Response• Limited High Frequency Sensitivity

i l S

Eddy Probe – Velocity Transducer - Accelerometer



• Require External Power Source

• Sensitive to observed material

• Often Measure Bearing Housings or Machinery Casing Vibration

• Effective in Low to Mid Frequency Range (10 Hz to around 1,500 Hz)

• Self Generating Devices

• Are Electro-Mechanical Devices With Moving Parts That Can Stick or Fail

• Rugged Devices

• Operate in Wide Frequency Range (Near 0 to above 400 kHz)

• Good High Frequency Response

• Some Models Suitable For High Temperature• Require Additional Electronics


Sensor Relationships

Amplitude



• The adjoining figure gives the

displacement and acceleration for a

velocity of 0.628 in/sec

• At 1 Hz, displacement x = v/p, i.e.,

mils

• Corresponding acceleration at 1 Hz

is a = vp

• Common machinery operating

range is shown by the rectangular

area, displacement and acceleration

at any frequency can be obtained in

terms of 0.628 in/sec velocity

Amplitude

(mils, in/sec, g’s)


Vibration Sensor Principle



• A vibration sensor utilizes the basic single degree system response.

• The sensor when placed on a vibrating member experiences x(t) that is to to be measured

• The mass responds with absolute displacement y(t)

• Relative displacement z(t) = y(t) - x(t) and Z/X is shown above used as a measure of x(t)

• The sensor range and application is depends on how we proportion its mass and spring,


Seismometer / Velocity Transducer



• A Seismometer or Velocity Transducer measures rate of

change of relative displacement z(t) = y(t) - x(t) using

the electromagnetic principle through the induced

voltage as illustrated in right side, as a measure of x(t)• The sensor range for Z=X is r >> w/p, therefore p

should as low as possible, i.e., the mass is relatively

heavier to the spring stiffness, then we measure the

velocity of the support and hence Velocity Transducer.

• Electromagnetic Principle

Variation of Permeance of magnetic

circuit causes a change in the flux FX

V

dt

dx

dt

d n

dt

d nV

8

8

10

10

−

−

×−=

×−=

φ

φ


Accelerometer Principle

• An Accelerometer uses Piezo electric principlegiven in next slide to measure the acceleration

directly



• The sensor range for Z to represent Xw2 the

acceleration of the support is r << w/p, therefore p

should as high as possible, i.e., the mass is

relatively lighter • Since p is a constant, we need the factor f to be as

close to unity as possible, then we measure

acceleration directly.

• When the damping ratio is 0.7, the factor f equals

unity to a large possible range of r, say up to 0.2.

Damping is therefore very crucial to an

accelerometer.

• Piezoelectric crystals have very low mass and

very

high frequencies, therefore are very good

candidates for measurement of acceleration. Theycan also be made very light and hence have useful

applications for measurements on small

components without effecting the basic structure.

( ) ( )

( ) ( )222

2

2

222

2

21

1

21

r r

f

f p

r r

r

X

Z

ξ

ω

ξ

+−=

=

+−=





Accelerometer Elements



Output impedance from accelerometer is very high - problems

matching, noise, cable length Charge amplifier is therefore used.

Elements of a measurement system




Sinusoid Signal - Characteristics



• The vibration signal in general is somewhat periodic with

several harmonics, the basic signal is therefore a sinusoid.

• A sinusoid, its squared sinusoid are shown above. Since the

energy in the system is proportional to velocity squared, we

prefer a root mean square value of the amplitude, rms value, to

judge the condition of the machine.


Calculation of RMS Value - Sinusoid



Let the pk to pk value of a sinusoid be 240 mils/s.

• The amplitude is 240/2 = 120 mils/s• Square of the signal varies from 0 to 1202 = 14400 mils/s

• Mean of the square = 7200 mils/s

• RMS = (7200)1/2 =120/1.414 = 84.85 mils/s• Pk to Pk RMS = 169.7 mils/s


Sum of Two Sinusoids in Phase



Both sinusoids amplitude = 1

Peak to peak of sum = 3.5203

Mean = 0

Root Mean Square of sum = 1

The amplitudes of both sinusoids that make

up the sum is obtained in frequency domain.


Sum of Two Sinusoids out of Phase



Peak to peak = 3.2506

Mean = 0

Root Mean Square = 1

• Frequency Domain remains same

• Phase Information lost


Consider a periodic signal with a sum (-0.1 sin 81.1t +0.05

sin 173.52t). Let us find the means, mean squares and root

mean squares of the components and the total signal.1. Let the two components be denoted by 1 and 2. The mean

values of individual components as well as the sum, being

Calculation of RMS with two components

0791.000125.0005.0

0354.000125.02

0707.0005.01

=+=

==

==

total RMS

RMS

RMS



harmonic signals, are zero

M1 = 0, M2 = 0, Mean of sum = 0

2. The time period of the lower frequency is large of the two

and is given by 2p /81.1 = 0.0775 s. We now follow the steps

given here to find MS (mean squares) and RMS [Integrate

and average over the period 0.0775s]

( ) ( )[ ]

( ) ( )[ ]

00125.0005.0

0775.02

0025.00775.0201.0

0775.01

52.173sin0025.01.81sin1.00775.0

1

52.173sin05.01.81sin1.00775.0

1

0775.0

0

22

0775.0

0

22

+=

×+

×=

+=

+=

∫

∫

dt t t

dt t t MS

units 0.1582 signal allover of pk pk to RMS =

2

2cos1sin 2 pt

pt −

=

Note

and integral of cosine term becomes zero




• Fourier analysis is a mathematical tool

th t h l i id tif i th f

Fourier Analysis

( ) ( )

( )T

n

nn t bt naat f 1

0

2

sincos21

∞

=++=

∫

∑ ω ω



that helps us in identifying the frequency

components of a periodic vibration signal

which is composed of several harmonics.• The periodic signal is assumed to consist

of several harmonics of the fundamental

frequency and an infinite series gives

accurate results. In practice, first few

components are considered. The harmoniccosine and sine components are

• The amplitude in each harmonic, n = 1, 2,

is given by

• The phase angle is

( )

( )

( ) ( )

n

nn

nnn

n

n

T

n

n

a

b

baA

nt nAat f

tdt nt F T

b

tdt nt F T

a

1

22

1

0

0

0

tan

cos2

1

sin2

cos2

−

∞

=

=

+=

−+=

=

=

∑

∫

∫

φ

φ ω

ω

ω


Fourier Analysis - Complex Form

• Fourier analysis is written in complex

form so that faster numerical tools can be

developed. The cosine and harmonicterms are written in exponential form as

given here

( )

( )

−=

+=

∞

−

−

111

2

1sin

2

1cos

t int in

t int in

eei

t n

eet n

ω ω

ω ω

ω

ω



g

• The harmonics can be written as one

sided series 0, 1, 2, … infinity

• Or as two sided series, - infinity, …, -2, -1, 0, 1, 2, … infinity.

• In this form we loose phase information,

the harmonics in two sided series have a

magnitude equal to half the values.

• A finite time signal is considered in thisprocess.

• Average power of the signal over a

period of time, T0 is

• Complex harmonic components directly

represent the average power in thecorresponding frequency term, this is

Parseval theorem.

( ) ( )

( )

( )

∑

∑

∫

∫

∑

∑

=

+

=

=

=

=−=

=

++−+=

∞

=

+

−

∞+

∞−

=

−

21

22

0

0

2

0

2

1

21

1

0

22

1

)(1

Complex

Components

Harmonic

2

1

2

1

)( series sided Two

2

1

2

1

2

1)(

0

n

n

n

T

av

T

T

t in

nn

ssnnn

t in

n

n

t in

nn

t in

nn

C

Aa

dt t xT

P

et f

T

C C

AC ibaC

eC t f

eibaeibaat f

ω

ω

ω ω

We will get back to this again


Saw Tooth Signal

( ) 04

2 0<<−= T t T

t t x



• To illustrate the usefulness of Fourier series, consider a periodic

saw tooth signal given above with its components.

• The top figure shows addition of terms one by one up to six.• The figure below shows the net result of sum of first six terms,

which is getting closer to the original saw tooth form.

( )

,...3,2,12

sin4

6sin

3

14sin

2

12sin

4

0

000

0

0

==

++=

k T

kt

k

T

t

T

t

T

t

T

π

π

π π π

π




• We can increase the number of terms in the summation to get more accurate result.

• Sixty terms sum is given in the figure below.• In practice, we are interested in the lower harmonics, or those harmonics which can be

excited to resonance from any per rev or nozzle excitations.




• Here, the sum up to 25 terms is given above with the harmonics in

frequency domain given below.• The first harmonic is one unit, followed by the second at ½ unit and

the higher ones decrease in magnitude rapidly.


Rectangular Signal

( )

...2

5sin5

123sin

3

12sin

42/01

2/01

000

0

0

+++=

<<−=

<<=

k

T

t

T

t

T

t T t

T t t x

π π π

π



• This and the next slide illustrate a rectangular periodic signal and its Fourier components.

,...3,2,12

sin4

0

== k T

kt

k

π

π





Octave Band Analysis



• 4-5 decades ago, a Fourier analysis was a tedious job, the signal is first recorded on a

recorder, e.g., a UV recorder, it was then enlarged and digitized manually to obtain the signal

as a function of time. A hand calculation or a main frame computer was then used to determine

the Fourier components in any diagnostics and trouble shooting exercise.

• Dedicated analog instruments are then developed using filter circuits, which are expensive – the accuracy was limited octave bands, e.g., 11 filters are common with center frequencies

beginning from 31.5 and doubling consecutively.

• A vibration chart thus developed is called octave band analysis, which gave the relative

energy levels in these bands to make a diagnostics analysis.


Sequential analysis

One filter at a time

A l i i



• Subsequently, we had 1/3 octave band analyzers which helped in

narrowing the frequency zones for a better diagnostics.

Analysis time

between each analysis

Significant transientdata may be lost


Fourier Transform – time to frequency domain

•Not limited to periodic functions alone

•Achieved through Fourier Integral•Aperiodic function repeats itself

after a large (infinite) time



•Inverse Transform – Conversion

from frequency to time domain

g ( )

•Associated frequency becomes negligibly small

•Replace w by w’, nw’ becomes continuous•T = 2 / pw’ approaches infinity

Forward Transform – Fourier Integral( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )∫

∫

∫

∞+

∞−

−−

∞+

∞−

−

∞+

∞−

−∞→

==

==

==

→

ω ω π

ω

ω

ω

ω ω

ω

ω

ω

d ef F F t f

dt et f t Ff F

dt et f TC F

n

t i

t i

t iT n

2

11


Discrete Fourier Transform (DFT)

• f(t) is analog should be convertedto a digital text file

• N intervals of a vibration record in

( )

( )

( )1 2

1,2,1,01

1,2,1,0

1,

2,,0

− −

−==

−==

−=

∑N

N

kr i

k

k

NrefF

N k N kT

N

T N

N

T

N

T t

K

K

KK

π



time T, time step T/N

• Frequency of the entire period = 2p /T

• Discrete frequencies 2p /0, 2pN /T , 2pN /2T , 2pN /3T, … 2pN /(N-1)TExample:

• Let 5000 samples be taken in 1 sec.

• Let us use N = 4096 out of these 5000 data points.

• T = 4096/5000

• Smallest discrete frequency = 2pN /(N-1)T = 2p(4096) /[4095(4096/5000)] rad/s

= 5000/4095 = 1.22 Hz.

( )

[ ] 11

0

1

1,2,1,0

×××

=

=

∑

N N N N

k

k r

F AF

N r ef N

F K


Fast Fourier Transform - FFT

[ ] 11

1××× = N N N N F A

NF

[ ] [ ]roweachin elements zeronon only two

A matrices log into factored i2 N A



N

N kr i

kr eaπ 2−

=

[ ]

→↓←↑

↓↑↓↑

←↓→↑

↑↑↑↑

=

3

2

1

0

A

[ ] [ ][ ] [ ]

↓↑

↓↑

↑↑

↑↑

←↑

→↑

↓↑

↑↑

=

=

00

00

00

00

00

00

00

00

1000

0010

0100

0001

21 AAP A

4NFor =

[P] Permutation matrix


• On the present day Desk top or lap top

computers, FFT can be performed in millisecs, thus making the conversion of time

domain data to frequency domain almost

instantaneous and in real time

Comparison - DFT/FFT Computation Requirements



instantaneous and in real time.

• With an A/D converter card, a lap top hasthus become a good diagnostics

instrument and tool.


Amplitude Modulated Signal

( )

( ) ( )aa

a

aa

xtox

x

t t xxt x ω ω

x-x Amplitude

Amplitude Modulation

coscos

+

+=



aω Frequency Modulation

( )

( ) ( )t xt xt x

t t xxt x

aaaa

aa

ω ω ω ω ω

ω ω

−+++=

+=

cos2

1cos

2

1cos

coscos

bandsside

- and

speed rotational

frequencycarrier

frequencymeshgear

gear mountedllyEccentrica

acac

a

c

=

+

=

ω ω ω ω

ω

ω


Amplitude Modulated Signal



Mean = 0

Mean square = 0.56

Root Mean Square = 0.75Side bands in Frequency domain


Frequency or Phase Modulated Signal

( ) ( )( ) ( )

modulation of frequency and magnitude - and'

Re

cos'cos

ω ω

ω ω ω ω

φ =

+=

a

t ia

Xet x

t t X t x



( ) ( )

Signal. ModulatedFrequency

a is signal resultig theand speed rotational

thetoingcorrespond mean value over the changes

frequencyangular the,vibrationalIn torsion

speedangular Variable

magnitudeconstant of Phasor

' to' variesFrequency

index modulation - /'

q yg

ω ω ω ω

ω ω

+−

a

a


Frequency or Phase Modulated Signal



Modulation Magnitude = 0.5

Modulation Freq = 4 Hz

Carrier Freq = 15 Hz

Mean = 0

Mean square = 0.50

Root Mean Square = 0.71


Torsional Vibration Measurement

Ω

vibrationtorsionalofamplitude

noscillatio torsionalof frequencyspeed rotational

θ

ω



Ω

−=

Ω+∞=Ω−∞=

ΩΩ=

Ω+=

2

to

range speedAngular

tsin -locityAngular ve

tcos t

positionshaft ousInstantane

vibrationtorsionalof amplitude

minmax

maxmin

ω ω θ

θ ω θ ω

θ ω φ

θ ω φ

θ

&




Power (Auto) Spectrum - Cross-Spectra

• Amplitude spectrum is given by the magnitude

and frequency of each harmonic component.• Power spectrum is obtained by squaring each

amplitude and halving them and plotted in

discrete form at each frequency.

( ) φ ω AF nn

1

SpectrumPower

,

Amplitude Spectrum

Discrete



discrete form at each frequency.

• Auto spectrum is same as power spectrum when

the frequency domain is expressed continuously,

this is obtained by taking the product of F(w)

and its conjugate marked by *

• Auto or Power spectrum is a measure of power

associated with corresponding frequency

component and hence important for vibration

engineers.• Cross spectrum involves two different functions.

Let us take the excitation and response functions

and their cross spectra are defined by

• Cross spectra are used in defining coherence,

which is a degree of linear dependence between

two signals, see next slide.

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )ω ω ω

ω ω ω

ω ω ω

F X

X F x(t)

f(t)

F F

An

*S

*S Response

Excitation

*S

Spectrum Auto2

1

xf

fx

ff

2

=

=

=Continuous

Cross spectra


Coherence

• Consider the frequency response function.

• Express this function in two different ways, first

using the conjugate of amplitude spectrum of

forcing function and next by using the conjugate

( )( )( )ω ω

ω F

X H =

Frequency Response Function



of amplitude spectrum of response function.

• Coherence is defined by the ratio of these two

response functions.

• If the measurement of forcing function and

response function are free from noise error and

that there is a perfect linear relationship between

them, Coherence will be unity.

• Coherence can be used to determine whether two

different signals are coming from the same

machine or if any of the signals are lost in

measurement due to some fault, coherence

between a good and a bad signal will be poor.This property can be used in identifying any

faulty sensor or problem in the transmitting path.

( )

( )

( )( ) ( )( ) ( )

( )

( )ω

ω

ω ω

ω ω

ω

ω

ω

ff

xf

S

S

F F

F X

F

X

H

=

=

=

,*

*

( )( )

( )( ) ( )( ) ( )

( )( )ω ω

ω ω

ω ω ω

ω ω

fx

xx

S

S

X F

X X F

X H

=

=

=

,*

*

2

( )( )( )

( )

( )

( )

( )ω

ω

ω

ω

ω

ω ω

xx

fx

ff

xf

S

S

S

S

H

H y

=

=2

12




• Definitions of Cepstrum• Terminology

Cepstrum for Spectrum

Quefrency (sec) for Frequency

Cepstrum - (Time Domain)

( ) ( )[ ] 2log

spectrumpower clogarithmiaof spectrumPower

ω τ xxxx S F C =

spectrumpower clogarithmi of ansformFourier tr Inverse



Q y ( ) q y

Rahmonics for Harmonics

Lifter for Filter Gamnitude (Hz) for Magnitude

Saphe for Phase

• Applications

• Side bands of a multi stage gear box,

they appear as distinct frequencies incepstrum domain.

( ) ( )[ ]ω τ xxxx S F C log1−=

sec/1and/1

frequencyat peaksdistinct Two

and sdifferencefrequency

bands side of families Two

cepstrumin peaksdistinct asAppear spectrumainstructures periodicIdentify

21

21

∆f ∆f

Hz ∆f ∆f

( ) ( )[ ]ω τ X F C xx log

spectrum amplitude of ansformFourier tr Inverse

1−=


Two Speed Gearbox with Defects




Advantage of Cepstrum Analysis




A/D Converter

• Analog to Digital converters and

modern high speed desk top computers

have revolutionized the way we record

and analyze vibration and other

process parameter time domain

signals The analog tape recorders



signals. The analog tape recorders

have disappeared and the storage has

become very simple and reliable. Theanalog instruments have also been

replaced by software and we can have

today a sensor and its conditioner

directly connected to a computer for

all the required analyses in real time.

• A 3 bit parallel A/D converter isillustrated here. Analog voltage

compared with each node’s voltage

• Output voltage high (on) when the

analog voltage is above ref. voltage

and low (off) when it is below.

• Binary encoder compares and gives a

3 bit binary output.


3-bit encoder binary output



An encoder circuit which reads the comparator outputs, high or low and produces a 3-bit binary

output corresponding to one of the eight possible

on/off conditions of the inputs 1 through 7


Resolution with

16 bit comparator

Examples

3 bit comparator

V10to10-:tageoutput volfor range Typical +8

5and

8

4between signalInput ref ref E E



range 10V to10-for 0.3mV resolution Voltage

increments 65,536 2 into divided Range

cardconverter A/Dbit-16

bits of number n range, scale full

2

1Ebitper resolution Voltage

2

1output sencoder' parallelbit-3

removaloffset dcor nattenuatio amplitude

viz.,ngconditioni signal eAppropriat

limitlower or upper theexceeds

signalinput theif saturatesConverter

16

nv

3

+

=

∆

∆=

fs

fs

ref

V

V

E

( )

0)(000 isoutput thelow, read scomparator All

7)(111 isoutput thehigh, read scomparator All

4020212100

outputbinary toingcorrespond

4 is stateInput

low read 7-5 andhigh read 4-1 scomparator

2

2

012

2

=

=

=×+×+×=


Signal Sampling

• Signal sampling is an important

criterion while acquiring the data

• The same signal above shows

t f s

∆=

1 rate Sampling



g

inadequate sampling rate, number of

points captured in a unit time is toolow and the original signal character

is lost.

• Decreasing sampling time or

increasing sampling rate improvesthe digitization process.


Aliasing – Nyquist Criterion



sNyq f f 2

1

=


Sampling Example

( )f N∆∆2

1 n∆∆f 2∆∆f ∆f, 0, resolved sFrequencie Nyq=KK



Hz 1.225000/4096 f samples theseof 4096 N

secondper samples 5000 f signalVibratory

4096 FFTin samples of Number

Hz 4000 rage sampling Minimum

rpm) (120,000 Hz 2000 -componentfrequencyHighest

s

==∆=

=

>



Documents

Vibration analysis lectures