Vibration analysis lectures

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.

1 DOF

6 DOFs


NATURAL FREQUENCY AND TIME RESPONSE

m&x&

Equation of motion : m&x& + kx = 0 Solution : x(t) = A cos ω nt + B sin ω nt = C sin(ω nt + φ ) k where ω n = is the natural frequency. m


DISPLACEMENT, VELOCITY & ACCELERATION - I

The constants ( A, B ) or (C , φ ) depend on the initial conditions. For instance, let us assume that x( 0 ) = x0 and x&( 0 ) = x&0 After some algebra, it can be shown that : x&0 x(t) = x0 cosω n t + sinω n t

ωn

In the general case : x(t) = A cos ω nt + B sin ω nt = C sin (ωnt + φ ) → x& max = Cωn x&(t) = Cωn cos (ωnt + φ) &x&(t) = −Cωn 2 sin (ωnt + φ)

→ &x& max = Cωn

2


DISPLACEMENT, VELOCITY & ACCELERATION - II

DISPLACEMENT x(t) = C sin (ωnt + φ ) VELOCITY x&(t) = Cωn cos (ωnt + φ ) ACCELERATION &x&(t) = −Cωn 2 sin (ωnt + φ )


SPRINGS - I

FORCE F

F=kx

F=kx0

Strain energy=1/2 kx02 x0 DISPLACEMENT x


SPRINGS - II

• What provides the spring (or restoring) force? • Gravity

• A spring • A float mechanism

• Beam flexibility


STIFFNESS FORMULAE


POSITION OF EQUILIBRIUM - I

IN ALL VIBRATION PROBLEMS, THE ORIGIN OF THE MOTION SHOULD BE TAKEN AS THE STATIC EQUILIBRIUM POSITION.


POSITION OF EQUILIBRIUM - II

Weight mg = Static deflection: δ = stiffnes k If the vibration amplitudeis x, total deflection will be : mg y = x +δ = x + k Newton's second law : m&x& = ∑ F m&x& = mg − ky = mg − k ( x + m&x& = −kx

mg ) k


TRANSLATIONAL, TORSIONAL & ROTATIONAL SDOF SYSTEMS


PENDULUM WITH SPRING

We will apply :

a

θ

L

m

Iθ&& = ∑ Moment

Spring restoring moment : Stiffness x displacement x distance = − k (a sin θ )a Gravity restoring moment : Effective weight x distance = −mgsinθ L Re - arranging : Iθ&& = −k (a sin θ )a − mgsinθ L

Re − arranging and assuming small θ mg 2 & & Iθ + (ka + mgL)θ = 0 (ka 2 + mgL) ωn = I Note: I=mL2


TORSIONAL SYSTEM

kT

θ

D=10 mm L= 300 mm G= 80 GPa

Iθ&& + kTθ = 0

I=0.16 kgm2

Torque T Torsional stiffness = kT = = Ang disp θ T Gθ πr 4 Also where J = = J L 2 Combining : kT = GJ / L = 261Nm / rad kT 261 ωn = = = 40.4 rad / s I 0.16


KINETIC ENERGY vs STRAIN ENERGY

1 2 KE = mv 2 1 2 SE = kx 2 x

v


ENERGY METHOD TO DETERMINE ωn

MAX KINETIC ENERGY = MAX STRAIN ENERGY Max velocity, zero displacement

Max displacement, zero velocity

Displacement : x(t ) = x0 sinωnt Velocity: x&(t ) = x0ωn cosωnt

Max = x0 Max = x0ωn

x&MAX → x0ωn AND xMAX → x0 FOR THE SDOF MASS-SPRING SYSTEM

1 1 2 2 m( x0ωn) = kx0 Hence:ωn = k m 2 2


MORE COMPLEX SYSTEMS

GENERAL LAGRANGE EQUATION

d  ∂T  ∂T ∂V ∂D   − = Qi + + dt  ∂x&i  ∂xi ∂xi ∂x&i


VERY COMPLEX SYSTEMS

Finite element model of a car


Simple Vibratory Phenomena External Force – Time dependent usually periodic One 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 circuit conditions … 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

d 2θ I 2 + kθ = 0 dt k p= rad/s I


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 at critical speeds • Bending vibrations effected by misalignment, loosely mounted parts, bearing stiffness, gears, asymmetry, instabilities due to oil film, etc… They cause most machinery problems


FREE MOTION OF A DAMPED SDOF SYSTEM

Equation of motion : m&x& + cx& + kx = 0 Dividing by m : &x& + 2ζωnx& + ω 2n x = 0 k where ωn = is the undamped natural frequency, m c c ζ = = is the viscous critical damping ratio. c0 2 km


SOLUTION OF THE EQUATION OF MOTION

EOM :

&x& + 2ζωnx& + ω 2n x = 0

α1t

Solution is of the form : x(t ) = Ae

α 2t

+ Be

A & B are two constants depending on initial conditions.

αt To find α1 & α 2 , insert x(t) = Ae into EOM. 2 t α Ae (α + 2ζωnα +

ω )=0 2 n

→ α1, 2 = (−ζ ± ζ − 1)ωn 2


POSSIBILITIES, DEPENDING ON THE VALUE OF ζ

Case 1. ζ < 1. Underdamped case with oscillatory motion. Both roots are complex : α1 = (−ζ + i 1 − ζ 2 )ωn

α 2 = (−ζ − i 1 − ζ 2 )ωn The general solution becomes : − ζωnt [ A cos( 1 − ζ 2 ωnt ) + B sin( 1 − ζ 2 ωnt )] x(t ) = e =e

− ζωnt

C sin( 1 − ζ 2 ωnt + φ ) = e

− ζωnt

C sin(ωdt + φ )

where ωd = 1 − ζ 2 ωn is the damped natural frequency.


TIME HISTORY FOR OSCILLATORY MOTION

Exponential term

e

− ζωnt t

Oscillatory term

C sin(ωdt + φ )


CASE 2

ζ > 1. Overdamped case with no oscillatory motion. Both roots are real : α1 = (−ζ + ζ 2 − 1)ωn

α 2 = (−ζ − ζ 2 − 1)ωn The solution becomes : x(t ) = Ae + Be

(−ζ + ζ 2 − 1)ωnt

(−ζ − ζ 2 − 1)ωnt


EFFECT OF OVERDAMPING


CASE 3

ζ = 1. Critically - damped motion → Max rate of decay Double root : α1 = α 2 = −ωn The solution becomes : − ω nt x(t ) = ( A + Bt )e


EXAMPLE: SYSTEM IDENTIFICATION FROM TIME RESPONSE

x(t) 1.0 s 2.5 s 4.0s

4000 kg

t

Initial T=2π/ωd=3 s ωd= 2π/3

Find k and c.

conditions

:

x& ( 0 ) = 0 x (1 ) = 0 , x ( 2 . 5 ) = 0 x(4) = 0


LOGARITHMIC DECREMENT - I

The amplitude ratio between two successive cycles : − ζωntm e C sin(ωdtm + φ ) x(tm) = x(tm + 1) e− ζωntm + 1C sin(ωdtm + 1 + φ ) x(tm) − ζωn(tm − tm + 1) =e x(tm + 1) AMP

ζωnT

=e

ζωn

=e

2π

ζωn

ωd = e

2π 1 − ζ 2 ωn


LOGARITHMIC DECREMENT - II

ζ x(tm) =e x(tm + 1)

2π 1−ζ 2

Taking logarithms of both sides : x(tm) 2πζ δ = ln = ≈ 2πζ x(tm + 1) 1−ζ 2 If the two amplitudes are separated by (N - 1) cycles : 1 x(tm) δ = ln N x(tm + N)


Viscous Damping Principle

τ =µ

• Damping force is proportional to velocity and

dv dz

F = (πDt ) µ

= Damping Coefficient C times Velocity dx/dt

v

δ

– dissipates energy

= cx&

πDtµ N - s/m c= δ

• 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


DAMPING IS NOT ALWAYS VISCOUS !

• Viscous damping ratio: ζ = c/c0

m&x& +cx& + kx = 0 • Material damping (what the material can dissipateDashpot in one cycle)

m &x& + k (1 +iη ) x = 0

•Aerodynamic damping: δ (e. g. gas pressure on a blade) • Friction damping (energy dissipation via contact mechanism) AT RESONANCE

ζ =η / 2 = δ / 2π


EXAMPLE OF DAMPED MOTION - PARACHUTE

Find the maximum compression in the spring if m=20 kg, k=10kN/m, c=540Ns/m and v=8 m/s.

Compression = static compression + dynamic compression

weight mg 20 x9.81 d1 = = = = 0.0196m stiffness k 10 x1000

d2


USE INITIAL CONDITIONS TO FIND A & B IN EOM:

d

t=0

Datum for motion

1

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

− ζωnt x(t ) = e ( Acosωdt + B sinωdt ) x(0) = A = x0 = −0.0196m x

− ζωnt Obtain velocity by differentiating : x(t ) = e ( A cos ωdt + B sin ωdt ) − ζωnt − ζωnt x& (t ) = −ζωne ( A cos ωdt + B sin ωdt ) + ωde (− A sin ωdt + B sin ωdt ) − ζωnt =e [( Bωd − Aωnζ ) cos ωdt − ( Aωd + Bωnζ ) sin ωdt ] → x& (0) = Bωd − Aωnζ = 8m / s Substitution for A, ωn , ωd and ζ gives B = 0.434


LAGRANGE’S EQUATION OF MOTION

0

d  ∂T  ∂T ∂V ∂D   − = Qi + + dt  ∂x&i  ∂xi ∂xi ∂x&i T : Kinetic energy

x is some general arbitrary coordinate.

V : Potential & spring energy D : Damping dissipation function Q : External force

i : Co - ordinate number (i = 1 for SDOFsystems) Free - vibration : Q = 0 No damping : D = 0

SDOF

d  ∂T  ∂V =0  + dt  ∂x&  ∂x


EXAMPLE 1 – SDOF SYSTEM

FIND THE EQUATION OF MOTION

1 2 ∂T d  ∂T  d T = mx& → = mx& →   = (mx& ) = m&x& 2 ∂x& dt  ∂x&  dt 1 2 ∂V V = kx → = kx 2 ∂x d  ∂T  ∂V = 0 → m&x& + kx = 0  + dt  ∂x&  ∂x


EX 2 - PENDULUM WITH SPRING

a

θ mgLcosθ

L

m mg L

1 &2 T = Iθ 2 D=0 V = ∆SE + ∆PE where ∆ : Change SE : Spring energy, PE : potential energy

1 2 SE = ky where y = aθ 2 PE = mgL(1 − cosθ ) 1 V = k (aθ ) 2 + mgL(1 − cosθ ) 2


OBTAIN DERIVATIVES

1 &2 T = Iθ 2 d  ∂T  d & ∂T & → & = Iθ →  &  = ( Iθ ) = Iθ&& dt  ∂θ  dt ∂θ 1 V = k (aθ ) 2 + mgL(1 − cosθ ) 2 ∂V 2 2 → = ka θ + mgL sin θ ≈ (ka + mgL)θ ∂θ


INSERT INTO LAGRANGE EQUATION

∂V d  ∂T    + = 0 dt  ∂θ&i  ∂θ i 2 & & Iθ + (ka + mgL)θ = 0

(ka + mgL) ωn = I 2


FORCED VIBRATION

EOM : m&x& + cx& + kx = F (t ) (2nd - order ODE with RHS)

Solution : x(t) = CF + PI xCF(t)

Transient

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

xPI(t)

Steady-state

Periodic solution


SOLUTION OF m&x& + cx& + kx = F (t ) If we consider ζ < 1 only , the CF is given by : x(t ) = e

− ζωnt

( A cos ωdt + B sin ωdt )

To obtain the PI, we must know the RHS, F(t). We will consider one type of excitation only : m&x& + cx& + kx = F (t ) = F0 sin ωt We now need to guess a PI.


PROPERTIES OF THE PI

When a linear system is subjected to a harmonic excitation of the form Fsinωt, •

It will respond harmonically at the same frequency.

•

There will be a phase lag between the force and the response.

Input :

F (t ) = F 0sin ωt

Output :

xPI (t ) = x0 sin(ωt − φ )

0<ω < ∞


PHASE LAG

φ

100

100

50

50

0

0

-50

-100 0

-50

0.1

Tim e

0.2

-100 0.3

F (t ) = F0 sin ωt

x(t ) = x0 sin(ωt − φ )


SOLUTION FOR THE STEADY-STATE VIBRATION

The solution for the steady − state vibration can be found by inserting the PI xPI (t ) = x0 sin(ωt − φ ) into the EOM m&x& + cx& + kx = F0 sin ωt x0 =

F0 (k − mω 2 ) 2 + (cω ) 2

ω Let r = ωn

c and ζ = 2 km 1

x0 = F0 k (1 − r 2 ) 2 + (2ζr ) 2

cω & tan φ = k − mω 2

2ζr & tan φ = 1− r 2


FREQUENCY RESPONSE FUNCTION (FRF)

Consider the expression : x0 1 1 H= = = 2 2 2 F0 k (1 − r ) + (2ζr ) (k − mω 2 ) 2 + (cω ) 2 It is of the form : Output = Function (k, m, c & ω ) = Function (system properties & ω ) Input Such a function is called Transfer Function in general It is called Frequency Response Function (FRF) in vibration analysis.


FRF PLOT

φ=900 at resonance I

RESONANCE


EFFECTS OF DAMPING

Reduces response at resonance. Has little effect elsewhere. Has relatively little effect on resonant frequency.


Another look at RESONANCE


Another look at RESONANCE……contd…








Q-FACTOR

Q=

Resonant amplitude  x0  =  Static deflection  d r = 1

We have : x0 =

F0 k (1 − r 2 ) 2 + (2ζr ) 2

F0 & d= k

     x0  1  x0    = Q=  =   (1 − r 2 ) 2 + (2ζr ) 2   d  r =1  F0   r =1  k  r =1  1 x  →Q = 0  = Inverse of Damping Ratio x 2  d  r =1 2ζ


HALF-POWER POINTS

Frequencies corresponding to

X0 /d

Half-power points Q

Q 2 are called half − power points. amplitude =

Q/1.414 1.0 r =1

ω


DAMPING ESTIMATION FROM AN FRF

H

Hres

x0 1 = Hres H = 2 2 2 H (ω1) = H (ω 2) = F − + k m c ω ω ( ) ( ) 0 2 1 Hres = at ω = ωn = k / m ω1 cωn ω n

ω2 Determine ω such that H (ω ) = Hres 2

ω

1 ω 2 → = → Yields a quadratic equation in ( ) 2 2 2 ωn 2cωn (k − mω ) + (cω ) 1

There are 2 frequencies ω 1 &ω 2 such that H (ω 1) = H (ω 2) = c 1 ω 2 − ω1 It can be shown that ≈ = 2ζ = Q ωn km

Hres 2


PHASOR DIAGRAM FOR F=F0sinωt

F0 is rotating with speed ω

50

F 0 sinωt

F0

-50

0

ωt = 2nπ

0.04

0.08

ωt = 2nπ + π

0.12

ωt = 2nπ + 2π

ωt


RELATIVE PHASE BETWEEN DISP, VEL & ACCEL

x = x0 sin(ωt − φ )

π

x& = ωx0 cos(ωt − φ ) = x0ω sin(ωt − φ + ) 2 ie x& leads x by

π

2 &x& = −ω 2 x0 sin(ωt − φ ) = −ω 2 x = ω 2 x0 sin(ωt − φ + π )

ie &x& leads x& by

π 2

, and x by π .

&x&

x

x& ωt-φ


3 CASES TO CONSIDER

mω2 x0

F0

kx0

cωx0 ω= ω n

kx0

-> Kx0 = mω2x0

Damping control

mω2 x0

F0

-> Kx0 > mω2x0

Stiffness control

mω 2 x0

F0

cωx0 ω < ω n

kx0

cωx0 ω> ω n -> Kx0 < mω2x0 Inertia control


EXCITATION BY ROTATING OUT OF BALANCE

(m + m' ) &x& + cx& + kx = F0 sin ωt where F0 = m' rω m’rω2cosωt

F0 (k − mω ) + (cω ) 2 2

Usually m >> m'

Example : Find x0 when ω >> ωn

m’rω2sinωt x0 =

2

2

→x0

m' rω2

=

m' rω

2

[k − (m + m' )ω 2 ]2 + (cω ) 2

m' rω2 m' r m' r = lim = ≈ lim x0 = lim 2 2 2 2 m + m' m ω →∞ ω →∞ [k − (m + m' )ω ] + (cω ) ω →∞ (m + m' )ω


VIBRATION ISOLATION


CAR ENGINE

•

Large forces over a wide frequency range.

•

They arise from the crankconnecting-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

UNDAMPED SPRING MOUNT

DAMPED SPRING MOUNT

PNEUMATIC RUBBER MOUNT


HIGH-SPEED PUNCH PRESS ON RUBBER MOUNTS

MOUNTS


BASIC THEORY

FT sin(ωt+θ)

Source of vibration

The force transmitted to ground is due to the spring and damper : FT = kx + cx& Force transmitted FT What we want to know is the ratio : = Excitation Force F 0


mω x0 2

F0

From the phasor diagram :

cωx0F

φ FT

T

kx0

ωt-φ

= x0 k 2 + (cω ) 2

From Lecture 4, we know that : x0 =

So, transmissibility, T, is given by :

F0 (k − mω 2 ) 2 + (cω ) 2

FT k 2 + (cω ) 2 1 + (2ζ r ) 2 = = T= 2 2 2 F0 (k − mω ) + (cω ) (1 − r 2 ) 2 + (2ζ r ) 2

ω where r = ωn

c and ζ = 2 km


HOW TO OBTAIN LOW TRANSMISSIBILITY ?

FT k 2 + (cω ) 2 T= = F0 (k − mω 2 ) 2 + (cω ) 2

•

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 2 m ω x0 ω<ω n

cωx0

FT

F0

ω=ωn

mω 2 x0

F0

kx0

cωx0

FT

kx0

ω>ωn

mω2 x0

F0

cωx0

FT

kx0

ω >> ω n

F0

mω2 x0

cωx0

FT

kx0


TRANSMISSIBILTY FT/F0

TRANSMISSIBILITY CURVES

AMPLIFICATION T> 1

ISOLATION T< 1


TRANSMISSIBILTY FT/F0

EFFECT OF DAMPING IN THE ISOLATION REGION

DECREASING T

LESS ISOLATION WITH INCREASING DAMPING


EXAMPLE: ISOLATION OF RAIL NOISE

RUBBER PADS

Before isolation After isolation


Lecture –3 Multi-Degree of Freedom Systems + Modal Analysis


MDOF VIBRATING SYSTEMS

• Two complications – Each node has 6 degrees of freedom – Many such nodes are needed to describe the geometry of representative engineering systems. FOR FE models that use 6 DOF/node elements Total no of DOFs = No of nodes x 6


EXAMPLES OF SDOF AND 2-DOF SYSTEMS

SDOF SYSTEM

2-DOF 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 in the 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 freedom systems. • 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

STEP 1: SPRING DEFLECTIONS REFERENC E k2 force

k2 force DEFLECTED

k1 force ELONGATION OF k1:

x1

COMPRESSION OF k2: (x1x2) DUE TO DISPLACEMENT OF BOTH ENDS


STEP 2: FREE-BODY DIAGRAMS

k1x1

k2(x1-x2) k2(x1-x2)

m1 x 1

m2 x2


STEP 3: OBTAIN THE EQUATIONS OF MOTION

k1x1

k2(x1-x2)

m1

k2(x1-x2)

m2

x1

x2

Newton' s 2nd law : m&x& = ∑ F Mass 1 : m1&x&1 = −k 1 x1 − k 2( x1 − x 2)

Mass 2 : m 2 &x&2 = k 2( x1 − x 2)

m1&x&1 + (k 1 + k 2) x1 − k 2 x 2 = 0

m 2 &x&2 − k 2 x1 + k 2 x 2 = 0


STEP 4: ASSUME SHM WRITE THE MATRIX EOM

We have :

m1&x&1 + (k 1 + k 2) x1 − k 2 x 2 = 0 m 2 &x&2 − k 2 x1 + k 2 x 2 = 0

Remembering that &x&1 = −ω 2 x1 & &x&2 = −ω 2 x 2 − m1ω 2 x1 + (k 1 + k 2 )x1 − k 2 x 2 = 0 -m 2ω 2 x 2 − k 2 x1 + k 2 x 2 = 0 m1 0  x1  k 1 + k 2 − k 2  x1   0    =     +  −ω    k 2  x 2   0   0 m 2  x 2   − k 2 2


STEP 4: CONTINUED

EIGENVALUE = (NATURAL FREQUENCY)2

([K]

−

k 1 + k 2 − k 2   − k2  k 2   STIFFNESS MATRIX

ω [M ]) {x} = {0} 2

m1 0   0 m2   MASS MATRIX

 x1     x2 

MODE SHAPE VECTOR

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.

k 1 + k 2 − k 2   − k2  k 2  

m1 0   0 m2  


STEP 6: OBTAIN THE NATURAL FREQUENCIES FOR LARGE N, THERE ARE MANY NUMERICAL SOLUTION TECHNIQUES. USE DET=0 FOR SMALL SYSTEMS.

([K] - ω 2 [ M ]) {x} = {0} → det ([K] - ω 2 [ M ]) = 0 or {x} = {0} For a non − trivial solution : k 1 + k 2 − ω m1 − k2 det ([K] - ω [ M ]) = 0 → =0 2 k 2 − ω m2 − k2 2

2

→ (k 1 + k 2 − ω 2 m1)(k 2 − ω 2 m 2) − k22 = 0 Quadratic in ω2 → 2 natural frequencies : ωn1 & ωn2. (ωn1 ≤ ωn2)


STEP 7: OBTAIN THE MODE SHAPES

Insert ωn1 into ([K]-ω2[M]){x}={0} By definition, det([K]- ωn1 2 [M])=0 x1 & x2 are linearly dependent, but we can obtain x1/x2 Using the previous result : − m1ω x1 + (k 1 + k 2 )x1 − k 2 x 2 = 0 2

x2 k 1 + k 2 − m1ω n21 = Hence : x1 ω =ωn1 k2 x2 k 1 + k 2 − m1ω = Similarly, for the 2nd mode : x1 ω =ωn 2 k2

2 n2


MODE SHAPE INTERPRETATION

Assume that, inserting values for m, k, ω gives : x2 =1 x1 ω =ωn1

&

MODE 1

The masses move in phase. X1 and X2 move by +1 unit each.

x2 = −1 x1 ω =ωn 2 MODE 2

The masses move out of phase. X1 moves by +1 unit, X2 moves by –1 unit.


GENERAL MDOF ANALYSIS

Ignore damping. Matrix equation of motion : [M] NxN {&x&}Nx1 + [K] Nx1{x}Nx1 = { 0 }Nx1 {x} = −ω 2{X} gives : ([K] − ω 2 M ]) Nx1{x}Nx1 = { 0 }Nx1 Eigenvalue problem in ω 2 and {x} N modes → (ω 2 ) Natural frequency {x} Mode shape

AN N-DOF SYSTEM HAS N MODES.


3 DOF SYSTEM

TIME

FREQUENCY


MODAL SUPERPOSITION MODE 1

MODE 3

MODE 2

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 excitation became 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 12 angular stiffnesses, therefore, we have a system with 13 degrees of freedom. • We will attempt to understand the behavior of such a system and study some important vibrational terms that are regularly used in routine testing and analysis. • It is suggested that the mathematical intricacies in this process may be ignored by an engineer in the field – we will emphasize the physical concepts that are of concern to field vibration engineers and just brush aside the mathematical stuff (unless, of course you are otherwise interested)


• These are governing differential

I1θ&&1 + k1θ1 − k1θ 2 = 0

I 2θ&&2 − k1θ1 + ( k1 + k 2 )θ 2 − K 3θ 3 = 0 equations of motion written from equilibrium conditions, 13 for 13 inertias ...

• These equations are written in a compact

I13θ&&13 − k12θ12 + K13θ13 = 0

matrix form

• Mass Matrix • Stiffness Matrix

[M ]{θ&&}+ [K ]{θ } = 0  I1    I2  [M ] =  I3    I13   − k1  k1  − k  k1 − k 2 1  [M ] =  − k 2 k 2 + k3 − k3 − k12  ...   − k12 k12  


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.

p = p1

{θ }1

θ1 = 1  θ   2  =  θ3   ...     θ13 1

{θ }2

θ1 = 1  θ   2  =  θ3   ...     θ13  2

θ i = θ i cos pt i = 1,2,...13

[[K ] − p [M ]]{θ } = 0 [[K ] − p [M ]] = 0 2 2

• 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  θ11 θ12 θ13 θ1N  θ θ 22 θ 23 θ 2 N  21   [θ ] =  θ 31 θ 32 θ 33 θ 3 N    θ N 1 θ N 2 θ N 3 θ NN 

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

p = p2

p = p13

{θ }13

θ1 = 1  θ   2  =  θ3   ...     θ13 13


Bending Vibration α11m1 &x&1 + α12 m2 &x&2 + ... + α1n mn &x&n + x1 = 0 α 21m1 &x&1 + α 22 m2 &x&2 + ... + α 2 n mn &x&n + x2 = 0 ...

α n1m1 &x&1 + α n 2 m2 &x&2 + ... + α1n mn &x&n + xn = 0

[α ][M ]{&x&} + [I ]{x} = 0 [M ]{&x&} + [α ]−1 [I ]{x} = 0 [M ]{&x&} + [K ]{x} = 0 • 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.


FORCED RESPONSE

THE RESPONSE OF THE SYSTEM TO SOME GIVEN HARMONIC EXCITATION CAN BE FOUND USING A TRANSFER FUNCTION APPROACH:

OUTPUT = SYSTEM FUNCTION x INPUT X = H(system properties, ω) x Force For forced response, we have : ([K]-ω2 [M]) {X} = {F} → { X } = ([K] - ω 2 [ M ]) −1{F } = [ H ]{F } 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 special and 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 with mass 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 shapes that 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.

{u } [M ]{u }= 0 r ≠ s {u } [M ]{u }= m r = s {u } [K ]{u }= 0 r ≠ s {u } [M ]{u }= k r = s {u~ } [M ]{u~ }= 1 s T

r

s T

r

s T

r

r

s T

r

r T

r

{u~ }= r

1 mr

T

{u } r


Modal Analysis We will write the multi-degree system equations first. • Recall, we stated that the natural frequencies and mode shapes are special solutions to the system. We haven’t yet been able to say what happens, if one just bumps, or raps the system, how does it respond? The system knows its mode shapes and try to respond in all modes, and adjust them in such a way they satisfy the type of displacement given in the test. That is it weighs how much amount in each mode it should respond, as decided by the generalized coordinates (weighing terms) and the result that we can obtain is called Modal Analysis. This expansion is called modal expansion. • We pre-multiply the equations with transpose of the modal matrix and use the orthogonal properties derived before to get the result given here. • All practical systems, therefore reduce to simple single degree of freedom models for any given mode of vibration. • We can introduce damping in the system for each mode and get the damped modal equations.

[M ]{q&&}+ [K ]{q} = 0

{q} = [U~ ]{η }

[M ][U~ ]{η&&}+ [K ][U~ ]{η } = 0

[U~ ] [M ][U~ ]{η&&}+ [U~ ] [K ][U~ ]{η} = 0 T

T

[I ]{η&&}+ [λ2 ]{η } = 0 η&&(t ) + p 2η (t ) = 0

r = 1,2, K N

η&& + 2ξpη + p 2η = 0

r = 1,2,K N


Proportional Damping

[C ] = α [M ] + β [K ] • 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 corresponding mode and thus use experimental value to write damped modal equations. • General finite element codes such as ANSYS, NASTRAN … adopt beta damping in place of viscous damping..

[I ]{η&&}+ [U~ ] (α [M ] + β [K ])[U~ ]{η&}+ [p 2 ]{η } = 0 T

[I ]{η&&}+ (α [I ] + β [p 2 ]){η&}+ [p 2 ]{η } = 0 2ξ r pr = α + β pr2

ξr =

α 2 pr

+

β pr 2

η&&r + 2ξ r prη&r + pr2η r = 0

r = 1,2, K


Modal Analysis


Modal Analysis


Modal Analysis


Modal Analysis - example


Modal Analysis – example.. contd…












VIBRATION ABSORBER


VIBRATION ABSORBER

…contd…


LECTURE 4

Continuous Systems Approach


CONTINUOUS SYSTEMS Strings under tension

Systems governed by Wave Equation


CONTINUOUS SYSTEMS Axial vibration of Bars

WAVE EQUATION

contd….


CONTINUOUS SYSTEMS Torsional vibration of RODS

WAVE EQUATION

contd….


Solution of WAVE EQUATION


Solution of WAVE EQUATION

contd….


Solution of WAVE EQUATION Free Vibration of Strings

contd….


Solution of WAVE EQUATION Free Vibration of Bars

contd….


Solution of WAVE EQUATION Free Vibration of Bars

contd….

contd….


BEAMS

Bending Vibrations


BEAMS

Bending Vibrations

contd….


BEAMS

Bending Vibrations

contd….


BEAMS

Bending Vibrations

contd….


BEAMS

Bending Vibrations

contd….


BEAMS

Bending Vibrations

contd….


BEAMS

Bending Vibrations

contd….


RAYLEIGH’s Method


RAYLEIGH’s Method

contd….


Example………..RAYLEIGH’s Method



…. contd…



…. contd…



…. contd…


RAYLEIGH – Ritz Method

undetermined


RAYLEIGH – Ritz Method …. contd….


RAYLEIGH – Ritz Method

contd….


Example………..RAYLEIGH – RITZ Method



…. contd…



…. contd…


GALERKIN Method


GALERKIN Method

…. contd…


GALERKIN Method

…. contd…


GALERKIN Method

…. contd…


Lecture 5

Stability Considerations


STABILITY


STABILITY Considerations

Phase-Plane



Phase-Plane

..contd..



Phase-Plane

..contd..



Phase-Plane

..contd..



Phase-Plane

..contd..


• other criterion to be discussed during Rotor Dynamics


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 Rotor Dynamics is different from Structural Dynamics, as we deal with a rotating structure. 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 the existence 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 the critical 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)


Ingenious Flexible Shaft

• Though, we have not understood clearly about the basics of rotor dynamics, Laval built the first turbine in 1883 which ran successfully at 40000 rpm!!! • From simple equilibrium conditions, he derived a correct relation for the whirl radius (though whirl and spin have not been clearly differentiated), in his own notation, it is

y=

stiffness

ω 2δ Fg W

eccentricity

−ω2

• When the denominator is zero, the whirl radius is infinity, defining the critical speed. • Laval proved that one can cross the critical speed and is stable at high speeds as y approaches -d


Jeffcott’s fundamental contribution

• Jeffcott in 1919 treated the problem as forced 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 as shaft. 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.


Equations of Motion - Solution

d2 dz M 2 ( z + a cos ω t ) + C + Kz = 0 dt dt d2 dy M 2 ( z + a sin ω t ) + C + Ky = 0 dt dt r = z + iy d 2r dr M 2 +C + Kr = Maω 2 e iω t dt dt O is the bearing center line, E is the disk geometric center, G is the mass center, EG=a is the eccentricity, OE=R is the whirl radius about the bearing center line. Disk rotates/spins about E with angular velocity w ccw direction and the whirl is assumed synchronous with spin. The whirl is lagging in phase by an angle f from the unbalance force vector in direction of EG. • Write down the inertia forces, stiffness forces and damping forces in the respective directions …

• Solution for the amplitude R of r, in terms of a and W = w/p

R R= = a

Ω2

(1 − Ω ) + (2ξ Ω ) 2 2

φ=

2ξ Ω 1 − Ω2

2


Jeffcott Rotor - Solution

• When the rotor is stationary initially, the rotor speed is zero and the response is also zero for all damping values. • As the speed increases, the whirl radius increases with the phase angle less than 90 deg until resonance. At resonance, the phase angle is 90 deg irrespective of damping. The peak value occurs slightly beyond resonance. • Damping limits the resonant response and the quality factor (magnification) is 1 over twice the damping ratio, 1/2x. • At high speeds beyond resonance, the system runs smoothly with response equal to eccentricity and a phase 180 deg, thus keeping the cg steady at the bearing center line.


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 r0 at

α0

d 2r dr M 2 + C + Kr = Maω 2 e iωt + Kre i ( ax + a0 ) dt dt R a

R= =

Ω2

(1 − Ω ) + (2ξ Ω) 2 2

R0 = 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 direction opposite to the mass center.

2

e

i (ω t −φ )

+

R0 e ia0

(1 − Ω ) + (2ξ Ω) 2 2

2

r0 bow factor a  2ξ Ω  phase angle of unbalance 2  1− Ω 

φ = tan -1 

φb = φ + a0 phase angle of bow

•Response due to conventional unbalance. Response due to bow unbalance.

e i (ω t −φb )


Response - Self Balancing Speed

[ ] ) = [A + Be ]e (

R = Ae − iφ + Be − i (φ + a0 ) eiω t

A=

B=

Ω

R R= = a

i ω t −ψ

− ia0

(1 − Ω ) + (2ξ Ω)

2

R0

(1 − Ω ) + (2ξ Ω)

(1 − Ω ) + (2ξ Ω ) 2 2

2

R R= = a

Ω 2 − R0

(1 − Ω ) + (2ξ Ω ) 2 2

2

Ω s = R0

2

 A sin φ + B sin φb   ψ = tan   A cos φ + B cos φb  −1

Ω 2 + R0

a0 = 180 0

2

2 2

2 2

a0 = 00

•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.


Response of a bowed rotor

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


Rotor with bow phase 1800

Phase relationship of a bowed rotor

Whirl Amplitude of a bowed rotor


Rigid Rotors in Flexible Anisotropic Bearings



..contd..



..contd..



..contd..


Rigid Rotors in Flexible Anisotropic Bearings with Cross-Coupling & Damping



.contd..



.contd..


.contd..


UNEQUAL MOMENTS OF INERTIA



…. contd…



…. contd…



…. contd…



…. contd…


Instability in Torsional Systems


Instability in Torsional Systems …. contd…


Hill’s Equation; Mathieu‘s Equation


Floquet’s Theory;

Strutt Diagram

Floquet’s Theory determines the behaviour of the Mathieu’s Equation and describes the Strutt Diagram


GRAVITATIONAL Effect



…. contd…



…. contd…



…. contd…



…. contd…


OIL WHIRL


ROTOR STABILITY IN FLUID FILM BEARINGS

contd..


ROUTH-HURWITZ CRITERION



contd..


Example



Example

ROUTH-HURWITZ CRITERION …. contd…


Holzer’s Method

for Torsional Systems


Holzer’s Method


…. contd…


Holzer’s Method


…. contd…


Example …… Holzer’s Method


Myklestad-Prohl Method

for Beams



for Beams

..contd…



for Beams

..contd…



for Beams

..contd…


GYROSCOPIC EFFECTS


GYROSCOPIC EFFECTS

Freely Spinning Disc


GYROSCOPIC EFFECTS

Freely Spinning Disc ..contd..


GYROSCOPIC EFFECTS

Additional term in place of 0

Disc on Shaft


GYROSCOPIC EFFECTS

Disc on Shaft

.. contd…


Rigid Rotor Imbalance Classification

•The imbalance in a rotor can be static – i.e., the principal inertia axis is shifted parallel to the shaft axis, a single correction mass is sufficient to balance the rotor in a plane containing cg. •A special case of imbalance can be a coupled unbalance, in which the principal inertia axis passes through the center of gravity, then a couple, two equal and opposite masses are needed for correcting the imbalance. •In a general dynamic unbalance case, sum of the above two special cases, we need two correction masses in any two convenient planes – we will discuss this further.


Dynamic Unbalance


Single Plane Balancing •Balancing of single discs is a fairly simple task as all the unbalance can be considered as confined to be in one plane. •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 relate the 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 positions taken 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 the response 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 the lengths of the vectors Tv and Ov.


Rigid Rotor Balancing

Distributed Unbalance •Very few rotors can be considered as single plane rotors, therefore we have to devise a method of balancing rotors with distributed unbalance. •When the rotors do not deform and retain the original shape, the rotor can be considered as rigid – in practice rotors running well below the critical speed are rigid rotors, then we can devise the following procedure. •Divide the rotor into several disks, as many as you want, say n in number. Let us consider the ith rotor at a distance zi from a plane which we will denote left plane marked L. In this plane, the unbalance is taken as miai at an angle fi from x axis. There are n such masses in these n disks chosen. •We also choose another convenient plane and denote this as plane R as shown.


Rigid Rotor – Two Plane Balancing •Let Fi be the unbalance force in ith plane. Introduce two equal and opposite forces Fi in plane L. Since the rotor is rigid these two equal and opposite forces 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 two unbalance 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 measure the phase. Choose the measurement planes to be, say left bearing and right bearing. Let the responses without any trial mass be as shown. •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

Run 1 : No Trial Mass Left Plane L : 9.144 microns at 90o

L1 = 9.144i

Right Plane R : 10.16 microns at 45o

R1 = 7.188 + 7.188i

Run 2 : Trial Mass 6.8g at 22.5o in Right Plane T R = 6.294 + 2.608i Left Plane L : 5.08 microns at 27 o

L 2 = 4.521 + 2.311i

Right Plane R : 6.35 microns at 99 o

R 2 = −0.9906 + 8.274i


Example – contd.

a bR =

•Influence coefficient denoting the response in plane b due to a unit force in plane R

=

(− 0.9906 + 6.274i ) − (7.188 + 7.188i ) 6.294 + 2.608i

= −1.160 + 0.336i

a aR = •Influence coefficient denoting the response in plane a due to a unit force in plane R

R 2 − R1 TR

=

L 2 − L1 TR

(4.521 + 2.311i ) − (9.144i ) 6.294 + 2.608i

= 0.229 + 1.181i


Example – contd. Final Run 3 : 6.8g at 36 o T L = 5.5 + 3.997i Left Plane L : 9.4 microns at 0 o

•Influence coefficient denoting the response in plane b due to a unit force in plane L

L 3 = 9.4 + 0i Right Plane R : 30.5 microns at 90 o R 3 = −4.775 + 30.1i R 3 − R1 TL (− 4.775 + 30.1i ) − (7.188 + 7.188i ) a bL = 5.5 + 3.997i = −0.558 + 3.76i

a bL =

•Influence coefficient denoting the response in plane a due to a unit force in plane L

L 3 − L1 TL 9.4 − 9.144i a aL = 5.5 + 3.997i = 0.327 − 1.9i a aL =


Example – contd.

•We need to eliminate the original response vectors 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 theoretically should 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.

− R1 = m R a bR + m L a bL − L1 = m R a aR + m L a aL L1 a bL − R1 a aL mR = a bR a aL − a aR a bL R1 a aR − L1 a bR mL = a bR a aL − a aR a bL m R = 10.94 + 1.56i m L = −2.81 − 1.9i


Shop Balancing set up


Flexible Rotor Balancing


Influence Coefficients


Balancing Masses

Trial Mass in Plane 2 1 1 v v − i ai12 = i 2 T Trial Mass in Plane j

aij1 =

vij1 − vi1 T

... k th Speed Tests aijk =

vij1 − vik T

{U }= [a ] {v} −1


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 maximum operating 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. •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 wheel and 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


ISO 1940: Rotor Classification


Balance Quality Grade G


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, the current 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 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 us in 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 it represents the energy associated in the system; for a given velocity and frequency, we can find the associated amplitudes of displacement and acceleration.



Transducers Selection Type of Measurement: 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


Eddy Probe – Velocity Transducer - Accelerometer • Measure Relative Distance Between Two Surfaces • Accurate Low Frequency Response • Limited High Frequency Sensitivity • 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 (mils, in/sec, g’s)

• 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


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

dφ × 10 −8 dt dφ dx = −n × 10 −8 V dt dt

V = −n


Accelerometer Principle

• An Accelerometer uses Piezo electric principle given 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. They can also be made very light and hence have useful applications for measurements on small components without effecting the basic structure.

r2

Z = X =

ω2 p

f =

2

(1 − r ) + (2ξ r ) 2 2

f 1

(1 − r ) + (2ξ r ) 2 2

2

2



Accelerometer Elements

Output impedance from accelerometer is very high - problems matching, noise, cable length Charge amplifier is therefore used. Elements of a measurement system


Proximity Probe – Non contact Measurement

Probe mounted in the proximity of the moving object High frequency field setup in the gap. •Can act as a key phasor •Two probes can give the orbit •Mounted in the bearing - relative displacement


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


Calculation of RMS with two components 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 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]

RMS1 = 0.005 = 0.0707 RMS 2 = 0.00125 = 0.0354 RMS total = 0.005 + 0.00125 = 0.0791

RMS pk to pk of over all signal = 0.1582 units

Note sin 2 pt =

MS = = =

1 0.0775

1 0.0775

∫ [(0.1sin 81.1t ) + (0.05 sin 173.52t )

0.0775

2

2

]dt

0

∫ [(0.1sin

0.0775

2

) (

0

1  0.01   0.0025  × 0.0775  × 0.0775  +    0.0775  2   2 

= 0.005 + 0.00125

)]

81.1t + 0.0025 sin 2 173.52t dt

1 − cos 2 pt 2

and integral of cosine term becomes zero


Fourier Analysis 3D View

View in time domain

View in frequency domain


Fourier Analysis

• Fourier analysis is a mathematical tool 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 harmonic cosine and sine components are • The amplitude in each harmonic, n = 1, 2, is given by • The phase angle is

∞ 1 f (t ) = a 0 + ∑ (a n cos nωt + bn sin ωt ) 2 n =1 T

2 a n = ∫ F (t ) cos nω tdt T 0 T

2 bn = ∫ F (t )sin nω tdt T 0 ∞ 1 f (t ) = a 0 + ∑ An cos(nωt − φ n ) 2 n =1

An = a n2 + bn2

φ n = tan −1

bn an


Fourier Analysis - Complex Form • Fourier analysis is written in complex form so that faster numerical tools can be developed. The cosine and harmonic terms are written in exponential form as given here • 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 this process. • Average power of the signal over a period of time, T0 is • Complex harmonic components directly represent the average power in the corresponding frequency term, this is Parseval theorem.

(

)

(

)

1 inω t e + e −inω t 2 1 inω t sin nωt = e − e −inω t 2i ∞ 1 1 1  f (t ) = a0 + ∑  (an − ibn )e inω t + (an + ibn )e −inω t  2 2  n =1  2 cos nωt =

Two sided series

+∞

f (t ) = ∑ Cn e inω t −∞

Cn = Harmonic Cn Components Complex 1 Pav = T0

T0

∫ {x(t )} dt 2

0

2

2

∞ A a  = 0  +∑ n  2  n=1 2

= ∑ Cn

2

Cn =

1 (an − ibn ) Cs = 1 As 2 2 1 T

1 + T 2

∫ f (t )e

inω t

1 − T 2

We will get back to this again


Saw Tooth Signal

x(t ) = 2 −

4t T0

0 < t < T0

4  2πt 1 4πt 1 6πt  + sin + sin sin  T0 3 T0  π  T0 2 4 2πkt = sin k = 1,2,3,... kπ T0 =

• 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.


• 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 x(t ) = 1 = −1

0 < t < T0 / 2 0 < t < T0 / 2

 4  2πt 1 2πt 1 2πt + sin 3 + sin 5 + ... sin T0 5 T0 π  T0 3  4 2πkt k = 1,2,3,... = sin kπ T0 =

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



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 Analysis time between each analysis Significant transient data may be lost

• Subsequently, we had 1/3 octave band analyzers which helped in narrowing the frequency zones for a better diagnostics.


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

nω → ω Forward Transform – Fourier Integral

F (ω ) = (TC n )T →∞ =

+∞

∫

f (t )e −iω t dt

−∞

F (ω ) = Ff (t ) =

+∞

∫

f (t )e −iω t dt

−∞

1 f (t ) = F F (ω ) = 2π −1

•Inverse Transform – Conversion from frequency to time domain

+∞

∫

−∞

f (ω )e −iω t dω


Discrete Fourier Transform (DFT) (N − 1)T 2T T ,K ,K N N N kT = k = 0,1,2, K ( N − 1) N

t k = 0,

• f(t) is analog should be converted to a digital text file • N intervals of a vibration record in time T, time step T/N

1 Fr = N

N −1

∑f k =0

{F }N ×1 =

k

e

−i

2πkr N

r = 0,1,2, K ( N − 1)

1 [A]N × N {F }N ×1 N

• Frequency of the entire period = 2p /T • Discrete frequencies 2p /0, 2pN /T , 2pN /2T , 2pN /3T, … 2pN /(N-1)T Example: • 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.


Fast Fourier Transform - FFT [A] factored into log 2 N

{F }N ×1

1 = [A]N × N {F }N ×1 N

a kr = e

−i

2πkr N

0 ↑ ↑ ↑ ↑    1 ↑ → ↓ ← [A] =  2 ↑ ↓ ↑ ↓   3 ↑ ← ↓ →

matrices [A ]i

only two non zero elements in each row For N = 4

[A] = [P][A]1 [A]2 1 0 = 0  0

0 0 0 ↑  0 1 0 ↑ 1 0 0  0  0 0 1  0

0  ↑  ↓ 0 0 0 0 ↑ → ↑  0 ↑ ←  0 ↑ 0

[P] Permutation matrix

0 ↑ 0  ↑ 0 ↑ 0 ↓ 0  ↑ 0 ↓


Comparison - DFT/FFT Computation Requirements • On the present day Desk top or lap top computers, FFT can be performed in milli secs, thus making the conversion of time domain data to frequency domain almost instantaneous and in real time. • With an A/D converter card, a lap top has thus become a good diagnostics instrument and tool.


Amplitude Modulated Signal x(t ) = x + x a cos ω a t  cos ω t Modulation Amplitude x a

Amplitude (x - x a ) to (x + x a ) Modulation Frequency ω a

x(t ) = x + x a cos ω a t  cos ω t = x cos ω t +

1 1 x a cos(ω + ω a )t + x a cos(ω − ω a )t 2 2 Eccentrically mounted gear

ω c gear mesh frequency = carrier frequency

ω a rotational speed ω c + ω a andω c - ω a = side bands


Amplitude Modulated Signal

Mean = 0 Mean square = 0.56 Root Mean Square = 0.75 Side bands in Frequency domain


Frequency or Phase Modulated Signal x (t ) = X cos (ω + ω ' cos ω a t )ω t x (t ) = Re Xe

i φ (t

)

and frequency of modulation ω ' and ω a - magnitude index ω ' / ω a - modulation Frequency varies (ω − ω ' ) to (ω + ω ' ) Phasor of constant magnitude Variable angular speed In torsion al vibration , the angular frequency changes over the mean value correspond ing to the rotational speed Frequency

and the resultig signal Modulated Signal.

is a


Frequency or Phase Modulated Signal

Mean = 0 Mean square = 0.50 Root Mean Square = 0.71

Modulation Magnitude = 0.5 Modulation Freq = 4 Hz Carrier Freq = 15 Hz


Torsional Vibration Measurement

ω rotational speed Ω frequency of torsional oscillation θ amplitude of torsional vibration Instantaneous shaft position

φ = ω t + θ cosΩt Angular velocity φ& = ω - θΩ sin Ωt Angular speed range

ω min = ∞ − Ωθ to ω max = ∞ + Ωθ ω − ω min θ = max 2Ω


ω min = θ=

2π 2π , ω max = Tmax Tmin

ω max − ω min 2Ω


Power (Auto) Spectrum - Cross-Spectra Amplitude Spectrum • 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. • 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.

Discrete

Continuous Cross spectra

F (ω )An , φ n Power Spectrum 1 2 An 2 Auto Spectrum

S ff (ω ) = F * (ω )F (ω ) Excitation f(t) Response x(t) S fx (ω ) = F * (ω )X (ω )

S xf (ω ) = X * (ω )F (ω )


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 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.

Frequency Response Function H (ω ) =

X (ω ) F (ω )

X (ω ) H (ω ) = F (ω ) X (ω )X * (ω ) = , = F (ω )X * (ω ) S (ω ) = xx S fx (ω ) =

H 2 (ω ) =

y 2 (ω ) = =

H 1 (ω ) H 2 (ω )

S xf (ω ) S fx (ω )

S ff (ω ) S xx (ω )

X (ω ) F (ω ) X (ω )F * (ω ) , F (ω )F * (ω ) S xf (ω ) S ff (ω )


Cross and Auto Correlations • Correlation is a measure of how well two functions relate each other, cross correlation is given by the average of the product x(t ) f (t + τ ) where t is the delay period. • Autocorrelation function is a measure of a function correlates with itself. • Autocorrelation has a maximum value when the delay period is zero, and it reduces to mean square value of the function. • Autocorrelation decreases as the delay period is increased. • When a signal is random, its mean square value will be negligible. • If a weak periodic signal is buried in a truly random noise, the spectrum reveals very little, however, autocorrelation reveals this weak periodic signal, since the random part has zero autocorrelation.

R fx (τ ) = lim T →∞

T /2

1 x(t ) f (t + τ )dt ∫ T −T / 2

− R ff (τ ) = lim T →∞

T /2

1 f (t ) f (t + τ )dt T −T∫/ 2


Cepstrum - (Time Domain)

• Definitions of Cepstrum • Terminology Cepstrum for Spectrum Quefrency (sec) for Frequency 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 in cepstrum domain.

Power spectrum of a logarithmic power spectrum C xx (τ ) = F [log S xx (ω )]

2

Inverse Fourier transform of logarithmic power spectrum C xx (τ ) = F −1 [log S xx (ω )]

Inverse Fourier transform of amplitude spectrum C xx (τ ) = F −1 [log X (ω )]

Identify periodic structures in a spectrum Appear as distinct peaks in cepstrum Two families of side bands frequency differences ∆f 1 and ∆f 2 Hz Two distinct peaks at frequency 1 / ∆f 1 and 1 / ∆f 2 sec


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 have disappeared and the storage has become very simple and reliable. The analog 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 is illustrated 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


Examples Resolution with 16 bit comparator

3 bit comparator

5 4 E ref and E ref 8 8 comparators 1 - 4 read high and 5 - 7 read low Input signal between Input state is 4 corresponding to binary output

(

100 2 = 2 2 × 1 + 21 × 0 + 2 0 × 0 = 4

)

All comparators read high, the output is 1112 (= 7)

Typical range for output voltage : - 10 to + 10 V Converter saturates if the input signal exceeds the upper or lower limit Appropriate signal conditioning viz., amplitude attenuation or dc offset removal

3 - bit parallel encoder' s output

All comparators read low, the output is 000 2 (= 0)

1 E ref 23

1 ∆V fs 2n ∆V fs full scale range, n number of bits

Voltage resolution per bit E v =

16 - bit A/D converter card Range divided into 216 = 65,536 increments Voltage resolution 0.3mV for - 10 to + 10V range


Signal Sampling • Signal sampling is an important criterion while acquiring the data Sampling rate f s =

1 ∆t

• The same signal above shows inadequate sampling rate, number of points captured in a unit time is too low and the original signal character is lost. • Decreasing sampling time or increasing sampling rate improves the digitization process.


Aliasing – Nyquist Criterion

f Nyq =

1 fs 2


Sampling Example

1 Frequencies resolved 0, ∆f, 2∆∆fK n∆∆fK N∆∆= (f Nyq ) 2 Highest frequency component - 2000 Hz (120,000 rpm) Minimum sampling rage > 4000 Hz Number of samples in FFT 4096 Vibratory signal f s = 5000 samples per second N = 4096 of these samples ∆f = 5000/4096 = 1.22 Hz


Multiplexer – De-multiplexer

• A multiplexer is an IC chip (Transistor Transistor Logic TTL 74150) which performs like a selectable commutator, in this case a 16 position – single pole switch, 1 of 16 selector. A desired input is chosen by the 4 digit binary number dcba (d is Most Significant Bit and a is Least Significant Bit). Saves costs of wiring while acquiring a large number of signals. The signals are monitored in sequence, one by one by a computer separated by a few seconds. • De-multiplexer or a 1 of 16 data distributor is similar but with reversed action (TTL 74154). Here a multiplexer – de-multiplexer is shown where a single line is used rather than separate lines to connect two positions. For the purpose of binary control the ports d, c, b and a should be connected. • Only 5 circuits instead of 16 to connect between two points

Vibration analysis lectures

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