Part 1: Discrete systems

•Introduction•Single degree of freedom oscillator•Convolution integral•Beat phenomenon•Multiple degree of freedom discrete systemsp g y•Eigenvalue problem•Modal coordinates•DampingDamping•Anti‐resonances

1

Why suppress vibrations ?

FailureFailureBuilding response to earthquakes (excessive strain)Wind on bridges (flutter instability)FatigueFatigue

ComfortCar suspensionsCar suspensionsNoise in helicoptersWind-induced sway in buildings

Operation of precision devicesDVD readersWafer steppersWafer steppersTelescopes & interferometers

2

How ?

Vibration damping:Reduce the resonance peaks

Vibration isolation: Prevent propagation of disturbances to sensitive payloads

3

d l Active damping in civil engineering structures

TMD: Tuned Mass Damper = DVA: Dynamic Vibration AbsorberAMD: Active Mass Damper

4

Single degree of freedom (d.o.f.) oscillator

Free body diagramy g

Free response:p

Characteristic equation:

Solution ?

Characteristic equation:Eigenvalues:

5

6

(A ,B, A1, B1 depend on initial conditions)

Impulse responseImpulse response

Spring and damping forces Have finite amplitudes

7

Impulse response for various damping ratios

8

Convolution IntegralConvolution Integral

Linear system

For a causal system:

9

Harmonic response

1. Undamped oscillator

Dynamic amplification

10

Harmonic response

2. Damped oscillator

Dynamic amplification

11

Bode plots Quality factor

12

Nyquist plot

13

Frequency Response Function (FRF)Frequency Response Function (FRF)

Harmonic excitation:

FRF:

The FRF is the Fourier transform of the impulse response

14

Fourier transform

Convolution integral (linear systems):

Parseval theorem:

= energy spectrum of f(t)

15

Transient response (Beat)Transient response (Beat)

Undamped oscillator starting from rest:

[ ]

Modulating functionModulating function

At resonance

16

Transient response Beat

Steady stateSteady-state amplitude:

The beat is a transientPhenomenon !

17

State Space form(system of first order differential equations)

Oscillator:

State variables:State variables:

Alternative choice Of state variables:Of state variables:

18

Problem 1: Find the natural frequency of the single story building

19

Problem 2: write the equation of motion of the hinge rigid bar

20

Multiple degree of freedom systems

In matrix form:

21

Mass matrix Stiffness matrix Damping matrix

Symmetric & semi positive definite

[]

22

Eigenvalue problemEigenvalue problem

Free response of the conservative system (C=0)

A non trivial solution exists if Eigenvalueproblem

The eigenvalues s are solutions of

Because M and K are symmetric and ysemi-positive definite, the eigenvaluesare purely imaginary:

Natural frequency Mode shape

Two-mass system:

23

Example: Two-mass system:

Natural frequencies:

Mode shapes:

24

Orthogonality of the mode shapesOrthogonality of the mode shapes

Upon permuting i and j,

Subtracting:

The mode shapes corresponding to distinct natural frequencies are gorthogonal with respect to M and K

Modal mass(or generalized mass)

[Can be selected freely]

Rayleigh quotient:25

Orthogonality relationships in matrix form: withOrthogonality relationships in matrix form: with

Notes:(1) Multiple natural frequencies:

If several modes have the same natural frequency they form a subspaceIf several modes have the same natural frequency, they form a subspaceand any vector in this subspace is also solution of the eigenvalue problem.

(2) Rigid body modes:

They have no strain energy:

They also satisfy which means that they are solutions of theThey also satisfy which means that they are solutions of the eigenvalue problem with

26

Free response from initial conditionsFree response from initial conditions

2n constants to determine from the initial conditions.Using the orthogonality conditions,g g y ,

27

If there are rigid body modes (i=0)

Rigid body modes Flexible modesRigid body modes Flexible modes

28

Problem: write the mass and stiffness matrices for the structures

29

Kinetic energy:

30Strain energy:

Building with n identical floors

Natural frequencies:

Mode h

31

shapes:

Modal coordinates

Orthogonality relationships

xx

Assumption of modal damping:

Set of decoupled equations of single d.o.f. oscillators:

Mode i:Mode i:Work of the external

forces on mode i

32

Modal truncation

Mode i:

The modes within the bandwidth of f respond dynamically; y yThose outside the bandwidth respond in a quasi-static manner.

33

Modal truncation

If

The response may be split into two groups of modes:

Responding dynamically

Responding in a quasi-static manner

34

Damping

35

36

Passive damping of very lightlydamped Structures (0.0002)

with shunted PZT patches

37

Rayleigh damping

and are free parameters that and are free parameters that Can be selected to match the

Damping of two modes.

38

Dynamic flexibility matrixDynamic flexibility matrix

Harmonic response of:

[ ]

Modal expansion of G( ):Modal expansion of G():

39

M d l t tiModal truncation

Dynamic flexibility matrix

[m<<n]

Dynamic flexibility matrix

Dynamicamplification

of mode i

40

Modal truncationModal truncation

Residual modesmodes

Dynamic part restricted tothe low frequency modes

41

Anti resonancesAnti-resonances

Diagonal terms of the dynamicflexibility matrix:

collocated

If the system is undamped,Gkk is purely real:

All the residues are positive and Gkk is a monotonously increasing function of

Gkk () = 0

42

Gkk () = 0

1. There is always one anti-resonance frequency between two resonance frequencies.2 Th ti d d th l ti f th ll t d t t / i2. The anti-resonances depend on the location of the collocated actuator/sensor pair.

43

Poles of a single degree of freedom oscillator

44

P l tt f t t ith ll t d t t / iPole-zero pattern of a structure with collocated actuator/sensor pair

With damping

45

Alternative form of the open-loop transfer function of collocated systems:Alternative form of the open loop transfer function of collocated systems:

undamped damped

46

N i t & B d l t f ll t d tNyquist & Bode plots of collocated systems

47

Anti resonances and constrained systemAnti-resonances and constrained system

48

Lagrangian dynamics

•Principle of virtual work•D’Alembert principle•Hamilton’s principle•Hamilton s principle•Lagrange’s equations•Examples•First integrals of Lagrange’s equationsFirst integrals of Lagrange s equations•Green strain tensor•Geometric strain energy (prestress)•BucklingBuckling

1

L i D iLagrangian Dynamics

Newton (1642‐1727) introduced the equation of dynamics in vector formHamilton (1805 –1865) wrote them in variational form which is more generalHamilton (1805  1865) wrote them in variational form, which is more generalBecause it can be extended to distributed and electromechanical systems.

Generalized coordinates qi: set of coordinates describing the kinematics of the system.If minimum, they are independent.  If not, they are connected by kinematic constraints.

2

Principle of virtual work (example)

Find the relationship between f and w at the static equilibrium

The static equilibrium problem is transformed into kinematics:

Principle of virtual work:

Virtual displacements:

0=0

3

D’Alembert’s principle(Extension of the principle of virtual work to dynamics)

The inertia forces are added:

The virtual work of the effective forces on the virtual displacementsCompatible with the constraints is zero.

D’Alembert’s principle is most general, but it is difficult to apply because it refersto vector quantities expressed in inertial frame; it cannot be transformeddirectly in generalized coordinates This achieved with Hamilton’s principledirectly in generalized coordinates. This achieved with Hamilton s principle.

4

Hamilton’s Principle xi does not measure the displacements on the truepath, but the separation between the true path anda perturbed one at a given time.

Lagrangian:Lagrangian:

The actual path is that which cancels the variational indicator V I with respectThe actual path is that which cancels the variational indicator V.I. with respect To all arbitrary variations of the path between t1 and t2, compatible with the Kinematic constraints, and such that

5

E lExample:

Hamilton:

is eliminated by integrating by parts

(differential equation of the pendulum)6

Lagrange’s equation

Hamilton’s principle contains only scalar work and energy quantities.Does not refer to any specific coordinate system and the system configuration may beexpressed In terms of generalized coordinates:expressed In terms of generalized coordinates: We assume an explicit dependency on time which is important for gyroscopic systems.

Kinetic energyKinetic energy:

General formsof T and V:

7

Lagrange’s equation

Example 1: Example 2:

8

Vibration of a linear discrete non‐gyroscopic systemVibration of a linear discrete non gyroscopic system

Lagrange

Dissipation function

All the forces non already included in Dalready included in D

Viscous damping: 

9

Example 3: Pendulum with a sliding massp g

Gravity Spring elasticenergy

Note: Try to obtain these results by writing the absolute acceleration in moving frame andapplying Newton’s law. Which way is easier ? 10

11

Example 4: Pendulum with a sliding disk

Example 3: The rod is a uniform bar of length l and mass M

Two additional terms:

Kinetic energy of the rod:

Potential energy of the rod:

[ ]

Potential energy of the rod:

12

Example 5: Rotating pendulumExample 5: Rotating pendulum

T2 T0

13

Example 6: Rotating spring‐mass system(constant rotation speed)

T2 T02 0

Lagrange:Lagrange:

The system becomes unstable when

14

Example 7: Two‐axis oscillator with anisotropic stiffness(constant rotation speed)(constant rotation speed)

Generalized coordinates: position (x,y) in the rotating frame.

Absolute velocity in rotating frame:

Kinetic energy:

Potential energy:

[T1 is responsible forNon conservative forces:

[T1 is responsible for the gyroscopic effects]

or

Note: This model is representative of a « Jeffcott rotor »  with anisotropic shaft 15

Coupling between x and y

Anti‐symmetric matrixof gyroscopic forcesof gyroscopic forces, which couples the motionin the two directions

Modified potential:

16

In the particular case:(i i & d d)(isotropic & undamped) 

[ ]

To study the stability, we assume a solution

Characteristic equationCharacteristic equation:

The solutions are purely imaginaryThe solutions are purely imaginary: 

Campbell diagram

17

Anisotropic stiffness, undamped

Characteristic equation:

This term is negative for:This term is negative for:The system is unstable (Routh‐Hurwitz)

18

Vibrating angular rate sensor

19

Vibrating angular rate sensor (2)

Assuming:

Harmonic excitation:

20

« First integrals » of the Lagrange equations:

1. Jacobi integral

If the system is conservative and if the Lagrangian does not depend explicitly on timeIf  the system is conservative and  if the Lagrangian does not depend explicitly on time

Total time derivative of L:

Lagrange equation:

21

Euler theorem on homogeneous functions:

If Tn is homogeneous of degree n in some variables qi,

It satisfies the identity:It satisfies the identity:

Since

Jacobi integral or Painlevé integral

If T=T2, it reduces to the Conservation of  the total energy:

22

2 Ignorable coordinate2. Ignorable coordinate

If the Lagrangian does not depend explicitly on some coordinate qs, the coordinate is ignorable:

Lagrange equation:

The generalized momentum associated with an ignorable coordinate is conserved.

23

Example: the spherical pendulump p p

is ignorable:

[Conservation of the angular momentum about Oz][ g ]

24

Green Strain Tensor

G TGreen Tensor:

Quadratic partQuadratic part

25

Global rigid body rotation:Global rigid body rotation:

26

27

Geometric stiffness

•Tension prestresses tend to rigidify the system and increase the natural frequencies

•Compression prestresses tend to soften the system and decrease the natural frequencies

For a linear elastic material:

Constitutive equations:

Strain energy density:

Alternative form of the Constitutive equations:

28

Geometric strain energy due to prestress

[Additional stress and strainFrom prestressed state]

It is impossible to account for thestrain energy associated withthe prestress if the linear straintensor is usedtensor is used

It can be shown that the strain energy can be written:

Geometric strain energy due to prestressgy p(it may be >0 or <0 depending on the prestress)If >0, it rigidify the system and increases the resonances

29

Di t t ith t

For a discrete sytem, Vg takes the form of a quadratic functionof the generalized coordinates:

Discrete system with prestresses

of the generalized coordinates:

Where Kg is the geometric stiffness (no longer positive definite)

The Lagrangian of the system is:

Leading to the equation of motion:

The natural frequencies are solutions of the eigenvalue problem:

Buckling occurs when the smallest natural frequency is reduced to 0

30

BucklingBuckling

If a loading produces a prestress and a geometric stiffness matrix

Then, a proportional loading produces a prestress

And a geometric stiffness matrix

The solution     of the eigenvalue problem

h l b kl l f f f h l d d b d f d bgives the critical buckling amplification factor for the load distribution defined by

And      gives the buckling mode.

31

Part 3•Continuous beams, bars and string•Rayleigh‐Ritz method•Beam with prestresses•Rayleigh quotient

1

Vibration of beams (Euler‐Bernoulli)

Vertical equilibrium:

Rotational equilibrium:

2

Kinematic assumptions:

The fibers are in a uniaxial state of stress and strain:

(No axial loading)Bending moment:

Partial differential equation

3

Alternative derivation from Hamilton’s principle

StrainEEnergy:

KineticEnergy: Virtual work:Energy:

Hamilton:

(The configuration is fixed at t and t )

4

(The configuration is fixed at t1 and t2 )

Finally:

PDE

BoundaryConditions:At x=0 , x=L

PDE

5

Beam with axial prestress

Vertical equilibrium:

Moment about P:

6

Beam with axial prestress (from Hamilton’s principle)

Green strain:One must add the Geometric energy of prestress:

Hamilton:

PDE:Boundary conditions:

7

Free vibration of a uniform beam

A solution of the form exists if:

Introducing: 

Eigenvalue Non dimensionalEigenvalueProblem:

Non‐dimensionalFrequency:

Characteristic equation:Characteristic equation:

General solution:General solution:

8

Decoupling the boundary conditionsDecoupling the boundary conditions

We define:

Alternative form of the general solution:

Decoupled !!9

Example 1: Simply supported beam

Solutions: (natural frequency)

(mode shape)

10

Simply supported beam (modes)

11

Example 2: Free‐free beamp

12

Double root at1. Rigid body modes:Double root at

Boundary conditions:

2 Flexible modes:2. Flexible modes:

13

For any given value of µFor any given value of µ,

14

Orthogonality relationships

Mode i satisfies:

and

15

Orthogonality relationships (2)

Modal massModal mass

Rayleigh quotient:Rayleigh quotient:

Compare to similar results for discrete systems:

16

Modal decomposition

(integrating by parts)

(using the orthogonality relations)

(work of p on mode k)

17

Modal truncation[Previous discussion 

About discrete systems]

Mode i:

The modes within the bandwidth of f respond dynamically; Those outside the bandwidth respond in a quasi‐static manner.

18

Modal truncation[Previous discussion 

About discrete systems]

If

The response may be split into two groups of modes:

Responding dynamically

Responding in a quasi‐static manner

19

Vibration of a string

Can you derive this fromHamilton’s principle ??

Free vibration: 

Assuming:

General solution:

Boundary conditions at x=0 and x=L:

20

Axial vibration of a bar

Free vibration: S d f dFree vibration: Speed of sound

General solution:

21

Boundary conditions: fixed at x=0, free at x=L

22

Rayleigh‐Ritz method(global assumed mode method)

Transform s PDE to ODE by making global assumptions on the displacement field

Shape functions (assumed modes): •Linearly independent, •Complete,•Satisfying the boundary conditionsSatisfying the boundary conditions

Example: simply supported beam:

Note: In this particular example, the assumed modesare orthogonal functions, which makes the coefficients of the expansion independent. 

23

Axial vibration of a bar (Rayleigh‐Ritz)

Strain energy:

The strain is uniform in the cross section

Stiffness matrix:

Positive definite

24

Similarly, the kinetic energy

Mass matrix:

Positive semi‐definite

25

Virtual displacements:Virtual displacements:

(work of the distributed loadsOn the shape function)

26

Planar vibration of a beam (Rayleigh‐Ritz)

Euler‐Bernoulli assumption:

Strain energy:

Rayleigh‐RitzAssumption:

Stiffness matrix

27

Similarly, the kinetic energy:

Mass matrix:

Virtual work of external forces:

GeneralizedGeneralizedforce:

28

With distributed damping:

Dampingmatrix:

29

Beam with axial preload

G t i d t t

Quadratic part of The  Green strain

Geometric energy due to pre‐stress:

Pre‐stress

Geometric stiffness matrix:

(not positive definite, depending on N)30

Simply supported beam with uniform axial load PNatural frequency ?q y

N(x)=‐PAssumed

mode:

Equation of motionEquation of motion:

Pcr = Critical buckling load

Analytical solution31

Rayleigh quotient1. Continuous beams

The mode shapes and the natural frequencies satisfy

Rayleigh quotient: where v(x) is any displacementy g qCompatible with the kinematics

Expanding v(x) in termsOf the mode shapesOf the mode shapes,

32

Rayleigh quotient2 Di t t2.Discrete systems

The mode shapes and the natural frequencies satisfy:

Rayleigh quotient: where x is any vector of generalizeddisplacements compatible with thekinematics.

Principle of stationarity: The Rayleigh quotient is stationary in the vicinity of the natural frequencies and an error of the first order on the mode shape producesan error of the second order on the natural frequency.an error of the second order on the natural frequency.

Consider: normalized such that

33

It follows that:

and

is called the HessianmatrixIf the error is expanded in the mode shapes:

34

Recursive search for eigenvectors based on the Rayleigh quotient:

The eigenvalue and the eigenvectorof order k  are such that

How to project a vector in a space orthogonal to a set of modes        ?

1. The  coefficients of the expansion are obtained from the orthogonality condition:

2. One removes the component along mode k:p g

The projection matrix Ak projects an arbitraryvector in the subspace orthogonal to mode k

35

Return to Problem 1: Find the natural frequency of the single story building

Based on the exact static solution:

36

Based on the exact static solution:

Single storey building with gravity loads

Assumption:

(one column)

>24 !!

Geometric stiffness:

Total stiffness:

37

Natural frequency:

Finite Elements:•Bar element•Plane truss•Beam element•Beam with geometric stiffness•Guyan reduction•Craig‐Bampton reduction

1

Rayleigh‐Ritz vs Finite ElementsRayleigh Ritz vs. Finite Elements

« local assumed modes »

On the contrary to the Rayleigh‐Ritz method, the shape functionswithin an element are selected once and for all for every type of element.

2

Bar element

Kinematic assumptions (within an element):

(uniform strain within the element)( )

Strain energy:

3

Generalized coordinates:

Strain energy:gy

Stiffness matrix:

Kinetic energy:

Mass matrix:

4

Summary:

This model is sufficient for the analysis of the axial vibration of a barHowever if the bar is part of a truss structure one must alsoHowever, if the bar is part of a truss structure, one must also

consider the kinetic energy associated with the transverse motion.

5

Generalized coordinates:

6

Truss structure (assembly)

6 nodes with 2 d.o.f. each: (ui , vi)

Gl b l di tGlobal coordinates:

Using the topology of the structure the localUsing the topology of the structure, the local coordinates of every element are related to theglobal coordinates:

y’

7

The total strain energy and kinetic energy are expressed in global cooordinates

Total strain energy =  sum of the strain energy of all the elements

Global stiffness matrix:

Total kinetic energy =  sum of the kinetic energy of all the elements

Global mass matrix:

8

Beam element (Euler Bernoulli)

1 Ki ti1. Kinematics

Shape functions:

9

Euler‐Bernoulli beam:

In the element:

« consistent » mass matrix(based on the same shape functions

10

( pas the stiffness matrix)

« Lumped » mass matrixThe inertia associated with the rotationThe inertia associated with the rotationis neglected, and one half of the total 

mass is lumped at both ends of the element

11

Beam structure (assemby)Beam structure (assemby)

A bl d tiff t iAssembled stiffness matrix:

12

Assembled mass matrix:

13

Boundary conditions:Boundary conditions:

Partition of the coordinates: where

14

After enforcing the boundary conditions:

15

Eigenvalue problem:Eigenvalue problem:

In the theory of beams,ReducedThe reduced frequencywas defined

Eigenvalue:

First mode: > (analytical result)Larger but quite close

>>>>

1. The FE method overestimates the natural frequencies (Rayleigh quotient)2 Good accuracy ofm natural frequencies requires N>>m degrees of freedom

(the approximation is poor)

16

2. Good accuracy of m natural frequencies requires N>>m degrees of freedom3. High frequency modes depend on the dicretization (no physical meaning).

Convergence of the F.E. model

Consistent massmodel

Lumped mass model

Number of elements

17

Geometric stiffness of a planar beam elementp

Geometric energy due to an axial load N(x):to an axial load N(x):

Positive intraction

18

Example: Single  storey building with gravity loads

Each column may be modeledby a single finite element

After enforcing the boundary conditions:

Geometric stiffness: N=‐mg/2

Total stiffness :

19Previous result obtained with the Rayleigh Ritz method:(section 5.6.1)

Single storey building with gravity loadsPrevious result obtained bythe Rayleigh‐Ritz method

(section 5.6.1)

Assumption:

(one column)

>24 !!

Geometric stiffness:

Total stiffness:

20

Natural frequency:

Guyan Reduction

1. The size of FE models is governed by the representation of the stiffness.2 Automatic mesh generators tend to produce very big models (N>105 d o f )2. Automatic mesh generators tend to produce very big models (N>10 d.o.f.).3. Guyan’s idea (‘60): quasi‐static condensation before solving the eigenvalue problem.

The d.o.f. are separated in two groups:Masters : x1 Slaves : x2 (will be eliminated)The d.o.f. are separated in two groups: Masters : x1 Slaves : x2  (will be eliminated)

Case 1: The slaves have no inertia and have no external forces applied:

Involves only the master d.o.f.Th i i i i hiThere is no approximation in this case

21

Guyan’s assumption:  The quasi‐static relationship between masters and slaves applies in all cases

CoordinateT f iTransformation:

Kinetic energy:Kinetic energy:

Strain energy:

Reduced mass andsiffness matrices :

Virtual work ofExternal forces:

Equation of motionEquation of motion after reduction:

22

Guidelines for selecting the master d.o.f.

1. The d.o.f. without inertia and external forces applied may becondensed without affecting the accuracycondensed without affecting the accuracy

2. The translation d.o.f. carry more information than the rotation d.o.f.

3. The master d.o.f. should be selected in order to maximize the first natural frequency i of the constrained system (x1 blocked)

[the error is an increasing function of the ratio: (i i)2 ]

4. The frequency i of the first constrained mode should be far abovethe frequency band where the model is expected to be accurate.

23

Example 1: Clamped beam modelled with a single finite element

Reduced eigenvalue : Eigenvalue problem:

F.E. (2 dof) Analytical

24

(second row of the stiffness matrix)

Static deflection

Mass and stiffnessAfter reduction:

Constrained system:

25

First row of the stiffness matrix)

Static deflection

Poor quality !Constrained sytem:

q y

26

Example 2: Comparison of various Guyan reductions

27

Craig‐Bampton reduction:X contains all the d o f of interestX1 contains all the d.o.f. of interest

Step 1: Guyan reduction:

St 2 C id th t i d t ( 0)Step 2: Consider the constrained system (x1=0):

Let 2 be a set of modes normalized according to 

The solution is enriched by adding a set of fixed boundary modes:

28

Finally, one gets the reduced equation:

hi h b d ith i i b f t i d dwhich may be used with an increasing number of constrained modes.

29

Example: dynamics of a segmented mirrorE‐ELT Telescope

Primary mirror

One segment

30

31

32

Seismic Response

1

2

Hamilton:

X is arbitraryX0 is arbitraryy satisfies the clampedcondition at the base

The kinetic energy depends The strain energy depends on the gy pOn the absolute velocities

gy prelative displacements

=1 for a point force=1 for a point force

eliminated by Integrating by parts

3

Equations of motion:

With viscous dampingand assuming f=0

Structure clamped at the baseStructure clamped at the base:

Orthogonalityconditions:conditions:

Assumption of modal damping:p g

4

Equations of motion in modal coordinates:

Modal participation factor of mode i

Modal participation vector

d b d lRigid body velocity:

Total massOf the structure

5

Seismic response

Modal participation factorModal response:

Acceleration in the structure:Acceleration in the structure:

6

Fig.2.11

Natural frequencies:

Mode shapes:

7

Modal participation factor: 

Figure 7 2Figure.7.2

8

Amplification of the response within the structure

Precision devices shouldalways be placedon lower floors !!!on lower floors !!!

9

Reaction force f0 and Dynamic mass:

f is expressed in terms off0 is expressed in terms of the inertia forces

Dynamic mass:

10

Alternative representationAlternative representation:f0 is expressed in terms of the elastic forces

(assuming no damping)

Effective modal mass of mode i

For =0: 

11

Effective Modal mass

Figure 7 2Figure.7.2

12

Truncation to mmodes:

Quasi‐static contribution of the high frequency modes

Statically correctStatically correct (no error on the static mass)

13

Vibration alleviation with a Dynamic Vibration Absorber (DVA)

=ma/mT=0.01

Mass ratio= ma /m1m1 ?

DVA parameters:

14

Dynamic Vibration Absorber (DVA)

Equal peak design:(Den Hartog)

15

Equal peak design

16

Effect of the DVA on the reaction force 

17

Response Spectrum

(pseudo velocity)(pseudo‐velocity)

18

19

20

21

22

23

Rotor Dynamics:Jeffcott rotorJeffcott rotor with damping; stabilityJeffcott rotor with damping; stabilityGyroscopic effectsCampbell diagramRigid rotor on elastic supportsRigid rotor on elastic supportsAnisotropic shaft and supports

1

Jeffcott rotor (1919)

Perfectly balanced: The point mass Pcoincide with the elastic center C 

Unbalanced: P has a smallExcentricity with respect to C

Jeffcott rotor:the stiffness k is isotropicthe disk remains aligned along z

In the fixed reference frame Oxyz, the coordinates of the elastic center C are taken as generalized coordinates: (xc,yc)

2

Lagrange equations:Lagrange equations:External forces in fixed frame

(if any) (1)

Unbalanced response (Fx=Fy=0):

A particular solution of the form: 

If

hResponse= synchronousWhirl with amplitude:

(critical velocity)

3

(critical velocity)

Subcritical region: Supercritical region:

(response 180° out of phasewith the excitation)

Self‐centering at high speed

Laval (1889) was the first to run a steam turbine at a supercritical velocity,Demonstrating the weakness of Rankine’s model. 4

Rankine’s model

Example 6: Rotating spring‐mass system(constant rotation speed)

T2 T02 0

Lagrange:Lagrange:

The system becomes unstable when

5

Complex coordinates:

with

(Non‐rotating external forces)

With these notations, the unbalanced response reads:, p

Free whirl: Free motion in complex coordinates: 

Forwardwhirl

Backwardwhirl 6

7

Jeffcott rotor with viscous damping

Two types of damping forces: 1. Stationary damping, associated with the non‐rotating parts (always stabilizing)2. Rotating damping, associated with the motion of the rotor (destabilizing at supercritical velocities)

(1)Stationary damping forces:

Rotating damping forces: (velocity in f )

g p gMoving frame)

Rotating damping forces Expressed in fixed frame:

8

Rotating damping forces expressed in fixed frameg p g p

Skew symmetricContribution to 

The stiffness matrix

Or, using the complex coordinates

The stiffness matrix

9

Stability analysis: Free whirling with damping

The characteristic equation depends on the spin velocity of the rotor:

Stable if: 

Routh‐Hurwitz (after some algebra!..): the system becomes unstable for 

At supercritical velocities, the stability iscontrolled by the ratio between thecontrolled by the ratio between the stationary and the rotating damping

10

Gyroscopic effectsGyroscopic effects

The Jeffcott model ignores the angular momentum of the disk. It cannot account for the gyroscopic effectswhich result from the interaction between the spin of the disk and the bending of the shaft.

Rayleigh‐Ritz approximation: The transverse displacements in inertial frame are assumed of the form:

The function f(z) is normalized in such way that f(a)=1.

11

Strain energy (assuming a uniform beam):

with

(same form as for the Jeffcott rotor)

12

Kinetic energy Since the disk is axisymmetric two rotations are needed to transform theKinetic energy: Since the disk is axisymmetric, two rotations are needed to transform theinertial frame {x,y,z} into a moving frame where the inertia tensor is diagonal and constant:

In this referential, the disk rotatesabout z2 at and the inertia tensorIs diagonal:

13

The rotation velocity of the disk is:

These contributions

Must be transformed intothe moving frame:

Belong to the sameframe  (moving frame)

The absolute rotation velocityis expressed in moving frame:

Rotational kinetic energy:

Responsible for the gyroscopic effects14

Total kinetic energyTotal kinetic energy:

translation rotation

KinematicAssumptions:

(Generalized mass)with

(Gyroscopic constant)

with

If there is a small excentricity of the center of mass, one must add (same as for Jeffcott rotor)

15

D i ith i ff t (i fi d f )Dynamics with gyroscopic effects (in fixed frame)

Lagrange:External forces in theExternal forces in the fixed reference frame

In complex coordinatesIn complex coordinates

16Reduces to the Jeffcott rotor in the angular momentum is neglected, 

Free whirl, Campbell diagram

Free motion in complex coordinates:

Solution

Campbell diagram

Forwardwhirl

Backwardwhirl

17

At point A:

Critical velocity:

18

Unbalanced response

Particular solution:

At the critical velocity, the unbalanced response is

tuned on the forward whirltuned on the forward whirl

19

Response to asynchronous force

The response is obtained by superposition:

20

Rigid rotor on elastic support

Potential energy:

Kinetic energy:

21In what follows, we assume a=b

Lagrange equations: for a=b, two decoupled sets of  equations:

With complex cordinates:

Cylindrical mode

Conical mode

22

Conical mode: 

The eigenvalues are solutions of 

Forward whirl (s j ):

With the notations: 

Forward whirl (s=j1): 

Backward whirl (s=‐j2): 2

23

Critical velocity : No critical velocity for a disk (Iz > Ix)

24

Rigid rotor on elastic support: Campbell diagramg pp p g

25

Jeffcott rotor with anisotropic shaft (the equations are written in moving framewhere the elastic properties are constant)

In moving frame:

26

Example 7: Two‐axis oscillator with anisotropic stiffness(constant rotation speed)

(From chapter on Lagrange’s equation)

(constant rotation speed)

Generalized coordinates: position (x,y) in the rotating frame.

Absolute velocity in rotating frame:

Kinetic energy:

Potential energy:

[T1 is responsible forNon conservative forces:

[T1 is responsible for the gyroscopic effects]

or

Note: This model is representative of a « Jeffcott rotor »  with anisotropic shaft 27

Stationary and rotatingDamping forces: In moving frame:p g g

Lagrange:

Stability of the anisotropic shaft:(without damping)(without damping)

28

Stability of the anisotropic shaft:

Characteristic equation:

Unstable if:

29

Stability in presence of rotating damping:y p g p g

Characteristic equation:

Unstable for all spin velocities above the critical speedsUnstable for all spin velocities above the critical speeds

30