M. Rades - Dynamics of Machinery 2

Mircea Rade Dynamics

of Machinery II

2009

Preface

This textbook is based on the second part of the Dynamics of Machinery lecture course given since 1993 to students of the English Stream in the Department of Engineering Sciences (D.E.S.), now F.I.L.S., at the University Politehnica of Bucharest. It grew in time from a postgraduate course taught in Romanian between 1985 and 1990 at the Strength of Materials Chair and continued within the master course Safety and Integrity of Machinery until 2007.

Dynamics of Machinery, as a stand alone subject, was first introduced in the curricula of mechanical engineering at D.E.S. in 1993. To sustain it, we published Dynamics of Machinery in 1995, followed by Dinamica sistemelor rotor-lagre in 1996 and Rotating Machinery in 2005.

The course aims to: a) increase the knowledge of machinery vibrations; b) further the understanding of dynamic phenomena in machines; c) provide the necessary physical basis for the development of engineering solutions to machinery problems; and d) make the students familiar with machine condition monitoring techniques and fault diagnosis.

As a course taught to non-native speakers, it has been considered useful to reproduce, as language patterns, full portions from English texts. For the students of F.I.L.S., the specific English terminology is defined and illustrated in detail.

Basic rotor dynamics phenomena, simple rotors in rigid and flexible bearings as well as examples of rotor dynamic analyses are presented in the first part. This second part is devoted to the finite element modeling of rotor-bearing systems, fluid film bearings and seals, and instability of rotors. The third part treats the analysis of rolling element bearings, gears, vibration measurement for machine condition monitoring and fault diagnosis, standards and recommendations for vibration limits, balancing of rotors as well as elements of the dynamic analysis of reciprocating machines and piping systems. No reference is made to the vibration of discs, impellers and blades.

July 2009 Mircea Rade

Prefa

Lucrarea se bazeaz pe partea a doua a cursului de Dinamica mainilor predat din 1993 studenilor Filierei Engleze a Facultii de Inginerie n Limbi Strine (F.I.L.S.) la Universitatea Politehnica Bucureti. Coninutul cursului s-a lrgit n timp, pornind de la un curs postuniversitar organizat ntre 1985 i 1990 n cadrul Catedrei de Rezistena materialelor i continuat pn n 2007 la cursurile de masterat n specialitatea Sigurana i Integritatea Mainilor. Capitole din curs au fost predate din 1995 la cursurile de studii aprofundate i masterat organizate la Facultatea de Inginerie Mecanic i Mecatronic.

Dinamica mainilor a fost introdus n planul de nvmnt al F.I.L.S. n 1993. Pentru a susine cursul, am publicat Dynamics of Machinery la U. P. B. n 1995, urmat de Dinamica sistemelor rotor-lagre n 1996 i Rotating Machinery n 2005, ultima coninnd materialul ilustrativ utilizat n cadrul cursului.

Cursul are un loc bine definit n planul de nvmnt, urmrind: a) descrierea fenomenelor dinamice specifice mainilor; b) modelarea sistemelor rotor-lagre i analiza acestora cu metoda elementelor finite; c) narmarea studenilor cu baza fizic necesar n rezolvarea problemelor de vibraii ale mainilor; i d) familiarizarea cu metodele de supraveghere a strii mainilor i diagnosticare a defectelor.

Fiind un curs predat unor studeni a cror limb matern nu este limba englez, au fost reproduse expresii i fraze din lucrri scrise de vorbitori nativi ai acestei limbi. Pentru studenii F.I.L.S. s-a definit i ilustrat n detaliu terminologia specific limbii engleze.

n prima parte se descriu fenomenele de baz din dinamica rotorilor, rspunsul dinamic al rotorilor simpli n lagre rigide i lagre elastice, precum i principalele etape ale unei analize de dinamica rotorilor. n aceast a doua parte se prezint modelarea cu elemente finite a sistemelor rotor-lagre, lagrele hidrodinamice i etanrile cu lichid i gaz, precum i instabilitatea rotorilor. n partea a treia se trateaz lagrele cu rulmeni, echilibrarea rotoarelor, msurarea vibraiilor pentru supravegherea funcionrii mainilor i diagnosticarea defectelor, standarde i recomandri privind limitele admisibile ale vibraiilor mainilor, precum i elemente de dinamica mainilor cu mecanism biel-manivel i vibraiile conductelor aferente. Nu se trateaz vibraiile paletelor, discurilor paletate i ale roilor centrifugale.

Iulie 2009 Mircea Rade

Contents

Preface i

Contents iii

5. Finite element analysis of rotor-bearing systems 1 5.1 Rotor component models 1

5.2 Kinematics of rigid body precession 3

5.2.1 Main reference frames 3

5.2.2 Rigid-body precession 3

5.2.3 Small rotations of the spin axis 4

5.3 Equations of motion for rotor components 7

5.3.1 Thin disks 7

5.3.2 Uniform shaft elements 13

5.3.3 Bearings and seals 31

5.3.4 Flexible couplings 35

5.4 System equations of motion 36 5.4.1 Second order configuration space form 37

5.4.2 First order state space form 40

5.5 Eigenvalue analysis 40

5.5.1 Right and left eigenvectors 41

5.5.2 Reduction to the standard eigenvalue problem 42

5.5.3 Campbell and stability diagrams 43

5.6 Unbalance response 47

5.6.1 Modal analysis solution 47

5.6.2 Spectral analysis solution 48

5.7 Kinematics of elliptic motion 49

5.7.1 Elliptic orbits 49

5.7.2 Decomposition into forward and backward circular motions 52

5.7.3 Variable angular speed along the ellipse 54

5.8 Model order reduction 56

DYNAMICS OF MACHINERY iv

5.8.1 Model condensation 56

5.8.2 Model substructuring 62

5.8.3 Stepwise model reduction methods 68

References 76

6. Fluid film bearings and seals 77 6.1 Fluid film bearings 77

6.2 Static characteristics of journal bearings 78

6.2.1 Geometry of a plain cylindrical bearing 79

6.2.2 Equilibrium position of journal center in bearing 80

6.3 Dynamic coefficients of journal bearings 83

6.4 Reynolds equation and its boundary conditions 84 6.4.1 General assumptions 86

6.4.2 Reynolds equation 87

6.4.3 Boundary conditions for the fluid film pressure field 89

6.5 Analytical solutions of Reynoldsequation 90 6.5.1 Short bearing (Ocvirk) solution 90

6.5.2 Infinitely long bearing (Sommerfeld) solution 99

6.5.3 Finite-length cavitated bearing (Moes) solution 99

6.6 Physical significance of the bearing coefficients 103

6.7 Bearing temperature 107

6.7.1 Approximate bearing temperature 108

6.7.2 Viscosity-temperature relationship 109

6.8 Common fluid film journal bearings 112

6.8.1 Plain journal bearings 112

6.8.2 Axial groove bearings 112

6.8.3 Pressure dam bearings 114

6.8.4 Offset halves bearings 116

6.8.5 Multilobe bearings 117

6.8.6 Tilting pad journal bearings 124

6.8.7 Rayleigh step journal bearings 128

6.8.8 Floating ring bearings 129

6.9 Squeeze film dampers 130 6.9.1 Basic principle 130

CONTENTS v

6.9.2 SFD design configurations 132

6.9.3 Squeeze film stiffness and damping coefficients 133

6.9.4 Design of a squeeze film damper 135

6.10 Annular liquid seals 137 6.10.1 Hydrostatic reaction. Lomakin effect 138

6.10.2 Rotordynamic coefficients 139

6.10.3 Final remarks on seals 146

6.11 Annular gas seals 147

6.12 Floating contact seals 150 6.12.1 Design characteristics 151

6.12.2 Seal ring lockup 154

6.12.3 Locked-up oil seal rotordynamic coefficients 155

References 159

7. Instability of rotors 161 7.1 Whirling of rotating shafts 161

7.2 Instability due to rotating damping 164

7.2.1 Planar rotor model 165

7.2.2 Qualitative effect of damping 166

7.2.3 Whirl speeds of rotor with rotating damping 168

7.2.4 Quantitative effects of damping 172

7.3 Whirl in hydrodynamic bearings 173

7.3.1 Oil-whirl and oil-whip phenomena 174

7.3.2 Half frequency whirl 176

7.3.3 Onset speed of instability 178

7.3.4 Crandalls explanation of journal bearing instability 179

7.3.5 Stability of linear systems 187

7.3.6 Instability of a simple rigid rotor 189

7.3.7 Instability of a simple flexible rotor 194

7.3.8 Instability of complex flexible rotors 199

7.4 Interaction with fluid flow forces 199

7.4.1 Steam whirl 199

7.4.2 Impeller-diffuser interaction 201

DYNAMICS OF MACHINERY vi

7.5 Dry friction backward whirl 203

7.5.1 Rotor-stator rubbing 203

7.5.2 Dry friction whirl 204

7.6 Instability due to asymmetric factors 206

7.6.1 Parametric excitation 207

7.6.2 Shaft anisotropy 207

7.6.3 Asymmetric inertias 219

7.6.4 Finite element analysis of asymmetric rotors 226

References 229

Index 233

5. FINITE ELEMENT ANALYSIS

OF ROTOR-BEARING SYSTEMS

This chapter presents the background of finite element techniques used for the prediction of rotordynamics behavior. The inertia, gyroscopic and stiffness matrices are established for axi-symmetric disks and uniform shaft segments. Journal bearings are described by linearized models with generally non-symmetric stiffness and damping matrices. Seals, dampers and couplings are also modeled by corresponding matrices. The forcing vectors due to unbalance are established for constant angular velocity. Model order reduction and condensation techniques are also considered.

5.1 Rotor component models

For the analytic prediction of the rotordynamic response, the main components of a machine with rotating assemblies must be first identified and modeled.

The actual system (Fig. 5.1, a) is replaced by a physical model (Fig. 5.1, b), usually a discrete-parameter model, comprising: a) shaft segments (Bernoulli/Timoshenko, uniform/conical); b) bearings (isotropic/orthotropic, undamped/damped); c) disks (rigid/flexible, thin/thick); d) seals; e) dampers (squeeze-film); f) couplings; and g) pedestals (plus the static support structure).

A set of mathematical equations consistent with the modeling assumptions is then generated for each component of the system.

In the following, the inertia, gyroscopic and stiffness matrices are established for circular disks and axi-symmetric shaft segments, starting from expressions of the kinetic and potential energies, and using Lagranges equations:

iiii

QqU

qT

qT

t=

+

&d

d , n,..,,i 21= , (5.1, a)

DYNAMICS OF MACHINERY 2

where T kinetic energy, U potential energy, iq generalized displacement, iq& generalized velocity, and iQ generalized force. The response analysis of general unsymmetrical rotors is beyond the aim of this introductory presentation.

a

b

Fig. 5.1 (from [5.1])

In actual situations, the generalized forces are derived by identifying physically a set of generalized coordinates and writing the virtual work in the form

=

=n

kkk qQW

1 . (5.1, b)

Rotary inertia and shear deformations are considered for axi-symmetrical shafts, while external and internal damping is neglected. Bearings are described by linearized models with generally non-symmetric stiffness and damping matrices. The forcing vectors due to unbalance are established considering constant angular velocity.

Then, the system equations of motion in fixed coordinates are set up. Results of eigenvalue analysis and unbalance response calculations are presented in a form useful for engineering analysis.

5. FEA OF ROTOR-BEARING SYSTEMS 3

5.2. Kinematics of rigid body precession

In order to write the equations of motion of a rigid disk attached to a flexible shaft, it is necessary first to determine the components of its angular velocity.

5.2.1 Main reference frames

The analysis is carried out with respect to two reference frames: X,Y,Z an inertial frame and x,y,z a frame rotating with the rotor (Fig. 5.2). The analysis is restricted to the case when the disk rotates with constant angular velocity (rad/sec) about the spin axis Ox.

Fig. 5.2

5.2.2 Rigid-body precession

Two possible motions of the rigid disk are shown in Fig. 5.3.

a b

Fig. 5.3


The motion in which the Ox axis traces out a cone (Fig. 5.3, a), with angular velocity , is called a precession when point M moves along a closed orbit, and whirling when it moves along a spiralling orbit. In forward precession

0> , while in backward precession 0


b) a rotation around the 1z -axis which transfers the Y -axis into the 1y -axis and the 1x -axis into the x -axis; the 11 z,Y,x frame becomes the 11 z,y,x frame;

c) a rotation about the Ox axis: the 11 z,y,x frame becomes the z,y,x frame.

a b c

Fig. 5.5

The axes X,Y,Z have unit vectors K,J,I , the axes 111 z,y,x have unit vectors 111 k,j,i , and the axes x,y,z have unit vectors k,j,i .

a) Rotation (angular velocity & around OY) (Fig. 5.5, a) The relation between the unit vectors is

=

1

1

1

1

11

cossinsincos

ki

ki

KI

.

In expanded form

[ ]

=

=

1

1

1

1

10010

01

kJi

TkJi

KJI

.

b) Rotation (angular velocity & around 1Oz ) (Fig. 5.5, b) The relation between the unit vectors is

=

11

1

11

cossinsincos

ji

ji

Ji

.

In expanded form


[ ]

=

=

1

1

1

1

1

1

1000101

kji

Tkji

kJi

.

c) Rotation (angular velocity =& around Ox) (Fig. 5.5, c) The relationship between the unit vectors is

=

kj

kj

cossinsincos

1

1 .

In expanded form

[ ]

=

=

kji

Tkji

kji

cossin0sincos0001

1

1 .

The relationship between the unit vectors in the stationary and rotating frames is of the form

[ ]

=

kji

TKJI

.

where the transformation matrix

[ ] [ ][ ][ ]

+

==

cossinsincos

sincoscossin1TTTT .

The vector of the instantaneous angular velocity is

1kJi && ++= . In terms of the unit vectors of the x,y,z system of coordinates

kjiJ sincos += , kjk cossin1 += , so that

( ) ( ) ( ) kji sincossincos &&&&& ++++= . The components of the angular velocity in the rotor-fixed (mobile) x,y,z reference system are


++

=

sincossincos&&&&&

z

y

x

. (5.2)

These will be used in the expression of the disk kinetic energy.

5.3 Equations of motion for rotor components

In this section, the equations of motion of disks, shaft segments, bearings, seals and couplings are set up, together with the corresponding element matrices.

5.3.1 Thin disks

In the following, only axi-symmetric rotors (hence disks) are considered and x,y,z are principal directions of inertia.

5.3.1.1 Inertia and gyroscopic matrices of rigid disks

In the fixed frame X,Y,Z, the disk position is defined by two translations ( v and w) and two rotations ( and ) (Fig. 5.6, a), the displacements of the shaft cross-section at the disk attachment. The angles and are approximately equal to the angular displacements collinear with the Y- and Z-axis, respectively.

Neglecting any unbalance, the kinetic energy of a rigid axi-symmetric disk is

( ) ( )2222221

21

zzyyxxdd JJJwmT ++++= &&v , (5.3)

where Px JJ = and Tzy JJJ == are principal moments of inertia with respect to the x,y,z coordinate frame fixed to the disk and dm is the mass of the disk, PJ is the axial (polar) mass moment of inertia and TJ is the diametral mass moment of inertia.

Substituting x , y , z from (5.2) into (5.3), the kinetic energy becomes

( ) ( )( ) ( )[ ]22

222

sincossincos21

21

21

&&&&

&&&

+++

++++=

T

Pdd

J

JwmT v

or


( ) ( ) ( )2222222212

21

21 &&&&&& ++++++= TPdd JJwmT v .

Neglecting the term 2221 &PJ , the kinetic energy becomes

( ) ( ) 2222221

21

21 PPTdd JJJwmT +++++= &&&&&v . (5.4)

In equation (5.4), the term ( )2221 wmd && +v accounts for rectilinear

translation, ( )2221 && +TJ - for rotary inertia, &PJ - for gyroscopic

coupling and 221 PJ - for rotation around the spin axis.

a b

Fig. 5.6

Applying Lagranges equations, the equations of motion of the rigid circular disk are obtained in the form

yd

dTmT

t==

vv

&&&dd , (5.5, a)

zPT

ddMJJTT

t==

&&&&d

d , (5.5, b)

zd

dTmT

t==

ww

&&&dd , (5.5, c)

( ) ( ) yPTd

MJJTt

==

&&&&d

d , (5.5, d)

0=v

dT , 0=w

dT , ( ) 0=

dT ,


where yT , zT and yM , zM are the components of the applied force and couple, respectively (Fig. 5.6, b).

In matrix form

=

+

y

z

z

y

P

P

T

d

T

d

MTMT

J

J

Jm

Jm

&&&&

&&&&&&&&

-w

v

-w

v

0000000

0000000

000000000000

.

Introducing the 12 state vectors in the X-Y and X-Z planes, respectively,

{ }

= vd

yu , { } = wdzu , and the corresponding 12 forcing vectors acting on the disk in the two planes

{ }

=

z

ydy M

Tf , { }

= y

zdz M

Tf ,

equation (5.5) becomes

[ ] [ ][ ] [ ] { }{ }[ ] [ ][ ] [ ] { }{ }

{ }{ }

=

+

dz

dy

dz

dy

d

d

dz

dy

d

d

f

f

u

u

g

g

u

u

m

m

&

&

&&

&&0

0

0

0 (5.6)

with the forcing right hand term including mass and static misalignment unbalance, interconnection forces and other external effects on the disk.

The submatrices

[ ] = Tdd Jmm 0 0 , [ ] = Pd Jg 0 00 , (5.7) form the inertia and gyroscopic submatrices, respectively, of an axi-symmetric rigid disk.

5.3.1.2 Unbalance force vectors

For the analysis of the synchronous response of a rotor system, it is important to determine the vectors of mass unbalance and static angular misalignment of a rigid disk.


Mass unbalance vectors

Consider the disk shown in Fig. 5.7, a. The mass center G of the disk is located at an eccentricity aCG = from the geometric center C. With respect to the rotating frame x,y,z it is located at cosaac = , sinaas = (Fig. 5.7, b). At 0=t (at rest), is the angle of CG with the Y-axis.

a b

Fig. 5.7

The unbalanced centrifugal force acting in G along CG is 2amd . The components along the (spinning) rotor-fixed reference system are 2cd am and

2sd am , and along the (inertial) fixed reference system are

.tamtamf

,tamtamf

cdsdz

sdcdy

sincos

sincos22

22

+==

(5.8)

The vector of unbalance forces (neglecting disk skewness) can be written

{ } { }{ } taa

mta

a

mfff

c

s

ds

c

ddz

dyd

sin

0

0cos

0

02

2

2

2

+

=

= . (5.9)

Permanent disk skew vector

A disk misaligned with its driving shaft receives active gyroscopic moments which force pitch changes in the disk analogous to the translations set up by centrifugal forces due to mass unbalance.

Consider a rigid disk (Fig. 5.8, a) which is not perpendicular to the shaft. There is a static angular misalignment between a principal axis of inertia of the disk and the shaft axis.


The disk skew can be defined by a rotating vector normal to the spin axis and to the line of maximum skew angle. At 0=t the vector makes an angle with the rotating z axis, so that the disk skew vector has components

cos=c and sin=s in the rotating frame x,y,z (see also eq.(2.84) in Ch. 2). The components of angular displacements (Fig. 5.8, b) become

( )( ) ,t

,t

z

y

++=+=

cos

sin

where and are elastic rotations.

a b Fig. 5.8

The corresponding angular velocities and accelerations are

( )( ) ,t

,t

z

y

+=+=

sin

cos&&&&

( )( ).t

,t

z

y

+=++=

cos

sin2

2

&&&&&&&&

(5.10)

From equations (5.5, b) and (5.5, d), replacing by y and by z , we obtain

( )

( ) .MJJ,MJJ

yzPyT

zyPzT

==+

&&&&&&

Substitution of (5.10) yields

( ) ( )

( ) ( ) ( ).tJJJJM,tJJJJM

TPPTy

TPPTz

++=

++=sin

cos 2

2

&&&&&&

The vector of gyroscopic moments acting on the disk is

( )

+

=

ttJJMM

PTy

z

sincossin

cossincos2 . (5.11)


The vector of unbalance forces and moments is

{ } ( )( )

( )( )

t

JJamJJam

t

JJamJJam

f

cPT

cd

sPT

sd

sPT

sd

cPT

cd

d

sincos 22

+

= . (5.12)

Note that the moment of inertia difference PT JJ can be both negative (thin disk) and positive (thick disk). When 0 PT JJ , the applied moment increases the angular unbalance by pitching the thick disk further in the direction of the initial skew.

5.3.1.3 Flexible disks

Flexible disks can be modeled as two rigid disks connected by springs with rotational stiffness Rk (Fig. 5.9).

Fig. 5.9

The inner disk has four degrees of freedom, two translatios v , w , and two rotations , . The inertia properties are m, PJ , TJ . The outer disk has only two rotational degrees of freedom, and . It has only polar and diametral mass moments of inertia PJ and TJ . Its mass is lumped in the inner disk.

The corresponding 66 matrices [5.1] can be reduced to 44 matrices by a static condensation, selecting the coordinates of the inner disk as active and those of the outer disk as omitted (see Section 5.8.1.2).


5.3.2 Uniform shaft elements

Shaft segments are considered isotropic and axi-symmetric about the spin axis of the rotor. This presentation is restricted to uniform cross-section elements.

5.3.2.1 Timoshenko vs. Bernoulli beam elements

Consider a two-noded uniform shaft segment (Fig. 5.10) modeled by a Bresse-Timoshenko beam element.

Dropping the element index, the following notations are used: l - length of shaft element, E Youngs modulus, G shear modulus, IE - flexural rigidity,

sAG - effective shear rigidity, sA - reduced shear area, - shear coefficient, - mass density of shaft material, A = mass per unit length, I second moment of area of cross section, I = rotary mass per unit length.

Fig. 5.10

At a given node, the shaft has four degrees of freedom, two transverse displacements v and w , and two rotations and , measured in the fixed coordinate system. Define two 14 sub-vectors of nodal displacements in the X-Y and X-Z planes, respectively,

{ }

=j

j

i

i

syu

v

v

, { }

=j

j

i

i

szu

w

w

, (5.13, a)

and the corresponding vectors of nodal forces


{ }

=jz

jy

iz

iy

sy

MTMT

f , { }

=jy

jz

iy

iz

sz

MTMT

f . (5.13, b)

Note the minus sign of rotations and moments about the Y-axis whose positive signs are in accordance with the right-hand rule. The sign convention used here is that positive internal forces act in positive (negative) coordinate directions on beam cross sections with a positive (negative) outward normal.

In the Bernoulli-Euler beam theory, deformations due to shear are neglected. The Bresse-Timoshenko beam theory is based on the following hypotheses: a) planar cross sections remain undistorted and plane; b) warping is neglected; and c) an average shear strain is considered, independent of Y.

The planar section hypothesis introduces additional fake stiffness against warping. In some points, distortions due to shear are underestimated. The relationship for T is not correct. The solution is to use an effective shear area

AAs < . From energy considerations

=

ll

xGATxA

G sA

d21dd

21 22 ,

so that

( )

AA

AAs

==

d

d2

2

.

For a hollow steel shaft, with outer diameter D and inner diameter d, the shear coefficient is

( )2

21033131

1

++=

DdDd..

.

Equations in the X-Y plane

At any point in the beam cross section, the axial displacement is proportional to the distance from the neutral axis

yu = . The average shear strain (Fig. 5.11) is


vv +=+

= xy

uxy .

The axial strain is

== y

xu

x .

The bending moment is

==A

zxz IEAyM d .

The equations in the X-Y plane are presented in Table 5.1.

Neglecting the && term, a convenient static relationship can be established between v and =

s

zGAEI-v , (5.14)

though, for the Bresse-Timoshenko beam, v and are kinematically independent.

Fig. 5.11

Introducing the dimensionless variable

lx= ,

so that

dd1

dd

dd

dd

l== xx ,

the static relationship between v and becomes [5.3]


Table 5.1

Bernoulli-Euler beam Bresse-Timoshenko beam

a) Equilibrium

.v 00

=+=+&&y

yz

T

,TM

.v 00

=+=+

&&&&

y

yz

T

,TM

b) Elasticity

xyzz IEM = .AGT,IEM

xysy

xyzz

==

c) Kinematics

.-v =

=0

,xy .-v

==

xy

xy ,

Elimination of xy Elimination of xy and xy

v == zzz IEIEM ( )-v==

sy

zz

AGT,IEM

Equations of motion

Elimination of zM and yT Elimination of zM and yT

0=+ vv v &&zIE ( )( ) .vv-v-

00

=+=

&&&&

s

sz

AG,AGIE


22

dd

12dd1

=

vl (5.14, a)

where

212 lsz

AGIE= . (5.15)

Equations in the X-Z plane

The axial displacement at any point is

zu = . The average shear strain (Fig. 5.12) is

ww +=+

= xz

uxz .

The axial strain is

== z

xu

x .

The bending moment is

==A

yxy IEAzM d .

The equations in the X-Z plane are presented in Table 5.2.

Fig. 5.12

Neglecting the && term, the static coupling between w and is

( ) ( )=+= s

y

s

y

GAEI

GAEI

w (5.16)


Table 5.2

Bernoulli-Euler beam Bresse-Timoshenko beam

a) Equilibrium

.w 00

==+&&zzy

T

,TM

.w 00

==+

&&&&

z

zy

T

,TM

b) Elasticity

xzyy IEM = .AGT,IEM

xzsz

xzyy

==

c) Kinematics

.w==

,xz

.w+==

xz

xz ,

Elimination of xz Elimination of xz and xz

w == yyy IEIEM ( )w+==

sz

yy

AGT

,IEM

Equations of motion

Elimination of yM and zT Elimination of yM and zT

0=+ ww v &&yIE ( )( ) .www

0

0

=+=+

&&&&

s

sy

AG

,AGIE


Introducing the dimensionless variable

lx= ,

the static relationship between w and becomes

( ) ( )22

dd

12dd1

=wl (5.16, a) where

212 lsy

AGIE= , zy II = .

5.3.2.2 Coordinates and shape functions

Consider third-degree polynomials as approximating functions for the displacements and rotations in the X-Y plane

( )( ) .BBBB

,AAAA

012

23

3

012

23

3

+++=+++=

v

Substitution in the static coupling condition (5.14, a)

( ) ( )230122331223 262231 BBBBBBAAA ++++=++ l yields

03 =B , 32 3 AB l= , 212 AB l= ,

+= 310 2

1 AAB l ,

so that ( ) is of second degree. The translation v and the rotation are approximated in terms of the

nodal displacements, using static shape functions:

( ) ( ) ( ) ( ) ( ) { }( ) ( ) ( ) ( ) ( ) { },uN~N~N~N~N~

,uNNNNNsyji

syji

=+++==+++=

4321

4321

ji

ji

vv

vvv (5.17)

where

4321 NNNNN = , 4321 N~N~N~N~N~ = .


An example of calculation is given here for ( )3N . According to the general properties of shape functions, 3N has a unit value at coordinate 3 and is zero at coordinates 1, 2 and 4. The four boundary conditions yield a set of four equations

( ) 00 =v 00 =A , ( ) 1=lv 121 23 =+

AA ,

( ) 00 = 31 2 AA= , ( ) 0=l 023 23 =+ AA ,

wherefrom the four constants are obtained as

+= 12

3A , += 13

2A , += 11A , 00 =A ,

so that

( ) ( ) +++= 233 321 1N . The shear-modified shape functions for displacements are

( ) ( )[ ]( ) ( ) ( )( ) ( )( ) ( ) ( ) .N

,N

,N

,N

++++=

++=

+++=

+++=

2324

323

2322

321

211

231

12

21

1

12311

1

ll

ll (5.18)

For 0= , the above shape functions become the third-degree Hermite polynomials used for Bernoulli-Euler beams.

The shape functions for rotations are [5.3]

( ) ( )( ) ( )[ ]( ) ( )( ) ( ) .N~

,N~

,N~

,N~

++=

++=

+++=

+=

231

1

6611

1

13411

1

6611

1

24

23

22

21

l

l

(5.19)

For 0=


( ) ,NN~ ii = l

1 ( )41,..,i = . It is useful to introduce a third set of shape functions defined by

= dd1 NN~N l

which will be used in the derivation of the stiffness matrix.

Their expressions are

( ) ( )( ) ( ) .NN

,NN

+==+==

121

11

42

31 l (5.20)

Similar relations hold for the displacements in the X-Z plane

{ } { },uN~,uN

sz

sz

==w

(5.21)

because of the similar coupling relationship (5.16, a) between w and .

5.3.2.3 Inertia and gyroscopic matrices

For an infinitesimal uniform shaft element of length xd , the kinetic energy can be obtained from the similar expression (5.4) derived for a thin disk

( ) ( ) ( ) xIIwAT Ps d2212121d 22222 +++++= &&&&&v , where

IIP 2= , A = , I = , sPPP JI l1== . (5.22)

Integrating, the kinetic energy for the shaft element becomes

( ) ( ) +++++=lll

&&&&&0

2

0

22

0

22 d21d

2d

2xJx

xwT P

sP

s v .

a) The translatory energy is

( ) +=l

&&0

221 d2

1 xwT s v .


Expressing velocities in terms of nodal coordinates and shape functions

{ } { } TTsysy NuuN &&&& === Tvv , { } { }syTTsy uNNu &&&&& == vvv T2 , { } { }szTTsz uNNu &&& =2w ,

this energy can be written

{ } [ ]

{ } { } [ ]

{ }szm

TTsz

sy

m

TTsy

s uxNNuuxNNuT

sT

sT

&444 3444 21

&&444 3444 21

&ll

+=00

1 d21d

21 .

The translational mass submatrix of the shaft element

[ ] ( ) ( ) = 10

d NNm TsT l (5.23)

is the same in the X-Y and X-Z planes.

b) The rotary energy is

( ) +=l

&&0

222 d2

1 xT s .

Expressing angular velocities in terms of nodal coordinates and shape functions

{ }syuN~ && = , { }szuN~ && = , { } { }syTTsy uN~N~u &&& =2 , { } { }szTTsz uN~N~u &&& =2 , this energy can be written

{ } [ ]

{ } { } [ ]

{ }szm

TTsz

sy

m

TTsy

s uxN~N~uuxN~N~uT

sR

sR

&44 344 21

&&44 344 21

&ll

+=00

2 d21d

21 .

The rotational mass submatrix of the shaft element


[ ] ( ) ( ) =1

0

d N~N~m TsR l (5.24)

is the same in the X-Y and X-Z planes.

c) The gyroscopic effect energy is

=l&

03 dxT Ps ,

{ } [ ]

{ }syg

TP

Tsz

s uxN~N~uT

s444 3444 21

&l

=0

3 d .

The gyroscopic submatrix of the shaft element

[ ] ( ) ( ) [ ]sRTPs mN~N~g 2d10

== l . (5.25) The total kinetic energy of the uniform shaft element is

{ } [ ] { } { } [ ]{ } { } [ ]{ } 221

21

21 sPsysTszszsTszsysTsys JuguumuumuT ++= &&&&&

where [ ] [ ] [ ]sRsTs mmm += . (5.26)

The total mass submatrix of the uniform shaft element is

[ ]

=

s

ss

sss

ssss

s

mmmmmmmmmm

m

44

3433

242322

14131211

sym

. (5.27)

where

( ) RTsm 36140294156 211 +++= , ( ) ( ) RTs ..m ll 15351753822 212 +++= , ( ) RTsm 367012654 213 ++= ,


( ) ( ) RTs ..m ll 15351753113 214 +++= , ( ) ( ) RTs .m 222222 10545374 ll +++++= , ( ) ( ) RTs .m 222224 5515373 ll +++= , ss mm 1423 = , ss mm 1133 = , ss mm 1234 = , ss mm 2244 = ,

( )21420 +=l

T , ( )22 130 += ll

R .

The gyroscopic submatrix of the uniform shaft element is

[ ]

=

s

ss

sss

ssss

s

gggggggggg

g

44

3433

242322

14131211

sym

. (5.28)

where

Rsg 7211 = , ( ) Rsg l153212 = , Rsg 7213 = ,

ss gg 1214 = , ( ) Rsg 2222 10542 l++= , ( ) Rsg 2224 5512 l+= , ss gg 1223 = , ss gg 1133 = , ss gg 2334 = , ss gg 2244 = . Substitution of sT in Lagranges equations yields

[ ] [ ][ ] [ ] { }{ }[ ] [ ][ ] [ ] { }{ }

{ }{ }

=

+

sz

sy

sz

sy

s

s

sz

sy

s

s

f

f

u

u

g

g

u

u

m

m

&

&

&&

&&0

0

0

0

or

[ ]{ } [ ]{ } { }sss fuGuM =+ &&& , (5.29) where the 88 inertia matrix [ ]sM is symmetric and the 88 gyroscopic matrix [ ]sG is skew-symmetric .

5.3.2.4 Stiffness matrix

For an infinitesimal shaft element, the strain energy is


( ) xAGxx

IEU xzxys ddd

dd

21d 22

22

++

+

= .

Integrating and substituting the shear strains we obtain

( ) ( ) ( )[ ] ++++=ll

0

22

0

22 d2

d2

xwAG

xIEU s v .

a) The flexural energy is

( ) ( ) +=+=lll

0

2

0

2

0

221 d2

d2

d2

xIExIExIEU .

Expressing rotations and curvatures in terms of nodal coordinates and shape functions

{ }syuN~= , { }syuN~ = l1 , where

( ) ( )dd= ,

{ } { }syTTsy uN~N~u = 21l2 , the bending energy can be written

{ } [ ]

{ } { } [ ]

{ }szk

TTsz

sy

k

TTsy uN

~N~IEuuN~N~IEuU

sB

sB

444 3444 21l

444 3444 21l +=

1

0

1

0

1 d21d

21 .

The stiffness submatrix for bending is

[ ] =1

0

ddd

dd

N~N~IEkT

sB l . (5.30)

b) The shear energy is

( ) ( ) ++=ll

0

2

0

22 d2

d2

xwAGxAGU ss v where


( ){ } { } { }sysysy uNuNN~uNN~ = == l1v , ( ){ } { }szsz uNuNN~ ==+ w ,

so that

{ } [ ]

{ } { } [ ]

{ }szk

Ts

Tsz

sy

k

Ts

Tsy uNNAGuuNNAGuU

sS

sS

444 3444 21l

444 3444 21l +=

1

0

1

0

2 d21d

21 .

The stiffness submatrix for shear is

[ ] =1

0

dNNAGk TssS l . (5.31)

The total strain energy of the uniform shaft element is

{ } [ ] { } { } [ ]{ }szsTszsysTsy ukuukuU 2121 += , where the stiffness submatrix is

[ ] [ ] [ ]sSsBs kkk += . (5.32) or

[ ]

+

+=2

22

2

22

3

sym00

00000

4sym612

264612612

11

l

ll

lllllll

l

IEk s .

Substitution of U in Lagranges equations yields

[ ] [ ][ ] [ ] { }{ }

{ }{ }

=

sz

sy

sz

sy

s

s

f

f

u

u

k

k

0

0

or

[ ]{ } { }sss fuK = , where [ ]sK is the 88 full stiffness matrix of the uniform shaft element.


5.3.2.5 Unbalance force vectors

Consider a linear distribution of the mass unbalance along the shaft element, between cLa , sLa at the left end, and cRa , sRa at the right end

( ) ( )( ) ( ) ,aaa

,aaa

sRsLss

cRcLsc

+=+=

1

1 (5.33)

where LLcL aa cos= , LLsL aa sin= , RRcR aa cos= , RRsR aa sin= ,

La , Ra are the unbalance radii, and L , R are the unbalance phase angles. The corresponding distributed unbalance forces are

( ) ( )t,at,p syy 2= , ( ) ( )t,at,p szz 2= , where

( ) ( ) ( )( ) ( ) ( ) .tatat,a

,tatat,asc

ss

sz

ss

sc

sy

sincos

sincos

+==

The virtual work of these forces is

( ) ( ) +=ll

00

dd xpxpW zT

yT wv ,

{ } { }

{ } { } 44 344 21

l44 344 21l

sz

sy f

zTTs

z

f

yTTs

y pNupNuW +=1

0

1

0

dd .

The kinematically equivalent nodal forces are

{ } =1

0

2 d syTsy aNf l , { } =1

0

2 d szTsz aNf l . The shaft mass unbalance vector has the expression

{ }{ }

+

=

t

aN

aN

t

aN

aN

f

f

sc

T

ss

T

ss

T

sc

T

sz

sy

sin

d

d

cos

d

d

1

0

1

021

0

1

02 ll .


An example of calculation is given below

( ) ( ) ( )[ ]

( ) ( ) ( )

( ) ( ) ( )[ ].aaNaNa

aaNaN

cRcL

cRcL

cRcLsc

2018404211201

dd1

d1d

1

01

1

01

1

01

1

01

++++=

=+=

=+=

The vector of unbalance forces can be written

{ } ( )[ ][ ]

[ ][ ]

+

+= taa

a

aa

at

aa

a

aa

af

cR

cL

sR

sL

sR

sL

cR

cL

s sincos1120

2l ,

where

[ ] ( ) ( )( ) ( )

++++++++

=ll

ll

565440422018

545620184042

a ,

so that

{ } { }{ } { }{ }

+

= t

QQ

tQQ

fy

z

z

ys sincos .

The unbalance force vector is of the form

{ } ( )

+

+= ttfs

sin

cos1120

4

3

2

1

8

7

6

5

8

7

6

5

4

3

2

1

2

l

l

l

l

l

l

l

l

l , (5.34)

where


( ) ( ) cRcL aa 201840421 +++= , ( ) ( ) cRcL aa 54562 +++= , ( ) ( ) cRcL aa 404220183 +++= , ( ) ( ) cRcL aa 56544 +++= , ( ) ( ) sRsL aa 201840425 +++= , ( ) ( ) sRsL aa 54566 +++= , ( ) ( ) sRsL aa 404220187 +++= , ( ) ( ) sRsL aa 56548 +++= .

5.3.2.6 Rotor modeling

Figure 5.13 illustrates an example of finite element modeling for a low pressure turbine rotor (M. L. Adams, 1980).

Fig. 5.13

For the disks integral with the shaft, note the way in which the outer diameter of each shaft element is established, in order to ensure a continuous stepwise variation along the rotor length, which is important in the calculation of shaft element stiffnesses. The remaining part of each disk is considered in the calculation of the disk mass and mass moments of inertia.


Fig. 5.14

For monoblock rotors with integral disks (Fig. 5.14), as used in high-pressure turbines, and for shrunk-on disks (Fig. 5.15), the portion of the disk which contributes to the shaft stiffness is sometimes determined using the empirical angle rule. The same approach is used to replace a portion of a shaft with a large diameter step with several steps of slowly increased diameter (Fig. 5.16).

Fig. 5.15

Very often different diameters are used for the stiffness matrix and the mass matrix of a rotor portion. For instance, the wirings of an electrical motor contribute only to the kinetic energy, but not to the potential energy. Therefore, the diameter used in the calculation of the mass matrix is greater than the diameter used in the stiffness matrix calculation.

Fig. 5.16


Conical shaft finite elements are available [5.4] but their presentation is outside the scope of this presentation. User supplied elements are also used for complicated geometries where the stiffness matrix is calculated by inverting the experimentally obtained flexibility matrix.

5.3.3 Bearings and seals

Radial bearings are usually modeled by translational stiffness and damping coefficients. Inertia effects are considered only for annular seals in centrifugal pumps.

5.3.3.1 Stiffness matrix of a radial bearing

The force-displacement relationship for a radial bearing (in the Y-Z plane) can be written

[ ]

=

=

wv

wv b

zzzy

yzyy

z

y kkkkk

TT

, (5.35)

where the elements of [ ]bk are stiffness coefficients, so that ijk is the spring force in the ith direction due to a unit displacement in the jth direction.

Fig. 5.17

Consider the ZY - reference frame (Fig. 5.17), rotated through an angle with respect to the ZY reference frame. The transformation of displacements can be written

[ ]

=

=

wv

wv

wv

R

cossinsincos

.

The transformation of forces is


[ ]

=

=

z

y

z

y

z

y

TT

RTT

TT

cossinsincos

.

Dropping the b index, the new force-displacement relation is

[ ] [ ][ ] [ ][ ][ ]

=

=

=

wv

wv 1RkRkR

TT

RTT

z

y

z

y .

Because [ ] [ ]TTT =1 , the transformed stiffness matrix is [ ] [ ][ ][ ]

== cs

sckkkk

cssc

RkRkzzzy

yzyyT , (5.36)

where cos=c and sin=s . The original non-symmetric stiffness matrix can be split into the sum of a

symmetric and a skew-symmetric component:

[ ] [ ] [ ]434214342143421

as k

a

a

k

zzs

syy

k

zzzy

yzyy

kk

kkkk

kkkk

+

=

00

, (5.37)

where

( )zyyzs kkk += 21 , ( )zyyza kkk = 21 . (5.38) The transformed stiffness matrix becomes

[ ] [ ] [ ] [ ]( ) [ ] [ ] [ ]asTas kkRkkRk +=+= where, denoting cos=c and sin=s ,

[ ] ( ) ( )( ) ( )

+++++=sckcksksckcskksckcskkcskskck

kszzyysyyzz

syyzzszzyys 2

22222

2222, (5.39)

and [ ] [ ]aa kk = .

These results imply that:

a) the skew-symmetric part [ ]ak is independent of the rotation [ ]T , hence of the angle ;

b) when

zzyy

skk

k=

22tan ,


the symmetric part [ ]sk is diagonal and the angles and 090+ define the principal directions of flexibility.

The symmetric matrix [ ]sk can be diagonalized through the coordinate transformation [ ]R , whereas the skew-symmetric matrix [ ]ak remains unchanged. This implies that a non-symmetric stiffness matrix can be transformed into a matrix in which the off-diagonal elements are skew-symmetrical.

In the special case when the diagonal elements are identical, two particular cases are of interest:

a) when the off-diagonal elements are skew-symmetrical, the transformed matrix is the same as the original matrix:

=

12

21

12

21

kkkk

cssc

kkkk

cssc

hence the bearing is isotropic.

b) when the off-diagonal elements are identical

[ ]

+=

=

2sin2cos

2cos2sin

212

221

12

21

kkkkkk

cssc

kkkk

cssc

k

and for 045= [ ]

+=

21

21

00

kkkk

k ,

so that the bearing is orthotropic, with principal axes of elasticity at 045 and 0135 .

The following six important conclusions can be drawn [5.5]:

a) when zyyz kk , the bearing is anisotropic; b) when zyyz kk = and zzyy kk = , the bearing is isotropic; c) when zyyz kk = and zzyy kk , the bearing is orthotropic, with principal

directions defined by and 090+ ; d) when zyyz kk = and zzyy kk = , the bearing is orthotropic, with principal

directions at 045 and 0135 ;

e) when 0== zyyz kk and zzyy kk , the bearing is orthotropic, with horizontal and vertical principal directions;


f) when 0== zyyz kk and zzyy kk = , the bearing is isotropic; Moreover, the condition zyyz kk (non-zero skew-symmetric coupling

element ak ) is the major cause of the instability of an anisotropic system.

5.3.3.2 Fluid film bearings

Linearized hydrodynamic fluid film bearings are represented by the incremental force components obtained by a Taylor-series expansion of the force-displacement and force-velocity relations about an equilibrium configuration of the journal in the bearing [5.6]:

[ ] [ ]

+

=

wv

wv

&&

4342143421bb c

bzz

bzy

byz

byy

k

bzz

bzy

byz

byy

z

y

cccc

kkkk

TT

. (5.40)

In expanded form

.

MTMT

cc

cc

kk

kk

y

z

z

y

bzz

bzy

byz

byy

bzz

bzy

byz

byy

=

+

&&&&

-w

v

-w

v

000000000000

000000000000

In partitioned form

[ ] [ ][ ] [ ]

{ }{ }

[ ] [ ][ ] [ ]

{ }{ }

{ }{ }

=

+

bz

by

bz

by

bzz

bzy

byz

byy

bz

by

bzz

bzy

byz

byy

f

f

u

u

cc

cc

u

u

kk

kk

&

& (5.41)

where

[ ] = 00 0bijb

ijkk , [ ] = 00 0

bijb

ijcc , ( )z,yj,i = . (5.42)

In the following, bearing/pedestal inertia effects are neglected.

5.3.3.3 Annular seals

Short annular seals in centrifugal pumps are usually considered isotropic. The dynamic seal forces are represented by a reaction force / seal motion of the form [5.7]


+

+

=

wv

wv

wv

&&&&

&&

MCccC

KkkK

TT

z

y . (5.43)

The diagonal elements of their stiffness and damping matrices are equal, while the off-diagonal terms are equal, but with reversed sign. The cross-coupled damping and the mass term arise primarily from inertial effects. Cross-coupled stiffnesses arise from fluid rotation, in the same way as in uncavitated plain journal bearings. The dynamic coefficients of equation (5.43) are defined in Section 6.10.2.

The impeller-volute/diffuser interaction forces are generally modeled by an equation of the form

+

+

=

wv

wv

wv

&&&&

&&

MmmM

CccC

KkkK

TT

c

c

z

y . (5.44)

Note the presence of a cross-coupled mass coefficient cm , in comparison with the liquid-seal model of equation (5.43). The sign of cm is negative, which implies that it is destabilizing for forward precession. The model with identical diagonal elements and skew-symmetric off-diagonal elements ensures radial isotropy.

The forces developed by centrifugal compressor seal labyrinths are at least one order of magnitude lower than their liquid counterparts. They have negligible added-mass terms and are typically modeled by a reaction force/motion model of the form

+

=

wv

wv

&&

CccC

KkkK

TT

z

y . (5.45)

Unlike the pump seal model of equation (5.43), the direct stiffness term is typically negligible and is negative in many cases.

Usually only translational coefficients are considered. Angular dynamic coefficients are used only for long annular clearance seals in multi-stage centrifugal pumps, where forces give rise to tilting shaft motions and couples produce linear displacements.

5.3.4 Flexible couplings

A flexible coupling can be modeled as an elastic element with isotropic translational stiffness Tk and rotational stiffness Rk between station i on one shaft and station j on the other, as shown in Fig. 5.18 [5.1].

The stiffness matrix of such a flexible coupling is of the form


[ ]

=

RRTT

RRTT

RRTT

RRTT

coupling

kkkk

kkkk

kkkk

kkkk

k . (5.46)

This simple type of coupling model allows modest amounts of relative transverse motion and angular misalignment between the centerlines of the components being coupled, but prevents any relative axial motion, so it does not apply to spline couplings.

Fig. 5.18

Inertia effects can be taken into account by including thin rigid disks at each of the connecting points.

5.4 System equations of motion

The equations of motion for the rotor-bearings-seals system are first obtained in second order form, by assembling the element matrices.


5.4.1 Second order configuration space form

It is convenient to use a global displacement vector whose upper half contains the nodal displacements in the Y-X plane, while the lower half contains those in the Z-X plane

{ } {{ }

}{ }

T

Z

nn

Y

nnT

TT

,,,,,,,,,,,,,x 444444 3444444 21 L44444 344444 21 L 11112211 = wwwvvv . (5.47)

Correspondingly, the upper half of the global forcing vector contains the nodal forces in the Y-X plane, and the lower half has the nodal forces acting in the Z-X plane

{ } {{ }

}{ }

T

F

nynzyzyz

F

nznyzyzyT

Tz

Ty

M,T,,M,T,M,T,M,T,,M,T,M,f4444444 34444444 21

L444444 3444444 21

L 22112211

= T .

(5.48)

One can write

{ } { }{ }

=

ZY

x , { } { }{ }

=

z

y

FF

f . (5.49)

By combining the component element equations, the assembled system equations of motion can be written

[ ]{ } [ ]{ } [ ]{ } { }fxKxCxM =++ &&& , (5.50) where

[ ] [ ] [ ][ ] [ ]

=m

mM

00

, [ ] [ ] [ ][ ] [ ]

+

+=zzzy

yzyy

kkkkkk

K ,

[ ] [ ] [ ][ ] [ ] [ ] [ ][ ] [ ] +

=0

0g

gcc

ccC

zzzy

yzyy , (5.51)

{ } { }{ }

=

ZY

x &&

& , { } { }{ }

=

ZY

x &&&&

&&

are of order nN 4= , where n is the number of nodes. The global inertia matrix is shown in Fig. 5.19. The global gyroscopic

matrix is shown in Fig. 5.20. The global stiffness matrix is shown in Fig. 5.21.


Fig. 5.19

Fig. 5.20

Fig. 5.21


Matrices [ ]yzk , [ ]zyk and [ ]zzk resemble [ ]yyk . Bearing matrices are similar.

Fig. 5.22

For the rotor model from Fig. 5.22, the equations of motion have the form

Fig. 5.23

Fig. 5.24


The vectors of mass unbalance forces are

{ } { } { } tFtFf sc sincos += (5.52) where the vectors in the right hand side are shown in Fig. 5.24.

5.4.2 First order state space form

For computational purposes, the equation of motion (5.50) is transformed into the first order state space form. Introducing an auxiliary equation

[ ]{ } [ ]{ } { }0= xMxM && , (5.53) equations (5.50) and (5.53) can be combined to give

[ ] [ ][ ] [ ]

{ }{ }

[ ] [ ][ ] [ ]

{ }{ }

{ }{ }

=

+

00

00

fxx

MK

xx

MMC

&&&&

(5.54)

or

[ ]{ } [ ]{ } { }pqBqA =+& , (5.55) where the NN 22 matrices [ ]A and [ ]B are real but non-symmetrical [ ] [ ] [ ][ ] [ ]

= 0M

MCA , [ ] [ ] [ ][ ] [ ]

=

MK

B0

0.

The resulting system of equations is non-self-adjoint.

Note that equations (5.50) and (5.53) can alternatively give

[ ] [ ][ ] [ ]

{ }{ }

[ ] [ ][ ] [ ]

{ }{ }

{ }{ }

=

+

fx

x

CK

M

x

x

M

M 00

0

0

&&&&

(5.56)

but in the following only the form (5.54) will be used.

5.5 Eigenvalue analysis

Solving the rotordynamics eigenvalue problem, natural frequencies, damping ratios, precession modal forms and stability thresholds can be determined for damped gyroscopic systems.


5.5.1 Right and left eigenvectors

The eigenvalues and right eigenvectors are obtained by solving the homogeneous form of equation (5.55)

[ ]{ } [ ]{ } { }0=+ qBqA & . (5.57) Assuming a solution of the form

{ } { } te Rq = , equation (5.57) can be written

[ ] [ ]( ){ } { }0=+ RBA . (5.58) There are Nr 2= eigenvalues r obtained from [ ] [ ]( ) 0det =+ BA , (5.59)

and N2 right eigenvectors { }Rr satisfying the generalized eigenvalue problem [ ]{ } [ ]{ }RrR AB = , ( )N,...,r 21= . (5.60) For the transposed equation

[ ] { } [ ] { } { }0=+ qBqA TT & , assume solutions of the form

{ } { } te Lq = . This yields

[ ] [ ]( ){ } { }0=+ LTT BA . (5.61) The eigenvalues are given by the equation

[ ] [ ]( ) 0det =+ TBA which has the same solutions as those of equation (5.59).

Because the transposed of equation (5.61) is


{ } [ ] [ ]( ) 0=+ BATL , { }L are referred to as left eigenvectors, solutions of the eigenproblem

[ ] { } [ ] { }LrTrLrT AB = , ( )N,...,r 21= . (5.62) The 12 N right and left eigenvectors satisfy the bi-orthogonality relations

{ } [ ] { }{ } [ ] { }

===

===

srsr

B

srsr

A

rrsr

Rr

TLs

rrsr

Rr

TLs

for0for

for0for

(5.63)

so that

r

rr

= . (5.64)

These bi-orthogonality relations between the modes of the original system and those of the transposed system can be used to uncouple the system equations.

5.5.2 Reduction to the standard eigenvalue problem

Equation (5.58) can be written in the form

[ ]{ } { }RRR 1= (5.65) where

[ ] [ ] [ ] [ ] [ ] [ ] [ ][ ] [ ]

==

0

111

IMKCKABR

is a NN 22 non-symmetric real matrix. This formulation has the drawback of giving the reciprocals of eigenvalues.

The upper half of the eigenvectors yields the 1N eigenvectors of the original problem.

An alternative is to use the form (5.56). For an assumed solution

{ } { } teux = , (5.66) the homogeneous form of equation (5.56) becomes


[ ] [ ][ ] [ ]

{ }{ }

[ ] [ ][ ] [ ]

{ }{ }

{ }{ }

=

+

000

00

uu

CKM

uu

MM

(5.67)

or

[ ] [ ]

[ ] [ ] [ ] [ ][ ] [ ][ ] [ ]

{ }{ } { }00

0011 =

u

uI

ICMKM

I .

Remember that nN 4= , where n is the number of nodes.

5.5.3 Campbell and stability diagrams

Equation (5.65) has N2 eigensolutions, where N is the order of the system global matrices. They are purely real for overdamped modes and appear in complex conjugate pairs for underdamped or undamped modes of precession.

In general, the complex eigenvalues are of the form

rrr i+= , rrr i= , (5.68) and they are functions of the rotating assembly spin speed .

The imaginary part r of the eigenvalue is the damped natural frequency of precession (whirl speed or precession speed), and the real part r is the damping constant. A stable mode requires a non-positive value for r .

Often, the damping is expressed in terms of the damping ratio

r

rr

,

or in terms of the logarithmic decrement

r

rrr

22 = .

It is common practice to plot both the precession natural frequencies and damping constants as a function of the rotor spin speed . These plots are called whirl speed maps [5.1], and one is shown in Fig. 5.25.

In most practical applications, if the system becomes unstable, it is usually the first forward precession mode which yields the instability, while the remaining modes remain stable. In Fig. 5.25, the first mode is illustrated to become unstable at an onset speed of instability, oi , also called a threshold speed, th .

Plots of only the precession natural frequencies versus spin speed are commonly referred to as Campbell diagrams. When points are marked on the


natural frequency curves, with the corresponding values of the logarithmic decrement, the plots are called Lund diagrams. Plots of only the real part of the eigenvalues versus spin speed are called stability diagrams. Examples of Campbell diagrams and stability diagrams are given for selected rotor models in Sections 3.4 and 4.5 of Part I.

Fig. 5.25

A critical speed of order of a single-shaft rotor system is defined as a spin speed for which a multiple of that speed coincides with one of the systems natural frequencies of precession. An excitation frequency line of equation

= is included in Fig. 5.25. When equals r , the excitation r creates a resonance condition.

One approach for determining critical speeds is to generate the Campbell diagram, include all excitation frequency lines of interest, and graphically note the intersections to obtain the critical speeds associated with each excitation.

The complex conjugate eigenvectors are of the form

{ } { } { }rrr bau i+= , { } { } { }rrr bau i= . (5.69) The free precession solution can be written as the sum of two complex

eigensolutions associated with the pair of eigenvalues and right eigenvectors

( ){ } { } { }

{ } { }( ) ( ) { } { }( ) ( ) .babauutx

trr

trr

tr

trr

rrrr

rr

ii eiei

ee+

++==+=

( ){ } { } { }

{ } { }( ) .tbtabatx

rrrrt

tt

r

tt

rt

r

r

rrrrr

sincose2

ii2eei

2eee2

iiii

=

=

++=

(5.70)

Figure 5.26 shows the nature of the motion in an undamped precession mode (node and mode indices are dropped out).


Fig. 5.26

The factor tre2 does not influence the mode shape, being a common factor for all coordinates. Therefore, the upper half of the modal vector can be approximated by

{ } { } { } tutux rsrcr sincos += , where

{ } { } { }rrc uReau == , { } { } { }rrs uImbu == . This way, the orbit at any station becomes an ellipse instead of a spiral,

which is taken into account by plotting incomplete (open) ellipses.

An element of the rth mode of precession has the form

( ) tututx rsrcj sincos += , defining a harmonic motion with frequency r .

The two components of the translational motion at any station are


,tt

,tt

rsrc

rsrc

sincossincoswwwvvv+=+=

(5.71)

defining the parametric equations of an ellipse.

Fig. 5.27

The resulting displacement vector is

( ) ( ) ( )tttr wv i+= . Connecting the end points of the displacement vectors at all stations along

a rotor (at a given time t), the mode shape is obtained, which is a curve in space (Fig. 5.27). Examples are given in Sections 3.4 and 4.5 of Part I.

The special case of conservative gyroscopic systems is worth mentioning. If a pair of bearing principal axes exists, for an undamped gyroscopic system, equation (5.50) becomes

[ ] [ ][ ] [ ]

{ }{ } [ ] [ ][ ] [ ] { }{ } [ ] [ ][ ] [ ] { }{ } { }{ }=+

+

00

00

00

00

ZY

kk

ZY

gg

ZY

mm

z

y&&

&&&&

.(5.72)

For an assumed solution of the form (5.66), because the system has pure imaginary eigenvalues, i= , so that equation (5.72) becomes [5.8]

[ ] [ ] [ ]

[ ] [ ] [ ]{ } { }{ } { }

{ }{ }

=

++

00

ii

ii

2

2

IR

IR

z

yzzyy

mkggmk

,

which gives four real coupled matrix equations. It can be shown that the solutions of the equations are proportional to each other

{ } { }RI yy = , { } { }IR zz = , where is a proportionality constant.

The modal vector can be written

{ } { }{ } { }

{ } { }{ } { }

( ){ }( ){ } ( )

{ }{ }

+=

++=

++=

++

I

R

I

R

II

RR

IR

IR

zy

zy

zzyy

zzyy

ii1

ii1

ii

ii


and can be divided by ( )i1+ . Hence, conservative gyroscopic systems are described by pure imaginary

eigenvalues, and eigenvectors are real in the X-Y plane and pure imaginary in the X-Z plane, respectively.

5.6 Unbalance response

The forced response of a rotor system can be determined either indirectly, in modal coordinates, or directly, in physical coordinates.

5.6.1 Modal analysis solution

In the case of synchronous excitation due to mass unbalance

{ } { } tFf ie= , { } { } tXx ie= , equation (5.45) becomes

[ ] [ ][ ] [ ]

[ ] [ ][ ] [ ]

{ }{ }

{ }{ }

=

+

0i0

00

iF

XX

MK

MMC

or

[ ] [ ]( ){ } { }PQBA =+i where

{ } { }{ }

=

XX

Q i , { }{ }{ }

=

0F

P .

Assume a solution of the form

{ } { } rNr

rRQ

==

2

1

, (5.73)

where r are modal coordinates. Multiplying to the left by { }TrR and considering the bi-orthogonality

relations (5.53), the rth decoupled equation is

{ } [ ] [ ]( ){ } { } { }PBA TrLrrRTrL =+i


or

( ) { } { }PTrLrrr =i . The rth modal coordinate is

{ } { }( )rr

Tr

L

rP

= i .

From equation (5.63), the vector of physical coordinates is

{ } { } { }( ) { }PQ rrTr

Lr

RN

r

== i

2

1

,

whose upper half is

{ } { } { }( ) { }FX rrTr

Lr

RN

r

== i

2

1

, (5.74)

where { }rR and { }rL are the upper halves of the corresponding modal vectors.

5.6.2 Spectral analysis solution

Consider again equation (5.50). For a synchronous excitation

{ } { } { } tFtFf sc sincos += , (5.75) the steady-state response has the form

{ } { } { } tXtXx sc sincos += . (5.76) Substitution of (5.75) and (5.76) into (5.50) yields

[ ] [ ]( ){ } [ ]{ } { }[ ]{ } [ ] [ ]( ){ } { }.FXMKXC ,FXCXMK ssc csc =+

=+2

2

(5.77)

This is a linear set of equations which can be solved with known routines. The cos and sin components are substituted back in equation (5.76). The two components of the translational motion at any station are given by equations of the form (5.71), wherefrom the elliptic orbits can be determined as shown in Section 5.7.


5.7 Kinematics of elliptic motion

In the following, the stationary rectangular coordinate frame is z,y,x .

5.7.1 Elliptic orbits

The simplest steady motion of a rotor (nodal) point is a planar motion with the precession frequency . Its displacement components in the y and z directions are

.tztzz,tytyy

sc

sc

sincossincos

+=+=

(5.78)

They define an elliptic orbit, as illustrated in Fig. 5.28.

Fig. 5.28

The ellipse results by composition of two harmonic motions of different amplitudes and phase angles


( )( ) ,tAtAtAz

,tAtAtAy

zzzzzz

yyyyyy

sinsincoscoscos

sinsincoscoscos

=+==+=

in two perpendicular directions.

It is seen that

( ) 2122 scy yyA += , c

sy y

y=tan ,

( ) 2122 scz zzA += , c

sz z

z=tan .

The parameters of the elliptic motion are generally a function of the spin speed . Two particular cases are of interest [5.1]: a) If zy = , the y and z motions are in phase, and the orbit reduces to a straight line (Fig. 5.29).

Fig. 5.29

b) If cs zy = ( )cz+ and sc zy += ( )sz the y motion leads (lags) the z motion by 090 , and the orbit reduces to forward (backward) circular motion (Fig. 5.30).

Fig. 5.30


The parametric equations (5.68) define an ellipse, which is inclined with respect to the y-z axes. Solving first for tcos and tsin as

( ) ( )( ) ( ) ,yzzyzyyzt

,yzzyyzzyt

scsccc

scscss

==

sincos

and then eliminating the time, the orbit equation is obtained as

( ) ( ) ( ) ( ) 2222222 2 sccsscssccsc zyzyzyyzyzyzyyzz =++++ . (5.79) Equation (5.79) is more often expressed in terms of the major and minor

semiaxes a and b, and the orbital inclination angle (Fig. 5.31).

Fig. 5.31

In an 11 zy (principal) coordinate system, with the axes along the ellipse axes, the motion is described as

( )( ),tbz

,tay

+=+=

sincos

1

1 (5.80)

where is a phase angle (inclination of radius vector at 0=t ) so that

12

12

1 =

+

bz

ay . (5.81)

The coordinate transformation

,zyz,zyy

cossinsincos

11

11

+==

(5.82)

leads to parametric equations of the form (5.78). By combining equations (5.78), (5.80) and (5.82), it is possible to obtain the four parameters a, b, and in terms of cy , sy , cz and sz . This will be done in the following by a different approach.


5.7.2 Decomposition into forward and backward circular motions

Elliptical orbits may be represented with the aid of rotating complex vectors, by treating the y-z plane as a complex plane with a real axis along the y-axis and the imaginary axis along the z-axis.

Fig. 5.32

The resultant of a forward rotating vector fr and a backward rotating vector br is a rotating vector whose end moves along an elliptical orbit (Fig. 5.32):

( ) ( )bf tbtftbtf rrrrr ++ +=+= iiii eeee . (5.83) From equations (5.78)

( ) ( )( ) ( ) ,zzz

,yyy

tts

ttc

tts

ttc

iiii

iiii

eei2

1ee21

eei2

1ee21

++=

++=

so that

( ) ( ) ( ) ( ) .zyzyezyzyezyr

cssct

cssct

+++

+++==+=

21i

21

21i

21

i

ii

It is seen that r is the sum of a forward rotating vector with complex amplitude fr , and a backward rotating vector with complex amplitude br , where


( ) ( )( ) ( ) .erzyzyr

,erzyzyr

b

f

bcsscb

fcsscf

i

i

21i

21

21i

21

=++=

=+++= (5.84)

The amplitudes and phase angles are given by

( ) ( )( ) ( ) .

zyzy,bazyzyr

,zyzy,bazyzyr

sc

csbcsscb

sc

csfcsscf

+==++=++=+=+++=

tan22

1

tan22

1

22

22

(5.85)

The semi-major axis is

bf rra += , (5.86)

( ) ( ) ( ) 2222222222241

21

csscscscscsc zyzyzzyyzzyya +++++++= .

The semi-minor axis is

==sc

scbf zz

yya

rrb det1 . (5.87)

The inclination of the major axis (attitude angle) is

( )bf += 21 . (5.88) From the condition of phase coincidence

bf tt +=+= , fbt =2 , where t is the time when P is in the point of maximum displacement amplitude.

The following useful relations can be established

( )222222tanscsc

scsc

zzyyzzyyt +

+= , (5.89)

( )222222tanscsc

sscc

zzyyzyzy+

+= , (5.90)

The precession is forward when

bf rr > , 0>b , cssc zyzy > .


The precession is backward when

bf rr < , 0


Point A on the y axis corresponds to 0=t . Point B corresponds to t, and point C to 2+t . Note that points B and C correspond to a time interval of

2 ( )090=MON while the angle BOC is larger than 090 .

Fig. 5.34

If point M moves along the circle of radius a with constant angular velocity , the point B moves along the ellipse with variable angular velocity.

Denoting AOB= , the angular position at time t is given by t

ab

yz tantan == . (5.92)

The variation of as a function of t is shown in Fig. 5.35. The deviation from the straight line indicates a variable angular velocity.

Fig. 5.35


Indeed, the angular velocity of the motion along the ellipse is

( )t

ab

tab

tt

22

2

tan1

sec

dd

dd

+===& . (5.93)

The variation of & as a function of t is plotted in Fig. 5.36. It is seen that & varies between ( ) ab= and ( ) ba= so that = .

Fig. 5.36

In conclusion, the angular speed of the precession motion is not & - that of the point B along the elliptic orbit, but the speed of the points M and P along the generating circles of radii a and b, respectively.

5.8 Model order reduction

The initial finite element discretization of a rotor system needs a relatively large number of degrees of freedom (DOFs) for satisfactory accuracy. A reduction of the number of DOFs is sometimes necessary because, in actual applications, only a few lower modes are of concern, giving little justification for solving the complete equation of motion in the dynamic analysis.

5.8.1 Model condensation

Static and dynamic condensation techniques can be used to produce reduced models possessing eigensystems that approximate that of the original full system model.


5.8.1.1 Formalism of coordinate reduction

Consider equation (5.50) in the form

[ ]{ } [ ] { } [ ]{ } { }1111

=++NNNNNNNNNNfxKxCxM &&& . (5.94)

A rectangular transformation matrix [ ]T is seeked, which relates the nN 4= elements (coordinates) of the vector { }x to a smaller number NL < of

elements (coordinates) of a vector { }u , so that { } [ ]{ }

11 =

LLNNuTx . (5.95)

The transformation is time independent, so that

{ } [ ]{ }uTx = , { } [ ]{ }uTx && = , { } [ ]{ }uTx &&&& = . (5.96) Substituting equations (5.96) in (5.94) and premultiplying by [ ]TT , the

energy equivalence yields the reduced equations of motion

[ ]{ } [ ] { } [ ]{ } { }1111

=++L

red

LLL

red

LLL

red

LLL

red fuKuCuM &&& , (5.97)

where

[ ] [ ] [ ][ ]TMTM Tred = , [ ] [ ] [ ][ ]TCTC Tred = , [ ] [ ] [ ][ ]TKTK Tred = , { } [ ] { }fTf Tred = .

A proper choice of [ ]T will drastically reduce the number of DOFs without altering the lower eigenfrequencies and the mode shapes of interest.

5.8.1.2 Guyan/Irons condensation

The basis for the Guyan/Irons reduction is to follow a standard procedure used in static structural analysis, namely, elimination of DOFs at which no forces are applied, whence the name of static condensation.

The coordinates (DOFs) are partitioned into two groups: a) the active (master, retained) coordinates, and b) the omitted (slave, discarded) coordinates, denoted by a and o, respectively.

Partitioning the equation (5.94) accordingly


[ ] [ ][ ] [ ]

{ }{ }

[ ] [ ][ ] [ ]

{ }{ }

[ ] [ ][ ] [ ]

{ }{ }

{ }{ }

=

+

+

o

a

o

a

oooa

aoaa

o

a

oooa

aoaa

o

a

oooa

aoaa

ff

xx

KKKK

xx

CCCC

xx

MMMM

&&

&&&&

.

(5.98)

Assuming { } { }0=of , the static force-deflection relationship reduces to

[ ] [ ][ ] [ ]

{ }{ }

{ }{ }

=

0a

o

a

oooa

aoaa fxx

KKKK

. (5.99)

The lower partition provides a static constraint equation

[ ]{ } [ ]{ } { }0=+ oooaoa xKxK (5.100) which can be written

{ } [ ] [ ]{ }aoaooo xKKx 1= . (5.101) The original set of coordinates { }x can be related to the subset of active

coordinates by the equation

{ } { }{ }[ ]

[ ] [ ] { } [ ]{ }aaoaooa

o

a xTxKK

Ixx

x =

=

= 1 . (5.102)

Equation (5.102) may be referred to as a Ritz transformation. The Ritz basis vectors, which are columns of the Ritz transformation matrix [ ]T , are the displacement patterns associated with unit-displacement of the respective a-coordinates while the o-coordinates are released

{ } [ ]{ } { } { } { }[ ] { }=

=

==L

jajj

aL

a

La xtx

xtttxTx

1

1

21 LL . (5.103)

The reduced set of equations has the form (5.97), where

{ } { }axu = , [ ] [ ] [ ][ ] [ ]oaooaoaared KKKKK 1= , (5.104)

[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ][ ] [ ] [ ] [ ] [ ],KKMKK

KKMMKKMM

oaooooooao

oaooaooaooaoaared

11

11

++= , (5.105)

and [ ]redC has an expression similar to (5.105).


Replacing the dynamic relationship between active and omitted DOFs by a static relationship, the Guyan/Irons reduction is an incomplete extension of the Static condensation, with inherent loss of accuracy.

One exception is worth mentioning: a lumped-mass model, consisting of point masses at the nodes where translational displacements are defined (mass moments of inertia neglected). With all the rotational DOFs as o-DOFs and all the translational DOFs as a-DOFs,

[ ] [ ]0=ooM , [ ] [ ]0=oaM , [ ] [ ]0=aoM , [ ] [ ]aared MM = and the Guyan reduction is accurate.

Drawbacks: a) misapplication can lead to serious modeling errors; b) destroys the banded form of matrices; and c) requires insight, experience and skill to partition the DOFs, though automatic procedures exist for the selection of active DOFs. In conclusion, the accuracy of the refined finite element model one tries very hard to achieve may be lost by the Guyan/Irons reduction.

5.8.1.3 Use of macroelements

Rotating shafts have variable cross sections and usually it is necessary to use a large number of finite elements to obtain a good model of the rotor. The number of elements can be reduced by introducing macro-elements [5.9]. Several short cylindrical elements can be treated as one element. Formally, this is done by a static condensation, treating the interior coordinates at the steps of the cross section as o-DOFs and the boundary coordinates as a-DOFs. This increases numerical economy without loss of accuracy in the results.

a b

Fig. 5.37

For the stepped shaft from Fig. 5.37, a, a macro-element is shown in Fig. 5.37, b. Considering only the motion in the Y-X plane, the 88 macro-element matrix has a banded form (Fig. 5.38). Reordering the nodal displacements, moving up the external DOFs selected as a-DOFs and moving down the internal DOFs selected as o-DOFs, destroys the banded form.


Fig. 5.38

Elimination of internal DOFs using the transformation (5.85) yields a condensed 44 matrix. This allows to maintain the banded structure of the system matrix (Fig. 5.39).

Fig. 5.39

5.8.1.4 Modal condensation

Consider the homogeneous part of the equation (5.50) of a damped anisotropic rotor system, written as

[ ]{ } [ ] [ ]( ) { } [ ] [ ]( ) { } { }0=++++ xKKxGCxM bsb &&& , (5.106) where [ ]bK and [ ]bC are the stiffness and damping matrices for bearings, respectively, [ ]sK is the shaft stiffness matrix, [ ]G is the (shaft+disks) gyroscopic matrix, [ ]M is the mass matrix, and { }x is the 14 n state vector.

The corresponding complex eigenvalue problem yields the damped eigenfrequencies and the complex eigenvectors. For rotor systems with a large


number of DOFs, the complex eigenvalue procedure may run into numerical difficulties and may be time consuming.

One approach to avoid some of these problems is the modal condensation method. One variant is based on the analysis of the isotropic undamped non-gyroscopic part of equation (5.106) in the Y-X plane:

[ ]{ } [ ] { } { }0=++ YkkYm yy&& . (5.107) The shaft is assumed symmetrical, rotor gyroscopic effects are neglected,

bearing damping is neglected, and only an average bearing principal stiffness term is considered, usually the symmetric component defined in equation (5.37). This (neighboring) associate conservative system has planar undamped modes entering as columns in the modal matrix

[ ] { } { } { }[ ]n221 L= . (5.108) A truncated modal matrix is used as the transformation matrix, using only

the L lower modes of the system described by (5.107).

Retaining the first L columns of the matrix (5.108)

[ ] { } { } { }[ ]L L21= . (5.109) The coordinate transformation is

{ } { }{ } [ ] { }{ } [ ] [ ][ ] [ ] { }{ } [ ]{ }uuuuuZYx zyzy =

=

=

=

0

0 (5.110)

where { }u is the reduced state vector. Substitution of (5.110) into (5.106) and pre-multiplication by [ ]T yields

the reduced set of equations of motion (5.97)


where

[ ] [ ] [ ] [ ]= MM Tred , { } [ ] { }ff Tred = , [ ] [ ] [ ] [ ]( ) [ ] += GCC bTred , (5.111) [ ] [ ] [ ] [ ]( ) [ ] += bsTred KKK .

After determining the modal coordinates { }u , the physical coordinates are calculated from equation (5.110).

One can say that up to 10-12 modal vectors { }j have to be used in order to accurately determine the first 2-3 eigenvalues whose imaginary parts fall within the operating speed range. It is possible to introduce diagonal modal damping values accounting for the external and internal/structural damping.

The modal method does not require any intuition on mass lumping, component mode selection, and iterative procedures to improve the transformation matrix. The only basic assumption is that the linear combination of the Ritz basis vectors obtained by considering the undamped isotropic rotor system constitutes a good approximation to the complex eigenvectors of the heavily damped anisotropic rotor system.

5.8.2 Model substructuring

The rotor-stator-foundation system can be subdivided into components or substructures, analyzing the components separately, and then coupling them to obtain the mathematical model of the full system. Methods of component mode synthesis (CMS) or substructure coupling for dynamic analysis are used to synthesize the system equations of motion from the characteristic displacement modes of the components.

In the CMS method, the displacement of any point in a component is represented as the superposition of two types of component modes: a) constrained normal modes, defining displacements relative to the fixed component boundaries, and b) constraint modes, produced by displacing the boundary coordinates. In addition, the coordinates of any component are classified as: a) boundary (junction) coordinates, if they are common to two or more components, and b) interior coordinates, if they do not interface with any other component.

The static constraint modes are determined by giving each of the boundary coordinates a unit displacement in turn, fixing all other boundary coordinates and allowing the internal coordinates to displace. The dynamic constrained normal modes are found by fixing all boundary coordinates and determining the free vibration modes of the constrained component.


The reduction of t

M. Rades - Dynamics of Machinery 2

Documents

Transcript of M. Rades - Dynamics of Machinery 2