A Professional Short Courseon

Elements of Torsional Vibrationin

Machinery Trains

Samuel Doughty, Ph.D.Registered Professional Engineer

(Retired)

© Copyright November 2013Samuel Doughty

Disclaimer & Copyright Notice

These notes are prepared for educational purposes only. Whilegreat care has been exercised in preparing them, they are notintended to applied directly to any real, professional engineeringsituations. The user must assume all liability for the validity andapplicability of this information in any particular situation.

All material used in these notes is either: (1) in the public domainby virtue of long public use and knowledge, or (2) the original,copyrighted property of the author.No part of this material maybe reproduced by any means whatsoever, without the explicit,written permission of the copyright holder.

Contents

I Nature of the Problem 31 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 System Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2.1 Parameter Evaluation . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2.1.1 Mass Moments of Inertia . . . . . . . . . . . . . . . . . . . . . 4

2.1.2 Shaft Sti¤ness . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.1.3 Excitations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

II Modal Approach 93 Mathematical Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

4 Free Vibration Analysis - Eigensolutions . . . . . . . . . . . . . . . . . . . 10

5 Modal Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

5.1 Modal Response Solutions . . . . . . . . . . . . . . . . . . . . . . . . 13

5.1.1 Zero Frequency Mode Forced Response . . . . . . . . . . . . . 14

5.2 Twisting Mode Forced Response . . . . . . . . . . . . . . . . . . . . . 14

5.2.1 Homogeneous Solution . . . . . . . . . . . . . . . . . . . . . . 14

5.2.2 Particular Solution . . . . . . . . . . . . . . . . . . . . . . . . 15

5.3 Summary of Modal Responses . . . . . . . . . . . . . . . . . . . . . . 16

6 Physical Response Reconstruction . . . . . . . . . . . . . . . . . . . . . . . 16

7 Discussion of the Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

III Direct Approach 198 A More Direct Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

8.1 Mathematical Model . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

8.2 Free Vibrations Transient Response . . . . . . . . . . . . . . . . . . 20

8.3 Steady State Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . 22

8.3.1 Non-sinusoidal Responses . . . . . . . . . . . . . . . . . . . . 23

1

8.3.2 Sinusoidal Responses . . . . . . . . . . . . . . . . . . . . . . . 24

8.4 Complete Physical Response . . . . . . . . . . . . . . . . . . . . . . . 25

IV Summary 27

List of Figures

1 Pictorial Presentation of a 2 Degree of Freedom, Torsionally VibratorySystem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Two Station System Schematic . . . . . . . . . . . . . . . . . . . . . 8

3 Twist Distribution Along the Shaft . . . . . . . . . . . . . . . . . . . 21

2

Part I

Nature of the Problem

1 Introduction

Torsional vibration of machine trains is a very common engineering problem, and yet,the whole area of torsional vibration is somewhat cloudy in the available literature.Much of that literature focuses on methods of computation, rather than being directedat understanding the nature of the motions involved. In recent years, it has becomepopular to use nite element models for torsional vibration analysis of machinery, butthis tends to further obscure the actual understanding of the motions while increasingthe modeling and computational power.

E¤orts to understand torsional vibration of machines are considerably complicatedby the inability of people, in most cases, to directly sense torsional vibration inmachines; it cannot be seen or felt. This immediately limits intuitive understandingof the phenomena, and forces a greater dependence on analysis.

The purpose of this note is to point to some of the phenomena displayed in the analysisof a very simple problem. In particular, the phenomena of (1) steady rotation, (2)rigid body oscillatory rolling response, (3) natural frequency twisting response, and(4) forced, twisting, vibratory response are all shown.

2 System Description

The system considered here is very simple, consisting of two disks, with mass momentsof inertia J1 and J2, joined by a shaft with torsional sti¤ness K, as shown in Figure1. There is assumed to be no damping in the system.

One might be inclined to ask, "Does such a simple model represent any real, physical,engineering system?" This simple system has been used e¤ectively in the past toanalyze such diverse systems as

1. A ships engine and propeller, coupled by a long drive shaft;

2. An airplane engine and propeller, coupled through a relatively short drive shaft;

3. A motor generator set, where two entirely separate electrical machines arejoined for converting electric power from one form to another (not all M-G setsare made this way; some eliminate the drive shaft, building both machines in asingle frame);

3

The element of mass is

dm =

gdv =

gr drd�dx

so that the total mass of the ring is

MRing =

Z h0

Z 2�0

Z RoRi

gr drd�dx

=

�h

g

�R2o �R2i

�The di¤erential (polar) mass moment of inertia is

dJ = r2dm =

gr3 drd�dx

JRing =

Z h0

Z 2�0

Z RoRi

gr3 drd�dx

=

�h

2g

�R2o �R2i

� �R2i +R

2o

�=

1

2Mring

�R2i +R

2o

�For a steel ring ( = 0:283 lb/in3), with radii Ri = 3:0 in. and Ro = 3:5 in., andheight h = 1:5, the mass and mass moment of inertia are

MRing =

�h

g

�R2o �R2i

�=

(0:283) (�) (1:5)

32:174 � 12�3:52 � 3:02

�= 1:1226 � 10�2 lb-s2/in

JRing =

�h

2g

�R2o �R2i

� �R2i +R

2o

�=

(0:283) (�) (1:5)

2 � 32:174 � 12�3:52 � 3:02

� �3:52 + 3:02

�= 0:11928 lb-s2-in

5

2.1.2 Shaft Sti¤ness

The shaft sti¤ness is a function of the shaft diameter and length. For a simple, cylin-drical shaft with no steps, the sti¤ness is calculated from the mechanics of materialsformula for the torsion of a cylinder, � = TL= (JAreaG). This gives an expression forthe torsional sti¤ness:

K =JAreaG

L(2)

where

JArea =

ZA

r2da = 132�D4 for a solid circular shaft of diameter D;

G = shear modulus for the material;

L = length of the shaft.

The question of determining a value for L is where things become a bit indenite,and the calculation becomes as much art as science.

Is it the length between the rotors? Is it the length from rotor center to rotor center? Is there a key and keyway involved? Is there a step in the shaft diameter? Are their llet radii involved?...

What should be used for L?

Much depends upon how the rotors are joined to the shaft, and it is beyond the scopeof the present discussion to go into this matter in depth. It can be said, however, thatin all cases, L is greater than the distance between the rotors. This is because thetwisting action does not stop immediately at the entrance to the rotor. For a rotorshrunk onto a shaft, it is common to say that the twisting action extends roughly1=3 of the way through the rotor, but this is only a rule of thumb that must becarefully evaluated for each situation. As an example, for two rotors with thicknessestR1 and tR2, and applying the rule of thumb just given, the e¤ective value of L is L =Distance between the rotors +1

3(tR1 + tR2).

The foregoing is all for a simple, straight-sided cylindrical shaft, but it is common forshafting to have many diameter changes, often step changes in diameters and some-times tapers as well. Let it su¢ ce to say here that these are much more complicatedsituations, but by applying the proper rules, a good estimate can be obtained for the

6

overall sti¤ness of a shaft segment. These matters are all discussed extensively in theclassic references by W. Ker Wilson1 and BICERA2.

2.1.2.1 Example: Shaft Sti¤ness Calculation Consider a hollow steel shaft(G = 79:3 GPa) with inner and outer diameters, ID = 32 mm and OD = 40 mm.The length of the shaft is L = 1:275 m. The sti¤ness is to be calculated.

JArea =1

32��OD4 � ID4

�=

1

32��0:0404 � 0:0324

�= 1:483 8 � 10�7m4

The shaft torsional sti¤ness is then computed as

K =JAreaG

L

=(1:483 8 � 10�7) (79:3 � 109)

1:275= 9228:7 N-m/rad

Note that all issues regarding the length to be used have simply been side-steppedfor this numerical example.

2.1.3 Excitations

The externally applied torques, T1 and T2 are the source of both the gross rotationalmotion and the angular vibration. If they were both steady torques, there would beonly the transient torsional vibration associated with start-up and shut-down of themachine, but this is very rare. Most prime movers provide an unsteady (variable)torque, such as that associated with

1. an internal combustion engine, either gasoline or diesel;

2. a reciprocating steam engine;

1Wilson, W. Ker, Practical Solution of Torsional Vibration Problems, Vol. I, 3rd. ed. Wiley,1956, Chapter 10, p. 562 ¤ .

2British Internal Combustion Engine Research Association (BICERA), A Handbook on TorsionalVibration, Cambridge University Press, 1958, Sect. 1.2, p. 35, ¤ .

7

Part II

Modal ApproachHere a highly systematic approach to the problem is presented, one that does notrely much at all on physical understanding. The modal approach is widely used in allareas of vibration analysis, both analytical and experimental. It is particularly usefulfor this study because of its systematic methodology and the insight that it provides.

3 Mathematical Model

The angular positions of the two disks are described by a pair of angular coordinates,�1 and �2, both positive in the same sense and measured from a stationary reference.These angles measure the sum of gross (rigid body) rotation and the vibratory rota-tion. When these angles are both zero, it is understood that there is no strain in theshaft. The torques acting on the two disks are T1 and T2, taken positive in the samesense as the two angles, �1 and �2:�

T1T2

�=

�T0�T0

�+

�TC1TC2

�cost+

�TS1TS2

�sint (3)

The value T0 is the useful torque, while all of the other terms have zero average valuebut still do drive torsional vibration in the system. The frequency is often the shaftspeed as will be assumed here, but that is not essential.

The system equation of motion is

�J1

J2

���1�2

�+

�K �K�K K

���1�2

�=

�T1T2

�=

�T0�T0

�+

�TC1TC2

�cost+

�TS1TS2

�sint

(4)

The notation d:c requires some comment. This is used in equation (4) to indicate adiagonal matrix (note also that the o¤-diagonal entries have not been lled in; this isa further indication that this is a diagonal matrix). This notation is used as neededin what follows.

9

4 Free Vibration Analysis - Eigensolutions

For the free vibration analysis, all of the exciting torques are ignored, and the systemnatural frequencies and mode shapes the eigensolutions are determined. The formassumed for the homogeneous solution is

f�g = fAg ej!t

When this is substituted into the equation of motion, the result is

�[K]� !2 dJc

�fAg = f0g (5)

This system of linear, homogeneous equations has nontrivial solutions, if and only if,the determinant of the coe¢ cients is zero. This requirement is expressed as����� K � !2J1 �K�K K � !2J2

����� = 0 (6)which is called the characteristic equation for this system. When the determinant isexpanded, this shows the characteristic polynomial

!2�!2J1J2 �K (J1 + J2)

�= 0 (7)

with solutions

!20 = 0 !21 =

K (J1 + J2)

J1J2(8)

The associated mode vectors are

fA0g =�11

�fA1g =

�1

�J1=J2

�(9)

Note the use of the subscripts 0 and 1 rather than 1 and 2 for the eigenval-ues and the associated eigenvectors. This departure from conventional mathe-matical convention is deliberate. It is to remind one and all that the rst eigenso-lution, the one with subscript 0, is signicantly di¤erent from all the others. Thisbecomes more evident below.

The modied equation of motion shown above, repeated here for convenience,

�[K]� !2 dJc

�fAg = f0g (5)

10

is called the generalized eigenproblem. It is generalized in that it involves two matrices,here denoted as [K] and dJc, which is more di¢ cult to deal with than the classic orsimple eigenproblem dened by

([C]� � dIc) = f0g (10)

where [C] is any matrix, dIc is the identity matrix of the same order as [C], and � iscalled the eigenvalue. There has been more e¤ort devoted to the development of solu-tion methods for the simple eigenproblem, although recently generalized eigensolversare becoming much more readily available. This is not a matter of great importanceto this presentation, and nothing more is said about it here.

5 Modal Transformation

The eigensolutions found in the previous section form the basis for the modal trans-formation. The modal matrix, [A], is composed of the mode vectors as follows:

[A] = [fA0g j fA1g] =�1 11 �J1=J2

�(11)

The modal transformation uses the modal matrix as a transformation matrix, suchthat

f�g = [A] f�g (12)

where f�g = col (�0 (t) ; �1 (t)) is the column vector of modal coordinates. Applyingthis transformation to the equation of motion gives

dJcn�o+ [K] f�g = fTg

dJc [A]n�o+ [K] [A] f�g = fTg

[A]T dJc [A]n�o+ [A]T [K] [A] f�g = [A]T fTg

where in the last line the previous equation has been multiplied through from the leftby [A]T . The triple matrix products are of particular interest, because each of themresults in a diagonal matrix:

11

dMc = [A]T [J ] [A]

=

�1 11 �J1=J2

��J1

J2

� �1 11 �J1=J2

�

=

�1 11 �J1=J2

� �J1 J1J2 �J1

�

=

�J1 + J2 J1 � J1J1 � J1 J1 (1 + J1=J2)

�

=

�J1 + J2

(J1=J2) (J1 + J2)

�= Diagonal Modal Mass Matrix

(13)

dSc = [A]T [K] [A]

=

�1 11 �J1=J2

� �K �K�K K

� �1 11 �J1=J2

�

=

�1 11 �J1=J2

� �0 K (1 + J1=J2)0 �K (1 + J1=J2)

�

=

�0 00 1 + J1=J2 + (J1=J2) (1 + J1=J2)

�

=

�0 0

0 K [(J1 + J2) =J2]2

�= Diagonal Modal Sti¤ness Matrix

(14)

The diagonalization accomplished just above is the primary reason for the modaltransformation. It accomplishes the decoupling of the two equations of motion, re-sulting in two equations, one in terms of �0 and the second in �1. Note that this resultis in no way unique to this system; it applies equally well for any number of degreesof freedom.

The unique character of torsional vibrations is immediately evident in the zero modalsti¤ness for the rst mode. This reects the fact that the system is positive semi-denite, rather than positive denite. A system is positive denite if any possibledisplacement results in positive strain potential energy; a system is positive semide-

12

nite if there exists some displacements that result in positive strain potential energybut there also exists other displacements for which the strain potential energy canremain zero. Because of the free-free nature of the shaft, a rigid body rotation ofthe entire system does not store any strain energy. If one end of the system wereanchored, tied to a nonrotating support, then any displacement of the free end wouldnecessarily involve twisting of the shaft and positive strain energy storage.

The term [A]T also multiplied the right side of the equation, so it is necessary to seewhat this does to the torque terms.

fEg = [A]T fTg

=

�1 11 �J1=J2

��T1T2

�

=

�T1 + T2T1 � (J1=J2)T2

�

=

�T0 + TC1 cost+ TS1 sint� T0 + TC2 cost+ TS2 sintT0 + TC1 cost+ TS1 sint� (J1=J2) (�T0 + TC2 cost+ TS2 sint)

�

=

�(TC1 + TC2) cost+ (TS1 + TS2) sintT0 [1 + (J1=J2)] + [TC1 � (J1=J2)TC2] cos t+ [TS1 � (J1=J2)TS2] sint

�= Modal Excitation Vector

(15)

It is signicant that the constant torque, T0, is not a part of the excitation in therst modal equation of motion; the only contributions there are from the sinusoidalterms. All of the original excitation terms appear in the excitation for the secondmode.

5.1 Modal Response Solutions

The solution proceeds by next solving the modal di¤erential equations, to obtain themodal responses as functions of time, �0 (t) and �1 (t). After this is complete, thephysical displacements will be reconstructed through the modal transformation.

13

5.1.1 Zero Frequency Mode Forced Response

The term zero frequency mode refers to the rst mode of a free-free system, a reminderof the fact that the modal sti¤ness in this mode is zero. The equation of motion forthe zero frequency mode is

M00�0 = (J1 + J2)�0 = (TC1 + TC2) cost+ (TS1 + TS2) sint (16)

This is clearly the equation of motion for the system considered as a rigid body, whichis exactly what the zero frequency mode represents. The equation can be solved bydirect integration,

_�0 =1

(J1+J2)

[(TC1 + TC2) sint� (TS1 + TS2) cost+ C1]

�0 = � 1(J1+J2)2 [(TC1 + TC2) cost+ (TS1 + TS2) sint+ C1t+ C2](17)

This is the full solution for the zero frequency (rst) mode, and it includes twoarbitrary constants, C1 and C2.

5.2 Twisting Mode Forced Response

For the simple system considered here, there is only one twisting mode, the secondmode of the system. The equation of motion governing this mode is

M11�1 + S11�1= T0 [1 + (J1=J2)] + [TC1 � (J1=J2)TC2] cos t+ [TS1 � (J1=J2)TS2] sint

(18)

where the modal mass and sti¤ness are

M11 = (J1=J2) (J1 + J2)

S11 = K [(J1 + J2) =J2]2

Although somewhat hidden in all of the detail regarding the excitation, this is simplyan undamped oscillator, subject to excitation by a constant, a cosine term, and a sineterm. As usual for such an equation, the complete solution consists of homogeneousand particular parts.

5.2.1 Homogeneous Solution

The homogeneous solution is determined without regard for the excitation terms, andis of the form

14

�1H (t) = �1 cos!1t+ �1 sin!1t (19)

where !21 = S11=M11. The coe¢ cients �1 and �1 must be determined to satisfy theboundary conditions of the problem, but that will not be carried out in detail here.Rather, the homogenous solution terms will simply be carried in the form shown.

5.2.2 Particular Solution

The particular solution is determined by the excitation, and since this is a lineardi¤erential equation, the full particular solution can be determined as the sum of theparticular solutions for each of the excitation terms.

5.2.2.1 Constant Solution The constant term is T0 [1� (J1=J2)], and the responseto this is simply a constant, say �1P1

S11�1P1 = T0 [1 + (J1=J2)]�1P1 = T0 [1 + (J1=J2)] =S11

= T0 � J1+J2J2 �J1J2

K(J1+J2)

= T0 � J2K(J1+J2)

(20)

5.2.2.2 Cosine Solution The solution for the cosine excitation is simply a cosineat the same frequency, say �1P2 = C3 cost. After di¤erentiation the assumed formand substitution it into the equation and canceling cosine factors, the result is

(�2 + !21)C3 = 1M11 [TC1 � (J1=J2)TC2]C3 =

[TC1�(J1=J2)TC2]M11(!21�2)

so that

�1P2 =[TC1 � (J1=J2)TC2]M11 (!21 � 2)

cost (21)

5.2.2.3 Sine Solution The sine solution is obtained by the same sort of processas the cosine solution, with the nal result

�1P3 =[TS1 � (J1=J2)TS2]M11 (!21 � 2)

sint (22)

15

5.2.2.4 Full Particular Solution As mentioned earlier, the full solution for thesecond mode is the sum of all of these terms

�1P (t) = T0 � J2K(J1+J2) +[TC1�(J1=J2)TC2]M11(!21�2)

cost+ [TS1�(J1=J2)TS2]M11(!21�2)

sint (23)

5,2.2.5 Complete Twisting Mode Solution The complete solution for the sec-ond mode is the sum of the homogeneous and particular solutions,

�1 (t) = �1 cos!1t+ �1 sin!1t+ T0 � J2K(J1+J2)+ [TC1�(J1=J2)TC2]

M11(!21�2)cost+ [TS1�(J1=J2)TS2]

M11(!21�2)sint

(24)

5.3 Summary of Modal Responses

The response in both modes has been determined, except for the various constantsthat are required to match the boundary conditions. The results are

�0 (t) = � 1(J1+J2)2 [(TC1 + TC2) cost+ (TS1 + TS2) sint+ C1t+ C2]�1 (t) = �1 cos!1t+ �1 sin!1t+ T0 � J2K(J1+J2)

+ [TC1�(J1=J2)TC2]M11(!21�2)

cost+ [TS1�(J1=J2)TS2]M11(!21�2)

sint(17. 24)

These two expressions contain all of the description of the motion, but they generallydefy interpretation. The one thing that can be said, however, is that all of the termsfound in each of the modal responses are generally expected to appear in the physicalangular displacement of each station. The exception to this occurs when some termdrops out by virtue of a zero coe¢ cient.

6 Physical Response Reconstruction

At this point, all that is required in order to reconstruct the physical response isto apply the modal transformation, equation (3.9), to the two expressions in thesummary just above. Thus,

��1 (t)�2 (t)

�= [A]

��1 (t)�2 (t)

�=

�1 11 �J1=J2

���0 (t)�1 (t)

�=

��0 (t) + �1 (t)�0 (t)� (J1=J2) �1 (t)

�The detailed expressions are then

16

�1 (t) = � 1(J1+J2)2 [C1t+ C2] Uniform, Steady Rotation

� 1(J1+J2)2

[(TC1 + TC2) cost+ (TS1 + TS2) sint] Rigid Body Rolling

+�1 cos!1t+ �1 sin!1t Free Vibration Response

+T0 � J2K(J1+J2) Steady Twist

+ [TC1�(J1=J2)TC2]M11(!21�2)

cost+ [TS1�(J1=J2)TS2]M11(!21�2)

sint Dynamic Twist

(25)

�2 (t) = � 1(J1+J2)2 [C1t+ C2] Uniform, Steady Rotation

� 1(J1+J2)2

[(TC1 + TC2) cost+ (TS1 + TS2) sint] Rigid Body Rolling

� (J1=J2) [�1 cos!1t+ �1 sin!1t] Free Vibration Response

�T0 � J1K(J1+J2) Steady Twist

� (J1=J2)�[TC1�(J1=J2)TC2]M11(!21�2)

cost+ [TS1�(J1=J2)TS2]M11(!21�2)

sint

�Dynamic Twist

(26)

7 Discussion of the Results

1. It is important to note that the terms identied as Uniform, Steady Rotationand Rigid Body Rolling are identical in the two expressions. This means thatthere is no twisting associated with these parts of the motion, only rigid bodyrotation.

2. The Steady Twist terms are identical, except for signs. This shows that thesame steady torque is applied to each station, transferring useful work from oneto the other.

3. Oscillatory motion at frequency appears in both the Rigid Body Rolling termsand in the Dynamic Twist terms. The rigid body rolling mode involves notwisting and does not contribute to fatigue damage. It may, however, be a source

17

of error in situations where precise angular positioning is critical. The fact oftwo di¤erent motion types at the same frequency means that experimentalmeasurements of the rotation at a single station (such as are made with aproximity probe sensing the passage of gear teeth), cannot distinguish betweenthe two motions. This fact makes it di¢ cult to interpret physical measurementsof torsional vibration, because the excitation frequencies show up as both rigidbody and twisting motions. This complicates the interpretation of torsionalvibration measurements. This has confused many experimental studies. Thissingle point may be the most important point in this entire presentation.

Just as a check, for reassurance, the di¤erence of the steady angular displacementterms is shown to add up to T0=K:

�1 Steady � �2 Steady = T0 �J2

K (J1 + J2)+ T0 �

J1K (J1 + J2)

=T0K

Note also that the useful torque, To, is transferred by a steady twist in the shaft ofthe amount that would be indicated by a static torque calculation, as seen in thecalculation just above.

18

Part III

Direct Approach

8 A More Direct Approach

Recall from Part II that there are two, essentially di¤erent, parts to the solution of thissystem. The rst is the homogeneous solution, which requires that the eigenproblembe solved. This is the solution that engineers usually refer to as the transient solution.The eigenproblem only needs to be solved if there is interest in the transient solution.In many cases, this is omitted because it is the second part of the solution that isreally of interest, the particular solution, or in engineering terms, the steady statesolution.

While it is certainly true that no damping has been considered in developing themodel, physical understanding dictates that the oscillatory terms in the transientsolution will eventually die away. Friction, although not included in the model, isalways present in reality; this is reason the transient part of the solution eventuallydisappears (after a long enough time) from the solution. Whether the transientsolution is of interest depends heavily on the machine type, the operating conditions,and, most importantly, the underlying purpose for the analysis. Many machinesundergo a brief transient during startup, then run at constant speed for a very longtime, and only experience another transient when they are shut down. For machinesof this type, usually it is the steady state solution that is required for fatigue or othervibration related analyses. On the other hand, there are machines that see theirmost severe stresses during startup, such as a large synchronous motor driven system(starting as an induction machine, with the damper (amortessuer) bars serving asrotor windings), may be lightly stressed during normal running but experience severefatigue damage due to the variable frequency excitations during startup.

In any case, it is not necessary to pursue the full modal solution as was done in theprevious part; there is a shorter, more direct approach available. The purpose for thispart is to show that alternate, more direct, approach. The results are fully equivalent.

8.1 Mathematical Model

In Part II, the system equation of motion was formulated as

19

�J1

J2

���1�2

�+

�K �K�K K

���1�2

�=

�T1T2

�=

�T0�T0

�+

�TC1TC2

�cost+

�TS1TS2

�sint

(3)

This same describing equation is used here, but the analysis to follow is somewhatdi¤erent.

8.2 Free Vibrations Transient Response

The transient response is understood to be that part of the oscillatory motion thatoccurs at the (non-zero) twisting natural frequency. This is true, even if no dampingis included in the model to cause the eventual decay of this motion.

Conservation of angular momentum follows in the absence of any torque to changethe angular momentum (remember, the discussion here is regarding free vibration, sothe external torques are all set to zero.) From conservation of momentum, it followsthat the vibrational motion of the two stations must be in opposition to each other.This leads quickly to the conclusion

J1�1 = �J2�2

This says, that as far as the free vibration goes, there is some point between the twostations that has no vibrational motion at all (it may still have the gross rotationalmotion of the whole system). Thus the angular motion may be visualized as shownin the following Figure 3.

The point c does not participate in the torsional free vibration. The shaft segmentof length a, extending to the left from c has twist in a positive sense, while the shaftsegment b to the right of c twists in the opposite (negative) sense, as indicated by thediagonal line. The point c may thus be considered as a xed point, with one motionoccurring to the left while another motion occurs to the right. The only requirementis that both motions have the same frequency, !1.

The sti¤ness to the left of c is inversely proportional to a, and the sti¤ness to theright of c is inversely proportional to b. Thus,

20

!21 =KaJ1

=KL

aJ1

=K

aJ1(a+ b)

=K

J1

�1 +

b

a

�!21 =

K (J1 + J2)

J1J2

Thus, it is now possible to write the expressions for the two transient motions as

�1�Trans (t) = �1 cos (!1t) + �1 sin (!1t)

�1�Trans (t) = �J1J2[�1 cos (!1t) + �1 sin (!1t)] (27)

where

!1 =

sK (J1 + J2)

J1J2(28)

This agrees with the transient solution found by the modal approach.

8.3 Steady State Solutions

The steady state solutions are all those that do not die away with time when dampingis included. These are typically terms for which there is an excitation term forcingthe continued response. This is not entire the case, as will be seen directly below.

The full equation of motion (again) reads

�J1

J2

���1�2

�+

�K �K�K K

���1�2

�=

�T1T2

�=

�T0�T0

�+

�TC1TC2

�cost+

�TS1TS2

�sint

(3)

22

8.3.1 Non-sinusoidal Responses

For those with a good physical feel for this problem, it is evident that there aresolutions other than the expected sinusoidal responses. Any term for which thesecond derivative is zero will satisfy this equation, so assume

f� (t)gns = fC1g t+ fC2g

where the components of fC1g and fC2g are yet to be determined. It is evident thatthe second derivative of the assumed form is the zero vector, so this form must satisfy

[K] fC1g t+ [K] fC2g = fT0g

This must be true at all times, including t = 0, so this implies two relations,

[K] fC2g = fT0g (29)[K] fC1g = f0g (30)

The rst is the result of requiring that the equation be valid when t = 0, and thenthe second is the result of subtracting the rst from the original equation. These areconsidered separately in what follows.

The rst relation, when written out in more detail, is�K �K�K K

��C21C22

�=

�T0�T0

�When this is expanded, it reads

K (C21 � C22) = T0K (C22 � C21) = �T0

so it is clear that there is really only one scalar equation present. From the secondform, it is evident that

C21 = C22 +T0K= C22 +

T0K

�J1 + J2J1 + J2

�

�C21 +T0K

J2(J1 + J2)

= �C22 �T0K

J1(J1 + J2)

23

The left side is the group of constants appearing in the solution for �1 (t) in the modalsolution. The right side is the group of constants appearing in the solution for �2 (t)in the modal solution. The fact that this can be demonstrated, starting with thedirect solution, shows that the constant terms are identical from both the modal anddirect solutions.

Now turn to the second relation above,�K �K�K K

��C11C12

�=

�00

�which, when expanded says

K (C11 � C12) = 0K (C12 � C11) = 0

Again, it is evident that there is really only one equation. This shows then that

C12 = C11

�C11C12

�= C1

�11

�The complete, non-sinusoidal, response is then:

f� (t)gns = fC1g t+ fC2g

= C1

�11

�t+ C2

�11

�+ C2

T0K

�10

�This is fully equivalent to both the constant solution and the solution linear in t asfound by the modal approach.

8.3.2 Sinusoidal Responses

The sinusoidal responses must satisfy the di¤erential equation written as

�J1

J2

���1�2

�+

�K �K�K K

���1�2

�=

�TC1TC2

�cos (t) +

�TS1TS2

�sin (t)

(31)

24

Assume a steady state solution of the form

f� (t)g =�c1c2

�cos (t) +

�s1s2

�sin (t)

This assumed form is then (1) di¤erentiated twice with respect to time, (2) substitutedinto the di¤erential equation, and (3) sine and cosine terms are separated and thesine and cosine factors are dropped. The result is a pair of equations,

�[K]� 2 [J ]

�� c1c2

�=

�TC1TC2

��[K]� 2 [J ]

�� s1s2

�=

�TS1TS2

�for which the solutions are

�c1c2

�=

�[K]� 2 [J ]

��1� TC1TC2

�=

1

2 (KJ2 + J1K � 2J1J2)

�K � 2J2 K

K K � 2J1

��TC1TC2

�=

1

2 (KJ2 + J1K � 2J1J2)

�(K � 2J2)TC +KT2CKTC + (K � 2J1)T2C

�(32)�

s1s2

�=

�[K]� 2 [J ]

��1� TS1TS2

�=

1

2 (KJ2 + J1K � 2J1J2)

�K � 2J2 K

K K � 2J1

��TS1TS2

�=

1

2 (KJ2 + J1K � 2J1J2)

�(K � 2J2)TS +KT2SKTS + (K � 2J1)T2S

�(33)

The required inverse (or the solution of the system of linear equations) exists, providedthat 6= !1.

8.4 Complete Physical Response

The complete physical response is then assembled from the sinusoidal and non-sinusoidal solution to give

25

f� (t)g = +C1�11

�t+ C2

�11

�+T0K

�10

�+

��1 cos!1t+ �1 sin!1t

�J1J2(�1 cos!1t+ �1 sin!1t)

�+�[K]� 2 [J ]

��1� TC1TC2

�cos (t) +

�[K]� 2 [J ]

��1� TS1TS2

�sin (t)

= C1

�11

�t+

�C21

C21 � T0K

�+

��1 cos!1t+ �1 sin!1t

�J1J2(�1 cos!1t+ �1 sin!1t)

�+

1

(K � 2J1) (K � 2J2)�K2

�K (TC1 + T2C)� TC12J2K (TC1 + T2C)� TC22J1

�cos (t)

+1

(K � 2J1) (K � 2J2)�K2

�K (TS1 + TS2)� TS12J2K (TS1 + TS2)� TS22J1

�sin (t)

or, when written separately,

�1 (t) = C1t+ C21 + �1 cos!1t+ �1 sin!1t

+K (TC1 + T2C)� TC12J2

(K � 2J1) (K � 2J2)�K2cos (t) +

K (TS1 + TS2)� TS12J2(K � 2J1) (K � 2J2)�K2

sin (t)

�2 (t) = C1t+ C21 �T0K� J1J2(�1 cos!1t+ �1 sin!1t) (34)

+K (TC1 + T2C)� TC22J1

(K � 2J1) (K � 2J2)�K2cos (t) +

K (TS1 + TS2)� TS12J2(K � 2J1) (K � 2J2)�K2

sin (t)

26

Part IV

SummaryThe reader will have gained a number of insights into important concepts in the areaof Torsional Vibration of Machine Trains. Specically:

� All of the important features of torsional vibration are evident in a system assimple as the two degree of freedom system.

� The standard model for analysis is a sequence of rigid rotor stations (havingmass) coupled by massless torsional spring elements.

� In this model, the mass moments of inertia for all of the stations are constantvalues, and the spring sti¤ness values are constants. This is a necessary simpli-cation; without it, the problem becomes intractable.

� Mass moments of inertia are determined by integration over the volume of eachstation. This integration may be relatively easy for simple geometries, but itmay be quite di¢ cult for more complex shapes.

� The basic torsional sti¤ness calculation is a simple mechanics of materials rela-tion, but the question regarding the appropriate length to be used may be quitecomplicated. This is as much an art as it is a science.

� The eigenproblem is introduced, leading to the determination of the naturalfrequencies for free vibration, and also the mode shape vectors associated withthose free vibrations.

� The characteristic equation is developed, and the zero natural frequency shownin that polynomial expression.

� The special zero frequency mode is noted, and its connection to the semidenitenature of the system sti¤ness matrix is noted. The associated mode shape forthe zero frequency mode is rigid body motion, denoted by the mode vector =col (1; 1).

� The concepts of positive denite and positive semidenite sti¤ness matrices areexplained.

� The modal transformation is employed to decouple the equations of motion,resulting in forms for which the solutions are much easier to express.

27

� The decoupled modal mass and modal sti¤ness matrices are developed, alongwith the modal excitation vector (the result of applying the modal transforma-tion to the excitations).

� The solution in modal coordinates is developed, with special emphasis on theunique modal solution for the zero frequency mode. This is simply a matterof developing the several parts required for the solution of each di¤erentialequation form.

� The reconstruction of the physical responses from the modal solutions is demon-strated.

� The physical signicance of each group of terms in the physical response ex-pressions is described.

� The existence of both rigid body and twisting motions at the fre-quency of every excitation term is shown. This is a most importantconcept, and one that is often overlooked.

� An alternative solution method, characterized as "direct" is also presented. It isevident that the modal solution is far more systematic while the direct solutionrequires more engineering insight into the nature of the problem.

� The results obtained by the modal solution and the direct solution are exactlyequivalent. A signicant bit of algebra is required to show this equivalence, butit can be done.

� The value of using matrix algebra in the formulation is evident. It makes theextension of these concepts to systems with greater number of degrees of freedomand more complicated excitations relatively self-evident. Matrix notation isreadily adapted to computer implementation for more complicated systems.

28