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Torsional Vibration of Machines with Gear Errors

AW Lees1 MI Friswell1 and G Litak2

1 College of Engineering Swansea University Singleton Park Swansea SA2 8PP UK2 Department of Applied Mechanics Technical University of Lublin PL-20-618 Lublin Poland

E-mail awleesswanseaacuk mifriswellswanseaacuk glitakpollubpl

Abstract Vibration and noise induced by errors and faults in gear meshes are key concernsfor the performance of many rotating machines and the prediction of developing faults Ofparticular concern are displacement errors in the gear mesh and for rigid gears these may bemodelled to give a linear set of differential equations with forced excitation Other faults suchas backlash or friction may also arise and give non-linear models with rich dynamics Thispaper considers the particular case of gear errors modelled as a Fourier series based on thetooth meshing frequency leading immediately to non-linear equations of motion even withoutthe presence of other non-linear phenomena By considering the perturbed response this systemmay be modelled as a parametrically excited system This paper motivates the analysis derivesthe equations of motion for the case of a single gear mesh and provides example responsesimulations of a boiler feed pump including phase portraits and power spectra

1 IntroductionGear systems are very common and practically impossible to replace in various applicationswhere mechanical power must be transferred The dynamic behaviour of systems involvinggears can be very complex and phenomena of interest include time varying mesh stiffness dueto multiple teeth contact a backlash between the teeth and excitation due to displacement errorsin the gears [1 2 3 4 5] In consequence under a dynamic load a typical gear system is anonlinear oscillator exhibiting a range of complex behaviour [6] As reported in [6] this includesfluctuating torque signals at a fixed frequency unrelated to shaft speed Such phenomena canonly be explained by non-linearities In the literature random noise is often included but in thepresent paper we seek a more rigorous approach

In practice it is important to minimise the effect of gear tooth errors to ensure machinestability and to minimise noise and gear wear But like all physical systems some error isinevitable and it is therefore important to understand the way in which a system will behavewith a real gearbox During operation wear will develop and at some stage the mating teeth maylose contact during some parts of the operational cycle this will give rise to further significantnon-linearities but in the paper we discuss the simpler situation in which the gears remain incontact

2 Modelling the Gear SystemWhen a torsional model of a complete shaft train is being prepared it is often necessary toinclude a model of a gearbox The effect of non-unity gear ratios on torsional vibration isexamined We focus on a relatively simple system with two shafts linked through a single gearalthough the derivation of the equations of motion is easily extended to the general case

9th International Conference on Damage Assessment of Structures (DAMAS 2011) IOP PublishingJournal of Physics Conference Series 305 (2011) 012020 doi1010881742-65963051012020

Published under licence by IOP Publishing Ltd 1

I1

I2 R2

I3 R3

I4

k1

k2

φ1 φ2

φ3 φ4

Figure 1 Four-inertia idealization of a drive system The ith gear inertia is Ii the workingradius of gear i is Ri the ith shaft stiffness is ki and the rotation of the ith gear is ϕi

21 Applying Constraints for Geared Systems [7]Consider the torsional behaviour in a system involving a gear drive depicted schematically inFigure 1 The system is modelled using four rotary inertias and two torsional springs Theangular displacements ϕ2 and ϕ3 of the inertias representing the gears are coupled throughthe gear mesh and are therefore not independent Let R2 and N2 be the working radius andthe number of teeth respectively for gear wheel 2 Similarly let R3 and N3 be the workingradius and the number of teeth respectively for gear wheel 3 If the gear pair is perfect thenthe tangential velocities of the two gears are identical For the present purposes it is necessary

only to recognise thatR2

N2=

R3

N3

Equating the tangential velocities yields R3ϕ3 = minusR2ϕ2 where ϕ2 and ϕ3 represent theinstantaneous angular velocities of gear wheels 2 and 3 respectively Defining γ to be the gear

ratio γ =N2

N3=

R2

R3 then ϕ3 = minusγϕ2 The negative sign indicates that a positive ϕ2 causes

negative ϕ3 With appropriate choice of initial conditions we can deduce that ϕ3 = minusγϕ2The four equations of motion of this simple system assuming viscous damping in the shaft

are

I1ϕ1 + c1

(ϕ1 minus ϕ2

)+ k1 (ϕ1 minus ϕ2) = T1

I2ϕ2 + c1

(ϕ2 minus ϕ1

)+ k1 (ϕ2 minus ϕ1) = T2 + R2F23

I3ϕ3 + c2

(ϕ3 minus ϕ4

)+ k2 (ϕ3 minus ϕ4) = T3 + R3F23

I4ϕ4 + c2

(ϕ4 minus ϕ3

)+ k2 (ϕ4 minus ϕ3) = T4

(1)

where F23 represents the instantaneous force acting between gears 2 and 3 along the commontangent and where T1 T2 T3 T4 are the instantaneous torques being exerted on inertias1 2 3 4 respectively from external sources It appears anomalous at first that F23 contributespositively to both gear wheels but inspection of the free body diagrams for disks 2 and 3 shownin Figure 2 reveals that this is correct

There are four rotation co-ordinates in Equation (1) but only three of these are independentbecause of the constraint between ϕ2 and ϕ3 We choose ϕ1 ϕ2 ϕ4 as independent coordinatesand ϕ3 is eliminated using ϕ3 = minusγϕ2 The forcing term F23 in Equation (1) is unknown andthis is removed by subtracting γ times the third equation from the second The resulting three


2

R2I3

k1( ) 2 1 k2( ) 3 4

I2R3

F23

F23

T2 T32 3

Disk 2 Disk 3

Figure 2 The free body diagrams for disks 2 and 3 of the four-to include a model of a gearboxinertia idealisation of a drive system

equations in ϕ1 ϕ2 ϕ4 are

I1ϕ1 + c1

(ϕ1 minus ϕ2

)+ k1 (ϕ1 minus ϕ2) = T1(

I2 + γ2I3

)ϕ2 + c1

(ϕ2 minus ϕ1

)+ c2

(γ2ϕ2 + γϕ4

)+k1 (ϕ2 minus ϕ1) + k2

(γ2ϕ2 + γϕ4

)= T2 minus γT3

I4ϕ4 + c2

(ϕ4 + γϕ2

)+ k2 (ϕ4 + γϕ2) = T4

(2)

These equations are equivalent to those obtained from a three-inertia model of a simplerotor but comprising only three inertias Suitable mass damping and stiffness matrices canbe extracted from Equations (2) and these may be used to find natural frequencies and modeshapes A similar approach may be applied to any gear arrangement consisting of multiple gearratios

Note that the angles ϕi used thus far are defined as the angular positions of the gears andinertias and hence are a combination of the steady state rotation and the torsional vibrationTo separate the two angles we define

ϕi = Ωit + qi (3)

where Ωi is the steady state rotation of inertia Ii and qi is the torsional vibration Thus Ω1 = Ω2Ω3 = Ω4 and Ω3 = minusγΩ2

Thus Equation (2) becomes in matrix form

Mq + Cq + Kq =

T1

T2 minus γT3

T4

(4)

where

q =

q1

q2

q4

M =

I1 0 00 I2 + γ2I3 00 0 I4

C =

c1 minusc1 0minusc1 c1 + γ2c2 γc2

0 γc2 c2

K =

k1 minusk1 0minusk1 k1 + γ2k2 γk2

0 γk2 k2


3

22 Displacement-driven Excitation of Torsional VibrationThus far each individual inertia initially has one independent rotation coordinate and at eachgear the constraint is of the form ϕ3 = minusγϕ2 With torsional models in particular it is veryoften necessary to model displacement-driven excitation The most important example of thisis gear geometry errors but other cases arise in misaligned coupled rotors Consider the twomeshing gears given in Figure 1 If no gear geometry errors are present then the ratio ϕ2ϕ3 isconstant and with suitable definition of angles such that ϕ2 = 0 when ϕ3 = 0 it follows thatN2ϕ2 + N3ϕ3 = 0 In the presence of gear geometry errors the ratio ϕ2ϕ3 will not be constantand the constraint equation is modified to

N2ϕ2 + N3ϕ3 = ε (ϕ2 ϕ3) (5)

where ε is the mesh error Most papers consider this error to be represented as a function oftime or frequency depending on the analysis being carried out However the mesh errors will inpractice be non-linear functions of the gear angles as discussed now

The factor N2ϕ22π represents the number of teeth on gear I2 passing through the mesh asthe gear rotates though an angle ϕ2 Suppose that the teeth on gear I2 are not equally spacedbut have errors Then this factor is modified to give (N2ϕ2minus e2(ϕ2))(2π) where e2 is a periodicfunction of ϕ2 The same analysis may be performed for gear I3 Hence Equation (5) becomesassuming no interaction between the mesh errors on the two gears

N2ϕ2 + N3ϕ3 = e2(ϕ2) + e3(ϕ3) (6)

or dividing by N3 we have

γϕ2 + ϕ3 =1

N3e2(ϕ2) + e3(ϕ3) (7)

If the functional form of the displacement errors is known then Equation (7) is a constraintequation linking the angles at the gear mesh ϕ2 and ϕ3 If we could find a closed form expressionfor ϕ3 in terms of ϕ2 or vice-versa then this constraint may be used directly in Equation (1) toobtain a three degree of freedom model in a similar approach to Section 21 However in generalthis is not possible and an alternative approach must be developed Here we consider twoapproaches namely an approximate constraint approach for small vibrations and an approachusing a finite tooth meshing stiffness

23 Defining Gear ErrorsTo proceed the gear errors e2(ϕ2) and e3(ϕ3) have to be specified One convenient option isto write the gear errors as Fourier series in ϕ2 and ϕ3 In particular the terms relating to thefundamental frequency and the harmonic term corresponding to the tooth meshing are likely tobe dominant Thus e2 is periodic in ϕ2 and hence

e2 =infinsum

n=1

α2n cos (nϕ2 + β2n) =infinsum

n=1

α2n cos (nΩ2t + nq2 + β2n) (8)

for constants α2n and β2n Similarly e3 is periodic in ϕ3 and hence

e3 =infinsum

n=1


n=1


for constants α3n and β3n


4

Of course defining the errors in this way means that it is impossible to solve for either ϕ2 or ϕ3

in closed form and therefore the constraints cannot be applied explicitly Often the non-linearityis ignored in displacement driven excitation of geared systems and in this case for example fore2

e2 =infinsum

n=1

α2n cos (nΩ2t + β2n) (10)

The gear errors are now solely a function of time and this merely leads to forcing terms in alinear equations of motion

24 Assuming Small VibrationsEquations (8) and (9) express the gear errors as non-linear functions of the gear angles Oneapproach is to assume that the vibration responses qi are small and ignore second and higherorder terms Thus the series for e2 may be expanded using trigonometric identities to give

e2 asympinfinsum

n=1

α2n cos (nΩ2t + β2n) minus nq2 sin (nΩ2t + β2n) (11)

which results in forcing and parametric terms at the frequency Ω2 and its harmonics Similarlyfor e3

e3 asympinfinsum

n=1


The constraints in terms of ϕ2 and ϕ3 can be written in terms of q2 and q3 using

N2ϕ2 + N3ϕ3 = (N2Ω2 + N3Ω3) t + N2q2 + N3q3 = N2q2 + N3q3 (13)

From Equation (6) substituting Equations (11) and (12) and collecting together terms in q2

and q3 we have[N2 + D2(t)] q2 + [N3 + D3(t)] q3 = E2(t) + E3(t) (14)

where

D2(t) =infinsum

n=1

α2nn sin (nΩ2t + β2n)

D3(t) =infinsum

n=1


E2(t) =infinsum

n=1

α2n cos (nΩ2t + β2n)

E3(t) =infinsum

n=1


Clearly these equations define the time dependent gear ratio as

γ(t) =N2 + D2(t)N3 + D3(t)

(15)

which results in a parametrically excited system There is also an excitation term

e(t) =E2(t) + E3(t)N3 + D3(t)

(16)


5

Thus the constraint equation isq3 = minusγ(t)q2 + e(t) (17)

Hence the equation of motion with no external excitation is

M(t)q+[C(t) + γ(t)C1(t)] q+[K(t) + γ(t)K1(t) + γ(t)K2(t)]q =

0I3γe + c2γe + k2γe

c2e + k2e

(18)

where the mass damping and stiffness matrices M C and K are defined in Equation (4)with the understanding that γ and hence the matrices are now time dependent The additionaldamping and stiffness matrix terms are given by

C1(t) =

0 0 00 2γ(t)I3 00 0 0

K1(t) =

0 0 00 γ(t)c2 00 c2 0

K2(t) =

0 0 00 γ(t)I3 00 0 0

(19)

Note that the model has a rigid body mode which can make a time simulation difficult sincethe absolute angles of the whole machine can drift together One option is to impose motion atsome point in the machines for example q1(t) equiv 0 which changes the boundary conditions ofthe machine and hence the natural frequencies A second option is to add a grounded spring oflow stiffness to one of the inertias which will also change the machine dynamics slightly Thealternative adopted here is to simulate the full system of equations and look at relative anglesfor example q2 minus q1

25 Assuming Finite Tooth Meshing StiffnessWe retain four degrees of freedom in Equation (1) and instead of imposing a rigid tooth meshstiffness constraint we impose the force as

F23 = minusktR2

N2(N2ϕ2 + N3ϕ3 minus e2(ϕ2) minus e3(ϕ3)) (20)

Note that kt is a translational tooth stiffness Note that if the teeth become rigid k2 rarr infin andsince the force between the teeth is finite we have that N2ϕ2 + N3ϕ3 = e2(ϕ2) + e3(ϕ3) whichrecovers Equation (6)

Then in Equation (1) the matrix equation of motion with a constant gear ratio γ is

Mq + Cq + Kq =γkt

N2

0γ10

(e2(ϕ2) + e3(ϕ3)) (21)

where

q =

q1

q2

q3

q4

M =

I1 0 0 00 I2 0 00 0 I3 00 0 0 I4

C =

c1 minusc1 0 0minusc1 c1 0 00 0 c2 c2

0 0 c2 c2

K =

k1 minusk1 0 0minusk1 k1 + γ2kt γkt 00 γkt kt + k2 minusk2

0 0 minusk2 k2

and kt = R23kt


6

3 Simulated ExampleThe parameters for the simulated system are

I1 = 7000 kg m2 I2 = 4000 kg m2 I3 = 13000 kg m2 I4 = 5000 kg m2

k1 = 16MNmrad k2 = 029MNmrad c1 = 36Nmsrad c2 = 58 Nmsrad

N2 = 111 N3 = 56 Ω2 = 45Hz

The natural frequencies of the linear system are 0 3974 and 1071Hz and hence the excitationfrequency is close to the first resonance frequency

The gear errors are assumed to be

α21 = 001 rad α2N2 = 0001 rad α2i = 00001 rad for 1 lt i lt N2 β2i = 0 for all i


Higher terms in the Fourier series are assumed to be zero The largest error is at the rotationspeed of the corresponding shaft for example due to errors in the location of the gear centre Thenext highest component is at the harmonic corresponding to the number of teeth and representstooth meshing errors The other harmonic terms are significantly smaller

From the Fourier series of the gear errors it is clear that the excitation contains manyfrequencies Furthermore the parametric excitation terms means that there will also be responsesat many combination frequencies Figure 3 shows the phase portrait of the twist in the firstshaft given by q2 minus q1 Figure 4 shows the FFT of the time response of the twist in the firstshaft It is clear that many frequencies are excited including many of the higher frequenciesThe interest in this paper is the effect on the low frequency dynamics Figure 5 shows a zoomedplot for the twist in the first shaft and clearly shows many combination frequencies There arelarge discrete responses at Ω2 = 45Hz and Ω3 = minus227Hz For comparison Figure 6 showsthe zoomed FFT plot when the parametric excitation is switched off Clearly the combinationfrequencies do not now appear Because the system has reached a steady state there is noexcitation and little response at the resonance frequencies of the system

4 ConclusionsThis paper has derived a model to simulate the torsional response of a machine due to gearerrors For errors on both gears in a gear mesh the constraint implicit in the displacementdriven excitation cannot be implemented exactly Approximate models were proposed assumingsmall vibrations that lead to parametric excitation The complicated nature of this parametricexcitation mean that a large range of frequencies are excited and this was demonstrated in asimulated example The broadband nature of the excitation and response mean that diagnosisof faults in geared systems is very difficult Fortunately the vibration amplitudes of the higherfrequency response background are relatively small compared to the main low frequency peakswhich could lead to the legitimate derivation of further approximate models The presence ofcombination response frequencies in the lower frequency range was also demonstrated Furtherinvestigation of the dynamics of this model are planned to understand properties of gear systemswith a view to deriving improved condition monitoring techniques

AcknowledgementsThe authors acknowledge the support of the Royal Society through the award of an InternationalJoint Project


7

minus5 0 5 10 15

x 10minus5

minus15

minus1

minus05

0

05

1

15

q2 minus q

1 (rad)

dq2

dt minus

dq 1

dt (

rad

s)

Figure 3 Phase portrait based on the twist in the first shaft (q2 minus q1)

0 1000 2000 3000 4000 5000 600010

minus12

10minus10

10minus8

10minus6

10minus4

Frequency (Hz)

Am

plitu

de (

rad)

Figure 4 Frequency response of the twist in the first shaft for the full model (q2 minus q1)


8

0 50 100 150 200 250 30010

minus10

10minus9

10minus8

10minus7

10minus6

10minus5

Frequency (Hz)

Am

plitu

de (

rad)

Figure 5 Low frequency response of the twist in the first shaft for the full model (q2 minus q1)

0 50 100 150 200 250 30010

minus12

10minus10

10minus8

10minus6

10minus4

Frequency (Hz)

Am

plitu

de (

rad)

Figure 6 Low frequency response of the twist in the first shaft for the full model (q2 minus q1)without parametric excitation

References[1] Ozguven HN and Houser DR 1988 Mathematical Models used in Gear Dynamics - A Review Journal of Sound

and Vibration 121 383-411[2] Theodossiades S and Natsiavas S 2000 Non-linear Dynamics of Gear-Pair Systems with Periodic Stiffness and

Backlash Journal of Sound and Vibration 229 287-310[3] Velex P and Maatar M 1996 A Mathematical Model for Analyzing the Influence of Shape Deviations and

Mounting Errors on Gear Dynamic Behaviour Journal of Sound and Vibration 191 629-660[4] Liu G and Parker RG 2008 Dynamic Modeling and Analysis of Tooth Profile Modification for Multimesh Gear

Vibration Journal of Mechanical Design 130 121402[5] Litak G and Friswell MI 2005 Dynamics of a Gear System with Faults in Meshing Stiffness Nonlinear Dynamics

41(4) 415-421


9

[6] Lees AW and Haines KA 1978 Torsional Vibrations of a Boiler Feed Pump Journal of Mechanical Design100 637-643

[7] Friswell MI Penny JET Garvey SD and Lees AW 2010 Dynamics of Rotating Machines (Cambridge UniversityPress)


10

Torsional Vibration of Machines with Gear Errors

AW Lees1 MI Friswell1 and G Litak2

1 College of Engineering Swansea University Singleton Park Swansea SA2 8PP UK2 Department of Applied Mechanics Technical University of Lublin PL-20-618 Lublin Poland

E-mail awleesswanseaacuk mifriswellswanseaacuk glitakpollubpl

Abstract Vibration and noise induced by errors and faults in gear meshes are key concernsfor the performance of many rotating machines and the prediction of developing faults Ofparticular concern are displacement errors in the gear mesh and for rigid gears these may bemodelled to give a linear set of differential equations with forced excitation Other faults suchas backlash or friction may also arise and give non-linear models with rich dynamics Thispaper considers the particular case of gear errors modelled as a Fourier series based on thetooth meshing frequency leading immediately to non-linear equations of motion even withoutthe presence of other non-linear phenomena By considering the perturbed response this systemmay be modelled as a parametrically excited system This paper motivates the analysis derivesthe equations of motion for the case of a single gear mesh and provides example responsesimulations of a boiler feed pump including phase portraits and power spectra

1 IntroductionGear systems are very common and practically impossible to replace in various applicationswhere mechanical power must be transferred The dynamic behaviour of systems involvinggears can be very complex and phenomena of interest include time varying mesh stiffness dueto multiple teeth contact a backlash between the teeth and excitation due to displacement errorsin the gears [1 2 3 4 5] In consequence under a dynamic load a typical gear system is anonlinear oscillator exhibiting a range of complex behaviour [6] As reported in [6] this includesfluctuating torque signals at a fixed frequency unrelated to shaft speed Such phenomena canonly be explained by non-linearities In the literature random noise is often included but in thepresent paper we seek a more rigorous approach

In practice it is important to minimise the effect of gear tooth errors to ensure machinestability and to minimise noise and gear wear But like all physical systems some error isinevitable and it is therefore important to understand the way in which a system will behavewith a real gearbox During operation wear will develop and at some stage the mating teeth maylose contact during some parts of the operational cycle this will give rise to further significantnon-linearities but in the paper we discuss the simpler situation in which the gears remain incontact

2 Modelling the Gear SystemWhen a torsional model of a complete shaft train is being prepared it is often necessary toinclude a model of a gearbox The effect of non-unity gear ratios on torsional vibration isexamined We focus on a relatively simple system with two shafts linked through a single gearalthough the derivation of the equations of motion is easily extended to the general case


Published under licence by IOP Publishing Ltd 1

I1

I2 R2

I3 R3

I4

k1

k2

φ1 φ2

φ3 φ4




N2=

R3

N3


ratio γ =N2

N3=

R2



are

I1ϕ1 + c1

(ϕ1 minus ϕ2

)+ k1 (ϕ1 minus ϕ2) = T1

I2ϕ2 + c1

(ϕ2 minus ϕ1

)+ k1 (ϕ2 minus ϕ1) = T2 + R2F23

I3ϕ3 + c2

(ϕ3 minus ϕ4

)+ k2 (ϕ3 minus ϕ4) = T3 + R3F23

I4ϕ4 + c2

(ϕ4 minus ϕ3

)+ k2 (ϕ4 minus ϕ3) = T4

(1)




2

R2I3

k1( ) 2 1 k2( ) 3 4

I2R3

F23

F23

T2 T32 3

Disk 2 Disk 3



I1ϕ1 + c1

(ϕ1 minus ϕ2

)+ k1 (ϕ1 minus ϕ2) = T1(

I2 + γ2I3

)ϕ2 + c1

(ϕ2 minus ϕ1

)+ c2

(γ2ϕ2 + γϕ4

)+k1 (ϕ2 minus ϕ1) + k2

(γ2ϕ2 + γϕ4

)= T2 minus γT3

I4ϕ4 + c2

(ϕ4 + γϕ2

)+ k2 (ϕ4 + γϕ2) = T4

(2)



ϕi = Ωit + qi (3)



Mq + Cq + Kq =

T1

T2 minus γT3

T4

(4)

where

q =

q1

q2

q4

M =

I1 0 00 I2 + γ2I3 00 0 I4

C =


0 γc2 c2

K =


0 γk2 k2


3


N2ϕ2 + N3ϕ3 = ε (ϕ2 ϕ3) (5)



N2ϕ2 + N3ϕ3 = e2(ϕ2) + e3(ϕ3) (6)


γϕ2 + ϕ3 =1

N3e2(ϕ2) + e3(ϕ3) (7)



e2 =infinsum

n=1


n=1



e3 =infinsum

n=1


n=1




4



e2 =infinsum

n=1




e2 asympinfinsum

n=1



e3 asympinfinsum

n=1






where

D2(t) =infinsum

n=1


D3(t) =infinsum

n=1


E2(t) =infinsum

n=1


E3(t) =infinsum

n=1



γ(t) =N2 + D2(t)N3 + D3(t)

(15)


e(t) =E2(t) + E3(t)N3 + D3(t)

(16)


5





c2e + k2e

(18)


C1(t) =

0 0 00 2γ(t)I3 00 0 0

K1(t) =

0 0 00 γ(t)c2 00 c2 0

K2(t) =

0 0 00 γ(t)I3 00 0 0

(19)



F23 = minusktR2




Mq + Cq + Kq =γkt

N2

0γ10

(e2(ϕ2) + e3(ϕ3)) (21)

where

q =

q1

q2

q3

q4

M =

I1 0 0 00 I2 0 00 0 I3 00 0 0 I4

C =


0 0 c2 c2

K =


0 0 minusk2 k2

and kt = R23kt


6




N2 = 111 N3 = 56 Ω2 = 45Hz










7

minus5 0 5 10 15

x 10minus5

minus15

minus1

minus05

0

05

1

15

q2 minus q

1 (rad)

dq2

dt minus

dq 1

dt (

rad

s)


0 1000 2000 3000 4000 5000 600010

minus12

10minus10

10minus8

10minus6

10minus4

Frequency (Hz)

Am

plitu

de (

rad)



8

0 50 100 150 200 250 30010

minus10

10minus9

10minus8

10minus7

10minus6

10minus5

Frequency (Hz)

Am

plitu

de (

rad)


0 50 100 150 200 250 30010

minus12

10minus10

10minus8

10minus6

10minus4

Frequency (Hz)

Am

plitu

de (

rad)







41(4) 415-421


9




10

I1

I2 R2

I3 R3

I4

k1

k2

φ1 φ2

φ3 φ4




N2=

R3

N3


ratio γ =N2

N3=

R2



are

I1ϕ1 + c1

(ϕ1 minus ϕ2

)+ k1 (ϕ1 minus ϕ2) = T1

I2ϕ2 + c1

(ϕ2 minus ϕ1

)+ k1 (ϕ2 minus ϕ1) = T2 + R2F23

I3ϕ3 + c2

(ϕ3 minus ϕ4

)+ k2 (ϕ3 minus ϕ4) = T3 + R3F23

I4ϕ4 + c2

(ϕ4 minus ϕ3

)+ k2 (ϕ4 minus ϕ3) = T4

(1)




2

R2I3

k1( ) 2 1 k2( ) 3 4

I2R3

F23

F23

T2 T32 3

Disk 2 Disk 3



I1ϕ1 + c1

(ϕ1 minus ϕ2

)+ k1 (ϕ1 minus ϕ2) = T1(

I2 + γ2I3

)ϕ2 + c1

(ϕ2 minus ϕ1

)+ c2

(γ2ϕ2 + γϕ4

)+k1 (ϕ2 minus ϕ1) + k2

(γ2ϕ2 + γϕ4

)= T2 minus γT3

I4ϕ4 + c2

(ϕ4 + γϕ2

)+ k2 (ϕ4 + γϕ2) = T4

(2)



ϕi = Ωit + qi (3)



Mq + Cq + Kq =

T1

T2 minus γT3

T4

(4)

where

q =

q1

q2

q4

M =

I1 0 00 I2 + γ2I3 00 0 I4

C =


0 γc2 c2

K =


0 γk2 k2


3


N2ϕ2 + N3ϕ3 = ε (ϕ2 ϕ3) (5)



N2ϕ2 + N3ϕ3 = e2(ϕ2) + e3(ϕ3) (6)


γϕ2 + ϕ3 =1

N3e2(ϕ2) + e3(ϕ3) (7)



e2 =infinsum

n=1


n=1



e3 =infinsum

n=1


n=1




4



e2 =infinsum

n=1




e2 asympinfinsum

n=1



e3 asympinfinsum

n=1






where

D2(t) =infinsum

n=1


D3(t) =infinsum

n=1


E2(t) =infinsum

n=1


E3(t) =infinsum

n=1



γ(t) =N2 + D2(t)N3 + D3(t)

(15)


e(t) =E2(t) + E3(t)N3 + D3(t)

(16)


5





c2e + k2e

(18)


C1(t) =

0 0 00 2γ(t)I3 00 0 0

K1(t) =

0 0 00 γ(t)c2 00 c2 0

K2(t) =

0 0 00 γ(t)I3 00 0 0

(19)



F23 = minusktR2




Mq + Cq + Kq =γkt

N2

0γ10

(e2(ϕ2) + e3(ϕ3)) (21)

where

q =

q1

q2

q3

q4

M =

I1 0 0 00 I2 0 00 0 I3 00 0 0 I4

C =


0 0 c2 c2

K =


0 0 minusk2 k2

and kt = R23kt


6




N2 = 111 N3 = 56 Ω2 = 45Hz










7

minus5 0 5 10 15

x 10minus5

minus15

minus1

minus05

0

05

1

15

q2 minus q

1 (rad)

dq2

dt minus

dq 1

dt (

rad

s)


0 1000 2000 3000 4000 5000 600010

minus12

10minus10

10minus8

10minus6

10minus4

Frequency (Hz)

Am

plitu

de (

rad)



8

0 50 100 150 200 250 30010

minus10

10minus9

10minus8

10minus7

10minus6

10minus5

Frequency (Hz)

Am

plitu

de (

rad)


0 50 100 150 200 250 30010

minus12

10minus10

10minus8

10minus6

10minus4

Frequency (Hz)

Am

plitu

de (

rad)







41(4) 415-421


9




10

R2I3

k1( ) 2 1 k2( ) 3 4

I2R3

F23

F23

T2 T32 3

Disk 2 Disk 3



I1ϕ1 + c1

(ϕ1 minus ϕ2

)+ k1 (ϕ1 minus ϕ2) = T1(

I2 + γ2I3

)ϕ2 + c1

(ϕ2 minus ϕ1

)+ c2

(γ2ϕ2 + γϕ4

)+k1 (ϕ2 minus ϕ1) + k2

(γ2ϕ2 + γϕ4

)= T2 minus γT3

I4ϕ4 + c2

(ϕ4 + γϕ2

)+ k2 (ϕ4 + γϕ2) = T4

(2)



ϕi = Ωit + qi (3)



Mq + Cq + Kq =

T1

T2 minus γT3

T4

(4)

where

q =

q1

q2

q4

M =

I1 0 00 I2 + γ2I3 00 0 I4

C =


0 γc2 c2

K =


0 γk2 k2


3


N2ϕ2 + N3ϕ3 = ε (ϕ2 ϕ3) (5)



N2ϕ2 + N3ϕ3 = e2(ϕ2) + e3(ϕ3) (6)


γϕ2 + ϕ3 =1

N3e2(ϕ2) + e3(ϕ3) (7)



e2 =infinsum

n=1


n=1



e3 =infinsum

n=1


n=1




4



e2 =infinsum

n=1




e2 asympinfinsum

n=1



e3 asympinfinsum

n=1






where

D2(t) =infinsum

n=1


D3(t) =infinsum

n=1


E2(t) =infinsum

n=1


E3(t) =infinsum

n=1



γ(t) =N2 + D2(t)N3 + D3(t)

(15)


e(t) =E2(t) + E3(t)N3 + D3(t)

(16)


5





c2e + k2e

(18)


C1(t) =

0 0 00 2γ(t)I3 00 0 0

K1(t) =

0 0 00 γ(t)c2 00 c2 0

K2(t) =

0 0 00 γ(t)I3 00 0 0

(19)



F23 = minusktR2




Mq + Cq + Kq =γkt

N2

0γ10

(e2(ϕ2) + e3(ϕ3)) (21)

where

q =

q1

q2

q3

q4

M =

I1 0 0 00 I2 0 00 0 I3 00 0 0 I4

C =


0 0 c2 c2

K =


0 0 minusk2 k2

and kt = R23kt


6




N2 = 111 N3 = 56 Ω2 = 45Hz










7

minus5 0 5 10 15

x 10minus5

minus15

minus1

minus05

0

05

1

15

q2 minus q

1 (rad)

dq2

dt minus

dq 1

dt (

rad

s)


0 1000 2000 3000 4000 5000 600010

minus12

10minus10

10minus8

10minus6

10minus4

Frequency (Hz)

Am

plitu

de (

rad)



8

0 50 100 150 200 250 30010

minus10

10minus9

10minus8

10minus7

10minus6

10minus5

Frequency (Hz)

Am

plitu

de (

rad)


0 50 100 150 200 250 30010

minus12

10minus10

10minus8

10minus6

10minus4

Frequency (Hz)

Am

plitu

de (

rad)







41(4) 415-421


9




10


N2ϕ2 + N3ϕ3 = ε (ϕ2 ϕ3) (5)



N2ϕ2 + N3ϕ3 = e2(ϕ2) + e3(ϕ3) (6)


γϕ2 + ϕ3 =1

N3e2(ϕ2) + e3(ϕ3) (7)



e2 =infinsum

n=1


n=1



e3 =infinsum

n=1


n=1




4



e2 =infinsum

n=1




e2 asympinfinsum

n=1



e3 asympinfinsum

n=1






where

D2(t) =infinsum

n=1


D3(t) =infinsum

n=1


E2(t) =infinsum

n=1


E3(t) =infinsum

n=1



γ(t) =N2 + D2(t)N3 + D3(t)

(15)


e(t) =E2(t) + E3(t)N3 + D3(t)

(16)


5





c2e + k2e

(18)


C1(t) =

0 0 00 2γ(t)I3 00 0 0

K1(t) =

0 0 00 γ(t)c2 00 c2 0

K2(t) =

0 0 00 γ(t)I3 00 0 0

(19)



F23 = minusktR2




Mq + Cq + Kq =γkt

N2

0γ10

(e2(ϕ2) + e3(ϕ3)) (21)

where

q =

q1

q2

q3

q4

M =

I1 0 0 00 I2 0 00 0 I3 00 0 0 I4

C =


0 0 c2 c2

K =


0 0 minusk2 k2

and kt = R23kt


6




N2 = 111 N3 = 56 Ω2 = 45Hz










7

minus5 0 5 10 15

x 10minus5

minus15

minus1

minus05

0

05

1

15

q2 minus q

1 (rad)

dq2

dt minus

dq 1

dt (

rad

s)


0 1000 2000 3000 4000 5000 600010

minus12

10minus10

10minus8

10minus6

10minus4

Frequency (Hz)

Am

plitu

de (

rad)



8

0 50 100 150 200 250 30010

minus10

10minus9

10minus8

10minus7

10minus6

10minus5

Frequency (Hz)

Am

plitu

de (

rad)


0 50 100 150 200 250 30010

minus12

10minus10

10minus8

10minus6

10minus4

Frequency (Hz)

Am

plitu

de (

rad)







41(4) 415-421


9




10



e2 =infinsum

n=1




e2 asympinfinsum

n=1



e3 asympinfinsum

n=1






where

D2(t) =infinsum

n=1


D3(t) =infinsum

n=1


E2(t) =infinsum

n=1


E3(t) =infinsum

n=1



γ(t) =N2 + D2(t)N3 + D3(t)

(15)


e(t) =E2(t) + E3(t)N3 + D3(t)

(16)


5





c2e + k2e

(18)


C1(t) =

0 0 00 2γ(t)I3 00 0 0

K1(t) =

0 0 00 γ(t)c2 00 c2 0

K2(t) =

0 0 00 γ(t)I3 00 0 0

(19)



F23 = minusktR2




Mq + Cq + Kq =γkt

N2

0γ10

(e2(ϕ2) + e3(ϕ3)) (21)

where

q =

q1

q2

q3

q4

M =

I1 0 0 00 I2 0 00 0 I3 00 0 0 I4

C =


0 0 c2 c2

K =


0 0 minusk2 k2

and kt = R23kt


6




N2 = 111 N3 = 56 Ω2 = 45Hz










7

minus5 0 5 10 15

x 10minus5

minus15

minus1

minus05

0

05

1

15

q2 minus q

1 (rad)

dq2

dt minus

dq 1

dt (

rad

s)


0 1000 2000 3000 4000 5000 600010

minus12

10minus10

10minus8

10minus6

10minus4

Frequency (Hz)

Am

plitu

de (

rad)



8

0 50 100 150 200 250 30010

minus10

10minus9

10minus8

10minus7

10minus6

10minus5

Frequency (Hz)

Am

plitu

de (

rad)


0 50 100 150 200 250 30010

minus12

10minus10

10minus8

10minus6

10minus4

Frequency (Hz)

Am

plitu

de (

rad)







41(4) 415-421


9




10





c2e + k2e

(18)


C1(t) =

0 0 00 2γ(t)I3 00 0 0

K1(t) =

0 0 00 γ(t)c2 00 c2 0

K2(t) =

0 0 00 γ(t)I3 00 0 0

(19)



F23 = minusktR2




Mq + Cq + Kq =γkt

N2

0γ10

(e2(ϕ2) + e3(ϕ3)) (21)

where

q =

q1

q2

q3

q4

M =

I1 0 0 00 I2 0 00 0 I3 00 0 0 I4

C =


0 0 c2 c2

K =


0 0 minusk2 k2

and kt = R23kt


6




N2 = 111 N3 = 56 Ω2 = 45Hz










7

minus5 0 5 10 15

x 10minus5

minus15

minus1

minus05

0

05

1

15

q2 minus q

1 (rad)

dq2

dt minus

dq 1

dt (

rad

s)


0 1000 2000 3000 4000 5000 600010

minus12

10minus10

10minus8

10minus6

10minus4

Frequency (Hz)

Am

plitu

de (

rad)



8

0 50 100 150 200 250 30010

minus10

10minus9

10minus8

10minus7

10minus6

10minus5

Frequency (Hz)

Am

plitu

de (

rad)


0 50 100 150 200 250 30010

minus12

10minus10

10minus8

10minus6

10minus4

Frequency (Hz)

Am

plitu

de (

rad)







41(4) 415-421


9




10




N2 = 111 N3 = 56 Ω2 = 45Hz










7

minus5 0 5 10 15

x 10minus5

minus15

minus1

minus05

0

05

1

15

q2 minus q

1 (rad)

dq2

dt minus

dq 1

dt (

rad

s)


0 1000 2000 3000 4000 5000 600010

minus12

10minus10

10minus8

10minus6

10minus4

Frequency (Hz)

Am

plitu

de (

rad)



8

0 50 100 150 200 250 30010

minus10

10minus9

10minus8

10minus7

10minus6

10minus5

Frequency (Hz)

Am

plitu

de (

rad)


0 50 100 150 200 250 30010

minus12

10minus10

10minus8

10minus6

10minus4

Frequency (Hz)

Am

plitu

de (

rad)







41(4) 415-421


9




10

minus5 0 5 10 15

x 10minus5

minus15

minus1

minus05

0

05

1

15

q2 minus q

1 (rad)

dq2

dt minus

dq 1

dt (

rad

s)


0 1000 2000 3000 4000 5000 600010

minus12

10minus10

10minus8

10minus6

10minus4

Frequency (Hz)

Am

plitu

de (

rad)



8

0 50 100 150 200 250 30010

minus10

10minus9

10minus8

10minus7

10minus6

10minus5

Frequency (Hz)

Am

plitu

de (

rad)


0 50 100 150 200 250 30010

minus12

10minus10

10minus8

10minus6

10minus4

Frequency (Hz)

Am

plitu

de (

rad)







41(4) 415-421


9




10

0 50 100 150 200 250 30010

minus10

10minus9

10minus8

10minus7

10minus6

10minus5

Frequency (Hz)

Am

plitu

de (

rad)


0 50 100 150 200 250 30010

minus12

10minus10

10minus8

10minus6

10minus4

Frequency (Hz)

Am

plitu

de (

rad)







41(4) 415-421


9




10




10

Torsional Vibration of Machines with Gear Errors - IOPscience

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Transcript of Torsional Vibration of Machines with Gear Errors - IOPscience