Effect of vibrations on friction in the context of

brake squeal

vorgelegt von

M.Sc. Nguyễn Thái Minh Tuấn

geboren in Hanoi, Vietnam

von der Fakultät V - Verkehrs- und Maschinensysteme

der Technischen Universität Berlin

zur Erlangung des akademischen Grades

Doktor der Ingenieurwissenschaften

-Dr.-Ing.-

genehmigte Dissertation

Promotionsausschuss:

Vorsitzender: Prof. Dr. rer. nat. Wolfgang H. Müller

Gutachter: Prof. Dr.-Ing. Utz von Wagner

Gutachter: Prof. Dr. rer. nat. Norbert Hoffmann

Tag der wissenschaftlichen Aussprache: 06. Februar 2019

Berlin 2019

i

Acknowledgements

This work has been done at the Chair of Mechatronics and Machine Dynamics (MMD), TU

Berlin. I acknowledge the financial support by Project 911 (from Vietnamese Government)

and a supplemental scholarship from DAAD.

First and foremost, I would like to thank Prof. Utz von Wagner for his firm yet kind

supervision and for all his patience and time spent in the last more than three years on

discussion and correction, from which I learnt a lot and found the way to complete this study.

I also would like to thank Prof. Norbert Hoffmann for evaluating my dissertation and giving

me valuable comments and Prof. Wolfgang Müller for chairing the doctoral committee. I am

indescribably grateful to Prof. Nguyễn Văn Khang for providing me not only the essential

knowledge in the field of Mechanics but also the guidance for my academic career since I was

a first-year student.

A special thank goes to Dr. Nils Gräbner, who worked most closely with me at MMD, and

spent much time discussing on my project. I would like to express my great appreciation to

Dr. Nils Gräbner, Dr. Sylwia Hornig, Dr. Holger Gödecker, Mr. Ronald Koll and Mr. Karl

Theet for their contribution in building and setting up the test rig and to Dr. Nguyễn Huy Thế

and Ms. Isabell Geier for their great help, without which it would be much more difficult for

me to finish my job. There are still other people that helped me at TU Berlin, both within and

out of MMD, and I am grateful to them even though I cannot name them individually.

I also received much help from Vietnam. I sincerely thank Dr. Nguyễn Thị Thu for helping

me in 3D drawing. I would like to express my gratitude to Dr. Nguyễn Phong Điền, Dr.

Nguyễn Quang Hoàng and the other colleagues at Department of Applied Mechanics, Hanoi

University of Science and Technology for their advices and supports during my study abroad.

I would like to thank my friends in Germany, Lương Huệ Trinh, Vũ Thị Thu Trà, Nguyễn

Phương Thảo, Nguyễn Thị Thu Hương, Claudia Krug, Tim Otto, Nguyễn Thái Chinh, Đoàn

Khánh Hoàng, just to name a few, for being nice to me. In just a few years, we had many

good times together and that helped me relieve some stress from work.

Finally, I would like to express my heartfelt thanks to my family – my parents, my younger

sister, my little son and especially my beloved wife Đỗ Thị Ái. It was hard for them when I

was abroad for a long time, but with their love, sympathy and effort, they are always my

reliable source of motivation and support.

ii

Contents

Acknowledgements .................................................................................................................... i Contents ..................................................................................................................................... ii 1 Introduction ........................................................................................................................... 1

1.1 Motivation 1 1.1.1 The compromise of modeling ................................................................................... 1

1.1.2 Brake squeal – a case of friction-induced vibration .................................................. 2 1.1.3 Dynamical identification of material parameters ...................................................... 3 1.1.4 Dynamical identification of contact parameters ........................................................ 3

1.2 Objectives of the research and structure of this dissertation 4

2 Some dry friction phenomena and models .......................................................................... 5 2.1 Dry solid contact 5 2.2 Static friction (stiction), break-away friction, and the Dahl’s effect 6 2.3 Steady-state sliding friction and the Stribeck curve 8

2.4 Frictional lag 8 2.5 Two friction models 9

2.5.1 A simple model of friction ........................................................................................ 9 2.5.2 A dynamic model of friction ..................................................................................... 9

2.6 Change of the coefficient of sliding friction due to vibration 13 2.7 Transverse friction-induced damping and tangential friction-induced damping 13

2.7.1 One-dimensional case ............................................................................................. 13 2.7.2 Two-dimensional case ............................................................................................. 18

3 Oscillating friction test rig .................................................................................................. 21 3.1 Objective of the test rig 21

3.2 General model of a test rig and requirements 22 3.3 A review on some tribotesters with different designs of normal force adaptors. 24

3.3.1 Some materials on friction measurement with elastic parts as normal force adaptor

.......................................................................................................................................... 24 3.3.2 Test rig with slide rail .............................................................................................. 24

3.4 The Oscillating Friction Test Rig (OFTR) with bending beams 26 3.4.1 The design of the OFTR .......................................................................................... 26

3.4.2 Basic measured and calculated values .................................................................... 31 3.4.3 Technical issues ....................................................................................................... 33

3.5 Some ideas for improving the OFTR 34

4 Experiments with the OFTR and results .......................................................................... 35 4.1 Dependence of the friction coefficient ratio on dynamical conditions 35

4.1.1 Measurement sequence ........................................................................................... 35 4.1.2 Dependence of the friction coefficient ratio on the vibration amplitude, the average

relative velocity and the excitation frequency .................................................................. 36 4.2 Qualitative observation of tangential friction-induced damping 44

5 Stability analysis of wobbling disk models with frictional point contacts ..................... 46 5.1 Constant coefficient of kinetic friction case 46

5.1.1 2-DOF model with in-plane stiffness ...................................................................... 46

5.1.2 8-DOF model ........................................................................................................... 59 5.2 Tangential friction-induced damping case 66

5.2.1 2-DOF model ........................................................................................................... 66

5.2.2 8-DOF model ........................................................................................................... 67

iii

6 Frequency-domain constants approach ............................................................................ 70

6.1 General idea 70

6.2 Determining the limit cycles by frequency-domain constants approach 70 6.2.1 Homogeneous linearized system ............................................................................. 70 6.2.2 Assumptions for frequency-domain constants ........................................................ 71 6.2.3 A procedure to find limit cycles with frequency-domain constants approach ........ 72

6.3 Discussion 75

7 Nonlinear analysis of wobbling disk models ..................................................................... 77 7.1 2-DOF model with in-plane stiffness 77 7.2 8-DOF model 82

7.2.1 8-DOF model without tangential friction-induced damping ................................... 82 7.2.2 8-DOF model with tangential friction-induced damping ........................................ 82

7.2.3 Subcritical behavior ................................................................................................. 86 8 Conclusions and outlook ..................................................................................................... 88

A Measurement results .......................................................................................................... 90 B Matrix form of Euler equations ...................................................................................... 103

B.1 Mathematical background 103 B.1.1 Kronecker product ................................................................................................ 103

B.1.2 Partial derivative of a matrix with respect to a vector .......................................... 103 B.1.3 Skew-symmetric matrix associated to cross product and its generalization ........ 104

B.2 Matrix form of Euler equations 104 Bibliography ......................................................................................................................... 107

1

1 Introduction

1.1 Motivation

Because of their important roles in the fundamentals of mechanics and in actual noise

problems, both friction and brake squeal are of great interest to mechanics scientists and

mechanical engineers. To develop the understandings of these two phenomena, a well-

implemented modeling process is needed. Therefore, the author’s opinion on modeling will be

presented before other motivations.

1.1.1 The compromise of modeling

To the author’s point of view, the most important functions of modeling real phenomena are:

i) Explanation: to help recognize the considered phenomena with words (a person may

witness the phenomena without knowing what they are, a verbal explanation makes it easier

for he or she to recognize them) and explain their nature in a logical manner;

ii) Knowledge conveyance: to pass the knowledge on the phenomena among humans or

artificial intelligence systems;

iii) Quantification and identification: to quantify the phenomena into quantities and identify

their values, usually with measurement devices;

iv) Reflection: to replicate the process of the phenomena, it can be done by building a model

physically or simulating with a mathematical model;

v) Prediction: to predict what will happen or the probability of certain phenomena under

certain conditions, it can be done qualitatively or quantitatively or both;

vi) Hypothesizing: to employ assumptions to compensate for the lack of knowledge so that the

process of studying a problem is not deadlocked.

These items can also be considered as the steps of modeling: a modeling process is a

permutation (repetition allowed) of this set of items. For example, to study a complicated

dynamical system, one can first break it down into simpler parts. Then, the behavior of each

part is predicted and explained and its parameters are identified with suitable available

theories and tools. In the next step, the parts are theoretically connected to form again the

original complicated system with all of their own models and parameters, so that a

mathematical model of the whole system can be created. Note that the word “modeling” is

usually used with a narrower meaning, referring to these early steps. Based on the complete

model, one can do simulations and compare the results with the observed phenomena. If the

model reflects well the behavior of the system, it can be used further in prediction and can be

taught to people who need to work with the system. In the process, simplifications and

assumptions may appear somewhere (for instance, in the steps of disassembling and

assembling) to reduce required time and effort and circumvent the lack of knowledge.

A question arises when putting smaller parts back to the origin system: Do they now undergo

the same conditions as those they do when they are separately studied? The answers in many

cases are no. Factor of safety is an example of compensation for the impracticalities in testing

(and for all the simplifications and assumptions). Another way to deal with the problem is to

2

bring the conditions in experiments closer to reality, which requires more time, effort, and

advanced technologies. This is the compromise between experimental identification and

hypothesizing.

There are more compromises within modeling. A logical and simple-enough explanation can

help a person learn the knowledge and do other functions of modeling faster and more easily

than a complex theory. For example, with some understanding on kinetic friction coefficient,

high-school pupils can predict that the car will stop if the motor is turned off and the brake

system is activated, and they can calculate the time required for the process; teaching them

anything about the nature of mechanical contacts without providing an appropriate

background carefully may lead to confusion*. Nevertheless, if the model is too simple, the

quality of reflection and prediction will become low in certain applications, e.g. rigid body

dynamics and Coulomb’s law of friction lead to Painlevé’s paradox [1], [2]. If the model is

too complicated, its parameters may not be identified relevantly due to the limitation of the

means of observation and measurement. Even if the parameters are sufficient, the difficulties

and errors in building and solving complicated mathematical models may hold the reflection

and prediction processes back.

Because of those conflicts, choosing a suitable model to serve a purpose or an application is a

critical issue. When the real system is well understood, the modeling process can be

standardized as a guideline (e.g. [3]). In contrast, when the knowledge is insufficient, for the

same reality, diverse models may be used, depending on which phenomena are examined,

which functions are focused, and the availabilities of observation and simulation tools. This is

perfectly true for the state-of-the-art studies on brake squeal.

1.1.2 Brake squeal – a case of friction-induced vibration

When a brake system is activated while the brake disk is still rotating, the contacts between

the brake pads and the brake disk are established; friction forces appear and slow down the

disk. While the whole system is desired to be calm, vibrations appear in different manners,

leading to noise, vibration, and harshness (NVH) in vehicles. This study is limited to the case

of brake squeal which has been described in numerous papers (e.g. [4], [5]). Since brake

squeal only happens with friction force, it is considered to be a case of friction-induced

vibration.

There are several ways to recognize or describe the existence of brake squeal. To a human

observer, brake squeal is the loud high-frequency (higher than 1000Hz) uncomfortable noise,

distinguished from other brake acoustic phenomena whose frequencies are lower than 1000Hz

such as brake moan [6], [7], creep groan [8], [9], and judder [10], [11]. By measuring, it is

found that when brake squeal starts to be heard, surges in the vibration amplitudes of the

brake pads and brake disk appear [12]-Fig. 56. The vibration amplitudes of the pads are in the

micrometer range, in which the vibrations in the circumferential and normal directions (with

respect to the disk) are dominant over that in the radial direction. More generally, the

circumferential direction can also be called longitudinal or tangential direction and the radial

* In a Vietnamese Physics textbook for high-school pupils, rolling resistant force is introduced within dynamics

of particles as a force appearing at the contact. This statement is not strictly wrong but is irrelevant in the

context: in dynamics of particles, rolling motion is not determined and the point of application of a force is also

unnecessary. If the pupils try to apply this taught knowledge into rigid body dynamics, which is introduced later

at universities, they will be confused.

The same problem can be seen at https://www.engineeringtoolbox.com/amp/rolling-friction-resistance-

d_1303.html (visited on 09/06/2018): if one tries to extend the provided figure with static friction forces between

the wheels and the surface, they will overlap the rolling resistant forces and cause confusion.

3

direction can also be called transverse direction. From the viewpoint of linear simulation,

where the brake system is modeled and then described by a set of linearized ordinary

differential equations (ODE), brake squeal occurs when the silent solution (the coordinate

system is usually chosen such that the silent solution is the trivial solution) representing the

silent motion is unstable. In nonlinear analysis, brake squeal is represented by limit cycles

which cannot be analyzed by linear models. Furthermore, nonlinearities may result in cases in

which a limit cycle may appear even when the linearized system is stable [13-15].

In some brake squeal tests (such as that in [12]) where the conditions are kept as steady as

possible, this phenomenon seems to occur uncertainly: sometimes it happens, sometimes not.

Nevertheless, its frequency is quite consistent for the same brake system, even when the disk

velocity and the brake pressure vary. The measurement result in [12]-Fig. 56 is depicted by a

curve representing subcritical Hopf bifurcation, a typical behavior of nonlinear system that

leads to the coexistence of solutions, which may be one reason why brake squeal is uncertain.

Some research does not consider the geometry constraints and the deformations of the

mechanical system and relates the instability to the negative slope of friction coefficient

versus relative velocity [16]. Meanwhile, some other research shows that adding, in an

appropriate way, springs and/or flexible bodies into a brake system model can also produce

instability and many of the analyzed brake squeal models, including a planar model

considering stiffness of joints and normal contact stiffness [17], planar models where the disk

is modeled as a visco-elastically supported rigid body [18] or as a bending beam [19], and

three-dimensional models with wobbling disk [20], [21], assume that the friction coefficient is

constant. The reasons for this assumption are probably the difficulties in both identification

and reflection.

1.1.3 Dynamical identification of material parameters

In order to simulate brake systems associated with brake squeal, the parameters of the systems

should be identified. Material parameters such as Young’s modulus and shear modulus can be

measured statically and then used in modeling. This approach comes from a simple

philosophy: when all the parameters are known, the vibration can be calculated. However,

since all those parameters are artificial concepts that may not reflect the true deep nature of

reality, it is questioned if they keep the same values under conditions different from which

they were identified under. In other words, do vibrations change the value of those

parameters?

The DCTR [22], [23] and DCSTR [12] were created at MMD to put the brake pad materials

under similar conditions they undergo in brake squeal and identify their parameters.

According to the results, vibration (as well as preload) seems to affect the materials

parameters: when the vibration amplitudes change, the parameters change. Lately, it is also

found that vibration amplitude affects the energy dissipation rates of brake shims made of

fibers [24-26]. These findings raise questions on the conventional models and demand new

ways of modeling in order to fulfill its functions.

1.1.4 Dynamical identification of contact parameters

Contact mechanics is a complicated field due to the complexity of the nature of mechanical

contacts and the difficulties in observing the phenomena within the contacts. Although such a

simple concept as the friction coefficient is useful in a wide range of engineering applications,

there are instances where it is irrelevant; therefore, the same questions as in the case of

material parameters arise.

4

Some authors already examine the friction behavior with the appearance of vibration and

notice the dependence of friction coefficient on vibration amplitude [27-39] (these literatures

will be discussed further in sections 2.6 and 3.1); a historical review on the studies of the

problem can be found in [34]. However, there has not been a universal explanation for this

behavior; therefore, it is still important to experimentally study the friction force for specific

material couples and conditions.

In [40], [41], a dynamical test setup is presented, focusing on studying the slope of friction

coefficient over relative velocity. The reported measurements are taken under different

frequencies, but the effect of vibration amplitude has not been examined. Though the

experimental results are interesting, the research has not drawn much attention (no citation

except for self-citations is found, according to Google Scholar and ResearchGate in July,

2018), which may be due partly to the fact that how to interpret the experimental results into

calculation models has not been discussed.

1.2 Objectives of the research and structure of this dissertation

The first objective of this research is to experimentally examine the frictional behavior in pad-

disk contacts under the conditions as close to the real conditions of brake squeal as possible.

These conditions include vibrations of the pads in all directions; however, due to the

limitation of time, only the case of tangential direction is investigated. The dynamical

parameters characterizing the frictional behaviors are then identified based on the

measurement results.

The second objective is to analyze the effects of those parameters on the behaviors of brake

system. The identified parameters are embedded into brake models to see if they can explain

the nonlinear brake squeal behavior better than the mere constant friction coefficient.

The main parts of this dissertation are structured as follows.

In chapter 2, some dry friction phenomena and models that may be useful in the analysis of

brake squeal are reviewed or derived. Important explanations and quantifications, for

instance, coefficient of sliding friction and tangential friction-induced damping, are presented

in this chapter.

In chapter 3, the design of the Oscillating Friction Test Rig (OFTR) developed within this

research to serve the first objective is presented. This chapter was partly published in [42].

Chapter 4 presents the measurement results and, by analyzing them, quantifies and identifies

the frictional behavior under tangential vibrations.

In chapter 5, some linearized models of brake squeal in which the disk is considered to have

wobbling motion are derived and their stability is analyzed.

Chapter 6 introduces frequency-domain constants approach, which, with appropriate

assumptions, is capable of using parameters measured in frequency domain to calculate the

size of limit cycles.

In chapter 7, the parameter identified in chapter 4, together with the results from DCTR are

embedded into the models in chapter 5 by using the newly-introduced frequency-domain

constants approach to achieve the second objective.

Chapter 8 is the conclusions and outlook.

5

2 Some dry friction phenomena and models

2.1 Dry solid contact

When two real dry solid surfaces are in contact, the actual contact area is dependent upon

their roughness and deformation (Fig. 2.1): there are points of contact where pairs of

asperities meet due to geometry constraints; the contact forces deform the materials from both

sides, making the points of contact become small surfaces (Fig. 2.2). These local contact

surfaces are scattered and vary in size and direction and so do the contact forces. When the

bodies are at rest, the total contact force acting on each body balances other forces. In the case

of plane-on-plane contact, the contact area is, in general, smaller than that of the macroscopic

contact surface. In contrary, when a sphere lies on a plane, also because of deformation, the

contact area is non-zero instead of a mere point. This explanation is widely accepted [43] but

is so complicated that some simplified models are used, such as the spring-like asperity

contact [44] (Fig. 2.3).

Figure 2.1: True contact between two solid surfaces, drawn based on [44]-Fig. 2.4

Figure 2.2: Contact between two asperities

Contact forces

and deformation Geometrical

contact

F

F

True contact Contact area

Asperities

Contact

points

Body 2

Body 1

6

Figure 2.3: Spring-like asperity model, drawn based on [44]-Fig. 2.6

2.2 Static friction (stiction), break-away friction, and the Dahl’s effect

A simple experiment to observe static friction is to put a rigid, box-shaped body on a

horizontal fixed plane and see how it reacts to external forces (Fig. 2.4a). At first, the total

contact force balances the gravitational force P (Fig. 2.4b). The contact force in this case is

called normal force N , as it is perpendicular to the contact surface. When a small force in

tangential direction F is applied to the body, no displacement is visible (Fig. 2.4c): the

contact force eliminates not only the gravitational force but also the tangential force (Fig.

2.4d). It means that there exists another component in the contact force, called static friction s

fF .

Figure 2.4: A simple experiment to prove the existence of static friction force

Then, increase the tangential force gradually: the body stands still, which means the static

friction also increases. Only when the tangential force reaches a critical value, ,maxs

fF , the

body slides. This force is called maximum static force or break-away friction and, according

to Coulomb, is determined by

,maxs

f sF N= (2.1)

where s is the coefficient of static friction.

Early research, such as that by Amontons and by Coulomb, claims that the coefficient of static

friction is independent of (macroscopic) contact area [44]. It is reasonable since, as mentioned

above, the real contact area is different from the macroscopic contact area and depends on the

Body 2

Body 1

b) Balanced force

system without

tangential forces

a) A box-shaped

body on a plane

c) The body

subjected to a small

tangential force

N

P

N

P

F F

s

fF

d) Balanced force

system with

tangential forces

7

normal force: the higher the normal force is, the more the asperities deform and the larger the

real contact area is. This relationship has already been included in calculation with the

appearance of the term N in (2.1).

While studying friction in machines with ball bearings, Dahl found a relationship between

friction force and displacement x [45]:

1 sgn( )

i

f f

C

dF Fx S

dx F

= −

(2.2)

where is the rest stiffness parameter, CF is the Coulomb friction level parameter, i is the

solid friction model exponent parameter and S is a stabilizing factor.

In static case, i.e. when 0x = , (2.2) becomes

fdF

dx= (2.3)

or

fF

x

= . (2.4)

Figure 2.5: Classical schematic diagram for the Dahl’s effect, drawn based on [44]-Fig. 2.7

Figure 2.6: Schematic diagram for the Dahl’s effect plus energy dissipation, considering only the

deformation in tangential direction

It can be seen that (2.4) represent a spring behavior: the displacement is proportional to

applied force [44] (Fig. 2.5). It means, though not visible, the body does change its rest

position when a small tangential force is applied to it and its displacement is called presliding

displacement [46]. This phenomenon is called the Dahl’s effect. In general, presliding

displacement is a function of static friction force, linear or nonlinear.

Body 2

Body 1

Body 2

Body 1

8

Moreover, friction is expected to dissipate energy, even in static friction case [44]. To meet

this assumption, there should be a damper parallel to the spring (Fig. 2.6).

2.3 Steady-state sliding friction and the Stribeck curve

Consider the case the movable body has a constant velocity, a simple formulation by

Coulomb to determine dry kinetic friction force k

fF in this case is:

k

f kF N= (2.5)

where k is the coefficient of kinetic friction.

Coulomb’s formulation does not clearly include the effect of sliding velocity. In reality, at

different steady-state sliding velocities, the kinetic friction force varies. The Stribeck curve

shows this dependence by the relation between k and relative velocity rv [44]

( )k k rv = . (2.6)

There are two important notes. The first is that the coefficient of kinetic friction is lower than

the coefficient of static friction, at least when rv is small enough.

k s . (2.7)

This can be explained by stress-strain curve of ductile material: breaking stress (rupture

stress) is lower than ultimate stress [45].

The second note is that, there is a region of relative velocity where the slope of friction

coefficient with respect to relative velocity is negative, which may result in instability.

Figure 2.7: Schematic diagram of a rigid body connected with a fixed plane through a stick-slip

element. The body either is stuck on the plane by stiction or slips and is subjected to sliding friction

To demonstrate the capability of both stick and slip, the contact can be depicted by a stick-slip

element (Fig. 2.7).

2.4 Frictional lag

When the movable body changes its velocity, i.e. accelerates or decelerates, it takes a short

period of time for the sliding friction force to reach a steady-state value as described by the

Stribeck curve. This phenomenon is called frictional lag or frictional memory [47].

9

2.5 Two friction models

2.5.1 A simple model of friction

The simplest way to model sliding friction is to use the coefficient of kinetic friction to

describe the ratio between the friction force and the normal force. This coefficient can be a

function of several parameters as follows

( )k k

f

N

F = =p (2.8)

where p is the vector of parameters, which can include state variables such as relative

velocity, qualitative values such as the material pair in contact, and vibration characteristics

such as amplitude and frequency.

To identify the parameters of this model, p should be controlled and monitored while N and

fF are measured. This is the way of experimental investigation in this research.

2.5.2 A dynamic model of friction

Since LuGre model, a dynamic model originally introduced in the field of machine control

[48-51], is used by some authors to explain the dependence of friction force on vibration, it is

worth mentioning some basic ideas of LuGre model. The original studies do not clearly show

how the model is built; therefore, a way to establish this model is presented here.

Figure 2.8: Schematic diagram of a dynamic friction model construction with the tangential contact

stiffness, the tangential contact damping, and a stick-slip element

1d

1k

2k

2d

1z

2z

x

ck

cd

z

x

a) Model with two deformations b) Model with combined deformations

10

Consider a one-dimension two-body problem in Fig. 2.8a: the displacement of the body is x ,

the average deformations of the asperities in the contact area on the surfaces of the fixed plane

and the body are respectively represented by two additional generalized coordinates 1z and

2z , the tangential contact stiffness and damping coefficient of the fixed plane are 1k and

1d ,

respectively, and the tangential contact stiffness and damping coefficient of the body are 2k

and 2d , respectively. The friction forces applied on the bodies are determined by

1 1 1 1 1fF k z d z= + , (2.9)

2 2 2 2 2fF k z d z= + . (2.10)

According to Newton’s third law of motion, one gets

1 2f fF F= (2.11)

With the denotations of combined tangential deformation

1 2z z z= + , (2.12)

combined tangential contact stiffness

1 2

1 2

c

k kk

k k=

+, (2.13)

and combined tangential contact damping coefficient

1 2

1 2

c

d dd

d d=

+, (2.14)

the friction force is now written as

f c cF k z d z= + . (2.15)

The model associated with (2.15) is shown in Fig. 2.8b: the contact is depicted by a stick-slip

element connected serially with a set of spring and damper in parallel. This model will be

used for further discussion.

In the case of steady-state velocity, in which x is constant and z is zero, it is expected that

( )fF s x= (2.16)

where (.)s is friction force as a function of velocity according to the Stribeck curve explained

in equation (2.6).

With the appearance of z , there should be an additional relation besides the equations of

motion of the mass. One choice for the additional relation is the differential of friction force

with respect to displacement (2.2) as in the Dahl’s model. Another way is to find an

appropriate function to determine z as follows

( , )z f x z= . (2.17)

To guarantee both (2.15) and (2.16), the zero point of (2.17) should be

( )

c

s xz

k= (2.18)

and the dimensions of both sides of (2.17) should match each other.

11

One choice for ( , )f x z is

( , ) 1( )

ckz f x z z x

s x

= = −

(2.19)

as in the LuGre model [48-51]. This model can be modified with an additional parameter as in

the following formulation

( , ) 1( )

ckz f x z z x

s x

= = −

(2.20)

where is a positive constant. LuGre model is a special case of this model when equals

one. In case the body is placed on a moving plane, the term x can be generally replaced by

the relative velocity between the two objects rv (with careful consideration of

positive/negative sign).

This model can simulate the following phenomena:

a) The Dahl’s effect

The static state is achieved if both x and z equal zero so (2.19) is satisfied without any

constraint on z .

The friction force is obtained from (2.15) as

f cF k z= (2.21)

which represents exactly a linear spring. It means that LuGre model captures the Dahl’s

effect.

b) Break-away friction

To observe the break-away phenomenon, let the body move at a low, positive steady state

velocity, i.e. 0, 0x const x= . Equation (2.20) becomes

1(0)

ckz z x

s

= −

. (2.22)

It can be drawn from (2.22) that as long as

0max

(0)

c

sz z

k = (2.23)

the value of z still increases together with the friction force. When z reaches 0maxz , it stops

increasing since z equals zero; therefore, the friction force also stops increasing and its

maximum value is governed by

,max

0max (0)s

f cF k z s= = . (2.24)

This value is also the break-away friction because if a horizontal external force with a higher

value is applied to the body, the displacement of the body will continually increase while the

deformation of the asperities will not, in other words, slip occurs.

c) Steady-state sliding friction – the Stribeck curve

Provided that the body is controlled to have a constant velocity x v= , (2.20) can be solved for

z analytically

12

( )( ) ( )( ) (0)

ckv t

s v

c c

s v s vz t z e

k k

− = − −

. (2.25)

When the time is high enough, the second term of (2.25) vanishes and z becomes a constant

( )

c

s vz

k= (2.26)

and the friction force agrees with the Stribeck curve

( )

( )f c c

c

s vF k z k s v

k= = = . (2.27)

Note that plays no role in (2.26) and (2.27); however, it can be seen from (2.25) that the

higher the value of is, the faster ( )z t reaches its steady value.

d) Frictional lag

Imagine an ideal experiment, where the body slides steadily with 1x v= , then, at time 1t , its

velocity changes suddenly to 2x v= . Solve (2.20) from time point 1t , one gets

2 1

2

( )( )2 2

1

( ) ( )( ) ( )

ckv t t

s v

c c

s v s vz t z t e

k k

− − = − −

. (2.28)

With the initial condition obtained from (2.26)

11

( )( )

c

s vz t

k= (2.29)

equation (2.28) can be rewritten as

2 1

2

( )( )2 2 1( ) ( ) ( )

( )

ckv t t

s v

c c c

s v s v s vz t e

k k k

− − = − −

(2.30)

or

2 1

2

( )( )

2 2 1( ) ( ) ( ) ( ( ) ( ))ck

v t ts v

f cF t k z t s v s v s v e− −

= = − − . (2.31)

It means that the friction force does not have a sudden change at 1t but it reaches its new

steady value after a frictional lag. The duration of this lag depends on parameter : the higher

is, the shorter the lag is. The LuGre model is without so that the lag only depends on k

and ( )s v , which makes the model less flexible.

It should be noted from the model construction that the stick-slip element has not been

handled directly, instead, an assumption has been made in the form of (2.19) or (2.20), which

does not have a clear physical meaning. Thus, though the model is good in machine control,

one should be careful when using it to investigate real friction behaviors.

It is much more difficult to identify the parameters of this model than those of the simple

model. Hence, the simple model is of more interest within this research.

13

2.6 Change of the coefficient of sliding friction due to vibration

Many authors have investigated the effect of vibration on both static friction and sliding

friction. Related to brake squeal, the case of sliding friction is of interest here. All tangential

vibration [27], [28], [32-38], transverse vibration [27], [32-35], and normal vibration [27],

[29-31] are observed to affect the coefficient of sliding friction. Usually, the higher the

vibration amplitude is, the lower the friction force is. [32] reports an exception in which

friction coefficient becomes higher as the vibration amplitude increases; however, unlike the

other studies, the excitation in this case is trapezoidal, not sinusoidal. In almost all the

experiments, the friction contacts are point or line contacts with low normal forces while the

real contacts in brake systems are large area contacts with high normal forces. Moreover, the

experimental results for different material pairs are different; therefore, it is necessary to

examine this phenomenon in specific cases.

In the case of in-plane vibration, when there are changes in the direction of relative velocity,

even Coulomb friction model produces the reduction in friction force, but dynamic models

such as Dahl’s model and LuGre model fit the experimental results better [33-39]. If there is

no change in the direction of the relative velocity, all those theoretical models show that there

is almost no change in the friction force. This characteristic will be tested later with the test

rig featured in this research.

2.7 Transverse friction-induced damping and tangential friction-induced damping

2.7.1 One-dimensional case

Consider a planar two-body model as in Fig. 2.9. In this model, both the deformations of body

1 and body 2 are neglected. Body 1 has a constant velocity dv while body 2 can only move

along body 1, so that the model is one-dimensional in terms of motion and has one degree-of-

freedom (DOF). The normal force N is constant while the applied tangential force tF has the

form

ˆ cos( )t t tF F F t= + (2.32)

where is the constant circular excitation frequency, tF and ˆtF also have constant values.

Figure 2.9: 1-DOF model with Coulomb’s friction law for a mass-damper explanation.

For intuitiveness, the friction force is drawn at the contact surface while the applied tangential force at

the middle height of the body. This creates a torque which is compensated by an offset between the

lines of action of the applied normal force and the reaction one. The underlying mechanism is out of

the scope of rigid body dynamics. If body 2 is considered as a point mass, it is no matter where the

points of action are but drawing them arbitrarily may lead to a feeling of falseness.

( )k rv

N

tF

dv

m

fFx

N

Body 2

Body 1

14

Under this excitation, body 2 is assumed to have the same vibration circular frequency

cos( )xx A t = − + . (2.33)

The relative velocity rv between body 2 and body 1 is determined by

r dv v x= + . (2.34)

When x is small enough and rv does not change its sign, the coefficient of kinetic friction k

is calculated by

( ) ( ) ( ) ( )k r k d k d k dv v x v v x = + + (2.35)

where

( )

( )

d

kk d

v v

vv

v

=

=

(2.36)

Hence, the friction force between body 2 and body 1 fF is determined by

( ) ( )f k d k dF v N v Nx = + . (2.37)

The equation of motion of body 2 is

t fmx F F= − (2.38)

where m is the mass of body 2.

Substituting (2.32) and (2.37) into (2.38) yields

ˆ cos( ) ( ) ( )t t k d k dmx F F t v N v Nx = + − − . (2.39)

Omitting the constant terms in (2.39)

( ) 0t k dF v N− = , (2.40)

we have

ˆ( ) cos( )k d tmx v Nx F t+ = . (2.41)

Expression (2.41) implies that the model in Fig. 2.9 is equivalent to a mass-damper model

ˆ cos( )tgt tmx d x F t+ = (2.42)

where

( ) ( )tgt tgt d k dd d v v N= = . (2.43)

The damping coefficient tgtd dissipates (or, if it has a negative value, feeds in) energy of

vibrations that make the relative velocity fluctuates along the tangential direction. Hence, this

phenomenon is called tangential friction-induced damping (or, for space-saving, “tangential

damping”) hereafter.

Substituting (2.33) into (2.42), one obtains

15

2

ˆ cos

ˆ sin

t

x

ttgt

x

Fm

A

Fd

A

=

=

(2.44)

or, if the vibration acceleration amplitude aA is used,

ˆ cos

ˆ sin

t

a

ttgt

a

Fm

A

Fd

A

=

=

. (2.45)

System (2.45) allows one to identify m and tgtd when the excitation force ˆtF , vibration

acceleration aA and the phase difference between the force and the acceleration (indicating

how long the force is delayed in comparison to the acceleration) can be measured.

Tangential friction-induced damping will not have much to do with modeling process if tgtd

really satisfies (2.43). However, the experimental results from [40], [41] and from this current

study show that it does not. The following example with LuGre model also illustrates how

(2.43) could be unsatisfied from a theoretical point of view.

Figure 2.10: 1-DOF model with LuGre friction law for a mass-damper explanation

Consider the case the friction law between the mass and the base is the modified LuGre model

(Fig. 2.10). The equations of motion are

ck

cd

z

x

dv

mˆ( ) cos( )k d tv N F t +

Body 2

Body 1

16

ˆ( ) cos( )

1 ( )( ) ( )

k d t f

d

k d k d

f c c

mx v N F t F

kz x v

v N v Nx

F k z d z

= + −

= − +

+

= +

. (2.46)

In this case, the system is nonlinear so that numerical integration and fast Fourier transform

(FFT) are used to find the acceleration amplitude and the phase difference between the

tangential force and the acceleration. Then, the damping coefficient is evaluated by (2.45).

The parameters for simulation are taken as follows (some parameters will be changed in

specific analyses).

61kg; 46 10 N/m; 100Ns/m;

1; 0.11m/s;

( ) 0.3; ( ) 0.1s/m;

ˆ1000N; 10N; 100 2 rad/s.

c c

d

k d k d

m k d

v

v v

N F

= = =

= =

= =

= = =

(2.47)

As can be seen in Fig. 2.11, ( )k dv N and the identified damping coefficient are correlated

but the identified damping coefficient is higher than ( )k dv N . While friction model

parameters such as cd (Fig. 2.12) and (Fig. 2.13) have little effect, the identified damping

coefficient increases much when the excitation angular frequency increases (Fig. 2.14). It

should be noted that while the modified LuGre model is designed to replicate several friction

phenomena, whether it is suitable in this case of periodic excitation or not is still an

unanswered question.

Figure 2.11: Dependence of the identified damping coefficient on ( )k dv N with LuGre model

17

Figure 2.12: Dependence of the identified damping coefficient on the contact damping coefficient with

LuGre model

Figure 2.13: Dependence of the identified damping coefficient on with modified LuGre model

Figure 2.14: Dependence of the identified damping coefficient on the excitation angular frequency

with LuGre model

18

Since (2.43) may be irrelevant, for the sake of simulation and identification, tgtd should be

considered as a standalone parameter rather than a parameter derived from the curve of

friction coefficient versus relative velocity. Introducing the tangential friction-induced

damping factor

( )

( )tgt d

d

d vv

N = , (2.48)

the friction force in the case of tangential excitation can be determined as follows

( ) ( )( )f k d d r dF v N v v v N + − . (2.49)

Equation (2.49) is more flexible than (2.37) because tangential friction-induced damping

factor ( )dv is used instead of ( )k dv . The change in relative velocity ( )r dv v− also replaces

x for generalization.

2.7.2 Two-dimensional case

In the one-dimensional case, the relative velocity does not change its direction, so that the

friction force vector can be determined easily with the coefficient of kinetic friction ( )k p

because the direction of the friction force is known. In contrast, when the relative velocity

changes its direction, it is more complicated to determine the direction of the friction force so

that an additional assumption may be needed.

Assumption of the alignment of sliding friction force and relative velocity: The sliding friction

force always opposes the relative velocity. This assumption works well with Coulomb’s law

of friction. In other friction laws, the friction force and the relative velocity can be unaligned.

a) Transverse vibration with a constant coefficient of kinetic friction

Consider the case of transverse vibration (Fig. 2.15) where body 1 has a constant velocity dv

while body 2 fluctuates around a fixed point so that the two bodies always touch each other

and their absolute velocities are perpendicular to each other. The mass of body 2 is m , its

velocity amplitude is v̂ , its frequency is , and the coefficient of kinetic friction between the

two bodies k is constant.

Figure 2.15: A 2-body system with transverse vibration

dv

Body 1

Body 2

Center position

of body 2

x

y

19

The two components of the relative velocity between body 2 and body 1 are

ˆcos( )rxv x v t= = , (2.50)

ry dv v= − (2.51)

where x is the coordinate of body 2.

The two components of the friction force applied on body 2 are

2 2

rxfx k k

r d

v xF N N

v v x = − = −

+, (2.52)

2 2

ry dfy k k

r d

v vF N N

v v x = − =

+. (2.53)

Considering the transverse motion of body 2, we have

2 2

fx k

d

xmx F N

v x= = −

+. (2.54)

If ˆdv v , (2.54) can be linearized as

k

d

Nmx x

v

= − . (2.55)

Again, (2.55) implies that the model in Fig. 2.15 can be re-modeled as a system of mass-

damper with the transverse damping coefficient

ktrv

d

Nd

v

= . (2.56)

This kind of damping phenomenon is introduced as friction induced damping in [52], [53].

Similar to its tangential counterpart, transverse damping coefficient trvd may not satisfy (2.56)

in reality. However, this will not be further investigated within this study.

b) General case

Figure 2.16: A 2-body system with vibrating relative velocity

dv

Body 1

Body 2

Center position

of body 2

20

Consider a two-body problem in Fig. 2.16, body 1 has a constant velocity dv while body 2

vibrates around a fixed point while moving on the surface of body 1. The maximum velocity

of body 2 is assumed to be much smaller than the velocity of body 1.

With the assumption of the alignment of sliding friction force and relative velocity and (2.49),

the friction force vector acting on body 2 is determined as follows

( ( ) ( ))rf k d r dN v v v − + −

vF t (2.57)

where rv is the magnitude of the relative velocity,

rvt is the unit vector of the relative

velocity vector of body 2 with respect to body 1, dv is the velocity of body 1 (Fig. 2.17).

Figure 2.17: Determination of the friction force acting on body 2 that opposes the relative velocity

a) Body 2 is at its leftmost position, its absolute velocity is zero so its relative velocity with respect to

body 1 rv equals dv− ; therefore, fF has the same direction as dv

b)-d) Body 2 is moving to the right, rv changes its magnitude so that fF changes its magnitude

accordingly, rv also changes its direction and so does fF so that fF always opposes rv

e)-h) The same as a)-d) but body 2 is now moving from its rightmost position to the left

dv dv

dv

0pv =

fFpv

dv−rv

fF

0pv =

fF

pv

dv− rv

fF

dv

pv

dv−rv

fF

dv

pv

dv−

rv

fF

a)

b)

c)

e)

f)

g)

dv

pv

dv− rv

fF

dv

dv

pv

dv−rv

fF

d)

h)

21

3 Oscillating friction test rig

3.1 Objective of the test rig

Consider a simple sketch representing a pad and a disk in a braking process (Fig. 3.1). The

normal direction of the disk plane, the tangential direction (or the circumferential direction of

the disk) and the transverse direction (or the radial direction of the disk) are called n , t and

p respectively. A normal force N is applied to the pad to keep it in contact with the disk

which is rotating and has a linear velocity dv at the contact point. The contact forces between

the pad and the disk are a normal force N and a dynamic friction force fF . Since the pad is

supported by a caliper system, this system applies to the pad a force dF , which prevents the

pad from moving far from its designed position due to fF .

Figure 3.1: Movement of the disk and forces acting on the pad in a brake system

x

dF

N

fF

n

t

p

t

dv

m

a) Top view

b) Front view

N

22

According to section 1.1.2, when break squeal occurs, the pad vibrates, and as discussed in

section 2.6, this vibration may affect the contact behavior which, in turn, may also have

influence on brake squeal. Thus, the effect of vibration on friction behavior should be

examined. The specific investigation in this research is limited to the case of tangential

vibration.

To approach the problem experimentally, a test rig which puts the pad-disk materials pair

under the desired conditions is to be built. While most of the existing tribometers for dry

friction usually measure friction of point or line contacts [28-34], [36], [38] due to the

advantage of self-aligning, this research would like to examine the friction behavior in the

condition as close to the condition in brake squeal as possible so a flat-to-flat contact is of

interest. During a test, constant values of N and dv should be kept and N is expected to

equals N . The disk should be controlled to have a constant angular velocity d . The

vibration of the pad can be characterized by its amplitude xA and frequency pf or circular

frequency 2p pf = and these parameters should also be controlled and kept unchanged

within a test.

The equation of motion of the pad is

d fmx F F= − (3.1)

where m is the vibrating mass and dF is now the driving force.

If the driving force and the acceleration of the pad are measured in time domain, then the

friction force can be calculated and the friction law can be found. However, as discussed in

the next sections, it is too difficult for the current technology, so the friction law has to be

estimated from measurable values.

In a real brake system, typical values for the brake pressure are e.g. 3 15bar− with respect to

squeal suspicious states. The normal force will therefore be 240 1200N− in order to get

similar pressure conditions in the contact. Squeal frequency lies in the kHz range and

amplitude xA is in the micrometer range.

3.2 General model of a test rig and requirements

According to the goal of the test, a theoretical simplified diagram of a test rig (Fig. 3.2) was

drawn. It consists of two main subsystems which can be built relatively independently: the

pad-subsystem and the disk-subsystem.

a) The disk-subsystem

The disk-subsystem includes:

- a disk,

- a motor and a transmission producing the constant velocity of the disk.

The task of this subsystem is simple: maintain the velocity of the disk at a desired value and

prevent the disk from in-plane vibrating as well as from wobbling. To fulfill this task, the

motor should be controlled and the shaft and disk should have high stiffness and high moment

of inertia.

b) The pad-subsystem

The pad-subsystem includes:

23

- a pad specimen,

- a pad specimen carrier,

- a clamping mechanism to apply normal force,

- a normal force adaptor allowing the carrier to move in tangential direction,

- an actuator making the carrier vibrate.

Figure 3.2: Theoretical block diagram of an oscillating friction test rig

The vibrating part should be light enough to lower the requirement for the power of the

actuator; hence, the use of gravity force to apply the normal force is not as suitable as the use

of the clamping mechanism and the normal force adaptor. The clamping mechanism and the

normal force adaptor have to be strong enough to withstand a high normal force. The normal

force adaptor should prevent the pad specimen from rotating to maintain flat-to-flat contact.

The rotation of the disk may result in vibration in normal direction of the pad specimen.

Therefore, the clamping mechanism or the normal force adaptor should be soft in this

direction to harmonize the normal force. Furthermore, the connection between the actuator

and the carrier is also a point to be considered.

As far as the author’s knowledge goes, there are two main methods to design the normal force

adaptor:

- Using elastic parts such as bending beams [54-57].

- Using a linear rail guide, presumably as a viscous damper [40], [41].

Clamping mechanism

Carrier

Pad specimen Actuator

Disk

Shaft and Transmission

Motor

Pad-

subsystem

Disk-

subsystem

N,r fv F

, dM

N

sinx px A t=

Normal force adaptor

N

24

In the next section, a review on different kinds of normal force adaptor is presented.

3.3 A review on some tribotesters with different designs of normal force adaptors.

3.3.1 Some materials on friction measurement with elastic parts as normal force adaptor

The following materials are on some tribotesters which measure sliding friction between a

pair of material specimens, the stationary one is kept relatively standstill and the moving one

is velocity-controlled. All of these setups employ bending beams to apply, adjust and measure

the reaction force between the surfaces. Although they do not deal with a tangentially

vibrating pad specimen, it is worth reviewing how the stationary part is pushed to the moving

part and can still vibrate in tangential direction, and how normal force and friction force are

measured.

a) Four-rod-device

A special design for the normal force adaptor is introduced in [54]. This adaptor includes two

highly stiff flanges and four identical rods. A rod has four pairs of notches, each pair form a

hinge that allows the rod to easily bend in a predetermined plane containing the axial axis of

the rod, and the bending planes of two consecutive pairs are perpendicular. The four rods are

placed in a parallel manner: their ends are mounted on the two flanges. One of the flanges is

fixed while the other carries the stationary material specimen. The parallel design keeps the

stationary part from rotating and its motion is pure translation; therefore, the flat-to-flat

contact between two specimens may be guaranteed during the test. Strain gauges are attached

to the notches on the rods and forces acting on the specimen will be calculated from the

measured values. This device needs to be preciously manufactured and the maximum allowed

forces (30N and 20N for normal force and friction force, respectively) are much lower than

the forces needed in this study (240-1200N for normal force as mentioned in section 3.1).

b) Two-bending-beam setup

The test setup in [55] Fig. 4.7 seems to be simpler than the four-rod-device. The stationary

material specimen is held by a series of two beams bending in two different perpendicular

planes so that it can move in both vertical and horizon directions. The forces are evaluated

with the values measured by the gauges attached on the beams. In this setup, the stationary

material specimen can rotate so that the contact between surfaces may not be flat-to-flat;

therefore, it is not suitable for a test in which a large contact area is desired.

c) Cantilever beam and laser

In this setup, the normal force adaptor contains only one cantilever beam which can undergo a

complex deformation including bending, lateral bending and torsion, which is detected and

measured by laser light and position-sensitive photodiode (PSPD) [56], [57]. The normal

force and friction force are then calculated from the measured deformation. This system was

designed for tests of friction in micro or nano scale. It is also not suitable for a flat-to-flat

contact test.

3.3.2 Test rig with slide rail

In the following, the test rig developed by Aoki et al. [40], [41] designed also for the

investigation of the pad/disk brake-problem is considered.

Flexible elements were not used; instead, they proposed a test rig including a slide rail and a

slide block which is simplified into a block diagram in Fig. 3.3. The slide rail mounting on the

25

moving base and the slide block attached to the carrier-pad block allow the pad to vibrate

freely in tangential direction. The interaction between the slide rail and slide block contains

not only normal force but also friction force. As can be seen in Fig. 3.3, this friction force is

included in the value measured by the load cell and, therefore, couples with the friction force

between the material pair being tested in the experiments.

Figure 3.3: Block diagram of the pad subsystem with slide rail used in [40], [41], drawn based on the

information provided by the literature

To overcome this problem, a pretest for the rail-block friction force was carried out with the

absence of normal force and resulted in a linear viscous damper. However, it is questionable

because the normal force in real test was so high that it may affect the damping coefficient,

not to mention the vibration of the pad. Additionally, the damping of the slide rail is also high,

so a change in it may cause a large error in the measured results.

This setup has an advantage that a high normal force can be applied, but the system is stiff in

normal direction so there is a risk of dramatic change in normal force during the tests.

Carrier

Pad

Actuator

Pad

subsystem

N

Load cell

Slide rail

Slide block

Load

cell

Accelero-

meter

,f viscousF

Moving base

26

3.4 The Oscillating Friction Test Rig (OFTR) with bending beams

The designing of OFTR is an important step of this study [42]. Elastic details are chosen to

comprise the normal force adaptor.

3.4.1 The design of the OFTR

Fig. 3.4 is a photo of the OFTR. The block diagrams of the disk-subsystem and pad-

subsystem are shown in Fig. 3.5 and Fig. 3.6, respectively.

Figure 3.4: The Oscillating Friction Test Rig

27

Figure 3.5: Block diagram of the disk-subsystem of the OFTR, which is a part of a regular brake test rig of MMD

Gearbox

fF

, dM

Brak

e disk

Receiver

dv

Driven shaft

Brass rin

g

anten

na

Strain

-gau

ge fo

r

torq

ue

measu

remen

t

Tran

smitter

Batteries

Driving shaft

Black-white stripes for

velocity measurement

Photocell

Frequency to voltage

converter

Power supply

Monitor Motor

controller

Motor N

28

Figure 3.6: Block diagram of the pad-subsystem of the OFTR

fF

N

sinx px A t= −

NThrust

screw Linear

spring

Carrier

Upper 3D

force cell

Pad

specim

en

Piezo actuator

Linear carrier

Linear rail guide

Threaded

plate

2 × 3D-

accelerometers

Upper beam

Lower beam

Lower beam

support

Between

beams support

Actuator adaptor

Back

plate Lower 3D

force cell

Vibrating

parts

Charg

e conv

erters

Charg

e conv

erters

Monitor Amplifier Signal generator

Base

Safety plate

N

29

Figure 3.7: The driving shaft

Figure 3.8: The driven shaft

Driven shaft Strain-gauge Cardan joint Transmitter Receiver Brass ring antenna

Batteries

Driven shaft Photocell Stripes

30

Figure 3.9: The pad-subsystem

The disk-subsystem, composed mostly of commercial devices, is part of a regular brake test

rig at MMD. The motor speed can be adjusted by the motor controller. A gearbox with ratio

of 16:1 lowers the speed transmitted from the driving shaft to the driven shaft. The driven

shaft is connected with the gearbox and the brake disk by two Cardan joints, allowing it to be

self-aligned.

To measure the velocity of the disk, a ring of black and white stripes is mounted on the

driving shaft and the number of the stripes passing by per revolution is counted by a photocell

containing a reflective optical sensor (Fig. 3.7). There are 8 white stripes and 8 black stripes.

A full-bridge strain-gauge placed at 45 degree is attached on the driven shaft (Fig. 3.8). The

voltage from the strain-gauge is converted into frequency by a transmitter. Both the

transmitter and its batteries are also mounted on the shaft. This frequency is read by a receiver

from a brass ring antenna and is then converted back to voltage by a converter.

The pad-subsystem has been designed and built for the special purpose of this study (Fig.

3.9). The base has an L shape, where a threaded plate, a linear carriage and a piezo actuator

are mounted on. A thrust screw and the threaded plate form the clamping mechanism since

the thrust screw can apply normal force on the normal force adaptor which is composed of a

linear spring and a system of two parallel beams with their support elements. The spring is to

soften the system in normal direction. The two parallel beams allow the vibrating parts to

move in tangential direction without large rotation. The linear rail guide and linear carriage

under the beams system are commercial devices by SKF. They support and allow the normal

force adaptor to move in normal direction in loading and unloading process as well as in

measuring period.

The vibrating part consists of a carrier, two 3D force cells, a back plate and the pad specimen.

The force cells are placed between the back plate (behind the pad specimen) and the carrier so

the values they measure do not include interaction forces between the carrier and the normal

force adaptor or the actuator.

Pad specimen Force cells Carrier Parallel beams Between beams support Accelerometers

Linear spring

Threaded

plate

Thrust

screw

Back plate

Linear rail guide

Linear carriage

Base

Safety plate

Actuator adaptor

Piezo actuator

31

The carrier is excited in tangential direction by the piezo actuator via a safety plate and an

actuator adaptor. The safety plate is fixed to the base but not to the actuator adaptor and the

actuator, so that the actuator is protected from any shear force when the actuator adaptor

moves. A signal generator is used to create a harmonic signal at a desired frequency which is

amplified by an amplifier before exciting the actuator.

The motion of the pad specimen is measured by two 3D accelerometers glued on the back

plate. The two force cells are connected to charge converters to convert electric charges into

voltages.

All the measured voltages are transmitted to the monitor and read by analyzer software.

3.4.2 Basic measured and calculated values

a) Angular velocity and linear velocity of the disk, and average relative velocity

The angular velocity d of the disk is determined as follows

0

2d

n

n t = (3.2)

where n is the number of black (or white) stripes counted in the time period t and 0n is the

total number of black (or white) stripes on the ring.

The linear velocity dv at the contact point between the brake and pad specimen is calculated

as

0

2d d

rnv r

n t= = (3.3)

where r is the distance between the contact point and the rotational axis of the disk. Here, the

contact point is assumed to be the middle point of the contact area.

In the OFTR, the parameters are 0 8n = and 0.14mr = .

The average relative velocity rv between the pad specimen and the disk approximately equals

the linear velocity of the disk at the contact point

r dv v . (3.4)

b) Normal force, static part

To measure the static part of the normal force, before the thrust screw is loaded, the two

outputs for normal direction of the 3D force cells are connected to two charge amplifiers

whose displayed values are set to zeros. Then, the thrust screw is loaded until the total value

displayed on the charge amplifiers reaches the desired normal force.

c) Average friction force

As the friction force of the axle bearing is close to zero, the equation of motion of the disk is

d fJ M rF = − (3.5)

where J is the moment of inertia of the disk and M is the torque on the driven shaft.

32

If the disk has a steady state angular velocity, the friction force is determined as

f

MF

r= . (3.6)

The torque is proportional to the torsion deformation of the driven shaft, which is measured

by the full-bridge strain-gauge. It should be noted that even when the disk does not rotate, the

voltage signal obtained from the system of the strain-gauge and the transmitter is not zero but

has a relaxation value.

( )MM k S R= − (3.7)

where S is the voltage signal measured when the friction force appears, R is the relaxation

value, and Mk is the conversion coefficient calibrated in a previous project. The measured

signals fluctuate during any measurement; therefore, the average values are considered. From

(3.6) and (3.7), we have

( )M

f

k S RF

r

−= (3.8)

where the over bars denote average values.

d) Normal force and friction force, dynamic parts

The dynamic parts, i.e. high-frequency components, of the interacting forces between the

carrier and the pad specimen are measured by the 3D force cells. The tangential interaction

force is simply calculated by adding the measured values from the upper force cell upper

tF and

the lower force cell lower

tF

ˆ ( ) ( ) ( )upper lower

t t tF t F t F t= + . (3.9)

It should be noted that, because of vibration, these forces include the inertial forces of the pad

specimen and the back plate. Different models for the pad specimen may lead to different

calculated contact forces.

e) Vibration amplitude

The acceleration of the pad specimen is considered to be the average value of the measured

value from the left accelerometer lefta and the right accelerometer righta .

( ) ( )

( )2

left righta t a ta t

+= . (3.10)

The average acceleration amplitude of the pad specimen can be estimated by the following

formulation

max min

1

1

2

periodn

i ia

iperiod

a aA

n =

−= (3.11)

where the measurement time is divided into periodn periods, maxia and minia are the maximum

and minimum measured accelerations in the i -th period.

An alternative way is to do fast-Fourier transform (FFT) of the measurement signal and take

the acceleration amplitude value fft

aA at the vibration frequency of the pad. This vibration

frequency is generally the excitation frequency. Note that the frequency of the signal

33

generator may be slightly shifted from the set value, so that fft

aA is determined as the highest

acceleration amplitude value in the vicinity of the considered frequency.

The respective velocity amplitudes and displacement amplitudes can be calculated from the

acceleration amplitudes as follows

2

,2 (2 )

a av x

p p

A AA A

f f = = , (3.12)

2

,2 (2 )

fft fftfft ffta a

v x

p p

A AA A

f f = = . (3.13)

In this research, the average values of amplitudes are used. The FFT values may be more

useful when there is more than one considered frequency.

3.4.3 Technical issues

a) Measuring the forces

As abovementioned, the low-frequency components and the high-frequency components of

the forces are measured in different ways. It is because none of the force measurement

devices currently available at MMD are able to measure both the low-frequency component

and the high-frequency component a dynamic force at once. It leads to difficulties in

designing the mechanical parts and in interpreting the measured values in time-domain. For

example, the static part of the normal force can only be measured before carrying out the

experiments. If this part changes in the course of the tests, the change cannot be observed.

b) The uncertainty in torque measurement

Even when the disk is at equilibrium, the voltage received from torque measurement is not

zero. This “relaxation value” changes overtime.

c) The limitation of the gain produced by the amplifier

To observe the influence of the vibration amplitude on the parameters in question, high

amplitudes are needed. However, the amplifier may be overloaded, restricting the amplitude

range.

d) Resonance and unwanted vibration shapes

The OFTR has a lot of flexible elements, which may lead to many resonance frequencies and

unwanted vibration shapes. The ideal vibration shape is where the pad specimen only has a

translational motion up and down. A finite element model of the test rig was studied. This

model is not good enough because of many unknown parameters, geometry simplifications

and a complicated contact between the linear rail and the carriage. It, however, can be seen

that the OFTR has lots of unwanted vibration shapes. To somehow avoid this problem,

different dimensions of the parallel beams are tested to choose the most appropriate one.

In certain cases, resonance can be helpful if it increases the vibration amplitude of the brake

pad while maintaining a good vibration shape. The frequency range used in this study is

actually chosen so that the maximum acceleration amplitude of the pad can go higher than

200m/s2.

34

e) Measurement noise

The measured results have many frequencies other than the excitation frequency so that the

dynamic values will be considered in frequency domain, not time domain.

f) The unreliability of motion measurement

Unlike the force, the two-point measurement is not enough to conclude about the motion of

the pad specimen. It may lead to false results of both the acceleration amplitude and,

especially, the acceleration phase.

g) Other issues

Due to the limited space, the chosen linear spring only works well under the normal force of

not more than 1000N.

Mechanical errors such as errors in the brake disk surface lead to errors in measured friction

force. Therefore, the time period for a particular test should be long enough and some parts of

the measured time period are excluded, if necessary, to prevent false results.

Despite the symmetry of the design, the normal force distributes unevenly between the upper

force cell and the lower force cell.

The temperature is not effectively controlled in the experiments.

3.5 Some ideas for improving the OFTR

An improvement can be made in the normal force adaptor. One possibility is to replace the

parallel beams with a better flexible system like a bigger version of the four-rod-device.

Another idea is to use one or more rail guides instead of the beams but still keep the force

cells between the pad specimen and the carrier. Another way is to have a moving base that

also carry the actuator like in [40], [41] but with force cells between the slide block and the

carrier and with a linear spring in normal direction. All of these ideas are to make the pad

specimen vibrate with a good shape.

For the motion measurement, there should be more measurement points or a better

measurement method should be employed.

35

4 Experiments with the OFTR and results

4.1 Dependence of the friction coefficient ratio on dynamical conditions

4.1.1 Measurement sequence

A few experimental studies use multiple-stage measurement with an alternate pattern of no-

vibration-stages and vibration-stages [27], [28], [37] while most of the others do not mention

any measurement sequence. The measurement sequence is considered here because it can

improve the measurement results.

There are some factors that may affect the friction force measurement:

- The relaxation value of the shaft deformation measurement is not zero and sometimes

changes over time.

- The dependence of friction force on temperature, wear and other unknown factors: A change

in the average friction force may occur when the same experiment is carried out shortly after

the first one.

They may lead to difficulty in drawing conclusions from the measurement results. For

example, the friction coefficient in a non-excitation case is measured first, then the actuator is

turned on and a lower friction coefficient is identified, but one cannot conclude that the

excitation reduces the friction coefficient because other factors should not be ignored.

Although increasing the number of repeated tests can make the evidence stronger, it can be

extremely costly and time-consuming.

Therefore, it is better to implement comparative measurements than to determine the friction

coefficient directly in each independent test. This research employs the sandwich

measurement sequence described in Table 4.1. This sequence includes five steps: (1) the disk

is stand still and there is no excitation, (2) the motor controlling the disk is turned on, (3) the

actuator is turned on so that the pad vibrates, (4) the actuator is turned off, (5) the motor is

turned off. Throughout this sequence, the normal force is expected to be unchanged; from step

2 to step 4, when measurements are performed, the velocity of the disk is expected to be

unchanged.

Table 4.1: Sandwich measurement sequence

Step Step name Measuring

time

Disk

rotating Excitation

Strain-gauge

signal

1 Relaxation Low (1 s) No No 1R

2 Steady state High (10 s) Yes No 1S

3 Excitation High (10 s) Yes Yes E

4 Steady state High (10 s) Yes No 2S

5 Relaxation Low (1 s) No No 2R

In table 4.1, a value of strain-gauge signal is the average voltage obtained from the strain-

gauge via the transmitter during a step. The following assumptions are adopted:

36

- The relaxation value of the shaft deformation measurement has a constant change rate over

time.

- The average friction force has a constant change rate over time due to the changes of

temperature and other factors.

- The average normal force is unchanged in all cases.

- The average relative velocity is unchanged in all cases where the disk rotates.

The friction coefficient ratio which is the ratio between the friction coefficient in excitation

case and that in steady state case is now of more interest than the friction coefficients

themselves. If the normal forces in the two cases are the same, the friction coefficient ratio

can be determined as

(with excitation)

(without excitation)

f

f

F

F = . (4.1)

Noting (3.8), one can write

1 2(with excitation)2

Mf

k R RF E

r

+= −

(4.2)

1 2 1 2(without excitation)2 2

Mf

k S S R RF

r

+ += −

. (4.3)

Hence, can be formulated as follows

1 2

1 2 1 2

2 ( )

( )

E R R

S S R R

− +=

+ − +. (4.4)

4.1.2 Dependence of the friction coefficient ratio on the vibration amplitude, the average

relative velocity and the excitation frequency

Figure 4.1: Pad specimens used in experiments. Pad specimen No. 1 (b) is cut from a commercial

brake pad (a) while pad specimen No. 2 (c) is made from another brake material

a) Commercial brake pad

b) Pad specimen No. 1

c) Pad specimen No. 2

37

Two pad specimens will be used in the experiments. Pad specimen No. 1 (Fig. 4.1b) was cut

from a commercial brake pad (Fig. 4.1a) and pad specimen No. 2 was made by gluing a piece

of another brake pad material on an aluminum plate (Fig. 4.1c).

Thereafter, “a test” will be used to refer to one five-step sandwich measurement and “a test

series” consists of tests with the same excitation frequency and average relative velocity but

different vibration amplitudes. Between two tests, there is a cool down period of three to five

minutes, so that the initial temperature of the disk is lower than 35 Celsius degree. There are

three “tries” for each test series (usually not consecutively) to get reliable results.

All the tests are performed with the normal force of approximately 1000N. This normal force

is applied with the thrust screw at the beginning of each test session (consisting of one or a

few test series) when the motor and actuator are turned off. It is measured by the 3D force

cells in static mode. The measured normal force may slowly and slightly change over time

which may be due to creep behavior. This error is small (less than 10N) compared to the

expected value and an exact value of normal force is also not needed here so this is not a

severe problem. The 3D force cells are then switched to dynamic mode.

In each test series for pad specimen No. 1, the generator for the actuator is fixed at one of four

different excitation frequencies: 1500Hz, 1600Hz, 1700Hz, and 1800Hz (outside this effective

frequency range, either the maximum acceleration amplitude of the pad specimen is too low

or its motion is too far from an up-and-down translational motion); the disk velocity is

controlled to get one of five different average relative velocities: 0.13m/s, 0.28m/s, 0.44m/s,

0.60m/s, and 1.26m/s; while the gain of the amplifier is varied among six levels to obtain

different vibration amplitudes.

The effective frequencies range for pad specimen No. 2 is different from, and narrower than,

the previous case. Two frequencies are considered in this case: 1100Hz and 1200Hz. Because

of time limitation, only two velocities are set for 1100Hz: 0.13m/s and 0.28m/s; and three for

1200Hz: 0.13m/s, 0.28m/s, and 0.44m/s.

Figure 4.2: Dependence of measured friction coefficient ratio on average acceleration amplitude for

pad specimen No. 1 at excitation frequency 1500Hz and average relative velocity 0.13m/s. The higher

the acceleration amplitude is, the lower the friction ratio (and, therefore, the friction force) is

The full measurement results for pad specimen No. 1 are represented in Table A.1-A.20. Fig.

4.2 is the visual display of Table A.1, a typical example of measurement results of three test

38

series: the measurements are implemented at 1500Hz and 0.13m/s and the values of the

friction coefficient ratio are calculated according to (4.4). When the acceleration amplitude is

low, the friction coefficient ratio may fluctuate around one. At higher amplitude, it can be

seen that the friction force decreases when the amplitude increases. Note that an acceleration

amplitude of 500m/s2 at 1500Hz means a displacement amplitude of about 5.6μm, which

belongs to actual brake squeal conditions. In all cases, the higher the average relative velocity

is, the higher the friction coefficient ratio is. It means that the friction reduction gets stronger

when the disk velocity is lowered. The same tendencies are also observed in the case of pad

specimen No. 2 (Table A.21-A.25).

The dependence of friction coefficient ratio on average vibration acceleration amplitude aA in

three tries of each test series is fitted into a line

,( , , ) 1 ( , )ka p r a p r aA f v f v A − (4.5)

where the friction reduction ratio over acceleration amplitude 2

, [s /m]k a is determined by

simple linear regression, the subscript a indicates that the acceleration amplitude is in use.

The quality of approximation is roughly assessed by coefficient of determination 2R [58]

2

2 1

2

1

( )

1

( )

m

m

n

i i

i

n

i

i

x y

R

x x

=

=

−

= −

−

(4.6)

where mn is the number of measurements, { }( 1, )i mx i n= are the measured values of ,

{ }( 1, )i my i n= are their fitted values, and x is the average value of { }ix . If 2R has a value on

the scale from 0 to 1 and near 1, the fitted line is probably good.

Table 4.2: Identified friction reduction ratio for pad specimen No. 1

Velocity

Frequency

0.13m/s 0.28m/s 0.44m/s 0.60m/s 1.26m/s

,k a

[ 2s /m ]

2R ,k a

[ 2s /m ]

2R ,k a

[ 2s /m ]

2R ,k a

[ 2s /m ]

2R ,k a

[ 2s /m ]

2R

1500Hz 20.9E-5 0.95 17.2E-5 0.93 13.6E-5 0.93 11.5E-5 0.94 7.91E-5 0.92

1600Hz 15.3E-5 0.96 12.9E-5 0.94 11.4E-5 0.97 8.77E-5 0.98 6.02E-5 0.92

1700Hz 13.0E-5 0.90 10.5E-5 0.95 7.77E-5 0.92 6.81E-5 0.93 4.92E-5 0.84

1800Hz 7.61E-5 0.89 6.00E-5 0.88 5.71E-5 0.85 4.41E-5 0.81 3.52E-5 0.70

Table 4.3: Identified friction reduction ratio for pad specimen No. 2

Velocity

Frequency

0.13m/s 0.28m/s 0.44m/s

,k a

[ 2s /m ]

2R ,k a

[ 2s /m ]

2R ,k a

[ 2s /m ]

2R

1100Hz 3.4E-5 0.29 1.5E-5 0.05 - -

1200Hz 28.0E-5 0.69 21.2E-5 0.69 15.3E-5 0.82

The identified results for pad specimen No. 1 shown in Table 4.2 are with quite good

coefficients of determination, almost all are more than 0.8. The fitted line for each frequency

is visually represented in each of Fig. 4.3-4.6. It can be confirmed again that, for the same

39

frequency and acceleration amplitude, the friction coefficient ratio is lower at lower average

relative velocity.

In contrast, the identified results for pad specimen No. 2 shown in Table 4.3 have worse

fitting quality, especially in the case of 1100Hz. It may be due to the lower maximum

acceleration amplitudes of only about 200-300m/s2, compared with 400-700m/s2 for pad

specimen No. 1. This problem emphasizes the urge of increasing the maximum vibration

amplitude in the experiments.

Figure 4.3: Fitted lines of friction coefficient ratio for pad specimen No. 1 at excitation frequency

1500Hz for different average relative velocities. The lower the average velocity is, the lower the

friction ratio (and, therefore, the friction force) is

Figure 4.4: Fitted lines of friction coefficient ratio for pad specimen No. 1 at excitation frequency

1600Hz for different average relative velocities

40

Figure 4.5: Fitted lines of friction coefficient ratio for pad specimen No. 1 at excitation frequency

1700Hz for different average relative velocities

Figure 4.6: Fitted lines of friction coefficient ratio for pad specimen No. 1 at excitation frequency

1800Hz for different average relative velocities

Besides average acceleration amplitude, average velocity amplitude vA and average

displacement amplitude xA can also be used as references for friction coefficient ratio

,( , , ) 1 ( , )kv p r v p r vA f v f v A − , (4.7)

,( , , ) 1 ( , )kx p r x p r xA f v f v A − (4.8)

where

2

av

p

AA

f= , (4.9)

41

2(2 )

ax

p

AA

f= . (4.10)

Thus, the friction reduction ratios over velocity amplitude , [s/m]k v and over displacement

amplitude , [1/m]k x are determined by

, ,( , ) 2 ( , )k kv p r p a p rf v f f v = , (4.11)

2

, ,( , ) (2 ) ( , )k kx p r p a p rf v f f v = . (4.12)

Note that, besides the average values, the FFT values of amplitudes can also be used.

Figure 4.7: Dependence of ,k a for pad specimen No. 1 at different frequencies on average relative

velocity.

Figure 4.8: Dependence of ,k v for pad specimen No. 1 at different frequencies on average relative

velocity

42

Figure 4.9: Dependence of ,k x for pad specimen No. 1 at different frequencies on average relative

velocity

Figure 4.10: Dependence of ,k a for pad specimen No. 2 at different frequencies on average relative

velocity.

43

Figure 4.11: Dependence of ,k v for pad specimen No. 2 at different frequencies on average relative

velocity

Figure 4.12: Dependence of ,k x for pad specimen No. 2 at different frequencies on average relative

velocity

To examine the effect of the excitation frequency, the dependence of ,k a at different

frequencies on average relative velocity for pad specimen No. 1 is plotted in Fig. 4.7, ,k v in

Fig. 4.8, and ,k x in Fig. 4.9. It is interesting to see that regardless whether the acceleration,

velocity or displacement amplitude is used, the friction force reduces more when the

frequency is lowered.

44

The results for pad specimen No. 2 shown in Fig. 4.10-4.12 reveal a reverse tendency: for the

same amplitude, the friction force reduction is much higher in the case of 1200Hz than in the

case of 1100Hz.

The maximum observed friction reduction with OFTR is more than 10%, which is

remarkable, noting that in all tests, the relative velocity does not change its sign, unlike most

of the experimental studies so far [27-29], [33-38]. It is probably that the large contact area

and the high normal force play a role in boosting friction reduction, as they are also the

differences between this study and the mentioned others. However, a theoretical explanation

is not available within this research.

4.2 Qualitative observation of tangential friction-induced damping

Due to the unreliability of the motion measurement (see section 3.4.3f), only qualitative

comments on tangential friction-induced damping coefficient are made here.

The pad specimen mass and the tangential friction-induced damping coefficient are identified

from the FFT of the measurement results according to (2.44) and presented in Table A.1-A.20

for pad specimen No. 1 and Table A.21-A.25 for pad specimen No. 2. Unlike the friction

coefficient ratio, the tangential friction-induced damping coefficient seems to be uncertain: in

some cases, the repeatability seems to be good while in the others cases, it does not, e.g. see

Fig. 4.9 and Fig. 4.10. This uncertainty is also observed in [40], [41]. Again, a reliable

explanation is still unavailable.

Figure 4.9: Dependence of identified tangential friction-induced damping coefficient on average

acceleration amplitude for pad specimen No. 1 at excitation frequency 1500Hz and average relative

velocity 0.13m/s

45

Figure 4.10: Dependence of identified tangential friction-induced damping coefficient on average

acceleration amplitude for pad specimen No. 1 at excitation frequency 1700Hz and average relative

velocity 0.60m/s

Despite the uncertainty, it seems that the tangential friction-induced damping coefficient

usually rises when the vibration amplitude increases. It may be somehow related to the rising

tendency of k

rv

under a growing vibration amplitude. This tendency can be proved,

knowing that the friction force reduces more at lower velocity, as follows.

We have

1 1 1 2 2 1 2 2( , , ) ( , , ) ( , , ) ( , , )k p a k p a k p a k p av f A v f A v f A v f A − − (4.13)

with 2 1 2 10; 0a av v A A .

Therefore,

2 2 1 2 2 1 1 1( , , ) ( , , ) ( , , ) ( , , )k a k a k a k av f A v f A v f A v f A − −

2 2 1 2 2 1 1 1

2 1 2 1

( , , ) ( , , ) ( , , ) ( , , )k a k a k a k av f A v f A v f A v f A

v v v v

− −

− −

2 1( , , ) ( , , )k a k av f A v f A

v v

. (4.14)

Taking v to zero, we get what we need to prove.

46

5 Stability analysis of wobbling disk models with frictional point

contacts

The models in this chapter are developed from the 2-DOF models in [20], [59] and the 8-DOF

model in [14]. The core idea of the models is that the brake disk exhibits wobbling motion,

which, under the effects of frictional contacts, can be destabilized.

5.1 Constant coefficient of kinetic friction case

5.1.1 2-DOF model with in-plane stiffness

5.1.1.1 Equations of motion

a) The model

Consider the system in Fig. 5.1. The brake disk is considered as a rigid circular disk fixed at

its center of symmetry O . It is supported by rotational springs tk and rotational dampers td

so that it lies horizontally at equilibrium, and it is driven by a torque AM . The two pads are

modeled as massless cone-headed elements. Each of them is supported in normal direction by

a linear spring (stiffness 1k on the upper half, 2k lower half) and a damper (damping

coefficient 1d on the upper half, 2d lower half) and in tangential direction by another linear

spring (stiffness 1ipk on the upper half,

2ipk lower half). The visco-elastic supports in normal

direction are preloaded by forces 0N . The coefficients of kinetic friction at the upper and

lower contact points are 1k and

2k , respectively. Compared with the model presented here,

the model in [20] does not consider the tangential spring and the model in [59] does not have

any dampers and both of them are absolutely symmetrical about the disk.

Figure 5.1: A 2-DOF brake system model with frictional point contacts

O

0 1x x0y

0z

2x3x

1 2y y

3y

1z

2 3z z

1k

1q

2q

3q

1 1,k d

2 2,k d

1ipk

2ipk

AM

1 1t tk q d q+

2 2t tk q d q+

2k

Preload 0N

Preload 0N

47

b) Coordinate systems and generalized coordinates

The fixed frame system 0 0 0Ox y z is chosen such that the 0z -axis is perpendicular to the disk at

equilibrium position, and the 0y -axis is chosen such that both the contact points lie on the

plane 0 0Oy z when the disk is at equilibrium. The body-fixed frame 3 3 3Ox y z coincides with

0 0 0Ox y z when the disk is at equilibrium. Position of the disk can be determined by three

angles 1q , 2q and 3q which are angles of rotation about 0x -, 1y - and 2z -axis respectively,

where 1 1 1Ox y z and 2 2 2Ox y z are intermediate frames.

The generalized coordinates are now defined as

1

2

3

q

q

q

=

q . (5.1)

Roughly commenting, small vibrations of 1q and 2q make the relative velocity vibrates

transversely and tangentially, respectively.

c) Kinematics of the disk

The direction cosine matrices of the frames are determined as

0

1 1 1 1

1 1

1 0 0

( , ) 0 cos sin

0 sin cos

x q q q

q q

= = −

A R , (5.2)

2 2

0

2 1 2 1 2 1 2 1

1 2 1 1 2

cos 0 sin

( , ) ( , ) sin sin cos cos sin

cos sin sin cos cos

q q

x q y q q q q q q

q q q q q

= = − −

A R R (5.3)

and

2 3 2 3 2

0

3 1 2 3 1 3 1 2 3 1 3 1 2 3 2 1

1 3 1 2 3 1 3 1 2 3 1 2

( , ) ( , ) ( , )

C C C S S

x q y q z q C S S S C C C S S S C S

S S C S C S C C S S C C

−

= = = + − − − +

A A R R R (5.4)

where the denotations

sin ; cos 1,2,3i i i iS q C q i= = =

were introduced and (*, )qR is the elementary rotation matrix around axis * by an angle q

[60], [61].

Each column of a direction cosine matrix is a unit vector in the corresponding frame. For

future use, denote

0 (0) (0) (0), ,i i i i = A x y z . (5.5)

The total angular velocity of the disk can be written in the form

(0) (0) ( )R=ω J q q (5.6)

on the fixed frame or

(3) (3) ( )R=ω J q q (5.7)

on the body-fixed frame, where (.)

RJ are rotational Jacobian matrices, which can be determined

as

48

2

(0) (0) (0) (0)

1 2 3 1 1 2

1 1 2

1 0 sin

( ) , , 0 cos sin cos

0 sin cos cos

R

q

q q q

q q q

= = −

J q x y z (5.8)

and

2 3 3

(3) 3 (0)

0 2 3 3

2

cos cos sin 0

( ) ( ) ( ) cos sin cos 0

sin 0 1

R R

q q q

q q q

q

= = −

J q A q J q . (5.9)

d) Euler equations

Euler equations for the wobbling disk can be written as (see B.2)

(0) *(0) (0)( ) ( )( ) k+ =M q q C q q q l (5.10)

where the mass matrix and the velocity-free Coriolis matrix are defined by

(0) (3) (3)( ) ( )R=M q AI J q , (5.11)

(3)

*(0) (3) (3) (3) (3)

3

( )( ) (( ( )) ))R

R R

= +

J qC q AI AJ I J q E

q (5.12)

in which (3)I is a matrix of constant representing the inertia tensor in body-fixed frame and

has the following form in the considered case

(3)

0 0

0 0

0 0

=

I (5.13)

and kE is the k k identity matrix.

Substitute (5.4), (5.7), and (5.13) into (5.11) and (5.12), one gets

2

2 2

(0)

1 2 2 1 1 2

1 2 2 1 1 2

( )sin 0 sin

( ) ( )sin sin cos cos sin cos

( )cos sin cos sin cos cos

q q

q q q q q q

q q q q q q

+ −

= − − − −

M q , (5.14)

1 2 2 1 2 2

2 2

2 2 1 2 1 2

1 1

*(0)

1 2 1 2

2 1 2 1 2

0 ( ) ( )

2( ) ( ) (2 1) ( ) (2 1)

0 0 0

0

( ) 0 0 0

0 0 0

0

0 0 0

TC S C S S C

S C S C C C

S C

C C S C

C S S C S

− − − −

− − − − − −

− = − − −

C q . (5.15)

The right-hand side of (5.10) is the total moment of external forces and torques acting on the

disk

1 2

(0) (0) (0) (0) (0) (0) (0) (0)

0 1 1 1 2 2 0 1 2( ) ( )k t t t t A p pk q d q k q d q M= − + − + + + +l x y z p f p f (5.16)

where each contact force is divided into two components

1 1 1 1

(0) (0) (0)

3p n t pF F= +f z t , (5.17)

2 2 2 2

(0) (0) (0)

3p n t pF F= − +f z t . (5.18)

49

The determination of the scalars inF and

itF , position vectors of the contact points (0)

ip and

friction force directions (0)

ipt will be discussed in the following.

e) The contact points

The position vectors can be determined on the trajectories of the pads

1

1

(0) (0) (0)

1 1 0 1

0

0

/ 2 1

I

l

r h h

h

= − + = + −

p p z , (5.19)

2

2

(0) (0) (0)

2 2 0 2

0

0

/ 2 1

I

l

r h h

h

= − + = +

p p z (5.20)

where 1h and 2h are respectively the 0z -direction displacement of the upper and lower pads

from the disk equilibrium, and 1l and 2l are their 0x -direction displacement.

Those two vectors can also be determined on the surfaces of the disk

1

1 1(0) (3) (3) (3)

1 3 3

1 1

0 1 0

0 0 1

/ 2 0 0

K

u u

v vh

= + = + −

p A A Ap A x y , (5.21)

2

2 2(0) (3) (3) (3)

2 3 3

2 2

0 1 0

0 0 1

/ 2 0 0

K

u u

v vh

= + = +

p A A Ap A x y . (5.22)

The equation pairs (5.19), (5.21) and (5.20), (5.22) form linear algebra systems of the

unknowns 1u , 1v , . 1h . and 2u , 2v , 2h .

1 1

1

(3) (3) (0) (0) (3)

3 3 0 1

1

I K

u

v

h

− = −

Ax Ay z p Ap , (5.23)

2 2

2

(3) (3) (0) (0) (3)

3 3 0 2

2

I K

u

v

h

− = −

Ax Ay z p Ap . (5.24)

Solving (5.23) and (5.24), we get

1 1

11

(3) (3) (0) (0) (3)

1 3 3 0

1

2 3 1 3 1 2 3 1 3 1 2 3

2 3 1 3 1 2 3 1 3 1 2 3 1

1 2

1 2 1 2 2

( )

2 ( ) 2 21

2 ( ) 2 22

2 2

I K

u

v

h

rC S h S S C S C C C S S S

rC C h S C C S S C S S S C lC C

rS C hC C h S

−

= − −

+ − − +

= − + + + + − +

Ax Ay z p Ap

, (5.25)

50

2 2

21

(3) (3) (0) (0) (3)

2 3 3 0

2

2 3 1 3 1 2 3 1 3 1 2 3

2 3 1 3 1 2 3 1 3 1 2 3 2

1 2

1 2 1 2 2

( )

2 ( ) 2 21

2 ( ) 2 22

2 2

I K

u

v

h

rC S h S S C S C C C S S S

rC C h S C C S S C S S S C lC C

rS C hC C h S

−

= − −

− − − +

= − − + + + + −

Ax Ay z p Ap

. (5.26)

The velocities of the pads are

2

2 1 2 1 1 2 2 1 2 1 1 21 1 2 12 2 2 2

1 2 1 2 1 2

2 2 2

2 2

rC hS C l S S C hC S l C Sh q q l

C C C C C C

+ + += − − − , (5.27)

2

2 1 2 2 1 2 2 1 2 2 1 22 1 2 22 2 2 2

1 2 1 2 1 2

2 2 2

2 2

rC hS C l S S C hC S l C Sh q q l

C C C C C C

− + − += − − − . (5.28)

The position vectors become

1

(0)

1

1 2 1 2

1 2

2 sin cos 2 sin

2cos cos

l

r

r q q h l q

q q

= −

+ + −

p , (5.29)

2

(0)

2

1 2 2 2

1 2

2 sin cos 2 sin

2cos cos

l

r

r q q h l q

q q

= −

− + −

p (5.30)

or in the body-fixed frame

2 3 1 3 1 2 3 1 1 3 1 1 2 3

1 2

(3) 2 3 1 3 1 2 3 1 1 3 1 1 2 31

1 2

2 2 2

2

2 2 2

2

2

rC S hS S hC S C l C C l S S S

C C

rC C hS C hC S S l C S l S S C

C C

h

+ − − + −

+ + + + = − −

p , (5.31)

2 3 1 3 1 2 3 2 1 3 2 1 2 3

1 2

(3) 2 3 1 3 1 2 3 2 1 3 2 1 2 32

1 2

2 2 2

2

2 2 2

2

2

rC S hS S hC S C l C C l S S S

C C

rC C hS C hC S S l C S l S S C

C C

h

− + − + −

− − + + = −

p . (5.32)

f) Relative velocities of the contact points

The relative velocity can be derived from the position vectors in the body-fixed frame.

1

(3) (3)

1r =v p , (5.33)

2

(3) (3)

2r =v p . (5.34)

51

Projecting (5.33) and (5.34) onto fixed frame yields

1 1 1 1

(0) (3)

1r r p p l= = +v Av J q j , (5.35)

2 2 2 2

(0) (3)

2r r p p l= = +v Av J q j (5.36)

with

1

1 2 2 1 1 1 2

2 1

2 2

1 2 1 2 1 1 2 1 2 2 2 11 1

1 2 1 2

2 2 2

1 2 1 1 1 2 1 1 2 2 12 1 1 22 2

1 2 1 2

2 2 20

2 2

2 2 2 2 2

2 2

2 2 2 2

2 2

p

h l S rC hS l S S

C C

rS C h l S l S C rS S C hS ll S

C C C C

rS C hS l S S l C C hS lrS l S C

C C C C

+ − − − − − − − − −

= − − − − − − −

J ,

1

2

1 2

1

0p

S

C C

= −

j ,

2

2 2 2 1 2 1 2

2 1

2 2

1 2 2 2 2 1 2 1 2 2 2 22 1

1 2 1 2

2 2 2

1 2 1 2 1 2 2 2 2 1 22 2 1 22 2

1 2 1 2

2 2 20

2 2

2 2 2 2 2

2 2

2 2 2 2

2 2

p

h l S rC hS l S S

C C

rS C h l S l S C rS S C hS ll S

C C C C

rS C hS l S S hS l l C CrS l S C

C C C C

− + − + − − + − − + −

= − − + − − + −

J ,

2

2

1 2

1

0p

S

C C

= −

j .

The directions of the friction forces acting on the disk are defined as

1

1

1

(0)

(0)

(0)

r

p

r

=v

tv

, (5.37)

2

2

2

(0)

(0)

(0)

r

p

r

=v

tv

. (5.38)

g) The contact forces

According to Coulomb’s law of friction, we have

1 1 1t k nF F= , (5.39)

2 2 2t k nF F= . (5.40)

The total force acting on each pad in 0z -direction must equal zero since the masses of the

pads are neglected.

1 1 1

(0) (0) (0) (0)

0 3 0 0 1 1 1 1 0T T

n t pF F N h k h d− − + − − =z z z t , (5.41)

52

2 2 2

(0) (0) (0) (0)

0 3 0 0 2 2 2 2 0T T

n t pF F N h k h d− − − − =z z z t . (5.42)

Solving (5.39)-(5.42), one gets

1

1 1

0 1 1 1 1

(0) (0) (0) (0)

0 3 0

n T T

k p

N h k h dF

− −=

+z z z t, (5.43)

1 1

1 1

0 1 1 1 1

(0) (0) (0) (0)

0 3 0

t k T T

k p

N h k h dF

− −=

+z z z t, (5.44)

2

2 2

0 2 2 2 2

(0) (0) (0) (0)

0 3 0

n T T

k p

N h k h dF

+ +=

−z z z t, (5.45)

2 2

2 2

0 2 2 2 2

(0) (0) (0) (0)

0 3 0

t k T T

k p

N h k h dF

+ +=

−z z z t. (5.46)

The total force acting on each pad in 0x -direction is also zero

1 1 1 1

(0) (0) (0) (0)

0 3 0 1 10( ) 0T T

n t p ipF F k l l− − − − =x z x t , (5.47)

2 2 2 2

(0) (0) (0) (0)

0 3 0 2 20( ) 0T T

n t p ipF F k l l− − − =x z x t (5.48)

where 10l and 20l are constants satisfying 1l and 2l are zeros when there is no vibration; for

example, when the disk rotates at a constant velocity.

h) A non-holonomic constraint

The 0z -component of the angular velocity of the disk, is assumed to have a constant value

d , which can be expressed as a non-holonomic constraint

(0) (0)

0 2 1 3 1 2sin cos cosT

dq q q q q= + =z ω (5.49)

or

2 13

1 2

sin

cos cos

d q qq

q q

−= . (5.50)

Note that the total angular velocity of the disk has more components, but for the simplicity of

communication, when there is no specific notice, d is called disk angular velocity or disk

velocity.

i) Minimal generalized coordinates

The constraint (5.50) can be rewritten as

1

1

2

2

12 1

1 21 21 2

0 1 0

0 0 1 ( ) ( )

sinsin0

cos coscos coscos cos

m m m

dd

qq

qq

qq q

q qq qq q

= = + = +

− −

q s q S q q (5.51)

where the vector of minimal generalized coordinates

1 2

T

m q q=q (5.52)

was introduced.

Taking derivative of (5.51) with respect to time variable, we get

3( ) ( )m m m m m m m m

m m m m

= + + = + +

s S s Sq q E q q Sq Sq q q q

q q q q. (5.53)

From the properties of Kronecker product (see B.1), the following equation can be proved

( ) ( ) ( ) ( )( )m m m m m = + + = + + + q q s Sq s Sq s s s S S s q S S q q . (5.54)

53

One important note is that all terms in (5.10) are independent on 3q because of the

geometrical shape of the disk. Therefore, by substituting (5.51), (5.53), (5.54), (5.14), (5.15)

and (5.16) into (5.10), one gets a equations system of minimal generalized coordinates

1 2

* (0) (0)

0 1 1 1 2 2

(0) (0) (0) (0) (0)

0 1 2

( ) ( ) ( )m m m m m m m m t t t t

A p p

k q d q k q d q

M

+ + + = − + − + +

+ + +

M q C q q D q d x y

z p f p f (5.55)

where

(3)( )m m =M q AM S , (5.56)

* (3) *(3)( ) ( )m m

m

= +

SC q AM AC S S

q, (5.57)

(3) *(3)( ) ( )m m

m

= + +

sD q AM AC s S S s

q, (5.58)

*(3)( ) ( )m m = d q AC s s , (5.59)

and any 3q appears in the last two terms of (5.55) is replaced by the right hand side of (5.50).

j) Cancellation of driving torque

To cancel out AM , multiply to the left of both sides of (5.55) with

1 0 0

0 1 0

=

G . (5.60)

The result is

1 2

1 1*

2 2 1

(0) (0) (0) (0)

1 2

( )( )cos

t t

m m m m m m m m

t t

p p

k q d q

k q d q q

+ + + + = − +

+

+ +

M q C q q GD q d

Gp f Gp f

(5.61)

or

* ( )m m m m m m m m m m m+ + + + =M q C q q D q K q d f (5.62)

where

2 1 22

1 2

2 2

1 11 2 2

1

( )

( )

( )

m m m

S SS

C C

S CS S C

C

+ − −

= = +− −

M q GM , (5.63)

* *

2 12 2 2 2

1 2 1 2

22 1

1 2 2 1 2 1 2

1

( )

0 2( )

( )( ) ( ) (2 1) 0

m m m

S SS C

C C C C

CC S C S C S

C

=

− − −

= + −− − − − −

C q GC

, (5.64)

1 2

2 2

1 2 1 2

1

12

1

0( )

0 cos

d dt

t

m m m

t dt

S Sd

C C C Cd

d qd C

C

+

= + =

−

D q GD , (5.65)

1

0( )

0 cos

t

m m

t

k

k q

=

K q , (5.66)

54

( )m m m= =d q Gd 0 , (5.67)

1 2

(0) (0) (0) (0)

1 2( )m m p p= +f q Gp f Gp f . (5.68)

k) Determination of driving torque

Multiplying to the left of both sides of (5.55) by (0)

0

Tz yields

1 2

2 2

1 2 2 1 1 2 1 2 2 1 1 2 1 2

(0) (0) (0) (0) (0) (0)

1 2 1 2 0 1 0 2

( ) ( ) ( ) 2( )A

T T

t t p p

M C S C q S q S S C q C S q q

d S q k S q

= − − − − − − − +

+ + − −z p f z p f. (5.69)

Hence, the driving torque AM that maintains the non-holonomic constraint (5.50) can be

determined provided that needed coordinates, velocities and accelerations are known.

l) Silent motion

For later linearization, a silent motion should be solved from (5.47), (5.48) and (5.62) such

that

1 2( ) ( ) ; 0m mt t l l= = = =q q 0 . (5.70)

However, it is a complicated system of nonlinear algebra equations of the four unknowns 1q ,

2q , 1l (if 10l is constant), and 2l (if 20l is constant) that can hardly be dealt with analytically.

Therefore, it is assumed that the silent motion is

* * *

1 2 1 2( ) ( ) ( ) ; 0m m mt t t l l l l= = = = = = =q q q 0 . (5.71)

This assumption should be rechecked after linearization.

m) Linearization of equations of motion

System (5.62) contains more than just mq and its time-derivatives: there are still unknowns 1l ,

1l , 2l , and

2l . Thus, there is a need for additional equations (5.47) and (5.48).

For later dimension reduction, those additional equations are rewritten as

1 1 1

1

(0) (0) (0) (0)

0 3 0

1 10

T T

n t p

ip

F Fl l

k

− −= +

x z x t, (5.72)

2 2 2

2

(0) (0) (0) (0)

0 3 0

2 20

T T

n t p

ip

F Fl l

k

−= +

x z x t. (5.73)

The two variables 10l and 20l are solved by substituting (5.71) into (5.72) and (5.73)

1

1

0

10

k

ip

Nl

k

= − , (5.74)

2

2

0

20

k

ip

Nl

k

= − . (5.75)

Then (5.72) and (5.73) are substituted into (5.62) and the result is rewritten as

* ( )m m m m m m m m m m m= + + + − =ψ M q C q q D q K q f 0 . (5.76)

To check for the validity of the silent motion, (5.71) is substituted into (5.76) which yields

1 20

0

( )

2

m m k khN

= = − −

0d ψ . (5.77)

The silent motion is valid only if md is a vector of zeros, which means, when the thickness of

the disk and the normal force are nonzero, the friction coefficients on two sides of the disk are

equal

1 2k k k = = . (5.78)

55

Hence, it is now the case to be considered.

Note that 1l ,

1l , 2l , and

2l still appear in (5.76), however

1 2 1 2[ , , , ]

m

Tl l l l

=

0

ψ0 (5.79)

so those variables will vanish in the linearized system.

Linearizing (5.76) around the silent motion, one gets

m m mm m m

m m m

+ + =

0 0 0

ψ ψ ψq q q 0

q q q. (5.80)

The linearization terms are

0

0

mm

m

= =

0

ψM

q, (5.81)

1 2

1 2

2 2 22 2 0 1 0 2 0

1 2

0 1 0 2 1 2

2 2 2

2 2

k k kt d

ip ip dmm

m k k k kd t

ip ip

N d h N d h N hd d r d r

k k r

N d r N d r d hr d hrd

k k

+ + + + +

= =

+ − − −

0

ψD

q, (5.82)

1

22

2 2

11

1

220 1 2

0

20 1 0 2 0 01 20 0

20 0 0

2

2 (1 )2

( )

2

12

mm

m

kt

ip ip

k t k

ip ip ip ip

kk

ip ip

NN h k kk N h r

k k

N k N k N Nk h k hr N k N h

k k k

h

k

N Nh

r k k

− + +

=

+ + + +

=

− − − + + + + − −

0

ψK

q

. (5.83)

The linearized equations are then

m m m m m m+ + =M q D q K q 0 . (5.84)

For simplicity of writing, character was cancelled from (5.84).

Linearizing (5.72) and (5.73) yields

1 1 1

2

1 1 0 2 1 11

( 1)k k k

ip ip ip

d rq N q k rql

k k k

+= − + , (5.85)

2 2 2

2

2 1 0 2 2 12

( 1)k k k

ip ip ip

d rq N q k rql

k k k

+= − + − . (5.86)

If the in-plane stiffnesses 1ipk and

2ipk are set to infinities and normal stiffness 1k equals 2k ,

the final equations of motion (5.84) reduce exactly to those in the case of no in-plane stiffness

[20].

Even when all the damping coefficients are zeros, the damping matrix is nonzero

2

0

(nodamper) 2

0

kd

m d

d

N h

r

= −

D . (5.87)

56

The skew-symmetric terms of (5.87) belong to gyroscopic terms and are a source of instability

[13], [62]. The first element of (5.87) is associated with transverse friction-induced damping

defined in chapter 2 and it appears as expected since the vibrating 1q causes transverse

vibration.

5.1.1.2 Stability analysis

Equations (5.84) can be rewritten as

=x Ax (5.88)

where

m

m

=

qx

q (5.89)

and

2 2 2 2

1 1

m m m m

− −

=

− −

0 EA

M K M D. (5.90)

The trivial solution of (5.88) is unstable if the maximum eigenvalue of A is positive [63].

Now define the critical speed crit as the disk angular velocity that makes the maximum real

part of the eigenvalues of A zero

max(Re(eig( ( )))) 0crit =A . (5.91)

Denote 1 1i as the complex-conjugate eigenvalues of A that have the highest real part

(the case of real eigenvalues is not considered here, as it does not lead to vibration). The

angular frequency of the dominant vibration mode is 1 .

Normally, the trivial solution becomes unstable when the disk angular velocity exceeds the

critical speed [20]. Hence, the influence of parameters on the stability of the disk can be

investigated through their effects on the critical speed.

The parameters used for calculation are taken from [20] (some parameters will be changed in

specific analyses).

1 2

2

7

6

1 2

1 2

6

0.02m; 0.13m;

0.16kg/m ; 2 ;

1.88 10 Nm; 0.1Nms;

6 10 N/m;

5Ns/m;

6 10 N/m;

0.6.

t t

ip ip ip

k

h r

k d

k k k

d d d

k k k

= =

= =

= =

= = =

= = =

= = =

=

The normal force is always chosen so that the braking torque is the same for every simulated

case.

0 0.13 0.6 3000Nmkr N = .

The stability map for varying k is shown in Fig. 5.2. It can be seen that the critical speed is

higher when k is lower. The critical speed also decreases with the increase in normal

57

stiffness of the pads (see Fig. 5.3). Fig. 5.4 and Fig. 5.5 show that high rotational damping

coefficient the disk or high damping coefficient of the pads can effectively stabilize the brake

system.

About the vibration frequency, the frequencies of the dominant vibration mode at critical

speed shown in Fig. 5.6 are close to the undamped natural frequency of the disk

1726Hztd

k

. (5.92)

Figure 5.2: Critical speed of the 2-DOF case for varying kinetic friction coefficient

Figure 5.3: Critical speed of the 2-DOF case for varying pad stiffness

58

Figure 5.4: Critical speed of the 2-DOF case for varying disk rotational damping coefficient

Figure 5.5: Critical speed of the 2-DOF case for varying pad damping coefficient

Figure 5.6: Frequency of the dominant vibration mode of the 2-DOF case at critical speed for varying

pad stiffness. This frequency is near the undamped natural frequency of the disk

59

5.1.2 8-DOF model

5.1.2.1 Equations of motion

The disk in this model is the same as that in the 2-DOF model. Thus, the schematic diagram

in Fig. 5.7 focuses on the pads, the carriers and the caliper [14]. The pads are now having the

masses 1pm and

2pm , normal stiffnesses 1k and 2k , and normal damping coefficients 1d and

2d . The carriers are represented by tangential stiffnesses 1ipk and

2ipk , and tangential

damping coefficients 1ipd and

2ipd . The caliper is modeled as two point masses 1c

m and 2cm

with the stiffness of the spring between them sk , the stiffnesses of the springs between them

and the pads 1bk and

2bk , and their corresponding damping coefficients 1bd and

2bd . Finally,

the caliper is connected with the ground by two springs (stiffnesses 1f

k and 2f

k ), and two

dampers (damping coefficients 1f

d and 2f

d ). The system composed of the pads, the carriers

and the caliper are now called in short the outer subsystem (the whole system minus the disk).

Figure 5.7: An 8-DOF brake system model with frictional point contacts

The nonlinear equations of motion of the disk in this case are established in the same way as

in the 2-DOF case and a system of equations similar to (5.62) is found

* ( )m m m m m m m m m m m+ + + + =M q C q q D q K q d f (5.93)

only with the differences in the in-plane coordinates and the contact forces. The variables

1l , 2l , 1l , and 2l are replaced by 4q , 5q , 4q , and 5q , respectively. The normal forces and

friction forces are determined as

1k1 1,k d

2 2,k d

1 1,ip ipk d

2kDisk

2 2,ip ipk d

1 1,b bk d

2 2,b bk d

2 2,f fk d

1 1,f fk d

sk

4q

5q

8q

9q

7q

6q

1pm

2pm

1cm

2cm

Preload 0N

Preload 0N

60

1

1 1

0 1 8 1 1 8 1

(0) (0) (0) (0)

0 3 0

( ) ( )n T T

k p

N h q k h q dF

− − − −=

+z z z t, (5.94)

1 1

1 1

0 1 8 1 1 8 1

(0) (0) (0) (0)

0 3 0

( ) ( )t k T T

k p

N h q k h q dF

− − − −=

+z z z t, (5.95)

2

2 2

0 2 9 2 2 9 2

(0) (0) (0) (0)

0 3 0

( ) ( )n T T

k p

N h q k h q dF

+ − + −=

−z z z t, (5.96)

2 2

2 2

0 2 9 2 2 9 2

(0) (0) (0) (0)

0 3 0

( ) ( )t k T T

k p

N h q k h q dF

+ − + −=

−z z z t. (5.97)

The outer subsystem is a linear system consisting of point masses, linear springs, and linear

dampers so that its equations of motion can be written quite easily as follows

0c c c c c c c+ + = +M q D q K q f f (5.98)

where

4 5 6 7 8 9[ , , , , , ]T

c q q q q q q=q , (5.99)

1 1 2 2 1 2 1 2

diag( , , , , , )c p c p c c c p pm m m m m m m m= + +M , (5.100)

1

2

1 1 1

2 2 2

1 1

2 2

0 0 0 0 0

0 0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

ip

ip

b f b

c

b f b

b b

b b

d

d

d d d

d d d

d d

d d

+ − = + −

−

−

D , (5.101)

1

2

1 1 1

2 2 2

1 1

2 2

0 0 0 0 0

0 0 0 0 0

0 0 0

0 0 0

0 0 0 0

0 0 0 0

ip

ip

b f s s b

c

s b f s b

b b

b b

k

k

k k k k k

k k k k k

k k

k k

+ + − − = − + + −

−

−

K , (5.102)

1 1 1

2 2 2

1 1 1

2 2 2

(0) (0) (0) (0)

0 3 0

(0) (0) (0) (0)

0 3 0

(0) (0) (0) (0)

0 3 0

(0) (0) (0) (0)

0 3 0

0

0

T T

n t p

T T

n t p

c

T T

n t p

T T

n t p

F F

F F

F F

F F

− −

− = − −

−

x z x t

x z x t

f

z z z t

z z z t

, (5.103)

61

0 c= −0

f f . (5.104)

Combining (5.93) and (5.98), we have

*

0

( ( ) )

( )

m m m m m m m m m m m

c c c c c c c c

= + + + − +

+ + + − − =

ψ E M q C q q D q K q f

E M q D q K q f f 0 (5.105)

where

2 2

6 2

m

=

EE

0, (5.106)

2 6

6 6

c

=

0E

E. (5.107)

System (5.105) will be linearized around

* * *; ;= = = = = =0 0 0

q q 0 q q 0 q q 0 (5.108)

where

m

c

=

qq

q. (5.109)

Like the 2-DOF case, to guarantee that the silent motion (5.108) is valid, or

=0

ψ 0 , (5.110)

the friction coefficients on both sides of the disk should be equal

1 2k k k = = (5.111)

The matrices of the linearized system are

1 1 2 2 1 2 1 2

diag( , , , , , , , )c p c p c c p pm m m m m m m m

= = + +

0

ψM

q, (5.112)

11 12

21 22

= = 0

D DψD

q D D, (5.113)

11 12

21 22

= = 0

K KψK

q K K (5.114)

with

2

1 2

11

1 2

2

0( )

(

2

)

2

k

d

t d

kd t

d d d r

d

N h

d hr

r

d

+ + +

= +− −

D ,

1 2

12 1 2

0 0 0 0

0 0 0 02 2

k k

d r d r

d h d h

= − −

D ,

62

1

2

21

1

2

0

0

0 0

0 0

0

0

k

k

d r

d r

d r

d r

−

=

D ,

1

2

1 1 1

2 2 2

1 1

2 2

1

2

22

1

2

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

ip k

ip k

b f b

b f b

b b

b b

d d

d d

d d d

d d d

d d d

d d d

− + − = + −

− +

− +

D ,

22 0

1 2 0

11

20 1 20

( )2

(4 )(1 )

2

kt

kt k

N hk k k r N h

r

N k h k h rk N h

+ + +

= + +

− + +

K ,

0 01 2

12

1 20 0

0 02 2

0 02 2

k k

k k

N h N hk r k r

r r

k h k hN N

−

= − − −

K ,

2

1 0

2

2 0

21

1

2

(1 )

(1 )

0 0

0 0

0

0

k k

k k

k r N

k r N

k r

k r

− +

− +

=

K ,

1

2

1 1 1

2 2 2

1 1

2 2

1

2

22

1

2

0 0 0 0

0 0 0 0

0 0 0

0 0 0

0 0 0 0

0 0 0 0

ip k

ip k

b f s s b

s b f s b

b b

b b

k k

k k

k k k k k

k k k k k

k k k

k k k

− + + − − = − + + −

− +

− +

K .

The linearized system is written as

+ + =Mq Dq Kq 0 . (5.115)

63

5.1.2.2 Converted stiffness and flexibility of the outer subsystem

Ignoring the damping coefficients, the in-plane vibrations and 2f

k (because an actual caliper

is usually mounted one side) and synchronizing the normal vibrations of the upper and lower

contact points, the outer subsystem can be simplified as in Fig. 5.8.

Figure 5.8: Simplified outer subsystem

The equations of motion of the simplified outer subsystem are

o o o o o+ =M q K q f (5.116)

where

6 7 8 9

T

o q q q q x=q , 1 2 1 2

diag( , , , ,0)o c c p pm m m m=M ,

1 1 1

2 2 2

1 1

2 2

1 1

2 2

1 2 1 2

0 0

0 0

0 0

0 0

0 0

b f s s b

s b f s b

o b b

b b

k k k k k

k k k k k

k k k k

k k k k

k k k k

+ + − − − + + − = − + − − + −

− − +

K ,

0 0 0 0T

o F=f .

Considering the case of harmonic excitation

ˆ ˆcos( ) 0 0 0 0 cos( )T

o o t F t = =

f f , (5.117)

the periodic excited motion has the form

ˆ cos( )o o t= q q . (5.118)

Substituting (5.118) into (5.116), we have

2 ˆˆ( )o o o o− =K M q f (5.119)

or

1fk

1bk

1k

1cm

1pm

2cm

2pm

sk

2bk

2k

6q

8q

7q

9q

F

x

64

2 1 ˆˆ ( )o o o o

−= −q K M f . (5.120)

Multiplying to the left of both sides of (5.120) by

5 0 0 0 0 1T =e , (5.121)

one gets the relation between the normal displacement amplitude of the contact points and the

amplitude of the excitation force

2 1

5 5ˆˆ ( )T

o ox F−= −e K M e . (5.122)

Denote

2 1

5 5

1( )

( )o T

o o

k−

=−e K M e

(5.123)

as the converted stiffness of the outer subsystem and its inverse

1 2 1

5 5( ) ( )T

o o ok − − = −e K M e (5.124)

as the converted flexibility.

5.1.2.3 Stability analysis

The parameters used for calculation are taken from [14] and [20] (some parameters will be

changed in specific analyses).

1 2 1 2

1 2 1 2 1 2 1 2

2 7

6

1 2 1 2

6

0.02m; 0.13m; 0.16kg/m ; 2 ; 1.88 10 Nm; 0.1Nms;

6 10 N/m; 5Ns/m;

6 10 N/m; 10Ns/m;

0.1kg; 0.7kg; 21Ns/m; 1.5Ns/m; 0;

1 10

t t

ip ip ip ip ip ip

p p c c b b f f

s

h r k d

k k k d d d

k k k d d d

m m m m d d d d

k

= = = = = =

= = = = = =

= = = = = =

= = = = = = = =

= 1 2 1 2

5 8 8 5N/m; 3.3 10 N/m; 3.3 10 N/m; 1 10 N/m; 0;

0.60.

b b f f

k

k k k k

= = = =

=

Figure 5.9: Critical speed of the 8-DOF case compared with that of the 2-DOF case for low varying

pad stiffness

65

a) Critical speed (the dashed line is for the 2-DOF case)

b) Converted flexibility at the disk’s natural frequency of the outer subsystem

c) Frequency of the dominant vibration mode at critical speed

Figure 5.10: Stability analysis of the 8-DOF case for high varying pad stiffness and different masses of

the caliper. Where the critical speed peaks, there is a point of discontinuity of the dominant frequency,

and it is near where the converted flexibility of the outer system reaches zero

66

The normal force is always chosen so that the braking torque is the same for every simulated

case.

0 0.13 0.60 3000Nmkr N = .

For low stiffnesses of the pads ( 6 71 10 1 10 /N m − ), it can be seen that there is only little

difference between the critical velocity of the disk in this case and that in the 2-DOF case

(Fig. 5.9). The frequency of the dominant vibration mode in this case is also close to the

undamped natural frequency of the disk determined by (5.92).

However, when the stiffnesses of the pads go higher, the critical velocity surges and reach a

maximum value before decreases again (Fig. 5.10). Around where the critical velocity peaks,

there are two notable phenomena. The first one is that the converted flexibility at the disk’s

natural frequency of the outer subsystem changes its sign. The second one is the point of

discontinuity of the dominant frequency. This jump indicates that there is a replacement of the

dominant vibration mode. It can be speculated that when the natural frequencies of the disk

and the outer subsystem become close, “inner resonance” happens: the two subsystems

interact with each other, changing the natural frequencies of the whole system and stabilizing

the system, two of the eigenvectors exchange the role of dominant vibration mode, making

the dominant frequency suddenly change.

5.2 Tangential friction-induced damping case

5.2.1 2-DOF model

The equations of motion in this case are generated with the same procedure as in the constant

friction coefficient case. There are two main differences. The first one is that the contact

forces satisfied (2.57) so that

1 1 1

( ( ) ( ))t n k d r dF F r v r = + − (5.125)

2 2 2

( ( ) ( ))t n k d r dF F r v r = + − . (5.126)

Here the coefficient of kinetic friction ( )k d r as well as the tangential friction-induced

damping factor of both contact points is considered to be the same. Solving (5.125),

(5.126), (5.41) and (5.42), one gets

1

1 1

0 1 1 1 1

(0) (0) (0) (0)

0 3 0( ( ) ( ))n T T

k d r d p

N h k h dF

r v r

− −=

+ + −z z z t, (5.127)

2

2 1

0 2 2 2 2

(0) (0) (0) (0)

0 3 0( ( ) ( ))n T T

k d r d p

N h k h dF

r v r

+ +=

− + −z z z t. (5.128)

The second difference is that the in-plane movements of the pads are ignored. If not, (5.79) is

not satisfied so that the linearized system is no longer 2-DOF.

The matrices of the final linearized system are

0

0m

=

M , (5.129)

22 2 0

1 2

2

1 2 0

2

2 2 2

kt d

m

k kd t

N hd d r d r

r

d hr d hr N hd

+ + +

= − − − +

D , (5.130)

67

2 2

1 2 0

220

2

20

0

102

2

2

(1 )2 2

kt

m

dk t k

k k r k r N h

N hk h k hN N h

N h

k

r

r

+ + +

=

− + + − + +

K . (5.131)

When there is no damper, the damping matrix reads

2

0

2

0

2

2

kd

m

d

N h

r

N h

= −

D . (5.132)

Unlike transverse friction-induced damping, tangential friction-induced damping factor

affects the second element of the diagonal of (5.132), so that it associates with the fluctuation

of 2q , which is also expected.

It can be predicted that an increasing may stabilize the brake system. This prediction is

verified by simulating with the same parameters as in the previous case (Fig. 5.11). The

stabilization is even stronger for lower kinetic friction coefficient (and, due to the assumption

of unchanged braking torque, higher normal force).

Figure. 5.11: Critical speed of the 2-DOF case for varying tangential friction-induced damping factor

and different coefficients of kinetic friction

5.2.2 8-DOF model

The contact forces in this case are

1

1 1

0 1 8 1 1 8 1

(0) (0) (0) (0)

0 3 0

( ) ( )

( ( ) ( ))n T T

k d r d p

N h q k h q dF

r v r

− − − −=

+ + −z z z t (5.133)

1 1 1

( ( ) ( ))t n k d r dF F r v r = + − (5.134)

2

2 1

0 2 9 1 2 9 1

(0) (0) (0) (0)

0 3 0

( ) ( )

( ( ) ( ))n T T

k d r d p

N h q k h q dF

r v r

+ − + −=

+ + −z z z t (5.135)

2 2 2

( ( ) ( ))t n k d r dF F r v r = + − . (5.136)

The matrices of the final linearized system are

68

1 1 2 2 1 2 1 2

diag( , , , , , , , )c p c p c c p pm m m m m m m m= + +M , (5.137)

11 12

21 22

= = 0

D DψD

q D D, (5.138)

11 12

21 22

= = 0

K KψK

q K K (5.139)

with

2

1 2

112

1 2 0

2

0

2( )

( )

2 2

t dk

kd t

d

d d d r

d

N h

r

d hr N hd

+ + +

= +− − +

D ,

1 2

12 0 0 1 2

0 0 0 0

0 02 2 2 2

k k

d r d r

N h N h d h d h

= − − −

D ,

1 2 1 2

21 0 0

0 0

0 0 0 02 2

T

k kd r d r d r d r

N h N h

− = −

D ,

1

2

1 1 1

2 2 2

1 1

2 2

0 1

0 2

22

1

2

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

ip k

ip k

b f b

b f b

b b

b b

d N d

d N d

d d d

d d d

d d d

d d d

+ −

+ + − = + −

− +

− +

D ,

22 0

1 2 0

11 220 1 2 0

0

( )2

(4 )(1 )

2 2

kt

k dt k

N hk k k r N h

r

N k h k h r N hk N h

+ + +

=+ +

− − + +

K ,

0 01 2

12

1 20 0

0 02 2

0 02 2

k k

k k

N h N hk r k r

r r

k h k hN N

−

= − − −

K ,

69

201 0

202 0

21

1

2

(1 )2

(1 )2

0 0

0 0

0

0

dk k

dk k

N hk r N

N hk r N

k r

k r

− − +

+ − + =

K ,

1

2

1 1 1

2 2 2

1 1

2 2

1

2

22

1

2

0 0 0 0

0 0 0 0

0 0 0

0 0 0

0 0 0 0

0 0 0 0

ip k

ip k

b f s s b

s b f s b

b b

b b

k k

k k

k k k k k

k k k k k

k k k

k k k

− + + − − = − + + −

− +

− +

K .

Similar to the 2-DOF model, the tangential friction-induced damping factor appears in the

second element of the diagonal of the damping matrix. And again, an increasing also helps

stabilizing the system (Fig. 5.12).

Figure. 5.12: Critical speed of the 8-DOF case for varying tangential friction-induced damping factor

and different coefficients of kinetic friction

70

6 Frequency-domain constants approach

6.1 General idea

When studying a vibratory system, it is ideal if the state-variable equations of motion of the

system are known. However, due to the lack of an appropriate model, some identified

parameters of the system may depend on frequency-domain variables (for instance, vibration

frequencies and amplitudes), so that the state-variable equations of motion of the system are

unavailable.

To overcome this problem without extending the model, those parameters, now called

frequency-domain constants, are considered to be unchanged if the frequency-domain

variables are unchanged, i.e. the system exhibits a stable vibration (periodic vibration, for

example). The equations of motion of the system become implicit (Fig. 6.1).

Figure 6.1: Implicit system with frequency-domain constants

The frequency-domain constants approach finds the stable solutions of the studied system by

solving this implicit system. One important kind of stable solution is limit cycle and the

simplest form of limit cycles, which will be considered here, is sinusoid.

Note that some calculation methods are also used to solve for sinusoidal limit cycle, such as

the averaging method [64] and the energy-balance method [65] [66], but they deal with

provided nonlinear state-variable equations of motion. In contrast, the frequency-domain

constants approach does not require state-variable equations of motion.

6.2 Determining the limit cycles by frequency-domain constants approach

6.2.1 Homogeneous linearized system

Consider a system of homogeneous linearized ODE representing a potentially self-excited

system

=q Aq (6.1)

where q is a vector of n state variables and A is a n n matrix.

As the system is linearized, A contains only constant parameters, and the stability of the

system depends solely on A . Let ( ) ( )j ji +A A and ( ) ( )j ji −A A ( 1,j m= ) be the

complex-conjugate pairs of eigenvalues of A , and ( ) ( )j ji+u A v A and ( ) ( )j ji−u A v A be

Vibratory

model

Frequency-domain

constants

Frequency-domain

variables

Stable

solution

71

their respective eigenvectors. Let ( )j A ( 2 1,j m n= + ) be the real eigenvalues of A and

( )ju A be their respective eigenvectors.

Consider the case ( )j A ( 1, ,2 1,j m m n= + ) are distinct real values, the general solution of

(6.1) is

1

2 1

( ) ( ( cos( ) sin( )) ( cos( ) sin( )))j

j

m

j j j j j m j j j j j

j

n

j j

j m

t e d t t d t t

e d

+

=

= +

= − + + +

+

q u v v u

u

(6.2)

where jd ( 1,j n= ) are constants.

If there exists at least an eigenvalue of matrix A that has a positive real part, the system is

unstable. In other words, system (6.1) is unstable if the maximum value of the set of real parts

of all eigenvalues of matrix A is positive

max ( ) max{ ( )} 0 1,j j n = =A A (6.3)

This value of max ( ) A is the maximal Lyapunov exponent of system (6.1). Consider the case

where max ( ) A is the real part of an eigenvalue maxmax ( ) ( )i +A A (the case max ( ) A is a

real eigenvalue is not considered here because it does not lead to a vibration mode). The

respective imaginary part max

( ) A is the angular frequency of the dominant vibration mode

of the system, which is of the form

max max

max max max max max max max1 2 1 2( ) ( )cos( ) ( )sin( )t t

t e c c t e c c t

= + + − +q u v v u (6.4)

where 1c and 2c are constants.

In case of instability, the vibration of system (6.1) increases to infinity at the rate determined

by max . However, in reality, some nonlinearities may restrict this vibration into a limit cycle.

Note that the dependences of some parameters on the vibration amplitude imply

nonlinearities. The linear system (6.1) alone cannot be used to find the limit cycle while the

frequency-domain constants approach can be used.

6.2.2 Assumptions for frequency-domain constants

Assume that the limit cycle is harmonic and is defined as

ˆ ˆ( ) cos( ) sin( )c st t t= + q q q (6.5)

where ˆcq and ˆ

sq are 1n vectors of coefficients of cosine and sine terms, respectively, and

is the vibration angular frequency. They are all frequency-domain variables.

Assume that matrix A is a matrix function of a vector of k parameters p , which in turn

depends on the frequency-domain variables ˆcq , ˆ

sq and

ˆ ˆ( , , )c s= p p q q , (6.6)

ˆ ˆ( ( , , ))c s= A A p q q . (6.7)

72

For a specific limit cycle, ˆcq , ˆ

sq and are determined so that the elements of p and,

therefore, A are constants. It means that the parameters do not depend on the state variables

but on the state of harmonic vibration and keep unchanged for the whole period.

The original system (6.1) is now rewritten as

ˆ ˆ( ( , , ))c s= q A p q q q , (6.8)

in which the dependence of parameters p on frequency-domain variables can be identified

through experiments

ˆ( , , )= p p x m (6.9)

where x̂ are the vibration amplitudes in the experiments and m is a vector of other

conditions. To get expression (6.6), a relationship between two cases should be established as

( )

ˆ

ˆ ˆ, ,c s

=

x

h q q

m

(6.10)

The limit cycle in (6.5) is valid and stable if the following conditions are satisfied.

i) Maximum value of the set of real parts of all eigenvalues of matrix ˆ ˆ( ( , , ))c s A p q q is zero

and the respective imaginary part is .

maxˆ ˆ( ( ( , , ))) 0c s =A p q q (6.11)

max

ˆ ˆ( ( ( , , )))c s =A p q q (6.12)

ii) (6.5) satisfies (6.8).

iii) At a close but “lower” amplitude, max is positive.

* * * *

maxˆ ˆ ˆ ˆ ˆ ˆ( ( ( , , ))) 0 ;c s c c s s A p q q q q q q (6.13)

iv) At a close but “higher” amplitude, max is negative.

* * * *

maxˆ ˆ ˆ ˆ ˆ ˆ( ( ( , , ))) 0 ;c s c c s s A p q q q q q q (6.14)

The first two conditions guarantee that equation (6.5) is not only a solution of (6.8) but also

the dominant vibration mode. The last two conditions mean that if the vibration of the system

deviates from (6.5), it will be pulled back to this solution: when the vibration amplitude is

“lower” than the amplitude of the limit cycle, a positive max increases the amplitude and

when the amplitude is “higher” than that, a negative max decreases the amplitude. However,

the definition of “lower” and “higher” with a type of norm is not unique and the transient

process cannot be described clearly with the abovementioned assumptions.

6.2.3 A procedure to find limit cycles with frequency-domain constants approach

Based on the presented idea and assumptions, the following procedure is constructed.

Step 1. Providing that the experimental results give relation (6.9), find (6.10) to get expression

(6.6)

73

Step 2. Establish functions or procedures to determine the eigenvalues ( ) ( )j ji +p p of

( )A p and their normalized eigenvectors ( ) ( )j ji+u p v p ( 1,j n= ) and then determine max ( ) p

and the respective imaginary part max

( ) p and eigenvector max max

( ) ( )i +u p v p .

Step 3. Solve the following system of 2 2n k+ + algebra equations

max

max

max

*

max

*

1

*

1

*

*

( ) 0

ˆ ( ) 0

ˆ ( ) 0

( ) 0

ˆ ˆ( , , ) 0

c

s

c s

c

c

=

− =

+ =− =

− =

p

q u p

q v p

p

p p q q

(6.15)

for ˆcq , ˆ

sq , , 1c , and *p . There is no need to determine 2c because the appearance of it will

only affect the initial phase of the limit cycle, which does not influence other parameters.

Step 4. Check if the following conditions are satisfied

maxˆ ˆ( ( ( , , ))) 0c s − − A p q ε q ε , (6.16)

maxˆ ˆ( ( ( , , ))) 0c s − − A p q ε q ε (6.17)

where ε is a vector of small positive constants.

If there is only one frequency-domain constant p , the equations in step 3 can be solved one-

by-one as follows.

Step 3’. Solve the following equation for *p

*

max ( ) 0p = , (6.18)

determine

max

*( ) = p , (6.19)

solve for 1c

max max

* * *

1 1( ( ), ( ), ) 0p p c c − − =u p v p , (6.20)

and calculate

max

*

1ˆ ( ),c c =q u p (6.21)

max

*

1ˆ ( ).s c = −q v p (6.22)

6.2.4 Verification of frequency-domain constants approach: A virtual experiment

The well-known Van der Pol oscillator has the following equation [67], [68]

2( 1) 0x x x x+ + − = (6.23)

where is a positive constant. If is small, the equation is weakly nonlinear.

The equation linearized around the trivial solution

0x x x+ − = (6.24)

74

has an unstable fixed point (0,0) because of the negative damping coefficient, so that (6.23) is

a self-excited oscillator.

Now consider an imaginary mechanical model of the Van der Pol oscillator (Fig. 6.2a) with a

mass of 1, a stiffness of 1 and a nonlinear damper whose coefficient, 2(1 )x− − , is assumed

to be unknown. To experimentally investigate the damper, it is embedded into a mass-spring-

damper system, where the mass can be harmonically excited (Fig. 6.2b). For simplicity, all the

quantities, including time, are considered dimensionless.

Figure 6.2: Applying frequency-domain constants approach to an imaginary mechanical model of the

Van der Pol oscillator

The excitation force and the displacement of the mass are measured and converted by FFT to

get the following values: the amplitude of the excitation force F̂ , the excitation angular

frequency , the amplitude of the displacement xA , and the phase difference , by which

the additive inverse of the displacement is shifted forward in comparison to the excitation

force. The total damping coefficient of the test setup is estimated according to (2.44), and the

identified nonlinear damping coefficient is a function of and xA .

ˆ sin

( , )identified x test

x

Fd A d

A

= −

. (6.25)

Applying the frequency-domain constant approach, the original oscillator is described by the

implicit equation

( , ) 0identified xx x d A x+ + = . (6.26)

Since the natural frequency of the mechanical model of the Van der Pol oscillator is 1, the

virtual experiments are carried out at 1= . The other parameters are

1; 0.1; 1; 0.5test test testm k d = = = = .

According to the virtual measurement results shown in Fig. 6.3, the solution of (6.26) found

by frequency-domain constants approach is

1m =

1k =

( )2( ) (1 )d x x= − −

x

unknown damper

a) An imaginary mechanical model

of the Van der Pol oscillator

testm

testk

( , )xd A

b) A virtual experiment to identify

the damping coefficient of the Van

der Pol oscillator

testd

measured displacement

( )cos( )xx A t − +

measured

excitation

force

( )ˆ cos( )F t

75

2.2cosx t , (6.27)

which agrees quite well with the approximate analytical solution of (6.23) [67]

2cosx t . (6.28)

Figure 6.3: Identified nonlinear damping coefficient of the Van der Pol damper for varying

displacement amplitude

6.3 Discussion

The frequency-domain constants approach has an advantage of keeping the linearized system

while doing nonlinear analysis to find limit cycles. Therefore, only a system of algebraic

equations is to be solved, no nonlinear ODE is involved, which may save much calculation

effort and time. Another significant advantage is that the experimental results, which are in

frequency domain, can be used in frequency domain, no artificial model of the parameters in

time domain is needed. This is well illustrated by the virtual experiment regarding Van der

Pol oscillator above: the vibration amplitude and frequency are estimated without knowing

the exact equation of motion. Although the result is less accurate than if the exact equation is

known, it is still, to a certain extent, acceptable. In reality, particularly within this research,

the instances where the exact models cannot be found with available knowledge are where

this approach can shine.

The disadvantage of a model treated with this approach is that it does not have an explicit

equation form. Thus, it is hard to use it to describe transient processes or to simulate in a

general case where the solution does not have a harmonic approximation. Moreover, the

errors (coming from identification and hypothesizing steps) are hard to control and there is no

guarantee that they do not lead to an unacceptable result.

On the calculation technique, when there is only one frequency-domain constant, step 3’ is

performed instead of step 3, only single-variable equations are to be solved; therefore, to get a

converged solution is not so hard. In contrast, for multiple frequency-domain constants, a set

of equations should be solved simultaneously in step 3, so that choosing appropriate starting

points (for Newton-Raphson algorithm, for example) that lead to converged solutions is the

central task of this step.

The converted stiffness or the converted flexibility (section 5.1.2.2) can be considered as

frequency-domain constants in a linear case: they only depend on the frequency, not the

amplitude. One may wonder whether frequency-domain constants approach can be used to

76

reduce the 8-DOF models in chapter 5 to 2-DOF. The answer is no because, in the linear

analysis of a linearized autonomous system, the system has only two states: stable and

unstable and there is no room for a limit cycle. However, the approach can be found helpful

for the case of a harmonically forced linear system, in which a harmonic motion is expected.

77

7 Nonlinear analysis of wobbling disk models

7.1 2-DOF model with in-plane stiffness

According to chapter 5, the linearized brake systems with wobbling disk are self-excited

systems. However, the linear analysis can only predict, to a certain extent, when the system

vibrates, but cannot predict how large the vibration would be. Here, the frequency-domain

constants approach in chapter 6 is used to estimate the size of the limit cycle produced by the

wobbling disk models.

Due to the way the parameters will be chosen, vibration amplitudes on two sides of the disk

will be the same. The amplitude of the displacement of in-plane vibration lA and the

amplitude of the acceleration of in-plane vibration l

A are respectively defined by

1 21 2

max( ) max( )max( ) max( )

2l

l lA l l

+= = = (7.1)

and

21 2 1 21 2

max( ) max( ) max( ) max( )max( ) max( )

2 2dl

l l l lA l l

+ += = = = (7.2)

where 1l and 2l are determined by (5.85) and (5.86), respectively.

The amplitude of the displacement of normal vibration of the contact points hA is calculated

as follows

1 2

1 2

max max2 2

max max2 2 2

h

h hh h

h hA h h

+ + −

= + = − =

(7.3)

where 1h and 2h are determined by (5.25) and (5.26), respectively. In this case, hA is also the

amplitude of the deformation of the pads.

The tilting angle of the disk is determined from the rotational matrix regarding 1q and 2q

(5.3) through Euler’s parameter as follows [60]

0

2 2 1 1 2

2 21 21 2

tr( ) 1 cos cos cos cos 12arccos 2arccos

4 4

2arccos cos cos2 2

d

q q q qq

q qq q

+ + + += =

= +

A

. (7.4)

The vibration amplitude of the disk dqA calculated as

( )2 2

1 2max( ) maxdq dA q q q= + (7.5)

is used to evaluate the size of the limit cycle.

The parameters used in calculation are kept the same as those from chapter 5 (some

parameters will be changed in specific analyses).

78

a) Vibration amplitude of the disk

b) Tangential vibration amplitude of the contact points

c) Normal vibration amplitude of the contact points

Figure 7.1: Vibration amplitude in the 2-DOF case with friction coefficient reduction for varying disk

angular velocity. dqA , lA , and hA are determined by Eq. (7.5), (7.1), and (7.3), respectively.

79

a) Vibration amplitude of the disk

b) Tangential vibration amplitude of the contact points

c) Normal vibration amplitude of the contact points

Figure 7.2: Vibration amplitude in the 2-DOF case with both friction coefficient reduction and pad

stiffness reduction for varying disk angular velocity

80

Firstly, the friction reduction due to vibration is taken into account. The frequency-domain

coefficient of kinetic friction is determined by

,( ) (1 ) (0)kk a kl l

A A = − . (7.6)

According to the results presented in Table 4.2, ,k a takes values between 40.35 10− and

42.09 10− s2/m. Fig. 7.1a shows the dependence of the size of the limit cycle on the angular

disk velocity with different friction reduction ratios ,k a : 40.5 10− , 41 10− and 42 10− s2/m.

It can be seen that the limit cycle is larger with a higher angular disk velocity, which agrees

with experimental observation [12]-Fig. 56: when the brake squeal appears, increasing the

disk velocity usually results in louder noise. In contrast, a higher value of ,k a lowers the

size of the limit cycle, which means that the more the friction coefficient reduces due to

vibration amplitude, the smaller the limit cycle is. The amplitudes of the displacement of in-

plane and normal vibration of the contact points under the same conditions are shown in Fig.

7.1b and Fig. 7.1c, respectively.

Next, together with the friction reduction ratio ,k a of 4 21 10 s /m− , the stiffnesses of the

pads are also considered to reduce due to normal vibration

,( ) (1 ) (0)h k x hk A A k= − (7.7)

where ,k x 2[N/m ] is the stiffness reduction ratio, which can be identified by the DCTR.

Based on [23]-Fig. 6.24, ,k x of a pad specimen with stiffness in the range of 107N/m is in the

range of 1012-1013N/m2, so it can be speculated that ,k x of a pad specimen with a stiffness of

66 10 N/m is in the range of 1011-1012N/m2.

Since two frequency-domain constants are considered, the starting point for searching for the

solution of (6.15) should be well chosen. Here, the starting point is determined as the solution

when the friction coefficient is fixed and k is the only frequency domain constant.

The vibration amplitude of the disk and the displacement amplitudes of in-plane and normal

vibration of the contact points for different values of ,k x : 120.2 10 , 120.4 10 and

120.6 10 N/m2 are shown in Fig. 7.2. The size of the limit cycle in this case is much smaller

than that in the case where only the nonlinearity of friction reduction is considered.

The stability of a limit cycle found by frequency-domain constants approach is verified by Eq.

(6.16) and (6.17). To visualize this verification with numerical integration, the system of

equations is linearized around the limit cycle and its smaller and larger scaled versions:

( 1,0,1)i i= = −x A x (7.8)

where

0 ( ( , , ))c s= A A p q q (7.9)

for the limit cycle,

1 ( (0.99 ,0.99 , ))c s− = A A p q q (7.10)

for the smaller scaled version, and

1 ( (1.01 ,1.01 , ))c s= A A p q q (7.11)

81

for the larger scaled version.

Each linearized system is solved with the appropriated initial values. Fig. 7.3 shows the time

history of 1q . The amplitude in the case of limit cycle is unchanged over time (Fig. 7.3b),

which is a sign of the validity of the limit cycle. The amplitude increases and decreases

respectively in the case of smaller and larger scaled version (Fig. 7.3a and Fig. 7.3c),

confirming the conditions for stability (6.16) and (6.17): the amplitude increases if it is

currently “lower” than the amplitude of the limit cycle and decreases if it is currently

“higher”.

a) b)

c)

Figure 7.3: Time history of 1q of system linearized around: (a) a smaller scaled version of the limit

cycle, (b) the stable limit cycle found by frequency-domain constants approach, and (c) a larger scaled

version of the limit cycle. The plots are the signs of the existence of the limit cycle and its stability.

82

7.2 8-DOF model

In this case, instead of lA and l

A , the displacement amplitude and acceleration amplitude of

in-plane vibration are respectively estimated by

45

4 5max( ) max( )

2q

q qA

+= (7.12)

and

45

4 5max( ) max( )

2q

q qA

+= . (7.13)

The amplitude of the deformation of the pads is estimated by

8 6 9 7max( ) max( )

2pq

q q q qA

− + −= . (7.14)

Note that due to asymmetry, the vibration amplitudes of the upper and lower parts may differ.

It may result in asymmetrical frequency-domain constants and, therefore, destruction of the

silent motion. For simplicity, symmetrical frequency-domain constants are assumed.

The frequency-domain coefficient of kinetic friction is determined by

45 45,( ) (1 ) (0)

kk q a q kA A = − . (7.15)

The frequency-domain stiffness of the pads is formulated as

,( ) (1 ) (0)p pq k x qk A A k= − . (7.16)

When a starting point is needed, one may follow the same strategy as in the 2-DOF case:

consider k as the only frequency-domain constant and choose the result as the starting point.

7.2.1 8-DOF model without tangential friction-induced damping

Similar to the previous section, two cases of nonlinearity will be considered: (a) friction

reduction and (b) both friction reduction and stiffness reduction. For the former case, the

vibration amplitude of the disk, the tangential vibration amplitude of the contact points, and

the deformation amplitude of the pads are shown in Fig. 7.4. The results for the latter case are

shown in Fig. 7.5. Again, stiffness reduction lowers the size of the limit cycle significantly.

7.2.2 8-DOF model with tangential friction-induced damping

In addition to the friction reduction ratio ,k a of 4 21 10 s /m− and the stiffness reduction ratio

,k x of 12 20.4 10 N/m , the tangential damping factor is assumed to increase due to vibration

and determined by

45 45,( ) (1 ) (0)q a qA A = + (7.17)

where ,a is the tangential damping factor addition ratio 3 2[s /m ] .

The calculation results in Fig. 7.6 show that increasing ,a decreases the size of the limit

cycle, which is expected since ,a help dissipate the vibration energy.

83

a) Vibration amplitude of the disk

b) Tangential vibration amplitude of the contact points

c) Deformation amplitude of the pads

Figure 7.4: Vibration amplitude in the 8-DOF case with friction coefficient reduction for varying disk

angular velocity. dqA ,

45qA , and pqA are determined by Eq. (7.5), (7.12), and (7.14), respectively.

84

a) Vibration amplitude of the disk

b) Tangential vibration amplitude of the contact points

c) Deformation amplitude of the pads

Figure 7.5: Vibration amplitude in the 8-DOF case with both friction coefficient reduction and pad

stiffness reduction for varying disk angular velocity

85

a) Vibration amplitude of the disk

b) Tangential vibration amplitude of the contact points

c) Deformation amplitude of the pads

Figure 7.6: Vibration amplitude in the 8-DOF case with tangential damping for varying disk angular

velocity

86

7.2.3 Subcritical behavior

According to Fig. 5.10a, the critical speed-pad stiffness curve in the linearized 8-DOF system

may have a local minimum, which means the stiffness reduction phenomenon may lead to the

coexistence of a stable fixed point, an unstable limit cycle, and a stable limit cycle (Fig. 7.7),

similar to what happens when there is a subcritical Hopf bifurcation in a brake squeal test

[12]-Fig. 56. However, the considered equations are not classical ODE (because with the

application of the frequency-domain constants approach, the system becomes implicit), so that

a true subcritical Hopf bifurcation should not be expected here, but a “subcritical behavior”.

Figure 7.7: Coexistence of a stable fixed point, an unstable limit cycle, and a stable limit cycle in the

8-DOF case with the reduction of pad stiffness due to vibration

Figure 7.8: The subcritical behavior in the 8-DOF case with the reduction of pad stiffness due to

vibration

stiffness reduction

unstable region

stable region

stable fixed point

unstable limit cycle

stable limit cycle

Pad stiffness

Dis

k a

ng

ula

r vel

oci

ty

87

Fig. 7.8 shows a subcritical behavior with the disk angular velocity as the control parameter.

The parameters used here are 1 2

0.3kgc cm m= = , k determined by (7.16) with

12 2

, 6 10 N/mk x = and 7(0) 1.5 10 N/mk = , and other parameters taken from chapter 5.

It can be seen that, different from Hopf bifurcation, the slope of the curve of the unstable limit

cycle is not infinite at the connecting point with the line of the stable fixed point. This may be

the result of the non-zero slope of the pad stiffness-dynamic amplitude curve at zero

amplitude [23]-Fig. 6.24, [12]-Fig. 64: if the pad stiffness-dynamic amplitude curve is

extended to the left, it should be symmetric about the vertical axis; therefore, the slope of the

curve is discontinuous at zero amplitude and the behavior of the whole system may become

abnormal around this point.

88

8 Conclusions and outlook

This research studies the brake squeal phenomenon by firstly experimentally investigating the

relationship between a more fundamental phenomenon – friction – and vibration, and then

embedding vibration-dependent nonlinearities into brake models for simulation. Throughout

the process, both experimental and theoretical contributions have been made.

In terms of experimental approach, this study contributes to the abundant, yet still insufficient,

collection of current tribometers with the OFTR. The OFTR is distinguished from normal

tribometers by its capacity of putting pairs of tested materials under complex dynamical

conditions. The testing conditions are desired to be close to the real conditions in the brake

system when brake squeal appears. Besides a high-normal-force pure-sliding large-area

contact between the pad specimen and the disk, the OFTR also succeeded in creating

tangential vibrations whose displacement amplitudes are in the micrometer range and

frequencies in the low kHz range. Though realistic, the available range of the conditions is

still narrow compared with the whole actual range for brake squeal. This problem and other

technical issues, especially the difficulties in measuring forces over time and measuring the

motion of the pad specimen as a whole, call for the improvement of the OFTR in the future

work. Vibration in normal and transverse directions (and maybe more complicated vibration

modes) should also be considered in some of the next designs.

The measurement results show some remarkable new findings on the friction reduction

phenomenon, which has already been described in plenty of literature sources. With the

sandwich measurement sequence, which is proposed to eliminate the interference of time-

dependent factors, it is carefully confirmed that the coefficient of kinetic friction drops as the

vibration amplitude goes up. The reduction ratio is significant even with no change in the

direction of the relative velocity. This characteristic contradicts all the theoretical results in

the literature, which relate the phenomenon to the varying relative velocity direction. On the

one hand, the disagreement between theory and experiment shows the lack of an appropriate

dynamic friction model regarding friction reduction. On the other hand, it is a big challenge to

nonlinear analysis and simulation for brake squeal.

It can be observed from the tests that tangential friction-induced damping also depends on the

vibration amplitude and frequency. Unfortunately, this phenomenon has not been successfully

quantified due to the inconsistency of the measurement results.

Regarding the stability analysis of linearized models for brake squeal, this study not only

replicates the 2-DOF and 8-DOF models with a wobbling disk, which are introduced in earlier

projects at MMD, but also gives new insights into them. Firstly, it is found that the ‘inner

resonance’ between the disk subsystem and outer subsystem in the 8-DOF model can stabilize

the whole system. Secondly, tangential friction-induced damping is added into the models and

proved to affect the stability maps.

To circumvent the lack of exact equations of motion for nonlinear analysis, the frequency-

domain constants approach is introduced. At the cost of surrendering accurate simulation over

time, the approach is dynamic-experiment friendly. It can employ the results from the DCTR

as well as the OFTR to calculate the size of the limit cycles for the brake models with a

wobbling disk. The result can be used to predict the effects of amplitude-dependent and

frequency-dependent nonlinearities on brake squeal.

89

The future work can be divided into four main branches. The first one is to improve the

OFTR, which has been mentioned above. The second one is to examine the friction-induced

damping with both tangential and transverse vibration, and maybe even normal vibration.

This requires not only experimental work but also theoretical effort to evaluate the effects of

the phenomenon on brake squeal. The third branch is to refine the brake system models and

parameters. No matter if the considered model is low-DOF or FEM, it should be adjusted to

match the real systems and phenomena. The last branch is to develop the new findings in

theory: the ‘inner resonance’ and the frequency-domain constants approach. They are

probably seen useful in NVH problems and beyond.

90

A Measurement results

Table A.1: Measurement results for pad specimen No. 1 with 1500Hzpf = , 0.13m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

23,6 1,003 0,225 92,13

22,3 1,006 0,231 144,93

26,7 1,002 0,234 147,88

2

42,9 0,996 0,226 102,9

36,1 1,003 0,22 125,84

42,2 0,995 0,23 159,25

3

73,9 0,982 0,226 120,5

68,8 0,989 0,218 116,28

76,3 0,98 0,226 168,09

4

240,7 0,937 0,22 203,52

186,3 0,961 0,215 148,99

254,1 0,937 0,223 203,38

5

317,3 0,926 0,223 305,7

297,7 0,94 0,216 211,07

343,6 0,924 0,227 276,76

6

464,8 0,905 0,208 338,02

510 0,912 0,218 348,47

436 0,901 0,218 394,03

Table A.2: Measurement results for pad specimen No. 1 with 1500Hzpf = , 0.28m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

25,7 0,998 0,233 30,6

23,4 0,999 0,225 100,3

26,7 1,002 0,228 111,92

2

41,7 0,992 0,234 40,26

40,1 0,994 0,221 70,13

44 0,993 0,225 110,41

3

80,7 0,981 0,237 69,21

75,8 0,991 0,223 71,13

72,8 0,984 0,223 110,38

4

259,1 0,933 0,227 188,97

200,8 0,962 0,221 133,47

279,9 0,94 0,222 160,52

5

368,6 0,926 0,227 263,82

377,4 0,926 0,214 178,97

422,6 0,923 0,22 180,89

6

640,3 0,912 0,213 211,17

560,4 0,912 0,214 259,02

530,6 0,911 0,218 274,82

91

Table A.3: Measurement results for pad specimen No. 1 with 1500Hzpf = , 0.44m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

26,3 1 0,229 38,72

24,9 1,004 0,222 113,49

24,9 0,998 0,218 143,71

2

42,4 0,996 0,232 27,36

39,9 1,004 0,217 82,51

40,4 0,996 0,219 140,4

3

85 0,986 0,237 50,9

71,7 0,992 0,219 80,43

72,2 0,99 0,215 119,02

4

287,7 0,947 0,227 174,48

208,1 0,953 0,214 147,48

237,7 0,959 0,22 80,79

5

420 0,942 0,228 213,32

421 0,932 0,211 152,02

445,8 0,94 0,225 133,98

6

684,4 0,923 0,21 144,29

637,5 0,918 0,212 185,75

574 0,922 0,221 230,82

Table A.4: Measurement results for pad specimen No. 1 with 1500Hzpf = , 0.60m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

26,3 0,998 0,231 66,87

24,2 0,996 0,21 141,54

23 1,001 0,21 341,95

2

41,7 0,997 0,229 70

41,3 1,007 0,21 154,29

36 0,996 0,204 287,68

3

78,1 0,989 0,231 89,62

68,4 0,994 0,205 143,81

65,7 0,993 0,201 188,82

4

330,4 0,957 0,222 149,51

186,7 0,97 0,209 93,24

203,6 0,975 0,208 63,25

5

343,5 0,952 0,23 245,56

405,2 0,941 0,213 106,81

371,9 0,961 0,227 163,19

6

687,8 0,937 0,209 127,67

678,6 0,921 0,216 130,72

512,5 0,939 0,221 243,44

92

Table A.5: Measurement results for pad specimen No. 1 with 1500Hzpf = , 1.26m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

27,6 1,002 0,226 149,46

28,1 1,003 0,23 182,61

28,2 1,006 0,216 118,33

2

43,5 0,995 0,23 143,12

41,6 1,001 0,228 166,41

42,5 1 0,215 125,02

3

77,2 0,993 0,23 181,78

73,3 0,992 0,228 179,42

71,9 0,996 0,216 130,98

4

177,3 0,984 0,216 131,16

169,5 0,979 0,215 144,15

182,3 0,981 0,213 120,21

5

286,6 0,976 0,217 93,35

260,4 0,974 0,215 83,19

268,4 0,973 0,211 82,09

6

491,6 0,967 0,225 247,55

479,8 0,96 0,23 270,34

710,6 0,948 0,223 128,87

Table A.6: Measurement results for pad specimen No. 1 with 1600Hzpf = , 0.13m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

36,8 0,994 0,208 259,06

34,5 1 0,203 163,56

34,9 1,014 0,197 150,75

2

67,4 0,991 0,208 209,41

62,4 1,002 0,202 137,62

66,8 1,003 0,2 133,83

3

143,1 0,974 0,211 297,93

180 0,971 0,209 191,22

183,9 0,974 0,202 137,58

4

434,7 0,929 0,195 275,93

304,1 0,946 0,217 460,76

470,7 0,922 0,204 266,03

5

532,5 0,918 0,195 243,7

378,7 0,946 0,207 484,72

542,8 0,913 0,205 315,77

6

602,5 0,91 0,199 221,01

545,1 0,915 0,197 434,82

562,2 0,922 0,213 434,19

93

Table A.7: Measurement results for pad specimen No. 1 with 1600Hzpf = , 0.28m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

30,7 0,997 0,199 296,66

41,4 0,982 0,208 191,75

41 0,996 0,206 166,57

2

55,4 0,996 0,198 262,3

83,3 1,011 0,21 182,47

83,4 0,993 0,211 154,02

3

121,1 0,984 0,201 206,78

209 0,976 0,218 232,83

210,8 0,97 0,224 258,01

4

408,5 0,944 0,208 267,64

310,3 0,958 0,216 443,1

288,5 0,962 0,218 490,68

5

525,1 0,932 0,204 179,57

372,7 0,945 0,202 527,49

386,5 0,947 0,201 492,9

6

588,6 0,928 0,205 135,05

604,7 0,922 0,198 306,1

599,5 0,928 0,199 311,75

Table A.8: Measurement results for pad specimen No. 1 with 1600Hzpf = , 0.44m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

39,5 0,999 0,209 236,76

41,8 0,999 0,205 179,11

43 1,002 0,203 80,8

2

79,4 0,994 0,211 222,03

75,1 0,992 0,208 167,07

100,8 0,994 0,211 81,34

3

220,6 0,976 0,217 211,52

233,6 0,971 0,217 211,43

222,7 0,975 0,223 203,92

4

341,4 0,963 0,214 406,49

320,1 0,957 0,214 397,96

312 0,963 0,217 398,77

5

419,2 0,955 0,205 399,13

436,9 0,942 0,204 347,46

410,9 0,947 0,208 391,98

6

584,9 0,94 0,206 162,66

636,1 0,927 0,206 205,09

630,3 0,933 0,21 215,66

94

Table A.9: Measurement results for pad specimen No. 1 with 1600Hzpf = , 0.60m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

68 0,998 0,184 97,36

39,4 1,002 0,213 504,3

44 1 0,204 117,62

2

178,1 0,984 0,197 128,16

66,9 0,999 0,2 270,33

82,8 0,997 0,208 95,37

3

493,1 0,955 0,212 -2,01

165,8 0,986 0,208 293,42

223,4 0,977 0,218 203,08

4

530,3 0,953 0,212 9,36

278,7 0,975 0,2 564,31

332,2 0,966 0,204 354,47

5

594,2 0,95 0,212 29,45

538,3 0,957 0,19 228,99

484 0,953 0,205 252,95

6

599,3 0,948 0,212 35,76

626,5 0,943 0,2 165,89

628,4 0,943 0,209 174,64

Table A.10: Measurement results for pad specimen No. 1 with 1600Hzpf = , 1.26m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

35,4 1,002 0,183 184,88

48,3 0,998 0,219 269,48

41,5 1,003 0,202 114,4

2

59,7 1,001 0,184 109,09

75 0,997 0,205 121,56

74,2 1,003 0,207 101,83

3

145,3 0,996 0,196 115,6

200,6 0,992 0,213 131,17

205,4 0,995 0,215 119,55

4

500,3 0,974 0,214 88,05

603,3 0,961 0,214 -56,73

482,2 0,966 0,223 210,13

5

577,7 0,972 0,213 73,31

621,1 0,96 0,219 -4,3

529,7 0,961 0,218 369,9

6

658 0,968 0,216 71,62

707,8 0,952 0,213 14,41

569,6 0,962 0,226 440,98

95

Table A.11: Measurement results for pad specimen No. 1 with 1700Hzpf = , 0.13m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

158,2 0,997 0,194 33,89

140,1 0,987 0,193 69,74

153,8 0,998 0,196 48,92

2

205,5 0,974 0,194 34,76

188,9 0,979 0,193 76,41

216,2 0,985 0,196 56,84

3

261,4 0,967 0,194 36,54

237,3 0,966 0,188 146,06

270,2 0,963 0,194 78,89

4

376,9 0,949 0,194 45,18

356,4 0,95 0,183 150,54

399,6 0,942 0,192 84,38

5

416,9 0,946 0,194 48,38

416,1 0,945 0,183 136,51

458,2 0,936 0,191 88,98

6

499,1 0,934 0,194 55,23

519,1 0,93 0,182 112,56

567,7 0,927 0,191 82,4

Table A.12: Measurement results for pad specimen No. 1 with 1700Hzpf = , 0.28m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

131,7 0,988 0,197 29,98

146,7 0,989 0,196 43,7

144,6 0,989 0,194 41,91

2

180,4 0,979 0,196 24,5

201 0,981 0,195 37,38

180,3 0,986 0,194 37,26

3

238,7 0,972 0,196 18,09

257,6 0,976 0,196 34,81

244,4 0,979 0,195 37,24

4

336,1 0,959 0,196 13,45

367,1 0,964 0,196 27,64

350,4 0,964 0,195 61,54

5

384,2 0,958 0,197 17,03

423,9 0,953 0,197 25,63

400,7 0,96 0,196 45,11

6

449,2 0,951 0,198 36,17

496,4 0,946 0,196 28,33

499,8 0,948 0,196 53,11

96

Table A.13: Measurement results for pad specimen No. 1 with 1700Hzpf = , 0.44m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

153 0,992 0,194 10,46

144,4 0,99 0,195 31,45

146,6 0,997 0,194 24,36

2

192 0,987 0,194 13,53

191 0,99 0,196 26,73

200,2 0,99 0,195 23,74

3

243,9 0,981 0,195 14,57

256,2 0,987 0,196 24,43

263,9 0,98 0,196 26,27

4

357,1 0,974 0,196 15,18

367,8 0,971 0,197 30,81

374,5 0,971 0,197 32,95

5

418,2 0,963 0,195 18,43

424,7 0,964 0,198 33,8

434,3 0,967 0,198 38,07

6

505,3 0,958 0,195 24,21

508,9 0,956 0,198 37

526,8 0,959 0,198 41,09

Table A.14: Measurement results for pad specimen No. 1 with 1700Hzpf = , 0.60m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

202,3 0,987 0,195 42,37

150,4 0,994 0,195 23,47

147 0,996 0,194 28,15

2

271,1 0,981 0,196 34,26

194 0,988 0,196 21,9

186,2 0,993 0,195 25,65

3

373,4 0,973 0,196 27,52

241,2 0,985 0,198 21,71

246,2 0,986 0,196 27,55

4

419,6 0,969 0,197 26,56

351,5 0,976 0,198 23,89

362,7 0,976 0,196 34,56

5

475,7 0,966 0,197 25,32

405 0,969 0,199 25,26

415,8 0,973 0,196 37,2

6

499,1 0,965 0,197 22,69

492,6 0,967 0,199 29,66

507,2 0,966 0,197 40,51

97

Table A.15: Measurement results for pad specimen No. 1 with 1700Hzpf = , 1.26m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

163,8 0,993 0,197 3,72

154,7 1 0,195 23,94

180,7 0,998 0,195 25,91

2

205,9 0,992 0,197 8,96

199,4 0,996 0,195 22,83

227,1 0,991 0,196 31,71

3

250,5 0,991 0,198 10,55

269,4 0,99 0,196 18,88

283,2 0,988 0,199 59,33

4

353,2 0,982 0,198 10,45

384,3 0,981 0,197 20,64

417,9 0,977 0,198 74,39

5

405,2 0,98 0,198 11,99

438,5 0,979 0,197 24,09

485,1 0,968 0,198 65,46

6

474,8 0,979 0,198 15,45

519,4 0,973 0,198 24,37

594,9 0,967 0,198 56,82

Table A.16: Measurement results for pad specimen No. 1 with 1800Hzpf = , 0.13m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

74,5 0,996 0,195 -9,1

86,6 0,997 0,195 2,68

80,4 1 0,194 5,17

2

105,1 0,997 0,196 -6,2

121,5 0,994 0,195 0,94

110,7 0,997 0,195 6,41

3

148,7 0,994 0,196 -2,19

170,4 0,987 0,196 2,14

160,1 0,991 0,196 7,67

4

238,7 0,983 0,197 8,97

269,5 0,979 0,196 10,25

262,6 0,981 0,196 14,54

5

299,2 0,977 0,197 13,75

321,2 0,971 0,196 13,9

304,6 0,979 0,196 17,52

6

362,3 0,971 0,197 20,97

400,8 0,962 0,196 20,74

385,5 0,971 0,196 24,34

98

Table A.17: Measurement results for pad specimen No. 1 with 1800Hzpf = , 0.28m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

73,8 0,998 0,195 -15,22

84,1 0,999 0,194 38,45

78,6 1 0,195 0,47

2

99,1 0,997 0,196 -10,51

122,6 0,994 0,194 32,47

110,3 0,997 0,196 -0,79

3

149 0,995 0,197 -6,91

167,8 0,993 0,195 27,3

162,6 0,997 0,196 1,32

4

251,2 0,986 0,197 0,5

271,1 0,986 0,196 22,79

261,3 0,982 0,197 4,83

5

294,5 0,982 0,197 3,74

320,8 0,98 0,196 20,71

307,5 0,976 0,197 7,83

6

368,9 0,976 0,197 7,42

397,8 0,974 0,197 20,91

387,4 0,975 0,197 12,43

Table A.18: Measurement results for pad specimen No. 1 with 1800Hzpf = , 0.44m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

75,2 0,998 0,196 -12,44

91,6 0,999 0,197 -6,94

80,2 1 0,194 -13,61

2

106,3 0,999 0,197 -9,04

129,5 0,996 0,197 -3,27

119,4 1 0,195 -17,52

3

150,3 0,995 0,197 -5,24

176,8 0,994 0,197 -2,69

165 0,994 0,196 -18,75

4

246,4 0,985 0,198 1,86

277 0,985 0,198 -0,74

263,4 0,985 0,197 -13,41

5

295,7 0,982 0,198 3,8

324,4 0,977 0,198 0,43

319,9 0,978 0,197 -10,68

6

371,2 0,98 0,198 5,94

407,8 0,976 0,198 3,67

398,9 0,973 0,198 -3,2

99

Table A.19: Measurement results for pad specimen No. 1 with 1800Hzpf = , 0.60m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

76,4 0,997 0,196 -12,19

81,5 1 0,197 -5,18

79,7 1,003 0,194 -45,28

2

105,9 0,996 0,197 -10,48

118,1 0,997 0,197 -3,45

112,5 0,998 0,196 -45,16

3

158 0,992 0,198 -7,62

168,2 0,996 0,198 -1,39

163,2 0,994 0,196 -38

4

246,4 0,988 0,198 -1,25

263,8 0,992 0,198 2,45

264,3 0,992 0,197 -27,21

5

307,2 0,985 0,198 1,11

314,5 0,99 0,198 4,11

316,5 0,982 0,198 -21,58

6

369,2 0,981 0,198 4,5

395,9 0,983 0,198 6,52

400,3 0,977 0,198 -14,92

Table A.20: Measurement results for pad specimen No. 1 with 1800Hzpf = , 1.26m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

79,2 0,999 0,197 -14,74

86,3 1,003 0,196 -16,5

95,1 0,999 0,194 46,52

2

111,5 0,998 0,198 -11,61

117,6 0,998 0,197 -16,95

133,7 0,998 0,194 64,34

3

162,4 0,998 0,199 -7,6

178,1 0,998 0,198 -13,05

188,7 0,996 0,193 69,18

4

259,4 0,995 0,199 -0,93

283,7 0,992 0,198 -8,14

301,5 0,984 0,193 62,42

5

300,4 0,993 0,199 1,14

335,7 0,989 0,198 -4,28

370,4 0,986 0,194 43,99

6

380 0,987 0,199 2,21

416,3 0,985 0,199 -1,92

453,7 0,973 0,195 34,95

100

Table A.21: Measurement results for pad specimen No. 2 with 1100Hzpf = , 0.13m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

14,4 0,997 0,212 100,08

12,5 1,003 0,208 37,51

11,9 1,005 0,216 47,02

2

25,8 1,002 0,212 95,67

21,5 0,998 0,208 30,52

20,7 0,999 0,218 55,05

3

45,8 0,999 0,212 81,43

40,3 0,999 0,209 20,89

40,4 1,002 0,219 53,26

4

109,7 1,004 0,212 45,62

101,6 1,002 0,21 -9,1

103,3 1,004 0,217 24,26

5

172,7 0,993 0,211 17,99

144,1 0,998 0,209 -35,22

142,6 1,004 0,215 4,04

6

249,2 0,976 0,208 -16,17

195,3 0,987 0,208 -77,55

219,7 0,997 0,211 -43,23

Table A.22: Measurement results for pad specimen No. 2 with 1100Hzpf = , 0.28m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

14,6 1,001 0,219 33,44

16,2 1 0,217 36,48

15,4 0,992 0,218 45,89

2

27,2 0,996 0,219 31,26

28,1 0,997 0,217 33,23

28,1 1,001 0,217 43,02

3

49,2 1,002 0,218 19,09

52,3 0,995 0,215 26,56

51,9 1,001 0,216 30,77

4

119,6 1,004 0,215 -28,74

130,5 1,004 0,21 -21,48

134 1,005 0,213 -24,54

5

160,1 1,006 0,212 -57,51

176,4 1,004 0,207 -60,36

170,3 0,998 0,21 -55,62

6

205,3 0,996 0,209 -114,42

217,1 0,987 0,206 -104,09

213,5 0,984 0,206 -95,5

101

Table A.23: Measurement results for pad specimen No. 2 with 1200Hzpf = , 0.13m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

36,3 1 0,261 52,68

28,5 1,003 0,247 32,34

29,1 1,007 0,254 29,17

2

66,2 0,999 0,26 45,19

66,2 1,006 0,253 22,45

68,2 1,004 0,259 26,68

3

107,1 0,988 0,256 50,43

129,8 0,994 0,254 12,99

118,8 0,994 0,258 21,69

4

194,7 0,935 0,247 61,09

216,5 0,965 0,252 26,16

212,4 0,955 0,254 40,36

5

221,3 0,905 0,246 71,87

244,9 0,947 0,251 41,93

244 0,935 0,252 55,08

6

265,5 0,866 0,242 83,44

293,9 0,922 0,248 80,07

293,8 0,905 0,248 78,7

Table A.24: Measurement results for pad specimen No. 2 with 1200Hzpf = , 0.28m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

45,2 1,004 0,26 4,76

41,6 1,009 0,261 21,34

45,1 1,014 0,261 2,89

2

74,5 1,008 0,263 21,76

60,8 1,004 0,265 37,18

70,2 1,005 0,264 14,15

3

117,8 1 0,263 23,76

106,5 1,002 0,265 32,57

105,3 1 0,265 17,02

4

203,5 0,966 0,26 18,57

197,3 0,973 0,258 16,91

196,6 0,969 0,259 6,39

5

226,2 0,951 0,26 25,15

232,3 0,956 0,255 13,27

227,7 0,955 0,256 6,86

6

288,9 0,898 0,245 38,47

295,3 0,923 0,248 11,24

292,4 0,914 0,244 17,65

102

Table A.25: Measurement results for pad specimen No. 2 with 1200Hzpf = , 0.44m/sdv =

Amplitude

level (three

tries)

Average

acceleration

amplitude [ 2m/s ]

Friction

coefficient

ratio

Identified

mass

[ kg ]

Identified tangential

damping coefficient

[ Ns/m ]

1

43,5 1 0,257 -10,69

46,4 1 0,254 -14,53

47,3 0,997 0,253 -16,55

2

66,1 0,996 0,259 -5,37

65,5 1,001 0,257 -2,51

64,5 1 0,257 -1,2

3

106,7 0,992 0,257 -1,15

108,9 0,993 0,257 2,65

109 0,993 0,257 1,92

4

194,7 0,97 0,251 -1,2

196,8 0,979 0,252 1,69

188 0,978 0,253 5,97

5

237 0,963 0,246 2,02

234,3 0,969 0,25 13,03

225 0,971 0,25 10,88

6

298,4 0,933 0,239 17,61

289,2 0,947 0,244 29,12

285,5 0,95 0,245 28,5

103

B Matrix form of Euler equations

B.1 Mathematical background

B.1.1 Kronecker product

Definition 1. The Kronecker product of two matrices [ ]r s ija =A and p qB is an rp sq

matrix defined by [69-71]

11 12 1

21 22 2

1 2

s

s

r r rs

a a a

a a a

a a a

=

B B B

B B BA B

B B B

. (12.1)

The following properties hold [69-71]

( ) ( ) = A B C A B C , (12.2)

( )T T T = A B A B , (12.3)

if B and C have the same size

( ) + = + A B C A B A C , (12.4)

( )+ = + B C A B A C A (12.5)

and if there exist matrix products AC and BD

( )( ) ( ) ( ) = A B C D AC BD . (12.6)

B.1.2 Partial derivative of a matrix with respect to a vector

Definition 2. The partial derivative of an r s matrix ( )A x which is a matrix function of a

1n vector x with respect to vector x is an r sn matrix given as [72], [73]

1 2( ) s =

aa aA x

x x x x (12.7)

where ia is the i -th column of matrix A

1 2 s=A a a a . (12.8)

Theorem 1. [72], [73]

( ) ( )

( ) ( )s

d

dt

= =

A x A xA x E x

x. (12.9)

Theorem 2. [72], [73]

( ( ) ( )) ( ) ( )

( ( ) ) ( )n

= +

A x B x A x B xB x E A x

x x x. (12.10)

Theorem 3.

1( ( ) ) ( )( ) ( )r s nd

dt

= +

J q q J qJ q q q q

q (12.11)

Proof.

104

1

1

( ( ) ) ( ( )) ( )( ) ( ) ( )( )

( ) ( )( ) ( ) ( ) ( ) ( )

n

n

d d

dt dt

= + = +

= + = +

J q q J q J qJ q q q J q q q E E q

q

J q J qJ q q qE E q J q q q q

q q

Theorem 4.

1 1( ) ( )( )p n p m m n = E x A d A E d x (12.12)

Proof.

1 1( ) ( )( ) ( ) ( ) ( ) ( )

( )( )

p p p n

n

= = =

=

E x Ad E x Ad E E Ad xE Ad E x

A E d x

B.1.3 Skew-symmetric matrix associated to cross product and its generalization

Definition 3. The skew-symmetric matrix corresponding to cross-product of the vector

1 2 3

Ta a a=a (12.13)

is defined as [60], [61]

3 2

3 1

2 1

0

0

0

a a

a a

a a

−

= − −

a . (12.14)

Definition 4. The block skew-symmetric matrix of the 3-row matrix

1

2

3

T

T

T

=

α

A α

α

(12.15)

is defined as

3 2

3 1

2 1

T T T

T T T

T T T

−

= − −

0 α α

A α 0 α

α α 0

. (12.16)

It is easy to prove that

3( )= Ax A E x . (12.17)

B.2 Matrix form of Euler equations

Consider a rigid body B moving in a fixed frame (0) . The body-fixed frame is denoted ( )b .

The direction cosine matrix of B is a three-by-three matrix ( )A q , in which 1nq is the vector

of independent general coordinates.

The angular velocity of B in fixed frame can be determined by [60], [61]

(0) (0)( ) ( )R=ω q J q q (12.18)

105

or in body-fixed frame

( ) ( )( ) ( )b b

R=ω q J q q (12.19)

where (.)

RJ are the rotational matrices, which satisfy

(0) ( )b

R R=J AJ . (12.20)

We have

( )

(0) ( ) ( ) ( ) ( )( )bb b b T bd

dt= = + = +

Aωω Aω Aω Aω AA Aω . (12.21)

Note that

(0) T=ω AA (12.22)

and

(0) ( )b=ω Aω , (12.23)

thus,

( ) ( ) ( )T b b b= =AA Aω ω ω 0 . (12.24)

Therefore,

(0) ( )b=ω Aω (12.25)

and

( ) (0)b T=ω A ω . (12.26)

The vector form of Euler equations reads [60]

. ( . )C C CI I l + = (12.27)

where CI is the moment of inertia tensor about the center of mass C of the body, is the

angular velocity of the body, Cl is total moment of the forces and torques acting on the body

about C .

Using matrix notations, one has [60], [61]

(0) (0) (0) (0) (0) (0)

C C C+ =I ω ω I ω l . (12.28)

Noting (12.26), we have

(0) (0) (0) (0) (0) (0)( )( ) ( )( )T T T T T

C C C+ =A A I A A ω A A ω A A I A A ω l

( ) ( ) ( ) ( ) ( ) (0)b b b b b

C C C+ =AI ω Aω I ω l (12.29)

Deriving (12.19), noting (12.11), yields

( )

( ) ( ) ( )( ) ( )

bb b R

R

= +

J qω J q q q q

q. (12.30)

Developing the second term of the left-hand side of (12.29) and considering (12.12), one gets

106

( ) ( ) ( ) ( ) ( ) ( )

3

( ) ( ) ( )

( )

(( ) )( )

b b b b b b

C R C R

b b b

R C R n

=

=

Aω I ω AJ E q I J q

AJ I J E q q. (12.31)

Substituting (12.30) and (12.31) into (12.29), one obtains

( )

( ) ( ) ( ) ( ) ( ) ( ) (0)(( ) ) ( )b

b b b b b bRC R C R C R n C

+ + =

JAI J q AI AJ I J E q q l

q. (12.32)

Equations (12.32) are a practical matrix form of Euler’s equations: all terms in the left hand

side depend explicitly on generalized coordinates, velocities and accelerations. They can also

be rewritten in a compact form as

* (0)( ) ( )( ) C+ =M q q C q q q l (12.33)

where

( ) ( )( ) b b

C R=M q AI J (12.34)

is the mass matrix and

( )

* ( ) ( ) ( ) ( )( ) (( ) )b

b b b bRC R C R n

= +

JC q AI AJ I J E

q (12.35)

is velocity-free Coriolis/Centripetal matrix.

107

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