Dynamics and Synthesis of Kinematic Chains with Impact and Clearance
Except where reference is made to the work of others, the work described in this
dissertation is my own or was done in collaboration with my advisory committee.
Eleonor D. Stoenescu
Certificate of Approval:
George T. Flowers
Professor
Department of Mechanical Engineering
Dan B. Marghitu, Chair
Professor
Department of Mechanical Engineering
David G. Beale
Professor
Department of Mechanical Engineering
Amnon J. Meir
Professor
Department of Mathematics
Stephen L. McFarland
Dean
Graduate School
Dynamics and Synthesis of Kinematic Chains with Impact and Clearance
Eleonor D. Stoenescu
A Dissertation
Submitted to
the Graduate Faculty of
Auburn University
in Partial Fulfillment of the
Requirements for the
Degree of
Doctor of Philosophy
Auburn, Alabama
May 13, 2005
Vita
Eleonor D. Stoenescu, son of Eleonor and Felicia Stoenescu, was born on January
15th, 1975, in Cisnadie, Romania. In 1993, he entered the University of Craiova, De-
partment of Mechanical Engineering, where he received his B.S. Diploma in 1998. He
received his M.S. Diploma from the same department in 1999. During 1999-2000 he
worked as an instructor at the Department of Electrical Engineering of the University of
Craiova. In August 2000 he joined the doctoral program of the Department of Mechanical
Engineering at Auburn University, Alabama.
iii
Dissertation Abstract
Dynamics and Synthesis of Kinematic Chains with Impact and Clearance
Eleonor D. Stoenescu
Doctor of Philosophy, May 13, 2005
(M.S., University of Craiova, Romania, 1999)
(B.S., University of Craiova, Romania, 1998)
162 Typed Pages
Directed by Dan B. Marghitu
In this dissertation, the dynamics and synthesis of open and closed kinematic chains
with frictional impact and joint clearance is studied.
First, the impact between rigid bodies with friction is investigated. A new model
of impact with friction is presented. The coefficient of rolling friction is defined and the
moment of rolling friction is introduced to the impact equations. The influence of the
moment of rolling friction and the geometrical characteristics of the links on the energy
dissipated by friction during the impact is analyzed.
Next, the effect of prismatic joint inertia on the dynamics of kinematic chains is
analyzed. The effect of the prismatic joint inertia on the position of the application
point of the joint contact forces is investigated. The influence of the joint inertia on
the dynamic response of a spatial robot arm with feedback control is analyzed. Also,
the influence of the joint inertia on the dynamic parameters of a planar mechanism is
exemplified using inverse dynamics.
Furthermore, a planar rigid-link mechanism with rotating prismatic joint and clear-
ance is modeled. The influence of the clearance gap size, crank speed, friction, and
iv
impact parameters on the dynamics of the system is analyzed. Nonlinear dynamics tools
are applied to analyze the data captured from the connecting rod of the mechanism.
Finally, a new structural synthesis of spatial mechanisms is developed based on the
system group classification. Spatial system groups of different families with one, two,
and three independent contours are presented. The advantage of the analysis of spatial
mechanisms based on the system group classification lies in its simplicity. The solution
of mechanisms can be obtained by composing the partial solutions of system groups.
For the previous models of impact with friction, the effect of the rolling friction was
neglected. In this dissertation, the moment of rolling friction is defined and introduced
to the impact equations. Prismatic joint inertia must be included for modeling high-
speed machine tools, manipulators, and robots. This problem is important, because in
some cases the moment of inertia of the prismatic joints is comparable to the moment
of inertia of the links and may significantly influence the dynamics of the system at high
speeds. Periodic motion is observed for the mechanism with rotating prismatic joint and
no clearance. The response of the mechanism with joint clearance is chaotic at relatively
high crank speeds. Also, a general method is presented in order to determine all the
configurations of complex spatial system groups and to automate the process.
v
Acknowledgments
The author would like to express his gratitude to his advisor Dr. Dan B. Marghitu,
Associate Professor of Mechanical Engineering, for his valuable guidance, support, pa-
tience, and encouragement towards the completion of this dissertation. The author also
wishes to acknowledge Dr. George T. Flowers, and Dr. David G. Beale, Professors of
Mechanical Engineering, and Dr. Amnon J. Meir, Professor of Mathematics, for their
helpful suggestions and serving as committee members.
ThefinancialsupportprovidedbytheDepartmentofMechanicalEngineering, Auburn
University, is gratefully appreciated.
The author would like to make a special acknowledgement to his parents, Eleonor
and Felicia Stoenescu, for their education, love, and constant support during this years.
vi
Style manual or journal used Journal of Sound and Vibration (together with the
style known as ?auphd?). Bibliography follows van Leunen?s A Handbook for Scholars.
Computer software used The document preparation package TEX (specifically
LATEX) together with the departmental style-file auphd.sty.
vii
Table of Contents
List of Tables x
List of Figures xi
1 Introduction 1
2 Impact with moment of rolling friction 3
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2.2 Rolling friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2.2.1 Pure rolling (no sliding) . . . . . . . . . . . . . . . . . . . . . . . . 8
2.2.2 Rolling with moment of friction . . . . . . . . . . . . . . . . . . . . 12
2.3 Impact with moment of rolling friction . . . . . . . . . . . . . . . . . . . . 13
2.3.1 Simple pendulum and energetic coefficient of restitution . . . . . . 13
2.3.2 Double pendulum and kinematic coefficient of restitution . . . . . 17
2.4 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.4.1 Simple pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.4.2 Double pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
2.5 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3 Effect of prismatic joint inertia 30
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
3.2 Mathematical background . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
3.2.1 Newton-Euler equations . . . . . . . . . . . . . . . . . . . . . . . . 35
3.2.2 Lagrange equations (unconstrained system) . . . . . . . . . . . . . 36
3.2.3 Lagrange equations (constrained system) . . . . . . . . . . . . . . 39
3.2.4 Kane equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
3.3 Kinematic chains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
3.3.1 Open kinematic chains . . . . . . . . . . . . . . . . . . . . . . . . . 43
3.3.2 Closed kinematic chains . . . . . . . . . . . . . . . . . . . . . . . . 48
3.4 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
3.4.1 Open kinematic chain . . . . . . . . . . . . . . . . . . . . . . . . . 51
3.4.2 Closed kinematic chain . . . . . . . . . . . . . . . . . . . . . . . . 53
3.5 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
4 Rigid body contact and impact 57
4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
4.2 R-RTR mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
viii
4.2.1 Newton-Euler?s method . . . . . . . . . . . . . . . . . . . . . . . . 65
4.2.2 Lagrange?s method . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
4.3 R-RTR mechanism with friction . . . . . . . . . . . . . . . . . . . . . . . . 71
4.3.1 Newton-Euler?s method . . . . . . . . . . . . . . . . . . . . . . . . 71
4.3.2 Kane?s method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
4.3.3 Kineto-static analysis . . . . . . . . . . . . . . . . . . . . . . . . . 76
4.4 R-RTR mechanism with friction and clearance . . . . . . . . . . . . . . . 79
4.4.1 Equations of motion . . . . . . . . . . . . . . . . . . . . . . . . . . 79
4.4.2 Simulation algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . 85
4.4.3 No contact . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
4.4.4 Contact on a single point . . . . . . . . . . . . . . . . . . . . . . . 89
4.4.5 Impact on a single point . . . . . . . . . . . . . . . . . . . . . . . . 92
4.4.6 Contact on two opposed points . . . . . . . . . . . . . . . . . . . . 95
4.4.7 Impact on two opposed points . . . . . . . . . . . . . . . . . . . . 97
4.4.8 Contact or impact on two points on the same side . . . . . . . . . 99
4.5 Working Model and Mathematica simulations . . . . . . . . . . . . . . . . 100
4.6 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
4.7 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
5 Structural synthesis of spatial mechanisms 108
5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
5.2 Degree of freedom and family . . . . . . . . . . . . . . . . . . . . . . . . . 110
5.3 Independent contours . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
5.4 Spatial system groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
5.4.1 System groups with one independent contour . . . . . . . . . . . . 119
5.4.2 System groups with two independent contours . . . . . . . . . . . 130
5.4.3 System groups with three independent contours . . . . . . . . . . . 133
5.5 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
6 Discussions and conclusions 142
Bibliography 144
ix
List of Tables
3.1 The maximum overshoot e2max and the settling time ts2 computed for
different values of IC3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
5.1 The number of configurations of system groups with one, two and three
independent contours (N = 1, 2, and 3) . . . . . . . . . . . . . . . . . . . 120
5.2 The configurations of system groups with one independent contour (N = 1)120
x
List of Figures
2.1 Homogeneous disk in motion on an inclined rough plane. . . . . . . . . . . 7
2.2 Impact of a rigid simple pendulum with a rough horizontal surface. . . . . 15
2.3 Impact of a rigid double pendulum with a rough horizontal surface. . . . . 19
2.4 The influence of the coefficient ?k and the angle ? on the ratio (?s/?a)/e?
for Mf = 0 and Mf negationslash= 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
2.5 The influence of the length L and the angle ? on the ratio (?s/?a)/e? for
Mf negationslash= 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.6 The influence of the radius r and the angle ? on the ratio (?s/?a)/e? for
Mf negationslash= 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
2.7 The influence of the coefficients s and e on the energy variation ?T. . . . 27
2.8 The influence of the coefficients s and ?k on the energy variation ?T. . . 29
3.1 a. Open kinematic chain with slider and friction; b. Force diagram for
the link 1; c. Force diagram for the link 2. . . . . . . . . . . . . . . . . . . 34
3.2 Three-link robot arm with prismatic joint. . . . . . . . . . . . . . . . . . . 45
3.3 Three-link mechanism with prismatic joint. . . . . . . . . . . . . . . . . . 49
3.4 The dynamic response of the co-ordinate q2 for the robot arm in the cases:
a. IC3 = 0, b. IC3 = 0.05 kg m2, c. IC3 = 0.1 kg m2, and d. IC3 = 0.15
kg m2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
3.5 The influence of the mass moment of inertia IC2 and angular speed ?1 on
the distance d. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
3.6 The influence of the mass moment of inertia IC2 and angular speed ?1 on
the motor torque Mm. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
4.1 Rigid body diagram for the R-RTR mechanism with rotating prismatic
joint. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
xi
4.2 The variation of the driver motor torque with respect to the angular speed
for the R-RTR mechanism. . . . . . . . . . . . . . . . . . . . . . . . . . . 66
4.3 Rigid body diagram for the R-RTR mechanism. . . . . . . . . . . . . . . . 67
4.4 Model of the R-RTR mechanism with rotating prismatic joint and clearance. 81
4.5 Geometry of the slider joint with clearance for: a. no contact; b. contact
or impact on a single point; c. contact or impact on two points on the
same side; d. contact or impact on two opposed points. . . . . . . . . . . 82
4.6 Rigid body diagram for the R-RTR mechanism with prismatic joint and
clearance in the case of no contact. . . . . . . . . . . . . . . . . . . . . . . 87
4.7 Geometry of the mechanism for: a. contact or impact on a single point;
b. contact or impact on two opposed points; c. contact or impact on two
points on the same side. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
4.8 Working Model simulation of the driver motor torque for the double pen-
dulum. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
4.9 Comparison of simulation results obtained using Mathematica and Work-
ing Model. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
4.10 Trajectory of the vertical coordinate yG2 in the state space for: a. zero
clearance (c = 0 mm); b. nonzero clearance (c = 1 mm). . . . . . . . . . . 105
4.11 Largest Lyapunov exponent computed for a set of values of the nominal
angular velocity ?10 and for the clearances: a. c = 0.5 mm; b. c = 1 mm;
c. c = 1.5 mm. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
5.1 Types of joints: a. Slider joint (class 5); b. Pin joint (class 5); c. Cylin-
drical joint (class 4); d. Spherical joint (class 3). . . . . . . . . . . . . . . 112
5.2 The cartesian spatial reference frame xOyz. . . . . . . . . . . . . . . . . . 113
5.3 Planar kinematic chains. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
5.4 Spatial kinematic chains. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
5.5 System groups with one independent contour (N = 1) of the family f = 0. 121
5.6 Spatial mechanism with one independent contour and a system group of
the family f = 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
xii
5.7 System groups with one independent contour (N = 1) of the family f = 1. 124
5.8 System groups with one independent contour (N = 1) of the family f = 2. 126
5.9 Spatial mechanism with one independent contour and a system group of
the family f = 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
5.10 System groups with one independent contour (N = 1) of the family f = 3. 128
5.11 System groups with one independent contour (N = 1) of the family f = 4. 129
5.12 System group with two independent contours (N = 2) of the family f = 1. 131
5.13 System groups with two independent contours (N = 2) of the family f = 2.132
5.14 System groups with two independent contours (N = 2) of the family f = 3.134
5.15 System group with two independent contours (N = 2) of the family f = 4. 135
5.16 Spatial mechanism with two independent contours and a system group of
the family f = 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
5.17 System group with three independent contours (N = 3) of the family f = 2.137
5.18 System group with three independent contours (N = 3) of the family f = 3.139
5.19 System group with three independent contours (N = 3) of the family f = 4.140
5.20 Spatial mechanism with three independent contours and a system group
of the family f = 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
xiii
Chapter 1
Introduction
Consideration of dynamic modeling is an important part in the analysis, design and
control of mechanical systems such as mechanisms, manipulators, robots, etc. In general,
mechanical systems have several desirable features relative to the coupling contact forces
such as higher speed, improved mobility and stability, and reduced power consumption.
The dynamics of mechanical systems with frictional contacts and impacts has been de-
veloped and applied to many industrial applications. One of the important factors that
influence the dynamic stability and the performance of machines is joint clearance. In
the last years, many researchers have been studied the effects of the clearance on the
motion of mechanical systems.
In this dissertation the nonlinear dynamics of a mechanism with rotating prismatic
joint and clearance is investigated. Nonlinear dynamics tools were applied in order to
study the behavior of the mechanism. Frictional contacts and impacts and prismatic
joint inertia have been considered to model the mechanism.
In Chapter 2 a new model of rigid body impact with friction is presented. The
coefficient of rolling friction and the moment of rolling friction are introduced.
In Chapter 3 the influence of the prismatic joint inertia on the position of the
application point of the joint reaction forces and the effect on the dynamics and control
of open and closed kinematic chains is analyzed.
1
2
In Chapter 4 the nonlinear dynamics of a planar, rigid-link mechanism with pris-
matic joint clearance is investigated. The influence of the clearance gap size, crank speed,
friction and impact parameters on the nonlinear behavior of the system are analyzed.
In Chapter 5 a new structural synthesis of spatial mechanisms is studied based on
the system group classification. Structural synthesis of mechanisms with the specified
number of contours and joint types is necessary in order to systematize the creative
design process.
Finally, in Chapter 6, general conclusions are recorded. For mechanical systems
with no clearance, the motion is periodic. Chaotic motion is observed for mechanical
systems with joint clearance.
Chapter 2
Impact with moment of rolling friction
The impact between rigid bodies and rough surfaces is studied. The rolling friction
moment and the coefficient of rolling friction are introduced, and an improved math-
ematical model of the planar impact with friction is presented. The influence of the
moment of rolling friction on the energy dissipated by friction during the impact is ana-
lyzed. For a simple pendulum, using the energetic coefficient of restitution, more energy
is dissipated for larger values of the coefficient of kinetic friction and contact radius, and
for smaller values of the length of the beam. For a double pendulum, using the kinematic
coefficient of restitution, in some cases one can obtain energetically inconsistent results.
If the moment of rolling friction is introduced, this problem can be solved for some values
of the coefficient of rolling friction.
2.1 Introduction
Newton [1] defines the coefficient of restitution e as a kinematic quantity that is used
to derive a relation between the normal impact velocities of approach and separation at
the contact point. Poisson [2] divides the collision period in two phases, compression
and restitution. Poisson defines e as a kinetic quantity that relates the normal impulses
at the contact point that occur during each phase. Routh [3] presented a graphical
method based on Poisson?s hypothesis [2] to treat collision problems. Later Whittaker [4]
expanded Newton?s method [1] considering the frictional impulse. Routh and Whittaker
presented different approaches in the treatment of motion on the tangential direction at
3
4
the point of contact. Routh solves for the slipping velocity during collision and introduces
the possibility of changes in slipping direction during contact. Considering Whittaker?s
method, slipping occurs when the ratio of the normal and tangential impulses are greater
that the coefficient of friction ?.
Kane and Levinson [5] observed that the classical solution of rigid body impact
problems using Newton?s theory produces energetically inconsistent results. Keller [6]
attributed this paradoxical behavior to slip reversals during collision subject to frictional
effects. The Newtonian approach ignores the changes in the direction of slip, leading to
the overestimation of the rebound velocity as a result of impact. Keller introduced a
revised formulation of rigid body collision equations based on Poisson?s hypothesis such
that impact never increases energy. Stronge [7, 8] divided the energy that is dissipated
during collision into two portions: dissipation due to frictional impulse and dissipation
due to normal impulse. He solved the impact with friction problem using an energetic
coefficient of restitution.
Brach [9, 10] has proposed a solution scheme based on revising Whittaker?s method
in order to avoid energy increases from resulting solutions. The approach treats the
tangential impulse as a constant fraction ? of the normal impulse. Then energy loss is
examined to determine the appropriate value of ? that can be used in the actual solution.
He has expanded his approach, to treat contacts that take place over finite areas and
introduced a moment coefficient em to solve the collision problem. The same author [11]
has introduced a moment due to peeling at the trailing edge of the contact surface during
rolling in the equations of a planar impact of a sphere in the presence of adhesion.
5
Marghitu [12] presented some impact friction versus impact angle plots for slender
steel beams with semi-spherical ends impacting the hard surface of a massive concrete
object. Similar impact friction plots were reported by Stoianovici and Hurmuzlu [13] for
a slender beam impacting a half-space. The classical coefficient of restitution is found to
depend strongly on the orientation of the bar, and the impacts are divided into a series
of micro-impacts. Calsamiglia et al. [14] observed that the coefficient of friction for
disks impacting a massive surface is found to depend on the inclination angle. Osakue
and Rogers [15] presented an experimental study of friction during planar low-speed
oblique impacts. Johansson and Kalbring [16] developed a numerical algorithm where
the impenetrability condition and Coulomb?s law of friction were formulated as equations
in terms of velocities and impulses rather than displacements and forces.
Pfeiffer and Glocker [17] presented theoretical and applied aspects of the dynamics
of multiple unilateral contacts in multibody mechanical systems. Kinetic restitution
was considered for the normal direction as well as the tangential restitution effects.
The problem of rigid body collision with multiple contact points was also studied by
Marghitu and Hurmuzlu [18]. Hurmuzlu [19] introduced a new method to solve collision
problems of slender bars with massive external surfaces on a revised energetic coefficient
of restitution that resolves the effect of impact induced vibrations on the post-collision
velocities of the impacting bars. The impulse-based rigid-body as well as the alternative
compliance-based approaches have failed to produce valid solutions to the problem of
predicting the post-impact velocities in multi-impact systems. Ceanga and Hurmuzlu
[20] considered the impulse-momentum-based rigid-body approach and solved the non-
uniqueness difficulty by introducing a new constant called the impulse transmission ratio.
6
In the new algebraic rigid-body collision model presented in Chatterjee and Ruina
[21], it is possible to predict partially sliding disk collisions for suitable choices of the tan-
gential coefficient of restitution. Lankarani [22] presented a general formulation for the
analysis of impact problems with friction in both open- and closed-loop multibody me-
chanical systems. Newton?s coefficient of friction and Poisson?s coefficient of restitution
were used as known quantities.
For the previous models of impact with friction, the effect of the rolling friction was
neglected. In this paper, the moment of rolling friction is defined and introduced to
the impact equations. For the simple pendulum, the energetic coefficient of restitution
and the coefficient of rolling friction are used to model the impact. The influence of
the coefficient of kinetic friction and the geometrical characteristics (the impact angle,
the length and the contact radius of the beam) on the energy dissipated by friction
during impact is analyzed. For the double pendulum, using the kinematic coefficient
of restitution, in some cases one can obtain energetically inconsistent results. If the
moment of rolling friction is introduced, this problem can be solved for some values of
the coefficient of rolling friction.
2.2 Rolling friction
A homogeneous circular disk in motion on an inclined plane is shown in Fig. 2.1.
The fixed cartesian reference frame xOyz is chosen with the origin at O. The angle
between the axis Ox and the horizontal is ?. The contact point between the disk and
the plane is B. The disk has the mass m, the radius r, and the center of mass at C. The
gravitational acceleration is g.
7
Figure 2.1: Homogeneousdiskinmotiononaninclined roughplane.
8
2.2.1 Pure rolling (no sliding)
The forces that act on the disk are the gravitational force G = ?mg? at the point
C, the normal reaction force N of the plane and the friction force Ff at the contact point
B. The rolling friction is considered negligible. The position vector rC of the center of
mass C of the disk is
rC = xC?+r?. (2.1)
The velocity vector vC of the center of mass C of the disk is
vC = ?rC = ?xC?+ ?r? = ?xC?. (2.2)
Denoting ?xC = v, the velocity of the center of mass C becomes
vC = v?. (2.3)
Thus, the acceleration vector of aC of the center of mass C of the disk is
aC = ?vC = ?v?. (2.4)
One can express the velocity vB of the contact point B as
vB = vC +??CB = v?+(??k)?(?r?) = (v ?r?)?. (2.5)
9
In order to find the equation of motion for the disk, one can write the Newton?s equation
maC = summationdisplayF. (2.6)
The sum of the external forces can be written as
summationdisplayF = mg+F
f +N = (mgsin??Ff)?+(N ?mgcos?)?, (2.7)
where Ff = ?Ff? is the friction force, and N = N? is the reaction force of the plane on
the disk. From Eqs. (2.4), (2.6), and (2.7) one can write the following equations
m?v = mgsin??Ff, (2.8)
N = mgcos?. (2.9)
The following moment equation can be written for the disk with respect to its center of
mass C
IC? = summationdisplayMC, (2.10)
where IC is the mass moment of inertia with respect to the point C, ? = ?? = ???k is
the angular acceleration, and ? = ??k is the angular velocity of the disk. The sum of
the external moments can be written as
summationdisplayM
C = CB?F = (?r?)?(?F?) = ?rFfk. (2.11)
10
From Eqs. (2.10) and (2.11) one can write the following equation
IC ?? = rFf. (2.12)
For no sliding, the velocity vB is zero (vB = 0). Thus, from Eq. (2.5) one can write
? = v/r and Eq. (2.12) becomes
IC ?vr2 = Ff. (2.13)
From Eqs. (2.8) and (2.13) the following equation of motion can be derived
parenleftbigg
m+ ICr2
parenrightbigg
?v = mgsin?. (2.14)
A homogeneous disk is considered in our case and the mass moment of inertia with
respect to its center of mass is IC = mr
2
2 . Thus, Eq. (2.14) can be written as
?v = 23gsin?. (2.15)
Condition for pure rolling
From Eq. (2.8) and Eq. (2.15) one can compute the friction force Ff as
Ff = m3 gsin?. (2.16)
The condition for the disk of rolling without sliding on the plane is
Ff ? ?kN, (2.17)
11
where ?k is the coefficient of kinetic friction. From Eq. (2.9), Eq. (2.16), and Eq. (2.17)
one can obtain
tan? ? 3?k, (2.18)
or
? ? ?, (2.19)
where the sliding friction angle ? can be determined from the equation
tan? = 3?k, where ?k = tan?. (2.20)
Eq. (2.19) represents the condition for rolling without sliding of the disk on the plane.
If the angle ? of the plane is smaller than the sliding friction angle ?, the disk rolls on
the plane without sliding. If the angle ? of the plane is greater than the sliding friction
angle ?, the disk rolls and slides on the plane simultaneously.
Moment of rolling friction
Experimentally one can observe that if the angle ? of the plane is small enough, the disk
does not move. The equilibrium conditions for the disk are v = 0, and ? = 0. The
rolling is stopped by a rolling resistant moment Mf that balances the active moment
rFf
Mf = rFf. (2.21)
The acceleration ?v is zero and from Eq. (2.8) one can express the friction force as Ff =
Gsin?. Thus, one can write
Mf = rmgsin? = rN tan?. (2.22)
12
If ?0 is the value of the angle ? when the rolling starts, the moment Mf is called the
rolling friction moment and has the value
Mf = rN tan?0. (2.23)
The constant rtan?0 is denoted by s and represents the coefficient of rolling friction
s = rtan?0. (2.24)
The rolling friction moment Mf become
Mf = sN. (2.25)
The rolling friction moment Mf is proportional to the normal reaction N and has the
expression
Mf = ?sN ?|?|. (2.26)
2.2.2 Rolling with moment of friction
In this case, Eq. (2.10) becomes
I ?? = rFf ?sN. (2.27)
From Eq. (2.8) and Eq. (2.27) one can write
?v = 23(sin??cos?tan?0)g = 2sin(???0)3cos?
0
g. (2.28)
13
In this case, the rolling condition is
sin(???0)
cos?0 ? 3
sin(???0)
cos? , (2.29)
or
tan? ? tan?+2(tan??tan?0). (2.30)
Eq. (2.30) can also be written as
? ? ?, (2.31)
where the angle ? can be obtained from
tan? = tan?+2(tan??tan?0). (2.32)
For the motion of the homogeneous disk on the plane of slope ?, the following three
cases are possible:
Case 1. ? < ?0 (Eq. (2.28)). The disk has no motion.
Case 2. ?0 ? ? < ? (Eq. (2.32)). The disk has pure rolling (no sliding) motion.
Case 3. ? ? ?. The disk has rolling and sliding motion simultaneously .
2.3 Impact with moment of rolling friction
2.3.1 Simple pendulum and energetic coefficient of restitution
The planar rigid pendulum with mass M and length L pivots around the frictionless
pin joint O, and the tip impacts an inelastic horizontal surface S at point C (Fig. 2.2).
The fixed cartesian reference frame xOyz is chosen. The inclination of the pendulum
14
with respect to the vertical axis Oy is the angle ?. At the impact point C, the coefficients
of kinetic and static friction are ?k and ?s, the coefficient of rolling friction is s, and
the energetic coefficient of restitution is e?. The ratio ?s/?a of the separation angular
speed ?s = ?(ts) and the approach angular speed ?s = ?(ta) at impact is calculated.
The kinetic energy of the pendulum is
T = 12I???, (2.33)
where I is the mass moment of inertia with respect to the joint O and ? = ?k is the
angular velocity of the pendulum.
The position of the contact point C relative to the center of axis O can be expressed as
rC = ?x??y?, (2.34)
where x = Lsin? and y = Lsin?. Thus, the velocity of the point C is
vC = ??rC. (2.35)
The differential of the impulse dp = dpt? + dpn? at the contact point C satisfies the
Amontons-Coulomb law
dpt = ?kdpn, vC ?? < 0,
??kdpn < dpt < ?kdpn, vC ?? = 0,
dpt = ??kdpn, vC ?? > 0,
(2.36)
15
Figure 2.2: Impact of arigid simple pendulum with arough horizontal surface.
16
where pt is the tangential impulse and pn is the normal impulse to the surface S at the
point C.
The impact equation for the pendulum can be written as
parenleftbigg?T
??
parenrightbigg
ts
?
parenleftbigg?T
??
parenrightbigg
ta
= ?vC?? ?(pt?+pn?)+ ???? ?Mf, (2.37)
where ta and ts are the approach and separation moments at the impact and Mf =
?spn ?|?| is the rolling friction moment at the point C.
The slip reverses in the direction of the compression impulse pc simultaneously with the
transition from compression to restitution. From Eq. (2.53) one can obtain the angular
velocity ? as a function of the normal impulse pn = p
?(p) =
??
??
???
?a ? x+?ky +sI p, 0 < p ? pc,
?x??ky ?sI (p?pc), pc < p < ps,
(2.38)
where ps is the separation impulse.
In order to have slip reversal, the following condition must be satisfied ?s < tan?.
Otherwise the pendulum sticks after compression (?s = 0).
The compression impulse pc can be calculated from ??vC(pc) = 0 as
pc = I?ax+?
ky +s
. (2.39)
The work of the normal impulse during compression Wn(pc) is
Wn(pc) =
integraldisplay pc
0
??vC(p)dp = ?x(x+?ky +s)2I p2c. (2.40)
17
The work of the normal impulse during restitution Wn(ps)?Wn(pc) is
Wn(ps)?Wn(pc) =
integraldisplay pf
pc
??vC(p)dp = x(x??ky ?s)2I (ps ?pc)2 . (2.41)
The energetic coefficient of restitution e? (Stronge [7]) can be written as
e2? = ?Wn(ps)?Wn(pc)W
n(pc)
= x??ky ?sx+?
ky +s
parenleftbiggp
s
pc ?1
parenrightbigg2
. (2.42)
From Eq. (2.42) one can compute the separation impulse ps as
ps = pc
parenleftBigg
1+e?
radicalBigg
x+?ky +s
x??ky ?s
parenrightBigg
. (2.43)
The ratio of the angular velocities of separation ?(ps) and approach ?(0) becomes
?s
?a =
?(ps)
?(0) = e?
radicalBigg
x??ky ?s
x+?ky +s. (2.44)
For no rolling friction moment (s = 0) the ratio ?s?
a
becomes
?s
?a = e?
radicalBigg
1??k cot?
1+?k cot?. (2.45)
2.3.2 Double pendulum and kinematic coefficient of restitution
In Fig. 2.3, two uniform rigid rods 1 and 2 with lengths L1 and L2 and masses
m1 and m2 are joined at point B by a frictionless pin joint in order to form a planar
double pendulum. The end of the rod 1 pivots around a frictionless pin joint at O. The
free end of the rod 2 strikes a rough horizontal surface S at the point C. The rods
18
have the angles of inclination from axis Oy denoted by ?1 and ?2 and angular speeds
of magnitudes ?1 = ??1 and ?2 = ??2, respectively. The fixed cartesian reference frame
xOyz is chosen. At the impact point C, the coefficients of kinetic and static friction
are ?k and ?s, the coefficient of rolling friction is s, and the kinematic coefficient of
restitution is e. The separation angular speeds ?s1 and ?s2 are calculated with respect
to the approach angular speeds ?a1 and ?a2 and the initial angles ?10 and ?20. Also, the
difference ?T = Ts ?Ta of the kinetic energies of separation and approach Ts and Ta at
the impact is computed. The positions of the center of masses C1 and C2 of the rods 1
and 2 are
rC1 = ?L12 sin?1?? L12 cos?1?, (2.46)
rC2 = ?(L1 sin?1 + L22 sin?2)??(L1 cos?1 + L22 cos?2)?. (2.47)
The velocities of the center of masses C1 and C2 are
vC1 = ?rC1, and vC2 = ?rC2. (2.48)
The position of the contact point C is
rC2 = ?(L1 sin?1 + L22 sin?2)??(L1 cos?1 + L22 cos?2)?. (2.49)
The velocity of the contact point C is vC = ?rC.
The kinetic energies T1 and T2 of the rods 1 and 2 are
Ti = 12(mivCi ?vCi +ICi?i ??i), i = 1,2, (2.50)
19
Figure 2.3: Impact of arigid doublependulum with arough horizontal surface.
20
where ICi is the mass moment of inertia with respect to the point Ci and ?i = ?ik is
the angular velocity of the rod i, for i = 1,2.
The kinetic energy T of the double pendulum is
T = T1 +T2. (2.51)
The differential of the impulse dp = dpt? + dpn? at the contact point C satisfies the
Amontons-Coulomb law
dpt = ?kdpn, vC ?? < 0,
??kdpn < dpt < ?kdpn, vC ?? = 0,
dpt = ??kdpn, vC ?? > 0,
(2.52)
where pt is the tangential impulse and pn is the normal impulse to the surface S at the
point C.
The impact equation for the rod i can be written as
parenleftbigg?T
??i
parenrightbigg
ts
?
parenleftbigg?T
??i
parenrightbigg
ta
= ?vC??
i
?(pt?+pn?)+ ??i??
i
?Mf, i = 1,2, (2.53)
where ta and ts are the approach and separation moments at the impact and Mf =
?spn ?2|?
2|
is the rolling friction moment at the point C.
The velocities of approach and separation vCa and vCs of the point C can be expressed
as
vCa = vC(ta), and vCs = vC(ts). (2.54)
21
In order to have slip reversal, the following condition must be satisfied ?s <
vextendsinglevextendsingle
vextendsinglevextendsinglept
pn
vextendsinglevextendsingle
vextendsinglevextendsingle, other-
wise the rod 2 sticks after compression (vCs ?? = 0).
Using the kinematic coefficient of restitution e (Newton [1]) the following equation can
be written
e = ?vCs ??v
Ca ??
. (2.55)
From Eqs. (2.53) and (4.107) one can compute the angular velocities of separation ?1s
and ?2s for the rods 1 and 2.
The kinetic energy dissipated by friction ?T is
?T = T(ts)?T(ta), (2.56)
where T(ta) and T(ts) are the kinetic energies before and after the impact for the double
pendulum.
2.4 Results
2.4.1 Simple pendulum
In this section, results from computer simulations are presented. The rigidpendulum
impacting a rough horizontal surface is shown in Fig. 2.2. The energetic coefficient of
restitution is e? = 0.3. Figures 2.4, 2.5, and 2.6 illustrate the ratio of the separation
angular speed ?s, the approach angular speed ?a, and the coefficient e? as a function of
the angle ? at the impact. The effect of the energy dissipated by friction at the impact
between the pendulum and the horizontal surface S is shown. At small values of the
angle ? the contact sticks (?s = 0), if the coefficient of static friction is sufficiently large
22
(?s ? tan?). Also, for small angles ? the work done by the friction force Ft is large in
comparison with the work done by the normal contact force Fn.
In Fig. 2.4 the pendulum with the length L = 0.2 m and the contact radius r =
0.002 m is considered. The ratio (?s/?a)/e?, that characterizes the energy dissipated
by friction at the impact, is plotted as function of the angle ? using different values of
the coefficient of kinetic friction ?k. The results for no friction moment (Mf = 0) are
represented with continuous line and the results for nonzero friction moment (Mf negationslash= 0)
are represented with dotted line. For larger values of ?k, larger differences between the
ratios considering zero and nonzero friction moment Mf are observed. For example, for
? = 1.044 rad and ?k = 0.3 the difference of ratios is dI = 0.010, and for ? = 1.044 rad
and ?k = 0.9 the difference of ratios is dIV = 0.025.
In Fig. 2.5 the length L of the pendulum is modified while all the other parameters
arekeptconstant. LessenergyisdissipatedbyfrictionforlargervaluesofL. Forexample,
for r = 0.002 m, ?k = 0.3 = constant, and ? = 1.044 rad, the ratio (?s/?a)/e? = 0.784
corresponds to L = 0.040 m and the ratio (?s/?a)/e? = 0.801 corresponds to L = 0.060
m.
A relation can also be established between the contact radius r of the pendulum and
the energy dissipated at impact (Fig. 2.6). Less energy is dissipated by friction for smaller
values of r. For example, for L = 0.200 m, ?k = 0.5 = constant, and ? = 1.044 rad, the
ratio (?s/?a)/e? = 0.729 corresponds to r = 0.001 m and the ratio (?s/?a)/e? = 0.706
corresponds to r = 0.004 m.
23
Figure 2.4: Thein uenceofthe coe?cient ?
k
andthe angle  on theratio (!
s
=!
a
)=e
?
for M
f
=0and M
f
6=0.
24
Figure 2.5: Thein uenceofthe length L andthe angle  on theratio (!
s
=!
a
)=e
?
for
M
f
6=0.
25
Figure 2.6: Thein uenceofthe radius r andthe angle  on theratio (!
s
=!
a
)=e
?
for
M
f
6=0.
26
2.4.2 Double pendulum
The rigid double pendulum impacting a rough horizontal surface is shown in Fig. 2.3.
For the rods 1 and 2, the following data are given: the masses m1 = m2 = 3 kg, the
lengths L1 = L2 = 2 m, the mass moments of inertia with respect to the axis Oz are
IC1 = IC2 = 1 kg m2, the inclination angles with respect to the axis Oy are ?1 = 20?
and ?1 = 30?, and the magnitudes of the angular speeds are ?1 = 1 rad/s and ?2 = 2
rad/s.
Figure 2.7 illustrates the energy variation ?T as function of the coefficient e, for different
values of the coefficient s and constant coefficient of kinetic friction ?k = 0.3 = constant.
For s = 0 (no moment of rolling friction) and 0.45 < e < 0.65, the energy variation ?T
is positive and a paradoxical increase of energy is observed. Using a moment of rolling
friction, negative energy variation ?T is obtained and energetically consistent results
for s > 0.16 and 0.45 < e < 0.65. For example, for e = 0.55 and s = 0, it results
?T = 2.455 J, and for e = 0.55 and s = 0.16, it results ?T = ?1.783 J. For s = 0.08,
the energy variation ?T is negative for 0.45 < e < 0.55 and positive for 0.55 < e < 0.65.
For example, for s = 0.08 and e = 0.5, it results ?T = ?0.779 J, and for s = 0.08 and
e = 0.6, it results ?T = 0.926 J.
Figure 2.8 illustrates the energy variation ?T as function of the coefficient ?k,
for different values of the coefficient s and constant kinematic coefficient of restitution
e = 0.5 = constant. For s = 0 (no moment of rolling friction) and 0.3 < ?k < 0.4, the
energy variation ?T is positive and a paradoxical increase of energy is observed. Using
a moment of rolling friction, negative energy variation ?T is obtained and energetically
consistent results for s > 0.25 and 0.3 < ?k < 0.4. For example, for ?k = 0.35 and s = 0,
27
Figure 2.7: Thein uenceofthe coe?cients s and e on theenergyvariation ?T.
28
it results ?T = 1.538 J, and for ?k = 0.35 and s = 0.25, it results ?T = ?2.354 J.
For s = 0.15, the energy variation ?T is negative for 0.3 < ?k < 0.35 and positive for
0.35 < ?k < 0.4. For example, for s = 0.15 and ?k = 0.3, it results ?T = ?2.344 J, and
for s = 0.15 and ?k = 0.4, it results ?T = 3.037 J.
Using the kinematic coefficient of restitution e to model the impact, for no rolling
friction moment (Mf = 0), the energy variation ?T is positive in the cases considered
above and paradoxical results are observed (Kane and Levinson [5]). Using a rolling
friction moment (Mf negationslash= 0), the energy variation ?T becomes negative for sufficiently
large values of s and energetically consistent results are obtained.
2.5 Conclusions
The results show the influence of the moment Mf on the energy dissipated by friction
during the impact for different values of the parameters ?, s, ?k, e, L, and r. More
energy is dissipated during impact for larger values of s. For the simple pendulum, using
the rolling friction moment, the ratios of the separation angular speed, approach angular
speed, and the energetic coefficient of restitution are compared for a constant value of
?k and different values of the length L and the radius r of the beam. For the double
pendulum, when the kinematic coefficient of restitution e is used to model the impact,
an energy increase is observed in some cases. One can partially solve this problem and
obtain energetically consistent results introducing the moment of rolling friction Mf to
the impact equations. In order to validate the analytical results, experimental data are
needed.
29
Figure 2.8: Thein uenceofthe coe?cients s and ?
k
on theenergyvariation ?T.
Chapter 3
Effect of prismatic joint inertia
The effect of prismatic joint inertia on dynamics of kinematic chains with friction
is investigated. The mathematical model of a planar kinematic chain consisting of a
prismatic joint sliding along a link that is connected to a revolute joint is developed.
The influence of the slider inertia on the position of the application point of the joint
forces is analyzed. The effect of the slider link inertia on the dynamic response of a spatial
robot arm with feedback control is analyzed using Kane?s formulation. Larger values of
the initial condition response characteristics are observed for larger values of the slider
link inertia. Also, the effect of the prismatic joint inertia on the dynamic parameters of
a planar mechanism is exemplified using inverse dynamics based on the Newton-Euler?s
method. Numerical results are obtained and compared for zero and larger values of the
prismatic joint inertia at different speeds. The numerical simulations reveal that the
effect of slider inertia may be negligible at low speeds, but becomes significant at high
speeds.
3.1 Introduction
The center of mass simplifies the translational motion of the body, but it gives no
information about the distribution of the mass on the body. The mass of the body rep-
resents the amount of matter contained in the body and the resistance of the body to
translational motion. The quantity that is dependent on how the mass is distributed and
30
31
describes the resistance of the body to rotation is the mass moment of inertia. Consid-
eration of dynamic modeling is an important part in the analysis, design and control of
mechanical systems such as mechanisms, robots, manipulators, etc. In general, mechan-
ical systems have several desirable features relative to the coupling contact forces such
as higher speed, improved mobility and control, and reduced power consumption. The
dynamics of mechanical systems with frictional contacts has been developed and applied
to many industrial applications. Examples in this area include fingered grippers [23] and
manipulation systems [24]. The contact normal and tangential forces can be determined
if the contacts are known for systems with independent constraints [25]. The contact
forces cannot be uniquely determined when the constraints are not all independent. It
has been shown that the initial value problem has no solution or multiple solutions for
some initial conditions [26].
The dynamics of elastic manipulators with prismatic joints has been investigated
by Kim [27] and Buffinton [28]. They examine the motion of a single beam moving
longitudinally over two distinct points. The equations of motion are formulated by
treating the beam?s supports as kinematical constraints imposed on an unrestrained
beam. Gordaninejad, Azhdari, and Chalhoub [29] examined the motion of a planar robot
arm consisting ofone revolute andone prismatic joint. Bensonand Talke [30]investigated
the dynamics of a magnetic recording slider of a rigid disk during its transition between
sliding and flying. The slider is modeled as a three degree-of-freedom system, capable of
lift, pitch, and roll. In addition to the load from the suspension arm and the impulsive
load arising from slider/disk collisions, they also considered the load due to inertia.
32
Do and Yang [31] solved the inverse dynamics of the Stewart platform manipulator
[32] assuming the joints are frictionless and the moment of inertia of the legs has not
been updated as a function of configuration in the simulation algorithm for path tracking.
Ji [33] considered the question of leg inertia and studied its effect on the dynamics of
the Stuart platform. The dynamic and gravity effects as well as the viscous friction
at the joints were considered for the inverse dynamic formulation of the general Stuart
platform presented by Dasgupta and Mruthyunjaya [34]. Important research related to
the subject of the present paper has been done by Xi, Sinatra, and Han [35]. The authors
investigated the effect of leg inertia on dynamic parameters of sliding-leg hexapods.
The theory presented in this study can be applied to the dynamic modeling of parallel
manipulators with prismatic joints [36].
In the present paper, the effect of slider link mass moment of inertia on the dynamics
of mechanical systems with friction is investigated. The mathematical model of an open
kinematic chain is developed using Lagrange?s method for unconstrained and constrained
systems. Also, a controlled three-link planar robot arm is modeled by using Kane?s
method. A conventional feedback control [37] is used for the robot. The three-link
planar mechanism and the controlled three-link planar robot arm with prismatic joint
are presented as applications. The influence of the prismatic joint mass moment of inertia
on dynamic parameters as the application point of the joint contact forces, angular speed
of the links, actuator torques and forces is analyzed.
In general, the effect of prismatic joint inertia may be negligible at low speeds, but
becomes significant at high speeds. Hence, prismatic joint inertia must be included for
33
modeling high-speed machine tools, manipulators, and robots. This problem is impor-
tant, because in some cases the moment of inertia of the prismatic joints is comparable
to the moment of inertia of the links and may significantly influence the dynamics of the
system at high speeds.
3.2 Mathematical background
The planar two-link mechanical system shown in Fig. 3.1 is considered. The carte-
sian reference frame xOOyO is chosen. The mobile reference frame xOy attached to the
link 1 is considered. The angle between the axis Ox and OxO is ?. For the links 1 and
2 the masses are m1 and m2, and the center of mass locations are designated by C1 and
C2. The length of the link 1 is L. The distance OC2 is denoted by r. The coefficient of
friction between the links 1 and 2 is ?. The gravitational acceleration g is considered.
The gravitational forces G1 and G2 that act on the links 1 and 2 are
G1 = ?m1g(sin??+cos??), G2 = ?m2g(sin??+cos??). (3.1)
The reaction force F12 and the friction force Ff12 exerted by the link 1 to the link 2 can
be written as
F12 = N?, Ff12 = ??Nsign(?r)?. (3.2)
The reaction force F21 and the friction force Ff21 exerted by the link 1 to the link 2 are
F21 = ?F12, Ff12 = ?Ff12. (3.3)
34
Figure 3.1: a. Open kinematicchain with slider andfriction; b. Forcediagram forthe
link1;c.Force diagram forthe link2.
35
3.2.1 Newton-Euler equations
The dynamic system presented above is considered (Fig. 3.1) and the equations of
motion are derived by using the Newton-Euler?s method.
The position rP of the point P is
rP = p?, (3.4)
where P is the application point of the reaction force N between the links 1 and 2.
The sum of the moments for the link 2 with respect to the center of mass C2 is zero
(Fig. 3.1.c)
(rP ?rC2)?F12 ?IC2? = 0. (3.5)
One can solve the linear equation Eq. (3.5) with respect to p
p = r + IC2N ??. (3.6)
The sum of the forces that act on the link 2 is zero (Fig. 3.1.c)
G2 +F12 +Ff12 ?m2aC2 = 0. (3.7)
One can project Eq. (3.7) on ? direction and find the reaction force N
N = m2(gcos? +2?r?? +r??). (3.8)
36
One can project Eq. (3.7) on ? direction
m2(gsin? ?r??2 + ?r)+?Nsign(?r) = 0. (3.9)
The sum of the moments for the link 1 with respect to the point O is zero (Fig. 3.1.b)
rC1 ?G1 +rP ?F21 ?IC1O? = 0. (3.10)
One can project Eq. (3.10) on k direction
1
6m1L1(3gcos? +2L
??)+Np = 0. (3.11)
From Eqs. (3.9) and (3.11) one can derive and solve the equations of motion with respect
to r and ?.
3.2.2 Lagrange equations (unconstrained system)
For the mechanical system shown in Fig. 3.1 the equations of motion for the uncon-
strained system are derived using Lagrange?s method. The polar coordinates q1 = r and
q2 = ? are chosen as generalized coordinates.
The positions rC1 and rC2 of the centers of mass C1 and C2 are
rC1 = L2?, rC2 = r?. (3.12)
37
The angular velocity ? and angular acceleration ? of the links 1 and 2 are
? = ??k, ? = ??k. (3.13)
The velocities vC1 and vC2 of the points C1 and C2 can be expressed as
vC1 = ?rC1 +??rC1, vC2 = ?rC2 +??rC2. (3.14)
The position rP of the point P is
rP = p?, (3.15)
where P is the application point of the reaction force F12.
The sum of the moments for the link 2 with respect to C2 is zero (Fig. 3.1.c)
(rP ?rC2)?F12 ?IC2? = 0. (3.16)
One can solve the linear equation Eq. (3.16) with respect to p
p = r + IC2N ??. (3.17)
The sum of the forces that act on the link 2 is zero
G2 +F12 +Ff12 ?m2aC2 = 0. (3.18)
38
One can project the constraint Eq. (3.18) on ? direction and find the reaction force N
N = m2(gcos? +2?r?? +r??). (3.19)
The Lagrange differential equations are
d
dt
parenleftbigg?T
? ?qi
parenrightbigg
? ?T?q
i
= Qi, i = 1,2, (3.20)
where T is the total kinetic energy and Qi is the generalized force corresponding to qi.
The kinetic energy T1 for the link 1 is
T1 = 12IO???, (3.21)
where IO is the mass moment of inertia of the link 1 with respect to the point O.
The kinetic energy T2 for the link 2 is
T2 = 12m2vC2 ?vC2 + 12IC2???, (3.22)
where IC2 is the mass moment of inertia of the link 2 with respect to the center of mass
C2.
The total kinetic energy is
T = T1 +T2. (3.23)
The velocity vP1 and vP2 of the point P attached to the links 1 and 2 can be written as
vP1 = ??rP, vP2 = ??rP + ?rP. (3.24)
39
The relative velocity vP21 of the link 2 with respect to the link 1 is
vP21 = vP2 ?vP1 = ?rP. (3.25)
The generalized force Qi for the link i is
Qi = ?vC1? ?q
i
?G1 + ?vC2?q
i
?G2 + ?vP1? ?q
i
?(F21 +Ff21)+
?vP2
? ?qi ?(F12 +Ff12), i = 1,2. (3.26)
From Eqs. (3.26), (3.25), and (3.3) one can write the generalized force Qi as
Qi = ?vC1? ?q
i
?G1 + ?vC2?q
i
?G2 + ?vP21? ?q
i
?(F12 +Ff12), i = 1,2. (3.27)
From Eq. (3.20) one can derive and solve the equations of motion with respect to q1 and
q2.
3.2.3 Lagrange equations (constrained system)
For the mechanical system shown in Fig. 3.1 the equations of motion for the con-
strained system are derived using Lagrange?s method. The rod (link 1) and the slider
(link 2) are considered separately. The motion of the slider is expressed using the polar
coordinates r and ?. To express the motion of the rod the angle ? is introduced. One
can chose the generalized coordinates q1 = r, q2 = ?, and q3 = ?.
The constraint equation is
? ?? = 0. (3.28)
40
The Lagrange differential equations are
d
dt
parenleftbigg?T
? ?qi
parenrightbigg
? ?T?q
i
= Qi, i = 1,2,3, (3.29)
where T is the total kinetic energy and Qi is the generalized force corresponding to qi.
The kinetic energy T1 for the link 1 is
T1 = 12IO?1 ??1, (3.30)
where ?1 = ??k.
The kinetic energy T2 for the link 2 is
T2 = 12m2vC2 ?vC2 + 12IC2?2 ??2, (3.31)
where ?2 = ??k and vC2 = ?rC2 +?2 ?rC2.
The total kinetic energy is
T = T1 +T2. (3.32)
The velocity vP1 of the point P attached to the link 1 can be written as
vP1 = ?1 ?rP1, (3.33)
where rP1 = p(cos??+sin??).
The velocity vP2 of the point P attached to the link 2 can be written as
vP2 = ?2 ?rP2 + ?p?, (3.34)
41
where rP2 = p(cos??+sin??).
The generalized force Qi for the link i is
Qi = ?vC1? ?q
i
?G1 + ?vC2?q
i
?G2 + ?vP1? ?q
i
?(F21 +Ff21)+
?vP2
? ?qi ?(F12 +Ff12), i = 1,2,3. (3.35)
The sum of the moments for the link 2 with respect to C2 is zero
(rP2 ?rC2)?F12 ?IC2?2 = 0, (3.36)
where ?2 = ??k.
One can solve Eq. (3.36) and find p.
p = r + IC2N ??. (3.37)
From Eq. (3.29), for i = 3, one can find the reaction force N.
From Eq. (3.29), for i = 1,2, and by using the constraint Eq. (3.28), one can derive and
solve the equations of motion.
3.2.4 Kane equations
For the system shown in Fig. 3.1, equations of motion are derived by using Kane?s
method. There are two generalized speeds u1 = ?r and u2 = ?? corresponding to the
generalized coordinates r and ?. To find the value of the reaction force N one can
introduce the third generalized speed u3 on Ox-axis. Thus, from Eq. (3.25) one can
42
write the relative velocity vP21 as
vP21 = ?rP +u3?. (3.38)
From Eqs. (3.24) and (3.38) the velocity vP2 becomes
vP2 = vP1 +vP21 = ??rP + ?rP +u3?. (3.39)
Furthermore, the velocity vC2 becomes
vC2 = ??rC2 + ?rC2 +u3?. (3.40)
The generalized forces Qj associated to the generalized speeds uj can be computed as
Qj = ?vC1?u
j
?G1 + ?vC2?u
j
?G2 + ?vP1?u
j
?(F21 +Ff21)+
?vP2
?uj ?(F12 +Ff12), j = 1,2,3. (3.41)
One can use Eq. (3.3) and rewrite Eq. (3.41) as
Qj = ?vC1?u
j
?G1 + ?vC2?u
j
?G2 + ?vP21?u
j
?(F12 +Ff12), j = 1,2,3. (3.42)
The generalized inertia forces F?j can be written as
F?j =
2summationdisplay
i=1
?vCi
?uj ?(?miaCi)+
2summationdisplay
i=1
??i
?uj ?(?ICi?i), j = 1,2,3. (3.43)
43
One can write Kane?s equations associated to the generalized speeds uj as
F?j +Qj = 0, j = 1,2,3. (3.44)
From Eq.(3.44), for j = 3, one can find the reaction force N.
From Eq.(3.44), for j = 1,2, one can derive and solve the equations of motion with
respect to r and ?.
3.3 Kinematic chains
The basic theory presented in the previous section can be applied for the study of
open and closed kinematic chains with prismatic joints.
3.3.1 Open kinematic chains
The equations of motion for the three-link spatial robot arm with prismatic joint
(Fig. 3.2) are derived by using Kane?s method. The cartesian reference frame xOOyOzO is
chosen. The mobile reference frame xiOyizi is attached to the link i, i = 1,2. The robot
arm has three degrees of freedom, those are the angles q1, q2, and the distance q3. The
link i has length Li, mass mi, center of mass Ci, and central inertia dyadic ?ICi, i = 1,2,3.
The coefficient of friction between the links 2 and 3 is ?. Friction is negligible for the
rotational joints. The gravitational acceleration g is considered. The initial conditions
at t = 0 are q1(0) = q10, q2(0) = q20, q3(0) = q30 m, and ?q1(0) = ?q2(0) = ?q3(0) = 0.
The feedback control is implemented using the actuator torques Mc01 and Mc12 applied
to the rotational joints O and A and the actuator force Fc23 acting to the translational
joint at A. The desired final state of the system is q1 = q1f, q2 = q2f, and q3 = q3f. The
44
Figure 3.2: Three-linkrobot armwithprismatic joint.
k
O
k
1
k
2
z
O
45
position rCi of the center of mass Ci, i = 1,2,3, is
rC1 = L12 ?1,
rC2 = L1?1 + L22 ?2, (3.45)
rC3 = L1?1 +q3?2.
The angular velocities and angular accelerations of the links 1 and 2 are
?1 = ?q1?1, ?1 = ?q1?1, (3.46)
?2 = ?q1?1 + ?q2k2, ?2 = ?q1?1 + ?q2k2. (3.47)
The velocities vCi and accelerations aCi of the points Ci, i = 1,2,3, can be expressed as
vCi = ?rCi +?i ?rCi, aCi = ?vCi +?i ?vCi. (3.48)
The gravitational forces G1, G2, and G3 that act on the links 1, 2 and 3 are
G1 = ?m1g?1,
G2 = ?m2g(sinq2?2 +cosq2?2), (3.49)
G3 = ?m3g(sinq2?2 +cosq2?2).
The reaction force F23 and the friction force Ff23 exerted by the link 2 to the link 3 can
be written as
F23 = N?2, Ff23 = ??Nsign(?q3)?2. (3.50)
46
The reaction force F32 and the friction force Ff32 exerted by the link 1 to the link 2 are
F32 = ?F23, Ff32 = ?Ff23. (3.51)
The feedback control of the arm is realized using the actuator torques and forces
Mc01 = ?[c11 ?q1 +c12(q1 ?q10)],
Mc12 = ?[c21 ?q2 +c22(q2 ?q20)]+
parenleftbiggm
2
2 L2 +m3q3
parenrightbigg
gcosq2 +NL1 cosq2 +
+[Fc23 ??Nsign(?q3)]L1 sinq2,
Fc23 = ?[c31 ?q3 +c32(q3 ?q30)]+m3gsinq2 +?Nsign(?q3), (3.52)
where c11, c12, c21, c22, c31, c32 are constants.
The position of the application point P of the reaction force F23 is
rP = L1?1 +p?2, (3.53)
where p = q3 + IC3N ?q2.
The velocities vP2 and vP3 of the point P attached to the links 2 and 3 can be written
as
vP2 = ?2 ?rP, vP3 = ?2 ?rP + ?rP. (3.54)
The relative velocity vP32 between the links 3 and 2 is
vP23 = vP3 ?vP2 = ?rP. (3.55)
47
One can define the generalized speeds ui = ?qi corresponding to the generalized coordi-
nates qi, i = 1,2,3. To find the reaction force N one can introduce the generalized speed
u4 on the direction Ox2 in the expression of the relative velocity vP23
vP23 = ?rP +u4?2. (3.56)
Thus, the velocities vP3 and vC3 become
vP3 = ?2 ?rP + ?rP +u4?2,
vC3 = ?2 ?rC3 + ?rC3 +u4?2. (3.57)
The generalized forces Qj associated to the generalized speeds uj can be written as
Qj = ?vC1?u
j
?G1 + ?vC2?u
j
?G2 + ?vC3?u
j
?G3 + ??1?u
j
?Mc01 +
?(?2 ??1)
?uj ?Mc12 +
?vP23
?uj ?(F23 +Ff23 +Fc23), j = 1,2,3. (3.58)
The generalized inertia forces Fj can be written as
Fj =
3summationdisplay
i=1
?vCi
?uj ?(?miaCi)+
3summationdisplay
i=1
??i
?uj ?(?
?IC
i ??i), j = 1,2,3. (3.59)
One can write Kane?s equations associated to the generalized speeds uj as
Fj +Qj = 0, j = 1,2,3. (3.60)
48
From Eq. (4.120), for j = 4, one can find the reaction force N.
From Eq. (4.120), for j = 1,2,3, one can derive and solve the equations of motion with
respect to the generalized co-ordinates q1, q2, and q3.
3.3.2 Closed kinematic chains
The three moving link planar mechanism shown in Fig. 3.3 is considered. The angle
between the link 1 and Ox-axis is the driver angle ?1. A motor torque Mm is acting
on the link 1 while an external torque Me is applied on the link 3. The distance from
the center of mass C2 of the link 2 to the application point P of the reaction force F23
between links 2 and 3 is d. The coefficient of friction between the links 2 and 3 is ?. The
link i has the length Li, the mass mi, the center of mass Ci, the mass moments of inertia
ICi, the linear acceleration aCi, and the angular acceleration ?i, for i = 1,2,3. The
gravitational force that acts on the link i is Gi = ?mig?, where g is the gravitational
acceleration. The distance d and the motor torque Mm can be computed using the
Newton-Euler?s equations. The Newton-Euler?s equations for the link 3 are
m3aC3 = F03 +F23 +Ff23 +G3, (3.61)
IC3?3 = C3C?F03 +C3P?F23 +Me, (3.62)
where F03 is the reaction force from the link 0 (the ground) on the link 3, and F23 =
F23(?sin?3?+cos?3?) is the reaction force from the link 2 on the link 3.
The friction force Ff23 has opposite direction to the relative translational velocity v23
49
Figure 3.3: Three-linkmechanism with prismaticjoint.
 
1
 
3
50
between the links 2 and 3
Ff23 = ? v23|v
23|
?F23. (3.63)
The application point P(xP,yP) of the reaction force F23 is not known but it is located
on the sliding direction
tan?3 = yPx
P ?OA
, (3.64)
where ?3 is the angle between the link 3 and the Ox-axis.
The Newton-Euler?s equations for the link 2 are
m2aC2 = F12 +F32 +Ff32 +G2, (3.65)
IC2?2 = C2P?F32, (3.66)
where F12 is the reaction force from the link 1 on the link 2, F32 = ?F23, and Ff32 =
?Ff23.
From Eqs. (4.70), (4.71), (4.64), (3.64), (4.68), and (4.69) one can compute the reaction
forces F03, F23, F12, the friction force Ff23, and the position of the point P(xP,yP).
The Newton-Euler?s equations for the link 1 are
m1aC1 = F01 +F21, (3.67)
IC1?1 = C1C2 ?F21 +C1O?F01 +Mm, (3.68)
where F01 is the reaction force from the link 0 (the ground) on the link 1, and F21 =
?F12.
51
From Eqs. (4.66) and (4.67) one can compute the reaction force F01 and the motor torque
Mm.
3.4 Results
3.4.1 Open kinematic chain
In this section, results from computer simulations are presented. Numerical data
captured from the three-link spatial robot arm shown in Fig. 3.2 is analyzed. The
lengths of the links 1 and 2 are L1 = L2 = 0.1 m. The masses of the links 1, 2, and 3
are m1 = m2 = 1 kg, and m3 = 0.2 kg. The mass moments of inertia of the links 1,
2, and 3 are IC1x = 0, IC1y = IC1z = 0.01 kg m2, IC2x = 0, IC2y = IC2z = 0.01 kg m2,
IC3x = IC3y = 0, and IC3z = IC3. The coefficient of friction between the links 2 and 3 is
? = 0.5. The gravitational acceleration is g = 9.807 m/s2. The initial conditions at t = 0
are q1(0) = pi/6 rad, q2(0) = pi/4 rad, q3(0) = 0.01 m, and ?q1(0) = ?q2(0) = ?q3(0) = 0.
The feedback control is implemented using the constants c11 = c12 = 10, c21 = c22 = 0.1,
and c31 = c32 = 1. The desired final state of the system is q1f = q2f = pi/3 rad,
and q3f = 0.1 m. The initial conditions response of the co-ordinate q2(t) for IC3 = 0,
IC3 = 0.05 kg m2, IC3 = 0.1 kg m2, and IC3 = 0.15 kg m2 is illustrated in Fig. 3.4.a-d.
One can define the error ei(t) = qi(t) ? qif for the co-ordinate qi, i = 1,2,3. The
maximum overshoot eimax = max|ei(t)| and the settling time tsi (ei(t) < ei0 for t > tsi),
can be computed for i = 1,2,3, where ei0 is a constant. The maximum overshoot e2max
and the settling time ts2 are computed for different values of IC3, where e20 = 10?3
(Table 1). For IC3 = 0, the maximum overshoot is approximately zero (e2max ? 0) and
the settling time is ts2 = 4.83 s. Larger values of e2max and ts2 are observed for larger
52
Figure 3.4: Thedynamic response of theco-ordinate q
2
forthe robotarm in thecases:
a. I
C
3
=0,b. I
C
3
=0:05 kg m
2
,c. I
C
3
=0:1kgm
2
,and d. I
C
3
=0:15 kg m
2
.
53
values of IC3 for the same control parameters values. For example, for IC3 = 1 kg m2,
the maximum overshoot is e2max = 0.049 and the settling time is ts2 = 12.66 s.
Similar results can be obtained for the generalized co-ordinates q1 and q3.
IC3[kg m2] 0 0.05 0.1 0.15
e2max ?0 0.02 0.049 0.068
ts2[s] 4.83 7.42 12.66 18.45
Table 3.1: The maximum overshoot e2max and the settling time ts2 computed for different
values of IC3.
3.4.2 Closed kinematic chain
Simulation results captured from the planar three-link mechanism shown in Fig. 3.3
are presented. The cartesian reference frame xOOyO is considered. The masses of the
links 1, 2, and 3 are m1 = 0.5 kg, m2 = 0.2 kg, and m3 = 0.8 kg, respectively. The
lengths of the links 1 and 3 are L1 = 0.5 m and L3 = 0.8 m. The distance between the
points O and A is OA = 0.2 m. The mass moments of inertia for the links 1 and 3 are
IC1 = 0.01 kg m2 and IC3 = 0.042 kg m2. An external torque Me = 500 Nm is applied
on the link 3. The angle ?1 = pi/3 is chosen for the simulations. The gravitational
acceleration is g = 9.807 m/s2.
Figures 3.5 and 3.6 illustrate the distance d and the torque Mm plotted for different
values of the mass moment of inertia IC2 and angular speeds ?1 = ??1, while the coefficient
of friction is ? = 0.5=constant. The distance d is zero (d = 0) for IC2 = 0. In Fig. 3.5
larger values of d are observed for larger values of IC2 and ?1. For example, for ?1 = 20
rad/s and IC2 = 0.004 kg m2 it results d = 0.0009 m, and for ?1 = 20 rad/s and
IC2 = 0.008 kg m2 it results d = 0.0019 m. Also, for IC2 = 0.006 kg m2 and ?1 = 10
54
rad/s it results d = 0.0003 m, and for IC2 = 0.006 kg m2 and ?1 = 30 rad/s it results
d = 0.0037 m. Simultaneously, smaller values of Mm are observed for larger values of
IC2 and ?1 (Fig. 3.6). For example, for ?1 = 20 rad/s and IC2 = 0.004 kg m2 it results
Mm = 725.145 Nm, and for ?1 = 20 rad/s and IC2 = 0.008 kg m2 it results Mm = 723.158
Nm. Also, for IC2 = 0.006 kg m2 and ?1 = 10 rad/s it results Mm = 789.958 Nm, and
for IC2 = 0.006 kg m2 and ?1 = 30 rad/s it results Mm = 614.474 Nm. The distance d
increases and significantly modifies the value of the torque Mm for relatively high values
of the angular velocity ?1.
3.5 Conclusions
The effect of prismatic joint inertia on the dynamic parameters of planar kinematic
chains is presented. The application point of the slider contact forces changes its position
for different values of the slider inertia. The effect of the slider inertia may be negligible
at low speeds but becomes significant at relatively high speeds. Dynamic response char-
acteristics of a planar robot arm are compared for different values of the slider inertia.
The maximum overshoot may be negligible for small values of the mass moment of iner-
tia and for some control parameters. Larger values of the maximum overshoot and the
settling time are observed for larger values of the mass moment of inertia for the same
control parameters. Experimental data are needed in order to validate the analytical
results.
55
Figure 3.5: Thein uenceofthe mass moment of inertia I
C
2
andangular speed !
1
on
thedistance d.
Graph I II III IV
!
1
[rad/s] 10 20 30 40
56
Figure 3.6: Thein uenceofthe mass moment of inertia I
C
2
andangular speed !
1
on
themotor torque M
m
.
Graph I II III IV
!
1
[rad/s] 10 20 30 40
Chapter 4
Rigid body contact and impact
In this study a planar rigid-link mechanism with a rotating slider joint and clearance
is investigated. The influence of the clearance gap size, crank speed, friction and impact
parameters on the nonlinear behavior of the system are analyzed. Periodic response is
observed for zero clearance and also at low crank speeds and low values of the coefficient
of restitution for the mechanism with clearance. Chaotic motion is observed for relatively
high crank speeds. The sliding joint with clearance is modelled using a kinematic coeffi-
cient of friction and a coefficient of restitution. Nonlinear dynamics tools are applied to
analyze the simulation data captured from the connecting rod of the mechanism. The
largest Lyapunov exponent is used as an index for studying the stability of the system
and a diagnostic tool.
4.1 Introduction
One of the important factors that influence the dynamic stability and the perfor-
mance of mechanisms is the joint clearance. In the last years, many researchers have been
studied the effects of the clearance on the motion of mechanical systems. Farahanchi
and Shaw [38] considered the model of a planar, rigid-link mechanism with clearance at
the slider joint. They observed that the response of the system appears to be chaotic,
although periodic motion become more common as dissipation effects are increased.
Abarbanel et al. [39, 40, 41] developed dynamic tools for analyzing observed chaotic
57
58
data. Nonlinear dynamic tools were presented by Nayfeh and Balachadran [42]. Re-
lated papers in the field of chaotic motion in mechanical systems are those of Kapitaniak
[43], and Ott et al. [44]. Deck and Dubowski [45] studied the problems encountered
in predicting the dynamic response of machines with clearance connections. Recent re-
search has contributed to the development of simulation methods for specific multibody
systems. Gilmore and Cipra [46] discussed a simulation method for planar dynamical
mechanical systems with changing topologies. The information provided by the rigid
bodies? boundary descriptions was used to automatically predict and detect impacts.
Conti et al. [47] described a unified method to predict the contact changes due to kine-
matics. Contact and friction constraints were used by Pfeiffer [48] to study the stick-slip
phenomena. The first trials to model impacts with friction can be found in Brach [49],
and Wang and Mason [50]. Brach [49] considered only single collisions and formulates
the impact equations using Newton?s law. Wang and Mason [50] applied a time-scaling
method and solve the impact equations by using Poisson?s Law and Coulomb?s law for a
single contact. Jean and Moreau [51] reformulated Newton?s law in an unilateral manner
for multiple impacts with friction. In this work, the models of rigid and flexible body
impacts described by Marghitu et al. [52, 53, 54] were used. In a closely related paper,
Marghitu and Stoenescu [55] developed a dynamic analysis of children gait.
In this section, the dynamic analysis of a planar rigid-link mechanism with prismatic
joint and clearance is investigated [56, 57]. Periodic motion is observed for the system
with no clearance. The response of the system with clearance is chaotic at relatively
high crank speeds and low values of the coefficient of restitution.
59
4.2 R-RTR mechanism
In this section, the planar R-RTR mechanism shown in Fig. 4.1 is considered. A fixed
reference frame xAyz is chosen. The lengths of the links are L1 = 0.001 m, L2 = 0.470 m,
and L3 = 0.047 m. The links 1 and 2 are rectangular prisms with the depth d = 0.001 m
and height h = 0.01 m. The link 3 has the height h3 = 0.02 m, and the depth d3 = 0.05
m. The mass density of the links is ? = 7850 Kg/m3. The center of mass locations of the
links i = 1, 2, 3 are designated by Gi(xGi,yGi,0). The initial conditions ?1(0) = pi/6
rad and ?1(0) = ??1(0) = 0 rad/s are given.
Generalized coordinates
The number of degrees of freedom for the mechanism can be computed using the relation
M = 3n?2c5 ?c4,
where n is the number of moving links, c5 is the number of pin joints or slider joints
with one degree of freedom, and c4 is the number of pin joints or slider joints with two
degrees of freedom.
In our case, n = 3, c5 = 4 (A(R), B(R), C(T), C(R)), c4 = 0, and the mechanism
has one degree of freedom (M = 1). One can choose the angle q1(t) = ?1(t) as the
generalized coordinate.
Kinematics
a. Position vectors
The position vector rG1 of the center of the mass G1 is
rG1 = xG1?+yG1?, (4.1)
60
Figure 4.1: Rigidbodydiagram forthe R-RTRmechanism with rotating prismaticjoint.
61
where xG1 and yG1 are the coordinates of G1
xG1 = L12 cosq1, yG1 = L12 sinq1. (4.2)
The position vector rG2 of the center of the mass G2 is
rG2 = xG2?+yG2?, (4.3)
where xG2 and yG2 are the coordinates of G2
xG2 = L1 cosq1 + L22 cos?2, yG2 = L1 sinq1 + L22 sin?2, (4.4)
where ?2 = arctan L1 sinq1L
1 cosq1 ?AC
.
The position vector rG3 of the center of the mass G3 is
rG3 = AC?. (4.5)
b. Velocity vectors
The velocity vector vG1 is the derivative with respect to time of the position vector rG1
vG1 = ?rG1 = ?xG1?+ ?yG1?, (4.6)
where
?xG1 = ?L12 ?q1 sinq1, ?yG1 = L12 ?q1 cosq1. (4.7)
62
The velocity vector vG2 is the derivative with respect to time of the position vector rG2
vG2 = ?rG2 = ?xG2?+ ?yG2?, (4.8)
where
?xG2 = ?L1 ?q1 sinq1 ? L22 ??2 sin?2,
?yG2 = L1 ?q1 cosq1 + L22 ??2 cos?2.
(4.9)
The velocity vector vG3 is zero
vG3 = 0. (4.10)
c. Acceleration vectors
The acceleration vector aG1 is the double derivative with respect to time of the position
vector rG1
aG1 = ?rG1 = ?xG1?+ ?yG1?, (4.11)
where
?xG1 = ?L12 ?q1 sinq1 ? L12 ?q21 cosq1,
?yG1 = L12 ?q1 cosq1 ? L12 ?q21 sinq1.
(4.12)
The acceleration vector aG2 is the double derivative with respect to time of the position
vector rG2
aG2 = ?rG2 = ?xG2?+ ?yG2?, (4.13)
where
?xG2 = ?L1?q1 sinq1 ?L1 ?q21 cosq1 ? L22 ??2 sin?2 ? L22 ??22 cos?2,
?yG2 = L1?q1 cosq1 ?L1 ?q21 sinq1 + L22 ??2 cos?2 ? L22 ??22 sin?2.
(4.14)
63
The acceleration vector of aG3 is zero
aG3 = 0. (4.15)
d. Angular velocity vectors
The angular velocity vectors of the links 1, 2, and 3 are
? = ?q1k,
?2 = ?3 = ??2k.
(4.16)
e. Angular acceleration vectors
The angular acceleration vectors of the links 1, 2, and 3 are
? = ?q1k,
?2 = ?3 = ??2k.
(4.17)
Force analysis
a. Masses
The masses of the links 1, 2 and 3 are
m1 = ?L1hd,
m2 = ?L2hd,
m3 = m3a ?m3b,
where m3a = ?L3h3d3, m3b = ?L3hd.
64
b. Gravitational forces
The gravitational forces of the link 1, 2, and 3 are
G1 = ?m1g?,
G2 = ?m2g?,
G3 = ?m3g?.
(4.18)
c. Mass moments of inertia
The mass moment of inertia of the link 1 with respect to the center of mass G1 is
IG1 = m112
parenleftBig
L21 +h2
parenrightBig
.
The mass moment of inertia of the link 2 with respect to the center of mass G2 is
IG2 = m212
parenleftBig
L22 +h2
parenrightBig
.
The mass moment of inertia of the link 3 with respect to the center of mass G3 is
IG3 = m3a12
parenleftBig
L23 +h23
parenrightBig
? m3b12
parenleftBig
L23 +h2
parenrightBig
.
d. Motor torque
The motor torque acts on the link 1
Mm = Mk. (4.19)
65
For a D.C. electric motor, M = M0
parenleftbigg
1? ?1?
0
parenrightbigg
, where M0 and ?0 are given in catalogues.
In our case, M0 = 1 Nm, and ?0 = 4 rad/s (Fig. 4.2).
4.2.1 Newton-Euler?s method
In this section the equation of motion for the mechanism is solved by using Newton-
Euler?s formulation. There are three rigid bodies in the system and one can write the
Newton-Euler equations for each link.
a. Link 1
The Newton-Euler equations for the link 1 are (see Fig. 4.3.a)
m1aG1 = F01 +F21 +G1, (4.20)
IG1?1 = G1A?F01 +G1B?F21 +Mm, (4.21)
where F01 is the joint reaction of the ground 0 on the link 1 at the point A, and F21 is
the joint reaction of the link 2 on the link 1 at the point B
F01 = F01x?+F01y?,
F21 = F21x?+F21y?.
(4.22)
b. Link 2
The Newton-Euler equations for the link 2 are (see Fig. 4.3.b)
m2aG2 = F12 +F32 +G2, (4.23)
66
Figure 4.2: Thevariation of thedrivermotor torque with respect to theangular speed
forthe R-RTRmechanism.
67
Figure 4.3: Rigidbodydiagram forthe R-RTRmechanism.
a.
b.
c.
68
IG2?2 = G2B?F12 +G2P?F32, (4.24)
where F12 is the joint reaction of the link 1 on the link 2 at the point B and F32 is the
joint reaction of the link 3 on the link 2 at the point P
F12 = ?F21,
F32 = F32x?+F32y?.
(4.25)
The application point P(xP,yP) of the reaction force F32 is not known but it is located
on the sliding direction
tan?2 = yPx
P ?AC
, (4.26)
where xP, yP are the coordinates of the point P.
The reaction force F32 is perpendicular to the sliding direction BD
F32 ?BD = 0. (4.27)
c. Link 3
The Newton-Euler equations for the link 3 are (see Fig. 4.3.c)
m3aG3 = F23 +F03 +G3 = 0, (4.28)
IG3?3 = CP?F23, (4.29)
where
F23 = ?F32,
F03 = F03x?+F03y?.
(4.30)
69
There are ten scalar equations with ten unknowns. From the Eqs. (4.39), (4.40), (4.44),
(4.45), (4.26), (4.42), and (4.47) one can find the unknown joint reaction forces F01x,
F01y, F21x, F21y, F32x, F32y, F03x, F03y, and the coordinates xP and yP of the point P.
Knowing the reaction forces and the position vector rP as functions of q1(t), ?q1(t), and
?q1(t), one can derive the equation of motion for the mechanism from Eq. (4.48).
4.2.2 Lagrange?s method
In this section the equation of motion for the mechanism is solved using Lagrange?s
formulation. The Lagrange differential equation for the mechanism with one degree of
freedom is
d
dt
parenleftbigg?T
? ?q1
parenrightbigg
? ?T?q
1
= Q, (4.31)
where T is the total kinetic energy of the system, and Q is the generalized force.
The kinetic energy T1 for the link 1 is
T1 = 12m1vG1 ?vG1 + 12IG1?1 ??1. (4.32)
The kinetic energy T2 for the link 2 is
T2 = 12m2vG2 ?vG2 + 12IG2?2 ??2. (4.33)
The kinetic energy T3 for the link 3 is
T3 = 12IG3?3 ??3. (4.34)
70
The total kinetic energy is
T =
3summationdisplay
i=1
Ti. (4.35)
The generalized force Qi for the link i is
Qi = ?rGi?q
1
?Gi. (4.36)
The generalized force Qm for the motor is
Qm = ??1? ?q
1
?Mm = M0
parenleftbigg
1? ?q1?
0
parenrightbigg
. (4.37)
The total generalized force Q for the mechanism is
Q =
3summationdisplay
i=1
Qi +Qm =
3summationdisplay
i=1
?rGi
?q1 ?Gi +
??1
? ?q1 ?Mm. (4.38)
For the link 1 some calculations are given
T1 = 12(IG1 + 14L21m1)?q21.
?T1
? ?q1 = (IG1 +
1
4L
2
1m1)?q1.
?rG1
? ?q1 = ?
1
2L1(sinq1?+cosq1?).
Q1 = ?rG1? ?q
1
?(?m1g?) = ?12m1gL1 cosq1.
71
From Eqs. (4.31), (4.35), and (4.38) one can derive and solve the equation of motion for
the mechanism.
Remark: Lagrange?s method does not require the calculation of the joint forces.
4.3 R-RTR mechanism with friction
The mechanism described in Section 4.2 is considered. The coefficient of kinetic
friction is ?k = 0.4. The mechanism has one degree of freedom. One can chose the angle
q1(t) = ?(t) as the generalized coordinate for the system.
4.3.1 Newton-Euler?s method
The equation of motion for the mechanism is solved using Newton-Euler?s formu-
lation. There are three rigid bodies in the system and one can write the Newton-Euler
equations for each link.
a. Link 1
The Newton-Euler equations for the link 1 are (see Fig. 4.3.a)
m1aG1 = F01 +F21 +G1, (4.39)
IG1?1 = G1A?F01 +G1B?F21 +Mm, (4.40)
72
where F01 is the joint reaction of the ground 0 on the link 1 at the point A, and F21 is
the joint reaction of the link 2 on the link 1 at the point B
F01 = F01x?+F01y?,
F21 = F21x?+F21y?.
(4.41)
b. Link 2
The reaction force F32 is perpendicular to the sliding direction BD
F32 ?BD = 0. (4.42)
The friction force Ff32 that acts on the link 2 is
Ff32 = Ff = ? vC2|v
C2|
?kF32. (4.43)
The Newton-Euler equations for the link 2 are (see Fig. 4.3.b)
m2aG2 = F12 +F32 +G2 +Ff32, (4.44)
IG2?2 = G2B?F12 +G2P?F32, (4.45)
where F12 is the joint reaction of the link 1 on the link 2 at the point B and F32 is the
joint reaction of the link 3 on the link 2 at the point P
F12 = ?F21,
F32 = F32x?+F32y?.
(4.46)
73
c. Link 3
The Newton-Euler equations for the link 3 are (see Fig. 4.3.c)
m3aG3 = F23 +F03 +G3 ?Ff32, (4.47)
IG3?3 = CP?F23, (4.48)
where
F23 = ?F32,
F03 = F03x?+F03y?.
(4.49)
There are eight scalar equations with eight unknowns. From the Eqs. (4.39), (4.40),
(4.44), (4.42), and (4.47) one can find the unknown joint reaction forces F01x, F01y, F21x,
F21y, F32x, F32y, F03x, F03y.
Knowing the reaction forces as functions of q1(t), ?q1(t), and ?q1(t), one can derive the
equation of motion for the mechanism from Eq. (4.45).
4.3.2 Kane?s method
The equation of motion for the mechanism is solved using Kane?s formulation.
The total kinetic energy is
T =
3summationdisplay
i=1
Ti, (4.50)
where Ti is the kinetic energy of the link i.
74
Generalized speeds
One can chose the generalized speed
u1 = ?q1. (4.51)
The velocity vector of the point C2 located on the link 2 can be written as
vC2 = vG2 +?2 ?G2C, (4.52)
where G2C = rC ?rG2.
In order to take in consideration the reaction force N between the links 2 and 3 one can
introduce a new generalized speed u2 in the expression of the relative velocity vector
vC32
vC23 = vC2 ?vC3 +u2e2n, (4.53)
where e2n = ?sin?2? + cos?2? and ?2 = arctan L1 sinq1L
1 cosq1 ?AC
. Thus, one can write
vG3 = vC3 = ?u2e2n.
Generalized forces
The reaction force F32 of the link 3 on the link 2 is
F32 = N = Ne2n. (4.54)
The reaction force F23 of the link 3 on the link 2 is
F23 = ?F32 = ?N. (4.55)
75
The friction force Ff32 that acts on the link 2 is
Ff32 = Ff = ? vC2|v
C2|
?kN. (4.56)
The friction force Ff23 that acts on the link 3 is
Ff23 = ?Ff32 = ?Ff. (4.57)
The generalized forces Qj associated to the generalized speeds uj, for j=1,2, can be
computed as
Qj =
3summationdisplay
i=1
?vGi
?uj ?Gi +
?vC2
?uj ?(N+Ff)+
?vC3
?uj ?(?N?Ff)+
??1
?uj ?Mm. (4.58)
Generalized inertia forces
The forces F?j can be written as
F?j =
3summationdisplay
i=1
?vGi
?uj ?(?miaGi)+
3summationdisplay
i=1
??i
?uj ?(?IGi?i). (4.59)
Kane?s equations
One can write two Kane equations associated to the generalized speeds u1 and u2
F?j +Qj = 0, j = 1,2. (4.60)
From Eqs. (4.113) and (4.120) one can find the the reaction force N and the equation
of motion for the mechanism.
76
One can write the angle ?2 as
?2 = pi2 ? q12 , (4.61)
the vector e2n as
e2n = ?cos q12 ?+sin q12 ?, (4.62)
and the reaction force N between the links 2 and 3 as
N = m3gsin q12 . (4.63)
4.3.3 Kineto-static analysis
In this section, the mechanism with no clearance (one degree of freedom) is con-
sidered (Fig. 4.1). Friction forces act at the rotational and translational joints. The
equation of motion is known and the reaction forces are computed. A Newtonian ap-
proach is used, that is, the method of consecutive approximations.
For the translational joint between the links i and j, the friction force Ffji acts on
the link i at the contact surface and is proportional to the coefficient of friction ?. The
force Ffji has opposite direction to the relative translational velocity vij between the
links i and j
Ffji = ? vij|v
ij|
?Fji, (4.64)
where vij = vi ?vj.
The friction forces induce a moment Mfji that acts at the rotational joint between the
links i and j. The moment Mfji is proportional to the coefficient of friction ?, the radius
r of the joint, and has opposite sense to the relative angular velocity ?ij between the
77
links i and j
Mfji = ?sign(?ij)?rFji, (4.65)
where ?ij = ?i ??j.
The Newton-Euler equations for the link 1 are (see Fig. 4.3.a)
m1aC1 = F01 +F21 +G1, (4.66)
IC1? = C1A?F01 +C1B?F21 +Mf21. (4.67)
The Newton-Euler equations for the link 2 are (see Fig. 4.3.b)
m2aC2 = F12 +F32 +G2 +Ff32, (4.68)
IC2?2 = C2B?F12 +C2Q?F32 +Mf12, (4.69)
where F12 = ?F21 and Mf12 = ?Mf21.
The Newton-Euler equations for the link 3 are (see Fig. 4.3.c)
0 = F23 +F03 +G3 +Ff23, (4.70)
IC3?3 = CQ?F23 +Mf03, (4.71)
where F23 = ?F32, and Ff23 = ?Ff32.
The application point Q(xQ,yQ) of the reaction force F32 is not known but it is located
78
on the sliding direction
tanq2 = yQx
Q ?AC
, (4.72)
where tanq2 = L1 sinq1L
1 cosq1 ?AC
, and xQ, yQ are the coordinates of the point Q.
The method of consecutive approximations consists of the following steps:
1. Initially, the friction forces and moments are considered zero. From Eqs. (4.66),
(4.67), (4.68), (4.69), (4.70), (4.71), and (4.72) one can calculate the unknown joint
reaction forces F01x, F01y, F21x, F21y, F32x, F32y, F03x, F03y, and the coordinates xQ and
yQ.
2. Using the values of the reaction forces computed at the step 1, one can calculate
the friction force Ff32 and the friction moments Mf21, Mf03 from Eqs. (4.64) and
(4.65).
3. Using the value of the friction forces and moments computed at the step 2, one
can recalculate the reaction forces from Eqs. (4.66), (4.67), (4.68), (4.69), (4.70), (4.71),
and (4.72).
4. Using the new values of the reaction forces computed at the step 3, one can
recalculate the friction forces and moments from Eqs. (4.64) and (4.65).
5. Step 3 and step 4 are repeated until
vextendsinglevextendsingle
vextendsingleFkij ?Fk+1ij
vextendsinglevextendsingle
vextendsingle < ?
for all reaction forces Fij. The error ? ? R+ is known a priori, and Fkij, Fk+1ij are the
magnitudes of the reaction forces Fij at two consecutive approximations.
79
The method converges after a finite number of steps and the reaction forces with
frictions are calculated.
4.4 R-RTR mechanism with friction and clearance
In order to study the effects of clearances on the motion of a connecting rod in a
slider crank mechanism, a simplified model is used, shown in Fig. 4.1. The following
basic assumptions are considered. (i) All components are rigid. (ii) All motions occur in
a fixed plane. (iii) A motor with a variable torque is used to crank the mechanism. (iv)
The clearances for the slider are symmetrically placed about the nominal slider path,
that is, without clearance, and have a fixed magnitude. (v) The impacts between the
connecting rod and slider are instantaneous and are modelled using a constant coefficient
of restitution, a coefficient of friction, and a moment coefficient.
4.4.1 Equations of motion
Various methods are used to derive the equations of motion for the mechanism. It
is assumed that during the impacts the system position does not change, because the
impact time is very small. It is also assumed that the effect of finite forces is neglected
during the impact. When two bodies impact against each other, an unknown impulsive
force acts between them. The friction between the impact bodies introduces an tangent
impactforce. Formulationofrigidbodycollisionproblemsarebasedontwophysicallaws,
Coulomb?s law of dry friction and balance of momentum. To solve the impact equations,
additional relations are obtained using a coefficient of restitution and a coefficient of
friction.
80
Figure 4.4 shows a planar slider joint where the backlash has been made very large
in order to make it clearly visible. Figure 4.5 illustrates a possible geometry for the slider
joint with clearance and the possible cases consist of: a) No contact (Fig. 4.5.a).
b) Contact or impact on a single point (Fig. 4.5.b).
c) Contact or impact on two opposed points (Fig. 4.5.c).
d) Contact or impact on two points on the same side (Fig. 4.5.d).
The conditions for switching from one case to a different one depend on the positions
of the links and the reaction forces at the contact points.
One can consider three lines (L, La and Lb) defined on the link 2, and four points (M,
N, P, and Q) defined on the link 3 (see Fig. 4.1).
The equations corresponding to the line L, can be expressed as a function of the coordi-
nates of the links 1 and 2 as
Line L: y ?m2x?n2 = 0,
where m2 = tan?2 is the slope and n2 = L1 sin?1 ?m2 cos?1 is the displacement of the
line L.
The equations corresponding to the lines La and Lb can be written as
Line La: y ?max?na = 0,
Line Lb: y ?mbx?nb = 0,
where ma = mb = m2 are the slopes and na = n2 ? l22cos?
2
, nb = n2 + l22cos?
2
are the
displacements of the lines La and Lb.
The coordinates of the points M, N, P, Q can be expressed as functions of the coordi-
nates of the link 3 as
81
Figure 4.4: Modelofthe R-RTRmechanism with rotating prismaticjoint andclearance.
82
Figure 4.5: Geometry of theslider jointwithclearance for: a. no contact; b. contact
or impact on asinglepoint;c.contact or impact on twopointsonthe same side;d.
contactorimpactontwo opposedpoints.
a.
b.
c.
d.
83
Point P: xP = xC +rcos?3, yP = yC +rsin?3,
Point Q: xQ = xC +rcos(?3 ??30), yQ = yC +rsin(?3 ??30),
Point M:xM = xC +rcos(?3 ??30 +pi), yM = yC +rsin(?3 ??30 +pi),
Point N: xN = xC +rcos(?3 +pi), yN = yC +rsin(?3 +pi),
where xC = AC, yC = 0 are the coordinates of the point C, r =
radicalBig
L23 +(l2 +c)2/2 is
the rotation radius of the link 3, and ?30 = 2arctan L3l
2 +c
is the angle negationslash MCN.
Using the expressions above, one can set the position conditions corresponding to the
four cases shown in Fig. 4.2 as following:
Case a) No conditions are necessary.
Case b) Point P is on the line La
yP ?maxP ?na ? 0. (4.73)
Case c) Point P is on the line La and point N is on the line Lb
yP ?maxP ?na ? 0,
yN ?mbxN ?nb ? 0. (4.74)
Case d) Points P and Q are on the line La
yP ?maxP ?na ? 0,
yQ ?maxQ ?na ? 0. (4.75)
84
Impacts can occur when the joint is in either case b), c) or d). The impact conditions
depend on the relative linear velocities of the contact points. For example, in case b),
one can write the following impact condition
vnP2 ?vnP3 ? 0. (4.76)
where vnP2 and vnP3 are the normal velocities to the collision surface of the contact point
P between the links 2 and 3.
The contact conditions also depend on the reaction forces between the links at the contact
points. For example, in case b), the force condition can be written as
NnP2 ?NnP3 ? 0. (4.77)
where NnP2 and NnP3 are the reaction forces between the links 2 and 3 at the contact
point P.
The motion of the contact point during the impact can be described by one of the
following two cases:
1. The contact point is slipping along surface while interacting with it in the normal
direction. Since contact is maintained and slipping occurs, the normal and tangential
components of the contact forces can be represented for dry friction as Ft = ??k Fn.
2. The contact point is not slipping along but interacting with it in the normal
direction. Thetangentialvelocityvt ofthecontactpointisvt = 0subjectto|Ft/Fn| ? ?s.
85
4.4.2 Simulation algorithm
Thesimulationalgorithmforthemechanismautomaticallydetermineswhenachange
in the topology occurs and reformulate the equations of motion to reflect the changes in
the system topology. The equations of motion depend on the contact and impact condi-
tions. Sets of nonlinear equations are solved for contact and sets of linear equations are
solved for impact.
The algorithm consisting of the following steps was written:
Step 1) Set up the input data, those are, the masses, the mass moments of inertia,
the dimensions of the links, and the coordinates of the mechanism.
Step 2) Set up the initial conditions: the initial coordinates, velocities and acceler-
ations of the links. Also, set up the initial time, the final time, and the step integration
time.
Step 3) Verify the position contact conditions (4.73), (4.75), and (4.74). If case a)
then go to step 4). If case b) then go to step 5). If case c) then go to step 6). If case d)
then go to step 7).
Step 4) Solve the equations of motion for no contact and go to step 8).
Step 5) Verify the impact condition. If impact is detected then solve the equation
of motion for impact on a single point and go to step 8). If no impact is detected then
verify the force contact condition. If case a) then go to step 4). If case b) then integrate
the equation of motion for contact on a single point and go to step 8).
Step 6) Verify the impact conditions. If impact is detected then solve the equation
of motion for impact on two points on the same side and go to step 8). If no impact is
detected then verify the force contact condition. If case a) then go to step 4). If case b)
86
then go to step 5). If case c) then integrate the equation of motion for contact on two
points on the same side and go to step 8).
Step 7) Verify the impact conditions. If impact is detected then solve the equation
of motion for impact on two opposed points and go to step 8). If no impact is detected
then verify the force contact condition. If case a) then go to step 4). If case b) then go
to step 5). If case d) then integrate the equation of motion for contact on two opposed
points and go to step 8).
Step 8) Increment the integration time with the step integration time. If the inte-
gration time is less than the final time then go to step 3).
Step 9) Export the output data for analyzing.
Next, the equations of motion for the previous cases are derived.
4.4.3 No contact
In this section, the mechanism with two degrees of freedom is considered (Fig. 4.6).
One can choose the generalized coordinates q1 = ?1 and q2 = ?2. The equation of motion
is derived using the Lagrange?s method
d
dt
parenleftbigg?T
? ?qi
parenrightbigg
? ?T?q
i
= Qi, i = 1,2 (4.78)
where T is the kinetic energy, qi is the generalized coordinate, Qi is the generalized force
associated with the coordinate qi. The kinetic energy T1 for the link 1 is
T1 = 12m1vG1 ?vG1 + 12IG1?1 ??1, (4.79)
87
Figure 4.6: Rigidbodydiagram forthe R-RTRmechanism with prismaticjoint and
clearanceinthe case of no contact.
88
where ?1 = ?q1k.
The kinetic energy T2 for the link 2 is
T2 = 12m2vG2 ?vG2 + 12IG2?2 ??2, (4.80)
where ?2 = ?q2k.
The kinetic energy T is
T = T1 +T2. (4.81)
One can write the generalized force Q1 as
Q1 = ?
parenleftbigg1
2m1 +m2
parenrightbigg
L1gcosq1. (4.82)
One can write the generalized force Q2 as
Q2 = ?12m2L2gcosq2. (4.83)
From Eqs. (4.78), (4.81), and (4.82), one can write
parenleftBigg
m1L21
4 +m2L
2
1 +IG2
parenrightBigg
?q1 + 12m2L1L2[?q2 cos(q2 ?q1)?
?q22 sin(q2 ?q1)] = ?L1
parenleftbigg1
2m1 +m2
parenrightbigg
gcosq1. (4.84)
From Eqs. (4.78), (4.81), and (4.83), one can write
parenleftBigg
m2L22
4 +IG2
parenrightBigg
?q2 + 12m2L1L2
bracketleftBig
?q1 cos(q2 ?q1)+ ?q21 sin(q2 ?q1)
bracketrightBig
=
89
= ?12L2m2gcosq2. (4.85)
Equations (4.84) and (4.85) are used and the equation of motion is derived.
4.4.4 Contact on a single point
In this case, the mechanism has two degrees of freedom (Fig. 4.7.a). One can chose
the generalized coordinates q1 = ?1 and q2 = ?2. Kane?s equations are used and the
equation of motion is derived. The kinetic energy T is
T = T1 +T2 +T3. (4.86)
One can find the position vector rP of the contact point P(xP,yP) solving the system
of equations
tanq2 = yB ?yPx
B ?xP
, (xC ?xP)2 +(yC ?yP)2 = r2. (4.87)
The angular velocity and acceleration vectors ?3 and ?3 of the link 3 are
?3 = ??3k, ?3 = ??3k, (4.88)
where ?3 = arctan yPx
P ?AC
.
One can chose the generalized speeds u1 and u2
u1 = ?q1, u2 = ?q2. (4.89)
90
Figure 4.7: Geometry of themechanism for: a. contactorimpactonasingle point; b.
contactorimpactontwo opposedpoints; c. contactorimpactontwo points on the
same side.
a.
b.
c.
91
In order to take in consideration the reaction force NP between the links 2 and 3 acting
at the point P one can introduce a new generalized speed u3 in the expression of the
relative velocity vP23
vP23 = vP2 ?vP3 +u3e2n, (4.90)
where vP2 = vG2 +?2 ?(rP ?rG2) and vP3 = ?3 ?(rP ?rC).
The reaction force NP of the link 3 on the link 2 is
NP = NPe2n. (4.91)
The friction force FfP that acts on the link 2 at the point P is
FfP = ? vP|v
P|
?kN. (4.92)
The generalized forces Qj, for j = 1,2,3, can be computed as
Qj =
3summationdisplay
i=1
?vGi
?uj ?Gi +
?vP2
?uj ?(NP +FfP)+
?vP3
?uj ?(?NP ?FfP)+
??1
?uj ?Mm. (4.93)
The generalized inertia forces F?j , for j = 1,2,3, can be written as
F?j =
3summationdisplay
i=1
?vGi
?uj ?(?miaGi)+
3summationdisplay
i=1
??i
?uj ?(?IGi?i). (4.94)
One can write three Kane equations associated to the generalized speeds uj, for j = 1,2,3
F?j +Qj = 0. (4.95)
92
From Eqs. (4.113) and (4.120) one can find the equation of motion for the mechanism
and the reaction force N.
4.4.5 Impact on a single point
Next, the mechanism with three generalized coordinates is considered (Fig. 4.7.a).
One can choose q3 = ?3 as the third generalized coordinate. To derive the equation of
motion for the impact, an integrated form of the Lagrange?s equations is used
parenleftbigg?T
? ?qi
parenrightbigg
ts
?
parenleftbigg?T
? ?qi
parenrightbigg
ta
= Pi, i = 1,2,3 (4.96)
where T is the kinetic energy, ?qi is the velocity associated with the generalized coordinate
qi, Pi is the generalized impulse associated with the coordinate qi, and ta, ts are the times
of approach and separation for the impact.
The kinetic energy T3 for the link 3 is
T3 = 12IG3?3 ??3 = 12IG3 ?q23, (4.97)
where ?3 = ?q3k.
The kinetic energy T is
T = T1 +T2 +T3. (4.98)
One can write the left hand sides of Eq. (4.96) as
parenleftbigg?T
? ?qi
parenrightbigg
ts
?
parenleftbigg?T
? ?qi
parenrightbigg
ta
= ?T? ?q
i
slashbigg
?qi=?i??i
, i = 1,2,3 (4.99)
93
where ?i = ?i(ta) = ?qi(ta) and ?i = ?i(ts) = ?qi(ts) are the angular velocities associated
with the coordinates qi before and after the impact.
One can express the position vector rP of the impact point P(xP,yP) solving the system
of equations
tanq2 = yB ?yPx
B ?xP
, tanq3 = yC ?yPx
C ?xP
. (4.100)
The velocity vector of the impact point P is vP = ?rP.
The generalized impulses (right-hand sides of Eq. (4.96)) can be written as
Pi = ?vP? ?q
i
?(Fne2n +Fte2t), i = 1,2,3 (4.101)
where e2n = ?sinq2? + cosq2? and e2t = cosq2? + sinq2? are the unit vectors normal
and tangential to the contact surface, and Fn,Ft are the normal and the tangential
components of the impulse momentum F.
For the link 1, one can write
parenleftBigg
m1L21
4 +m2L
2
1 +IG1
parenrightBigg
(?1 ??1)+ 12m2L1L2(?2 ??2)
cos(q2 ?q1) = P1. (4.102)
For the link 2 one can write
parenleftBigg
m2L22
4 +IG2
parenrightBigg
(?2 ??2)+ 12m2L1L2(?1 ??1)cos(q2 ?q1) = P2. (4.103)
For the link 3 one can write
IG3(?3 ??3) = P3. (4.104)
94
The velocities vP2 and vP3 of the contact points P2 and P3 located on the links 2 and 3
can be expressed as
vP2 = vB + ?q2k?BP,
vP3 = ?q3k?CP, (4.105)
where vB = ?rB is the linear velocity of the joint B, and CP = rP ?AC?.
One can write the velocity of approach va and separation vs for the impact as
va = vP2(ta)?vP3(ta), vs = vP2(ts)?vP3(ts). (4.106)
From the definition of the coefficient of restitution e, one can write
e = ?vsnv
an
, (4.107)
where van = va ? e2n and vsn = vs ? e2n are the projections of the linear velocities of
approach and separation va and vs on the normal direction e2n.
The tangential component vst of the velocity of separation vector vs can be expressed
as
vst = (vs ?e2t)e2t. (4.108)
There are two cases of impact with friction at the point P:
1. No slipping. The following condition must be satisfied
vextendsinglevextendsingle
vextendsinglevextendsingleFt
Fn
vextendsinglevextendsingle
vextendsinglevextendsingle < ?s. (4.109)
95
In this case, the velocity vector vst is zero
vst = 0. (4.110)
From Eqs. (4.102), (4.103), (4.104), (4.107), and (4.110) one can find the unknown
variables Fn, Ft, and ?i, i = 1,2,3.
2. Slipping. The following condition must be satisfied
vextendsinglevextendsingle
vextendsinglevextendsingleFt
Fn
vextendsinglevextendsingle
vextendsinglevextendsingle > ?s. (4.111)
In this case, the following relation can be written
Fte2n = ? vst|v
st|
?k |Fn|. (4.112)
From Eqs. (4.102), (4.103), (4.104), (4.107), and (4.112) one can find the unknown
variables Fn, Ft, and ?i, i = 1,2,3.
4.4.6 Contact on two opposed points
In this case, the mechanism has one degree of freedom. One can chose q1 = ?1 as
the generalized coordinate (Fig. 4.7.b). Kane?s equations are used and the equation of
motion is derived.
One can chose the generalized speed
u1 = ?q1. (4.113)
96
One can write the position vector rN of the contact point N(xN,yN) as
rN = rP +2re2t. (4.114)
There are two contact points between the link 2 and the link 3, those are P and N. In
order to take in consideration the reaction forces NP and NN acting at the points P and
N one can introduce the generalized speeds u2 and u3 in the expressions of the relative
velocities vP23 and vN23
vP23 = vP2 ?vP3 +u2e2n, vN23 = vN2 ?vN3 +u3e2n, (4.115)
where vP2 = vG2 +?2?(rP ?rG2), vP3 = ?3?(rP ?rC), vN2 = vG2 +?2?(rN ?rG2),
and vN3 = ?3 ?(rN ?rC).
The reaction forces NP and NN of the link 3 on the link 2 at the points P and N are
NP = NPe2n, NN = NNe2n. (4.116)
The friction forces FfP and FfN that act on the link 2 at the points P and N are
FfP = ? vP|v
P|
?kNP, FfN = ? vN|v
N|
?kNN. (4.117)
The generalized forces Qj, for j = 1,2,3, can be computed as
Qj =
3summationdisplay
i=1
?vGi
?uj ?Gi +
?vP2
?uj ?(NP +FfP)+
?vP3
?uj ?(?NP ?FfP)
+?vN2?u
j
?(NN +FfN)+ ?vN3?u
j
?(?NN ?FfN)+ ??1?u
j
?Mm. (4.118)
97
The generalized inertia forces F?j , for j = 1,2,3, can be written as
F?j =
3summationdisplay
i=1
?vGi
?uj ?(?miaGi)+
3summationdisplay
i=1
??i
?uj ?(?IGi?i). (4.119)
One can write three Kane equations associated to the generalized speeds uj, for j = 1,2,3
F?j +Qj = 0. (4.120)
From Eqs. (4.113) and (4.120) one can find the equation of motion for the mechanism
and the reaction forces NP and NN.
4.4.7 Impact on two opposed points
In this case, the mechanism has three generalized coordinates (Fig. 4.7.b). The link
2 impacts the link 3 simultaneously in two points, those are P(xP,yP) and N(xN,yN).
One can express the position vectors rP and rN of the impact points P(xP,yP) and
N(xN,yN) solving the system of equations
tanq2 = yB ?yx
B ?x
, (xC ?x)2 +(yC ?y)2 = r2. (4.121)
The velocity vectors of the impact points P and N are vP = ?rP and vN = ?rN.
The generalized impulses can be written as
Pi = ?vP? ?q
i
?(FIne2n +FIte2t)+ ?vN? ?q
i
?(FIIne2n +FIIte2t), i = 1,2,3, (4.122)
98
where FIn,FIt and FIIn,FIIt are the normal and the tangential components of the im-
pulse momenta FI and FII.
The velocities vN2 and vN3 of the contact points N2 and N3 located on the links 2 and
3 can be expressed as
vN2 = vB + ?q2k?BN, vN3 = ??3k?CN, (4.123)
where CN = rN ?AC?.
One can write the velocities of approach vIa, vIIa and separation vIs, vIIs for the impact
points P and N as
vIa = vP2(ta)?vP3(ta), vIIa = vN2(ta)?vN3(ta),
vIs = vP2(ts)?vP3(ts), vIIs = vN2(ts)?vN3(ts). (4.124)
From the definition of the coefficient of restitution e, one can write
e = ?vIsnv
Ian
, e = ?vIIsnv
IIan
, (4.125)
where vIan = vIa ?e2n, vIsn = vIs ?e2n, vIIan = vIIa ?e2n, and vIIsn = vIIs ?e2n.
The tangential components vIst and vIIst of the velocity of separation vectors vIs and
vIIs can be expressed as
vIst = (vIs ?e2t)e2t, vIIst = (vIIs ?e2t)e2t. (4.126)
There are two cases of impact with friction:
99
1. No slipping. The following two conditions must be satisfied
vextendsinglevextendsingle
vextendsinglevextendsingleFIt
FIn
vextendsinglevextendsingle
vextendsinglevextendsingle < ?s and
vextendsinglevextendsingle
vextendsinglevextendsingleFIIt
FIIn
vextendsinglevextendsingle
vextendsinglevextendsingle < ?s. (4.127)
In this case, the velocity vectors vIst and vIIst are zero
vIst = vIIst = 0. (4.128)
From Eqs. (4.102), (4.103), (4.125), and (4.128) one can find the unknown variables FIn,
FIt, FIIn, FIIt, and ?i, i = 1,2,3.
2. Slipping. One of the following conditions must be satisfied
vextendsinglevextendsingle
vextendsinglevextendsingleFIt
FIn
vextendsinglevextendsingle
vextendsinglevextendsingle > ?s or
vextendsinglevextendsingle
vextendsinglevextendsingleFIIt
FIIn
vextendsinglevextendsingle
vextendsinglevextendsingle > ?s. (4.129)
In this case, the following two relations can be written
FIte2n = ? vIst|v
Ist|
?k |FIn|, FIIte2n = ? vIIst|v
IIst|
?k |FIIn|. (4.130)
From Eqs. (4.102), (4.103), 4.125), and (4.130) one can find the unknown variables FIn,
FIt, FIIn, FIIt, and ?i, i = 1,2,3.
4.4.8 Contact or impact on two points on the same side
Considering the endpoints, the line contact is kinematically equivalent to two point
contact along a line segment [46]. Thus, this case is similar to the case c) for contact or
impact on two opposite points (Fig. 4.7.c).
100
4.5 Working Model and Mathematica simulations
For this section, the mechanism shown in Fig. 4.6 is considered. The numerical
results obtained solving the equation of motion for the mechanism using Mathematica
and data captured from the Working Model simulation are compared.
In order to compare the results, the initial conditions used in Mathematica are used
as input data in Working Model. For example, one can set up the variable torque M
of the motor in the ?Properties? window from ?Window? menu by choosing the motor
type ?Torque? and introducing the value M = M0(1??/?0), where M0, ?0 are constant
and ? is the rotational velocity of the driver link (see Fig. 4.8). The graph of the torque
M can be visualized by selecting the command ?Torque transmitted? from ?Measure?
menu.
Data from Working Model graphs can be exported choosing the command ?Ex-
port...? from ?File? menu. The data is exported in a plain text file for the desired
interval of time and accuracy. This way, the data can be imported and analyzed using
various tools. In this case, the simulated data for the motor torque M is exported from
Working Model and imported in a Mathematica program. To import a file in a Mathe-
matica program, one can use the command Import["file", "format"] which imports
data in the specified file format from a file and converts it to a Mathematica expression.
For example, if the Working Model exported file name is torque.dta, the command
to import the file in Mathematica can be Import["torque.dta", "Table"]. Also, the
motor torque M can be computed solving the equation of motion using Mathematica. In
our case, the Lagrange method was used. For a given interval of time, the torque M is
computed using Mathematica (Mc), and captured from the Working Model simulation
101
Figure 4.8: WorkingModel simulation of thedrivermotor torque forthe doublependu-
lum.
102
(Ms). The graphs of the torques Mc and Ms are compared in Fig. 4.9. The error err(t)
between the numerical data Mc(t) and Ms(t) at the time t can be computed as
err(t) = |Mc(t)?Ms(t)|.
The relative maximal error errrelmax can be calculated as
errrelmax = maxt?[ti,tf] err(t)M
0
?100,
where [ti,tf] is the time interval used for the analysis.
In our case, the value of the relative maximal error is computed with the Mathematica
program shown in Appendix A as errrelmax = 4.5%.
4.6 Results
In this section, results from computer simulations are presented using analysis tools.
In Fig. 4.4 the mechanism with slider clearance is shown. The masses of the links are
m1 = 0.008 kg, m2 = 0.038 kg, and m3 = 0.015 kg. The mass moments of inertia for the
links are IG1 = 6.733?10?6 Kg m3, IG2 = 6.925?10?4 Kg m3, and IG3 = 2.220?10?6
Kg m3. The lengths of links are L1=0.1 m, L2=0.47 m, and L3 = 0.047 m. The nominal
width of the slider (link 3) is l3 = 0.025 m. The distance between the pin joints A and
C is AC = 0.28 m. The kinetic coefficient of friction ?k=0.3, the static coefficient of
friction ?s=0.35, and the coefficient of restitution e = 0.4 are used. These values are
constant through the investigation. The analysis is performed for different values of the
103
Figure 4.9: Comparison of simulation resultsobtainedusing Mathematicaand Working
Model.
104
clearance c, varying the nominal angular velocity of the link 1, ?10. The torque of the
motor acting at joint A is chosen as Mm = M0(1??1/?10), where M0 = 1 Nm.
Figure 4.10.a shows the vertical trajectory for the center of mass G2 of the link 2,
yG2, in the state space for zero clearance (c=0 mm). On the three-dimensional graphic,
the coordinate of the position yG2(t) is plotted along the coordinate yG2(t+T) and the
coordinate yG2(t+2T), where T = 3 is the time lag. The trajectory is a closed loop and
the motion is periodic. In this case, the largest Lyapunov exponent is ? = 0 and all the
other exponents are less than zero, that is, a periodic orbit.
Figure 4.10.b shows the vertical trajectory yG2 in the state space for nonzero clear-
ance c=1 mm, and ?10 = 200 rpm. The curve is not closed, that is, an unstable orbit.
The largest Lyapunov exponent calculated is positive, denoting the chaotic behavior of
the system.
Next the largest Lyapunov exponent is computed for a set of simulation results for
different values of the nominal angular velocity of the crank: ?10=50 rpm, ?10=100 rpm,
?10=150 rpm, and ?10=200 rpm. Figure 4.11 shows the results for the clearances: c=0.5
mm (Fig. 4.11.a), c=1 mm (Fig. 4.11.b), and c=1.5 mm (Fig. 4.11.c). For constant
clearance (c=constant), and for larger values of the nominal angular velocity ?10 one
can obtain larger values of the Lyapunov exponent ?. For c=0.5 mm, ?10=50 rpm, it
results ?=20.54, and for c=0.5 mm, ?10=200 rpm, it results ?=26.55. Also, for constant
nominal angular velocity (?10=constant), and for larger values of the clearance c one
can obtain larger values of the Lyapunov exponent ?. For ?10=100 rpm, c=0.5 mm, it
results ?=24.66, and for ?10=100 rpm, c=1.5 mm, it results ?=28.90.
105
Figure 4.10: Trajectoryofthe vertical coordinate y
G
2
in thestate spacefor:a. zero
clearance(c =0mm);b.nonzeroclearance (c =1mm).
b.
a.
106
Figure 4.11: LargestLyapunovexponentcomputed forasetofvaluesofthe nominal
angularvelocity !
10
andfor theclearances:a. c =0:5 mm; b. c =1mm; c. c =1:5
mm.
c.
b.
a.
107
4.7 Conclusions
The dynamic analysis of a planar mechanism with clearance at the sliding joint is
presented. The mathematical model shows that either of four possible contact modes
can occur during motion, and the conditions for switching from one case to another.
Either contacts or impacts are detected at the contact points between the connecting
rod and the slider.
The results present the influence of the slider clearance and the crank speed on
the stability of the system. The Lyapunov exponents are computed for the simulated
data and used as a diagnostic tool. For the mechanism with no clearance, the motion
is periodic. Chaotic motion is observed for the mechanism with slider clearance. The
largest Lyapunov exponents are compared for different crank speeds at different values
of the clearance. For a constant value of the clearance, larger Lyapunov exponents
correspond to higher crank speeds.
Chapter 5
Structural synthesis of spatial mechanisms
A new structural synthesis of spatial mechanisms is studied based on the system
group classification. New spatial system groups of different families with one, two, and
three independent contours are presented. Several structure configurations of system
groups with the same number of independent contours can be obtained for a given fam-
ily. The advantage of the analysis of spatial mechanisms based on the system group
classification lies in its simplicity. The solution of mechanisms can be obtained by com-
posing the partial solutions of system groups.
5.1 Introduction
Structural synthesis of mechanisms with the specified number of contours and joint
types is necessary in order to systematize the creative design process. The structural
synthesis of mechanisms was accomplished using the graph theory [58, 59]. Tsai [60]
applied the graph theory, combinatorial analysis, and computer algorithms to systemat-
ically enumerate all possible mechanism topologies having same degrees of freedom and
joint types. Belfiore and Pennestri [61] elaborated a method for automatically drawing
kinematic chains with specified number of contours using the graph theory. Sen and
Mruthynjaya [62] studied the singularities in the workspace of planar closed-loop manip-
ulators. The singularities are determined using the centers of rotation for closed kine-
matic chains with two degrees of freedom. A classification of Assur groups with multiple
joints is presented by Jinkui and Weiqing [63]. Designers also generated collections of
108
109
mechanisms classified according to their functional characteristics[64]. Using the theory
of symmetric groups, Tuttle and Peterson [65] generated planar linkages by contraction
and expansion on a base structure. Huang and Huang [66] described a computer-aided
method to generate planar kinematic chains using the approach of contracted link ad-
jacency matrix. The methodology developed by Chiou and Kota [67] systematically
generates alternate mechanism concepts using symbolic matrices and constraint vectors
representing a library of mechanisms building blocks. Rao and Deshmukh [68] presented
a method to generate distinct kinematic chains that does not require the test of iso-
morphism. Shen, Ting and Yang [69] offers a general and versatile method to identify
the possible configurations up to twenty-nine types of basic kinematic chains contain-
ing up to four independent contours. A method of computer-aided structure synthesis
of multi-loop three-dimensional kinematic chains was presented by Shujun [70]. Struc-
tural synthesis of planar and spatial mechanisms with bars was studied by Popescu and
Ungureanu [71].
Given the required inputs to any single or multiple degree of freedom mechanism,
the mechanism can always be decomposed into system groups. The advantage of the
system group classification lies in the fact that the global solution can be obtained by
composing the partial solutions. Using subroutines for the system groups the spatial
mechanisms can be analyzed in a systematic way. The purpose of this article is to offer
a general method to determine all the configurations of complex spatial system groups
and to automate the process. Spatial mechanisms can always be decomposed into system
groups. The solution of mechanisms can be obtained by composing the partial solutions
of system groups.
110
5.2 Degree of freedom and family
The number of independent coordinates that uniquely determine the relative po-
sition of two links connected by a joint is called the degree of freedom of the joint.
Alternatively, the term joint class is introduced. A joint is called of the j-th class if it
diminishes the relative motion of a rigid linked bodies by j degrees of freedom (i.e., j
scalar constraint conditions correspond to the given kinematic pair). It follows that such
a joint has (6j) independent coordinates.
The family f of a mechanism is the number of degrees of freedom that are eliminated
from all the links of the system. A free body in space has six degrees of freedom. A
system of family f consisting of n movable links has (6?f)n degrees of freedom. Each
joint of class j diminishes the freedom of motion of the system by (j ? f) degrees of
freedom. Denoting the number of joints of class k as ck, it follows that the number of
degrees of freedom M of a particular system is
M = (6?f)n?
5summationdisplay
j=f+1
(j ?f)cj. (5.1)
In the literature this is referred to as the Dobrovolski formula.
For the general case of planar mechanisms, mechanisms of family f = 3, the number
of degrees of freedom M is calculated as
M = 3n?
5summationdisplay
j=4
(j ?3)cj = 3n?2c5 ?c4, (5.2)
where n is the number of moving links, c5 is the number of full joints, and c4 is the
number of half joints. The most common types of planar and spatial joints are shown
111
in Fig. 5.1. Two planar joints of the class 5 (one degree of freedom joints) are shown, a
slider joint in Fig. 5.1.a and a pin joint in Fig. 5.1.b. Figure 5.1.c represents a cylindrical
joint of the class 4 (two degrees of freedom joints). A spherical joint of the class 3 (three
degree of freedom joints) is shown in Fig. 5.1.d.
Of special interest are the kinematic chains which do not change their degree of
freedom after being connected to an arbitrary system. Kinematic chains defined this
way are called system groups. A structurally new system can be created connecting
them to or disconnecting them from a given system. For the case of planar systems,
from Eq. (5.2) one can obtain
3n?2c5 = 0, (5.3)
according to which the number of system group links n is always even. The simplest
fundamental kinematic chain is the binary group with two links (n = 2) and three full
joints (c5 = 3). This binary group is called dyad.
The cartesian reference frame xOyz is chosen for the mechanical systems (Fig.5. 2).
The rotations about the axis are represented by R and the translations along the axis
are represented by T.
5.3 Independent contours
A contour or loop is a configuration described by a polygon consisting of links
connected by joints. The presence of loops can be used to determine the type of kinematic
chains. Closed kinematic chains have one ore more loops so that each link and each joint
is contained in at least one loop. A closed kinematic chain has no open attachment points.
112
Figure 5.1: Typesofjoints: a. Slider joint(class5); b. Pinjoint (class 5);c.Cylindrical
joint(class4); d. Sphericaljoint (class 3).
b.
d.
c.
a.
113
Figure 5.2: Thecartesian spatialreference frame xOyz.
114
An open kinematic chain contains no loops. Mixed kinematic chains are a combination
of closed and open kinematic chains.
A contour with at least one link that is not included in any other contour of the chain
is called independent contour. The number of independent contours N of a kinematic
chain can be computed as
N = c?n, (5.4)
where c is the number of joints, and n is the number of moving links.
Planar kinematic chains are presented in Fig.5. 3. The kinematic chain shown in
Fig. 5.3.a has two moving links 1 and 2 (n = 2), three joints (c = 3), and one independent
contour (N = c?n = 3?2 = 1). This kinematic chain is a dyad. In Fig. 5.3.b, a new
kinematic chain is obtained by connecting the free joint of the link 1 to the ground (link
0). In this case, the number of independent contours is also N = c ? n = 3 ? 2 = 1.
The kinematic chain shown in Fig. 5.3.c has three moving links 1, 2, and 3 (n = 3), four
joints (c = 4), and one independent contour (N = c ? n = 4 ? 3 = 1). A closed chain
with three moving links 1, 2, and 3 (n = 3), and one fixed link 0, connected by four
joints (c = 4) is shown in Fig. 5.3.d. This is a four-bar mechanism. In order to find the
number of independent contours, only the of moving links are considered. Thus, there
is one independent contour (N = c ? n = 4 ? 3 = 1). The kinematic chain presented
in Fig. 5.3.e has four moving links 1, 2, 3, and 4 (n = 4), and six joints (c = 6). There
are three contours: 1-2-3, 1-2-4, and 3-2-4. Only two contours are independent contours
(N = 6?4 = 2).
Spatial kinematic chains are depicted in Fig. 5.4. The kinematic chain shown in
Fig. 5.4.a has five links 1, 2, 3, 4, and 5 (n = 5), six joints (c = 6), and one independent
115
Figure 5.3: Planar kinematicchains.
e.
c. d.
b.
a.
116
contour (N = c?n = 6?5 = 1). For the spatial kinematic chain shown in Fig. 5.4.b,
there are six links 1, 2, 3, 4, 5, and 6 (n = 6), eight joints (c = 8), and three contours
1-2-3-4-5, 1-2-3-6, and 5-4-3-6. In this case, two of the contours are independent contours
(N = c?n = 8?6 = 2).
5.4 Spatial system groups
One can determine the system groups for spatial mechanisms by analogy to the
system groups for the planar mechanisms. The system groups have the degree of freedom
M = 0. All possible system groups can be determined for each family of chains.
For the family f = 0, for system groups, from Eqs. (5.1) and (5.4) the mobility is
M = 6n?5c5 ?4c4 ?3c3 ?2c2 ?c1 = 0, (5.5)
and the number of moving links is
n = c?N. (5.6)
From Eqs. (5.5) and (5.6) one can express the number of joints of class 5 as
c5 = 6N ?5c1 ?4c2 ?3c3 ?2c4, (5.7)
and the number of moving links as
n = ?N +c1 +c2 +c3 +c4 +c5. (5.8)
117
Figure 5.4: Spatial kinematicchains.
a.
b.
118
For the family f = 1, c1 = 0 and it results
c5 = 5N ?4c2 ?3c3 ?2c4, n = ?N +c2 +c3 +c4 +c5. (5.9)
For the family f = 2, c1 = 0, c2 = 0 and it results
c5 = 4N ?3c3 ?2c4, n = ?N +c3 +c4 +c5. (5.10)
For the family f = 3, c1 = 0, c2 = 0, c3 = 0 and it results
c5 = 3N ?2c4, n = ?N +c4 +c5. (5.11)
For the family f = 4, c1 = 0, c2 = 0, c3 = 0, c4 = 0 and it results
c5 = 2N, n = ?N +c5. (5.12)
Using the above conditions, all the possible solutions for spatial system groups can be
determined. The number of joints c1, c2, ,c3, and c4 are cycled from 0 to w, where w
is a positive integer, for system groups with one or more independent contours (N ? 1).
The number of joints c5 and the number of moving links n are computed for each system
group. An acceptable solution has to verify the conditions n > 0 and c5 > 0. In Table 1,
the number of possible solutions is presented for some values of w between 0 and 40
and for kinematic chains with one contour (N = 1), two contours (N = 2), and three
contours (N = 3). For N = 1 and w ? 3, there are 23 possible solutions. For N = 2,
there are 85 solutions for w ? 6, and for N = 3 there are 220 solutions for w ? 9.
119
5.4.1 System groups with one independent contour
The combinations of spatial system groups with one independent contour (N = 1)
are presented in Table 2. The number of joints c1, c2, c3, and c4 are cycled from 0 to
3, and the number of joints c5 and the number of moving links n are computed. System
groups from Table 2 are exemplified next for each of the families f = 0, 1, 2, 3, and 4.
For the family f = 0, four system groups are illustrated in Fig. 5.5. The values c5
and n are computed from Eqs. (5.7) and (5.8), respectively. A spatial system group with
no joints of class 1, 2, 3, and 4 (c1 = c2 = c3 = c4 = 0) is shown in Fig. 5.5.a. The system
group has six joints of class 5 (c5 = 6(1) = 6), and five moving links (n = ?1 + 6 = 5).
A system group with one joint of class 4 (c4 = 1) and no joints of class 1, 2, and 3
(c1 = c2 = c3 = 0) is shown in Fig. 5.5.b. The system group has four joints of class 5
(c5 = 6(1) ? 2(1) = 4), and four moving links (n = ?1 + 1 + 4 = 4). A system group
with two joints of class 4 (c4 = 2) and no joints of class 1, 2, and 3 (c1 = c2 = c3 = 0) is
shown in Fig. 5.5.c. The system group has two joints of class 5 (c5 = 6(1)?2(2) = 2),
and four moving links (n = ?1 + 2 + 2 = 3). A system group with one joint of class 3
(c3 = 1) and no joints of class 1, 2, and 4 (c1 = c2 = c4 = 0) is shown in Fig. 5.5.d. The
system group has three joints of class 5 (c5 = 6(1)?3(1) = 3), and three moving links
(n = ?1+1+3 = 3).
The spatial mechanism presented in Fig. 5.6 is built from the system group shown in
Fig. 5.5.b. The mechanism has one degree of freedom (M = 6n?5c5?4c4?3c3?2c2?c1 =
6(5) ? 5(5) ? 4(1) = 1). The driver link is the link 5. The relative linear velocities are
symbolized by vij and the relative angular velocities are symbolized by ?ij, where i and j
120
w 0 1 2 3 4 5 6 7 8 9 10 20 30
N = 1 5 18 22 23 23 23 23 23 23 23 23 23 23
N = 2 5 30 62 76 82 84 85 85 85 85 85 85 85
N = 3 5 31 100 158 190 205 214 218 218 220 220 220 220
Table 5.1: The number of configurations of system groups with one, two and three
independent contours (N = 1, 2, and 3)
Index f c1 c2 c3 c4 c5 n
1 0 0 0 0 0 6 5
2 0 0 0 0 1 4 4
3 0 0 0 0 2 2 3
4 0 0 0 0 3 0 2
5 0 0 0 1 0 3 3
6 0 0 0 1 1 1 2
7 0 0 0 2 0 0 1
8 0 0 1 0 0 2 2
9 0 0 1 0 1 0 1
10 0 1 0 0 0 1 1
11 1 0 0 0 0 5 4
12 1 0 0 0 1 3 3
13 1 0 0 0 2 1 2
14 1 0 0 1 0 2 2
15 1 0 0 1 1 0 1
16 1 0 1 0 0 1 1
17 2 0 0 0 0 4 3
18 2 0 0 0 1 2 2
19 2 0 0 0 2 0 1
20 2 0 0 1 0 1 1
21 3 0 0 0 0 3 2
22 3 0 0 0 1 1 1
23 4 0 0 0 0 2 1
Table 5.2: The configurations of system groups with one independent contour (N = 1)
121
Figure 5.5: System groups with oneindependent contour(N =1)ofthe family f =0.
d.
c.
b.
a.
122
are the numbered links. Only one relative velocity is represented on the reference frame
in order to show that the respective motion exists.
For the family f = 1, three systems groups are depicted in Fig. 5.7. The values c5
and n are computed from Eq. (5.9). The missing translations and rotations with respect
to the axis of the reference frame xOyz are specified further on for each system group.
A spatial system group with no joints of class 1, 2, 3, and 4 (c1 = c2 = c3 = c4 = 0) is
shown in Fig. 5.7.a. The system group has five joints of class 5 (c5 = 5(1) = 5), and four
moving links (n = ?1+5 = 4). There are no rotations Rx for the links. A system group
with no joints of class 1, 2, and 3 (c1 = c2 = c3 = 0) and one joint of class 4 (c4 = 1) is
shown in Fig. 5.7.b. The system group has three joints of class 5 (c5 = 5(1)?2(1) = 3),
and three moving links (n = ?1+1+3 = 4). There are no translations Tz for the links.
A system group with one joint of class 3 (c3 = 1) and no joints of class 1, 2, and 4
(c1 = c2 = c4 = 0) is shown in Fig. 5.7.c. The system group has two joints of class 5
(c5 = 5(1)?3(1) = 2), and two moving links (n = ?1+3 = 2). There are no translations
Ty for the links.
For the family f = 2, four system groups are presented in Fig. 5.8. The values c5
and n are computed from Eq. (5.10). Two spatial system groups with no joints of class
1, 2, 3, and 4 (c1 = c2 = c3 = c4 = 0) are shown in Figs. 8.a and 8.b. The system groups
have four joints of class 5 (c5 = 4(1) = 4), and three moving links (n = ?1+4 = 3). For
the system group in Fig. 5.8.a, there are no translations Tx and no rotations Ry for the
links. For the system group in Fig. 5.8.b, there are no translations Ty and no rotations
Rx for the links. A system group with no joints of class 1, 2, and 3 (c1 = c2 = c3 = 0)
and one joint of class 4 (c4 = 1) is shown in Fig. 5.8.c. The system group has two joints
123
Figure 5.6: Spatial mechanismwithone independent contourand asystemgroup of the
family f =0.
124
Figure 5.7: System groups with oneindependent contour(N =1)ofthe family f =1.
c.
b.
a.
125
of class 5 (c5 = 4(1) ? 2(1) = 2), and two moving links (n = ?1 + 1 + 2 = 2). There
are no translations Tz and no rotations Ry for the links. A system group with one joint
of class 3 (c3 = 1) and no joints of class 1, 2, and 4 (c1 = c2 = c4 = 0) is shown in
Fig. 5.8.d. The system group has one joints of class 5 (c5 = 4(1) ? 3(1) = 1), and one
moving link (n = ?1+1+1 = 1). There are no translations Tx and Tz for the links.
The spatial mechanism presented in Fig. 5.9 is built from the system group shown
in Fig. 5.8.b. The mechanism has one degree of freedom (M = 4n ? 3c5 ? 2c4 ? c3 =
4(4)?3(5) = 1). The link 4 is the driver link.
For the family f = 3, three system groups are presented in Fig. 5.10. The values c5
and n are computed from Eq. (5.11). Three system groups with no joints of class 1, 2, 3,
and 4 (c1 = c2 = c3 = c4 = 0) are shown in Fig. 5.10. The system groups have three
joints of class 5 (c5 = 3(1) = 3), and two moving links (n = ?1 + 3 = 2). There are no
translations Tx and no rotations Ry and Rz for the system group in Fig. 5.10.a. There are
no translations Tx and no rotations Rx and Rz for the system group in Fig. 5.10.b. There
are no translations Tz and no rotations Rx and Ry for the system group in Fig. 5.10.c.
For the family f = 4, two planar system groups with no joints of class 1, 2, 3, and
4 (c1 = c2 = c3 = c4 = 0) are shown in Fig. 5.11. The values c5 and n are computed
from Eq. (5.12). for each system group, there are two joints of class 5 (c5 = 2(1) = 2),
and one moving link (n = ?1 + 2 = 1). Also, there are two planar translations for the
links and thus the family of the systems is f = 6?2 = 4.
126
Figure 5.8: System groups with oneindependent contour(N =1)ofthe family f =2.
d.
c.
b.
a.
127
Figure 5.9: Spatial mechanismwithone independent contourand asystemgroup of the
family f =2.
128
Figure 5.10: System groups with oneindependent contour(N =1)ofthe family f =3.
a.
b.
c.
129
Figure 5.11: System groups with oneindependent contour(N =1)ofthe family f =4.
a.
b.
130
5.4.2 System groups with two independent contours
Spatial system groups with two independent contours (N = 2) are presented. The
number of joints c1, c2, c3, and c4 are cycled, and the number of joints c5 and the number
of moving links n are computed. Examples of system groups with N = 2 are described
next for each of the families f = 1, 2, 3, and 4.
For the family f = 1, a system group is depicted in Fig. 5.12. The system group
has no joints of class 1, 2, 3, and 4 (c1 = c2 = c3 = c4 = 0). There are ten joints of class
5 (c5 = 5(2) = 10), and eight moving links (n = ?2+10 = 8). There are no translations
Tx for the links.
For the family f = 2, two system groups are illustrated in Fig. 5.13. A system group
with no joints of class 1, 2 and 3 (c1 = c2 = c3 = 0) and one joint of class 4 (c4 = 1) is
shown in Fig. 5.13.a. The system group has six joints of class 5 (c5 = 4(2)?2(1) = 6),
and five moving links (n = ?2+1+6 = 5). There are no translations Tx and no rotations
Rx for the links. A system group with no joints of class 1 and 2 (c1 = c2 = 0), one joint
of class 3 (c3 = 1), and one joint of class 4 (c4 = 1) is shown in Fig. 5.13.b. The system
group has three joints of class 5 (c5 = 4(2) ? 3(1) ? 2(1) = 3), and three moving links
(n = ?2+1+1+3 = 3). There are no translations Tx and Ty for the links.
For the family f = 3, three system groups are presented in Fig. 5.14. A system
group with no joints of class 1, 2, 3, and 4 (c1 = c2 = c3 = c4 = 0) is shown in
Fig. 5.14.a. The system group has six joints of class 5 (c5 = 3(2) = 6), and four moving
links (n = ?2 + 6 = 4). There are no translations Tx, Ty, and Tz for the links. A
spatial system group and a planar system group with no joints of class 1, 2 and 3
(c1 = c2 = c3 = 0) and one joint of class 4 (c4 = 1) are shown in Fig. 5.14.b and Fig.14.c,
131
Figure 5.12: System groupwithtwo independent contours (N =2)ofthe family f =1.
132
Figure 5.13: System groups with twoindependent contours (N =2)ofthe family f =2.
b.
a.
133
respectively. The system groups have four joints of class 5 (c5 = 3(2) ? 2(1) = 4), and
three moving links (n = ?2 + 1 + 4 = 3). There are no translations Ty, Tz and no
rotations Rz for the spatial system in Fig. 5.14.b.
For the family f = 4, a planar system group with no joints of class 1, 2, 3, and 4
(c1 = c2 = c3 = c4 = 0) is shown in Fig. 5.15. The system group has four joints of class
5 (c5 = 2(2) = 4), and two moving links (n = ?2+4 = 2).
The spatial mechanism shown in Fig. 5.16 contains a system group of the family
f = 0 that has c1 = c2 = 0, c3 = 1, c4 = 2, c5 = 6(2) ? 3(1) ? 2(2) = 5, and
n = ?2 + 1 + 2 + 5 = 6. The mechanism has two degrees of freedom M = 6n ? 5c5 ?
4c4 ?3c3 ?2c2 ?c1 = 6(8)?5(7)?4(2)?3(1) = 2. The links 7 and 8 are driver links.
5.4.3 System groups with three independent contours
Spatial system groups with three independent contours (N = 3) are presented. The
number of joints c1, c2, c3, and c4 are cycled and the number of joints c5 and the number
of moving links n are computed. System groups with N = 3 are exemplified next for
each of the families f = 2, 3, and 4.
For the family f = 2, a spatial system group with no joints of class 1 and 2 (c1 =
c2 = 0), one joint of class 3 (c3 = 1), and one joint of class 4 (c4 = 1) is shown in
Fig. 5.17. The system group has seven joints of class 5 (c5 = 4(3) ? 3(1) ? 2(1) = 7),
and six moving links (n = ?3+1+1+7 = 6). There are no translations Tx and Tz for
the links.
For the family f = 3, a planar system group with no joints of class 1, 2, and 3
(c1 = c2 = c3 = 0) and one joint of class 4 (c4 = 1) is depicted in Fig. 5.18. The
134
Figure 5.14: System groups with twoindependent contours (N =2)ofthe family f =3.
a.
b.
c.
135
Figure 5.15: System groupwithtwo independent contours (N =2)ofthe family f =4.
136
Figure 5.16: Spatial mechanismwithtwo independent contours andasystem groupof
thefamily f =0.
137
Figure 5.17: System groupwiththree independent contours (N =3)ofthe family f =2.
138
system group has seven joints of class 5 (c5 = 3(3) ? 2(1) = 7), and five moving links
(n = ?3+1+7 = 5).
For the family f = 4, a planar system group with no joints of class 1, 2, 3, and 4
(c1 = c2 = c3 = c4 = 0) is shown in Fig. 5.19. The system group has six joints of class 5
(c5 = 2(3) = 6), and three moving links (n = ?3+6 = 3).
The spatial mechanism presented in Fig. 5.20 contains a system group of the family
f = 0 that has c1 = c2 = 0, c3 = 3, c4 = 4, c5 = 6(3) ? 3(3) ? 2(4) = 1, and
n = ?3 + 3 + 4 + 1 = 5. The mechanism has three degrees of freedom M = 6n?5c5 ?
4c4?3c3?2c2?c1 = 6(8)?5(4)?4(4)?3(3) = 3. The links 6, 7 and 8 are driver links.
5.5 Conclusions
The method of computer-aided structural synthesis of spatial mechanisms presented
in this paper is based essentially on system group formation using the number of inde-
pendent contours and joints as inputs. The number of joints of different classes are cycled
for different families and several structures of spatial system groups with one, two, or
more independent contours are obtained. For a given family, different configurations of
system groups with the same number of independent contours can be obtained. Spatial
mechanisms can be structured based on spatial system groups.
139
Figure 5.18: System groupwiththree independent contours (N =3)ofthe family f =3.
140
Figure 5.19: System groupwiththree independent contours (N =3)ofthe family f =4.
141
Figure 5.20: Spatial mechanismwiththree independent contours andasystem groupof
thefamily f =0.
Chapter 6
Discussions and conclusions
In this dissertation the nonlinear dynamics of a mechanical system with joint clear-
ance is investigated. Modeling this system leads to modeling contacts and impacts that
occur between the links and the joints. Also, at high speeds, the joint moment of inertia
may significantly influence the behavior of the system.
In Chapter 2, the influence of the moment of rolling friction on the energy dissipated
byfrictionduringtheimpactfordifferentvaluesofthegeometricalparametersofthelinks
is analyzed. More energy is dissipated during impact for larger values of the coefficient
of rolling friction. An energy increase is observed in some cases when the kinematic
coefficient of restitution is used to model the impact. One can partially solve this problem
and obtain energetically consistent results introducing the moment of rolling friction to
the impact equations.
In Chapter 3, the effect of joint inertia on the dynamics of kinematic chains is
presented. The application point of the joint contact forces changes its position for
different values of the slider inertia. Dynamic response characteristics of a planar robot
arm are compared for different values of the prismatic joint inertia.
In Chapter 4, the influence of the prismatic joint clearance and the crank speed
on the stability of mechanisms is studied. The Lyapunov exponents are computed for
simulated data and used as a diagnostic tool. The largest Lyapunov exponents are
compared for different crank speeds at different values of the clearance.
142
143
In Chapter 5, a computer-aided structural synthesis of spatial mechanisms is pre-
sented. The method is based essentially on system group formation using the number
of independent contours and joints as inputs. For each family, the number of joints are
cycled and several structures of spatial system groups are obtained. The solution of a
spatial mechanism can be obtained by composing the partial solutions of the system
groups.
Future work includes experimental approaches of dynamics and control of robotic
arms with joint clearance. The influence of the clearance at joints and the effect of joint
inertia on the control parameters of the systems should be investigated.
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