Machine Dynamics and System Dynamics - uni-siegen.de · 1 INTRODUCTION 1.1tasks of systems and machine dynamics Insystemdynamicsweareconcernedwiththepredictionandanalysisofthe evolution

MACHINE DYNAMICS AND SYSTEM DYNAMICS

prof. dr.-ing. c.-p. fritzen

Lecture NotesUniversität Siegen

SS 2014 – 1. Edition

Prof. Dr.-Ing. C.-P. Fritzen:Machine Dynamics and System Dynamics, LectureNotes, © SS 2014 -- 1. Edition

CONTENTS

i lecture notes 11 introduction 3

1.1 Tasks of Systems and Machine Dynamics . . . . . . . . . . . . 31.2 Modelling of Mechanical Systems . . . . . . . . . . . . . . . . . 6

1.2.1 Degree Of Freedom . . . . . . . . . . . . . . . . . . . . 61.2.2 Different Categories of Models . . . . . . . . . . . . . . 7

2 kinematics 112.1 Kinematics of Particles . . . . . . . . . . . . . . . . . . . . . . 11

2.1.1 Motion on a Straight Path . . . . . . . . . . . . . . . . 122.1.2 Description of the Motion Using Generalized Coordinates 13

2.2 Kinematics of Rigid Bodies . . . . . . . . . . . . . . . . . . . . 152.2.1 Transformation Matrices . . . . . . . . . . . . . . . . . 182.2.2 Velocity in the Inertial System . . . . . . . . . . . . . . 232.2.3 Relation Between Matrix and Vector Representation

of the Velocity . . . . . . . . . . . . . . . . . . . . . . 252.2.4 Velocity in the Body-Fixed Reference Frame . . . . . . 262.2.5 Accelerations in the Inertial System . . . . . . . . . . . 272.2.6 Acceleration in the Body-Fixed Reference Frame . . . . 282.2.7 Angular Accelerations . . . . . . . . . . . . . . . . . . 292.2.8 Systems with Constraints . . . . . . . . . . . . . . . . 29

2.3 Relative Motion of a Particle . . . . . . . . . . . . . . . . . . . 342.3.1 Relation Between Absolute and Relative Velocity . . . 352.3.2 Relation Between Absolute and Relative Acceleration . 372.3.3 Summary of the Formula for Relative Kinematics . . . 39

3 kinetics 413.1 Kinetics of a Single Particle . . . . . . . . . . . . . . . . . . . . 41

3.1.1 Momentum and Angular Momentum, Newton’s Law . 413.1.2 Rotation of a Body About a Fixed Axis . . . . . . . . 433.1.3 Kinetics of a Particle for Relative Motion . . . . . . . 453.1.4 Work and Work-Energy Principles . . . . . . . . . . . 463.1.5 Power . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

3.2 Kinetics of Rigid Bodies . . . . . . . . . . . . . . . . . . . . . 513.2.1 Momentum of a Rigid Body and the Momentum The-

orem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513.2.2 Angular Momentum and Moments of Inertia . . . . . 523.2.3 Angular Momentum Theorem . . . . . . . . . . . . . . 573.2.4 Change of the Reference Frame . . . . . . . . . . . . . 57

iii

iv contents

3.2.5 Eulerian Equations, Angular Momentum Theorem ina Rotating Coordinate Frame . . . . . . . . . . . . . . 61

3.2.6 Angular Momentum Theorem in a Guided CoordinateSystem . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

3.3 Kinetic Energy of a Rigid Body . . . . . . . . . . . . . . . . . 663.4 Lagrange’s Equations of Motion of 2nd. Kind . . . . . . . . . . 68

3.4.1 Conservative Systems . . . . . . . . . . . . . . . . . . . 693.4.2 Conservative and Non-conservative Forces, Rayleigh

Energy Dissipation Function . . . . . . . . . . . . . . . 703.5 Equations of Motion of a Mechanical System . . . . . . . . . . 71

3.5.1 Linearization of the Equations of Motion . . . . . . . . 723.5.2 Equation of Motion of a Linear Time-Variant and Time-

Invariant Mechanical System . . . . . . . . . . . . . . . 733.6 State Space Representation of a Mechanical System . . . . . . 74

3.6.1 The General Non-linear Case . . . . . . . . . . . . . . 743.6.2 The Linear Time-Invariant Case . . . . . . . . . . . . . 753.6.3 The Linear Time-Invariant Case in Discrete Time . . . 76

4 linear vibrations of systems with one degree offreedom 794.1 General Classification of Vibrations . . . . . . . . . . . . . . . 794.2 Free Undamped Vibrations of the Linear Oscillator . . . . . . . 82

4.2.1 Equation of Motion . . . . . . . . . . . . . . . . . . . . 824.2.2 Solution of the Equation of Motion . . . . . . . . . . . 834.2.3 Complex Notation . . . . . . . . . . . . . . . . . . . . 844.2.4 Relation Between Complex and Real Notation . . . . . 854.2.5 Further Examples of Single Degree of Freedom Systems 864.2.6 Approximate Consideration of the Spring Mass . . . . 87

4.3 Free Vibrations of a Viscously Damped Oscillator . . . . . . . 894.3.1 Equation of Motion . . . . . . . . . . . . . . . . . . . . 894.3.2 Solution of the Equation of Motion . . . . . . . . . . . 91

4.4 Forced Vibrations From Harmonic Excitation . . . . . . . . . . 954.4.1 Excitation with Constant Force Amplitude . . . . . . . 964.4.2 Harmonic Force from Imbalance Excitation . . . . . . . 1014.4.3 Support Motion / Ground Motion . . . . . . . . . . . . 102

4.5 Excitation by Impacts . . . . . . . . . . . . . . . . . . . . . . . 1044.5.1 Impact of Finite Duration . . . . . . . . . . . . . . . . 1044.5.2 DIRAC-Impact . . . . . . . . . . . . . . . . . . . . . . 106

4.6 Excitation by Forces with Arbitrary Time Functions . . . . . . 1074.7 Periodic Excitations . . . . . . . . . . . . . . . . . . . . . . . . 108

4.7.1 Fourier Series Representation of Signals . . . . . . . . . 1084.7.2 Forced Vibration Under General Periodic Excitation . 110

4.8 Vibration Isolation of Machines . . . . . . . . . . . . . . . . . 114

contents v

4.8.1 Forces on the Environment Due to Excitation by Iner-tia Forces . . . . . . . . . . . . . . . . . . . . . . . . . 115

4.8.2 Tuning of Springs and Dampers . . . . . . . . . . . . . 1175 vibration of linear multiple-degree-of-freedom

systems 1215.1 Equation of Motion . . . . . . . . . . . . . . . . . . . . . . . . 1215.2 Influence of the Weight Forces and Static Equilibrium . . . . . 1225.3 Ground Excitation . . . . . . . . . . . . . . . . . . . . . . . . . 1245.4 Free Undamped Vibrations of the Multiple-Degree-of-Freedom

System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1265.4.1 Eigensolution, Natural Frequencies and Mode Shapes

of the System . . . . . . . . . . . . . . . . . . . . . . . 1265.4.2 Modal Matrix, Orthogonality of the Mode Shape Vectors1275.4.3 Free Vibrations, Initial Conditions . . . . . . . . . . . 1305.4.4 Rigid Body Modes . . . . . . . . . . . . . . . . . . . . 130

5.5 Forced Vibrations of the Undamped Oscillator under HarmonicExcitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

ii appendix 135a einleitung - ergänzung 137b kinematics - appendix 139

b.1 General 3-D Motion in Cartesian Coordinates . . . . . . . . . . 139b.2 Three-Dimensional Motion in Cylindrical Coordinates . . . . . 140b.3 Natural Coordinates, Intrinsic Coordinates or Path Variables . 141

c fundamentals of kinetics - appendix 145c.1 Special cases for the calculation of the angular momentum . . 145

d vibrations - appendix 149d.1 Excitation with constant amplitude of force - Complex Approach149d.2 Excit. with constant amp. of force - Alternative Complex App. 151d.3 Fourier Series - Alternative Real Representation . . . . . . . . 152d.4 Fourier Series - Alternative Complex Representation . . . . . . 152d.5 Magnification Functions . . . . . . . . . . . . . . . . . . . . . . 154

e literature 157f formulary 159

ABBREVIAT IONS

DOF Degree Of Freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6SDOF Single Degree Of Freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79MDOF Multiple Degrees Of Freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79EOM Equation Of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89MBS Multi-body system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8FEM Finite-Element Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

vi

Part I

LECTURE NOTES

1INTRODUCTION

1.1 tasks of systems and machine dynamics

In system dynamics we are concerned with the prediction and analysis of theevolution of the system’s state. This state can be expressed in terms of tem-perature pressure, chemical concentrations, electrical currents, wear or damageetc.

Especially in machine and structural dynamics, we deal with the motion interms of displacements, velocities and accelerations as well as with dynamicinternal forces and moments in machine and structures. What we are interestedin are e.g. the precision with which a robot can follow a given trajectory, theoccurrence of unstable motion, amplitudes of vibration in a stationary stateor the transient behaviour following to a disturbance of the system

Some typical fields of applications can be found in

• automotive engineering (vehicle dynamics, vibration and noise),

• railway systems (high speed train ICE) and magnetic levitation systems(Transrapid),

• space vehicles and satellites,

• airplanes and helicopters,

• robotic systems,

• milling machines,

• printing machines,

• internal combustion engines,

3

4 introduction

• turbomachinery (steam turbines, water turbines, wind turbines, pumpsand turbo compressors → rotor dynamics is a special field of machinedynamics,

• biomechanical systems: walking and mobility prosthetics,

• . . . .

Mechatronics

Other names are used for this special class of problems, including controlledMechatronicsmachines, smart machines, smart structures, and intelligent machines. Theterm mechatronics is mainly in use in Europe and Japan where mechatronicdevices such as magnetic bearings or automated cameras have been pioneered.A large-scale application is the mag-lev train that has been developed in Japanand Germany. Active magnetic bearings for small and large rotating machinessuch as pipeline pumps and machine tool spindles have been developed inSwitzerland, France or Japan. In the USA a significant amount of researchand development has been directed forward toward Micro-Electro-MechanicalSystems or MEMS. Other fields of interest are self-diagnosis of machines andstructures using built-in diagnostic devices and computational intelligence aswell as vibration and noise control.

The distinguishing feature of most of these systems compared to classical con-trolled machines has been the incorporation of sensing, actuation, and intelli-gence in producing and controlling motion in machines and structures. Thismeans that we have to integrate control and intelligence into the mechanicaldesign from the very beginning and not as an add-on after the machine isdesigned.

Dynamic Failures1

While dynamic analysis in engineering is often used to create motions in physi-Dynamic Failurescal systems, in many unwanted dynamic failures are to be avoided. Such failuresinclude:

• large deflections,

1 F.C. Moon, Applied Dynamics

1.1 tasks of systems and machine dynamics 5

• fatigue of materials from high or low amplitude vibrations,

• motion-induced fracture,

• dynamic instability, e.g. flutter or chatter,

• impact-induced local damage (e.g. delaminations of plies in carbon-fibrereinforced plastics),

• motion-induced noise,

• instability about a steady motion, e.g. wheels on rails,

• thermal heating due to dynamic friction.

Avoidance of Dynamic Failure

• understand the dynamics before the design becomes a product, usingsimulation tools and/or measurements,

• choose materials with enhanced properties to resist fatigue, fracture orwear, or choose materials with higher damping to minimize resonance,

• use passive damping,

• use active control,

• use internal diagnostics, sensors, limit switches , etc. to detect imminentfailure and avoid catastrophe (Structural Health Monitoring (SHM)).

Of great importance is the knowledge of the sources and phenomena of un-wanted vibrations. Vibrations may be induced by:

• oscillating and rotating machine parts with mass unbalance,

• periodic variations of the torque in internal combustion engines,

• interaction of mechanical machine parts with a fluid (turbulent windloads, self-excited vibrations , flutter),

• earthquakes (important in civil engineering but also in mechanical engi-neering in safety relevant areas like nuclear power plants),

• road roughness (road vehicles),

6 introduction

• . . . .

1.2 modelling of mechanical systems

1.2.1 Degree Of Freedom

The expression Degree Of Freedom (DOF) plays a basic role in the modelling of’A model should beas simple as possi-ble - not simpler’

A. Einstein

a system. It is an important quantity to describe the complexity of an analyticalor numerical model. In general, an increasing number of DOFs increases themodel accuracy, but also computation time and computer memory is increased.

The engineer’s art is to find a compromise between accuracy and computa-tional cost. For a mechanical system, the lower limit is given by the numberof DOFs of the rigid body motion, the upper limit is infinity (in the case ofa continuous distributed parameter system) and can be very high in the caseof a very detailed finite-element mesh (e. g.100,000 DOFs). In practice, for adynamic analysis, it is recommended to add to the rigid body DOFs as manyelastic DOFs so that the highest excitation frequency which is of interest forthe technical problem is included in the analysis. If the lowest eigenfrequencyis much higher than the highest excitation frequency, then the machine orstructure can be modeled as a rigid system and only the DOFs of the rigidbodies are considered.

From “Engineering Mechanics” we know that a free rigid body has 6 DOFs,3 translational and 3 rotational DOFs, in 3D space, and 3 DOFs in 2D space2 translational and 1 rotational DOF. A multibody-system of n rigid bodies,which can move freely without constraints, have

ffree =n∑i=1

ffree,i (1.1)

where ffree,i is the number of DOFs of the i-th free body and

ffree,i =

6 in 3D space

3 in 2D space(1.2)

No machine would properly work, if the different parts of the machine wouldnot be connected in a certain way, e.g. by hinges, joints etc. These constraints,whose overall number is nc, reduce the number of the DOFs so that

f = ffree − nc (1.3)

1.2 modelling of mechanical systems 7

The number f of the constrained system corresponds to the number of thecoordinates which are necessary to describe uniquely the position of the system.

To realize a desired motion in a machine, we need supports, joints, etc. Allthese elements introduce constraints. According to these constraints we getconstraint forces (e.g. in a hinge, holding the two parts of the machine to-gether).

The constraints can be divided into several classes. They can depend on thepositions, velocities and sometimes explicitly on time.

Therefore, we distinguish: see section 2.2.8for further illustra-tionholonomic constraints: The constraint equation can be formulated by the

positions and time only.

non-holonomic constraints: Besides the positions and time also the ve-locities appear in the constraint equations.

rheonomic constraints: Time t?appears explicitly in the constraint equa-tion.

scleronomic constraints: Time t does not appear explicitly.

1.2.2 Different Categories of Models

Mechanical systems are characterized by

• inertia effects due to the mass of the single elements of the system,

• elasticity of the elements, in the case that we can neglect the deformationswe use a rigid body model.

In addition, we can have effects from:

• damping/friction,

• external forces (from motors, hydraulic elements, or disturbances fromthe environment).

In some applications we might have also contact or play between some bodies.This leads to nonlinear behavior of the system.

8 introduction

The inertia effects are determined by the mass density distribution and thegeometry of the body. In all considerations we keep the mass constant, however,the mass moment of inertia can depend on the actual configuration of thesystem.

The modelling of elastic elements depends on the ratio of elastic forces/inertiaforces, that means whether they can be considered as massless spring elementsor as bodies with mass, e.g. the ratio is very large for suspension spring of cars,the vertical motion of the car is very slow compared to the vibrations of thespring. With an idealization of a massless spring, modelling errors will not betoo large.

The same arguments hold true for damping elements where the ratio of damp-ing forces/inertia forces play an important role.

The properties of the real technical systems have to be described by idealizedmodels. Basically we distinguish between models with concentrated parametersan distributed parameters. That can be divided into different categories.

Multibody Systems

A Multi-body system (MBS) consists of rigid bodies, which are subjected toloads in discrete points. Forces and moments can be generated by springs,dampers, external forces (e.g. gravitation) , magnetic forces, devices like mo-tors.

A MBS-model has always a finite number of degrees of freedom. The discretestructure of the model is due to the physical discretization.

Finite-Element Systems

A model which is generated by means of the finite-element method is composedof single finite elements which are connected in the so-called nodal points. Theproperties of the elements are based on distributed elastic and inertial effects.By means of internal functions for each element the properties of the interiorof an element can be concentrated on the nodal points.

Standard elements in finite elements programs are e.g. truss, beam, plate, 2D-,or 3D solid elements.

A model based on Finite-Element Method (FEM) has a finite number of DOFs(which can become very large). In contrast to MBS, the discretization is reachedby mathematical discretization.

1.2 modelling of mechanical systems 9

Distributed Systems

The system is considered to have a distributed mass and elasticity. The equilib-rium of forces and moments are formulated for an infinitesimal small elementwhich leads to partial differential equations which may be difficult to be solvedanalytically. Distributed systems have an infinite number of DOFs.

There is no discretization and the solution is exact in the sense of the assump-tions which are made in continuum mechanics. Here, the solution is a function,not a vector.

The advantage is: the solution is exact, however as disadvantage, the geometrieswhich can be treated usually are very simple (such as beams or plates withconstant thicknesses or constant cross sections).

However, it is possible to get approximate solutions using a the classical Ritz-approach or difference methods.

Elastic Multibody Systems

This is a connection of rigid MBS and flexible elastic elements modeled byFEM. We can consider the large motions of MBS which are super imposed bysmall motions of the elastic parts of the structure. This closes the gap betweenthe two worlds of MBS and FEM.

2KINEMATICS

The following chapter is partly a repetition of the material learned in “En-gineering Mechanics” and partly a presentation of new materials which arenecessary to solve more complicated technical problems, such as the motion ofrobotic systems.

Kinematics in general deals with the motion of a single particle, rigid bodies ora system of bodies. In “Kinematics” we do not ask how this motion is generated,how large the forces and moments are in links between several bodies etc. Thisis a task of “Dynamics”. Here, we look only at the geometric relations of singlebodies or multibody systems and its behavior in time.

In kinematics the motion of a particle or a rigid body is described by positionvectors, as well as the velocities (German: Geschwindigkeit) and accelerations(German: Beschleunigung). We investigate the relations between these quan-tities which are given by the time derivatives. In general these quantities arevectors.

2.1 kinematics of particles

During the motion of a particle with a position P this point is moving onthe so-called path (German: Bahnkurve or Bahn). The actual position of Pis described uniquely by the position vector r (t) in a fixed reference coor-dinate system (German: Inertialsystem). The first derivative of the positionvector r (t) with respect to time yields the velocity vector v (t) (German:Geschwindigkeitsvektor)

v (t) =dr (t)

dt = r (t) . (2.1)

From the second derivative we get the acceleration vector a (t) (German: Beschle-unigungsvektor):

a (t) =dv (t)

dt = v (t) =d2r (t)

dt2 = r (t) . (2.2)

11

12 kinematics

x

y

z

P

r(t)

Path

Figure 2.1: Position vector and path

2.1.1 Motion on a Straight Path

The most simple case is the particle motion along a straight line. s = s (t)is the position coordinate along the straight path. We do not need a vectorrepresentation for the description for the velocity v (t) and the accelerationa (t) we immediately obtain

v (t) = s (t) (2.3)a (t) = v (t) = s (t) (2.4)

Given the position depending on time t the velocity and the acceleration canbe calculated by differentiation of the function s (t) .

In the inverse way, given the acceleration a (t), we obtain the velocity and theposition of P by integration in the following way:

v (t) = v0 +∫ t

t0a (t∗) dt∗ (2.5)

s (t) = s0 +∫ t

t0v (t∗) dt∗ (2.6)

where s0 (initial position) and v0 (initial velocity) are called the initial condi-tions.

In cases where the velocity and the acceleration are not given as a function oftime, e. g. v = v (s) or a = a (v); a = a (s) we refer to the literature1.

1 H.G. Hahn: Technische Mechanik

2.1 kinematics of particles 13

2.1.2 Description of the Motion Using Generalized Coordinates

Often we face the problem that the position, the velocity and the accelerationof a point P has to be expressed in terms of another set of variables, thegeneralized coordinates q (t), e. g. in the 3D-space, where the components qican be longitudes and angles:

q (t) =[q1 (t) q2 (t) q3 (t)

]T. (2.7)

A simple example is the motion of a particle on a circular path. We look forthe position, the velocity and the acceleration vector, respectively, in Cartesiancoordinates depending on the angle q (t) = q1 (t) = α (t) (example givenduring the lecture). The solution of this problem can be formalized and solvedusing computer algebraic systems2. Given a set of generalized coordinates

q (t) mit q (t) ∈ <m; 1 ≤ m ≤ 3,

we first have to establish the relation between the position vector and thesegeneralized coordinates (when treating multibody problem, m can be largerthan 3) which has the general form

r = r(q (t)

)(2.8)

First, we calculate the velocity and acceleration vector expressed by the gen-eralized coordinates qi. The purpose of this transformation is to express com-plicated motions of mechanisms or systems like a robot by the rotations of adrive motor, represented by an angle α (t).

2.1.2.1 Velocities

In order to get the velocity v (t) we differentiate the position vector in eq. (2.8),with respect to time, applying the chain rule. With m = 3 we get:

v = r =∂r

∂q1

dq1dt +

∂r

∂q2

dq2dt +

∂r

∂q3

dq3dt

=∂r

∂q1q1 +

∂r

∂q2q2 +

∂r

∂q3q3

=m∑i=1

∂r

∂qiqi (2.9)

2 Program systems for PCs: Mathematica, Maple, MathCad, Maple is also available as Matlab-toolbox

14 kinematics

orv (t) = r (t) = Jrq

(q (t)

)q (t) (2.10)

where Jrq is the Jacobian-Matrix containing the derivatives with respect toJacobian-Matrixthe generalized coordinates:

Jrq =

∂r1∂q1

∂r1∂q2

∂r1∂q3

∂r2∂q1

∂r2∂q2

∂r2∂q3

∂r3∂q1

∂r3∂q2

∂r3∂q3

(2.11)

so that v1

v2

v3

=

∂r1∂q1

∂r1∂q2

∂r1∂q3

∂r2∂q1

∂r2∂q2

∂r2∂q3

∂r3∂q1

∂r3∂q2

∂r3∂q3

q1

q2

q3

Note:When we have a nested relation e.g. given by r (t) = r

(p(q (t)

)), the proce-

dure is almost identical. The only thing is to use the chain rule one more timeto get the velocity v (t) = JrpJpq q (t).

If the position vector depends not only on the coordinates q (t) but directlyon the time t:

r = r(q (t) , t

)(2.12)

then the velocity is

v = r =∂r

∂t+

∂r

∂q1

dq1dt +

∂r

∂q2

dq2dt +

∂r

∂q3

dq3dt

=∂r

∂q1q1 +

∂r

∂q2q2 +

∂r

∂q3q3

=∂r

∂t+

m∑i=1

∂r

∂qiqi (2.13)

orv (t) = r (t) =

∂r

∂t+ Jrq

(q (t)

)q (t) (2.14)

2.2 kinematics of rigid bodies 15

2.1.2.2 Accelerations

To derive expressions for the accelerations we continue to differentiate thevelocity in eq. (2.10) with respect to time:

v (t) = r (t) = Jrq

(q (t)

)q (t)

We obtain a general relation a = a(q (t) , q (t) , q (t)

). Using the chain rule of

differentiation we get (it is a good exercise to derive this expression)

a (t) = r (t) = Jrq q (t) +Krq qQ (2.15)

Again, the Jacobian plays an important role in the first term of the right-handside which is coupled with the second derivatives of the generalized coordinates.The second term contains quadratic expressions of the first derivatives of theqi. The matrix Krq is built up by the 2nd order derivatives of the positionvector with respect to the generalized coordinates:

Krq =

[∂2r

∂q21

∂2r

∂q1∂q2

∂2r

∂q1∂q3

∂2r

∂q22

∂2r

∂q2∂q3

∂2r

∂q23

](2.16)

The order corresponds to the vector qQof the squares of velocities Index "‘Q"’ indi-

cates quadratic ex-pression of the ve-locitiesq

Q=[q2

1 2q1q2 2q1q3 q22 2q2q3 q2

3

]T(2.17)

The factor of 2 is due to the mixed derivatives which appear twice and whichcan put together in one term. All we have to do to get the accelerations isto calculate the 2nd order derivatives with respect to the qi. The Jacobian isalready known from the calculation of the velocities.

2.2 kinematics of rigid bodies

In treating the motion of a simple particle we had to consider only translations.Now, the rotation about an arbitrary axis has to be considered, too. In the3D space the rotation of the rigid body is given by the ω (t) vector with3 components representing the 3 DOFs of rotation.

We investigate the motion of an arbitrary point P of the rigid body. First,we have to describe the position vector and subsequently, the velocity and theacceleration vector. Knowing this for any point P , the state of motion is knownfor the whole rigid body.

16 kinematics

0

A

P

rr

rA

PAP

w

Figure 2.2: Representation of a rotating rigid body

We consider a fixed reference frame with origin O. The position vector in fig. 2.2for a point P

rP = rA + rAP (2.18)can be composed of the position of the reference point A and the position vectorpointing from A to P . The choice of A is arbitrary, too. In some applications,it makes sense to choose the center of gravity, but also a point where tworigid bodies are connected by a hinge can be chosen. The well-known Eulerianformula describes the velocity of the point P of a rigid body:Eulerian Formal

vP = vA + ω× rAP (2.19)

where ω is the vector of the angular velocity (German: Winkelgeschwindigkeits-vektor). The velocity vector vA with it’s 3 components characterizes the 3translations and the cross product deliver the 3components of the rotationwhich in sum correspond to 6 DOFs of the rigid body in the 3D-space. Furtherdifferentiation with respect to time t lead to the acceleration of point P :

aP = vA + ω× rAP + ω× rAP= aA + ω× rAP + ω× (ω× rAP )

(2.20)

In the plane (the motion takes place in the x-y-plane) the equations can besimplified:

vP = vA + ω (ez × rAP ) (2.21)aP = aA + ω (ez × rAP )− ω2rAP (2.22)

Now, the vector of the angular velocity ω is perpendicular to the x-y-plane andpoints in z-direction (or in negative z-direction). Introducing a reference frame


w

0

A

P

rr

rA

PAP

eej

r

Figure 2.3: Motion of a rigid body in a plane

which is fixed on the body (and rotates with the body), and where the unitvector er always points in the direction of rAP we get a very representation ofthe velocity and the acceleration:

rP = rA + rer (2.23)vP = vA + rωeϕ (2.24)aP = aA + rωeϕ − rω2er (2.25)

withr = |rAP | . (2.26)

In the next section, we want to consider the kinematics of a rigid body asfollows:

The motion of a point P is described by means of the motion of a referencepoint A in a fixed reference frame, or a so-called inertial (I) system (in German“Inertialsystem” (I)). As seen earlier, we can write again: Position Vector in

Inertial FrameIrP = IrA + IrAP (2.27)

The subscript I indicates that we consider the vectors in the fixed referenceframe (inertial system). It turns out that it is very useful to describe the motionof point A in the (I)-system while we describe the motion of P relative to Ain a the body-fixed coordinate system (K). ((K) stands for the german word“körperfest” which means body-fixed). The reason for this is that when we deal The distance be-

tween two points ofthe rigid body isconstant

with rigid bodies the distance and orientations relative to the reference pointA and another arbitrary point P always remains constant. Hence, we expressthe vector IrAP in the body-fixed reference frame (K) by Kr

′P , where we must

keep in mind that the (K)-frame has it’s origin in point A. Furthermore, wemust consider that the instantaneous orientation of the (K)-frame in generalis not identical with the fixed frame (I). Due to the 3D-motion of the body,the (K)-frame is rotated against the fixed (I)-frame.

18 kinematics

I Pr

x

y

z y’

x’

z’

I Ar

(I)

(K)

K PrA

P

Figure 2.4: Different reference frames: fixed ref. frame (I) and moving frame (K)fixed on the moving body

We express this rotation by a rotation matrix AIK which has dimension 3× 3.We can also say: AIK represents the transformation of a vector from the (K)-into (I)-coordinates:

IrP = IrA +AIKKr′P (2.28)

Matrix AIK can characterized by 3 rotation parameters, e.g. 3 angles. We shallFrom K → I:AIK

From I → K:AKI

see later how the rotation matrix is built up .

2.2.1 Transformation Matrices

For the determination of the transformation matrix AIK by three rotationalangles defined by three elementary rotations about the x−axis with angle α,about y−axis with angle β and about z−axis with angle γ. The elementaryrotation matrix Aα maps the vector Ir in the fixed reference frame to the vectorKr. The (K)-frame is rotated about the x− axis and the angle of rotation isα.

x y

z

α

Aα =

1 0 00 cosα sinα0 − sinα cosα

(2.29)


In the same way the rotations about the y and the z axes are carried out. Therotation matrices which correspond to these rotations are denoted by Aβ andAγ , respectively:

x y

z

β Aβ =

cos β 0 − sin β

0 1 0sin β 0 cos β

(2.30)

x y

z

γ

Aγ =

cos γ sin γ 0− sin γ cos γ 0

0 0 1

(2.31)

Now, we perform all three rotations subsequently. The first rotation about thex axis yields:

Kr(1) = AαIr (2.32)

after that we carry out the rotation about the y axis

Kr(2) = AβKr

(1) = AβAαIr (2.33)

and after the third rotation we get the final position:

Kr(3) = AγAβAαIr (2.34)

Leaving the superscript away we get:

Kr = AγAβAαIr (2.35)

We can see that the transformation matrix is a product of elementary rotations.From matrix calculus we know that we are not allowed to change the order ofmultiplication of the single matrices. This leads to a different matrix product.Physically, this means that we are not allowed to change the order of rotation.This would lead to a different final position.

AKI = AγAβAα (2.36)

For our application we also need the inverse mapping AIK . We simply have totake the inverse of the matrix AKI :

AIK = A−1KI =

(AγAβAα

)−1= A−1

α A−1β A−1

γ (2.37)

20 kinematics

Because the transformation matrices are orthogonal, we can replace the inverseOrthogonalMatrix:B−1 = BT

by the much simpler transpose of the matrix

AIK = A−1KI = ATαA

TβA

Tγ =

(AγAβAα

)T(2.38)

so thatIr = AIKKr (2.39)

can be calculated explicitly:

AIK =

cos β cos γ − cos β sin γ sin β

cosα sin γ + sinα sin β cos γ cosα cos γ − sinα sin β sin γ − sinα cos βsinα sin γ − cosα sin β cos γ sinα cos γ + cosα sin β sin γ cosα cos β

(2.40)

This final matrix AIK is valid only for the pre-defined order of rotation aboutx−, y−, z−axis.

The order of rotation can be arbitrary choosen, in such cases, the transforma-tion matrix AIK is to be recalculated for each choosen order. Two differentorders of rotation are mostly used in the machine dynamic studies and arecalled Cardinian and Eular angles.

2.2.1.1 Cardanian Angles

The so-called Cardanian angles represent one possibility to describe the rota-Change the orderof the rotations isnot allowed

tion of a rigid body in a unique way. What we are doing is to carry out the 3Drotation by 3 subsequent rotations about the 3 axes of the body-fixed referenceframe. It is important to note, that is not allowed to change the order of thethree rotations! A change of the order of the subsequent rotation usually willlead to a different position at the end of the rotations3.

The Cardanian angles are defined in a way to carry-out the three subsequentCardanian anglesare carried-outabout x-y-z axes

rotations about the x-y-z axes.

ACardanianIK =(AγAβAα

)T(2.41)

3 Finite angles of rotation do not have vector character, they do not obey the commutativelaw. The order of the rotations is absolutely important. Contrary to this, infinitesimal smallrotations and angular velocities (which are based on infinitesimal small angles) have allproperties of a vector.


Elementary rotationsa

b

g b

a

g

zK

zI

xI

xK

yI

yK

Aα =

1 0 00 cosα sinα0 − sinα cosα

Aβ =

cos β 0 − sin β

0 1 0sin β 0 cos β

Aγ =

cos γ sin γ 0− sin γ cos γ 0

0 0 1

Rotation matrix

AIK =

cos β cos γ − cos β sin γ sin β

cosα sin γ + sinα sin β cos γ cosα cos γ − sinα sin β sin γ − sinα cos βsinα sin γ − cosα sin β cos γ sinα cos γ + cosα sin β sin γ cosα cos β

Angular velocities

Iω =

1 0 sin β0 cosα − sinα cos β0 sinα cosα cos β

α

β

γ

; Kω =

cos β cos γ sin γ 0− cos β sin γ cos γ 0

sin β 0 1

α

β

γ

Kinematic equation

α

β

γ

= 1cosβ

cos γ − sin γ 0

sin γ cos β cos γ cos β 0− sin β cos γ sin γ sin β 1

Kωx

Kωy

Kωz

= 1cosβ

cos β sin β sinα − sin β cosα

0 cos β cosα cos β sinα0 −. sinα cosα

Iωx

Iωy

Iωz

Table 1: Cardanian angles

This means: at first we rotate about the x-axis with an angle α (t). This leadsto a new position of the rotated coordinate system x

′-y′-z′ , where the x′-axisand the old x-axis are still identical. After that we rotate about the new y

′-axiswith an angle β (t) leading to the new orientation x′′-y′′-z′′ (with y′ = y

′′) andfinally the last rotation about the z′′-axis with an angle γ (t) leading to thefinal orientation x′′′-y′′′-z′′′ of the reference frame. An alternative possibility isto use the so-called Euler ian angles. Here, the rotations are also carried out in

22 kinematics

3 subsequent steps but the order is different: the 3 elementary rotation havethe sequence “z-x-z”4.

2.2.1.2 Eulerian Angles

The transformation we discussed before was based on the Cardanian anglesEulerian Angles ro-tate as following:

z-x-z

(see chapter 2.2.1.1), which are characterized by a special order of the rotationsabout the x-y-z axes:

AIK = ACardanIK (2.42)

Basically, we are free to choose any combination of rotations. We mentionedalready the Euler ian angles which are based on rotations about the differentaxes in the order z-x-z.

AKI = AEulerKI = AϕAϑAψ (2.43)

where we perform the following rotations

1. Rotation about the z-axis:

Aψ =

cosψ sinψ 0− sinψ cosψ 0

0 0 1

(2.44)

2. Rotation About x-axis:

Aϑ =

1 0 00 cosϑ sinϑ0 − sinϑ cosϑ

(2.45)

3. One more rotation about the z-axis (which now of course has a differentorientation in the 3D space):

Aϕ =

cosϕ sinϕ 0− sinϕ cosϕ 0

0 0 1

(2.46)

4 In the English literature the expression “Eulerian angles” is used in a more general wayincluding the Eulerian angles as described above (Type A) and the Cardanian angles (asEulerian angles, Type B), see F.C. Moon, Applied Dynamics.


In general we can choose the order of the rotations, but once we have selectedone, we have to retain this order unchanged. As mentioned already the for-malism we derived can be applied to any rotation transformations AIK . Bythis, it is possible to match the transformation to a special application withoutchanging the general formula.

2.2.1.3 Small Rotations

If the angles α, β and γ are small, we can simplify the trigonometric functionsin the rotation matrix. With the approximation for small angles

cosα ≈ 1 (2.47)

and

sinα ≈ α . (2.48)

The elementary rotations become

Aα =

1 0 00 1 α

0 −α 1

Aβ =

1 0 −β0 1 0β 0 1

Aγ =

1 γ 0−γ 1 00 0 1

(2.49)

leading to the rotation matrix eq. (2.40)

AKI = AγAβAα =

1 γ −β−γ 1 α

β −α 1

AIK = ATKI =(AγAβAα

)T=

1 −γ β

γ 1 −α−β α 1

where we have also neglected products of small angles like e.g. αβ ≈ 0 etc.

2.2.2 Velocity in the Inertial System

The velocity of point P in the inertial system can be obtained by differentiationof

IrP = IrA +AIKKr′P (2.50)

24 kinematics

with respecto to time so that

I rP = I rA + AIKKr′P (2.51)

Now, we can use the fact that the derivative of Kr′P with respect to time is

zero, because this vector is constant in the fixed-body reference frame. Andbecause Kr = AKIIr for any position vector, especially for

Kr′P = AKIIr

′P (2.52)

we get velocity in the inertial system

IvP = I rP = I rA + AIKAKIIr′P . . (2.53)

This turns out to be a skew-symmetric important formula allows to determineA skew-symmetricmatrix B is definedby:BT = −B

the velocity based on the rotation matrix. The general formulation is indepen-dent of the special rotation parameters (which were Cardanian angles in thiscase).

The matrix product AIKAKI matrix. It contains the components of the vectorof the angular velocity (see also the next section).

I ωKI = AIKAKI (2.54)

Formal replacemnt in eq. (2.53) yields

I rP = I rA + I ωKIIr′P (2.55)

orIvP = IvA + I ωKIIr

′P . (2.56)

where

I ωKI =

0 −ωz ωy

ωz 0 −ωx−ωy ωx 0

. (2.57)

The corresponding vector of the angular velocity is

Iω =

I

ωx

ωy

ωz

(2.58)

This allows us to calculate the three components of the angular velocities fromthe product I ωKI = AIKAKI


Note: Proof of skew-symmetryWe start with the relation that forward and subsequent backward transforma-tion leads to the identity, where I is the identity matrix:

AIKAKI = I

Differentiation with respect to time using the product rule leads tod

dt

(AIKAKI

)= AIKAKI +AIKAKI = 0

The derivative of the constant identity matrix is zero. This gives

AIKAKI = −AIKAKIMaking use of the orthogonality property of the transformation matricesAKI =A−1IK = ATIK we get

AIKAKI = −AIKAKI = −ATKIA

T

IK = −(AIKAKI

)Twhich shows us the desired skew-symmetry.

2.2.3 Relation Between Matrix and Vector Representation of the Velocity

Comparing the relation

IvP = IvA + I ωKIIr′P (2.59)

with the Eulerian equation discussed in section 2.2:

vP = vA + ω× rAP (2.60)

we can see immediately the similarity of both representations of the velocity.The matrix representation is very useful as we could see when we derive theangular velocity from the rotation matrix (in this case based on Cardanianangles).

Note:A vector product a× b of two vectors a and b can also be written as a matrix-vector-multiplication

a× b = a b (2.61)with

a =

ax

ay

az

und b =

bx

by

bz

. (2.62)

26 kinematics

We can show easily that the components of vectors a have to be arranged inthe matrix scheme as follows:

a =

0 −az ay

az 0 −ax−ay ax 0

(2.63)

This matrix is skew-symmetric. The symbol “ ˜ ” indicates that the vector com-ponents of a vector a have to be arranged according to the scheme of eq. (2.63).We can see that

a× b = a b =

0 −az ay

az 0 −ax−ay ax 0

bx

by

bz

=

aybz − azbyazbx − axbzaxby − aybx

. (2.64)

We know that a× b = − (b× a) so that we can derive that

a× b = − (b× a) (2.65)a b = −b a . (2.66)

2.2.4 Velocity in the Body-Fixed Reference Frame

The velocity expressed in inertial system was:

IvP = IvA + I ωKIIr′P

We know that we can transform a vector from the I into the K-system bymeans of the rotation matrix AKI and vice versa. This is valid also for thevelocity vector:

KvP = AKIIvPeq. (2.56)

= AKIIvA +AKII ωKIIr′P (2.67)

eq. (2.54)= AKIIvA +AKIAIKAKIIr

′P (2.68)

eq. (2.52)= KvA +AKIAIKKr

′P (2.69)

orKvP = KvA +AKIAIKKr

′P (2.70)

In analogy to the considerations of chapter. 2.2.2 we can derive the angularvelocity, but now expressed by the components of the K-system

K ωIK = AKIAIK (2.71)


Performing some elementary steps

K ωIK = AKIAIK

= AKIAIK

(AKIAIK

)= AKI

(AIKAKI

)AIK

= AKI

(I ωKI

)AIK

= ATIK

(I ωKI

)AIK

we get the well-known transformation law

K ωIK = ATIK

(I ωKI

)AIK . (2.72)

This allows us to express the velocity in terms of the coordinates of the bodyfixed coordinate system:

KvP = KvA + K ωIK Kv′P . (2.73)

Note:We want to note that the velocity of point A is not identical to the timederivative of KrA: KvA 6= K rA.

However we obtain the absolute velocity of A in components of theK-referenceframe:

KvA = AKI IvA = AKI I rA . (2.74)

The absolute velocity expressed in the moving, body-fixed reference frame canbe obtained only by differentiating the position vector in the inertial referenceframe I. Only the time derivative with respect to an inertial system yields anabsolute velocity (otherwise it is only a relative velocity). So, the procedure is:

• first calculate the velocity in the I-System,

• then transform it to the K-System using the rotation matrix.

2.2.5 Accelerations in the Inertial System

In order to derive an expression for the accelerations we have to differentiateonce more with respect to time t:

I rP = I rA + AIKKr′P (2.75)

28 kinematics

withKr′P = AKIIr

′P (2.76)

we get

I rP = I rA + AIKAKIIr′P . (2.77)

With these relations we already can calculate the accelerations.

We investigate the product of the transformation matrix

ddt(AIKAKI

)︸︷︷︸

I ˙ωKI

= AIKAKI + AIKAKI (2.78)

= AIKAKI + AIK

(AKIAIK

)︸︷︷︸

I

AKI (2.79)

The product AIKAKI is nothing else but the matrix of the angular velocitiesI ωKI , so that it follows that

AIKAKI = I ˙ωKI + I ωKII ωKI . (2.80)

This yields the absolute accelerations

I rP = I rA +(I ˙ωKI + I ωKII ωKI

)Ir′P (2.81)

or shorter (indices)

I rP = I rA + I

(˙ω+ ω ω

)KII

r′P . (2.82)

orIaP = IaA + I

(˙ω+ ω ω

)KII

r′P . (2.83)

We compare this result to the vector formula which was presented earlier eq. (2.20):

aP = aA + ω× rAP + ω× (ω× rAP )

and we can see that there is a perfect analogy.

2.2.6 Acceleration in the Body-Fixed Reference Frame

The transformation into the K-frame is identical as with the velocities. Wemultiply by AKI :

AKI IaP = AKI IaA +AKI I(

˙ω+ ω ω)KII

r′P (2.84)

= AKI IaA +AKI I(

˙ω+ ω ω)KI

AIK Kr′P (2.85)


KaP = AKII rP = AKIIaP . (2.86)

The acceleration of the reference point A is

KaA = AKII rA = AKIIaA (2.87)

2.2.7 Angular Accelerations

To get a transformation rule for the vector of the angular accelerations in theI and the K-frame, respectively, we start with:

I ω =ddt (Iω) (2.88)

The transformation leads to

AKI I ω = AKIddt (Iω)

= AKIddt(AIKKω

)= AKI

(AIKKω+AIKK ω

)= AKIAIKKω+AKIAIKK ω

= K ωIKKω+ K ω

Because the cross product of two identical vectors is zero:

K ωIKKω = Kω×Kω = 0 (2.89)

we finally obtainK ω = AKII ω (2.90)

2.2.8 Systems with Constraints

In section 1.2.2 we already discussed the reduction of the number of DOF byconstraints. The number of DOFs for a motion in the 3D-space for n rigidbodies was given by

f = 6n− c (2.91)

where c denotes the number of constraints. For a set of n particles (no rotations)we only get

f = 3n− c (2.92)

30 kinematics

The position of a particle can be described by the position vector r. For nparticles, we have n position vectors ri

ri = ri(q1, q2,K, q, f) i = 1, 2,K,n (2.93)

The description using the position vectors of some (n) representative points isknown as the configuration space. However as can be seen in fig. 2.5, the de-scription in the configuration space with 3 quantities (x, y, z) is over-specified,the number of DOFs is only 2. Thus, two generalized coordinates q1 and q2 aresufficient to describe the position of the particle in a unique way.

Figure 2.5: Example for a system with constraints: the particle can only move in theq1-q2-plane

2.2.8.1 Holonomic Constraints

The constraints are formulated mathematically as an vector equation:HolonomicConstraints

Φ(y) = 0 (2.94)

where y represents all the coordinates necessary to describe the constraint. Thevector Φ has as many components as mathematical constraints are available.If time t does not appear explicitly, the constraint is said to be scleronomic.

If the time appears explicitly

Φ(y, t) = 0 (2.95)

we call the constraint rheonomic.

Examples:


1. A particle which can only move along a parabolic curve:

Parabola : y = x2 leads to Φ(x, y) = y− x2 = 0 (2.96)

This constraint is holonomic-scleronomic because the time does not ap-pear explicitly.

2. A particle is fixed with a thread of constant length L .The other endof the thread is fixed to origin of the coordinate system. The particle isconstrained to move on a spherical surface with radius L:

sphere : x2 + y2 + z2 = L2 or Φ(x, y, z) = x2 + y2 + z2 −L2 = 0(2.97)

The particle has 2 DOFs. The constraint is also holonomic-scleronomic.The description using the coordinates x, y, z is over-specified, only twoquantities e.g. the two angles of spherical coordinates are required (2generalized coordinates). If the length of the thread is depending on time:L = L(t), the constraint becomes holonomic-rheonomic, the constraintis now

Φ(x, y, z, t) = x2 + y2 + z2 −L(t)2 = 0e.g. L(t) = L0 + ∆L cos(ωt)(2.98)

In spherical coordinates the motion of the particle can be described ina simple way because only the radial component r is influenced by theconstraint:

Φ = r−L(t) = 0 (2.99)so that

r =

L(t) cosψ sinϑL(t) sinψ sinϑL(t) cosϑ

= r(q1, q2, t) (2.100)

If the velocities and accelerations have to be calculated, we proceed asdescribed in chapter 2.1.2.

For holonomic-scleronomic constraints the velocities can be calculated by meansof the Jacobian

r =f∑i=1

∂r

∂qiqi = Jrq q (2.101)

For holonomic-rheonomic constraints time t appears explicitly in the constraintequations so that the direct derivative with respect to t has to be considered:

r =∂r

∂t+

f∑i=1

∂r

∂qiqi = Jrq q (2.102)

The calculation of the accelerations also follows chapter 2.1.2.

32 kinematics

2.2.8.2 Non-holonomic Constraints

While holonomic constraints only deal with the coupling of geometric quan-tities, non-holonomic constraints couple also velocities. Hence, the implicitmathematical description of a non-holonomic constraints has the form:

Ψ = Ψ(y, y, t) = 0 (2.103)

In some special cases the non-holonomic constraints can be integrated in orderto obtain equivalent geometric constraints, but in general non-holonomic con-straints are not integrable. This means that the position coordinates cannotbe derived from the integration of the constraint equations. This means thatwe can get a certain position on different paths.

As an example for a non-holonomic system we consider a rolling coin or wheelwhich moves along a curved path in the x-y-plane without sliding. For a motionwith pure rolling the velocity of the center point of the wheel is coupled withthe rotational speed.

x

y

z

Path

Coin

Figure 2.6: A rolling wheel as an example for a non-holonomic system

The wheel as a free rigid body has 6 DOFs.

y = (x, y, z,α, β, γ)T (2.104)

The center of the wheel has a constant height described by the holonomicconstraint z = R (where R is the radius of the wheel).

Φ1 = z −R = 0, (2.105)

If we further assume that the wheel is always in a vertical position, a secondholonomic constraint can be added:

Φ2 = α− α0 with α0 = 0 (2.106)

The state of the wheel can be described by 4 DOFs (4 generalized coordinates,f = 4)

q = (x, y, β, γ)T (2.107)


where the angle γ describes the tangent of the path and hence the actualdirection of the rolling wheel in the x-y-plane, the angle β describes the rotationof the wheel along the path, see fig. 2.7.

z

y

x

a

b Path

Figure 2.7: Rolling wheel

The relation between the two angles β and γ and the translation of the centerof the wheel given by the coordinates (x, y, z = R) can be derived by geometricconsiderations when we rotate the wheel about an infinitesimal small angle dβ.With ds = −Rdβwe get

dx = ds cos γ = −Rdβ cos γ (2.108)dy = ds sin γ = −Rdβ sin γ (2.109)

If we relate the infinitesimal small displacements dx and dy to an infinitesimalsmall time interval dt we get the velocities

x =dx

dt= −Rdβ

dtcos γ = −Rβ cos γ (2.110)

y =dy

dt= −Rdβ

dtsin γ = −Rβ sin γ (2.111)

Now we have found the coupling between the different velocities: the velocitiesx and y are determined by the non-holonomic constraints:

Ψ1(x, β, γ) = x+Rβ cos γ = 0 (2.112)Ψ2(y, β, γ) = y+Rβ sin γ = 0 (2.113)

In the special case that the angle γ = γ0 = const., (that means the path is astraight line) the constraint can be integrated:

x− x0 = −R cos γ0(β − β0) (2.114)y− y0 = −R sin γ0(β − β0) (2.115)

(2.116)

34 kinematics

y

xg

Path

ds ds sing

ds cosg

Figure 2.8: Geometrical quantities for the rolling constraint in the x-y-plane

Now, we have a unique relation between the angle β and the two coordinatesx, y. This means that we have only 1 DOFs: the motion can be fully describedby the angle β. Presetting the angle γ = γ0 = const. changes the problemfrom a non-holonomic to a holonomic system.

In technical systems holonomic constraints occur more frequently than non-holonomic ones.

2.3 relative motion of a particle

As is it well-known, Newton’s law is only valid in an inertial system. But oftenit is more convenient to describe the motion of a particle in a moving referenceframe. This is the case when the interest lies in examining the motion ofparticles relative to spinning bodies or the motion of spinning bodies relativeto a fixed reference frame. (Examples for this are the motion of a body onthe rotating earth or the motion of gyroscopes or similar devices with respectto the inertial space). Other examples are friction where we must know therelative velocity between the two machine parts which are in contact, or themotion of car occupants in an accelerated car during a car crash. We have todistinguish between the coordinate frame in which the vector components arewritten and the coordinate frame in which the time derivatives are taken.

We consider two reference frames: one is the inertial system (I) which is fixedInertial system (I)is fixed.Reference frame(R) can translateand rotate.

and the second is amoving reference frame (R). We wish to describe the motionof the particle P with respect to the (R)-frame. The reference frame (R) canperform a translation and a rotation.

2.3 relative motion of a particle 35

relative path of P

PMoving system r

0P

Inertial system

0

(I)

(R)

r0

Figure 2.9: Motion of particle P relative to the moving reference frame

2.3.1 Relation Between Absolute and Relative Velocity

If we look at point P from the origin of the inertial system (I) we can describeit by the position vector

IrP = Ir0 + Ir0P (2.117)

Now we replace IrP (as is done in the chapter on kinematics of rigid bodies) bythe corresponding vector in the moving reference frame (R) and the rotationmatrix:

IrP = Ir0 +AIRRr0P (2.118)

Contrary to the case where we considered a point P on the rigid body which Rr0P is not con-stant in relativesystem

had always a fixed distance from the origin of the moving frame, now we haveto consider that also the position IrP of point P relative to frame (R) maychange. Hence, we have to look at the temporal change in the reference frame(R) which we denote by the symbol d

′

dt which is the derivative with respect totime but in the moving frame, while the derivative d

dt is the absolute changewith respect to the fixed system.

The absolute velocity now is:

IvP = I rP = I r0 + Ir′0P = I r0 + AIRRr0P +AIR

d′

dt(Rr0P ) (2.119)

with the relative velocity: relative velocity

Rvrel =d′

dt(Rr0P ) (2.120)

36 kinematics

w

Relative pathof P

P

Moving system

wxr 0PR

RvFü

R 0v

R 0v

R Pv

Rvrel

R 0Pr

Inertial system

0

(I)

(R)

I 0r

Figure 2.10: Components of the Velocity

We use the rotation matrix ARI to transform the vector in (I)-components tothe (R) system.

RvP = ARIIvP (2.121)= ARII r0 +ARIAIRRr0P +ARIAIR︸︷︷︸

I

Rvrel (2.122)

RvP = Rv0 +ARIAIRRr0P + Rvrel (2.123)

We already know the expression from earlier considerations ARIAIR as theskew-symmetric matrix of the angular velocities:

RωIR = ARIAIR (2.124)

so that the velocity is

Rv = Rv0 + RωIRRr0P + Rvrel (2.125)

This expression for the absolute velocity has three different terms which wecan explain intuitively: The last term is the relative velocity of P with respectto the moving reference frame (R); the first two terms represent the velocity ofpoint P when we consider this point to be fixed in the (R)-frame. They are thevelocity of the origin of (R)-frame and the rotation of the (R)-frame. We callthese two terms guidance velocity vg (German: “Führungsgeschwindigkeit”).guidance velocity

With the guidance velocity we can write

Rv = Rvg + Rvrel . (2.126)


A person sitting at the origin of the moving (R)-frame only sees the relativevelocity, while a person sitting in the inertial system sees all guidance velocityplus relative velocity.

Using the rotation matrix it is very easy to express the velocity in coordinatesof the fixed coordinate system:

Iv = AIRRv (2.127)= AIRRvg +AIRRvrel (2.128)= Ivg + Ivrel (2.129)

Note:In the classical notation of the relative kinematics the vector product is usedwhich we have already shown to be equivalent to the matrix notation. In theclassical notation of the relative kinematics the vector product is used whichwe have already shown to be equivalent to the matrix notation.

v = v0 + ω× rOP + vrel = vg + vrel . (2.130)

2.3.2 Relation Between Absolute and Relative Acceleration

The acceleration can be obtained by further differentiation:

IaP = I rP = I r0 + I r0P (2.131)

orIaP = Ia0 +

(AIRRr0P +AIRRr

′0P).

(2.132)This gives:

IaP = Ia0 + AIRRr0P + 2AIRRr′0P +AIRRr

′′0P (2.133)

The derivatives with the prime indicate again that these express the changes relative accelera-tionwith respect to the moving frame. With the relative velocity of the last section

and the relative acceleration

Rarel = Rr′′0P , (2.134)

we can write

IaP = Ia0 + AIRRr0P + 2AIRRr′0P +AIRRarel (2.135)

Transformation as we have done it with the velocities now yields:

RaP = ARIIaP = ARIIa0 +ARIAIRRr0P . . .

. . .+ 2ARIAIRRvrel +ARIAIR︸︷︷︸I

Rarel (2.136)

38 kinematics

RaP = Ra0 +ARIAIRRr0P + 2ARIAIRRvrel + Rarel (2.137)

The product terms ARIAIR and ARIAIR can be expressed by the matrix ofthe angular velocities:

RωIR = ARIAIR

and using eq. (2.80)

ARIAIR = R ˙ωIR + RωIRRωIR

we get

RaP = Ra0 + R ˙ωIRRr0P + RωIRRωIRRr0P + 2RωIRRr′0P + Rr

′′0P . (2.138)

We can see that the absolute acceleration expressed in terms of the R-frameconsists of five terms. The acceleration terms now appear to be less intuitiveas the velocities.

The last one is the relative acceleration of point P in the moving coordinatesystem which the observer sitting in the moving R-frame can see. We get thisquantity by differentiating the position vector in the moving frame twice withrespect to time, regardless how the R-frame is moving.

The term with the factor of 2 is the Coriolis accelerationCoriolisacceleration

RaCor = 2RωIRRr′0P = 2RωIRRvrel (2.139)

which appears when a point P moves in the R-frame with a relative velocityvrel in a rotating reference frame. (This happens always when we move on thesurface of the earth.)

The first three terms are the guidance acceleration (German: “Führungsbeschle-Guidanceacceleration unigung”):

Rag = Ra0 + R ˙ωIRRr0P + RωIRRωIRRr0P (2.140)

This expression has three terms: the acceleration of the origin of the R-frame, aCentripetalacceleration term describing the part of the acceleration resulting from angular acceleration

and the last one with quadratic terms of the angular velocities is the centripetalacceleration. We can write

Ra = Rag + RaCor + Rarel (2.141)


If we are interested to get the acceleration in components of the inertial systemall have to do is multiplying by the rotation matrix:

Ia = AIRRa (2.142)

Note:In classical notation using the cross product the acceleration have the form:

ag = a0 + ω× r0P + ω× (ω× r0P ) (2.143)aCor = 2ω× vrel (2.144)

arel =d′2dt2 r0P (2.145)

where we can recognize the corresponding terms of the matrix notation.

2.3.3 Summary of the Formula for Relative KinematicsSummary

1. Absolute velocity in moving frame

RvP = Rv0 + RωIRRr0P︸︷︷︸Rvg

+ Rr′0P︸︷︷︸

Rvrel

(2.146)

with

Rvg = Rv0 + RωIRRr0P guidance velocity (2.147)

Rvrel = Rr′0P relative velocity (2.148)

The observer in the moving system gets aware of the relative velocitywithout knowing anything about the motion of the reference system.

• Classical notation with cross product:

vP = v0 + ω× r0P︸︷︷︸vg

+d′r0P

dt︸︷︷︸vrel

(2.149)

Derivation d′dt means derivation with respect to time in the moving

system.

2. Absolute acceleration expressed in the R-system:

RaP = Ra0 + R ˙ωIRRr0P + RωIRRωIRRr0P︸︷︷︸Rag

+ . . .

. . .+ 2RωIRRr′0P︸︷︷︸

RaCor

+ Rr′0P︸︷︷︸

Rarel

(2.150)

40 kinematics

with

Rag = Ra0 + R ˙ωIRRr0P + . . . Relative acceleration. . .+ RωIRRωIRRr0P (2.151)

RaCor = 2RωIRRr′0P = 2RωIRRvrel Coriolis-acceleration (2.152)

Rarel = Rr′′0P Guidance acceleration (2.153)

• Classical notation

aP = a0 + ω× r0P + ω× (ω× r0P )︸︷︷︸ag

+ 2ω× vrel︸︷︷︸aCor

+d′2dt2 r0P︸︷︷︸arel

(2.154)Derivation d′2

dt2 means derivation with respect to time in the movingsystem.

3KINET ICS

Here, we investigate the interaction of motion of a body or system of bodies Definition ofKineticsand the forces and moments causing this motion.

3.1 kinetics of a single particle

The idealization of a “real world body” as a particle is admissible, if only thetranslation of the body is of interest. The mass of the body is concentrated tothe centre of gravity (CG).

3.1.1 Momentum and Angular Momentum, Newton’s Law

Newton’s Axioms1 belong to the most important fundamentals in mechanics 2. Newton’s Ax-ioms(Basic Kinetic con-cept)

and had a great impact on the further scientific development at that time.In applied mechanics (where we are far away from the speed of light) thesebasic laws are still valid today. Especially, the second axiom is one of the mostimportant foundations of kinetics. It says that: “the temporal change of themotion2 is proportional to the impressed force . . . ”. This implies that, if noimpressed force is present, the motion remains unchanged. The momentum p Momentumis defined by

p = mv (3.1)In mathematical terms Newton’s 2. axiom reads as: Basic Kinetic con-

cept(Differential Form)F =

dpdt =

ddt (mv) (3.2)

which is valid in an inertial system. For many applications the mass can be Called also:Impulse lawconsidered constant. For constant mass m we get:

F = mdvdt = ma für m = const. (3.3)

1 Philosophiae Naturalis Prinzipia Mathematica, 16872 Today we would say “momentum”

41

42 kinetics

where a is the absolute acceleration. In integrated form Newton’s law isBasic Kinetic con-cept(Integrale Form)

p− p0 =∫ t

t0F (t∗) dt∗ (3.4)

The integral of the force over time is call impulse. So, we see that change ofmomentum = impulse. Especially, for constant m we get

m (v (t)− v (t0)) = p− p0 =∫ t

t0F (t∗) dt∗ (3.5)

For a particle we can define the angular momentum which can be derived byAngularMomentum the cross product of the position vector and the momentum:

L0 = r× p (3.6)

The angular momentum is related to the origin of the reference frame (as wewill see later, any other reference point can be chosen). If we multiply eq. (3.7)

p =ddt (mv) = F (3.7)

by the position vector from the left hand side we get:

r× p = r× ddt (mv) = r× F (3.8)

On the right hand side of this equation, we see the moment of the impressedforce with respect to the origin of the reference frame. The left hand side leadsto

ddt (r×mv)︸︷︷︸

r×p

= r× ddt (mv) +

drdt ×mv︸︷︷︸

0

yx

z

m

(I)

v

F

r

0

htaP

Figure 3.1: Particle under the influence of an impressed force

3.1 kinetics of a single particle 43

and

drdt ×mv = v×mv = 0

we obtain the relation for the angular momentum and the moment of the Conservationof angularmomentum fora mass point(Differential Form)

impressed force called angular impulse-momentum principle:

ddtL0 = M0 with M0 = r× F (3.9)

Note: While we could derive this relation for a particle from Newton’s law the Conservationof angularmomentum fora mass point(Integrale Form)

relation between angular momentum and moment is an independent axiom forrigid bodies. The integrated form is:

L0 (t)−L0 (t0) =∫ t

t0M0 (t

∗) dt∗ (3.10)

Instead of cross product in vector notation, in matrix notation we can writethe angular momentum in the following form:

L0 = r p (3.11)

where the skew-symmetric matrix is built up from the components of the po-sition vector

r =

0 −z y

z 0 −x−y x 0

(3.12)

Note: Compare this to angular velocities in vector and matrix notation inchapter 2.

3.1.2 Rotation of a Body About a Fixed Axis

Although we mainly deal with particles here, the rotational motion of a spin-ning rigid body of mass m about a fixed axis can be easily derived from New-ton’s law. We can find rotational motion about a fixed axis in many machinessuch as turbines, gears etc.

If we consider a small mass element dm of the rigid body (fig. 3.2), we can seethat it perform a circular motion about the rotation axis (with constant radiusr). From the previous kinematic considerations we know the circular motioncauses acceleration (also for constant angular speed). In cylindrical coordinateswe got (please compare to chapter B.2):

44 kinetics

a =

ar

aϕ

0

=

−rω2

rω

0

=

−rϕ2

rϕ

0

(3.13)

The (infinitesimal) moment we need to accelerate a single element dm in cir-cumferential direction (ϕ-direction) is

dM0 = rdFϕ (3.14)

According to Newton’s law eq. (3.3) for an infinitesimal small element dm wecan write

dFϕ = dmaϕ (3.15)dM0 = r2ϕdm (3.16)

The integration over the whole rigid body yields:

M0 = ϕ∫(m)

r2dm (3.17)

Or:M0 = J0ϕ (3.18)

with the mass moment of inertia

J0 =∫(m)

r2dm (3.19)

with respect to the rotation axis through the origin of the reference frame.

For an arbitrary axis x through a point A we get (see fig. 3.3):

MA = JAϕ

z

r

er

dm

j

ej

x

Figure 3.2: Rotational motion of a rigid body about a fixed axis


withJA =

∫(m)

r2dm . (3.20)

In many cases the mass moment of inertia for a fixed axis x′-x′ through centerof gravity S is known (e.g. from tables). The transformation for a parallel axisx-x through A can be done by means of Steiner’s theorem Steiner theorem

JA = JS + r2Sm (3.21)

As we can see, the mass moment of inertia with respect to an axis through theCG is always the smallest possible one, because the additional Steiner term isalways positive.

3.1.3 Kinetics of a Particle for Relative Motion

As stated earlier, Newton’s law is valid only when applied in an inertial system.For constant mass and indicating the inertial system by the subscript I we canwrite:

IF = mIa

Now we can replace the absolute acceleration in the I-system by the compo-nents of the moving R-system by using the rotation matrix:

RF = ARIIF = mARIIa = mRa (3.22)

In a next step we express the acceleration by means of the guidance, theCoriolis and the relative acceleration

a = ag + aCor + arel

dm

r

S

A

x

x’

x

x’

rs

Figure 3.3: On the rotational motion about different fixed axes: explanation of geo-metric quantities

46 kinetics

where the subscript R has been omitted for simplicity. Now we introduce thisinto eq. (3.22):

F = ma = m(ag + aCor + arel

)(3.23)

or after re-arranging the last equation:

F rel = marel = ma−mag −maCor (3.24)

orF rel = marel = F −mag −maCor (3.25)

What can be seen clearly is that we are not allowed to write Newton’s law init’s simple form in the moving R-system. However, we have to “correct” theformula by the inertia forces: theCoriolis force

Coriolis force FCor = −maCor

and theGuidance forceguidance force F g = −mag

(which consists of three terms and among them the centrifugal force as the mostprominent representative of the inertia forces). If we introduce these forces intothe last equation, we see that:

F rel = marel = F + F g + FCor (3.26)

The appearance of the inertia forces is directly coupled with the shift fromthe inertial system (where we have no inertia forces) to the moving referenceframe R. The inertia forces play an important role in D’Alembert’s principle.

Finally we consider the special case that the reference frame is rigidly connectedwith our mass particle m. Then per definition of the R-frame we cannot haveany relative motion. The relative acceleration and the relative velocity both arezero. The latter implies that the Coriolis force is zero. So we end up formallyin an dynamic equilibrium where the sum of all forces including the inertiaforces is zero:

0 = F + F g (3.27)

3.1.4 Work and Work-Energy Principles

3.1.4.1 Translational Motion of a Particle

The force F moves the particle m along the path. If we consider a small dis-placement dr we get an infinitesimal contribution of work

dW = Fdr = Fdr cosα = FSds (3.28)


yx

z

m

(I)

dr

F

r

0

htaP

as

Figure 3.4: Quantities to explain the concept of work of a force F

The work is calculated by means of the scalar product of force and displace-ment, α is the angle between the vectors F and dr . The magnitude of theinfinitesimal displacement is dr = ds. FS is the component of the force in direc-tion of the tangent to the path. Obviously, the component which is orthogonalto the path tangent cannot contribute to the work of the force.

Integration along the path from s0 to s1 yields the work

W =∫ s1

s0F (s) cos (α (s)) ds (3.29)

Next we set dr = vdt in eq. (3.28) and F is replaced by ma (Newton’s law)

F = ma = mdvdt (m=const.)

so thatdW = Fdr = mvdv (3.30)

The integration now yields the very important relation

W = m∫ v1

v0vdv = m

2(v2

1 − v20)

(3.31)

where the right hand side can be identified as change of the kinetic energy of Kinetic Energythe particle (mass m). The kinetic energy3 is

Ekin =12mv

2 (3.32)

From eq. (3.31) follows the work-energy theorem:

W = Ekin,1 −Ekin,0 (3.33)

3 In the anglo-american literature, frequently the symbol T is used for the kinetic energy.

48 kinetics

The work of the external force causes a change of the kinetic energy (increaseor decrease depending on the angle between the force and the path).

The left hand side is a path integral which considers what happens along thepath, while the energies only take the initial and final state into account.

3.1.4.2 Conservative and Non-Conservative Forces, Potential, Energy Theo-rem

We can distinguish between conservative and non-conservative forces. For con-Conservativeforces are

independent ofintegral path

servative forces the value of the path integral eq. (3.29) for a fixed starting andfinal point is independent of the special shape of the path between these points.This is a very important property of conservative forces.

For conservative forces a scalar function Π, the potential, exist which allowsthe calculation of the force vector from the negative gradient of the potential:

F = − grad Π (3.34)

where the negative sign is by definition. For example, in cartesian coordinateswe get

Fx = −∂Π∂x

Fy = −∂Π∂y

Fz = −∂Π∂z

(3.35)

example: The potential of the gravitational field of the earth (close toExamplethe surface) is

Π (x, y, z) = mgz , (3.36)

where the z-axis is orthogonal to the surface and points away from the center ofthe earth. The calculation of the gradient immediately yields: Fx = 0, Fy = 0und Fz = −mg.

The total differential of Π is

dΠ =∂Π∂x

dx+ ∂Π∂y

dy+ ∂Π∂z

dz (3.37)

and on the other hand from eq. (3.28) and eq. (3.29) (in cartesian coordinates)we get

W =∫ 1

0(Fxdx+ Fydy+ Fzdz) (3.38)


If we replace the force components by the derivatives of the gradient an usingeq. (3.37), we obtain

W = −∫ 1

0

(∂Π∂x

dx+ ∂Π∂y

dy+ ∂Π∂z

dz)= −

∫ 1

0dΠ = − (Π1 −Π0) , (3.39)

which shows us that the work W does not depend on the path but only on Concervativeforces does notdepend on path

the values of the potential Π0 and Π1 at the initial and final point of the pathwhich proves the independence of the path. The condition that a force field isconservative is the existence of a potential!

In the special case that initial and final point coincide, it follows from eq. (3.39),that W = 0: if we walk along a closed loop in a conservative force the workresulting from this process is zero!

For non-conservative forces such a potential does not exist. The work integral Non-conservativeforces depend onpath

is path-dependent and the integration along a closed loop yields (in general) avalue W 6= 0.

• gravitational forces,

• elastic forces (e.g. of elastic springs, bending of beams, etc.),

• magnetic forces

Non-conservative forces are

• friction forces or

• forces and moments from external sources.

Finally, for forces having a potential, from eq. (3.33) and eq. (3.39) we can Energy theoremderive the important energy theorem:

W = Ekin,1 −Ekin,0 = − (Π1 −Π0) (3.40)

from which follows:Ekin,1 + Π1 = Ekin,0 + Π0 (3.41)

With the potential energy Epot which is corresponding to the value of the Πwe get

Ekin,1 +Epot,1 = Ekin,0 +Epot,0 (3.42)

This describes the invariance of the total energy (which is the sum of kineticand potential energy) for a process in conservative fields.

50 kinetics

3.1.4.3 Rotation About a Fixed Axis

In analogy to the translational motion of a particle we obtain for the rotationof a rigid body about a fixed axis:

dW = Mdϕ (3.43)

W =∫ ϕ1

ϕ0M (ϕ) dϕ (3.44)

Again, the work changes the kinetic energy. For the rotation about a fixed axiswe get:

Ekin =12Jω

2 (3.45)

where ω is the angular velocity and J is the moment of inertia with respect tothe spinning axis. We obtain (according to eq. (3.31) and (3.33)):

W = Ekin,1 −Ekin,0 =12J

(ω2

1 − ω20)

(3.46)

A more general analysis of the kinetic energy for the rotation about a free axiswill be given later.

3.1.5 Power

Relating the work to time, we come to the concept of power which is veryPower unit is 1 W(Watt).

Metric horsepower:1 hp = 735.5 W

important in all technical disciplines.

The power P is mathematically defined by the gradient:

P =dWdt (3.47)

For a translational motion we get

P = Fdrdt = Fv cos (α) = FSv (3.48)

where Fs is again the tangential component of the force F . For the rotationalmotion with a fixed axis, the power is

P = Mdϕdt = Mω (3.49)

3.2 kinetics of rigid bodies 51

(I)

S

Pdm

P

SP

S

dF

rr

r

x

z

y0

Figure 3.5: On the kinetics of a rigid body

3.2 kinetics of rigid bodies

3.2.1 Momentum of a Rigid Body and the Momentum Theorem

The total mass of a body results from integration over the whole body

m =∫(m)

dm =∫(V )

ρdV (3.50)

where ρ is the mass density and dV is a small volume element. In fig. 3.5, P isan arbitrary point of the body, while S is the center of gravity4. In the sameway, the resultant of the external forces dF acting on a small mass elementdm can be obtained by integration over the rigid body K:

F =∫(K)

dF (3.51)

If we consider the contribution of the momentum of a single mass element dm(at point P ):

dp = vdm (3.52)

As the rigid body performs translational and rotational motion as well, weapply Euler ’s formula eq. (2.19) and use the center of gravity S as referencepoint

v = vP = vS + ω× rSP (3.53)If we put this into eq. (3.52) we get

dp = (vS + ω× rSP ) dm = vSdm+ (ω× rSP ) dm (3.54)

Again the total momentum of the rigid body is obtained by integration:

p =∫(m)

vSdm+∫(m)

(ω× rSP ) dm = vSm+ ω×∫(m)

rSPdm (3.55)

4 German: “Schwerpunkt” or “Massenschwerpunkt”

52 kinetics

The second integral vanished because we have chosen the reference point asthe mass center of gravity (good choice!) 5 The momentum then is

p = mvS . (3.56)

It is the product of the mass m which is concentrated in the center of gravitymultiplied by the velocity vS of point S. FromNewton’s law eq. (3.3) formulatedfor a small element dm

rdm = dF (3.57)we obtain after integration∫

(m)rdm =

∫(m)

dF = F . (3.58)

In chapter 2 we derived an expression for the acceleration of a point P

r = rP = aP = aS + ω× rSP + ω× (ω× rSP ) (3.59)

putting eq. (3.59) into eq. (3.58) we get (the integrals∫(m) rRPdm are zero):∫

(m)rdm =

∫(m)

aSdm = aSm (3.60)

The yields the momentum theorem:Momentumtheorem

F = aSm = mddtvS = p (3.61)

where m = const. was assumed. We see that the center of gravity (CG) movesas if the force resultant F would act directly on the CG where the total massm is concentrated. The resulting force F determines the acceleration as of theCG. In general there is an additional rotation of the rigid body, because Fdoes not act directly on the CG which causes a moment about S.

While we could derive the angular momentum - moment relation from themomentum principle for a mass particle this is not possible for a rigid body.In the latter case we need additional assumptions about the internal forces.Instead the angular momentum theorem was formulated as an independentlaw for the rigid body by Euler.

3.2.2 Angular Momentum and Moments of Inertia

A small element dm delivers a contributioncompare to.fig. 3.5 5 Recall that the formula rRS = 1

m

∫(m) rRPdm yields the position of the center of gravity

S as vector pointing from the reference point R to S. If the reference point R is alreadyidentical with S it follows immediately that rRS = rSS = 0.


dL0 = r× dp = r× vdm (3.62)

to the angular momentum (compare to our considerations with respect to aparticle) which after integration leads to

L0 =∫r× vdm =

∫(rS + rSP )× (vS + ω× rSP ) dm (3.63)

and multiplying all the terms in brackets of the cross product:

L0 =∫rS × vSdm+

∫rSP × vSdm

+∫rS × (ω× rSP ) dm+

∫rSP × (ω× rSP ) dm (3.64)

The vectors rS , vs and ω are constant with respect to the integration over thebody and can be moved outside the integral. Now, the second and third termvanish because the integral ∫

rSPdm = 0 (3.65)

as we showed before so that

L0 = rS × vSm+∫rSP × (ω× rSP ) dm (3.66)

remainsL0 = rS × p+

∫rSP × (ω× rSP ) dm . (3.67)

The first term is the translational part, the second part is the rotational partof the angular momentum:

L0,trans = rS × p (3.68)

L0,rot =∫rSP × (ω× rSP ) dm (3.69)

L0 = L0,trans + L0,rot (3.70)

We see that also if the body does not rotate we have a contribution to theangular momentum from the translation.

The rotational part can be further investigated and expressed as :

L0,rot =∫(m)

rSP × (ω× rSP ) dm = −∫(m)

rSP × (rSP × ω) dm (3.71)

Splitting the integral terms and the angular velocities, the integral yields the Moment of inertiasymmetric tensor of the moments of inertia:

JS =

Jxx Jxy Jxz

Jxy Jyy Jyz

Jxz Jyz Jzz

(3.72)

54 kinetics

S

dm

r

x

z

y

Figure 3.6: Rigid body with reference frame in the center of gravity

The subscript ’S ’ indicates that the inertia tensor is related to center of gravity.The matrix is called moment of inertia matrix or simply inertia matrix. Thediagonal elements are called the moments of inertia (with respect to the x-,y- or z-axis) and the off-diagonal terms are the products of inertia or crossproducts of inertia:

• Moments of Inertia

Jxx =∫ (

y2 + z2)

dm (3.73a)

Jyy =∫ (

x2 + z2)

dm (3.73b)

Jzz =∫ (

x2 + y2)

dm (3.73c)

• Products of inertia:

Jxy = Jyx = −∫xydm (3.73d)

Jxz = Jzx = −∫xzdm (3.73e)

Jyz = Jzy = −∫yzdm (3.73f)

where the coordinates x = xSP , y = ySP and z = zSP . For a change ofthe reference frame to another reference frame which has axes parallel to thefirst one Steiner ’s theorem is valid. While the coordinates always appear inquadratic terms in the moments of inertia, the sign of the coordinates are notimportant, however, in the products of inertia the sign of the coordinates haveto be considered.

Now we could write the rotational part of the angular momentum as

L0,rot = JSω (3.74)


then, the total angular momentum is given as

L0 = rS × p+ JSω (3.75)

And in matrix notation, the angular momentum is given as

IL0 = I rSIp+ IJSIω (3.76)

3.2.2.1 Principle Axes

If the x-, y- amd z-axes are principle axes, the products of inertia (off-diagonal Principal inertiasterms) vanish

JS =

J1 0 00 J2 00 0 J3

(3.77)

and J1 and J3 are the maximum and minimum values of the moments of iner-tia, respectively. The three diagonal moments of inertia are the eigenvalues ofmatrix in eq. (3.72) and are called principal inertias. In a basis with symmetricplanes of a body, the off-diagonal terms are zero.

Due to the fact that the integration is a linear operator, the inertia-matrixcalculation is an additive operation. The inertia matrix of a body with a volumeV1 + V2 is the sum of the inertia matrix of volume V1 and the inertia matrixof V2.

3.2.2.2 Steiner’s Theorem or Parallel Axes Theorem

If we know the inertia matrix with respect to the center of gravity S we can Steiner’s theoremeasily calculate also the inertia matrix for any point, say A, which is the originof a reference frame which has axes (x′-y′-z′-axes) which are parallel to thex-y-z-axes through S. Steiner’s theorem says that

Jx′x′ = Jxx +(y′

2S + z′

2S

)(3.78a)

Jy′y′ = Jyy +(x′

2S + z′

2S

)(3.78b)

Jz′z′ = Jzz +(x′

2S + y′

2S

)(3.78c)

and

Jx′y′ = Jxy − x′Sy′S (3.78d)Jx′z′ = Jxz − x′Sz′S (3.78e)Jy′z′ = Jyz − y′Sz′S (3.78f)

56 kinetics

S

dm

r

x

z

yS

r‘

x‘

z‘

y‘

Figure 3.7: Rigid body with reference frame in the center of gravity

where the coordinates x′S , y′S and z′S define the position of S in the x′-y′-z′coordinate system in A.

Note: For completeness we note that we can also find the definition of theproducts of inertia without the negative sign (eq. (3.73d) - eq. (3.73f)) in theliterature. But in this case the negative sign is included directly in the JS ,eq. (3.73), where all Jkl, k, l = x, y, z, k 6= l having negative signs.

The rotational part of the angular momentum is

L0,rot = JSω (3.79)

The components of this vector in Cartesian coordinates are

L0,rot,x = Jxxωx + Jxyωy + Jxzωz (3.80a)L0,rot,y = Jyxωx + Jyyωy + Jyzωz (3.80b)L0,rot,z = Jzxωx + Jzyωy + Jzzωz (3.80c)

or for the principle axes:

L0,rot,1 = J1ω1 (3.81a)L0,rot,2 = J2ω2 (3.81b)L0,rot,3 = J3ω3 (3.81c)

The total angular momentum finally is

L0 = rS × p+ JSω (3.82)

In matrix notation the angular momentum is

IL0 = I rSIp+ IJSIω (3.83)


3.2.3 Angular Momentum Theorem

The theorem of angular momentum introduced by Euler is -as already mentioned-an independent axiom, that cannot be derived from the momentum theoremwithout further assumption about the internal forces.

The angular momentum theorem is AngularMomentumTheoremL0 = M0 (3.84)

The external moment changes the angular momentum. For a rigid body theinternal forces do not have any influence on the momentum and the angularmomentum.

The result of all external moments follows from the integration of all contribu-tions rdF over the whole body

M0 =∫r× dF (3.85)

It is important to note that the reference point 0 of the coordinate frameis arbitrary, however, it must be a fixed point. Angular momentum L0 andmoment M0 must be related to the same point (0). Finally, the general formof the angular momentum theorm could be written as

L0 = (rs ×mvs) + (Jsω) = M0 (3.86)

The product rule has to be applied. The term

(rs × p) = (rS ×mvS ) = (vS ×mvS) + (vS ×maS) (3.87)

and because vS × vS = 0 only the second term remains

L0 = (rS ×maS) + (JSω) = M0 (3.88)

3.2.4 Change of the Reference Frame

If we move from the fixed reference “0”-system to a moving one “0∗”-systemas shown in fig. 3.8, additional terms describing the motion of the referenceframe have to be added. Starting with the position vector

rP = r = r0∗ + r∗P (3.89)

and the same applies for the velocity vector

vP = v = v0∗ + v∗P (3.90)

58 kinetics

then, the angular momentum could be derived from the integration of eq. (3.62)with eq. (3.90)as follows

L0 =∫r× vdm =

∫(r0∗ + r∗P )× vdm =

∫(r0∗ + r∗P )× (v0∗ + v∗P ) dm

=∫r0∗ × vdm+

∫r∗P × v0∗dm+

∫r∗P × v∗Pdm

= r0∗ × p+∫r∗P × v0∗dm+ L0∗

as (static moment) ∫r∗Pdm = r∗Sm

we getL0 = r0∗ × p+mr∗S × v0∗ + L0∗

finally the angular momentum in the moving frame is given as

L0∗ = L0 − r0∗ × p−mr∗S × v0∗ (3.91)

with the momentum of the rigid body given as: p = mvS .

If the moving reference frame origin is the same as the center of gravity of themore special casesof reference frameare further illus-trated in the ap-pendix

moving rigid body “0∗ == S” such as r∗S = 0, the angular momentum couldbe written as

LS = L0 − r0∗ × p (3.92)

Using the angular momentum L0 given in eq. (3.82), the angular momentumwriten at the center of gravity of the moving rigid body is

LS = JSω (3.93)

(I)

S

Pdm

P

P

SP

S

P

S

v

r

r*

rv

r

rr*

x

x*

z

z*y

y*

0

0*

0*

0*

v0*

v*P

Figure 3.8: Change of the reference system


As the moment of inertia of the rigid body depends in general on the currentposition of the body (e.g. angles). This became the main disadvantage of theeasy application of the theory. Therefor, the angular momentum is written interms of body-fixed reference frame, where the dependency of the moments ofinertia on time is removed, and the moments of inertia stay constant.

3.2.4.1 Angular Momentum Theorem for Pure Rotation

For a fix-point A only rotation of the body about this point is possible. Euler’sequation for the velocities was

rP = rA + ω× rAP (3.94)

If A is a fix-point, then rA = 0 so that rP = ω × rAP . In the case that pointP is the CG S:

vS = rS = ω× rS (3.95)

Putting this into eq. (3.86) with 0 = A:

LA = (rS ×mω× rS ) + (JSω) = MA (3.96)

The double cross product is

(rS ×mω× rS) =

(y2S + z2

S)mωx − ySxSmωy − zSxSmωz−ySxSmωx + (x2

S + z2S)mωy − zSySmωz

−zSxSmωx − zSySmωy + (y2S + x2

S)mωz

(3.97)

=

(y2S + z2

S)m− ySxSm− zSxSm−ySxSm+ (x2

S + z2S)m− zSySm

−zSxSm− zSySm+ (y2S + x2

S)m

ωx

ωy

ωz

(3.98)

The matrix can be identified from Steiner’s theorem, see eq. (3.78), so that wecan write

JA = JS +

(y2S + z2

S)m− ySxSm− zSxSm−ySxSm+ (x2

S + z2S)m− zSySm

−zSxSm− zSySm+ (y2S + x2

S)m

(3.99)

which is the inertia matrix with respect to fix-point A. The angular momentumtheorem then is

LA = (JAω) = MA (3.100)

As we can see the translational part of the angular momentum is eliminated.

60 kinetics

3.2.4.2 Angular Momentum Theorem with Respect to the Center of Gravity

The center of gravity is fixed on the body, but moving in space. Until now,we considered the angular momentum theorem only for an inertial system.Starting our considerations with eq. (3.88):

L0 = (rS ×maS) + (JSω) = M0

we express the moment according to eq. (3.85)

M0 =∫r× df =

∫(rS + r′)× dF = rS × F +MS (3.101)

where r′ is the position vector from point S to the actual mass element. Ascan be seen, M0 can be split into a part of the resultant force (1.part) and amoment about the center of gravity MS .

L0 = (rS ×maS) + (JSω) = rS × F +MS (3.102)

Now we use the momentum theorem (Newton’s law for a rigid body):

maS = F

and we see that eq. (3.102) simplifies to

LS = (JSω) = MS (3.103)

As can bee seen the choice of the center of gravity as reference point turns outto be a good choice because only the rotational part of the angular momentumremains.

We should keep in mind that although the reference frames moves with S, thedirections of x-y-z-axes are always constant. This means also that if the body

S

x

z

y

Figure 3.9: Rigid body (with reference frame in the center of gravity) under externalforce and moment loading


is rotating the moments and products of inertia change with time during therotational motion. This is a serious disadvantage of this formulation and wehave to think about the possibility of a reference frame which is fixed on thebody (in S).

3.2.5 Eulerian Equations, Angular Momentum Theorem in a Rotating Coor-dinate Frame

As we saw, the rotational motion of a rigid body can be described by theangular momentum theorem. The derivative of the angular momentum withrespect to time had to be calculated in a fixed coordinate system, the inertialsystem.

The extremely big disadvantage of this procedure is that if a body rotates butthe coordinate system is fixed you always have to calculate the mass momentsof inertia and the products of inertia with respect to this fixed coordinatesystem but while the body is rotating you may have different moments andproducts of inertia at every time instant. This makes the practical use of theangular momentum theorem very difficult.

The way out of this problem is to find a moving coordinate system (e. g. abody-fixed coordinate frame) in which the moments and products of inertiaremain constant during rotation.

The angular momentum theorem says that:

L = M (3.104)

To indicate which coordinate frame is used, we introduce the index I:

I L = IM (3.105)

The components in the (I)-frame can be transformed to the rotating frame (R)using the rotation matrix AIR (as we did many times before):

ddt(AIRRL

)= AIRRM (3.106)

orAIRRL+ AIRRL = AIRRM (3.107)

Premultiplying by ARI from the left side: ARIAIR

= I

RL+ARIAIRRL = RM (3.108)

62 kinetics

In the second expression we recognize the matrix of the angular speeds (com-pare to chap. 2, kinematics of the rigid body):

RL+ RωIRRL = RM (3.109)

The angular speed are the rotational speeds of the rotating frame in compo-nents of the R-frame. The product of the angular velocities and the angularmomentum can be written in classical formulation as a cross product.

d′dt (RL) + Rω× RL = RM (3.110)

The first term d′dt (RL) is marked by the prime and indicates that the derivative

is the temporal change in the rotating frame.

3.2.5.1 Coordinate System in the Center of Gravity

As we have seen, the expression for the angular momentum will be muchsimpler if the reference point is the center of gravity S. In this special case wegot:

LS = JSω

Because the moments and products of inertia are constant with respect to abody-fixed rotating coordinate system, the derivative of L with respect to timeyields:

RLS = RJSRω . (3.111)

and putting this into eq. (3.109) leads to

RJSRω+ RωIRRJSRω = RMS (3.112)

or shortly, leaving the R-subscript away:

JSω+ ω JSω = MS . (3.113)

In classical notation this equation writes as

JSd′ωdt + ω× JSω = MS (3.114)

3.2.5.2 Body-fixed Principal Axes

In body-fixed principal axes; these equation become very simple in structure.In principle axes;the moment ofinertia matrix hasalways diagonalform.


The inertia matrix then becomes diagonal because the products of inertia arezero:

JS =

J1 0 00 J2 00 0 J3

(3.115)

It follows thatJ1ω1 − (J2 − J3)ω2ω3 = M1

J2ω2 − (J3 − J1)ω3ω1 = M2

J3ω3 − (J1 − J2)ω1ω2 = M3

(3.116)

These are the so-called Eulerian equations for gyroscopic systems6. They are Eulerian equationsthree non-linear coupled differential equations. All components of J ,ω and mo-mentM have to be written in components of the rotating (R)-coordinate frame.These equations have a great relevance for the derivation of the equations ofmotion of mechanical systems with rotating components.

3.2.6 Angular Momentum Theorem in a Guided Coordinate System

As already mentioned it can be more convenient to use neither a body fixednor an inertial system to describe the kinematics of a rotating body.

The approach can be clarified with an example. A cylindrical roll is rotatingabout its rotation axis, which corresponds to the x-axis (fig. 3.10). The wholesystem is rotating again about the vertical z-axis. The guided coordinate sys-tem has its origin in the z-axis. The y-axis is always horizontal and does not-in contrast to the body fixed x′-y′-z′-system- rotate with the cylinderical roll,whereas x- and x′-axis are identical. The coordinate system in the Center ofGravity is -in contrast to the former chapter- not used here.

The angular momentum theorem expressed in the guided coordinate systemusing the point of origin of the R-system as centre of reference is

d′dt (RL0) + RωGKS × RL0 = RM0 (3.117)

whereas the first term is analogue to eq. (3.109) and eq. (3.110) and describesthe relative change of the angular momentum in the guided coordinate system GCS: Guided Co-

ordinate System(GCS). The second term of the left hand side is the change of angular momen-tum due to the rotation of the coordinate system. The expression on the righthand side is the moment expressed in coordinates of the guided R-system.

6 Euler, 1765

64 kinetics

S

z

z’

x

yy’

R

ωz

a

rl

Homogeneous cylinder roll of mass m

ωx

Figure 3.10: System with rotating cylindrical roll

wx

wwz

L

L

Lz

Figure 3.11: Angular velocity, and angular momentum vector

In this example the rotation of the guided coordinate system about the verticalrotation axis is expressed by following angular velocity vector

RωGCS =

00ωz

(3.118)

The mass moments of inertia are:

JR =

Jxx 0 00 Jyy 00 0 Jzz

(3.119)


Due to the symmetry of the cylindrical rolls Jzz is equal to Jyy, the off-diagonalterms are zero. The mass moment of inertia of a homogeneous cylinder (massm, radius r, length l) is:

Jxx =12mr

2 (3.120a)

Jzz = Jyy =112m

(l2 + 3r2

)+ma2 (3.120b)

The last term in eq. (3.120b) is the parallel axis part. The angular velocityvector of the rotor consists of two components: rotation about x-axis androtation about z-axis:

Rω =

ωx

0ωz

(3.121)

The angular momentum is:

RL0 = JRω =

Jxxωx

0Jyyωz

(3.122)

With constant angular velocity in terms of the guided coordinate systemd′dt (RL0) = 0

the moment can be determined to be

RM0 =

0

Jxxωxωz

0

= −RM0,gyroscopic . (3.123)

z

x

y

R

ωx

ωz

Mgyroscopic

Figure 3.12: Gyroscopic moment due to the rotation of the cylindrical roll

66 kinetics

The gyroscopic moment on the right hand side is the action of the rotor tothe environment. It loads the bearings as well as other machine parts. In thepresent example the gyroscopic moment has only a y-component. Comparedto the guided coordinate system the gyroscopic moment has negative sign,assuming that the two values of the angular velocities are positive.

3.3 kinetic energy of a rigid body

The kinetic energy of a particle is

Ekin =12mv

2 (3.124)

For a rigid body the kinetic energy is obtained by integration over all infinites-imal element of mass dm :

Ekin =12

∫(m)

v2dm (3.125)

Using the Eulerian kinematic equation (see eq. (2.19))

v = vP = vA + ω rAP ,

where A is an arbitrary reference point. We express the square of the velocitiesby:

v2 = vT v =(vA + ω rAP

)T (vA + ω rAP

)(3.126)

and put it into the integral. Further manipulations lead to

Ekin =12mv

2A +mvTA

(ω rAS

)+

12ω

TJAω (3.127)

rAS is the position vector pointing from point A to the center of gravity S.This quantity results from the integral∫

(m)rAPdm = rASm

and is the static moment which defines the position of the CG.

In vector notation the kinetic energy is

Ekin =12mv

2A +mvTA (ω× rAS) +

12ω

TJAω (3.128)

We see that kinetic energy has three terms:

3.3 kinetic energy of a rigid body 67

• The first is a pure translational part resulting from the velocity of point A,

Etrans =12mv

2A

• The third describes the pure rotation related to point A,

Erot =12ω

TJAω

• And the middle term is a coupling term of translation and rotation. Don’tforget this part of the kinetic energy.

Ecoupling = mvTA (ω× rAS)

Important Special Cases

1. If the reference point is the center of gravity: A = S, then,

rAS = rSS = 0

the coupling term vanishes, and we get:

Ekin =12mv

2S +

12ω

TJSω (3.129)

or simplyEkin = Etrans +Erot (3.130)

In cartesian coordinates we get

Etrans =12(v2x + v2

y + v2z

)(3.131)

and

Erot =12[ωx ωy ωz

] Jxx Jxy Jxz

Jyx Jyy Jyz

Jzx Jzy Jzz

ωx

ωy

ωz

(3.132)

Erot =12(Jxxω

2x + Jyyω

2y + Jzzω

2z + 2Jxyωxωy + . . .

. . .+ 2Jxzωxωz + 2Jyzωyωz) (3.133)

For principal axes the rotational part of the energy is

Erot =12(J1ω

21 + J2ω

22 + J3ω

23)

(3.134)

68 kinetics

2. Rotation about a fix-point A: vA = 0

Ekin =12ω

TJAω (3.135)

kinetic energy for pure rotation about point A.

3. Rotation about a fixed axis through A.The vector of the angular velocity is identical with the spinning axis. Thewhole expression reduced to one term

Ekin =12JAω

2 (3.136)

where JA is the moment of inertia with respect to the axis throughpoint A.

3.4 lagrange’s equations of motion of 2nd . kind

The Lagrangian equations of motion or the Lagrangian equations of the 2nd.kind (LE2)7 play a major role in dynamics. They belong to the class of ana-lytical methods which analyze the equations of motions from a global kineticenergy and potential energy consideration. Usually, reaction forces do not ap-pear. They are a standard tool to obtain the equations of motion of dynamicsystem. The LE2 can be derived from D’Alembert’s principle.

Our system consists of n rigid bodies and has f DOFs. We need the relationbetween the position vector r and the f generalized coordinates qi. As shownin chapter eq. (2.8) we can write

r i = r i (q1 , q2 , . . . , qf ) i = 1, . . . , n

where i is the subscript of the i-th rigid body. The Jacobian-matrix (eq. (2.11))contains the first derivatives of the position vector with respect to the gener-alized coordinates

J i =

[∂r i∂q1

∂r i∂q2

. . .∂ r i∂qf

]

The LE2 have the general form:

ddt

(∂Ekin∂ qk

)− ∂Ekin

∂qk= Qk k = 1, . . . , f (3.137)

7 There are also Lagrange’s equations of the first kind, which we do not treat here. Theseequations contain the constraints explicitely, while the equations of the second kind workwith the generalized coordinates which already contain the constraints implicitely.

3.4 lagrange’s equations of motion of 2nd . kind 69

where Ekin is the total kinetic energy of the whole system which is the sumof the kinetic energies of all n rigid bodies:

Ekin =n∑i=1

Ekin,i (3.138)

The eq. (3.137) yields f differential equations according to the number of thedegrees of freedom f , respectively the number of generalized coordinates. TheQk are generalized forces. Which result from the external impressed forces fromall n rigid bodies.

The generalized force Qk follows from the forces and moments (l-th force/mo-ment at the i-th body)

Qk =n∑i=1

Li∑l=1

(∂ril∂qk

)TF(e)il +

Li∑l=1

(∂ϕ

il

∂qk

)TM

(e)il

(3.139)

by a projection of a force or moment with the help of the Jacobian matrix. Thisis done in such a way that only the part of the force/moment is considered thatcontributes to the motion of the system (to the k-th gen. coordinate) accordingto the constraints8.

3.4.1 Conservative Systems

Generally, for conservative forces a potential exists (see chapter 3.1.4.2), whichis not the case for non-conservative forces. Thus, if the generalized force resultsfrom conservative forces/moments we can derive them directly from the poten-tial energy

Qk = Qk,kons = −∂Epot∂qk

(3.140)

As shown earlier, conservative forces can be calculated from the potential bydifferentiation of Epot = Epot (q1, . . . , qf ) but now with respect to the k-thgeneralized coordinate. Introducing eq. (3.140) into eq. (3.137) we get

ddt

(∂Ekin∂qk

)− ∂Ekin

∂qk= −∂Epot

∂qk(3.141)

with the total energies

Ekin =n∑i=1

Ekin,i

8 example is given in the class room

70 kinetics

and

Epot =n∑i=1

Epot,i .

Introducing the Lagrangian function L9 (also called kinetic potential)

L = Ekin −Epot (3.142)

we see thatThe equationsof conservativesystem areexpressed usingLE2

ddt

(∂L

∂qk

)− ∂L

∂qk= 0 k = 1, . . . , f . (3.143)

which are the famous Langrangian equations of the 2. kind.

3.4.2 Conservative and Non-conservative Forces, Rayleigh Energy Dissipa-tion Function

In this case a potential exists only for the conservative forces/moments andonly these forces/moments can be treated by potential energy terms. The non-conservative forces/moments have to be considered by the generalized forcesQk = Qk,nc:

ddt

(∂L

∂qk

)− ∂L

∂qk= Qk,nc k = 1, . . . , f (3.144)

However, some types of non-conservative forces/moments (like viscous damp-Lagrangeequation fornon-conservativesystems

ing forces) can be expressed by a energy dissipation function D which has thedimension of a power. This dissipation function is called Rayleigh dissipationfunction and can be defined by a general expression

D =12∑i

∑j

cij qiqj (3.145)

where the the q’s are the time derivatives of the generalized coordinates q andthe c’s are coefficients. The non-conservative forces can be derived from D

in a similar way as the conservative forces, namely by differentiating a scalarfunction:

Qk,nc = −∂D

∂qk(3.146)

9 A confusion of the Lagrangian function L with the angular momentum L is not expectedhere

3.5 equations of motion of a mechanical system 71

but here we differentiate with respect to a velocity. This leads to an extensionof the Lagrangian equations

ddt

(∂L

∂qk

)− ∂L

∂qk= Qk,nc −

∂D

∂qkk = 1, . . . , f (3.147)

Here the Q’s only contain those forces which cannot be expressed by D. Asan example we can consider a viscous damper with damping constant c. Themotion of the left attachment point of the damper is described by ql and theright attachment point by qr, both having the same direction. The dissipationfunction for this case is

D =12cv

2rel =

12c (qr − ql)

2

As we can see this example also leads to the general form of eq. (3.145):

D =12c (qr − ql)

2 =12c (qrqr − 2qlqr + qlql)

The damping force is obtain by differentiation

Fdamper,r = −∂D

∂qr= −c (qr − ql)

Fdamper,l = −∂D

∂ql= c (qr − ql)

3.5 equations of motion of a mechanical system

From the Lagrangian equations or Newton/Eulers equations we get the differ-ential equations describing the motion of a mechanical system with f DOF.They have the general form

M(q)q+ h

(q, q, t

)= Qnc (t) (3.148)

72 kinetics

where:M(q)

f × f mass matrix (which in general is depending on the gen-eralized coordinates q)

Qnc (t) vector of time dependent external forces and moments (dimen-sion f)

h(q, q

)vector of forces/moments (dimension f) depending on the gen-eralized coordinates and/or the generalized velocities and/ortime t, e.g.

• Conservative elastic forces, depending on q.

• Dissipative forces, e.g. from viscous damping, fric-tion, depending on q.

• Gyroscopic moments, depending on velocities q.

• . . .

3.5.1 Linearization of the Equations of Motion

In many applications, the motion described by q(t) can be splitted into areference motion q

r(t) which is the desired motion of the system and a more

or less small disturbance ∆q(t) (e.g. a vibration about this reference motionqr(t)).

q(t) = qr(t) + ∆q(t) (3.149)

The same can be done with the external forces which are split into referenceforces and disturbances

Q(t) = Qr(t) + ∆Q(t) (3.150)

withF(q, q, q, t

)= M(q)q+ h

(q, q, t

)= Q(t) (3.151)

we can derive a linearized differential equation for the motion around the de-sired trajectory q

r(t) by using the Taylor series which we truncate after the

linear part:

F(qr+ ∆q, q

r+ ∆q, q

r+ ∆q, t

)= F

(qr, qr, qr, t)+∂F

∂q

∣∣∣∣∣r

∆q+ . . .

. . .+∂F

∂q

∣∣∣∣∣r

∆q+∂F

∂q

∣∣∣∣∣r

∆q (3.152)

Splitting the last equation into reference motion (which can also be a staticdisplacement) and disturbance (which is a small motion around the referencetrajectory), we get

3.5 equations of motion of a mechanical system 73

• Reference motion

M(qr)qr+ h

(qr, qr, t)= Q

r(t) (3.153)

• Disturbance

M(qr)∆q+

∂h

∂q

∣∣∣∣∣r

∆q+∂h

∂q

∣∣∣∣∣r

∆q+∂(M(q)q

)∂q

∣∣∣∣∣∣r

∆q = ∆Q(t) (3.154)

3.5.2 Equation of Motion of a Linear Time-Variant and Time-Invariant Me-chanical System

If the linearization is done with repect to a reference trajectory qr(t) the result-ing system matrices are time dependent. We call this a time-variant system.

M(qr)∆q+

(C(t) +G(t)

)∆q+

(K +N

)∆q = ∆Q(t) (3.155)

If the linearization is done with repect to a reference point qr(e.g. the static

equilibrium position due to the gravitational forces) the resulting system matri-ces are time independent and the system is called time-invariant. The systemmatrices are

M∆q+(C +G

)∆q+

(K +N

)∆q = ∆Q(t) (3.156)

The f × f matrices of the equation of motion are:

M mass matrixC symmetric damping matrix (describing velocity dependent damping

forces/moments)G skew-symmetric gyroscopic matrix (describing velocity dependent gyro-

scopic moments)K symmetric stiffness matrix (describing position dependent restoring

forces/moments)N skew-symmetric matrix of non-conservative position dependent forces/-

moments)

74 kinetics

3.6 state space representation of a mechanical system

3.6.1 The General Non-linear Case

The equation of motion

M(q)q+ h(q, q, t

)= Q(t) (3.157)

is in general non-linear and contains the second order derivatives with respectto time as highest order (due to the fact the acceleration appears in the equa-tions) . However, many numerical algorithms10 for ordinary differential equa-tions (ODE) and theories in control make use of first-order formulations.

z = f (z, t) (3.158)

which is a first-order differential equation. The second-order equation of motioneq. (3.157) has to be converted into a first-order equation which is done by thefollowing intermediate step. We introduce the state space vector z:

z =

qq

=

z1

z2

(3.159)

which contains the generalized coordinates and the velocities as well. The stateof a mechanical system is defined by the generalized coordinates (displacementsor angles) and the velocities!

The acceleration comes into the play when the derivative of eq. (3.159) isformulated:

z =

qq

=

z1

z2

(3.160)

Comparing eq. (3.160) and eq. (3.159), we see that

z1 = z2 (3.161)

If we use z instead of the q’s, eq. (3.157) becomes

M(z1)z2 + h (z1, z2, t) = Q(t) (3.162)

so that we only have first-order time derivatives. However we have to pay theprice that the number of the differential equations doubled and has dimen-sion 2f now.

10 Examples are the Euler- (very simple but inaccurate algorithm), the Runge-Kutta- or theAdams-method

3.6 state space representation of a mechanical system 75

Converting eq. (3.161) and eq. (3.162) into eq. (3.158) we get:

z =

z1

z2

=

z2

M−1(z1)[Q(t)− h (z1, z2, t)

] = f (z, t) (3.163)

For the solution of the problem we also have to define the initial positions andthe initial velocities:

z (t = 0) = z0 =

q0q0

(3.164)

Usually, in a state-space representation, a second equation called the mea-surement equation, is used. The measurement equation links the measuredquantities y with state-space variables in z:

y = g (z, t) (3.165)

This is a very general (non-linear) formulation. As an example a measuredquantity y1 could be a strain which has to be related to displacements q.

3.6.2 The Linear Time-Invariant Case

In many practical cases it is possible to linearize the non-linear equationsof motion or they are linear from the beginning. The linear theory is wellexamined and many important theorems are available for linear systems.

If the mechanical system behaves linearly, the equation of motion is (the ∆ isleft away here for simplicity)

Mq+(C +G

)q+

(K +N

)q = Q(t) (3.166)

In this case the state-space equation z = f (z, t) becomes

z =

z1

z2

=

z2

M−1[Q(t)−

(C +G

)q−

(K +N

)q] = f (z, t) (3.167)

This can be re-organized in the following way:z1

z2

=

0 I

−M−1(K +N

)−M−1

(C +G

) z1

z2

+ 0M−1Q(t)

(3.168)

or qq

=

0 I

−M−1(K +N

)−M−1

(C +G

) q

q

+ 0M−1Q(t)

(3.169)

76 kinetics

The general formulation of a linear state space model in control theory has theform

z = Az +Bu (3.170)

whereA 2f × 2f system matrixB 2f × f input matrixu input (here the generalized forces u(t) = Q(t))

The two matrices can be identified as:

A =

0 I

−M−1(K +N

)−M−1

(C +G

) (3.171)

B =

0M−1

(3.172)

Assume we have measured ny quantities which we arrange in the vector y. Thecorresponding linear measurement equation is

y = Cmeasz +Dmeasu(t) (3.173)

withCmeas: measurement or ouput matrix (size ny × 2f)Dmeas: transmission matrix (size ny × f)

The eigenvalues λ of the system matrix A are the system poles and they containimportant information about the resonant frequencies and damping behaviour.If one of the real parts of the λ’s is positive, the system is unstable and largeamplitude can occur so that the system can be destroyed. More about theserelations will be given in the chapters dealing with vibrations of mechanicalsystems.

3.6.3 The Linear Time-Invariant Case in Discrete Time

Dealing with digital techniques, we have sampled time series and the valuesare only available at time instants tk, k = 1, 2, 3, . . .

Assuming that u is constant between two sample times and transformingeq. (3.170) from discrete time to continuous time yields:

z(k+ 1) = Adisz(k) +Bdisu(k) (3.174)

3.6 state space representation of a mechanical system 77

The measurement equation can be written as

y(k) = Cmeasz(k) +Dmeasu(k) (3.175)

where the discrete-time matrices can be calculated from the continuous-time ∆t = tk − tk−1

matrices A and B:

Adis = eA.∆t (3.176)

Bdis =∫ ∆t

0eA.(∆t−τ )Bdτ (3.177)

This can be shown by solving the differential equation11. The matrices of themeasurement equations remain unchanged, because the measurement equationis an algebraic equation and contains no differential expression.

11 J.-N. Juang: "Applied System Identification", Prentice Hall, 1994.

4L INEAR VIBRATIONS OF SYSTEMS WITH ONEDEGREE OF FREEDOM

4.1 general classification of vibrations

Different possibilities for a classification exist1:

• Classification with respect to the number of degrees of freedom: f

– 1 DOF or known as Single Degree Of Freedom (SDOF)

– n DOFs or known as Multiple Degrees Of Freedom (MDOF)

– infinite numbder of DOFs (continuous system: beams, plates, . . .)

Spatially discrete vibration systems with f = n DOFs can result fromdiscrete multi-body systems but from finite-element systems, where thecontinuum was discretized to some representative nodal points leadingto a finite number of DOFs.

• Classification with respect to the character of the describing equation ofmotion:

– linear vibrations

– non-linear vibrations (e.g. oscillators with non-linear spring or dampercharacteristics, contact or play, etc.)

Even for an apparently simple non-linear system with one DOF the ana-lytical solution can be very complicated, if not impossible. In the lattercase, a solution is only possible by numerical methods.

1 S. z.B. Magnus, K., Popp, K. : Schwingungen, Teubner, 1997.

79

80 linear vibrations of systems with one degree of freedom

As shown in the last chapter non-linear systems can be linearized abouta reference trajectory or a reference point in many cases. The descriptionby linear differential equations has many methodological advantages.

In the context of non-linear differential equations we can get new, verycomplex dynamic behaviour, which does not occur with linear systems.An interesting phenomenon is deterministic chaos. Although the motionof the system is described by deterministic equations and the input exci-tation is deterministic (e.g. a sine-excitation) the response of the systemseems to be random (which we call chaotic in this context). Starting withtwo adjacent initial conditions which differ only slightly we may end upin totally different system states. This phenomenon is called the "but-terfly effect". However, order can be found when we observe the systemover a very long time span. The system converges to a so-called "strangeattractor".

• Classification with respect to the origin of the vibration

– free vibrations

– self-excited vibrations

– parameter-excited vibrations

– forced vibrations

– coupled vibrations

If the system performs free vibrations (linear or non-linear) it starts from aninitial state and is left alone, there is no other influence from outside thesystem.

self-excited vibrations can occur if the system has access to an exter-nal reservoir of energy other than forced vibrations where the rhythm ofthe excitation is prescribed, here the system itself determines the rhythmof the energy transfer into the system. The system excites itself. It takesas much energy as it needs to maintain the vibration (e. g.as in the caseof a pendulum clock where the energy storage is an elastic spring). Thesesystems are called autonomous systems. If more energy per cycle flowsinto the system than energy is dissipated due to friction etc. then insta-ble vibration can occur leading to strongly increasing amplitudes as inthe case of unstable feedback control loops. This must be avoided underany conditions.

4.1 general classification of vibrations 81

This may also occur if there is an interaction of a structure with a fluidas in the case of an airplane wing which tends to unstable vibrationwhen reaching a critical speed. This phenomenon is called "wing flutter".Usually, this speed is beyond the travelling speed of the airplane, but hasto be checked for every new prototype.

Also the destruction of the Tacoma-Bridge in the 40th in the US is a"good" example how dangerous self-excited vibrations by fluid-structureinteraction can be. Other examples are unstable vibrations of turbo ma-chinery coming from or friction induced vibrations in brakes (squealnoise) or slip-stick phenomena with tool machines leading to marks ofthe cutting tool at the surface of the workpiece. Also unstable motionof the bogie of rail vehicles at very high speeds belong to the group ofself-excited vibrations.

parameter-excited vibrations occur if one or more coefficients ofthe differential equations are not constant but periodically time-varying.The frequency of the parameter change is prescribed explicitely as a func-tion of time, e.g. by the rotational speed of a shaft. Examples are: pendu-lum with periodically varying length, rotating shaft with unsymmetriccross-section, periodically varying stiffness of gear-wheels.

forced vibrations emerge from external disturbances (e.g. periodicaldisturbances from unbalance). The rhythm of the vibration here is notprescribed by varying parameters but by a time-dependent disturbanceterm on the right hand side of the equation of motion.

coupled oscillations occur by the fact that two oscillators can influ-ence and excite each other.

All these origins for vibrations can occur also in combined form.

Another possibility of classification is due to the character of the excitation:

• deterministic

– harmonic / periodic

– transient (e.g. vibrations after an impact)

• stochastic (or random) vibrations

– stationary


– instationary

4.2 free undamped vibrations of the linear oscillator

4.2.1 Equation of Motion

Newton’s resp. Euler ’s equations or the Lagrange’s formalism can be used herein order to derive the equation of motion. The example shows a mass m withelastic foundation (total stiffness k)

Newton’s law:

mx =∑i

Fi = −2k2x (4.1)

mx+ kx = 0 (4.2)

Division by the mass yields:

x+ ω20x = 0 (4.3)

withNatural circularfrequency[ω0] = s−1 ω2

0 =k

m(4.4)

ω0 is the natural circular frequency of the free undamped vibration.

xm

k

2k

2

xm

k

2x

k

2x

k

2x

k

2x

Figure 4.1: Vibration system with one DOF

4.2 free undamped vibrations of the linear oscillator 83

Tx

A

t

Figure 4.2: Example for a solution x (t) of the undamped free oscillator

4.2.2 Solution of the Equation of Motion

This differential equation has the solution: Basic approach tosolve a vibrationequationx (t) = Ac cosω0t+As sinω0t (4.5)

The constants Ac und As follow from the initial conditions:

x0 = x (t = 0) (4.6)v0 = x (t = 0) (4.7)

For t = 0 this leads directly to Ac = x0 and after differentiation of eq. (4.5) toget the velocities, the constant As = v0

ω0so that we can express the constants

in terms of the initial displacement and the initial velocity:

x (t) = x0 cosω0t+v0ω0

sinω0t (4.8)

Further important quantities are the time of a cycle or period: Period[T ] = s

T0 =2πω0

(4.9)

And the natural frequency : Natural frequency[f0] = s−1 = Hz

f0 =1T0

=ω02π (4.10)

Another kind of representation of the solution is:

x (t) = A sin (ω0t+ ϕ) (4.11)


Using trigonometric theorems:

sin (α+ β) = sinα cos β + sin β cosα . (4.12)

Here:

x (t) = A (sinω0t cosϕ+ sinϕ cosω0t) (4.13)= A cosϕ︸︷︷︸

As

sinω0t+A sinϕ︸︷︷︸Ac

cosω0t (4.14)

with

Ac = A sinϕ (4.15)As = A cosϕ (4.16)

Furthermore:A2c +A2

s = A2(sin2 ϕ+ cos2 ϕ

)(4.17)

andAcAs

=A sinϕA cosϕ (4.18)

so thatA =

√A2c +A2

s (4.19)and

tanϕ =AcAs

. (4.20)

x

j

A

2p

w t

Figure 4.3: Phase Angle ϕ

4.2.3 Complex Notation

Starting with the equation of motion (eq. (4.3))

x+ ω20x = 0 (4.21)


and solve it using an exponential approach: Exponentialapproach tosolve a vibrationequation

x (t) = eλt (4.22)x (t) = λeλt (4.23)x (t) = λ2eλt (4.24)

we get

λ2eλt + ω20eλt = 0 (4.25)

λ2 + ω20 = 0 (4.26)λ2 = −ω2

0 . (4.27)

which has the two conjugate complex solutions λ:

λ1,2 = ±iω0 (4.28)

The general solution for x (t) is:

x (t) = A+eiω0t +A−e−iω0t (4.29)

where the constants A+, A− are conjugate complex, too, so that the solu-tion x (t) becomes real again. The constants can be derived from the initialdisplacement and velocity, respectively.

4.2.4 Relation Between Complex and Real Notation

We start with the well-known general formula:

e±iλ = cosλ± i sinλ (4.30)

which we can use to express sin and cos by:

cosλ =eiλ + e−iλ

2 (4.31)

sin λ =eiλ − e−iλ

2i (4.32)

Putting this into eq. (4.29) yields:

x =Ac2(eiω0t + e−iω0t

)+As2i(eiω0t − e−iω0t

)(4.33)

=Ac2(eiω0t + e−iω0t

)− iAs2

(eiω0t − e−iω0t

)(4.34)

=Ac − iAs

2︸︷︷︸A+(=A−

∗)

eiω0t +Ac + iAs

2︸︷︷︸A−

e−iω0t (4.35)


where (...)∗ denotes the conjugate complex number of (...).complex conjugateof A is A∗

This delivers the relation between the real constants Ac, As and the complexconstants A+ and A−:

A+ =Ac − iAs

2 = A∗− (4.36)

A− =Ac + iAs

2 (4.37)

Furthermore, it follows that :

ReA+eiω0t

=

12 (Ac cosω0t+As sinω0t) (4.38)

ReA−e−iω0t

=

12 (Ac cosω0t+As sinω0t) (4.39)

so that

ReA+eiω0t +A−e−iω0t

= (Ac cosω0t+As sinω0t) (4.40)

ImA+eiω0t +A−e−iω0t

= 0 (4.41)

4.2.5 Further Examples of Single Degree of Freedom Systems

There are many applications, where the elastic element of a vibration systemis not a simple spring but e.g. an elastic beam. The lowest natural frequency ofsuch a system can be calculated by a relatively simple approximation, wherewe assume that the deflection shape of the vibrating beam or shaft etc. isapproximately equal to the static deflection shape (which is approximatelytrue for the lowest vibration mode).

example Flexural Vibration of a Beam:

The mass of the beam is assumed to be small: m >> mB

Figure 4.4: Flexural vibration of a beam


The stiffness constant is k = 48EIl3 where E is the Young’s modulus, I is the

area moment of inertia. The constant follows from the elementary theory ofbeam bending, where the deflection x of the beam due to a static force F in themiddle of the beam is x = Fl3

48EI . The stiffness coefficient k follows immediatelyfrom F = kx. The natural circular frequency is:

ω0 =

√k

m=

√48EIml3

(4.42)

example Torsion Vibration of a Disc:

The equation of motion is:Jϕ+ k

_

ϕ = 0 (4.43)Here the stiffness coefficient of the shaft is k

_

= GIp

l , where G is the shearmodulus and Ip is the polar moment of inertia. A static torque MT at theposition of the disc will cause a twist angle ϕ of the shaft:

ω0 =

√k_

J=

√GIpJl

f0 =ω02π (4.44)

Here, again the mass moment of inertia of the bar has been neglected.

4.2.6 Approximate Consideration of the Spring Mass

If the spring mass (which can be the mass of a beam, bar or shaft dependingon the actual case) cannot be neglected compared to the main mass m, we canconsider the effect of the spring mass on the natural frequency.

Figure 4.5: Torsion vibration of a disc


example Mass m connected to a bar:

Figure 4.6: Mass m connected to a bar

The bar has the mass mS , length l, a cross section area A and Young’s modu-lus E. The stiffnes coefficient is k = EA

l .

The real deflection shape is approximated by the static deflection shape. At acertain position z of the bar we have a static deflection:

u (z,x) = z

lx (4.45)

where x is the deflection of the mass m. At the top z = 0 and hence u = 0, atz = l we get u = x. Now we consider the kinetic energy of both masses:

Ekin =12mx

2 +12

∫ l

0u2dm =

12mx

2 +12

∫ l

0

(z

lx)2ρAdz (4.46)

=12mx

2 +12ρA

l2x2∫ l

0z2dz = 1

2mx2 +

12ρA

l2x2 1

3 l3 (4.47)

The mass of the bar is mS = ρAl with ρ as the mass density so that

Ekin =12mx

2 +12

(13mS

)x2 (4.48)

Finally, we get:

Ekin =12

(m+

mS

3

)x2 =

12(meff )x

2 (4.49)

As we can see (for this example) that the mass of the bar is considered by onethird of its total mass.

The natural frequency is:

f0 =1

2π

√√√√ k(m+ mS

3

) =1

2π

√√√√ EA/l(m+ mS

3

) (4.50)

4.3 free vibrations of a viscously damped oscillator 89

xm

k

2k

2

xm

k

2x

k

2x

k

2x

k

2x

ccx

cx

Figure 4.7: SDOF oscillator with viscous damping

which is smaller as in the case when we neglect the mass of the bar.

The factor 13 applies also in the case of mass connected with a helical spring, or

in the case of a disc mounted on a torsion bar. In the case of flexural bendingof beams the static deflection shape is more complicated as in this examplebut can be calculated by means of the beam theory or taken from tables. Thegeneral procedure to calculate the kinetic energy is identical to the example.

4.3 free vibrations of a viscously damped oscillator

4.3.1 Equation of Motion

We consider viscous damping which means that the damper force is propor-tional to the relative velocity of the two connection pins of the damper. Theforce represents fairly well the conditions of damping due to the oil in a dash-pot. Other, more complicated damping laws can be considered which lead tonon-linear equations of motion. Non-linear Equation Of Motion (EOM) (EOM)require a much more complicated mathematical treatment. In most cases onlya numerical solution is possible.

The free body diagram and application of Newton’s law yields:

mx =∑i

Fi = −2k2x− cx (4.51)


ormx+ cx+ kx = 0 (4.52)

The mass m, the viscous damping coefficient c and the stiffness k are constants(as in the previous case) andm, c, k > 0. The viscous damping reduces the totalenergy of the vibrating system. Division by m yields:

x+ 2δx+ ω20x = 0 (4.53)

whereδ =

c

2m (4.54)

and

ω0 =

√k

m

Transition to the dimensionless time (or an angle, depending on the interpre-tation)

τ = ω0t (4.55)

leads to the following derivatives (using the chain rule):

d (. . .)

dt =d (. . .)

dτdτdt =

d (. . .)

dτ ω0 . (4.56)

If we distinguish the derivatives with respect to time t by the dot, and thex = dxdt

x′= dx

dτ2nd derivativesanalogically!

derivatives with respect to τ by the prime, we find that:

ω20x′′+ 2δω0x

′+ ω2

0x = 0 . (4.57)

Division by the square of the natural circular frequency:

x′′+

2δω0x′+ x = 0 (4.58)

orx′′+ 2Dx

′+ x = 0 (4.59)

The parameter D is the only quantity which describes the behavior of the freeDimensionlessdamping damped oscillator. It is called the dimensionless damping (in Germany also

called Lehr’s damping):

D =δ

ω0=

c

2mω0=

c

2√km

(4.60)

As we can see, D contains all three parameters m, c and k.


Rel

Rel

Rel

Rel

Iml

Iml

Iml

Iml

D = 1

D = 0 0 < D < 1

D > 1

No Damping Weak Damping

Critical Damping Strong Damping

Figure 4.8: Solutions of the characteristic equation depending on D

4.3.2 Solution of the Equation of Motion

The solution to eq. (4.59) can be found by the assumption that the solutionhas the form:

x (τ ) = Aeλτ (4.61)where A and λ are in general complex. Putting this into eq. (4.59) we get the Characteristic

equationcharacteristic equationλ2 + 2Dλ+ 1 = 0 (4.62)

which has two solutions: i =√−1

λ1,2 = −D±√D2 − 1 = −D± i

√1−D2 (4.63)

Eq. (4.63) shows that the oscillator changes its behaviour depending whetherDleads to real or complex solutions of λ (see Fig. 4.8). The general mathematicalsolution for x (τ ) resp. x (t) for λ1 6= λ2 is:

x (τ ) = A1eλ1τ +A2eλ2τ (4.64)

and if λ1 = λ2 = λ:

x (τ ) = A1eλτ + τA2eλτ = (A1 + τA2) eλτ (4.65)


The constants A1,2 are determined from the initial conditions (initial displace-ment and velocity, respectively). They can be real or complex.

For different D we get different types of solutions, which we will discuss in thefollowing sections.

1. Overdamped Case: Strong Damping:D > 1D > 1 represents strong damping, we do not get a vibration, but the masscreeps back to the equilibrium position. The mathematical solution forthis case is:

λ1,2 = −D±√D2 − 1

we obtain two real (and different) values. The solution is:

x (τ ) = A1e(−D+√D2−1)τ +A2e(−D−

√D2−1)τ (4.66)

or, if we go back to the t-time scale:

x (t) = A1e(−D+√D2−1)ω0t +A2e(−D−

√D2−1)ω0t (4.67)

Because the real values of λ are always negative we find the creep motionback to the static equilibrium.

2. Critical Damping:D = 1For D = 1 we get the transition from vibration to creep motion (hencewe call this state critical). The two solutions of λ are identical and realwhich follows immediately from eq. (4.63):

λ1,2 = −D±√D2 − 1 = −1 = λ

The resulting motion is

x (τ ) = A1e−τ + τA2e−τ = (A1 + τA2) e−τ (4.68)

and on the t-time scale:

x (t) = A1e−ω0t + ω0tA2e−ω0t = (A1 + ω0A2t) e−ω0t (4.69)

3. Weak Damping, Damped Vibrations:0 < D < 1For 0 < D < 1 we get:

λ1,2 = −D± i√

1−D2

which are conjugate complex. The solution is:

x (τ ) = A1e(−D+i√

1−D2)τ +A2e(−D−i√

1−D2)τ (4.70)

On the t-time scale we obtain:

x (t) = A1e(−D+i√

1−D2)ω0t +A2e(−D−i√

1−D2)ω0t (4.71)


0 10 20 30 40 50 60 700

0.05

0.1

0.15

0.2

0.25

Time t

Am

plit

ude

x

(a) D = 2

0

0.1

0.2

0.3

0.4

Am

plit

ude

x

0 10 20 30 40 50 60 70

Time t

(b) D = 1

0 10 20 30 40 50 60 70-1.5

-1

-0.5

0

0.5

1

1.5

Time t

Am

plit

ude

x

(c) D = 0.5

0 10 20 30 40 50 60 70-1.5

-1

-0.5

0

0.5

1

1.5

Time t

Am

plit

ude

x

(d) D = 0.1

0 10 20 30 40 50 60 70-1.5

-1

-0.5

0

0.5

1

1.5

Time t

Am

plit

ude

x

(e) D = 0.01

0 10 20 30 40 50 60 70-1

-0.5

0

0.5

1

Time t

Am

plit

ude

x

(f) D = 0

Figure 4.9: Free vibrations of a viscously damped oscillator for different D


If we separate the real- and imaginary parts in the exponential function (eq. (4.71))we get:

x (t) = e−Dω0t(A1e+i

√1−D2ω0t +A2e−i

√1−D2ω0t

)(4.72)

The exponential term standing left of the bracket describes the decaying be-havior of the vibration (for D = 0 we obtain the undamped case withoutreduction of the amplitudes).

The expression with the imaginary exponent show us the oscillation because

e±ωDt = cosωDt± sinωDt

and using eq. (4.72) the circular frequency of the damped oscillation is:

ωD =√

1−D2ω0 (4.73)

The frequency is reduced by the influence of the damping. However, if thevalues for D are in the range 0.01 - 0.1, the change of the frequency comparedto the undamped case is negligible.

The period (from one maximum to the next maximum) then is:

T =2πωD

(4.74)

Using trigonometric functions instead of the exponential function eq. (4.72)becomes:

x (t) = e−Dω0t[A cos

(√1−D2ω0t

)+B sin

(√1−D2ω0t

)](4.75)

Introducing the initial displacement x0 and initial velocity v0 at time instantt = 0 yields the constants:

A = x0 (4.76)

andB =

v0 +Dω0x0

ω0√

1−D2 . (4.77)

or alternativelyx (t) = Ce−Dω0t sin (ωDt+ ϕ) (4.78)

The maximum amplitude C and the phase angle ϕ can be determined fromthe following equations:

C =√A2 +B2 =

√(x0ωD)

2 + (v0 +Dω0x0)2

ωD(4.79)

4.4 forced vibrations from harmonic excitation 95

andtanϕ =

A

B(4.80)

An important quantity to describe the damping behavior is the logarithmic Logarithmicdecrementdecrement:

ϑ =1n

ln x1xn+1

= 2π D√1−D2 (4.81)

The xi are maxima of the damped vibration and can be determined frommeasured curves of the free damped oscillations of a vibrating system. Becausethe use of two subsequent maxima can lead to inaccurate results we observe thedecaying process over a longer time (e.g. n = 5− 10 periods, if the damping issmall enough). From the logarithmic decrement we can immediately determineD. From the nT on the time axis from maximum xi to the maximum xn+i wecan calculate the damped natural circular frequency ωD.

4.4 forced vibrations from harmonic excitation

As discussed earlier, forced vibrations are one very important practical mech-anism for the occurrence of vibrations. The equation of motion of the dampedlinear SDOF oscillator with an external force (see fig. 4.10) is:

mx+ cx+ kx = F (t) (4.82)

The general solution of this differential equation is: Approach to theequation of motionin case of forced vi-bration

x (t) = xhom (t)︸︷︷︸free vibrations

+ xpart (t)︸︷︷︸results from external force

(4.83)

x

m

F(t)

ck

Figure 4.10: SDOF oscillator with viscous damping and external force


Figure 4.11: Homogeneous and particular part of the solution and superposition

which consists of the homogeneous part resulting from the free vibration andthe particular part resulting from the external disturbance F (t) (see fig. 4.11).The homogeneous solution has already been treated in the last chapter.

While the homogeneous part of the solution will decay to zero with time weare especially interested in the stationary solution.

4.4.1 Excitation with Constant Force Amplitude

The excitation function is harmonic, Ω is the frequency of excitation:In the script onlythe real approachis presented,however, thereis an equivalentcomplex approachthat could befollowed. Complexapproach isillustrated inappendix D.1.

F (t) = F cos Ωt (4.84)

Eq. 4.82 becomes:mx+ cx+ kx = F cos Ωt (4.85)

Dividing by the mass m:

x+c

mx+

k

mx =

F

mcos Ωt . (4.86)

Introducing again the dimension less damping (eq. (4.60)) and the naturalcircular frequency (eq. (4.4)):

2D =c

mω0, (4.87)

andω2

0 =k

m(4.88)

and the amplitude

f =F

m(4.89)


This yields:x+ 2Dω0x+ ω2

0x = f cos Ωt . (4.90)

To solve this differential equation, we make an approach with harmonic func-tions

x (t) = A cos Ωt+B sin Ωt (4.91)

This covers also a possible phase lag due to the damping in the system.Differentiating eq. (4.91) to get the velocity and the acceleration and puttingthis into eq. (4.90) leads to:

−Ω2A cos Ωt−Ω2B sin Ωt+ . . .

. . .+ 2Dω0 (−ΩA sin Ωt+ ΩB cos Ωt) + . . .

. . .+ ω20 (A cos Ωt+B sin Ωt) = f cos Ωt (4.92)

After separating the coefficients of the sin- and cos-functions and comparingthe coefficients we get:

−Ω2A+ 2Dω0ΩB + ω20A = f (4.93)

−Ω2B − 2Dω0ΩA+ ω20B = 0 (4.94)

From the second equation we see that

−Ω2B + ω20B = 2Dω0ΩA , (4.95)

which leads toB =

2Dω0Ωω2

0 −Ω2A (4.96)

and we put this result into eq. (4.93):

−Ω2A+ 2Dω0Ω2Dω0Ωω2

0 −Ω2A+ ω20A = f (4.97)[(

ω20 −Ω2

)+

4D2ω20Ω2

ω20 −Ω2

]A = f (4.98)[(

ω20 −Ω2

)2+ 4D2ω2

0Ω2]A = f

(ω2

0 −Ω2)

(4.99)

This yields the solution for A and B:

A =f(ω2

0 −Ω2)

(ω20 −Ω2)

2+ 4D2ω2

0Ω2(4.100)

andB =

f (2Dω0Ω)

(ω20 −Ω2)

2+ 4D2ω2

0Ω2(4.101)


Introducing the dimensionless ratio of frequencies:

η =Excitation frequenzNatural frequenz =

Ωω0

(4.102)

leads to:

A =

(fω2

0

) (1− η2

)(1− η2)2 + 4D2η2

(4.103)

and

B =

(fω2

0

)(2Dη)

(1− η2)2 + 4D2η2. (4.104)

with A and B we have found the solution for x (t) = A cos Ωt+B sin Ωt.

Another possibility is to present the solution with amplitude and phase angle:

x (t) = C cos (Ωt− ϕ) (4.105)

The amplitude is:

C =√A2 +B2 =

1√(1− η2)2 + 4D2η2

f

ω20

(4.106)

Considering that f = Fm und ω2

0 = km we get:

C =1√

(1− η2)2 + 4D2η2

F

k(4.107)

Introducing the dimensionless magnification factor V 1 which only depends onthe frequency ratio and the damping D:

V1 (η,D) =1√

(1− η2)2 + 4D2η2(4.108)

we get the amplitude as

C = V1F

k(4.109)

and the phase angle (using trigonometric functions similar as in 4.2.2 on p. 84):

tanϕ =B

A=

2Dη1− η2 (4.110)


In fig. 4.12(a) we can see that as η approaches 1 the amplitude grows rapidly,and its value near or at the resonance is very sensitive to changes of the damp-ing D. The maximum of the magnification curve for a given D can be foundat

ηres =√

1− 2D2 =Ωresω0

(4.111)

If D is very small then ηres ≈ 1. The maximum amplitude for this D then is:

Cmax = V1 (ηres,D)F

k=

12D√

1−D2F

k(4.112)

η → 0 V1 ≈ 1 The system behaves quasi-statically.η →∞ V1 → 0 The vibrations are very small.


0 0,5 1 1,5 2 2,5 3 3,5 4 4,5 50

1

2

3

4

5

6

7

8

9

10

η ω= /Ω0

V1

D=0,05

D=0,1

D=0,2

D=0,3

D=0,5

D=0,7071

√(1- ) +42 2D

2 2η η

1V1=

(a) Amplitude

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50°

20°

40°

60°

80°

100°

120°

140°

160°

180°

φ

η ω= /Ω0

D=0,05D=0,1

D=0,2D=0,3

D=0,5D=0,7071

1- 2η

φ =2Dη

arctan

D=0

(b) Phase

Figure 4.12: Magnification factor V1 and phase angle to describe the vibration behav-ior of the damped oscillator under constant force amplitude excitation


c

Ω Ω

mu

2

k

2k

2

ε ε

mM

xmu

2

Figure 4.13: SDOF oscillator with unbalance excitation

4.4.2 Harmonic Force from Imbalance Excitation

The total mass of the system in fig. 4.13 consists of the mass mM and the tworotating unbalance masses mu:

m = mM + 2mu

2 (4.113)

The disturbance force from the unbalance is depending on the angular speedΩ, ε the excentricity:

FUnbalance (t) = Ω2εmu cos Ωt (4.114)

Now, following the same way as before (real or complex) leads to the solution:

x (t) = C cos (Ωt− ϕ) (4.115)

with Amplitude

x = C =η2√

(1− η2)2 + 4D2η2εmu

m= V3 (η,D) ε

mu

m, (4.116)

Phasetanϕ =

2Dη1− η2 (4.117)

and Magnification factor

V3 (η,D) =η2√

(1− η2)2 + 4D2η2. (4.118)

The phase has the same expression as in the previous case (eq. (4.110)), how-ever, the magnification factor is different (see fig. 4.14), because the forceamplitude is increasing with increasing angular speed.


V3

0 0,5 1 1,5 2 3 3,5 4 4,5 50

1

2

3

4

5

6

7

8

9

10

D=0,05

D=0,1

D=0,2

D=0,3

D=0,5

D=0,7071

η ω= /Ω0

2,5

√(1- ) +42 2D

2 2η η

η2

V3=

Figure 4.14: Magnification factor V3 for the case of imbalance excitation

As can be seen, for:η → 0 V3 ≈ 0 There is no force if the system is not rotating or

rotates only slowlyη →∞ V3 → 1 The Mass mM is vibrating with an amplitude

(εmum ), but the common center of gravity of total

system mM and mu does not move.

4.4.3 Support Motion / Ground Motion

4.4.3.1 Case 1

The equation of motion for this system depicted in fig. 4.15 is:

mx+ cx+ kx = ku (t) (4.119)

Under harmonic excitation:

u (t) = u cos Ωt (4.120)


u(t)

c

k

m

x

Figure 4.15: Excitation of the SDOF oscillator by harmonic motion of one springend

The mathematical treatment is nearly identical to the first case, only the exci-tation function is different: the excitation F

k is replaced by u here. This leadsto the result for the amplitude of vibration:

x =1√

(1− η2)2 + 4D2η2︸︷︷︸mag. function V1

u = V1 (η,D) u (4.121)

The magnification factor again is V1 (eq. (4.108)). Also, the phase relation isidentical as before (eq. (4.110)):

tanϕ =2Dη

1− η2 (4.122)

4.4.3.2 Case 2

The equation of motion now (see fig. 4.16) also contains the velocity u:

mx+ cx+ kx = cu+ ku (4.123)

Amplitude of vibration and phase shift become:

x =

√1 + 4D2η2√

(1− η2)2 + 4D2η2︸︷︷︸mag. function V2

u = V2 (η,D) u (4.124)

and

tanϕ =2Dη3

1− η2 + 4D2η2 . (4.125)


ck

mx

u(t)

Figure 4.16: Excitation of the SDOF oscillator by harmonic motion of the spring/-damper combination

As can be seen the phase now is different due to the fact that the damper forcedepending on the relative velocity between ground motion and motion of themass plays a role. The amplitude behaviour is described by the magnificationfactor:

V2 =

√1 + 4D2η2√

(1− η2)2 + 4D2η2(4.126)

Notice that all curves have an intersection point at η =√

2 (see fig. 4.17 whichmeans that for η >

√2 higher damping does not lead to smaller amplitudes but

increases the amplitudes. This is due to the fact that larger relative velocities(due to higher frequencies η) make the damper stiffer and hence the dampingforces.Further cases of ground motion excitation are possible.

4.5 excitation by impacts

4.5.1 Impact of Finite Duration

We consider an impact of finite length Ti and constant force level during theimpact (see fig. 4.18). The impact duration Ti is much smaller than the periodof vibration T .

Ti << T =2πωD

(4.127)

4.5 excitation by impacts 105

η ω= /Ω0

V2

0 0,5 1 1,5 2 2,5 3 3,5 4 4,5 50

1

2

3

4

5

6

7

8

9

10

D=0,05

D=0,1

D=0,2

D=0,3

D=0,5

D=0,7071

√(1- ) +42 2Dη η

2 2

√1+4D2 2η

V2=

√2Figure 4.17: Magnification factor V2 for the case of ground excitation via spring and

damper

With the initial condition that there is no initial displacement x0 = 0 we cancalculate the velocity by means of the impulse of the force:

p = mv0 =∫ Ti

0Fdt = F Ti (4.128)

This leads to the initial velocity:

v0 =F Tim

(4.129)

Ti

F(t)

t

F(t)

kc

x

m

F(t)^

Figure 4.18: SDOF oscillator under impact loading


Using the results of the viscously damped free oscillator for D < 1 (eq. (4.75)),

x (t) = e−Dω0t

A cos

√1−D2ω0︸︷︷︸ωD

t

+B sin

√1−D2ω0︸︷︷︸ωD

t

(4.130)

we can immediately find the result with the initial conditions x0 und v0:

x0 = 0 ⇒ A = 0 (4.131)

and

v0 =F Tim

⇒ B =v0

ω0√

1−D2 =v0ωD

(4.132)

so that the system response to the impact is a decaying oscillation where wehave assumed that the damping D < 1:

x (t) =v0ωD

e−Dω0t sin (ωDt) (4.133)

4.5.2 DIRAC-Impact

The DIRAC-Impact is defined by

F (t) = F δ (t)→ δ (t) =

0 t 6= 0

∞ t = 0but

∫ ∞−∞

δ (t) dt = 1 (4.134)

δ is the Kronecker symbol. The duration of this impact is infinitely shortbut the impact is infinitely large. However, the integral is equal to 1 or F ,

F

t

Figure 4.19: DIRAC-Impact

4.6 excitation by forces with arbitrary time functions 107

respectively. For the initial displacement x0 = 0 and calculation of the initialvelocity following the previous chapter, we get:

x (t) =F

mωDe−Dω0t sin (ωDt) (4.135)

For F = 1, the response x (t) is equal to the impulse response function (IRF)h (t):

h (t) =1

mωDe−Dω0t sin (ωDt) (4.136)

The IRF is an important characteristic of a dynamic system in control theory.

4.6 excitation by forces with arbitrary time functions

F( )τ

τ τ τ+Δ

F

t

t

x

τ

Figure 4.20: Interpretation of an arbitrary time function as series of DIRAC-impacts

Using the results of the previous chapters we can solve the problem of an arbi-trary time function F (t) (see fig. 4.20) as subsequent series of Dirac-impacts,where the initial conditions follow from the time history of the system.


The solution is given by the Duhamel-Integral or convolution integral:

x (t) =∫ t

0

1mωD

e−Dω0(t−τ ) sin (ωD (t− τ ))F (τ ) dτ (4.137)

=∫ t

0h (t− τ )F (τ ) dτ (4.138)

As can be seen, the integral contains the response of the SDOF oscillator withrespect to a DIRAC-impact multiplied with the actual force F (τ ), which isintegrated from time 0 bis t.

4.7 periodic excitations

4.7.1 Fourier Series Representation of Signals

Periodic signals can be decomposed into an infinite series of trigonometricfunctions, called Fourier series (see fig. 4.21). The period of the signal is T andthe corresponding fundamental frequency is:

ω =2πT

(4.139)

Now, the periodic signal x (t) can be represented as follows:

x (t) =a02 +

∞∑k=1

ak cos (kωt) + bk sin (kωt) (4.140)

Figure 4.21: Scheme of signal decomposition by trigonometric functions

4.7 periodic excitations 109

The Fourier-coefficients a0, ak and bk must be determined. They describe howstrong the corresponding trigonometric function is present in the signal x (t).The coefficient a0 is the double mean value of the signal in the interval 0 . . . T :

a0 =2T

∫ T

0x (t) dt (4.141)

and represents the off-set of the signal. The other coefficients can be determinedfrom

ak =2T

∫ T

0x (t) cos (kωt) dt (4.142)

andbk =

2T

∫ T

0x (t) sin (kωt) dt (4.143)

The individual frequencies of this terms are

ωk = kω =2πkT

(4.144)

For k = 1 we call the frequency ω1 fundamental frequency or basic harmonicand the frequencies for k = 2, 3, . . . the second, third, . . . harmonic (or generallyhigher harmonics).

The real trigonometric functions can also be transformed into complex expo-nential expression:

x (t) =∞∑

k=−∞Xke

ikωt (4.145)

TheXk are the complex Fourier coefficients which can be determined by solvingthe integral:

Xk =1T

∫ T

0x (t) e−ikωtdt (4.146)

or

Xk =1T

∫ T

0x (t) [cos kωt− i sin kωt] dt (4.147)

which clearly shows the relation to the real Fourier coefficients series given byeq. (4.142) and eq. (4.143):

Re Xk =ak2 ; Im Xk = −

bk2 (4.148)

The connection to the other real representation (chap. 4.7.1) is:

|Xk| = ck tanϕk =(

Im XkRe Xk

)(4.149)

The coefficients with negative index are the conjugate complex values of thecorresponding positive ones:

X−k = X∗k (4.150)


4.7.2 Forced Vibration Under General Periodic Excitation

Let us use once more the SDOF oscillator (see fig. 4.22) but now the force isa periodic function which can be represented by a Fourier series

F (t) =F02 +

∞∑k=1

Fck cos (kΩt) + Fsk sin (kΩt) (4.151)

The Fck and Fsk are the Fourier coefficients which can be determined ac-

x

m

F(t)

ck

Figure 4.22: SDOF oscillator with viscous damping and external force

cording to the last chapter (eqns. 4.141 - 4.143). The response due to such anexcitation is:Here, the stiffness

of the system iscalled k∗. x (t) =

F02k∗ +

∞∑k=1

V1 (ηk,D)Fckk∗

cos (kΩt− ϕk) + . . .

. . .+ V1 (ηk,D)Fskk∗

sin (kΩt− ϕk)(4.152)

with the frequency ratio:

ηk =kΩω0

, k = 1, 2, . . .∞ (4.153)

Each individual frequency is considered with its special amplification factor Vand individual phase shift ϕ, which in the present case can be calculated from(see eq. (4.108)):

V1 (ηk,D) =1√

(1− η2k)

2+ 4D2η2

k

tanϕk =2Dηk1− η2

k

For the other cases of mass unbalance excitation or ground excitation theprocedure works analogously. The appropriate V -functions have to be usedand the correct pre-factors (which is in the present case 1

k ) have to be used.


Figure 4.23: Perturbation function

example: Graphical determination of the complete solution for a givenperturbation function F (t) (see fig. 4.23)

F (t) = F1 (t) + F3 (t) = F1 sin (Ω1t) + F3 sin (Ω3t)

Fig. 4.24 shows the graphical representation of both partial functions of theperturbation function (only the first two not vanishing parts of the Fourierseries) where:

F1 (t) = F1 sin (Ω1t) F3 (t) = F3 sin (Ω3t)

F1 = 1 F3 =13 F1

Ω1 = Ω Ω3 = 3Ω

The frequency-dependent amplitude (magnification factor) including the val-ues V1 and V3 which are determined for η1 and η3 is depicted in a resonancediagram shown in fig. 4.25(a) (dimensionless damping amounts D = 0.2). Fig.4.25(b) shows the according phase angle diagram. The graphical representa-tions of the partial solutions

x1 (t) = A1 sin (Ω1t− ϕ1)

x3 (t) = A3 sin (Ω3t− ϕ3)

are illustrated in fig. 4.25(c). In fig. 4.25(d) one can study the complete solution

x (t) = x1 (t) + x3 (t)

with A1 > A3.


The difference in case of resonance is shown by the graphical representation ofthe frequency-dependent amplitude (magnification factor) including the values

Figure 4.24: Graphical representation of both sub-functions

(a) Magnification factor

0 1

h

φ3

φ1

h1

h3

p/2

Pha

se a

ngle

(b) Phase angle

Time t

Dis

plac

emen

t x

(c) Partial solution

Time t

Dis

plac

emen

t x

(d) Complete solution

Figure 4.25: Graphical determination of the complete solution for a given perturba-tion function F (t) without resonance


Amplitude

(a) Magnification factor

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50°

20°

40°

60°

80°

100°

120°

140°

160°

180°

φ

η = Ω / ω0

D=0,05

D=0,1

D=0,2

D=0,3D=0,5

D=0,7071

21-ηφ =

2Dηarctan

D=0

(b) Phase angle

Time t

Dis

plac

emen

t x

(c) Partial solution

Time t

Dis

plac

emen

t x

(d) Complete solution

Figure 4.26: Graphical determination of the complete solution for a given perturba-tion function F (t) in case of resonance

V1 and V3 which are determined for η1 and η3. This is depicted in a resonance di-agram in fig. 4.26(a). Here we have resonance, dimensionless damping amountsD = 0.2. Fig. 4.26(b) shows the according phase angle diagram. The graphicalrepresentations of the partial solutions

x1 (t) = A1 sin (Ω1t− ϕ1)

x3 (t) = A3 sin (Ω3t− ϕ3)

are illustrated in fig. 4.26(c). In fig. 4.26(d) one can study the complete solution

x (t) = x1 (t) + x3 (t)

with A1 < A3.


4.8 vibration isolation of machines

In dynamics of machines and systems the linear vibrations of SDOF systemsintroduced at the beginning of chapter 4 are often applied to isolate machineryfrom vibrations.

• Requirements:

– machine can be idealized as rigid

– vibration process can be considered to be a SDOF system

• Vibration ssolation:

– important assignment of dynamics of machines and systems

– concerning machine manufactures, project planers, operators

– any installation site whether ground, floor or another carrying struc-ture is elastic. That is why excitation by imbalanced inertia forcescan occur

• Purpose:

– installing the machine in order to minimize forces (inertia forces, ...)diverted into the foundation / structure

– mounting sensitive devices in a way so that they are well shieldedfrom their vibrating environment

We differ between:

1. Low tuning:The highest natural frequency of the ground vibration is less than thelowest excitation frequency.

2. High tuning:The natural frequencies of the ground vibration are above the range ofthe excitation frequencies.

3. Mixed tuning:The range of natural frequencies and the range of excitation frequenciespartially overlap each other, but there is no resonance.

4.8 vibration isolation of machines 115

c

Ω

k

2k

2

ε

x

mu

Figure 4.27: Imbalanced machine - substituted mechanical system

These terms are generally valid for MDOF systems too.

4.8.1 Forces on the Environment Due to Excitation by Inertia Forces

In order to achieve vibration isolation we initially have to calculate the forceson the environment in order to use encountered dependencies for the purposeof minimizing these forces. The determination of these forces is performed bymeans of an example.

example: Imbalanced machine (see fig. 4.27)

Figure 4.28: Imbalanced machine - forces acting on the foundation


Amplitude of the excitation force:

Fu = muεΩ2 = UΩ2

with total mass m and imbalance mass mu.

Forces acting on the foundation like in fig. 4.28:

FFu = kx+ cx

In case of harmonic excitation x (t):

X (t) = XeiΩt

˙X (t) = iΩXeiΩt

¨X (t) = −Ω2XeiΩt

FFu (t) = kXeiΩt + ciΩXeiΩt = (k+ iΩc) XeiΩt

ˆFFu (t) = (k+ iΩc) η2

(1− η2) + 2Dηiεmu

m

=(ω2

0 + 2Dω0iΩ) η2

(1− η2) + 2Dηiεmu

Referring the complex amplitude to the amplitude of the excitation force:

ˆFFuFu

=

(ω2

0 + 2Dω0iΩ)

η2

(1−η2)+2Dηiεmu

Ω2εmu=

1 + 2Dηi1− η2 + 2Dηi

Real amplitude ratio (see. eq. (4.126)):

FFu

Fu=

√√√√ 1 + 4D2η2

(1− η2)2 + 4D2η2= V2 (η,D) (4.154)

Absolute amplitude of the foundation force:

FFu = V2 (η,D) Fu = V2Ω2muε = V2η2muεω

20 = V2η

2mu

mεk (4.155)

FFu = V2η2mu

mεk = V4

mu

mεk (4.156)

V4 = η2V2 (4.157)


η ω= /Ω0

V2

0 0,5 1 1,5 2 2,5 3 3,5 4 4,5 50

1

2

3

4

5

6

7

8

9

10

D=0,05

D=0,1

D=0,2

D=0,3

D=0,5

D=0,7071

√(1- ) +42 2Dη η

2 2

√1+4D2 2η

V2=

√2Figure 4.29: Magnification factor V2

4.8.2 Tuning of Springs and Dampers

From the real amplitude ration (see eq. (4.154))

FFu

Fu= V2 (η,D) , (4.158)

we can deduce, η should attain higher values so that V2 becomes as small aspossible (see fig. 4.29).

High values of η mean:

Ω = ηω0 ⇒ Ω > ω0

This is called low or "soft" tuning.

Soft springs, heavy foundation block ⇒ small ω0.

Using soft springs, it is necessary to mind the static lowering under dead load,because it can increase to high values (too high!).

Moreover it shows: In order to have V2 as small as possible damping should besmallest possible at fixed Ω. But this brings along disadvantages for passingthe case of resonance (while starting up and shutting down).


Remedy:

• passing resonance frequency very fast (sufficient torque required!)

• connecting a damper while passing resonance frequency

By comparison: rigid installation would lead to FF u

Fu= 1.

numerical example D = 0; Tuning target: we want to tune the foun-dation in a way, so that only 5% of the excitation force is transferred to theenvironment!

FFu

Fu< 5% =

120

V2 (η,D = 0) = 11− η2

η =Ωω0

⇒ 120 = ±

∣∣∣∣∣ 11− η2

∣∣∣∣∣⇒ 1− η2 = −20⇔ −η2 = −21⇔ η =

√21 ≈ 4, 58

that meansΩ ≥ 4, 58ω0 or ω0 ≤

(1

4, 58

)Ω (4.159)

In case of constant force amplitudes we obtain the following isolation effect:Isolation effect incase of excitation

with constantforce amplitude

F (t) = F sin Ωt

or

F (t) = F eiΩt .

Applying the same approach as in section 4.8 we get (see eq. (4.154)):

FFu

Fu= V2 (η,D)

Dealing with machinery installed in way that it is excited by ground motionVibration isolationin case of groundmotion

we have (see eq. (4.126)):x

xa= V2 (η,D)


η ω= /Ω 0

V2

0 0,5 1 1,5 2 2,5 3 3,5 4 4,5 50

1

2

3

4

5

6

7

8

9

10

Hard foundation Soft foundation

Figure 4.30: Magnification factor V2 for tuning the spring

x

m

ck

xa

Figure 4.31: Single-mass oscillator

We notice the most suitable way to manage a vibration isolated installation ofsensitive devices is to use low tuning. The same applies to soundproof roomsetc.

5VIBRATION OF L INEARMULTIPLE -DEGREE -OF -FREEDOM SYSTEMS

5.1 equation of motion

The equation of motion can be derived by using the principles we have learnedsuch as Newton’s/Euler’s laws or Lagrange’s equation of motion. For a generallinear system mDOF system we found that we can write in matrix form:

M x+(C +G

)x+

(K +N

)x = F (5.1)

with the matrices: Note:A general matrix Bcan be decomposedinto the symmetricpart and the skew-symmetric part Band B:

B = B + B

where:

B = 12(B +BT

)B = 1

2(B −BT

)

• M : Mass matrix (symmetric) M = MT

• C: Damping matrix (symmetric) C = CT

• K: Stiffness matrix (symmetric) K = KT

• G: Gyroscopic matrix (skew-symmetric) G = −GT

• N : Matrix of non-conservative forces (skew-symmetric) N = −NT

• F : External forces

In the standard case that we have no gyroscopic forces and no non-conservativedisplacement dependent forces but only inertial forces, damping forces andelastic forces the last equation reduces to:

M x+C x+K x = F (5.2)

121

122 vibration of linear multiple-degree-of-freedom systems

example The system shown in fig. 5.1, where the masses can slide withoutfriction (µ = 0), has them1 0 00 m2 00 0 m3

x1

x2

x3

+c1 0 00 0 00 0 0

x1

x2

x3

+ . . .

. . .+

k1 + k2 + k3 −k2 −k3

−k2 k2 0−k3 0 k3

x1

x2

x3

=

F1 (t)

F2 (t)

F3 (t)

(5.3)

5.2 influence of the weight forces and static equilib-rium

The static equilibrium displacements are calculated by (xstat = xstat = 0):

K xstat = F stat (5.4)

which means for the case of the example shown in fig. 5.2:

K xstat = F stat =

m1g

m2g

(5.5)

F (t)1

F (t)2

F (t)3

x2

k1

c1

x1

x3

k2

k3

m1

m3

m2

m=0

m=0

x1x1x1x1x1

F (t)1

m1

FC1

FK1FK3

FK2

FK3

FK2 m2

F (t)3m3

F (t)2

x1x1x1 x2

x3

Figure 5.1: Multi-Degree-of-Freedom (MDOF) system with free body diagram

5.2 influence of the weight forces and static equilibrium 123

g

k2

k1

m1

m1

m2

m2

k1

k2

x stat1

x stat2

Figure 5.2: Static equilibrium position of a two DOFs system

The dynamic problem for this example is:

M x+K x = F stat︸︷︷︸static force

+F (t) (5.6)

x = xstat + xdyn︸︷︷︸part of the motion

describing the vibrationabout the static equilibrium

(5.7)

From the last equation also follows that

x = xdyn ⇒ x = xdyn (5.8)

so thatM xdyn +K

(xdyn + xstat

)= F stat + F (t) (5.9)

and after rearrangement

M xdyn +K xdyn = F stat −K xstat︸︷︷︸=0

+F (t) (5.10)

M xdyn +K xdyn = F (t) (5.11)

As can be seen the static forces and static displacements can be eliminatedand the equation of motion describes the dynamic process about the staticequilibrium position.

In cases where the weight forces influences the dynamic behavior a simple elim- Noteination of the static forces and displacements is not possible. In the exampleof an inverted pendulum shown in fig. 5.3 the restoring moment is mgl sinφ,where l is the length of the pendulum.


k k

mg

m

Figure 5.3: Case where the static force also influences the dynamics

5.3 ground excitation

Fig. 5.4 shows a MDOF system.

Without ground motion x0 = 0 the equation of motion is:m1 0

0 m2

x1

x2

+c1 0

0 0

x1

x2

+ . . .

. . .+

k1 + k2 −k2

−k2 k2

x1

x2

=

F1

F2

(5.12)

Now, if we include the ground motion, the differences (x1 − x0) and the rel-ative velocity d(x1 − x0)/dt determine the elastic and the damping force, re-

k2

k1

m1

m2

c1

x2

x1

xo

F2

F1

Figure 5.4: 2DOF system with excitation by ground motion x0

5.3 ground excitation 125

spectively at the lower mass. This can be expressed by adding x0 to the lastequation in the following manner:m1 0

0 m2

x1

x2

+c1 0

0 0

x1

x2

+k1 + k2 −k2

−k2 k2

x1

x2

= . . .

. . . =

F1

F2

+k1

0

x0 (t) +

c10

x0 (t) (5.13)

The dynamic force f0 of the vibrating system on the foundation is:

F0 = k1 (x1 − x0) + c1 (x1 − x0) (5.14)

Figure 5.5: Example for ground motion excitation of a building structure (earth-quake excitation)

Figure 5.6: Excitation of a vehicle by rough surface


5.4 free undamped vibrations of the multiple-degree-of-freedom system

5.4.1 Eigensolution, Natural Frequencies and Mode Shapes of the System

The equation of motion of the undamped system is:

M x+K x = 0 (5.15)

To find the solution of the homogeneous differential equation, we make theharmonic solution approach as in the SDOF case. However, now we have toconsider a distribution of the individual amplitudes for each coordinate. Thisis done by introducing (the unknown) vector ϕ:

x = ϕeiωt (5.16)x = −ω2ϕeiωt (5.17)

Putting this into eq. (5.15) yields:(K − ω2M

)ϕ = 0 (5.18)

This is a homogeneous equation with unknown scalar ω and vector ϕ. If we setλ = ω2 we see that this is a general matrix eigenvalue problem1:(

K − λM)ϕ = 0 (5.19)

where λ is the eigenvalue and ϕ is the eigenvector. Because the dimension ofthe matrices is f by f we get f pairs of eigenvalues and eigenvectors:

λi = ω2i for i = 1, 2, . . . f (5.20)

ωi is the i-th natural circular frequency andϕi

the i-th eigenvector which has the physical meaning of a vibrationmode shape.

The solution of the characteristic equation

det(K − ω2M

)= 0 (5.21)

1 The well-known special eigenvalue problem has the form(A− λI

)x = 0, where I is the

identity matrix, x the eigenvector and λ the eigenvalue.

5.4 free undamped vibrations of the multiple-degree-of-freedom system 127

yields the eigenvalues and natural circular frequencies λ = ω2, respectively.The natural frequencies are:

fi =ωi2π (5.22)

The natural frequencies are the resonant frequencies of the structure.

The eigenvectors can be normalized arbitrarily, because they only represent avibration mode shape, no absolute values. Commonly used normalizations are

• normalize ϕiso that

∣∣∣ϕi

∣∣∣ = 1

• normalize ϕiso that the maximum component is 1

• normalize ϕiso that the modal mass (the generalized mass) is 1.

Generalized mass or modal mass:

Mi = ϕTiM ϕ

i(5.23)

Generalized stiffness:Ki = ϕT

iK ϕ

i(5.24)

whereKi = ϕT

iK ϕ

i= ω2

i if Mi = 1 (5.25)

The so-called Rayleigh ratio is: Rayleigh ratio

ω2i =

Ki

Mi=

ϕTiK ϕ

i

ϕTiM ϕ

i

(5.26)

It allows the calculation of the frequency if the vectors are already known.

5.4.2 Modal Matrix, Orthogonality of the Mode Shape Vectors

If we order the natural frequencies so that

ω1 ≤ ω2 ≤ ω3 ≤ . . . ≤ ωf

and put the corresponding eigenvectors columnwise in a matrix, the so-calledmodal matrix, we get: Modal matrix

Φ =[ϕ1,ϕ2, . . . ,ϕ

f

]=

ϕ11 ϕ12 . . . ϕ1f

ϕ21 ϕ22 . . . ϕ2f

. . . . . . . . . . . .

ϕf1 ϕf2 . . . ϕff

(5.27)


The first subscript of the matrix elements denotes the number of the vectorcomponent while the second subscript characterizes the number of the eigen-vector.

The eigenvectors are linearly independent and moreover they are orthogonal.This can be shown by a pair i and j(

K − ωi2M)ϕi= 0 and

(K − ωj2M

)ϕj= 0 (5.28)

Premultplying by the transposed eigenvector with index j and i respectively:

ϕjT(K − ωi2M

)ϕi= 0 and ϕ

iT(K − ωj2M

)ϕj= 0 (5.29)

If we take the transpose of the second equation

ϕjT(KT − ωj2MT

)ϕi= 0 (5.30)

and consider the symmetry of the matrices: M = MT and K = KT andsubtract this equation ϕ

jT(K − ωi2M

)ϕi= 0 from the first eq. (5.28) we get(

ωj2 − ωi2

)ϕjT(M)ϕi= 0 (5.31)

which means that if the eigenvalues are distinct ωi 6= ωj for i 6= j the secondscalar product expression must be equal to zero:

ϕjTMϕ

i= 0 (5.32)

That means that the two distinct eigenvectors i 6= j are orthogonal with respectto the mass matrix. For all combinations we can write:Kronecker

symbol δijϕjTMϕ

i= δijMi

ϕjTKϕ

i= δijωi

2Mi

δij =

0, for i = j

1, for i 6= j(5.33)

or with the modal matrix:

ΦTMΦ = diag Mi =

M1 0 . . . 00 M2

. . . ...... . . . . . . 00 . . . 0 Mf

ΦTKΦ = diagωi

2Mi

=

ω12M1 0 . . . 0

0 ω22M2. . . ...

... . . . . . . 00 . . . 0 ωf

2Mf

(5.34)


Figure 5.7: Mode shapes and natural frequencies of a two storey structure


5.4.3 Free Vibrations, Initial Conditions

The free motion of the undamped system x (t) is a superposition of the modesvibrating with the corresponding natural frequency:

x (t) =f∑i=1

ϕi(Aci cosωit+Asi sinωit) (5.35)

Each mode is weighted by a coefficient Aci and Asi which depend on theinitial displacement shape and the velocities. In order to get these coefficients,we premultiply eq. (5.35) by the transposed j-th eigenvector:

ϕjTMx (t) =

f∑i=1

ϕjTMϕ

i(Aci cosωit+Asi sinωit)︸︷︷︸for i6=j⇒=0

(5.36)

= ϕjTMϕ

j︸︷︷︸Mj

(Acj cosωjt+Asj sinωjt) (5.37)

All but one of the summation terms are equal to zero due to the orthogonalityconditions. With the initial conditions for t = 0 we can derive the coefficients:

t = 0x (t = 0) = x0

ϕjTMx0 = MjAcj

⇒ Acj =ϕjTMx0

Mj(5.38)

x (t) =f∑i=1

ϕiωi (−Aci sinωit+Asi cosωit) (5.39)

t = 0x (t = 0) = v0

ϕjTMx0 = MjωjAsj

⇒ Asj =ϕjTMv0

Mjωj(5.40)

which we have to calculate for modes j.

5.4.4 Rigid Body Modes

As learned earlier the constraints reduce the DOFs of the rigid body motion.If the number of constraints is not sufficient to suppress rigid body motion thesystem has also zero eigenvalues. The number of zero eigenvalues correspondsdirectly to the number of rigid body modes. In the example shown in fig. 5.8 theExample of a rigid

body mode


two masses which are connected with a spring can move with a fixed distanceas a rigid system. This mode is the rigid body mode, while the vibration ofthe two masses is a deformation mode. The equation of motion of this systemis: m1 0

0 m2

x1

x2

+ k −k−k k

x1

x2

=

00

(5.41)

The corresponding eigenvalue problem is:k− λm1 −k−k k− λm2

ϕ1

ϕ2

=

00

(5.42)

The eigenvalues follow from the determinant which is set equal to zero:

det [. . .] =[(k− λm1) (k− λm2)− k2

]= 0 (5.43)

λ2m1m2 − λ (km1 + km2) = 0 (5.44)

Obviously, this quadratic equation has the solution

λ1 = ω21 = 0 (5.45)

andλ2 = ω2

2 = km1 +m2m1m2

(5.46)

The corresponding (unnormalized) eigenvectors are

ϕ1 =

11

(5.47)

which is the rigid body mode: both masses have the same displacement, nopotential energy is stored in the spring and hence no vibration occurs. Thesecond eigenvector, the deformation mode is

ϕ2 =

1−m1m2

(5.48)

which is a vibration of the two masses. Other examples for systems with rigidbody modes are shown in the following figures.

x1x2

m1 m2

k

Figure 5.8: A two-DOFs oscillator which can perform rigid body motion


5.5 forced vibrations of the undamped oscillator underharmonic excitation

The equation of motion for the type of system depicted in fig. 5.12 is:

M x+K x = F (t) (5.49)

Figure 5.9: Examples for systems with torsional and transverse bending motion withrigid body motion

Figure 5.10: Flying airplane (Airbus A318) as a system with 6 rigid body modes anddeformation modes

Figure 5.11: Commercial communication satellite system (EADS) with 6 rigid bodymodes and deformation modes

5.5 forced vibrations of the undamped oscillator under harmonic excitation 133

For a harmonic excitation we can make an exponential approach to solve theproblem as we did with the SDOF system.

F (t) = F︸︷︷︸complexamplitudevector

eiΩt (5.50)

We make a complex harmonic approach for the displacements with Ω as theexcitation frequency:

x = XeiΩt (5.51)The acceleration vector is the second derivative

x = −Ω2XeiΩt (5.52)

Putting both into the equation of motion (eq. (5.49)) and eliminating theexponential function yields (

K −Ω2M)X = F (5.53)

which is a complex linear equation system that can be solved by hand for asmall number of DOFs or numerically. The formal solution is:

X =(K −Ω2M

)−1F (5.54)

which can be solved if determinant of the coefficient matrix:

det(K −Ω2M

)6= 0 (5.55)

If the excitation frequency Ω coincides with one of the natural frequenciesωi we get resonance of the system with infinitely large amplitudes (in theundamped case).

k1

k2

m2

m1

k2

x1

x2

F (t)2

F (t)1

2 2

Figure 5.12: Example for a system under forced excitation


Resonance:det

(K −Ω2M

)= 0⇔ Ω = ωi (5.56)

Part II

APPENDIX

AE INLE ITUNG - ERGÄNZUNG

137

BKINEMATICS - APPENDIX

b.1 general 3-d motion in cartesian coordinates

The position vector in cartesian coordinates with time invariant unit vectorsex, ey und ez:

r (t) = x (t) ex + y (t) ey + z (t) ez (B.1)

The coordinates x (t), y (t) and z (t) are scalar functions of time t. Because theunit vectors are time invariant in the cartesian system, we immediately get thevelocity vector (first derivative) and the acceleration vector (second derivative)of the position vector with respect to time:

v (t) = r (t)

= x (t) ex + y (t) ey + z (t) ez (B.2)a (t) = v (t) = r (t)

= x (t) ex + y (t) ey + z (t) ez (B.3)

with components vx = x, vy = y and vz = z, also ax = x, ay = y and az = z.

The velocity vector is always tangential to the path of the particle P . In general,this is not true for the acceleration vector.

x

y

z

P

r

e

e

e

z

x y x

y

z

Path

Figure B.1: Cartesian Coordinate System

139

140 kinematics - appendix

The magnitude of the position, velocity and acceleration vector is:

r = |r| =√x2 (t) + y2 (t) + z2 (t) (B.4)

v = |v| =√x2 (t) + y2 (t) + z2 (t) (B.5)

a = |a| =√x2 (t) + y2 (t) + z2 (t) (B.6)

b.2 three-dimensional motion in cylindrical coordinates

The description of motion on a curved path might be easier described in cylin-drical coordinates than in cartesian coordinates.

The position vector is in cylindrical coordinates

r (t) = r (t) er + z (t) ez (B.7)

whereas the unit vector er depends on the angle ϕ? and this on his part againdepends on the time, which can be written as er = er (ϕ (t)) . Due to thistime dependence of the unit vector the derivations with respect to time for thevelocity and the acceleration are not as simple expressions as in cartesian coor-dinates. In fact the following expression can be obtained (without derivation):

v (t) = r (t) = r (t) er + r (t) ϕ (t) eϕ + z (t) ez (B.8)

with the velocity components:

vr = r Radial velocityvϕ = rϕ Circular velocityvz = z Axial velocity

Path

y

x

z

ejez

erj

r

r

z

P

Figure B.2: Cylindrical Coordinates

B.3 natural coordinates, intrinsic coordinates or path variables 141

and the acceleration is:

a (t) = r (t) =[r (t)− r (t) ϕ2 (t)

]er

+ [2r (t) ϕ (t) + r (t) ϕ (t)] eϕ + z (t) ez (B.9)

with the components:

ar = r− rϕ2 Radial acceleration (B.10)aϕ = 2rϕ+ rϕ Circular acceleration (B.11)az = z Axial acceleration (B.12)

Whereas the component −rϕ2 = −rω2 is the centripetal acceleration and ω isthe angular velocity. The component 2rϕ = 2vrω is the Coriolis acceleration.The magnitude of velocity and acceleration can be obtained using the rule ofvector operation:

|v| =√v2r + v2

ϕ + v2z (B.13)

|a| =√a2r + a2

ϕ + a2z (B.14)

The magnitude of the position vector may not be mistaken with the radialcoordinate r

|r| =√r2 + z2 (B.15)

b.3 natural coordinates, intrinsic coordinates or pathvariables

If the path and the velocity with which the particle P moves on the path aregiven, it is of benefit to use path coordinates (so called “natural” coordinates).The coordinate along the path is denoted by s = s (t) and the path velocity by

ej

er

ez

r

j

rj

r

z

ej

er

ez

r

j

rj

r

zrj2

2rj

Figure B.3: Visualisation of the single velocity and acceleration components

142 kinematics - appendix

yx

z

eteb

r

P

Path

s

M

en

0

r

Figure B.4: Natural coordinates / Path variables

v (t) = s (t). The accompanying coordinate system is moving with the particleP on the path. The time variant tangent unit vector et always points in thedirection of the current motion and is always tangential to the curve. Since thevelocity vector is always tangent to the path, the velocity vector can only beexpressed in terms of the tangent vector:

v (t) = v (t) et = set (B.16)

Approximating the path through a circle in the point P (the radius of thecircle ρ is variable, in general), the normal unit vector points to the center ofthis circle. The binormal ~eb is perpendicular to the osculating plane. It can beobtained using the cross product of the tangent unit vector and the normalunit vector:

eb = et × en . (B.17)

The acceleration can be obtained by differentiation of v (t) with respect totime. Note that the tangent unit vector is time variant and has to be taken inconsideration for the derivative of v (t). With the aid of the Frenet’s Formulaof the circle theory, the acceleration is:

a (t) = vet +v2

ρen (B.18)

with the components

at = v = s tangential acceleration, path acceleration (B.19)

an =v2

ρ=s2

ρnormal acceleration (B.20)

B.3 natural coordinates, intrinsic coordinates or path variables 143

You may recognize that for a linear motion with a circle of infinite large radiusthe equations derived earlier can be obtained.

CFUNDAMENTALS OF KINET ICS - APPENDIX

c.1 special cases for the calculation of the angularmomentum

In general, The angular momentum in a ∗-reference frame is given by eq. (3.91)

L0∗ = L0 − r0∗ × p−mr∗S × v0∗

We distinguish the following special cases

1. 0∗ is a fixed point:then the vlocity of this point isv0∗ = 0 which results:

L0∗ = L0 − r0∗ × p

2. 0∗ coincids with the point A which belongs to the moving rigid body, using EularequationvA = vS +ω× rSAand the velocityof the center ofgravity of the rigidbody is includedin the angularmomentumthrough the termvA.

and moves with it:

L0∗ = LA = L0 − rA × p−mrAS × vA

mit r0∗ = rA und p = mvS .

3. 0∗ coincids with the center of gravity of the rigid body A = S:then rAS = 0 and vA = vS which means:

L0∗ = LS = L0 − rS × p

As the angular momentum L0 has two terms, translation and rotation,such as

L0 = rS ×mvS + JSω

this will reduce the angular momentum at the center of gravity of therigid body S to just the rotational term

LS = JSω .

145

146 fundamentals of kinetics - appendix

4. 0∗ is ths instantaneous center of rotation A:at the instantaneous center of rotation, the velocity is given as vA = 0.and this point is not fixed in space, neither fixed to the body, but itsmotion is described by the path of the instantaneous center of rotation.

Using the translational and rotational terms of the angular momentumas in 3, and using vA = 0, we get:

LA = JAω

This final result is also valid if the point A is fixed point.

It is now important to keep in mind what are the terms that change in theEffects on the An-gular momentumtheorem

angular mometum theorem with the change of reference point.

Starting with the angular momentum theorm with reference point 0 (see eq. (3.84)):

L0 = M0

With the change of reference point to an arbitrary moving point 0* the angularmomentom is written according to eq. (3.105) as

L0∗ = L0 − r0∗ × p−mr∗S × v0∗

and the moment with regard to the new reference point is:

M0∗ = r∗ × F = (r− r0∗)× FM0∗ = M0 − r0∗ × F

which leads to the following angular momentum theorm applied at referencepoint 0:

ddt[L0∗ + r0∗ × p+mr∗S × v0∗

]= M0∗ + r0∗ × F (C.1)

Calculating the time derivative and noticing that p = F and

r0∗ × p = r0∗ × F ,

which will result

L0∗ + r0∗ × p+mr∗S × v0∗ +mr∗S × v0∗ = M0∗ . (C.2)

Moreover, we havevS = v0∗ + v∗S

andr∗S = v∗S = vS − v0∗ ,

C.1 special cases for the calculation of the angular momentum 147

so the thrid term on the left side becomes:

mr∗S × v0∗ = m (vS − v0∗)× v0∗

= mvS × v0∗ −mv0∗ × v0∗

= mvS × v0∗

= −v0∗ ×mvS= −v0∗ × p= −r0∗ × p

Finally, we have:L0∗ +mr∗S × v0∗ = M0∗ (C.3)

It is clear from this last relation, that a new term is introduced to the angularmomentum theorm when it is written in term of the moving point ’0∗’, andthe new term, the second term in the left side in eq. (C.3), should always beget in mind. At this point, different special cases could be observered

1. The reference point 0∗ is moving with constant velocity:as v∗0 = konst then v∗0 = 0 and

L0∗ = M0∗

2. The reference point 0∗ is the center of gravity: 0∗ = S:S could also perform accelerated motion. As r∗S = 0, then

LS = MS

3. The point 0∗ is fixed: 0∗ = A or it is the instantanuous center of rotation:for a fixed point or the instantanuous center of rotation we have: vA = 0which results

LA = MA

Normally the center of gravity or the inatatanuous center of rotation is usedin solving problems of classical mechanics, which seems to be a good choice asthe second term in the left side in eq. (C.3) will disappear.

As we have seen, the angular momentum has two terms, the translational andthe rotational. Choosing the center of gravity, or a fixed point ( i.e. instan-tanuous center of rotation) as reference point, this will reduce the angulatmomentum to the rotational part, and the angular momentum theorm basedon such choice takes the following form

LS =ddt(JSω

)= MS (C.4)

148 fundamentals of kinetics - appendix

similarly,LA =

ddt(JAω

)= MA (C.5)

The term ddt

(JSω

)does not simply equal to JSω, as the moment of inertiaNote

of a rotating rigid body in the a fixed reference frame can be time dependent.An exception to this is, for example, a rotating cylinder about its axis that iscoicides with one of the axis of the reference frame, in this situation, we couldwrite that

ddt(JSω

)= JSω

The dependency of the moment of inertia on the actual position (angle) is adeciding factor to the easy and direct application of the angular momentumtheorm. Because of this body-fixed reference frame is used, as the moment ofinertia in the body-fixed frame stays constant and the time derivative of themoment of inertia is cenceled!

DVIBRATIONS - APPENDIX

d.1 excitation with constant amplitude of force - com-plex approach

Let us first recall that we can represent a real harmonic functions by a complexexponential function using

eiΩt = cos Ωt+ i sin Ωt (D.1)

From this we can derive that

cos Ωt =eiΩt + e−iΩt

2 (D.2)

andsin Ωt =

eiΩt − e−iΩt

2i (D.3)

The harmonic force is:

F (t) = F cos Ωt =F

2(eiΩt + e−iΩt

)=F

2 eiΩt +

F

2 e−iΩt (D.4)

This means that we have to solve the equation of motion twice, for the eiΩtand the e−iΩt term. For the first step we make the approach:

x1 (t) = x1eiΩt (D.5)

x2 (t) = x2e−iΩt (D.6)

Putting both approaches into the equation of motion yields:

(−Ω2m+ iΩc+ k)x1eiΩt =

F

2 eiΩt (D.7)

(−Ω2m− iΩc+ k)x2eiΩt =

F

2 e−iΩt (D.8)

Dividing by k and introducing the frequency ratio η (eq. (4.102))[(1− η2) + 2Dηi

]x1 =

F

2k (D.9)[(1− η2)− 2Dηi

]x2 =

F

2k (D.10)

149

150 vibrations - appendix

The solution for x1 and x2 are:

x1 =(1− η2)− 2Dηi

(1− η2)2 + 4D2η2F

2k (D.11)

x2 =(1− η2) + 2Dηi(1− η2)2 + 4D2η2

F

2k (D.12)

As we can see, the solution of one part is the conjugate complex of the other:

x1 = x∗2 (D.13)

The solution for x (t) is combined from the two partial solutions, which we justhave found:

x (t) = x1eiΩt + x2e

−iΩt (D.14)

This can be resolved:

x (t) = x1 cos Ωt+ ix1 sin Ωt+ x2 cos Ωt+ ix2 sin Ωt (D.15)

and using eq. (D.13), we finally get:

x (t) = 2 Re x1 cos Ωt− 2 Im x1 sin Ωt (D.16)

The factor of 2 compensates the factor 12 associated with the force amplitude.

All the information can be extracted from x1 only so that only this part of thesolution has to be solved:

x (t) =(1− η2)

(1− η2)2 + 4D2η2F

kcos Ωt+

2Dη(1− η2)2 + 4D2η2

F

ksin Ωt (D.17)

which is the same result as eq. (4.91) with eq. (4.103) and eq. (4.104).Also the magnitude x = |x1| and phase ϕ can be obtained in the same wayand yield the previous results:

Magnitude: x =1√

(1− η2)2 + 4D2η2︸︷︷︸magnification factor V1

1kF = V1 (η,D)

1kF (D.18)

Phase: tanϕ = −Im x1Re x1

=2Dη

1− η2 (D.19)

D.2 excit. with constant amp. of force - alternative complex app. 151

d.2 excitation with constant amplitude of force - al-ternative complex approach

Instead of eq. (D.4), we can write

F (t) = F cos Ωt =F

2(eiΩt + e−iΩt

)= 2 Re

F

2 eiΩt

(D.20)

orF (t) = F cos Ωt = Re

F eiΩt

(D.21)

According to this approach, we formulate the steady state response as

x (t) = ReXeiΩt

(D.22)

The complex amplitude X is determined from the equation of motion, solving

Re(−Ω2m+ iΩc+ k

)XeiΩt

= Re

F eiΩt

(D.23)

The real parts are equal if the complex expression is equal:(−Ω2m+ iΩc+ k

)XeiΩt = F eiΩt (D.24)

Elimination of the time function yields:(−Ω2m+ iΩc+ k

)X = F (D.25)

The expression in brackets is also called the dynamic stiffness

kdyn (Ω) =(k−Ω2m+ iΩ

)(D.26)

Now we solve eq. (D.25) to get the complex amplitude:

X =F

(−Ω2m+ iΩc+ k)(D.27)

The expression

H (Ω) =1

(−Ω2m+ iΩc+ k)=X

F=

OutputInput (D.28)

is the complex Frequency Response Function (FRF). Introducing the dimen-sionless frequency η as before yields:

X =1

(1− η2 + i2Dη)F

k(D.29)


Because

x cos (Ωt− ϕ) = xReei(Ωt−ϕ)

= Re

xe−iϕeiΩt

= Re

XeiΩt

(D.30)

we take the magnitude x and phase lag ϕ of this complex result

X (Ω) = xe−iϕ (D.31)

which leads to the same result as before, see eq. (D.18) and eq. (D.19):

x =1√

(1− η2)2 + 4D2η2︸︷︷︸magnification factor V1

1kF = V1 (η,D)

1kF (D.32)

tanϕ = −Im x1Re x1

=2Dη

1− η2 (D.33)

d.3 fourier series - alternative real representation

We can write the Fourier series as a sum of cosine functions with amplitude ckand a phase shift ϕk:

x (t) = c0 +∞∑k=1

ck cos (kωt+ ϕk) (D.34)

ck =√ak2 + bk

2 and ϕk = arctan (− bkak

) (D.35)

d.4 fourier series - alternative complex representation

The real trigonometric functions can also be transformed into complex expo-nential expression:

x (t) =∞∑

k=−∞Xke

ikωt (D.36)

TheXk are the complex Fourier coefficients which can be determined by solvingthe integral:

Xk =1T

∫ T

0x (t) e−ikωtdt (D.37)

D.4 fourier series - alternative complex representation 153

or

Xk =1T

∫ T

0x (t) [cos kωt− i sin kωt] dt (D.38)

which clearly shows the relation to the real Fourier coefficients series given byeq. (4.142) and eq. (4.143):

Re Xk =ak2 ; Im Xk = −

bk2 (D.39)

The connection to the other real representation (chap. 4.7.1) is:

|Xk| = ck tanϕk =(

Im XkRe Xk

)(D.40)

The coefficients with negative index are the conjugate complex values of thecorresponding positive ones:

X−k = X∗k (D.41)


d.5 magnification functions

0 0,5 1 1,5 2 2,5 3 3,5 4 4,5 50

1

2

3

4

5

6

7

8

9

10

η ω= /Ω0

V1

D=0,05

D=0,1

D=0,2

D=0,3

D=0,5

D=0,7071

√(1- ) +42 2D

2 2η η

1V1=

Figure D.1: Magnification function V1

η ω= /Ω0

V2

0 0,5 1 1,5 2 2,5 3 3,5 4 4,5 50

1

2

3

4

5

6

7

8

9

10

D=0,05

D=0,1

D=0,2

D=0,3

D=0,5

D=0,7071

√(1- ) +42 2Dη η

2 2

√1+4D2 2η

V2=

√2Figure D.2: Magnification function V2

D.5 magnification functions 155

V3

0 0,5 1 1,5 2 3 3,5 4 4,5 50

1

2

3

4

5

6

7

8

9

10

D=0,05

D=0,1

D=0,2

D=0,3

D=0,5

D=0,7071

η ω= /Ω0

2,5

√(1- ) +42 2D

2 2η η

η2

V3=


V4

0 0,5 1 1,5 2 3 3,5 4 4,5 5

1

2

3

4

5

6

7

8

9

10

η ω= /Ω0

2,5

D=0,05

D=0,1

D=0,2

D=0,3

D=0,5

D=0,7071 D=0,05

D=0,7071

√(1- ) +42 2D

2 2η η

V4=η

2√1+4D2 2η

√2



0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50°

20°

40°

60°

80°

100°

120°

140°

160°

180°

φ

η ω= /Ω0

D=0,05D=0,1

D=0,2D=0,3

D=0,5D=0,7071

1- 2η

φ =2Dη

arctan

D=0

Figure D.5: Phase ϕ - for magnification function V1 and V3

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50°

20°

40°

60°

80°

100°

120°

140°

160°

180°

φ

η = Ω / ω0

D=0,05

D=0,1

D=0,2

D=0,3D=0,5

D=0,7071

21-ηφ =

2Dηarctan

D=0

Figure D.6: Phase ϕ - for magnification function V2 and V4

EL ITERATURE

Die hier vorliegende Zusammenstellung maschinendynamischer Themen um-fasst grundlegende Prinzipien und Arbeitsgebiete. Die dargestellten Inhaltelassen sich in vielen Lehrbüchern wiederfinden. Im besonderen sei jedoch aufdie folgenden Werke hingewiesen:

• Moon, F.C., Applied Dynamics, John Wiley & Sons, 1998.

• Ginsberg, J.H., Advanced Engineering Dynamics, 2nd edition, CambridgeUniv. Press, 1998.

• Weaver, W., Timoshenko, S.P., Young, D.H., Vibration Problems in En-gineering, John Wiley, 1990.

• Inman, D.J., Engineering Vibrations, Prentice Hall, 1994.

• Ginsberg, J.H., Mechanical and Structural Vibrations - Theory and Ap-plications, John Wiley, 2001.

157

FFORMULARY

159

160 formulary

sin(α± β) = sinα cos β ± cosα sin βcos(α± β) = cosα cos β ∓ sinα sin β

sin(2ϕ) = 2 sinϕ cosϕcos(2ϕ) = cos2(ϕ)− sin2(ϕ) = 1− 2 sin2(ϕ)

Table 2: Trigonometric functions

Configuration Equivalent stiffness(bending stiffness EI = const.)

l

l/2

F

wmax

F = keqwmax

kers = 48EIl3

l

F

wmax

F = keqwmax

kers = 12EIl3

l

F

wmax

F = keqwmax

kers = 3EIl3

Table 3: Equivalent stiffnesses for different loaded beams

formulary 161

(1± x)m = 1±mx+ m(m−1)2! x2 ± m(m−1)(m−2)

3! x3 + . . .

+ (±1)n m(m−1)...(m−n+1)n! xn + . . .

m > 0

√1− x = 1− 1

2x−18x

2 − . . . m = 12

Table 4: Reihenentwicklung: Binomische Reihe mit positiven Exponenten (|x| ≤ 1)

Springs connection Equivalent stiffness

k1

k2

Fkn

Parallelschaltungkeq =

n∑i=1

ki

n = 2:

keq = k1 + k2

F

Reihenschaltung

k1

k2

kn

1keq

=n∑i=1

1ki

n = 2:

1keq

= 1k1

+ 1k2

keq =k1k2k1+k2

Table 5: Equivalent stiffness of springs in parallel and serie connection

162 formulary

Rigid body Mass moment of inertia

x y

zl

b

hS

QuaderJxx =

112m

(b2 + h2

)Jyy =

112m

(h2 + l2

)Jzz =

112m

(l2 + b2

)

Stab (b,h<<l)

S

xy

zl Jyy = Jzz =112ml

2

x y

z

l

rZylinder

S

Jxx = Jyy =112m

(l2 + 3r2

)Jzz =

12mr

2

Scheibe ( )h r<<

x y

z

hr

S

Jxx = Jyy =14mr

2

Jzz =12mr

2

x y

z

l

RHohlzylinder r

S

Jxx = Jyy =14m

(R2 + r2 + h2

3

)Jzz =

12m

(R2 + r2

)

KreisringscheibeR r h R≈ , <<

x y

z

R

r

h

S

Jxx = Jyy =12mR

2

Jzz = mR2

Kugel

x y

z

r

SS

Jxx = Jyy = Jzz =25mr

2

Table 6: Mass moment of inertia (given at the center of gravity S)

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