Third Year Dynamics Lecture Notes

Michael Zaiser

Contents

1 Introduction 1

1.1 Prerequisites . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Course outline . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.3 Use of these notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.4 Basic concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 The theory of systems with one degree of freedom 5

2.1 Lumping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2 Free vibration of a SDF system with no damping . . . . . . . . . . . . . . . . . . 6

2.2.1 Equation of motion approach . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.2.2 Energy approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.3 Free vibration of a SDF system with damping . . . . . . . . . . . . . . . . . . . . 10

2.3.1 Viscous damping and solution of the damped motion equation . . . . . . . 10

2.3.2 Canonical form of the damped free vibration equation . . . . . . . . . . .14

2.3.3 Rotational vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.4 Periodically forced vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . .19

2.4.1 Steady-state response of a periodically forced system . . . . . . . . .. . . 19

2.4.2 Force transmission and vibration damping . . . . . . . . . . . . . . . . . . 22

i

2.4.3 Moving boundary condition and amplitude transmission . . . . . . . . . . 23

2.5 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.5.1 Accelerometer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.5.2 Rotating out of balance rotor . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.5.3 Out of balance in reciprocating internal combustion engines . . . . . . .. 29

2.6 Non-periodic forcing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .36

2.6.1 Shock damping and shock response spectrum . . . . . . . . . . . . . . .. 39

3 The theory of systems with many degrees of freedom 42

3.1 Free vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

3.2 Eigenvalue problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

3.3 Forced vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

3.4 Orthogonal modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

3.5 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

3.5.1 Shaft whirling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

3.5.2 Beating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

3.5.3 Anti-resonance and vibration absorbers . . . . . . . . . . . . . . . . . .. 58

4 Self-excited vibration 61

4.1 Positive Feedback . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .61

4.2 Stick-slip motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

ii

1 Introduction

1.1 Prerequisites

You have already studied some freely vibrating systems last year, and we will be building on thisknowledge. You will need to draw on the material you studied both in dynamics,mathematics andsolid mechanics. As the course starts, make sure you know how to draw free body diagrams (FBDs)and extract the equations of motion from these diagrams. You have studied the theory of secondorder ordinary differential equations with constant coefficients in second year maths. We will makea lot of use of the results, so revise them! I will be going over some of the material in the first fewlectures, but it will only be to refresh your memory and to define my own notations, so don’t rely onlearning it for the first time here. As a test, before you start the course you should be able to solve

x+5x+4x = exp(−3t) (1)

with initial conditionsx(0) = 0 , x(0) = 1 . (2)

Remember you will need to find the two complementary functions and the particularintegral andthen use the initial conditions to determine the constants in the general solution. Sound familiar?Could you solve the same equation if the right hand side was sin(3t)? You will also need to knowhow to use matrices to represent systems of equations and know the meaning and use of eigenvaluesand eigenvectors. To deal with the problems in tutorial and exam questions,you also should recallwhat you have learned about rotational motion, moments of inertia and beam flexure.

1.2 Course outline

The focus of the course is dynamic vibration. You can find more details and alecture by lecturebreakdown in the syllabus, available from the dynamics web page. We will study the theory ofvibration of single and multiple degree of freedom systems with and without forcing. With thistheoretical work, we will be able to look at a range of applications. These will include shaft whirling,vibration dampers, balancing in internal combustion engines and vibration measuring devices.

We will look first at single degree of freedom (SDF) systems in some detail because they exhibitmany of the features of more complicated systems while remaining relatively easyto analyze math-ematically. After reviewing free body diagrams and clarifying notation, we willconsider a freelyvibrating SDF system with and without damping. The equation of motion for the system followsfrom the free body diagram. It is a second order, linear, homogeneousordinary differential equationwith constant coefficients and it can be tackled by methods which will alreadybe familiar from pre-vious courses. We will look at the physical significance of the solution andsee how the parametersof the system determine its vibrational characteristics. Having understood the behaviour of freely

1

vibrating systems, we will introduce a forcing term and consider its effects.Some new mathemat-ical techniques will be introduced here. We will first look at systems forced by a periodic forceand then at the more difficult general case. At this point we will develop a general solution for allSDF systems, free or forced and damped or undamped, and discuss the different properties of eachsystem and the methods used to analyse them.

The second part of the module will address multiple degree of freedom (MDF) systems. We willsee that the problem is similar to an SDF system but with some added complications.We find thatthe mathematical analysis is similar in SDF and MDF systems if we replace the mass andstiffnessconstants with matrices. We will look at how to obtain solutions of the matrix equations and whatthese solutions tell us about the behaviour of the system. We will see that in many cases we canmake a coordinate transformation that will allow us to solve the MDF system as a series of SDFsystems. The equations for MDF systems are often very hard to solve. In practice they are usuallysolved numerically, and we will look at some of the methods that are used to do this.

Along with the mathematical theory, we will look at some applications to engineeringsystems. Wewill find in most instances that we must simplify the true situation but that the answers we get fromthe theory agree quite well with experiments. We will look at shaft whirling, ICengine balanc-ing, rotating out-of-balance systems, vibration absorbers, anti-resonance devices, accelerometers,torsional and beam systems. Numerical methods are becoming the dominant method of analysingvibrating systems and we will spend some time using MATLAB to explore how the methods workand what we need to be able to do to use them. It is a common misconception that numerical meth-ods are easy to use. We will see that it is indeed almost too easy to use numerical methods to get“answers”, but rather more difficult to use them properly and get valid answers.

At the end of the course we will look at what we have achieved and hopefully conclude that we cananalyse many interesting systems that are important in engineering. We will alsodiscuss what wehave not yet studied and see the limitations of our theories.

1.3 Use of these notes

I suggest that you will find the material easier to follow if you look at the relevant sections of thesenotes before the lectures. Do the exercises in the notes as we go along; they will help your under-standing. You need to be active in your study by trying out the techniques on problems yourself.Passively reading the notes will not get you very far.

It should really go without saying that these notes are in no way a replacement for the lectures, andnot only because they are incomplete. The notes complement the lectures andrelieve you of thetedious, error-prone task of copying chunks of algebra from the board.

2

1.4 Basic concepts

We will be building on the work from previous courses. You should already be familiar with thebasic ideas of freely-vibrating systems. In particular, you ought to knowthe meaning of period andfrequency of vibration, inertial forces, spring stiffness, moments of inertia, beam bending, comple-mentary functions and particular integrals. You should also be able to draw free body diagramscorrectly. We will revise free body diagrams before starting the course proper.

A free body diagram shows the real and inertial forces that act on each mass in a system. Once afree body diagram has been constructed, the equation of motion can oftenbe written down. Whenconstructing the free body diagram, it is important to bear in mind the direction ofeach force in thesystem.

x 1 x 2

k 1 m 1 k 2 m 2 k 3Figure 1: An illustrative two-degree of freedom system. The masses are free to move in the hori-zontal direction only. Whenx1 = 0 andx2 = 0 the system is in static equilibrium

The best way to see how this works is to consider an example. In Figure 1 thetwo objects withmassesm1 andm2 are free to move in the horizontal direction. However, they are not free tomovein the vertical direction, an example of a constraint. The masses can move independently, so twocoordinates are needed to specify their position. The system therefore has two degrees of freedom.

The three springs in Figure 1 have stiffness constantsk1,k2 and k3 and like all springs we willconsider they are linear. If a linear spring is extended by a distancex, it exerts a forcekx wherek is its stiffness. From this information, we can draw the free body diagram for the system. As areference configuration we always chose the static equilibrium configuration of the system. Figure 2shows the system after the masses have been moved by (arbitrary) distances ofx1 andx2 out of thisconfiguration. If the distance between the masses in the equilibrium configuration isl , the distancebetween them after they have moved isl −x1 +x2. The extension of the central spring is thereforex2−x1. The forces due to the springs can now be deduced. Figure 3 shows theforces acting on eachmass, including the inertial forces. Note that the final spring is compressedwhenx2 is positive andso the force must be directed to the left. We choose to show this by drawing thearrow to the rightand including a minus sign. Equivalently, we could have kept the force as positive and directed thearrow to the left.

3

x 1 x 2l

l - x 1 + x 2Figure 2: Distances when the masses have moved.

k 1 x 1 k 2 ( x 2 - x 1 ) - k 3 x 2

11 xm && 22 xm &&

Figure 3: The free-body diagram for the illustrative system.

The direction of the inertial forces is important, and can be determined by considering the effectof the inertia of the mass. Ifx1 is positive, as it is shown in Figure 2, then the massm1 has beenmoved to the right. If we accelerate a mass in the direction ofx1 to the right, it’s inertia acts tooppose this and so the inertial force is directed to the left. The same argumentapplies to the massm2. Hence, the direction of the inertial forces must always be taken oppositeto the direction of therespective coordinate. The equations of motion for the two mass system cannow be written downusing D’Alembert’s principle that the sum of the forces on each body must be zero in dynamicequilibrium. We obtain one equation for each mass, and they are given by

m1x1 +k1x1−k2(x2−x1) = 0 ,

m2x2 +k2(x2−x1)+k3x2 = 0 . (3)

Exercise: 1It is a useful exercise at this point to redraw the system of Figure 1 with a differentcoordinate system. Label the system with coordinatesx3 andx4 so thatx3 is justthe same asx1 but x4 runs in the opposite direction tox2. Draw the free bodydiagram and deduce the equations of motion. When you have done this, substitutex3 = x1 andx4 = −x2 and check you have the same equation of motion as derivedabove. Tip: You may find you need to multiply one of the equations by -1 to makeit the same as equation (3).

4

What can we conclude from this brief exercise? Minus signs can be a headache when we come tofigure out the directions of forces, but actually the mathematics takes care of all this for us! If weset up a coordinate system so that things work properly for positive displacements, everything willalso work for negative displacements. Therefore, when setting up a FBD, just specify a direction forthe displacement of a body (and stick to it!), and draw in the correctly directed spring force whichresults from this displacement. If in reality the body moves the other way, it doesn’t matter becausethe maths will sort out the appropriate sign on the force.

2 The theory of systems with one degree of freedom

Systems with one degree of freedom are very important and will be the starting point for our studyof vibration. We will find that there are many interesting engineering problemsthat can be modelledwell by single degree of freedom (SDF) systems. Other more complicated systems will need extradegrees of freedom, but we will discover that many of the ideas and techniques we develop forSDF systems will be easily extendible to many degree of freedom (MDF) systems, although themathematical effort needed to solve them will be greater.

2.1 Lumping

Before we start to analyse a SDF system, let us consider for a moment whatwe are really thinkingabout when we draw our initial diagram for a system. In reality, we can never cover all the compli-cated aspects of a real system. For instance, consider a mass supportedby a spring in the absenceof gravity. This appears to be a very simple dynamic system to study, and we might draw a diagramlike that shown in Figure 4.

Remember there is no gravity so the equilibrium position of the mass is at the unextended lengthof the string. If we look at the details, the dynamic behavior of this system may contain manycomplexities: The mass of the object supported by the spring is in reality distributed over the objectrather than concentrated at a point. The stiffness of the spring is similarly spatially extended. Thespring has a finite mass and its response is unlikely to be exactly linear. However, if we disregardthese complexities and treat the system as a point mass supported by a massless spring, the behaviorof our model system and the true physical system turn out to be very similar.We have captured theessential features of the system in our simple model, without including all the complex details thathave only small effects.

The process whereby we consider the mass to be concentrated at a pointrather than distributedin space is calledlumpingand we refer to the model as alumped parametermodel. Lumping isa very powerful idea because it greatly simplifies the mathematical treatment ofthe system whilemanaging to retain the essentials. Masses and springs can both be lumped, i.e.replaced by a single

5

mx k

Figure 4: The simplest SDF system.

element. We will see later that another major component of our modelled systems,the damping,can also be lumped. In all cases, the essential point to bear in mind is that we are modelingthe realphysical situation. We know that the mass is not concentrated at a point in reality, but that we canmodel it as being so and obtain a good approximation of the real situation.

2.2 Free vibration of a SDF system with no damping

We now start our mathematical analysis of the system shown in Figure 4. We willstart off in thenormal way by drawing the free body diagram, deriving the equation of motion and finding solutionsto this equation. We will also look at the energy of the system and see how this too can be used toderive an equation of motion. Of course, the two methods give the same results.

2.2.1 Equation of motion approach

This is the way we will tackle most problems in the course. First we draw the free body diagram asshown in Figure 5.

6

mx k x

xm &&

Figure 5: The free body diagram for the system of Figure 4.

Now we can write down the equation of motion

mx+kx= 0 . (4)

Exercise: 2Consider the same system but with the effects of gravity included. Draw the freebody diagram and write down the equation of motion. Show that by appropriatelyshifting thex coordinate one obtains the same equation of motion (4) as in thegravity-free case. (Chose asx = 0 the static equilibrium configuration of thesystem with gravity)

Before we try to solve this equation, we will classify it. The equation is secondorder in time becausethe inertial term involves a second order derivative. The equation is linear, since the coefficients donot depend onx, and it has constant coefficients becausem andk do not depend ont. It is alsohomogeneous because there is no term independent ofx. In full then, the equation is a second orderhomogeneous ordinary differential equation with constant coefficients.

To solve this equation, we need to find two complementary functions. We will notneed a particularintegral because the equation is homogeneous. The equation is second order, so we expect to obtaintwo complementary functions. The sum of these two functions is the general solution. We can seeby inspection that either of

x = sin(ωt) ,

7

x = cos(ωt) , (5)

is a solution of the equation. The constantω must be chosen so that them andk constants areaccounted for correctly. The value needed to do this isω =

√

k/m.

Exercise: 3Verify the value given forω. To do this, consider sin(ωt) first. Compute thesecond derivative with respect to time and substitute into equation (4). Fromtheresulting equation you can calculate the relation betweenω andk andm. Repeatthe calculation for the cosine solution.

From the complementary functions, we can now construct the general solution. It is

x = Asin(ωt)+Bcos(ωt) . (6)

It is easier to see what is going on if we rewrite the general solution forx in a different form. Wecan write the solution as

x = Csin(ωt +φ) . (7)

where the constants are given byC =√

A2 +B2 and tanφ = B/A.

Exercise: 4Show this is the case. Hint: start withx = Csin(ωt +φ) and expand the sine usingthe formula sin(X +Y) = sin(X)cos(Y)+sin(Y)cos(X).

From this expression we can see more clearly what is happening. The constantC is the amplitudeof the vibration andφ is the phase. These two constants are undetermined in the general solutionand must be calculated from the initial conditions.

Let’s now put some numbers into the formula, specify initial conditions, and see what the solutionlooks like. Supposem= 1 kg andk = 25 N/m. If we set the mass off at time zero at its equilibriumposition with a velocity of 1 cm/s, what does the subsequent motion look like? We calculate firstωand findω = 5 rad/s. The motion of the mass is therefore given by

x = Csin(5t/s+φ) . (8)

From the initial conditions, we know thatx(0) = 0 which impliesφ = 0. The velocity at any timetcan be found by differentiating the above equation to get

v(t) = x(t) = [5/s]Ccos(5t/s+φ) (9)

and we now knowφ = 0 so at timet = 0 the velocity is given by

v(0) = [5/s]C . (10)

The initial conditions tell us thatv(0) = 10−2 m/s soC = 2×10−3 m. The final solution then is

x(t) = [2×10−3m] sin(5t/s) . (11)

This solution is shown in Figure 6 for the first three seconds of motion.

8

0 1 2 3

-2

0

2

disp

lace

men

t [m

m]

time[s]

Figure 6: The displacement of the mass in the system of Figure 4 as a functionof time. Theparameters and initial conditions are given in the text.

2.2.2 Energy approach

We can tackle the same problem discussed in Section 2.2.1 using an alternativeapproach based onthe energy of the system. We will find that we can actually derive the equationof motion once weknow the energy of the system, so all the results from the previous section can be re-obtained. Theadvantage is that we don’t need to consider the forces in the problem, which in some cases canbe difficult. On the other hand, the energy approach as discussed hereworks only if there is nodamping.

Consider the energy stored in the spring. The force exerted by a springis F = −kx wherex is theextension from the equilibrium length andk is the stiffness. The work done in extending the springby a lengthx is

W =∫

Fds= −kx2

2. (12)

The energy stored in the spring, its potential energy, is the negative of thework done on the springand we will call thisU(x). We have

U(x) =kx2

2. (13)

9

The kinetic energy of the mass we will callT. It is given by

T =mx2

2, (14)

so the total energy of the systemE = U +T is

E =mx2 +kx2

2. (15)

The principle of conservation of energy tells us thatE must be the same at all times, so dE/dt = 0.We can use this information now to deduce the equation of motion. Differentiatingthe expressionfor E with respect to time gives us

E = mxx+kxx= 0 . (16)

We can factor this equation intox(mx+kx) = 0. (17)

so either ˙x = 0, which tells us that the system may be at rest, ormx+kx= 0 which is the equationof motion derived above. The situation in which the system never moves is notof great interest tous in a course on dynamics. The other result is very useful, however, because it shows us how toobtain an equation of motion for a system once we know the total energy of thesystem. Note thatwe have assumed that the energy in the system is conserved. In some cases, friction and other forcesdissipate energy to the surrounding. In these cases, the energy of the system decreases in the courseof time and so the approach we have used here cannot be applied.

2.3 Free vibration of a SDF system with damping

In reality, many systems that we meet have damping. The damping can be caused by friction insliding parts or by viscous effects in a fluid, for example. The mathematical specification of adamping force can be quite tricky because of the different physical mechanisms that can causethe damping. In all cases, however, the effects are similar: energy in the system is removed anddissipated to the surroundings, often in the form of heat or noise. We will be primarily concernedwith viscous damping for which the mathematical specification is relatively simple. Just as with themasses and springs, we will use lumped dampers to model the effects of damping in a system. Thismeans that although in reality the damping effects may be distributed throughoutthe system, in ourmodel the damping will occur at a well defined point.

2.3.1 Viscous damping and solution of the damped motion equation

The idea of viscous damping is that a system moving with a velocityv is slowed down by a forceproportional tov. Fast moving objects therefore encounter large forces, while slowly moving objects

10

encounter only small forces. The direction of the force is always to oppose the velocity so that thesystem is always slowed by the damping, rather than speeded up. Mathematically, we can say thatthe viscous damping forceF is given byF =−cvwherec is a constant which determines the amountof damping present in the system. Lets look at a simple system acting under the effects of viscousdamping.

x k m c

Figure 7: A spring-mass-damper system.

Figure 7 shows how we represent symbolically a spring-mass system retarded by a viscous damper(’dashpot’). Recalling that the force due to the viscous damper is−cv= −cx we can draw the freebody diagram for this system as shown in Figure 8.

xm &&

xc &

x m

k xFigure 8: The free body diagram for the system of Figure 7.

11

Notice the direction of the forces. The positivex direction is from left to right in the figure, so ifthe mass is accelerating from left to right both ˙x and x are directed left to right also. As we havealready discussed, the inertial force acts to oppose the acceleration we attempt to give the mass, soacts right to left. Also, the damping force acts to oppose the motion, and since the mass moves fromleft to right for positive ˙x, the damping force must act right to left. The equation of motion can nowbe written down for this system. It is

mx+cx+kx= 0 . (18)

This is a linear, homogeneous second order equation. The solutions of thisequation are of the formexp(λt). We can see why this is the case by thinking about the properties needed for a function tobe a solution of equation (18). The coefficients in the equation are constants, so if we can choosea function that under differentiation is equal to itself multiplied by a constant, then this is a goodcandidate for a solution to (18). The function that satisfies this need is of course the exponentialfunction. The constantλ must now be chosen so that equation (18) is satisfied. To this end, wesubstitutex = exp(λt), x = λexp(λt) andx = λ2exp(λt) into (18). We get

λ2 +cm

λ+km

= 0 . (19)

This quadratic equation (characteristic equation) has roots

λ± =−c±

√c2−4km

2m, (20)

and the corresponding solutions are given by

x+(t) = exp(λ+t) ,

x−(t) = exp(λ−t) .

These are two complementary functions which solve equation (18). Since either of these functionsis a solution of (18), and the equation is linear, any combinationAx+ +Bx− is also a solution (A andB are arbitrary constants). The most general solution is therefore

x(t) = Aexp(λ+t)+Bexp(λ−t) . (21)

It is worth recognizing that the rootsλ+ andλ− and therefore also the complementary functionsx+(t) andx−(t) are not always real. Of course, the final solutionx(t) must be real, and this can beensured by taking, in a last step of the calculation, the real part of equation (21),

x(t) = Re[Aexp(λ+t)+Bexp(λ−t)] . (22)

The behaviour of this solution depends on whether the rootsλ± of the characteristic equation arereal or complex.

12

If c2 > 4km both roots given by equation (20) and, hence, also the complementary functions arereal. Inserting the values forλ+ andλ− into equation (21) gives

x(t) = Aexp

([

− c2m

+

√

( c2m

)2− k

m

]

t

)

+Bexp

([

− c2m

−√

( c2m

)2− k

m

]

t

)

. (23)

This expression can be re-arranged andx(t) can written in terms of hyperbolic functions:

x(t) = exp(

− c2m

t)

[

Ccosh

(√

( c2m

)2− k

mt

)

+Dsinh

(√

( c2m

)2− k

mt

)]

(24)

whereC = A+B andD = A−B.

If on the other handc2 < 4km then the roots are complex and the values of exp(λ+t) and exp(λ−t)will be complex quantities. The roots will be complex conjugates of each other,so the real part ofthe roots is the same while the imaginary part is the same magnitude for each root, but of oppositesign. We can therefore write

λ+ = λR + iλI ,

λ− = λR− iλI

where the real and imaginary partsλR andλI are

λR = − c2m

, λI =

√

km−( c

2m

)2.

As for the real roots case, the general solution is given by equation (21), but now the exponentialsand the constantsA,B must be envisaged as complex numbers,A = AR + iAI andB = BR + iBI .Inserting in Eq. (21) gives

x(t) = Aexp(λ+t)+Bexp(λ−t)

= exp(λRt) [(AR + iAI)exp(iλIt)+(BR + iBI)exp(−iλIt)]

= exp(λRt) [(AR +BR)cos(λIt)− (AI −BI)sin(λIt)

+i((AI +BI)cos(λIt)+(AR−BR)sin(λI t))]

= exp(− c2m

t)

[

Ecos

(√

km−( c

2m

)2t

)

+F sin

(√

km−( c

2m

)2t

)]

(25)

where in the last line I have taken the real part and used new constantsE = AR + BR and F =−AI +BI to replace the old ones. Note the similarity to equation (24).

The form of the solutions (23) and (25) is very interesting. Equation (23)is the sum of two expo-nential decays. This means that the motion of a system for which this equation isapplicable, i.e.whenc2 > 4km, is just a decay towards zero. No oscillations can be seen, and we referto the system

13

Figure 9: An example of the solution for anover-damped system. The solution exhibits nooscillations.

Figure 10: An example of the solution for anunder-damped system. The envelope of the os-cillations is a decaying exponential.

as being over-damped. On the other hand, whenc2 < 4km, equation (25) shows that the motionconsists of an oscillatory part given by the sinusoidal functions in the brackets, multiplied by an ex-ponential decay. The system therefore oscillates, but the amplitude of the oscillations decays. Thesystem is referred to as being under-damped. Figures 9 and 10 show plots of what these solutionslook like.

The two different types of behaviour, over-damped and under-damped oscillation, can be identifiedusing the ratioc2/4km. Forc2/4km> 1 the system is over-damped and forc2/4km< 1 the systemis under-damped. Forc2/4km= 1 the system is said to be critically damped. We won’t look at thiscase because in practice the ratio is never precisely equal to one.

2.3.2 Canonical form of the damped free vibration equation

We have seen how to solve the damped free vibration equation for a mass-spring damper system.We considered the equation

mx+cx+kx= 0 . (26)

When we are dealing with other vibrating systems, the coefficients in this differential equation maychange, and even their physical meaning may be different. We will see an example of this in thenext section. However, the basic structure of the equation of motion and themethod of solutionare always the same. Rather than finding the solution separately for each case, we write equationsof this type in a general form. This is usually referred to as canonical form. We can then use thesolution to the canonical equation straight away, rather than having to redoall the algebra. To thisend, we use two fundamental parameters. The first parameter is the frequency of vibration of thesystem without damping. We also refer to this as thenatural frequencyω0 of the system. For the

14

mass-spring system described by equation (26), we find that the naturalfrequency is

ω0 =

√

km

. (27)

To characterize the damping, we introduce thedamping ratioδ which determines whether the sys-tem is under-damped or over-damped. For our mass-spring-damper system the damping ratio isdefined by

δ2 =c2

4km. (28)

As we have seen in the preceding section,δ > 1 gives over-damping, andδ < 1 gives under-damping.We can rearrange equation (28) to give a useful alternative expression for the damping ratio:

δ =c

2√

km=

c2mω0

. (29)

We can now write equation (26) in terms ofδ andω0 instead ofk, m andc. Dividing through bymwe have

x+cm

x+km

= 0 . (30)

and from the relations forω0 andδ

km

= ω20 ,

cm

= 2δω0 , (31)

we find that this can be written as

x+2ω0δx+ω20x = 0 . (32)

By writing our equations in thiscanonical form, we can see at once the value of the dampingparameter and the undamped natural frequency. The true frequency of vibration is the coefficient oft in the sine and cosine terms in equation (25). It is called the damped frequency ωd and is given by

ωd =

√

km−( c

2m

)2(33)

We can express this in terms ofω0 andδ:

ωd = ω0

√

1−δ2 . (34)

This gives us the result that if damping is very small thenωd ≈ ω0. We can also write the solution,equation (25), using these new parameters:

x(t) = exp(−ω0δt)[Ecos(ωdt)+F sin(ωdt)] . (35)

Just as we did with the undamped system, we can rewrite the sum of the sine andcosine term as asine term with a phase shift. We get an equivalent formula to the above:

x(t) = H exp(−ω0δt)sin(ωdt +φ) . (36)

15

The constantsH andφ can be determined from the initial conditions.

Another parameter which is often used to characterize damped vibration is thelogarithmic decre-mentη of the oscillation. The logarithmic decrement is defined as the natural logarithmof the ratioof any two successive maxima of the oscillation. It is related to the damping ratio by

η =2πδ√1−δ2

(37)

Exercise: 5Use the solution (36) to derive this expression from the ratio between two succes-sive maxima of the oscillating solution.

Despite all the mathematics above, we can see that the effect of damping on thevibrating systemis actually quite straightforward. If the damping is larger that a critical amount,given byδ = 1,the system is over-damped and simply relaxes to its equilibrium position without any oscillation.The more interesting situation from our point of view is when the damping is less than the criticaldamping. In this case, the addition of the damping alters the frequency of the oscillations, andmakes their amplitude decay gradually to zero.

To use the canonical form to best advantage, we first write down the equation of motion for a systemusing the free body diagram. We can then identify the parametersω0 andδ, and from these deducewhether the system is under or over-damped, the frequency of vibrationand the rate of decay of theoscillations (if present).

2.3.3 Rotational vibration

Until now we have considered systems where masses move along a given fixed direction, i.e. themotion is translational. However, in many important cases vibration is related to therotation ofparts. In this case our procedure of setting up the equations of motion must be modified. Figure11 shows a system which may exhibit rotational vibrations: A square block of massm can rotatearound the axis A and is attached to the walls by a spring of constantk and a damper with dampingconstantc. The spring constant is such that the block is at rest in the position shown.

Let’s see what happens when we rotate the block to the right by a small angleφ. The top right cornerof the block (where the damper is attached) moves to the right byl sinφ and the bottom right cornerwhere the spring is attached moves downward by the same amount. This leads toa compression ofthe spring and, if the rotation occurs at finite speed, to a viscous damping force from the damper.To work out the spring force and the damping force, we uselinearizedrelations by noting that, forsmall anglesφ, sinφ ≈ φ. Hence the spring and the damper are compressed byx = lφ. The velocityof compression of the damper is obtained by taking the time derivative, ˙x = l φ.

The corresponding forces are drawn in Figure 12. Now we have to keep in mind that the block

16

c k

m

l

l

f

A

Figure 11: A system undergoing rotational vibration.

rotatesaround the axis A, so we have to consider the sum of all moments with respectto this axis.The spring and damper forces areklφ andclφ, and their moments with respect to A arekl2φ andcl2φ. In addition we have to consider the effect of inertia. This leads to a momentIAα whereIA isthe moment of inertia of the block with respect to the axis A andα = φ is the angular acceleration.

Exercise: 6Calculate the moment of inertia of the block around the axis A. Assume that theblock is homogeneous. (Result:IA = 2ml2/3)

The sum of the moments around A must be zero. Hence we end up with the equation of motion

IAφ+cl2φ+kl2φ = 0 . (38)

or, after insertingIA = 2ml2/3 and dividing by 2l2/3

mφ+(3c/2)φ+(3k/2)φ = 0 . (39)

This equation is very similar to the equation of motion for the translational motion of amass-spring-

17

A

f

mf &c l

fk l

f

fff

&&

&

Jk lc l

2

2

Figure 12: (Left) forces and (right) moments acting in the system of Fig. 11

damper system, equation (18), and can be written in the same canonical form.To this end, we divideby the pre-factor of the term with the highest derivative to obtain

φ+3c2m

φ+3k2m

φ = 0 . (40)

We identify this with the general canonical form as given by equation (32)- here we have the angleφ instead ofx as the dependent variable, but everything else remains the same:

φ+2ω0δφ+ω20φ = 0 . (41)

By comparing coefficients, we find

ω20 =

3k2m

, 2ω0δ =3c2m

, (42)

and, hence, the natural frequency and damping ratio for this system aregiven by

ω0 =

√

3k2m

, δ =

√

3c2

8km. (43)

We can now use the general solution of the canonical vibration equation byinserting these param-eters into equation (35) or (36) and determining the remaining unknown parameters from initialconditions.

18

xc m f ( t )k

Figure 13: An example of a simple system subjected to forcing

2.4 Periodically forced vibration

We now consider the vibration of a system which involves an external force. We first considerperiodic forcing, where we will make use of the complex exponential method once again. You mayhave found it possible to work through the problems so far without using complex exponentials. Youwill struggle to do this for forced vibration. Please make sure you are familiarwith the workings ofthe complex exponential method, and make sure you can do the questions I’ve set on it. Once youare happy with the method, you’ll probably find that the material that follows isnot too complicated.

2.4.1 Steady-state response of a periodically forced system

Consider the system shown in Figure 13. We assume that the forcingf (t) is a periodic function, sof (t) = f0cos(Ωt). The equation of motion for the system shown in Figure 13 is

mx+cx+kx= f (t) = f0cos(Ωt) . (44)

Before we solve this equation we consider the simplest case where the applied force is a constant(Ω = 0). A constant force leads to a constant elongation of the spring byx0 = f0/k. We callx0 thestatic responseof the system.

Now we put the equation into canonical form. We first divide bym to get

x+2ω0δx+ω20x =

f0m

cos(Ωt) . (45)

The right-hand side of this equation can be written in terms of the static response and the naturalfrequency:

x+2ω0δx+ω20x = ω2

0x0cos(Ωt) . (46)

19

This is the form which we will always use in the following. Looking at the problem from a math-ematical point of view, I can see that the equation of motion is an inhomogeneous equation and sothe solution will be given by the sum of two complementary functions and a particular integral. Thecomplementary functions are the solutions of the homogeneous problem and have been discussed inthe previous section, they will both be damped oscillations, so after a reasonable time they will havedecayed to small values. At this stage, I’m interested in how the system responds after quite sometime so that these initial effects will have died away. That’s what we call thesteady-state responseof the system. I will therefore ignore the complementary functions in what follows.

Now let us take a closer look at equation (46). The left-hand side (LHS) basically involvesx mul-tiplied by various constants and differentiated. The solutions to this sort of equation are things likex∼ exp(λt), whereλ may be complex. The right-hand side (RHS) is a cosine, but I can write thatalso in complex exponential form

x+2ω0δx+ω20x = ω2

0x0exp(iΩt) (47)

and regardx as a complex variable. I can recover the original equation of motion by looking at justthe real part of this equation. Since the equation is linear, I can say that ifx = xR + ixI is a solutionthenxR is a solution to the real part of the equation andxI is a solution to the imaginary part. I cantherefore go ahead and solve for the complex variablex and take the real part ofx at the end of thecalculation. This real part will be a solution to the original equation of motion.

I expect the steady-state response to be

x(t) = X exp(iΩt) , (48)

wherex andX are complex. Substituting this expression gives

X =x0ω2

0

−Ω2 +2iΩω0δ+ω20

. (49)

To bring this complex amplitude into a more useful form, I use that any complex numberA+ iB canalso be written in the form of a complex exponential:

A+ iB = r exp(iφ) (50)

wherer2 = A2 +B2 and tanφ = B/A. When we writeX in equation (49) in this form, we get

X =x0

√

(

1−Ω2/ω20

)2+4δ2Ω2/ω2

0

exp(iφ) , (51)

where

φ = −arctan

(

2δΩ/ω0

1−Ω2/ω20

)

. (52)

20

We now insert insertX into the solution (48):

x(t) =x0

√

(

1−Ω2/ω20

)2+4δ2Ω2/ω2

0

exp(i[Ωt +φ]) , (53)

As I explained above, this is the solution to the complex form of the equation of motion. Thephysical solution is simply the real part of it. I use that

cos(Ωt) = Re[exp(iΩt)] . (54)

and findxR =

x0√

[

1−Ω2/ω20

]2+4δ2Ω2/ω2

0

cos(Ωt +φ) . (55)

What does this mean? There are several interesting points to extract fromthis solution. Firstly, it isbasically another sinusoidal function of the form

x = Acos(Ωt +φ) , (56)

where the amplitude of the vibration is

A =x0

√

(

1−Ω2/ω20

)2+4δ2Ω2/ω2

0

(57)

and the frequency is the same as that of the forcing. However, there is a phase shift between theforcing and the response. The system therefore responds to the forcing by vibrating at the samefrequency but with a delay,

φ = −arctan

(

2δΩ/ω0

1−Ω2/ω20

)

. (58)

The phase lagφ depends on the frequency of forcing. For smallΩ, φ is small. ForΩ = ω0,φ = −π/2, and forΩ → ∞, φ →−π.

Exercise: 8What does a phase shift of−π radians mean physically?

Exercise: 9Write down the values ofφ for δ = 0.1 andΩ = 0,ω0 + ε,ω0− ε,∞ whereε is avery small number. Verify that the values of the phase shift for smallΩ areφ ≈ 0,for Ω = ω0, φ = −π/2, and forΩ → ∞, φ →−π.

When you use your pocket calculator to get the phase shiftφ, you must be careful when taking theinverse tangent of a negative quantity. The inverse tangent is only defined up to a factor ofnπ wheren is an integer number, and you may have to subtractπ to get the phase shift right.

21

We now look how the amplitude behaves as a function ofΩ. Take the expression forA,

A =x0

√

(

1−Ω2/ω20

)2+4δ2Ω2/ω2

0

(59)

and consider the case when the damping ratioδ is small. Now, the squared terms inside the squareroot are always positive, so since they appear in the denominator, the amplitude is largest when theyare small. The first term is zero whenΩ = ω0. So when the damping is small and the forcing isthe same frequency as the frequency of free vibration, the amplitude is a maximum. This is calledresonance. For small values of the damping ratioδ we can often use the approximate expression

A =x0

|1−Ω2/ω20|

. (60)

This expression shows that the resonance peak is located at approximately ω0 (this follows fromthe small damping ratioδ assumed in equation (60)). The amplitude of the dynamic response atresonance for smallδ is 1/(2δ).

Exercise: 10Calculate the frequency and amplitude of resonance (the frequency andamplitudewhere the response has its maximum) for the case whereδ is not small.

Most of what we’ve been dealing with here can be neatly expressed in twographs; one for the am-plitude and one for the phase as functions of the frequency of forcing.The graph for the amplitudeis probably the most useful form of the frequency-response graph.It tells us a great deal about thedynamics of a system, all on a single plot. If we plot on(x,y) axes, then we let

y = A/x0 , x = Ω/ω0 , (61)

and so from equation (59) we have

y =1

√

(1−x2)2 +4δ2x2(62)

which determines the shape of the frequency-response graph. Because at resonance the response isusually very large, you will often find these graphs plotted in logarithmic coordinates (i.e. plot lnyagainst lnx).

Exercise: 11Use MATLAB or Mathcad to deduce the shape of these graphs for a number ofvalues ofδ and add sketches to your notes. Deduce the height and width of theresonance peak and label it.

2.4.2 Force transmission and vibration damping

We now consider the following question: Given the amplitudef0 of a periodic forcing acting on themassm in Figure 13, what is the amplitude of the force transmitted to the support?

22

Adding the spring and damper forces, we see that the transmitted force is given by

fs(t) = kx(t)+cx(t) (63)

Inserting the solution given by equation (56), we find

fs(t) = kAcos(Ωt +φ)−cΩAsin(Ωt +φ) (64)

This is a periodic function, and at the moment we are interested only in the amplitude. If we shiftthe time axis according tot ′ = t −φ/Ω we find that

fs(t′) = fs,0sin(Ωt ′ +φ′) (65)

where tanφ′ = −k/(cΩ) and fs,0 =√

k2A2 +c2Ω2A2. Hence the amplitude of the transmitted forceis

fs,0 = A√

k2 +Ω2c2 =x0√

k2 +Ω2c2√

(

1−Ω2/ω20

)2+4δ2Ω2/ω2

0

(66)

and using the relationsx0 = f0/k, c2 = 4kmδ2 andω20 = k/m this can be written as

fs,0 = f0

√

√

√

√

1+4δ2Ω2/ω20

(

1−Ω2/ω20

)2+4δ2Ω2/ω2

0

= f0×TR (67)

The ratio between the force acting on the mass and the force transmitted to the surroundings iscalled thetransmissibility ratio

TR =

√

√

√

√

1+4δ2Ω2/ω20

(

1−Ω2/ω20

)2+4δ2Ω2/ω2

0

. (68)

Exercise: 12Sketch the curves of TR plotted againstx = Ω/ω0 for x = 0. . .5 in intervals of 0.5.Use values ofδ = 0.1,0.2,0.5 and compare the different dampings. Plot the samegraph on logarithmic scales.

The form of the transmissibility ratio shows that at the resonant frequency, large values of dampingare good because they reduce force transmission. However, at high frequencies the opposite is thecase and the most efficient way to reduce force transmission is to makeω0 as small as possible (softsprings, large masses). In practical circumstances it is vital to be aware of this compromise.

2.4.3 Moving boundary condition and amplitude transmission

Until now we have considered forced vibration where an external force is acting directly on a vi-brating mass. A different type of excitation is shown in Figure 14.

23

x c m k y ( t )Figure 14: A system with moving boundary excitation

We assume that the support of the system moves periodically according toy(t) = y0cos(Ωt) and askfor the amplitude of the steady-state response of the system.

The equation of motion ismx+c(x− y)+k(x−y) = 0. (69)

We collect the terms involvingzyon the right-hand side and bring the equation into canonical formby dividing throughm

x+2ω0δx+ω20x = −2ω0δΩy0sin(Ωt)+ω2

0y0cos(Ωt). (70)

We now bring also the right-hand side into canonical form. To this end, we write

−2ω0δΩy0sin(Ωt)+ω20y0cos(Ωt)

= ω20y0[cos(Ωt)− (2δΩ/ω0)sin(Ωt)]

= ω20[y0

√

1+4δ2Ω2/ω20]sin(Ωt +ψ) . (71)

We finally shift the time axis to get rid of the phaseψ and find that the equation of motion assumesthe canonical form

x+2ω0δx+ω20x = ω2

0x0cos(Ωt ′) (72)

with x0 = y0

√

1+4δ2Ω2/ω20 . We can now use the solution (56),x(t) = Acos(Ωt + φ) with the

amplitudeA given by (59). As a result we find that the ratio of the vibration amplitudes ofthe massand of the support is given by

A/y0 = TR (73)

with the same transmissibility TR as in the previous section. Hence, equation (68)gives us bothforce and amplitude transmission, and what has been said about the influence of the parametersω0

andδ on vibration isolation also applies to the present case.

24

2.5 Applications

2.5.1 Accelerometer

As an application of the theory of forced vibration with moving boundary excitation, we consider adevice known as an accelerometer. As the name suggests, it is used to measure the acceleration ofa vibrating surface. Essentially, the device consists simply of a “seismic” mass, basically a lump ofmetal, that is attached by a spring damper system to the surface whose vibration is to be measured.A sensor is able to measure the distance between the seismic mass and the surface. Our task is torelate the extension of the spring, which we can measure, to the accelerationof the surface. Figure15 shows a diagram of the situation.

x c m k y ( t )

Figure 15: A schematic diagram of an accelerometer.x measures distance from a fixed point inspace to the mass,y measures distance from a fixed point in space to the vibrating surface.

The equation of motion ismx+c(x− y)+k(x−y) = 0 . (74)

We now introduce the new variablez= x−y. We can differentiate to find

z= x− y , (75)

z= x− y , (76)

and then eliminatex from the equation of motion in favour ofz. We get

mz+cz+kz= −my , (77)

which looks quite familiar. Let us assume that the surface is vibrating with frequencyΩ so that

y(t) = Ycos(Ωt) (78)

25

and soy = −Ω2Ycos(Ωt) , (79)

which we can use in our equation of motion to give

mz+cz+kz= mΩ2Ycos(Ωt) , (80)

which is in the form which we are used to solving, with the amplitude of the forcinggiven bymΩ2Y.Dividing through bymwe have

z+2ω0δz+ω20z= Ω2Ycos(Ωt) , (81)

We know how to solve this sort of equation. The canonical form is

z+2ω0δz+ω20z= ω2z0cos(Ωt) , (82)

By comparison, we find that the static responsez0 is

z0 = YΩ2/ω20 (83)

and the amplitude of the dynamic response is

Z =z0

√

(1− (Ω/ω0)2)2 +4δ2(Ω/ω0)2=

Y(Ω/ω0)2

√

(1− (Ω/ω0)2)2 +4δ2(Ω/ω0)2. (84)

Hence, the ratio of the amplitudes of vibration of the accelerometer and the surface is

ZY

=(Ω/ω0)

2√

(1− (Ω/ω0)2)2 +4δ2(Ω/ω0)2. (85)

If we make the free vibration frequencyω0 very high by a suitable choice ofk andm, thenΩ/ω0

will be small andZY

=Ω2

ω20

(86)

to a good approximation. Therefore we can write

Ω2Y = ω20Z . (87)

Now, the surface is moving sinusoidally according to

y = Ycos(Ωt) (88)

and so the acceleration of the surface is

y = −Ω2Ycos(Ωt) (89)

and the amplitude of this acceleration isA = Ω2Y (90)

26

Therefore,A = ω2

0Z . (91)

Since we can measure the amplitudeZ and we already know the frequencyω0, we can measure theacceleration amplitudeA. Since we are far below the free vibration frequency, our measured signalz(t) will be in phase with the term on the right-hand side of Eq. (81) and therefore in antiphase withthe acceleration.

A further degree of sophistication is to add damping to the accelerometer in order to improve itsaccuracy. The whole idea in going from Eq. (85) to (86) is that the term under the square rootshould be as close as possible to 1. This is achieved by makingΩ/ω0 as small as possible. To doeven better, we retain an extra term of orderΩ2/ω2

0:

ZY

=(Ω/ω0)

2√

(1− (Ω/ω0)2)2 +4δ2(Ω/ω0)2

≈ Ω2

ω20

[1− (2δ2−1)(Ω/ω20)] . (92)

This time we have kept extra terms to which is why our original result (86) is not exact. These termsbecome significant whenΩ2/ω2

0 is not so small. However, we can get rid of them! If we setδ =1/√

2 then the above equation reduces to equation (86). We can therefore expect our accelerometerto be more accurate if we apply damping so thatδ ≈ 0.7.

Many accelerometers work on this principle. Low frequency accelerometers use a damping ratioof 0.7 as described above. This also improves the phase distortion; you can read a description ofphase distortion in Thompson’s book if you’re interested. The ones you will meet in the lab use apiezoelectric crystal which has a very high natural frequency and therefore there is no need for anydamping (can you explain why?). These accelerometers are more suited to high frequency work.

2.5.2 Rotating out of balance rotor

Many machines have rotating parts which can vibrate. The system we will lookat are the vibrationsof a machine which are caused by a rotor which is out of balance. Figure 16 shows a machine ofmassM which contains a rotor of massm with an eccentricity ofe. The machine is mounted ona solid floor by a spring of stiffnessk and a damper with damping constantc and can move in thevertical (x) direction only.

From the free-body diagram shown in Figure 17, we find that the equationof motion for the machineand rotor in thex direction is

(m+M)x+cx+kx= meω2cos(ωt) . (93)

27

w t G e O x ( t ) m M

c k

Figure 16: Schematic diagram of a machine with an out of balance rotor. Thecentre of mass of therotor is located at G.

We follow our usual procedure of casting this into canonical form,

x+2ω0δx+ω20x = ω2

0x0cos(ωt) . (94)

where now

ω0 =

√

km+M

, δ =

√

c2

4k(m+M), x0 =

mem+M

ω2

ω20

. (95)

It follows from the results of Section 2.4.1 that the steady-state solution is given byx(t)= X cos(ωt+φ) where the amplitudeX is given by

X =me

m+Mω2/ω2

0√

(1−ω2/ω20)

2 +4δ2ω2/ω20

(96)

and the phase is given by

tanφ = − 2δω/ω0

1−ω2/ω20

. (97)

If we look at the amplitude and the phase of the motion for low, resonant and high frequencies wesee that

ω ≪ ω0 X → 0 φ → 0 ,

ω = ω0 X = me/[2δ(m+M)] φ = −π/2 ,

ω ≫ ω0 X → me/(m+M) φ →−π . (98)

28

m e m x ( t )

c M k xx & x &&

2w

x &&

Figure 17: Free body diagram for the system of Figure 16.

The highest possible amplitude occurs at resonance in a system for whichm≫ M. In this case, themaximum amplitude isX = e/δ. If the damping is small this can be very large.

Exercise: 13Use the results of the previous sections to determine the amplitude of the forcewhich is transmitted to the ground. What is the force transmitted at resonance?

2.5.3 Out of balance in reciprocating internal combustion engines

The inertial forces due to the motion of the piston and con rod in an engine mustbe balanced byforces acting on the engine and these can cause vibration. In this section we will deduce what theseforces are and what measures can be taken to keep them to a minimum. Figure 18 shows a schematicdiagram of a piston in a cylinder a distancex from bottom dead centre attached by a con rod AB toa crank shaft BC which rotates about C. The length of the shaft isr and the length of the con rod is

A f q C

Bl r

x l + r - xFigure 18: Schematic diagram of a piston, con rod AB and crank BC.

29

l . The distance between the piston and the crank axis is thereforer + l −x as shown.

To find the inertial force due to the motion of the piston, con rod and crank shaft, we will seek torelatex to θ, the angle through which the crank shaft has rotated. If the engine runsat a constantspeed thenθ, the angular velocity of the crankshaft, is a constantΩ.

The problem is that because of the geometry the motionx(t) of the piston is not simply a sinusoidalfunction. To work out what it is, we project the lengths of AB and BC onto AC to give

ABcosφ+BCcosθ = AC (99)

and putting in the various terms gives us

l cosφ+ r cosθ = l + r −x (100)

From the sine law, we havel sinφ = r sinθ (101)

therefore we can eliminateφ because

cosφ =

√

1−sin2 φ =

√

1− r2sin2 θl2 . (102)

hence

x = r + l − r cosθ− l

√

1− r2sin2 θl2 . (103)

If we introduce a parameterλ = r/l then we have

x = r

(

1+1λ−cosθ− 1

λ

√

1−λ2sin2 θ)

. (104)

In an engine, the value ofλ is usually around 1/3 to 1/4 and so we can expand the square root as apower series and ignore higher powers ofλ because they will be small. For smallλ, we have fromthe Taylor series that

√

1−λ2sin2 θ ≈ 1− λ2sin2 θ2

, (105)

so we can write

x = r

(

1−cosθ+λsin2 θ

2

)

. (106)

where I have ignored powers ofλ higher than one. This approximation is usually good enough. Wenow want to differentiate twice with respect tot in order to find the acceleration. We assume thatthe engine is running at constant speed,θ = Ωt. We insert this into x(t) and use that

sin(Ωt)cos(Ωt) =12

sin(2Ωt) . (107)

30

Taking the derivative ofx(t) gives

x = r[sin(Ωt)+λsin(Ωt)cos(Ωt)]Ω = r[sin(Ωt)+λ2

sin(2Ωt)]Ω , (108)

andx = rΩ2[cos(Ωt)+λcos(2Ωt)] . (109)

Now we have calculated the acceleration of the piston, let us consider the forces acting in the systemof Figure 18. To make the analysis easier, we will model the con rod in an approximate way. Thetrue con rod shape is rather complicated, and we will replace it in our “lumped” model with twomasses,m1 andm2, one at each end of the real rod. Figure 19 shows the con rod and its model.

Figure 19: Schematic diagram of the con rod and its model. G is the centre of gravity.

We choose the lengths a and b and the massesm1 andm2 so thatm1 +m2 = MCR, whereMCR is themass of the con rod, andm1a = m2b so the the centre of gravity of the con rod and the model of thecon rod are the same. Figure 20 shows the forces acting on the system with the model for the conrod inserted. In order to balance the system, the size and shape of the crank is chosen so the rotationof the mass labelledm3 causes the same centrifugal force as the rotation of the massm2 from thecon rod. These forces balance, but the remaining inertial forceM1x along the axis of motion of thepiston, whereM1 = MP +m1 andMP is the mass of the piston, causes a forceF on the crankshaftand hence on the engine.

To summarise this section, we have looked at the motion of the piston, con rod and crankshaftsystem at a constant angular velocity and deduced the inertial force dueto the motion of the pistonand the con rod. The crankshaft shape is chosen to eliminate the non-axial component of the conrod inertial force and the resulting force on the engine is given by

F = (MP +m1)x = (M1)rΩ2[cos(Ωt)+λcos(2Ωt)] (110)

31

M 1 x

m 2 r W 2

m 2 r W 2

m 3M p m 1

:

Figure 20: Forces acting on the piston, con rod and crank.

The termM1rΩ2cos(Ωt) is called the primary unbalance force and the termM1rΩ2λcos(2Ωt) isthe secondary unbalance force. To understand the origin of the secondary unbalance force, it isuseful to have a look at Figure 21. The motionx(θ) of the piston as calculated from Eq. (104) is notstrictly sinusoidal, and the difference with a pure cosine can be well approximated by a cosine withtwice the frequency. The corresponding acceleration gives rise to the secondary unbalance force.

0 2 4 6 8 100.0

0.5

1.0

1.5

2.0

Θ

x/r

Figure 21: Upper full line: motion of the piston as a function of the angle of rotation of thecrankshaft calculated from Eq. (104) forλ = 1/4. Dotted line: approximation by a cosine. Lowerfull line: Difference between the actual motion and a cosine.

32

We now look at the assembly of cylinders in an engine, and the resultant forces which act on the bodyof the engine. The goal will be to devise an engine that is balanced, so the unbalance forces fromeach cylinder cancel out. We will consider a four cylinder inline arrangement. Other arrangementscan be analysed by the same method.

Figure 22: Four cylinder inline arrangement. The cranks are attached atangles of 0, 180, 180 and 0degrees.

Figure 22 illustrates a four cylinder engine where the pistons are connected to the crankshaft atangles of 0,π, π and 0 radians (1-3-4-2 firing sequence). We want to know what the forces andmoments acting on the engine are. We will consider a general case, then seehow this applies to thefour cylinder engine.

Suppose that there areN cylinders in a line, and that thei-th crank is attached at an angle ofαi .After timet, the shaft rotates through an angleΩt and the crank is at an angleΩt +αi to the vertical.The force due to the piston and con rod is

Fi = M1rΩ2[cos(Ωt +αi)+λcos2(Ωt +αi)] (111)

The total force is the sum over all the cylinders, so

F = ∑i

Fi (112)

and is composed of a primary unbalance force oscillating at angular frequencyΩ, and a secondaryforce oscillating at 2Ω. By adding the contributions from all cylinders, we find that the total primary

33

unbalance forceFP is given by

FP = M1rΩ2∑i

cos(Ωt +αi) = M1rΩ2[cosΩt ∑i

cosαi −sinΩt ∑i

sinαi ] (113)

and the total secondary unbalance force is

FS = M1rΩ2λ∑i

cos2(Ωt +αi) = M1rΩ2λ[cos2Ωt ∑i

cos2αi −sin2Ωt ∑i

sin2αi ] . (114)

We also want to know the moments acting on the engine. If we take moments about apoint O (thecenter of gravity of the engine), and each cranki is a vertical distancezi from O, the primary andsecondary moments are

MP = M1rΩ2∑i

zi cos(Ωt +αi) = M1rΩ2[cosΩt ∑i

zi cosαi −sinΩt ∑i

zi sinαi ] (115)

and

MS = M1rΩ2λ∑i

zi cos2(Ωt +αi) = M1rΩ2λ[cos2Ωt ∑i

zi cos2αi −sin2Ωt ∑i

zi sin2αi ] . (116)

We can now evaluate the primary and secondary moments and forces from these expressions forany inline arrangement of cylinders. For the four cylinder example, we have i = 1,2,3,4 andz1 − b,z2 = −a,z3 = a andz4 = b. We can evaluate the sums of the various terms as shown inTable 1. The terms in the columns are the terms in the sums in the equations above.

crank αi zi sinαi cosαi sin2αi cos2αi

1 0 -b 0 1 0 12 π -a 0 -1 0 13 π a 0 -1 0 14 0 b 0 1 0 1

total 0 0 0 4

crank αi zi zi sinαi zi cosαi zi sin2αi zi cos2αi

1 0 -b 0 -b 0 -b2 π -a 0 a 0 -a3 π a 0 -a 0 a4 0 b 0 b 0 b

total 0 0 0 0

Table 1: Sums for the primary and secondary unbalance forces and moments in a four cylinderinline engine as shown in Figure 22

34

Using the table we can see that the primary and secondary unbalance momentsw as well as theprimary unbalance forces cancel, but the secondary unbalance force remains. The amplitude of thisis given by

F0S = 4M1rΩ2 . (117)

This force can in principle be countered using out-of-balance rotating shafts. This adds to thecomplexity of the engine, but ensures that it does not vibrate excessively.

Exercise: 14Another possible solution to the problem of minimising vibration of a 4-cylinderinline engine could be to use angles of 0o,90o,270o,180o. Draw your own versionof Table 1 and confirm that in this case the primary and secondary unbalance forcesand primary unbalance moments cancel. Why is this design not commonly used?

Exercise: 15Demonstrate that for 6 inline cylinders attached at angles of0o,120o,240o,240o,120o and 0o and at distances of 2.5,1.5,0.5,−0.5,−1.5,−2.5from the centre of the shaft, the primary and secondary unbalance forces andmoments all cancel.

35

2.6 Non-periodic forcing

In certain cases we wish to analyse the response of a system to non-periodic forcing. One exampleof this might be a sudden shock. We will consider a shock first of all, and then show that morecomplicated forcings can be considered as a amalgamation of shocks one after the other. This willgive us a formula with which we can calculate the response to an arbitrary forcing.

F o r c e ~ 1 / e

e T i m e

Figure 23: A short-lasting force.

Consider a spring-damper system such as that shown above in Figure 13, with the forcing functionshown in Figure 23. The duration of the force shown is given byε which we will take to be small.This means that the force lasts only briefly, but is large ( 1/ε). Such a shock force can be envisagedas a heavy kick. The strength of the kick can be characterized by itsimpulsewhich is defined as theintegral

I =∫ ∞

0f (t)dt . (118)

What is the response of the system to an impulsive shock like this? Let’s be clear about the problemwe wish to solve. Given the spring-damper system of Figure 13 subjected toan impulse of sizeI ,what is it’s response? We will assume that initially the system is in its equilibrium state.

It’s actually quite straight-forward to work out. We will find that the effect of the forcing is just togive the system an initial kick, and from then on it will behave like a freely vibrating system. If wecan work out the size of this kick, this will allow us to set the initial condition for the subsequentfreely vibrating behaviour.

36

All we need to do is to recognize that Newton’s second law can be written as

Fm =dpdt

(119)

wherep= mvis the momentum, andFm is the force acting on the massm. Therefore, the momentumof the mass is given by

p(t)− p(0) =∫ t

0Fmdt . (120)

When the initial kick is very strong and very short, the effect of the springs and dampers during thekick can be ignored because the external force is much larger than the spring and damper forces.Initially, v(0) = 0 because the system starts at rest, sop(0) = 0 and the momentum of the massmjust after the shock is given by

p(t = ε) = mv(ε) = I , (121)

and sov(t = ε) = I/m . (122)

Now, if ε is very small (the ’kick’ is very short), we can consider the motion of the system to startat time 0 with the velocityv = I/m, and since the forcing is zero after timeε we need only considerthe motion of the freely vibrating system. We found above that the solution for afreely vibratingsystem is

x(t) = exp(−ω0δt)[Acos(ωdt)+Bsin(ωdt)] (123)

and if we set the initial conditionx(0) = 0, it follows thatA = 0, i.e.,

x(t) = Bexp(−ω0δt)sin(ωdt) . (124)

The other initial condition is ˙x(0) = I/m. We find:

x(t) = Bexp(−ω0δt)[−ω0δsin(ωdt)+ωdcos(ωdt)] , (125)

sox(0) = Bωd = I/m , (126)

hence

B =I

mωd, (127)

and the solution we require is

x(t) =I

mωdexp(−ω0δt)sin(ωdt) . (128)

This expression gives us the response of a system to an impulse of sizeI delivered at timet = 0. Itis usual to define the unit impulse response functionh(t) to be

h(t) =1

mωdexp(−ω0δt)sin(ωdt) , (129)

37

and so the response to an impulseI is

x(t) = Ih(t) . (130)

We often find that we can approximate the behaviour of other forces as shocks, even if they last forquite a long time. The spring-mass system has a time scale for its vibrations ofT0 = 2π/ω0 and ifthe time over which a force acts is much smaller thanT0 we can usually treat it as a shock and writedown the response using the unit impulse response functionh(t).

For complicated non-periodic forcing, we can generalise the above method. Consider the forcingfunction shown in Figure 24:

F o r c e t t - t

t t + d t t T i m e

Figure 24: A general non-periodic forcing.

We can imagine that the forcingF(τ) is split up into lots of pieces defined over intervals of time dτ(note that for clarity I’m using the variableτ to denote time rather thant - the reason will becomeclear later). Each of these can be thought of as a little impulsive shock. If we consider the responseat timet, we can think of this response as being built up of many responses to the previous shocks- remember, we are allowed to build up solutions like this because the equation is linear and so thesuper-position principle holds.

Let’s consider the response at timet due to the shock shown at timeτ. The impulse due to this shockis

Iτ = F(τ)dτ (131)

Now, the effect of this shock at timet depends on the timet − τ that is elapsed after the shock. It isgiven by

xτ = h(t − τ)Iτ = h(t − τ)F(τ)dτ , (132)

where I’ve used the subscriptτ to denote that this is the part of the response due to the shock at time

38

τ. The full response is the total of all the shocks from time 0 to timet, i.e.,

x(t) =∫ t

0xτdτ =

∫ t

0F(τ)h(t − τ)dτ , (133)

and so we can calculate the response using this expression, provided wecan do the integral. In full,the formula for the response is

x(t) =∫ t

0

F(τ)mωd

exp(−ω0δ(t − τ))sin(ωd(t − τ))dτ (134)

which is called the convolution formula. For the case of small or no damping, thisexpressionsimplifies a bit to

x(t) =∫ t

0

F(τ)mω0

sin(ω0(t − τ))dτ (135)

which is in general easier to integrate. This I will refer to as the convolution formula withoutdamping.

2.6.1 Shock damping and shock response spectrum

To assess whether a vibrating system is capable of effectively damping shocks, we consider theshock response spectrum. This is a plot relating the response to the forcing duration, but plottedin a clever choice of variables: On the y-axis, the ratio of the peak response x and the peak staticresponse ˆx0 is plotted. The peak response is the highest maximum of the response to the shock,which is also sometimes called themaximax. The peak static response is just the response thatwould be caused by a force equal to the peak value of the forcing appliedstatically. On the x-axis,the frequency of free vibration divided by the frequency scale of the forcing is plotted. For a forceof durationTs, the frequency scale is simply 2π/Ts. Therefore, if the peak value of the shock forceis F0 and the system has a massm, the peak static response is

F0/k = F0/(mω20) . (136)

If the shock force lasts for a timeTs then the frequency of free vibration of the system divided bythe frequency scale of the forcing isω0Ts/(2π). On the shock response spectrum, we plot ˆxmω2

0/F0

againstω0Ts/(2π) .

The shock response spectrum depends in general on the shape of theforce pulse. As an example,we consider the response of an undamped mass-spring system to a rectangular pulse,

F(τ) =

F0 , 0≤ τ ≤ Ts

0 else .(137)

The response of an undamped spring-mass system to this pulse is obtained from equation (135). Weconsider separately the two casest ≤ Ts andt > Ts:

39

0.0 0.5 1.0 1.5 2.00

1

2

Pea

k am

plitu

de [

F0/

(mω

02 )]

Shock duration [2π/ω0]

Figure 25: Shock response spectrum for a rectangular pulse. Full line: Primary SRS; Dashed line:Residual SRS

(i) For t ≤ Ts the response is

x(t) =F0

mω0

∫ t

0sin(ω0(t − τ))dτ =

F0

mω20

[1−cos(ω0t)] . (138)

The maximum of the response is attmax= (2nmax−1)π/ω0 with the peak value ˆx= 2F0/mω20. nmax

is a positive integer number. Hence, this maximum is reached only ifTs > π/ω0.

(ii) For t > Ts we find

x(t) =F0

mω0

∫ Ts

0sin(ω0(t − τ))dτ =

F0

mω20

[cos(ω0(t −Ts))−cos(ω0t)] . (139)

The maxima are in this case given by

x =2F0

mω20

|sin(ω0Ts/2)| . (140)

The shock response spectrum (SRS) is plotted in Figure 25. Theprimary SRSshows the absolutemaximum of the response which occurs during the time of the shock, and theresidual SRSshowsthe absolute maximum of the response occurring after the end of the shock.It is interesting to notethat for shock durations which are integer multiples of the period of vibration, there is no residualresponse at all.

40

We now determine the conditions under which the maximum of the shock response falls below thestatic response, i.e. where we haveshock damping. This requires that

x = sin(ω0Ts/2) < 1/2 . (141)

Hence, shock damping is achieved if the pulse is sufficiently short such that ω0Ts < π/3, i.e. theratio between the shock duration and the period of free vibration must beTs/T < 1/6. We maycompare this with the criterion for vibration damping which follows from equation(67), where wehave observed that vibration damping is achieved ifω0/Ω <

√2, i.e., the ratio between the period

of the forcing and the period of free vibration must be less than 1/√

2. In either case an efficientstrategy to achieve damping is to make the natural frequency of the system assmall as possible(large masses and/or soft springs).

41

x 1 x 2 k k k m 2 m

Figure 26: A two degree of freedom system consisting of two masses connected by springs.

3 The theory of systems with many degrees of freedom

The generalisation of what we have done for one degree of freedom systems to many degrees offreedom is straight-forward provided we do it in the right way. What we will find is that if we usematrices to represent our equations, then thefree body diagram→ equation of motion→ complexexponentialmethod we have used works just as well for the many degree of freedom systems. Themain complication is the maths, which can get rather lengthy. In fact, we will oftenhave to resortto computers and numerical methods in order to solve the equations. We will alsofind there are oneor two extra things going on in these systems that simply do not occur in a single degree of freedomsystem.

3.1 Free vibration

We will start our study of multiple degree of freedom systems with a relatively simple exampleshown in Figure 26.

Exercise: 16We have discussed in the Introduction how to obtain equations of motion for thiskind of system. Draw a free body diagram and write down the equation of motionfor each mass.

If we write down the equations of motion for the two masses in the system shown and collect similarterms together we get a set of two coupled linear differential equations:

42

mx1 +2kx1−kx2 = 0 ,

2mx2−kx1 +2kx2 = 0 . (142)

I can represent these equations of motion in matrix form:

(

m 00 2m

)(

x1

x2

)

+

(

2k −k−k 2k

)(

x1

x2

)

= 0 . (143)

Exercise: 17Multiply out the matrix equation above to check that you do recover the two equa-tions of motion.

I could write this matrix equation symbolically as

M~x+K~x = 0 , (144)

where bold quantities represent matrices and the arrows represent vectors. M andK are the massand the stiffness matrices. They are given by

M =

(

m 00 2m

)

,

K =

(

2k −k−k 2k

)

. (145)

Written in this form, you can see that the equation looks really similar to the single degree offreedom equations we are used to solving. I can solve the equation using the complex exponentialmethod. I write

~x = ~Aexp(iωt) , (146)

where the usualx and A quantities I get in the single degree of freedom systems have becomevectors. Alternatively, I can write out this equation in long hand as the pair

x1 = A1exp(iωt) ,

x2 = A2exp(iωt) . (147)

I calculate the derivatives and substitute into the equation of motion in the usualway to get

−ω2M~Aexp(iωt)+K~Aexp(iωt) = 0 . (148)

Cancelling out the exponentials and collecting terms gives me

(−ω2M+K)~A = 0 , (149)

43

which I can write asB~A = 0 (150)

withB = −ω2M+K . (151)

It is important to understand that this is just a set of two coupled linear equations for the unknowncomponents of the vector~A, i.e., for the amplitudes of vibration of the two masses. Now a generalresult in the theory of linear algebra is that for the system of equationsB~A = 0 to have nontrivialsolutions, it must be true that the determinant ofB must be zero,

|B| = 0 . (152)

If this does not ring a bell, please take the time to go over your maths notes to refresh you memory.ReplacingB, we have deduced at this point in the calculation that

|−ω2M+K| = 0 (153)

If I now put in the expressions for the matrices I get the determinant∣

∣

∣

∣

2k−mω2 −k−k 2k−2mω2

∣

∣

∣

∣

= 0 . (154)

Now I can expand this determinant out to get

(2k−mω2)(2k−2mω2)−k2 = 0 (155)

which rearranges to2m2ω4−6kmω2 +3k2 = 0 , (156)

which is a quadratic equation inω2. I can solve the quadratic equation to find two roots. I get

ω2± =

km

(

32±

√3

2

)

. (157)

Negative frequencies don’t really mean anything. I therefore have twosolutions which I will labelω+ andω−. They are given by

ω+ =

√

√

√

√

km

(

32

+

√3

2

)

, (158)

ω− =

√

√

√

√

km

(

32−

√3

2

)

. (159)

This result is important. I have managed to calculate the frequencies of freevibration of the systemof two masses. Interestingly, there are two possible frequencies for free vibration. Either can occur

44

i.e. the system will resonate at either of these frequencies. More generally, a system withn degreesof freedom hasn frequencies of free vibration.

Now I have evaluatedω, I can go back to equation (149) and put in eitherω+ or ω−. For each valueI can use the matrix equation to calculate a ratioA1/A2.

Exercise: 18Use the values I have obtained forω+ andω− to write out equation (149) in full,using the definitions ofM andK in equation (145). Expand out the matrix equationinto two ordinary equations, and solve to find the ratioA1/A2.

For the frequencyω+, the ratioA1/A2 is -2.73 and forω−, the ratioA1/A2 is 0.73. What does allthis mean? Let’s go back to the single degree of freedom system, and consider what happened there.We made a substitution of the form

x = Aexp(iωt) (160)

in pretty much the same way as we have here, calculated the allowed value ofω, and that was it.We couldn’t determine the value ofA, the amplitude of vibration, without more information. Thisextra information came in the form of the initial conditions. For this two degrees of freedom system,we have found two possible values ofω, and for each a value of the ratioA1/A2. Now, from thesubstitution we made

~x = ~Aexp(iωt) , (161)

we can see that the first componentA1 of the vector~A represents the amplitude of vibration ofthe left mass since it is measured by coordinatex1, and the second componentA2 represents theamplitude of vibration of the right mass. Once again, we cannot deduce the values of bothA1 andA2 without knowing the initial conditions. However, we can deduce the ratio of the two, and this isinteresting. For the frequencyω+, we knowA1/A2 = -2.73. In this case, the amplitude of the leftmass is more than twice that of the right. The minus sign indicates that the direction of oscillationof the left mass is opposite to that of the right, i.e. they are in antiphase. For thesecond frequencyω−, A1/A2=0.73, and so both masses are in phase, with the amplitude of vibration of the right massa little larger than that of the left one.

We see that the system can oscillate at one of its natural frequencies only at if the amplitudes ofthe two masses have the right ratioA1/A2. We call such an oscillation afundamental modeof thesystem. The fundamental modes are often characterized by vectors~Swhere the componentsS1 andS2 are chosen such that they have the right ratioS1/S2 = A1/A2 and the vector has length 1, i.e.,S2

1 + S22 = 1. Such a vector is referred to as themode shape. For our example system, the mode

shapes are forω = ω+

~S+ =

(

−0.940.34

)

(162)

and forω = ω−

~S− =

(

0.590.81

)

(163)

45

Each of

~x(t) = Re

[

A+

(

−0.940.34

)

exp(iω+t)

]

(164)

and

~x(t) = Re

[

A−

(

0.590.81

)

exp(iω−t)

]

(165)

are valid solutions of the equation of motion. The constantsA+ andA− are in general complex, soby taking the real part of the complex exponential we get a combination of sines and cosines, oralternatively a sine with a phase shift.

The general solution can be written as a sum of both solutions,

x(t) = A+

(

−0.940.34

)

sin(ω+t +φ+)+A−

(

0.590.81

)

sin(ω−t +φ−) , (166)

where themode amplitudes A+ andA− are now real and together with the phasesφ+ andφ− mustbe determined from initial conditions. So we need four initial conditions. In general, for a systemwith n degrees of freedom 2n initial conditions are required.

3.2 Eigenvalue problems

We found in the above section that we could write our equations of motion in matrixform as

M~x+K~x = 0 , (167)

and indeed this is always the case, even for many more degrees of freedom. Using the substitution

~x = ~Aexp(iωt) , (168)

we can transform the equation to−ω2M~A+K~A = 0 (169)

or rearranging terms(M−1K)~A = ω2~A . (170)

Now this last form is called the matrix eigenvalue problem, and it arises from many different phys-ical situations. Because it occurs so often, it has been studied very thoroughly. Moreover, there aremany computer packages that will quickly solve it for us. Let’s consider what the solutions are likefirst, then see how a computer can be used to help in the calculations.

If we writeY = M−1K (171)

46

andω2 = λ then the problem we wish to study is

Y~A = λ~A (172)

The mathematics tells us that ifY is a n× n matrix then there aren values ofλ that allow thisequation to have non-trivial solutions, and each of these has associatedwith it a vector~A. Thevalues ofλ are called eigenvalues and the associated vectors are eigenvectors.

We have seen that we can take our vibration problem, cast it in matrix form and then manipulate thematrices so that it is a eigenvalue problem. Using a package such as MATLAB, the solution of thiseigenvalue problem is very easy. In this way, we can tackle systems with quitea large number ofdegrees of freedom that would be very hard indeed if we did not have access to a computer.

3.3 Forced vibration

Just as with the single degree of freedom systems, we would like to be able to calculate what happenswith a forced system. The path from the free body diagram to the equations of motion is no morecomplicated than we have met already. Once again, we can write the equationsin matrix form sowe end up with something like

M~x+K~x = ~F cos(Ωt) , (173)

where the new part is the RHS consisting of a forcing term proportional to cos(Ωt). Many timesonly one component of~F will be non-zero, but that need not concern us now. Using what we knowabout matrices, solving this really quite difficult problem is surprisingly easy. Just as with the singledegree of freedom systems, we introduce a “guessed” solution of the form

~x = ~Acos(Ωt) (174)

where you should note that we have been able to use cos(Ωt) directly rather than the complexexponential because there is no damping and so we know that the response is either in phase or inantiphase with the forcing. Substituting this into the equation of motion gives

−Ω2M~Acos(Ωt)+K~Acos(Ωt) = ~F cos(Ωt) . (175)

and we can cancel the cosines from each side so we have an equation ofthe form

Z~A = ~F (176)

whereZ = −Ω2M+K . (177)

The matrix equation (177) corresponds to system of equations for the components of the vector~X.For small matrices (systems with only a few degrees of freedom) this can be solved by hand. For

47

larger systems, one uses a formal solution which can be computed numerically. To this end, wehave simply to determine the inverse matrix ofZ:

~A = Z−1~F . (178)

The inverse of a matrix can be calculated using MATLAB. Hence, we may simplyuse MATLAB tocalculate the inverse matrixZ−1 and multiply this with the vector~F which contains the amplitudesof the external forces.

Exercise: 19Go back to the two spring system we defined above and add a forcing term of theform F0cos(Ωt) to the left mass. Use the above method to deduce an expressionfor the frequency response curve.

3.4 Orthogonal modes

I want to conclude our consideration of the theory of multiple degree of freedom systems with somemore matrix algebra. Make sure your understanding of matrix algebra is up toscratch before westart. The following considerations are rather formal, but they lead to a surprising result: A vibratingsystem withn degrees of freedom can always be transformed in such a manner that itbehaves liken independentsingle degree of freedom systems.

The matrix equation of motion for an degree of freedom system is

M~x+K~x = 0 . (179)

If we now assume that the system is vibrating at one of its fundamental modes,then

~x =~Si exp(iωt) , (180)

and we haveK~Si = ω2

i M~Si , (181)

where we know that there should ben possible values ofω and I’ve acknowledged this in theequation by labelling these different possible frequencies and the corresponding mode shapes withan indexi which can run from 1 up ton.

Now I’m going to prove an important result concerning the vectors~Si . Firstly I will multiply theabove equation by the transpose vector~ST

j to give me

~ATj K~Si = ω2

i~ST

j M~Si . (182)

It will become clear later just why I want to do this. For the moment, just make sure you understandwhat this means - write out example matrices for yourself if you like. I will also make use of exactlythe same equation written out with the i and j subscripts interchanged,

~STi K~Sj = ω2

j~ST

i M~Sj . (183)

48

Now, the theory of matrix algebra shows that

(AB)T = BTAT . (184)

Using this result, we can also prove quite quickly that

(ABC)T = CTBTAT . (185)

If I take my equation (183) above and transpose it, then I get

(~STi K~Sj)

T = ω2j (~S

Ti M~Sj)

T , (186)

and using the above theorem of matrix algebra gives

~STj K~Si = ω2

j~ST

j M~Si , (187)

remembering that the transpose operation twice just goes back to the originalmatrix and that sinceM andK are symmetric the transpose operation has no effect on them. Subtracting equations (182)and (187) gives me the interesting result

(ω2j −ω2

i )~STj M~Si = 0 . (188)

It follows that~ST

j M~Si = 0 and ~STj K~Si = 0 ifi 6= j . (189)

These equations describe a property of the vectors~Sj called orthogonality with respect to the matri-cesM andK. This property turns out to be very useful.

For the casei = j, we define by~ST

i M~Si = Mii (190)

and~ST

i K~Si = Kii (191)

the generalised mass and stiffness. We will now make use of these ideas in a very powerful method.Before we can do so, we must make one last definition; an object called the modal matrixP. Wedefine it to be the sequence of all the column vectors~Si , i.e.

P = [~S1~S2 . . .~Sn] , (192)

so thatP is an×n matrix formed by the eigenvectors~Si . What is the use of this matrix? Considerthe matrix

PTMP = [~S1~S2 . . .~Sn]TM[~S1~S2 . . .~Sn] . (193)

Because of the orthogonality relations above, most of the terms when we multiplyout this expres-sion are zero. We are left with

PTMP =

M11 0 . . . 00 M22 . . . 0. . .0 0 . . . Mnn

(194)

49

which is a diagonal matrix. The matrixPTKP is also diagonal. The result we have proved is thatthe modal matrix diagonalises the mass and stiffness matrices. This is useful! Ifwe consider theequation of motion

M~x+K~x = ~F0exp(iΩt) , (195)

then if we define a new set of variables by~x = P~y then

MP~y+KP~y = ~F0cos(Ωt) , (196)

and multiplying this equation byPT gives

PTMP~y+PTKP~y = PT~F0cos(Ωt) . (197)

Since the matrices are diagonal, thisuncouplesthe problem: This equation is justn single degree offreedom problems

Mii yi +Kii yi = [PT~F0]i cos(Ωt) (198)

which we know how to solve. Hence, our coupled n-degree of freedomsystem is in fact equivalentto n uncoupled spring-mass systems. Each of these represents one of the fundamental modes of thesystem. The effective force amplitude exciting this mode is given by[PT~F0]i = ~Si~F0, i.e., it is equalto the projection of the force vector on the respective mode shape. This is important in practicesince it allows us to assess whether a particular forcing is ’dangerous’ inexciting a certain vibrationmode of the system.

The above mathematics seems quite complicated, but don’t be intimidated. You cantry the methodfor yourself. In brief, to solve any forced multiple degree of freedom problem you need to:

• Draw a free body diagram

• Write down the equation of motion for each component, and put the equations inmatrix form

• Set the forcing to zero temporarily and solve the freely vibrating system to find the frequenciesof free vibration

• Use these frequencies to solve the matrix equation and determine the mode shapes

• Construct the modal matrix P from the mode shapes

• Use the modal matrix to define a new set of coordinatesyi (this is conceptual - you don’t needto calculate anything here)

• Calculate the productsPTMP andPTKP and write down the new matrix equation fory

• Expand out the matrix equation to given single degree of freedom systems

• Solve the single degree of freedom systems

• Use~x = P~y to put the solution back into the original coordinates

50

It’s quite a lengthy procedure, but the steps are quite simple to carry out if you have a computer athand - and it does give a general method to solve any system, however complicated, provided youcan calculate the eigenvalues and vectors. In particular, the method is suitedto programs such asMATLAB where you can define and handle matrices easily.

3.5 Applications

3.5.1 Shaft whirling

The next system we will look at is the vibrations of a rotating out-of-balanceshaft. For simplicity,let’s consider a massless shaft with a disc of massm in the centre. Figure 27 shows what this lookslike.

A C w B

G C e

Figure 27: Schematic diagram of a disc rotating on a shaft. The shaft AB meets the disc at C. Thecentre of mass of the disc is located at G.

Figure 28 shows the coordinate system we will use. The origin of the coordinates has been chosento be at the equilibrium position of the shaft centre. The shaft centre is attached to the disc at C, soin equilibrium C rests at O. The centre of mass G is rotated about the shaft centre C at the angularvelocity of the rotation of the shaft, which isω. Because the shaft is rotating, C can be displacedfrom its equilibrium position and will then be rotated about O. We want to know how far the centreof the shaft C is displaced from its equilibrium position.

51

y G w t e

z C q O x

Figure 28: Motion of the shaft centre C and the centre of mass G. G moves around C due to therotation of the shaft. C moves around O because of the inertial forces caused by rotating G.

Our first step in analyzing this problem is to draw the free body diagram. Thisis shown in Figure29.

y O x

m e w 2

m x G m y k z C

:

:

Figure 29: Free body diagram for the system of Figure 28, showing the inertial forces and the forcedue to the elastricty of the shaft

52

The force acting on C is due to the displacement of the centre of the shaft from its equilibriumposition. If the force for a unit displacement isk then the force here iskzdirected towards O. Thevalue ofk is a property of the shaft and the way it is mounted and can be calculated using solidmechanics. The force acting on G is the centrifugal force due to its rotation about C. This is theforcing in our problem, which causes C to be displaced from O. The other two forces shown actingon G are the inertial forces due to its motion. Note that we must give two of thesesince there aretwo independent coordinates needed to specify the position of G. G and C are rigidly linked so theequation of motion can be deduced by summing all the forces in the system. We get two equations,by resolving in thex andy directions:

mx+kzcos(θ)−meω2cos(ωt) = 0 , (199)

my+kzsin(θ)−meω2sin(ωt) = 0 . (200)

Now x = zcos(θ) andy = zsin(θ), so we can write the two equations of motion

mx+kx= meω2cos(ωt) , (201)

my+ky= meω2sin(ωt) . (202)

This is a most simple two-degree of freedom system since the two equations are not coupled. There-fore each of them can be solved using the usual method for forced SDF systems. We find

x(t) = eω2/ω2

0

1−ω2/ω20

cos(ωt) , (203)

y(t) = eω2/ω2

0

1−ω2/ω20

sin(ωt) . (204)

These can be added together to calculatez2 = x2 +y2. We find

z= eω2/ω2

0

1−ω2/ω20

. (205)

whereω0 is the natural frequency of free vibration for the shaft. Ifω is small, i.e. the shaft ro-tates slowly, thenk ≫ mω2 and z = e(ω/ω0)

2, which shows C is not far displaced from O. Athigher speeds of rotation, however, C does move a long way from O and at resonance (ω = ω0) theexpression goes to infinity. At high speeds, for whichω is large,

z≈−e , (206)

so the centre of mass sits at the origin and the shaft centre C rotates about it.

This simple example shows that shaft whirling occurs at the natural frequency of vibration of theshaft. This is a very useful result because now we can calculate when whirling will occur by ana-lyzing the free vibrational properties of the shaft.

53

3.5.2 Beating

We will look at the beating behaviour of two pendulums. This will allow us a bit ofpractice oninserting the initial conditions into a MDF system as well as providing an interesting study of thephenomena of beating.

q 1 q 2

a k l

m mFigure 30: Two weakly coupled pendulums.

q 1 q 2

k a ( q 2 - q 1 )

m g m gm l 2 q 1 m l 2 q 2

: :

Figure 31: Free body diagram for the system of Figure 25.

Beating occurs when we have a weak link between two systems with natural frequencies that areclose together. I’m going to consider two identical pendulums joined by a weak spring. In realitythe coupling between the pendulums might occur because of the motion of the mounting plate, butwe will just take it as given here, and model it using the weak spring. Figure 30 shows the two

54

pendulums and Figure 31 is the free body diagram.

From the free body diagram, we can write down the equation of motion by takingthe moments ofthe forces about the pivots of the pendulums.

ml2θ1 +mglsinθ1−ka2(θ2−θ1) = 0 , (207)

ml2θ2 +mglsinθ2 +ka2(θ2−θ1) = 0 . (208)

In the approximation of small oscillations, we can put sinθ ≈ θ and write the equations of motionin matrix form:

(

ml2 00 ml2

)(

θ1

θ2

)

+

(

mgl+ka2 −ka2

−ka2 mgl+ka2

)(

θ1

θ2

)

= 0 . (209)

We substitute~θ = ~Aexp(iωt). If we carry out the differentiation ont and insert the result in theequation of motion we get

(

−ml2ω2 +mgl+ka2 −ka2

−ka2 −ml2ω2 +mgl+ka2

)(

A1

A2

)

= 0 . (210)

For the equation to have solutions, the determinant of the matrix must be zero which gives us ourusual quadratic equation forω2. If we do the algebra we get

−ml2ω2 +mgl+ka2 = ±ka2 . (211)

It follows that the values ofω are

ω+ =

√

gl

, ω− =

√

gl+

2ka2

ml2. (212)

Putting these values back into equation (210) gives two equations for the AmplitudesA1 andA2

which are

ka2(

1 −1−1 1

)(

A1

A2

)

= 0 , ω = ω+ ,

ka2(

−1 −1−1 −1

)(

A1

A2

)

= 0 , ω = ω− . (213)

We therefore find thatA1/A2 = 1 for ω = ω+ andA1/A2 = −1 for ω = ω−. The mode shapes aretherefore

~S+ =1√2

(

11

)

, ~S− =1√2

(

−11

)

. (214)

Now we will construct the general solution.

The two solutions~θ =~S+ exp(iω+t) , ~θ =~S−exp(iω−t) (215)

55

are just the complementary functions. The general solution is just the sum ofthe complementaryfunctions multiplied with arbitrary (complex) amplitudes. It is

~θ(t) = Re

[

A+√2

(

11

)

exp(iω+t)+A−√

2

(

1−1

)

exp(iω−t)

]

. (216)

At this point in the calculation, I will take the real part. If I define

− ImA+√2

= B ,ReA+√

2= C , − ImA−√

2= D ,

ReA−√2

= E , (217)

this gives(

θ1

θ2

)

=

(

Bsin(ω+t)+Ccos(ω+t)+Dsin(ω−t)+Ecos(ω−t)Bsin(ω+t)+Ccos(ω+t)−Dsin(ω+t)−Ecos(ω+t)

)

. (218)

This then is the general solution, and it contains four unknownsB,C,D,E that are determined by theinitial conditions. For the initial condition, I will specify that both pendulums areat rest, pendulumone is at an angleα to the vertical and pendulum two is vertical. If we put these conditions into theabove equations forθ1 andθ2, we findC = E = 0 andB = D = α/2. Please check this for yourself.The final solution is therefore

θ1 = (α/2)[cos(ω1t)−cos(ω2t)] (219)

andθ2 = (α/2)[cos(ω1t)+cos(ω2t)] (220)

We can make use of the trig identities to write this in a more convenient form. We have

cosA+cosB = 2cosA+B

2cos

A−B2

cosA−cosB = 2sinA+B

2sin

A−B2

so our solution can be written as

θ1 = αcos((ω1 +ω2)t/2)cos((ω1−ω2)t/2) (221)

andθ2 = αsin((ω+ +ω−)t/2)sin((ω+−ω−)t/2) (222)

Finally we can put in the values we calculated forω+ andω−. If the coupling is very weak,k issmall and so

ω+ +ω−2

≈√

gl

(223)

56

-1.0

-0.5

0.0

0.5

1.0

2π/ω 2π/∆

Θ2/

α

-1.0

-0.5

0.0

0.5

1.0

Θ1/

α

Figure 32: Plot of the beating behaviour in the coupled pendulum system. The outer line shows thesinusoidal envelope curve

57

and

ω+−ω−2

=12

√

gl

1−√

1+2ka2

mgl

≈−√

gl

ka2

2mgl(224)

If we write

∆ =

√

gl

ka2

2mgl, ω =

√

gl

, (225)

then the solution can be written as

θ1 = αcos(∆t)cos(ωt) , (226)

θ2 = αsin(∆t)sin(ωt) . (227)

A plot of this form is shown in Figure 32. As you can see, the vibration amplitude is periodicallymodulated, and the energy of vibration oscillates between the two pendulums witha period ofπ/∆.This phenomenon is called beating.

3.5.3 Anti-resonance and vibration absorbers

We have seen in previous sections that the vibration of a forced system can be always suppressed ifone makes the natural frequency low enough, i.e. by designing a system with large mass and smallstiffness. However, this may often not be a feasible option due to other design constraints.

In this case, another strategy may be chosen to suppress unwanted oscillations. Consider the systemin Figure 33 (left) with a sinusoidal forceF(t) = F0cos(Ωt). The equation of motion of this systemis

m1x1 +k1x1 = F0cos(Ωt) (228)

and the steady-state response of the system is

x(t) =x0

1−Ω2/ω20

cos(Ωt) =F0

k1−m1Ω2 cos(Ωt) . (229)

If the forcing frequency is not much larger than the system’s natural frequencyω0 =√

k1/m1 theresponse amplitude may be significant, and at resonance it becomes infinite.

How can we suppress the vibration of the massm1 without changing the mass or the stiffness of thesystem? We will see that we can deal with this problem by adding another mass-spring subsystem(m2 andk2) as shown in Figure 33 (right). This gives a two-degree of freedom system

m1x1 +k1x1 +k2(x1−x2) = F0cos(Ωt) (230)

m2x2 +k2(x2−x1) = 0 (231)

58

F ( t )

x 1 m 1

k 1

x 2 m 2 F ( t ) k 2

x 1 m 1

k 1

Figure 33: Left: a forced oscillator. Right: the same oscillator with an additional spring-masssystem acting as anti-vibration device.

or, in matrix form(

m1 00 m2

)(

x1

x2

)

+

(

k1 +k2 −k2

−k2 k2

)(

x1

x2

)

=

(

F0

0

)

cos(Ωt) . (232)

We solve this by assuming(

x1

x2

)

=

(

A1

A2

)

cos(Ωt) . (233)

whereA1 andA2 are the vibration amplitudes of the massesm1 andm2. This leads to the system ofequations

(−m1Ω2 +k1 +k2)A1−k2A2 = F0 , (234)

−k2A1 +(−m2Ω2 +k2)A2 = 0 . (235)

By solving this system we obtain the vibration amplitudesA1 andA2. The result is

A1 =(k2−m2Ω2)F0

K(Ω), (236)

A2 =k2F0

K(Ω), (237)

59

whereK(Ω) = (k1 +k2−m1Ω2)(k2−m2Ω2)−k2

2 . (238)

We see that the amplitude of vibration of the massm1 becomes zero ifk2 andm2 fulfil the relationk2/m2 = Ω2, i.e., if the resonant frequency of the added subsystem matches the excitation frequency.In particular, we can suppress the resonant response of the originalsystem atΩ = ω0 =

√

k1/m1

completely by chosingk2/m2 = k1/m1.

60

4 Self-excited vibration

We have seen how vibrations can be excited by a force acting from outsideon a system. In thepresent section, on the other hand, we will be concerned with vibrations that emerge without suchexternal influence through a mechanism called positive feedback.

4.1 Positive Feedback

In our vibrating systems we have always been dealing with situations where the damping forceopposes the motion. Positive feedback, on the other hand, means a situationwhere the motion ofthe system leads to a netforward force which accelerates the motion. This leads to afeedback loop:The accelerated motion increases the forward force, this further accelerates the system motion, andso on.

To understand how this can happen, we consider the example in Figure 34.This is a fairly commonsituation: A slider is moving on a surface, with a friction force that decreases as the velocity of mo-tion increases. This includes the well-known fact that static friction is largerthan kinetic friction,but also more complicated situations e.g. in the friction of snow and ice: Because of frictional heat-ing, the motion of a slider creates a water film which lubricates the sliding interface. This reducesthe friction force – the larger the velocity, the larger the heating and the betterthe lubrication, hencethe lower the friction.

x c

y m k F f

Figure 34: Left: A slider pulled across a surface with a velocity-dependent friction forceFf andconstant pulling speed ˙y = v0.

Let us now analyse a situation where we try to slide a massmacross a surface at a constant velocity

61

y = v0. The equation of motion is given by

mx+c(x− y)+k(x−y)−Ff(x) = 0 . (239)

It is convenient to envisage the motion of the slider in a coordinate systemz= x− y which movesat the constant velocityv0. Since ¨y = 0, we have

mz+c(z)+kz= −Ff(v0 + z) . (240)

If the massmslides at the constant velocity ˙x = v0, thenz= 0 and this equation reduces to

kz0 = −Ff(v0) , (241)

i.e., the spring is pulled by such an amount to the right that the spring force offsets the friction forcethat has to be overcome for the mass to slide at the imposed velocityv0. This is simple, but can sucha situation be sustained?

To investigate this, we assume that the position of the massm deviates by a small amountz− z0

from the force equilibrium position. We get forz′ = z−z0:

mz′ +[c+F ′f (v0)]z′ +kz′ = 0 . (242)

We see that the damping is replaced by an ’effective damping’[c+ F ′f ] which depends on the ac-

tual damping constant, but also on the way how the friction force changes with velocity (F ′f is the

derivative of the friction force with respect to velocity, taken at the pointv0.)

If friction decreases strongly with velocity,[c+F ′f ] < 0 and the effective damping becomes negative!

Recalling the solution of the free vibration equation, we find that this impliesδ < 0 and the vibrationgrows exponentially. That is what we call vibration arising from a positivefeedback mechanism.

In machining, several physical mechanisms may lead to positive feedback and self-excited vibration(’chattering vibration’). These include

• Thermomechanical instability of the cutting process: During cutting, the workpiece is locallyheated and thereby softened. If this effect is pronounced, it may reduce the cutting force andlead to accelerated cutting which produces even more heat. Ice lubrication as discussed abovemay be considered a special case of this ’thermomechanical’ feedback mechanism.

• Negative velocity sensitivity of friction forces and stick-slip vibration. It isa fairly generalobservation even in dry friction that the friction force decreases with increasing velocity.An example is grinding: If the grinding tool slides faster, this may decrease the experiencedfriction force and, if the damping of the tool is not sufficiently large, causefurther accelerationof the motion.

62

• Regenerative Vibration: Suppose a drill vibrates a bit at its natural frequency. This smallvibration leads to uneven cutting of the workpiece and leaves a pattern on thesurface of thedrilled hole. Subsequently the tool cuts into this uneven surface and therefore experiencesa time-dependent force. If the force is such that it increases the vibration, we again have apositive feedback mechanism that will lead to ’chattering’ vibration.

4.2 Stick-slip motion

We analyse in some more detail the dynamical behavior which arises if the friction force decreaseswith increasing velocity. In particular, we consider the case of dry frictionwhere the friction forcein Figure 34 can be described by a piecewise constant force law:

|Ff | =

µsFN = Fs , x− y = 0 ,µdFN = Fd , x− y 6= 0 .

(243)

If the slider is at rest, we have static friction with the friction coefficientµs, otherwise we havedynamic friction with the friction coefficientµd < µs. In either case the friction force is proportionalto the normal forceFN, and is directed such that it opposes the sliding.

Let’s now assume that the slider is initially at rest on the surface, and we start to move the supportto the right at constant velocity ˙y. The force on the mass acting to the left is thency+ kz, and thisis balanced by the static friction force. At a critical elongationzc, Fs = cy+kzc, and if the spring isfurther elongated, the static friction is overcome and the system starts to move against the surface.This happens at

zc = −Fs−cyk

. (244)

As soon as the mass starts to slide on the surface, its equation of motion becomes

mz+cz+kz= −Fd . (245)

We know how to solve this equation: We definez′ = z+Fd/k to get rid of the constant term on theright-hand side and bring the equation into canonical form. The solution is (see Section 2.3)

z′ = Aexp(−ω0δt)sin(ωdt +φ) . (246)

The amplitudeA and phaseφ follow from the initial conditions at the onset of sliding, which weidentify with the momentt = 0. These are given by

z′(0) = zc +Fd

k= −Fs−Fd−cy

k, z′(0) = −y = −v0 . (247)

From these we can work out the amplitudeA and phaseφ. For simplicity, we focus on the casewhere damping is small,δ ≪ 1. In this case we find approximately thatωd ≈ ω0 and

z′ ≈ Aexp(−ω0δt)sin(ω0t +φ) ,

z′ ≈ Aω0exp(−ω0δt)cos(ω0t +φ) . (248)

63

Inserting the initial conditions we get

tanφ = −z(0)ω0

v0≈ Fs−Fdω0

kv0, A =

√

[z′(0)]2 +[v0/ω]2 (249)

We find that the mass undergoes damped oscillation. Now everything depends on whether thevelocity of the mass becomes large enough to allow it to ’catch up’ again with the sliding surface.This occurs if ˙x = z+v0 = 0, i.e., z=−v0. As soon as this happens, we switch from sliding to staticfriction, i.e., the mass gets stuck and is carried along with the surface. It will then again be releasedat the positionzc, and this ’stick-slip cycle’ will repeat itself.

The first and deepest minimum of ˙z occurs atω0t + φ = π. Inserting this we find the condition forstick-slip to be given by

v0 < Aω0exp(−δ[π−φ]) , (250)

which after some transformations can be written as

1 < (1+ tan2 φ)exp(−2δ[π−φ]) . (251)

In the limit of large velocities, large spring stiffnesses, or small friction force difference,φ ≈ 0 andthis condition can never be met. In this case, the mass does not catch up againwith the slidingsurface, the oscillation will decay and the system ultimately reaches a regime ofsteady-state slidingat the imposed velocity ˙y.

64