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

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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

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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 dont 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 particular integral 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 a lecture 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 easy to analyze math-ematically. After reviewing free body diagrams and clarifying notation, we will consider 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, homogeneous ordinary differential equationwith constant coefficients and it can be tackled by methods which will already be familiar from pre-vious courses. We will look at the physical significance of the solution and see how the parametersof the system determine its vibrational characteristics. Having understood the behaviour of freely

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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 and stiffnessconstants 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 engineering systems. 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, IC engine 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 getanswers, 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 also discuss 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 and relieve you of thetedious, error-prone task of copying chunks of algebra from the board.

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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 know the 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 often be written down. Whenconstructing the free body diagram, it is important to bear in mind the direction of each 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. When x1 = 0 and x2 = 0 the system is in static equilibrium

The best way to see how this works is to consider an example. In Figure 1 the two objects withmasses m1 and m2 are free to move in the horizontal direction. However, they are not free to movein 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 constants k1,k2 and k3 and like all springs we willconsider they are linear. If a linear spring is extended by a distance x, it exerts a force kx 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 of x1 and x2 out of thisconfiguration. If the distance between the masses in the equilibrium configuration is l, the distancebetween them after they have moved is l x1 + x2. The extension of the central spring is thereforex2x1. The forces due to the springs can now be deduced. Figure 3 shows the forces acting on eachmass, including the inertial forces. Note that the final spring is compressed when x2 is positive andso the force must be directed to the left. We choose to show this by drawing the arrow to the rightand including a minus sign. Equivalently, we could have kept the force as positive and directed thearrow to the left.

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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. If x1 is positive, as it is shown in Figure 2, then the mass m1 has beenmoved to the right. If we accelerate a mass in the direction of x1 to the right, its inertia acts tooppose this and so the inertial force is directed to the left. The same argument applies to the massm2. Hence, the direction of the inertial forces must always be taken opposite to the direction of therespective coordinate. The equations of motion for the two mass system can now be written downusing DAlemberts 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 coordinates x3 and x4 so that x3 is justthe same as x1 but x4 runs in the opposite direction to x2. Draw the free bodydiagram and deduce the equations of motion. When you have done this, substitutex3 = x1 and x4 = 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).

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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 doesnt 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 problems that 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 what we 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 supported by 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 point rather than distributedin space is called lumping and we refer to the model as a lumped parameter model. Lumping isa very powerful idea because it greatly simplifies the mathematical treatment of the system whilemanaging to retain the essentials. Masses and springs can both be lumped, i.e. replaced by a single

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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 modeling the 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 will start 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.

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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 the x coordinate one obtains the same equation of motion (4) as in thegravity-free case. (Chose as x = 0 the static equilibrium configuration of thesystem with gravity)

Before we try to solve this equation, we will classify it. The equation is second order in time becausethe inertial term involves a second order derivative. The equation is linear, since the coefficients donot depend on x, and it has constant coefficients because m and k do not depend on t. It is alsohomogeneous because there is no term independent of x. 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 not need 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(t) ,

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x = cos(t) , (5)is a solution of the equation. The constant must be chosen so that the m and k constants areaccounted for correctly. The value needed to do this is =

k/m.

Exercise: 3Verify the value given for . To do this, consider sin(t) first. Compute thesecond derivative with respect to time and substitute into equation (4). From theresulting equation you can calculate the relation between and k and m. Repeatthe calculation for the cosine solution.

From the complementary functions, we can now construct the general solution. It isx = Asin(t)+Bcos(t) . (6)

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

x = C sin(t +) . (7)where the constants are given by C =

A2 +B2 and tan = B/A.

Exercise: 4Show this is the case. Hint: start with x = C sin(t +) 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 constant C is the amplitudeof the vibration and is the phase. These two constants are undetermined in the general solutionand must be calculated from the initial conditions.

Lets now put some numbers into the formula, specify initial conditions, and see what the solutionlooks like. Suppose m = 1 kg and k = 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 first and find = 5 rad/s. The motion of the mass is therefore given by

x = C sin(5t/s+) . (8)From the initial conditions, we know that x(0) = 0 which implies = 0. The velocity at any time tcan be found by differentiating the above equation to get

v(t) = x(t) = [5/s]C cos(5t/s+) (9)and we now know = 0 so at time t = 0 the velocity is given by

v(0) = [5/s]C . (10)The initial conditions tell us that v(0) = 102 m/s so C = 2103 m. The final solution then is

x(t) = [2103m] sin(5t/s) . (11)This solution is shown in Figure 6 for the first three seconds of motion.

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0 1 2 3

-2

0

2

disp

lace

me

nt [m

m]

time[s]

Figure 6: The displacement of the mass in the system of Figure 4 as a function of 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 alternative approach based onthe energy of the system. We will find that we can actually derive the equation of motion once weknow the energy of the system, so all the results from the previous section can be re-obtained. Theadvantage is that we dont need to consider the forces in the problem, which in some cases canbe difficult. On the other hand, the energy approach as discussed here works only if there is nodamping.

Consider the energy stored in the spring. The force exerted by a spring is F = kx where x is theextension from the equilibrium length and k is the stiffness. The work done in extending the springby a length x is

W =

Fds =kx2

2. (12)

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

U(x) =kx22

. (13)

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

T =mx2

2, (14)

so the total energy of the system E = U +T is

E =mx2 + kx2

2. (15)

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

E = mxx+ kxx = 0 . (16)We can factor this equation into

x(mx+ kx) = 0. (17)so either x = 0, which tells us that the system may be at rest, or mx + kx = 0 which is the equationof motion derived above. The situation in which the system never moves is not of 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 the system. 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 throughout the 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 velocity v is slowed down by a forceproportional to v. Fast moving objects therefore encounter large forces, while slowly moving objects

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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 force F is given by F =cv where c 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.

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Notice the direction of the forces. The positive x 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 this equation are of the formexp(t). 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 constant must now be chosen so that equation (18) is satisfied. To this end, wesubstitute x = exp(t), x = exp(t) and x = 2 exp(t) into (18). We get

2 + cm

+ km

= 0 . (19)

This quadratic equation (characteristic equation) has roots

=cc24km

2m, (20)

and the corresponding solutions are given by

x+(t) = exp(+t) ,x(t) = exp(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 combination Ax+ +Bx is also a solution (A andB are arbitrary constants). The most general solution is therefore

x(t) = Aexp(+t)+Bexp(t) . (21)

It is worth recognizing that the roots + and and therefore also the complementary functionsx+(t) and x(t) are not always real. Of course, the final solution x(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(+t)+Bexp(t)] . (22)

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

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

x(t) = Aexp

([ c

2m+

( c2m

)2 k

m

]t

)+Bexp

([ c

2m( c

2m

)2 k

m

]t

). (23)

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

x(t) = exp( c

2mt)[

C cosh

(( c2m

)2 k

mt

)+Dsinh

(( c2m

)2 k

mt

)](24)

where C = A+B and D = AB.

If on the other hand c2 < 4km then the roots are complex and the values of exp(+t) and exp(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

+ = R + iI , = R iI

where the real and imaginary parts R and I are

R = c2m , I =

km( c

2m

)2.

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

x(t) = Aexp(+t)+Bexp(t)= exp(Rt) [(AR + iAI)exp(iIt)+(BR + iBI)exp(iIt)]= exp(Rt) [(AR +BR)cos(It) (AIBI)sin(It)

+i((AI +BI)cos(It)+(ARBR)sin(I t))]

= exp( c2m

t)

[E cos

(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 constants E = 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 is applicable, i.e.when c2 > 4km, is just a decay towards zero. No oscillations can be seen, and we refer to the system

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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, when c2 < 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 ratio c2/4km. For c2/4km > 1 the system is over-damped and for c2/4km < 1 the systemis under-damped. For c2/4km = 1 the system is said to be critically damped. We wont 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 the method 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 redo all 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 the natural frequency 0 of the system. For the

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

0 =

km. (27)

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

2 = c2

4km . (28)As we have seen in the preceding section, > 1 gives over-damping, and < 1 gives under-damping.We can rearrange equation (28) to give a useful alternative expression for the damping ratio:

= c2

km=

c

2m0. (29)

We can now write equation (26) in terms of and 0 instead of k, m and c. Dividing through by mwe have

x+c

mx+

km

= 0 . (30)and from the relations for 0 and

km

= 20 ,c

m= 20 , (31)

we find that this can be written as

x+20x+20x = 0 . (32)By writing our equations in this canonical 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 d and is given by

d =

km( c

2m

)2(33)

We can express this in terms of 0 and :

d = 0

12 . (34)This gives us the result that if damping is very small then d 0. We can also write the solution,equation (25), using these new parameters:

x(t) = exp(0t)[E cos(dt)+F sin(dt)] . (35)Just as we did with the undamped system, we can rewrite the sum of the sine and cosine term as asine term with a phase shift. We get an equivalent formula to the above:

x(t) = H exp(0t)sin(dt +) . (36)

15

The constants H and can be determined from the initial conditions.

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

= 2pi12 (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 the vibrating systemis actually quite straightforward. If the damping is larger that a critical amount, given by = 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 parameters 0 and , and from these deducewhether the system is under or over-damped, the frequency of vibration and 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 the rotation 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 mass m can rotatearound the axis A and is attached to the walls by a spring of constant k and a damper with dampingconstant c. The spring constant is such that the block is at rest in the position shown.

Lets see what happens when we rotate the block to the right by a small angle . The top right cornerof the block (where the damper is attached) moves to the right by l sin and the bottom right cornerwhere the spring is attached moves downward by the same amount. This leads to a 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 use linearized relations by noting that, forsmall angles , sin . Hence the spring and the damper are compressed by x = l. The velocityof compression of the damper is obtained by taking the time derivative, x = l .

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.

rotates around the axis A, so we have to consider the sum of all moments with respect to this axis.The spring and damper forces are kl and cl , and their moments with respect to A are kl2 andcl2 . In addition we have to consider the effect of inertia. This leads to a moment IA where IA isthe moment of inertia of the block with respect to the axis A and = 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

IA + cl2 + kl2 = 0 . (38)

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

m+(3c/2)+(3k/2) = 0 . (39)

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

17

A

f

mf

&c l

fk l

f

f

f

f

&&

&

Jk lc l2

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

+ 3c2m

+ 3k2m

= 0 . (40)

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

+20+20 = 0 . (41)By comparing coefficients, we find

20 =3k2m

, 20 =3c2m

, (42)

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

0 =

3k2m

, =

3c28km . (43)

We can now use the general solution of the canonical vibration equation by inserting 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 familiar with the workings ofthe complex exponential method, and make sure you can do the questions Ive set on it. Once youare happy with the method, youll probably find that the material that follows is not too complicated.

2.4.1 Steady-state response of a periodically forced system

Consider the system shown in Figure 13. We assume that the forcing f (t) is a periodic function, sof (t) = f0 cos(t). The equation of motion for the system shown in Figure 13 is

mx+ cx+ kx = f (t) = f0 cos(t) . (44)

Before we solve this equation we consider the simplest case where the applied force is a constant( = 0). A constant force leads to a constant elongation of the spring by x0 = f0/k. We call x0 thestatic response of the system.

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

x+20x+20x =f0m

cos(t) . (45)

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

x+20x+20x = 20x0 cos(t) . (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, Im interested in how the system responds after quite sometime so that these initial effects will have died away. Thats what we call the steady-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 involves x mul-tiplied by various constants and differentiated. The solutions to this sort of equation are things likex exp(t), where may be complex. The right-hand side (RHS) is a cosine, but I can write thatalso in complex exponential form

x+20x+20x = 20x0 exp(it) (47)

and regard x 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 if x = xR + ixI is a solutionthen xR is a solution to the real part of the equation and xI is a solution to the imaginary part. I cantherefore go ahead and solve for the complex variable x and take the real part of x 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(it) , (48)

where x and X are complex. Substituting this expression gives

X =x020

2 +2i0+20. (49)

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

A+ iB = r exp(i) (50)

where r2 = A2 +B2 and tan = B/A. When we write X in equation (49) in this form, we get

X =x0(

12/20)2

+422/20exp(i) , (51)

where =arctan

(2/0

12/20

). (52)

20

We now insert insert X into the solution (48):

x(t) =x0(

12/20)2

+422/20exp(i[t +]) , (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(t) = Re[exp(it)] . (54)

and findxR =

x0[12/20

]2+422/20

cos(t +) . (55)

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

x = Acos(t +) , (56)

where the amplitude of the vibration is

A =x0(

12/20)2

+422/20(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,

=arctan(

2/012/20

). (58)

The phase lag depends on the frequency of forcing. For small , is small. For = 0, =pi/2, and for , pi.Exercise: 8What does a phase shift of pi radians mean physically?

Exercise: 9Write down the values of for = 0.1 and = 0,0 + ,0 , where is avery small number. Verify that the values of the phase shift for small are 0,for = 0, =pi/2, and for , pi.When you use your pocket calculator to get the phase shift , you must be careful when taking theinverse tangent of a negative quantity. The inverse tangent is only defined up to a factor of npi wheren is an integer number, and you may have to subtract pi to get the phase shift right.

21

We now look how the amplitude behaves as a function of . Take the expression for A,

A =x0(

12/20)2

+422/20(59)

and consider the case when the damping ratio 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 when = 0. 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 ratio we can often use the approximate expression

A =x0

|12/20|. (60)

This expression shows that the resonance peak is located at approximately 0 (this follows fromthe small damping ratio assumed in equation (60)). The amplitude of the dynamic response atresonance for small is 1/(2).

Exercise: 10Calculate the frequency and amplitude of resonance (the frequency and amplitudewhere the response has its maximum) for the case where is not small.Most of what weve been dealing with here can be neatly expressed in two graphs; 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 = /0 , (61)and so from equation (59) we have

y =1

(1 x2)2 +42x2 (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 of 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 amplitude f0 of a periodic forcing acting on themass m 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(t +) cAsin(t +) (64)This is a periodic function, and at the moment we are interested only in the amplitude. If we shiftthe time axis according to t = t/ we find that

fs(t ) = fs,0 sin(t +) (65)where tan =k/(c) and fs,0 =

k2A2 + c22A2. Hence the amplitude of the transmitted force

is

fs,0 = A

k2 +2c2 = x0

k2 +2c2(12/20

)2+422/20

(66)

and using the relations x0 = f0/k, c2 = 4km2 and 20 = k/m this can be written as

fs,0 = f0 1+422/20(

12/20)2

+422/20= f0TR (67)

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

TR =

1+422/20(12/20

)2+422/20

. (68)

Exercise: 12Sketch the curves of TR plotted against x = /0 for x = 0 . . .5 in intervals of 0.5.Use values of = 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 make 0 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 to y(t) = y0 cos(t) 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 involving zy on the right-hand side and bring the equation into canonical formby dividing through m

x+20x+20x =20y0 sin(t)+20y0 cos(t). (70)We now bring also the right-hand side into canonical form. To this end, we write

20y0 sin(t)+20y0 cos(t)= 20y0[cos(t) (2/0)sin(t)]= 20[y0

1+422/20]sin(t +) . (71)

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

x+20x+20x = 20x0 cos(t ) (72)with x0 = y0

1+422/20 . We can now use the solution (56), x(t) = Acos(t + ) with the

amplitude A given by (59). As a result we find that the ratio of the vibration amplitudes of the 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 parameters 0and 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 acceleration of 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 variable z = x y. We can differentiate to find

z = x y , (75)z = x y , (76)

and then eliminate x from the equation of motion in favour of z. We get

mz+ cz+ kz =my , (77)

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

y(t) = Y cos(t) (78)

25

and soy =2Y cos(t) , (79)

which we can use in our equation of motion to give

mz+ cz+ kz = m2Y cos(t) , (80)

which is in the form which we are used to solving, with the amplitude of the forcing given by m2Y .Dividing through by m we have

z+20z+20z = 2Y cos(t) , (81)

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

z+20z+20z = 2z0 cos(t) , (82)

By comparison, we find that the static response z0 is

z0 = Y 2/20 (83)

and the amplitude of the dynamic response is

Z =z0

(1 (/0)2)2 +42(/0)2=

Y (/0)2(1 (/0)2)2 +42(/0)2

. (84)

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

ZY

=(/0)2

(1 (/0)2)2 +42(/0)2. (85)

If we make the free vibration frequency 0 very high by a suitable choice of k and m, then /0will be small and

ZY

=2

20(86)

to a good approximation. Therefore we can write

2Y = 20Z . (87)

Now, the surface is moving sinusoidally according to

y = Y cos(t) (88)

and so the acceleration of the surface is

y =2Y cos(t) (89)

and the amplitude of this acceleration isA = 2Y (90)

26

Therefore,A = 20Z . (91)

Since we can measure the amplitude Z and we already know the frequency 0, we can measure theacceleration amplitude A. 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 making /0 as small as possible. To doeven better, we retain an extra term of order 2/20:

ZY

=(/0)2

(1 (/0)2)2 +42(/0)2

2

20[1 (221)(/20)] . (92)

This time we have kept extra terms to which is why our original result (86) is not exact. These termsbecome significant when 2/20 is not so small. However, we can get rid of them! If we set =1/

2 then the above equation reduces to equation (86). We can therefore expect our accelerometerto be more accurate if we apply damping so that 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 Thompsons book if youre 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 look at are the vibrationsof a machine which are caused by a rotor which is out of balance. Figure 16 shows a machine ofmass M which contains a rotor of mass m with an eccentricity of e. The machine is mounted ona solid floor by a spring of stiffness k and a damper with damping constant c and can move in thevertical (x) direction only.

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

(m+M)x+ cx+ kx = me2 cos(t) . (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. The centre of mass of therotor is located at G.

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

x+20x+20x = 20x0 cos(t) . (94)

where now

0 =

k

m+M, =

c2

4k(m+M) , x0 =me

m+M2

20. (95)

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

X =me

m+M2/20

(12/20)2 +422/20(96)

and the phase is given bytan = 2/0

12/20. (97)

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

0 X 0 0 , = 0 X = me/[2(m+M)] =pi/2 , 0 X me/(m+M) pi . (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 which mM. In this case, themaximum amplitude is X = e/. 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 must be 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 distance x from bottom dead centre attached by a con rod AB toa crank shaft BC which rotates about C. The length of the shaft is r 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 therefore r + l x as shown.

To find the inertial force due to the motion of the piston, con rod and crank shaft, we will seek torelate x to , the angle through which the crank shaft has rotated. If the engine runs at a constantspeed then , the angular velocity of the crankshaft, is a constant .

The problem is that because of the geometry the motion x(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

ABcos+BC cos = AC (99)and putting in the various terms gives us

l cos+ r cos = l + r x (100)From the sine law, we have

l sin = r sin (101)therefore we can eliminate because

cos =

1 sin2 =

1 r2 sin2

l2 . (102)

hence

x = r + l r cos l

1 r2 sin2

l2 . (103)

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

x = r

(1+

1 cos

1

12 sin2 )

. (104)

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

12 sin2 1 2 sin2

2, (105)

so we can writex = r

(1 cos+ sin

2 2

). (106)

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

sin(t)cos(t) = 12

sin(2t) . (107)

30

Taking the derivative of x(t) gives

x = r[sin(t)+sin(t)cos(t)] = r[sin(t)+ 2

sin(2t)] , (108)and

x = r2[cos(t)+cos(2t)] . (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 and m2, 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 masses m1 and m2 so that m1 +m2 = MCR, where MCR is themass of the con rod, and m1a = 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 labelled m3 causes the same centrifugal force as the rotation of the mass m2 from thecon rod. These forces balance, but the remaining inertial force M1x along the axis of motion of thepiston, where M1 = MP + m1 and MP is the mass of the piston, causes a force F 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 due to 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)r2[cos(t)+cos(2t)] (110)

31

M 1 x

m 2 r W 2

m 2 r W 2m 3

M p m 1:

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

The term M1r2 cos(t) is called the primary unbalance force and the term M1r2cos(2t) isthe secondary unbalance force. To understand the origin of the secondary unbalance force, it isuseful to have a look at Figure 21. The motion x() 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

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) for = 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 at angles of 0, 180, 180 and 0degrees.

Figure 22 illustrates a four cylinder engine where the pistons are connected to the crankshaft atangles of 0, pi, pi 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 see how this applies to thefour cylinder engine.

Suppose that there are N cylinders in a line, and that the i-th crank is attached at an angle of i.After time t, the shaft rotates through an angle t and the crank is at an angle t +i to the vertical.The force due to the piston and con rod is

Fi = M1r2[cos(t +i)+cos2(t +i)] (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 frequency , and a secondaryforce oscillating at 2. By adding the contributions from all cylinders, we find that the total primary

33

unbalance force FP is given by

FP = M1r2 i

cos(t +i) = M1r2[cost i

cosi sint i

sini] (113)

and the total secondary unbalance force is

FS = M1r2i

cos2(t +i) = M1r2[cos2t i

cos2i sin2t i

sin2i] . (114)

We also want to know the moments acting on the engine. If we take moments about a point O (thecenter of gravity of the engine), and each crank i is a vertical distance zi from O, the primary andsecondary moments are

MP = M1r2 i

zi cos(t +i) = M1r2[cost i

zi cosi sint i

zi sini] (115)

and

MS = M1r2i

zi cos2(t +i) = M1r2[cos2t i

zi cos2i sin2t i

zi sin2i] . (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 and z4 = 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 i zi sini cosi sin2i cos2i1 0 -b 0 1 0 12 pi -a 0 -1 0 13 pi a 0 -1 0 14 0 b 0 1 0 1

total 0 0 0 4

crank i zi zi sini zi cosi zi sin2i zi cos2i1 0 -b 0 -b 0 -b2 pi -a 0 a 0 -a3 pi 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 = 4M1r2 . (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.

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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 eFigure 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 by which we will take to be small.This means that the force lasts only briefly, but is large ( 1/). Such a shock force can be envisagedas a heavy kick. The strength of the kick can be characterized by its impulse which is defined as theintegral

I =

0f (t)dt . (118)

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

Its 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.

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All we need to do is to recognize that Newtons second law can be written as

Fm =dpdt (119)

where p = mv is the momentum, and Fm is the force acting on the mass m. 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, so p(0) = 0 and the momentum of the mass mjust after the shock is given by

p(t = ) = mv() = I , (121)and so

v(t = ) = I/m . (122)Now, if is very small (the kick is very short), we can consider the motion of the system to startat time 0 with the velocity v = I/m, and since the forcing is zero after time we need only considerthe motion of the freely vibrating system. We found above that the solution for a freely vibratingsystem is

x(t) = exp(0t)[Acos(dt)+Bsin(dt)] (123)and if we set the initial condition x(0) = 0, it follows that A = 0, i.e.,

x(t) = Bexp(0t)sin(dt) . (124)

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

x(t) = Bexp(0t)[0sin(dt)+d cos(dt)] , (125)

so

x(0) = Bd = I/m , (126)hence

B =I

md, (127)

and the solution we require is

x(t) =I

mdexp(0t)sin(dt) . (128)

This expression gives us the response of a system to an impulse of size I delivered at time t = 0. Itis usual to define the unit impulse response function h(t) to be

h(t) = 1md

exp(0t)sin(dt) , (129)

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and so the response to an impulse I 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 of T0 = 2pi/0 and ifthe time over which a force acts is much smaller than T0 we can usually treat it as a shock and writedown the response using the unit impulse response function h(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 eFigure 24: A general non-periodic forcing.

We can imagine that the forcing F() is split up into lots of pieces defined over intervals of time d(note that for clarity Im using the variable to denote time rather than t - the reason will becomeclear later). Each of these can be thought of as a little impulsive shock. If we consider the responseat time t, 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.

Lets consider the response at time t due to the shock shown at time . The impulse due to this shockis

I = F()d (131)Now, the effect of this shock at time t depends on the time t that is elapsed after the shock. It isgiven by

x = h(t )I = h(t )F()d , (132)where Ive used the subscript to denote that this is the part of the response due to the shock at time

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

x(t) = t

0xd =

t0

F()h(t )d , (133)

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

x(t) = t

0

F()md

exp(0(t ))sin(d(t ))d (134)

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

x(t) = t

0

F()m0

sin(0(t ))d (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 x0 is plotted. The peak response is the highest maximum of the response to the shock,which is also sometimes called the maximax. The peak static response is just the response thatwould be caused by a force equal to the peak value of the forcing applied statically. On the x-axis,the frequency of free vibration divided by the frequency scale of the forcing is plotted. For a forceof duration Ts, the frequency scale is simply 2pi/Ts. Therefore, if the peak value of the shock forceis F0 and the system has a mass m, the peak static response is

F0/k = F0/(m20) . (136)

If the shock force lasts for a time Ts then the frequency of free vibration of the system divided bythe frequency scale of the forcing is 0Ts/(2pi). On the shock response spectrum, we plot xm20/F0against 0Ts/(2pi) .

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

F() ={

F0 , 0 Ts0 else . (137)

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

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0.0 0.5 1.0 1.5 2.00

1

2

Pea

k a

mpl

itude

[F0/(

m

02)]

Shock duration [2pi/0]

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

m0

t0

sin(0(t ))d = F0m20

[1 cos(0t)] . (138)

The maximum of the response is at tmax = (2nmax1)pi/0 with the peak value x = 2F0/m20. nmaxis a positive integer number. Hence, this maximum is reached only if Ts > pi/0.

(ii) For t > Ts we find

x(t) =F0

m0

Ts0

sin(0(t ))d = F0m20

[cos(0(tTs)) cos(0t)] . (139)

The maxima are in this case given by

x =2F0m20

|sin(0Ts/2)| . (140)

The shock response spectrum (SRS) is plotted in Figure 25. The primary SRS shows the absolutemaximum of the response which occurs during the time of the shock, and the residual SRS showsthe 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.

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

x = sin(0Ts/2) < 1/2 . (141)

Hence, shock damping is achieved if the pulse is sufficiently short such that 0Ts < pi/3, i.e. theratio between the shock duration and the period of free vibration must be Ts/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 if 0/