Dynamics Lecture Notes

8/11/2019 Dynamics Lecture Notes

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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 Forced vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.4.1 Periodic forcing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

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

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

2.4.4 Rotating out of balance rotor . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.4.5 Non-periodic forcing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.4.6 Shock damping and shock response spectrum . . . . . . . . . . . . . . . . 30

3 The theory of systems with many degrees of freedom 33

3.1 Free vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3.2 Eigenvalue problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.3 Forced vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.4 Orthogonal modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

4 Applications 42

4.1 Accelerometer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

4.2 Shaft whirling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

4.3 Beating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

4.4 Anti-resonance and vibration absorbers . . . . . . . . . . . . . . . . . . . . . . . 52

4.5 Out of balance in reciprocating internal combustion engines . . . . . . . . . . . . . 53

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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 this

knowledge. You will need to draw on the material you studied both in dynamics, mathematics and

solid 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 second

order ordinary differential equations with constant coefficients in second year maths. We will make

a lot of use of the results, so revise them! I will be going over some of the material in the first few

lectures, but it will only be to refresh your memory and to define my own notations, so dont rely on

learning it for the first time here. As a test, before you start the course you should be able to solve

x + 5 x + 4x=exp(3t) (1)

with initial conditions

x(0) =0 , x(0) =1 . (2)

Remember you will need to find the two complementary functions and the particular integral and

then 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 eigenvalues

and eigenvectors. To deal with the problems in tutorial and exam questions, you also should recall

what 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 lecture

breakdown in the syllabus, available from the dynamics web page. We will study the theory of

vibration of single and multiple degree of freedom systems with and without forcing. With this

theoretical 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 freely

vibrating SDF system with and without damping. The equation of motion for the system follows

from the free body diagram. It is a second order, linear, homogeneous ordinary differential equation

with 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 parameters

of 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 all

SDF systems, free or forced and damped or undamped, and discuss the different properties of each

system and the methods used to analyse them.

The second part of the module will address multiple degree of freedom (MDF) systems. We will

see that the problem is similar to an SDF system but with some added complications. We find that

the mathematical analysis is similar in SDF and MDF systems if we replace the mass and stiffness

constants with matrices. We will look at how to obtain solutions of the matrix equations and what

these solutions tell us about the behaviour of the system. We will see that in many cases we can

make a coordinate transformation that will allow us to solve the MDF system as a series of SDF

systems. The equations for MDF systems are often very hard to solve. In practice they are usually

solved 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. We

will find in most instances that we must simplify the true situation but that the answers we get from

the 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 analysing

vibrating systems and we will spend some time using MATLAB to explore how the methods work

and 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 can

analyse many interesting systems that are important in engineering. We will also discuss what we

have 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 these

notes 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, and

not only because they are incomplete. The notes complement the lectures and relieve you of the

tedious, 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 the

basic ideas of freely-vibrating systems. In particular, you ought to know the meaning of period and

frequency 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 diagrams

correctly. 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 a

free body diagram has been constructed, the equation of motion can often be written down. When

constructing the free body diagram, it is important to bear in mind the direction of each force in the

system.

Figure 1: An illustrative two-degree of freedom system. The masses are free to move in the hori-

zontal direction only. When x1=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 the two objects with

massesm1 and m2are free to move in the horizontal direction. However, they are not free to move

in the vertical direction, an example of a constraint. The masses can move independently, so two

coordinates 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 where

k is its stiffness. From this information, we can draw the free body diagram for the system. As a

reference configuration we always chose the static equilibrium configuration of the system. Figure 2

shows the system after the masses have been moved by (arbitrary) distances ofx1and x2out of this

configuration. 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 whenx2is positive and

so the force must be directed to the left. We choose to show this by drawing the arrow to the right

and including a minus sign. Equivalently, we could have kept the force as positive and directed the

arrow to the left.

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Figure 2: Distances when the masses have moved.

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 effect

of the inertia of the mass. Ifx1 is positive, as it is shown in Figure 2, then the mass m1 has been

moved to the right. If we accelerate a mass in the direction ofx1 to the right, its inertia acts to

oppose this and so the inertial force is directed to the left. The same argument applies to the mass

m2. Hence, the direction of the inertial forces must always be taken opposite to the direction of the

respective coordinate. The equations of motion for the two mass system can now be written down

using DAlemberts principle that the sum of the forces on each body must be zero in dynamic

equilibrium. We obtain one equation for each mass, and they are given by

m1 x1+ k1x1 k2(x2x1) =0 ,m2 x2+ k2(x2x1) + k3x2=0 . (3)

Exercise: 1

It is a useful exercise at this point to redraw the system of Figure 1 with a differentcoordinate system. Label the system with coordinatesx3 and x4 so that x3 is just

the same as x1 but x4 runs in the opposite direction to x2. Draw the free body

diagram and deduce the equations of motion. When you have done this, substitute

x3= 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 make

it 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 to

figure 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 will

also work for negative displacements. Therefore, when setting up a FBD, just specify a direction for

the displacement of a body (and stick to it!), and draw in the correctly directed spring force which

results from this displacement. If in reality the body moves the other way, it doesnt matter because

the 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 modelled

well by single degree of freedom (SDF) systems. Other more complicated systems will need extra

degrees of freedom, but we will discover that many of the ideas and techniques we develop for

SDF systems will be easily extendible to many degree of freedom (MDF) systems, although the

mathematical 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 thinking

about 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 absence

of gravity. This appears to be a very simple dynamic system to study, and we might draw a diagram

like that shown in Figure 4.

Remember there is no gravity so the equilibrium position of the mass is at the unextended length

of the string. If we look at the details, the dynamic behavior of this system may contain many

complexities: The mass of the object supported by the spring is in reality distributed over the object

rather than concentrated at a point. The stiffness of the spring is similarly spatially extended. The

spring has a finite mass and its response is unlikely to be exactly linear. However, if we disregard

these complexities and treat the system as a point mass supported by a massless spring, the behavior

of 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 that

have only small effects.

The process whereby we consider the mass to be concentrated at a point rather than distributed

in space is called lumping and we refer to the model as a lumped parametermodel. Lumping is

a very powerful idea because it greatly simplifies the mathematical treatment of the system while

managing to retain the essentials. Masses and springs can both be lumped, i.e. replaced by a single

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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 real

physical situation. We know that the mass is not concentrated at a point in reality, but that we can

model 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 the

normal way by drawing the free body diagram, deriving the equation of motion and finding solutions

to this equation. We will also look at the energy of the system and see how this too can be used to

derive 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 as

shown in Figure 5.

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

Now we can write down the equation of motion

m x + kx=0 . (4)

Exercise: 2Consider the same system but with the effects of gravity included. Draw the free

body diagram and write down the equation of motion. Show that by appropriately

shifting the x coordinate one obtains the same equation of motion (4) as in the

gravity-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 because

the inertial term involves a second order derivative. The equation is linear, since the coefficients do

not depend on x, and it has constant coefficients because m and kdo not depend on t. It is also

homogeneous because there is no term independent ofx. In full then, the equation is a second order

homogeneous ordinary differential equation with constant coefficients.

To solve this equation, we need to find two complementary functions. We will not need a particular

integral because the equation is homogeneous. The equation is second order, so we expect to obtain

two complementary functions. The sum of these two functions is the general solution. We can see

by 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 kconstants areaccounted for correctly. The value needed to do this is=

k/m.

Exercise: 3

Verify the value given for . To do this, consider sin(t) first. Compute thesecond derivative with respect to time and substitute into equation (4). From the

resulting equation you can calculate the relation between and kand m. Repeat

the calculation for the cosine solution.

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

x=

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. We

can write the solution as

x= Csin(t+) . (7)

where the constants are given by C=

A2 +B2 and tan=B/A.

Exercise: 4

Show this is the case. Hint: start with x = Csin(t)and expand the sine using theformula sin(X+ Y) =sin(X) cos(Y) + sin(Y) cos(X).

From this expression we can see more clearly what is happening. The constant Cis the amplitude

of 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 solution

looks 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 first

and find=5 rad/s. The motion of the mass is therefore given by

x= Csin(5t/s +) . (8)

From the initial conditions, we know thatx(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]Ccos(5t/s +) (9)and we now know=0 so at timet= 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 isx(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

displacement[mm]

time[s]

Figure 6: The displacement of the mass in the system of Figure 4 as a function of time. The

parameters and initial conditions are given in the text below.

2.2.2 Energy approach

We can tackle the same problem discussed in Section 2.2.1 using an alternative approach based on

the energy of the system. We will find that we can actually derive the equation of motion once we

know the energy of the system, so all the results from the previous section can be re-obtained. The

advantage is that we dont need to consider the forces in the problem, which in some cases can

be difficult. On the other hand, the energy approach as discussed here works only if there is no

damping.

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

by a lengthx is

W=

Fds=kx22

. (12)

The energy stored in the spring, its potential energy, is the negative of the work done on the spring

and we will call this U(x). We have

U(x) =kx2

2 . (13)

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

T=m x2

2 , (14)

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

E=m x2 + kx2

2 . (15)

The principle of conservation of energy tells us thatEmust be the same at all times, so dE/dt= 0.We can use this information now to deduce the equation of motion. Differentiating the expression

forEwith respect to time gives us

E= m x x + kxx=0 . (16)

We can factor this equation into

x(m x + kx) =0. (17)

so either x=0, which tells us that the system may be at rest, or m x + kx=0 which is the equationof motion derived above. The situation in which the system never moves is not of great interest to

us in a course on dynamics. The other result is very useful, however, because it shows us how to

obtain an equation of motion for a system once we know the total energy of the system. Note that

we have assumed that the energy in the system is conserved. In some cases, friction and other forces

dissipate energy to the surrounding. In these cases, the energy of the system decreases in the course

of 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 in

sliding parts or by viscous effects in a fluid, for example. The mathematical specification of a

damping force can be quite tricky because of the different physical mechanisms that can cause

the damping. In all cases, however, the effects are similar: energy in the system is removed and

dissipated to the surroundings, often in the form of heat or noise. We will be primarily concerned

with viscous damping for which the mathematical specification is relatively simple. Just as with the

masses and springs, we will use lumped dampers to model the effects of damping in a system. This

means that although in reality the damping effects may be distributed throughout the system, in our

model 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 force

proportional 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 the

system is always slowed by the damping, rather than speeded up. Mathematically, we can say thatthe viscous damping force Fis given by F=cv where cis a constant which determines the amountof damping present in the system. Lets look at a simple system acting under the effects of viscous

damping.

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= c xwe can draw the freebody diagram for this system as shown in Figure 8.

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

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Ifc2 >4km both roots given by equation (20) and, hence, also the complementary functions are

real. Inserting the values for+ and into equation (21) gives

x(t) =A exp

c

2m+

c2m

2 k

m

t

+B exp

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

Ccosh

c2m

2 k

mt

+D sinh

c2m

2 k

mt

(24)

where C= (A +B)/2 andD= (AB)/2.

If on the other hand 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 an

over-damped system. The solution exhibits nooscillations.

Figure 10: An example of the solution for an

under-damped system. The envelope of the os-cillations is a decaying exponential.

as being over-damped. On the other hand, whenc2 1 the system is over-damped and for c2/4km


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

0=

k

m. (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 is

defined 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:

=

c

2km = c

2m0 . (29)

We can now write equation (26) in terms of and 0instead ofk,m and c. Dividing through bym

we have

x + c

mx +

k

m=0 . (30)

and from the relations for0and

k

m=20 ,

c

m=20 , (31)

we find that this can be written as

x + 20 x +2

0x=0 . (32)

By writing our equations in this canonical form, we can see at once the value of the damping

parameter and the undamped natural frequency. The true frequency of vibration is the coefficient of

tin the sine and cosine terms in equation (25). It is called the damped frequencydand is given by

d=

k

m c

2m

2(33)

We can express this in terms of0and :

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)[Ecos(dt) + Fsin(dt)] . (35)Just as we did with the undamped system, we can rewrite the sum of the sine and cosine term as a

sine term with a phase shift. We get an equivalent formula to the above:

x(t) =Hexp(0t) sin(dt+) . (36)

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The constantsHand can be determined from the initial conditions.

Another parameter which is often used to characterize damped vibration is the logarithmic decre-

mentof the oscillation. The logarithmic decrement is defined as the natural logarithm of the ratio

of any two successive maxima of the oscillation. It is related to the damping ratio by

= 2

12 (37)

Exercise: 5

Use 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 system

is 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 critical

damping. In this case, the addition of the damping alters the frequency of the oscillations, and

makes their amplitude decay gradually to zero.

To use the canonical form to best advantage, we first write down the equation of motion for a system

using the free body diagram. We can then identify the parameters0and , and from these deduce

whether the system is under or over-damped, the frequency of vibration and the rate of decay of the

oscillations (if present).

2.3.3 Rotational vibration

Until now we have considered systems where masses move along a given fixed direction, i.e. the

motion is translational. However, in many important cases vibration is related to the rotation of

parts. In this case our procedure of setting up the equations of motion must be modified. Figure

11 shows a system which may exhibit rotational vibrations: A square block of mass m can rotate

around the axis A and is attached to the walls by a spring of constant kand a damper with damping

constantc. 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 corner

of the block (where the damper is attached) moves to the right byl sinand the bottom right cornerwhere the spring is attached moves downward by the same amount. This leads to a compression of

the 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, for

small angles, sin . Hence the spring and the damper are compressed byx=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

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Figure 11: A system undergoing rotational vibration.

rotatesaround 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 and

cl2. In addition we have to consider the effect of inertia. This leads to a moment IAwhereIA is

the moment of inertia of the block with respect to the axis A and = is the angular acceleration.

Exercise: 6

Calculate the moment of inertia of the block around the axis A. Assume that the

block 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 insertingIA=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-

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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 divide

by the pre-factor of the term with the highest derivative to obtain

+ 3c

2m+

3k

2m=0 . (40)

We identify this with the general canonical form as given by equation (32) - here we have the angle

instead ofx as the dependent variable, but everything else remains the same:

+ 20+20=0 . (41)

By comparing coefficients, we find

20= 3k

2m, 20=

3c

2m, (42)

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

0= 3k

2m, =3c2

8km. (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 initial

conditions.

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

2.4 Forced vibration

We will now consider the vibration of a system which involves an external force. This will we

call the forcing. We will consider both periodic and non-periodic forcing. We will make use of

the complex exponential method once again. You may have found it possible to work through

the problems so far without using complex exponentials. You will struggle to do this for forced

vibration. Please make sure you are familiar with the workings of the complex exponential method,

and make sure you can do the questions Ive set on it. Once you are happy with the method, youll

probably find that the material that follows is not too complicated.

2.4.1 Periodic forcing

Consider the system shown in Figure 13. The forcing f(t) is quite general; we will first considerthe case where f(t)is a periodic function, so f(t) = f0 cos(t).

Exercise: 8

Draw the free body diagram for Figure 13 and deduce the equation of motion.

The equation of motion for the system shown in Figure 13 is

m x + c x + 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 responseof the system.

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

x + 20 x +20x=

f0

mcos(t) . (45)

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

frequency:

x + 20 x +20x=

20x0 cos(t) . (46)

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 so

the solution will be given by the sum of two complementary functions and a particular integral. The

complementary functions are the solutions of the homogeneous problem and have been discussed in

the previous section, they will both be damped oscillations, so after a reasonable time they will have

decayed to small values. At this stage, Im interested in how the system responds after quite some

time so that these initial effects will have died away. Thats what we call thesteady-state response

of 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 like

x exp(t), where may be complex. The right-hand side (RHS) is a cosine, but I can write thatalso in complex exponential form

x + 20 x +20x=

20x0 exp(it) (47)

and regardx as a complex variable. I can recover the original equation of motion by looking at just

the real part of this equation. Since the equation is linear, I can say that ifx=xR+ ixIis a solution

thenxRis 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 the

calculation. This real part will be a solution to the original equation of motion.

I expect the steady-state response to be

x(t) =Xexp(it) , (48)

wherex and Xare complex. Substituting this expression gives

X= x0

20

2 + 2i0+20. (49)

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

A + iB=rexp(i) (50)

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

X= x0

12/202

+ 422/20

exp(i) , (51)

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where

=arctan 2/012/20 . (52)

We now insert insert Xinto the solution (48):

x(t) = x0

12/202

+ 422/20

exp(i[t+]) , (53)

As I explained above, this is the solution to the complex form of the equation of motion. The

physical solution is simply the real part of it. I use that

cos(t) =Re[exp(it)] . (54)

and find

xR= x012/20

2+ 422/20

cos(t+) . (55)

What does this mean? There are several interesting points to extract from this solution. Firstly, it is

basically another sinusoidal function of the form

x=A cos(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 the

forcing and the response. The system therefore responds to the forcing by vibrating at the same

frequency but with a delay,

=arctan

2/012/20

. (58)

The phase lag depends on the frequency of forcing. For small , is small. For = 0,=/2, and for,.Exercise: 9

What does a phase shift ofradians mean physically?

Exercise: 10Write 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,=/2, and for,.When you use your pocket calculator to get the phase shift, you must be careful when taking the

inverse tangent of a negative quantity. The inverse tangent is only defined up to a factor ofnwhere

nis an integer number, and you may have to subtract to get the phase shift right.

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

A= x012/20

2+ 422/20

(59)

and consider the case when the damping ratio is small. Now, the squared terms inside the square

root are always positive, so since they appear in the denominator, the amplitude is largest when they

are 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 called

resonance. 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 from

the small damping ratio assumed in equation (59)). The amplitude of the dynamic response at

resonance for smallis 1/(2).

Exercise: 11

Calculate the frequency and amplitude of resonance (the frequency and amplitude

where the response has its maximum) for the case whereis 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 amplitude

is probably the most useful form of the frequency-response graph. It tells us a great deal about the

dynamics 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

(1x2)2 + 42x2 (62)

which determines the shape of the frequency-response graph. Because at resonance the response is

usually very large, you will often find these graphs plotted in logarithmic coordinates (i.e. plot lny

against lnx).

Exercise: 12

Use MATLAB to deduce the shape of these graphs for a number of values ofand

add sketches to your notes. Deduce the height and width of the resonance peak andlabel it.

2.4.2 Force transmission and vibration damping

We now consider the following question: Given the amplitude f0of a periodic forcing acting on the

massm in Figure 13, what is the amplitude of the force transmitted to the support?

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

fs(t) =kx(t) + c x(t) (63)

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

fs(t) =kA cos(t+)cA sin(t+) (64)Now we use that this can be written as

fs(t) = fs,0 cos(t+) (65)

where fs,0=

k2A2 + c22A2. Hence the amplitude of the transmitted force is

fs(t) =A

k2 +2c2 = x0

k2 +2c2

12/202

+ 422/20(66)

and using the relationsx0= f0/k,c2 =4km2 and20=k/mthis can be written as

fs(t) = f0

1 + 422/2012/20

2+ 422/20

= f0TR (67)

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

called the transmissibility ratio

TR= 1 + 422/20

12/202

+ 422/20. (68)

Exercise: 13

Sketch the curves of TR plotted againstx= /0for 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 damping

are good because they reduce force transmission. However, at high frequencies the opposite is the

case and the most efficient way to reduce force transmission is to make 0as small as possible (soft

springs, 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.

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Figure 14: A system with moving boundary excitation

We assume that the support of the system moves periodically according to z(t) =z0 cos(t)and askfor the amplitude of the steady-state response of the system.

The equation of motion is

m x + c( x z) + k(xz) =0. (69)We collect the terms involving z on the right-hand side and bring the equation into canonical form

by dividing throughm

x + 20 x +20x=20z0 sin(t) +20z0 cos(t). (70)

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

20z0 sin(t) +20z0 cos(t)=20z0[cos(t) (2/0) sin(t)]=20[z0

1 + 422/20] sin(t+

) . (71)

We finally shift the time axis to get rid of the phase (we are interested in the long-time responseso the choice of the start time really doesnt matter) and find that the equation of motion assumes

the canonical formx + 20 x +

20x=

20x0 cos(t) (72)

with x0 = z0

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

amplitudeA given by (59). As a result we find that the ratio of the vibration amplitudes of the mass

and of the support is given by

A/z0=TR (73)

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with the same transmissibility TR as in the previous section. Hence, equation (68) gives us both

force and amplitude transmission, and what has been said about the influence of the parameters0and on vibration isolation also applies to the present case.

2.4.4 Rotating out of balance rotor

Figure 15: Schematic diagram of a machine with an out of balance rotor. The centre of mass of the

rotor is located at G.

Many machines have rotating parts which can vibrate. The system we will look at are the vibrations

of a machine which are caused by a rotor which is out of balance. Figure 15 shows a machine of

mass Mwhich contains a rotor of mass m with an eccentricity ofe. The machine is mounted on

a solid floor by a spring of stiffness kand a damper with damping constant c and can move in the

vertical (x) direction only.

From the free-body diagram shown in Figure 16, we find that the equation of motion for the machine

and rotor in the x direction is

(m +M) x + c x + kx=me2 cos(t) . (74)

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

x + 20 x +20x=

20x0 cos(t) . (75)

where now

0=

k

m +M, =

c2

4k(m +M) , x0=

me

m +M

2

20. (76)

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Figure 16: Free body diagram for the system of Figure 15.

It follows from the results of Section 2.4.1 that the steady-state solution is given byx(t) =Xcos(t+)where the amplitudeXis given by

X= me

m +M

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

(77)

and the phase is given by

tan=

2/0

12/20

. (78)

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

see that

0 X 0 0 ,=0 X= me/[2(m +M)] =/2 , 0 Xme/(m +M) . (79)

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

Exercise: 14Use the results of the previous sections to determine the amplitude of the force

which is transmitted to the ground. What is the force transmitted at resonance?

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2.4.5 Non-periodic forcing

In certain cases we wish to analyse the response of a system to non-periodic forcing. One example

of this might be a sudden shock. We will consider a shock first of all, and then show that more

complicated forcings can be considered as a amalgamation of shocks one after the other. This will

give us a formula with which we can calculate the response to an arbitrary forcing.

Consider a spring-damper system such as that shown above in Figure 13 two pages above, with the

forcing function shown in Figure 17.

Figure 17: A short-lasting force.

The force shown contains a parameter which we will take to be small. This means that the force

lasts only briefly, but is large - it can be envisaged as a heavy kick.

We define the impulse to be the integral

I=

0 f(t)dt . (80)

What is the response of the system to an impulsive shock like this? Lets be clear about the problem

we 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 to

give the system an initial kick, and from then on it will behave like a freely vibrating system. If we

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can work out the size of this kick, this will allow us to set the initial condition for the subsequent

freely vibrating behaviour.

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

Fm=dp

dt(81)

wherep = mv is the momentum, and Fmis the force acting on the mass m. Therefore, the momentumof the mass is given by

p(t)p(0) = t

0Fmdt . (82)

When the initial kick is very strong and very short, the effect of the springs and dampers during the

kick 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 m

just after the shock is given by

p(t=) =mv() =I , (83)

and so

v(t=) =I/m . (84)

Now, if is very small (the kick is very short), we can consider the motion of the system to start

at time 0 with the velocityv=I/m, and since the forcing is zero after timewe need only considerthe motion of the freely vibrating system. We found above that the solution for a freely vibrating

system is

x(t) =exp(0t)[A cos(dt) +B sin(dt)] (85)

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

x(t) =B exp(0t) sin(dt) . (86)

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

x(t) =B exp(0t)[0sin(dt) +d cos(dt)] , (87)

so

x(0) =Bd=I/m , (88)

hence

B= Imd

, (89)

and the solution we require is

x(t) = I

mdexp(0t) sin(dt) . (90)

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This expression gives us the response of a system to an impulse of size Idelivered at timet= 0. It

is usual to define the unit impulse response functionh(t)to be

h(t) = 1

mdexp(0t) sin(dt) , (91)

and so the response to an impulseIis

x(t) =Ih(t) . (92)

We often find that we can approximate the behaviour of other forces as shocks, even if they last for

quite a long time. The spring-mass system has a time scale for its vibrations ofT0= 2/0 and ifthe time over which a force acts is much smaller thanT0we can usually treat it as a shock and write

down the response using the unit impulse response functionh(t).

Exercise: 17

Sketch the behaviour of a mass-spring-damper system after an impulse is applied

For complicated non-periodic forcing, we can generalise the above method. Consider the forcing

function shown in Figure 18:

Figure 18: A general non-periodic forcing.

We can imagine that the forcing F(u)is split up into lots of pieces defined over intervals of time du(note that for clarity Im using the variable u to denote time rather than t- the reason will become

clear later). Each of these can be thought of as a little impulsive shock. If we consider the response

at time t, we can think of this response as being built up of many responses to these shocks -

remember, we are allowed to build up solutions like this because the equation is linear and so the

super-position principle holds.

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Lets consider the response at time tdue to the shock shown at time u. The impulse due to this

shock isI=F(u)du (93)

At times after the application of a shock of impulse I, we know that the response due to the shock

is

x=h(s)I (94)

Now, the time after the shock is shown by the arrows and is tuso the response at timetdue to theshock at timeu is

xu=h(tu)F(u)du , (95)where Ive used the subscript u to denote that this is the part of the response due to the shock at time

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

x(t) = t

0xudu=

t

0F(u)h(tu)du , (96)

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(u)

mdexp(0(tu)) sin(d(tu))du (97)

which is called the convolution formula. For the case of small or no damping, this expression

simplifies a bit to

x(t) =

t

0

F(u)

m0 sin(0(tu))du (98)which is in general easier to integrate. This I will refer to as the convolution formula without

damping.

2.4.6 Shock damping and shock response spectrum

To assess whether a vibrating system is capable of effectively damping shocks, we consider the

shock response spectrum. This is a plot relating the response to the forcing duration, but plotted

in a clever choice of variables: On the y-axis, the ratio of the peak response x and the peak static

response 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 that

would 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 force

of duration, the frequency scale is simply 2/. Therefore, if the peak value of the shock force isF0and the system has a mass m, the peak static response is

F0/k=F0/(m20) . (99)

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0.0 0.5 1.0 1.5 2.00

1

2

Pe

akamplitude

[F0

/(m0

2)]

Shock duration [2/0]

Figure 19: Shock response spectrum for a rectangular pulse. Full line: Primary SRS; Dashed line:

Residual SRS

If the shock force lasts for a time then the frequency of free vibration of the system divided by

the frequency scale of the forcing is0/(2). On the shock response spectrum, we plot xm20/F0

against0/(2).

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(u) =

F0 , t 0 else .

(100)

The response to this pulse is obtained from equation (98). We consider separately the two cases

t and t>:

(i) Fort

the response is

x(t) = F0

m0

t

0sin(0(tu))du= F0

m20[1 cos(0t)] . (101)

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

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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 of

freedom is straight-forward provided we do it in the right way. What we will find is that if we use

matrices to represent our equations, then the free body diagram equation of motion complexexponential method we have used works just as well for the many degree of freedom systems. The

main complication is the maths, which can get rather lengthy. In fact, we will often have to resort

to computers and numerical methods in order to solve the equations. We will also find there are one

or two extra things going on in these systems that simply do not occur in a single degree of freedom

system.

3.1 Free vibration

We will start our study of multiple degree of freedom systems with a relatively simple example

shown in Figure 20.

Figure 20: A two degree of freedom system consisting of two masses connected by springs.

Exercise: 15

We have discussed in the Introduction how to obtain equations of motion for this

kind of system. Draw a free body diagram and write down the equation of motion

for each mass.

If we write down the equations of motion for the two masses in the system shown and collect similar

terms together we get

mx1+ 2kx1kx2=0 ,

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2mx2kx1+ 2kx2=0 . (105)

I can represent these equations of motion in matrix form:

m 0

0 2m

x1

x2

+

2k kk 2k

x1x2

=0 . (106)

Exercise: 16

Multiply 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

Mx + Kx=0 , (107)

where bold quantities represent matrices and the arrows represent vectors. M and K are the mass

and the stiffness matrices. They are given by

M =

m 0

0 2m

K =

2k kk 2k

(108)

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 exponential

method. I write

x= A exp(it) , (109)

where the usual x and A quantities I get in the single degree of freedom systems have become

vectors. Alternatively, I can write out this equation in long hand as the pair

x1=A1 exp(it) ,

x2=A2 exp(it) . (110)

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

2MA exp(it) + KA exp(it) =0 (111)

Cancelling out the exponentials and collecting terms gives me

(2M + K)A=0 (112)

which I can write as

BA=0 (113)

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with

B =2

M + K (114)Now a general result in the theory of linear algebra that you may be aware of is that for the system

of equationsBA=0 to have nontrivial solutions, it must be true that the determinant ofB must bezero,

|B|=0 . (115)If this is not familiar then take the time to go over your maths notes to refresh you memory. Replac-

ingB, we have deduced at this point in the calculation that

|2M + K|=0 (116)If I now put in the expressions for the matrices I get the determinant 2km2 kk 2k2m2

=0 . (117)Now I can expand this determinant out to get

(2km2)(2k2m2) k2 =0 (118)which rearranges to

2m246km2 + 3k2 =0 , (119)which is a quadratic equation in2. I can solve the quadratic to find two roots. I get

= km32

3

2

. (120)

Negative frequencies dont really mean anything. I therefore have two solutions which I will label

+ and . They are given by

+=

km

3

2+

3

2

, (121)

=

km

3

2

3

2

. (122)

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

i.e. the system will resonate at either of these frequencies. It turns out, as we shall see, that a system

withndegrees of freedom hasnfrequencies of free vibration. It is usual to refer to vibration at these

frequencies as modes of vibration or vibration modes.

Now I have evaluated, I can go back to equation (100) and put in either+or . For each valueI can use the matrix equation to calculate the ratioA1/A2.

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

x(t) =c+

2.731

sin(+t++) + c

0.73

1

sin(t+) , (129)

where themode amplitudes c+and care now real and together with the phases +and must bedetermined from initial conditions. So we need four initial conditions. In general, for a system with

ndegrees of freedom 2ninitial conditions are required.

3.2 Eigenvalue problems

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

Mx + Kx=0 , (130)

and indeed this is always the case, even for many more degrees of freedom. The matrices are square,

and usually symmetric although there are occasions when they are not. Using the substitution

x= A exp(it) , (131)

we can transform the equation to

2MA + KA=0 (132)or rearranging terms

(M1K)A=2A . (133)

Now this last form is called the matrix eigenvalue problem, and it arises from many different phys-

ical situation. Because it occurs so often, it has been studied very thoroughly. Moreover, there are

many computer packages that will quickly solve it for us. Lets consider what the solutions are like

first, then see how a computer can be used to help in the calculations.

If we write

Y = M1K (134)

and2 =then the problem we wish to study is

YA=A (135)

The mathematics tells us that ifY is a n n matrix then there are n values of that allow thisequation to have non-trivial solutions, and each of these has associated with it a vector A. Thevalues of 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 the

matrices so that it is a eigenvalue problem. Using a package such as MATLAB, the solution of this

eigenvalue problem is very easy. In this way, we can tackle systems with quite a large number of

degrees of freedom that would be very hard indeed if we did not have access to a computer.

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3.3 Forced vibration

Just as with the single degree of freedom systems, we would like to be able to calculate what happens

with a forced system. The path from the free body diagram to the equations of motion is no more

complicated than we have met already. Once again, we can write the equations in matrix form so

we end up with something like

Mx + Kx=Fcos(t) , (136)

where the new part is the RHS consisting of a forcing term proportional to cos (t). Many timesonly one component ofFwill 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 single

degree of freedom systems, we introduce a guessed solution of the form

x= Xcos(t) (137)

where you should note that we have been able to use cos (t) directly rather than the complexexponential because there is no damping and so we know that the response is either in phase or in

antiphase with the forcing. Substituting this into the equation of motion gives

2MXcos(t) + KXcos(t) =Fcos(t) . (138)

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

ZX= F (139)

whereZ =2M + K . (140)

The matrix equation (139) 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

larger systems, one uses a formal solution which can be computed numerically. To this end, we

have simply to determine the inverse matrix ofZ:

X= Z1F . (141)

The inverse of a matrix can be calculated using MATLAB. Hence, we may simply use MATLAB to

calculate the inverse matrix Z1 and multiply this with the vectorFwhich contains the amplitudes

of the external forces.

Exercise: 18

Go back to the two spring system we defined above and add a forcing term of the

form F0 cos(t) to the left mass. Use the above method to deduce an expressionfor the frequency response curve.

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3.4 Orthogonal modes

I want to conclude our consideration of the theory of multiple degree of freedom systems with some

more matrix algebra. Make sure your understanding of matrix algebra is up to scratch before we

start. The following considerations are rather formal, but they lead to a surprising result: A vibrating

system withn degrees of freedom can always be transformed in such a manner that it behaves like

n independentsingle degree of freedom systems.

The matrix equation of motion for a n degree of freedom system is

Mx + Kx=0 . (142)

If we introduce the vector of coordinates Asuch that

x= A exp(it) , (143)

then we have

KAi=2i M

Ai , (144)

where we know that there should be n possible values of and Ive acknowledged this in the

equation by labelling these different possible s and the correspondingAs with an index i whichcan run from 1 up ton. Now Im going to prove an important result concerning the vectors . Firstly

I will multiply the above equation by the transpose vector ATj to give me

ATjKAi=2iATjMAi . (145)

It will become clear later just why I want to do this. For the moment, just make sure you understand

what this means - write out example matrices for yourself if you like. I will also make use of exactly

the same equation written out with the i and j subscripts interchanged,

ATi KAj=2jATi MAj . (146)

Now, the theory of matrix algebra shows that

(AB)T = BTAT . (147)

Using this result, we can also prove quite quickly that

(ABC)T

= CT

BT

AT

. (148)

If I take my equation (146) above and transpose it, then I get

(ATi KAj)T =2j (A

Ti M

Aj)T , (149)

and using the above theorem of matrix algebra gives

ATjKAi=2jATjMAi , (150)

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remembering that the transpose operation twice just goes back to the original matrix and that since

Mand Kare symmetric the transpose operation has no effect on them. Subtracting equations (145)and (150) gives me the interesting result

(2j 2i)ATjMAi=0 . (151)

It follows thatATjMAi=0 and A

TjK

Ai=0 ifi = j . (152)These equations describe a property of the vectorsAj called orthogonality with respect to the ma-tricesM and K. This property turns out to be very useful.

For the casei= j, we define byAT

iMA

i=M

ii (153)

andATi KAi=Kii (154)

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 matrix P. We

define it to be the sequence of all the column vectors Ai, i.e.

P = [A1A2 . . .An] , (155)

so thatP is annmatrix formed by the eigenvectorsAi. What is the use of this matrix? Considerthe matrix

P

T

MP = [A1

A2 . . .

An]

T

M[A1

A2 . . .

An] . (156)

Because of the orthogonality relations above, most of the terms when we multiply out this expres-

sion are zero. We are left with

PTMP =

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

(157)

which is a diagonal matrix. The matrix PTKPis also diagonal. The result we have proved is that

the modal matrix diagonalises the mass and stiffness matrices. This is useful! If we consider the

equation of motion

Mx + Kx=F0 exp(it) , (158)

then if we define a new set of variables by x= Pythen

MPy + KPy=F0 cos(t) , (159)

and multiplying this equation by PT gives

PTMPy + PTKPy= PTF0 cos(t) . (160)

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Since the matrices are diagonal, thisuncouplesthe problem: This equation is just nsingle degree of

freedom problemsMii yi+ Kiiyi= [P

TF0]i cos(t) (161)

which we know how to solve.

The above mathematics seems quite complicated, but dont be intimidated. You can try the method

for 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

Collect the equations together in matrix form Set the forcing to zero temporarily and solve the freely vibrating system to find the eigenvalues

(i.e. frequencies of free vibration)

Use the eigenvalues to solve the matrix equation and determine the eigenvectors Construct the modal matrix P from the eigenvectors Use the modal matrix to define a new set of coordinatesyi(this is conceptual - you dont need

to calculate anything here)

Calculate the products PTMPand PTKPand 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= Pyto put the solution back into the original coordinates

Its quite a lengthy procedure, but the steps are quite simple to carry out if you have a computer at

hand - and it does give a general method to solve any system, however complicated, provided you

can calculate the eigenvalues and vectors. In particular, the method is suited to programs such as

MATLAB where you can define and handle matrices easily.

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4 Applications

4.1 Accelerometer

As a first application of the theory of vibration, we consider a device known as an accelerometer.

As the name suggests, it is used to measure the acceleration of a vibrating surface. Essentially,

the device consists simply of a seismic mass, basically a lump of metal, 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 to relate the vibration of the

mass, which we can measure, to the acceleration of the surface. Figure 21 shows a diagram of the

situation.

Figure 21: A schematic diagram of an accelerometer. x measures distance from a fixed point in

space to the mass,y measures distance from a fixed point in space to the vibrating surface.

The equation of motion is

m x + c( x y) + k(xy) =0 . (162)We now introduce the new variablez=xy. We can differentiate to find

z= x y , (163)

z= x y , (164)and then eliminatex from the equation of motion in favour ofz. We get

mz + cz + kz= m y , (165)

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

y(t) = Ycos(t) (166)

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and so

y=2

Ycos(t) , (167)which we can use in our equation of motion to give

mz + cz + kz=m2Ycos(t) , (168)

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

Dividing through bym we have

z + 20z +20z=

2Ycos(t) , (169)

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

z + 20z +

2

0z=

2

z0 cos(t) , (170)

By comparison, we find that the static responsez0is

z0= Y2/20 (171)

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. (172)

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 . (173)

If we make the free vibration frequency0 very high by a suitable choice ofkand m, then/0will be small and

Z

Y=2

20(174)

to a good approximation. Therefore we can write

2Y=20Z . (175)

Now, the surface is moving sinusoidally according to

y= Ycos(t) (176)

and so the acceleration of the surface is

y=2Ycos(t) (177)and the amplitude of this acceleration is

A=2Y (178)

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Therefore,

A=20Z . (179)

Since we can measure the amplitudeZand we already know the frequency 0, we can measure the

acceleration amplitudeA. Since we are far below the free vibration frequency, our measured signal

z(t) will be in phase with the term on the right-hand side of Eq. (169) and therefore in antiphasewith the acceleration.

A further degree of sophistication is to add damping to the accelerometer in order to improve its

accuracy. The whole idea in going from Eq. (173) to (174) is that the term under the square root

should 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 order2/20:

ZY

= (/0)2

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

2

20[1 (221)(/20)] . (180)

This time we have kept extra terms to which is why our original result (174) is not exact. These

terms become significant when 2/20 is not so small. However, we can get rid of them! If weset = 1/

2 then the above equation reduces to equation (174). We can therefore expect our

accelerometer to be more accurate if we apply damping so that 0.7.

Many accelerometers work on this principle. Low frequency accelerometers use a damping ratio

of 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 a

piezoelectric crystal which has a very high natural frequency and therefore there is no need for any

damping (can you explain why?). These accelerometers are more suited to high frequency work.

4.2 Shaft whirling

The next system we will look at is the vibrations of a rotating out-of-balance shaft. For simplicity,

lets consider a massless shaft with a disc of massm in the centre. Figure 22 shows what this looks

like.

Figure 23 shows the situation we will analyse. The origin of the coordinates has been chosen to

be at the equilibrium position of the shaft centre. The shaft centre is attached to the disc at C, so

in equilibrium C rests at O. The centre of mass G is rotated about the shaft centre C at the same

angular velocity as the rotation of the shaft, which is.

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Figure 22: Schematic diagram of a disc rotating on a shaft. The shaft AB meets the disc at C. The

centre of mass of the disc is located at G.

Because the shaft is rotating, C can be displaced from its equilibrium position and will then be

rotated about O. We want to know how far the centre of the shaft C is displaced from its equilibrium

position.

Figure 23: Motion of the shaft centre C and the centre of mass G. G moves around C due to the

rotation of the shaft. C moves around O because of the inertial forces caused by rotating G.

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

24.

Figure 24: Free body diagram for the system of Figure 22

The force acting on C is due to the displacement of the centre of the shaft from its equilibrium

position. If the force for a unit displacement is kthen the force here is kzdirected towards O. The

value ofkis 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 the

forcing in our problem, which causes C to be displaced from O. The other two forces shown acting

on G are the inertial forces due to its motion. Note that we must give two of these since there are

two independent coordinates needed to specify the position of G. G and C are rigidly linked so the

equation of motion can be deduced by summing all the forces in the system. We get two equations,

by resolving in thex and y directions:

m x + kz cos()me2 cos(t) =0 , (181)m y + kz sin()me2 sin(t) =0 . (182)

Nowx=z cos()and y=z sin(), so we can write the two equations of motion

m x + kx=me2 cos(t) , (183)

m y + ky=me2 sin(t) . (184)

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) =e2/20

12/20cos(t) , (185)

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y(t) =e2/20

12/20

sin(t) . (186)

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

z=e2/20

12/20. (187)

where 0 is the natural frequency of free vibration for the shaft. If is small, i.e. the shaft ro-

tates slowly, then k m2 and z= e(/0)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 (=0) theexpression goes to infinity. At high speeds, for whichis large,

ze , (188)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 the

shaft. 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.

4.3 Beating

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

phenomena of beating.

Figure 25: Two weakly coupled pendulums.

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above equations for1and 2, we find C=E=0 andB=D= /2. Please check this for yourself.

The final solution is therefore

1= (/2)[cos(1t) cos(2t)] (201)

and

2= (/2)[cos(1t) + cos(2t)] (202)

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

cosA + cosB=2 cosA +B

2 cos

AB2

cosA

cosB=2sin

A +B

2

sinAB

2so our solution can be written as

1=cos((1+2)t/2) cos((12)t/2) (203)

and

2=sin((++)t/2) sin((+)t/2) (204)Finally we can put in the values we calculated for + and . If the coupling is very weak,kissmall and so

++2

g

l(205)

and+

2 =

1

2

g

l

1

1 +

2ka2

mgl

g

l

ka2

2mgl(206)

If we write

=

g

l

ka2

2mgl, =

g

l, (207)

then the solution can be written as

1=cos(t) cos(t) , (208)

2= sin(t) sin(t) . (209)

A plot of this form is shown in Figure 27. As you can see, the vibration amplitude is periodically

modulated, and the energy of vibration oscillates between the two pendulums with a period of/.This phenomenon is called beating.

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-1.0

-0.5

0.0

0.5

1.0

2/

2/

2

/

-1.0

-0.5

0.0

0.5

1.0

1

/

Figure 27: Plot of the beating behaviour in the coupled pendulum system. The outer line shows the

sinusoidal envelope curve

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Figure 28: Left: a forced oscillator. Right: the same oscillator with an additional spring-mass

system acting as antivibration device.

4.4 Anti-resonance and vibration absorbers

We have seen in previous sections that the vibration of a forced system can be always suppressed if

one makes the natural frequency low enough, i.e. by designing a system with large mass and small

stiffness. 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 system

in Figure 28 (left) with a sinusoidal forceF(t) =F0 cos(t). The equation of motion of this systemis

m1x1+ k1x1=F0 cos(t) (210)and the steady-state response of the system is

x(t) = x0

12/20cos(t) =

F0

k1m12cos(t) . (211)

If the forcing frequency is not much larger than the systems natural frequency 0=

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

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How can we suppress the vibration of the massm1without changing the mass or the stiffness of the

system? We will see that we can deal with this problem by adding another mass-spring subsystem(m2and k2) as shown in Figure 28 (right). This gives a two-degree of freedom system

m1 x1+ k1x1+ k2(x1x2) =F0 cos(t) (212)m2 x2+ k2(x2x1) =0 (213)

or, in matrix form m1 0

0 m2

x1

x2

+

k1+ k2 k2k2 k2

x1x2

=

F00

cos(t) . (214)

We solve this by assuming x1x2

=

X1X2

cos(t) . (215)

whereX1 and X2are the vibration amplitudes of the massesm1and m2. This leads to the system of

equations

(m12 + k1+ k2)X1k2X2 = F0 , (216)k2X1+ (m22 + k2)X2 = 0 . (217)

By solving this system we obtain the vibration amplitudesX1and X2. The result is

X1 = (k2m22)F0

K()

, (218)

X2 = k2F0

K() , (219)

where

K() = (k1+ k2m12)(k2m22) k22 . (220)We see that the amplitude of vibration of the massm1 becomes zero ifk2 and m2 fulfil the relation

k2/m2=2, i.e., if the resonant frequency of the added subsystem matches the excitation frequency.

In particular, we can suppress the resonant response of the original system at = 0=

k1/m1completely by chosingk2/m2=k1/m1.

4.5 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 by

forces acting on the engine and these can cause vibration. In this section we will deduce what these

forces are and what measures can be taken to keep them to a minimum. Figure 29 shows a schematic

diagram of a piston in a cylinder a distance x from bottom dead centre attached by a con rod AB to

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Figure 29: Schematic diagram of a piston, con rod AB and crank BC.

a crank shaft BC which rotates about C. The length of the shaft is rand the length of the con rod isl. The distance between the piston and the crank axis is thereforer+ lxas shown.

To find the inertial force due to the motion of the piston, con rod and crank shaft, we will seek to

relate x to , the angle through which the crank shaft has rotated. If the engine runs at a constant

speed then, the angular velocity of the crankshaft, is a constant.

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

AB cos+BCcos=AC (221)

and putting in the various terms gives us

l cos+ rcos=l + rx (222)

From the sine law, we havesin

l=

sin

r(223)

therefore we can eliminate because

cos=

1 sin2=

1 r

2 sin2

l2 . (224)

hence

x=r+ l rcos l

1 r2 sin2

l2 . (225)

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

x=r

1 +

1

cos 1

12 sin2

. (226)

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In an engine, the value of is usually around 1/3 to 1/4 and so we can expand the square root as a

power 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 , (227)

so we can write

x=r

1 cos+ sin

2

2

. (228)

where I have ignored powers ofhigher than one. This approximation is usually good enough. We

now want to differentiate twice with respect to tin order to find the acceleration. We use the chain

rule sod

dt

sin= cos=cos , (229)

whereis the constant angular velocity described above. Carrying out this procedure onx gives

x=r(sin+sincos)=r(sin+

2sin2) , (230)

and

x=r(cos+cos2)2 . (231)

Now we have calculated the acceleration of the piston, let us consider the forces acting in the system

of Figure 29. To make the analysis easier, we will model the con rod in an approximate way. The

true con rod shape is rather complicated, and we will replace it in our lumped model with two

masses,m1and m2, one at each end of the real rod. Figure 31 shows the con rod and its model.

Figure 30: Schematic diagram of the con rod and its model. G is the centre of gravity.

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We choose the lengths a and b and the massesm1and m2so thatm1+ m2=MCR, whereMCRis the

mass of the con rod, andm1a=m2bso the the centre of gravity of the con rod and the model of thecon rod are the same. Figure 30 shows the forces acting on the system with the model for the con

rod inserted. In order to balance the system, the size and shape of the crank is chosen so the rotation

of the mass labelled m3 causes the same centrifugal force as the rotation of the mass m2 from the

con rod. These forces balance, but the remaining horizontal component of the inertial forceM1 x,

whereM1=MP + m1andMPis the mass of the piston, causes a force Hon the crankshaft and henceon the engine.

Figure 31: Forces acting on the piston, con rod and crank.

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 piston

and the con rod. The crankshaft shape is chosen to eliminate the non-horizontal component of the

con rod inertial force and the resulting force on the engine is given by

H= (MP+ m1) x= (M1)r(cos+cos2)2 (232)

The termM1r2 cosis called the primary unbalance force and the termM1r

2cos2is the sec-

ondary unbalance force. To understand the origin of the secondary unbalance force, it is useful

to have a look at Figure 32. The motionx() of the piston is not strictly sinusoidal, and the dif-ference with a pure cosine can be well approximated by a cosine with twice the frequency. The

corresponding acceleration gives rise to the secondary unbalance force.

We now look at the assembly of cylinder in an engine, and the resultant forces which act on the body

of the engine. The goal will be to devise an engine that is balanced, so the unbalance forces from

each cylinder cancel out. We will consider a four cylinder inline arrangement. Other arrangements

can be analysed by the same method.

Figure 32 illustrates a four cylinder engine where the pistons are connected to the crankshaft at

angles of 0,,and 0 radians.

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