chapter 2 2dof systems fall 2013

8/14/2019 Chapter 2 2DOF Systems Fall 2013

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CHAPTER 2: TWO DEGREE OF FREEDOM SYSTEMS

The number of degrees of freedom (DOF) of a system is the number of independent coordinates

necessary to define motion. Also, the number of DOF is equal to the number of masses multiplied

by the number of independent ways each mass can move. Consider the 2 DOF system shown below.

From Newtons law, the equations of motion are:

1 1 1 1 1 1 2 2 1 2 2 1 1

2 2 3 2 3 2 2 2 1 2 2 1 2

m x k x c x k x x c x x F

m x k x c x k x x c x x F

(1)

Rearranging gives:

1 1 1 2 1 1 2 1 2 2 2 2 1

2 2 2 3 2 2 3 2 2 1 2 1 2

m x c c x k k x c x k x F

m x c c x k k x c x k x F

(2)

These equations can be written in matrix form:

1 2 2 1 2 21 1 1 1 1

2 2 3 2 2 32 2 2 2 2

0

0

c c c k k k m x x x F

c c c k k k m x x x F

(3)

Defining:

1 1

2 2

1 2 2 1 2 21

2 2 3 2 2 32

0

0

x Fx fx F

c c c k k k mM C K

c c c k k k m

we get:

M x C x K x F (4)

Free Undamped Vibration

Setting the damping [C]and forcing {F}terms to zero, we get:

1 1 1 2 1 2 2

2 2 2 3 2 2 1

0

0

m x k k x k x

m x k k x k x

(5)

Solution: we can assume that each mass undergoes harmonic motion of the same frequency andphase. This is proved on page 4. The solution is thus written as:

1 1

2 2

cos

cos

x X t

x X t

(6)

k1

c1

m1

k2

c2

m2

k3

c3

x1 x2

F1 F2


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or:

cosx X t (7)where

1

2

XX

X

(8)

Substituting into the equation of motion yields:

2 20 0M x K x K M x

(9)

This is an eigenvalue problem. For a non-trivial solution, the determinant must vanish so we have:

2 0K M (10)

Or2

1 1 2 2

2

2 2 2 3

0m k k k

k m k k

(11)

For our problem, this results in:

2 21 1 2 2 2 3 2 2 0m k k m k k k k (12)combining terms we get:

2

2 21 2 1 2 3 2 1 2 1 2 2 3 1 3 0m m m k k m k k k k k k k k (13)

which is a quadratic equation in terms of 2. From this we can get:

2

1 2 3 2 1 2 1 2 3 2 1 2 1 2 1 2 2 3 1 32

1 2

4

2

m k k m k k m k k m k k m m k k k k k k

m m

(14)

These values of are the natural frequenciesof the system. The values ofX1andX2remain to be

determined. To simplify the analysis, let m1=m2=mand k1=k2=k3=k. The determinant will be:2

2

20

2

k m k

k k m

(15)

yielding the characteristic equation:

2

2 22 0k m k (16)

which has the solutions:

1 2

3,

k k

m m (17)

Note that these values are the solutions to this particular case (masses are identical, springs are

identical).To determineX1andX2, we need to substitute into

2 0K M x

with the values of 1 2and just obtained. Hence at 1k

m

we have:

21 11

22 21

20 0

2

X Xk kk m k

X Xk kk k m

This has infinite number of solutions, but they must satisfy a certain ratio, namely:

1

1

2

1

1

X

X

(18)

Similarly, at 2we have:


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21 12

22 22

20 0

2

X Xk kk m k

X Xk kk k m

(19)

so

2

1

2

1

1

X

X

(20)

The ratio of amplitudes1

2

X

X

defines a certain pattern of motion called the normal mode of

vibration. The vectors

1 2

1 2

1 1

2 2

andX X

X XX X

are called the modal vectorsor eigenvectors. They define the mode shapes of the system. In this

particular case, if the system vibrates in its first mode, the masses will move in phase with the sameamplitudes, while in the second mode of vibration the masses move out of phase also with the same

amplitudes.

The solution for the vibration of the system at the first mode is:

1

1 1

1 1 1

2 2

cosx t X

A tx t X

(21)

and for the second mode:

2

1 1

2 2 2

2 2

cosx t X

A tx t X

(22)

so the general solution is:

1 2

1 1 1

1 1 1 2 2 2

2 2 2

cos cosx t X X

A t A t

x t X X

(23)

whereA1,A2, 1and 2are 4 constants to be determined from the initial conditions.


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Proof: It was noted that the solution of the equations

1 1 1 2 1 2 2

2 2 2 3 2 2 1

0

0

m x k k x k x

m x k k x k x

takes the form:

1 1 2 2sin , sinx X t x X t

meaning the masses undergo harmonic motions of the same frequencywith no phase difference

between them. In order to justify this, let us re-write the equations of motions in a more general

form:

1 11 1 12 2

2 21 1 22 2

0

0

x a x a x

x a x a x

Now assume a general solution in the form:

1 1 1 2 2 2sin , sin( )x X t x X t

where 2is taken to be different from 1.There is no loss of generality in assuming no phase forx1

and only a phase difference between the two motions. We wish to prove that 1 2 and 0 .

Substituting into the equations of motion yields:

2

11 1 1 1 12 2 2

221 1 1 22 2 2 2

sin sin 0

sin sin 0

a X t a X t

a X t a X t

These relations must be valid for all t. Setting t=0 in the first equation gives:

12 2sin 0a X

since a12andX2cannot be zero, we must have 0 . Thus there can be no phase difference between

the harmonic motions of the two parts.

The first expression may then be written as:

211 1 1 1 12 2 2sin sin 0a X t a X t

or

2

1 11 12

1 12 2

sin

constantsin

a Xt

t a X

Since the left hand side must be constant for all values of t, we must have 2=1and consequentlythe harmonic motions occur at the same frequency.


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Example

For various initial conditions, obtain the free response of the previous system having m = 1 and k =1.

Solution

Recall the natural frequencies were

1 2

31 , 3

k k

m m

and the mode shapes were

1 2

1 1

2 2

1 1,

1 1

X X

X X

and the general solution is:

1 2

1 1 1

1 1 1 2 2 2

2 2 2

sin sinx t X X

A t A tx t X X

hence

1 1 1 2 22

1 1sin sin 3

1 1

x tA t A t

x t

Differentiating w.r.t. time we get:

1 1 1 2 22

1 1cos 3 cos 3

1 1

x tA t A t

x t

For the initial conditions

1 1

2 2

(0) 5 (0) 0

(0) 0 (0) 0

x x

x x

we have:

1 1 2 2

1 1 2 2

1 1 2 2

1 1 2 2

5 sin sin

0 sin sin

0 cos 3 cos

0 cos 3 cos

A A

A A

A A

A A

from which we get:

1 2

1 2

2

5 2A A

hence the solution is:

1 25 5

cos cos 3 , cos cos 32 2

x t t t x t t t


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Modes of vibration contribute equally to the solution.


1 1 2 2(0) 1 (0) 0 (0) 1 (0) 0x x x x

we get

1 2 1 22, 1, 0A A


1 2cos , cosx t t x t t i.e. the masses move in-phase with the same amplitude and frequency 1 rad /s (mode 1)


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1 1 2 2(0) 1 (0) 0 (0) 1 (0) 0x x x x

we get

1 2 1 22, 0, 1A A


1 2cos 3 , cos 3x t t x t t

i.e. the masses move out-of-phase with the same amplitude and frequency 3 rad /s (mode 2)


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Forced vibration analysis

Consider the shown system

The equations of motion are:

1 1 1 1 0

2 2 2 2

0sin

0 0

m x k k k x Ft

m x k k k x

(24)

or:

sinM x K x F t (25)Seeking a steady-state solution in the form

sinx X t (26)yields:

2 sin sinK M X t F t (27)

hence:2

1 01 1

222 2 0

X Fk k m k

Xk k k m

(28)

which can be solved for the unknown amplitudes. For our special case where m1=m2=m and

k1=k2=k, we have:2

1 0

22

2

02

X Fk m k

Xk k m

(29)

thus:1

21 0

22

2

02

X Fk m k

X k k m

(30)

This results in:

2 01

2 2 22

0

21

3

k m FX

k kXm kFm m

(31)

or:

2

0

01 22 2 2 2 2 2 2 2 2 2

1 2 1 2

2

,

k m F kFX Xm m

(32)

where

1 2

3,

k k

m m

are the natural frequencies obtained earlier. Plotting the amplitudes of the masses reveals that

resonance occurs when the frequency of excitation coincides with either of the two naturalfrequencies of the system.

k1 k k2m1 m2

F0sint


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1 2


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Dynamic vibration absorber

Consider the primary system shown, which is a model of a single DOF

vibrating structure, acted upon by a harmonic force.

Suppose that the exciting frequency, , is constant and is equal to the natural frequency of thesystem, i.e.

1

1

k

m

We wish to reduce the vibrations of m1at the exciting frequency .We can do this by adding a secondary system, consisting of a mass

m2and spring k2as shown.

If we derive the equations of motion, we will get:

1 1 1 2 2 1 0

2 2 2 2 2

0sin

0 0

m x k k k x Ft

m x k k x

Substituting into

1

2 2K M X F X K M F

yields:

12

1 01 2 1 2

22 2 2 2

0

X Fk k m k

X k k m

or:

21 02 2 2

22 2 22 2 1 2 11 2 1 2 2 2

1

0

X Fk m k

X k k k mk k m k m k

hence:

2

0 2 2

1 2 2 2

1 2 1 2 2 2

0 22 2 2 2

1 2 1 2 2 2

F k mX

k k m k m k

F kX

k k m k m k

k2

m2

m1F0sint

k1

k1

m1F0sint


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now define

1 211 22

1 2

,k k

m m

For the primary system (without absorber), resonance occurs when

111

1

k

m

ForX1to be zero at this frequency, we must have2 2

2 2 22

2

0 k

k mm

Therefore if k2and m2are chosen such that

1 2

1 2

k k

m m

thenX1will be zero at =11. This is what we call a tuned dynamic absorber, in which

11 22

At this frequency, the displacement ofX2will be:

0 2 02

2 21 21 2 1 2 2 2

1 2

F k FX kk k

k k m k m k m m

Adding the secondary system (dynamic absorber) will result in zero vibrations of the primary mass

at 11=22. However, two resonant frequencies n1and n2 are introduced at which the amplitudeof X1 becomes significantly large. Thus the dynamic absorber can only be useful when the

disturbing frequency is constant.

X1

n1 n2

No

absorberWith

absorber


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How to design the vibration absorber?

1. Based on space limitation. Choose k2and m2such that

1 2

1 2

k k

m m

noting that the amplitude of m2will be

02

2

FX

k

2. Based on how far apart should the natural frequencies be . The two new natural frequencies of

the system can be obtained by setting the determinant equal to zero:

2

1 2 1 2

2

2 2 2

0k k m k

k k m

which gives:

2 2 21 2 1 2 2 2 0k k m k m k This gives:

4 2 1 2 2 1 2

1 1 2 1 2

0k k k k k m m m m m

which can be put in the form:

4 2 2 2 2 2 211 22 22 11 22 0

where 2

1

mass ratiom

m

The roots of this equation n1and n2satisfy the relations:

2 2 2 2

1 2 11 11

2 2 2 2

1 2 11 22 1

n n

n n

But for a tuned absorber we have 11 22 hence2 2

1 2

2 2

22 22

2 2

1 2

2 2

22 22

1

2

n n

n n

As you increase the mass ratio (), the natural frequencies n1and n2will grow further apart. Note

that n1is always closer to 11than n2.


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Systems with 2 or More Degrees Of Freedom

Systems having more than 2 degrees of freedom are easily studied using the same principles that are

applied for 2 DOF systems. Let us investigate 2 cases.

Unrestrained 2 DOF systemConsider an unrestrained system like the one shown below.

m1k

m2

x1 x2

From Newtons law, the equations of motion are:

1 1 1 2 2 2 2 1( ) 0 , ( ) 0m x k x x m x k x x (33)

Which can be written in matrix form as:

1 1 1

2 2 2

0 0

0 0

m x xk k

m x xk k

(34)

as before, we have:2

11

222

0

0

Xm k k

Xk m k

(35)

to get the natural frequencies, we impose the determinant to zero:2

1

2

2

0m k k

k m k

(36)

which gives:2 2

1 2 1 2[ ( )] 0m m k m m (37)

or:

2 2 1 21 2

1 2

( )0,

k m m

m m

(38)

When = 0, use Eq. (3) to get:Hence

1

2

0

0

Xk k

Xk k

(39)

or:

1 2X X (40)This mode is called a rigid body mode, since it resembles rigid body motion, and is typical of

unrestrained systems.

Now recall that:

2 1 22

1 2

( )k m m

m m

(41)

If we let m1= m2= mthen:

2

2

2k

m (42)


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so we get:

1

2

0

0

Xk k

Xk k

(43)

Or

1 2X X (44)

i.e. the masses move out of phase in this mode of vibration.

Systems with more than 2 DOFThe system shown below has 3 masses on a taught string. Let us study the free transverse vibration.

l l l l

x1

x2

x3

To get the equations of motion, we can draw free body diagrams as shown, assuming the tension to

be uniform.

l l l l

x1

x2

x3T

T

TT

T

T

The equations of motion are then:

1 2 1

1 1

3 22 1

2 2

2 3 3

3 33

x x xT T m x

l l

x xx xT T m x

l l

x x xT T m x

l l

(45)

which can be written in matrix form as:

1 1 1

2 2 2

3 3 3

0 0 2 0 0

0 0 2 0

0 0 0 2 0

m x T l T l x

m x T l T l T l x

m x T l T l x

(46)


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If we let m1= m2= m, then the equations become:

1 1

2 2

3 3

1 0 0 2 1 0 0

0 1 0 1 2 1 0

0 0 1 0 1 2 0

x xT

m x xl

x x

(47)

We can get the natural frequencies and mode shapes from MATLAB using the command eig.

m = 3;

M = m*eye(3);

T = 100; L = 0.8;K = T/L*[2 -1 0;-1 2 -1;0 -1 2]

[vec,val]=eig(inv(M)*K)

vec =

0.5000 -0.7071 -0.5000

0.7071 0.0000 0.7071

0.5000 0.7071 -0.5000

val =

24.4078 0 0

0 83.3333 00 0 142.2589

And the mode shapes can be plotted as shown below.


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Recall that the equation of motion governing the free response of a multi-DOF system is written as:

0M x K x (48)Imposing harmonic motion in the form:

sinx X t (49)

we get:

2K X M X (50)or:

1

M K X X

(51)

which is a standard eigenvalueproblem. Solving Eq. (52) using MATLAbs eigcommand yields

the system eigenvaluesi

and associated eigenvectors iX for 1,2, ,i n , where nis the system

size, i.e. degrees of freedom.


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Active Vibration Control

We have earlier seen how vibration can be passivelyisolated or absorbed through proper choice of

m, c and k or by adding a dynamic absorber. In active vibration control, we use external active

devices called actuators (e.g. hydraulic cylinder, electric motor, piezoelectric device, etc.) to

provide a force to the structure or machine whose vibration properties are to be changed. The figureshows a feedback control system illustrating the required sensor and actuator setup for active

vibration control.

m

k

x

c

F

Actuator

u

Control law

electronics

Power supplyControl signal

Motion sensor

The equation of motion is:

mx cx kx F u (52)A simple control law is called state-feedback, which for the case of a single degree of freedom

system utilizes the position and velocity measurements. The controller is designed to provide asignal to the actuator proportional to the displacement and velocity. This is also called PD

(proportional plus derivative) control. The actuator is hence commanded to provide a force for the

mass in the form:

1 2u g x g x (53)

whereg1andg2are calledgains. These gains are determined by the sensor and actuator properties

and by the designers choice. In control system terminology,g1is calledproportional gainand g2iscalled derivative gain. The equation of motion with PD control becomes:

2 1mx c g x k g x F (54)We immediately see that the gaing1acts like an artificial stiffness and the gaing2like an artificial

damping. In this way, we can change the effective stiffness and damping by properly designing the

gains.

Example. Examine the free response of the system shown by varying the control gainsg2andg2.

10 kg

200 N/m 50 Ns/m u


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% Active vibration control

clear; %close all;

tspan = 0:0.05:10; % time span

x0 = [5;0]; % initial conditions

[t,x] = ode23('odefun3',tspan,x0);

plot(t,x(:,1));

xlabel('Time [s]')

ylabel('Displacement [m]')

grid

functionf = odefun3(t,x)

f = zeros(2,1);

g1 = 30;

g2 = 5;

f(1) = x(2);

f(2) = -(200+g1)/10*x(1)-(5+g2)/10*x(2);

0 1 2 3 4 5 6 7 8 9 10-5

-4

-3

-2

-1

0

1

2

3

4

5

Time [s]

Displacement[m]

Effect of derivative control (g1= 0,g2= 5)

0 1 2 3 4 5 6 7 8 9 10-5

-4

-3

-2

-1

0

1

2

3

4

5

Time [s]

Displacem

ent[m]

Effect of proportional control (g1= 30,g2= 0)


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0 1 2 3 4 5 6 7 8 9 10-5

-4

-3

-2

-1

0

1

2

3

4

5

Time [s]

Displacem

ent[m]

Effect of PD control (g1= 30,g2= 5)

chapter 2 2dof systems fall 2013

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