multiple degree of freedom (mdof) systems

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http://WikiCourses.WikiSpaces.com Multiple Degree of Freedom Systems

Mohammad Tawfik

Multiple Degree of Freedom

Systems

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Mohammad Tawfik

Multiple Degrees of Freedom

VibrationClips/Mechanical Resonance.WMV

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Mohammad Tawfik

Objectives

• What is a multiple degree of freedom system?

• Obtaining the natural frequencies of a multiple

degree of freedom system

• Interpreting the meaning of the eigenvectors of a

multiple degree of freedom system

• Understanding the mechanism of a vibration

absorber

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Mohammad Tawfik

Two Degrees of Freedom

Systems

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Mohammad Tawfik

Two Degrees of Freedom Systems

• When the dynamics of the system can be

described by only two independent

variables, the system is called a two

degree of freedom system

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Two Degrees of Freedom

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Free-Body Diagram

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Equations of Motion

)()()(

)()()()(

12222

1221111

txtxktxm

txtxktxktxm

0)()()(

0)()()()(

221222

2212111

txktxktxm

txktxkktxm

Rearranging:

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Mohammad Tawfik

Initial Conditions

• Two coupled, second -order, ordinary

differential equations with constant

coefficients

• Needs 4 constants of integration to

solve

• Thus 4 initial conditions on positions

and velocities

202202101101 )0(,)0(,)0(,)0( xxxxxxxx

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Mohammad Tawfik

In Matrix Form

)()(

)(,)()(

)(,)()(

)(2

1

2

1

2

1

txtx

ttxtx

ttxtx

t

xxx

22

221

2

1 ,0

0kkkkk

Km

mM

0xx KMWhere:

With initial conditions:

20

10

20

10 )0( ,)0(xx

xx

xx

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Mohammad Tawfik

Recall: For SDOF

• The ODE is

• The proposed

solution:

• Into the ODE you get

the characteristic

equation:

• Giving:

0)()( tkxtxmtaetx )(

02 tt aemk

ae

mk

2mk

j

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Mohammad Tawfik

Solving the system

• The ODE is

• The proposed

solution:

• Into the ODE you get

the characteristic

equation:

• Giving:

tjet ax )(

02 tjtj ee KaMa

0xx KM

02 tje aKM

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Mohammad Tawfik

Giving:

02

1

aa

aEither:

Trivial solution;

No motion!

02 KMOR:

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Mohammad Tawfik

Giving:

022

22

22112

kmk

kkkm

0)( 212

2212214

21 kkkmkmkmmm

Which can be solved as a quadratic equation in 2.

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Mohammad Tawfik

NOTE!

• For spring mass systems, the resulting

roots are always positive, real, and distinct

• Which give two couples of distinct roots.

224,3

212,1 &

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Example

• m1=9 kg,m2=1kg,

k1=24 N/m and k2=3

N/m

• In Matrix form:

033327

1009

xx

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Mohammad Tawfik

Example (cont’d)

• The proposed solution:

• Into the ODE you get the characteristic equation:

4-62+8=(2-2)(2-4)=0

• Giving:

2 =2 and 2 =4

tjet ax )(

Each value of 2 yields an expression for a:

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Mohammad Tawfik

Calculating the corresponding

vectors a1 and a2

0a

0a

222

121

)(

)(

KM

KM

A vector equation for each square frequency

And:

4 equations in the 4 unknowns (each

vector has 2 components, but...

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Mohammad Tawfik

Computing the vectors a

let 2,=For 12

111

21

aa

a

2 equations, 2 unknowns but DEPENDENT!

03 and 039

00

)2(333)2(927

)(-

12111211

12

11

21

aaaa

aa

KM 0a

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Mohammad Tawfik

0a0a

a0a

u

1211

21

1121

1

21

121112

11

)()(

:arbitrary , does so ,)(

satisfies Suppose arbitrary. is magnitude The

.0 :because is This

!determined becan magnitude not the direction, only the

:equationsboth from 31

31

cKMcKM

ccKM

KM

aaaa

continued

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Mohammad Tawfik

For the second value of 2:

31

aor 039

00

)4(333)4(927

)(-

have then welet 4,=For

22212221

22

21

21

22

212

22

aaa

aa

KM

aa

0a

a

Note that the other equation is the same

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Mohammad Tawfik

What to do about the

magnitude!

11

11

31

222

31

112

a

a

a

a

Several possibilities, here we just fix one element:

Choose:

Choose:

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Mohammad Tawfik

Thus the solution to the

algebraic matrix equation is:

1 ,2

1 ,2

31

24,2

31

13,1

a

a

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Mohammad Tawfik

Return now to the time

response:

nintegratio of constants are and ,,, where

)sin()sin(

)(

)(

,,,)(

2121

22221111

21

2211

2211

2211

2211

2211

AA

tAtA

decebeaet

edecebeat

eeeet

tjtjtjtj

tjtjtjtj

tjtjtjtj

aa

aax

aaaax

aaaax

We have four solutions:

Since linear we can combine as:

determined by initial conditions

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Mohammad Tawfik

Physical interpretation of all that

math! • Each of the TWO masses is oscillating at

TWO natural frequencies 1 and 2

• The relative magnitude of each sine term,

and hence of the magnitude of oscillation

of m1 and m2 is determined by the value of

A1a1 and A2a2

• The vectors a1 and a2 are called

mode shapes

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Mohammad Tawfik

What is a mode shape?

• First note that A1,A2, 1 and 2 are

determined by the initial conditions

• Choose them so that A2 = 1 = 2 =0

• Then:

• Thus each mass oscillates at (one)

frequency 1 with magnitudes proportional

to a1 the1st mode shape

taa

Atxtx

t 112

111

2

1 sin)()(

)(

x

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Mohammad Tawfik

Multiple Degrees of Freedom

VibrationClips/Mechanical Resonance.WMV

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Mohammad Tawfik

Things to note

• Two degrees of freedom implies two

natural frequencies

• Each mass oscillates at these two

frequencies present in the response

• Frequencies are not those of two

component systems

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Mohammad Tawfik

Eigenvalues and Eigenvectors

• Can connect the vibration problem with the

algebraic eigenvalue problem

• This will give us some powerful

computational skills

• And some powerful theory

• All the codes have eigensolvers so these

painful calculations can be automated

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Compound Pendulum

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Pendulum Video

VibrationClips/compound_pendulum.wmv

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Frequency Response

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Frequency Response

• Similar to SDOF systems, the frequency

response of a MDOF system is obtained by

assuming harmonic excitation.

• An analytical relation between all the possible

input forces and output displacements may be

obtained, called transfer function

• For our course, we will pay more attention to the

plot of the relation.

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Mohammad Tawfik

Dynamic Stiffness

• The system of equations we obtain for an undamped vibrating system is always in the form

fKxxM • For harmonic excitation harmonic

response, we may write

fxKM 2fxKD

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Mohammad Tawfik

Dynamic Stiffness

• Now, we have a system of algebraic

equations that may be solved for the

amplitude of vibration of each DOF as a

response to given harmonic excitation at a

certain frequency!

fKx D1

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Example

• For the three DOF system given in the

sketch, consider all stiffness values to be 2

and m1=2, m2=1, m3=3

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Mohammad Tawfik

Example

• The equations of motion may be written in

the form:

3

2

1

3

2

1

3

2

1

420242

024

300010002

fff

xxx

xxx

FKxxM

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Mohammad Tawfik

Example

• Getting the eigenvalues, and frequencies

796.0295.1241.2

,633.0000677.1000023.5

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Getting the Frequency Response

300010002

420242

0242DK

010

3

2

1

fff

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Mohammad Tawfik

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Mohammad Tawfik

Notes:

• For all degrees of freedom, as the frequency reaches one of the natural frequencies, the amplitudes grows too much

• For some frequencies, and some degrees of freedom, the response becomes VERY small. If the system is designed to tune those frequencies to a certain value, vibration is absorbed: “Vibration absorber”

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Mohammad Tawfik

Vibration Absorber

The first passive damping

technique we will learn!

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Mohammad Tawfik

For a 2-DOF System

• For the shown 2-DOF

system, the equations

of motion may be

written as:

• Where:

fxx KM

2

1

ff

f

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Mohammad Tawfik

For Harmonic Excitation

• We may write the

equation for each of

the excitation

frequency in the form

of:

• Then we may add

both solutions!

0

11 tCosfKM

xx

tCosf

KM22

0

xx

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Mohammad Tawfik

Consider the first force

• We may write the

equation in the form:

• And the solution in

the form:

• Which will give:

tCosfKM 101

xx

tCosxx

2

1x

xx 2

2

12

tCosxx

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Mohammad Tawfik

The equation of motion becomes

• Get x1() and find out when does it equal

to zero!

000 1

2

1

22

221

22

12 f

xx

kkkkk

mm

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Mohammad Tawfik

Using the Dynamic Stiffness

Matrix • Writing down the dynamic stiffness matrix:

Getting the inverse:

01

2

1

222

2

22112 f

xx

KmKKKKm

01

2222

2211

2211

22

2222

2

1 f

KKmKKm

KKmKKKm

xx

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Mohammad Tawfik

Obtaining the Solution

• Multiply the inverse by the right-hand-side

• For the first degree of freedom:

12

1222

212

212124

212

1 1fK

fKmKKKmKKmmmx

x

0

212

212124

21

1222

1

KKKmKKmmmfKm

x

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Mohammad Tawfik

Vibration Absorber

• For the first degree of freedom to be

stationary, i.e. x1=0

• The excitation frequency have to satisfy:

• Note that this frequency is equal to the

natural frequency of the auxiliary spring-

mass system alone

2

2

mK

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Mohammad Tawfik

Vibration absorber

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Mohammad Tawfik

Vibration absorber

VibrationClips/vibration_isolation.wmv

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Homework #2

• Repeat the example of this lecture using

f2=f3=0 and f1=1 AND f1=f2=0 and f3=1

• Plot the response of each mass for each

of the excitation functions

• Comment on the results in the lights of

your understanding of the concept of

vibration absorber

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Mohammad Tawfik

Homework #2 (cont’d)

• Use modal decomposition

(diagonalization) to obtain the same

results.

multiple degree of freedom (mdof) systems

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