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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Multiple Degrees of Freedom

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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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Two Degrees of Freedom

Systems

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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

)()()(

)()()()(

1221111

txtxktxm

txtxktxktxm

0)()()(

0)()()()(

221222

2212111

txktxktxm

txktxkktxm

Rearranging:

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Initial Conditions

• Two coupled, second -order, ordinary

differential equations with constant

coefficients

• Needs 4 constants of integration to

• Thus 4 initial conditions on positions

and velocities

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

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In Matrix Form

)(,)()(

0kkkkk

0xx KMWhere:

With initial conditions:

10 )0( ,)0(xx

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Recall: For SDOF

• The ODE is

• The proposed

solution:

• Into the ODE you get

the characteristic

equation:

• Giving:

0)()( tkxtxmtaetx )(

02 tt aemk

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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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Giving:

aEither:

Trivial solution;

No motion!

02 KMOR:

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Giving:

0)( 212

2212214

21 kkkmkmkmmm

Which can be solved as a quadratic equation in 2.

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• For spring mass systems, the resulting

roots are always positive, real, and distinct

• Which give two couples of distinct roots.

212,1 &

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Example

• m1=9 kg,m2=1kg,

k1=24 N/m and k2=3

• In Matrix form:

033327

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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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Calculating the corresponding

vectors a1 and a2

A vector equation for each square frequency

4 equations in the 4 unknowns (each

vector has 2 components, but...

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Computing the vectors a

let 2,=For 12

2 equations, 2 unknowns but DEPENDENT!

03 and 039

)2(333)2(927

12111211

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121112

:arbitrary , does so ,)(

satisfies Suppose arbitrary. is magnitude The

.0 :because is This

!determined becan magnitude not the direction, only the

:equationsboth from 31

cKMcKM

continued

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For the second value of 2:

aor 039

)4(333)4(927

have then welet 4,=For

22212221

Note that the other equation is the same

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What to do about the

magnitude!

Several possibilities, here we just fix one element:

Choose:

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Thus the solution to the

algebraic matrix equation is:

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Return now to the time

response:

nintegratio of constants are and ,,, where

)sin()sin(

22221111

decebeaet

edecebeat

tjtjtjtj

We have four solutions:

Since linear we can combine as:

determined by initial conditions

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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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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

1 sin)()(

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Multiple Degrees of Freedom

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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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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

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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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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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Example

• The equations of motion may be written in

the form:

420242

300010002

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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

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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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Vibration Absorber

The first passive damping

technique we will learn!

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For a 2-DOF System

• For the shown 2-DOF

system, the equations

of motion may be

written as:

• Where:

fxx KM

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For Harmonic Excitation

• We may write the

equation for each of

the excitation

frequency in the form

• Then we may add

both solutions!

11 tCosfKM

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Consider the first force

• We may write the

equation in the form:

• And the solution in

the form:

• Which will give:

tCosfKM 101

tCosxx

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The equation of motion becomes

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

to zero!

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Using the Dynamic Stiffness

Matrix • Writing down the dynamic stiffness matrix:

Getting the inverse:

22112 f

KmKKKKm

KKmKKm

KKmKKKm

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Obtaining the Solution

• Multiply the inverse by the right-hand-side

• For the first degree of freedom:

212124

fKmKKKmKKmmmx

212124

KKKmKKmmmfKm

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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

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Vibration absorber

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Vibration absorber

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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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Homework #2 (cont’d)

• Use modal decomposition

(diagonalization) to obtain the same

results.

Multiple Degree of Freedom (MDOF) Systems

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