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Multiple Degree of Freedom SystemsMaged Mostafa

Multiple Degree of Freedom Systems



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



Two Degrees of Freedom Systems



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



Two Degrees of Freedom



Free-Body Diagram



Equations of Motion

)()()(

)()()()(

12222

1221111

txtxktxmtxtxktxktxm

0)()()(0)()()()(

221222

2212111

txktxktxmtxktxkktxm

Rearranging:



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



In Matrix Form

)()(

)(,)()(

)(,)()(

)(2

1

2

1

2

1

txtx

ttxtx

ttxtx

t

xxx

22

221

2

1 ,0

0kkkkk

Km

mM

0xx KM Where:

With initial conditions:

20

10

20

10 )0( ,)0(xx

xx

xx



Recall: For SDOF• The ODE is• The proposed

solution:• Into the ODE you get

the characteristic equation:

• Giving:

0)()( tkxtxm taetx )(

02 tt aemkae

mk

2mkj



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



Giving:

02

1

aa

aEither:

Trivial solution;No motion!

02 KMOR:



Giving:

022

22

22112

kmk

kkkm

0)( 212

2212214

21 kkkmkmkmmm

Which can be solved as a quadratic equation in 2.



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 &



Example

• m1=9 kg,m2=1kg, k1=24 N/m and k2=3 N/m

• In Matrix form:

033327

1009

xx



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:



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 (eachvector has 2 components, but...



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

aaaaaa

KM 0a



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



For the second value of 2:

31aor 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



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:



Thus the solution to the algebraic matrix equation is:

1 ,2

1 ,2

31

24,2

31

13,1

a

a



Return now to the time response:

nintegratio of constants are and ,,, where

)sin()sin( )(

)(

,,,)(

2121

22221111

21

2211

2211

2211

2211

2211

AA

tAtAdecebeaet

edecebeat

eeeet

tjtjtjtj

tjtjtjtj

tjtjtjtj

aaaax

aaaax

aaaax

We have four solutions:

Since linear we can combine as:

determined by initial conditions



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



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



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



Multiple Degrees of Freedom



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



Compound Pendulum



Pendulum Video



Frequency Response



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.



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



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



The Decibel or dB scaleIt is often useful to use a logarithmic scale to plot vibration levels (or noise

levels). One such scale is called the decibel or dB scale. The dB scale is always relative to some reference value x0. It is define as:

dB 10 log10xx0

2

20 log10xx0

(1.22)

For example: if an acceleration value was 19.6m/s2 then relative to 1g (or 9.8m/s2) the level would be 6dB,

10 log1019.69.8

2

20 log10 2 6dB



Example

• For the three DOF system given in the sketch, consider all stiffness values to be 2 and m1=2, m2=1, m3=3



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



Example

• Getting the eigenvalues, and frequencies

796.0295.1241.2

,633.0000677.1000023.5



Getting the Frequency Response

300010002

420242

0242DK

010

3

2

1

fff



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”



Vibration Absorber

The first passive damping technique we will learn!



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



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



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



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



Using the Dynamic Stiffness Matrix

• Writing down the dynamic stiffness matrix:

Getting the inverse:

01

2

1

222

2

22112 f

xx

KmKKKKm

0

12

2222

2112

2112

2

2222

2

1 fKKmKKm

KKmKKKm

xx



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



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



Vibration absorber



Damped Vibration



Generic Example:• If the damping

mechanisms are known then

• Sum forces to find the equations of motion

c1 Ý x 1

c2( Ý x 2 Ý x 1)

c2( Ý x 2 Ý x 1)

Free Body Diagram:

)()()()(

12212222

122111221111

xxkxxcxmxxkxkxxcxcxm



Matrix form of Equations of Motion:

00

)()(

)()(

)()(

00

2

1

22

221

2

1

22

221

2

1

2

1

txtx

kkkkk

txtx

ccccc

txtx

mm

The C and K matrices have the same form.

It follows from the system itself that consisted damping and stiffness elements in a similar manner.



Homework #3

• 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



Homework #3 (cont’d)

• Use modal decomposition (diagonalization) to obtain the same results.

03 multipledof

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