Multiple Degree of Freedom Systems Explained

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Multiple Degree of Freedom Systems
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Multiple Degree of Freedom
Systems

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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
 
 )()()(
)()()()(
12222
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
solve
• Thus 4 initial conditions on positions
and velocities
202202101101 )0(,)0(,)0(,)0( xxxxxxxx  

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In Matrix Form



















)(
)(
)(,
)(
)(
)(,
)(
)(
)(
2
1
2
1
2
1
tx
tx
t
tx
tx
t
tx
tx
t





 xxx















22
221
2
1
,
0
0
kk
kkk
K
m
m
M
0xx  KM
Where:
With initial conditions:













20
10
20
10
)0(,)0(
x
x
x
x


xx

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Recall: For SDOF
• The ODE is
• The proposed
solution:
• Into the ODE you get
the characteristic
equation:
• Giving:
0)()(  tkxtxm 
t
aetx 
)(
02
 tt
ae
m
k
ae 

m
k
2

m
k
j

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Solving the system
• The ODE is
• The proposed
solution:
• Into the ODE you get
the characteristic
equation:
• Giving:
tj
et 
ax )(
02
 tjtj
ee 
 KaMa
0xx  KM
  02
 tj
e 
 aKM

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Giving:
0
2
1








a
a
a
Either:
Trivial solution;
No motion!
02
 KM
OR:

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Giving:
0
22
2
2
2211
2



kmk
kkkm


0)( 21
2
221221
4
21  kkkmkmkmmm 
Which can be solved as a quadratic equation in 2.

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NOTE!
• For spring mass systems, the resulting
roots are always positive, real, and distinct
• Which give two couples of distinct roots.
2
24,3
2
12,1 &  

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Example
• m1=9 kg,m2=1kg,
k1=24 N/m and k2=3
N/m
• In Matrix form:
0
33
327
10
09














xx

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Example (cont’d)
• The proposed solution:
• Into the ODE you get the characteristic equation:
4-62+8=(2-2)(2-4)=0
• Giving:
2 =2 and 2 =4
tj
et 
ax )(
Each value of 2 yields an expression for a:

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Calculating the corresponding
vectors a1 and a2
0a
0a


2
2
2
1
2
1
)(
)(
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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Computing the vectors a
let2,=For
12
11
1
2
1 






a
a
a
2 equations, 2 unknowns but DEPENDENT!
03and039
0
0
)2(33
3)2(927
)(-
12111211
12
11
2
1






















aaaa
a
a
KM 0a

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0a0a
a0a
u




1
2
11
2
1
11
2
1
1
2
1
1211
12
11
)()(
:arbitrary,doesso,)(
satisfiesSupposearbitrary.ismagnitudeThe
.0:becauseisThis
!determinedbecanmagnitudenot thedirection,only the
:equationsbothfrom
3
1
3
1
cKMcKM
ccKM
KM
aa
a
a



continued

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For the second value of 2:
3
1
aor039
0
0
)4(33
3)4(927
)(-
havethen welet4,=For
22212221
22
21
2
1
22
21
2
2
2
aaa
a
a
KM
a
a





























0a
a


Note that the other equation is the same

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What to do about the
magnitude!















1
1
1
1
3
1
222
3
1
112
a
a
a
a
Several possibilities, here we just fix one element:
Choose:
Choose:

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Thus the solution to the
algebraic matrix equation is:















1
,2
1
,2
3
1
24,2
3
1
13,1
a
a



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Return now to the time
response:
   
nintegratioofconstantsareand,,,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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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
t
a
a
A
tx
tx
t 1
12
11
1
2
1
sin
)(
)(
)( 











x

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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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Multiple Degrees of Freedom

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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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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  2

fxKD 

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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 D
1


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The Decibel or dB scale
It 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  10log10
x
x0




2
 20log10
x
x0



 (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,
10log10
19.6
9.8




2
 20log10 2  6dB

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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:























































3
2
1
3
2
1
3
2
1
420
242
024
300
010
002
f
f
f
x
x
x
x
x
x



FKxxM 

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Example
• Getting the eigenvalues, and frequencies






















796.0
295.1
241.2
,
633.000
0677.10
00023.5


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Getting the Frequency Response

























300
010
002
420
242
024
2
DK





















0
1
0
3
2
1
f
f
f

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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







2
1
f
f
f

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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 tCosf
KM

xx
 






tCosf
KM
22
0

xx

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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 1
0
1






 xx
 tCos
x
x








2
1
x
  xx 2
2
12
 






 tCos
x
x


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The equation of motion becomes
• Get x1() and find out when does it equal
to zero!





































00
0 1
2
1
22
221
2
2
1
2
f
x
x
kk
kkk
m
m



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Using the Dynamic Stiffness
Matrix
• Writing down the dynamic stiffness matrix:
Getting the inverse:




















0
1
2
1
22
2
2
2211
2
f
x
x
KmK
KKKm


   




















0
1
2
222
2
211
2
211
2
2
222
2
2
1 f
KKmKKm
KKmK
KKm
x
x




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Obtaining the Solution
• Multiply the inverse by the right-hand-side
• For the first degree of freedom:
  
 





 







12
122
2
21
2
21212
4
212
1 1
fK
fKm
KKKmKKmmmx
x 

 
  
0
21
2
21212
4
21
122
2
1 



KKKmKKmmm
fKm
x



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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
2
2
m
K


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Vibration absorber

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Damped Vibration

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Generic Example:
• If the damping
mechanisms are
known then
• Sum forces to find
the equations of
motion
c1
Ýx1
c2(Ýx2  Ýx1) c2(Ýx2  Ýx1)
Free Body Diagram:
)()(
)()(
12212222
122111221111
xxkxxcxm
xxkxkxxcxcxm





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Matrix form of Equations of
Motion:















































0
0
)(
)(
)(
)(
)(
)(
0
0
2
1
22
221
2
1
22
221
2
1
2
1
tx
tx
kk
kkk
tx
tx
cc
ccc
tx
tx
m
m




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.

#WikiCourses
Maged Mostafa
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

#WikiCourses
Maged Mostafa
Homework #3 (cont’d)
• Use modal decomposition
(diagonalization) to obtain the same
results.

Multiple Degree of Freedom Systems Explained

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