Multiple Degree of Freedom (MDOF) Systems

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Multiple Degree of Freedom Systems
Mohammad Tawfik
Multiple Degree of Freedom
Systems

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Mohammad Tawfik
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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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
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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Mohammad Tawfik
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
kae 

m
k2

m
kj

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

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Mohammad Tawfik
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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Mohammad Tawfik
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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Mohammad Tawfik
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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Mohammad Tawfik
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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Mohammad Tawfik
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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Mohammad Tawfik
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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Mohammad Tawfik
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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Mohammad Tawfik
For the second value of 2:
3
1aor039
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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Mohammad Tawfik
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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Mohammad Tawfik
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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Mohammad Tawfik
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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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
t
a
a
A
tx
tx
t 1
12
11
1
2
1
sin
)(
)(
)( 











x

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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

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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  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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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
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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Mohammad Tawfik
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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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
f
f
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 tCosf
KM

xx
 






tCosf
KM
22
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 1
0
1





 xx
 tCos
x
x








2
1
x
  xx 2
2
12
 






 tCos
x
x


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Mohammad Tawfik
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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Mohammad Tawfik
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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Mohammad Tawfik
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
fKmx



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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
m
K

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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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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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