Mechanical Vibrations
Chapter 5
Peter Avitabile
Mechanical Engineering Department
University of Massachusetts Lowell
22.457 Mechanical Vibrations - Chapter 5
1
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Multiple Degree of Freedom Systems
Systems with more than one DOF:
• Referred to as a Multiple Degree of Freedom
• An NDOF system has ‘N’ independent degrees of
freedom to describe the system
• There is one natural frequency for every DOF in
the system description
22.457 Mechanical Vibrations - Chapter 5
2
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Multiple Degree of Freedom Systems
Properties of a MDOF system:
• Each natural frequency has a displacement
configuration referred to as a ‘normal mode’
• Mathematical quantities referred to as
‘eigenvalues’ and ‘eigenvectors’ are used to
describe the system characteristics
• While the resulting motion appears more
complicated, the system set of equations can
always be decomposed into a set of equivalent
SDOF systems for each mode of the system.
22.457 Mechanical Vibrations - Chapter 5
3
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Multiple Degree of Freedom Systems
Assumptions
f2
• lumped mass
m2
• stiffness proportional
k2
to displacement
• damping proportional to
velocity
• linear time invariant
f1
c2
x1
m1
k1
• 2nd order differential
x2
c1
equations
22.457 Mechanical Vibrations - Chapter 5
4
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Multiple Degree of Freedom Systems
Free body diagram
f2
k 2 ( x2 - x1 )
f1
c 2 ( x2 - x1 )
x1
m1
k 1x 1
22.457 Mechanical Vibrations - Chapter 5
x2
m2
c 1x 1
5
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Multiple Degree of Freedom Systems
Newton’s Second Law
m1&x&1=f1 (t ) −c1x& 1+c 2 (x& 2 − x& 1 )− k1x1+k 2 (x 2 − x1 )
m 2 &x& 2 =f 2 (t )−c 2 (x& 2 − x& 1 )−k 2 (x 2 − x1 )
Rearrange terms
m1&x&1+(c1 + c 2 )x& 1−c 2 x& 2 +(k1 + k 2 )x1−k 2 x 2 =f1 (t )
m 2 &x& 2 −c 2 x& 1+c 2 x& 2 −k 2 x1 +k 2 x 2 =f 2 (t )
f
2
k 2 ( x2 - x1 )
f1
c 2 ( x2 - x1 )
6
x1
m1
k 1x 1
22.457 Mechanical Vibrations - Chapter 5
x2
m2
c 1x 1
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Multiple Degree of Freedom Systems
Matrix Formulation
  &x&1 
 

m 2  &x& 2 
(c1 + c 2 ) − c 2   x& 1 
+
 x& 
−
c
c

2
2  2 
m1


Matrices and
Linear Algebra
are important !!!
(k1 + k 2 ) − k 2   x1   f1 ( t ) 
+

 =

k 2  x 2  f 2 ( t )
 − k2
22.457 Mechanical Vibrations - Chapter 5
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
MDOF - Frequencies and Mode Shapes
Example 5.1.1
m&x&1 = − kx1 + k (x 2 − x1 )
2m&x& 2 = − k (x 2 − x1 ) − kx 2
22.457 Mechanical Vibrations - Chapter 5
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(5.1.1)
MDOF - Frequencies and Mode Shapes
For normal mode type of oscillation, we can write
x1 = A1 sin ωt or A1eiωt
x 2 = A 2 sin ωt or A 2eiωt
(5.1.2)
substituting into the differential equation yields
(2k − ω2m)A1 − kA 2 = 0
− kA1 + (2k − 2ω2 m )A 2 = 0
22.457 Mechanical Vibrations - Chapter 5
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(5.1.3)
MDOF - Frequencies and Mode Shapes
In matrix form this is
(
 2 k − ω2 m

−k

)
  A1  0
−k
=
2  A  0 
2 k − 2ω m   2   
(
)
(5.1.4)
and the determinant of the matrix is
(2k − ω2m)
−k
−k
=0
2
2 k − 2ω m
(
)
(5.1.5)
whose solution yields the eigenvalues
2
2
3 k 
4 3k 2 3  k 
2 3k
ω − ω +   =λ − λ+   =0
m
2m
m
2m
22.457 Mechanical Vibrations - Chapter 5
10
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(5.1.6)
MDOF - Frequencies and Mode Shapes
The frequencies of the system are
k
k
λ1 = 0.634
⇒ ω1 = 0.634
m
m
k
k
λ 2 = 2.366
⇒ ω2 = 2.366
m
m
and the general ratio of response is
A1
k
2 k − 2ω 2 m
=
=
2
A 2 2k − ω m
k
22.457 Mechanical Vibrations - Chapter 5
11
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(5.1.7)
(5.1.8)
MDOF - Frequencies and Mode Shapes
The ratio for the first frequency , ω1, is
 A1 


 A2 
(1)
k
=
= 0.731
2
2k − ω m
(5.1.8a)
The ratio for the second frequency, ω2, is
 A1 


 A2 
( 2)
22.457 Mechanical Vibrations - Chapter 5
=
k
= −2.73
2
2k − ω m
12
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(5.1.8b)
MDOF - Frequencies and Mode Shapes
The mode shape for the two different modes is
0.731
− 2.73
φ1 ( x ) = 
 ; φ2 ( x ) = 

 1 
 1 
(5.1.8)
Each mode oscillates according to
 x1 
 
x 2 
 x1 
 
x 2 
(1)
( 2)
0.731
= A1 
 sin (ω1t + ψ1 )
 1 
− 2.73
= A1 
 sin (ω2 t + ψ 2 )
 1 
22.457 Mechanical Vibrations - Chapter 5
13
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(5.1.8b
MDOF - Initial Conditions
The general description of the system
(for the example considered) is
0.731
k
ω1 = 0.634
; φ1 ( x ) = 

m
 1 
− 2.73
k
; φ2 ( x ) = 
ω2 = 2.366

m
 1 
and the initial conditions are specified as
 x1 
 
x 2 
(i )
= ci φi sin (ωi t + ψ i )
22.457 Mechanical Vibrations - Chapter 5
14
i = 1,2
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(5.2.1)
MDOF - Initial Conditions
The displacement is written as
 x1 
0.731
  = c1 
 sin (ω1t + ψ1 )
 1 
x 2 
− 2.732
+ c2 
 sin (ω2 t + ψ 2 )
 1 
(5.2.2)
The velocity is written as
 x& 1 
0.731
  = ω1c1 
 cos(ω1t + ψ1 )
 1 
x& 2 
− 2.732
+ ω2 c 2 
 cos(ω2 t + ψ 2 )
 1 
22.457 Mechanical Vibrations - Chapter 5
15
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(5.2.3)
MDOF - Initial Conditions
Example 5.2.1 - Initial conditions are:
 x 1 ( 0 )   2 .0 
 x& 1 (0)  0.0

=  ; 
= 
 x 2 ( 0 )   4 .0 
x& 2 (0) 0.0
which correspond to
2 
0.731
− 2.732
  = c1 
 sin (ψ1 ) + c 2 
 sin (ψ 2 )
4 
 1 
 1 
(5.2.2a)
0
0.731
− 2.732
  = ω1c1 
 cos(ψ1 ) + ω2c 2 
 cos(ψ 2 ) (5.2.3a)
0
 1 
 1 
22.457 Mechanical Vibrations - Chapter 5
16
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
MDOF - Initial Conditions
Example 5.2.1 - upon solving these equations for
the response due to the specified initial conditions
yields:
 x1 
0.731
− 2.732
  = 3.732
 cos(ω1t ) + 0.268
 cos(ω2 t )
 1 
 1 
x 2 
22.457 Mechanical Vibrations - Chapter 5
17
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
MDOF - Coordinate Coupling
Coordinate coupling exists in many problems.
Either static coupling, dynamic coupling or both
static and dynamic coupling can exist.
The equations of motion are:
m11&x&1 + m12 &x& 2 + k11x1 + k12 x 2 = 0
m 21&x&1 + m 22 &x& 2 + k 21x1 + k 22 x 2 = 0
(5.3.1)
and can be cast in matrix form as:
 m11 m12   &x&1   k11 k12   x1  0
 + 
 = 
m


 21 m 22  &x& 2  k 21 k 22  x 2  0
22.457 Mechanical Vibrations - Chapter 5
18
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(5.3.2)
MDOF - Coordinate Coupling
Coordinate coupling can be eliminated through a
transformation to a different coordinate system
wherein the independent variables are not coupling
either statically or dynamically.
These coordinates are referred to as
‘principal coordinates’
or
‘normal coordinates’
22.457 Mechanical Vibrations - Chapter 5
19
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
MDOF - Coordinate Coupling
For systems with general damping, this is not
easily possible unless the damping is of a special
form or the system is first converted to the state
space formulation of the system equations
0   &x&1 
m11
 0 m  &x& 

22   2 
 c11 c12   x& 1 
+ 
 x& 
c
c
 21 22   2 
(5.3.3)
k11 0   x1  0
+
 = 

 0 k 22  x 2  0
22.457 Mechanical Vibrations - Chapter 5
20
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
MDOF - Coordinate Coupling
Example 5.3.1 Static Coupling
m 0 &x&   (k1 + k 2 )
 0 J  &θ&  + (k l − k l )

   2 2 1 1
22.457 Mechanical Vibrations - Chapter 5
21
(k 2l 2 − k1l1 ) x  0
(
k1l12
+ k 2l 22
)
 θ  = 0
   
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
MDOF - Coordinate Coupling
Example 5.3.1 Dynamic Coupling
 m me &x& c  (k1 + k 2 )
me J   &θ&  + 
0

c   
22.457 Mechanical Vibrations - Chapter 5
22
0
k1l32 + k 2l 24
(
 x c  0
  θ  = 0
   
)
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
MDOF - Coordinate Coupling
Example 5.3.1 Static & Dynamic Coupling
 m ml1  &x&1  (k1 + k 2 ) k 2l  x1  0
 &&  + 
2   =  
ml

k 2l   θ  0
 1 J   θ   k 2l
22.457 Mechanical Vibrations - Chapter 5
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
MDOF - Forced Harmonic Vibration
Consider a system excited by a harmonic force
m11
  &x&1   k11 k12   x1  F1 
 +
 =  sin ωt



m 22  &x& 2  k 21 k 22  x 2   0 

(5.4.1)
which has a solution assumed to be
 x1   X1 
 =  sin ωt
 x 2  X 2 
22.457 Mechanical Vibrations - Chapter 5
24
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
MDOF - Forced Harmonic Vibration
Substituting into the differential equation yields
(
 k11 − m11ω2

k 21

)
k12
k 22 − m 22ω2
(
  X1  F1 
  = 
 X 2   0 
)
(5.4.2)
which is generally written in terms of the
impedance matrix as
 X1  F1 
[Z(ω)] = 
X 2   0 
22.457 Mechanical Vibrations - Chapter 5
25
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
MDOF - Forced Harmonic Vibration
Solving this yields
 X1 
−1F1  Adj[Z(ω)] F1 
 =[Z(ω)]   =
 
 0  det Z(ω)  0 
X 2 
(5.4.3)
where the adjoint matrix and determinant are
(
 k 22 − m 22ω2
Adj[Z(ω)] = 
− k 21

(
)

− k12
2 
k11 − m11ω 
(
)
)(
det Z(ω) = m1m 2 ω12 − ω2 ω22 − ω2
22.457 Mechanical Vibrations - Chapter 5
26
)
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
MDOF - Forced Harmonic Vibration
The general equation becomes
 k 22 − m 22ω2
− k12

k11 − m11ω2
− k 21
 X1  
 =
m1m 2 ω12 − ω2 ω22 − ω2
X 2 
(
)
(
(
)(
)


 F1 
 
0
)
(5.4.3)
and the amplitudes of response are
(
k 22 − m 22ω2 )F1
X1 =
m1m 2 (ω12 − ω2 )(ω22 − ω2 )
− k 21F1
X2 =
m1m 2 ω12 − ω2 ω22 − ω2
(
)(
)
(5.4.6)
Example 5.4.1 and 5.4.2 are good examples
22.457 Mechanical Vibrations - Chapter 5
27
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
MDOF - Vibration Absorber
A very common, practical application of a 2 DOF
system is that of the ‘tuned absorber’. This is
commonly used to minimize objectionable resonance
Recall
ω12 =
k1
k
; ω22 = 2
m1
m2
(5.6.1)
The amplitude of response for X1 is
  ω 2 
1 −   
  ω2  
X1k1
=
F0
 k  ω 2    ω 2  k
1 + 2 −    1 −    − 2
 k1  ω1     ω2   k1
22.457 Mechanical Vibrations - Chapter 5
28
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(5.3.2)
MDOF - Tuned Absorber
  ω 2 
1 −   
  ω2  
X1k1
=
F0
 k  ω 2    ω 2  k
1 + 2 −    1 −    − 2
 k1  ω1     ω2   k1
22.457 Mechanical Vibrations - Chapter 5
29
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
MDOF - Tuned Absorber
  ω 2 
1 −   
  ω2  
X1k1
=
F0
 k  ω 2    ω 2  k
1 + 2 −    1 −    − 2
 k1  ω1     ω2   k1
22.457 Mechanical Vibrations - Chapter 5
30
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
MATLAB Examples - VTB4_2
VIBRATION TOOLBOX EXAMPLE 4_2
>> clear
>> m=[1 0;0 1];k=[2 -1;-1 1];x0=[1;0];v0=[0;0];tf=5;plotpar=1;
>> [x,v,t]=VTB4_2(m,k,x0,v0,tf,plotpar);
P res s any key to continue
1
0.8
0.6
P res s any key to continue
Dis placement of X1
0.4
0.8
0.2
0.6
0
0.4
Dis placement of X2
-0.2
-0.4
-0.6
-0.8
0
0.5
1
1.5
2
2.5
time (s ec)
3
3.5
4
0.2
0
-0.2
4.5
5
-0.4
-0.6
-0.8
22.457 Mechanical Vibrations - Chapter 5
0
31
0.5
1
1.5
2
2.5
time (s ec)
3
3.5
4
4.5
5
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
MATLAB Examples - VTB5_1
VIBRATION TOOLBOX EXAMPLE 5_1
>> clear
>> m=1; c=.1; k=2;
>> VTB5_1(m,c,k)
>>
Trans mis s ibility plot for ζ = 0.035355 ω = 1.4142 rad/s
15
Trans mis s ibility Ratio
10
5
0
0
22.457 Mechanical Vibrations - Chapter 5
0.2
0.4
0.6
32
0.8
1
1.2
Dimens ionles s Frequency
1.4
1.6
1.8
2
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
MATLAB Examples - VTB5_4
VIBRATION TOOLBOX EXAMPLE 5_4
>> beta=1
beta =
Mas s ratio vers us s ys tem natural frequency for β = 1
1
1
0.9
>> VTB5_4(beta)
>>
0.8
0.7
mas s ratio - µ
0.6
0.5
0.4
0.3
0.2
0.1
0
0.6
22.457 Mechanical Vibrations - Chapter 5
0.8
1
33
1.2
1.4
normalized frequency ω/ωa
1.6
1.8
2
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory