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1 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 6
Mechanical VibrationsChapter 6
Peter AvitabileMechanical Engineering DepartmentUniversity of Massachusetts Lowell
2 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 6
MDOF Equations of Motion
Equation of Motion for 2 DOF
can be written in compact matrix form as
( )
( )
=
−
−++
−
−++
)t(f)t(f
xx
kkkkk
xx
ccccc
xx
mm
2
1
2
1
22
221
2
1
22
221
2
1
2
1
&
&
&&
&&
[ ] [ ] [ ] FxKxCxM =++ &&&
3 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 6
MDOF Equations of Motion
This coupled set of equations can be uncoupled byperforming an eigensolution to obtain ‘eigenpairs’for each mode of the system, that is
‘eigenvalues’ and ‘eigenvectors’
or
‘frequencies’ (poles) and ‘mode shapes’
4 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 6
MDOF Equations of Motion
Equation of Motion[ ] [ ] [ ] )t(FxKxCxM =++ &&&
Eigensolution
Frequencies (eigenvalues) andMode Shapes (eigenvectors)
[ ] [ ][ ] 0xMK =λ−
[ ] [ ]L2122
21
2 uuUand\\
\=
ω
ω=
Ω
5 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 6
MDOF - Orthogonality of Eigenvectors
Normal modes are ‘orthogonal’ with respect to thesystem mass and stiffness matrices.The eigen problem can be written as
For the ith mode of the system, the eigenproblemcan be written as
[ ][ ] [ ][ ][ ]2UMUK Ω=
[ ] [ ] iii uMuK λ= (6.6.1)
6 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 6
MDOF - Orthogonality of Eigenvectors
Premultiply that equation by the transpose of adifferent vector uj
and write a second equation for uj and premultiplyit by the transpose of vector ui
(6.6.2)
(6.6.3)
[ ] [ ] iTjii
Tj uMuuKu λ=
[ ] [ ] jTijj
Ti uMuuKu λ=
7 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 6
MDOF - Orthogonality of Eigenvectors
Subtracting 6.6.3 from 6.6.2, yields
Since the eigenvalues for each mode are different,
When i=j, then the values are not zero
(6.6.2)
(6.6.8)
( ) [ ] 0uMu jT
iji =λ−λ
[ ] [ ]
jikuKumuMu
iiiT
i
iiiT
i =
==
[ ] ji0uMu jT
i ≠= (6.6.6)
8 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 6
Modal Matrix and Modal Space Transformation
Define the modal matrix as the collection ofmodal vectors for each mode organized incolumn fashion in the modal matrix
This modal matrix is then used to define themodal transformation equation with a newcoordinate with ‘p’ as the principal coordinate
[ ] [ ]L321 uuuU =
[ ] [ ]
==
M
L 2
1
21 pp
uupUx
9 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 6
Modal Space Transformation
Substitute the modal transformation into theequation of motion
In order to put the equations in normal form, thisequation must be premultiplied by the transpose ofthe projection operator to give
[ ] [ ][ ] [ ] [ ][ ] [ ] [ ][ ] [ ] FUpUKUpUCUpUMU TTTT =++ &&&
[ ][ ] [ ][ ] [ ][ ] FpUKpUCpUM =++ &&&
10 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 6
Modal Space Transformation
The first term of the modal acceleration canbe expanded as
[ ] [ ][ ]
[ ] ( ) [ ] ( ) [ ] ( ) [ ] ( ) [ ] ( ) [ ] ( ) [ ] ( ) [ ] ( ) [ ] ( )
=
OMMM
L
L
L
3T
3T
31T
3
3T
22T
21T
2
3T
12T
11T
1
T
uMu2uMuuMuuMuuMuuMuuMuuMuuMu
UMU
11 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 6
Modal Space Transformation
Recall from orthogonality that
[ ] ji0uMu jT
i ≠=
so that
[ ] [ ][ ]
[ ] ( ) [ ] ( )
[ ] ( )
=
OMMM
L
L
L
3T
3
2T
2
1T
1
T
uMu000uMu000uMu
UMU
12 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 6
Modal Mass, Modal Damping, Modal Stiffness
The mass becomes
The stiffness becomes
The damping becomes(under special conditions)
[ ] [ ][ ]
=
OMMM
L
L
L
33
22
11
T
m000m000m
UMU
[ ] [ ][ ]
=
OMMM
L
L
L
33
22
11
T
k000k000k
UKU
[ ] [ ][ ]
=
OMMM
L
L
L
33
22
11
T
c000c000c
UCU
13 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 6
Modal Space Transformation
The physical set of highly coupled equations aretransformed to modal space through the modaltransformation equation to yield a set of uncoupledequations
Modal equations (uncoupled)
=
+
+
MMM
&
&
M
&&
&&
FuFu
pp
\k
kpp
\c
cpp
\m
mT
2
T1
2
1
2
1
2
1
2
1
2
1
2
1
14 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 6
Modal Space Transformation
Diagonal Matrices - Modal Mass Modal Damping Modal Stiffness
Highly coupled system
transformed intosimple system
[ ] FUp\
K\
p\
C\
p\
M\
T=
+
+
&&&
m
k
1
1 c1
mp
3
3
c3
m
k
2
2 c2
f3
k 3
p 2 f2f1p 1
MODE 1 MODE 2 MODE 3
15 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 6
Modal Space Transformation
It is very clear to see that these modal spaceequations result in a set of independent SDOFsystemsThe modal transformation equation uncouples thehighly coupled set of equationsThe modal transformation appropriates the forceto each modal oscillator in modal spaceThe modal transformation equation combines theresponse of all the independent SDOF systems toidentify the total physical response
16 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 6
Modal Space Response Analysis
Since the MDOF system is reduced to equivalentSDOF systems with appropriate force, theresponse of each SDOF system can be determinedusing SDOF approaches discussed thus far.The total response due to each of the SDOFsystems can be determined using the modaltransformation equation
17 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 6
Modal Space Transformation
[ M ]p + [ C ]p + [ K ]p = [U] F(t)T
m
k
1
1 c1
f1p1
MODE 1
m
k
2
2 c2
p2 f2
MODE 2
mp
3
3
c3
f3
k3
MODE 3
MODAL
[M]x + [C]x + [K]x = F(t)
x = [U]p = [
SPACE
++ u p 33u p 22u p 11
u 3u 2u 1 ]p
.. .
.. .
+
+
Σ=
18 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 6
Modal Space - Modal Matrices
[ ] [ ][ ]
=
\M
\UMU 111
T1
[ ] [ ][ ]
=
\C
\UCU 111
T1
[ ] [ ][ ]
=
\K
\UKU 111
T1Modal Stiffness
Modal Damping
Modal Mass
The mode shapes are linearly independent andorthogonal w.r.t the mass and stiffness matrices
TRUE !!!
???????
TRUE !!!
19 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 6
Proportional Damping
[ ] [ ] [ ][ ][ ]
β+α=β+α
\KM
\UKMU 1
T1
The damping matrix is only uncoupled for a specialcase where the damping is assumed to beproportional to the mass and/or stiffness matrices
Many times proportional damping is assumed sincewe do not know what the actual damping isThis assumption began back when computationalpower was limited and matrix size was of criticalconcern. But even today we still struggle with thedamping matrix !!!
20 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 6
Non-Proportional Damping
If the damping is not proportional then a statespace solution is required (beyond our scope here)
[ ] [ ][ ] [ ]
[ ] [ ][ ] [ ]
=
−
−
F0
xx
K00M
xx
CMM0
1
1
11
1 &
&
&&
[ ] [ ] QYAYB 11 =−&
[ ] [ ] [ ][ ] [ ]
=
11
11 CM
M0B [ ] [ ] [ ]
[ ] [ ]
−
=1
11 K0
0MA
[ ] [ ][ ] 0YBA 11 =λ− [ ] 11 pY Φ=
The equation of motion can be recast as
The eigensolution and modal transformation is then
21 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 6
MATLAB Examples - VTB4_1
VIBRATION TOOLBOX EXAMPLE 4_1
>> clear>> m=[1 0;0 1];k=[2 -1;-1 1];>> [P,w,U]=VTB4_1(m,k)
P =0.5257 0.8507
0.8507 -0.5257w =
0.6180 1.6180U =
0.5257 0.8507 0.8507 -0.5257>> mm =
1 0 0 1>> kk =
2 -1 -1 1
>>
22 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 6
MATLAB Examples - VTB4_3VIBRATION TOOLBOX EXAMPLE 4_3
>> clear>> m=[1 0;0 1];d=[.1 0;0 .1];k=[2 -1;-1 1];>> [v,w,zeta]=VTB4_3(m,d,k)Damping is proportional, eigenvectors are real.
v = -0.5257 -0.8507 -0.8507 0.5257w = 0.6180 1.6180zeta = 0.0809 0.0309>> mm = 1 0 0 1>> dd = 0.1000 0 0 0.1000>> kk =
2 -1 -1 1