1 Dr. Peter Avitabile Modal Analysis & Controls Laboratory System Modeling Concepts System Modeling Concepts Peter Avitabile Modal Analysis & Controls Laboratory Mechanical Engineering Department University of Massachusetts Lowell [ K ] n [ M ] n [ M ] a [ K ] a [ E ] a [ ω ] 2 Structural Dynamic Modeling Techniques & Modal Analysis Methods m m k k c c 1 1 2 2 1 2 4. 0 0. 1 1. 0 Li near Fr equency ( Hz) 3. 0 2. 0 5. 0 1.0 E+1 1.0 E+2 m k 2 c 2 2 4. 0 0. 1 1. 0 Li near Fr equency ( Hz) 3. 0 2. 0 5. 0 1.0 E+3 1.0 E+4 m k 1 1 c 1
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
1 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
System Modeling Concepts
Peter AvitabileModal Analysis & Controls LaboratoryMechanical Engineering DepartmentUniversity of Massachusetts Lowell
19 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
Component Mode Synthesis
We can represent the physical coordinates x in terms of component generalized coordinates p:
where Ψ = a matrix of component modes.For the Craig-Bampton method, constraint modes and fixed-interface normal modes are used as component modes.
px Ψ=
20 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
Component Mode Synthesis
Fixed-Interface normal modes Um (where “m” refers to kept modes)All juncture coordinates are constrainedObtain the normal modes by solving the eigenvalueproblem:
It is assumed that the modes are scaled to unit modal mass.
( ) 0xmk 2 =ω−
21 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
Component Mode Synthesis
Constraint modesPartition physical coordinates x into two sets: c, the constrained coordinates—the coordinates relative to which the constraint modes will be defined; and i, the remaining (interior) coordinatesStatically impose a unit displacement on one constrained coordinate, and maintain a zero displacement on the other constrained coordinates. The remaining i coordinates are free to deform.
22 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
Component Mode Synthesis
Constraint modesThe constraint mode matrix is therefore
Where Tc = constraint mode transformation
−
=
= −
ic1ii
cc
ic
cccKK
ICI
T
23 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
Component Mode Synthesis
Applying the Craig Bampton method:Let for each component, whereUm = kept fixed-interface normal modes, andC = constraint modes.
cmm CppUx +=
24 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
Component Mode Synthesis
Since the normal modes are fixed-interface modes, and , this can be written in partitioned form aswhereIcc = identity associated with unit displacement at connection DOF (for constraint modes)Cic = Resulting constraint modes, equal to Uim = Normal modes of system with c constraints applied (fixed-interface normal modes)
Note: n – c = i n = all points (as usual)c = constrained points i = interior points
cc xp ≡
=
m
c
imic
cc
i
c
pp
UC0I
xx
ic1ii KK−−
25 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
Component Mode Synthesis
In general, for each component, the mass and stiffness matrices are
where Ψ is the matrix of component modes.
ΨΨ
ΨΨ
K
MT
T
=κ
=µ
26 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
Component Mode Synthesis
Using the Craig Bampton method, the mass is given by
where
µµµµ
=µmmmc
cmcc
( )( ) ccicciicicii
Ticcc
ijiciiTim
Tcmmc
mmmm
mCmmCmC
mCMU
I
+++=µ
+=µ=µ
=µ
27 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
Component Mode Synthesis
Using the Craig Bampton method, the stiffness is given by
Where
Note that κcc is the Guyan reduced stiffness matrix
κκκκ
=κmmmc
cmcc
ic1iicicccc
cmTmc
2mmmm
KKKK
0−−=κ
=κ=κ
==κ ΩΛ
28 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
Component Mode Synthesis
To assemble system matrices, first write the equation of interface displacement compatibility (with pc
α being the dependent coordinate):
Written in the form therefore
this is
D = dependent
I = linearly independent
0pp cc =− αβ
0Dp =
[ ] 0
pppp
0I0I
m
c
m
c
=
−
β
β
α
α
[ ]0I0ID −=
[ ]0I0DID
DI
DD
=−=
29 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
Component Mode Synthesis
Then define S so , wherep = original generalized coordinates, andq = new generalized coordinates.
In this case
Sqp =
=
β
α
β
β
β
α
α
m
m
c
m
c
m
c
ppp
I0000I0I000I
pppp
30 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
Component Mode Synthesis
To get synthesized system M and K matrices:
SSK
SSMT
T
κ=
µ=αβ
αβ
µµµµ
µµµµ
=
ββ
ββ
αα
αα
αβ
I0000I0I000I
0000
0000
I0I00I00000I
M
mmmc
cmcc
mmmc
cmcc
31 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
System Modeling Concepts
Then
where
=ββ
αα
βα
αβ
mmmc
mmmc
cmcmcc
M0M0MMMMM
M
( )( )
βα
βββ
ααα
ββ
αα
µ+µ=
µ==
µ==
=
=
cccccc
mcT
cmmc
mcT
cmmc
mmmm
mmmm
M
MM
MM
IM
IM
32 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
System Modeling Concepts
And
Where= diagonal matrix, modal stiffness of α= diagonal matrix, modal stiffness of β
= full matrix of stiffness terms for reduced α and β
=
β
ααβ
kk
mm
cc
K000K000K
K
αα = mmmmK Λββ = mmmmK Λ
βα κ+κ= ccccccK
33 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
System Modeling Concepts
Then solve the equation of motion for the assembled system:
Transform from q to p coordinates using
and then from p to u (physical coordinates) using
0qKqM =+ αβαβ &&
Sqp =
=
m
c
imic
cc
i
c
pp
UC0I
xx
34 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
Impedance Modeling Techniques
Consider a cantilever beam. It is desired to estimate the FRF between point c and b when the tip of the beam is pinned to ground.In particular, the FRF hcb when xa = 0
abc
35 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
Impedance Modeling Techniques
The response at "a" is related to the force at "a" and "b" through
where xa is the vertical translation at the tip of the beam. With the constraint xa = 0, the force at point "a" becomes
abc
aaababa fhfhx +=
bab1-aaa fhhf −=
36 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
Impedance Modeling Techniques
The response at "c" due to an excitation at “a” and "b" is
In order to include the effects of the constraint at “a” , the force at point "a" with the constraint xa = 0, changes this equation to
which are obtained from the unconstrained system
abc
bcbacac fh+fhx =
ab1-aacacb
b
ccb~
hhh-h=fxh =
37 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
Summary of Impedance Modeling
Frequency Response Functions can also be used to investigate structural modifications. The FRF can be written as
Using force balance and compatibility equations, the effects of a modification can be written in terms of the unmodified system as
)pj(
uuq)pj(
uuq)j(H *
k
*jk
*ikk
m
1k k
jkikkij
−ω+
−ω=ω ∑
=
ab1aacacb
b
ccb HHHH
FxH~ −−==
aaababa FHFHx +=
bab1aaa FHHF −−=
bcbacac FHFHx +=
c b a
38 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
Frequency Based System Modeling Techniques
Consider combining two systems together
COMPONENT (A)
a-DOFs b-DOFsc-DOFs c-DOFs
SYSTEM (S)
COMPONENT (B)
39 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
Frequency Based System Modeling Techniques
The equation of motionfor each component is
where n = a + c for component (A) n = b + c for component (B).
Note that the number of "c" coordinates are the same on component (A) and (B)
COMPONENT (A)
a-DOFs b-DOFsc-DOFs c-DOFs
SYSTEM (S)
COMPONENT (B)
[ ] nnnn FHX =
40 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
FBS Modeling Techniques
Component A can be partitioned as
Component B can be partitioned as
[ ] [ ][ ] [ ]
ncAa
A
nnccA
caA
acA
aaA
ncAa
A
FF
HHHH
XX
=
[ ] [ ][ ] [ ]
nbBc
B
nnbbB
bcB
cbB
ccB
nbBc
B
FF
HHHH
XX
=
COMPONENT (A)
a-DOFs b-DOFsc-DOFs c-DOFs
SYSTEM (S)
COMPONENT (B)
(4-8)(4-8)(4-8)
(4-8)
(4-9)
41 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
FBS Modeling Techniques
When rigidly connecting Component A to Component B,compatibility implies that
and equilibrium at the "c" DOFs requires that
where "S" superscript is used to represent system comprised of Component A rigidly coupled to Component B at the connection DOFs "c"
COMPONENT (A)
a-DOFs b-DOFsc-DOFs c-DOFs
SYSTEM (S)
COMPONENT (B)
cScB
cA XXX ==
cScB
cA FFF =+
(4-10)
(4-11)
42 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
FBS Modeling Techniques
The FRFs of the uncoupled system can be defined as
From the partitioned equations for Component A and Component B, the connection DOF are
COMPONENT (A)
a-DOFs b-DOFsc-DOFs c-DOFs
SYSTEM (S)
COMPONENT (B)
[ ] [ ] [ ][ ] [ ] [ ][ ] [ ] [ ]
nbSc
Sa
S
nnbbS
bcS
baS
cbS
ccS
caS
abS
acS
aaS
nbSc
Sa
S
FFF
HHHHHHHHH
XXX
=
(4-12)
(4-13)
(4-14)
[ ] [ ] cAcc
Aa
Aca
Ac
A FHFHX +=
[ ] [ ] [ ] bBcb
Bc
Bcc
Bc
B FHFHX +=
43 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
FBS Modeling Techniques
These two equations can be equated and used to solve for the connection force as
From these equations derived above, the coupled system FRFs can be determined in terms of the uncoupled FRFs of the individual components. Equations (4-8), (4-9), (4-12) and (4-15) are used in the development of the coupled system.
COMPONENT (A)
a-DOFs b-DOFsc-DOFs c-DOFs
SYSTEM (S)
COMPONENT (B)
(4-15)
[ ] [ ][ ] [ ] [ ] [ ] [ ]cSccB
aA
caA
bB
cbB1
ccB
ccA
cA FHFHFHHHF~ +−+=
−
44 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
FBS Modeling Techniques
As an example,will be derived The first equation of (4-12) of the coupled system is
and the first equation of (4-8) of the uncoupled system is
COMPONENT (A)
a-DOFs b-DOFsc-DOFs c-DOFs
SYSTEM (S)
COMPONENT (B)
(4-16)
(4-17)
[ ]aaSH
[ ] [ ] [ ] bSabS
cS
acS
aS
aaS
aS FHFHFHX ++=
[ ] [ ] cAac
Aa
Aaa
Aa
A FHFHX +=
45 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
FBS Modeling Techniques
When the systems are coupled, the force on "A" from "B" is given by from (4-15) and the corresponding response associated withthat coupling force is which then becomes
then (4-16) and (4-18) combine to give
COMPONENT (A)
a-DOFs b-DOFsc-DOFs c-DOFs
SYSTEM (S)
COMPONENT (B)
(4-18)
(4-19)
cAF~
aSX [ ] [ ] cA
acA
aA
aaA
aS F~HFHX +=
[ ] [ ] [ ] [ ] [ ] cA
acA
aA
aaA
bS
abS
cS
acS
aS
aaS
F~HFH
FHFHFH
+=
++
46 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
FBS Modeling Techniques
The FRF, is developed realizing that are zero
Substituting (4-15) into (4-19) and simplifying, allows for the calculation of in terms of the uncoupled FRF matrices as
COMPONENT (A)
a-DOFs b-DOFsc-DOFs c-DOFs
SYSTEM (S)
COMPONENT (B)
(4-20)
[ ]aaSH
cSF bSF bBF
[ ]aaSH
[ ] [ ] [ ] [ ] [ ][ ] [ ]caA1cc
Bcc
Aac
Aaa
Aaa
S HHHHHH−
+−=
47 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts
FBS Modeling Techniques
Schematically this is shown as [ ] [ ][ ] cjA1
ccB
ccA
icAA
ijSij HHHHhh
−+−=
COMPONENT A
COMPONENT B
CONNECTION POINTS
ji
FRFsdescribingconnection
points
FRFsdescribinginput force
points
FRFsdescribing
output responsepoints
COMPONENTEVALUATION
48 Dr. Peter AvitabileModal Analysis & Controls LaboratorySystem Modeling Concepts