sdvc.kaist.ac.krsdvc.kaist.ac.kr/course/ce514/ch15.pdf · 1 chapter15. dynamic response of mdof...
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![Page 1: sdvc.kaist.ac.krsdvc.kaist.ac.kr/course/ce514/ch15.pdf · 1 Chapter15. DYNAMIC RESPONSE OF MDOF SYSTEMS: MODE-SUPERPOSITION METHOD Dynamic Response - Direct Integration Methods: efficient](https://reader030.vdocument.in/reader030/viewer/2022040120/5e7a3e6205f2eb2c896161dc/html5/thumbnails/1.jpg)
1
Chapter15. DYNAMIC RESPONSE OF MDOF SYSTEMS: MODE-SUPERPOSITION METHOD
Dynamic Response
- Direct Integration Methods: efficient for the system under loads of short duration such as impact loads or for non-linear
systems - Vector Superposition Methods: efficient for the system under
loads of long duration or harmonic loads or various loads -
Vector Superposition Methods/ a set of coupled equations are transformed into a set of uncoupled equations through use of the orthogonal vectors (Normal Modes) of the system
Orthogonal Vectors
- Normal Modes: orthogonal wrt km and Mode Superposition (Normal Mode) Methods
- Ritz Vectors: orthogonal wrt m - Lanczos Vectors: orthogonal wrt m -
Mode Superposition Methods - Mode Displacement Method - Mode Acceleration Method
§15.1 General Solution for Dynamic Response: Normal Mode Method Initial Value Problem
)(pkuucum t=++ &&& (15.1) 00 u)0(u,u)0(u && == (15.2)
)()()( tututu ph +=
where
solutionparticulartu
solutionmogeneous htusolutiongeneraltu
p
h
:)(o :)(
:)(:
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2
0)mk( 2 =− rr φω (15.3)
)()()(11
ttutu r
N
rr
N
rr ηφ∑∑
==
== )()( ttu ηΦ= (15.1)
[ ]Nφφφ L21=Φ modal matrix (15.6) )(tη principal coordinates.
)()()()(111
tpkcmN
rrrr
N
rrr
N
rr =++ ∑∑∑
===
ηφηφηφ &&&
)()()()( tptktctm =Φ+Φ+Φ ηηη & Pre-multiply by T
sφ and use the orthogonality of φ 0=r
Ts mφφ for sr ≠ ),..,,( 21 N
T MMMdiagm =ΦΦ 0=r
Ts cφφ for sr ≠ ),..,,( 21 N
T CCCdiagc =ΦΦ 0=T
rTs kφφ for sr ≠ ),..,,( 21 N
T KKKdiagk =ΦΦ Note that .and,wrt orthogonalis kcmφ Then
+)(m trTr ηφφ && +)(m tr
Tr ηφφ && )()(k tpt T
rrTr φηφφ =
)(tmt η&&ΦΦ + )(tct η&ΦΦ )()( tptk TT Φ=ΦΦ+ η Let
)()(mcm
r
r
tptPKCM
Trr
rTrr
rTr
rTr
φφφφφφφ
=
=
=
=
)()(),...,,(
),..,,(),..,,(
21
21
21
tptPKKKdiagkK
CCCdiagcCMMMdiagmM
Tr
NT
NT
NT
Φ=
=ΦΦ=
=ΦΦ=
=ΦΦ=
force modal)(stiffness modaldamping modal
mass modal
====
tPKCM
r
r
r
r
vector force modal)(matrix stiffness modalmatrix damping modal
matrixmassmodal
====
tPKCM
(15.9)
Then
)(tPKCM rrrrrrr =++ ηηη &&& )(tPKCM =++ ηηη &&& (15.8)
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3
r
rrrtrrr M
tP )(2 2 =++ ηωηωζη &&&
The transformation to principle coordinates has uncoupled the equations of motion (compare Eq. 15.8 with the original equation of motion, Eq. 15.1). initial conditions
)0()0( r
N
rru ηφ∑
=
= )0()0(u ηΦ=
)0()0( r
N
rru ηφ && ∑
=
= )0()0(u η&& Φ= (15.10)
)0()0( r
N
rr
Ts
Ts mmu ηφφφ ∑
=
= )0()0(muT ηΦΦ=Φ mT
=)0(muTrφ )0(rr
Tr m ηφφ )0(muTΦ )0(ηM=
Therefore
=)0(muTrφ )0(rrM η
Similarly =)0(umT
r &φ )0(rrM η& modal initial conditions
Nr
Mmum
Mmmu
rrTr
Tr
r
Tr
rrTr
Tr
r
,,2,1(0)um1)0()0(
mu(0)1)0()0(
Tr
K
&&
&
=
==
==
φφφ
φη
φφφ
φη (15.12)
:)0(),0( uu & Known
)(tPKCM rrrrrrr =++ ηηη &&& )(tPKCM =++ ηηη &&& (15.8) Uncoupled equations
)()()( ttt prhrr ηηη += Solution Mehods
- Duhamel integration methods for simple loads - Direct integration methods for complex loads
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4
§15.2 Mode-Displacement Solution for Response of Undamped MDOF Systems
∑∑==
==N
rrr
N
rr ttutu
11)()()( ηφ
≈)(tu ∑∑==
==N
rrr
N
rr ttutu
ˆ
1
ˆ
1)()(ˆ)(ˆ ηφ )(ˆˆ)(ˆ ttu ηΦ= (15.13)
[ ]N21
ˆ φφφ L=Φ (15.14) where NN <<ˆ (ex. )1000 ,50ˆ == NN To determine the number of modes used in the analysis, engineering experience and judgment are required.
)(ˆˆˆˆˆ tPKM rrrrr =+ ηη&& )(ˆˆˆˆˆ tPKM =+ ηη&& (15.15) By the Duhamel integration method
ττωτω
ωηω
ωηη
dtPM
ttt
r
t
rrr
rrr
rrr
)sin()(1
)sin()0(1)cos()0()(
0−
+
+=
∫
&
(15.18)
The process of computing modal responses and substituting these, )(trη , back into Eq. 15.13 to obtain the approximate system response
will be called the mode-displacement method. Note The contribution of the higher modes is small. •Free Vibration )(tuh
0p(t) = (15.19)
)sin()0(1)cos()0()( ttt rrr
rrr ωη
ω
+ωη=η & (15.18)
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5
Nr
M
M
rr
Tr
rr
,,2,1
(0)um1)0(
mu(0)1)0(
Tr
K
&&
=
=
=
φη
φη
(15.12)
Example 13.5 Free Vibration
Initial Conditions
=
02
1 000
u)(u)(u
=
00
00
2
1
)(u)(u
&
&
=
=
11
, 1
2/1
1 φωmk
−
=
=
11
,32
2/1
2 φωmk
Solution a. r
TrrM φφ m= (1)
k k km m
1u 2u
=
−
−+
00
22
00
2
1
2
1
uu
kkkk
uu
mm
&&
&&
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6
[ ] mm
mM 2
11
00
111 =
= (2)
[ ] mm
mM 2
11
00
112 =
−
−=
b.
rr
Tr
r
r
Tr
r
M
M
ωφη
φη
(0)um)0(
mu(0))0(
&& =
= (3)
=
=
=
=
00
00
00
(0)um
000
0mu(0)
00
mm
muumm
&
(4)
[ ]
[ ]2
011
21mu(0))0(
20
1121mu(0))0(
0
02
22
0
01
11
umumM
umumM
T
T
−=
−
==
=
==
φη
φη (5)
−
=1
12
)0( 0uη
[ ]
[ ] 000
1121mu(0))0(
000
1121mu(0))0(
2
22
1
11
=
−
==
=
==
mM
mMT
T
φη
φη
&
&
(5)
=00
)0(η&
)sin()0(1)cos()0()( ttt rrr
rrr ωηω
ωηη &
+= (15.18)
)cos()0()( tt rrr ωηη =
)(tη =
tt
22
11
cos)0(cos)0(
ωηωη
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7
=
− t
tu
2
10
coscos
2 ωω
)()( ttuh ηΦ=
=
−
− t
tu
2
10
coscos
21111
ωω
[ ]
[ ])cos()cos(2
)(
)cos()cos(2
)(
210
2
210
1
ttutu
ttutu
ωω
ωω
+
=
−
=
•Particular Solution )(tup for harmonic load IF tΩ= cosPp(t) Harmonic Loads (15.19)
tP T Ω= cosP)((t) φ (15.20a) tFP rr Ω= cos(t) (15.20b)
PrTrF φ= (15.20c)
)(tPKCM rrrrrrr =++ ηηη &&& Assuming tYt rr Ω= cos)(η
Ω−
=
2)/(11
rr
rr K
FYω
(15.21)
tKFt
rr
rr Ω
Ω−
= cos
)/(11)( 2ω
η (15.21)
tKFt
rr
rN
rr Ω
ωΩ−
φ= ∑
=
cos)/(1
1)(u2
ˆ
1p (15.22)
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8
rr
rr KK
FD PTrφ=≡ modal static deflection (15.23)
§12.4 Response of an Undamped 2-DOF System to Harmonic Excitation: Mode-Superposition Example 12.6 the steady-state response )(tup
.
tP
uu
kuu
m Ω
=
−
−+
cos
03112
2001 1
2
1
2
1
&&
&&
the natural frequencies and modes are
Mode 1 mk /21 =ω
Mode 2 )/(252
1 mk=ω
k k k2
mm2
1u2u
1p
tPp Ω= cos11
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9
Solution
[ ]
−==Φ
211
1121 φφ (1)
[ ] )()(2
121 ttu ηηη
φφ Φ=
= (2)
−
+=
−=
21
21
2
1
2
1
21
211
11
ηη
ηη
ηη
uu
(3)
)]()()([ tpkmT =Φ+ΦΦ ηη&& (4) pkm TTT Φ=ΦΦ+ΦΦ ηη )()( &&
PKM =+ ηη&& (5) pPkKmM TTT Φ=ΦΦ=ΦΦ= ,, (6)
=
−
−=
23003
211
11
200
211
11m
mm
M (7)
=
−
−
−
−=
415003
211
11
32
211
11k
kkkk
K (8)
tPP
tP
P Ω
=Ω
−= coscos
0211
11
1
11 (9)
tPP
km Ω
=
+
cos
415003
23003
1
1
2
1
2
1
ηη
ηη&&
&&Uncoupled Equations (10)
tPkm Ω=+ cos33 111 ηη&& (11a) tPkm Ω=+ cos)4/15()2/3( 122 ηη&& (11b)
tYY
tt
Ω
=
cos)()(
2
1
2
1
ηη
tY Ω= cos11η (12a) tY Ω= cos22η (12b)
21
12
11 )/(1
)3/1(33 ωΩ−
=Ω−
=Pk
mkPY (13a)
![Page 10: sdvc.kaist.ac.krsdvc.kaist.ac.kr/course/ce514/ch15.pdf · 1 Chapter15. DYNAMIC RESPONSE OF MDOF SYSTEMS: MODE-SUPERPOSITION METHOD Dynamic Response - Direct Integration Methods: efficient](https://reader030.vdocument.in/reader030/viewer/2022040120/5e7a3e6205f2eb2c896161dc/html5/thumbnails/10.jpg)
10
22
12
12 )/(1
)15/4()2/3()4/15( ωΩ−
=Ω−
=Pk
mkPY (13b)
[ ] )()(2
121 ttu ηηη
φφ Φ=
= (2)
=
−=
2
1
2
1
211
11
ηη
uu
tYY
Ω
− cos
211
11
2
1
tYY
YY
uu
Ω
−
+=
cos21
21
21
2
1
tUU
uu
Ω
=
cos2
1
2
1
Ω−
−
Ω−
=
Ω−
+
Ω−
=
22
12
1
12
22
12
1
11
)/(115/4
21
)/(13/
)/(115/4
)/(13/
ωω
ωω
kPkPU
kPkPU
(15a)
(15b)
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11
Example 15.1
tP Ωcos1
3u
4u
2u
1u
23 =m
34 =m
22 =m
./inseck1 21 −=m
.k/in8001 =k
16002 =k
24003 =k
32004 =k
![Page 12: sdvc.kaist.ac.krsdvc.kaist.ac.kr/course/ce514/ch15.pdf · 1 Chapter15. DYNAMIC RESPONSE OF MDOF SYSTEMS: MODE-SUPERPOSITION METHOD Dynamic Response - Direct Integration Methods: efficient](https://reader030.vdocument.in/reader030/viewer/2022040120/5e7a3e6205f2eb2c896161dc/html5/thumbnails/12.jpg)
12
−−−−−
−−−
=Φ
=×
=
=
−−−
−−−
=
63688.070797.043761.023506.000000.115859.053989.049655.044817.000000.109963.077910.0
15436.090145.000000.100000.1
882.55079.41660.29294.13
,10
12279.368746.187970.017672.0
3000020000200001
,
73003520
02310011
32 ωω
mk
Solution a.
rrrrTr MKM 2
r ,m ωφφ == (1)
=
23506.049655.077910.00000.1
3000020000200001
23506.049655.077910.00000.1
1
T
M
695.507)87288.2(78.176
87288.2
1
1211
1
===
=
KMK
Mω
4.374,1164239.343.736836658.439.191517732.2
695.50787288.2
44
33
22
11
==
====
==
KMKMKMKM
(2)
b.
=
000
P
1P
(3)
![Page 13: sdvc.kaist.ac.krsdvc.kaist.ac.kr/course/ce514/ch15.pdf · 1 Chapter15. DYNAMIC RESPONSE OF MDOF SYSTEMS: MODE-SUPERPOSITION METHOD Dynamic Response - Direct Integration Methods: efficient](https://reader030.vdocument.in/reader030/viewer/2022040120/5e7a3e6205f2eb2c896161dc/html5/thumbnails/13.jpg)
13
PrTrF φ= (4)
14
13
12
11
15436.090145.0
PFPF
PFPF
=
−==
=
(5)
c.
)1(cos)/()( 2r
rrr r
tKFt−
Ω=η (6)
rrr ω
Ω= (7)
d.
∑=
=N
rrr ttu
ˆ
111 )()( ηφ (8)
e.
)]79.3122/(1)[4.11374()cos15436.0)(15436.0(
3ˆ
)]46.1687/(1)[43.7368()cos90145.0)(90145.0(
2ˆ
)]70.879/(1)[39.1915()cos)(0.1(
1ˆ)]72.176/(1)[695.507(
)cos)(0.1()(
21
21
21
21
1
Ω−Ω
+
=
Ω−Ω−−
+
=
Ω−Ω
+
=
Ω−
Ω=
tP
N
tP
NtP
NtPtu
(9)
80.2851,402.53,3.1179.44,6486.6,5.0
23
21
=Ω=Ω=Ω
=Ω=Ω=Ω
ωω
Constant C in tCPtu Ω= cos)( 11
)10(987.4)10(228.5)10(630.3)10(301.13.1)10(291.3)10(289.3)10(176.3)10(626.25.0)10(604.2)10(602.2)10(492.2)10(970.10
4ˆ3ˆ2ˆ1ˆ
33333
33331
3333
−−−−
−−−−
−−−−
−−−−=Ω
=Ω
=Ω
====
ωω
NNNN
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14
Example 15.2: Example 15.1 Compute rF and rD
=
=
1111
,
0001
ba PP
Solution a.
00000.1
0001
23506.049655.077910.000000.1
P11 =
==
T
aT
aF φ (1a)
51071.2
1111
23506.049655.077910.000000.1
P11 =
==
T
bT
bF φ (1b)
06931.0,15436.076801.0,90145.007713.0,00000.151071.2,00000.1
44
33
22
11
==
−=−=
−==
==
ba
ba
ba
ba
FFFFFFFF
(2)
b.
r
rr K
FD = (3)
)10(6094.0),10(3571.1)10(0423.1),10(2234.1)10(4027.0),10(2209.5)10(9453.4),10(9697.1
34
34
33
33
32
32
31
31
−−
−−
−−
−−
==
−=−=
−==
==
ba
ba
ba
ba
DDDDDDDD
(4)
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15
Example 12.3 assumed-modes model (Ex. 11.9) Solve for the natural frequencies and modes of this model, and sketch the modes. Use the notation 2211 , uquq →→ .
Solution Assumed modes
)()()()(),( 2211 tuxtuxtxu ψψ +=
Lxx =)(1ψ
2
2 )(
=
Lxxψ
)()(),( 2
2
1 tuLxtu
Lxtxu
+
= (12)
=
+
00
4333
312151520
60 2
1
2
1
uu
LEA
uuAL&&
&&ρ (1)
)cos(2
1
2
1 αωφφ
−
=
tuu
(2)
=
−
00
12151520
4333
2
12
φφ
µ i (3)
22
2
20 ii EL ωρµ
=
0)153()124)(203( 2222 =−−−− iii µµµ 03)(26)(15 222 =+− ii µµ (5)
6090.1,1243.030
496262 =±
=iµ (6)
Lx
),( txu
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16
=
=
=
=
ρµ
ρω
ρµ
ρω
EEL
EEL
18.3220)(
486.220)(
22
22
21
21
=
ρω EL exact 467.2)( 2
1 (8a)
=
ρω EL exact 21.22)( 2
2 (8b)
0153203 2
21
2 =−+− )i(i
)i(i )()( φµφµ (9)
)()(
i
i
)i(
i 2
2
1
2
153203µµ
φφβ
−−
−=
≡
(10)
381.1
453.0136.1514.0
2
1
−=
−=−=
β
β
−
=
453011
2
1
.
)(
φφ
−
=
381112
2
1
/
)(
φφ
)cos()cos( 22
)2(
2
111
)1(
2
1
2
1 αωφφ
αωφφ
−
+−
=
ttuu
[ ]
−−
=)cos()cos(
22
1121 αω
αωφφ
tt
)()(),( 2
2
1 tuLxtu
Lxtxu
+
= (12)
=
)()(
2
12
tutu
Lx
Lx
= [ ]
−−
)cos()cos(
22
1121
2
αωαω
φφtt
Lx
Lx
(11a)
(7a)
(7b)
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17
=
−−
)cos()cos(11
22
11
21
2
αωαω
ββ tt
Lx
Lx
=
+
+
2
2
2
1 Lx
Lx
Lx
Lx ββ
−−
)cos()cos(
22
11
αωαω
tt
=[ ])()( 21 xx φφ
−−
)cos()cos(
22
11
αωαω
tt
)cos()cos()( 222111 αωφαωφ −+−= ttx = ),(),( )2()1( txutxu +
2
)(
+
=
Lx
Lxx ii βφ (15)
)cos()(),()(iii
i txtxu α−ωφ= (14)
)cos(),(2
)(iii
i tLx
Lxtxu α−ω
β+
= (13)
§15.3 Mode-Acceleration Solution for Response of Undamped MDOF Systems
)(pkuum t=+&& (15.24) The mode-acceleration solution is based on the following.
)ump(ku 1 &&−= − (15.25) )ump(ku~ 1 &&−= − (15.26)
![Page 18: sdvc.kaist.ac.krsdvc.kaist.ac.kr/course/ce514/ch15.pdf · 1 Chapter15. DYNAMIC RESPONSE OF MDOF SYSTEMS: MODE-SUPERPOSITION METHOD Dynamic Response - Direct Integration Methods: efficient](https://reader030.vdocument.in/reader030/viewer/2022040120/5e7a3e6205f2eb2c896161dc/html5/thumbnails/18.jpg)
18
rr
N
rηφ &&∑
=
−− −=ˆ
1
11 mkpk (15.27)
rr
N
rηφ
ω&&∑
=
−
−=
ˆ
12r
1 1pk (15.28)
The first term in the above equation is the pseudo static response, while the second term gives the method its name, the mode-acceleration method. Example 15.3: Example 15.1 a. b. c.
)10(
31250.031250.031250.031250.031250.072917.072917.072917.031250.072917.035417.135417.131250.072917.035417.160417.2
ka 31 −−
==
Solution a.
rr
N
rtpatu ηφ
ω&&1
ˆ
12r
1111
1cos)(~ ∑=
−Ω= (1)
rr
N
rtpatu ηφ
ω&&1
ˆ
12r
2
1111 cos)(~ ∑=
Ω+Ω= (2)
)]79.3122/(1)[4.11374()cos15436.0)(15436.0)(79.3122/(
3ˆ)]46.1687/(1)[43.7368(
)cos90145.0)(90145.0)(46.1687/(
2ˆ)]70.879/(1)[39.1915()cos)(0.1)(70.879/(
1ˆ)]72.176/(1)[695.507()cos)(0.1)(72.176/(
)cos)(10(60417.2)(~
21
2
21
2
21
2
21
2
13
1
Ω−ΩΩ
+
=
Ω−Ω−−Ω
+
=
Ω−ΩΩ
+
=
Ω−ΩΩ
+
Ω= −
tP
NtP
NtP
NtPtPtu
(3)
b. 80.2851and,197.44,0 222 =Ω=Ω=Ω
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19
Constant C in tCPtu Ω= cos)( 11
)10(987.4)10(207.5)10(506.2)10(044.53.1)10(291.3)10(291.3)10(288.3)10(261.35.0)10(604.2)10(604.2)10(604.2)10(604.20
4ˆ3ˆ2ˆ1ˆ
33333
33331
3333
−−−−
−−−−
−−−−
−−−=Ω
=Ω
=Ω
====
ωω
NNNN
c. Use Eq. 15.18 to obtain a general expression for )(trη&& to substitute Into the mode-acceleration equation, Eq. 15.28. Thus
ττωτωωηωωηωη
dtPMM
tPttt
r
t
rr
r
r
r
rrrrrrr
)(sin)()(sin)0(cos)0()(
0
2
−−+
−−=
∫
&&&
(15.29)
Alternative form
ττωττ
ω
ωηωωηωη
dtPddtP
M
ttt
r
t
rrrr
rrrrrrr
)(cos)]([cos)0(1sin)0(cos)0()(
0
2
−−+
−−=
∫
&&&
(15.30).
§15.4 Mode-Superposition Solution for Response of Certain Viscous damped Systems
srsTr ≠= ,0cφφ proportional damping (15.31)
NrtPM r
rrrrrrr ,,2,1),(12 2 K&&& =
=++ ηωηωζη (15.32)
rTr
rrrr
rr MM
C φφωω
ζ c2
12
== (15.33)
te
te
dtePM
t
drt
rrrrdr
drt
r
drtt
rdrr
r
rr
rr
rr
ωηωζηω
ωη
ττωτω
η
ωζ
ωζ
τωζ
sin)]0()0([1
cos)0(
)(sin)(1)( )(
0
−
−
−−
+
+
+
−
= ∫
&
(15.34)
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20
21 rrdr ζωω −= (15.35) mode-displacement solution
∑∑==
==N
rrr
N
rr ttutu
ˆ
1
ˆ
1)()(ˆ)(ˆ ηφ
ignores completely the contribution of the modes from )1ˆ( +N to N . mode-accerleration solution
)umucp(ku 1 &&& −−= − (15.36)
∑∑=
−
=
−− −=N
rrr
N
rrr ttt(t
ˆ
1
1ˆ
1
11 )(mk-)(ck)p(k)u~ ηφηφ &&& (15.37)
The resulting mode-accerleration solution is
∑∑==
−
−=
N
rrr
r
N
rrr
r
r ttt(tˆ
12
ˆ
1
1 )(1-)(2)p(k)u~ ηφω
ηφωζ
&&& (15.38)
tFM
tPpFor
rr
rrrrrr Ω
=++
Ω=
cos12
cos
2ηωηωζη &&& (15.39)
PrTrF φ= (15.20c)
ti
r
rrrrrrrr
ti
eMF
PepFor
Ω
Ω
=++
=
222
ωηωηωζη &&& (15.40)
steady-state response (see Eq. 4.30) ti
rFr eFH rr
ΩΩ= )(/ηη (15.41) where )(/ Ωrr FH η is the complex frequency response function for principal coordinates, given by
)2()1(/1)()( 2/
rrr
rFr
rirKHH rr ζ
η+−
=Ω≡Ω (15.42)
)cos()2()1(
/)(
solutionofpart real the take ,cos
222 r
rrr
rrr t
rrKFt
tPpFor
αζ
η −Ω+−
=
Ω=
(15.43a)
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21
)1(2tan 2
r
r
rr
−=
ζα (15.43b)
∑=
=Φ=N
rrr tt(t
1)()()u ηφη (15.44)
tiN
r rrrr
Trr e
rirKt Ω
=∑
+−
=
12 )2()1(
1P)(uζ
φφ (15.45)
∑=
+−
=Ω≡Ω
N
r rrrr
jrirPuij
rirKHH jir
12/
)2()1(1)()(
ζφφ
(15.46)
)cos()2()1(
1P)(u
cosFor
122 r
N
r rrrr
Trr t
rrKt
tPp
αζ
φφ−Ω
+−
=
Ω=
∑=
(15.47)
A plot of Eq. 15.46 on the complex plane is referred to as a complex frequency response plot.
∑=
+−−
=
N
r rrr
r
r
jririj
rrr
KHR
1222
2
)2()1()1()(ζ
φφ (15.48a)
∑=
+−
−
=
N
r rrr
rr
r
jririj
rrr
KHI
1222 )2()1(
2)(ζ
ζφφ (15.48b)
Complex frequency response functions (sometimes called transfer function) as given by Eq. 15.46 are frequently employed in determining the vibrational characteristics of a system experimentally.
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22
Example 15.4
0628.0,6284.01,217,987
cos11
=′===′=
Ω=
ccmkk
tPp
Solution a.
=
′+′−
′−′++
′+′−
′−′++
0)()(
)()(
00
1
2
1
2
1
2
1
puu
kkkkkk
uu
cccccc
uu
mm
&
&
&&
&&
(1)
=
′+′−
′−′++
00
)()(
00
2
1
2
1
uu
kkkkkk
uu
mm
(2)
tωφ cosu = (3)
=
−
′+′−
′−′+00
00
)()(
2
12
φφ
ωm
mkkk
kkk (4)
mkk
mk ′+
==2, 2
221 ωω (5)
−
=
=1
1,
11
21 φφ (6)
70.37,42.31
14211
1421,9871
987
21
22
21
==
====
ωω
ωω
1u 2u
c
c
c′ k
k
k′ 1p
m
![Page 23: sdvc.kaist.ac.krsdvc.kaist.ac.kr/course/ce514/ch15.pdf · 1 Chapter15. DYNAMIC RESPONSE OF MDOF SYSTEMS: MODE-SUPERPOSITION METHOD Dynamic Response - Direct Integration Methods: efficient](https://reader030.vdocument.in/reader030/viewer/2022040120/5e7a3e6205f2eb2c896161dc/html5/thumbnails/23.jpg)
23
HzfHzf 00.62
,00.52
22
11 ====
πω
πω
(7)
b.
−
=Φ11
11 (8a)
=
−
−
=ΦΦ=2002
1111
1001
1111
mTM (8b)
=
−
−
=
−
−
−
−
=ΦΦ=
5080.1002568.1
7540.06284.07540.06284.0
1111
1111
6912.00628.00628.06912.0
1111
cTC
2842)2(14211974)2(987
2222
1211
===
===
MKMK
ωω
=
2842001974
K (8c)
c.
rr
rr M
Cω
ζ2
= (9)
0100.0)70.37)(2(2
5080.1
0100.0)42.31)(2(2
2568.1
2
1
==
==
ζ
ζ (10)
d.
∑=
Ω+Ω−
=Ω
2
12 )/2())/(1(
1)(r rrrr
jririj
iKH
ωζωφφ
Ω+Ω−
+
Ω+Ω−
=Ω
]70.37/)01.0(2[)70.37/(11
2842)1)(1(
]42.31/)01.0(2[)42.31/(11
1974)1)(1()(
2
211
i
iH
(11)
![Page 24: sdvc.kaist.ac.krsdvc.kaist.ac.kr/course/ce514/ch15.pdf · 1 Chapter15. DYNAMIC RESPONSE OF MDOF SYSTEMS: MODE-SUPERPOSITION METHOD Dynamic Response - Direct Integration Methods: efficient](https://reader030.vdocument.in/reader030/viewer/2022040120/5e7a3e6205f2eb2c896161dc/html5/thumbnails/24.jpg)
24
]70.37/)01.0(2[)70.37/(1)10519.3(
]42.31/)01.0(2[)42.31/(1)10066.5(
2
4
2
4
11
Ω+Ω−×
+
Ω+Ω−×
=
−
−
i
iH
(12a)
Ω+Ω−
−
+
Ω+Ω−
=Ω
]70.37/)01.0(2[)70.37/(11
2842)1)(1(
]42.31/)01.0(2[)42.31/(11
1974)1)(1()(
2
221
i
iH
]70.37/)01.0(2[)70.37/(1)10519.3(
]42.31/)01.0(2[)42.31/(1)10066.5(
2
4
2
4
21
Ω+Ω−×
−
Ω+Ω−×
=
−
−
i
iH
(12b)
e.
∑=
Ω+Ω−
Ω−
=
2
1222
2
))/(2())/(1())/(1()(
r rrr
r
r
jririj
KHR
ωζωωφφ
(13a)
∑=
Ω+Ω−
Ω−
=
2
1222 ))/(2())/(1(
)/(2)(r rrr
rr
r
jririj
KHI
ωζωωζφφ
(13b)
![Page 25: sdvc.kaist.ac.krsdvc.kaist.ac.kr/course/ce514/ch15.pdf · 1 Chapter15. DYNAMIC RESPONSE OF MDOF SYSTEMS: MODE-SUPERPOSITION METHOD Dynamic Response - Direct Integration Methods: efficient](https://reader030.vdocument.in/reader030/viewer/2022040120/5e7a3e6205f2eb2c896161dc/html5/thumbnails/25.jpg)
25
f.
π2Ω
=f
![Page 26: sdvc.kaist.ac.krsdvc.kaist.ac.kr/course/ce514/ch15.pdf · 1 Chapter15. DYNAMIC RESPONSE OF MDOF SYSTEMS: MODE-SUPERPOSITION METHOD Dynamic Response - Direct Integration Methods: efficient](https://reader030.vdocument.in/reader030/viewer/2022040120/5e7a3e6205f2eb2c896161dc/html5/thumbnails/26.jpg)
26
![Page 27: sdvc.kaist.ac.krsdvc.kaist.ac.kr/course/ce514/ch15.pdf · 1 Chapter15. DYNAMIC RESPONSE OF MDOF SYSTEMS: MODE-SUPERPOSITION METHOD Dynamic Response - Direct Integration Methods: efficient](https://reader030.vdocument.in/reader030/viewer/2022040120/5e7a3e6205f2eb2c896161dc/html5/thumbnails/27.jpg)
27
Example 15.5
0031.0,6284.0,10,987 =′==′= cckk Solution a.
mkk
mk ′+
==2, 2
221 ωω (1)
73.31,42.31
10071
1007,9871
987
21
22
21
==
====
ωω
ωω
HzfHzf 05.52
,00.52
22
11 ====
πω
πω
(2)
b.
−
=Φ11
11,
=
2002
M (3a,b)
=
−
−
=
−
−
−
−
=ΦΦ=
2692.1002568.1
6346.06284.06346.06284.0
1111
1111
6315.00031.00031.06315.0
1111
cTC (3c)
2014)2(10071974)2(987
2222
1211
===
===
MKMK
ωω
=
2014001974
K (3d)
c.
rr
rr M
Cω
ζ2
= (4)
0100.0)73.31)(2(2
2692.1
0100.0)42.31)(2(2
2568.1
2
1
==
==
ζ
ζ (5)
d.
![Page 28: sdvc.kaist.ac.krsdvc.kaist.ac.kr/course/ce514/ch15.pdf · 1 Chapter15. DYNAMIC RESPONSE OF MDOF SYSTEMS: MODE-SUPERPOSITION METHOD Dynamic Response - Direct Integration Methods: efficient](https://reader030.vdocument.in/reader030/viewer/2022040120/5e7a3e6205f2eb2c896161dc/html5/thumbnails/28.jpg)
28
Ω+Ω−
+
Ω+Ω−
=Ω
]73.31/)01.0(2[)73.31/(11
2014)1)(1(
]42.31/)01.0(2[)42.31/(11
1974)1)(1()(
2
211
i
iH
]73.31/)01.0(2[)73.31/(1)10659.4(
]42.31/)01.0(2[)42.31/(1)10066.5(
2
4
2
4
11
Ω+Ω−×
+
Ω+Ω−×
=
−
−
i
iH
(6a)
]73.31/)01.0(2[)73.31/(1)10965.4(
]42.31/)01.0(2[)42.31/(1)10066.5(
2
4
2
4
21
Ω+Ω−×
−
Ω+Ω−×
=
−
−
i
iH
(6b)
e.
![Page 29: sdvc.kaist.ac.krsdvc.kaist.ac.kr/course/ce514/ch15.pdf · 1 Chapter15. DYNAMIC RESPONSE OF MDOF SYSTEMS: MODE-SUPERPOSITION METHOD Dynamic Response - Direct Integration Methods: efficient](https://reader030.vdocument.in/reader030/viewer/2022040120/5e7a3e6205f2eb2c896161dc/html5/thumbnails/29.jpg)
29
f.
![Page 30: sdvc.kaist.ac.krsdvc.kaist.ac.kr/course/ce514/ch15.pdf · 1 Chapter15. DYNAMIC RESPONSE OF MDOF SYSTEMS: MODE-SUPERPOSITION METHOD Dynamic Response - Direct Integration Methods: efficient](https://reader030.vdocument.in/reader030/viewer/2022040120/5e7a3e6205f2eb2c896161dc/html5/thumbnails/30.jpg)
30
Figure 15.2. Inertance Bode plot for a complex structure
Figure 15.2 is an inertance Bode plot, that is, Acceleration/Force as a function of frequency, of the response of a complex system with excitation at one point and response measured at another point.
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31
§15.5 Dynamic Stresses by Mode-Superposition Mode Displacement Method
∑=
=N
rrr tt
ˆ
1)(s)(ˆ ησ (15.49)
Mode Acceleration Method
∑=
−=
N
rrr
r
ttˆ
12icpseudostat )(s1)(~ η
ωσσ && (15.50)
Example 15.6 Solution
)4~1( =iiσ : Story Shear
444
4333
3222
2111
)()()(
ukuukuukuuk
=
−=
−=−=
σσσσ
(1)
−−
−
=
4
3
2
1
4
33
22
11
4
3
2
1
00000
0000
uuuu
kkk
kkkk
σσσσ
(2)
][)(][
)()(
φηφσηφ
kstk
ttu
===
−−
−
=
4
3
2
1
4
33
22
11
4
3
2
1
00000
0000
φφφφ
kkk
kkkk
ssss
(3)
![Page 32: sdvc.kaist.ac.krsdvc.kaist.ac.kr/course/ce514/ch15.pdf · 1 Chapter15. DYNAMIC RESPONSE OF MDOF SYSTEMS: MODE-SUPERPOSITION METHOD Dynamic Response - Direct Integration Methods: efficient](https://reader030.vdocument.in/reader030/viewer/2022040120/5e7a3e6205f2eb2c896161dc/html5/thumbnails/32.jpg)
32
−
−=
−
−
=
−−
=
=
02.203851.392807.2317
02.482
s,
50.226551.131874.185316.1521
s
35.140047.245
42.70470.879
s,
19.75258.62708.45272.176
s
43
21
(4)
§15.6 Mode-Superposition for Undamped Systems with Rigid-Body Modes skip
)()()(u)(u)(u ttttt EERRER ηη Φ+Φ=+= (15.51) pT
RRRM Φ=η&& (15.52a) pT
EEEEE KM Φ=+ ηη&& (15.52b)
R
rr
t Tr
rr
Nrt
ddM
t
,,2,1for )0()0(
)(p1)(0 0
K
&
=++
= ∫ ∫
ηη
τξξφητ
(15.53)
Mode-Displacement Method )(ˆˆ)()(u ttt EERR ηη Φ+Φ= (15.54)
∑=
==EN
rrrEE ttt
ˆ
1)(s)(ˆS)(ˆ ηησ (15.55)
Mode-Acceleration Method EEEE
TEEERR MKK ηη &&)(-p)(u 11 −− ΦΦΦ+Φ= (15.56)
EEEEERR MK ηη &&)ˆˆˆ-pu 1−Φ+Φ= a (15.57) One expression for the elastic flexibility matrix Ea is
TEEE K ΦΦ= −1a (15.58)
EEE pkuum =+&& (15.59a) RE umpp &&−= (15.59b)
pu 1R
TRRRRR M ΦΦ=Φ= −η&&&& (15.60)
pp R=E (15.61a) TRRR MmI ΦΦ−= −1R (15.61b)
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33
Epw a= (15.62) RRE cΦ−= ww (15.63)
0)w(m =Φ−Φ RRTR c (15.64)
mwMc 1 TR
-RR Φ= (15.65)
wm)wMI(w 1 TTR
-RRE R=ΦΦ−= (15.66)
pw EE a= (15.67) RR aa T
E = (15.68)
∑=
−=
EN
rrr
r
ttˆ
12icpseudostat )(s1)(~ η
ωσσ && (15.69)
Example 15.7 Solution a.
=
−−−
−+
000
110121
011
100010001
3
2
1
3
2
1
uuu
uuu
&&
&&
&&
(1)
Let tωφ cosu = . (2)
3u2u1u
11 =k
)(3 tp
12 =m
13 =m11 =m
12 =k
)(3 tp
t
0p
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34
=
−−−−−
−−
000
)1(101)2(1
01)1(
3
2
1
2
2
2
φφφ
ωω
ω (3)
0)34( 242 =+− ωωω (4) 3,1,0 2
322
21 === ωωω (5)
−=
−=
=
12
1,
101
,111
321 φφφ (6)
rTrr
TrrM φφφφ == m (7)
rrr MK 2ω= (8)
18,2,06,2,3
321
321
===
===
KKKMMM
(9)
b. 3 2, 1,,0)0()0( === rrr ηη && (10)
c. )(p)( ttP T
rr φ= (11) 033032031 )(,)(,)( ptpPptpPptpP ==−=−=== (12)
d.
018622
3
033
022
01
≥
=+
−=+
=
tpp
p
ηηηη
η
&&
&&
&&
(13)
)cos1(18
)cos1(2
6
30
3
20
2
20
1
tp
tp
tp
ωη
ωη
η
−
=
−
−=
=
(14)
tp
tp
30
3
20
2
cos6
cos2
ωη
ωη
=
−=
&&
&&
(15)
e.
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35
rr
r ss
−
−=
=
3
2
1
2
1
110011
sφφφ
(16)
−−
=
−−
=33
s,11
s 32 (17)
f. )(s)(ˆ 22 tt ησ = (18a)
)(s)(s)(ˆ 3322 ttt ηησσ +=≡ (18b)
)cos1(11
2
)cos1(21
1ˆˆ
ˆ
20
20
2
1
tp
tp
ω
ωσσ
σ
−
=
−
−
−−
=
=
(19a)
)cos1(11
6
)cos1(11
2
30
20
2
1
tp
tp
ω
ωσσ
σ
−
−
+
−
=
=
(19b)
g.
=
32
30
0
icpseudostat2
1
p
p
σσ
(20)
2222
icpseudostat s1)(~ ηω
σσ &&
−=t (21a)
3323
2222
icpseudostat s1s1~ ηω
ηω
σσσ &&&&
−
−=≡ (21b)
tpp2
00
2
1 cos11
221
3~~
~ ωσσ
σ
−
=
=
0p
3/0p
3/0p
3/0p
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36
tptpp3
02
00
2
1
cos1
16
cos11
221
3
~
ωω
σσ
σσ
−
+
−
=
=≡
(22b)
Comparison of Maximum Spring Forces Computed by Mode-Displacement (M-D) Method and Mode-Acceleration (M-A) Method
333241.1166667.1000000.1/999933.0833333.0000000.1/
02
01
pp
Exactmode1mode1AMDM
σσ
−−
end • Review of Dynamic Response by Normal Mode Method
Dynamic Response Analysis )(][][][ tpukucum =++ &&&
Free vibration Analysis
]][][[]][[or ][][)(][)(or )()(
0][][
ΛΦ=Φ=Φ==
=+
mkmkttuttu
ukum
φλφηηφ
&&
Premultiply by T][Φ ][]][[][)(]][[][])][[][ pktcm TTTT Φ=ΦΦ+ΦΦ+ΦΦ ηηη &&&
][],][[][][ ],][[][][],][][][][ pPkKcCmM TTTT Φ=ΦΦ=ΦΦ=ΦΦ=
][Φ : Normal Mode If ][c is proportional damping matrix
iii
iiiiii
iiiiiiiiii
PM
PKCMPKdiagonalCdiagMdiag
12
][][][
2 =++
=++=++
ηωηωζη
ηηηηηη
&&&
&&&
&&&
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37
§15.7 Dynamic Response by Ritz Vectors
Combination of the Gram-Schmidt orthogonalization and inverse Iteration
- Reduction of System Order For m =1 to q
][][ 1 mm xmxk =+
1
1*
1 i
m
iimm xrxx ∑
=++ −= (1)
Let m1,2,..,i 0][ *
1 ==+mT
i xmx Premultiply (1) by ][ mx T
i , then ][ 1+= m
Tii xmxr
2/1*1
*1
*1
1 )][(
++
++ =
mm
mm xmx
xx
Go to Eq.(1)
],..,x ,[][ 21 qxxX = : Ritz vectors Use ][for ][ ΦX
][]][[][]][[][]][[][][ by yPremultipl
)(][][][][
pXqXKXqXCXqXMXX
tqXupuKuCuM
TTTT
T
=++
==++
&&&
&&&
][][][ pqKqCqI =++ &&&
Note
][]][[][][ IXMXM T == ][X are orthonormal wrt ][M
][ ,][ KC : full matrix, small order qxq matrix
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38
- Solution of the Reduced System
Free vibration analysis
]][[]][[or ][
0][][
Λ=
=
=+
ZZKzzK
qKqIλ
&&
][]][[][]][[][]][[][[Z] by Pr
)(][
pZZKZZCZZIZemultiply
tZq
TTTT
T
=++
=
ηηη
η
&&&
][][][ ** pCI =Λ++ ηηη &&&
][ - ][ ][ ][ *** ηηηη &&&& ijii CpdiagCdiagI =Λ++ Modify Numerical Solution If C is proportional damping matrix
][ ][ ][ ** pdiagCdiagI ii =Λ++ ηηη &&&
)(][)(]][[)(][)( ttZXtqXtu ηη Φ=== ]][[][ ZX=Φ
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39
§15.8 Dynamic Response by Lanczos Vectors
Combination of the Gram-Schmidt orthogonalization and inverse Iteration
- Reduction of System Order For i = 2,3,…N
It can be shown that 01 =−iγ
],..,,[][ 21 NXXXX = Lanczos Vectors
and )(][
][][][][][][][][][
111
tqXuLet
pKuIuCKuMKpuKuCuM
=
=++
=++−−− &&&
&&&
Premultiply by MX T
Where
( ) 1
][
1
2/1
XX
XmX
XAssumeT
γ
γ
=
=
1][][ −= ii XmXk
iTii XmX ][ 21 −− =β
( ) 1
][
*
2/1**
ii
T
XX
XmX
γ
γ
=
=
)1(..... 212111* −−−−= −−−−−− iiiiiiii XXXXX γβα
iTii XmX ][ 11 −− =α
Nmorder small Matrix, lTridiagona that show can We
)MKX()MXX()XMKX(ˆ)MXMKX(
1
1T1T1T
<<=
=++
−
−−−
TMMKX
pqqCq
T
T&
System Reduced ][][][ pqIqCqT =++ &&&
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40
][]][[][][ IXMXM T == ][X are orthonormal wrt ][M
of order of m
Free vibration analysis
1]][[]][[or 1][
0][][
−Λ==
=+
ZZTzzT
qIqT
λ
&&
][][][]][[][]][[][[Z] by Pr
)(][
pZZZZCZZTZemultiply
tZq
TTTT
T
=++
=
ηηη
η
&&&
][][][ **1 pICdiag =++Λ − ηηη &&& Modify
If C is proportional damping matrix ][][][ **1 pICdiagdiag =++Λ − ηηη &&&
)(][)(]][[)(][)( ttZXtqXtu ηη Φ=== ]][[][ ZX=Φ
• Comparison of Vector Superposition Methods
- Both Ritz method and Lanczos method reduce the system
order using the inverse iteration and Rayleigh-Ritz method,
and compute the eigenpairs of the reduced system,
which are different from the eigenpairs of the original system.
- Both methods save computational time, while the accuracy of
pMKXpCXMKXC
T
T
1
1 Matrix Full −
−
=
=
![Page 41: sdvc.kaist.ac.krsdvc.kaist.ac.kr/course/ce514/ch15.pdf · 1 Chapter15. DYNAMIC RESPONSE OF MDOF SYSTEMS: MODE-SUPERPOSITION METHOD Dynamic Response - Direct Integration Methods: efficient](https://reader030.vdocument.in/reader030/viewer/2022040120/5e7a3e6205f2eb2c896161dc/html5/thumbnails/41.jpg)
41
response is good enough for engineering purpose
- For the free vibration and dynamic response analysis,
the Rayleigh-Ritz method uses the Ritz vectors
],...,,[][ 21 NXXXX = , while Lanczos method uses ii βα and which
form the tridiagonal matrix ][T
- ir in Rayleigh-Ritz method and ii βα and in the Lanczos
methods are same. Theoretically we should have three non-zero
coefficients )1(,),1( +−= mmmiri for the thm Ritz vector, the
other coefficients are all zero.
![Page 42: sdvc.kaist.ac.krsdvc.kaist.ac.kr/course/ce514/ch15.pdf · 1 Chapter15. DYNAMIC RESPONSE OF MDOF SYSTEMS: MODE-SUPERPOSITION METHOD Dynamic Response - Direct Integration Methods: efficient](https://reader030.vdocument.in/reader030/viewer/2022040120/5e7a3e6205f2eb2c896161dc/html5/thumbnails/42.jpg)
42
1. Cantilever beam
g(t) = sin7t
[email protected] m = 5 m
0 5 10 15 20time (sec)
-2
-1
0
1
2
g(t)
(kN
)
Geometry and loading configuration
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43
0 5 10 15 20Number of vectors
1e-7
1e-6
1e-5
1e-4
1e-3
1e-2
1e-1
1e0
Erro
r
Ritz vector methodLanczos vector method
Mode displacement methodMode acceleration method
0 5 10 15 20Number of vectors
0
1
2
3
4
5
Com
putin
g tim
e (s
ec) Ritz vector method
Lanczos vector method
Mode displacement methodMode acceleration method
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44
2. Two-dimensional frame
[email protected] m = 61 m
10@
3.05
m =
30.
5 m
g(t)
0 5 10 15 20time (sec)
0
1
2
g(t)
(kN
)
Geometry and loading configuration
![Page 45: sdvc.kaist.ac.krsdvc.kaist.ac.kr/course/ce514/ch15.pdf · 1 Chapter15. DYNAMIC RESPONSE OF MDOF SYSTEMS: MODE-SUPERPOSITION METHOD Dynamic Response - Direct Integration Methods: efficient](https://reader030.vdocument.in/reader030/viewer/2022040120/5e7a3e6205f2eb2c896161dc/html5/thumbnails/45.jpg)
45
0 6 12 18 24 30Number of vectors
1e-3
1e-2
1e-1
1e0Er
ror
Ritz vector methodLanczos vector method
Mode displacement methodMode acceleration method
0 6 12 18 24 30Number of vectors
0
20
40
60
80
Com
putin
g tim
e (s
ec) Ritz vector method
Lanczos vector method
Mode displacement methodMode acceleration method
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46
3. Multi-span continuous bridge(Dong-Jin bridge)
Connection element
Element for pot bearing
0 5 10 15 20 25 30 35 40Time (sec)
-4
-2
0
2
4
Acc
eler
atio
n (m
/sec
2 )
Geometry and loading configuration(El Centro earthquake)
![Page 47: sdvc.kaist.ac.krsdvc.kaist.ac.kr/course/ce514/ch15.pdf · 1 Chapter15. DYNAMIC RESPONSE OF MDOF SYSTEMS: MODE-SUPERPOSITION METHOD Dynamic Response - Direct Integration Methods: efficient](https://reader030.vdocument.in/reader030/viewer/2022040120/5e7a3e6205f2eb2c896161dc/html5/thumbnails/47.jpg)
47
0 30 60 90Number of vectors
1e-12
1e-10
1e-8
1e-6
1e-4
1e-2
1e0Er
ror
Ritz vector methodLanczos vector method
Mode displacement methodMode acceleration method
0 30 60 90Number of vectors
0
500
1000
1500
2000
Com
putin
g tim
e (s
ec) Ritz vector method
Lanczos vector method
Mode displacement methodMode acceleration method
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48
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49
§15.8 State-Space Form of Differential Equations For structures with proportional and non-proportional damping Useful for the system with non-proportional damping
)(][][][ tPUKUCUM =++ &&&
=
−
+
0
)(0
00
tP
UU
MK
UU
MMC
&&&
& or
=
+
− 0
)(00
0
tPUU
KKC
UU
KM &
&
&&
][][ ***** PyKyM =+& State-Space Form of Dynamic Equations
=
=
−
=
=
0)(
00
][ 0
][M
**
**
tPP
UU
y
MK
KM
MC
&
Free vibration analysis
0][][ **** =+ yKyM &
)( * tyy η=
**
*
yUU
UU
y
UU
y
UUUUeUU
eUUt
t
λλ
λλλλ λ
λ
=
=
=
=
==
==
=
&&&
&&
&
&&&
&
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50
0])[*][( * =+ yKMλ
0])[][( ** =+ yKMλ eigenvalue problem Solution for y and λ
- Lee’s Method - Lanczos Method
Dynamic response analysis
)(][)( * tYty η=
],...,,[][ 221 NyyyY =
)()(]][[)(]][[ *** tPtYKtYM =+ ηη& )(][)(][][)(]][[][ *** tPYtYKYtYMY TTT =+ ηη&
)()(][)(][ tPtKdiagtMdiag =+ ηη&
ii
iiii M
tP )(=+ ηλη&
![Page 51: sdvc.kaist.ac.krsdvc.kaist.ac.kr/course/ce514/ch15.pdf · 1 Chapter15. DYNAMIC RESPONSE OF MDOF SYSTEMS: MODE-SUPERPOSITION METHOD Dynamic Response - Direct Integration Methods: efficient](https://reader030.vdocument.in/reader030/viewer/2022040120/5e7a3e6205f2eb2c896161dc/html5/thumbnails/51.jpg)
51
• Solution of Eigenvalue Problem
0)()()( =++ txKtxCtxM &&& (
where M : Mass matrix, Positive definite C : Damping matrix K : Stiffness matrix )(tx : Displacement vector
Eigenanalysis Proportional Damping( CMKKMC 11 −− = ) (
low in cost straightforward Non-Proportional Damping
very expensive
CURRENT METHODS
Transformation methods QR method(Moler and Stewart) LZ method(Kaufman) Jacobi method(Veselic)
Lanczos methods Unsymmetric Lanczos method(Kim & Craig) Symmetric Lanczos method(Chen & Taylor)
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52
Vector iteration method(Gupta)
Subspace iteration method(Leung ) : efficient method • Lee’s Method
General eigenproblem
02 =++ φφλφλ KCM (3)
Eigenproblem of order 2n zBzA λ= (4)
where
−=
MK
A0
0,
=
0MMC
B (5)
=λφφ
z
ijj
Ti zBz δ= (6)
Newton-Raphson technique
( )( ) 1
0)1()1(
)1()1(
=
=−++
++
kj
Tkj
kj
kj
zBz
zBA λ (7)
)()()1(
)()()1(
kj
kj
kj
kj
kj
kj
zzz ∆+=
∆+=+
+ λλλ (8)
where )(k
jλ∆ , )( kjz∆ : unknown incremental values
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53
Introducing Eq.(8) into Eq.(7) and
Neglecting nonlinear terms ( ) )()()()()( k
jk
jkj
kj
kj rzBzBA −=∆−∆− λλ (9)
( ) 0)()( =∆ kj
Tkj zBz (10)
where
)()()()( kj
kj
kj
kj zBzAr λ−= (11)
( )(kjr : residual vector)
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54
Matrix form of Eqs.(9) and (10)
( )
−=
∆∆
−
−−
00
)(
)(
)(
)(
)()( kj
kj
kj
Tkj
kj
kj rz
zBzBBA
λλ
(12)
Note that coefficient matrix is - Symmetric - Nonsingular
Modified Newton-Raphson technique
( )
−=
∆∆
−
−−
00
)(
)(
)(
)(
)()0( kj
kj
kj
Tkj
kjj rz
zBzBBA
λλ
(13)
)()()1(
)()()1(
kj
kj
kj
kj
kj
kj
zzz ∆+=
∆+=+
+ λλλ (8)
NONSINGULARITY
Let jj λλ =)0( and jk
j zz =)( in Eq.(13), and consider
12,,2,1 +== niuFuE iii Kγ (14)
where
( )
−
−−=
0Tj
jj
zBzBBA
Eλ
, (15)
(2n) (1)
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55
=
=
100000
100
MMC
BF (16)
Eigensolution of Eq.(14)
)(,1,1
,2,,2,1,0
,1
,1
jki
jjji jknk
zzzu
λλγ −−=
≠=
−
= K (17)
[ ] [ ] )(detdet2
1j
n
jkk
kFE λλ∏≠=
−−= (18)
≠ 0
[ ] [ ] [ ] [ ]MMIF detdetdetdet −=Q (19)
≠ 0
NUMERICAL EXAMPLES
Structures Cantilever beam with multi-lumped dampers Framed structure with a lumped damper
Analysis methods
Proposed method Subspace iteration method(Leung, 1995)
Comparisons
Solution time(CPU) Convergence
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56
2
)(2
)()()(
Error Normk
j
kj
kj
kj
zA
zBzA λ−= (20)
Computer
CONVEX with 100MIPS, 200MFLOPS
CANTILEVER BEAM WITH MULTI-LUMPED DAMPERS
Fig 1. Cantilever beam with multi-lumped dampers
TANGENTIAL DAMPER : c = 0.1 RAYLEIGH DAMPING : βα , = 0.001 YOUNG’S MODULUS : 1000 MASS DENSITY : 1 CROSS-SECTION INERTIA : 1 CROSS-SECTION AREA : 1 NUMBER OF EQUATIONS : 200 NUMBER OF MATRIX ELEMENTS : 696 MAXIMUM HALF BANDWIDTHS : 4 MEAN HALF BANDWIDTHS : 4
2 3
5
100 101
c
1
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Table 1. The Results of the Proposed Method for Cantilever Beam
Proposed Method Mode
Number
Error Norm of Starting Eigenpair (Lanczos method)
Number of Iterations
Eigenvalue Error Norm
1 2 3 4 5
0.872989E-04 0.763146E-03 0.437867E-04 0.605684E-02 0.420530E-00
1 1 1 1 1
-1.02232 ± i 3.95028
-1.18011 ± i 18.3991
-1.79640 ± i 39.6535
-2.87171 ± i 60.9945
-4.40255 ± i 82.2930
0.183316E-07 0.189217E-09 0.373318E-10 0.371279E-11 0.983166E-07
Table 2. CPU Time for the First Five Eigenpairs of Cantilever Beam
Method CPU time(in seconds) Ratio
Subspace Iteration Method Proposed method
(Lanczos method + Iteration scheme)
96.10 76.75
(10.55 + 66.20)
1.25 1.00
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Proposed Method
Subspace Iteration Method
0 1 2 3 4 5
Iteration Number
1E-8
1E-7
1E-6
1E-5
1E-4
1E-3
1E-2
1E-1
1E+0
Erro
r Nor
m
Error Limit
Fig. 2. Error norm versus iteration number of
the first eigenpair
0 1 2 3 4 5 6 7
Iteration Number
1E-10
1E-9
1E-8
1E-7
1E-6
1E-5
1E-4
1E-3
1E-2
1E-1
1E+0
Erro
r Nor
m
Proposed Method
Subspace Iteration Method
Error Limit
Fig. 3. Error norm versus iteration number of
the second eigenpair
Lanczos method
Lanczos method
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0 2 4 6 8 10
Iteration Number
1E-11
1E-10
1E-9
1E-8
1E-7
1E-6
1E-5
1E-4
1E-3
1E-2
1E-1
1E+0
Erro
r Nor
mProposed Method
Subspace Iteration Method
Error Limit
Fig. 4. Error norm versus iteration number of
the third eigenpair
0 2 4 6 8 10 12 14
Iteration Number
1E-121E-111E-10
1E-91E-81E-71E-61E-51E-41E-31E-21E-11E+0
Erro
r Nor
m
Proposed Method
Subspace Iteration Method
Error Limit
Fig. 5. Error norm versus iteration number of
the fourth eigenpair
Lanczos method
Lanczos method
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0 2 4 6 8 10 12 14 16
Iteration Number
1E-8
1E-7
1E-6
1E-5
1E-4
1E-3
1E-2
1E-1
1E+0
Erro
r Nor
m
Proposed Method
Subspace Iteration Method
Error Limit
Fig. 6. Error norm versus iteration number of
the fifth eigenpair
Lanczos method
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FRAMED STRUCTURE WITH A LUMPED DAMPER
Fig 7. Framed structure with a lumped damper
HORIZONTAL DAMPER : c = 10.0 RAYLEIGH DAMPING : α β, = 0.001 YOUNG’S MODULUS : 500 MASS DENSITY : 1 CROSS-SECTION INERTIA : 1 CROSS-SECTION AREA : 1
NUMBER OF EQUATIONS : 267 NUMBER OF MATRIX ELEMENTS : 1326 MAXIMUM HALF BANDWIDTH S: 6
c
5
2
1
2
30
62
90
91
31 60
61
32
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Table 3. The Results of the Proposed Method for Framed Structure
Proposed Method Mode
Number
Error Norm of Starting Eigenpair (Lanczos method)
Number of Iterations
Eigenvalue Error Norm
1 2 3 4 5
0.169894E-05 0.274663E-04 0.193503E-01 0.732792E-01 1.000000E-00
1 1 1 1 2
-0.06543 ± i 7.44209
-0.39695 ± i 8.40284
-0.07532 ± i 11.7071
-0.71155 ± i 13.9090
-0.46457 ± i 20.0691
0.483552E-10 0.165798E-09 0.200729E-10 0.447537E-10 0.300293E-10
Table 4. CPU Time for the First Five Eigenpair of Framed Structure
Method CPU time(in seconds) Ratio
Subspace Iteration Method Proposed method
(Lanczos method + Iteration scheme)
204.74 173.51
(14.24 + 159.27)
1.18 1.00
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0 1 2 3 4 5 6 7 8 9
Iteration Number
1E-11
1E-10
1E-9
1E-8
1E-7
1E-6
1E-5
1E-4
1E-3
1E-2
1E-1
1E+0
Erro
r Nor
mProposed Method
Subspace Iteration Method
Error Limit
Fig. 8. Error norm versus iteration number of
the first eigenpair
0 2 4 6 8 10
Iteration Number
1E-10
1E-9
1E-8
1E-7
1E-6
1E-5
1E-4
1E-3
1E-2
1E-1
1E+0
Erro
r Nor
m
Proposed Method
Subspace Iteration Method
Error Limit
Fig. 9. Error norm versus iteration number of
the second eigenpair
Lanczos method
Lanczos method
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0 2 4 6 8 10 12
Iteration Number
1E-11
1E-10
1E-9
1E-8
1E-7
1E-6
1E-5
1E-4
1E-3
1E-2
1E-1
1E+0
Erro
r Nor
mProposed Method
Subspace Iteration Method
Error Limit
Fig. 10. Error norm versus iteration number of
the third eigenpair
0 2 4 6 8 10 12 14
Iteration Number
1E-11
1E-10
1E-9
1E-8
1E-7
1E-6
1E-5
1E-4
1E-3
1E-2
1E-1
1E+0
Erro
r Nor
m
Proposed Method
Subspace Iteration Method
Error Limit
Fig. 11. Error norm versus iteration number of
the fourth eigenpair
Lanczos method
Lanczos method
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0 2 4 6 8 10 12 14 16 18
Iteration Number
1E-11
1E-10
1E-9
1E-8
1E-7
1E-6
1E-5
1E-4
1E-3
1E-2
1E-1
1E+0
Erro
r Nor
mProposed Method
Subspace Iteration Method
Error Limit
Fig. 12. Error norm versus iteration number of
the fifth eigenpair
Lanczos method
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§ 15.9 Estimation of Damping Ratio and Damping Matrix
Ref: Text §18.1 Damping in MDOF Systems Definition of orthogonal, classical, modal, or proportional damping
srsTr ≠= ,0Cφφ (18.1)
Rayleigh Damping a particular form of proportional damping, defined by
KMC βα += (a) The above C satisfies eq.(18.1) Let
( ) iiiTi KM ζωφβαφ 2=+
Since 1=i
Ti Mφφ 2
iiTi K ωφφ =
We get
iii ζωβωα 22 =+ (b) Using 11 and ζω , and 22 and ζω , which are known we compute βα and
1121 2 ζωβωα =+
2222 2 ζωβωα =+
In matrix form
=
22
11
22
21 2
11
ζωζω
βα
ωω
Solve for βα and and compute
KMC βα += From (b), for other damping ratios
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i
ii ω
βωαζ2
2+= i=3,4,..,N
Disadvantage of Rayleigh damping
It does not permit realistic damping to be defined for all the modes of interest
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68
Example 9.9 Compute βα and for Rayleigh damping in order that a direct step-by-step Integration can be carried out.
10.0 302.0 2
222
111
======
ξζωξζω
iii ξωβωα 22 =+ (b)
60.0)10,0)(3(2908.0)02,0)(2(24
==+==+
βαβα
(c)
0.104 336.0 =−= βα KMKMC 104.0336.0 +−=+= βα (d)
i
ii ω
ωξ2
104.0336.0 2+−= Ni ,..,3,2=
Example 9.10 Assume that the approximate damping to be specified for a MDOF System is as follows. Choose appropriate Rayleigh damping parameters .and βα
19;14.015;10.0
;7;04.03;03.0
;2;002.0
55
44
33
22
11
==
==
======
ωξωξωξωξωξ
iii ξωβωα 22 =+ (a)
Only two pairs of values determine .andβα Considering the spacing of the frequencies(see the figure below), we use
17;12.04;03.0
22
11
==
==
ωξωξ
(b)
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69
Figure 9.5 Damping as a function of frequency
08.428924.016
=+=+βα
βα
KMC βα += KM 01405.001498.0 += (c)
i
ii ω
ωξ2
01405.001498.0 2+−= Ni ,..,3,2=
Proportional Damping
)2(C rrrT ωζdiagC M=ΦΦ= (18.8)
1C −− ΦΦ= CT (18.9)
)(M rT diagM M=ΦΦ= (18.10a)
ΦΦ=ΦΦ== −−− 111 )M(I TMMM (18.10b)
MΦΦ 11 TM −− = (18.10c)
)M()M( C
11
1
T
T
MCMC
ΦΦ=
ΦΦ=−−
−−
(18.11)
MMdiagdiagMdiagM Trrrrr ΦΦ= −− 11 )( )2()( Mωζ
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70
∑=
=
N
r
Trr
r
rrωζ1
)M)(M(2C φφM
(18.12)
ssss
Ts ωζ M2C =φφ (18.13)
∑=
=
cN
r
Trr
r
rr
1)M)(M(2C φφωζ
M (18.14)
:cN No of specified damping ratios
∑−
=
+=
1
11 )M)(M(
ˆ2KCcN
r
Trr
r
rra φφωζM
Home Work (18.15)
where
c
c
N
Naωζ2
1 = (18.16a)
−=
c
c
N
rNrr ω
ωζζζ (18.16b)
++=
=
= NNNs
Ns
ccN
sN
c
s
c
c,),2(),1(,
,,2,1 value, specified
K
K
ωωζζ (18.17)
Example 18.1
01.021 == ζζ
660.29 294.13
2
1
==
ωω
Determine 43 and ζζ , and C Solution
882.55079.41
4
3
=
=
ωω
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71
a. From Eq. 18.15 Ta )M)(M(
ˆ2KC 111
111 φφωζ
+=
M (1)
where
2
21
2ωζ
=a (2a)
−=
2
1211
ˆωωζζζ (2b)
Thus,
41 107431.6
660.29)01.0(2 −×==a (3a)
31 105179.5
660.29294.1301.001.0ˆ −×=
−=ζ (3b)
=
=
70518.099310.055820.100000.1
23506.049655.077910.000000.1
3000020000200001
M 1φ (4)
−−−
−−−
=
73003520
02310011
800K (5)
−−−
−−−
=
80153.358258.105611.003601.058258.174760.299987.005071.0
05611.099987.074233.145988.003601.005071.045988.059051.0
C (6)
b. From Eq. 18.17b,
4,3,2
2 =
= ss
s ωωζζ (7)
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72
0188.0660.29882.5501.0
0138.0660.29079.4101.0
4
3
=
=
=
=
ζ
ζ (8)
Assume Rayleigh Damping 01.021 == ζζ 660.29 294.13 21 == ωω
882.55079.41
4
3
=
=
ωω
iii ξωβωα 22 =+
0.18359 )294.13(04656.0)01.0)(294.13(2
0004656.042.954
2
)294.1366.29)(294.1366.29()01.0)(366.16(2
)01.0)(660.29(2)660.29()01.0)(294.13(2)294.13(
2
2
2
=−=
==
+−=
=+
=+
α
β
βαβα
0.0138) with (compare 0118.0 2(41.079)
41.079)0.0004656(0.18359
20004656.018359.0
2
3
23
3
=
+=
+=
ωωξ
• Caughey series
srsTr ≠= ,0Cφφ Proportional Damping (18.1)
If CKMKCM 11 −− = , (18.2) then (18.1) is satisfied
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73
Assume that the p damping ratios )~1( pii =ξ are given to define C. Then a damping matrix that satisfies the relation is obtained using the Caughey series,
∑−
=
−=1
0
1 ][p
k
kk KMaMC (9.57)
where the coefficients )~1( pkk =α are calculated from the simultaneous equations. Note that with p=2, (9.57) reduces to Rayleigh damping, as specified as
KMC βα += and
++++= −
−32
13
210
21 p
ipiii
i aaaa ωωωω
ξ L (9.58)
Note that (9.57) satisfies (18.2)
Approximation of [C] Example 9.11 Approximation of [C]
)(][][][ tRUKUCUM =++ &&&
)(R210141
012
5.00
1.0
21
121
tUUU =
−−−
−+
+
&&& (a)
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74
=Ω
−−
−
=Φ6
42
;
211
21
210
21
211
21
2
Let ][ XU Φ=
)(][]][[][]][[][]][[][ tRXKXCXM TTTT Φ=ΦΦ+ΦΦ+ΦΦ &&&
=
+
−−
−−+
3
2
1
)X(6
42
)(X3.022.03.0
22.06.022.03.022.03.0
)(XRRR
ttt &&& (b)
where
)(R
21
21
21
10121
21
21
3
2
1
tRRR
−−
−=
Note that ]][[] ΦΦ CT is not diagonal Neglecting the off-diagonal elements of the coefficient matrix of )(tX&
=
+
+
3
2
1
)X(6
42
)(X3.000
06.00003.0
)(XRRR
ttt &&&
Note: Uncoupled equations