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 :)(
:)(:
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)
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
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)
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
&&
&&
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
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)
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
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)
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)
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
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)
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
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)
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
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)
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)
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
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)
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)
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.
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
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)
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)
25
f.
π2Ω
=f
26
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.
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.
29
f.
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.
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)
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)
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
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.
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
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
&&&
&&&
&&&
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
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
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 =++ &&&
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 −
−
=
=
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.
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
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
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
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
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)
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
48
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
=
=
=
=
==
==
=
&&&
&&
&
&&&
&
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&
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)
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
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)
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)
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
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
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
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
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
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
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
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
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
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
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
§ 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
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
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)
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
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
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)
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
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)
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