sdvc.kaist.ac.krsdvc.kaist.ac.kr/course/ce514/ch15.pdf · 1 Chapter15. DYNAMIC RESPONSE OF MDOF...

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

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


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


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



0 6 12 18 24 30Number of vectors

0

20

40

60

80

Com

putin

g tim

e (s




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



0 30 60 90Number of vectors

0

500

1000

1500

2000

Com

putin

g tim

e (s




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

&


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


Error Limit


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


Error Limit


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


Error Limit


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


Error Limit


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


Error Limit


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


Error Limit


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


Error Limit


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


Error Limit


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


Error Limit


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

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