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

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 +=

solutionparticulartu

solutionmogeneous htusolutiongeneraltu

:)(o :)(

0)mk( 2 =− rr φω (15.3)

)()()(11

ttutu r

rr ηφ∑∑

== )()( ttu ηΦ= (15.1)

[ ]Nφφφ L21=Φ modal matrix (15.6) )(tη principal coordinates.

)()()()(111

tpkcmN

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

tptPKCM

φφφφφφφ

)()(),...,,(

),..,,(),..,,(

tptPKKKdiagkK

CCCdiagcCMMMdiagmM

=ΦΦ=

force modal)(stiffness modaldamping modal

mass modal

vector force modal)(matrix stiffness modalmatrix damping modal

matrixmassmodal

(15.9)

)(tPKCM rrrrrrr =++ ηηη &&& )(tPKCM =++ ηηη &&& (15.8)

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

rru ηφ∑

= )0()0(u ηΦ=

)0()0( r

rru ηφ && ∑

= )0()0(u η&& Φ= (15.10)

)0()0( r

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

,,2,1(0)um1)0()0(

mu(0)1)0()0(

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

§15.2 Mode-Displacement Solution for Response of Undamped MDOF Systems

∑∑==

rr ttutu

11)()()( ηφ

≈)(tu ∑∑==

rr ttutu

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

)sin()(1

)sin()0(1)cos()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

+ωη=η & (15.18)

(0)um1)0(

mu(0)1)0(

(15.12)

Example 13.5 Free Vibration

Initial Conditions

u)(u)(u

)(u)(u

1 φωmk

2 φωmk

Solution a. r

TrrM φφ m= (1)

k k km m

[ ] mm

ωφη

(0)um)0(

mu(0))0(

0mu(0)

21mu(0))0(

1121mu(0))0(

φη (5)

)0( 0uη

[ ] 000

1121mu(0))0(

)0(η&

)sin()0(1)cos()0()( ttt rrr

rrr ωηω

ωηη &

+= (15.18)

)cos()0()( tt rrr ωηη =

)(tη =

cos)0(cos)0(

ωηωη

coscos

2 ωω

)()( ttuh ηΦ=

coscos

[ ])cos()cos(2

)cos()cos(2

•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

2)/(11

(15.21)

)/(11)( 2ω

η (15.21)

ωΩ−

φ= ∑

cos)/(1

1p (15.22)

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

2001 1

the natural frequencies and modes are

Mode 1 mk /21 =ω

Mode 2 )/(252

1 mk=ω

k k k2

tPp Ω= cos11

Solution

−==Φ

1121 φφ (1)

[ ] )()(2

121 ttu ηηη

φφ Φ=

)]()()([ tpkmT =Φ+ΦΦ ηη&& (4) pkm TTT Φ=ΦΦ+ΦΦ ηη )()( &&

PKM =+ ηη&& (5) pPkKmM TTT Φ=ΦΦ=ΦΦ= ,, (6)

415003

−= coscos

11 (9)

415003

ηη&&

&&Uncoupled Equations (10)

tPkm Ω=+ cos33 111 ηη&& (11a) tPkm Ω=+ cos)4/15()2/3( 122 ηη&& (11b)

cos)()(

tY Ω= cos11η (12a) tY Ω= cos22η (12b)

11 )/(1

)3/1(33 ωΩ−

=Ω−

mkPY (13a)

12 )/(1

)15/4()2/3()4/15( ωΩ−

=Ω−

mkPY (13b)

[ ] )()(2

121 ttu ηηη

φφ Φ=

− cos

)/(115/4

)/(13/

)/(115/4

)/(13/

Example 15.1

tP Ωcos1

./inseck1 21 −=m

.k/in8001 =k

16002 =k

24003 =k

32004 =k

−−−−−

−−−

63688.070797.043761.023506.000000.115859.053989.049655.044817.000000.109963.077910.0

15436.090145.000000.100000.1

882.55079.41660.29294.13

12279.368746.187970.017672.0

3000020000200001

73003520

02310011

32 ωω

Solution a.

rrrrTr MKM 2

r ,m ωφφ == (1)

23506.049655.077910.00000.1

3000020000200001

23506.049655.077910.00000.1

695.507)87288.2(78.176

87288.2

4.374,1164239.343.736836658.439.191517732.2

695.50787288.2

KMKMKMKM

PrTrF φ= (4)

15436.090145.0

)1(cos)/()( 2r

tKFt−

Ω=η (6)

rrr ω

Ω= (7)

rrr ttu

111 )()( ηφ (8)

)]79.3122/(1)[4.11374()cos15436.0)(15436.0(

)]46.1687/(1)[43.7368()cos90145.0)(90145.0(

)]70.879/(1)[39.1915()cos)(0.1(

1ˆ)]72.176/(1)[695.507(

)cos)(0.1()(

Ω−Ω

Ω−Ω−−

Ω−Ω

80.2851,402.53,3.1179.44,6486.6,5.0

=Ω=Ω=Ω

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

−−−−

−−−−=Ω

Example 15.2: Example 15.1 Compute rF and rD

Solution a.

00000.1

23506.049655.077910.000000.1

aF φ (1a)

51071.2

23506.049655.077910.000000.1

bF φ (1b)

06931.0,15436.076801.0,90145.007713.0,00000.151071.2,00000.1

−=−=

FFFFFFFF

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

−−

−=−=

DDDDDDDD

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

1 tuLxtu

= (12)

312151520

uuAL&&

&&ρ (1)

)cos(2

1 αωφφ

12151520

µ i (3)

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)

),( txu

18.3220)(

486.220)(

ρω EL exact 467.2)( 2

1 (8a)

ρω EL exact 21.22)( 2

2 (8b)

0153203 2

2 =−+− )i(i

)i(i )()( φµφµ (9)

153203µµ

φφβ

−−

453.0136.1514.0

−=−=

453011

381112

)cos()cos( 22

1 αωφφ

αωφφ

−−

=)cos()cos(

1121 αω

αωφφ

)()(),( 2

1 tuLxtu

= (12)

−−

)cos()cos(

αωαω

φφtt

−−

)cos()cos(11

αωαω

ββ tt

Lx ββ

−−

)cos()cos(

αωαω

=[ ])()( 21 xx φφ

−−

)cos()cos(

αωαω

)cos()cos()( 222111 αωφαωφ −+−= ttx = ),(),( )2()1( txutxu +

Lxx ii βφ (15)

)cos()(),()(iii

i txtxu α−ωφ= (14)

)cos(),(2

Lxtxu α−ω

= (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)

rηφ &&∑

−− −=ˆ

11 mkpk (15.27)

ω&&∑

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.

31250.031250.031250.031250.031250.072917.072917.072917.031250.072917.035417.135417.131250.072917.035417.160417.2

ka 31 −−

Solution a.

rtpatu ηφ

1cos)(~ ∑=

−Ω= (1)

rtpatu ηφ

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)(~

Ω−ΩΩ

Ω−Ω−−Ω

Ω−ΩΩ

Ω= −

NtPtPtu

b. 80.2851and,197.44,0 222 =Ω=Ω=Ω

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

−−−−

−−−=Ω

c. Use Eq. 15.18 to obtain a general expression for )(trη&& to substitute Into the mode-acceleration equation, Eq. 15.28. Thus

ττωτωωηωωηωη

rrrrrrr

)(sin)()(sin)0(cos)0()(

−−+

−−=

(15.29)

Alternative form

ττωττ

ωηωωηωη

dtPddtP

rrrrrrr

)(cos)]([cos)0(1sin)0(cos)0()(

−−+

−−=

(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)

C φφωω

== (15.33)

rrrrdr

ωηωζηω

ττωτω

τωζ

sin)]0()0([1

cos)0(

)(sin)(1)( )(

−−

(15.34)

21 rrdr ζωω −= (15.35) mode-displacement solution

∑∑==

rr ttutu

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 ttt(t

11 )(mk-)(ck)p(k)u~ ηφηφ &&& (15.37)

The resulting mode-accerleration solution is

∑∑==

r ttt(tˆ

1 )(1-)(2)p(k)u~ ηφω

ηφωζ

&&& (15.38)

tPpFor

rrrrrr Ω

2ηωηωζη &&& (15.39)

PrTrF φ= (15.20c)

rrrrrrrr

PepFor

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/

rirKHH rr ζ

η+−

=Ω≡Ω (15.42)

)cos()2()1(

solutionofpart real the take ,cos

tPpFor

η −Ω+−

(15.43a)

)1(2tan 2

ζα (15.43b)

rrr tt(t

1)()()u ηφη (15.44)

r rrrr

rirKt Ω

12 )2()1(

1P)(uζ

φφ (15.45)

=Ω≡Ω

r rrrr

jrirPuij

rirKHH jir

)2()1(1)()(

ζφφ

(15.46)

)cos()2()1(

cosFor

r rrrr

φφ−Ω

(15.47)

A plot of Eq. 15.46 on the complex plane is referred to as a complex frequency response plot.

+−−

jririj

)2()1()1()(ζ

φφ (15.48a)

jririj

1222 )2()1(

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.

Example 15.4

0628.0,6284.01,217,987

=′===′=

Solution a.

′+′−

′−′++

′+′−

′−′++

kkkkkk

cccccc

′+′−

′−′++

kkkkkk

tωφ cosu = (3)

′+′−

′−′+00

kkk (4)

mk ′+

==2, 2

221 ωω (5)

21 φφ (6)

70.37,42.31

1421,9871

c′ k

k′ 1p

HzfHzf 00.62

,00.52

11 ====

11 (8a)

=ΦΦ=2002

mTM (8b)

=ΦΦ=

5080.1002568.1

7540.06284.07540.06284.0

6912.00628.00628.06912.0

2842)2(14211974)2(987

2842001974

K (8c)

0100.0)70.37)(2(2

5080.1

0100.0)42.31)(2(2

2568.1

ζ (10)

Ω+Ω−

12 )/2())/(1(

1)(r rrrr

jririj

ωζωφφ

Ω+Ω−

]70.37/)01.0(2[)70.37/(11

2842)1)(1(

]42.31/)01.0(2[)42.31/(11

1974)1)(1()(

]70.37/)01.0(2[)70.37/(1)10519.3(

]42.31/)01.0(2[)42.31/(1)10066.5(

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()(

]70.37/)01.0(2[)70.37/(1)10519.3(

]42.31/)01.0(2[)42.31/(1)10066.5(

Ω+Ω−×

Ω+Ω−

))/(2())/(1())/(1()(

jririj

ωζωωφφ

Ω+Ω−

1222 ))/(2())/(1(

)/(2)(r rrr

jririj

ωζωωζφφ

Example 15.5

0031.0,6284.0,10,987 =′==′= cckk Solution a.

mk ′+

==2, 2

221 ωω (1)

73.31,42.31

1007,9871

HzfHzf 05.52

,00.52

11 ====

M (3a,b)

=ΦΦ=

2692.1002568.1

6346.06284.06346.06284.0

6315.00031.00031.06315.0

cTC (3c)

2014)2(10071974)2(987

2014001974

K (3d)

0100.0)73.31)(2(2

2692.1

0100.0)42.31)(2(2

2568.1

ζ (5)

Ω+Ω−

]73.31/)01.0(2[)73.31/(11

2014)1)(1(

]42.31/)01.0(2[)42.31/(11

1974)1)(1()(

]73.31/)01.0(2[)73.31/(1)10659.4(

]42.31/)01.0(2[)42.31/(1)10066.5(

Ω+Ω−×

]73.31/)01.0(2[)73.31/(1)10965.4(

]42.31/)01.0(2[)42.31/(1)10066.5(

Ω+Ω−×

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.

§15.5 Dynamic Stresses by Mode-Superposition Mode Displacement Method

rrr tt

1)(s)(ˆ ησ (15.49)

Mode Acceleration Method

12icpseudostat )(s1)(~ η

ωσσ && (15.50)

Example 15.6 Solution

)4~1( =iiσ : Story Shear

)()()(

ukuukuukuuk

−=−=

σσσσ

−−

σσσσ

][)(][

φηφσηφ

−−

φφφφ

−−

02.203851.392807.2317

02.482

50.226551.131874.185316.1521

35.140047.245

42.70470.879

19.75258.62708.45272.176

§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)

,,2,1for )0()0(

)(p1)(0 0

= ∫ ∫

τξξφητ

(15.53)

Mode-Displacement Method )(ˆˆ)()(u ttt EERR ηη Φ+Φ= (15.54)

rrrEE ttt

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)

TRRRRR M ΦΦ=Φ= −η&&&& (15.60)

pp R=E (15.61a) TRRR MmI ΦΦ−= −1R (15.61b)

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)

12icpseudostat )(s1)(~ η

ωσσ && (15.69)

Example 15.7 Solution a.

−−−

110121

100010001

Let tωφ cosu = . (2)

3u2u1u

)(3 tp

13 =m11 =m

)(3 tp

−−−−−

−−

)1(101)2(1

φφφ

ω (3)

0)34( 242 =+− ωωω (4) 3,1,0 2

21 === ωωω (5)

321 φφφ (6)

TrrM φφφφ == m (7)

rrr MK 2ω= (8)

18,2,06,2,3

KKKMMM

b. 3 2, 1,,0)0()0( === rrr ηη && (10)

c. )(p)( ttP T

rr φ= (11) 033032031 )(,)(,)( ptpPptpPptpP ==−=−=== (12)

018622

ηηηη

)cos1(18

)cos1(2

110011

sφφφ

−−

s 32 (17)

f. )(s)(ˆ 22 tt ησ = (18a)

)(s)(s)(ˆ 3322 ttt ηησσ +=≡ (18b)

)cos1(11

)cos1(21

ωσσ

−−

)cos1(11

ωσσ

icpseudostat2

icpseudostat s1)(~ ηω

σσ &&

−=t (21a)

icpseudostat s1s1~ ηω

σσσ &&&&

−=≡ (21b)

1 cos11

~ ωσσ

tptpp3

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/

Exactmode1mode1AMDM

−−

end • Review of Dynamic Response by Normal Mode Method

Dynamic Response Analysis )(][][][ tpukucum =++ &&&

Free vibration Analysis

]][][[]][[or ][][)(][)(or )()(

ΛΦ=Φ=Φ==

mkmkttuttu

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

iiiiii

iiiiiiiiii

PKCMPKdiagonalCdiagMdiag

][][][

=++=++

ηωηωζη

ηηηηηη

§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 =+

iimm xrxx ∑

=++ −= (1)

Let m1,2,..,i 0][ *

1 ==+mT

i xmx Premultiply (1) by ][ mx T

i , then ][ 1+= m

Tii xmxr

1 )][(

mm xmx

Go to Eq.(1)

],..,x ,[][ 21 qxxX = : Ritz vectors Use ][for ][ ΦX

][]][[][]][[][]][[][][ by yPremultipl

)(][][][][

pXqXKXqXCXqXMXX

tqXupuKuCuM

][][][ pqKqCqI =++ &&&

][]][[][][ IXMXM T == ][X are orthonormal wrt ][M

][ ,][ KC : full matrix, small order qxq matrix

- Solution of the Reduced System

Free vibration analysis

]][[]][[or ][

ZZKzzK

qKqIλ

][]][[][]][[][]][[][[Z] by Pr

pZZKZZCZZIZemultiply

ηηη

][][][ ** 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

§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 )(][

][][][][][][][][][

tqXuLet

pKuIuCKuMKpuKuCuM

=++−−− &&&

Premultiply by MX T

XAssumeT

1][][ −= ii XmXk

iTii XmX ][ 21 −− =β

)1(..... 212111* −−−−= −−−−−− iiiiiiii XXXXX γβα

iTii XmX ][ 11 −− =α

Nmorder small Matrix, lTridiagona that show can We

)MKX()MXX()XMKX(ˆ)MXMKX(

1T1T1T

−−−

System Reduced ][][][ pqIqCqT =++ &&&

][]][[][][ IXMXM T == ][X are orthonormal wrt ][M

of order of m

1]][[]][[or 1][

−Λ==

ZZTzzT

][][][]][[][]][[][[Z] by Pr

pZZZZCZZTZemultiply

ηηη

][][][ **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

1 Matrix Full −

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.

1. Cantilever beam

g(t) = sin7t

100@0.05 m = 5 m

0 5 10 15 20time (sec)

Geometry and loading configuration

0 5 10 15 20Number of vectors

Ritz vector methodLanczos vector method

Mode displacement methodMode acceleration method

0 5 10 15 20Number of vectors

ec) Ritz vector method

Lanczos vector method

2. Two-dimensional frame

10@6.1 m = 61 m

0 5 10 15 20time (sec)

Geometry and loading configuration

0 6 12 18 24 30Number of vectors

3. Multi-span continuous bridge(Dong-Jin bridge)

Connection element

Element for pot bearing

0 5 10 15 20 25 30 35 40Time (sec)

Geometry and loading configuration(El Centro earthquake)

0 30 60 90Number of vectors

§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 =++ &&&

][][ ***** PyKyM =+& State-Space Form of Dynamic Equations

0][][ **** =+ yKyM &

)( * tyy η=

UUUUeUU

λλλλ λ

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&

iiii M

tP )(=+ ηλη&

• 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)

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)

=λφφ

Ti zBz δ= (6)

Newton-Raphson technique

( )( ) 1

0)1()1(

)1()1(

=−++

zBA λ (7)

)()()1(

zzz ∆+=

∆+=+

+ λλλ (8)

where )(k

jλ∆ , )( kjz∆ : unknown incremental values

Introducing Eq.(8) into Eq.(7) and

Neglecting nonlinear terms ( ) )()()()()( k

kj rzBzBA −=∆−∆− λλ (9)

( ) 0)()( =∆ kj

Tkj zBz (10)

)()()()( kj

kj zBzAr λ−= (11)

( )(kjr : residual vector)

Matrix form of Eqs.(9) and (10)

∆∆

−−

)()( kj

zBzBBA

Note that coefficient matrix is - Symmetric - Nonsingular

Modified Newton-Raphson technique

∆∆

−−

)()0( kj

kjj rz

zBzBBA

)()()1(

zzz ∆+=

∆+=+

+ λλλ (8)

NONSINGULARITY

Let jj λλ =)0( and jk

j zz =)( in Eq.(13), and consider

12,,2,1 +== niuFuE iii Kγ (14)

−−=

zBzBBA

, (15)

(2n) (1)

100000

BF (16)

Eigensolution of Eq.(14)

)(,1,1

,2,,2,1,0

jjji jknk

λλγ −−=

= K (17)

[ ] [ ] )(detdet2

kFE λλ∏≠=

−−= (18)

[ ] [ ] [ ] [ ]MMIF detdetdetdet −=Q (19)

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

)()()(

Error Normk

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

100 101

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

Error Limit

Fig. 2. Error norm versus iteration number of

the first eigenpair

0 1 2 3 4 5 6 7

Iteration Number

Proposed Method

Error Limit

the second eigenpair

Lanczos method

0 2 4 6 8 10

Iteration Number

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

Proposed Method

Error Limit

the fourth eigenpair

Lanczos method

0 2 4 6 8 10 12 14 16

Iteration Number

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

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

mProposed Method

Error Limit

the first eigenpair

0 2 4 6 8 10

Iteration Number

Proposed Method

Error Limit

the second eigenpair

Lanczos method

0 2 4 6 8 10 12

Iteration Number

mProposed Method

Error Limit

the third eigenpair

0 2 4 6 8 10 12 14

Iteration Number

Proposed Method

Error Limit

the fourth eigenpair

Lanczos method

0 2 4 6 8 10 12 14 16 18

Iteration Number

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

ζωζω

Solve for βα and and compute

KMC βα += From (b), for other damping ratios

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

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

======

ξζωξζω

iii ξωβωα 22 =+ (b)

60.0)10,0)(3(2908.0)02,0)(2(24

==+==+

βαβα

0.104 336.0 =−= βα KMKMC 104.0336.0 +−=+= βα (d)

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

======

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

ωξωξ

Figure 9.5 Damping as a function of frequency

08.428924.016

=+=+βα

KMC βα += KM 01405.001498.0 += (c)

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

ΦΦ=−−

−−

(18.11)

MMdiagdiagMdiagM Trrrrr ΦΦ= −− 11 )( )2()( Mωζ

rrωζ1

)M)(M(2C φφM

(18.12)

Ts ωζ M2C =φφ (18.13)

1)M)(M(2C φφωζ

M (18.14)

:cN No of specified damping ratios

∑−

11 )M)(M(

ˆ2KCcN

rra φφωζM

Home Work (18.15)

Naωζ2

1 = (18.16a)

rNrr ω

ωζζζ (18.16b)

= NNNs

c,),2(),1(,

,,2,1 value, specified

ωωζζ (18.17)

Example 18.1

01.021 == ζζ

660.29 294.13

Determine 43 and ζζ , and C Solution

882.55079.41

a. From Eq. 18.15 Ta )M)(M(

ˆ2KC 111

111 φφωζ

=a (2a)

ˆωωζζζ (2b)

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

b. From Eq. 18.17b,

s ωωζζ (7)

0188.0660.29882.5501.0

0138.0660.29079.4101.0

ζ (8)

Assume Rayleigh Damping 01.021 == ζζ 660.29 294.13 21 == ωω

882.55079.41

iii ξωβωα 22 =+

0.18359 )294.13(04656.0)01.0)(294.13(2

0004656.042.954

)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(

βαβα

0.0138) with (compare 0118.0 2(41.079)

41.079)0.0004656(0.18359

20004656.018359.0

ωωξ

• Caughey series

srsTr ≠= ,0Cφφ Proportional Damping (18.1)

If CKMKCM 11 −− = , (18.2) then (18.1) is satisfied

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,

∑−

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

++++= −

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

tUUU =

−−−

&&& (a)

−−

Let ][ XU Φ=

)(][]][[][]][[][]][[][ tRXKXCXM TTTT Φ=ΦΦ+ΦΦ+ΦΦ &&&

−−

−−+

)(X3.022.03.0

22.06.022.03.022.03.0

)(XRRR

ttt &&& (b)

−−

Note that ]][[] ΦΦ CT is not diagonal Neglecting the off-diagonal elements of the coefficient matrix of )(tX&

)(X3.000

06.00003.0

)(XRRR

ttt &&&

Note: Uncoupled equations

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

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