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 k m 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 ) ( p ku u c u m t = + + & & & (15.1) 0 0 u ) 0 ( u , u ) 0 ( u & & = = (15.2) ) ( ) ( ) ( t u t u t u p h + = where solution particular t u solution mogeneous h t u solution general t u p h : ) ( o : ) ( : ) ( :
74
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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 -
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
- 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)
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)
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.
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
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
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
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
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]