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Fundamentals ofFundamentals of Structural Vibration
Speaker:Speaker:
Prof. FUNG Tat Ching
Date & Time: Wed 20 August 2014, 1:30 - 5:30 pm
School of Civil and Environmental EngineeringN T h l i l U i it
Venue: CEE Seminar Room D (N1-B4C-09B)
1Nanyang Technological University
Topics in Fundamentals of Structural Vibration (1.5 hrs)
SDoF SystemsDynamic Equilibrium
MDoF SystemsMode Shapesy q
Natural Freq/Period
Damping ratio
p
Modal decomposition
Modal responsesDamping ratio
Phase lag
Disp Resp Factor
Modal responses
Disp Resp Factor
Response Spectrum
2
Course Outline for CV6103Lecture Course Content Chapters in
Textbook 1 Si l D f F d S t
(21 hours)1 Single-Degree-of-Freedom Systems
Equations of motionFree vibration
1.2 – 1.6 2.1 – 2.2
2 Single-Degree-of-Freedom SystemsResponse to harmonic and periodic excitations
3.1 – 3.4 3.12 – 3.13
3 Single-Degree-of-Freedom Systemsg g yResponse to arbitrary, step and pulse excitations 4.1 – 4.11
4 Multi-Degree-of-Freedom SystemsEquations of motion 9.1 – 9.2Equations of motionNatural vibration frequencies and modes
9.1 9.2 10.1 – 10.7
5 Multi-Degree-of-Freedom SystemsFree vibration response 10 8 – 10 15Free vibration response 10.8 – 10.15
6 Multi-Degree-of-Freedom SystemsForced vibration response 12.1 – 12.7
7 S t ith G li d D f F d3
7 Systems with Generalized Degrees of FreedomGeneralized coordinates and their applications 14.3, 17.1
Textbook and References
● Main TextCh A K D i f St t Th● Chopra, A. K., Dynamics of Structures: Theory and Applications to Earthquake Engineering, Prentice Hall 4rd Edition 2011Prentice Hall, 4rd Edition, 2011.
● References● References● Clough, R. W., and Penzien, J.,
Dynamics of Structures,Dynamics of Structures, McGraw-Hill, 1993.
● Meirovitch, L. Fundamentals of ,Vibrations, McGraw-Hill, 2001.
4
Why Is There A Need To Do Dynamic Analysis?
● Static analysis● External Load = Internal Force● External Load Internal Force
Magnitude of loading & stiffness
Dynamic analysis● Dynamic analysis● External Load ≠ Internal Force
f & ffMagnitude of loading & stiffness
Frequency characteristics of loading, and the dynamic properties of structures (mass stiffness damping)properties of structures (mass, stiffness, damping)
5
Examples of SDOF Systems
Water tank
Mass concentrated at one locationMass concentrated at one location
Supports assumed to be massless
C b d l d SDOF tCan be modeled as a SDOF system
Pendulum
Rod is assumed to be massless
Only allowed to rotate about hinge
Can be modelled as a SDOF system
6
Can be modelled as a SDOF system
Equation of Motion
Damping force Restoring force
pffum SD =++&&External force
Newton’s Second Law of Motion:
External force
umffp DS&&=−−
D’Alembert’s Principle (Dynamic equilibrium):
Inertia force
(Dynamic equilibrium):
umffffp IIDS&&==−−− with0
7
Single-degree-of-freedom
cu(t)
(SDoF) Systems
m
(mass) p(t)
Typical representation:
(mass) p( )
k
Mass-spring-damper system
External force p(t)Assumptions:
External force p(t)
Restoring force kud
Linear elastic restoring force
Li i d iDamping force
2dut
uc
d
dLinear viscous damping
I ti f)(tk&&& 2dtm
8
Inertia force)(tpkuucum =++
Undamped Free Vibration
● Equation of Motion: )(tpkuucum =++ &&&
● with c = 0, p(t) = 0 kω0+ kuum && 02+ uu ω&&
● Initial conditions: )0( ),0( uuuu && == mn =ω0=+ kuum 0=+ uu nωor
● Exact Solution:u )0(&
tu
tutu nn
n ωω
ω sin)0(
cos)0()( +=
2⎞⎛
See Page 46 in Chopra( )θω −= tutu ncos)( maxor
9( )2
2max
)0()0( ⎟⎟
⎠
⎞⎜⎜⎝
⎛+=
n
uuu
ω&
)0(
)0(tan 1
u
u
nωθ
&−=
Free Vibration of a System without yDamping
umax
umaxumax
10
Periods of Vibration of Common Structures
Common Structures Period20-story moment resisting frame 1.9 sec10-story moment resisting frame 1.1 sec1-story moment resisting frame 0.15 sec
20-story braced frame 1.3 sec10-story braced frame 0.8 sec1-story braced frame 0.1 sec
Gravity dam 0.2 sec
11Suspension bridge 20 sec
Viscously Damped Free Vibration
0=++ kuucum &&&
0kc &&&
Equation of Motion:
0=++ um
um
uDivided by m:
ccLet
crn c
c
m
c ==ω
ζ2
(reasons will be clear later)zeta
02 2 =++ uuu nn ωζω &&&Hencem
kn =ω
Note:as before
kmmc ncr 22 == ωc
12critical damping ratio
Type of Motion
ζ :damping ratio
Three scenarios
ζ
ccr : critical damping coefficient
ζ < 1, i.e. c < ccr ⇒ under-damped (oscillating)
ζ = 1, i.e. c = ccr ⇒ critically damped
13ζ > 1, i.e. c > ccr ⇒ over-damped
Typical Damping RatioStructure ζWelded steel frame 0.010e ded stee a e 0 0 0Bolted steel frame 0.020
Uncracked prestressed concrete 0.015Uncracked reinforced concrete 0.020Cracked reinforced concrete 0 035Cracked reinforced concrete 0.035
Glued plywood shear wall 0.100p yNailed plywood shear wall 0.150
Damaged steel structure 0.050Damaged concrete structure 0.075
14Structure with added damping 0.250
Effects of Damping in Free Vibration
⎟⎞
⎜⎛ + uut ζωζω i
)0()0()0()(
&
21 ζωω =
⎟⎟⎠
⎞⎜⎜⎝
⎛ ++= − tuu
tuetu DD
nD
tn ωωζωωζω sin
)0()0(cos)0()(
1 ζωω −= nD
T
1521 ζ−= n
DT
Tmaxu=ρ
Decay of MotionOne way to measure damping is from rate of decay from free vibration
TD
decay from free vibration
TD
( )θωζω −= − tuetu Dtn cos)( max
(exactly)
( ) ⎟⎞
⎜⎛− 2
expexpπζζω
ζω ti T
eu n
Since peaks are separated by TD , ( y)
( ) ( )⎟⎟
⎠⎜⎜
⎝ −=== +−
+2
1 1expexp
ζζζωζω DnTt
i
i Teu Dn
16
Type of Excitations
● Harmonic / Periodic Excitations● Commonly encountered in engineering
Steady-state Responses
● Commonly encountered in engineeringUnbalanced rotating machinery
Wave loadingWave loading
● Basic components in more general periodic excitationsexcitations
Fourier series representation
More General Excitations T i t● More General Excitations● Step/Ramp Forces
TransientResponses
● Pulses Excitations17
Equation of Motion)(tpkuucum =++ &&& Resonance
● Linear ● ⇒u(t) can be replaced● ⇒u(t) can be replaced
by u(t) + any free vibration responsesp
● For example, c = 0 u(t) = u sinωt
2/ >nωω
SlowlyRapidlyloaded
● c = 0, u(t) = u0sinωt
● (-mω2 + k) u0 sinωt = p(t)
loadedloaded
● ⇒ p(t) = p0 sinωt
0 1pk
18( )20
0/1 n
k
pu
ωω−=
m
kn =ω
Excitation tp ωsin tp ωsin0
un
0)0(
2.0ωω
=
=
k
pu
u
n 0)0(
0)0(
ω=
=
&
Response
k
( ) tk
ptup ω
ωωsin
/1
1)(
20=
19
( )knωω /1−
Undamped Resonant Systems● For resonance, ω = ωn
( )tttp
tu ωωω sincos1
)( 0=
Derivation: See Page 70 & 72 in Chopra
● Response grows indefinitely
( )tttk
tu nnn ωωω sincos2
)( −−=
0)0( ,0)0( == uu &p g y
● Becomes infinite after infinite duration
20
Harmonic Vibration of Viscous DampingEquation of Motion
Sinusoidal forcetpkuucum ωsin0=++ &&&
Equation of Motion
m
kn =ω
Excitation frequencyAmplitude of forceS
tpkuucum ωsin0++ m
Amplitude of forceParticular SolutiontDtCtup ωω cossin)( += Derivation: See
Page 73 in Chopra
222
20
]/2[])/(1[
)/(1 n
k
pC
ζωω+
−=
p Page 73 in Chopra
222 ]/2[])/(1[ nnk ωζωωω +−0 /2 np
Dωζω−=
222 ]/2[])/(1[ nnkD
ωζωωω +−=
21
Steady-State SolutionThe particular solution can also be written as:
( )φω= tutu sin)(
220 DCu +=
( )φω −= tutup sin)( 0
whereD
tan 1 −= −φ0 DCu +
Using previously derived results for C and D,
whereC
tanφ
[ ]20
0max1
k
puu ==
( )[ ] ( )[ ]2220max
/2/1 nnk ωωζωω +−
( )( )2
1
/1
/2tan n
ωωωωζφ = −
22
( )/1 nωω−
2.0=ζ
Static response exactly in phase with forceStatic response exactly in-phase with force
Dynamic response has a time lag, φ/ω 23
General Solution
Complementary solution is the free damped vibration response:vibration response:
( )tBtAetu DDtn ωωζω sincos)( += −
21 ζωω −= nD
( )tBtAetu DDc ωω sincos)( +
Complete solution: Recall: A, B derived by satisfying initial conditions
)()()( tututu pc +=satisfying initial conditions
( ) tDtCtBtAe DDtn ωωωωζω cossinsincos +++= −
Transient Steady state
24Derivation: See Page 73 in Chopra
Example 1
ω /ωn= 0.2, ζ = 0.05, u(0) = 0, kpu n /)0( 0ω=&
Observe how the transient response decays due to damping, leaving only the steady state part
25
Example 2
ω = ωn (resonant response), ζ = 0.05
With d i h stuWith damping, response approach
ζ2maxstu =
26But can still be larger for another value of ω!
Example 3
27
Significance of Steady-state Solutions
● In certain problems, e.g. wave loads on an offshore structure, the load is assumed to beoffshore structure, the load is assumed to be in place for a sufficiently long time, so that the transient response has completely decayedtransient response has completely decayed.
● The interest is in the steady-state solution.
28
Maximum Response and Phase Angle
( )tt i)(St d t t l ti i ( )φω −= tutup sin)( 0Steady state solution is
1
( )[ ] ( )[ ]222
00
/2/1
1
k
pu
ωωζωω +=
( )[ ] ( )[ ]/2/1 nn ωωζωω +−
( )/2 ωωζCalled Deformation Response Factor (DRF) Rd in Chopra
( )( )2
1
/1
/2tan
n
n
ωωωωζφ
−= −
( ) d ptextbook (Page 76)
( )n
φ is the phase lag
29Time lag = φ/ω
DRF and PhaseR
For ω /ωn << 1
Slowly varying
Resonance
y y g
DRF ≅ 1
u0 ≅ p0/k
Displacement in-phase with force
Response dominated by stiffness
DR
F
For ω /ωn >> 1
R idl i
D
Slowly Rapidlyloaded Rapidly varying
DRF → 0
φ 02
0 ω pp n ⎟⎞
⎜⎛
loaded loaded
Displacement anti-phase with force
Pha
se φ
200
0 ωωω
m
p
k
pu n =⎟
⎠⎞
⎜⎝⎛≅
Response dominated by massP
30
When ω ≈ ωn
● Forcing freq ≈ natural freq
● Resonance
DAF is very large close to max at ω● DAF is very large, close to max at ωn
● pkpu 00 / =≅ ( ) 0
0/= kp
u
● Response dominated by dampingnc
uωζ0 2
=≅ ( )2max0
12 ζζ −u
● Response dominated by damping
● Displacement is 90° out of phase with force when ω = ωn
● This is the scenario we want to avoid! (butThis is the scenario we want to avoid! (but not always possible) 31
Periodic Excitation
p
t
)()( tjTt
T0 T0 T0 00 /2 Tπω =
Separate into harmonic components using Fourier series
)()( 0 tpjTtp =+ j : integer in (-∞, ∞ )
Separate into harmonic components using Fourier series
)sin()cos()( 000 tjbtjaatp jj ωω ∑∑∞∞
++= )()()( 01
01
0 jjpj
jj
j ∑∑==
Note: Arbitrary excitations can also be transformed into Fourier
32
Note: Arbitrary excitations can also be transformed into Fourier
series with appropriate technique, such as FFT.
Response To Arbitrary Time-Varying ForcesE ti f M ti
)(tpkuucum =++ &&&
Equation of Motion
Initial conditions:
)(p
0)0( ,0)0( == uu &
p(t): varying arbitrarily with timep(t): varying arbitrarily with timee.g. step forces (with finite rise time), pulses etcpulses etc.
Interested in the max response.
33
p
Max response Response/Shock spectrum
Simple Examples
p
pStep Force
p00 )( 0 ≥= tptp
0 t
Ramp or li l
p
plinearly increasing force
p0
0 )( 0 ≥= tt
tptp
r
Step force
0 ttr
pStep force with finite rise time
p0 ( )⎩⎨⎧
≥≤
= /
)( 0 rr
ttp
ttttptp
0 ttr
⎩ ≥ 0 rttp
34
Dynamic Response to Step Forces
⎤⎡ ⎞⎛
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−+−= − tte
k
ptu DD
tn ωζ
ζωζω sin1
cos1)(2
0
⎦⎣ ⎠⎝ ζ1
35
Dynamic Response of Ramp tp0
or Linearly Increasing Force⎞⎛ ttp sin
rtp0
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
rn
n
r t
t
t
t
k
ptu
ωωsin
)( 0
0p
0
)( 00
=
=st k
pu
ζ
5.2=n
r
T
t
ζ
n
36
Dynamic Response to Step
pForce With Finite Rise Time
p0( )⎩⎨⎧
≥≤
=
/)(
0
0
r
rr
ttp
ttttptp
0 ttr
⎩ 0 rttp
Phase 1 Phase 2Consider undamped response: Phase 1 Phase 2
1. Ramp phase: rn tt
t
t
t
t
k
ptu ≤⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
sin)( 0
ωω
Consider undamped response:
rnr ttk ⎟⎠
⎜⎝ ω
p ⎫⎧ 12. Constant phase: ( )[ ] rrnn
rn
ttttttk
ptu >
⎭⎬⎫
⎩⎨⎧
−−−= ωωω
sinsin1
1)( 0
37
Step Force With Finite Rise Time
38
Maximum Deformation
Tt )/sin(π
nr
nrd Tt
TtR
/
)/sin(1
ππ
+=
39Response Spectrum
Single Pulse ExcitationsF
● Example of pulse excitations: ● underground explosions
ForceBlast overpressure● underground explosions
● Idealized by simple shapesE
Time
● E.g.
40
Response Spectrum
⎨⎧ <
== 21
0//sin2 ndnd TtTtu
Rπ
⎩⎨ ≥
==21
0, /2 ndstd Ttu
R
Also called Shock Spectrum for single pulse
41
Also called Shock Spectrum for single pulse
Multi-Degree-of-Freedom Systems
p (t)
kc
(t)
u (t)p (t)
p (t) 44
44
3
kc
u (t)
u (t)p ( )
p (t) 33
2
33
2
kc
u (t)
u (t)
p (t) 22
1
2
1
kc 11
1
42
General Approach for Complex Structures
● Elastic resisting forces● Same as static analysis (i e ku=p)● Same as static analysis (i.e. ku p)
● fS = ku
D i f ll th i l● Damping forces: usually rather simple● ucfD &=
● Inertia forces: usually simple ● umfI &&=
● Equations of motionumfI
)()()()( tttt k&&&43
)()()()( tttt pkuucum =++ &&&
Arbitrary u(0)• NOT Simple Harmonic Motion (SHM)• Frequency of motion cannot be defined
Modal coordinators
• u1 and u2 are not proportional (⇒deflected shapes varies with time)
Modal coordinators
44
When u(0) = φ1
• can be SHM with appropriate initial conditions• u1 proportional to u21 2
• φ1 is a natural mode
45
When u(0) = φ2
• can be SHM with appropriate initial conditions• u1 proportional to u21 2
• φ2 is a natural mode
46
Natural FrequencyThe natural period of vibration Tn = the time required for one cycle of the harmonic motion in q yone of these natural modes.
π21
nnT
ωπ2=
nn Tf
1=
f = natural cyclic frequency of vibrationfn = natural cyclic frequency of vibrationωn = natural circular frequency of vibration
An N-DOF system has N number of natural periods and N number of natural modesand N number of natural modes.
47
How to Find the natural periods and natural modes?
● EoM:
● Natural frequency ω and mode shape φ can
)()()()( tttt pkuucum =++ &&&
● Natural frequency ωn and mode shape φn can be obtained by solving the following eigenvalue problemeigenvalue problem
0mk =− nn φ][ 2ω● Different modes can be shown to be
orthogonal wrt the m and k matrices, i.e.orthogonal wrt the m and k matrices, i.e.
0=rTn φφ m 0=r
Tn φφ k (n ≠ r)
48nTnnn
Tnn KΜ φφφφ km == ,
Vibration Analysis of MDOF Systems
● EoM: with
● Time Stepping Methods
)()()()( tttt pkuucum =++ &&& )0( and )0( uu &
● Time Stepping Methods● E.g. Central Difference, Newmark’s method etc.
● Modal Decomposition● MDoF problem ⇒ a number of SDoF problems
u
⎫⎧
+++= NNqqq L2211 φφφ uΦq 1−=
Φq=⎪⎬
⎫⎪⎨
⎧== ∑ n
N
rr
q
q ML1
1 ][ φφφ T
Tn
nqφφ
φm
mu=
49
q⎪⎭
⎬⎪⎩
⎨∑=
N
nr
rr
q
q 11
][ φφφnn φφ m
Uncoupled Equations
● EoM:
with
● Transformed into N SDoF Systems, each
)()()()( tttt pkuucum =++ &&& )0( and )0( uu &
y ,
)()()()( tPtqKtqCtqM nnnnnnn =++ &&&classical damping
tP )(
)( , , tPKΜ Tnnn
Tnnn
Tnn pkm φφφφφ === n
TrnrC φφ c=
K
With initial conditionsn
nnnnnnn M
tPtqtqtq
)()()(2)( 2 =++ ωωζ &&&
n
nn Μ
K=ω
● With initial conditionsTnφ mu )0( T
nφ um )0(&&50n
Tn
nnq
φφφ
m
)(0, =
nTn
nnq
φφφ
m
um )0(0,
& =
Displacement ResponsesOnce q1(t), …, qN(t) are determined, the response u1(t) uN(t) in u(t) can be obtainedresponse u1(t), …, uN(t) in u(t) can be obtained from
∑=N
tqt )()( φu ∑=
=n
nn tqt1
)()( φu
and subsequently the internal forces can alsoand subsequently the internal forces can also be calculated if required
Caution: The expression could be very lengthy.
51
Modal Contribution
It is useful to define the contribution of the nth mode to u(t) as
)()( tqt nnn φ=u
mode to u(t) as
Then internal force due to un(t) can be evaluated first and then sum up for all the modes laterfirst and then sum up for all the modes later.
Further ImprovementpSince qn(t) is a scalar function, the internal force due to φn can be evaluated first (static analysis) and φn ( y )then times qn(t) before sum up for all the modes later.
52
How To Calculate the Internal Forces?
● Directly from u(t) or un(t) or φn
Alternatively the same internal forces can be● Alternatively, the same internal forces can be obtained by considering the same structure s bjected to the eq i alent static forces k (t)subjected to the equivalent static forces ku(t) or kun(t) or kφn
53
Equivalent Static Force)()()()( tttt k&&&
)()()( ttt ucump &&& −− )()( ttS kuf =)()()()( tttt pkuucum =++
p5(t)m5
mk5
u5(t)fS5(t)
k5
)(55 tum &&
p4(t)m4
mk4
u4(t)fS4(t)
k4
)(44 tum &&
p3(t)
p (t)
m3
m2k3
u3(t)fS3(t)
(t)f (t)
k3
)(33 tum &&
)(t&&p2(t)
p (t)m1
k2
u2(t)fS2(t)
u (t)f (t)
k2
)(22 tum
)(tum &&p1(t)
k1
u1(t)fS1(t)
k1
)(11 tum
54V(t) V(t)
)()()()( 2 tqtqtt nnnnnnn φφ mkkuf ω===
1211 or:Forces φφ mk ω 111 or :Forces φφ mk ω
q1(t)M1
P1(t)r (t) = r st q (t)
K1, ζ1r1st
r1(t) = r1 q1(t)
2NNN φφ mk 2 or :Forces ω
qN(t)MN
PN(t)
KN, ζNrNst
rN(t) = rNst qN(t)
55
Recap
SDOF SystemsDynamic Equilibrium
MDOF SystemsMode Shapesy q
Natural Freq/Period
Damping ratio
p
Modal decomposition
Modal responsesDamping ratio
Phase lag
DAF/DRF
Modal responses
DAF/DRF
Response Spectrum
56