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