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1

Structural Dynamics

Prof. Dr. Andrei V. Metrikine

Lecture 1

CT 4140 Structural Mechanics

Faculty of Civil Engineering and Geosciences2009 Delft, The Netherlands

CT 4140

Section CM, Room 6.59, Tel.: 84749

[email protected]://www.mechanics.citg.tudelft.nl

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Course Objectives

CT 4140 Structural MechanicsFaculty of Civil Engineering and Geosciences2009 Delft, The Netherlands

n-mass-spring

system

2-mass-spring

system

1-mass-spring

systemrigid

body

string

string

bar

bar

beam

beam

plate

plate

Main: Modelling of the dynamic behavior of civil engineering structures

Models to be considered

Lecture 1 General information about the course 6

Earthquake-Induced Vibrations of Buildings


Lecture 1 Significance of dynamics for civil engineering

Cause Result Model

F(t)

F

t

Earthquakes are extremely dangerous for

metropolitans not to mention that they can causetsunami

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Earthquakes are also dangerous because of possible liquefaction(indirect effect)

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Banda Aceh (Indonesia 2004) Debris Detail(indirect effect through tsunami)

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Wind-Induced Vibrations of Structures



Vibrating Tacoma Narrows Bridge Skyscrapers Chimney

Wind-farm

( ) ( )Wind force: ,WIND MEAN FLUCT F f X f X t = +

F

t

Wind can cause intense vibrations of relatively flexiblestructures

10

Traffic-Induced Vibrations



Bullet train in Kyoto City Bullet train passing a bridge

High-speed trains cause significant amplification of the dynamic response of railways

Dynamic rail bucklingis a serious danger

11

Soil Vibration from a High-Speed Train


Lecture 1 Significance of dynamics for civil engineering 12

Super-fast Train



Transatlantic Underwater Tunnel New-York -London

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Super-fast Train



Final.avi

14

Construction- and Production-Induced Vibrations



Pile driving Construction of a machine foundation

The vibration level in the surrounding buildings must be low!

15

Human-Induced Vibrations



Millennium bridge experienced severe walking-induced lateral vibrations

16

Wave- and Flow-Induced Vibrations



Semi-submersible offshore platform Mega-Float in Tokyo Bay

Floating offshore facility Installation of a riser

Waves and currents are capableof causing violent vibrations of

offshore structures

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Cross-Flow-Induced Vibrations



Floating offshore facility

Vortex Induced Vibration ofRisers

18

Axial-Flow-Induced Vibrations



Floating LNG factoryInstability of a submerged

cantilever

Guido 1.avi

19

Ice-Induced Vibrations



Light-house surrounded by ice

The Molikpaq

IIV tests

20

Ice-Induced Vibrations



ice sheet

ice

rubblescracks

rigidrigid

struct.struct.

ice sheet

ice

rubblescracks

rigidrigid

struct.struct.

Ice failure against cylinderIce failure against square

shaped cross-section

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Classification of Dynamic Loads



harmonicF

t

F

t

F

t

pulseF

t

F

t

periodical

Machines

Vortex / Galopping

Walking / Dancing

Waves

Wind

Traffic

Piling

Earthquakes

Explosions

Collisions

random

transient

22

Examples of One-Degree-of-Freedom Idealizations


Lecture 1 Recollection of knowledge regarding undamped 1 DOF systems

m

x(t )

F (t )

Idealization of bending motion of an offshore platform in waves

m

x(t )

F (t )

Unloaded

static

position

Idealization of vertical vibration of a lorry on a bridge

23

1 DOF Systems without Damping

CT 4140 Structural MechanicsFaculty of Civil Engineering and Geosciences

2009 Delft, The Netherlands

Equation of motion Model:and initial conditions:

( )

( ) ( )0 00 0

mx+kx= F t

x x , x v= =

&&

&

Free Vibrations

m

x(t )

F (t )

x

k

( )

( )

0

0

0 0

0

0

2

n

n

mx + kx = x + x =

x x k m

x v

ω

ω

⇔

= =

=

&& &&

&

Lecture 1 Recollection of knowledge regarding undamped 1 DOF systems 24

The General Solution for 1 DOFS without Damping



Equation of motion: The General Solution:

02n

x + x =ω && ( ) ( )2

1

exp , , complexk k k k

k

x t X s t X s=

= −∑

Equation of motion of N DOFS: The General Solution:

( ) ( ) ( ) ( )1 (1)1 11

... 0 exp N

N N

N N k k

k

x a x a x a x x t X s t −

−

=

+ + + = = ∑

Begin Generalization

Characteristic Equation:

2 2

1,20 i eigenvaluesk n ns sω ω + = ⇒ = ± −

Characteristic Equation:

( ) ( )11 1... 0

N N

k k N k N s a s a s a

−

−+ + + =

End Generalization


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The General Solution in Real Form and Free Vibrationsof 1 DOFS without Damping


The General Solution:

( ) ( ) ( ) ( ) ( )1 2exp i exp i cos sinn n n n x t X t X t A t B t ω ω ω ω = + − = +

( ) ( ) ( ) ( )

( )

00 0 0

22

0 0 0

1 00

0

cos sin cos

amplitude

tan initial phase

n n n

n

n

n

v x t x t t A t

A x v

v

x

ω ω ω ϕ ω

ω

ϕ ω

−

= + = −

= + −

⎛ ⎞= −⎜ ⎟

⎝ ⎠

Begin Important Formula

Free Vibration:

( ) ( ) ( )exp i cos isinn n nt t t ω ω ω = +

End Important Formula

t x

x0 A0

2π / ωn

v0


Contents of Lecture 2

Lecture 2


1. Recollection of knowledge regarding undamped 1 DOF

systems• vibrations under harmonic force, amplitude-frequency and phase-

frequency characteristics, resonance, vibrations under generaldisturbing force

2. Dirac delta-function and derivation of Duhamel’s integral

3. Recollection of knowledge regarding 1 DOF systems withviscous damping

• Equation of motion, characteristic equation, aperiodic and dampedfree vibrations, forced vibration under harmonic and general-typeforce

4. Frequency-Domain Analysis of 1 DOFS

• Integral Fourier transform, spectrum, frequency response function

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Forced Vibration of 1 DOFS without Damping underHarmonic Excitation




Equation of motion Beating:

( )

( ) ( )

0

0 0

cos

0 , 0

mx kx F t

x x x v

ω + =

= =

&&

&

( ) ( ) ( )

( ) ( )( )

00

0

2 2

cos sin

1cos cos

1

n n

n

n

n

v x t x t t

F t t

k

ω ω ω

ω ω ω ω

= +

+ −−


The Steady-State Solution: ( ) ( )02 2

1cos

1n

F x t t

k ω

ω ω =

−

0 50 100 150 200 250time t [s]

-30

-20

-10

0

10

20

30

x ( t ) k / F 0

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Amplitude-Frequency and Phase-FrequencyCharacteristics of 1 DOFS without Damping



Steady-state solution:

Amplitude-Frequency Characteristic(Magnification Factor)

( ) ( ) ( )0 2 21

cos cos1

steady n

n

F x t t x A t

k ω ω ϕ

ω ω = ⇒ = +

−

0 1 2 3

ω/ωn

0

1

2

3

4

5

A / x

s t a t i c

02 21

1n

F Ak ω ω = −

0 1 2 3

ω/ωn

0

90

180

ϕ

Phase-Frequency Characteristic

0, 1

, 1

n

n

ω ω ϕ

π ω ω

⎧ ⎪⎩


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Resonance in 1 DOFS without Damping


Frequency-domain representation Time-domain representation

of resonance : of resonance

0 1 2 3

ω/ωn

0

1

2

3

4

5

A / x

s t a t i c

0

2 21

1 n

F Ak ω ω

=−

0 5 10 15 20 25t [s]

-20

-10

0

10

20

x ( t ) k / F

0

( ) ( )0 sin2

F x t t t k

ω ω =


Vibrations of 1 DOFS without Damping underGeneral Disturbing Force


Equation of motion and initial conditions:

( )

( ) ( )0 00 0

mx+kx= F t

x x , x v= =

&&

&

Solution in the form of Duhamel’s integral:

( ) ( ) ( ) ( ) ( )( )000

1cos sin sin

t

n n n

n n

v x t x t t F t t t dt

mω ω ω

ω ω ′ ′ ′= + + −∫

The idea behind the Duhamel’s integral is representationof the force by a sequence of short pulses:

t

F

t ’ dt ’

( ) ( ) ( )( )

1sin ,

p nn

x t F t t t dt t t m

ω ω

′ ′ ′ ′= − >


31

Dirac Delta-Function as Mathematical Representationof Impulsive Forces



How do we describe mathematically an impulsive force (force of short duration),which applies a momentum I = 1?

t

F

t ’

2dt

2

I

dt 22

I I F t dt I

dt = Δ = =

( )2 ,

0, otherwise

I dt t dt t t dt F

⎧ − < < +⎪= ⎨

⎪⎩

Let us make the force shorter and shorter keeping I = 1.

t

F

t ’

( )0

0,lim

,dt

t t F t t

t t δ

→

′≠⎧′= − = ⎨

′∞ =⎩

Dirac Delta-Function

( )

( ) ( ) ( )

1t t dt

F t t t dt F t

δ

δ

∞

−∞

∞

−∞

′ ′− =

′ ′ ′− =

∫

∫

Properties of Dirac Delta-Function

Lecture 2 Dirac delta-function and derivation of Duhamel’s integral 32

Derivation of Duhamel’s Integral



Representation of the external force using Dirac delta-function:

( ) ( ) ( ) ( ) ( ) ( )0 0

1s

mx + kx = F t F t t t dt F t t t dt t δ δ δ

∞ ∞

′ ′ ′ ′ ′ ′ ′⎡ ⎤= − = − =⎣ ⎦∫ ∫&&

Response to Dirac delta-function:

( )

( ) ( ) ( ) ( )

01N s

0 0 00 0 0 0

p p p p

p p p p

m x + k x =mx + kx = I t I

x , x I m x , x

δ ⎧ ⎧⎪ ⎪⇔ =⎨ ⎨

= == = ⎪⎪ ⎩⎩

&&&&

&&

( ) ( ) ( ) ( )( )1 1

sin , 0 sin , 0 p n p n

n n

x t t t x t t t t t t m m

ω ω ω ω

′ ′ ′= ≥ ⇒ − = − − ≥

Response to the original force:

( ) ( ) ( ) ( ) ( )( )0 0

1sin

t

p p n

n

x t F t x t t dt F t t t dt m

ω ω

∞

′ ′ ′ ′ ′ ′= − = −∫ ∫

Lecture 2 Dirac delta-function and derivation of Duhamel’s integral

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1 DOFS with Viscous Damping


Equation of motion Model:

and initial conditions:( )

( ) ( )0 00 0

mx+cx+kx= F t

x x , x v= =

&& &

&

Free Motion

22 0,

the natural frequency of undamped vibration

2 damping factor

n

n

x nx x

k m

c m n

ω

ω

+ + =

= −

= −

&& &

Lecture 1 Rec ol lec tion o f knowledge regarding 1 DOFS with v iscous damping

m

x(t )

F (t )

x

k

c

The general Solution:

( ) ( ) ( ) ( ) ( )( )2

2 2 2 2

1 2

1

exp exp exp expn n n n

n

x t X s t nt X t n X t nω ω =

= = − − + − −∑

34

Free Motion of 1 DOFS with Viscous Damping


Characteristic Equation: Eigenvalues:

2 2 2 2 2 2

1 22 0 ,n n ns ns s n n s n nω ω ω + + = = − + − = − − −

Critical Damping:

Lecture 2 Rec ol lect ion o f knowledge regarding 1 DOFS with v iscous damping

( )2 22crit crit nc km c c n ω = = ⇔ = Aperiodic Motion: Damped Vibration:

t

x

x0

( ) ( ) ( )( )0 1 2 2 11 2

exp exp x

x t s s t s s t s s

= −−

( ) ( ) ( ) ( )( )1 12 2

1

exp cos sin ,

n

x t nt A t B t

n

ω ω

ω ω

= − +

= −

35

Free Vibrations of 1 DOFS with Viscous Damping



Damped Vibration:

Lecture 2 Rec ol lec tion o f knowledge regarding 1 DOFS with v iscous damping

t

x

x0

ϕ0

ω1

A0

A0exp(-nt )

-A0exp(-nt )

T 1=

2π

ω1

( ) ( ) ( ) ( )( )1 12 2

1

exp cos sin ,

n

x t nt A t B t

n

ω ω

ω ω

= − +

= −

( ) ( ) ( )0 1 02

2 0 0 0 00 0 0

1 1 0 1

exp cos

, arctan

x t A nt t

v nx v nx A x

x

ω ϕ

ϕ ω ω ω

= − −

⎛ ⎞⎛ ⎞ += + + = ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠

( )0

1 12 2

0

Amplitude: exp

2Period: 2

Phase angle:

n

A nt

T n

π π ω

ω

ϕ

−

= =−

36

Forced Vibration of 1 DOFS with Viscous Dampingunder Harmonic Excitation




Equation of motion

( )

( ) ( )

0

0 0

cos

0 , 0

mx cx kx F t

x x x v

ω + + =

= =

&& &

&

( ) ( ) ( ) ( )( ) ( )1 1exp cos sin part x t nt A t B t x t ω ω = − + +


Real form of particular solution:

( ) ( )

( )

( ) ( )

2 2

0 0

2 22 2 2 2 2 2 2 2

cos sin ,

2,

4 4

part c s

n

c s

n n

x X t X t

F F n X X

m mn n

ω ω

ω ω ω

ω ω ω ω ω ω

= +

−= =

− + − +

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Complex Form of Particular Solution for Dynamic Systemsunder Harmonic Excitation



( )

( ) ( )

auxiliarysystem

0 0

cosexp i

sin

t mx cx kx F mx cx kx F t

t

ω ω

ω

⎧⎪+ + = ⇒ + + =⎨

⎪⎩

&& &&& & % % %

Particular solution of the auxiliary problem is found as:

Begin Generalization

( ) ( )( ) ( )2 00

2 2

exp i exp i i exp i

1

2i

substitution

part

n

x X t X t m c k F t

F X

m n

ω ω ω ω ω

ω ω ω

= ⇒ − + + =

=− + +

%

( )( ) ( )

( )( ) ( )0

0

Re exp i , cos

Im exp i , sin part

X t F F t

x X t F F t

ω ω

ω ω

⎧ =⎪

= ⎨ =⎪⎩End Generalization

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Steady-State Vibrations of 1 DOFS with Viscous Dampingunder Harmonic Excitation



Equation of motion

( ) ( ) ( ) ( )( ) ( )1 1exp cos sin cos , arg( ) x t nt A t B t X t X ω ω ω ϕ ϕ = − + + − =The Steady-State Solution:

( ) ( ) ( )lim cossteadyt

x t x t X t ω ϕ →∞

= = −

0.0 0.5 1.0 1.5 2.0 2.5

ω/ωn

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

X / x

s t a t i c

γ = 0

γ = 2

γ = 1 γ = 0.5

γ = 0.3

γ = 0.2

Magnification factor (dynamicamplification factor):

1/ 22

2 22

2 21

2 2

static n n

n crit

X

x

n c

c

ω ω γ

ω ω

γ ω

−⎛ ⎞⎛ ⎞⎜ ⎟= − +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= =

39

ibrations of 1 DOFS with Viscous Damping underGeneral Disturbing Force



Equation of motion and initial conditions:

( )

( ) ( )0 00 0

mx+cx+kx= F t

x x , x v= =

&& &

&

The general solution in the form of Duhamel’s integral:

( ) ( ) ( ) ( )

( ) ( )( ) ( )( )

0 00 1 1

1 1

1

1 0

exp cos sin

1exp sin

t

v nx x t nt x t t

F t n t t t t dt m

ω ω ω ω

ω ω

⎛ ⎞⎛ ⎞= − + +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

′ ′ ′ ′+ − − −∫

Reminder: the main idea behind this solution isthe representation of the external force by asequence of short pulses

t

F

t ’ dt ’


Frequency-Domain Analysis of Forced Vibrations of1 DOFS under General Disturbing Force



Main idea: represent the external force as the superposition of harmonic vibrations.

( )mx+cx+kx= F t && &

Lecture 2 Frequency-Domain Analysis of 1 DOFS

Mathematical tool: Integral Fourier Transform

Equation of motion: Decomposition of the force and displacement:

( )( )

( )( )

( )1 exp i2

F t F t d x t x

ω ω ω π ω

∞

−∞

⎧ ⎫⎧ ⎫⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

∫%

%

Equation in the frequency domain and its solution:

( ) ( ) ( ) ( ) ( ) ( )2 2i im c k x F x F m c k ω ω ω ω ω ω ω ω − + + = ⇒ = − + +% %% %

( ) ( ) ( ) ( )and are frequency spectrums of andF x F t x t ω ω % %

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The Frequency Response Function (FRF)


Frequency Response Function of 1 DOFS with Viscous Damping:

( ) 21i

G im c k

ω ω ω

=− + +

Physical Significance: G (i ω) is the complex amplitude of the steady-state responseof a dynamic system to a harmonic force of unit amplitude

Mathematical definition: G (i ω) is the spectrum of the response of a dynamicsystem to the pulse load (Dirac delta function)

Solution in the Frequency Domain:

( ) ( ) ( )i x G F ω ω ω = %%

Transformation of the Frequency Domain Solution to the Time Domain:

( ) ( ) ( ) ( )1

i exp i2

x t G F t d ω ω ω ω π

∞

−∞

= ∫ %

Lecture 2 Frequency-Domain Analysis of 1 DOFS 42

The Integral Fourier Transform and Typical Spectrums


The Forward Fourier Transform: The Inverse Fourier Transform:

( ) ( ) ( ) ( ) ( ) ( )1exp i , exp i2

f = f t t dt f t f t d ω ω ω ω ω π

∞ ∞

−∞ −∞

− =∫ ∫% %

( ) f t ( ) f ω %

( )t δ 1

t

0 f

0t

( ) ( )0 0 f t f H t t = −

( )... is the Heaviside (step) Function H

( ) ( )00

sin2

t f f

ω ω

ω =%

0t −

0 02 f t

ω

( )0 f t t − ( ) ( )0exp i f t ω ω −%

Lecture 2 Frequency-Domain Analysis of 1 DOFS

43


Lecture 3



1. Unstable 1 DOFS – Galloping in Wind

2. Unstable 1 DOFS under Harmonic Disturbing Force

3. Recollection of knowledge regarding undamped 2 DOFS

• equations of motion, matrix form of equations of motion, free

vibrations, characteristic equation, natural frequencies and normalmodes, forced vibration under harmonic force, the undamped vibrationabsorber

44

Unstable 1 DOFS – Galloping in Wind



Lecture 3 Unstable 1 DOFS – Galloping in Wind

Prismatic bar subject to a steady wind blowing in x-direction:

The Lift (across-wind) and Drag (along-wind) forces on the bar (small attack angles):

Visco-elastically supported rigid prismatic bar in steady wind;

the bar may move in the vert ical direction only.

B

B L

2k v

2k

z

y x

v

z&

zF

zF cosv α

z− &

α

2c

2 21 1,

2 2 L L D D

F v C BL F v C BL ρ ρ = =

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Total force in z-direction:

Approximation of C z:

( ) ( )0 0 0 0

, 0 z z z L z DdC dC dC dC z

C C d v d d d α α α α α α α α α α = = = =

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞≈ ≈ − = +⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

&

( ) ( ) ( ) ( ) ( ) ( ) ( )21cos sin 2 z L D L D Z F F F F F v BLC α α α α α α α ρ α = + ≈ + =

zF L

F D

F

xF α

v0

zC

α

∂⎡ ⎤⎢ ⎥∂⎣ ⎦

zC

α

v

z& z− &

α

46




Equation of vertical motion of the bar:

Characteristic Equation and Eigenvalues:

0

0 0

1- 02

1can be negative since 0

2

zeff

z zeff

dC mz cz kz vBLz m z c z kzd

dC dC c c vBL

d d

α

α α

ρ α

ρ α α

=

= =

⎛ ⎞+ + = ⇔ + + =⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞= + − % %&& & &

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Unstable 1 DOFS under External Harmonic Force


Lecture 3 Unstable 1 DOFS under External Harmonic Force

Linearized Equation of Motion and Initial Conditions

( ) ( ) ( ) ( )( ) ( )

( )

1, 1,

2 2

1,

exp cos sin cos arg( ) ,

2 ,

eff eff eff

eff eff eff eff n

x t n t A t B t X t X

n c m n

ω ω ω

ω ω

= − + + −

= = −

( )( ) ( )

0

0 0

cos , 0

0 , 0

eff eff mx c x kx F t c

x x x v

ω + + = <= =

&& &

&


The Steady-State Solution

( ) ( ) ( )lim cossteadyt

x t x t X t ω ϕ →∞

≠= −

The steady-state solution of the linearized problem does not exist if ceff < 0 !

The Fourier Transform is not applicable!

50

Recollection of Knowledge Regarding Undamped 2 DOFS


Lecture 3 Recollection of knowledge regarding undamped 2 DOFS

Equations of motion: Model:

( ) ( )

( ) ( )

1 1 1 1 3 1 2 1

2 2 2 2 3 2 1 2

m x k x k x x F t

m x k x k x x F t

+ + − =

+ + − =

&&

&&

x

m1

x1(t )

F 1(t )

m2

F 2(t )

x2(t )

k 1

k 2

k 3

Equations of motion in matrix form:

, x x F + =&& M K [ ]

[ ]

1 2

1 2

1

2

1 3 3

3 2 3

, Displacemetn Vector

, Force Vector

0Mass Matrix

0

Stiffness Matrix

T

T

x x x

F F F

m

m

k k k

k k k

=

=

⎡ ⎤= ⎢ ⎥

⎣ ⎦

+ −⎡ ⎤= ⎢ ⎥− +⎣ ⎦

M

K

51

Free Vibrations of Undamped 2 DOFS




Equations of motion:

( )

( )

1 1 1 1 3 1 2

2 2 2 2 3 2 1

00

0

m x k x k x x x x

m x k x k x x

⎧ + + − =⎪⇔ + =⎨

+ + − =⎪⎩

&&

&&

&&

M K


( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )4 4 4

1 2

1 2

1 1 1

exp , exp expn n n n n n

n n n

x t X s t x t X s t x t X s t = = =

⎧ ⎫= = ⇔ =⎨ ⎬

⎩ ⎭∑ ∑ ∑

Substitution of the General Solution into Equation of Motion gives:

The Characteristic Equation:

( ) ( ) ( )1 22 0, ,T

n n ns X X X X ⎡ ⎤+ = = ⎣ ⎦ M K

( )2

2 1 1 3 3

2

3 2 2 3

det 0nn

n

m s k k k s

k m s k k

+ + −+ = =

− + + M K

52

Free Vibrations of Undamped 2 DOFS




The Characteristic Equation as Polynomial:

The Frequency Equation:

( ) ( )2 2 2 4 2 1 3 2 3 1 2 1 3 2 31 1 3 2 2 3 31 2 1 2

0 0n n n nk k k k k k k k k k

m s k k m s k k k s sm m m m

⎛ ⎞+ + + ++ + + + − = ⇔ + + + =⎜ ⎟

⎝ ⎠ All eigenvalues are imaginary!

( ) ( )

4 2 2 2 2 2 4 0a b a b ab

ω ω ω ω ω ω ω − + + − =

1 3

1

2 3

2

3

1 2

, - frequency

partial frequencies

'coupling frequency'

a

b

ab

s i

k k

m

k k

m

k

m m

ω ω

ω

ω

ω

=

⎫+= ⎪

⎪⎬

+ ⎪= ⎪⎭

=

m1

x1(t )

F 1(t )

x

m2

F 2(t )

x2(t )

k 1

k 2

k 3

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Natural Frequencies and Normal Modesof Undamped 2 DOFS



Natural Frequencies (both real):

The General Solution of the Governing Equations:

( ) ( )2 2

2 2 2 2 4 2 2 2 2 4

1 2

1 14 , 4

2 2n a b a b ab n a b a b abω ω ω ω ω ω ω ω ω ω ω ω = + − − + = + + − +

We have 4 initial conditions and 8 unknown constants to determine???

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

1 1 1 1 1 1 2 1 2

2 2 1 2 1 2 2 2 2

cos sin cos sin

cos sin cos sin

n n n n

n n n n

x t A t B t C t D t

x t A t B t C t D t

ω ω ω ω

ω ω ω ω

= + + +

= + + +

Mode 1 Mode 2

54

Mathematics behind Normal Modes



Matrix Form of the General Solution with 8 unknowns:

( ) ( )4

1

expn nn

x t X s t =

= ∑

Thus, only 4 unknowns remain in the general solution, which can be determined fromthe initial conditions.

4 pairs of 2 algebraic equations (each pair is linearly dependent):

( )2 0, 1..4n ns X n+ = = M K

From these equations, one can find:

( )

( )

2 2 2

11 11 21 21

2 2112 12 22 22

n n n

n nn

X m s k m s k

m s k m s k X

+ += − = −

+ +

55

Physical Significance of Normal Modes




The General Solution with 4 unknowns in Real Form:

Particular case of symmetrical system

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( ) ( )( ) ( )

1 1 1 2 2

1 2 1 22 2 2 2

2 1 1 1 2 2 22 2

cos sin cos sin

cos sin cos sin

n n n n

n n a n n n a n

ab ab

x t A t B t C t D t

m m m m x t A t B t C t D t

ω ω ω ω

ω ω ω ω ω ω ω ω ω ω

= + + +

= + − + + −

m

x1(t )

x

m

x2(t )

k k

k 3

Symmetrical system

1 2 1 2,k k k m m m= = = =

2 2

a bω ω =

x

m

x1(t )

m

x2(t )

k k

2 2

1 2 1 2 1, 1n a ab k m A A B Bω ω ω = − = = =

Mode 1

m

x1(t )

x

m

x2(t )

k k

32k 32k

Mode 2

( )2 22 3 2 1 2 12 1n a ab k k m C C D Dω ω ω = + = + = = −

56

Undamped 2 DOFS under Harmonic Loading




Equations of Motion:

The Steady-State Solution:

( ) ( ) ( )

( ) ( ) ( )

1

1 1 1 1 3 1 2 0

2

2 2 2 2 3 2 1 0 0

cos

cos

m x k x k x x F t

m x k x k x x F t

ω

ω ϕ

+ + − =

+ + − = +

&&

&&

( ) ( )

( ) ( )

1 1 1 0

2 2 2 0

cos cos

cos cos

x A t B t

x A t B t

ω ω ϕ

ω ω ϕ

= + +

= + +

Algebraic equations with respect to the amplitudes:

( )

( )

2121 1 1 1 3 1 3 21 1 1 1 3 1 3 2 0

22 2

2 2 2 2 3 2 3 1 2 2 2 2 3 2 3 1 0

0

0

m B k B k B k Bm A k A k A k A F

m A k A k A k A m B k B k B k B F

ω ω

ω ω

⎧ ⎧− + + − =− + + − =⎪ ⎪⎨ ⎨

− + + − = − + + − =⎪ ⎪⎩ ⎩

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The Undamped Dynamic Absorber



Model:

Examples - a bridge and a vibrating machine:

M

x1(t )

x

m

x2(t )

K

k

M

K /2

k

m

Bridge beam

Auxiliary mass

(a)m

M

K /2

k

(b)

58





The Steady-State Solution:

( ) ( )( )

1 1 3 1 2 0

2 2 1

cos

0

Mx Kx k x x F t

mx k x x

ω + + − =+ − =

&&

&&

( )

( )

1 1

2 2

cos

cos

x A t

x A t

ω

ω

=

=

Algebraic equations with respect to the amplitudes:

2

1 1 1 1 3 1 3 2 0

2

2 2 2 2 3 2 3 1

0

m A k A k A k A F

m A k A k A k A

ω

ω

⎧− + + − =⎪⎨

− + + − =⎪⎩

M

x1

(t )

x

m

x2(t )

K

k

59





Model:

The Magnification Factors:

M

x1(t )

x

m

x2(t )

K

k

( )( )( )

( )( )

2 2 2

1

2 2 2 2

1 2

2 2

2

2 2 2 2

1 2

n b

static n n

n ab

static n n

A

x

A

x

ω ω

ω ω ω ω

ω

μ ω ω ω ω

Ω −=

− −

Ω=

− −

ω

A / x

s t a t i c

A1

A2

ωa

ωb

ωn1

ωn2

, ,a b a b

k K k k

M m m M ω ω ω

+= = =

60

Contents of Lectures 4&5



1. Unstable 2 DOFS – Flutter

2. Formulation of equations of motion of a 3 DOFS using thedisplacement method

3. Representation of a structure by N DOFS

4. A numerical recipe for finding the stiffness matrix for a nodeof N DOFS.

Lecture 4 & 5

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Unstable 2 DOFS – Flutter


Lecture 4 & 5 Unstable 2 DOFS – Flutter

Prismatic rigid bar of rectangular shape subject to a uniform flow:


vertical motion

angular motion

z

y

mz cz kz F

I c k M ϕ ϕ

ϕ ϕ ϕ

+ + = −

+ + = −

&& &

&& &

x

z

v

2k ϕ

2 z

k

L

B

H

y

ϕ

z

62




The vertical force and the moment:

Resulting equations of motion:

21 2 3

2 2

4 5 6

12

1

2

z

y

z BF v BL C C C v v

z B M v B L C C C

v v

ϕ ρ ϕ

ϕ ρ ϕ

⎛ ⎞= + +⎜ ⎟⎝ ⎠

⎛ ⎞= + +⎜ ⎟

⎝ ⎠

&&

&&

ϕ

zv

2 2

1 2 3

3 2 2 2

5 6 4

1 1 10

2 2 2

1 1 10

2 2 2

mz c vBLC z kz vB LC v BLC

I c vB LC k v B LC vB LC zϕ ϕ

ρ ρ ϕ ρ ϕ

ϕ ρ ϕ ρ ϕ ρ

⎛ ⎞+ − + − − =⎜ ⎟

⎝ ⎠

⎛ ⎞ ⎛ ⎞+ − + − − =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

&&& &

&& & &

The motion of the bar is governed by 2 coupled second-order differential equations

63

2

2

2

4

3

6

1

2

3

5

2 2

11 1

2

1

2

2

1

2

1

2

0

0

2v BLc

k v B

vB Lmz z kz

I z

C

vB LC

vBLC

c vB

C

L L C C ϕ ϕ

ϕ ϕ

ϕ

ρ

ρ

ρ

ϕ ϕ ρ ρ

ρ + + − − =

+ ⎛ ⎞

−⎜ ⎟+ −⎝ ⎠

⎛ ⎞−⎜ ⎟

⎝ ⎠

⎛ ⎞−⎜ ⎟

⎝ ⎠ =

&&& &

&& & &





The meaning of the coefficients in the equations of motion:

ϕ

z

Effective damping of thevertical motion

Effective damping of theangular motion

Effective stiffness ofthe angular motion

Velocity couplingcoefficients

‘Displacement’ coupling

64





Matrix form of the equations of motion:

[ ]

2 2

1 2 3

2 3 2 2

4 5 6

00, , ,

0

1 1 1

2 2 2,

1 1 10

2 2 2

T m x x x x z

I

c vBLC vB LC k v BLC

vB LC c vB LC k v B LC ϕ ϕ

ϕ

ρ ρ ρ

ρ ρ ρ

⎡ ⎤+ + = = = ⎢ ⎥

⎣ ⎦

⎡ ⎤ ⎡ ⎤− − −⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥= =⎛ ⎞ ⎛ ⎞⎢ ⎥ ⎢ ⎥− − −⎜ ⎟ ⎜ ⎟

⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

&& & M C K M

C K

The characteristic equation:

( )2 4 3 21 2 3 4det 0n n n n n ns s s a s a s a s a+ = + + + + = M C + K

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Unstable 2 DOFS – Flutter – Eigenvalue Analysis



Suppose, that the vertical and angular motions are uncoupled, and the effectivedamping and stiffness coefficients are positive (C2= C3=C4=0, C1< 0, C5< 0, C6< 0):

( )

( ) ( )1 2 3

2

5 6 4

0

0, 2

mz c QC z kz QBC QvC

I c QB C k QvBC QBC z Q vBLϕ ϕ

ϕ ϕ

ϕ ϕ ϕ ρ

+ − + − − =

+ − + − − = =

&&& &

&& & &

Wind velocity Wind velocity

V i b r a t i o n f r e q u e n c y – I m ( s )

V

i b r a t i o n d e c r e m e n t – R e ( s )

Stable vibrations!

66




Suppose, that the velocity coupling is active and the velocity couplingcoefficients are positive (C3=0, C2 >0, C4 >0, C1< 0, C5< 0, C6< 0):

( )

( ) ( )1 2 3

2

5 6 4

0

0, 2



ϕ ϕ

ϕ ϕ ϕ ρ

+ − + − − =

+ − + − − = =

&&& &

&& & &



V


Instability!

67





Suppose, that the velocity coupling is active and the velocity couplingcoefficients are negative (C3=0, C2

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Suppose, that both the velocity and displacement couplings are active and allcoupling coefficients are positive (C3>0,C2 >0, C4 >0, C1< 0, C5< 0, C6< 0):

( )

( ) ( )1 2 3

2

5 6 4

0

0, 2



ϕ ϕ

ϕ ϕ ϕ ρ

+ − + − − =

+ − + − − = =

&&& &

&& & &



V


Instability!

70

Formulation of equations of motion of a 3 DOFSusing the displacement method


Lecture 4 & 5 Demonstration of the displacement method on 3 DOFS

A rigid foundation block:

1k

3k

2k

a b

e

h

g

2F

1F

1 x 3 x

2 x

3F

This block has 3 DOF: 2 translational (horizontal and vertical) and 1 rotational

71





Three positions of the block, in each of which only one degree of freedom is activated:


1 1k x 2 1k x

1 x

3 2k x

2 x

1 3k ax

2 3k b x

3 3k h x

3ax

3hx

2 3

k bx

3b x

1 1 1 2 1 1 3 2 3 1

2 3 2 3 3 2

3 1 1 2 1 3 2 3 3

1 3 2 3 3 1 2

m x k x k x k ax k bx F

m x k x k hx F

J x k x a k x b k x h k hx h

k a x a k bx b F eF gF

= − − − + +

= − − +

= − ∗ + ∗ − ∗ − ∗

− ∗ − ∗ + + −

&&

&&

&&

1k

3k

2k

a b

e

h

g

2F 1

F

1 x 3 x

2 x

3F

72





Matrix form of the equations of motion:

[ ]

[ ]

1 2 3

1 2 3

1 2 1 2

3 3

2 2 2

1 2 3 1 2 3

, , generalized displacement vector

, , generalized force vector

0 0

0 0 mass matrix

0 0

0

0 stiffness matrix

T

T

x x x x

F F F F

m

m

J

k k a k b k

k h k

a k b k h k a k b k h k

= −

= −

⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦

+ −⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥− + +⎣ ⎦

M

K

, x x F + =&& M K

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Representation of a structure by N DOFS


Lecture 4 & 5 Representation of a structure by N DOFS

A road sign: Discrete Element or Finite Element mesh:

2F

1F

1 x 3 x

2 x3F

1 x

2 x3 x

7 x

8 x9 x

4 x

5 x6 x

10 x

11 x12 x

13 x

14 x15 x

16 0 x =

17 0 x =18 0 x =

1F

2F

Internodal connection

74

A numerical recipe for finding the stiffness matrix for anode of N DOFS


Lecture 4 & 5 Nodal Stiffness Matrix

Three consecutive calculations giving a nodal stiffness matrix:

1 11F k =

1

2

3

1

0

0

x

x

x

===

3 31F k =

2 21F k =

1 12F k =

1

2

3

0

1

0

x

x

x

===

3 32F k =

2 22F k =

1 13F k =

1

2

3

0

0

1

x

x

x

===

3 33F k =

2 23F k =

[ ] [ ]

11 12 13

nodal nodal nodal nodal 21 22 23 nodal 1 2 3 nodal 1 2 3

31 32 33

, , , , , , ,T T

k k k

x F k k k x x x x F F F F

k k k

⎡ ⎤

⎢ ⎥= = = =⎢ ⎥⎢ ⎥⎣ ⎦

K K

75




1. Free Vibrations of Undamped N DOFS - General

2. Example: 3 DOF Foundation Block

3. The Orthogonality Property

4. The Modal Mass Matrix and Modal Stiffness Matrix

5. The Matrix of Natural Frequencies

Lecture 6 76

Free Vibrations of Undamped N DOFS - General



Lecture 6 Free Vibrations of Undamped N DOFS


0 x x+ =&& M K


( ) ( ) ( )2 for undamped systems only

1 1

ˆexp sin N N

i i i i i

i i

x t X s t X t ω ϕ = =

= = +∑ ∑The corresponding system of N linear algebraic equations (the eigenvalue and the

eigenfrequency problems):

( ) ( )2 2det 0 or det 0i is ω + = − + = M K M K

( ) ( )2 2 ˆ0 or 0i i i is X X ω + = − + = M K M K

The Characteristic Equation: The Frequency Equation:

The natural frequencies are the positive roots of the frequency equation (this definition will begeneralized to the case of systems with damping later)

complexreal

20

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Free Vibrations of Undamped N DOFS - General


Lecture 6 Free Vibrations of Undamped N DOFS

The General Solution can be written as:

( ) ( ) ( ) ( )1 1 1

ˆ ˆ ˆsin sin ,

ˆ eigenvectors (independent of initial conditions)

natural frequencies (independent of initial conditions)

phase angles (to be determined by

N N N

i i i i i i i i i

i i i

i

i

i

x t X t x A t x u t

x

ω ϕ ω ϕ

ω

ϕ

= = == + = + =

−

−

−

∑ ∑ ∑

)

unknown constants (

initial conditions

to be determined by initial condit n )io si

A −

( ) ( ) 2sin these satisfy 0i i i i i i iu t A t u uω ϕ ω = + + =&&

The Eigenfunctions:

The Eigenmatrix:

1 2ˆ ˆ ˆ...

N x x x⎡ ⎤= ⎣ ⎦ E

78

Example: 3 DOF Foundation Block


Lecture 6 Example: 3 DOF Foundation Block


1k

3k

2k

a b

e

h

g

2F 1F

1 x 3 x

2 x

3F

1 11 2 1 2

2 3 3 2

2 2 2

1 2 3 1 2 33 3

0 0 0

0 0 0 0

0 0

x xm k k a k b k

m x k h k x

J a k b k h k a k b k h k x x

+ −⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥+ =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− + +⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦

&&

&&

&&

Assume that the parameters are such that:

1

111

1 1 21 1

31

12

2 2 22 2

32

fundamantal natural frequencyˆ 2.0 2.0

therad ˆ ˆ2.3 ; 1.6 1.6

and the eigenvector s1.0 1.0ˆ

ˆ 0.8rad

ˆ ˆ5.7

firs

; 1.0s

1 0

t

.ˆ

x

x x

x

x

x x

x

α

ω α

ω α

=⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢⎢ ⎥ ⎢−⎣ ⎦⎣ ⎦

2

3

1

13 1/2

3 3 23 3

33

0.8the natural frequency

1.0and the eigenvector

1.0

ˆ 1.0 0.5 the natural frequencyrad

ˆ ˆ19.8 ; 1.0 0.5s

2.

s

0 1

third

econd

s c nd

0ˆ

e o

.

x

x x

x

α

α

ω α

=−

=

−⎡ ⎤⎢ ⎥= − −⎥ ⎢ ⎥

⎥ ⎢ ⎥⎣ ⎦

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = − = − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦third and the eigenvector

79

Example: 3 DOF Foundation Block



Lecture 6 Example: 3 DOF Foundation Block


( ) ( ) ( )3 3

1 1

ˆ ˆsini i i i i ii i

x t x A t x u t ω ϕ = =

= + =∑ ∑

The Eigenmatrix:

11 12 13

21 22 23

31 32 33

ˆ ˆ ˆ 2.0 0.8 0 .5

ˆ ˆ ˆ 1.6 1.0 0.5

ˆ ˆ ˆ 1.0 1 .0 1.0

x x x

x x x

x x x

⎡ ⎤ −⎡ ⎤⎢ ⎥ ⎢ ⎥= = − −⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

E

12ˆ 0.8 x =

22ˆ 1 x =

32ˆ 1 x− =

2ω

2 RC

13ˆ 1 x =

33ˆ 2 x =

23ˆ 1 x− =

3ω

3 RC

11ˆ 2 x =

21ˆ 1.6 x =

31ˆ 1 x =

1ω

1 RC

80

The Orthogonality Property



Lecture 6 The Orthogonality Property

The Eigenfrequency Problem:

( )2 2ˆ ˆ ˆ0 or i i i i i X X X ω ω − + = = M K M K

Consider two different solutions . These satisfy:( ) ( )2 2ˆ ˆ, and ,r r s s X X ω ω 2 2ˆ ˆ ˆ ˆ,r r r s s s X X X X ω ω = = M K M K

Pre-multiplying these equations by , respectively, we obtainˆ ˆand T T s r

X X

2 2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ,T T T T s r r s r r s s r s X X X X X X X X ω ω = = M K M K

The following equality holds for any symmetric matrix: . Thus, asare symmetric, we may rewrite the above equations as

T T x y y x= A A and M K

( )2 2 2 2ˆ ˆ ̂ ̂ˆ ˆ ˆ ˆ ˆ, 0ˆT T T s s r T T r r s r s r s r s s r X X X X X X X X X X ω ω ω ω = = ⇒ − = M K M K M

21

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81

The Modal Mass Matrix and Modal Stiffness Matrix



The orthogonality with respect to the mass matrix:

ˆ ˆ 0 if T

s r X X r s= ≠ M

The orthogonality with respect to the stiffness matrix:

2ˆ ˆ ˆ ˆ ˆ ˆ0 if (follows from )T T T s r s s r s r X X r s X X X X ω = ≠ = K M K

Definitions:

the Modal Mass Matrix (diagonal)

the Modal Stiffness Matrix (diagonal)

where is the eigenmatrix

T

T

= −

= −

*

*

M E M E

K E K E

E

82



Demonstration that the modal mass matrix is diagonal:

11 1 1 2 1

2 2 1 2 2 2

1 2

1 2

1 1

2 2

ˆ ˆ ˆ ˆ ˆ ˆ ˆ.

ˆ ˆ ̂ ̂ ̂ ˆ ̂.ˆ ˆ ˆ...

. . . ..

ˆ ˆ ˆ ˆ ˆ ˆ.ˆ

ˆ ˆ 0 . 0

ˆ ˆ0 . 0

. . . .

ˆ ˆ0 0 .

T T T T

N

T T T T

N T

N

T T T T

N N N N N

T

T

T

N N

x x x x x x x

x x x x x x x x x x

x x x x x x x

x x

x x

x x

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎡ ⎤= = =⎢ ⎥ ⎣ ⎦ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

⎡ ⎤⎢ ⎥⎢ ⎥= =⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

*

M M M

M M M M E M E = M

M M M

M

M

M

*

11

*

22

*

0 . 0

0 . 0

. . . .

0 0 . NN

m

m

m

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

The Modal Mass Matrix and Modal Stiffness Matrix

83

The Matrix of Natural Frequencies



Lecture 6 The Matrix of Natural Frequencies

The Eigenfrequency Problem:

( )2 2 2ˆ ˆ ˆ ˆ ˆ0 or or r r r r r r r r X X X x xω ω ω − + = = = M K M K M K

Pre-multiplying by , we obtain which can be written asˆT

s x

2 ˆ ˆ ˆ ˆT T r s r s r x x x xω = M K

2 2 2

1 1 1 2 1 2 1 1 1 1 2 1

2 2 2

1 2 1 2 2 2 2 2 1 2 2 2

2 2 2

1 1 2 2 1 2

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ. .

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ. .

. . . . . . . .ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ. .

T T T T T T

N N N

T T T T T T

N N N

T T T T T T

N N N N N N N N N

x x x x x x x x x x x x



ω ω ω

ω ω ω

ω ω ω

⎡ ⎤⎢ ⎥⎢ ⎥ =⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

M M M K K K

M M M K K K

M M M K K K

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

Using the orthogonality,this simplifies to:

2 * *

1 11 11

2 * *

2 22 22

2 * *

0 . 0 0 . 0

0 . 0 0 . 0

. . . . . . . .

0 0 . 0 0 . N NN NN

m k

m k

m k

ω

ω

ω

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

84

The Matrix of Natural Frequencies



Introducing the matrix of natural frequencies as:

We can write:

2

1

2

2 2

2

0 . 0

0 . 0

. . . .

0 0 . N

ω

ω

ω

⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

2 * *

1 11 11

2 * *

2 22 22

2 * *

0 . 0 0 . 0

0 . 0 0 . 0

. . . . . . . .0 0 . 0 0 . N NN NN

m k

m k

m k

ω

ω

ω

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥=

⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Instead of

2 * *

1 11 11

2 * *

2 22 22

2 * *

0 . 0 0 . 0 0 . 0

0 . 0 0 . 0 0 . 0

. . . . . . . . . . . .

0 0 . 0 0 . 0 0 . N NN NN

m k

m k

m k

ω

ω

ω

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

2 * *= M K

Lecture 6 The Matrix of Natural Frequencies

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1. Modal Analysis of Forced Vibrations of Undamped N DOFS

• General

• Use of Initial Conditions

• The Steady-State Response to a Harmonic Load

2. Finding the Forced Response of N DOFS using IntegralFourier Transform

Lecture 7 86

Modal Analysis of Forced Vibrations of UndampedN DOFS - General


Lecture 7 Modal Analysis of Forced Vibrations of Undamped N DOFS


( ) x x F t + =&& M K

The main idea of the Modal Analysis is to represent the forced motion as thesuperposition of the normal modes of free vibrations multiplied by unknownfunctions of time:

( ) ( ) ( )1

ˆ N

i i

i

x t x u t u t =

= =∑ E

Substitution of the above into the equations of motion gives:

u u F + =&& M E K E

Pre-multiplying this by E T , we obtain:

* *T T T T u u F u u F + = ⇒ + =&& && E M E E K E E M K E

87




Uncoupled System of Equations:

( ) ( )

2 * *

* * * * 2 * * 2

2 * * *

ˆ

or

ˆ, where

T T T

ii i ii i i i

T

i i i i ii i i

u u F u u F m u m u x F

u u F t m F t x F

ω

ω

=

+ = ⇒ + = ⇔ + =

+ = =

&& && &&

&&

M K

M K E M M E

Once the modal displacements are known, the structural displacements can becalculated as:

The general solution to the modal equation of motion (Duhamel’s integral):

( ) ( ) ( ) ( )( )**0

1sin sin

where and are to be determined from the initial conditions

t

i i i i i i

i ii

i i

u t A t F t d m

A

ω ϕ τ ω τ τ ω

ϕ

= + + −∫

( )iu t

( ) ( ) ( )1

ˆ N

i i

i

x t x u t u t =

= =∑ E

Modal Analysis of Forced Vibrations of UndampedN DOFS- General

88

Modal Analysis of Forced Vibrations of UndampedN DOFS – Use of Initial Conditions




Explicit form for the vector-displacement and vector-velocity:

( ) ( ) ( ) ( ) ( )( )

( ) ( ) ( ) ( ) ( )( )

*

*1 1 0

*

*1 1 0

1ˆ ˆ sin sin

1ˆ ˆ cos cos

t N N

i i i i i i i i

i i i ii

t N N

i i i i i i i i i

i i ii

x t x u t x A t F t d m

x t x u t x A t F t d m

ω ϕ τ ω τ τ ω

ω ω ϕ τ ω τ τ

= =

= =

⎛ ⎞= = + + −⎜ ⎟⎜ ⎟

⎝ ⎠

⎛ ⎞= = + + −⎜ ⎟

⎝ ⎠

∑ ∑ ∫

∑ ∑ ∫& &

To determine the unknown constants, we consider the above expressions at t=0:

( ) ( ) ( ) ( ) ( ) ( )1 1 1 1

ˆ ˆ ˆ ˆ0 0 sin , 0 0 cos N N N N

i i i i i i i i i i i

i i i i

x x u x A x x u x Aϕ ω ϕ = = = =

= = = =∑ ∑ ∑ ∑& &

Pre-multiplying these equations by , we obtainˆT j x M

( ) ( ) ( ) ( )1 1

ˆ ˆ ˆ ˆ ˆ ˆ0 sin , 0 cos N N

T T T T

j j i i i j j i i i i

i i

x x x x A x x x x Aϕ ω ϕ = =

= =∑ ∑& M M M M

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Using the orthogonality with respect to the mass matrix, we obtain:

( ) ( ) ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ0 sin , 0 cosT T T T j j j j j j j j j j j x x x x A x x x x Aϕ ω ϕ = =& M M M M

Squaring the above equations and subsequently summing them up we obtain thefollowing expression for the amplitudes:

( ) ( )22

* *

ˆ ˆ0 0T T j j j

jj j jj

x x x x A

m mω

⎛ ⎞⎛ ⎞= + ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

& M M

The phases are found by dividing the first above equation by the second:

( )

( )

ˆ 0arctan

ˆ 0

T

j

j j T j

x x

x xϕ ω

⎛ ⎞= ⎜ ⎟

⎜ ⎟⎝ ⎠&

M

M

90

Modal Analysis of Undamped N DOFS - The

Steady-State Response to a Harmonic Load


Lec ture 7 The Steady-State Response of Undamped N DO FS to a Harmonic Load

Consider a N DOFS under a synchronous harmonic load:

( )ˆ sin x x F t + = Ω&& M K

Rewriting these equations in the form of the uncoupled system yields:

( )2*

ˆˆsin

T

i

i i i

ii

x F u u t

mω + = Ω&&

In the steady-state regime, the N DOFS has to vibrate on the frequency of theexternal load, i.e.:

( ) ( )2 2 * 2 2 *

2 2 *

ˆ ˆˆ ˆ1ˆ ˆ ˆsin

ˆ1ˆor

T T

i i

i i i i i

ii i ii

T

i ii

x F x F u u t u u

m m

F u

m

ω ω

ω

= Ω ⇒ −Ω + = ⇒ =− Ω

=− Ω

E

91






The Modal Frequency Response Function of the mode i to the harmonic force

applied to the degree of freedom p:

Varying i and p, one can compose the

following (not symmetric) ModalFrequency Response Matrix:

( )2 2 *

2 2 * 2 2 *

ˆ FRF of mode to the force applied toˆ 1-

ˆ degree of freedo

ˆ ˆˆ ˆ1 1ˆ ˆ ˆ0 0 ...

0

mi p

T T

i pi p

p i

i ii i i

piiu F

i ii p

i

x F x F F F

x iu H

pmF

um mω ω

ω

⎡ ⎤= ⇒ = =⎣ ⎦ − Ω − Ω

Ω = =

− Ω

1 1 1 2 1

2 1 2 2 2

1 2

.

.

. . . .

.

N

N

N N N N

u F u F u F

u F u F u F

uF

u F u F u F

H H H

H H H

H H H

⎡ ⎤⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

H

92






General expression for the modal amplitudes:

ˆˆuF

u F = H

The original vector-displacement is given as:

( ) ( )( ) ( ) ( ) ( )ˆ ˆsin , sin

ˆ ˆˆ ˆ

ˆˆ , where

x t x t u t u t

xF xF uF

uF xF x

x

t u t x u F

F

F = Ω = Ω

= ⇒ = =

= =

= E E E

H

H

H E H

H

The Frequency Response Matrix:1 1 1 2 1

2 1 2 2 2

1 2

.

.

. . . .

.

N

N

N N N N

x F x F x F

x F x F x F

xF

x F x F x F

H H H

H H H

H H H

⎡ ⎤⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

H

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Expressions for the amplitude of vibrations of a particular degree of freedom:

1 1 1 2 1

1

1

2

2

1

1.

.. . .

ˆ ˆ .ˆ ˆ :

. . .

ˆ

ˆ.

.

.

ˆ

.

q q q

N

q p

N N N N

N x F x F x F q

N

x F x F x F

N

xF q x F p

p

x F x F x F

H H H

x F x H F

H H H

F

H H H F

F

=

⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥= = ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

∑ H

Elaborating further we obtain:

1

2 2 * 2 2 *1 1 1

ˆˆ , where

ˆ ˆ ˆ1 1ˆ ˆ

q p

q p i p

N

q x F p

p

N N N

pi qi pi x F qi u F qi

i i ii ii i ii

x H F

x x x H x H x

m mω ω

=

= = =

=

= = =− Ω − Ω

∑

∑ ∑ ∑

FRF of degree of freedom to the force

applied to degree of freedom

i

p

94

Modal Analysis of Undamped N DOFS –The




A representative plot of FRF for 5 DOFS:q p x F H

1ω 2ω 3ω 4ω 5ω Ω

q p x F H

95

Finding the Forced Response of N DOFS by usingIntegral Fourier Transform




( ) x x F t + =&& M K

( )

( )

( )

( ) ( )

1exp

2

F t F i t d

x t x

ω ω ω

π ω

∞

−∞

⎧ ⎫⎧ ⎫⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬

⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

∫%

%

Fourier Transform: Inverse Fourier Transform

( )

( )

( )

( ) ( )exp

F t F i t dt

x t x

ω ω

ω

∞

−∞

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪= −⎨ ⎬ ⎨ ⎬

⎪ ⎪⎪ ⎪ ⎩ ⎭⎩ ⎭ ∫

%

%

Equation of Motion and its Solution in the Frequency Domain:

( ) ( ) ( ) ( ) ( ) ( )1

2 2 x F x F ω ω ω ω ω ω −

− + = ⇒ = − +% %% % M K M K

Solution in the Time Domain:

( ) ( ) ( ) ( )1

21 exp2

x t F i t d ω ω ω ω π

∞−

−∞

= − +∫ % M K

96



1. Modal Analysis of Forced Vibrations of N DOFS with ViscousDamping

• General

• Use of Initial Conditions

• The Steady-State Response to a Harmonic Load

Lecture 8

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101



Explicit form for the vector-displacement and vector-velocity:

( ) ( ) ( ) ( ){

( ) ( )( ) ( )( )

( ) ( )

( ) ( ) ( )( ){

( )

2

1 1

* 2

* 20

1

2 2 2

1

*

* 20

ˆ ˆ exp sin 1

1sin 1 exp

1

ˆ

ˆ exp sin 1 1 cos 1

sin 11

N N

i i i i i i i i i

i i

t

i i i i i

ii i i

N

i i

i

N

i i i i i i i i i i i i

i

t

ii i i

ii i

x t x u t x A t t

F t t d m

x t x u t

x t t t

F m

ξ ω ω ξ ϕ

τ ω ξ τ ξ ω τ τ ω ξ

ω ξ ω ξ ω ξ ϕ ξ ω ξ ϕ

ξ τ ω ξ

ξ

= =

=

=

= = − − +

⎫⎪+ − − − − ⎬

− ⎪⎭

=

= − − − + + − − +

− −−

∑ ∑

∫

∑

∑

∫

& &

( )( ) ( )( )

( ) ( )( ) ( )( )

2

* 2

*

0

exp

1cos 1 exp

i i

t

i i i i i

ii

t t d

F t t d m

τ ξ ω τ τ

τ ω ξ τ ξ ω τ τ

− − −

⎫+ − − − − ⎬⎭

∫

Lecture 8 Modal Analysis of Forced Vibrations of Damped N DOFS 102



To determine the unknown constants, we consider the above expressions at t=0:

( ) ( ) ( )

( ) ( ) ( ) ( )( )1 1

2

1 1

ˆ ˆ0 0 sin ,

ˆ ˆ0 0 sin 1 cos

N N

i i i i i

i i

N N

i i i i i i i i i

i i

x x u x A

x x u x A

ϕ

ω ξ ϕ ξ ϕ

= =

= =

= =

= = − + −

∑ ∑

∑ ∑& &

Pre-multiplying these equations by , and using the orthogonality, we obtainˆT

j x M

( ) ( ) ( )

( ) ( ) ( )( )

( ) ( )( )

*

1

2

1

* 2

ˆ ˆ ˆ0 sin sin

ˆ ˆ ˆ0 sin 1 cos

sin 1 cos

N T T

j j i i i jj j j

i

N T T

j j i i i i i i i

i

jj j j j j i j

x x x x A m A

x x x x A

m A

ϕ ϕ

ω ξ ϕ ξ ϕ

ω ξ ϕ ξ ϕ

=

=

= =

= − + −

= − + −

∑

∑&

M M

M M

Lecture 8 Modal Analysis of Forced Vibrations of Damped N DOFS

103

( ) ( )

( ) ( )

( ) ( ) ( )

( )

2

2

* 2

ˆ1 0tan ,

ˆ ˆ0 0

ˆ ˆ ˆ0 0 01

ˆ1 0

T

i i j

j T T

j i j j

T T T

j j i j j

jT

jj i i j

x x

x x x x

x x x x x x A

m x x

ω ξ ϕ

ω ξ

ω ξ

ω ξ

−=

+

⎛ ⎞+⎜ ⎟= +⎜ ⎟

−⎝ ⎠

&

&

M

M M

M M M

M



From the above two equations, the following expressions can be derived:

Lecture 8 Modal Analysis of Forced Vibrations of Damped N DOFS

Mathematically, the problem has been solved. But … what about our assumptionconcerning the diagonality of the Modal Damping Matrix?

104

Proportional (Rayleigh) Damping


The proportional damping is defined as follows:

Lecture 8 Proportional Damping

To identify the unknown constants, one has to know the Modal Damping Ratios forat least two modes. Suppose, these are known for the first two modes. Then

0 1 0 1, with and unknown constantsa a a a+ −C = M K

The modal damping takes the form:

( )* * *0 1 0 1 0 1T T T T a a a a a a= + +C E C E = E M K E = E ME + E KE = M K

The modal Modal Damping Ratio takes the form:

* * *

0 1 0 1* *2 2 2 2ii ii ii i

i

ii i ii i i

c a m a k a am m

ω ξ ω ω ω += = = +

( ) ( )1 2 1 2 2 1 2 2 1 10 12 2 2 2

2 1 2 1

2 2,a a

ω ω ξ ω ξ ω ξ ω ξ ω

ω ω ω ω

− −= =

− −

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Proportional Damping


Three Cases(assuming ω1 > < <

> ≤ ≤

≤ > ≤

1

2

0 1a a

ξ

ξ

1ω

2ω

0 1a a

1

2 i

aω

0

2i

a

ω

case A

iω

iξ

1 2 2

2 1 1

ω ξ ω

ω ξ ω < <

1

2

ξ

ξ

iξ

0 1a a−

1ω 2ω

0

2i

a

ω

1

2 i

aω

iω

2 1

1 2

ξ ω

ξ ω ≤ case B

2

1

ξ

ξ

0 1a a−

1ω

iξ

2ω

0

2i

a

ω

1

2 i

aω

iω

case C

2 2

1 1

ξ ω

ξ ω ≥

106

Modal Analysis of Damped N DOFS - The Steady-

State Response to a Harmonic Load


Lec ture 8 The Steady-State Response of Damped N DOF S to a Harmonic Load

Consider a N DOFS under a synchronous harmonic load:

( )ˆ sin x x x F t + + = Ω&& & M C K

Rewriting these equations in the form of the uncoupled system yields:

( )2*

ˆˆ2 sin

T

i

i i i i i i

ii

x F u u u t

mξ ω ω + + = Ω&& &

In the steady-state regime, the N DOFS has to vibrate on the frequency of theexternal load but with a certain phase shift, i.e.:

( )

( )( ) ( ) ( )22 *22 2

ˆ sin

ˆˆ 21 1

ˆ , tan 11 2

substitution

i i i

T

i i i

i i

i ii ii i i

u u t

x F

u m

ϕ

ξ ω

ϕ ω ω ω ξ ω

= Ω − ⇒

Ω

= = − Ω− Ω + Ω

107





The Modal Amplitude-Frequency Response Characteristic:

Varying i and p, one can compose the

following (not symmetric) Modal Amplitude-Frequency Response Matrix:

( )The ratio of the amplitude of the displacement of modeˆ

-ˆ to that of the force applied to degree of freedomi p

A iu F

p

iu H

pF Ω =

1 1 1 2 1

2 1 2 2 2

1 2

.

.

. . . .

.

N

N

N N N N

u F u F u F

u F u F u F A

uF

u F u F u F

H H H

H H H

H H H

⎡ ⎤⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

H

( )

( )( ) ( )2 *2

2 2

ˆ1 1

1 2i p

pi A

u F

i ii

i i i

x H

mω ω ξ ω

Ω =− Ω + Ω

108





The typical shape of the Modal Amplitude-Frequency Response Characteristic andthe Modal Phase-Frequency Response Characteristic:

i p

A

u F H

( ) ( )22 22

ˆ1 1ˆ ˆ

1 2

i p

pi

u F T

i i i

iii

x H

x M xω ξ

ω ω

=⎧ ⎫ ΩΩ− +⎨ ⎬⎩ ⎭

2

ˆ

ˆ ˆ2

pi

T

i i i i

x

x M xξ ω

2

ˆ

ˆ ˆ

pi

T

i i i

x

x M xω

0 iω Ω

( )2

2

arctan

1

ii

i

i

ξ ω

φ

ω

Ω⎛ ⎞⎜ ⎟

= ⎜ ⎟Ω⎜ ⎟−⎜ ⎟

⎝ ⎠

0 iω Ω

2π

π

0

iφ

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Modal Analysis of Damped N DOFS - The Ste

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