aero 7sem ae2403nol

7/30/2019 Aero 7sem Ae2403nol

1/51

AE 2403 VIBRATIONS ANDELEMENTS OF AEROELASTICITY

BYMr. G.BALAJIDEPARTMENT OF AERONAUTICAL ENGINEERINGREC,CHENNAI


2/51

Fundamentals of Linear Vibrations

1. Single Degree-of-Freedom Systems

2. Two Degree-of-Freedom Systems

3. Multi-DOF Systems

4. Continuous Systems


3/51

Single Degree-of-Freedom Systems

1. A spring-mass systemGeneral solution for any simple oscillator

General approach

Examples

1. Equivalent springsSpring in series and in parallel

Examples

1. Energy MethodsStrain energy & kinetic energyWork-energy statement

Conservation of energy and example


4/51

A spring-mass system

General solution for any simple oscillator:

Governing equation of motion:0=+kxxm

)sin()cos()( tv

txtx nn

ono

+=

2

n

o2

o

n

n

n

ooo

v

xamplitudeC;2

T

1

Hz)orc.(cycles/sefrequencyf

vibrationofperiodT;T

2)(rads/sec.frequencynatural

m

k

(sec.)timet;xvelocityinitialvnt;displacemeinitialx

+=====

====

====

where:


5/51

Any simple oscillator

General approach:

1. Select coordinate system

2. Apply small displacement3. Draw FBD4. Apply Newtons Laws:

)(

)(

Idt

dM

xmdt

dF

=

=


6/51

Simple oscillator Example 1

22

mlmdI

inertiaofmomentmassI

cg =+=

=

IK

IM

=

=

02 =+ Kml 2ml

Kn =

+


7/51


=

=+=

l

a

m

k

mlmdII

n

cg

22

2)( mlaak

IM oo

=

=

022 =+ kaml

+

(unstable),l

a

As

m

k,

l

aWhenits:limNote

n

n

00

1

==


8/51


9/51


Lma

GJ

L

JGK:stiffnessEquivalent

TL

JG

JG

TL

maI:tableFrom

n 2

2

2

2

2

=

=

=

=

=

IT

IMz

==

02

2

=+ L

GJma

+


10/51

Equivalent springs

Springs in series:

same force - flexibilities add

Springs in parallel:

same displacement - stiffnesses add

21 kkkeq +=

=+= += eqkkkkkP)( 21

21

PfPff

Pkk

eq=+=

+=+=)(

11

21

2121

21 fffeq +=


11/51

Equivalent springs Example 1

0=+ xKxm eq

0312

3

2

3

1

=

++ xL

EI

L

EIxm


12/51


)a(

ml

Wlka

nn

n

=

=

2

22

2mllWa)ak(

IM oo

=+

=

022 =+ )Wlka(ml

+

Consider:

ka2 > Wl n2 is positive - vibration is stable

ka2 = Wl statics - stays in stable equilibrium

ka2 < Wl unstable - collapses


13/51


02

2

=+

=

=

sinmglml

mlsinWl

IM oo

0=+ sin

l

g

l

g

l

g

n =

=+ 0

+We cannot definen

since we havesin term

If< < 1, sin:


14/51


15/51

Work-Energy principles

Work done = Change in kinetic energy

Conservation of energy for conservative systems

E = total energy = T + U = constant

12

2

1

2

1

TTdTrdFT

T

r

r

==


16/51

Energy methods Example

0

0

=+

=

xxmxkx

)E(dt

d

0=+ kxxm 22

2

2

2

1

2

1

2

1

2

1

xmkxTUE

xmT

kxU

+=+=

=

=

Same as vector mechanics

Work-energy principles have many

uses, but one of the most useful is

to derive the equations of motion.

Conservation of energy: E = const.


17/51

Two Degree-of-Freedom Systems

1. Model problemMatrix form of governing equation

Special case: Undamped free vibrations

Examples

1. Transformation of coordinatesInertially & elastically coupled/uncoupled

General approach: Modal equations

Example

1. Response to harmonic forcesModel equationSpecial case: Undamped system


18/51


19/51

Undamped free vibrationsZero damping matrix [C] and force vector

{P}

)cos(2

1

2

1

=

tA

A

x

xAssumed general solutions:

Characteristic polynomial (for det[ ]=0):

021

212

2

2

1

214

=+

+

+ mm

kk

m

k

m

kk

+

+

+

+==

21

21

21

2

2

2

1

21

2

2

1

212

21

21

4

2

1

mm

kk

m

k

m

kk

m

k

m

kk

Eigenvalues (characteristic values):

Characteristic equation:

=

+

0

0

)(

)(

2

1

2

222

22121

A

A

mkk

kmkk


20/51

Undamped free vibrationsSpecial case when k

1

=k2

=k and m1

=m2

=m

Eigenvalues and frequencies:

periodlfundamenta

frequencylfundamenta

==

==

T

m

k.

2

61801

m

k

=

=618.2

3819.02

1

2

1

21

Two mode shapes (relative participation of each mass in the motion):

1

618.12 2

1

2 ==k

mk

A

A shapemode1st

1

618.02

1

2 =

=mk

k

A

Ashapemode2nd

The two eigenvectors are orthogonal:

=

618.1

1

)1(2

)1(

1

A

A

=

618.0

1

)2(2

)2(

1

A

AEigenvector (1) = Eigenvector (2) =


21/51

Undamped free vibrations (UFV)

For any set of initial conditions:We know {A}(1) and {A}(2), 1 and 2

Must find C1, C2, 1, and 2 Need 4

I.C.s

{ } )cos()cos()(

)(22)2(

2

)2(

1

211)1(

2

)1(

1

1

2

1 +

++

=

= tA

ACt

A

AC

tx

txx

Single-DOF:

For two-DOF:

)cos()( += tCtx n


22/51

UFV Example 1

{ } )cos(618.0

0.1)cos(

618.1

0.12211

2

1tCtC

x

xx

+

=

=

Given:

No phase angle since initial velocity is 0:

{ } { } { }

==

618.1

0.10 oxx and

{ }

+

=

=618.0

0.1

618.1

0.1

618.1

0.121 CCxo

From the initial displacement:

1

1

21

2;0;

===

T

CC


23/51

UFV Example 2

{ } )cos(618.0

1

)171.0()cos(618.1

1

)171.1( 21 ttx

+

=

Now both modes are involved:

Solve for C1 and C2:

{ } { } { }

==

2

10 oxx and

{ }

=

+

=

=2

1

21618.0618.1

11

618.0

1

618.1

1

2

1

C

CCCxo

From the given initial displacement:

=

=

171.0

171.1

2

1

1618.1

1618.0

618.1618.0

1

2

1

C

C

Hence,

or

Note: More contribution from mode 1

)cos()618.0(171.0)cos()618.1(171.1)(

)cos()1(171.0)cos()1(171.1)(

212

211

tttx

tttx

==


24/51

Transformation of coordinates

Introduce a new pair of coordinates that represents spring stretch:

=

++

0

0)(

0

0

2

1

22

221

2

1

2

1

x

x

kk

kkk

x

x

m

m

UFV model problem:

inertially uncoupled

elastically coupled

z1(t) = x1(t) = stretch of spring 1

z2(t) = x2(t) - x1(t) = stretch of spring 2

or x1(t) = z1(t) x2(t) = z1(t) + z2(t)Substituting maintains symmetry:

=

+

+

0

0

0

0)(

2

1

2

1

2

1

22

221

z

z

k

k

z

z

mm

mmm

inertially coupled elastically uncoupled


25/51


We have found that we can select coordinates so that:

1) Inertially coupled, elastically uncoupled, or

2) Inertially uncoupled, elastically coupled.

Big question: Can we select coordinates so that both are uncoupled?

Notes in natural coordinates:

The eigenvectors are orthogonal w.r.t [M]:

The modal vectors are orthogonal w.r.t [K]:

Algebraic eigenvalue problem:

{ } { }

=

=

=

=618.0

1

618.1

1

:vectors)(modalrsEigenvecto

)2(

2

)2(

1

2)1(

2

)1(

1

1A

Au

A

Au

{ } [ ] { }{ } [ ] { } 0

0

12

21

==

uMu

uMuT

T

{ } [ ] { }

{ } [ ] { } 0

0

12

21

=

=

uKu

uKu

T

T

[ ]{ }

[ ]{ }

[ ]{ }

[ ]{ }

222111

uMuKuMuK ==


26/51


Governing equation:

Modal equations:

Solve for these using initial conditions then substitute into (**).

[ ]{ } [ ]{ } 0=+ xKxM

{ } { } { }

)()()(

)(

)()(

2

22

12

1

21

11

2

1

2211

tqu

u

tqu

u

tx

tx

tqutqux

+

=

+= (**)

General approach for solution

We were calling A - Change to u to match Meirovitch

{ }{ }

=+

=+

0)()((*)

0)()((*)

2

2

222

1

2

111

tqtqu

tqtqu

T

T

[ ] { } { }( ) [ ] { } { }( ) 0)()()()( 22112211 =+++ tqutquKtqutquM (*)Substitution:

Let

or

Known solutions


27/51

Transformation - Example

{ } )cos()171.0(618.0

1)cos(171.1

618.1

121 ttx

+

=

2) Transformation:

=

=

=

=618.0

1;618.1

618.1

1;618.0

22

12

2

21

11

1u

u

u

u and

1) Solve eigenvalue problem:

=

=

=

+

=

)cos()0()(

)cos()0()(

171.0

171.1

)0(

)0(

)0(618.0

1)0(

618.1

1

2

1

222

111

2

1

21

tqtq

tqtq

q

q

qq

and

So

As we had before.

More general procedure: Modal analysis do a bit later.

Model problem with: { } { }

=

=0

0

2

1oo xx and

{ } { } { } =+ =++= 0)()(0)()(

)()(2

2

22

1

2

112211tqtq

tqtqtqutqux

and


28/51

Response to harmonic forces

Model equation:

[M], [C], and [K] are full but symmetric.

[ ]{ } [ ]{ } [ ]{ } { } tieFFtFxKxCxM

==++

2

1)(

{F}

not function of time

Assume:

{ } { }ti

eiX

iXiXx

== )()(

)(2

1

Substituting gives:[ ] [ ] [ ]( ){ } { }FiXKCiM =++ )(2

[ ] matriximpedance2x2=)( iZ

[ ] [ ] { } [ ] { }FiZiXiZiZ 11 )()()()( =

{ }

=

=2

1

1112

1222

2

1222112

1 1

F

F

zz

zz

zzzX

XX

Hence:

( )212 ,ji,kcimz ijijijij =++=

:)(ioffunctionarezAll ij


29/51


30/51

Multi-DOF Systems

1. Model EquationNotes on matrices

Undamped free vibration: the eigenvalue problem

Normalization of modal matrix [U]

1. General solution procedureInitial conditions

Applied harmonic force


31/51

Multi-DOF model equation

Model equation:

Notes on matrices:

They are square and symmetric.

[M] is positive definite (since T is always positive)

[K] is positive semi-definite: all positive eigenvalues, except for some potentially 0-eigenvalues which

occur during a rigid-body motion. Ifrestrained/tied down positive-definite. All positive.

[ ]{ } [ ]{ } [ ]{ } { }Q=++ xKxCxM

1) Vector mechanics (Newton or D Alembert)

2) Hamilton's principles

3) Lagrange's equations

We derive using:

Multi-DOF systems are so similar to two-DOF.

{ } [ ]{ }

{ } [ ] { }xKxU

xMxT

T

T

21

21

=

=

:springinenergyStrain

:energyKinetic


32/51

UFV: the eigenvalue problem

Matrix eigenvalue problem

Equation of motion:

{ } { } titi eAeAtftfuq +== 21)()(

[ ]{ } [ ]{ } { }0=+ qKqM

Substitution of

in terms of the generalized D.O.F. qi

leads to

[ ] { } [ ] { }uMuK 2=

For more than 2x2, we usually solve using computational techniques.

Total motion for any problem is a linear combination of the naturalmodes contained in {u} (i.e. the eigenvectors).


33/51

Normalization of modal matrix [U]

Do this a row at a time to form [U].

This is a common technique

for us to use after we have solved

the eigenvalue problem.

We know that: [ ] { } { } [ ] { } ijjT

iji CuMuuMu ==

{ }

=1

ku

=

=

=

ji

ji

ij

if

if

deltaKronecker

:where

0

1So far, we pick our

eigenvectors to look like:

Instead, let us try to pick

so that:

{ } { }

==1

knewk uu

{ } [ ] { } { } [ ] { } 12 == kT

knewk

T

newkuMuuMu

Then: [ ] [ ] [ ] [ ]IUMUT = [ ] [ ] [ ] [ ]=UKU Tand

[ ]

=

2

2

2

2

1

..0

....

..0

0.0

n

:where

Let the 1st

entry be 1


34/51

General solution procedure

For all 3 problems:

1. Form [K]{u} = 2

[M]{u} (nxn system)Solve for all 2 and {u} [U].

2. Normalize the eigenvectors w.r.t. mass matrix (optional).

Consider the cases of:

1. Initial excitation

2. Harmonic applied force

3. Arbitrary applied force

{ } { }oo qq and


35/51

Initial conditions

2n constants that we need to determine by 2n conditions

General solution for any D.O.F.:

Alternative: modal analysis

{ } { } { } { } )cos()cos()cos()( 22221111 nnnn tCutCutCutq +++=

Displacement vectors:

{ } { }ioio

qq andon

{ } [ ]{ }{ } { } { } { } )()()()( 2211 tutututq

Uq

nn

+++==

UFV model equation: [ ]{ } [ ]{ } { }

[ ] [ ] [ ]{ } [ ] [ ] [ ]{ } { }{ } [ ]{ } { }00

0

=+ =+

=+

UKUUMU

qKqM

TT

n modal equations:

=+

=+

=+

0

0

0

2

2

2

22

1

2

11

nnn

Need initial conditions on ,

not q.


36/51

Initial conditions - Modal analysis

Using displacement vectors: { } [ ] { }

[ ] [ ] { } [ ] [ ] [ ] { }UMUqMU

Uq

TT =

=

As a result, initial conditions:

Since the solution of

{ } [ ] [ ] { }

{ } [ ] [ ] { } == o

T

o

o

T

o

qMU

qMU

)sin()(

)cos()()(

)sin()(

)cos()()( 11

1111

ttt

ttt

n

n

nonnon

oo

+=

+=

And then solve

hence we can easily solve for

{ } [ ] [ ] { }qMU T=or

02 =+ is:

)sin()cos()(

)cos(

ttt

tC

oo

+=

or

{ } [ ]{ }Uq =


37/51

Applied harmonic force

Driving force {Q} = {Qo}cos(t)

Equation of motion:

{ } [ ]{ }

[ ]{ } unknownknownU

Uq =

[ ]{ } [ ]{ } { }Q=+ qKqM

Substitution of

leads to

[ ] [ ] [ ]{ } [ ] [ ] [ ] { } [ ] { } { }NtQUUKUUMU oTTT ==+ )cos(

{ } { } requencydriving ftQQ o

==)cos(

and

Hence, { } { }

{ } { }

.

)cos(

)cos(

22

2

22

22

1

11

etc

tQu

tQu

o

T

oT

=

= then

{ } [ ]{ }Uq =


38/51

Continuous Systems

1. The axial barDisplacement field

Energy approach

Equation of motion

1. ExamplesGeneral solution - Free vibration

Initial conditions

Applied force

Motion of the base

1. Ritz method Free vibrationApproximate solution

One-term Ritz approximation

Two-term Ritz approximation


39/51


40/51


41/51

Axial bar - Equation of motion

2

22

2

2

x

u

t

u

=

Hamiltons principle leads to:

If area A = constant

( ) 0=

+

x

uEA

xuA

t

Since x and t are independent, must have both sides equal to a constant.

Separation of variables: )()(),( tTxXtxu =

)sin()cos(

02

tpBtpAT

TpT

+==+ ( )

( ) ( )

xpDxpCX

XpX

sincos

02

+==+

Hence[ ] ( ) ( )[ ]

=

++=1

sincos)sin()cos(),(

i

iiiiiiii xpDxpCtpBtpAtxu

=

3

22

LM

LFE

:where

( ) ( ) 222222contant p-

T

dtTd

X

dxXd===


42/51

Fixed-free bar General solution

0cos0 == LpD ii orsolution)(trivialEither

= wave speed

E

=

For any time dependent problem:

=

+

=,5,3,1 2

sin

2

cos

2

sin),(i

ii

L

tiB

L

tiA

L

xitxu

Free vibration:

[ ] ( ) ( )[ ]

= ++= 1 sincos)sin()cos(),( iiiiiiiii xpDxpCtpBtpAtxu

EBC:

NBC:

0)0( =u

00 =

=

== LxLx

x

u

x

uEA

General solution:

EBC [ ]

=

=+=1

0)sin()cos(),0(i

iiiii tpBtpACtu

( ) [ ]

== =+=

1

0)sin()cos(cosi

iiiiiii

Lx tpBtpALppD

x

u

0=iC

25

23

2

oror=Lpi

),5,3,1(2

== iL

ipi

NBC


43/51


44/51

Fixed-free bar Initial conditions

or

=

=,3,1

22

)1(

2

2

cos

2

sin1

)1()(8

),(i

i

o

L

ti

L

xi

i

LLtxu

Give entire bar an initial stretch.Release and compute u(x, t).

0)0,( 0 =

= =to

t

ux

L

LLxu and

Initial conditions:

Initial velocity:

Initial displacement:

=

=

== 0

2sin

2,3,10

i

itLxiB

Li

tu 0=iB

22sin2sin2sin

2sin

,3,100

,3,1

L

AdxL

xi

L

xi

AdxL

xi

xL

LL

L

xiAx

L

LL

ii

L

i

Lo

i

io

=

=

=

=

=

),3,1()1()(8

2sin

)(22

)1(

2202==

=

iiLL

dxL

xix

L

LLA

i

oL

oi

Hence


45/51

Fixed-free bar Applied force

or

( )txL

EA

Ftxu o

sinsinsec),(

=

Now, B.Cs:

=

=

= )sin(

0),0(

tFxuEA

tu

oLx

From

B.C. at x = 0:

B.C. at x = L:

= 0),0( tu 01 =A

=

L

EA

FA o sec2

Hence

2

22

2

2

x

u

t

u

=

)sin()(),( txXtxu n=we assume:

Substituting: ( )tx

Ax

Atxu

sinsincos),( 21

+

=

)sin()sin(cos2 tFtL

L

AEAx

u

EA oLx

=

=

=


46/51


47/51

Ritz method Free vibration

Start with Hamiltons principle after I.B.P. in time:

Seek an approximate solution to u(x, t):In time: harmonic function cos(t) ( = n)

In space: X(x) = a1 1(x)

where: a1 = constant to be determined

1(x) = known function of position

( ) ( ) dtdxuxx

uEAuuA

t

t

t

L

= 2

1 00

1(x) must satisfy the following:

1. Satisfy the homogeneous form of the EBC.

u(0) = 0 in this case.

2. Be sufficiently differentiable as required by HP.


48/51

One-term Ritz approximation 1

Ritz estimate is higher than the exact

Only get one frequency

If we pick a different basis/trial/approximation function 1, we would get a different result.

)cos()cos()(

)cos()cos()(),()(

1

1111

txtxu

txatxatxuxx

==

===

:eapproximatAlso

:Pick

[ ] dttdxEAxxAatt

L

)(cos)1)(1())((02

0

2

1

2

1

=Substituting:

2

22

23

2 33

3

L

E

LLEA

LA =

==

LLRITZ

732.1

3==

LLEXACT

571.12 ==

10

10

22adxEAadxxA

LL

=

Hence

[ ] { } [ ] { }( )aKaM =2:formmatrixin

=

L

xEXACT

2sin1

xRITZ =1


49/51


50/51

Two-term Ritz approximation

221)( xaxaxX +=:Let

[ ] dtdxxaaEAxxaxaAt

t

L

++= 21 0

21

2

21

2)1()2()(0

where:

:1 xu == eapproximatIf

xaadxdX 21 2+=

:2

xu =eapproximatIf [ ] dtdxxxaaEAxxaxaAt

t

L

++= 2

1 0 21

22

21

2 )2()2()(0

=

2

1

2221

1211

2

1

2221

12112

a

a

KK

KKE

a

a

MM

MM

==

===

==

5))((

4))((

3))((

5

0

22

22

4

0

2

2112

3

011

LdxxxM

LdxxxMM

L

dxxxM

L

L

L

In matrix form:

==

===

==

3

4)2)(2(

)1)(2(

)1)(1(

3

022

2

02112

011

LdxxxK

LdxxKK

LdxK

L

L

L


51/51

aero 7sem ae2403nol

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