AE 2403 VIBRATIONS AND ELEMENTS OF AEROELASTICITY

BYMr. G.BALAJIDEPARTMENT OF AERONAUTICAL ENGINEERINGREC,CHENNAI

Fundamentals of Linear Vibrations

1. Single Degree-of-Freedom Systems

2. Two Degree-of-Freedom Systems3. Multi-DOF Systems4. Continuous Systems

Single Degree-of-Freedom Systems

1. A spring-mass systemGeneral solution for any simple oscillatorGeneral approachExamples

2. Equivalent springsSpring in series and in parallelExamples

3. Energy MethodsStrain energy & kinetic energy Work-energy statementConservation of energy and example

A spring-mass system

General solution for any simple oscillator:

Governing equation of motion: 0 kxxm

)sin()cos()( tv

txtx nn

ono

2

n

o2o

nn

n

ooo

ωv

xamplitudeC;2ω

T1

Hz)or c.(cycles/sefrequencyf

vibrationofperiodT;T2

)(rads/sec.frequencynaturalmk

ω

(sec.)timet;xvelocityinitialvnt;displacemeinitialx

π

π

where:

Any simple oscillator

General approach:

1. Select coordinate system2. Apply small displacement3. Draw FBD4. Apply Newton’s Laws:

)(

)(

Idt

dM

xmdt

dF

Simple oscillator – Example 1

22 mlmdI

inertiaofmomentmassI

cg

IK

IM

02 Kml 2ml

Kωn

+

Simple oscillator – Example 2

l

a

m

kω

mlmdII

n

cg22

2)( mlaak

IM oo

022 kaml

+

(unstable)ω,l

aAs

m

kω,

l

aWhenits:limNote

n

n

00

1

Simple oscillator – Example 3

l

b

m

kω

mlmm

ml

mdII

mllA

AdxxdmrI

n

cgo

cg

l

3

3212

1212

2

222

2

23

22 2/

0

3)(

2mlbbk

IM oo

03

22

kbml

+

Simple oscillator – Example 4

Lma

GJL

JGK:stiffnessEquivalent

TL

JG

JG

TL

maI:tableFrom

n 22

2

2

2

IT

IM z

02

2

L

GJma

+

Equivalent springs

Springs in series:same force - flexibilities add

Springs in parallel:same displacement - stiffnesses add

21 kkkeq

eqkkk

kkP

)( 21

21

PfPff

Pkk

eq

)(

11

21

2121

21 fffeq

Equivalent springs – Example 1

0 xKxm eq

0312

32

31

x

L

EI

L

EIxm

Equivalent springs – Example 2

)a(ml

Wlkaω

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

Equivalent springs – Example 3

02

2

sinmglml

mlsinWl

IM oo

0 sinl

gl

gω

l

g

n

0

+We cannot define n

since we have sin term

If < < 1, sin :

Energy methods

Strain energy U:energy in spring = work done

Kinetic energy T:

Conservation of energy:work done = energy stored

PkU2

1

2

1 2

Tenergy kinetic ofincrement

done work ofIncrement

dT) rrm d(dt) r()r (m

rdF

21

rrmT

2

1

Work-Energy principles

Work done = Change in kinetic energy

Conservation of energy for conservative systems

E = total energy = T + U = constant

122

1

2

1

TTdT rdFT

T

r

r

Energy methods – Example

0

0

xxmxkx

)E(dt

d

0 kxxm 22

2

2

2

1

2

12

12

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.

Two Degree-of-Freedom Systems

1. Model problemMatrix form of governing equation Special case: Undamped free vibrationsExamples

2. Transformation of coordinates

Inertially & elastically coupled/uncoupledGeneral approach: Modal equationsExample

3. Response to harmonic forcesModel equationSpecial case: Undamped system

Two-DOF model problem

Matrix form of governing equation:

2

1

2

1

22

221

2

1

22

221

2

1

2

1 )()(

0

0

P

P

x

x

kk

kkk

x

x

cc

ccc

x

x

m

m

where:[M] = mass matrix; [C] = damping matrix;[K] = stiffness matrix; {P} = force vector

Note: Matrices have positive diagonals and are symmetric.

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

2222

22

121

A

A

mkk

kmkk

Undamped free vibrationsSpecial case when k1=k2=k and m1=m2=m

Eigenvalues and frequencies:

period lfundamenta

frequency lfundamenta

ω

π T

m

k.ω

2

61801

m

k

618.2

3819.021

21

21

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

1

618.12 2

1

2 k

mk

A

A shape mode 1st

1

618.02

1

2

mk

k

A

Ashape mode 2nd

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) =

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

12

1

tA

ACt

A

AC

tx

txx

Single-DOF:

For two-DOF:

)cos()( tCtx n

UFV – Example 1

)cos(618.0

0.1)cos(

618.1

0.12211

2

1 tCtCx

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:

11

21

2

;0;

T

CC

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

121 618.0618.1

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

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”

Transformation of coordinates

We have found that we can select coordinates so that:1) Inertially coupled, elastically uncoupled, or2) 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)(modal rsEigenvecto

)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

uKuT

T

222111 uMuKuMuK

Transformation of coordinates

Governing equation:

Modal equations:

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

0 xKxM

)()()(

)(

)()(

222

121

21

11

2

1

2211

tqu

utq

u

u

tx

tx

tqutqux

(**)

General approach for solution

We were calling “A” - Change to u to match Meirovitch

0)()((*)

0)()((*)

22222

12111

tqtqu

tqtquT

T

0)()()()( 22112211 tqutquKtqutquM (*)

Substitution:

Let

or

Known solutions

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

122

21

111 u

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)()()()(

2222

1211

2211tqtq

tqtqtqutqux

and

Response to harmonic forces

Model equation:

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

tieF

FtFxKxCxM

2

1)(

{F}not function of

timeAssume: tie

iX

iXiXx

)(

)()(

2

1

Substituting gives: FiXKCiM )(2

matrix impedance 2x2)( iZ

FiZiXiZiZ 11 )()()()(

2

1

1112

1222

21222112

1 1

F

F

zz

zz

zzzX

XX

Hence:

212 ,ji,kciωmωz ijijijij

:)(i of function are z All ij

Special case: Undamped system

Zero damping matrix [C]Entries of impedance matrix [Z]:

For our model problem (k1=k2=k and m1=m2=m), let F2 =0:

212

2222

2111

22

11111222

122

2222

111

21212

2221 ))((

)(;

))((

)(

kmkmk

FmkFkX

kmkmk

FkFmkX

Notes:1) Denominator originally (-)(-) =

(+). As it passes through 1, changes

sign.2) The plots give both amplitude and phase angle (either 0o or

180o)

Substituting for X1 and X2:

12122

222222

11111 )(;)(;)( kzmkzmkz

)()(;

)()(

)(22

221

221

222

221

221

2

1

m

FkX

m

FmkX

Multi-DOF Systems

1. Model EquationNotes on matrices Undamped free vibration: the eigenvalue problemNormalization of modal matrix [U]

2. General solution procedureInitial conditionsApplied harmonic force

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.

If restrained/tied down positive-definite. All positive.

Q xKxCxM

1) Vector mechanics (Newton or D’ Alembert)

2) Hamilton's principles3) Lagrange's equations

We derive using:

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

xKxU

xMxTT

T

21

21

:spring inenergy Strain

:energy Kinetic

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 natural modes contained in {u} (i.e. the eigenvectors).

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

j i

j i

δij

if

if

deltaKronecker

:where

0

1So far, we pick our eigenvectors to look like:

Instead, let us try to pickso that:

1

knewk uu

12 kT

knewkT

newk uMuuMu

Then: IUMU T UKU Tand

2

22

21

..0

....

..0

0.0

n

:where

Let the 1st entry be

1

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 force3. Arbitrary applied force

oo qq and

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 and on

)()()()( 2211 tutututq

Uq

nn

UFV model equation:

0

0

0

ηUKUηUMU

qKqMTT

n modal equations:

0

0

0

2

2222

1211

nnn

Need initial conditions on ,

not q.

Initial conditions - Modal analysis

Using displacement vectors:

ηUMUqMU

UqTT

As a result, initial conditions:

Since the solution of

oT

o

oT

o

qMUη

qMUη

)sin()(

)cos()()(

)sin()(

)cos()()( 11

1111

ttt

ttt

nn

nonnon

oo

And then solve

hence we can easily solve for

qMUη T or

02 is:

)sin()cos()(

)cos(

ttt

tC

oo

or

ηUq

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

Equation of motion:

unknownη

known U

ηUq

Q qKqM

Substitution of

leads to

NtQUηUKUηUMU oTTT )cos(

requency driving fω

tQQ o

)cos(

and

Hence,

.

)cos(

)cos(

222

22

221

11

etc

tQu

tQu

oT

oT

then

ηUq

Continuous Systems

1. The axial barDisplacement field Energy approachEquation of motion

2. ExamplesGeneral solution - Free vibrationInitial conditionsApplied forceMotion of the base

3. Ritz method – Free vibrationApproximate solution One-term Ritz approximationTwo-term Ritz approximation

The axial bar

Main objectives:1. Use Hamilton’s Principle to derive the equations of

motion.2. Use HP to construct variational methods of solution.

A = cross-sectional area = uniform

E = modulus of elasticity (MOE)u = axial displacement = mass per volume

Displacement field: u(x, y, z) = u(x, t)v(x, y, z) =

0w(x, y, z) =

0

Energy approach

L tt

t

t

L L

t

t

L

dxuuAdtux

uEAdxu

x

uEA

xuudxA

t

dtdxuxx

uEAuudxA

00 0

0

2

1

2

1

2

1

0

0

221

21

21

2

1

um

x

u

x

uE)εε(Eε σ xx

energy kinetic T

U energy strain energy potentialV

densityenergy strainUo

For the axial bar:

Hamilton’s principle:

dtux

uEAdxu

x

uEA

xuA

t

t

t

L L

2

1 0 0

0

2

1

)(0t

tdtVT

221 u(Adx)ρ

V odVU

2

2

x

uE

Axial bar - Equation of motion

2

22

2

2

x

u

t

u

Hamilton’s 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

22222

2 contant p-T

dtTd

X

dxXd

Fixed-free bar – General solution

0cos0

Lp

D ii or solution) (trivial Either

= wave speed

E

For any time dependent problem:

,5,3,1 2sin

2cos

2sin),(

iii L

tiB

L

tiA

L

xitxu

Free vibration:

1

sincos)sin()cos(),(i

iiiiiiii 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

0iC

2

5

2

3

2

ororLpi

),5,3,1(2

iL

ipi

NBC

Fixed-free bar – Free vibration

E

L

in 2

are the eigenfunctions

L

xi

2sin

For free vibration:

General solution:

Hence

)cos()(),( txAtxu n

are the frequencies (eigenvalues)

2

22

2

2

x

u

t

u

),5,3,1( i

Fixed-free bar – Initial conditions

or

,3,12

2

)1(

2 2cos

2sin

1)1(

)(8),(

i

io

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

iit L

xiB

L

i

t

u 0iB

22sin

2sin

2sin

2sin

,3,100

,3,1

LAdx

L

xi

L

xiAdx

L

xix

L

LL

L

xiAx

L

LL

ii

L

i

Lo

ii

o

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

2sin

)(2 2

)1(

2202

ii

LLdx

L

xix

L

LLA

io

Lo

i

Hence

Fixed-free bar – Applied force

or

txL

EA

Ftxu o

sinsinsec),(

Now, B.C’s:

)sin(

0),0(

tFx

uEA

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 nwe assume:

Substituting:

txA

xAtxu

sinsincos),( 21

)sin()sin(cos2 tFtL

LAEA

x

uEA oLx

Fixed-free bar – Motion of the base

)sin()sin(),0( 1 tUtAtu o

2

22

2

2

x

u

t

u

Using our approach from before:

Resonance at:

txLxUtxu o

sinsintancos),(

oUA 1

L

UA o tan2

Hence

txA

xAtxu

sinsincos),( 21

0sincossin 2 tLALU

x

u oLx

0

Lxx

uEA

From

B.C. at x = 0:

B.C. at x = L:

or,2

3,

2

L.,

2

3,

2etc

LL

Ritz method – Free vibration

Start with Hamilton’s 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) = a11(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.

One-term Ritz approximation 1

Ritz estimate is higher than the exactOnly get one frequencyIf we pick a different basis/trial/approximation function 1, we would get a different result.

)cos()cos()(

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

1

1111

txtxu

txatxatxuxx

:eapproximat Also

:Pick

dttdxEAxxAat

t

L)(cos)1)(1())((0 2

0

21

2

1

Substituting:

222

23

2 33

3

L

E

LLEA

LA

LLRITZ

732.13

LLEXACT

571.12

1010

22 adxEAadxxALL

Hence

aKaM 2:formmatrix in

L

xEXACT

2sin1

xRITZ 1

One-term Ritz approximation 2

Both mode shape and natural frequency are exact.But all other functions we pick will never give us a frequency lower than the exact.

L

xx

2sin)(1

:pick we ifWhat

dttdxL

x

LEA

L

xAa

dtdxuxx

uEAuuA

t

t

t

L

t

t

L

)(cos2

cos22

sin0

0

2

0

2

2

221

0

2

1

2

1

Substituting:

EXACTRITZ L

E

L

22Hence

)cos(2sin)cos()(

)cos(2sin)cos()(),(

1

111

tLxtxu

tLxatxatxu

:eapproximat Also

L

x

Ldx

d

2cos

21

Two-term Ritz approximation

221)( xaxaxX :Let

dtdxxaaEAxxaxaAt

t

L

2

1 0 212

212 )1()2()(0

where:

:1 xu eapproximat If

xaadx

dX21 2

:2xu eapproximat If dtdxxxaaEAxxaxaAt

t

L

2

1 0 2122

212 )2()2()(0

2

1

2221

1211

2

1

2221

12112

a

a

KK

KKE

a

a

MM

MM

5))((

4))((

3))((

5

0

2222

4

0

22112

3

011

LdxxxM

LdxxxMM

LdxxxM

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

Two-term Ritz approximation (cont.)

E

22 and

LaaLaL 4526.00)3785.01713.0( 2212

0

0

)534()4(

)4()3(

2

1

532422

42232

a

a

LLLL

LLLL

leads to

Solving characteristic polynomial (for det[ ]=0) yields 2 frequencies:

LL RITZRITZ 67.5)(5767.1)( 21 and

Substitution of:

LL EXACTEXACT 7123.4)(5708.1)( 21 and

Mode 1:

Let a1 = 1:

LxxxX 21 4526.0)(

:1 shape Mode

LaaLaL 38.10)10.5043.7( 22122

Mode 2:

LxxxX 22 38.1)(

:2 shape Mode