Vibrations Tutorial2

Elementary Tutorial

Prepared by Dr. An Tranin collaboration with Professor P. R. Heyliger

Department of Civil EngineeringColorado State University

Fort Collins, ColoradoJune 2003

Fundamentals of Linear Vibrations

Developed as part of the Research Experiences of Undergraduates Program on “Studies of Vibration and Sound” , sponsored by National Science Foundation

and Army Research Office (Award # EEC-0241979). This support is gratefully

acknowledged.

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()( tvtxtx nn

ono

2

n

o2o

nn

n

ooo

ωvxamplitudeC;2

ω T1Hz)or c.(cycles/sefrequencyf

vibrationofperiodT;T2)(rads/sec.frequencynaturalm

kω

(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:

)(

)(

IdtdM

xmdtdF

Simple oscillator – Example 1

22 mlmdI

inertiaofmomentmassI

cg

IKIM

02 Kml 2ml

Kωn

+


la

mkω

mlmdII

n

cg22

2)( mlaak

IM oo

022 kaml

+

(unstable)ω,laAs

mkω,

laWhenits:limNote

n

n

00

1


lb

mkω

mlmmml

mdII

mllA

AdxxdmrI

n

cgo

cg

l

33212

1212

2

222

2

23

22 2/

0

3)(

2mlbbk

IM oo

03

22

kbml

+


LmaGJ

LJGK:stiffnessEquivalent

TL

JGJGTL

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

eqkkkkkP

)( 21

21

PfPff

Pkk

eq

)(

11

21

2121

21 fffeq

Equivalent springs – Example 1

0 xKxm eq

031232

31

x

LEI

LEIxm


)a(ml

Wlkaω

nn

n

2

22

2mllWa)ak(

IM oo

022 )Wlka(ml

+

Consider: ka2 > Wl n

2 is positive - vibration is stableka2 = Wl statics - stays in stable equilibriumka2 < Wl unstable - collapses


02

2

sinmglml

mlsinWl

IM oo

0 sinlg

lgω

lg

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

PkU21

21 2

Tenergy kinetic ofincrement

done work ofIncrement

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

21

rrmT

21

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(dtd

0 kxxm 22

2

2

21

21

2121

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 coordinatesInertially & 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

0PP

xx

kkkkk

xx

ccccc

xx

mm

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

t

AA

xxAssumed general solutions:

Characteristic polynomial (for det[ ]=0):

021

212

2

2

1

214

mmkk

mk

mkk

21

21

212

2

2

1

21

2

2

1

212

21

21

421

mmkk

mk

mkk

mk

mkk

Eigenvalues (characteristic values):

Characteristic equation:

00

)()(

2

12

222

22

121

AA

mkkkmkk

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

Eigenvalues and frequencies:

period lfundamenta

frequency lfundamenta

ωπ T

mk.ω

2

61801

mk

618.2

3819.021

21

21

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

1618.12 2

1

2 kmk

AA shape mode 1st

1618.0

21

2

mk

kAAshape mode 2nd

The two eigenvectors are orthogonal:

618.11

)1(2

)1(1

AA

618.01

)2(2

)2(1

AAEigenvector (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

tAA

CtAA

Ctxtx

x

Single-DOF:

For two-DOF:

)cos()( tCtx n

UFV – Example 1

)cos(618.00.1

)cos(618.10.1

22112

1 tCtCxx

x

Given:No phase angle since initial velocity is 0:

618.10.1

0 oxx and

618.00.1

618.10.1

618.10.1

21 CCxo

From the initial displacement:

11

21

2;0;

T

CC

UFV – Example 2

)cos(618.01

)171.0()cos(618.11

)171.1( 21 ttx

Now both modes are involved:

Solve for C1 and C2:

21

0 oxx and

2

121 618.0618.1

11618.01

618.11

21

CC

CCxo

From the given initial displacement:

171.0171.1

21

1618.11618.0

618.1618.01

2

1 CC

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

tttxtttx

Transformation of coordinates

Introduce a new pair of coordinates that represents spring stretch:

00)(

00

2

1

22

221

2

1

2

1

xx

kkkkk

xx

mm

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:

00

00)(

2

1

2

1

2

1

22

221

zz

kk

zz

mmmmm

“inertially coupled” “elastically uncoupled”


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:

618.01

618.11

: vectors)(modal rsEigenvecto

)2(2

)2(1

2)1(2

)1(1

1 AA

uAA

u

0

0

12

21

uMu

uMuT

T

0

0

12

21

uKu

uKuT

T

222111 uMuKuMuK


Governing equation:

Modal equations:

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

0 xKxM

)()()()(

)()(

222

121

21

11

2

1

2211

tquu

tquu

txtx

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

618.11

21 ttx

2) Transformation:

618.01

;618.1618.11

;618.022

122

21

111 u

uuu

and

1) Solve eigenvalue problem:

)cos()0()()cos()0()(

171.0171.1

)0()0(

)0(618.0

1)0(

618.11

21

222

111

2

1

21

tqtqtqtq

qq

qq

and

So

As we had before.More general procedure: “Modal analysis” – do a bit later.

Model problem with:

00

21

oo xx and

0)()(0)()(

)()(2

222

1211

2211 tqtqtqtq

tqutqux

and

Response to harmonic forces

Model equation:

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

tieFF

tFxKxCxM

2

1)(

{F}not function of time

Assume: tie

iXiX

iXx

)()(

)(2

1

Substituting gives: FiXKCiM )(2

matrix impedance 2x2)( iZ

FiZiXiZiZ 11 )()()()(

2

1

1112

122221222112

1 1FF

zzzz

zzzXX

X

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 =0:2

122

2222

111

22

11111222

122

2222

111

21212

2221 ))((

)(;))((

)(kmkmk

FmkFkXkmkmk

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 X2:1212

222222

211111 )(;)(;)( kzmkzmkz

)()(;

)()()(

22

221

221

222

221

221

2

1

m

FkXm

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

ifif

deltaKronecker :where

01So far, we pick our

eigenvectors to look like:

Instead, let us try to pickso that:

1

knewk uu

12 kT

knewkTnewk uMuuMu

Then: IUMU T UKU Tand

2

22

21

..0......00.0

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:

00

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 UηUq

Q qKqM Substitution of

leads to NtQUηUKUηUMU o

TTT )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 = uniformE = 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

dxuuAdtuxuEAdxu

xuEA

xuudxA

t

dtdxuxx

uEAuudxA

00 0

0

2

1

2

1

2

1

0

0

221

21

21

21

um

xu

xuE)εε(Eε σ xx

energy kinetic TU energy strain energy potentialV

densityenergy strainUo

For the axial bar:

Hamilton’s principle:

dtuxuEAdxu

xuEA

xuA

tt

t

L L

2

1 0 00

2

1

)(0t

tdtVT

221 u(Adx)ρ

V odVU

2

2

xuE

Axial bar - Equation of motion

2

22

2

2

xu

tu

Hamilton’s principle leads to:

If area A = constant

0

xuEA

xuA

t

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

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

)sin()cos(02

tpBtpATTpT

xpDxpCXXpX

sincos02

Hence

1

sincos)sin()cos(),(i

iiiiiiii xpDxpCtpBtpAtxu

3

22

LM

LFE

:where

22222

2 contant p-T

dtTdX

dxXd

Fixed-free bar – General solution

0cos0

LpD i

i or solution) (trivial Either

= wave speed

E

For any time dependent problem:

,5,3,1 2sin

2cos

2sin),(

iii L

tiBL

tiALxitxu

Free vibration:

1

sincos)sin()cos(),(i

iiiiiiii xpDxpCtpBtpAtxu

EBC:

NBC:

0)0( u

00

LxLx xu

xuEA

General solution:

EBC

1

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

iiiii tpBtpACtu

1

0)sin()cos(cosi

iiiiiii

Lx tpBtpALppDxu

0iC

2

52

32

ororLpi

),5,3,1(2

iL

ipi

NBC

Fixed-free bar – Free vibration

EL

in 2

are the eigenfunctions

Lxi

2sin

For free vibration:

General solution:

Hence

)cos()(),( txAtxu n

are the frequencies (eigenvalues)

2

22

2

2

xu

tu

),5,3,1( i

Fixed-free bar – Initial conditions

or

,3,12

2)1(

2 2cos

2sin1)1()(8),(

i

io

Lti

Lxi

iLLtxu

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

0)0,( 0

to

tux

LLLxu and

Initial conditions:

Initial velocity:

Initial displacement:

0

2sin

2,3,10

iit L

xiBL

itu

0iB

22sin

2sin

2sin

2sin

,3,100

,3,1

LAdxLxi

LxiAdx

Lxix

LLL

LxiAx

LLL

ii

L

i

L o

ii

o

),3,1()1()(82

sin)(2 2)1(

2202

ii

LLdxLxix

LLLA

io

Lo

i

Hence

Fixed-free bar – Applied force

or

txLEA

Ftxu o

sinsinsec),(

Now, B.C’s:

)sin(

0),0(

tFxuEA

tu

oLx

From

B.C. at x = 0:

B.C. at x = L:

0),0( tu 01 A

L

EAFA o sec2

Hence

2

22

2

2

xu

tu

)sin()(),( txXtxu nwe assume:

Substituting:

txAxAtxu

sinsincos),( 21

)sin()sin(cos2 tFtLLAEA

xuEA oLx

Fixed-free bar – Motion of the base

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

2

22

2

2

xu

tu

Using our approach from before:

Resonance at:

txLxUtxu o

sinsintancos),(

oUA 1

LUA o tan2

Hence

txAxAtxu

sinsincos),( 21

0sincossin 2 tLALUxu o

Lx

0

LxxuEA

From

B.C. at x = 0:

B.C. at x = L:

or,2

3,2

L.,

23,

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 determined1(x) = known function of position

dtdxuxx

uEAuuAt

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

txtxutxatxatxuxx

:eapproximat Also:Pick

dttdxEAxxAat

t

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

0

21

2

1

Substituting:

222

23

2 333

L

EL

LEALA

LLRITZ 732.13

LLEXACT 571.1

2

1010

22 adxEAadxxALL

Hence

aKaM 2:formmatrix in

LxEXACT

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.

Lxx

2sin)(1

:pick we ifWhat

dttdxLx

LEA

LxAa

dtdxuxx

uEAuuAt

t

t

L

t

t

L

)(cos2

cos22

sin0

0

2

0

22

221

0

2

1

2

1

Substituting:

EXACTRITZ LE

L

22Hence

)cos(2sin)cos()(

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

1

111

tLxtxutLxatxatxu

:eapproximat Also

Lx

Ldxd

2cos

21

Two-term Ritz approximation2

21)( xaxaxX :Let

dtdxxaaEAxxaxaAt

t

L

2

1 0 212

212 )1()2()(0

where:

:1 xu eapproximat If

xaadxdX

21 2

:2xu eapproximat If dtdxxxaaEAxxaxaAt

t

L

2

1 0 2122

212 )2()2()(0

2

1

2221

1211

2

1

2221

12112

aa

KKKKE

aa

MMMM

5))((

4))((

3))((

5

0

2222

4

0

22112

3

011

LdxxxM

LdxxxMM

LdxxxM

L

L

L

In matrix form:

34)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

00

)534()4()4()3(

2

1532422

42232

aa

LLLLLLLL

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 2

2 38.1)(

:2 shape Mode

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