Post on 27-Aug-2014
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
+
Simple oscillator – Example 2
la
mkω
mlmdII
n
cg22
2)( mlaak
IM oo
022 kaml
+
(unstable)ω,laAs
mkω,
laWhenits:limNote
n
n
00
1
Simple oscillator – Example 3
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
+
Simple oscillator – Example 4
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
Equivalent springs – Example 2
)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
Equivalent springs – Example 3
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”
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:
618.01
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
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
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
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