Elementary Tutorial Prepared by Dr. An Tran in collaboration with Professor P. R. Heyliger Department of Civil Engineering Colorado State University Fort Collins, Colorado June 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
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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
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.
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.