Post on 04-Jan-2016
description
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
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