aero 7sem ae2403nol
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AE 2403 VIBRATIONS ANDELEMENTS OF AEROELASTICITY
BYMr. G.BALAJIDEPARTMENT OF AERONAUTICAL ENGINEERINGREC,CHENNAI
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Fundamentals of Linear Vibrations
1. Single Degree-of-Freedom Systems
2. Two Degree-of-Freedom Systems
3. Multi-DOF Systems
4. Continuous Systems
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Single Degree-of-Freedom Systems
1. A spring-mass systemGeneral solution for any simple oscillator
General approach
Examples
1. Equivalent springsSpring in series and in parallel
Examples
1. Energy MethodsStrain energy & kinetic energyWork-energy statement
Conservation of energy and example
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A spring-mass system
General solution for any simple oscillator:
Governing equation of motion:0=+kxxm
)sin()cos()( tv
txtx nn
ono
+=
2
n
o2
o
n
n
n
ooo
v
xamplitudeC;2
T
1
Hz)orc.(cycles/sefrequencyf
vibrationofperiodT;T
2)(rads/sec.frequencynatural
m
k
(sec.)timet;xvelocityinitialvnt;displacemeinitialx
+=====
====
====
where:
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Any simple oscillator
General approach:
1. Select coordinate system
2. Apply small displacement3. Draw FBD4. Apply Newtons Laws:
)(
)(
Idt
dM
xmdt
dF
=
=
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Simple oscillator Example 1
22
mlmdI
inertiaofmomentmassI
cg =+=
=
IK
IM
=
=
02 =+ Kml 2ml
Kn =
+
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Simple oscillator Example 2
=
=+=
l
a
m
k
mlmdII
n
cg
22
2)( mlaak
IM oo
=
=
022 =+ kaml
+
(unstable),l
a
As
m
k,
l
aWhenits:limNote
n
n
00
1
==
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Simple oscillator Example 4
Lma
GJ
L
JGK:stiffnessEquivalent
TL
JG
JG
TL
maI:tableFrom
n 2
2
2
2
2
=
=
=
=
=
IT
IMz
==
02
2
=+ L
GJma
+
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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 +=
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Equivalent springs Example 1
0=+ xKxm eq
0312
3
2
3
1
=
++ xL
EI
L
EIxm
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Equivalent springs Example 2
)a(
ml
Wlka
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
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Equivalent springs Example 3
02
2
=+
=
=
sinmglml
mlsinWl
IM oo
0=+ sin
l
g
l
g
l
g
n =
=+ 0
+We cannot definen
since we havesin term
If< < 1, sin:
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Work-Energy principles
Work done = Change in kinetic energy
Conservation of energy for conservative systems
E = total energy = T + U = constant
12
2
1
2
1
TTdTrdFT
T
r
r
==
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Energy methods Example
0
0
=+
=
xxmxkx
)E(dt
d
0=+ kxxm 22
2
2
2
1
2
1
2
1
2
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.
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Two Degree-of-Freedom Systems
1. Model problemMatrix form of governing equation
Special case: Undamped free vibrations
Examples
1. Transformation of coordinatesInertially & elastically coupled/uncoupled
General approach: Modal equations
Example
1. Response to harmonic forcesModel equationSpecial case: Undamped system
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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
2
222
22121
A
A
mkk
kmkk
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Undamped free vibrationsSpecial case when k
1
=k2
=k and m1
=m2
=m
Eigenvalues and frequencies:
periodlfundamenta
frequencylfundamenta
==
==
T
m
k.
2
61801
m
k
=
=618.2
3819.02
1
2
1
21
Two mode shapes (relative participation of each mass in the motion):
1
618.12 2
1
2 ==k
mk
A
A shapemode1st
1
618.02
1
2 =
=mk
k
A
Ashapemode2nd
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) =
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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
1
2
1 +
++
=
= tA
ACt
A
AC
tx
txx
Single-DOF:
For two-DOF:
)cos()( += tCtx n
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UFV Example 1
{ } )cos(618.0
0.1)cos(
618.1
0.12211
2
1tCtC
x
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:
1
1
21
2;0;
===
T
CC
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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
1
21618.0618.1
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
==
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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
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Transformation of coordinates
We have found that we can select coordinates so that:
1) Inertially coupled, elastically uncoupled, or
2) 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)(modalrsEigenvecto
)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
uKu
T
T
[ ]{ }
[ ]{ }
[ ]{ }
[ ]{ }
222111
uMuKuMuK ==
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Transformation of coordinates
Governing equation:
Modal equations:
Solve for these using initial conditions then substitute into (**).
[ ]{ } [ ]{ } 0=+ xKxM
{ } { } { }
)()()(
)(
)()(
2
22
12
1
21
11
2
1
2211
tqu
u
tqu
u
tx
tx
tqutqux
+
=
+= (**)
General approach for solution
We were calling A - Change to u to match Meirovitch
{ }{ }
=+
=+
0)()((*)
0)()((*)
2
2
222
1
2
111
tqtqu
tqtqu
T
T
[ ] { } { }( ) [ ] { } { }( ) 0)()()()( 22112211 =+++ tqutquKtqutquM (*)Substitution:
Let
or
Known solutions
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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
12
2
21
11
1u
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)()(
)()(2
2
22
1
2
112211tqtq
tqtqtqutqux
and
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Response to harmonic forces
Model equation:
[M], [C], and [K] are full but symmetric.
[ ]{ } [ ]{ } [ ]{ } { } tieFFtFxKxCxM
==++
2
1)(
{F}
not function of time
Assume:
{ } { }ti
eiX
iXiXx
== )()(
)(2
1
Substituting gives:[ ] [ ] [ ]( ){ } { }FiXKCiM =++ )(2
[ ] matriximpedance2x2=)( iZ
[ ] [ ] { } [ ] { }FiZiXiZiZ 11 )()()()( =
{ }
=
=2
1
1112
1222
2
1222112
1 1
F
F
zz
zz
zzzX
XX
Hence:
( )212 ,ji,kcimz ijijijij =++=
:)(ioffunctionarezAll ij
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Multi-DOF Systems
1. Model EquationNotes on matrices
Undamped free vibration: the eigenvalue problem
Normalization of modal matrix [U]
1. General solution procedureInitial conditions
Applied harmonic force
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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. Ifrestrained/tied down positive-definite. All positive.
[ ]{ } [ ]{ } [ ]{ } { }Q=++ xKxCxM
1) Vector mechanics (Newton or D Alembert)
2) Hamilton's principles
3) Lagrange's equations
We derive using:
Multi-DOF systems are so similar to two-DOF.
{ } [ ]{ }
{ } [ ] { }xKxU
xMxT
T
T
21
21
=
=
:springinenergyStrain
:energyKinetic
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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 naturalmodes contained in {u} (i.e. the eigenvectors).
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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
=
=
=
ji
ji
ij
if
if
deltaKronecker
:where
0
1So far, we pick our
eigenvectors to look like:
Instead, let us try to pick
so that:
{ } { }
==1
knewk uu
{ } [ ] { } { } [ ] { } 12 == kT
knewk
T
newkuMuuMu
Then: [ ] [ ] [ ] [ ]IUMUT = [ ] [ ] [ ] [ ]=UKU Tand
[ ]
=
2
2
2
2
1
..0
....
..0
0.0
n
:where
Let the 1st
entry be 1
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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 force
3. Arbitrary applied force
{ } { }oo qq and
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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 andon
{ } [ ]{ }{ } { } { } { } )()()()( 2211 tutututq
Uq
nn
+++==
UFV model equation: [ ]{ } [ ]{ } { }
[ ] [ ] [ ]{ } [ ] [ ] [ ]{ } { }{ } [ ]{ } { }00
0
=+ =+
=+
UKUUMU
qKqM
TT
n modal equations:
=+
=+
=+
0
0
0
2
2
2
22
1
2
11
nnn
Need initial conditions on ,
not q.
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Initial conditions - Modal analysis
Using displacement vectors: { } [ ] { }
[ ] [ ] { } [ ] [ ] [ ] { }UMUqMU
Uq
TT =
=
As a result, initial conditions:
Since the solution of
{ } [ ] [ ] { }
{ } [ ] [ ] { } == o
T
o
o
T
o
qMU
qMU
)sin()(
)cos()()(
)sin()(
)cos()()( 11
1111
ttt
ttt
n
n
nonnon
oo
+=
+=
And then solve
hence we can easily solve for
{ } [ ] [ ] { }qMU T=or
02 =+ is:
)sin()cos()(
)cos(
ttt
tC
oo
+=
or
{ } [ ]{ }Uq =
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Applied harmonic force
Driving force {Q} = {Qo}cos(t)
Equation of motion:
{ } [ ]{ }
[ ]{ } unknownknownU
Uq =
[ ]{ } [ ]{ } { }Q=+ qKqM
Substitution of
leads to
[ ] [ ] [ ]{ } [ ] [ ] [ ] { } [ ] { } { }NtQUUKUUMU oTTT ==+ )cos(
{ } { } requencydriving ftQQ o
==)cos(
and
Hence, { } { }
{ } { }
.
)cos(
)cos(
22
2
22
22
1
11
etc
tQu
tQu
o
T
oT
=
= then
{ } [ ]{ }Uq =
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Continuous Systems
1. The axial barDisplacement field
Energy approach
Equation of motion
1. ExamplesGeneral solution - Free vibration
Initial conditions
Applied force
Motion of the base
1. Ritz method Free vibrationApproximate solution
One-term Ritz approximation
Two-term Ritz approximation
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Axial bar - Equation of motion
2
22
2
2
x
u
t
u
=
Hamiltons 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
( ) ( ) 222222contant p-
T
dtTd
X
dxXd===
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Fixed-free bar General solution
0cos0 == LpD ii orsolution)(trivialEither
= wave speed
E
=
For any time dependent problem:
=
+
=,5,3,1 2
sin
2
cos
2
sin),(i
ii
L
tiB
L
tiA
L
xitxu
Free vibration:
[ ] ( ) ( )[ ]
= ++= 1 sincos)sin()cos(),( iiiiiiiii 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
0=iC
25
23
2
oror=Lpi
),5,3,1(2
== iL
ipi
NBC
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Fixed-free bar Initial conditions
or
=
=,3,1
22
)1(
2
2
cos
2
sin1
)1()(8
),(i
i
o
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
i
itLxiB
Li
tu 0=iB
22sin2sin2sin
2sin
,3,100
,3,1
L
AdxL
xi
L
xi
AdxL
xi
xL
LL
L
xiAx
L
LL
ii
L
i
Lo
i
io
=
=
=
=
=
),3,1()1()(8
2sin
)(22
)1(
2202==
=
iiLL
dxL
xix
L
LLA
i
oL
oi
Hence
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Fixed-free bar Applied force
or
( )txL
EA
Ftxu o
sinsinsec),(
=
Now, B.Cs:
=
=
= )sin(
0),0(
tFxuEA
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 n=we assume:
Substituting: ( )tx
Ax
Atxu
sinsincos),( 21
+
=
)sin()sin(cos2 tFtL
L
AEAx
u
EA oLx
=
=
=
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Ritz method Free vibration
Start with Hamiltons 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) = a1 1(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.
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One-term Ritz approximation 1
Ritz estimate is higher than the exact
Only get one frequency
If we pick a different basis/trial/approximation function 1, we would get a different result.
)cos()cos()(
)cos()cos()(),()(
1
1111
txtxu
txatxatxuxx
==
===
:eapproximatAlso
:Pick
[ ] dttdxEAxxAatt
L
)(cos)1)(1())((02
0
2
1
2
1
=Substituting:
2
22
23
2 33
3
L
E
LLEA
LA =
==
LLRITZ
732.1
3==
LLEXACT
571.12 ==
10
10
22adxEAadxxA
LL
=
Hence
[ ] { } [ ] { }( )aKaM =2:formmatrixin
=
L
xEXACT
2sin1
xRITZ =1
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7/30/2019 Aero 7sem Ae2403nol
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Two-term Ritz approximation
221)( xaxaxX +=:Let
[ ] dtdxxaaEAxxaxaAt
t
L
++= 21 0
21
2
21
2)1()2()(0
where:
:1 xu == eapproximatIf
xaadxdX 21 2+=
:2
xu =eapproximatIf [ ] dtdxxxaaEAxxaxaAt
t
L
++= 2
1 0 21
22
21
2 )2()2()(0
=
2
1
2221
1211
2
1
2221
12112
a
a
KK
KKE
a
a
MM
MM
==
===
==
5))((
4))((
3))((
5
0
22
22
4
0
2
2112
3
011
LdxxxM
LdxxxMM
L
dxxxM
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
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