AAE 556
Aeroelasticity
The V-g method
g
k decreasing
V/b
mode 1
mode 2
flutter
point
1
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Airfoil dynamic motion
Ma
e
P=-L
x
(t)
V
aero K 
center T
Kh h
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This is what we’ll get when we use the V-g
method to calculate frequency vs. airspeed and
include Theodorsen aero terms
1.6
1.4
Frequency Ratio ( /  )

1.2
1
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5
3
Velocity (V/  b)
3.5
4
4.5
5
3
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When we do the V-g method here is
damping vs. airspeed
1
0.8
0.6
0.4
flutter
0.2
g
0
-0.2
divergence
-0.4
-0.6
-0.8
-1
0
0.5
1
1.5
2
2.5
3
Velocity (V/  b)
3.5
4
4.5
5
4
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To create harmonic motion at all airspeeds we
need an energy source or sink at all airspeeds
except at flutter
 Input
energy when the aero damping
takes energy out (pre-flutter)
 Take away energy when the aero forces
put energy in (post-flutter)
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2D airfoil free vibration with everything
but the kitchen sink


h
Mh  Mx  K h  g h  g   h   P   Leit
 

  M  K
2
h

1  i  g h  g  h   Mx  P
2



I  Mx h  K  g  g      M a  M a eit



  I
2


 K 1  i  g  g      Mx h  M a
2
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We will still get matrix equations
that look like this
A B h / b  0 
  



D E 
   0 
…but have structural damping that
requires that …
A(k, , g)E(k, , g)  B(k)D(k)  0
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Here is how the equations are slightly
different
A B h / b  0 

  


D E    0 
Each term contains inertial,
structural stiffness, structural
damping and aero
information
A  {1 ( /  )[1  i(gh  g)]} Lh
2
h
2
B   x   L - Lh (1 / 2  a)
D = x   Mh  Lh(1/ 2  a)
2
E   r {1 (  /  )[1  i(g  g)]}
2
2
Mh (1 / 2  a)  M  L (1 / 2  a)  Lh (1 / 2  a) 2
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One approximation and one definition allows
us to construct an eigenvalue problem
  h 2

A   1  
1  i  g h  g    Lh



 

We change the eigenvalue from a pure frequency term to a
frequency plus fake damping term. So what?
  h 2   2

A   1  
1  ig   Lh




    



  ( /  )(1  ig)    i
2
2
2
2
R
2
I
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The three other terms can also be
modified
A B h / b  0 

  


D E    0 
Each term contains inertial,
structural stiffness, structural
damping and aero
information
B   x   L - Lh (1 / 2  a)
D =  x   Mh  Lh(1/ 2  a)
2



2



E   r 1  
 1  ig  


 

1

1

1

 M h   a   M   L   a   Lh   a 
2

2

2

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2
10
We input k and compute 
  
  
     ig     2R  i 2I
 
 
2
2
2
2
1
2
R 1
2
I 1
  ( )  i( )
2
2
2
R 2
2
I 2
  ( )  i( )
1   / (R )1
2
I 1
2
R 1
g1  ( ) / ( )
2   / (R ) 2
The value of g represents the amount of
damping that would be required to keep
the system oscillating harmonically. It
should be negative for a stable system
2
2
g2  (I )2 / (R )2
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Now compute airspeed using
the definition of k
V1  b1 / k
 1    / ( R )1
Remember that we always input k so the same
value of k is used in both cases. One k, two
airspeeds and damping values
V2  b2 / k
 2    / ( R ) 2
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Typical V-g Flutter Stability Curve
g '  gh  g  g  g
gh  g
k decreasing
g
V/b
mode 1
flutter
point
mode 2
  ( /  )(1  ig)
2
2
2
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Now compute the eigenvectors
V1  b1 / k
h
2
2
(b / h)1  D / E(1 ) ;
 1 (   1 )
b
V2  b2 / k
(h / b  )2  B / A(2 ) ;   1 (  2 )
2
2
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Example
Two-dimensional airfoil
mass ratio,  = 20
quasi-static flutter speed VF = 160 ft/sec
g  gh  0.03
b  3.0 ft
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Example
k  0.32
1 / k  3.1250
h 10 rad / sec
  25 rad / sec.
L  13.4078  i3.7732
Lh  0.10371 i40973
M  0.37500  i3.1250
Mh  0.50000
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The determinant
k  0.32
2
A  19.896 i4.0973  3.2
B  11.3767  i2.5440
D  2.5311 i1.22919
2
E  9.2380  i2.3618  5.0
4
2
AE  BD  16()  (129.043  i28.044) 199.794  i64. 418  0
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Final results for this k value – two
g’s and V’s
b  3.0 ft
2
  4.0326  i0.87638  3.0067  i3.0420
2
  4.0326  i0.87638  (1.9084  i0.79702)
2
1  5.9410  i1.67340
1 10.257 rad / sec (h 10 rad / sec)
2
2  2.1242  i0.07936
V1  96.157 ft / sec
g1  g  g  0.2817
2 17.153 rad / sec (  25 rad / sec) V2  160.810 ft / sec
g2  g  g  0.0374
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Final results
Flutter
g = 0.03
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