AAE556 v g flutter lecture 32 The Vg method

AAE 556
Aeroelasticity
The V-g method
g

k decreasing
V/bωθ

mode 1

mode 2

flutter
point


Airfoil dynamic motion

Ma

e
P=-L

θ(t)
V

xθ

aero K θ
center T
Kh h

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 (ω / ω )
θ

1.2
1
0.8
0.6
0.4
0.2
0

0

0.5

1

1.5

2
2.5
3

Velocity (V/ ω b)
θ

3.5

4

4.5

5


When we do the V-g method here is
damping vs. airspeed
1
0.8
0.6
0.4

flutter

0.2

g

0

divergence

-0.2

-0.4
-0.6
-0.8
-1
0

0.5

1

1.5

2
2.5
3
Velocity (V/ ω b)
θ

3.5

4

4.5

5


To create harmonic motion at all airspeeds we
need an energy source or sink at all airspeeds
except at flutter

i
i

Input energy when the aero damping
takes energy out (pre-flutter)
Take away energy when the aero forces
put energy in (post-flutter)


2D airfoil free vibration with everything
but the kitchen sink
&


h
&&
&&
Mh + Mxθ θ + K h ( g h + g ) + h  = P = − Leiωt
ω



( −ω M + K
2

h

)

1 + i ( g h + g )  h − ω Mxθ θ = P

2

&


θ
&&
&&
Iθ θ + Mxθ h + Kθ ( gθ + g ) + θ  = M a = M a eiωt
ω



( −ω I
2

θ

)

+ Kθ 1 + i ( gθ + g )  θ − ω Mxθ h = M a
2


We will get matrix equations that
look like this

 A B  h / b  0 
= 



 D E   θ  0 

m
µ=
2
πρ b

…but have structural damping that
requires that …

A(k, ω , g)E(k, ω , g) − B(k)D(k) = 0


The EOM’s are slightly different from those
before (we also multiplied the previous
equations by µ)
B  h / b  0  Each term contains inertial,
structural stiffness, structural
=




E   θ  0  damping and aero
information

A
D

A = µ{1− (ω / ω )[1 + i(gh + g)]} + Lh
2
h

2

B = µ x θ + Lα =- Lh (1 / 2 + a)

2
θ

1

D =µxθ +M h −Lh  +a ÷
2


E = µ r {1 − ( ω / ω )[1 + i(gθ + g)]}
2
θ

2

− Mh (1 / 2 + a) + Mα − Lα (1 / 2 + a) + Lh (1 / 2 + a) 2


Look at the “A” coefficient and identify the
eigenvalue – artificial damping is added to keep
the system oscillating harmonically

  ωh  2

A = µ 1 − 
÷ 1 + i ( g h + g fake )  + Lh
ω

 


(

)

We change the eigenvalue from a pure frequency term to a
frequency plus fake damping term. So what?

  ωh 2  ωθ  2

A = µ 1 − 
1 + ig fakier )  + Lh
(
÷

÷
ωθ   ω 




Ω = (ω / ω )(1 + ig) = Ω + iΩ

2

2
θ

2

2
R

2
I


The three other terms are also
modified

 A B   h / b  0 
D E   θ  = 0 

Each term contains inertial,
structural stiffness, structural
damping and aero
information

B = µ x θ + Lα =- Lh (1 / 2 + a)
D =+ µ x θ + Mh − L h (1/ 2 + a)
2



ω
2


θ
E = µ rθ 1 − 
÷ ( 1 + ig ) 
ω

 

2

1

1

1

− M h  + a ÷+ M α − Lα  + a ÷+ Lh  + a ÷
2

2

2



To solve the problem we input k and
compute the two values of Ω2

2

2

 ωθ 
 ωθ 
Ω =  ÷ + ig  ÷ = Ω 2R + iΩ 2I
ω 
ω 
2

Ω = (Ω ) + i(Ω )
2
1

2
R 1

2
I 1

Ω = (Ω ) + i(Ω )
2
2

2
R 2

2
I 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

ω 1 = ω θ / (Ω R )1
g1 = (Ω ) / (Ω )
2
I 1

2
R 1

ω 2 = ω θ / (Ω R ) 2
g2 = (Ω 2I )2 / (Ω 2R )2


Now compute airspeeds
using the definition of k

V1 = bω 1 / k

ω 1 = ω θ / (Ω 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

V 2 = bω 2 / k

ω 2 = ω θ / (Ω R ) 2


Typical V-g Flutter Stability Curve
g ' = g h + g = gθ + g

gh ≈ gθ

k decreasing

g

V/bωθ

mode 1

flutter
point

mode 2

Ω = (ω / ω )(1 + ig′ )
2

2
θ

2

Now compute the eigenvectors

V1 = bω 1 / k
h
2
2
(bθ / h)1 = −D / E(Ω1 ) ;
= 1 (Ω = Ω 1 )
b

V 2 = bω 2 / k
(h / b θ )2 = − B / A(Ω 2 ) ; θ = 1

(Ω 2 = Ω 22 )


Example
Two-dimensional airfoil
mass ratio, µ = 20
quasi-static flutter speed VF = 160 ft/sec

gθ = g h = 0.03
b = 3.0 ft


Example
k = 0.32
1 / k = 3.1250

ω h = 10 rad / sec

ωθ = 25 rad / sec.

Lα = −13.4078− i3.7732

Lh = −0.10371− i40973
Mα = 0.37500 − i3.1250

Mh = 0.50000


The determinant
k = 0.32
A = 19.896 − i4.0973 − 3.2Ω

2

B = −11.3767 − i2.5440
D = 2.5311+ i1.22919
2
E = 9.2380 − i2.3618 − 5.0Ω
A E − BD = 16(Ω) 4 + (−129.043+ i28.044)Ω2 + 199.794 − i64. 418= 0


Final results for this k value – two
g’s and V’s
b = 3.0 ft

Ω = 4.0326 − i0.87638± 3.0067 − i3.0420

2

Ω − 4.0326 − i0.87638± (1.9084− i0.79702)
2

Ω12 = 5.9410 − i1.67340
ω 1 = 10.257 rad / sec (ω h = 10 rad / sec)
Ω 22 = 2.1242− i0.07936

V1 = 96.157 ft / sec
g1 = g + gθ = −0.2817

ω 2 = 17.153 rad / sec (ω θ = 25 rad / sec) V 2 = 160.810 ft / sec
g 2 = g + gθ = −0.0374


Final results
Flutter
g = 0.03