Lectures 22, 23 Typical dynamic instability problems and test review

AAE 556
Aeroelasticity
Lectures 22, 23
Typical dynamic instability
problems and test review

Purdue Aeroelasticity

22-1


How to recognize a flutter problem in the making

Given: a 2 DOF system with a parameter Q that
creates loads on the system that are linear functions of
the displacements
 M1
 0


0   &&x1   K1
 +
M 2   &&x2   0

0   x1 
 0
=
Q
 

K 2   x2 

 p21


 x1 
 
 x1 

  =  ei ωt

 x 2
 
x 2 


p12   x1 
 
0   x2 

(

)

K
= 1
M1

ω 22 =

ω 4 − ω 2 ω 12 + ω 22 + ω12ω 22 = 0

Q is a real number
If p12 and p21
have the same sign
(both positive or
both negative) can
flutter occur?

ω12


 −ω 2 + ω 12



 Q
 − M p21 


2

(

(

)

∆ = −ω +ω
2

2

1

)(

K2
M2

 Q

 −
p12     
 M1
  x 1  = 0 
 x  0 
−ω 2 + ω 22  2   


(

)

)

Q=0

Q not zero

Q2
−ω +ω −
p12 p21 = 0

M 1M 2
2

2
2


The modified determinant
 M1
−ω 
 0
2

0   x1   K1
 + 

M 2   x2   0

(

∆ = − ω 2 + ω12
ω12 =
2
2
ω
+
ω
1
2
1

2
ωn =
±
2
2

)(

0   x1 
 0
 −Q

K 2   x2 
 p21

p12   x1  0 
r  =  

0   x2  0 

)

2
Q
− ω 2 + ω22 −
p12 p21 = 0
M 1M 2

K1
M1

ω 22

(ω

2
1

−ω

K2
=
M2

)

2 2
2

Q2
+4
p12 p21
M 1M 2


If flutter occurs two frequencies must merge
2
2
ω
+

ω
1
2
1
2
ωn =
±
2
2

(ω

2
1

−ω

)

2 2
2

Q2
+4
p12 p21
M 1M 2

FLUTTER – Increasing Q must cause the term under the radical
sign to become zero.
K1

2
ω1 =
ω 22 =

(ω

M1

2
1

K2
M2
2

Q =−

−ω

)

Q2
=− 4
p12 p21
M 1M 2

???
2 2
2

(

M 1 M 2 ω 12 − ω 22
4 p12 p21

)
2

p12 p21 = −

(

M 1 M 2 ω 12

− ω 22

)

2

4Q 2

For frequency merging flutter to occur, p12 and p21 must have opposite
signs.


If one of the frequencies can be driven
to zero then we have divergence

(

∆ = −ω +ω
2

2
1

ωn = 0

)(

)

Q2
−ω +ω −
p12 p21 = 0
M 1M 2
2

2
2

( )( )
ω 12

ω 22

( )( )
2

Q2
=
p12 p21
M 1M 2

KK
Q = 1 2
p12 p21
2

2

∆ = 0 = ω1 ω2

Q2
−
p12 p21
M 1M 2

M 1 M 2ω 12ω 22
Q =
p12 p21
2

p12 p21 =

M 1 M 2ω12ω 22
Q2

Divergence requires that the cross-coupling terms have the same sign

Aero/structural interaction model
TYPICAL SECTION
What did we learn?

L = qSCL α (αo + θ )
V

lift
e

θ

GJ
KT ∝
span

torsion spring
KT


qScCMAC 
 αo +

K
T

L = qSCLα 
 1 − qSeC Lα 



K


T


Divergence-examination vs.
perturbation
L=

1−

qSCLα
qSeCLα

αo +
KT
Kh
 0

1−

qSCLα
qSeC Lα

KT

 qScC 

MAC 


K


T

0 h   − L 
 =


KT θ  MSC 

∞
1
= 1 + q + q 2 + q 3 +...= 1 + ∑ q n
1− q
n=1


Perturbations & Euler’s Test
V

KT (∆θ ) > (∆L)e

lift
e

θ

torsion spring
KT

...result - stable - returns -no static equilibrium in perturbed state

KT ( ∆θ ) < ( ∆L)e
...result - unstable -no static equilibrium - motion
away from equilibrium state

KT ( ∆θ ) = ( ∆L)e
...result - neutrally stable - system stays - new static equilibrium point


Stability equation is original
equilibrium equation with R.H.S.=0.

∆θ ≠ 0

V
θ

lift
e

torsion spring
KT

(KT − qSeC Lα )= KT = 0
The stability equation is an equilibrium equation that represents an

equilibrium state with no "external loads" –
Only loads that are deformation dependent are included
The neutrally stable state is called self-equilibrating


Multi-degree of freedom systems
A

2KT

3KT

panel 2

panel 1
e

b/2

V
5
KT 
 −2

A

b/2

αο + θ2
αο + θ1

shear
centers

aero
centers

From linear algebra,
we know that there is
a solution to the
homogeneous
equation only if the
determinant of the
aeroelastic stiffness
matrix is zero

view A-A

−2 θ1 
 −1 0  θ1 
1
  + qSeC Lα 
  = qSeC Lα αo  

2 θ2 
0
−1

 θ 2 
1

MDOF stability
Mode shapes? Eigenvectors and eigenvalues.

[KT ]{∆θ i }= {0}
KT = 0

Kij − qAij = 0

System is stable if the aeroelastic stiffness matrix
determinant is positive. Then the system can absorb
energy in a static deformation mode. If the stability
determinant is negative then the static system, when
perturbed, cannot absorb all of the energy due to work done
by aeroelastic forces and must become dynamic.


Three different definitions of roll
effectiveness
• Generation of lift – unusual but the only game in town
for the typical section
• Generation of rolling moment –
• contrived for the typical section – reduces to lift
generation
• Multi-dof systems – this is the way to do it
• Generation of steady-state rolling rate or velocity-this
is the information we really want for airplane
performance
• Reversal speed is the same no materr which way you

do it.


Control effectiveness

q  c  CM δ 
 1+



qD e CLδ 

L = qSCL δ δ o
=0
q
1−
qD

q  c  CM δ
1+
=0


qD e CLδ

KT  CL δ 

qR = −

ScCLα  CMδ 

Lift
α0+ θ

V

MAC t orsion spring KT
shear center
e

δ0

reversal is not an
instability - large
input produces
small output
opposite to
divergence
phenomenon


Steady-state rolling motion
 qScCMδ
v 

L = 0 = qSCLα
δo −
+ qSC Lα δ o
 KT
V

Lift
α0+ θ

V

MAC t orsion spring KT
shear center
e

δ0


Swept wings
α structural= θ − φ tan Λ

qn = qcos2 Λ

K1
d

f

K2

αo

Λ

V

C
V cosΛ

A

b

c

B

A

  Kφ
 0


 −tb
0 
− Q 2
Kθ 
 −te

b  φ
b
  Qα o  
2   θ  = cosΛ  2 
e  
e

B

C


Divergence

bt


∆ = Kθ Kφ + Q Kθ − Kφe


2
Kθ

 e   c   Kφ 
tan Λ crit = 2   
c b  Kθ 

2.0

nondimensional divergence
dynamic pressure

Seao
qD =
  b  K  tan Λ 
2
cos Λ  1−   θ 

 e K  2 

φ


nondimensional divergence dynamic
pressure vs. wing sweep angle
1.5

sweep back

sweep f orward

1.0

5.72 degrees

0 .5
0 .0
-0 .5
-1.0

b/c=6
e/c=0.10
Kb/Kt=3

-1.5
-2.0
-90 -75 -60 -45 -30 -15

0

15

30

sweep angle
(degrees)

45

60

75

90


Lift effectiveness
lift eff ect iveness
vs.
dynamic pressure

2.0

lif t ef f ect iveness

unswept
wing
1.5

unswept wing
divergence

1.0

15 degrees
sweep

0 .5

30 degrees
sweep
0 .0
0

50

10 0

150

20 0

250

dynamic pressure (psf )

30 0

350


Flexural axis

x

ref e
r enc
e ax
is

Λ

θ E = θ − φ tanΛ

β
y

Flexural axis - locus of points where a concentrated force creates no
stream-wise twist (or chordwise aeroelastic angle of attack)

θE = 0

The closer we align the
airloads with the flexural
axis, the smaller will be
aeroelastic effects.

How to recognize a flutter problem in the making

Given: a 2 DOF system with a parameter Q that
creates loads on the system that are linear functions of
the displacements
 M1
 0


0   &&x1   K1
 +
M 2   &&x2   0

0   x1 
 0
=
Q
 

K 2   x2 
 p21


 x1 
 
 x1 

  =  ei ωt

 x 2

 
x 2 


p12   x1 
 
0   x2 

(

)

K
= 1
M1

ω 22 =

ω 4 − ω 2 ω 12 + ω 22 + ω12ω 22 = 0

Q is a real number
If p12 and p21
have the same sign
(both positive or
both negative) can
flutter occur?

ω12



 −ω 2 + ω 12



 Q
 − M p21 


2

(

(

)

∆ = −ω +ω
2

2
1

)(

K2
M2

 Q

 −

p12     
 M1
  x 1  = 0 
 x  0 
−ω 2 + ω 22  2   


(

)

)

Q=0

Q not zero

Q2
−ω +ω −
p12 p21 = 0
M 1M 2
2

2
2


If flutter occurs two frequencies must merge
2
2

ω
+
ω
1
2
1
2
ωn =
±
2
2

(ω

2
1

−ω

)

2 2
2

Q2
+4
p12 p21
M 1M 2

FLUTTER – Increasing Q causes the term under the radical sign to

be zero.
K1
2
ω1 =
ω 22 =

(

M1

ω 12

K2
M2
2

Q =−

− ω 22

)
2

Q2
= −4
p12 p 21
M1 M 2

(

M 1 M 2 ω 12 − ω 22
4 p12 p21

)
2

p12 p21 = −

(

M 1 M 2 ω 12

− ω 22

)

2

4Q 2

For frequency merging flutter to occur, p12 and p21 must have opposite
signs.


If one of the frequencies is driven to
zero then we have divergence

ωn = 0

 M1

 0


0   &&x1   K1
 +

M 2   &&x2   0

0   x1 
 0
=
Q
 

K 2   x2 
 p21

( )( )
2

2

∆ = 0 = ω1 ω2

( )( )
ω 12

ω 22

Q2

=
p12 p21
M 1M 2

KK
Q = 1 2
p12 p21
2

p12   x1 
 
0   x2 

Q2
−
p12 p21
M 1M 2

M 1 M 2ω 12ω 22
Q =
p12 p21
2

p12 p21 =

M 1 M 2ω12ω 22
Q2

Divergence requires that the cross-coupling terms are of the same sign

Fuel line flutter
A hollow, uniform-thickness, flexible tube has a mass per unit length of m
slugs/ft. and carries liquid fuel with density ρ to a rocket engine. The fuel flow
rate is U ft/sec. through a pipe cross-section of A. The tube is straight and has
supports a distance L apart, the tube bending displacement is approximated to
be

 πy 
 2π y 
w (y,t ) = φ1 sin  + φ 2 sin 
L
L 

φ1
φ2

Unknown amplitudes of vibrational motion

The free vibration frequencies when the
fluid is not flowing are:

4
 π  EI
ω = 
L mo
2
1

 2π EI

ω 22 =  
L mo
4

mo = m + ρA


Fluid flow creates system coupling, but through
the velocity, not the displacement

8  ρ AU
&&
φ1 − 
3  mo L

2
 &  2 ρ AU 2  π  
φ1 = 0
÷φ2 +  ω1 −
 ÷÷
÷
mo  L  



8  ρ AU
&&
φ2 − 
3  mo L

2
 &  2 ρ AU 2  2π  
φ2 = 0
÷φ1 +  ω2 −

÷÷
mo  L  ÷




1. Find the divergence speed
2. Estimate the flow speed that flutter occurs, if it occurs


Divergence is found by computing the
determinant of the aeroelastic stiffness matrix

∆ aesm

8  ρ AU
φ&&1 − 
3  mo L

2
 &  2 ρ AU 2  π  
φ1 = 0
÷φ2 +  ω1 −
 ÷÷
÷

m
L
  
o



8  ρ AU
φ&&2 − 
3  mo L

2
 &  2 ρ AU 2  2π  
φ2 = 0
÷φ1 +  ω2 −

÷÷
÷
m
L

 
o



 2 ρAU 2  π  2  2 ρAU 2  2π  2 
 ω2 −
=0
=  ω1 −








m
L
m
L




o
o




2


L
2
 
U1 =
ρA  π 

mo ω12

2


L
2


U2 =

ρA  2π 

mo ω 22

 π  EI
U div =  
 L  ρA


Assume that coupling leads to flutter and
find an estimate of the merging point

 2 ρ AU 2  π  2 
φ&&1 +  ω1 −
φ1 = 0
 ÷÷

÷
mo  L  


 2 ρ AU 2  2π 2 
φ&&2 +  ω2 −
φ2 = 0

÷÷

÷
mo  L  

Harmonic motion?

2
2   2 
2

ρ
AU
π
ρ
AU
2
π
 2
 2

2
2
 −ω + ω 1 − m  L   −ω + ω 2 − m  L   = 0


o   
o 

