AAE 556
Aeroelasticity
Lecture 4
Reading: notes assignment from
Lecture 3
[email protected]
Armstrong 3329
765-494-5975
Purdue Aeroelasticity
4-1
Summary to-date
 Development
of simple models of wing aeroelastic
behavior with pitch (torsion) only and pitch and
plunge (bending)
 Models show that torsional deformation creates
additional lift, deflection (and stress).
 Models identify an aeroelastic parameter that
defines a dynamic pressure at which lift and
torsional deflection approach infinity
– Models are linear so this will never really happen
– This special dynamic pressure is called the “divergence
dynamic pressure.”
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4-2
Today and next week’s
agenda
 Define
and discuss static stability
– Concept of perturbations
– Distinguish stability from response
 Learn
how to do a stability analysis
 Find the divergence dynamic pressure
using a “perturbation” analysis
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4-3
The perturbed structure

Static stability analysis considers
what happens to a flexible system
that is in static equilibrium and is
then disturbed.
– If the system tends to come back to its
original, undisturbed position, it is stable
- if not - it is unstable.

We need to apply these above words
to equations so that we can put the
aeroelastic system to a mathematical
test
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4-4
Stability investigation
 Given
a system that we know is in static
equilibrium (forces and moments sum to zero)
 Add a disturbance to perturb the system to move it
to a different, nearby position (that may or may not
be in static equilibrium)
 Is this new, nearby state also a static equilibrium
point?
 Write static equilibrium equations and see if forces
and moments balance
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4-5
Perturbed airfoil
 In
flight this airfoil is in static equilibrium at the
fixed angle  but what happens if we disturb
(perturb) it?
L  qSCL   
lift + perturbation lift

o
MS=KT(+)
torsion spring
KT
V
 There
are three possibilities
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4-6
Perturbation possibilities
 KT()>(L)e
– statically stable because it tends to return
– no static equilibrium in the perturbed state
 KT()<(L)e
– statically unstable
– motion away from original position
 KT()=(L)e
– system stays perturbed but static
– we have found new static equilibrium point
– Euler test has found neutral stability
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4-7
Example
 Perturb
the airfoil when it is in static
equilibrium
 To be neutrally stable in this new
perturbed position this equation must be
an true
K
T
 
 qSeCL   KT  qSeCL

Purdue Aeroelasticity

     qSeC
L
4-8
o
Static stability investigation is
“stiffness based”
KT      L  e  0
Neutral stability
means this relationship must
be zero (2 states)
 KT  qSeCL     0
so...
KT  0
Not zero
condition at neutral stability
static equilibrium displacement () is not unique
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4-9
Observations
 The
equation for neutral stability is simply the
usual static equilibrium equation with right-handside (the input angle o) set to zero.
 The neutral stability equation describes a special
case
– only deformation dependent external (aero) and
internal (structural) loads are present
– these loads are “self-equilibrating” without any other
action being taken
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4-10
Stability investigation

Take a system that we know is in static equilibrium (forces and
moments sum to zero)
Kh
 0


 Kh
 0


Perturb the system to move it to a different, nearby position (that may
or may not be in static equilibrium)
0   h  h 

  qSCL
KT     
0
0

1  h  h  (?)
1

  qSCL  o    qScCMAC
e     
e
Is this new, nearby state also a static equilibrium point?
 Kh
 0


 1 h 
 1
0
   qSCL  o    qScCMAC  
e   
e
1
0  h 
0
   qSCL 
KT   
0
0   h 

  qSCL

KT   
0
0

1  h  (?) 0

 

e    0
Static equilibrium equations for stability are those for a selfequilibrating system
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11
0
 
1 
More observations
 At
neutral stability the deformation is not
unique ( is not zero but can be plus or
minus)
 At neutral static stability the system has
many choices (equilibrium states) near
its original equilibrium state.
– airfoil position is uncontrollable - it has no
displacement preference when a load is
applied.
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4-12
The 1 DOF divergence condition
 Neutral
stability
 KT  qSeCL     0
KT  qD SeCL
 or
KT
qD 
SeCL
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4-13
System stiffness, not strength, is
important
M shear
center
M structure  KT
Structural
resistance
Aero overturning
M aero  qSeCL o   
Slope depends on qSCL
Equilibrium point
twist 
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4-14
Stable perturbed system
M shear
M structure  KT

center
M aero  qSeCL o   
Equilibrium point
twist 
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4-15
Perturbed system-neutral stability
M aero  qSeCL o   
Lines are parallel
M shear
center
M structure  KT
Equilibrium point at infinity
twist 
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4-16
Unstable system
M aero  qSeCL o   
M shear
center
M structure  KT
Equilibrium point?
twist 
Purdue Aeroelasticity
4-17
Aeroelastic stiffness
K
T

 qSeCL   Le  M AC  M SC
M SC  M SC

 K effective  K e


M sc
Keffective  KT  qSeCL

twist 
Aeroelastic stiffness decreases as q increases
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4-18
Aeroelastic divergence
 Look
at the single degree of freedom
typical section and the expression for
twist angle with the initial load
 neglect wing camber
previous result
"twist amplification"

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qSeCL  o

KT 1  q 
4-19
Twist amplification
qo

1 q

1
2
3
n
 1  q  q  q  ...  1   q
1 q
n 1
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4-20
Example corrections

qSeCL  o

KT
1  q  q
2

 ...
1
relative sizes
of terms
the sum of the
terms is 2
q bar = 0.5
0.75
0.5
0.25
0
1
2
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3
4
5
6
7
4-21
Aeroelastic feedback
process

qSeCL  o
KT
o is the twist angle
with no aero
load/structural
response "feedback"
1  q  q
o 
2

 ...
qSeCL  o
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
KT
4-22
More terms
1  q  o 
qSeCL  o

KT
the response to angle of attack o
instead of o
…and, the third term
 2  q  o  q 1
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2
4-23
Conclusion
Each term in the series represents a feedback
"correction" to the twist created by load
interaction

n
   o 1   q 
 n 1 


   0   n
n 1
Series convergence
q 1
Series divergence
q 1
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4-24
Summary
 Divergence
condition is a neutral
stability condition
 Divergence condition can be found
using the original equilibrium conditions
 Stability does not depend on the value
of the applied loads
Purdue Aeroelasticity
4-25