AAE 556
Aeroelasticity
Lecture 5 –
1) Compressibility;
2) Multi-DOF systems
Reading: Sections 2-13 to 2-15
Purdue Aeroelasticity
5-1
Homework for Monday?


Prob. 2.1
– Uncambered (symmetrical sections) MAC = 0
– Lift acts at aero center (AC) a distance e ahead to
the shear center
Problem 2.3 – wait to hand in next Friday
Purdue Aeroelasticity
5-2
Aeroelasticity matters
Reflections on the feedback process
Purdue Aeroelasticity
5-3
Topic 1 - Flow compressibility (Mach number) has an
effect on divergence because the lift-curve slope
depends on Mach number
Approximate the effect of Mach number by adding the
Prandtl-Glauert correction factor for sub-sonic flow
KT
qD 
SeC L
q Do
KT

SeC L
CL 
0
CL
0
1 M 2
KT 1  M 2
qD 
 qDo 1  M 2
SeCL
0
Plots as a curve vs. M
Purdue Aeroelasticity
5-4
But wait! – there’s more!
Mach number depends on altitude and airspeed
so two expressions must be reconciled
qatmosphere  qa
Physics
1
1 2 2
2
qa  V   a M
2
2
M=V/a
Speed of sound, “a," depends on temperature
and temperature depends on altitude
KT 1  M
qD 
 qDo 1  M 2
SeCL
2
0
Purdue Aeroelasticity
5-5
The divergence equation which contains
Mach number must be consistent with the
“physics” equation
qDo 1  M D
2
1 2
2
2
 a M D  q1M D
2
1. Choose an altitude
2. Find the speed of sound
3. Square both sides of the above
equation and solve for MD
1 2
q1  a
2
2


 qDo 
2
4

 1  M D  M D
 q1 
Purdue Aeroelasticity
5-6
Determining MD requires solving a
quadratic equation
2
MD
4
2
 q Do 
 qDo 
2
 M D  
  0
 
 q1 
 q1 
qatmosphere  qa
250
dynamic pressure
(lb/sq. ft)
1
1 2 2
2
qa  V  a M
2
2
qDo 1  M 2
sea level q
200
divergence q
150 20,000 ft.
100
40,000 ft.
50
0
0.00
0.25
0.50
0.75
Mach number
Purdue Aeroelasticity
1.00
5-7
If we want to increase the divergence Mach
number we must increase stiffness (and
weight) to move the math line upward
dynamic pressure
(lb/sq. ft)
400
350
sea level q
divergence q
300
250
200
20,000 ft.
150
100
40,000 ft.
50
0
0.00
0.25
0.50
0.75
1.00
Mach number
Purdue Aeroelasticity
5-8
Summary
 Lift
curve slope is one strong factor that
determines divergence dynamic pressure
– depends on Mach number
 Critical
Mach number solution for divergence
dynamic pressure must be added to the
solution process
Purdue Aeroelasticity
5-9
Topic 2 – Multi-degree-of-freedom
(MDOF) systems
 Develop
process for analyzing MDOF
systems
 Define theoretical stability conditions for
MDOF systems
 Reading - Multi-degree-of-freedom
systems – Section 2.14
Purdue Aeroelasticity
5-10
Here is a 2 DOF, segmented, aeroelastic finite wing model - two
discrete aerodynamic surfaces with flexible connections used to
represent a finite span wing (page 57)
A
2KT
3KT
fuselage
Torsional
springs
panel 2
panel 1
e
A
b/2
V
wing root
shear
centers
b/2
aero
centers
 + 2
 + 1
view A-A
wing tip
Torsional degrees of freedom
Purdue Aeroelasticity
5-11
Introduce “strip theory” aerodynamic modeling
to represent twist dependent airloads

Strip theory assumes that lift depends only on local
angle of attack of the strip of aero surface
– why is this an assumption?
L1  qSCL o  1 
L2  qSCL  o  2 
Purdue Aeroelasticity
q twist angles are
measured from a
common reference
5-12
The two twist angles are unknowns - we have to
construct two free body diagrams to develop
equations to find them
Structural restoring torques depend on the difference
between elastic twist angles
Wing root
Wing tip
Double arrow vectors
are torques
Internal shear forces are present,
but not drawn
Purdue Aeroelasticity
5-13
This is the eventual lift re-distribution
equation due to aeroelasticity – let’s see
how we find it
L1 
W 2  q 
 


L2  24  q   2 
Observation - Outer wing panel carries more of the total load
than the inner panel as q increases
Purdue Aeroelasticity
5-14
Torsional static equilibrium is
a special case of dynamic equilibrium


M

I

 1 1 1  0  L1e  3KT1  2 KT  2  1 


M

I

 2 2 2  0  L2e  2 KT  2  1 
Arrange these two simultaneous equations in matrix form
 5  2 1 
1 0 1 
1
KT 
   qSeCL 
   qSeCL  o  


 2 2   2 
0 1  2 
1
Purdue Aeroelasticity
5-15
Summary


The equilibrium equations are written in terms of
unknown displacements and known applied loads due to
initial angles of attack. These lead to matrix equations.
Matrix equation order, sign convention and ordering of
unknown displacements (torsion angles) is important
 5  2 1 
 1 0  1 
1
KT 
   qSeCL 
   qSeCL  o  


 2 2   2 
 0  1  2 
1
Purdue Aeroelasticity
5-16
Problem solution outline
 5  2 1 
 1 0  1 
1
KT 
   qSeCL 
   qSeCL  o  


 2 2   2 
 0  1  2 
1
Combine structural and aero stiffness matrices on the left hand side
1 
1
KT    qSCL  o 1

 2
The aeroelastic stiffness matrix is
 K T 
Invert matrix and solve for 1 and 2
 i   qSeCL  o KT  1
1
Purdue Aeroelasticity
5-17
The solution for the ’s requires inverting
the aeroelastic stiffness matrix
 K11
 KT   
 K21
 K11

 K 21
K12 

K22 
1
K12 
1  K 22  K12 
  

   K 21 K11 
K 22 
  K11 K 22  K12 K 21
Purdue Aeroelasticity
5-18
The aeroelastic stiffness matrix
determinant is a function of q
 The
determinant is   K11 K 22  K12 K 21
  q  7q  6
2
where
  1  q  6  q 
q
qSeCL
KT
When dynamic pressure increases, the determinant
 tends to zero – what happens to the system then?
Purdue Aeroelasticity
5-19
Plot the aeroelastic stiffness
determinant  against dynamic
pressure (parameter)
6
4
2
STABLE
0
UNSTABLE
-2
-4
-6
-8
0
2
4
6
8
Dynamic pressure parameter
The determinant of the stiffness matrix is always
positive until the air is turned on
Purdue Aeroelasticity
5-20
Solve for the twist angles created by
an input angle of attack o
1  q o
 

 2 
4  q 


7  q 
q
qSeCL
KT
  1  q  6  q 
Purdue Aeroelasticity
5-21
Twist deformation vs. dynamic
pressure parameter
A
2KT
3KT
panel 2
panel 1
shear
centers
e
b/2
V
b/2
 + 2
 + 1
view A-A
divergence
panel twist, i/o
A
aero
centers
Unstable q region
Outboard
panel (2)
determinant  is zero
Purdue Aeroelasticity
5-22
Panel lift computation on each segment
gives:

q 4  q  


1
L1 


qSC

 

L o 
q
7

q
 
L2 
1  

 
L flex
Lrigid
L1  L2
L1  L2


q  2S  CL  o qStotal CL  o
Note that
Stotal  2S
Purdue Aeroelasticity
5-23
More algebra - Flexible system lift
L flex
Lrigid
q
 1   5.5  q 

L flex
 q

 qStotal CL  o 1   5.5  q  
 

Set the wing lift equal to half the airplane weight
L flex
 q
 Weight
 qStotal CL  o 1   5.5  q   
2
 

Purdue Aeroelasticity
5-24
Lift re-distribution due to aeroelasticity
(originally presented on slide 13)
W
o 
2qStotal CL    q  5.5  q  

1  q 4  q  

L1 
 
  qSCL  o 
L2 
1  q 7  q  

 
L1 
W 2  q 
 


L2  24  q   2 
Observation - Outer wing panel carries more of the total load
than the inner panel as q increases
Purdue Aeroelasticity
5-25