AAE 556
Aeroelasticity
Lecture 8
Multi-degree-of-freedom systems
with feedback control
Purdue Aeroelasticity
8-1
Goals
 Demonstrate
how to increase
divergence q of MDOF systems by
adding a feedback “control loop”
 Define stability conditions for controlled
MDOF systems
 Reading - Multi-degree-of-freedom
systems – Section 2.21
Purdue Aeroelasticity
8-2
Add an aileron surface to outboard panel
2 of previous example in Section 2.14
V
3 KT
aero
centers
2 KT
e
panel 2
panel 1
aileron
b/2
shear
centers
b/2
added aileron
Purdue Aeroelasticity
8-3
Static equilibrium equations change
aileron deflection adds lift and pitching (torsional) moment
added moment
due to aileron
panel 1
panel 2
added lift
due to
aileron
1/4 chord
aileron
V
shear
center
The end view, looking inboard
How do lift and pitching moment depend on aileron movement?
Purdue Aeroelasticity
8-4
Review - The flap-to-chord ratio
determines the aileron aero derivative
values
Big letter, little letter?
What’s the diff?
1.0
parameters
0.8
0.4
center-of-pressure
distance behind the
1/4 chord
-0.2
0.0
All-movable section
center of pressure behind ¼ chord
0.2
0.0
Nose-down pitch


cl 
c
c L delta
l divided by
lift curve slope
0.6
c m delta/lift curve slope
0.2
0.4
0.6
0.8
1.0
E = flap/chord ratio
Purdue Aeroelasticity
8-5
Compute changes in lift and pitching moment
on outboard panel 2 aileron due to aileron
deflection
L2  qSCL  o   2   qSCL  o
M AC2  qScCMAC  o
about aero center @ ¼ chord of panel 2
Purdue Aeroelasticity
8-6
Write equilibrium equations in matrix form
Two aileron torque terms are added
 5  2 1 
 1 0  1  Q1 
KT 
   qSeC L 
  


 2 2   2 
 0  1  2  Q2 
0
0
1
 qSeC L  o    qSeC L    qScC MAC  

 
1
 o
 o 
Aileron input
Purdue Aeroelasticity
8-7
Three applied aero torsion loads
Nondimensionalize to identify aeroelastic
terms
0
0
1
Qi   qSeCL  o    qSeCL    qScCMAC  
1
 o 
 o 
Divide load matrix terms by KT
Q 
i
so that
qSeCL
KT
1 qSeCL
o   
KT
1
 0  qScCMAC
 
KT
 o 
0
 
 o 
CL 0
CMAC c 0
1
Qi  q o    q  o    q
o  
CL 1
CL e 1
1
Purdue Aeroelasticity
8-8
Combine the aileron load terms
1  CL CMAC c  0
Qi  q  o    q  
o  

CL e  1
1  CL
1   CMAC c  0
Qi  q  o    q  
o  

CL e  1
1  
CMsc
Purdue Aeroelasticity
0.8
parameters
  CMAC c 



 

C
e
L



1.0
c L delta divided by
lift curve slope
0.6
0.4
center-of-pressure
distance behind the
1/4 chord
0.2
0.0
-0.2
0.0
c m delta/lift curve slope
0.2
0.4
0.6
0.8
1.0
E = flap/chord ratio
8-9
The final equilibrium equation set
 5  q 

 2
2  1 
1
0 
    q o    qCMsc  o  
 2  q   2 
1
1 
inputs
outputs
Purdue Aeroelasticity
8-10
The divergence condition still comes
from the determinant of the aeroelastic
stiffness matrix on the left hand side
5  q 
2
2
2  q 
0
We have applied an aileron deflection
Nothing about divergence has changed - why?
Purdue Aeroelasticity
8-11
Feedback control laws
what are they?
 Nature
already has designed a feedback
relationship between aerodynamic loads
and structural deflection - that is why part
of the aero load is on the left-hand side of
the equilibrium equation
 o  G11
 Let’s put in an artificial feedback
relationship between aileron deflection on
the outer panel and twist of the inner panel
Purdue Aeroelasticity
8-12
What difference does this make?
Look at the equilibrium equations
 5  q 

 2
2  1 
1
    q o    qCMsc
 2  q   2 
1
 0 


G11 
This term is out of place.
How did it get here?
The aileron term belongs over here with these guys.
Purdue Aeroelasticity
8-13
Get the aileron control vector into the
“correct” form
 0   a11 a12  1   0 0 1 


 
 


 g11   a21 a22  2  G1 0 2 
qCM 
 5  q 

 2
 0 

  qCMsc
 g11 
 0 0 1 
G 0  
 1
 2
2  1 
    qCMsc
 2  q   2 
0
G
 1
q
qSeC L
KT
0  1 
1
   q o  

0   2 
1
notice the minus sign
Purdue Aeroelasticity
8-14
Reduce the equations
to nondimensional form
G1  CMsc g1
 5  q 

 2  qk 
2  1 
1
    q o  
 2  q   2 
1
Purdue Aeroelasticity
8-15
The divergence dynamic pressure has changed
 5  q 

 2  qk 
2  1 
1
    q o  
 2  q   2 
1
Compute the Determinant
5  q 
2
 2  qk   2  q 
Purdue Aeroelasticity
0
8-16
Expand the stability
determinant
  q  7q  6  2qk  0
2
Polynomial
2nd order
   o  2qk  0
Purdue Aeroelasticity
8-17
G=1
G=0.5
baseline
G=-1
Crossing points are different
G=-0.5
Plot the Stability Determinant vs.
dynamic pressure parameter for different k (=G)
values
aeroelastic stiffness determinant
6
4
divergence
2
0
-2
Positive values of k
mean that the aileron
increases load in
response to positive

-4
-6
-8
0
2
4
6
8
10
dynamic pressure parameter, q
Purdue Aeroelasticity
8-18

aeroelastic stiffness determinant
A close look at the first crossing point
6
STABLE
REGION
4
Negative aileron action,
load reduction
divergence
2
G=-1
0
-2
divergence
-4
-6
G=-0.5
no aileron
UNSTABLE
REGION
G=1
-8
0.0
0.5
1.0
1.5
2.0
G=0.5
2.5
dynamic pressure parameter
q
qSeC L
KT
Purdue Aeroelasticity
8-19
Summary
 When
a control surface is added, its
deflection creates just another load - unless…
– the control surface deflection responds to surface
deflection – using a control law that we choose.
A
feedback control law changes Mother
Nature’s aeroelastic feedback process and
the divergence dynamic pressure changes
Purdue Aeroelasticity
8-20
Homework for next Friday?
 Five
problems handed out in class and
posted on-line
 Watch for updates
 Helpful hints for maximizing points
– FBD’s
– Definitions
– Stability is a perturbation event
Purdue Aeroelasticity
8-21