AAE 556
Aeroelasticity
Lecture 6-Control effectiveness
Purdue Aeroelasticity
6-1
Today’s goals
 Review
what we have done with the 2
DOF model and draw some conclusions
 Begin the study of control effectiveness
Purdue Aeroelasticity
5-2
Reading review
 Sections
2.1-2.18
– Some of these sections are painfully
worked example problems – work through
them to understand principles discussed in
class
– Skip 2.19 for now (next week)
– Read 2.20, 2.20.1 and 2.20.2
Purdue Aeroelasticity
6-3
Summary-Develop 2 DOF segmented aeroelastic
finite wing model
A
Torsional
springs
2KT
3KT
fuselage
panel 2
panel 1
e
A
b/2
V
wing root
shear
centers
b/2
aero
centers
 + 2
 + 1
view A-A
Torsional degrees of freedom
Purdue Aeroelasticity
wing tip
6-4
Lift re-distribution due to aeroelasticity
Wing sections support ½ the total W
1 

 L1  W  2 
   
 L2  2  1 
 2
 L1 
W
 
 L2  2  q

1  q 
2

 1 
2

q
qSeCL
KT
Observation - Outer wing panel carries more of the total load
than the inner panel as q increases
Purdue Aeroelasticity
6-5
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 – divergence occurs
Purdue Aeroelasticity
6-6
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
  q 2  7q  6
panel twist, i/o
A
aero
centers
Unstable q region
Outboard
panel (2)
  1  q  6  q 
determinant  is zero
Purdue Aeroelasticity
6-7
How do we determine divergence for
an MDOF system?
The general form of the aeroelastic static equilibrium equations is
 KT   i   Qi 
n degrees of freedom
Perturb the system by an amount
i 
Euler question – “Is there an equilibrium solution to the
following relationship?”
 KT   i    i   Qi 
Purdue Aeroelasticity
6-8
Static stability
The static stability test reduces to the
existence of a homogeneous matrix
equation that must hold if the system is
neutrally stable
 KT   i    i   Qi 
 KT   i   0
?
Purdue Aeroelasticity
6-9
Linear algebra says...
only if … K
or
ij
 qAij  0
KT    0
This nth order determinant is called the stability
determinant or the characteristic equation
Purdue Aeroelasticity
6-10
Our next goal
control effectiveness



Demonstrate the
aeroelastic effect of
deflecting aileron surfaces
to increase lift or rolling
moment
Examine the ability of an
aileron or elevator to
produce a change in lift,
pitching moment or rolling
moment
Reading – Sections 2.202.20.2
Purdue Aeroelasticity
6-11
Setting the stage

Many of the
uncertified minimum
ultralights, and
perhaps some of the
certificated aircraft,
have low torsional
wing rigidity. This will
not only make the
ailerons increasingly
ineffective with speed
(and prone to flutter),
but will also place
very low limits on g
loads.
– http://www.auf.asn.a
u/groundschool/flutt
er.html#flutter
Purdue Aeroelasticity
6-12
The ability of an aileron or elevator to produce a change in
lift, pitching moment or rolling moment is changed by
aeroelastic interaction
L  qSCL  o  qSCL 
no  o
M AC  qScCMAC
Lift
0 
V
aileron
MAC torsion spring KT
shear center deflection
e
Purdue Aeroelasticity
0
6-13
Herman Glauert’s estimators for CL
and CMAC
The flap-to-chord ratio is
CL 
CL

cos
CMAC  
1
E
1  2E   2
CL

cf
c
E 1  E 

1  E  1  E  E
Purdue Aeroelasticity
6-14
1 DOF idealized model – no camber
Sum moments about the shear center
L
Linear problem (what
does that mean?)
e
M
sc
 0  Le  M AC  KT
Remember  o  0
Purdue Aeroelasticity
6-15
Solve for the twist angle
due only to aileron deflection 
c

qSe C L  C MAC
e


KT  qSeC L
Lift



o
L  qSCL  o  qSCL 

Purdue Aeroelasticity
6-16
The aeroelastic lift due to
deflection

CMAC
q
c


1 
 
 qD  e  CL


L  qSC L  o
q
1
qD
Compare answer to
the lift computed
ignoring aeroelastic
interaction




Lrigid  qSCL  o

Purdue Aeroelasticity
6-17
One definition for
the reversal condition
L flex  0
Is this possible?
We usually use an aileron to produce a rolling
moment, not just lift. What is the dynamic
pressure to make the lift or rolling moment zero
even if we move the aileron?
Purdue Aeroelasticity
6-18
How do I make the numerator term in
the lift expression equal to zero?

q  c  CMAC
L=0, reversal
1   
 qD  e  C L


L  qSCL  o
q
1
qD



 0
L=infinity, divergence
Purdue Aeroelasticity
6-19
Solve for the q at the reversal
condition
qR
1
qD
 c  CMAC
 0 numerator=0
 
 e  CL
q  qreversal  qR
KT
e C L
qR  qD
or qR   ScC
c CMAC
L
 CL


 CMAC


Why the minus sign?
Purdue Aeroelasticity
6-20




Summary
 Control
surfaces generate less lift because
the control deflection creates a nose-down
pitching moment as it generates lift.
 At a special dynamic pressure (a combination
of airspeed and altitude) the deflection of an
aileron creates more downward lift due to
nose-down deflection than upward lift
Purdue Aeroelasticity
6-21