AAE 556
Aeroelasticity
Lecture 7-Control effectiveness
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Purdue Aeroelasticity
Reading
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Sections 2.15-2.18
– These sections are painfully worked example problems – read through
them to understand principles discussed in class
– Section 2.18.2 has a virtual work example – wait to read this until next
week
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Skip 2.19 for now (will do next week)
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Read 2.20, 2.20.1 and 2.20.2
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Our next goal
learn about control effectiveness
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Demonstrate the aeroelastic effect of
deflecting aileron surfaces to increase lift or
rolling moment
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Examine the ability of an aileron or elevator to
produce a change in lift, pitching moment or
rolling moment
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Reading – Sections 2.20-2.20.2
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Ailerons are required for lateral stability
They become increasingly ineffective at high speeds
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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.
–
/>lutter.html#flutter
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Purdue Aeroelasticity
The ability of an aileron or elevator to produce a change in lift, pitching moment or rolling moment is
changed by aeroelastic interaction
L = qSCLδ δ o + qSCLα θ
no α o
M AC = qScCMACδ
Lift
α0 + θ
V
MAC torsion spring KT
shear center
δ0
e
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aileron
deflection
Herman Glauert’s estimators for CLd and CMACd
The flap-to-chord ratio is
C Lδ =
C Lα
π
(cos
CMACδ = −
−1
E=
(1 − 2 E ) + 2
CLα
π
cf
c
E (1 − E )
( 1− E ) ( 1− E ) E
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)
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 − KTθ
Remember
αo = 0
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Purdue Aeroelasticity
Solve for the twist angle
due only to aileron deflection d
c
qSe C Lδ + CMACδ
e
θ=
KT − qSeC Lα
Lift
δ
o
L = qSC L δ o + qSC Lα θ
δ
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The aeroelastic lift due to deflection
CMACδ
q
c
1 +
qD e C Lδ
L = qSC Lδ δ o
q
1−
qD
Compare answer to the lift computed
ignoring aeroelastic interaction
Lrigid = qSC L δ o
δ
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The aileron deflection required to generate a fixed increases as q increases
1− q
qR
Lo = qSC L δ o
δ
1 − q
qD
The required control input is …
Aileron deflection increases as q
approaches reversal
Lo
δo =
qSC L
δ
1− q
qD
1 − q
qR
Is aileron reversal an instability?
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The most common definition for
the reversal condition
L flex = 0
Is it possible that I deflect and aileron and get no lift?
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?
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How do I make the numerator term in the lift expression equal to
zero?
q c CMACδ
L=0, reversal
1 + ÷
qD e C L
δ
L = qSCLδ δ o
q
1−
qD
L=infinity, divergence
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÷
÷
=0
Solve for the q at the reversal condition
qR
1+
qD
c CMACδ
=0
÷
e CLδ
numerator=0
q = q reversal = q R
e C Lδ
qR = −qD
c CMACδ
or
KT
qR = −
ScC Lα
CL
δ
CMAC
δ
Why the minus sign?
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Understanding what the aileron does
Two different ways to compute pressure distribution resultants due to aileron
deflection
aerodynamic
center
aileron lift = qSC L δ δo
aileron lift = qSCL δδo
d
MAC = qScC MAC δ δo
δ0
e
(a) aerodynamic center
representation
δ0
e
(b) aileron center of
pressure
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Force equivalence
the same moment at the AC with 2 different models
+
− Lδ d = qScCMACδ δ o
Solve for the distance d to find the CP distance from
=
d
the AC
A lift force at d produces the
same result at the AC as a lift
force and moment at the AC
CM δ
d
=−
c
CLδ
e
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Lδ
δ0
Aileron center of pressure depends on the aileron chord
mid-chord
midchord position
Dist ance f rom aileron ce nt e r of pre ssure
t o airf oil ae rodynamic ce nt e r
(Glaue rt pre dict ion)
Distance
distance
ngths
aftinofchord
1/4 le
chord
0.20
AC
e xample
she ar
ce nte r
location
All-movable surface
0.10
quarte r chord location
0.00
0.0
0.2
0.4
0.6
0.8
1.0
f lap to chord ratio
Aileron flap to chord ratio, E
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Summary
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Control surfaces generate less lift because the control deflection creates a
nose-down pitching moment as it generates lift.
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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
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