Then the system can absorb energy in a static deformation mode.. If the stability determinant is negative then the static system, when perturbed, cannot absorb all of the energy due to w
Trang 1AAE 556
Aeroelasticity
Lecture 22 Typical dynamic instability problems and test
review
ARMS 3326 6:00-8:00 PM
Trang 2How to recognize a flutter problem in the making
Q
&&
&&
Q is a real number
If p12 and p21 have the same
sign (both positive or both
negative) can flutter occur?
x1
x2
=
x 1
x 2
e
iωt
2 2
2 2
−
=
Given: a 2 DOF system with a parameter Q that creates loads on the system that are linear functions of the displacements
ω 4 − ω 2 ( ω 1 2 + ω 2 2 ) + ω 1 2 ω 2 2 = 0
ω12 = K1
M2
Q=0
−ω2 +ω12
M1 p12
M2 p21
−( ω2 +ω22)
x 1
x 2
=
0 0
Q not zero
Trang 3If flutter occurs two frequencies must merge
2 1
2 2
2 2
2 1
2 2
2 1
2
1
Q
ω
( ω 1 2 − ω 2 2 ) 2
= − 4 Q
2
M 1 M 2 p 12 p 21
For Flutter – Increasing Q must cause the term under the radical sign to become zero and then go negative The zero
condition is:
Q2 = − M1M2( ω12 − ω22)2
4 p12p21 p12p21 = − M1M2( ω12 − ω22)2
For frequency merging flutter to occur, p12 and p21 must have opposite signs.
ω12 = K1
M1
ω22 = K2
M2
Trang 4If one of the frequencies is driven to zero then we have divergence
ω n = 0
∆ = 0 = ( ) ω 1 2 ( ) ω 2 2 − Q 2
M 1 M 2 p 12 p 21
( ) ω 1 2 ( ) ω 2 2 = Q 2
M 1 M 2 p 12 p 21
Q 2 = M 1 M 2 ω 1 2 ω 2 2
p 12 p 21
p12p21= M1M2ω12ω22
Q2
Q 2 = K 1 K 2
p 12 p 21
Divergence requires that the cross-coupling terms are of the same sign
2 2 21
2
2
0 0
Q
K x p
M
x
x
M x
ω =
Trang 5Aero/structural interaction model
TYPICAL SECTION What did we learn?
torsion spring
KT
V
θ
lift e
K T ∝ GJ
span
L = qSC L α
α o + qScC MAC
K T
1 − qSeC L α
K T
Trang 6Divergence-examination vs perturbation
K h 0
h
θ
=
− L
M SC
L = qSC Lα
1 − qSeC Lα
K T
α o + qSC Lα
1 − qSeC Lα
K T
qScC MAC
K T
1
1 − q = 1 + q + q 2 + q 3 + = 1 + q n
n = 1
∞
∑
Trang 7Perturbations & Euler’s Test
K T ( ) ∆ θ > ∆ ( ) L e
K T ( ) ∆ θ < ∆ ( ) L e
K T ( ) ∆ θ = ∆ ( ) L e
result - stable - returns -no static equilibrium in perturbed state
result - unstable -no static equilibrium - motion away from equilibrium state
result - neutrally stable - system stays - new static equilibrium point
torsion spring
KT
V
θ
lift
e
Trang 8Stability equation is original equilibrium equation with R.H.S.=0.
K T − qSeC L α
( ) = K T = 0
torsion spring
KT
V
θ
lift
e
∆ θ ≠ 0
The stability equation is an equilibrium equation that represents an equilibrium state with no "external loads" –
Only loads that are deformation dependent are included
The neutrally stable state is called self-equilibrating
Trang 9Multi-degree of freedom systems
e
shear centers
aero centers
panel 1
panel 2
V
A
A
view A-A
αο+ θ1
αο+ θ2
− 2 2
θ1
θ2
+ qSeCLα
− 1 0
θ1
θ2
= qSeCLααo 1
1
From linear algebra, we know that there is a solution to the
homogeneous equation only if the determinant of the aeroelastic stiffness matrix is zero
Trang 10MDOF stability
K T = 0 K ij − qA ij = 0
System is stable if the aeroelastic stiffness matrix determinant is positive Then the system can absorb energy in a static deformation mode If the stability determinant is negative then the static system, when perturbed, cannot absorb all of the energy due to work done by aeroelastic forces and must
become dynamic
Mode shapes? Eigenvectors and eigenvalues
K T
[ ] { } ∆ θ i = { } 0
Trang 11Three different definitions of roll effectiveness
• Generation of lift – unusual but the only game in town for the typical section
• Generation of rolling moment –
• contrived for the typical section – reduces to lift generation
• Multi-dof systems – this is the way to do it
• Generation of steady-state rolling rate or velocity-this is the information we really want for airplane performance
• Reversal speed is the same no materr which way you do it
Trang 12Control effectiveness
V
α0+ θ
Lift
δ0
e
shear center
t orsion spring
L = qSCLδδo
1 + q
qD
c e
CM
δ
CLδ
1 − q
qD
q D
c e
C M δ
C L δ = 0
qR = − KT
ScCLα
CLδ
CMδ
reversal is not an instability - large input produces small output
opposite to divergence phenomenon
Trang 13Steady-state rolling motion
L = 0 = qSC L α qScC M δ
K T δ o − v
V
+ qSC L α δ o
V
α0+ θ
Lift
δ0
e
shear center
t orsion spring
Trang 14Swept wings
K2
K1
f
V
V cosΛ
b
αo
C C
α structural = θ − φ tan Λ
q n = qcos 2 Λ
Kφ 0
− Q
− tb
2
b
2
− te e
φ θ
=
Q αo
b
2
e
Trang 15∆ = K θ K φ + Q K θ bt
2 − K φ e
qD =
Kθ Seao
cos2 Λ 1 − b
e
K Kθφ
2
tan Λ crit = 2 e
c
c b
K φ
K θ
90 75 60 45 30 15 0 -15 -30 -45 -60 -75 -90 -2.0 -1.5 -1.0 -0 5
0 0
0 5 1.0 1.5 2.0
nondimensional divergence dynamic pressure vs wing sweep angle
sweep angle (degrees)
sweep f orward sweep back
b/c=6 e/c=0.10 Kb/Kt=3
Trang 16Lift effectiveness
350 300
250
20 0 150
10 0 50
0
0 0
0 5
1.0
1.5
2.0
lift effectiveness vs
dynamic pressure
dynamic pressure (psf)
divergence
30 degrees
sweep
15 degrees sweep unswept wing
Trang 17Flexural axis
β
Λ
x
y
Flexural axis - locus of points where a concentrated force creates no stream-wise twist (or chordwise aeroelastic angle
of attack)
θ E = θ − φ tan Λ
θ E = 0
The closer we align the airloads with the flexural axis, the smaller will be aeroelastic effects