AAE556
Lecture 33
V-g Method revisited
Purdue Aeroelasticity
1
Final EOM’s for forced response
h
Fo
A + B θ =
mb
b
ω is known because we pre-select it
ω Lh
2
2
A=−
− ω + ωh
µ
2
ω
B = −ω xθ −
µ
2
2
1
Lα − 2 + a Lh
2
Purdue Aeroelasticity
Moment equilibrium equation
h
D ÷+ Eθ = 0
b
2
2
ω
ω
1
2
D = −ω xθ −
Mh +
Lh + a
µ
µ 2
2
2
ω
1
ω
E = −ω 2 rθ2 + ωθ2 rθ2 +
Mh + a −
Mα
µ
2
µ
2
ω2 1
ω
1
+
Lα + a −
Lh + a
µ
2
µ
2
2
3
Purdue Aeroelasticity
Theodorsen’s method
The system is self-equilibrating
h
Fo
A ÷+ Bθ =
=0
mb
b
ω Lh ω ω
=− 2 − 2 +
ω µ ω ω
2
ATM
BTM
2
ω2
ω2
= − 2 xθ −
2
ω
µω
2
h
2
1
Lα − 2 + a ÷Lh
4
Purdue Aeroelasticity
Moment equilibrium equation
h
D ÷+ Eθ = 0
b
ω
ω
ω
1
= − 2 xθ −
Mh +
L
+ a÷
2
2 h
ω
µω
µω
2
2
DTM
ETM
2
2
2
ω 2 2 ωθ2 2 ω 2
1
ω
= − 2 rθ + 2 rθ + 2 M h + a ÷− 2 M α
ω
ω
ω µ
2
ω µ
2
ω
1
ω
1
+ 2 Lα + a ÷− 2 Lh + a ÷
ω µ 2
ω µ 2
2
2
5
Purdue Aeroelasticity
Eigenvalue Equation of Motion #1
2
h
h
h
ω
1
2
2
2
−ω
− ω xθθ + ωh −
Lh + Lα − + a ÷Lh θ
b
b µ b
2
Divide by ω
÷= 0
2
ωh2 h 1 h
h
1
− − xθ θ + 2 − Lh + Lα − + a ÷Lh θ
b
ω b µ b
2
÷= 0
Include structural damping
ωh2
h
h 1 h
1
− − xθ θ + 2 ÷( 1 + ig ) − Lh + Lα − + a ÷Lh θ
b
b µ b
2
ω
6
Purdue Aeroelasticity
÷= 0
Equation #2, moment equilibrium
2
h
ω
h
2
2 2
2 2
−ω xθ ÷− ω rθ θ + ωθ rθ θ −
M θθ θ + M θ h ÷ = 0
µ
b
b
2
M θθ
1
1
= M α − + a ( Lα + M h ) + + a Lh
2
2
Divide by ω
Mθ h
1 1
= − + a ÷Lh
2 2
2
h 2
ωθ2 2
1
h
− xθ ÷− rθ θ + 2 rθ θ − M θθ θ + M θ h ÷ = 0
ω
µ
b
b
Include structural damping
h 2
ωθ2
1
h
2
− xθ ÷− rθ θ + 2 ( 1 + ig ) rθ θ − M θθ θ + M θ h ÷ = 0
ω
µ
b
b
7
Purdue Aeroelasticity
Matrix equations
ωh2 h 1 h
ωθ2
h
1
− − xθ θ + 2 ÷( 1 + ig ) 2 ÷ − Lh + Lα − + a ÷Lh θ
b
2
ω
ωθ b µ b
÷= 0
h 2
ωθ2
1
h
2
− xθ ÷− rθ θ + 2 ( 1 + ig ) rθ θ − M θθ θ + M θ h ÷ = 0
ω
µ
b
b
2
ω
h
2
0
ωθ
2 ÷
h b 1
− 2 ÷( 1 + ig ) ωθ
+ x
ω
2
0
θ θ
r
θ
1 Lh
+
µ
M θ h
xθ h
b
2
rθ θ
1
h
L
−
+
a
÷Lh b 0
α 2
=
0
θ
M θθ
8
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The eigenvalue problem
ωh2
ω
2 ÷ 0 h b 1
−
+
÷( 1 + ig ) ωθ
ω
θ xθ
2
0
rθ
2
θ
2
1 Lh
+
µ
M θ h
ωθ2
0
h ω 2 ÷
1
Ω 2 b = h
x
1
θ 0
θ
rθ 2
xθ h
b
2
rθ θ
1
h
L
−
+
a
÷Lh b = 0
α 2
0
θ
M θθ
xθ 1 Lh
+
rθ 2 µ
M θ h
1
h
Lα − 2 + a ÷Lh b
θ
M θθ
9
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Another look at it
This should be easy for a 6
ωθ2
0
h ω 2 ÷
1
Ω 2 b = h
x
1
θ 0
θ
rθ 2
xθ 1 Lh
+
rθ 2 µ
M θ h
th
grader with MATLAB
1
h
Lα − 2 + a ÷Lh b
θ
M θθ
ωθ2 Lh ωθ2
Lα Lh 1
− + a ÷÷
1 + ÷ 2 ÷ xθ +
h ωh 2 ÷
µ ωh
µ µ 2
h
2
b
Ω b=
1
Mθ h
1 2 M θθ
θ
θ
x
+
r
+
÷
2 θ
rθ 2 θ
µ ÷
r
µ
θ
10
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The flutter problem – a complex eigenvalue with flutter frequency and airspeed
unknown
a = elastic axis location (shear center)
ωh
= bending-torsion frequency ratio
ωθ
S
xθ = θ
= dimensionless static unbalance
mb
rθ = dimensionless radius of gyration about SC
µ =density ratio
ω = frequency
k=reduced frequency
11
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