Aeroelastic Stability: Divergence

Analytical ExactLater: Approximate

Structural Analysis: Aeroelastic Divergence© 2021 Mayuresh Patil. Licensed under a Creative Commons Attribution 4.0 license https://creativecommons.org/licenses/by-nc-sa/4.0/ mpatil@gatech.edu

Aeroelasticity

• Aeroelastic phenomena– Stability: Divergence (static) and Flutter (Dynamic)– Lift redistribution– Control reversal

AerodynamicsStructural Analysis

Undeformed AiplaneFlight Condition

Airloads DeformationStresses

Aeroelastic Coupling

Torsion (review)

• Equations of Equilibrium

• Internal Twisting Moment

• Strain and Stress (Axisymmetric)

• Strain and Stress (Thin-Walled, Single Cell)

mx(x)

d

dx

GJ(x)

d�(x)

dx

�= �mx(x)

Tx(x) = GJ(x)d�(x)

dx

⌧xs =Tx

2Act✏xs =

Ac

tRc1t ds

d�

dxJ =

4A2cR

c1t ds

⌧x✓ = Grd�

dx=

Txr

J✏x✓ =

1

2rd�

dxJ = Ip

Torsion of Wing

• Aerodynamic pitching moment:

– Twist of the wing due to aerodynamic loads

a.c.s.c.

e

mx = Mxsc = Mxac + Ly ⇥ e

Mxac =1

2⇢V 2c2Cm0

Ly =1

2⇢V 2cCl =

1

2⇢V 2c(Cl0 + Cl↵↵)

d

dx

GJ(x)

d�(x)

dx

�= �1

2⇢V 2c(cCm0 + eCl0 + eCl↵↵)

�(0) = 0 GJd�

dx|x=L = 0

Aeroelasticity:Fluid-Structure Interaction (coupling)

• Aerodynamic loads lead to structural deformation (φ)• Does structural deformation affect the aerodynamic loads?

– where:

• Aeroelastic equation for torsion of wing↵ = ↵0 + �(x)

d

dx

GJ(x)

d�(x)

dx

�= �1

2⇢V 2c(cCm0 + eCl0 + eCl↵(↵0 + �(x)))

d

dx

GJ(x)

d�(x)

dx

�+

1

2⇢V 2ceCl↵�(x) = �1

2⇢V 2c(cCm0 + eCl0 + eCl↵↵0)

Ly =1

2⇢V 2c(Cl0 + Cl↵↵)

Wing twist under aerodynamic loads• We can calculate the twist at a given flight condition

• Possibility of Instability? Yes • Large twist at certain airspeed: Divergence– Consider solutions of homogenous equation

– Trivial solution:– Is there a non-trivial solution?

d

dx

GJ(x)

d�(x)

dx

�+

1

2⇢V 2ceCl↵�(x) = �1

2⇢V 2c(cCm0 + eCl0 + eCl↵↵0)

d

dx

GJ(x)

d�(x)

dx

�+

1

2⇢V 2ceCl↵�(x) = 0

�(x) ⌘ 0

Stability Solution

• Uniform wing (constant GJ, c, e, Clα)

– Differential eigenvalue problem

• Solution:• Example: Cantilevered wing

GJd2�(x)

dx2+

1

2⇢V 2ceCl↵�(x) = 0

d2�(x)

dx2+ �2�(x) = 0 �2 =

⇢V 2ceCl↵

2GJ

�(x) = A sin�x+B cos�x

�(0) = 0 �0(L) = 0

Algebraic Eigenvalue problem• Matrix form

• Nontrivial solution only if matrix is singular

• Lowest divergence speed:

– No divergence if e < 0

=)

=)

0 1

cos�L � sin�L

�⇢AB

�= 0

cos�L = 0 �L =(2n� 1)⇡

2

�2 =(2n� 1)2⇡2

4L2Vdiv =

(2n� 1)⇡

2

s2GJ

⇢ceCl↵L2

Vdiv =⇡

2

s2GJ

⇢ceCl↵L2

qdiv =⇣⇡2

⌘2 GJ

ceCl↵L2

Uniform Wing:twist at a given flight conditions

• Cantilevered• Solution:

– What happens when– Note: if no aeroelastic coupling:

�(0) = 0 GJd�

dx|x=L = 0

�(x) =qc(cCm0 + e(Cl0 + Cl↵↵0))

2GJ(2L� x)x

q !⇣⇡2

⌘2 GJ

ceCl↵L2

�(x) =cCm0 + e(Cl0 + Cl↵↵0)

eCl↵

"�1 + cos

rqceCl↵

GJx

!+ sin

rqceCl↵

GJx

!tan

rqcL2eCl↵

GJ

!#

GJd2�(x)

dx2+ qceCl↵�(x) = �qc(cCm0 + eCl0 + eCl↵↵0)

Aeroelastic Wing Twist

• GJ = 0.26e9 N-m2, L = 25 m, ρ=1 kg/m3, c = 4 m, e = 1 m, Clα = 2 π, Cl0= 0.25, Cm0 = 0, α0 = 3 deg

• Vdiv = 285.8 m/s

50 100 150 200 250 300Vm

s

-20

-10

10

20

30

�tip(deg)

Change in Lift due to Flexibility