L3_Static Aeroelasticity of Slender Wing

8/10/2019 L3_Static Aeroelasticity of Slender Wing

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Xie Changchuan2014 Autumn

3rd Static Aeroelasticityof Slender Wing


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2

Contents

1Model of slender wing

2Method of divergence

3Load redistribution

4Control efficiency and reversal

Main AimsUnderstand the basic mechanic principles and

analysis methods of static aeroelasticity fromthe simple slender wing model.


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Aiefoil: symmetry thin Spanl

Chord

C(y) Air force axis : ChordSlope of lifting curve Zero Moment coefficient:

Elastic axis: behind air force axis

Torsion stiffness: Twist angle: + nose up

Linear density of weight: Overload factor: NWeight center before elastic axis( )d y

( )y

( )e y

( )GJ y

( )LC y

0 ( )mC y

( )gm y

dy

TT dy

y

+

M

Nmg

e

w

y

y

l

d

C

V V

T mydy

Air force center line

Wei ht center

Elastic axis

L

Model of slender wing without swept


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Forces on slender wingAssumption of Small deformation

Wing bending would not produce air force

0ydT

T dy m dy T dy

+ + =

( )d

T GJ y

dy

=

Aerodynamics flat airfoil strip theory

Equilibrium equation

of small element

Elastic torque

( ) ( )L yC y C y

=

External Torque = Air force moment + weight moment

y

dTm

dy=

[ ] 20 0( ) ( ) ( ) ( ) ( ) ( ) ( )y L mm C y qC y e y C y qC y Ngm y d y = + +

dy

TT dy

y

+

V

T mydy

T Internal Torque

External Torqueym


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Differential equation of slender wing

20 0( ) ( )L L m

d dGJ C qCe C qCe C qC Ngmd

dy dy

+ = +

Boundarycondition

0, 0

, 0

y

dy l

dy

= =

= =

Twist divergence ---- Static stability problem

Static stability of equilibrium state:A system at it s staticequilibrium stategets an arbitrary external disturbance. When

the disturbance is eliminated, the system can go back to the

original static equilibrium state. Then, the static equilibriumstate of system is said to be static stable.

Unstable StableStable, but not

asymptotic


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Static stability of slender wing

= +

For general form of these functions,

the equation can not solved explicitly.

Obtain ahomogeneous

equation

L b u = =Note the equation as

operator form

0L 0 = =

( ) 0

L

d dGJ C qCe

dy dy

+ =

0, 0

, 0

y

dy l

dy

= =

= =

( ), ( ), ( ), ( )LGJ y C y C y e y

Attention

Let

L b u = = Equilibrium solution

Disturbed solution


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If there are any nonzero solutions

No! The disturbed solution is always zero. The system goes back

to its original state, then the system is stable.

Yes! The system can not go back to its original state, then the

system is stable.

Supposing the wing is uniform, the disturbed

equation is simplified as

2 2

2

2

0

L

ddy

qCe

CGJ

+ =

=

0, 0

, 0

y

dy l

dy

= =

= =

0 =Zero solution must exist



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22

2

0

d

dy

+ =

It means the equation has nonzero solutions

Equation have general solutions

sin cosA y B y = +

Substitute them into

boundary conditions s 0

0

co l

B

=

=

cos 0l = (2 1) ( 0,1,2, , )2

il i i = + =

2

L

qCeC

GJ

=2

24div L

GJ

q l CeC

=


When cos 0l =sin 0A y = Then for ally

From


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Eigenvalue problem in linear algebra

Discussion 1) Comparing with characteristic equation in linear algebra

22

2

d

dy

= Rewrite the

disturbed equation2 D u = =Abstract form

2

2D y

=

0, 0

, 0

y

d

y l dy

= =

= =

Ax px=,T T TA A y Ax x Ay= = 0Tx Ax >

Eigen vectors constitute a basement of linear space.

The homogeneous solution is combined linearly by the Eigen vectors.


is a linear partial difference operator

If A is symmetry and positive definite, then the

equation has nonnegative eigenvalues.

Symmetry: Positive definite


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Discussion 2) How to solve a non-uniform wing

( ) ( ) ( ) ( ) 0Ld dGJ y qC y C y e y

dy dy

+ =

[ ( ) ]d d

D GJ y

dy dy

=

0, 0

, 0

y

dy l

dy

= =

= =

0

( )( , ) ( ) ( )

l d dw yDw v GJ y v y dy

dy dy

=

Selected homework: Discuss the properties of the partial

differential operator in disturbed equation.

Inner production

2

20

( )( , ) ( )

ld w yDw v v y dy

dy

=

The eigenvalue theory of symmetric operator in Hilbert space


is a linear partial difference operator

Inner production


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Load redistribution of slender wing

The general

solutions

Considering an uniform wing, the un-homogeneous equation is

simplified as 22

2

2L

dK

dy

qCeC

GJ

+ =

=

0, 0

, 0

y

dy ldy

= = = =

2

24 Ldiv

Gq

Jq

l CeC

< = the equation has

unique stable solution

2

0 0

1( )L mK C qCe C qC Ngmd GJ

= +

2sin cos

KA y B y

= + +

2 unknows are determined by 2 boundary conditions

When


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Lift of rigid wing

2 2tan

K KA l B

= =

0( ) (cos tan sin 1) LL y y l y C qC = +

Then

2 (1 tan sin cos )

K

l y y =

2l

The condition of divergence:

At initial AOA , the lift distribution of uniform straight slender wing is0

[ ]0( ) ( ) ( )r

LL y C y qC L L y

= + = +

0r

LL C qC

=

cos tan sinr

r r

L L Ly l y

L L

+

= = +


Elastic increment


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Lift distribution of the wing with different stif fness

at given AOA and dynamic pressure

0 0.25 0.5 0.75 1.0

y

l

L/Lr

2.0

1.75

1.5

1.0

1.25

0l =8

l

=

4l

=

3l

=



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Model of swept slender wing

Swept angle:

Effective span:

Effective chord:

Aero axis1/4 chord line

Effective lifting slope:

Elastic axis: behind aero axis

Elastic twist: + nose upWeight axis: before elastic axisd

eL

C

cl

dw

dy

w

x

x

V

Vcos

A

B

l

l

y,

de

y,(elastic axis)

EffectiveRoot

Aero axis

Weight axis

Vsin

sin dw

Vdy

c

B

r +

Vcos sin dw

Vdy

tanr dw

dy +

tanidw

dy

=


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Divergence characters of swept slender wing

Lift of wing segment2( ) cosLL y qcC

=

Effective AOA

without bending r = +

r

tanr w

y

= + Bending induced AOA

2tan( ) ( ) cosrLL y qc w

Cy

= +

For sweep back wing, the elastic deformation decreases theeffectiveAOA and the lift coefficient. So the divergencedynamic pressure is increased.

For sweep forward wing, the elastic deformation increases the

effectiveAOA and the lift coefficient. So the divergencedynamic pressure is decreased.

Effective AOA

of rigid wing

Effective AOA

with bending

Nose down effect


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Divergence dynamics pressure changing

with sweep angle of slender wing

0-45

1.0

qD/qD|=0

Divergence characters of swept slender wing


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Factors to promote divergence speed of wings

Increase stiffness weight increasing

light structure with high stiffness

Increase sweep angle affects the aerodynamic

decrease the aerodynamic efficiency, especially in low air speed

Decrease the distance of elastic and aero axis

limitation on structure assignment

the aero axis is different in sub/supersonic Decrease aspect ratio, increase taper ratio

limitation on aerodynamic configuration and performance

decrease lift-drag ratio of the wing

Aeroelastic tailoring optimized design, utilize elastic

change the spanwise distribution of elastic and aero axis

change the stiffness coupling e.g. For sweep forward wing,

make it nose down when bending upward


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1Please give out divergence dynamic pressure ofuniform slender wing without sweep using energy

method.

2(selective) Considering the effective bending

stiffness EI of slender swept backward wing, please

solve the divergence dynamic pressure and load

distribution. Then discuss the possibility to eliminatethe divergence.

Load and divergence of slender wing

HOMEWORK


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py

V

B

d(y)e(y)

x

V

A

l1

l

y()

py

V

A -B

Rolling efficiency and reversal of

slender wing with aileron

Span: Chord:Aero axis:1/4 chord line Effective slope of lift:

Elastic axis behind aero axis elastic twist: + nose up

Weight axis before elastic axis

aileron position: from to wing tip

C

d

eL

Cl

1l


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2

0, 0

, 0

y

dy ldy

= =

= =

Boundary

conditoins

( )2 0 1mL L L

CC C Cd d pyGJ qCe qCe qC qCe mpyd y l

dy dy V

+ = +

( )1L LC Cd d pyGJ qCe qCe mpyd y l

dy dy V

+ = +

Simplify: Uniform rolling rate 0p =

1

1

1 ( )1 ( )

0 ( )

a

y ly

y l

>=


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22

)(12222

2

yKyV

p

dy

da

=+Amalgamate

equations 2

0

L

mL L

CqeC

GJ

CC CCK

e

=

= +

Solution: )()()( 12 yCyCV

ply +=

( ) 11 1sin ( )

( ) 1 ( ) 1 cos ( ) sincosa

l lC y K y y l yl

=

ll

y

l

yyC

cos

sin)(2 =




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23

10

20

( ) 1 ( )

( )

lL

a

L

a l

CplC y y ydy

CV

yC y ydyl

+ = =

Rolling

efficiency

pl/V

l

Rl=1.44

1.61.20.80.40

pl

VSpiral angle

of wing tip

/pl

V

10

0.5,

0.6, 0.6

L L

m L

C C

lC C C

e l

=

= =



Spiral angle by

unit deflection


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Reversal

Let Numerator be zero

10

20

( ) 1 ( )

( )

lL

a

L

a l

CplC y y ydy

CV

yC y ydy

l

+ = =

Rolling rate and acc. are both zerowhen the aileron deflects.

( ) 0cos2coscos0

21

222

1

0 =

+

+

l

C

e

Cllll

C

e

CC mmL

Transcendental equation Calculate the smallest satisfyingthe equation numerically. Note it as

rev

Reversal

dynamic pressure

2R rev

L

GJqC

eC

=




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Thinking and trying

When there is connecting stiffness between

the wing and aileron, try to solve the control

efficiency and reversal dynamic pressure. Justconsider a wing segment with control surface.

k



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L3_Static Aeroelasticity of Slender Wing