Aeroelasticity & Experimental Aerodynamics (AERO0032-1)

Lecture 4 Practical Aircraft Aeroelasticity

T. Andrianne

2015-2016

3 assumptions by Theodorsen: •  Attached flow à small motion

•  Airfoil = flat plate •  Flat wake à Unsteady circulatory lift :

Assuming sinusoidal motion à à

From Lecture 3

2

lc = πρUcC k( ) Uα + !h+ 3c4− x f

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α =α0ejωt

h = h0ejωt

Non-sinusoidal motion

•  Theodorsen analysis assumes sinusoidal motion à Equations of motion are only valid at zero airspeed or at the flutter condition.

•  But they are also valid in the case of forced sinusoidal excitation.

à Objective: Calculate the response of an aeroelastic system with Theodorsen aerodynamics to any excitation force

3

Forced response

•  Until now, unforced motion only:

•  Consider an external loads (e.g. control input)

applied on one of the degrees of freedom – Force on plunge:

– Moment on pitch:

4

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Sinusoidal force

•  Using Theodorsen analysis, the force must be sinusoidal:

•  Once all the transients have died out, the steady-state response of the pitch-plunge wing will also be sinusoidal:

à Replace Fh (or Mα) and h (or α) in the EoM

5

Fh = F0ejωt

Mα =M0ejωt

h = h0ejωt

α =α0ejωt

Forced equations of motion

Force in plunge:

6

Kh −ω2m+πρUcC k( ) jω

−ω 2ρπb

−ω 2S + ρπb2Ujω + ρπb2 x f −c2

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+πρUcC k( ) U +34c− x f

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−ω 2S −πρUecC k( ) jω

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&'ρπb2ω 2

Kα −ω2Iα +

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− x f −c2

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with k=ωb/U

FRF Matrix

This equation is of the form H(ω)q0=F

where H-1(ω) is = Frequency Response Function Matrix

à Knowing H-1(ω) and F à q0=H(ω)-1F

•  H is is a function of ω à  Response amplitude q0 is also a function of ω •  The response is still sinusoidal. •  What happens if F is also a function of ω? 7

Fourier Transforms

•  Using Fourier analysis, any force time signal Fh(t) can be transformed to the frequency domain and written as:

F0(ω)ejωt

•  The signal that is of particular interest here is the impulse. Its Fourier Transform is

F0(ω)=1 à If we set F0(ω) = 1 and then calculate

q0(ω)=H-1(ω)F = H-1(ω) The resulting response amplitude q0(ω), will be the Frequency Response Function (FRF).

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FRF for pitch-plunge system

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FRF of h The two modes are clearly present FRF of α The first mode is present as an anti-resonance

Impulse Response Function

•  Impulse Response Function (IRF) = inverse Fourier Transform of the FRF

•  From the knowledge of the FRF, q0(ω), we can simply apply the inverse Fourier Transform to it and calculate the IRF.

•  The IRF is a time domain signal and it is not sinusoidal: –  For a dynamic system, the IRF is typically an

exponentially decaying sinusoid. –  For a fluttering aeroelastic system, the IRF is an

exponentially increasing sinusoid.

10

Impulse response of pitch-plunge airfoil

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Damped sinusoidal motion

•  Until now, we showed: –  Theodorsen theory is only valid for sinusoidal motion. –  But it can also be used to calculate impulse responses. –  The form of the impulse response can tell is if the

system is stable or unstable.

•  Stability analysis is slow and can be less accurate when performed on impulse responses.

à We need a method for calculating the damping at all airspeeds directly from the EoM

12

The p-k Method

13

= most popular technique for obtaining aeroelastic solutions from Theodorsen-type aeroelastic systems •  Proposed in the 80’s and since then has

become the industrial standard

•  Virtually all aircraft flying today have been designed using the p-k method

Basics

14

•  The p-k method uses the structural equations of motion in the standard form

•  Coupled with Theodorsen aerodynamic forces of the form

with k=ωb/U

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Time-Frequency domain

•  Coefficients of EoM are frequency dependent •  EoM = Time-frequency domain equations.

•  General form:

15

Ms!!q+Ksq−12ρU 2Q k( )q = 0

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−4πC k( ) jk + 2πk2 −2πcC k( )− 2πbjk − 4πC k( ) 34c− x f

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4πecC k( ) jk − 2π x f −c2

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•  Solving time-frequency equations of motion is a nonlinear eigenvalue problem.

•  The basis of the p-k method is to define so that

•  Re-writing the equations of motion as

p2Ms +Ks −12ρU 2Q k( )

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p ≡ ddt

!!q = p2q

Ms!!q+Ksq−12ρU 2Q k( )q = 0

Time-Frequency domain

•  These equations of motion are clearly an eigenvalue problem. For a non-trivial solution:

= 1 equation with 2 unknowns (p and k) à  Need a second condition.

•  Note that p is an eigenvalue of the matrix

p2I+Ms−1 Ks −

12ρU 2Q k( )

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Eigenvalue estimation (1)

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Ms−1 Ks −

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(1)

Eigenvalue estimation (2)

•  k is a reduced frequency: •  Recall that the magnitude of the eigenvalue

is the frequency : à The second equation is:

à  The p-k method consists in: Finding an eigenvalue and a frequency that satisfy both equations (1) and (2).

18

p = kUb

(2)

p =ω

k = ωbU

The p-k solution

The solution of these equations is iterative: 1.  Guess a value for the frequency k 2.  Calculate p from the resulting eigenvalue

problem (1) :

3.  p and k should satisfy equation (2) :

4.  If they do not, we change the value of k and re-calculate p until equation (2) is satisfied.

19

p2I+Ms−1 Ks −

12ρU 2Q k( )

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p = kUb

= Frequency matching

Frequency matching

20

p2I+Ms−1 Ks −

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p-k method characteristics

21

•  Converges very quickly to the correct eigenvalue.

•  Suitable for large computational problems. •  Calculates complete eigenvalues and therefore

damping ratios and natural frequencies. •  Valid at all airspeeds, not just the flutter speed. •  Estimated flutter speeds are very similar to

those obtained from the flutter determinant solution.

Sample result

22

Sample result

23

Roger’s Approximation

= Alternative to solve the p-k equations à Transformation of EoM into the time domain using Roger’s Approximation. The frequency-dependent part of equations of motion, Q(k), is approximated as:

where nl is the number of aerodynamic lags γn are aerodynamic lag coefficients.

24

Q k( ) =A0 +A1 jk +A2 jk( )2 + A2+njk

jk +γnn=1

nl

∑ (3)

Ms!!q+Ksq−12ρU 2Q k( )q = 0

Roger’s EOMs

•  The equations of motion of the complete aeroelastic system become:

where

•  Usually:

25

!q =

−M−1C −M−1K −M−1A3 " −M−1Anl+2

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M = Ms −12ρb2A 2, C = Cs −

12ρUbA1, K = K s −

12ρU 2A 0, A j = −

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nl = 4, γ n = −1.7kmaxn

nl +1( )2 , kmax = maximum k of interest

Roger details

•  Matrices A0, A1, .. are obtained from equation (3) by performing a least-squares curve fit of Q(k) for several values of k.

•  If the aeroelastic system has n degrees of freedom,

Roger’s equations of motion will have (nl+2)n states. Of those, nln states are aerodynamic states.

•  There are similar but more efficient schemes for improving the curve fit of Q(k) and reducing nl, e.g. the minimum state method.

26

Practical Aeroelasticity

•  For an aircraft, the matrix Q(k) is obtained using a panel method-based aerodynamic model.

•  Usually performed by means of commercial packages, such as MSC.Nastran or Z-Aero.

•  For a chosen set of k values, e.g. k1, k2, …, km, the corresponding Q matrices are returned.

•  Values of Q at intermediate k values are obtained by interpolation.

•  p-k method à flutter solution, time-domain responses

27

3D wings

•  All the methods described until now concern airfoil sections (2D)

•  These results must now be extended to 3D wings because all wings are 3D

•  Two methods for 3D wing aeroelasticity: – Strip theory – Panel methods

28

Strip theory

The instantaneous lift and moment acting on each strip 2D sectional lift and moment theories (quasi-steady, unsteady, ..)

29

S

y

dy

à Dividing the wing into spanwise small strips

Wake Panels

i+1,j+1

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Panel methods

30

à Wing is replaced by its camber surface.

Camber surface itself is replaced by panels of mathematical singularities, solutions of Laplace’s equation

Hancock Model

= Simple 3D wing model

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•  Rigid flat plate •  Dimensions

o  span s o  chord c o  thickness t,

•  Suspended through

an axis xf by two torsional springs:

o  in roll (Kγ) o  in pitch (Kθ)

Equations of motion

•  As with the 2D pitch plunge wing, the equations of motion are derived using energy considerations.

•  The kinetic energy of a small mass element dm of the wing is given by

•  The total kinetic energy of the wing is:

32

€

dT =12

˙ z 2dm =12

dm y ˙ γ + x − x f( ) ˙ θ ( )2

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T =m12

2s2 ˙ γ 2 + 3s c − 2x f( ) ˙ γ ̇ θ + 2 c 2 − 3x f c + 3x f2( ) ˙ θ 2( )

Structural equations

•  The potential energy of the wing is simply

•  The full structural equations of motion are then:

33

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Iγ = ms2 /3, Iγθ = m c − 2x f( )s /4, Iθ = m c 2 − 3x f c + 3x f2( ) /3

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Aerodynamic model

Using the strip-theory –  2D aerodynamics : quasi-steady or unsteady

approximations for the lift and moment around the flexural axis are applied to infinitesimal strips of wing

–  Integration of lift and moment over the entire span of the wing

à Quasi-steady pseudo-3D lift and moment acting on the Hancock wing:

34

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Quasi-steady strip theory

Notations: θ = α (pitch) and h = yγ (flap) à Strip theory integrations will yield the total moments around the y = 0 and x = xf axes.

35

l

mxf

3D Quasi-steady equations of motion

Full 3D quasi-steady equation of motion:

36

Natural frequencies and damping ratios

37

Theodorsen function aerodynamics

Theodorsen function aerodynamics (unsteady frequency domain) can be implemented directly using strip theory:

38

l

mxf

Flutter determinant

Flutter determinant for the Hancock model

To solve similarly than the 2D pitch-plunge airfoil (flutter determinant)

39

p-k solution

40

Instead of solving the flutter determinant, we can apply the p-k method. The process is exactly as presented in the 2D case.

Comparison of flutter speeds

41

Wagner and Theodorsen solutions are identical. Quasi-steady solution is the most conservative (ignore the other solutions)

Comparison of 3D and strip theory, static case

42

The 3D lift distribution is completely different to the strip theory result !

Strip theory: Lifting line method:

CL = 2πα

CL =2πARAR+ 2

α

•  Strip theory is only exact when the wing’s aspect ratio is infinite •  It becomes completely unsatisfactory at moderate and small

aspect ratios (less than 8-10).

i+1,j+1

i+1,j

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y

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Vortex Lattice Method

Vortex ring = •  Rectangle made up of

four straight line vortex segments.

•  Elementary solution of Laplace’s equation.

•  nij, normal to the ring,at its midpoint

•  Circulation, Γ, is constant over all the 4 segments of a vortex ring.

43

Dividing the wing planform into panels on which lie vortex rings, usually called a vortex lattice

•  Each segment induces a velocity [u v w] at a general point P. •  Self-induced velocity = 0

If point P lies on a vortex ring segment, the velocity is 0.

Panelling up and solving •  The wing can be swept, tapered and twisted.

•  It cannot have thickness; the panels lie on the wing’s camber surface.

•  The wake must also be panelled up.

•  Objective of VLM: calculation of the values of the circulations Γ on each wing panel at each instance in time.

•  The circulation of the wake panels does not change in time. Only the vorticities of the wing panels are unknowns.

44

Wake Panels

i+1,j+1

s

i,j+1

i+1,ji,j

c

y

x

z

0

Boundary conditions Two boundary conditions:

–  Impermeability ‘No flow normal to the panel at its collocation point’ à This conditions gives all the Γ values.

– Kutta condition ‘Flow separates at the trailing edge’ à Automatically satisfied by placing the leading edge of each vortex ring on the quarter-chord of its corresponding geometric panel.

45

Calculating forces

•  Once the vorticities on the wing panels are known, the lift and moment acting on the wing can be calculated

•  These are calculated from the pressure difference acting on each panel

•  Summing the pressure differences of the entire wing yields the total forces and moments

46

Panels for static wing

47

Even if the wing is not moving à 3D wing tip effects (wing tip vortices, induced drag) à Wake must be modelled

Wake shapes

48

Wake shape behind a rectangular wing (AR=4, AOA=5°) that underwent an impulsive start from rest.

Effect of Aspect Ratio on lift coefficient

49

3D results approach Wagner’s function (2D result) as the AR increases.

Free wakes

•  Free wake behind a flapping goose wing. •  Airspeed: 18m/s.

50

Industrial use

•  Unsteady wakes are beautiful but expensive to calculate.

•  For practical purposes, a fixed wake is used with unsteady vorticity

à Theodorsen’s method. •  The wake propagates at the free stream

airspeed and in the free stream direction. •  Only a short length of the wake is simulated (a

few chord-lengths). •  The result is a linearized aerodynamic model

51

Commercial packages

•  There are two major commercial packages that can calculate 3D unsteady aerodynamics using panel methods: – MSC.Nastran (MSC Software) – ZAERO (ZONA Technology)

•  They can both deal with complex aircraft geometries.

52

ZAERO AFA example

•  The examples manual of ZAERO features an Advanced Fighter Aircraft (AFA) model.

53

Structural model

Aerodynamic model

BAH Example

•  Bisplinghoff, Ashley and Halfman wing

•  FEM with 12 nodes and 72 dof

54

First 5 modes of BAH wing (US àinversed signs !)

55

BAH Example

‘hump’ mode

GTA Example

= Very simple aeroelastic model of a Generic Transport Aircraft (GTA)

56

Finite element model: Bar elements with 678 degrees of freedom

Aerodynamic model: 2500 doublet lattice panels

Flutter plots for GTA

57

First 7 flexible modes: Clear flutter mechanism between first (first wing bending) and third mode (aileron deflection)

Time domain plots for the GTA

58

V<VF V=VF

SuperSonic Transport

•  SST = Concorde and Tupolev Tu-144 •  The aeroelastic model is a half-model •  The aerodynamics features the wing and a rectangle

for the wall

59

Flutter plots for SST

60

First 9 flexible modes: Clear flutter mechanism between first and third mode.

Summary

•  Aeroelastic design in industry is almost exclusively carried out using lattice methods (vortex or doublet) combined with a Finite Element model with few retained modes (generally fewer than 100).

•  Panel methods (source and doublet) are also used for more detailed geometric representation, including thickness.

•  The aerodynamic forces are written in the form of Aerodynamic Force Coefficient Matrices that depend on frequency.

•  The flutter solution is obtained using the p-k method = Frequency matching 61