Aeroelasticity

Lecture 2: Dynamic Aeroelasticity

G. Dimitriadis

1 Aeroelastic EOM

• In the previous lecture we developed the aeroelastic for a pitching and plunging flat plate:

2 2nd Order ODEs • The equations are 2nd order linear ODEs of the form

(A + ρB)q!! +(C + ρUD)q! +(E + ρU 2F)q = 0

! ! $ $ # c & • where 1 # − x f & ! $ # " % & ! $ ! $ m S 2 2 0 0 h A = # &, B = πb , C = # &, q = # & # & # 2 2 & S Iα ! $ ! $ " 0 0 % " α % " % # c − c − + b & # # x f & # x f & & " " 2 % " 2 % 8 % ! $ !3c $ c # 1 # − x &+ & " f % ! $ ! $ # 4 4 & Kh 0 0 1 D = cπ , E = # &, F = cπ # & # 2 & # & ! $ ! $ 0 Kα " 0 −ec % # − c − + 3c − c & " % # ec # x f & # x f & & " " 2 % " 4 % 4 %

3 First order form • The second order equations can be easily written in first order form:

! $ ! −1 −1 2 $! $ q!! −M C + ρUD −M E + ρU F q! # & = # ( ) ( ) &# & # ! & # &# & " q % " I 0 %" q % • where M=A+ρB • The first order ODEs are of the form z! = Qz • where ! $ ! −1 −1 2 $ q! −M (C + ρUD) −M E + ρU F z = # &, Q = # ( ) & # & # & " q % " I 0 %

4 Analytical solution • Recall from last year’s Flight Mechanics course that first order linear ODEs have an analytical solution: z(t) = eQtz(0) • or, after decomposing the matrix exponential:

n λit z(t) = ∑vie ci i=1 • where c=V-1z(0), n is the number of states, V is the eigenvector matrix of Q and l are the eigenvalues of Q.

5 Frequency and • The absolute values of the eigenvalues are the natural frequencies, ωn=|λ| • The damping ratios are defined as: ζ=-Re(λ)/ωn • The damping ratios are measures of the amount of damping present in each mode of • It must be kept in mind that both natural frequencies and damping ratios are functions of airspeed and air density because the matrix Q is a function of these two quantities.

6 Variation with airspeed

As the airspeed increases, the two natural frequencies approach each other. One of the damping ratios increases while the other first increases and then decreases. The critical damping ratio becomes zero and then negative. Instability ensues. This phenomenon is called flutter and the zero damping speed is the flutter speed.

7 Subcritical System response

Solve the equations of motion for the time responses of the system from initial conditions (α(0)=5o). Time responses for U=30m/s. Both pitch and plunge decay with time.

8 Critical System Response

Solve the equations of motion for the time responses of the system from initial conditions (α(0)=5o). Time responses for U=35.9m/s. Both pitch and plunge amplitudes remain constant.

9 Supercritical Responses

Solve the equations of motion for the time responses of the system from initial conditions (α(0)=5o). Time responses for U=38m/s. Both pitch and plunge oscillation amplitudes increase with time.

10 Stability criteria

• The stability of the system can be estimated directly from the eigenvalues of the system matrix: – If all eigenvalues have negative real parts, the system is stable – If at least one real eigenvalue is positive, the system has undergone static divergence – If at least one pair of complex conjugate eigenvalues has positive real part, the system has undergone flutter.

11 Determining the flutter speed

• The flutter speed can be determined by trial and error: – Choose an air density (i.e. flight altitude) – Calculate the system eigenvalues for a starting airspeed – Keep increasing the airspeed until at least one pair of complex eigenvalues has positive real part – Continue to try different airspeeds until the real part is almost zero.

12 Routh-Hurwitz (1)

• The static divergence and flutter speeds can also be obtained directly from the characteristic polynomial • This can be achieved using the Routh- Hurwitz stability criterion. • The criterion applies to a polynomial of the form 4 3 2 a4λ + a3λ + a2λ + a1λ + a0 = 0

13 Routh-Hurwitz (2) • The system is unstable if

– any of the coefficients ai is zero or negative while at least one is positive – There is at least one sign change in the first column of the matrix H • The matrix H is given by

14 Routh-Hurwitz (3)

• The condition a0<0 gives the static divergence 2 2 condition, Kα<ρU ec π • The condition c1<0 yields

• Which, when expanded, yields a 4th order polynomial in U. • Two of the solutions are U=+0 and U=-0

• The other two solutions are U=+UF and U=- U=-UF

15 Numerical searches

• Routh-Hurwitz can be easily applied to a 2- DOF system. • aeroelastic models can have more than 100 DOFs. Routh-Hurwitz is totally impractical for such large systems. • Numerical methods can be used instead. • These are generally divided into two categories – Directed searches, e.g. Newton-Raphson – Indirect searches, e.g. trial and error

16 Newton-Raphson • Newton-Raphson is a very widely used method for solving nonlinear problems. • Suppose we need to solve the nonlinear equation f(U)=0. • We start with a first guess Ui. This is a guess so f(Ui)=0. However, we want to calculate a correction DU, such that f(Ui+DU)=0. • We expand f(Ui+DU) in a Taylor series around Ui: df f (Ui + ΔU) = f (Ui ) + ΔU = 0 dU Ui

17 Newton-Raphson

• Solving for DU we get: −1 # df & ΔU = −% ( f (Ui ) $ dU Ui ' • Now we can calculate a better approximation for the solution of f(U)=0, which is Ui+1=Ui+DU. • This value is still not exact. We need to re-apply the procedure in order to calculate Ui+2, which will be an even better approximation. • We keep iterating until |DU|

18 Flutter test functions

• For flutter determination we need to define a suitable function f(U)=0. • Several different test functions work well. The simplest is: n f (U) = ∏ℜ(λ j (U)) j=1 • Where n is the number of states. • This test function is equal to 0 when the real part of any of the eigenvalues is equal to 0. • If we want to detect only flutter and not static divergence, then we can choose to include only the complex eigenvalues in the product.

19 Flutter derivative • As the calculation of the eigenvalues is numerical, it is not possible to evaluate the derivative analytically. • We can use a forward difference scheme to calculate the derivative numerically: f U +δU − f U df = ( i ) ( i ) δ dU Ui U • Where dU is a very small user-defined speed increment.

20 Starting guess

• The starting guess for the flutter speed should not be close to 0. Aeroelastic systems without structural damping flutter at U=0. • Aeroelastic systems with structural damping can flutter at negative airspeeds. • Choose an airspeed within the flight envelope but far from 0. • Some aeroelastic systems may have many flutter airspeeds. Only the lowest flutter airspeed is of interest.

21 Effect of flexural axis

The position of the flexural axis has a significant effect on both flutter and static divergence. For this aeroelastic system the flutter speed is always lower than the static divergence speed, unless xf/c>0.75. Also note that placing the flexural axis in front of the aerodynamic center is bad for the flutter speed!

22 Unsteady

• As mentioned in the first lecture, quasi-steady aerodynamics ignores the effect of the wake on the flow around the airfoil. • The effect of the wake can be quite significant, it effectively reduces the magnitude of the aerodynamic forces acting on the airfoil. • This reduction can have a significant effect on the values of the flutter.

23 Kelvin’s theorem

• One of the bases of unsteady aerodynamics is Kelvin’s theorem. • It states that the total circulation in a flow cannot change in time; this includes circulation on the wing, , and in the wake, . • If the circulation over a wing increases at a particular Γtime" instance (for exampleΓ# because the angle of attack increases) then equal and opposite circulation must be shed into the wake. • In equation form:

' Γ" + Γ# = 0 • so that any change in'( bound circulation must be accompanied by a change in wake circulation . ΔΓ" # ΔΓ = 24 − ΔΓ" Starting Vortex (1) • The simplest unsteady flow is a flat plate at 0o angle of attack in a steady flow of airspeed . • At a particular instance in time, , the angle of attack is increased impulsively to, say, 5o. ! • This impulsive change causes the"# shedding of a strong vortex in the wake, known as the starting vortex. • The starting vortex induces a significant amount of local velocity around the airfoil. However, it travels downstream because of the steady flow . • As the starting vortex distances itself from the wing,! its effect decreases • After a while it has no effect at all and the flow becomes steady 25 Starting Vortex (2)

Wake shape of an airfoil whose angle of attack was impulsively increased to 5o. The starting vortex is clearly seen

26 Wagner’s model

• The first solution of the impulsively started flat plate problem was obtained by Wagner in the 1920s. • He modelled the wake as a flat horizontal line containing a continuous vorticity distribution. • The flat plate is also horizontal while the free stream is inclined to the angle by a small angle . ! " / Δ% Δ%

" 0 % ΔΓ+ ΔΓ- ΔΓ, ΔΓ. !• The vortices are ejected at the th time instance and then travel at the free 0stream airspeed,0 such that . • Their strength does not change# as they travel downstream. • The vorticity distribution becomes continuousΔ% = !Δ' as .

Δ' → 0 27 Vortex summary

• Recall from the Aerodynamics course that a point vortex induces a potential around it. • The potential induced at a point by a vortex of strength lying at point is given by !, # Γ !%,(1)#% Γ ./ # − #% & !, # = − tan • Then the horizontal and2* vertical! − !% velocities induced at the same point are

(2) 1& Γ # − #% 0 !, # = = 2 2 1! 2* ! − !% + # − #% (3) 1& Γ ! − !% 28 4 !, # = = − 2 2 1# 2* ! − !% + # − #% Circle plane

• The flow solution is obtained using a conformal transformation from a plane containing a circle of radius to a plane containing the flat plate of chord . • The complex coordinate of the circle! plane is defined by . • Vortices are placed2! in pairs of equal but opposite strength and at positions and respectively. Kelvin’s theorem# = is% + '( automatically satisfied since. the sum of circulation is zero. −*Γ • *ThereΓ is an infinityΞ- of vortices! /Ξ- ranging from to and from to .

! ∞ ! 0 ( !

*Γ −*Γ

. % !- /Ξ- 29 Ξ- Flat plate plane

• The complex coordinate of the flat plate plane is defined by . • The conformal transformation between the two planes is given by ! = # + %& (4) 1 1 ! = 0 + 2 0

&

' (Γ, −(Γ

# ,- 30 Circle transformation

• The equation of the circle in the plane is

! + + + • which means that on the$ circle+ - = )

+ + • Applying the transformation!./0.12 = $of+ equation3 ) − $ (1) to the circle we obtain:

7 7 7 7 = 5 6 5 6 8:/ 6 :8 + + 7 7 + + 7 7 • or4 = + $ + 3 ) −. $As+ 89on/ 6 the:8 circle= + $ takes+ 3 ) values− $ + between896 :8 $ and , this means that the circle of radius does indeed collapse" = $, &onto= 0 a flat$ plate with chord . −) ) ) 2) 31 Transformation of vortices • The vortices on the circle plane lie at and . $ • Applying) the transformation of equation! = (4)Ξ to, & the= 0 first vortex we! = obtain( /Ξ$, & = 0 ) 1 ( • Applying the same transformation+ = Ξ$ + to the second vortex we have 2 Ξ$ 1 1 1 / 0 0 / 0 1 • This means that both+ = )vortices23 + 0 /2 from3 = ) theΞ$ +circle23 plane transform to the same point on the flat plate plane

) 1 ( • Solving for we see that4$ = Ξ$ + 2 Ξ$ or (5) ) ) ) ) $ $ $ $ $ $ Ξ − 24 Ξ + ( Ξ = 4 + ( − 4 32 Velocity induced by the wake • In the circle plane, the velocity induced by the wake vortices at position is calculated from equation (3) as ξ = #, % = 0 'Γ 'Γ ', = − 0 + # 2/ Ξ+ − # • Substituting from equation (5)2/ and# − simplifying we obtain Ξ+

'Γ *+ + # • The total velocity induced by', all= −the vortices+ is obtained by integrating dv, such that 2/# * − #

6 1 *+ + # • Finally, we can write , = − 45, where+ is' vortexΓ strength per unit length, such that 2/# * − # 'Γ = Γ) *+ '*+ Γ) (6) 6 1 *+ + # , = − 4 Γ) *+ '*+ 2/# 5 *+ − # 33 Kutta condition

• The position in the circle plane is actually the trailing edge in the flat plate plane. • The Kutta conditionξ = #, % says= 0 that all velocities must be equal to zero at the trailing edge. There are two vertical velocity components , induced by the vortices , the vertical component of the free stream. • The– ( Kutta condition then means that . Substituting from– ) *equation (3): )* + ( = 0(7) 5 1 ./ + # 3 Γ, ./ 7./ = )* • This is the fundamental22# 4 equation./ − # of Wagner theory. It relates the strength of the vortices in the wake to the vertical free stream component. • It is one equation with one unknown, the vortex strength .

/ Γ, . 34 Solution

• Equation (7) is an integral equation so it does not have an easy solution. • It can be solved numerically very easily. • Consider the flat plate one time step after the impulsive start. There is a single vortex of strength lying at . Equation (7) becomes Δ" ΔΓ$ %&' = ) + +Δ" 1 +Δ" + 2) +Δ" ΔΓ$ = +1 ⟹ ΔΓ$ = 20)+1 • At the second20) time+Δ" instance, there are two vortices,+Δ" + 2) lying at and lying at . Equation (7) becomes ΔΓ$ %&, = ) + 2+Δ" ΔΓ. %&' = ) + +Δ" 1 +Δ" + 2) 1 2+Δ" + 2) ΔΓ. + ΔΓ$ = +1 • Which can2 be0) easily+Δ" solved for20) .2 And+Δ" so on for the next time instances. 35 ΔΓ. Vortex strength

• The strengths of the 0.3 vortices are thus easily calculated for all the time steps of interest. 0.25 • The figure plots against non-dimensional 0.2 time . NoteΔΓ that , i.e. the

" ! 0.15 vortex# =strengths%&/( are ΔΓplotted#) =againstΔΓ) the non- dimensional time at 0.1 which they were ejected. • The results were obtained for a flat plate 0.05 with m, and m/s. . 0 • All that( = remains0.5 - is= to5 0 0.5 1 1.5 2 calculate% = 10 the lift cause by = these vortices!

36 Unsteady lift

• The lift is obtained from the pressure difference between the upper and lower side of the plate. • In steady conditions, the pressure is calculated from the Bernoulli equation

constant • In unsteady conditions, the0 pressure1 1 is calculated from the unsteady Bernoulli equation / + 1 2 ! + ) = constant 0 1 1 34 • The velocity is given/ +by2 1 ! + ) + .3 5 = • On the surface, and , i.e. the potential is equal and opposite! on the two sides! = #of +,the%& plate:/%( . ) = 0 & (, 0 = −& (, 0 + constant 1 , 0 34 34 / (, 0 = −2 1 # + 36 + 35 + constant 1 . 0 34 34 • Calculate the pressure/ (, 0 difference= −2 1 so# −that,36 after− 35 linearising (8) 37 . , %& %& Δ/ = / (, 0 − / (, 0 = 22 # + %( %9 Potential • The potential induced by a pair of vortices on the circle is obtained from equation (1) as

+Γ 34 ! 34 ! +, &, ! = tan − tan $ • Recall that on the circle2/ & − Ξ6 and& − that# /Ξ the6 same potential applies to the flat plate’s$ surface,$ where . The potential becomes ! = # − & ' = &, ) = 0 $ $ $ $ +Γ 34 # − ' 34 # − ' +, ', 0 = tan − tan $ • Substituting for from2/ equation' (5)− Ξ 6and simplifying' − # /Ξ we6 obtain

$ $ $ $ 6 +Γ 34 # − ' 8 − # +, ', 0 = tan $ • Finally, the total potential2/ due to all #the− 8pairs6 of vortices is

$ $ $ $ (9) < 6 38 1 34 # − ' 8 − # , ', 0 = : tan $ Γ= 86 +86 2/ ; # − 86 Lift equation

• We now substitute the potential of equation (9) into the pressure difference of equation(8). • First we recall that the wake travels with the free stream airspeed, so that and, therefore,

!" = $ + &' () () *!" () • Equation (8) gives (after =a conveniently= & neglected integration session) (' (!" *' (!"

/& - + !" Δ, -, !" = Γ3 !" *!" 1 1 1 1 • Then, the lift due to a pair0 of vortices" is (after an integration marathon) $ − - ! − $

8 9: 78 " ; ; 3 " " • And the total lift*5 = ∫ Δ, -, ! *- = /& 9: 78 Γ ! *! (10) = " ! 39 5 = /& < Γ3 !" *!" 8 1 1 !" − $ Calculating the lift

• Again, equation (10) is difficult to evaluate analytically but easy to calculate numerically. • At the first time instance there is a single vortex of known strength lying at . Equation (10) becomes ΔΓ# $%& = ( + *Δ+ 12345 # , , # • At the second time/ = instance0* 12345 6there1 ΔΓ are two vortices, lying at and lying at . ΔΓ# %, . %& $ = ( + 2*123Δ4+5 ΔΓ 12.345 $ = ( + *Δ+ . , , . , , # • As , / etc= 0*are1 2already345 61 ΔΓ known+ 0* 1 2we.34 5can61 ΔcalculateΓ the lift at all time instances. # . ΔΓ ΔΓ 40 Lift result

• The figure plots 0.6 the lift against non-dimensional 0.5 time. • At the first time 0.4 instance the lift l jumps to half the c 0.3 steady-state value. 0.2 • It then increases cl(t) cl =2 0.1 1 asymptotically cl towards the 1 steady-state 0 value. 0 10 20 30 40 = We can write where is a function increasing asymptotically from 0.5 to 1 " ! # = 2&'Φ # Φ # 41 Wagner Function (1) • The effect of the starting vortex on the aerodynamic forces around the airfoil can be modeled by the Wagner function . • The Wagner function states that the instantaneous lift at the start of theΦ motion" is equal to half the value of the steady lift (i.e. the value of the lift if the flow had been steady) • The instantaneous lift then slowly increases to reach its steady value as time tends to infinity

42 Wagner Function (2)

The Wagner function is equal to 0.5 when 1 t=0. It increases asymptotically to 1. 0.8 It describes the growth of lift after the )

impulsive start of a = 0.6 flat plate airfoil. ) ( It is independent of 0.4 the angle of attack. If it is calculated as a function of non- 0.2 dimensional time, it is also independent 0 of the airspeed and 0 10 20 30 40 chord. =

43 Wagner Function (3) • An approximate expression for the Wagner function is given by

−ε 1Ut / b −ε 2Ut / b Φ(t) = 1− Ψ1e − Ψ2e • where Ψ1=0.165, Ψ2=0.335, ε1=0.0455, and ε2=0.3. • The lift coefficient variation with time after a step change in incidence is given by

cl (t) = 2παΦ(t) • So that the lift force variation becomes l(t) = ρπU 2cαΦ(t) = ρπUcwΦ(t)

w=Uα is the downwash velocity 44

Unsteady Motion

• Unsteady motion can be modeled as a superposition of many small impulsive changes in angle of attack • The increment in lift due to a small change in pitch angle at time t0 dw(t ) dl(t) = ρπUcΦ(t − t )dw(t ) = ρπUcΦ(t − t ) 0 dt 0 0 0 dt 0 • So that the lift variation at all times can be0 obtained by integrating from time -∞ to time t, i.e. t dw(t ) l(t) = ρπUc Φ(t − t ) 0 dt ∫-∞ 0 0 dt0

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Unsteady Motion (2) • Using the thin airfoil theory result obtained in the first lecture, the downwash velocity can be written as $ 3 & w(t) = Uα (t) = Uα(t) + h (t) + c − x α (t) tot % 4 f ' • For a motion starting at t=0, w=0 for t<0 and w=w(0) at t=0. • The lift generated at negative times is given by 0 dw(t0 ) l(t) = ρπUc Φ(t − t0 ) dt0 = ρπUcΦ(t)dw(0) = ρπUcΦ(t)w(0) t0 <0 ∫-∞ dt0

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Unsteady Motion (3) • Then, the lift at all times is " % ! " 3 % l(t) = ρπUc$Uα (0) + h(0) +$ c − x f 'α! (0)'Φ(t) + # # 4 & &

t " " % % ! !! 3 !! ρπUc ∫ Φ(t − t0 )$Uα (t0 ) + h(t0 ) +$ c − x f 'α (t0 )'dt0 0 # # 4 & & • Use integration by parts to get rid of acceleration terms inside the integral (these are very difficult to deal with) • Result is:

47 Unsteady Motion (4)

& & 3 ( ( l(t) = ρπUc* Uα(t) + h (t) + c − x α (t)+ Φ(0) − ' ' 4 f ) )

t ∂Φ − & ( (t t0 )  & 3 ( ρπUc * Uα(t0) + h (t0) + c − x f α (t0)+ dt0 ∫0 ' ) ∂t0 ' 4 )

• This equation is the basis of Wagner function aerodynamics • It includes the effect of the entire motion history of the system in the calculation of the current lift force

48 Moment

• The aerodynamic moment around the flexural axis due to the unsteady lift force is simply mxf(t)=ec l(t) • However, for a complete representation of the aerodynamic force and moment, the added mass effects must be superimposed, exactly as was done in the quasi-steady case. • The complete equations of motion become

49 Unsteady equations of motion

This type of equation is known as integro-differential since it contains both integral and differential terms.

50 Integro-differential equations

• Integro-differential equations cannot be readily solved in the manner of Ordinary Differential Equations. • A numerical solution can be applied, based on finite differences, e.g. Houbolt’s Method • However, numerical solutions are not very good for conducting stability analysis • The equations must be transformed to ODEs in order to perform stability analysis

51 Transform to ODEs (1)

• Use the following substitutions:

The wi variables are known as the aerodynamic states. They arise from the substitution of the approximate form of the Wagner function, , in the equations of motion. Φ

52 Transform to ODEs (2)

• The integral in the lift equation can be expanded by parts. Then, substituting for the aerodynamic states we obtain

53 Transform to ODEs (3) • The integrals have been absorbed by the aerodynamic states. The full equations of motion are " % + ρπ 2 − ρπ 2 − $ m b S b (x f c / 2) '" !! % M $ '$ h '+ 2 2 2 2 $ α!! ' $ S − ρπb (x f − c / 2) Iα + ρπb (x f − c / 2) + b / 8 '# & # ( ) & " % Φ 0 c / 4 + Φ 0 3c / 4 − x " % C $ ( ) ( )( f ) ' h! πρUc$ '$ '+ $ − Φ − − Φ '# α! & # ec (0) (3c / 4 x f )(c / 4 ec (0)) & " % (1) + πρ Φ! πρ Φ + − Φ! " % $ Kh Uc (0) Uc(U (0) (3c / 4 xf ) (0)) ' h K $ '$ '+ 2 ! 2 ! α $ −πρUec Φ 0 Kα − πρUec UΦ 0 + 3c / 4 − xf Φ 0 '# & # ( ) ( ( ) ( ) ( )) & " w % " %$ 1 ' −Ψ ε 2 −Ψ ε 2 Ψ ε −ε − Ψ ε −ε − $ 1 1 / b 2 2 / b 1 1 (1 1 (1 2e)) 2 2 (1 2 (1 2e)) '$ w ' W 2πρU 3 2 $ 2 2 '$ ' $ ecΨ ε b ecΨ ε b −ecΨ ε −ε − e −ecΨ ε −ε − e ' w3 # 1 1 / 2 2 / 1 1 (1 1 (1 2 )) 2 2 (1 2 (1 2 )) &$ ' $ ' # w4 & " ! % $ πρUcΦ(t)(h(0) +(3c / 4 − xf )α (0)) + p(t) ' = $ ' $ −πρ 2Φ! + − α + ' 54 # Uec (t)(h(0) (3c / 4 xf ) (0)) r(t) & Transform to ODEs (4)

• There are two equations with 6 unknowns; 4 more equations are needed. • These can be obtained by noting that the definitions of wi are of the form

• Differentiating this equation with time:

55 Leibniz Integral Rule

• E.g for w1(t):

• For all wi(t):

(2)

56 Complete Equations • Equations (1) and (2) make up the complete aeroelastic system of equations. • Equations (1) are 2nd order Ordinary Differential Equations (ODEs). They describe the dynamics of the system states. • Equations (2) are 1st order ODEs. They describe the dynamics of the aerodynamic states.

57 Complete Equations (2) • Here is the form of the complete equations # h  & % ( u = Qu % α  ( % h ( • where # −M−1C −M−1K −M−1 W& % ( % ( α = = % ( Q % I 0 0 (, u % ( % ( w1 $ 0 W0 ' % ( % w2 ( % w ( % 3 ( $ w4 ' # 1 0 −ε U /b 0 0 0 & % 1 ( −ε % 1 0 0 2U /b 0 0 ( W0 = % 0 1 0 0 −ε1U /b 0 ( % ( $ 0 1 0 0 0 −ε 2U /b'

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Aerodynamic States • The aerodynamic states are mathematical constructs that are used to represent history effects. • As already mentioned several times, the aerodynamic forces depend not only the current state of the system but also on the history of the motion. • This history is stored in the aerodynamic states. After all they are integrals.

59 Solution of the ODEs

• Now the unsteady aeroelastic equations are in complete ODE form (6 equations with 6 unknowns) and can be solved as usual, by injecting a harmonic component

λt u = u0e • A 8th order characteristic polynomial is obtained of the form

8 7 6 5 4 3 2 a8λ + a7λ + a6 λ + a5λ + a4λ + a3λ + a2λ + a1λ + a0 = 0

60 Natural Frequencies and damping ratios

61 Hard flutter • There are significant differences between the unsteady and quasi-steady natural frequencies and damping ratios. • The unsteady flutter speed is much higher than the quasisteady one: 50.7 m/s instead of 35.9 m/s. • The bad news is that the unsteady flutter mechanism is much more abrupt: the damping drops very quickly to zero. • This phenomenon is known as hard flutter and can be very dangerous.

62 Effect of Flexural Axis

The divergence speed is the same as in the quasi-steady case. The flutter speeds obtained from Wagner’s method is always higher than that obtained from quasi-steady calculations.

63 Discussion

• Wagner function aerodynamics leads directly to time domain equations of motion. • The application of this approach has been mostly limited to simple systems, such as the pitch-plunge airfoil or the pitch-plunge-control airfoil. • Commercial aeroelastic packages calculate the aerodynamic forces in the frequency domain and then transform to the time domain, if needed.

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