Aeroelasticity_A_R_10241445

MEng - Aeroelasticity - 7AAD0039 School of Engineering and Technology University of Hertfordshire SWEPT WING DIVERGENCE AND FLUTTER

Report by Mr A R Tutor DR ANDREW LEWIS Date 7th January 2015

Mr A R - 10241445 – A5 – University of Hertfordshire

Aeroelasticity – Swept Wing Divergence and Flutter 2

Introduction This text is separated into 2 sections, namely Part 1 and Part 2. Part 1 considers Aeroelastic divergence of forward swept wings, and Part 2 considers Aeroelastic flutter.

1 Assignment – Part 1

a If the UAV is required to achieve the divergence speed requirement indicated in the Table for your case number, determine the minimum magnitude of K required to do this. Is K positive or negative? Given that K values between ±4500 Nm are achievable, is your value of K feasible?

Distance of aerofoil aerodynamic center from shear center, e = 0.075m Aerofoil Chord, c = 0.3m Aerofoil Lift curve slope, a = 5.9 / Rad Wing Span, l = 2.5m Wing sweep angle (+ve for Aft sweep, -ve for Fwd sweep), Λ = −25° Density at SL, ρISA_SL =1.226kg m3 CASE NUMBER 16: EI =15000Nm2 GJ =18000Nm2 Minimum Divergence speed, V =140m s

qD =12ρV 2 =

12×1.226×1402 =12014.8Pa

qD =π2GJ 1−κ 2( )

4ecal2 cos2 Λ( ) 1−κ GJEItan Λ( )− 3π

2

76leGJEI

tan Λ( )−κ EIGJ

#

$%

&

'(

)*+

,+

-.+

/+

Where, κ = KEI ⋅GJ

Substituting:

qD =π2GJ 1− K

EI ⋅GJ#

$%

&

'(2#

$%%

&

'((

4ecal2 cos2 Λ( ) 1− KEI ⋅GJ

#

$%

&

'(GJEItan Λ( )− 3π

2

76leGJEI

tan Λ( )− KEI ⋅GJ

#

$%

&

'(

EIGJ

*

+,

-

./

012

32

452

62

Substituting variables:

qD =π2 ×18000 1− K

15000×18000#

$%

&

'(2#

$%%

&

'((

4×0.075×0.3×5.9×2.52 × cos2 −25( ) 1− K15000×18000

#

$%

&

'(1815tan Λ( )− 3π

2

76le1815

tan Λ( )− K15000 ⋅18000

#

$%

&

'(1518

+

,-

.

/0

123

43

563

73



Substituting and rearranging:

qD =18000π2 −18000π

2K 2

2.7×108#

$%%

&

'((

2.726 K 8.96842×10−4( )+8.26678{ } qD =

177652.8792− 6.57974×10−4K 2( )K 2.44479×10−3( )+ 22.53524

K 2 +12014.8

6.57974×10−4K 2.44479×10−3( )+ 22.53524( )− 177652.87926.57974×10−4

= 0

K 2 +K × 4.46426×104 +1.415×108 = 0

Using quadratic equation: K1,2 =−b± b2 − 4ac

2a Where

a =1b = 4.46426×104

c =1.415×108

K1,2 =−4.46426×104 ± 4.46426×104( )

2− 4 1.415×108( )

2= −2.23213×104 ±1.88876×104

∴K1 = −41208.873Nm∴K2 = −3433.7Nm

The stiffness K2 is feasible and is within±4500Nm , the negative magnitude dictates the direction of flexure is opposed.



b Carry out a short parametric study to look at how divergence speed varies with: (i) Variation to EI Aerodynamic centre to shear centre = e (m) [Constant] 0.075 Aerofoil chord = c (m) [Constant] 0.3 Aerofoil lift curve slope = a [Constant] 5.9 Wing span = L (m) [Constant] 2.5 Sweep angle = Λ (deg) [Constant] -25 GJ (Nm^2) [Constant] 18000

EI (Nm^2) [Variable] 15000 14000 16000

Divergent Dynamic pressure = 𝑞! (Pa) for EI=15000 -12951.85 Divergent Dynamic pressure = 𝑞! (Pa) for EI=14000 -11852.55 Divergent Dynamic pressure = 𝑞! (Pa) for EI=16000 -14088.19 k for EI=15000 -0.20897 k for EI=14000 -0.21630 k for EI=16000 -0.20233 K stiffness for existing Case, EI=15000 [Constant] -3433.7 Divergence Speed = V (m/s) when EI=15000 145.36 Divergence Speed = V (m/s) when EI=14000 139.05 Divergence Speed = V (m/s) when EI=16000 151.60

It can be seen from the figure above how the divergence speed increases as the Flexural rigidity, EI is increased. This observation is in agreement with elementary dynamics, where one can observe that higher stiffness structures have a higher natural frequency. In the case of aeroelastic divergence the same principles apply, the stiff system is less prone to divergence at low velocity and excitations.

138.00

140.00

142.00

144.00

146.00

148.00

150.00

152.00

154.00

13500 14000 14500 15000 15500 16000 16500

V (m

/s)

EI (Nm^2)

Divergence Speed varia9on of Variable Flexural rigidity EI



(ii) Variation to GJ Aerodynamic centre to shear centre = e (m) [Constant] 0.075 Aerofoil chord = c (m) [Constant] 0.3 Aerofoil lift curve slope = a [Constant] 5.9 Wing span = L (m) [Constant] 2.5 Sweep angle = Λ (deg) [Constant] -25 EI (Nm^2) [Constant] 15000

GJ (Nm^2) [Variable] 18000 17000 19000

k for GJ=18000 -0.20897 k for GJ=17000 -0.21503 k for GJ=19000 -0.20339 K stiffness for existing Case, GJ=18000 [Constant] -3433.7 Divergent Dynamic pressure = 𝑞! (Pa) for GJ=18000 -12951.85 Divergent Dynamic pressure = 𝑞! (Pa) for GJ=17000 -12560.37 Divergent Dynamic pressure = 𝑞! (Pa) for GJ=19000 -13321.46 Divergence Speed = V (m/s) when GJ=18000 145.36 Divergence Speed = V (m/s) when GJ=17000 143.14 Divergence Speed = V (m/s) when GJ=19000 147.42

It can be seen from the figure above how the divergence speed increases as the Torsional Rigidity, GJ is increased. This observation is in agreement with elementary dynamics, where one can observe that higher stiffness tubular structures have a higher natural torsional frequency. In the case of aeroelastic divergence the same principles apply, the stiff system is less prone to divergence at low velocity and torsional excitations.

142.50 143.00 143.50 144.00 144.50 145.00 145.50 146.00 146.50 147.00 147.50 148.00

16500 17000 17500 18000 18500 19000 19500

V (m

/s)

GJ (Nm^2)

Divergence Speed varia9on of Variable Torsional Rigidity GJ



(iii) Variation to stiffness, K Aerodynamic centre to shear centre = e (m) [Constant] 0.075 Aerofoil chord = c (m) [Constant] 0.3 Aerofoil lift curve slope = a [Constant] 5.9 Wing span = L (m) [Constant] 2.5 Sweep angle = Λ (deg) [Constant] -25 EI (Nm^2) [Constant] 15000 GJ (Nm^2) [Constant] 18000 K Stiffness for Existing case [Variable] -3433.7 K Stiffness -500 from existing K [Variable] -3933.7 K Stiffness +500 to existing K [Variable] -2933.7 K Stiffness +1000 to existing K [Variable] -2433.7 k for existing, K=-3433.7 -0.20897 k for K=-3933.7 -0.23940 k for K=-2933.7 -0.17854 k for K=-2433.7 -0.14811 Divergent Dynamic pressure = 𝑞! for existing, K=-3433.7 -12951.85 Divergent Dynamic pressure = 𝑞! for K=-3933.7 -11538.31 Divergent Dynamic pressure = 𝑞! for K=-2933.7 -14674.27 Divergent Dynamic pressure = 𝑞! for K=-2433.7 -16830.97 Divergence Speed = V (m/s) for existing, K=-3433.7 145.36 Divergence Speed = V (m/s) for K=-3933.7 137.20 Divergence Speed = V (m/s) for K=-2933.7 154.72 Divergence Speed = V (m/s) for K=-2433.7 165.70

It can be seen from the figure above how the divergence speed increases as the Spring Stiffness, K is increased. This observation is in agreement with dynamics principals, where one can observe that higher stiffness systems having a higher natural frequency under torsion. In the case of aeroelastic divergence the same principles apply, the stiff system is less prone to divergence at low velocity and torsional excitations. The importance for greater torsional stiffness for high speed aircraft and Fwd sweep wing angles is significant as a result.

130.00

135.00

140.00

145.00

150.00

155.00

160.00

165.00

170.00

-‐4500 -‐4000 -‐3500 -‐3000 -‐2500 -‐2000

V (m

/s)

K (Nm/rad)

Divergence Varia9on of Variable S9ffness K



c It is decided that it would be cheaper to use an isotropic material and so K will now be zero, but the required divergence speed is still as shown in the case table. If EI and GJ are unaltered, estimate the possible sweep angles which would enable the divergence speed to be achieved.

Minimum Divergence speed,

qD =GJπ2 1+ tan2 Λ( )( )

4ecal2 1− 3π2

76leGJEItan Λ( )

#$%

&'(

Rearranging:

qD4ecal2

GJπ2=

1+ tan2 Λ( )

1− 3π2

76leGJEItan Λ( )

qD4ecal2

GJπ21− 3π

2lGJ76eEI

tan Λ( )#

$%

&

'(=1+ tan2 Λ( )

qD4ecal

2

GJπ2−1= tan2 Λ( )+ qD4ecal

2

GJπ23π 2lGJ76eEI

tan Λ( )

0 = tan2 Λ( )+ 3qDcal3

19EItan Λ( )− qD4ecal

2

GJπ2+1

Substituting: 0 = tan2 Λ( )+3.497729605tan Λ( )+ 0.775550401

Solving as quadratic:

tan Λ( ) =−3.497729605± 3.4977296052 − 4 0.775550401( )

2∴= −1.748865±1.510968

∴= −0.237897 or −3.259833 Hence: Λ1 = tan

−1 −0.237897( ) = −13.38°Λ2 = tan

−1 −3.259833( ) = −72.95°

Where both are forward sweep angles required for divergence to be achieved with the same parameters as calculated above.

EI =15000Nm2

GJ =18000Nm2

V =140m s

qD =12ρV 2 =

12×1.226×1402 =12014.8Pa



2 Assignment – Part 2

A ridged wing is mounted flexibly so that both vertical translational and pitching oscillations are

possible. A Simplified form of the flutter equation for this particular wing is:

0.23s4 + 0.3264− 0.05333V 2( )s2 − 0.0144V 2 + 0.0864 = 0

Where V is a non-dimensionalised air speed related to the actual air speed by: V =Ubωθ

And s is a non-dimensionalised root of the flutter equation such that the heave and pitch motions

are:

h = hesωθ t

θ =θesωθ t

a Investigate whether flutter occurs for speeds V = 1.0, 1.2, 1.4 and 1.6.

Set V=1.0 as shown:

0.23s4 + 0.3264− 0.05333V 2( )s2 − 0.0144V 2 + 0.0864 = 0

0.23s4 + 0.3264− 0.05333 1( )2( )s2 − 0.0144 1( )2 + 0.0864 = 0

Let x = s 2 ∴0.23x2 + 0.27307x + 0.072 = 0

Using quadratic equation: x1,2 =−b± b2 − 4ac

2a Where

a = 0.23b = 0.27307c = 0.072

x1,2 =−0.27307± 0.27307( )2 − 4 0.23( ) 0.072( )

2 0.23( )∴x1 = −0.39525∴x2 = −0.79201

∴s2 = −0.39525 Or −0.79201

∴s = ±0.628691i Or ±0.889948i

If s =ℜe+ Img = a1 + b1i

The both values above are imaginary and contain no real part, hence system is stable and flutter

does not exist.



Set V=1.2 as shown:

0.23s4 + 0.3264− 0.05333V 2( )s2 − 0.0144V 2 + 0.0864 = 0

0.23s4 + 0.3264− 0.05333 1.2( )2( )s2 − 0.0144 1.2( )2 + 0.0864 = 0

Let x = s 2 ∴0.23x2 + 0.249604x + 0.065664 = 0

Using quadratic equation as shown in the V=1.0 section: x1,2 =−b± b2 − 4ac

2a

∴x1 = −0.44806825∴x2 = −0.63717001

∴s2 = −0.44806825 Or −0.63717001

∴s = ±0.669379i Or ±0.798229i

If s =ℜe+ Img = a1 + b1i

The both values above are imaginary and contain no real part, hence system is stable and flutter

does not exist.

Set V=1.4 as shown:

0.23s4 + 0.3264− 0.05333V 2( )s2 − 0.0144V 2 + 0.0864 = 0

0.23s4 + 0.3264− 0.05333 1.4( )2( )s2 − 0.0144 1.4( )2 + 0.0864 = 0

Let x = s 2 ∴0.23x2 + 0.2218732x + 0.058176 = 0


2a

∴x1 = −0.48233304+ 0.14245689i∴x2 = −0.48233304− 0.14245689i

s12 = −0.48233304+ 0.14245689i ; ∴s1 = ± a1 + b1i( )

s22 = −0.48233304− 0.14245689i ; ∴s2 = ± a2 − b2i( )

If s =ℜe+ Img = a+ bi

The values above contain both real and imaginary parts, hence system is unstable and flutter can

occur.

For V=1.2, no flutter exists and system is stable, hence flutter occurs when speed changed

between V=1.2 and V=1.4. The exact velocity will be shown in part 2(b) below; V=1.6 is also

considered for completeness.



Set V=1.6 as shown:

0.23s4 + 0.3264− 0.05333V 2( )s2 − 0.0144V 2 + 0.0864 = 0

0.23s4 + 0.3264− 0.05333 1.6( )2( )s2 − 0.0144 1.6( )2 + 0.0864 = 0

Let x = s 2 ∴0.23x2 + 0.1898752x + 0.049536 = 0


2a

∴x1 = −0.41250434+ 0.21211564i∴x2 = −0.41250434− 0.21211564i

s12 = −0.41250434+ 0.21211564i ; ∴s1 = ± a1 + b1i( )

s22 = −0.41250434− 0.21211564i ; ∴s2 = ± a2 − b2i( )

If s =ℜe+ Img = a+ bi

The values above contain both real and imaginary parts, hence system is unstable and flutter can

occur. The transition was established between V=1.2 and V=1.4. The exact velocity will be shown

in part 2(b) below.

Figure 1 – Flutter being approached - from Fig 4.1 of (Hodges and Pierce, 2002)



b By trial and error, or by assuming that the onset of flutter is coincident with frequency coalescence, calculate the flutter speed VF.

Using frequency coalescence and using Eqn 6.36 (Lewis, 2014):

( ) 02222222

222222

222422 =−−+⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−−−++−

µσ

µσ

σµµµ

σ θθ

aVVrsVxaVVrrsxr

Eqn 6.37 (Lewis, 2014):

024 =++ CBsAs where

( )

µσ

µσ

σ

µµµσ θ

θ

aVVrC

VxaVVrrB

xrA

222222

222222

22

2

22

−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−+=

−=

Eqn 6.39 (Lewis, 2014): for coalescence;B2 = 4AC

Divergence is also identifiable if required and occurs when C=0, giving Eqn 6.41 and 6.42

(Lewis, 2014):

σ 2r2 −σ2V 2

µ− 2σ

2V 2aµ

= 0 a

rVD 21+=

µ

From the above however, for coalescence:

0.23s4 + 0.3264− 0.05333V 2( )s2 − 0.0144V 2 + 0.0864 = 0

A = 0.23

B = 4ac = 0.3264− 0.05333V 2

C = 0.0864− 0.0144V 2

From above, the simple equality is evaluated for frequency coalescence condition for B:

B = 4×0.23× 0.0864− 0.0144V 2( ) = 0.3264− 0.05333V 2

0.079488− 0.013248V 2 = 0.3264− 0.05333V 2( )2

0.079488− 0.013248V 2 = 0.10653696− 0.034813824V 2 + 0.0028440889V 4

0.0028440889V 4 − 0.021565824V 2 + 0.02704896 = 0

Let x =V 2 ∴0.0028440889x2 − 0.021565824x + 0.02704896 = 0



x1,2 =−b± b2 − 4ac

2a= x1,2 =

+0.021565824± −0.021565824( )2 − 4 0.0028440889( ) 0.02704896( )2 0.0028440889( )

x1 = 3.791341403+ 2.205375443= 5.9967x2 = 3.791341403− 2.205375443=1.58596

∴V 2 = 5.9967 or 1.58596

We will be interested on the smaller value of V2

∴VF = 1.58596 =1.25935

The non-dimensional flutter speed is 1.25935, and is between that calculated in section 2(a)

above, where it was observed that the wing became unstable between V=1.2 and V=1.4.

The physical speed corresponding to this non-dimensional flutter speed can be calculated if the

following was known:

UF =VFbωθ where b is the semi-chord and ωθ is the uncoupled natural frequency for torsion.

c Calculate the corresponding non-dimensional flutter frequency. The calculation for non-dimensional transition flutter frequency is completed by substituting the non-dimensionalised velocity back into the wing flutter equation. 0.23s4 + 0.3264− 0.05333V 2( )s2 − 0.0144V 2 + 0.0864 = 0

Where: VF =1.25935 0.23s4 + 0.3264− 0.05333 VF( )2( )s2 − 0.0144 VF( )2 + 0.0864 = 0

0.23s4 + 0.3264− 0.05333 1.25935( )2( )s2 − 0.0144 1.25935( )2 + 0.0864 = 0

0.23s4 + 0.241821s2 − 0.063562 = 0

S1,22 =

−b± b2 − 4ac2a

=−0.241821± 0.241821( )2 − 4 0.23( ) −0.063562( )

2 0.23( )S12 = 0.2177S22 = −1.2691

∴S1 = ±0.46658&∴S2 = ±1.12656i



REFERENCES

HODGES, D. H. & PIERCE, G. A. 2002. Introduction to Structural Dynamics and Aeroelasticity, Cambridge University Press.

LEWIS, A. 2014. Introduction to Flutter of a Typical Section. Aeroelasticity, University of Hertfordshire, Chapter 6.

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