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
Mr A R - 10241445 – A5 – University of Hertfordshire
Aeroelasticity – Swept Wing Divergence and Flutter 3
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.
Mr A R - 10241445 – A5 – University of Hertfordshire
Aeroelasticity – Swept Wing Divergence and Flutter 4
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
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Aeroelasticity – Swept Wing Divergence and Flutter 5
(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
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Aeroelasticity – Swept Wing Divergence and Flutter 6
(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
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Aeroelasticity – Swept Wing Divergence and Flutter 7
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
Mr A R - 10241445 – A5 – University of Hertfordshire
Aeroelasticity – Swept Wing Divergence and Flutter 8
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.
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Aeroelasticity – Swept Wing Divergence and Flutter 9
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
Using quadratic equation as shown in the V=1.0 section: x1,2 =−b± b2 − 4ac
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.
Mr A R - 10241445 – A5 – University of Hertfordshire
Aeroelasticity – Swept Wing Divergence and Flutter 10
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
Using quadratic equation as shown in the V=1.0 section: x1,2 =−b± b2 − 4ac
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)
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Aeroelasticity – Swept Wing Divergence and Flutter 11
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
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Aeroelasticity – Swept Wing Divergence and Flutter 12
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
Mr A R - 10241445 – A5 – University of Hertfordshire
Aeroelasticity – Swept Wing Divergence and Flutter 13
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.