This document summarizes a student's analysis of aeroelastic divergence and flutter for a swept wing. In part 1, the student calculates the required stiffness (K) to achieve a specified divergence speed for a given wing configuration. A parametric study shows that increasing flexural rigidity (EI), torsional rigidity (GJ), or stiffness (K) increases divergence speed as expected. With K=0, the student estimates possible forward sweep angles that would achieve the required divergence speed. In part 2, the student considers flutter of the flexible wing and calculates flutter speeds for different configurations.

1. 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

2. 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 =
1
2
ρV2
=
1
2
×1.226×1402
=12014.8Pa
qD =
π2GJ 1−κ2
( )
4ecal2
cos2
Λ( ) 1−κ
GJ
EI
tan Λ( )−
3π 2
76
l
e
GJ
EI
tan Λ( )−κ
EI
GJ
#
$
%
&
'
(
)
*
+
,+
-
.
+
/+
Where, κ =
K
EI ⋅GJ
Substituting:
qD =
π2GJ 1−
K
EI ⋅GJ
#
$
%
&
'
(
2#
$
%%
&
'
((
4ecal2
cos2
Λ( ) 1−
K
EI ⋅GJ
#
$
%
&
'
(
GJ
EI
tan Λ( )−
3π 2
76
l
e
GJ
EI
tan Λ( )−
K
EI ⋅GJ
#
$
%
&
'
(
EI
GJ
*
+
,
-
.
/
0
1
2
32
4
5
2
62
Substituting variables:
qD =
π2 ×18000 1−
K
15000×18000
#
$
%
&
'
(
2#
$
%%
&
'
((
4×0.075×0.3×5.9×2.52
×cos2
−25( ) 1−
K
15000×18000
#
$
%
&
'
(
18
15
tan Λ( )−
3π 2
76
l
e
18
15
tan Λ( )−
K
15000⋅18000
#
$
%
&
'
(
15
18
+
,
-
.
/
0
1
2
3
43
5
6
3
73

3. Mr A R - 10241445 – A5 – University of Hertfordshire
Aeroelasticity – Swept Wing Divergence and Flutter 3
Substituting and rearranging:
qD =
18000π2 −
18000π2K2
2.7×108
#
$
%
%
&
'
(
(
2.726 K 8.96842×10−4
( )+8.26678{ }
qD =
177652.8792 − 6.57974×10−4
K2
( )
K 2.44479×10−3
( )+ 22.53524
K2
+
12014.8
6.57974×10−4
K 2.44479×10−3
( )+ 22.53524( )−
177652.8792
6.57974×10−4
= 0
K2
+ K × 4.46426×104
+1.415×108
= 0
Using quadratic equation: K1,2 =
−b± b2
− 4ac
2a
Where
a =1
b = 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.

4. 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

5. Mr A R - 10241445 – A5 – University of Hertfordshire
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

6. Mr A R - 10241445 – A5 – University of Hertfordshire
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

7. Mr A R - 10241445 – A5 – University of Hertfordshire
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−
3π 2
76
l
e
GJ
EI
tan Λ( )
#
$
%
&
'
(
Rearranging:
qD 4ecal2
GJπ2
=
1+ tan2
Λ( )
1−
3π 2
76
l
e
GJ
EI
tan Λ( )
qD 4ecal2
GJπ2
1−
3π 2
lGJ
76eEI
tan Λ( )
#
$
%
&
'
( =1+ tan2
Λ( )
qD 4ecal2
GJπ2
−1= tan2
Λ( )+
qD 4ecal2
GJπ2
3π 2
lGJ
76eEI
tan Λ( )
0 = tan2
Λ( )+
3qDcal3
19EI
tan Λ( )−
qD 4ecal2
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 =
1
2
ρV2
=
1
2
×1.226×1402
=12014.8Pa

8. 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 − 0.05333V2
( )s2
− 0.0144V2
+ 0.0864 = 0
Where V is a non-dimensionalised air speed related to the actual air speed by: V =
U
bωθ
And s is a non-dimensionalised root of the flutter equation such that the heave and pitch motions
are:
h = hesωθt
θ =θesωθt
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 − 0.05333V2
( )s2
− 0.0144V2
+ 0.0864 = 0
0.23s4
+ 0.3264− 0.05333 1( )
2
( )s2
− 0.0144 1( )
2
+ 0.0864 = 0
Let x = s2
∴0.23x2
+ 0.27307x + 0.072 = 0
Using quadratic equation: x1,2 =
−b± b2
− 4ac
2a
Where
a = 0.23
b = 0.27307
c = 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.

9. Mr A R - 10241445 – A5 – University of Hertfordshire
Aeroelasticity – Swept Wing Divergence and Flutter 9
Set V=1.2 as shown:
0.23s4
+ 0.3264 − 0.05333V2
( )s2
− 0.0144V2
+ 0.0864 = 0
0.23s4
+ 0.3264− 0.05333 1.2( )
2
( )s2
− 0.0144 1.2( )
2
+ 0.0864 = 0
Let x = s2
∴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 − 0.05333V2
( )s2
− 0.0144V2
+ 0.0864 = 0
0.23s4
+ 0.3264− 0.05333 1.4( )
2
( )s2
− 0.0144 1.4( )
2
+ 0.0864 = 0
Let x = s2
∴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
s1
2
= −0.48233304+ 0.14245689i ; ∴s1 = ± a1 + b1i( )
s2
2
= −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.

10. 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 − 0.05333V2
( )s2
− 0.0144V2
+ 0.0864 = 0
0.23s4
+ 0.3264− 0.05333 1.6( )
2
( )s2
− 0.0144 1.6( )
2
+ 0.0864 = 0
Let x = s2
∴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
s1
2
= −0.41250434+ 0.21211564i ; ∴s1 = ± a1 + b1i( )
s2
2
= −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)

11. Mr A R - 10241445 – A5 – University of Hertfordshire
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):
( ) 0222
2222
222
222
222422
=−−+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−−−++−
µ
σ
µ
σ
σ
µµµ
σ θ
θ
aVV
rs
VxaVV
rrsxr
Eqn 6.37 (Lewis, 2014):
024
=++ CBsAs where
( )
µ
σ
µ
σ
σ
µµµ
σ θ
θ
aVV
rC
VxaVV
rrB
xrA
2222
22
222
222
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 2
r2
−
σ 2
V2
µ
− 2
σ 2
V2
a
µ
= 0
a
rVD
21+
=
µ
From the above however, for coalescence:
0.23s4
+ 0.3264 − 0.05333V2
( )s2
− 0.0144V2
+ 0.0864 = 0
A = 0.23
B = 4ac = 0.3264− 0.05333V2
C = 0.0864− 0.0144V2
From above, the simple equality is evaluated for frequency coalescence condition for B:
B = 4× 0.23× 0.0864 − 0.0144V2
( ) = 0.3264 − 0.05333V2
0.079488− 0.013248V2
= 0.3264− 0.05333V2
( )
2
0.079488− 0.013248V2
= 0.10653696 − 0.034813824V2
+ 0.0028440889V 4
0.0028440889V 4
− 0.021565824V2
+ 0.02704896 = 0
Let x = V2
∴0.0028440889x2
− 0.021565824x + 0.02704896 = 0

12. Mr A R - 10241445 – A5 – University of Hertfordshire
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= 5.9967
x2 = 3.791341403− 2.205375443=1.58596
∴V2
= 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 − 0.05333V2
( )s2
− 0.0144V2
+ 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
S1,2
2
=
−b± b2
− 4ac
2a
=
−0.241821± 0.241821( )
2
− 4 0.23( ) −0.063562( )
2 0.23( )
S1
2
= 0.2177
S2
2
= −1.2691
∴S1 = ±0.46658
&∴S2 = ±1.12656i

13. 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.