Fluttering aerofoil with trailing edge flap

1. Simulation of the dynamics of a two-dimensional fluttering aerofoil
with a trailing edge flap
Coursework 1
DENM032
Saqib Shah

2. Introduction
In Aero elasticity we try to understand the effects of the aerodynamic and structural loads on
many different structures including Aircraft. Prior to this study we have assumed that Aircraft
and most structures are perfectly rigid, however this is not the case in real life and structures
can buckle and hence fail if not engineered properly. With respect to Aircraft we experience
phenomena called “fluttering” which will be studied in this report. Flutter can be described as
“an unstable oscillation which can lead to destruction. Flutter can occur on fixed surfaces,
such as the wing or the stabilizer, as well as on control surfaces such as the aileron or the
elevator for instance.” (Aviation) Flutter doesn’t necessarily have to be on associated with
wings of an Aircraft; however it can also be associated with other large structural components
like Bridges and Buildings. As the Aircraft increases with speed, we need dampers within the
Aircraft to minimise the energy for stability; however the damping process is not always
successful hence we need to understand the dynamics and characteristics of flutter for a
successful design of an Aircraft, whether it is undergoing sub-sonic speeds or super-sonic
speeds.
Aims
The aim of the simulation is to investigate and understand the factors governing the response
of a three degree of freedom aerofoil model with a trailing edge flap and hence relate these to
basic analytical techniques. (Vepa) By using Mat lab/Simulink we will simulate the dynamics
of a three degree of freedom aerofoil with a trailing edge flap. This will be done by inputting
the complex equations of motion which will eventually give us results which can be analysed.
Background theory
The Aircraft wing which we modelled was assumed to be a flat plate with mass 𝑚. The
aerofoil section slides across a plane, where the three degrees of freedom are described by the
torsion of the aerofoil, the angle 𝛼 which is inevitably the angle of attack between the aerofoil and
the horizontal positive axis. We have two springs each with a different stiffness which can be denoted
as 𝐾 𝐴 and 𝐾 𝐵. The Chord length of the wing is 2b and the origin of the coordinate system can
be taken from the elastic Centre of the area which is assumed to be at EA. The figure below
shows the coordinate system of the aerofoil.

3. Figure 1: Section of aerofoil with the coordinate systems
Firstly we need to derive the potential and kinetic energies for the above system. After doing
this we need to use the energy equation, by using the Lagrangian approach to compute the
difference in kinetic and potential energies.
𝐿 = 𝑇 − 𝑉
Where 𝐿 is the Lagrangian, 𝑇 is the sum of all the kinetic energies and 𝑉 is sum of all the potential
energies of the system. For simplicity we will compute the potential energy first.
The translational potential energy of the two springs is given by:
𝑉1 =
1
2
𝑘ℎℎ2
As the aerofoil rotates, the potential energy of the rotating aerofoil needs to be taken into
account also, hence:
𝑉2 =
1
2
𝑘 𝛼 𝛼2
And the potential energy for the trailing edge flap is given by:
𝑉3 =
1
2
𝑘 𝛽 𝛽2
Hence by adding up all the potential energy terms, we obtain the following:
𝑉 =
1
2
(𝑘ℎℎ2 + 𝑏2𝑘 𝛼 𝛼2 + 𝑘 𝛽 𝛽2)
The next step is to compute the kinetic energies of the above system.

4. The translational vertical motion of the wing with the trailing edge flap locked gives us the
kinetic energy at the centre of mass, hence:
𝑇1 =
1
2
𝑚(ℎ̇ + 𝑏𝑥 𝛼 𝛼̇ )
2
As we have a rotational effect on the aerofoil, we need to take into account the rotational component
of the kinetic energy, hence:
𝑇2 =
1
2
𝐼 𝐶𝑀 𝛼̇ 2
Where 𝐼 𝐶𝑀 is the moment of inertia relative to the centre of mass.
The mass moment of inertia due to the parallel axis theorem about the centre of mass is
related to the mass moment of inertia about the elastic axis. The relationship can be shown
below as follows:
𝐼 𝐶𝑀 = ( 𝐼 𝛼 − 𝑚( 𝑏𝑥 𝛼)2)
The trailing edge flap has a great effect on the total kinetic energy. The vertical displacement
of the trailing edge flap can be worked out as:
ℎ 𝛽 = ℎ + 𝛼( 𝑐 − 𝑎 + 𝑥 𝑏) 𝑏
The deflection of the trailing edge can be given as:
𝑏𝑥 𝛽 𝛽
Hence the kinetic energy due to the trailing edge flap can be expressed as:
𝑇3 =
1
2
𝑚 𝛽[(ℎ̇ 𝛽 + 𝑏𝑥 𝛽 𝛽̇ )
2
− ℎ̇ 𝛽
2
]
Now we can add up all the components of the kinetic energy and end up with the total kinetic
energy as follows:
𝑇 =
1
2
𝑚(ℎ̇ + 𝑏𝑥 𝛼 𝛼̇)2 +
1
2
𝐼 𝐶𝑀 𝛼̇2 +
1
2
𝑚 𝛽[(ℎ̇ 𝛽 + 𝑏𝑥 𝛽 𝛽̇)
2
− ℎ̇ 𝛽
2
]
Now the equations of motion can be obtained using the expressions for the total energy into
the Lagrangian equation.
𝑑
𝑑𝑡
(
𝜕𝑇
𝜕𝑞̇
) −
𝜕𝐿
𝜕 𝑞
= 0
By substituting the values for the differential above with 𝜕𝑞̇ being equal to 𝜕ℎ̇ and 𝜕 𝑞 being
equal to 𝜕ℎ, we obtain the following expression.

5. 𝑑
𝑑𝑡
𝜕 [[
1
2
𝑚(ℎ̇ + 𝑏𝑥 𝛼 𝛼̇ )2
+
1
2
𝐼 𝐶𝑀 𝛼̇2
+
1
2
𝑚 𝛽 [(ℎ̇ 𝛽 + 𝑏𝑥 𝛽 𝛽̇ )
2
− ℎ̇ 𝛽
2
]] + [
1
2
𝑘ℎℎ2
+
1
2
𝑘 𝛼 𝛼2
+
1
2
𝑘 𝛽 𝛽2
]]
𝜕ℎ̇
−
𝜕[[[
1
2
𝑚(ℎ̇ + 𝑏𝑥 𝛼 𝛼̇ )2
+
1
2
𝐼 𝐶𝑀 𝛼̇ 2
+
1
2
𝑚 𝛽 [(ℎ̇ 𝛽 + 𝑏𝑥 𝛽 𝛽̇)
2
− ℎ̇ 𝛽
2
]]] + [
1
2
𝑘ℎℎ2
+
1
2
𝑏
2
𝑘 𝛼 𝛼2
+
1
2
𝑏
2
𝑘 𝛽 𝛽2
]]
𝜕ℎ
= 0
By simplifying the equation above we obtain the expressions below which are in the inertia
coupled form.
𝑚ℎ̈ + 𝑚𝑏𝑥 𝛼 𝛼̈ + 𝑘ℎℎ = 0
𝑚𝑏𝑥 𝛼ℎ̈ + 𝐼 𝛼 𝛼̈ + 𝑘 𝛼 𝛼 = 0
Given that the aerodynamic restoring force and restoring moment about the elastic axis, are 𝐿
and 𝑀 respectively, the disturbance force and moment are 𝐿 𝐺 and 𝑀 𝐺, show that the
equations of motion in inertia coupled form are: (Vepa)
𝑚ℎ̈ + 𝑚𝑏𝑥 𝛼 𝛼̈ + 𝑘ℎℎ + 𝐿 = 𝐿 𝐺
And
𝑚𝑏𝑥 𝛼ℎ̈ + 𝐼 𝛼 𝛼̈ + 𝑘 𝛼 𝛼 + 𝑀 = 𝑀 𝐺
Normally the restoring lift and moment may be shown as convolution integrals. However by
making certain constraining assumptions that the motion is purely simple harmonic, we can
show that the equations are. (Vepa)
[
𝐿
𝑀
] = 𝑀 𝑎 [ℎ̈
𝛼̈
] + 𝐶 𝑎 [ℎ̇
𝛼̇
] + 𝐾 𝑎 [
ℎ
𝛼
]
Where T. Theodoresen showed that:
𝑀 𝑎 = 𝜋𝜌𝑏3
[
1
𝑏
−𝑎
−𝑎 𝑏 ( 𝑎2
+
1
8
)
]

6. 𝐶 𝑎 = 𝜋𝜌𝑏3
𝑈[
2𝐶( 𝑘)
𝑏
1 + 𝐶( 𝑘)(1 − 2𝑎)
−𝐶( 𝑘)(1 + 2𝑎) 𝑏 (
1
2
− 𝑎)(1 − 𝐶( 𝑘)(1+ 2𝑎))
]
𝐾 𝑎 = 𝜋𝜌𝑏𝑈2
𝐶( 𝑘)[
0 2
0 −𝑏(1 + 2𝑎)]
The constant 𝐶( 𝑘) which is shown in the above equations is known as the “Theodoresen
function.” The non-dimensional parameter 𝑘 =
𝜔𝑏
𝑈
is known as the reducing velocity, where
𝑈 is the speed of the airflow relative to the aerofoil. As the As the aerodynamic stiffness matrix alone
is a function of the square of the velocity𝑈2
, one may ignore the effects of aerodynamic inertia and
damping in the first instance. Thus approximating C(k) as equalto unity, one may express the
equations of motion as a set of coupled second order matrix equations representing a vibrating system;
i.e. in the form, (Vepa)
Mẍ + Cẋ + Kx = F(t)
In the first instance one should identify the matrices, M, C, K, x and F(t).
Dividing each of the equations by 𝑚𝑏2 and it follows that in non-dimensional matrix form,
the equations of motion in the presence of an external disturbance force and an external
disturbance moment are: (Vepa)
[
1 𝑥 𝑎
𝑥 𝑎 𝐼 𝑎 𝑚𝑏2⁄
][ℎ̈ 𝑏⁄
𝛼̈
] + [
𝑘ℎ 𝑚⁄ 0
0 𝑘 𝛼 𝑚𝑏2⁄
][
ℎ 𝑏⁄
𝛼
] + [
𝐿̅ 𝑏
𝑀̅
] = [
𝐿̅ 𝐺 𝑏
𝑀̅ 𝐺
]
The aerodynamic restoring moments are,
[ 𝐿̅ 𝑏
𝑀̅] =
𝜋𝜌𝑏2
𝑚
(
𝑈
𝑏
)
2
( 𝑀̃ 𝑎 [
ℎ̈
𝑏
𝛼̈
] + 𝐶̃ 𝑎 [
ℎ̇
𝑏
𝛼̇
] + 𝐾̃𝑎 [
ℎ
𝑏
𝛼
])
Where:
𝑀̃ 𝑎 = (
𝑏
𝑈
)
2
[
1 −𝑎
−𝑎 ( 𝑎2
+
1
8
)
]
𝐶̃ 𝑎 =
𝑏
𝑈
[
0 1
0 (
1
2
− 𝑎)
] +
𝑏
2𝑈
𝐶( 𝑘)[
4 2(1 − 2𝑎)
−2(1 + 2𝑎) −(1 − 2𝑎)(1 + 2𝑎)
]
𝐾̃ 𝑎 = 2𝐶( 𝑘)[
0 1
0 −(
1
2
+ 𝑎)
]

7. 𝐿̅ 𝐺 and 𝑀̅ 𝐺 are an external non-dimensional disturbance vertical force and an external
disturbance anti-clockwise moment due to a typical gust. The equations of motion of the
aerofoil including the trailing edge flap may be expressed as: (Vepa)
[
𝑚𝑏2 𝑚𝑏2 𝑥 𝛼 𝑚 𝛽 𝑏2 𝑥 𝛽
𝑚𝑏2 𝑥 𝛼 𝐼 𝛼 𝑚 𝛽 𝑏2 𝑥 𝛽( 𝑐 − 𝑎) + 𝐼𝛽
𝑚 𝛽 𝑏2 𝑥 𝛽 𝑚 𝛽 𝑏2 𝑥 𝛽( 𝑐 − 𝑎) + 𝐼𝛽 𝐼𝛽
]
[
ℎ̈
𝑏
𝛼̈
𝛽̈]
+ [
𝑘ℎ 𝑏2 0 0
0 𝑘 𝛼 0
0 0 𝑘 𝛽
][
ℎ
𝑏
𝛼
𝛽
][
𝐿𝑏
−𝑀
−𝑀𝛽
] = [
0
0
0
]
The aerodynamic equations for the restoring moment can hence be shown as:
[
𝐿𝑏
−𝑀
−𝑀 𝛽
] = 𝜋𝜌𝑏3
𝑈2
( 𝑀̃ 𝑎 [
ℎ̈
𝑏
𝛼̈
𝛽̈
] + 𝐶̃ 𝑎 [
ℎ
𝑏
̇
𝛼̇
𝛽̇
] + 𝐾̃ 𝑎 [
ℎ
𝑏
𝛼
𝛽
])
Where the values of 𝑀̃ 𝑎, 𝐶̃ 𝑎 and 𝐾̃ 𝑎 are shown as:
𝑴̃ 𝒂 = (
𝑏
𝑈
)
2
[
1 −𝑎
𝑇1
𝜋
−𝑎 ( 𝑎2
+
1
8
) 2
𝑇13
𝜋
𝑇1
𝜋
2
𝑇13
𝜋
−
𝑇3
𝜋2 ]
, 𝐶̃ 𝑎 = 𝐶̃ 𝑎𝑛𝑐 + 𝐶̃ 𝑎𝑐−𝑛𝑡𝑙 + 𝐶̃ 𝑎𝑐−𝑡𝑙 𝐶( 𝑘),
𝑪̃ 𝒂𝒏𝒄 =
𝑏
𝑈
[
0 1 −
𝑇4
𝜋
−1 0
𝑇15
𝜋
𝑇4
𝜋
−
𝑇15
𝜋
0 ]
, 𝐶̃ 𝑎𝑐−𝑛𝑡𝑙 =
𝑏
𝑈
[
0
1
−
𝑇4
𝜋
][1 (
1
2
− 𝑎)
𝑇11
2𝜋
]
𝑪̃ 𝒂𝒄−𝒕𝒍 =
𝑏
𝑈
[
2
−(1 + 2𝑎)
𝑇12
𝜋
] [1(
1
2
− 𝑎)
𝑇11
2𝜋
]
And
𝑲̃ 𝒂 = 𝑲̃ 𝒂𝒏𝒄 + 𝑲̃ 𝒂𝒄−𝒏𝒕𝒍 + 𝑲̃ 𝒂𝒄−𝒕𝒍 𝐶( 𝑘), 𝑲̃ 𝒂𝒏𝒄 =
[
0 0 0
0 −1
𝑇4
𝜋
0
𝑇4
𝜋
𝑇5
𝜋2]

8. 𝑲̃ 𝒂𝒄−𝒏𝒕𝒍 = [
0
1
−
𝑇4
𝜋
][0 1
𝑇10
𝜋
], 𝐾̃ 𝑎𝑐−𝑡𝑙 = 2
[
1
− (
1
2
+ 𝑎)
𝑇12
𝜋 ]
= [0 1
𝑇10
𝜋
]
The equations of motion can therefore finally be expressed as:
(M + 𝑆𝑞M̃ 𝑎)ẍ + (C + 𝑆𝑞C̃ 𝑎)ẋ + (K + 𝑆𝑞K̃ 𝑎)x = −[ 𝐿̅ 𝐺 𝑀̅ 𝐺 𝑀̅ 𝛽𝐺] 𝑇
Where:
𝑆 = 2𝜋𝑏2, 𝑞 =
1
2
𝜌𝑈2and x = [ℎ/𝑏 𝛼 𝛽] 𝑇
Results and Discussion
As the equations of motion have now been completely derived we can use this to simulate our
problem. The equations of motion can first be expressed as shown below for the two degree
of freedom system:
[ℎ̈ 𝑏⁄
𝛼̈
] = [
1 𝑥 𝑎
𝑥 𝑎 𝐼 𝑎 𝑚𝑏2⁄
]
−1
+ {[
𝐿̅ 𝐺 𝑏
𝑀̅ 𝐺
] − [
𝑘ℎ 𝑚⁄ 0
0 𝑘 𝑎 𝑚𝑏2⁄
][
ℎ 𝑏⁄
𝛼
]}
Which can also be noted as:
[ℎ̈/𝑏
𝛼̈
] = [
1 𝑥 𝛼
𝑥 𝛼 𝑟 𝛼
2 ]
−1
([
𝐿̅ 𝐺 𝑏
𝑀̅ 𝐺
]− [
𝜔ℎ0
2
0
0 𝑟 𝛼
2
𝜔 𝛼0
2 ][
ℎ/𝑏
𝛼
])
And:
[ℎ̈ 𝑏⁄
𝛼̈
] =
1
𝑟 𝑎
2 − 𝑥 𝑎
2
[
𝑟 𝑎
2
−𝑥 𝑎
−𝑥 𝑎 1
]
−1
{[
𝐿̅ 𝐺 𝑏
𝑀̅ 𝐺
] − [
𝜔ℎ0
2
0
0 𝑟 𝑎
2
𝜔 𝑎0
2 ][
ℎ 𝑏⁄
𝛼
]}
≡
1
𝑅 − 𝑆2 [
𝑅 −𝑆
−𝑆 1
] {[
𝐿̅ 𝐺 𝑏
𝑀̅ 𝐺
] − [
𝐾 0
0 𝑃
][ℎ 𝑏⁄
𝛼
]}