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.
simple pendulum and compound pendulum | vaibration | u.o.b | Saif al-din ali
saif aldin ali madi
سيف الدين علي ماضي
s96aif@gmail.com
In this laboratory practice the simulation of a simple pendulum was
carried out with the objective of determining the acceleration of gravity
and its uncertainty, through the data obtained in the simulation. In this
one was made the assembly of a simple pendulum through a rope, a
weight, a grader and the base for pendulums, which allowed us to
obtain through the following instruments rule and timer, data as the time
in which 20 oscillations are completed and the length of the pendulum,
taking into account the uncertainties of each instrument, these data
were organized into tables and then used in the realization of graphs
expressing the time as a function of the length of the pendulum, in
addition to calculating the acceleration of gravity and its uncertainty.
Troubleshooting and Enhancement of Inverted Pendulum System Controlled by DSP...Thomas Templin
An inverted pendulum is a pendulum that has its center of mass above its pivot point. It is often implemented with the pivot point mounted on a cart that can move horizontally and may be called a cart-and-pole system. A normal pendulum is always stable since the pendulum hangs downward, whereas the inverted pendulum is inherently unstable and trivially underactuated (because the number of actuators is less than the degrees of freedom). For these reasons, the inverted pendulum has become one of the most important classical problems of control engineering. Since the 1950s, the inverted-pendulum benchmark, especially the cart version, has been used for the teaching and understanding of the use of linear-feedback control theory to stabilize an open-loop unstable system.
The objectives of this project are to:
• Focus on hardware and software troubleshooting and enhancement of an inverted-pendulum system controlled by a DSP28355 microprocessor and CCSv7.1 software.
• Use the swing-up strategy to move the pendulum into the unstable upward position (‘saddle’). The cart/pole system employs linear bearings for back-and-forward motion. The motor shaft has a pinion gear that rides on a track permitting the cart to move in a linear fashion. Both rack and pinion are made of hardened steel and mesh with a tight tolerance. The rack-and-pinion mechanism eliminates undesirable effects found in belt-driven and free-wheel systems, such as slippage or belt stretching, ensuring consistent and continuous traction.
• The motor shaft is coupled to a high-resolution optical encoder that accurately measures the position of the cart. The angle of the pendulum is also measured by an optical encoder, and the system employs an LQR controller to stabilize the pendulum rod at the unstable-equilibrium position.
• Addition of real-time status reporting and visualization of the system.
For the project, the Quanser High Frequency Linear Cart (HFLC) was used. The HFLC system consists of a precisely machined solid aluminum cart driven by a high-power 3-phase brushless DC motor. The cart slides along two high-precision, ground-hardened stainless steel guide rails, allowing for multiple turns and continuous measurement over the entire range of motion.
Our team implemented a control strategy that consists of a linear stabilizing LQR controller, proportional-integral swing-up control, and a supervisory coordinator that determines the control strategy (LQR or swing-up) to be used at any given time. The function of the linear stabilizer is to stabilize the system when it is in the vicinity of the unstable equilibrium. When the pendulum is in its natural state (straight-down stable-equilibrium node), the swing-up controller provides the cart/pendulum system with adequate energy to move the pendulum to the unstable equilibrium inside the “region of attraction” in which the linearized LQR controller is functional.
Analytic Model of Wind Disturbance Torque on Servo Tracking AntennaIJMER
International Journal of Modern Engineering Research (IJMER) is Peer reviewed, online Journal. It serves as an international archival forum of scholarly research related to engineering and science education.
Mechanics of Aircraft Structures solution manual C.T. Sun 2nd edDiego Fung
Designed to help students get a solid background in structural mechanics and extensively updated to help professionals get up to speed on recent advances This Second Edition of the bestselling textbook Mechanics of Aircraft Structures combines fundamentals, an overview of new materials, and rigorous analysis tools into an excellent one-semester introductory course in structural mechanics and aerospace engineering. It's also extremely useful to practicing aerospace or mechanical engineers who want to keep abreast of new materials and recent advances. Updated and expanded, this hands-on reference covers: * Introduction to elasticity of anisotropic solids, including mechanics of composite materials and laminated structures * Stress analysis of thin-walled structures with end constraints * Elastic buckling of beam-column, plates, and thin-walled bars * Fracture mechanics as a tool in studying damage tolerance and durability Designed and structured to provide a solid foundation in structural mechanics, Mechanics of Aircraft Structures, Second Edition includes more examples, more details on some of the derivations, and more sample problems to ensure that students develop a thorough understanding of the principles.
simple pendulum and compound pendulum | vaibration | u.o.b | Saif al-din ali
saif aldin ali madi
سيف الدين علي ماضي
s96aif@gmail.com
In this laboratory practice the simulation of a simple pendulum was
carried out with the objective of determining the acceleration of gravity
and its uncertainty, through the data obtained in the simulation. In this
one was made the assembly of a simple pendulum through a rope, a
weight, a grader and the base for pendulums, which allowed us to
obtain through the following instruments rule and timer, data as the time
in which 20 oscillations are completed and the length of the pendulum,
taking into account the uncertainties of each instrument, these data
were organized into tables and then used in the realization of graphs
expressing the time as a function of the length of the pendulum, in
addition to calculating the acceleration of gravity and its uncertainty.
Troubleshooting and Enhancement of Inverted Pendulum System Controlled by DSP...Thomas Templin
An inverted pendulum is a pendulum that has its center of mass above its pivot point. It is often implemented with the pivot point mounted on a cart that can move horizontally and may be called a cart-and-pole system. A normal pendulum is always stable since the pendulum hangs downward, whereas the inverted pendulum is inherently unstable and trivially underactuated (because the number of actuators is less than the degrees of freedom). For these reasons, the inverted pendulum has become one of the most important classical problems of control engineering. Since the 1950s, the inverted-pendulum benchmark, especially the cart version, has been used for the teaching and understanding of the use of linear-feedback control theory to stabilize an open-loop unstable system.
The objectives of this project are to:
• Focus on hardware and software troubleshooting and enhancement of an inverted-pendulum system controlled by a DSP28355 microprocessor and CCSv7.1 software.
• Use the swing-up strategy to move the pendulum into the unstable upward position (‘saddle’). The cart/pole system employs linear bearings for back-and-forward motion. The motor shaft has a pinion gear that rides on a track permitting the cart to move in a linear fashion. Both rack and pinion are made of hardened steel and mesh with a tight tolerance. The rack-and-pinion mechanism eliminates undesirable effects found in belt-driven and free-wheel systems, such as slippage or belt stretching, ensuring consistent and continuous traction.
• The motor shaft is coupled to a high-resolution optical encoder that accurately measures the position of the cart. The angle of the pendulum is also measured by an optical encoder, and the system employs an LQR controller to stabilize the pendulum rod at the unstable-equilibrium position.
• Addition of real-time status reporting and visualization of the system.
For the project, the Quanser High Frequency Linear Cart (HFLC) was used. The HFLC system consists of a precisely machined solid aluminum cart driven by a high-power 3-phase brushless DC motor. The cart slides along two high-precision, ground-hardened stainless steel guide rails, allowing for multiple turns and continuous measurement over the entire range of motion.
Our team implemented a control strategy that consists of a linear stabilizing LQR controller, proportional-integral swing-up control, and a supervisory coordinator that determines the control strategy (LQR or swing-up) to be used at any given time. The function of the linear stabilizer is to stabilize the system when it is in the vicinity of the unstable equilibrium. When the pendulum is in its natural state (straight-down stable-equilibrium node), the swing-up controller provides the cart/pendulum system with adequate energy to move the pendulum to the unstable equilibrium inside the “region of attraction” in which the linearized LQR controller is functional.
Analytic Model of Wind Disturbance Torque on Servo Tracking AntennaIJMER
International Journal of Modern Engineering Research (IJMER) is Peer reviewed, online Journal. It serves as an international archival forum of scholarly research related to engineering and science education.
Mechanics of Aircraft Structures solution manual C.T. Sun 2nd edDiego Fung
Designed to help students get a solid background in structural mechanics and extensively updated to help professionals get up to speed on recent advances This Second Edition of the bestselling textbook Mechanics of Aircraft Structures combines fundamentals, an overview of new materials, and rigorous analysis tools into an excellent one-semester introductory course in structural mechanics and aerospace engineering. It's also extremely useful to practicing aerospace or mechanical engineers who want to keep abreast of new materials and recent advances. Updated and expanded, this hands-on reference covers: * Introduction to elasticity of anisotropic solids, including mechanics of composite materials and laminated structures * Stress analysis of thin-walled structures with end constraints * Elastic buckling of beam-column, plates, and thin-walled bars * Fracture mechanics as a tool in studying damage tolerance and durability Designed and structured to provide a solid foundation in structural mechanics, Mechanics of Aircraft Structures, Second Edition includes more examples, more details on some of the derivations, and more sample problems to ensure that students develop a thorough understanding of the principles.
A paper which analyses the motion of a satellite launch vehicle, a rocket, from the moment it is launched till when it is placed into orbit. The paper contains derivations for equations for thrust, mass, mass loss, distance, velocity, burnout time and burnout velocity
Finite Element Analysis of the Aeroelasticity Plates Under Thermal and Aerody...Mohammad Tawfik
At the flight condition of panel-flutter (usually supersonic flight conditions), the phenomenon is associated with elevated temperatures, produced from the aerodynamic heating through the boundary layer friction and the presence of shock waves. This heating adds to the complexity of the problem by introducing panel stiffness reduction and thermal loading, which might also be associated with post-buckling deflection. In the following, a literature review of panel-flutter analysis and control topics is presented.
#WikiCourses
https://wikicourses.wikispaces.com/TopicX+Nonlinear+Solid+Mechanics
https://eau-esa.wikispaces.com/Topic+Nonlinear+Solid+Mechanics
Tractor-trailers are used all across America to transport cargo, but are not designed with fuel efficiency in mind. Therefor, there is an incentive for companies to invest in making their tractor-trailers more aerodynamic in order to save on fuel costs. I go into the testing and methodology of how my team and I decided to tackle the problem of reducing the coefficient of drag on tractor-trailers by implementing air channeling devices (ACDs). Then, I cover the results from our experiments and our ACD recommendation.
Presentation Acme engineering- Two Stage Turbo Shaft Engine- Pratt and WhittneySiddharth Salkar
Design and Analysis of Two Stage Turbofan Gas Turbine Engine-Pratt and Whitney
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Pipe insulation efficiency study unit |HEAT TRANSFER LaboratorySaif al-din ali
SAIF ALDIN ALI MADIN
سيف الدين علي ماضي
S96aif@gmail.com
1. Experiment Name: Pipe Insulation Efficiency Study Unit
2. Experiment Aim: The study unit for the assessment of thermal
insulation efficiency permits investigating the effect of thermal
insulation of steam pipes, the unit consist of a set of four pipes, three of
which are covered with insulating materials, placed vertically. Steam is
fed by means of an overhead manifold
3. Composition:
The lagging of piping unit includes:
1. - 3 test pipes covered with materials which thermal conductivity
coefficient is different.
2. One pipe without covering,
3. Manual control valve on steam inlet,
4. Bourdon pressure gauge on steam inlet
5. Thermometer in steam inlet
6. Condenser discharge system,
7. Graduated containers of glass to measure the condensate,
Free heat convection |HEAT TRANSFER LaboratorySaif al-din ali
SAIF ALDIN ALI MADIN
سيف الدين علي ماضي
S96aif@gmail.com
Experiment Name: Free Heat Convection from a Horizontal Heated
Cylinder
2. Experiment Aim:
1. Calculating the free heat convection coefficient (ℎ푁퐶) for a
horizontal heated solid cylinder.
2. Find the relationship between RaNo. And NuNo. for fluid flow
around a cylinder
Solution manual for water resources engineering 3rd edition - david a. chinSalehkhanovic
Solution Manual for Water-Resources Engineering - 3rd Edition
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This solution manual include all problems (Chapters 1 to 17) of textbook. in second section of solution manual, Problems answered using mathcad software .
Calibrating a CFD canopy model with the EC1 vertical profiles of mean wind sp...Stephane Meteodyn
For some projects, applying the basic rules of EC1 is not sufficient, and it is required to get a more accurate estimation of the wind speed on the construction site. This can be done by using computational fluid dynamics codes which have the advantage, both to take into account of the terrain inhomogeneity and to calculate 3D orographic effects. In this way, the orography and roughness effects are coupled as they are in the real world. However, applying CFD computations must be in coherence with EC1 code. Then it is necessary to calibrate the ground friction for low roughness terrains as well as the drag force and turbulence production in case of high roughness lengths due to the presence of a canopy (forests or built areas). That is the condition for such methods to be commonly used and agreed by Building Control Officers. In this mind, TopoWind has been developed especially for wind design applications and can be a very useful, practical and objective tool for wind design engineers. The canopy model implemented in TopoWind has been calibrated in order to get the mean wind and turbulence profiles as defined in the EC1 for standard terrains. In this way, TopoWind computations satisfy the continuity between the EC1 values for homogeneous terrains and the more complex cases involving inhomogeneous roughness or orographic effects
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.