skma4253 aircraft design ii group two · 2020. 7. 16. · ahmad amirul amin bin abdul wahab...
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1
SKMA4253 AIRCRAFT DESIGN II
GROUP TWO
RC-PLANE GROUP PROJECT (FINAL DRAFT)
LEADER NAME:
MOHAMAD ASHRAF BIN AB HAN
LECTURERS NAME:
DR. ING. MOHD. NAZRI BIN MOHD. NASIR
DR. WAN ZAIDI BIN WAN OMAR
i
GROUP MEMBERS
1. AHMAD AMIRUL AMIN BIN ABDUL WAHAB (A16KM0017)
2. AHMAD FAKHRURRAZI BIN ABDUL LATIF (A16KM0022)
3. ALEXANDER JONG JIUN WEI (A16KM0033)
4. CHENG WEI QIN (A16KM0062)
5. LEONG JOE YEE (B17KM0011)
6. MOHAMAD ASHRAF BIN AB HAN (A16KM0176)
7. MOHAMAD AZMIL BIN BAHJAM (A16KM0178)
8. MUHAMAD AZAMUDDIN BIN MOHD TAHIR (A16KM0222)
9. MUHAMAD HANIF SHAH BIN KHAIRUL ANUAR (A16KM0224)
10. MUHAMMAD HAFIZ AKMAL BIN HARIS FADZILAH (A16KM0289)
11. MUHAMMAD NASRUL BIN YAZI (A16KM0496)
12. MUHAMMAD SHAFRIEZ BIN MD SHAIFUL (A16KM0330)
13. NURUL ‘AFIFAH BINTI ZULKEFLI (A16KM0390)
14. SAMALADEWI A/P MURUKAPPAN (A16KM0417)
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TABLE OF CONTENTS
TITLE PAGE
TABLE OF CONTENTS i
CHAPTER 1 INTRODUCTION 1
1.1 Objectives 1
1.2 Gantt chart 2
1.3 Flowchart 3
CHAPTER 2 CONCEPTUAL DESIGN 5
2.1 What is Conceptual Design? 5
2.2 Design Selection 5
2.2.1 Wing Configuration 5
2.2.2 Fuselage Configuration 7
2.2.3 Tail Configuration 8
2.2.4 Landing Gear 9
2.2.5 Propulsion System 10
2.3 Material Selection For Rc Plane 11
2.4 Weight Estimation 12
2.5 Cost Estimation 13
2.5.1 Design Sketches 14
CHAPTER 3 RC PLANE CONFIGURATION 16
CHAPTER 4 PROPULSION AND POWER PLANT SYSTEM 18
4.1 Introduction 18
4.2 Objective 20
4.3 Methodology 20
4.4 Preliminary Design 20
4.5 Preliminary Selection of Motor 20
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4.6 Power Plant Configuration 24
4.7 Aerodynamic Drag 26
4.8 Thrust Required 28
4.9 Power Available 29
4.10 Power Required 31
4.11 Rate of Climb 32
4.12 Endurance 33
4.13 Range 34
4.14 Takeoff Performance 34
4.15 Landing Performance 36
4.16 Flight Path 37
4.17 Flight Power Consumption 38
CHAPTER 5 AVIONICS 40
5.1 Electronic Components Selection 40
5.1.1 Transmitter and Receiver 40
5.1.2 TX-RX Communication Signal 43
5.1.3 Electronic Speed Controller (ESC) 49
5.1.4 Battery 52
5.1.5 Servo 54
5.1.6 Wiring System Schematic Diagram 56
CHAPTER 6 WING 57
6.1 Introduction 57
6.1.1 Function of Wing 57
6.1.2 Aircraft Wing Design 58
6.1.3 Wing Load Analysis 60
6.1.4 Aircraft’s Airworthiness 62
6.1.5 Assumptions on Wing Design 63
6.2 Flight Envelope and Wing Loadings 64
6.2.1 Flight Envelope (V-n Diagram) 64
6.2.2 Wing Loading without Aileron 66
6.2.3 Wing Loading with Aileron 69
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6.2.4 Reaction force of Wing Strut 72
6.2.5 Shear Force Diagram and Bending Moment
Diagram 77
6.2.6 Discussion 89
6.3 Structural Analysis 91
6.3.1 Material Selection 91
6.3.2 Rib Analysis 92
6.3.3 Spar Analysis 100
6.3.4 Skin Analysis 104
6.3.5 Stress Analysis of Multi Cell Sections 108
6.3.5.1 Shear stress due to shear force
(monocoque) 108
6.3.5.2 Shear stress due to shear force (semi-
monocoque) 112
6.3.5.3 Shear stress due to torsion
(monocoque) 116
6.3.5.4 Shear stress due to torsion (semi-
monocoque) 118
6.3.5.5 Shear Stress Summary 121
6.3.6 Part Attachment 123
6.3.6.1 Joint of the part attachment 123
6.3.6.2 Structural Analysis of Wing Strut 128
6.3.6.3 Part attachment between aerofoil and
aileron 130
6.4 Summary of Safety Factor 134
CHAPTER 7 FUSELAGE AND TAIL 142
7.1 Introduction 142
7.2 Monocoque Structure 143
7.3 Material Properties 144
7.4 Shear and Bending Moment Diagram for Fuselage 145
7.4.1 𝑮 = 𝟏 145
7.4.2 𝑮 = 𝟑 146
7.4.3 𝑮 = −𝟏. 𝟓 147
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7.5 Flow Simulation 150
7.6 Stress Analysis 151
7.6.1 Former structure 151
7.6.2 Bulkhead structure 153
7.7 Compressive-Buckling Analysis 155
7.7.1 Former structure 156
7.7.2 Bulkhead structure 157
7.8 Shear-Buckling Analysis 158
7.8.1 Former structure 159
7.8.2 Bulkhead structure 160
7.9 Shear Flow of Fuselage due to Torsion 161
7.10 Structure Analysis on Horizontal Tail 164
7.11 Shear and Bending Moment Diagram for Tail 167
7.11.1.1 G = 1 167
7.11.2 G = 3 168
7.11.3 G = -1.5 169
CHAPTER 8 LANDING GEAR 173
8.1 Introduction 173
8.2 Static Load Analysis 173
8.3 Introduction of FEM 177
8.3.1 Software Used 177
8.3.2 Material of Landing Gear 178
8.3.2.1 Material Property for Landing Gear 178
8.3.3 Applied Load on Front Landing Gear 178
8.3.4 Boundary Conditions for Front Landing Gear 179
8.3.5 Meshing for Front Landing Gear 180
8.4 Results of Finite Element Analysis for Front Landing
Gear 182
8.5 Discussion 184
CHAPTER 9 FLIGHT STABILITY AND CONTROL 185
9.1 Center of Gravity 185
v
9.1.1 Empty Weight cg 186
9.1.2 Maximum Weight cg 187
9.1.3 Forward cg 188
9.1.4 Aft cg 189
9.2 Stability Control System 191
9.2.1 Contribution of the Wing to Mcg 191
9.2.2 Contribution of the Wing-Body to Mcg 192
9.2.3 Contribution of the the Tail to Mcg 193
9.2.4 Total Pitching Moment about the centre of
gravity 194
9.2.5 Equations for Longitudinal Static Stability 194
9.3 Static Lateral Roll Stability 205
9.3.1 Influence of roll rate on the rolling moment 206
9.3.2 Influence of deflection angle of aileron on the
roll rate 209
9.3.3 Influence of deflection angle of aileron on the
rolling moment 210
9.3.4 Influence of airspeed on rolling moment 212
9.4 Static Lateral Directional Stability 214
9.4.1 Pure Yawing Motion 217
9.5 Dynamic Longitudinal Stability System 220
9.6 Dynamic Lateral Stability System 226
CHAPTER 10 FLIGHT TESTING PLANNING AND
PREPARATION 232
10.1 Flight test and preparation 232
10.2 Checklist 234
REFERENCES 236
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LIST OF TABLES
TABLE NO. TITLE PAGE
Table 1-1 : List of activities from week 1 to week 14 2
Table 2-1 : Properties of EPP Foam 11
Table 2-2 : Weight & Area estimation (based on SolidWorks): 15
Table 3-1 : The configuration of RC plane 16
Table 4-1 General characteristics or our RC plane 20
Table 4-2 Thrust-to-Weight Ratio for different type of aircraft 20
Table 4-3 The values of airspeed according to the motor Kv 21
Table 4-4 : Different brand of brushless motor 23
Table 4-5 : The time and altitude at which the RC plane will takeoff,
cruising and landing 38
Table 4-6 : Percentage of thrust needed for Takeoff, Cruising and Landing
38
Table 5-1: Transmitter specification 41
Table 5-2: Receiver specification 42
Table 5-3: Sunnysky X2216 Motor Performance Chart 50
Table 5-4: ESC Specifications 51
Table 5-5: Comparison of RC Battery 52
Table 5-6: Battery Specification 54
Table 5-7: Servo Specification 55
Table 6-1 Specification of the wing 60
Table 6-2 Federal Aviation Regulations, Part 23 design requirements and
specifications 63
Table 6-3 Project Requirement on Wing Design 63
Table 6-4 Wing loading in each station for point A (Aileron deflects
downwards) 77
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Table 6-5 Wing loading in each station for point D (Aileron deflects
downwards) 78
Table 6-6 Wing loading in each station for point E (Aileron deflects
downwards) 79
Table 6-7 Wing loading in each station for point G (Aileron deflects
downwards) 80
Table 6-8 Wing loading in each station for point A (Aileron deflects
upwards) 81
Table 6-9 Wing loading in each station for point D (Aileron deflects
upwards) 82
Table 6-10 Wing loading in each station for point E (Aileron deflects
upwards) 83
Table 6-11 Wing loading in each station for point G (Aileron deflects
upwards) 84
Table 6-12 Maximum Shear force and Bending moment in each position 88
Table 6-13 Material Selection of each Wing Component 91
Table 6-14 The chordwise load distribution for each bay 92
Table 6-15 Material properties of plywood 93
Table 6-16 Material Properties of EPP Foam 100
Table 6-17 Factor of safety of upper and lower wing 107
Table 6-18 Dimensions of cell 1 and cell 2 wing section 109
Table 6-19 Integral of cell 1 and cell 2 110
Table 6-20 Total shear flow for monocoque and semi-monocoque wing
structure 122
Table 6-21 Total Shear Stress of Monocoque and Semi-monocoque 122
Table 6-22 Safety Factor of Monocoque and Semi-monocoque 122
Table 6-23 Material Properties of AN Steel (hinge pin) 124
Table 6-24 Properties of AN Standard Steel bolt/pin 124
Table 6-25 Material Properties of Mica Plastic (fixed hinge) 126
Table 6-26 Average limit control surface loading 131
Table 6-27 Factor of Safety of Wing Components 134
Table 7-1: Material properties 144
viii
Table 7-2: Maximum values of shear force and bending moment for
fuselage 148
Table 7-3: Stress result for former 152
Table 7-4: Stress result for bulkhead 154
Table 7-5: Fuselage parameter 161
Table 7-6: Stress on surface 164
Table 7-7: Maximum values of shear force and bending moment 170
Table 8-1 Material used 178
Table 8-2 Properties of the material used 178
Table 9-1 Empty weight cg calculation 186
Table 9-2 Maximum weight cg calculation 187
Table 9-3 Forward cg calculation 188
Table 9-4 Aft weight cg calculation 189
Table 9-5 Pitching moment at various angle of attack for empty weight 197
Table 9-6: Pitching moment at various angle of attack for maximum
weight 199
Table 9-7: Pitching moment at various angle of attack for forward cg 200
Table 9-8: Pitching moment at various angle of attack for aft cg 201
Table 9-9 Torque varies with the aileron deflection angle 211
Table 9-10 Torque varies with the airspeed 212
Table 9-11: Longitudinal dimensionless derivatives (Nelson, 1998) 220
Table 9-12: Longitudinal dimensional derivatives (Nelson, 1998) 221
Table 9-13: Longitudinal stability coefficients 221
Table 9-14: Longitudinal dimensional derivatives 222
Table 9-15 Parameters of Cessna 172 226
Table 9-16 Flying Qualities 231
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LIST OF FIGURES
FIGURE NO. TITLE PAGE
Figure 1-1 : Flow chart for the project 3
Figure 2-1 : Cessna 172 with rectangular wing design 6
Figure 2-2 : Semi-monocoque configuration 7
Figure 2-3 : Conventional tail on RC plane 8
Figure 2-4 : Tail wheel design on light aircraft 9
Figure 2-5 : Expanded Polypropylene Foam (EPP) 11
Figure 2-6 : Isometric View of RC Plane 14
Figure 2-7 : Front, Side and Upward View of the RC plane 15
Figure 4-1 Aerodynamics forces acting on an airfoil (Anderson Jr, 2001) 18
Figure 4-2 Graph of W/g versus kv 22
Figure 4-3: Detail specification of Sunnysky X2216 23
Figure 4-4 : SunnySky X2216 23
Figure 4-5 Three common methods to mount propellers (and engines). A
is a tractor configuration, B is a pusher configuration, and
C is a tractor mounted on a nacelle configuration. 24
Figure 4-6 : Graph of CL/CD against Velocity 27
Figure 4-7 : Graph of Drag against Velocity 28
Figure 4-8 Dimension of motor 30
Figure 4-9 Propeller Specification 30
Figure 4-10 : Graph or PR, PA against velocity 31
Figure 4-11 : Illustration of climbing flight 32
Figure 4-12 : Rate of Climb against Velocity 33
Figure 4-13 : Flight path for the RC plane 37
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Figure 4-14: Flight Current/Thrust Consumptions 38
Figure 5-1: Flysky FS-iA6 RC Controller 41
Figure 5-2: Flying mode 41
Figure 5-3: Flysky FS-i6 Receiver 42
Figure 5-4: PWM Pulse 44
Figure 5-5: PPM Pulse 44
Figure 5-6: High gain omnidirectional antenna radiation pattern 47
Figure 5-7: Horizontally Polarized RC Radio Antenna (Left) 48
Figure 5-8: Forward TX Antenna Position (Right) 48
Figure 5-9: 90 Degree Polarization Diversity 48
Figure 5-10: Skywalker 30A ESC 51
Figure 5-11: Maxpower Graphene LiPo Battery 54
Figure 5-12: PWM Signal in servo control 54
Figure 5-13: Tower Pro 9g SG90 Servo Motor 55
Figure 6-1 Aerodynamic forces, airspeeds and pressure acting on wing
aerofoil 58
Figure 6-2 Dimension of the rectangular wing 59
Figure 6-3 Wing components 62
Figure 6-4 Flight Envelope 66
Figure 6-5 Resultant aerodynamic force and the components into which it
splits (Anderson, 2010) 67
Figure 6-6 the parameters c, cf and δ in aerofoil with plain aileron
(McCormick, 1995) 69
Figure 6-7 Position and dimension of ailerons 70
Figure 6-8 Aerodynamic characteristics of NACA 2412 airfoil section
(Abbott, 1945) 71
Figure 6-9 Front View of the Aircraft 72
Figure 6-10 Free Body Diagram (before simplification) 72
Figure 6-11 Free Body Diagram (after simplification) 73
Figure 6-12 Release System 74
Figure 6-13 Unit load system 74
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Figure 6-14 Reaction force of wing strut 75
Figure 6-15 Graph of Normal Lift across Wingspan (Zoom in) 85
Figure 6-16 Graph of Normal Lift across Wingspan (Zoom out) 85
Figure 6-17 Graph of Shear Force across Wingspan when the aileron
deflects downwards 86
Figure 6-18 Graph of Shear Force across Wingspan when the aileron
deflects upwards 86
Figure 6-19 Graph of Bending Moment across Wingspan when aileron
deflects downwards 87
Figure 6-20 Graph of Bending Moment across Wingspan when aileron
deflects upwards 87
Figure 6-21 Graph of Torsion across Wingspan 88
Figure 6-22 Location of ribs along the wingspan 92
Figure 6-23 The load distribution on rib 93
Figure 6-24 Cell 1 of Rib 2 94
Figure 6-25 Cell 2 of Rib 2 96
Figure 6-26 Shear force across the chord 98
Figure 6-27 Bending Moment across the chord 99
Figure 6-28 Dimension and location of spar in wing 100
Figure 6-29 Cross section of spar 101
Figure 6-30 Wing’s skin 104
Figure 6-31 Shear flow diagram of 2 cells wing 108
Figure 6-32 Vertical distance from centroid, z of cell 1 and cell 2 109
Figure 6-33 Line integral of wing section 112
Figure 6-34 Part attachment (wing strut) 123
Figure 6-35 Dimension of the connection 124
Figure 6-36 The moment arm for bending moment on hinge pin 125
Figure 6-37 Graph of kt against d/w 127
Figure 6-38 Shear stress on the contact area of the glue 127
Figure 6-39 Combined compression, bending and shear (Bruhn, 1973) 130
Figure 6-40 Average limit control surface loading 131
xii
Figure 6-41 Chordwise lift distribution of the aerofoil with deployed
aileron 132
Figure 6-42 Wing loading on the deployed aileron 132
Figure 6-43 Aileron attachment area 133
Figure 7-1: Fuselage monocoque structure 143
Figure 7-2: Cross section of maximum bending moment location for
fuselage (cm) 148
Figure 7-3: Pressure contour 150
Figure 7-4: Flow visualization 151
Figure 7-5: Load applied on former 152
Figure 7-6: Pressure contour on former structure for (a) equivalent (Von-
Mises) stress (b) shear stress 153
Figure 7-7: Load applied on bulkhead 154
Figure 7-8: Pressure contour on bulkhead structure for (a) equivalent
(Von-Mises) stress (b) shear stress 155
Figure 7-9: Shear flow of the fuselage 161
Figure 7-10: Flow simulation with 33.6m/s and 860RPM 164
Figure 7-11: Shear-Buckling-Stress Coefficient of Plates as a Function of
a/b 165
Figure 7-12: Chart of Nondimensional Compressive Buckling Stress 166
Figure 8-1 Diagram for nose landing gear load calculation 174
Figure 8-2 Centre of gravity for nose landing gear (front landing gear) 174
Figure 8-3 Centre of gravity for main landing gear (rear landing gear) 175
Figure 8-4 Overview of ABAQUS on Front Landing Gear being modelled
177
Figure 8-5 Loading on Front Landing Gear 179
Figure 8-6 Boundary Condition of Front Landing Gear 179
Figure 8-7 Meshing of Front Landing Gear 180
Figure 8-8 Incrementation of the analysis 181
Figure 8-9 Plot Undeformed Shape of Rear Landing Gear 182
Figure 8-10 Contour Plot of Front Landing Gear 183
Figure 8-11 Magnitude displacement of Front Landing Gear 183
xiii
Figure 9-1 CG location of each weight on the aircraft 185
Figure 9-2 Location of cg empty and payloads 188
Figure 9-3 Location of all cg on the aircraft 190
Figure 9-4 Wing geometry for pitching moment. 191
Figure 9-5 Geometry of wing tail combination. 193
Figure 9-6: Graph of moment coefficient vs angle of attack for empty
weight 198
Figure 9-7: Graph of moment coefficient vs angle of attack for maximum
weight 199
Figure 9-8: Graph of moment coefficient vs angle of attack for forward cg
200
Figure 9-9: Graph of moment coefficient vs angle of attack for aft cg 201
Figure 9-10 Static Lateral Roll Stability 205
Figure 9-11 Wing planform undergoing a rolling motion. 207
Figure 9-12 Flap effectiveness parameter 208
Figure 9-13 Transient reponse of roll rate 209
Figure 9-14 Graph of roll rate vs deflection angle 209
Figure 9-15 Ailerons for roll control 210
Figure 9-16 Graph of torque vs deflection angle 211
Figure 9-17 Graph of torque vs airspeed 213
Figure 9-18 Static directional stability 214
Figure 9-19: Wing Body Interference Factor 215
Figure 9-20: Reynolds number correction factor 216
Figure 9-21: Time response open-loop for longitudinal stability 223
Figure 9-22: Time response for short period 224
xiv
LIST OF APPENDICES
APPENDIX TITLE PAGE
Appendix I Graph to determine aerodynamic coefficients 238
Appendix II McCormick’s method on wing aileron analysis with aileron
239
Appendix III Bruhn’s method on structural analysis 241
Appendix IV EPP foam material properties 244
Appendix V MATLAB coding on dynamic stability and control 245
Appendix VI Minutes of meetings 247
1
CHAPTER 1
INTRODUCTION
1.1 Objectives
Upon receiving this project, there are several objectives that we have to achieve. The
objectives are as follows:
i. Design, build and fly unmanned remote control aircraft
ii. Fixed wing
iii. Maximum takeoff weight is less than 5kg
iv. There will be payload with the mass of 500g
v. The flying time is around 5 minutes
2
1.2 Gantt chart
The Gantt chart shows the planning for our group activities from week 1 to week 14.
Table 1-1 : List of activities from week 1 to week 14
Project SEMESTER 2 SESSION 2019/2020
W1 W2 W3 W4 W5 W6 W7 W8 W9 W10 W11 W12 W13 W14
Conceptual
Design
Preliminary
Design
Detailed Design
Construction
and
Manufacturing
Flight Testing
and
Competition
Report
3
1.3 Flowchart
No
No
No
Yes
Yes
Yes
Performance data
O.K?
Select team members Design
• Fuselage
• Power Plant
• Wing
• Tail
• Landing gear
Calculation and
analysis Is design
O.K?
Construction
• Fuselage
• Power Plant
• Wing
• Tail
• Landing gear
Report Funding
Test Flight
O.K?
Practice Flight
Competition
Yes
Figure 1-1 : Flow chart for the project
4
The figure above shows the flow chart for our RC plane project. It started with
the selection of members into subgroups, which will design the fuselage, wing, tail and
the landing gear. The design of the parts must be compatible with other parts according
to some regulations. Some subgroups will select the appropriate type of power plant
and electronics systems so that it will suit the RC plane.
Detailed calculations are made by each groups to avoid any unnecessary errors
when assembling or fabrication of the RC plane. The funding for this RC plane project
is by collecting RM50 from each member to buy the necessary parts. If the theoretical
performance data is acceptable, then we proceed to the flight test for our RC plane and
move on to practice flying and finally compete with the other groups.
5
CHAPTER 2
CONCEPTUAL DESIGN
2.1 What is Conceptual Design?
Aircraft conceptual design involves sketching a variety of possible
configurations that meet the required design specifications. By drawing a set of
configurations, designers seek to reach the design configuration that satisfactorily
meets all requirements as well as go hand in hand with factors such as aerodynamics,
propulsion, flight performance, structural and control systems. This is called design
optimization. Fundamental aspects such as fuselage shape, wing configuration and
location, engine size and type are all determined at this stage. Constraints to design
like those mentioned above are all taken into account at this stage as well. The final
product is a conceptual layout of the aircraft configuration on paper or computer
screen, to be reviewed by engineers and other designers.
2.2 Design Selection
In this section, we will discuss the configuration of our RC plane.
2.2.1 Wing Configuration
Aircraft wings are airfoils that create lift when moved rapidly through the air.
Aircraft designers have created a variety of wings with different aerodynamic
properties. Attached to the body of an aircraft at different angles, these wings come in
different shapes. These different types of angle and shape of wing need to be evaluated
to determine the most suitable wing configuration of our aircraft design.
6
The wing configuration are choose based on the analysis between different wing types
of aircraft that have been produced. From this, the best design will be used to apply on
our RC plane design. The design that we chose is the rectangular wing design. Below
shows the justification for our selection.
Rectangular Wing (Selected Design)
Figure 2-1 : Cessna 172 with rectangular wing design
The rectangular wing design is one of the most wing configuration among
aircrafts. Besides being the simplest wing to manufacture, it is also a non-tapered,
straight wing that being used in small aircrafts. The wing shape extends out from the
aircraft’s fuselage at right angles approximately.
Rectangular wing is also easy to build, and easy to design. They are also easy
to maintain and can carry more fuel. It can be designed to have one single airfoil profile
from root to tip (no aerodynamic twist) and evenly thick.
7
2.2.2 Fuselage Configuration
There are several types of fuselage configuration namely Truss/framework,
monocoque and semi-monocoque. In this project, we chose semi-monocoque
configuration. This is because:
• It relies largely on the strength of the skin to carry the primary loads.
• This construction uses formers, frame assemblies, and bulkheads to
give shape to the fuselage.
• The heaviest of these structural members are located at intervals to
carry concentrated loads and at points where fittings are used to attach
others units such as wings, power plants, and stabilizers.
• The design is simple and ease to fabricate.
Figure 2-2 : Semi-monocoque configuration
8
2.2.3 Tail Configuration
For the tail configuration, we had to choose between the conventional tail, T-tail and
the V-tail. After comparison, the tail that we are using is the conventional type. This
is because conventional type has its own advantages, which are:
• Most common form.
• Provides adequate stability and control with lowest structural weight.
• Low risk and ease of control and manufacturability.
• Most efficient for the speed RC planes are expected to fly it.
Figure 2-3 : Conventional tail on RC plane
9
2.2.4 Landing Gear
Landing gear is one of the most important part for RC plane because we have to fly it
from the ground and land it to the ground too. Upon consideration, the suitable landing
gear for our RC plane is the tail wheel because we can connect it to our rudder for
ground movement control. The other advantages are:
• Helps slow the aircraft upon landing and provides directional stability.
• Advantageous when operating in and out of non-paved runways.
• A steerable tail wheel, connected by cables to the rudder or rudder
pedals, is also a common design.
Figure 2-4 : Tail wheel design on light aircraft
10
2.2.5 Propulsion System
Based on the calculation in motor selection, the power required by the RC
aircraft is 273W. The motor that will be selected should have higher power compared
to required power. For the propeller, it depends on the motor used because different
motor have different recommended propeller. Since there is only one type of motor
that our lecturer recommended hence we chose Sunnysky X2216.
Sunnysky X2216
RPM 1100kv
Battery 2~4S
Max Current 18A
Max Power 385W
Motor Resistance 72mΩ
ESC 30A
Weight 72g
11
2.3 Material Selection For Rc Plane
For the fuselage, tail and wing. We use only one material to fabricate it which
is the EPP Foam. Expanded Polypropylene (EPP) is a highly versatile closed-cell bead
foam that provides a unique range of properties, including outstanding energy
absorption, multiple impact resistance, thermal insulation, buoyancy, water and
chemical resistance, exceptionally high strength to weight ratio and 100%
recyclability. EPP can be made in a wide range of densities, from 15 to 200 grams per
litre, which are transformed by moulding into densities ranging from 18 to 260 grams
per litre. Individual beads are fused into final product form by the steamchest moulding
process resulting in a strong and lightweight shape.
Figure 2-5 : Expanded Polypropylene Foam (EPP)
Table 2-1 : Properties of EPP Foam
EPP density range 15 g/l to 200 g/l
Tensile strength (kPa) 270 to 1930
Tensile elongation (%) 21 to 7.5
Compressive strength (kPa)
25% strain 80 to 2000
50% strain 150 to 3000
75% strain 350 to 9300
12
2.4 Weight Estimation
Below shows the weight estimation for our RC plane. The weight estimation is based
on the information that we got from the supplier and from a reliable source in the
internet. The total estimated weight for our RC plane is 1.467 kg.
i. Propulsion & Engine
Motor Sunnysky X2216 100g
Propeller Blade 35g
ii. Electronics
Servo 9g (x4) 36g
Flysky FS-i6 Receiver 6.4g
Lipo Battery (4S) 200g
ESC 30A 25g
Wires 5g
Pushrod (x4) 60g
iii. Fuselage
Polyfoam (x2 boards) 200g
Plywood (x3) 300g
iv. Payload
Plasticine 500g
13
2.5 Cost Estimation
After contacting all the supplier and surveying all the components cost. This is our
cost estimation for our RC plane.
i. Propulsion & Engine
Motor Sunnysky X2216 + Propeller
blade RM120.00
ii. Electronics
Servo 9g (x4) RM27.60
Flysky FS-i6 Receiver RM66.76
Lipo Battery 3S 2200mah RM59.60
ESC 30A RM27.00
wires RM6.00
Pushrod (x4) RM16.00
iii. Fuselage
Polyfoam (x2 boards) RM40.00
Plywood (x3) RM15.00
iv. Payload
Plasticine RM4.90
Total estimated cost of aircraft = RM382.86
14
2.5.1 Design Sketches
By using the software SolidWorks, we design the 3D sketches for our RC plane. The
purpose of these sketches is to obtain the weight estimation based on the software since
we can change the material for the RC plane based on our preferences and to have the
clear view of the RC plane. We can also calculate the surface are of the RC plane by
using SolidWorks.
Figure 2-6 : Isometric View of RC Plane
15
Figure 2-7 : Front, Side and Upward View of the RC plane
Table 2-2 : Weight & Area estimation (based on SolidWorks):
Fuselage & Tail (Foam only) = 20.97𝑔
Wing (include rib & spar) = 34.46𝑔
Surface area for polyfoam:
Wing Skin + Aileron = 4167.620 𝑐𝑚2
Fuselage = 2083.44 𝑐𝑚2
Horizontal Tail + Elevon = 307.44 𝑐𝑚2
Vertical Tail + Rudder = 124.41 𝑐𝑚2
Total Foam Area = 6682.91 𝑐𝑚2
16
CHAPTER 3
RC PLANE CONFIGURATION
The configuration of our RC plane is discussed in this chapter. Since the estimated
weight is approximately 14.715N (1.5kg) including the payload, then the empty weight
is without the 500g payload which makes it 9.81N.
Table 3-1 : The configuration of RC plane
𝐺𝑟𝑜𝑠𝑠 𝑤𝑒𝑖𝑔ℎ𝑡 14.715 𝑁
𝐸𝑚𝑝𝑡𝑦 𝑤𝑒𝑖𝑔ℎ𝑡 9.81 𝑁
𝑊𝑖𝑛𝑔 𝑎𝑖𝑟𝑓𝑜𝑖𝑙 𝑠𝑒𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑁𝐴𝐶𝐴 2412
𝑊𝑖𝑛𝑔 𝑠𝑝𝑎𝑛, 𝑏 1.2 𝑚
𝑊𝑖𝑛𝑔 𝑝𝑙𝑎𝑡𝑓𝑜𝑟𝑚 𝑎𝑟𝑒𝑎, 𝑆 0.204 𝑚2
𝐴𝑠𝑝𝑒𝑐𝑡 𝑟𝑎𝑡𝑖𝑜, 𝐴𝑅 7.0588
𝐶ℎ𝑜𝑟𝑑 𝑙𝑒𝑛𝑔𝑡ℎ, 𝑐 0.17 𝑚
𝑇𝑎𝑝𝑒𝑟 𝑟𝑎𝑡𝑖𝑜, 𝜆 5.8824
𝐹𝑢𝑠𝑒𝑙𝑎𝑔𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 0.73 𝑚
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑙𝑖𝑓𝑡 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 1.29
Limit factors
Our lecturer had given limiting factors to us and the values are as shown.
𝑛𝑙𝑖𝑚𝑝𝑜𝑠= 3; 𝑛𝑙𝑖𝑚𝑛𝑒𝑔
= −1.5
Stall speed
Stall speed is the slowest speed a plane can fly to maintain level flight. In our case,
by using the formula with the known value, we got:
𝑉𝑠 = √2𝑊𝑀𝑇𝑂𝑊
ρ𝐶𝐿𝑚𝑎𝑥𝑆
= √2(1.5 𝑥 9.81)
(1.225)(1.29)(0.204) = 9.5179 𝑚/𝑠
17
Takeoff speed
Takeoff speed is the speed needed for the plane to take off. Based on our
calculations, the takeoff speed for our RC plane is:
𝑉𝑇𝑂 = 1.2𝑉𝑠𝑡𝑎𝑙𝑙 = 1.2√2𝑊
ρ∝𝑆𝐶𝐿𝑚𝑎𝑥
= 11.4215 𝑚/𝑠
Landing speed
Landing speed is the speed when the aircraft land. The landing speed of our plane is
shown below:
𝑉𝐿0 = 1.3𝑉𝑠𝑡𝑎𝑙𝑙 = 1.3√2𝑊
ρ∝𝑆𝐶𝐿𝑚𝑎𝑥
= 1.3(8.6293) = 12.3733 𝑚/𝑠
18
CHAPTER 4
PROPULSION AND POWER PLANT SYSTEM
4.1 Introduction
According to NASA, the propulsion means to push forward or drive an object
forward while a propulsion system is a machine that produces thrust to push an object
forward. The foundations of powered flight can be set in 1799 when Sir George
Cayley, the father of Aerodynamics, first identified the four basic forces of flying:
weight, lift, drag, and thrust and set forth the concept of the modern aeroplane as a
fixed-wing flying machine (no flapping wings like in bird flight), with separate
systems for lift (tilted planes), propulsion (engine), and controls, designing the first
successful glider to carry a human being aloft. The basic forces of flying or known as
aerodynamic forces can be shown as in Figure 1.
Figure 4-1 Aerodynamics forces acting on an airfoil (Anderson Jr, 2001)
On airplanes, thrust is usually generated through some application of Newton's
third law of action and reaction. A gas, or working fluid, is accelerated by the engine,
and the reaction to this acceleration produces a force on the engine. A general
19
derivation of the thrust equation shows that the amount of thrust generated depends on
the mass flow through the engine and the exit velocity of the gas. Aircraft propulsion
has a wide variety of systems available. Different propulsion systems generate thrust
in slightly different ways. The choice depends on the speed or Mach number required
and the role for which the aircraft is to be used.
Aircraft electric propulsion is not a new, the idea of an all-electric air vehicle
has been researched and implemented since the mid-70’s (Newcome, 2004). Even
though piston engine is one of the most efficient internal combustion engines when
compared to others but electric motor has the potential to reach 80-90% of efficiencies.
This is due to the output of electric motor is relatively not affected by altitude,
temperatures and humidity factors which bring a lot of impact to the power output of
a combustion engine. Other than that, electrical propulsion could increase the
sustainability in aviation since it can be powered by renewable sources such as wind
and solar. The instantaneous torque and silent operation also the advantages of using
electric motor.
There are few things need to be considered when selecting electric motor
depending the application for example Alternating Current (AC) against Direct
Current (DC) and others. Even though the AC motor has a cost and a maximum torque
advantage over the DC motor, but the efficiency loss of converting the DC battery
power to AC current for an aircraft application is a primary reason that a DC motor is
selected for this project. The DC motor has a brushed and brushless construction but
brushless motor offered less maintenance, more controllable speed, no voltage drop
across brushes, high output power, small frame size and high speed range. Thus, for
electric motor, these are the following motor characteristics:
1. Direct Current (DC)
2. Brushless
20
4.2 Objective
The objective is to determine the aircraft’s motor and propeller with the power
available and required to meet the requirement of aircraft in order to produce thrust.
4.3 Methodology
4.4 Preliminary Design
Table 4-1 General characteristics or our RC plane
Type RC aircraft
Maximum Take-off Weight 1.5 kg
Engine type DC motor engine
4.5 Preliminary Selection of Motor
An important criterion in selecting a motor is a thrust to weight ratio, which
depends on the type of the aircraft.
Table 4-2 Thrust-to-Weight Ratio for different type of aircraft
Type of Aircraft Thrust- to- Weight Ratio
Glider/Trainer 0.35 to 0.55
Scale Flight 0.6 to 0.7
Sport and Slow Acrobatic 0.7 to 0.8
Acrobatic Fast 0.8 to 1.00
Jets and 3D 1.00 to 2.5
Estimate weight of
aircraft
Select motor for
desired
thrust/weight ratio
Select propeller
blade that suitable
for the motor
21
Based on the table, since the aircraft is a scale flight, the suitable thrust-to-weight ratio
that suits our aircraft is from 0.6 to 0.7.
To calculate the necessary thrust of the aircraft, the following formula is used:
𝑇ℎ𝑟𝑢𝑠𝑡 = 𝑊𝑒𝑖𝑔ℎ𝑡 ×𝑇ℎ𝑟𝑢𝑠𝑡
𝑊𝑒𝑖𝑔ℎ𝑡
Since the maximum weight of the aircraft would be not more than 5000g, then the
necessary thrust is:
𝑇ℎ𝑟𝑢𝑠𝑡 = 1500 × 0.6 = 900𝑔
Kv is simply the revolutions per minute (rpm) an electric motor will spin at per volt
when under no load.
To choose the Kv of the motor, the required speed of the aircraft must be determined:
Table 4-3 The values of airspeed according to the motor Kv
Kv Air Speed (Km/h) Air Speed (mph)
1000 70 43
2000 140 87
3000 210 131
4000 280 175
Based on the table above, it can be conclude that the higher the value of Kv, the higher
the air speed. Since the aircraft will be flying in a low air speed, the suitable value of
Kv will be around 1000.
22
Victor (2017) to determine the suitable power/thrust ratio with has done an experiment
regarding the power/thrust ratio given value of kv.
Figure 4-2 Graph of W/g versus kv
Based on the graph plotted, the equation to obtain the power/thrust ratio is :
𝑊
𝑔= 0.17 ×
𝐾𝑣
1000+ 0.09
For this case, the required thrust is 900g and by interpolation, the power/thrust ratio is
approximately 0.26. From there, the needed power can be calculated by:
𝑃𝑜𝑤𝑒𝑟(𝑊) = 900𝑔 ×0.26𝑊
𝑔= 234𝑊
With these calculated values, the motor is chosen from manufacturer’s catalogue. A
higher value of the motor power with 40% to 50% more to fly easily at 50 to 60%
throttle. The power value in parenthesis reflex this consideration.
Currently, there are 2 motors that can be used for this aircraft:
0
0.2
0.4
0.6
0.8
1
0 1000 2000 3000 4000 5000 6000
W/g
kv (rpm/volt)
Watts/gram vs Kv
23
Table 4-4 : Different brand of brushless motor
Motor Turnigy D2836/8 Sunnysky X2216
RPM 1100kv 1100kv
Max Power 336W 385W
Weight 70g 72g
The most suitable motor that the aircraft can use is the Sunnysky X2216 Brushless
Outrunner Motor. The specifications for the motor is as shown in the figure below:
Figure 4-3: Detail specification of Sunnysky X2216
Figure 4-4 : SunnySky X2216
24
4.6 Power Plant Configuration
Propellers can be mounted in a number of ways to an airplane. Three common
methods are shown in Figure x, a tractor (A), pusher (B), and a configuration featuring
the engine and propeller mounted on the wing in nacelles (C). Configuration (C) is a
variation of (A) or (B). The advantages and disadvantages of these installation methods
are listed in Table 14-1. Note that so-called “inline” configurations, which consist of
a tractor and pusher, can simply be treated as a combination of configurations (A) and
(B). Tractor configuration will be use in this propject.
Figure 4-5 Three common methods to mount propellers (and engines). A is a tractor
configuration, B is a pusher configuration, and C is a tractor mounted on a nacelle
configuration.
The incoming air of Tractor Configuration is undisturbed. Ground clearance is
not an issue during rotation at T-O, or flare before touch-down. A Placing the fuselage
behind the propeller allows for a reduced streamtube inflow distortion and less
asymmetric disc loading that would increase blade stresses. There is less chance that
the propeller suffers damage due to FOD, in particular when the aircraft is moving.
Propeller is not subject to excessive heat from the exhaust. Propwash can help with T-
O rotation during soft- or short-field take-offs, by increasing the dynamic pressure at
the horizontal tail.
25
While in pusher configuration, Ground clearance may be an issue during
rotation at T-O, or flare before landing. Fuselage ahead of the propeller may distort
flow inside the streamtube, causing asymmetric disc loading and increased blade
stresses. This distortion may affect the propeller’s performance. Propeller may suffer
FOD, in terms of both pebbles shot by tires and ingestion by ice shedding off a
fuselage. Propeller may be subject to excessive heat from the exhaust. Special
regulatory requirements for pushers are stipulated in 14 CFR 23.905. Results in a
higher fly-over noise and possible propeller corrosion issues.
The nacelle Configuration has many of the same disadvantages as the tractor
configuration; however, since the nacelle usually extends far forward from the wing’s
structural support, the airframe may be subject to structural oscillations that, if ignored,
may cause premature structural failure, and even whirl flutter phenomena.
26
4.7 Aerodynamic Drag
In this section, the theoretical calculation of our aerodynamic drag is shown below;
𝐶𝐿 =𝑊
0.5𝜌𝑉2𝑆
𝐶𝐷 = 𝐶𝐷,0 + 𝛽𝐶𝐿2 = 𝐶𝐷,0 +
𝐶𝐿2
𝜋𝑒𝐴𝑅
𝑤ℎ𝑒𝑟𝑒 𝐶𝐷0= 𝑝𝑎𝑟𝑎𝑠𝑖𝑡𝑒 𝑑𝑟𝑎𝑔 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑎𝑡 𝑧𝑒𝑟𝑜 𝑙𝑖𝑓𝑡 = 0.01
𝑒 = 𝑜𝑠𝑤𝑎𝑙𝑑 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 =1
1.05 + 0.007𝜋𝐴𝑅= 0.8297
𝐷𝑟𝑎𝑔 =𝐶𝐷
0.5𝜌𝑉2𝑆
Velocity
(m/s)
Lift coefficient,
CL
Induced drag,
CDi
Drag
coefficient, CD CL/CD
Drag /
Thrust
required
4 8.742536 4.661796 14.00549 0.624222 27.9998
6 3.885572 0.920849 2.782646 1.396359 12.5169
8 2.185634 0.291362 0.894187 2.44427 7.1506
10 1.398806 0.119342 0.378126 3.699312 4.7247
12 0.971393 0.057553 0.192759 5.039414 3.4683
14 0.713676 0.031066 0.113297 6.299159 2.7747
16 0.546409 0.01821 0.07473 7.311728 2.3904
18 0.43173 0.011369 0.054206 7.964693 2.1944
20 0.349701 0.007459 0.042477 8.232798 2.1230
22 0.289009 0.005095 0.035384 8.167908 2.1398
24 0.242848 0.003597 0.030891 7.861406 2.2233
26 0.206924 0.002612 0.027935 7.407422 2.3595
28 0.178419 0.001942 0.025925 6.882174 2.5396
30 0.155423 0.001473 0.02452 6.338597 2.7574
32 0.136602 0.001138 0.023514 5.809296 3.0086
27
34 0.121004 0.000893 0.022779 5.312046 3.2903
Figure 4-6 : Graph of CL/CD against Velocity
Based on the graph, we can obtain the L/D max for our RC plane;
(𝐿
𝐷)𝑚𝑎𝑥
= 6.7347
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑡 (𝐿
𝐷)𝑚𝑎𝑥
= 18 𝑚/𝑠
0
1
2
3
4
5
6
7
8
0 5 10 15 20 25 30 35 40
CL/
CD
Velocity (m/s)
28
4.8 Thrust Required
Consider an airplane in steady and level flight at a given altitude and a given
velocity. Forces acting:
𝑇 = 𝐷 = 𝑞∞𝑆𝐶𝐷
𝑊 = 𝐿 = 𝑞∞𝑆𝐶𝐿
The thrust required for an airplane to fly at a given velocity in level, unaccelerated
flight is:
𝑇𝑅 = 𝐷 = 𝑞∞𝑆(𝐶𝐷,0 + 𝐶𝐷,𝑖) = 𝑞∞𝑆𝐶𝐷
Figure 4-7 : Graph of Drag against Velocity
At minimum 𝑇𝑅,
𝐶𝐷,0 = 𝐶𝐷,𝑖
𝑧𝑒𝑟𝑜 − 𝑙𝑖𝑓𝑡 𝑑𝑟𝑎𝑔 = 𝑑𝑟𝑎𝑔 𝑑𝑢𝑒 𝑡𝑜 𝑙𝑖𝑓𝑡
This yields an interesting aerodynamic result that at minimum thrust required, zero-
lift drag equals drag due to lift.
𝑡ℎ𝑟𝑢𝑠𝑡 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑚𝑖𝑛 = 𝑑𝑟𝑎𝑔 𝑚𝑖𝑛 = 2.5952
𝑉𝑚𝑖𝑛 𝑎𝑡 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑡ℎ𝑟𝑢𝑠𝑡 = 18 𝑚/𝑠
0
2
4
6
8
10
12
14
0 5 10 15 20 25 30 35 40 45
Dra
g
Velocity (m/s)
29
4.9 Power Available
Motor SunnySky X2216 1100kv
Propeller APC9045 (9x4.5)
The maximum power available based on the datasheet of the motor is 143.63W with
the efficiency of 4.628 g/W and 1000gf thrust.
To find the actual power available of the motor, the calculations are based on the
website (STARLINO) where they provide the relevant equations in obtaining the
actual power.
Based on the propeller diameter expressed in inches, the theoretical thrust can be
calculated:
𝑇𝑔𝑟𝑎𝑚 = (𝑃𝐷𝑖𝑛𝑐ℎ
𝐶) = (
143.63(9)
0.0278)
23
= 1316.987 𝑔𝑓
Where C is the air density dependent coefficient 30𝑜𝐶, 1 𝑎𝑡𝑚
= 1
0.0127√
𝑔3
2𝜋𝑄𝑎𝑖𝑟 = 0.0278
The efficiency of the motor and the propeller is given by :
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =𝑡ℎ𝑟𝑢𝑠𝑡
𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑡ℎ𝑟𝑢𝑠𝑡=
924
1316.897= 70%
Hence, the actual power available supplied by the motor is:
𝑃𝐴 = 143.63 × 0.7 = 100.8𝑊
30
Figure 4-8 Dimension of motor
Figure 4-9 Propeller Specification
Efficiency (g/W) Amps (A) Thrust (gf) Watts (W)
Voltage (V)
31
4.10 Power Required
Consider an airplane in level, unaccelerated flight at a given altitude and with
velocity 𝑉∞
The power required:
𝑃𝑅 = 𝑇𝑅𝑉∞
But as 𝑇𝑅 = 𝐷 (for an unaccelerated flight)
𝑃𝑅 = 𝑞∞𝑆(𝐶𝐷,0 + 𝐶𝐷,𝑖)𝑉∞
Or in other word,
𝑃𝑅 = 𝑧𝑒𝑟𝑜 𝑙𝑖𝑓𝑡 𝑝𝑜𝑤𝑒𝑟 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 + 𝑙𝑖𝑓𝑡 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑝𝑜𝑤𝑒𝑟 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑
Figure 4-10 : Graph or PR, PA against velocity
At minimum power required,
𝑃𝑅𝑚𝑖𝑛 = 42.24
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑡 𝑃𝑅𝑚𝑖𝑛 = 14 𝑚/𝑠
The maximum flight velocity is determined by the intersection of the maximum 𝑃𝐴
and the 𝑃𝑅 curves.
𝑉𝑚𝑎𝑥 = 28 𝑚/𝑠
0
50
100
150
200
250
0 5 10 15 20 25 30 35
Po
wer
(W
Att
)
Velocity (m/s)
Power Required
Power Available
32
4.11 Rate of Climb
Consider an airplane in steady, unaccelerated and climbing flight,
Figure 4-11 : Illustration of climbing flight
Summing forces parallel to the flight path:
𝑇 = 𝐷 + 𝑊𝑠𝑖𝑛휃
And perpendicular to the flight path:
𝐿 = 𝑊𝑐𝑜𝑠휃
The vertical velocity is called the rate of climb, R/C:
𝑅/𝐶 = 𝑉∞𝑠𝑖𝑛휃
As we know,
𝑃𝐴 = 𝑇𝑉∞ (𝑤ℎ𝑒𝑟𝑒 𝑇 = 𝑇𝐴)
𝑃𝑅 = 𝐷𝑉∞ (𝑓𝑜𝑟 𝑎 𝑙𝑒𝑣𝑒𝑙 𝑓𝑙𝑖𝑔ℎ𝑡)
𝑠𝑖𝑛𝑐𝑒 𝐷 = 𝑇𝑅
Hence,
𝑇𝑉∞ − 𝐷𝑉∞ = 𝑒𝑥𝑐𝑒𝑠𝑠 𝑝𝑜𝑤𝑒𝑟
And for the rate of climb,
33
𝑅
𝐶= 𝑒𝑥𝑐𝑒𝑠𝑠 𝑝𝑜𝑤𝑒𝑟/ 𝑊
Figure 4-12 : Rate of Climb against Velocity
For the maximum R/C:
(𝑅
𝐶)𝑚𝑎𝑥 = 𝑒𝑥𝑐𝑒𝑠𝑠 𝑝𝑜𝑤𝑒𝑟𝑚𝑎𝑥/ 𝑊 =
58.56
19.62= 2.9847
From graph power vs V
max 𝑒𝑥𝑐𝑒𝑠𝑠 𝑝𝑜𝑤𝑒𝑟 =58.56 𝑎𝑡 14 𝑚/𝑠
4.12 Endurance
RC airplane can’t fly with a LiPo battery until it is completely flat. This is because
there is a limit on how many amp-hours can discharge before the battery isn’t able to
be recharged. This means that the effective capacity of a LiPo is only 80% of your
total amp-hours.
2200𝑚𝐴ℎ 𝑏𝑎𝑡𝑡𝑒𝑟𝑦 ∶ 4 𝑥 (2200𝑚𝐴ℎ
1000)𝑥 0.8 𝑥 60 / 35𝐴 = 12 𝑚𝑖𝑛𝑢𝑡𝑒𝑠
-8
-6
-4
-2
0
2
4
0 5 10 15 20 25 30 35 40
Rat
e o
f cl
imb
(R
/C)
Velocity (m/s)
34
4.13 Range
To calculate the range, we multiply the endurance by the velocity of our RC plane.
𝑅𝑎𝑛𝑔𝑒 = 𝐸𝑛𝑑𝑢𝑟𝑎𝑛𝑐𝑒 𝑥 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦
= 12.0686 𝑥 60 𝑥 14 = 10137.624 𝑚𝑒𝑡𝑒𝑟𝑠
4.14 Takeoff Performance
The takeoff performance will estimate the distance that our aircraft needs in before
takeoff. The general formula of takeoff performance is:
𝑆𝐿𝑂 ≈1.44W2
gρ∞SCLmax[T − [D + μr(W − L)]0.7𝑉𝐿𝑂]
Where,
𝑉𝑠𝑡𝑎𝑙𝑙 = √2𝑊
𝜌∞𝐶𝐿𝑀𝑎𝑥𝑆
𝑉𝑠𝑡𝑎𝑙𝑙 = 9.5179 𝑚/𝑠
VLO = 1.2(Vstall) = 1.2(9.5179) = 11.4215 𝑚/𝑠
0.7VLO = 0.7(11.4215) = 7.9951𝑚/𝑠
𝜑 = (16ℎ𝑏
)2
1 + (16ℎ𝑏
)2
)
= (16(0.045)
1.2 )2
1 + (16(0.045)
1.2)2
= 0.2647
L = 1
2𝜌𝑉2𝑆CL
14.715 = 0.5(1.225)(7.99512)(0.204)CL
CL = 1.8424
CDi = 𝜑 (𝐶𝐿
2
𝜋𝑒𝐴𝑅)
35
= 0.2647 (1.84242
𝜋(0.8297)(7.06))
= 0.0488
DLO = 0.5𝜌𝑉0.7𝑉𝐿𝑜2𝑆(CDo + CDi)
= 0.5(1.225)(7.9951)2(0.204)(0.01 + 0.0488)
= 0.4696 𝑁
𝐿 = 0.5𝜌𝑉0.7𝑉𝐿𝑜2𝑆CLmax
= 0.5(1.225)(7.9951)2(0.204)(1.3)
= 10.3831 𝑁
g = 9.81
𝜌 = 1.225
S = 0.204
During take-off, the angle of attack of the airplane is restricted by the
requirement that the tail not drag the ground, and therefore CLmax is assumed
to be limited to 1.3 during ground roll.
CLmax = 1.3
T = 9.8 𝑁
μr = 0.02
W = 14.715 𝑁
𝑆𝐿𝑂 ≈1.44W2
gρ∞SCLmax[T − [D + μr(W − L)]0.7𝑉𝐿𝑂]
𝑆𝐿𝑂
=1.44(14.715)2
(9.81)(1.225)(0.204)(1.3)[9.8 − [0.4696 + 0.02(14.715 − 10.3831)]0.7𝑉𝐿𝑂]
𝑆𝐿𝑂 ≈ 10.58 m
36
4.15 Landing Performance
The landing performance will estimate the landing distance that our RC plane needed.
The formula for calculating the landing performance is given by:
𝑆𝐿 =1.69W2
gp∞SCLmax[T𝑟𝑒𝑣 + [D + μr(W − L)]0.7𝑉𝑇𝐷]
μr = 0.02
W = 14.715 𝑁
g = 9.81
p∞ = 1.225
S = 0.204 m2
VT = 1.3(Vstall) = 1.3(9.5179) = 12.3733 𝑚/𝑠
𝑉0.7𝑉𝑇= 0.7(12.3733) = 8.6613 𝑚/𝑠
𝐿 = 0.5𝜌𝑉0.7𝑉𝐿𝑜2𝑆CLmax
= 0.5(1.225)(8.6613)2(0.204)(1.3)
= 12.1856 𝑁
𝑊 = 𝑊1
L = 1
2𝜌𝑉2𝑆CL
14.715 = 0.5(1.225)(8.6613 2)(0.204)CL
CL = 1.5698
CDi = 𝜑 (𝐶𝐿
2
𝜋𝑒𝐴𝑅)
= 0.2647 (1.56982
𝜋(0.8297)(7.06)) = 0.0354
37
DT = 6.2462N
T𝑟𝑒𝑣 = 0 𝑙𝑏
𝑆𝐿 =1.69W2
gp∞SCLmax[T𝑟𝑒𝑣 + [D + μr(W − L)]0.7𝑉𝑇𝐷]
𝑆𝐿 =1.69(14.715)2
(9.81)(1.225)(0.204)(1.3)[0 + [6.2462 + 0.02(14.715)]0.7𝑉𝑇𝐷]
𝑆𝐿 ≈ 17 m
4.16 Flight Path
The figure below depicts the flight path for our RC plane; the maximum altitude we
estimated that our RC plane could fly is up to 20m. Based on the previous calculations,
we estimated the flight path of our RC plane for the whole time.
Figure 4-13 : Flight path for the RC plane
0
5
10
15
20
25
0 50 100 150 200 250 300 350 400 450
Alt
itu
de
(met
re)
Time(sec)
Flight Path
38
Table 4-5 : The time and altitude at which the RC plane will takeoff, cruising and
landing
Condition Time (sec) Altitude (metre)
Takeoff 0 − 11 0 to 20
Cruising 11 − 186 20
Landing 186 − 300 20 to 0
4.17 Flight Power Consumption
The figure below shows the flight current consumption for the RC plane. The grey line
indicates the thrust while the blue line indicates the current.
Figure 4-14: Flight Current/Thrust Consumptions
Table 4-6 : Percentage of thrust needed for Takeoff, Cruising and Landing
0
100
200
300
400
500
600
700
800
900
1000
0
2
4
6
8
10
12
14
0 50 100 150 200 250 300 350 400 450
Thru
st (
gf)
Cu
rren
t (a
mp
ere)
Time(sec)
Current Thrust
39
Condition Thrust (gf)
*Max thrust=1320gf
Current (A)
Takeoff speed 70% 𝑥 1320 = 924 13.04
Cruising speed 50% 𝑥 1320 = 660 9.314
Landing Speed 20% 𝑥 1320 = 264 2.34
We estimated the flight current and thrust consumption based on the percentage that
we need from the maximum thrust available from the motor. Since we required more
thrust and current during takeoff, we estimated 70% of the thrust, which is equivalent
to 924gf and 13.04A, are needed for our RC to takeoff from the ground. In cruising,
only 50% of the thrust is needed and for the landing it could be less than 20% of the
thrust because we can save more battery consumption if we lessen the thrust needed
for landing.
40
CHAPTER 5
AVIONICS
5.1 Electronic Components Selection
The selection of electronics components is mainly based on the suitability of
project, compatibility with the other components, price and avaibility of components
(preferable to buy them locally than overseas). The main electronic components to
powered an UAV are transmitter, receiver, electronic speed controller (ESC), battery
and servo. Detail description on the selection, including the specifications of the
products is explained below.
5.1.1 Transmitter and Receiver
Transmitter is an electronic device that uses radio signals to transmit
commands wirelessly via a set radio frequency over to the receiver. (All About
Multirotor Drone Radio Transmitters and Receivers, 2018) Pilot control the RC
airplane remotely through transmitter. The chosen transmitter model is the Flysky FS-
i6 RC Controller. The specification of the Flysky is as shown in Table 5-1. 6 channels
are chosen because it is needed to control the aileron, rudder, transmitter and landing
gears. It is recommended for beginners, as it is one of the reputable brand, where all
the tutorials on how to use the transmitter can be easily obtained online. In addition, it
has 20 models memory, which means it can works with 20 different receivers. Then,
the price is consider affordable for a 6 channels transmitter and receiver set (RM165).
41
Table 5-1: Transmitter specification
Figure 5-1: Flysky FS-iA6 RC
Controller
Besides, mode 2 is set as default mode on the transmitter, the stick movement
is as shown in Figure 5-2. Also, this transmitter adapt a 2.4GHz frequency, which is a
newer technology that offers “frequency hopping”. It manages a multiple user
frequency transmitting at the same time by scanning the frequency band and finding
the best available channel during the transmission. (All About Multirotor Drone Radio
Transmitters and Receivers, 2018)
Figure 5-2: Flying mode
Receivers are electric devices that works closely with transmitter. It has built
in antennas that intercept the radio signals from the transmitters, and convert them into
Model Flysky FS-i6 RC Controller
Channel 6
Suitable RC Type Fixed wing / helicopter / glider
Weight 392g
Size 174 × 89 × 190mm
Channel resolution 1024 steps
Battery Voltage 6V 1.5AA × 4
Bandwidth 500 kHz
DSC port PS2
Output PPM, PWM, iBUS
Control range 500m
42
alternating current pulses. (Perlman, 2016) The receiver then relays these signals to
the servos and sends out to the respective channels through Pulse Position Modulation
(PPM) or S-BUS commands. It is important to take note on the frequencies and
protocol during the selection of receiver. Frequencies must be the same on both
receiver and transmitter. For instance; a 2.4GHz transmitter can only work with
2.4GHz receiver. Also, a receiver must be compatible with the transmitter in order to
establish a good communication, thus a same brand of receiver and transmitter is
always recommended.
Therefore, we decided to buy the same brand receiver, which is the Flysky FS-
iA6 6 Channel receiver, where the specifications are shown in Table 5-2.
Table 5-2: Receiver specification
Model Flysky FS-i6 Receiver
Channel 6
RF receiver sensitivity -105dBm
Transmitting Power ≤20dBm
Frequency 2.4GHz
Weight 6.4g
Size 40.4 × 21.2 × 7.35mm
Channel resolution 1024 steps
Antenna length 26 × 2mm (Dual Antenna)
Modulation GFSK
System type AFHDS2A / AFHDS
Blind port Yes
Power port Yes (VCC)
Input Power 4V – 6.5V DC
Figure 5-3: Flysky FS-i6
Receiver
43
5.1.2 TX-RX Communication Signal
Frequency Band
Some commonly use frequencies are 27MHz, 72MHz, 433MHz, 900MHz,
1.3GHz and 2.4Ghz. 433Mhz, 900Mhz and 1.3GHz are typically used in long range
FPV and RC systems, while 27Mhz and 72Mhz are older frequencies which were
being used for many years in RC. Equipment operating on 27Mhz and 72Mhz used
crystals to bind the Transmitter with a Receiver, which can found on a few older
version of RC toys in the market.
2.4GHz is most popular frequency band, which adapted on Flysky FS-iA6 set
as well. It is a newer technology and it offers “frequency hopping” which does the job
of managing multiple users frequency transmitting at the same time. This is done by
scanning the frequency band and finding the best available channel during the
transmission. 2.4GHz antennas are very compact as well. Generally speaking the lower
the frequency, the larger the antenna. For that reason, 2.4GHz quickly became the “go
to” frequency.
Communication Protocol
Different Transmitter and Receiver might be capable of different protocols. There
are 2 major type of communication protocol, which is:
• TX Protocols: Communication between Radio Transmitter and Radio
Receiver.
• RX Protocols: Communications between Radio Receiver and Flight
Transmitter.
For TX Protocols, FS-iA6 transmitter capable to switch between AFHDS and
AFHDS 2A. AFHDS stands for Automatic Frequency Hopping Digital System, and is
a digital protocol that ensures 2 or more radios can operate at the same time without
interfering each other’s respective aircraft (or receiver). AFHDS 2A is the 2nd
generation of the system that added 2-way communication capability and allows for
telemetry.
44
For RX Protocols, FS-iA6 receiver support the universal PWM and PPM, as
well as the brand specific iBUS. PWM stands for Pulse Width Modulation. In PWM
each RC channel has own cable. The value of each channel is represented as a 1
millisecond (ms) to 2ms “ON” signal and this signal repeats (or updates) every 20
milliseconds. When PWM command is selected on TX, the receiver will output each
channel 1-6 respectively. . (FS-i6 Instruction Manual, 2015)
Figure 5-4: PWM Pulse
On the other hand, PPM stands for Pulse Position Modulation.Think of PPM
as several PWM signals lined up back to back. In PPM, the same signalling is used but
each channel is sent successively, then a delay, then it loops back to channel 1. In
normal PWM there are 50 updates sent per second (50Hz) which means each update
takes 20 milliseconds. So if each channel takes up to 2ms, then 10 channels can be
performed within that 20ms before which loop back to channel 1. PPM has less wiring
compared to PWM. When PPM is selected on transmitter, the receiver will output a
standard PPM signal. (Liang, 2013)
Figure 5-5: PPM Pulse
Similar to SBUS, iBUS is a digital serial protocol that allows for up to 18
channels to be supported on a single wire (compared to the 16 channels offered by
SBUS). These digital protocols have less latency than PPM, providing an overall better
solution to traditional analogue RX signals. Another point in favour of iBUS over
45
SBUS is that the signal is un-inverted, so you are not required to invert the signal when
using F1 and F4 boards as you do with SBUS. (Liang, FLYSKY TRANSMITTER &
RECEIVER BUYER’S GUIDE, 2019)
TX and RX Special Features
As the AFHDS 2A protocol is developed on the transmitter, thus the systems features
mentioned below are thus available:
• Bidirectional Communication
Capable of sending and receiving data, each transmitter is capable of receiving data
from temperature, altitude and many other types of sensors, servo calibration and
iBUS support.
• Multi-channel Hopping Frequency
This systems bandwidth range from 2.4055GHz to 2.475GHz. This band is divided in
140 channels. Each transmitter hips between 16 channels in order to reduce
interference from other transmitters.
• Omni-directional Gain Antenna
The high efficiency Omni-directional high gain antenna cuts down on interference,
while using less power and maintaining a strong reliable connection.
• ID Recognition System
Each transmitter and receiver has it’s own unique ID. Once the transmitter and
receiver have been paired, they will only communicate with each other, preventing
other systems accidentally connecting to or interfering with the system operation.
• Low Power Consumption
The system is built using highly sensitive low power components, maintaining high
receiver sensitivity, while consuming as little as little as one tenth the power of a
standard FM system, extending battery life.
Besides, safety features such as Failsafe is available to protect the models and
users if the receiver loses signal and therefore is no longer controllable. In the case of
a loss of signal, the corresponding servo will keep its last received position. If it
displays a percentage, the servo will instead move to the selected position.
46
TX-RX Binding Procedures
The transmitter and receiver have been pre-bound before delivery. The
following steps show the the binding procedures of transmitter and receiver (FS-i6
Instruction Manual, 2015):
1. Connect the supplied bind cable to the B/VCC port on the receiver.
2. Insert power into any other port.
3. Hold the ‘BIND KEY’ while powering on the transmitter to enter the bind mode.
4. Remove the power and bind cable from the receiber. Then connect the power cable
to the B/VCC port.
5. Check the servos’ operation. If not working as expected, restart this procedure from
the beginning.
Range Test Procedures
The limit of range is normally where the receiver can no longer clearly hear
what the transmitter telling it and typically falls in the 500m-600m range in normal
conditions.
In order to know the effective range, range test can be performed. Basically
range test can be done once the binding procedures are completed. The following
procedures shows the range test procedures (FS-i6 Instruction Manual, 2015):
1. Operator/pilot holds the transmitter.
2. Another helper moves the model away from the transmitter.
3. Check the model and mark the distance from where the model starts to lose
control.
47
Antenna Gain
In electromagnetics, an antenna’s power gain is a key performance number
which combines the antenna’s directivity and electrical efficiency. As transmitting
antenna, gain is describes as how well the antenna converts input power into radio
waves headed in a specified direction. As receiving antenna, gain is described as how
well the antenna converts radio waves
arriving from a specified direction into
electrical power.
Figure 5-6: High gain omnidirectional
antenna radiation pattern
High-gain omnidirectional antennas are
built for this Flysky FS-iA6 set. "Higher gain" means that the antenna radiates less
energy at higher and lower elevation angles and more in the horizontal directions.
High-gain omnidirectional antennas are generally realized using collinear dipole
arrays. The gain is increased which in turn, reduces the beamwidth, this can reeduce
interference as well. (mpnxt1, 2019) (Salt, 2019)Figure 5-6 represents the high gain
omnidirectional antenna radiation pattern.
Next, we investigate on the best antenna orientation for both transmitter and
receiver to get the best radio frequency energy. If we observe the 3D pattern from
Figure 5-6, we can see a big round doughnut shaped area of radio waves radiating
outward all around the vertical antenna, which also referred as “RF Doughnut”. The
strongest radio waves will be out to the sides all the way around the antenna (red), and
the weakest (the null zone) in line with both top and bottom tips of the antenna (green).
Therefore, it is advice to avoid having the point/tip of the TX antenna, pointing
directly at your aircraft. In that orientation it's radiating the least amount of RF energy
toward the model. The solution is to tilt the TX antenna from the vertical to the
horizontal (Figure 5-7) RF doughnut is now on its side and no worries about the
doughnut hole null zone out to the front and above. Another even better solution is
48
tilted the antenna up and forward (Figure 5-8) RF doughnut hole null zones are directed
up and out behind the pilot and down forward in front.
Figure 5-7: Horizontally Polarized RC Radio Antenna (Left)
Figure 5-8: Forward TX Antenna Position (Right)
Next, move on to the receiver antenna. Most RC manufacturers suggested to
orientate a 90 Polarization Diversity with vertical and horizontal polarized RC
antenna as shown in Figure 5-8. Polarization Diversity improves the RX's "listening"
ability in both horizontal and vertical planes as well. Bounced and reflected signals
that become polarized can also be detected in some instances by having more than one
plane of antenna polarization. (Salt, 2019)
Figure 5-9: 90 Degree Polarization Diversity
49
5.1.3 Electronic Speed Controller (ESC)
An electronic speed control or ESC is an electronic devices that controls and
regulates the speed of the brushless motor. (Electronic speed control, 2020) It may also
provide reversing of the motor and dynamic braking. The 3 main considerations when
selecting the right ESC are:
• Electric Motor: The size of the motor will greatly dictate the amount of amperage the
ESC must be able to handle.
• Propeller: Choice of propeller will also dictate what amp rating the ESC should be.
Will your drone be spinning 3, 4, 5, or 6 inch propellers? What kind of performance
will you want, and therefore what kind of propeller pitch will you use?
• Battery: ESCs are rated for amperage, along with battery cells.
The battery, motor and propeller selected are the 4S LiPo battery, SunnySky
X2216 1100kV Brushless Motors and APC 9045 Propeller. Therefore, to decide the
most suitable ESC, the motor performance chart (Table 5-3) from official SunnySky
website is referred.
From the chart, we found that with 9 inch APC propeller and 4S 14.8V LiPo
battery, the expected maximum current drawn at maximum throttle is approximately
21A. A suitable ESC need to have a 20% higher amp rating than the maximum current
drawn, which is 25.2A. Thus, a 30A rating ESC would be sufficient to support the
system.
Secondly, generally there are two type of ESC in the market, the programmable
and non programmable ESC. Programmable speed controls generally have user-
specified options which allow setting low voltage cut-off limits, timing, acceleration,
braking and direction of rotation. Reversing the motor's direction may also be
accomplished by switching any two of the three leads from the ESC to the motor.
Skywalker ESC is one of the trusted brand for programmable ESC, thus 30A
Skywalker ESC is selected.
50
Next, we found that a battery eliminator circuit (BEC) built in with the ESC is
a better option than an external BEC. A ESC with BEC onboard is cheaper, lighter
and less complicated wiring. The downside of built-in BEC is most of them has linear
circuit. The linear power supplys step down voltage by turning the extra wattage into
heat, which reduce the efficiency. Also, it is incapable of driving larger loads without
burning up, however 4 servos is still within the BEC capability, thus it is safe. Table
5-4 shows the specification of the selected 30A Skywalker ESC chosen.
Table 5-3: Sunnysky X2216 Motor Performance Chart (SunnySky X2216 Brushless Motors, n.d.)
Prop(inch) Voltage (V)
Amps (A)
Thrust (gf)
Watts (W)
Efficiency (g/W)
Load temperature in 100% throttle
APC9045 14.8
0.7 100 10.36 9.652509653
54°
1.7 200 25.16 7.949125596
2.7 300 39.96 7.507507508
4.2 400 62.16 6.435006435
5.7 500 84.36 5.926979611
7.3 600 108.04 5.553498704
9 700 133.2 5.255255255
10.6 800 156.88 5.099439062
12.5 900 185 4.864864865
14.6 1000 216.08 4.627915587
16.4 1100 242.72 4.531970995
21 1320 310.8 4.247104247
51
Table 5-4: ESC Specifications
Figure 5-10: Skywalker 30A
ESC
Model Skywalker 30A ESC
Continuous Current 30A
Burst Current 40A (10s)
BEC included
BEC output 5V/ 2A
BEC output capability 4 servos
Size 70 × 25 × 8 mm
Weight 38g
Programmable Yes
52
5.1.4 Battery
Battery plays an essential role in powering the motor, receiver and servo. The
common battery for RC plane are the NiMH (Nickel Metal Hydride), Li-ion (Lithium-
Ion), LiPo (Lithium Polymer) battery. Table 5-5 below are the comparison in term of
properties and capability. (Monti, 2010) (Differences Between NiMH and LiPo
Batteries, n.d.)
Table 5-5: Comparison of RC Battery NiMH battery Li-ion battery LiPo battery
Rechargable Yes
Memory effect exist
Yes
No Memory effect
Yes
No memory effect
Weight Heavy Medium Light
Energy density Least High Medium
Self discharge High High Low
Discharge rate Low High High
Capacitance Low High High
Suitability Simple project More complicated RC
project
More complicated
RC project
Price Lowest Medium Highest
Safety Reliable, common
battery type, safe
choice for beginners.
Not fully mature -
metals and chemicals
are changing on a
continuing basis.
higher resistant to
overcharge and
electrolyte leakage.
*Memory effect: Batteries must be fully discharged in order to keep full capacity
available.
* No memory effect: Batteries don’t have to be fully discharged before recharging.
Next, the major considerations when choosing the suitable battery include the
following:
• Voltage: The higher the voltage, the more power and higher rpm of motor will
achieve.
• Capacity: the current that could be discharged for an hour.
53
• Weight: High power/weight ratio is always preferable.
• Dimension: Able to fit in the compartment (fuselage).
• Connector: Compatible with ESC.
• Discharge rate: The rating of how many amps of power the battery can deliver.
• Safety: Trusted and reliable brand.
By comparing the type of batteries with above factors, we notice that LiPo
battery is the best option, although it comes with a higher price, but it is rechargable,
higher power/weight ratio, higher discharge rate and capacitance, and more safer
option for a RC airplane project. Therefore, a Maxpower LiPo battery is chosen. The
specifications is as below (Table 5-6).
Next, a 4S 14.8V of LiPo battery is chosen as it can produce a relatively more
power required than a 3S LiPo battery for a stable take off. Next, for a 4S 2200maH
35C LiPo pack we obtained:
(2200mAh × 35C) / 1000 = 77A
In other words, a 2200mAh / 35C LiPo can only handle an continuous current
draw of 77A at max. By referring the motor performance chart again (Table 3),
maximum current draw is only 21A, which means a 35C of discharge rate is sufficient.
(Salt, 11 Things to Know About LiPo Batteries to Get the Best Performance, Life,
Value & Fun Out of Them, Whatever You Fly., 2020)
As the requirement for the flight test is a 5 minutes of endurance, thus the
capacity of battery (2200mAh) has ensured to meet the requirement using formula as
below:
((2200mAh/1000) × 0.8 × 60) / 20A = 5.28 minutes
54
Table 5-6: Battery Specification
Figure 5-11: Maxpower Graphene
LiPo Battery
5.1.5 Servo
Servos provide the ability to move control surfaces. A typical servo consists of
an electric motor that moves a rotary actuator via a reduction gearset, normally
providing increased torque as well as clutch mechanism. Theoretically, each servo
output generates a standard PWM signal: the peak of the signal lasts for a period of
between one and two milliseconds, with a resolution control of 1 μs. So when the on-
time is 1ms the motor will be in 0° and when 1.5ms the motor will be 90°, similarly
when it is 2ms it will be 180°. So, by varying the on-time from 1ms to 2ms the motor
can be controlled from 0° to 180°. (Introduction to Servos, 2020) The figure represents
these parameters graphically.
Figure 5-12: PWM Signal in servo control
Model Maxpower Graphene Lipo
2200mAh 4S 14.8V 35C
Capacity 2200 mAh
Voltage 4S 14.8V
Discharge rate 35C
Size 106 × 34 × 33 mm
Weight 233g
Connector Dean Plug
55
The specifications of the servo selected is as shown in table 5-7. The servo
chosen is Arduino Tower Pro 9g SG90 Servo Motor. A total of 4 servos are purchased
for aileron, rudder and elevator respectively. The most important parameter, is the
torque at which the motor operates. The Towerpro SG90 Motor provides a 1.8kg/cm
torque at 4.8V voltage. In other words, this 1.8kg/cm torque means that the motor can
pull a force of 1.8kg when it is suspended at a distance of 1cm through pushrod. So if
the pushrod is 0.5cm long then the motor can pull a load of 3.6kg similarly if the
pushrod is 2cm long then can pull only 0.9kg.
Also, it is important to familiar with the wire configuration. There are 3 wires
with different colour and purposes respectively. To make this motor rotate, we have to
power the motor with +5V using the Red wire, while the Brown wire as ground wire
and PWM signal is given in through this Orange wire to drive the servo motor. (Servo
Motor SG-90, 2017)
Table 5-7: Servo Specification
Figure 5-13: Tower Pro 9g
SG90 Servo Motor
Model Arduino Tower Pro 9g SG90
Servo Motor
Modulation Analog
Operating voltage 4.8V
Stall torque (4.8V) 1.8 kg/cm
Operating speed (4.8V) 0.12 sec/ 60
Size 23 × 12.2 × 29 mm
Weight 9g
Gear type POM gear set
Operating Temperature 0C - 55C
Servo wire length 25cm
56
5.1.6 Wiring System Schematic Diagram
57
CHAPTER 6
WING
6.1 Introduction
6.1.1 Function of Wing
A wing is a type of fin that produces lift, while moving through air or some other fluid. As
such, wings have streamlined cross-sections that are subject to aerodynamic forces and act as
aerofoils. A wing's aerodynamic efficiency is expressed as its lift-to-drag ratio. The lift a wing
generates at a given speed and angle of attack can be one to two orders of magnitude greater than
the total drag on the wing. A high lift-to-drag ratio requires a significantly smaller thrust to propel
the wings through the air at sufficient lift.
The design and analysis of the wings of aircraft is one of the principal applications of the
science of aerodynamics, which is a branch of fluid mechanics. In principle, the properties of the
airflow around any moving object can be found by solving the Navier-Stokes’ equations of fluid
dynamics. However, except for simple geometries these equations are notoriously difficult to solve
and simpler equations are used.
58
Figure 6-1 Aerodynamic forces, airspeeds and pressure acting on wing aerofoil
For a wing to produce lift, it must be oriented at a suitable angle of attack. When this occurs,
the wing deflects the airflow downwards as it passes the wing. Since the wing exerts a force on
the air to change its direction, the air must also exert an equal and opposite force on the wing,
resulting in different air pressures over the surface of the wing. A region of lower-than-normal air
pressure is generated over the top surface of the wing, with a higher pressure on the bottom of the
wing. These air pressure differences can be measured directly using instrumentation or can be
calculated from the airspeed distribution using basic physical principles such as Bernoulli's
principle, which relates changes in air speed to changes in air pressure.
It is possible to calculate lift from: the pressure differences, the different velocities of the
air above and below the wing, or from the total momentum change of the deflected air. Debates
over which mathematical approach is the most convenient to use can be mistaken as differences
of opinion about the basic principles of flight.
6.1.2 Aircraft Wing Design
In our RC place design, we implemented the high wing design by considering some of the
factors. Each wing configuration is beneficial in its unique way for training, performance,
maintenance, and everyday use. High wing aircraft are generally less aerodynamic, slightly easier
to train in for the new pilot, and easier to access for routine maintenance than low wing aircraft.
59
For high and low wing aircraft, major differences will occur in the following regimes:
lateral (roll) stability, low speed handling characteristics, and general aircraft performance (cruise
speed, take-off and landing distances etc.).
For the high wing aircraft, the centre of gravity sits below the wing, meaning the fuselage
of the aircraft acts as a pendulum to increase roll stability relative to the low wing aircraft, whose
centre of gravity is balanced above the wing.
Figure 6-2 Dimension of the rectangular wing
Rectangular wing design is chosen as it can bring a lot of advantages such as act as wing-
root stallers. For the high wing aircraft, the centre of gravity sits below the wing, meaning the
fuselage of the aircraft acts as a pendulum to increase roll stability relative to the low wing aircraft,
whose centre of gravity is balanced above the wing.
According to Part 23 of Federal Aviation Regulations, the criteria of wing design that are
required to abide by are as follows,
60
1. A main wing located closer to the airplane's center of gravity than to the aft, fuselage-
mounted, empennage;
2. A main wing that contains a quarter-chord sweep angle of not more than 15 degrees
fore or aft;
3. A main wing that is equipped with trailing-edge controls (ailerons or flaps, or both);
4. A main wing aspect ratio not greater than 7.
Table 6-1 Specification of the wing
Type High Wing
Airfoil Selection NACA2412
Aspect Ratio 7.06
Chord Length 0.17m
Wing Area 0.20387m
2
Wingspan 1.2m
6.1.3 Wing Load Analysis
In this study, our RC plane is designed and analyzed. The design process starts with a
sketch of how the airplane is envisioned. Weight is estimated based on the sketch and a chosen
design mission profile. A more refined method is conducted based on calculated performance
parameters to achieve a more accurate weight estimate which is used to acquire the external
geometry of the airplane.
In order to obtain a more refined estimation of the airplane weight several performance
parameters must be estimated such as the maximum lift-to-drag ratio, the maximum lift coefficient,
the wing loading and the thrust-to-weight ratio.
According to Federal Aviation Regulations (FAR), aircraft loads are those forces and
loadings applied to the aircraft structural components to establish the strength level of the
61
completed aircraft. Those are the loads that an aircraft structure must be designed to withstand. A
large part of the forces that make up design loads are the forces resulting from the flow of air about
the aircraft surfaces – the same forces that enables flight and control of the aircraft. The loads that
must be determined early in the structural design.
In normal straight and level flight, the wing lift supports the weight of the aircraft.
However, the greatest air loads on an aircraft usually come from the generation of lift during high-
g manoeuvres. The amount of additional loads depends on the severity or turbulence (during gust),
and its magnitude is measures in terms of load factor, n. load factor is a multiplying factor which
defines a load in terms of weight.
At lower speeds, the highest load factor an aircraft may experience is limited by the lift
available, and it is limited to some arbitrary value based upon the operation of the aircraft. At high
angle of attack, the load direction may actually be forward of the aircraft body-axis vertical,
causing a forward load component on the wing structure.
The aircraft maximum speed, or dive speed 𝑉𝐷𝑖𝑣𝑒 represents the maximum dynamic
pressure, q. The point representing maximum q and maximum load factor is vital for structural
sizing. At this condition, the aircraft is at a fairly low angle of attack because of the high dynamic
pressure, so the load is approximately vertical in the body axis.
Compliance with the strength requirements must be shown at any combination of airspeed
and load factor on and within the boundaries of a flight envelope, that represents the flight loading
conditions specified by the manoeuvring (excluding gust) criteria in our case. In manoeuvring
envelope, the aircraft is assumed to be subjected to symmetrical manoeuvres resulting in the
following limit load factors.
1. The positive manoeuvring load factor (+3.0) at speeds up to 𝑉𝐷𝑖𝑣𝑒.
2. The negative manoeuvring load factor (-1.5) at 𝑉𝐶 (Design cruising speed).
62
The process is then followed by the analysis of the wing parts such as spar, skin, rib and
also shear flow of the wing.
Figure 6-3 Wing components
In a fixed-wing aircraft, the spar is often the main structural member of the wing, running
span wise at right angles (or thereabouts depending on wing sweep) to the fuselage. The spar
carries flight loads and the weight of the wings while on the ground. The wing rib of an aircraft is
a very critical part. It provides airfoil contour to the wing. Its principal role in wing structure is
that to transfer load from skin to stringers or other parts of wing.
6.1.4 Aircraft’s Airworthiness
To maintain a high level of safety, the aviation industry is heavily regulated. The design
process of an aircraft is regulated by airworthiness authority to ensure the aircraft’s suitability for
safe flight. In this project, the Federal Aviation Regulation (FAR), Part 23 – Airworthiness
Standards: Normal Category Airplanes, is applied in the design process and summarized in Table
6-2, for utility aircraft. Also, our project requirement is align with the FAR as shown in Table 6-
3.
63
Table 6-2 Federal Aviation Regulations, Part 23 design requirements and specifications
Table 6-3 Project Requirement on Wing Design
Project Requirement
Maximum take-off weight <5kg
Positive load factor, n +3.0
Negative load factor -1.5
6.1.5 Assumptions on Wing Design
There are a few assumptions outlined on the wing design, which can be listed as follows,
1. The wing is assumed to be a 2D wing, at which the factors of 3D wing design will not be
taken into account.
2. The value of maneuvering speed 𝑉𝐴 need not exceed the value of cruising speed 𝑉𝐶 used
in design, as stated in FAR Section 23.335.
Utility Aircraft: FAR 14
Maximum take-off weight 12,500 lbs (5700kg) 14 CFR23.3
Positive load factor, n +4.4 14 CFR23.337
Negative load factor -1.76 (> 0.4 times positive n)
Design airspeeds 𝑉𝐷𝑖𝑣𝑒 > 1.5 𝑉𝑐𝑚𝑖𝑛 14 CFR23.335
64
6.2 Flight Envelope and Wing Loadings
6.2.1 Flight Envelope (V-n Diagram)
In aerodynamics, the flight envelope defines operational limits for an aerial platform with
respect to maximum speed and load factor given a particular atmospheric density. The flight
envelope is the region within which an aircraft can operate safely. If an aircraft flies 'outside the
envelope' it may suffer damage; the limits should therefore never be exceeded.
To construct the flight envelope, the load factor limits are +3.0 and -1.5 according to the
project requirement, as per FAR Part 23. Also, there are various important features of the V-n
diagram at point A, D, E, and G.
The intersection of the positive limit of the load factor and the line of maximum lift (point
A) defines the maximum airspeed that allows full manoeuverability. This point is called the
manoeuver speed or corner speed. At lower speeds, the structure cannot be overstressed as it will
stall before reaching the limit load factor. At the manoeuver airspeed the aircraft's limit load factor
will be reached at the lowest possible airspeed. At higher speeds, possible structural damage may
be caused.
Besides, the intersection of the negative limit load factor and line of maximum negative
lift capability (point G) defines the maximum airspeed that allows full manoeuverability in a
negative lift situation. Airspeeds greater than point G provide sufficient negative lift to damage
the structure.
Additionally, the diagram defines the never exceed speed or diving speed (point D and E).
This is the maximum speed before the aircraft enters the region where structural failure is possible.
65
Positive load factor
𝑛𝑚𝑎𝑥 = 3
Maneuver speed (Positive Limit)
𝑉𝐴 = √𝑛𝑚𝑎𝑥𝑝𝑜𝑠
𝑊
0.5𝜌𝑆𝐶𝐿𝑚𝑎𝑥
𝑉𝐴 = √3(14.715)
0.5(1.225)(0.203871)(1.29) = 16.5545 𝑚/𝑠
Design Dive Speed
𝑉𝐷 = 1.2𝑉𝑚𝑎𝑥
𝑉𝐷 = 1.2(28)
𝑉𝐷 = 33.6 𝑚/𝑠
Maximum Negative Load Factor
𝑛𝑚𝑎𝑥𝑛𝑒𝑔= −1.5
Maneuver Speed (Negative Limit)
𝑉𝐺 = √𝑛𝑚𝑎𝑥𝑛𝑒𝑔
𝑊
0.5𝜌𝑆𝐶𝐿𝑚𝑎𝑥
𝑉𝐺 = √−1.5(14.715)
0.5(1.225)(0.203871)(−0.6647)
𝑉𝐺 = 16.3073 𝑚/𝑠
For the curve part of the V-n diagram, the curve is drawn by varying airspeed before maneuvering
speed with equation:
𝑛 =0.5𝜌𝑉2𝑆𝐶𝐿𝑚𝑎𝑥
𝑊 (for positive V − n)
66
For positive V-n region, airspeed varies from 0 to 16.5545 ft/s.
𝑛 =0.5𝜌𝑉2𝑆(−𝐶𝐿𝑚𝑎𝑥
)
𝑊 (for negative V − n)
For negative V-n region, airspeed varies from 0 to 16.3073 m/s.
Figure 6-4 Flight Envelope
6.2.2 Wing Loading without Aileron
Normal Lift Force
Firstly, the lift coefficient, 𝐶𝐿 is calculated from the equation:
𝐶𝐿 =𝑛𝑊
12⁄ 𝜌𝑉2𝑆
where,
𝑊 : maximum take-off weight
𝑛 : load factor at points A, D, E and G of V-n diagram
𝐶L : lift coefficient
67
𝜌 : air density
𝑉 : stall velocity at points A, D, E and G of V-n diagram
𝑆 : wing area
Since assumption was made that the wing was a 2D wing, all the aerodynamic coefficient
are obtained, based on 2D wing analysis. the section aerodynamic angle of attack, 𝛼 is then
determined from the graph of 𝐶𝐿 𝑣𝑠 𝛼 for NACA 2412 wing section in Appendix I. Then, the
aerofoil section lift coefficient, 𝐶𝑙 and drag coefficient, 𝐶𝑑 are determined from graphs in Appendix
I and substituted into the following equations, which are derived from resolving aerodynamic
forces in Figure 6-5, to find the lift force acting perpendicular to aerofoil, at each station for points
A, D, E and G of V-n diagram.
Figure 6-5 Resultant aerodynamic force and the components into which it splits (Anderson, 2010)
𝐶𝑛=𝐶𝑙𝑐𝑜𝑠𝛼+𝐶𝑑𝑠𝑖𝑛𝛼
𝑁 =1
2𝐶𝑛𝜌𝑉2𝑆
68
Shear Force
The changes in shear force, Δ𝑉𝑛 at each station is given by,
Δ𝑉𝑛 =𝑁𝑛+1 + 𝑁𝑛
2(𝑦𝑛+1 − 𝑦𝑛)
and the corresponding shear force, 𝑉𝑛 at each station is,
𝑉𝑛 = ∆𝑉20 + ∆𝑉19 + ⋯+ ∆𝑉𝑛
Bending Moment
The changes in bending and moment distribution, Δ𝑀𝑛 at each station is given by,
Δ𝑀𝑛 =𝑉𝑛+1 + 𝑉𝑛
2(𝑦𝑛+1 − 𝑦𝑛)
and the corresponding bending moment, 𝑀𝑛 at each station is,
𝑀𝑛 = ∆𝑀20 + ∆𝑀19 + ⋯+ ∆𝑀𝑛
Torsional Moment
The value of moment coefficient (𝐶𝑚) can be obtained in similar way as how angle of
attack is determined. At each station, the 𝐶𝑚 value can be determined from the graph in Appendix
I for points A, D, E and G of V-n diagram.
where,
T : Torsion
𝜌 : air density
𝑉 : stall velocity at points A, D, E and G of V-n diagram
𝑆=cdy : wing area at each station, y
c : chord length at each station, y
𝐶𝑚 : moment coefficient
𝑇 =1
2𝐶𝑚𝜌𝑉2𝑆𝑐
69
The calculated results are tabulated in the following tables for all points A, D, E and G while the
graphs are plotted in figures in section 6.2.5.
6.2.3 Wing Loading with Aileron
Change in Lift Coefficient
Roll control can be achieved by the differential deflection of ailerons. The deflection of
aileron controls the increment of lift and the incremental change in roll moment. Downward aileron
deflection will increase the lift coefficient value while upward aileron deflection will decrease the
lift coefficient value. The deflection of aileron, either upward or downward, will affect the resultant
normal lift acting on the aircraft’s wing. McCormick (1995) gives the following formula for the
section lift coefficient increment due to deflection of ailerons.
Where
𝐶𝑙𝛼 :Airfoil section lift-curve slope
τ :Flap effectiveness factor
η :Correction factor to flap effectiveness factor
δ :Deflection angle
Figure 6-6 the parameters c, cf and δ in aerofoil with plain aileron (McCormick, 1995)
∆𝐶𝐿 = 𝐶𝑙𝛼𝜏η𝛿
70
Figure 6-7 Position and dimension of ailerons
Sample calculation is done for aircraft’s wing at position A. From Figure 1 in Appendix II,
τ = 0.64 for cf/c = 0.24, and from Figure 2 in Appendix II, η = 0.53 for a plain flap deflected 30o.
Hence, from Equation, ∆𝐶𝐿 is equal to
∆𝐶𝐿 = 𝐶𝑙𝛼𝜏η𝛿
∆𝐶𝐿 = (0.2623)(0.64)(0.53)(30)(𝜋
180)
∆𝐶𝐿 = 0.04659
From Figure 3 in Appendix II, the ratio of ∆𝐶𝐿𝑚𝑎𝑥 to ∆𝐶𝐿 is obtained as 0.75. Hence,
∆𝐶𝐿𝑚𝑎𝑥 = 0.03494
Sample calculation is done for aircraft’s wing at position D. From Figure 1 in Appendix II,
τ = 0.64 for cf/c = 0.24, and from Figure 2 in Appendix II, η = 0.8 for a plain flap deflected 10o.
Hence, from Equation, ∆𝐶𝐿 is equal to
∆𝐶𝐿 = 𝐶𝑙𝛼𝜏η𝛿
∆𝐶𝐿 = (0.2623)(0.64)(0.8)(10)(𝜋
180)
∆𝐶𝐿 = 0.02344
From Figure 3 in Appendix II, the ratio of ∆𝐶𝐿𝑚𝑎𝑥 to ∆𝐶𝐿 is obtained as 0.75. Hence,
∆𝐶𝐿𝑚𝑎𝑥 = 0.01758
71
Position A and position G will have the same maximum aileron deflection while the
maximum aileron deflection at position D and E is the same. Based on the roll control analysis, to
control the roll, when the critical speed is too high, the approximated maximum aileron deflection
is 10o, while when the critical speed is lower, the maximum aileron deflection is approximated to
be 30o. When the aileron deflects downwards, the total 𝐶𝐿 value for the wing section with aileron
is the addition of value of ∆𝐶𝐿𝑚𝑎𝑥 and the value of 𝐶𝐿 at the angle of attack in position A, D, E and
G respectively. On the other hand, when the aileron deflects upwards, the total 𝐶𝐿 value for the
wing section with aileron is the subtraction of value of ∆𝐶𝐿𝑚𝑎𝑥 and the value of 𝐶𝐿.
Change in Moment Coefficient
The roll control of the aileron also affects the value of moment coefficient 𝐶𝑚. The
estimated change in 𝐶𝑚 value can be determined from Figure 6-8. The figure shows that the graph
shifts down when the aileron deflects with an angle of 60o. As our maximum aileron deflection
angle is 30o and 10o for position A and G as well as position D and E respectively, interpolation
method is used to obtain the approximated change in moment coefficient value.
Figure 6-8 Aerodynamic characteristics of NACA 2412 airfoil section (Abbott, 1945)
72
6.2.4 Reaction force of Wing Strut
The reaction force of the wing strut is necessary when considering the wing loading, as this
affects the shear force diagram and bending moment diagram. To determine the reaction force of
the wing strut, the analysis of the wing loading must first be conducted for half of the wingspan.
The simplification of the problem is shown as follows,
Figure 6-9 Front View of the Aircraft
Figure 6-10 Free Body Diagram (before simplification)
The positive convention of the reaction force and moment is indicated in Figure 6-10. The
aircraft’s wing is assumed to be a cantilevered beam. The reaction force of the wing strut is
indicated at point B, which is located at the middle of the wingspan. The horizontal reaction forces,
RxA and RxB will cancel out each other at the end of the calculation. During initial stage of design,
since the wing weight is smaller, as compared to other component weight of the aircraft, the weight
73
of the aircraft can be assumed to be concentrated at a point A. The applied load, which is indicated
with the orange line, shows the total concentrated weight of the whole aircraft.
Figure 6-11 Free Body Diagram (after simplification)
The simplified problem now is a statically indeterminate structure. The reaction force of
the wing strut at point B can be determined by using the flexibility method, according to the
Castigliano’s Theorem. Since the system has three reaction forces, RyA, RyB and MA, the degree
of statically indeterminate structures can be determined through the equation below,
Q = [Total number of reactions] – [Total number of Static Equations]
Where, Q = Redundancy
The general equations of the flexibility method is shown as follows,
𝛿𝑖𝑛 = 𝛿𝑖𝑛0 + ∑ 𝑋𝑖𝑚
𝑘𝑚=1 ∙ 𝑓𝑚𝑛 n=1,2,3, … k
𝛿𝑖𝑛0 = ∫
𝑀𝐿(𝑥)𝑀𝑢𝑛(𝑥)
𝐸𝐼𝑑𝑥
𝑓𝑚𝑛 = ∫𝑀𝑢𝑚(𝑥)𝑀𝑢𝑛(𝑥)
𝐸𝐼𝑑𝑥
Where,
𝑋𝑖𝑚 = Parameter referring to redundancy
𝛿𝑖𝑛0 = Displacement of statically determinate system
𝑓𝑚𝑛 = flexibility
74
𝑀𝐿(𝑥) = Bending moment from actual load
𝑀𝑢𝑚(𝑥) = Bending moment for unit load in reference to redundancy
𝑀𝑢𝑛(𝑥) = Bending moment for unit load due to deflection
Sample Calculation at Position A
𝑄 = 3 − 2 = 1 𝑟𝑒𝑑𝑢𝑛𝑑𝑎𝑛𝑐𝑦
Take 𝑅𝐵 as the redundancy,
For release system,
Figure 6-12 Release System
𝑀𝐿 = −𝑞𝑥2
2
For unit load,
Figure 6-13 Unit load system
𝑀𝑢1 = 0 𝑓𝑜𝑟 0 < 𝑥 <𝐿
2
𝑀𝑢1 = (𝑥 −𝐿
2) 𝑓𝑜𝑟
𝐿
2< 𝑥 < 𝐿
The general equation,
𝛿𝐵10 = ∫
𝑀𝐿(𝑥)𝑀𝑢𝑛(𝑥)
𝐸𝐼𝑑𝑥
𝐿
𝐿2
75
𝛿𝐵10 = ∫
−𝑞𝑥2
2(𝑥 −
𝐿2)
𝐸𝐼𝑑𝑥
𝐿
𝐿/2
𝛿𝐵10 = −
𝑞𝐿2
8𝐸𝐼+
𝑞 (𝐿2)2
8𝐸1+ [
𝑞𝐿4
12𝐸𝐼−
𝑞𝐿 (𝐿2)3
12𝐸𝐼]
𝛿𝐵10 = −
17𝑞𝐿4
384𝐸𝐼
𝑓𝑚𝑛 = ∫(𝑥 −
𝐿2) 2
𝐸𝐼𝑑𝑥
𝐿
𝐿/2
𝑓𝑚𝑛 =1
𝐸𝐼∫ (𝑥2 − 𝑥𝐿 +
𝐿2
4)𝑑𝑥
𝐿
𝐿/2
𝑓𝑚𝑛 =𝐿3
24𝐸𝐼
Therefore,
𝛿𝐵1 = 𝛿𝐵10 + 𝑥1 ∙ 𝑓11 = 0
−17𝑞𝐿4
384𝐸𝐼+
𝐿3
24𝐸𝐼𝑥1 = 0
𝑥1 = 𝑅𝐵 = 1.0625𝑞𝐿
At position D, q = 44.18931 N/m, L = 0.6m
𝑅𝐵 = 28.1707 𝑁 (𝑎𝑐𝑡𝑖𝑛𝑔 𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑)
To determine the reaction force of the wing strut,
Figure 6-14 Reaction force of wing strut
𝑅𝑠𝑡𝑟𝑢𝑡 =𝑅𝐵
sin(19.46)
76
𝑅𝑠𝑡𝑟𝑢𝑡 =27.7721
sin(19.46)= 84.5589
77
6.2.5 Shear Force Diagram and Bending Moment Diagram
Table 6-4 Wing loading in each station for point A (Aileron deflects downwards)
y (m) y/b α (o) Cl Cd Cn Va (N) N (N) Delta V Vn (N) Delta M M (Nm) Cm T (Pa)
0 0.0000 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.3941 -1.3938 -0.0669 0.0179 -0.1 -0.58184
0.032 0.0533 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.3505 -2.7879 -0.1074 0.0848 -0.1 -0.58184
0.063 0.1050 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.3941 -4.1384 -0.1547 0.1921 -0.1 -0.58184
0.095 0.1583 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.3505 -5.5324 -0.1924 0.3469 -0.1 -0.58184
0.126 0.2100 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.3941 -6.8829 -0.2426 0.5393 -0.1 -0.58184
0.158 0.2633 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.3505 -8.2770 -0.2775 0.7819 -0.1 -0.58184
0.189 0.3150 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.3941 -9.6275 -0.3304 1.0594 -0.1 -0.58184
0.221 0.3683 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.3941 -11.0215 -0.3750 1.3898 -0.1 -0.58184
0.253 0.4217 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.3505 -12.4156 -0.4058 1.7648 -0.1 -0.58184
0.284 0.4733 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 0.0436 -13.7661 -0.0138 2.1706 -0.1 -0.58184
0.285 0.4750 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 -13.8425 -13.8096 -0.0069 2.1844 -0.1 -0.58184
0.286 0.4767 12.0179 1.2898 0.0543 1.2728 16.5545 -27728.53 -13.8425 0.0329 0.0070 2.1912 -0.1 -0.58184
0.287 0.4783 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.2634 13.8754 0.3841 2.1843 -0.1 -0.58184
0.316 0.5267 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.3505 12.6120 0.3700 1.8002 -0.1 -0.58184
0.347 0.5783 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.3941 11.2615 0.3381 1.4302 -0.1 -0.58184
0.379 0.6317 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.4128 9.8675 0.2932 1.0921 -0.1 -0.58184
0.411 0.6850 12.0179 1.3248 0.0543 1.3070 16.5545 44.7338 1.3867 8.4547 0.2406 0.7990 -0.54 -3.14194
0.442 0.7367 12.0179 1.3248 0.0543 1.3070 16.5545 44.7338 1.4315 7.0679 0.2033 0.5584 -0.54 -3.14194
0.474 0.7900 12.0179 1.3248 0.0543 1.3070 16.5545 44.7338 1.3867 5.6365 0.1532 0.3551 -0.54 -3.14194
0.505 0.8417 12.0179 1.3248 0.0543 1.3070 16.5545 44.7338 1.4315 4.2497 0.1131 0.2019 -0.54 -3.14194
0.537 0.8950 12.0179 1.3248 0.0543 1.3070 16.5545 44.7338 1.3867 2.8182 0.0659 0.0888 -0.54 -3.14194
0.568 0.9467 12.0179 1.3248 0.0543 1.3070 16.5545 44.7338 1.4315 1.4315 0.0229 0.0229 -0.54 -3.14194
0.6 1.0000 12.0179 1.3248 0.0543 1.3070 16.5545 44.7338 0.0000 0.0000 0.0000 0.0000 -0.54 -3.14194 0.0000 -1.3938 0.0179
78
Table 6-5 Wing loading in each station for point D (Aileron deflects downwards)
y (m) y/b α (o) Cl Cd Cn Va (N) N (N) Delta V Vn (N) Delta M M (Nm) Cm T (Pa)
0 0.0000 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.4141 -1.1490 -0.0594 0.1497 -0.10 -2.3969
0.032 0.0533 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.3699 -2.5631 -0.1007 0.2091 -0.10 -2.3969
0.063 0.1050 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.4141 -3.9329 -0.1485 0.3098 -0.10 -2.3969
0.095 0.1583 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.3699 -5.3470 -0.1870 0.4583 -0.10 -2.3969
0.126 0.2100 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.4141 -6.7168 -0.2376 0.6453 -0.10 -2.3969
0.158 0.2633 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.3699 -8.1309 -0.2733 0.8828 -0.10 -2.3969
0.189 0.3150 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.4141 -9.5008 -0.3266 1.1561 -0.10 -2.3969
0.221 0.3683 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.4141 -10.9148 -0.3719 1.4828 -0.10 -2.3969
0.253 0.4217 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.3699 -12.3289 -0.4034 1.8547 -0.10 -2.3969
0.284 0.4733 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 0.0442 -13.6988 -0.0137 2.2581 -0.10 -2.3969
0.285 0.4750 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 -14.0412 -13.7429 -0.0067 2.2718 -0.10 -2.3969
0.286 0.4767 0.3376 0.3131 0.0543 0.3134 33.6000 -28126.51 -14.0412 0.2982 0.0073 2.2786 -0.10 -2.3969
0.287 0.4783 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.2815 14.3394 0.3973 2.2712 -0.10 -2.3969
0.316 0.5267 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.3699 13.0579 0.3836 1.8740 -0.10 -2.3969
0.347 0.5783 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.4141 11.6880 0.3514 1.4904 -0.10 -2.3969
0.379 0.6317 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.4537 10.2740 0.3055 1.1390 -0.10 -2.3969
0.411 0.6850 0.3376 0.3307 0.0543 0.3310 33.6000 46.6679 1.4467 8.8202 0.2510 0.8335 -0.27 -6.4716
0.442 0.7367 0.3376 0.3307 0.0543 0.3310 33.6000 46.6679 1.4934 7.3735 0.2121 0.5825 -0.27 -6.4716
0.474 0.7900 0.3376 0.3307 0.0543 0.3310 33.6000 46.6679 1.4467 5.8802 0.1599 0.3705 -0.27 -6.4716
0.505 0.8417 0.3376 0.3307 0.0543 0.3310 33.6000 46.6679 1.4934 4.4335 0.1180 0.2106 -0.27 -6.4716
0.537 0.8950 0.3376 0.3307 0.0543 0.3310 33.6000 46.6679 1.4467 2.9401 0.0687 0.0926 -0.27 -6.4716
0.568 0.9467 0.3376 0.3307 0.0543 0.3310 33.6000 46.6679 1.4934 1.4934 0.0239 0.0239 -0.27 -6.4716
0.6 1.0000 0.3376 0.3307 0.0543 0.3310 33.6000 46.6679 0.0000 0.0000 0.0000 0.0000 -0.27 -6.4716
0.0000 -1.1490 0.1497
79
Table 6-6 Wing loading in each station for point E (Aileron deflects downwards)
y (m) y/b α (o) Cl Cd Cn Va (N) N (N) Delta V Vn (N) Delta M M (Nm) Cm T (Pa)
0 0.0000 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.7146 1.3452 0.0545 0.3042 -0.10 -2.3969
0.032 0.0533 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.6923 2.0598 0.0746 0.2497 -0.10 -2.3969
0.063 0.1050 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.7146 2.7521 0.0995 0.1752 -0.10 -2.3969
0.095 0.1583 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.6923 3.4667 0.1182 0.0757 -0.10 -2.3969
0.126 0.2100 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.7146 4.1590 0.1445 -0.0425 -0.10 -2.3969
0.158 0.2633 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.6923 4.8736 0.1618 -0.1871 -0.10 -2.3969
0.189 0.3150 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.7146 5.5659 0.1895 -0.3489 -0.10 -2.3969
0.221 0.3683 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.7146 6.2806 0.2124 -0.5384 -0.10 -2.3969
0.253 0.4217 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.6923 6.9952 0.2276 -0.7508 -0.10 -2.3969
0.284 0.4733 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.0223 7.6875 0.0077 -0.9784 -0.10 -2.3969
0.285 0.4750 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 7.0959 7.7098 0.0042 -0.9861 -0.10 -2.3969
0.286 0.4767 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 14214.17 7.0959 0.6139 -0.0029 -0.9903 -0.10 -2.3969
0.287 0.4783 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.6476 -6.4820 -0.1786 -0.9873 -0.10 -2.3969
0.316 0.5267 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.6923 -5.8344 -0.1701 -0.8087 -0.10 -2.3969
0.347 0.5783 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.7146 -5.1421 -0.1531 -0.6386 -0.10 -2.3969
0.379 0.6317 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.6750 -4.4275 -0.1309 -0.4855 -0.10 -2.3969
0.411 0.6850 -2.0480 -0.1390 0.0543 -0.1408 -33.6000 -19.8547 -0.6155 -3.7525 -0.1068 -0.3546 -0.25 -5.9923
0.442 0.7367 -2.0480 -0.1390 0.0543 -0.1408 -33.6000 -19.8547 -0.6354 -3.1370 -0.0902 -0.2478 -0.25 -5.9923
0.474 0.7900 -2.0480 -0.1390 0.0543 -0.1408 -33.6000 -19.8547 -0.6155 -2.5017 -0.0680 -0.1576 -0.25 -5.9923
0.505 0.8417 -2.0480 -0.1390 0.0543 -0.1408 -33.6000 -19.8547 -0.6354 -1.8862 -0.0502 -0.0896 -0.25 -5.9923
0.537 0.8950 -2.0480 -0.1390 0.0543 -0.1408 -33.6000 -19.8547 -0.6155 -1.2508 -0.0292 -0.0394 -0.25 -5.9923
0.568 0.9467 -2.0480 -0.1390 0.0543 -0.1408 -33.6000 -19.8547 -0.6354 -0.6354 -0.0102 -0.0102 -0.25 -5.9923
0.6 1.0000 -2.0480 -0.1390 0.0543 -0.1408 -33.6000 -19.8547 0.0000 0.0000 0.0000 0.0000 -0.25 -5.9923
0.0000 1.3452 0.3042
80
Table 6-7 Wing loading in each station for point G (Aileron deflects downwards)
y (m) y/b α (o) Cl Cd Cn Va (N) N (N) Delta V Vn (N) Delta M M (Nm) Cm T (Pa)
0 0.0000 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.7075 1.0646 0.0454 0.1684 -0.10 -0.5646
0.032 0.0533 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.6854 1.7720 0.0656 0.1230 -0.10 -0.5646
0.063 0.1050 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.7075 2.4574 0.0900 0.0575 -0.10 -0.5646
0.095 0.1583 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.6854 3.1649 0.1087 -0.0325 -0.10 -0.5646
0.126 0.2100 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.7075 3.8502 0.1345 -0.1412 -0.10 -0.5646
0.158 0.2633 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.6854 4.5577 0.1519 -0.2757 -0.10 -0.5646
0.189 0.3150 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.7075 5.2431 0.1791 -0.4276 -0.10 -0.5646
0.221 0.3683 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.7075 5.9506 0.2017 -0.6067 -0.10 -0.5646
0.253 0.4217 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.6854 6.6580 0.2170 -0.8085 -0.10 -0.5646
0.284 0.4733 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.0221 7.3434 0.0074 -1.0255 -0.10 -0.5646
0.285 0.4750 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 7.0249 7.3655 0.0039 -1.0329 -0.10 -0.5646
0.286 0.4767 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 14071.99 7.0249 0.3406 -0.0032 -1.0367 -0.10 -0.5646
0.287 0.4783 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.6411 -6.6844 -0.1846 -1.0335 -0.10 -0.5646
0.316 0.5267 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.6854 -6.0432 -0.1767 -0.8490 -0.10 -0.5646
0.347 0.5783 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.7075 -5.3579 -0.1601 -0.6723 -0.10 -0.5646
0.379 0.6317 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.6891 -4.6504 -0.1378 -0.5121 -0.10 -0.5646
0.411 0.6850 -8.0000 -0.6297 0.0543 -0.6311 -16.3073 -20.9594 -0.6497 -3.9613 -0.1127 -0.3743 -0.505 -2.8512
0.442 0.7367 -8.0000 -0.6297 0.0543 -0.6311 -16.3073 -20.9594 -0.6707 -3.3116 -0.0952 -0.2616 -0.505 -2.8512
0.474 0.7900 -8.0000 -0.6297 0.0543 -0.6311 -16.3073 -20.9594 -0.6497 -2.6409 -0.0718 -0.1664 -0.505 -2.8512
0.505 0.8417 -8.0000 -0.6297 0.0543 -0.6311 -16.3073 -20.9594 -0.6707 -1.9911 -0.0530 -0.0946 -0.505 -2.8512
0.537 0.8950 -8.0000 -0.6297 0.0543 -0.6311 -16.3073 -20.9594 -0.6497 -1.3204 -0.0309 -0.0416 -0.505 -2.8512
0.568 0.9467 -8.0000 -0.6297 0.0543 -0.6311 -16.3073 -20.9594 -0.6707 -0.6707 -0.0107 -0.0107 -0.505 -2.8512
0.6 1.0000 -8.0000 -0.6297 0.0543 -0.6311 -16.3073 -20.9594 0.0000 0.0000 0.0000 0.0000 -0.505 -2.8512
0.0000 1.0646 0.1684
81
Table 6-8 Wing loading in each station for point A (Aileron deflects upwards)
y (m) y/b Cl Cn N (N) Delta V Vn (N) Delta M M (N) Cm T (Pa)
0 0.0000 1.2898 1.2728 43.5641 1.3941 -1.8734 -0.0823 -0.2204 -0.100 -0.5818
0.032 0.0533 1.2898 1.2728 43.5641 1.3505 -3.2674 -0.1222 -0.1382 -0.100 -0.5818
0.063 0.1050 1.2898 1.2728 43.5641 1.3941 -4.6179 -0.1701 -0.0159 -0.100 -0.5818
0.095 0.1583 1.2898 1.2728 43.5641 1.3505 -6.0120 -0.2073 0.1541 -0.100 -0.5818
0.126 0.2100 1.2898 1.2728 43.5641 1.3941 -7.3625 -0.2579 0.3614 -0.100 -0.5818
0.158 0.2633 1.2898 1.2728 43.5641 1.3505 -8.7565 -0.2924 0.6193 -0.100 -0.5818
0.189 0.3150 1.2898 1.2728 43.5641 1.3941 -10.1070 -0.3457 0.9117 -0.100 -0.5818
0.221 0.3683 1.2898 1.2728 43.5641 1.3941 -11.5011 -0.3903 1.2575 -0.100 -0.5818
0.253 0.4217 1.2898 1.2728 43.5641 1.3505 -12.8951 -0.4207 1.6478 -0.100 -0.5818
0.284 0.4733 1.2898 1.2728 43.5641 0.0436 -14.2456 -0.0143 2.0685 -0.100 -0.5818
0.285 0.4750 1.2898 1.2728 43.5641 -13.8425 -14.2892 -0.0074 2.0828 -0.100 -0.5818
0.286 0.4767 1.2898 1.2728 -27728.5359 -13.8425 -0.4467 0.0065 2.0901 -0.100 -0.5818
0.287 0.4783 1.2898 1.2728 43.5641 1.2634 13.3958 0.3702 2.0836 -0.100 -0.5818
0.316 0.5267 1.2898 1.2728 43.5641 1.3505 12.1324 0.3552 1.7135 -0.100 -0.5818
0.347 0.5783 1.2898 1.2728 43.5641 1.3941 10.7819 0.3227 1.3583 -0.100 -0.5818
0.379 0.6317 1.2898 1.2728 43.5641 1.3753 9.3879 0.2784 1.0356 -0.100 -0.5818
0.411 0.6850 1.2549 1.2387 42.3945 1.3142 8.0126 0.2280 0.7572 0.340 1.9783
0.442 0.7367 1.2549 1.2387 42.3945 1.3566 6.6983 0.1926 0.5292 0.340 1.9783
0.474 0.7900 1.2549 1.2387 42.3945 1.3142 5.3417 0.1452 0.3365 0.340 1.9783
0.505 0.8417 1.2549 1.2387 42.3945 1.3566 4.0275 0.1072 0.1913 0.340 1.9783
0.537 0.8950 1.2549 1.2387 42.3945 1.3142 2.6709 0.0624 0.0841 0.340 1.9783
0.568 0.9467 1.2549 1.2387 42.3945 1.3566 1.3566 0.0217 0.0217 0.340 1.9783
0.6 1.0000 1.2549 1.2387 42.3945 0.0000 0.0000 0.0000 0.0000 0.340 1.9783
0.0000 -1.8734 -0.2204
82
Table 6-9 Wing loading in each station for point D (Aileron deflects upwards)
y (m) y/b Cl Cn N (N) Delta V Vn (N) Delta M M (N) Cm T (Pa)
0 0.0000 0.3131 0.3134 44.1893 1.4141 -2.1652 -0.0919 -0.3552 -0.1000 -2.3969
0.032 0.0533 0.3131 0.3134 44.1893 1.3699 -3.5793 -0.1322 -0.2633 -0.1000 -2.3969
0.063 0.1050 0.3131 0.3134 44.1893 1.4141 -4.9492 -0.1810 -0.1311 -0.1000 -2.3969
0.095 0.1583 0.3131 0.3134 44.1893 1.3699 -6.3632 -0.2185 0.0499 -0.1000 -2.3969
0.126 0.2100 0.3131 0.3134 44.1893 1.4141 -7.7331 -0.2701 0.2684 -0.1000 -2.3969
0.158 0.2633 0.3131 0.3134 44.1893 1.3699 -9.1471 -0.3048 0.5385 -0.1000 -2.3969
0.189 0.3150 0.3131 0.3134 44.1893 1.4141 -10.5170 -0.3592 0.8433 -0.1000 -2.3969
0.221 0.3683 0.3131 0.3134 44.1893 1.4141 -11.9311 -0.4044 1.2024 -0.1000 -2.3969
0.253 0.4217 0.3131 0.3134 44.1893 1.3699 -13.3451 -0.4349 1.6068 -0.1000 -2.3969
0.284 0.4733 0.3131 0.3134 44.1893 0.0442 -14.7150 -0.0147 2.0418 -0.1000 -2.3969
0.285 0.4750 0.3131 0.3134 44.1893 -14.0412 -14.7592 -0.0077 2.0565 -0.1000 -2.3969
0.286 0.4767 0.3131 0.3134 -28126.5107 -14.0412 -0.7180 0.0063 2.0643 -0.1000 -2.3969
0.287 0.4783 0.3131 0.3134 44.1893 1.2815 13.3231 0.3678 2.0579 -0.1000 -2.3969
0.316 0.5267 0.3131 0.3134 44.1893 1.3699 12.0416 0.3521 1.6902 -0.1000 -2.3969
0.347 0.5783 0.3131 0.3134 44.1893 1.4141 10.6718 0.3189 1.3381 -0.1000 -2.3969
0.379 0.6317 0.3131 0.3134 44.1893 1.3744 9.2577 0.2743 1.0192 -0.1000 -2.3969
0.411 0.6850 0.2955 0.2958 41.7107 1.2930 7.8833 0.2243 0.7450 0.0700 1.6778
0.442 0.7367 0.2955 0.2958 41.7107 1.3347 6.5903 0.1895 0.5206 0.0700 1.6778
0.474 0.7900 0.2955 0.2958 41.7107 1.2930 5.2555 0.1429 0.3311 0.0700 1.6778
0.505 0.8417 0.2955 0.2958 41.7107 1.3347 3.9625 0.1054 0.1882 0.0700 1.6778
0.537 0.8950 0.2955 0.2958 41.7107 1.2930 2.6278 0.0614 0.0828 0.0700 1.6778
0.568 0.9467 0.2955 0.2958 41.7107 1.3347 1.3347 0.0214 0.0214 0.0700 1.6778
0.6 1.0000 0.2955 0.2958 41.7107 0.0000 0.0000 0.0000 0.0000 0.0700 1.6778
0.0000 -2.1652 -0.3552
83
Table 6-10 Wing loading in each station for point E (Aileron deflects upwards)
y (m) y/b Cl Cn N (N) Delta V Vn (N) Delta M M (N) Cm T (Pa)
0 0.0000 -0.1565 -0.1584 -22.3318 -0.7146 0.3296 0.0220 -0.2004 -0.1000 -2.3969
0.032 0.0533 -0.1565 -0.1584 -22.3318 -0.6923 1.0442 0.0431 -0.2224 -0.1000 -2.3969
0.063 0.1050 -0.1565 -0.1584 -22.3318 -0.7146 1.7365 0.0670 -0.2655 -0.1000 -2.3969
0.095 0.1583 -0.1565 -0.1584 -22.3318 -0.6923 2.4511 0.0867 -0.3325 -0.1000 -2.3969
0.126 0.2100 -0.1565 -0.1584 -22.3318 -0.7146 3.1434 0.1120 -0.4192 -0.1000 -2.3969
0.158 0.2633 -0.1565 -0.1584 -22.3318 -0.6923 3.8580 0.1303 -0.5312 -0.1000 -2.3969
0.189 0.3150 -0.1565 -0.1584 -22.3318 -0.7146 4.5503 0.1570 -0.6616 -0.1000 -2.3969
0.221 0.3683 -0.1565 -0.1584 -22.3318 -0.7146 5.2649 0.1799 -0.8186 -0.1000 -2.3969
0.253 0.4217 -0.1565 -0.1584 -22.3318 -0.6923 5.9796 0.1961 -0.9985 -0.1000 -2.3969
0.284 0.4733 -0.1565 -0.1584 -22.3318 -0.0223 6.6718 0.0067 -1.1946 -0.1000 -2.3969
0.285 0.4750 -0.1565 -0.1584 -22.3318 7.0959 6.6942 0.0031 -1.2013 -0.1000 -2.3969
0.286 0.4767 -0.1565 -0.1584 14214.1682 7.0959 -0.4017 -0.0039 -1.2044 -0.1000 -2.3969
0.287 0.4783 -0.1565 -0.1584 -22.3318 -0.6476 -7.4977 -0.2080 -1.2005 -0.1000 -2.3969
0.316 0.5267 -0.1565 -0.1584 -22.3318 -0.6923 -6.8500 -0.2016 -0.9924 -0.1000 -2.3969
0.347 0.5783 -0.1565 -0.1584 -22.3318 -0.7146 -6.1578 -0.1856 -0.7908 -0.1000 -2.3969
0.379 0.6317 -0.1565 -0.1584 -22.3318 -0.7543 -5.4431 -0.1621 -0.6052 -0.1000 -2.3969
0.411 0.6850 -0.1741 -0.1760 -24.8089 -0.7691 -4.6889 -0.1334 -0.4431 0.0500 1.1985
0.442 0.7367 -0.1741 -0.1760 -24.8089 -0.7939 -3.9198 -0.1127 -0.3097 0.0500 1.1985
0.474 0.7900 -0.1741 -0.1760 -24.8089 -0.7691 -3.1259 -0.0850 -0.1969 0.0500 1.1985
0.505 0.8417 -0.1741 -0.1760 -24.8089 -0.7939 -2.3568 -0.0627 -0.1120 0.0500 1.1985
0.537 0.8950 -0.1741 -0.1760 -24.8089 -0.7691 -1.5630 -0.0365 -0.0492 0.0500 1.1985
0.568 0.9467 -0.1741 -0.1760 -24.8089 -0.7939 -0.7939 -0.0127 -0.0127 0.0500 1.1985
0.6 1.0000 -0.1741 -0.1760 -24.8089 0.0000 0.0000 0.0000 0.0000 0.0500 1.1985
0.0000 0.3296 -0.2004
84
Table 6-11 Wing loading in each station for point G (Aileron deflects upwards)
y (m) y/b Cl Cn N (N) Delta V Vn (N) Delta M M (N) Cm T (Pa)
0 0.0000 -0.6646 -0.6657 -22.1085 -0.7075 0.5934 0.0303 -0.0657 -0.1000 -0.5646
0.032 0.0533 -0.6646 -0.6657 -22.1085 -0.6854 1.3009 0.0510 -0.0960 -0.1000 -0.5646
0.063 0.1050 -0.6646 -0.6657 -22.1085 -0.7075 1.9863 0.0749 -0.1469 -0.1000 -0.5646
0.095 0.1583 -0.6646 -0.6657 -22.1085 -0.6854 2.6937 0.0941 -0.2218 -0.1000 -0.5646
0.126 0.2100 -0.6646 -0.6657 -22.1085 -0.7075 3.3791 0.1195 -0.3159 -0.1000 -0.5646
0.158 0.2633 -0.6646 -0.6657 -22.1085 -0.6854 4.0866 0.1373 -0.4354 -0.1000 -0.5646
0.189 0.3150 -0.6646 -0.6657 -22.1085 -0.7075 4.7719 0.1640 -0.5727 -0.1000 -0.5646
0.221 0.3683 -0.6646 -0.6657 -22.1085 -0.7075 5.4794 0.1867 -0.7367 -0.1000 -0.5646
0.253 0.4217 -0.6646 -0.6657 -22.1085 -0.6854 6.1869 0.2024 -0.9234 -0.1000 -0.5646
0.284 0.4733 -0.6646 -0.6657 -22.1085 -0.0221 6.8722 0.0069 -1.1258 -0.1000 -0.5646
0.285 0.4750 -0.6646 -0.6657 -22.1085 7.0249 6.8944 0.0034 -1.1327 -0.1000 -0.5646
0.286 0.4767 -0.6646 -0.6657 14071.9915 7.0249 -0.1306 -0.0036 -1.1361 -0.1000 -0.5646
0.287 0.4783 -0.6646 -0.6657 -22.1085 -0.6411 -7.1555 -0.1982 -1.1324 -0.1000 -0.5646
0.316 0.5267 -0.6646 -0.6657 -22.1085 -0.6854 -6.5144 -0.1913 -0.9342 -0.1000 -0.5646
0.347 0.5783 -0.6646 -0.6657 -22.1085 -0.7075 -5.8290 -0.1752 -0.7429 -0.1000 -0.5646
0.379 0.6317 -0.6646 -0.6657 -22.1085 -0.7259 -5.1215 -0.1523 -0.5677 -0.1000 -0.5646
0.411 0.6850 -0.6995 -0.7003 -23.2576 -0.7210 -4.3957 -0.1251 -0.4154 0.3050 1.7220
0.442 0.7367 -0.6995 -0.7003 -23.2576 -0.7442 -3.6747 -0.1057 -0.2903 0.3050 1.7220
0.474 0.7900 -0.6995 -0.7003 -23.2576 -0.7210 -2.9305 -0.0797 -0.1846 0.3050 1.7220
0.505 0.8417 -0.6995 -0.7003 -23.2576 -0.7442 -2.2095 -0.0588 -0.1049 0.3050 1.7220
0.537 0.8950 -0.6995 -0.7003 -23.2576 -0.7210 -1.4652 -0.0342 -0.0462 0.3050 1.7220
0.568 0.9467 -0.6995 -0.7003 -23.2576 -0.7442 -0.7442 -0.0119 -0.0119 0.3050 1.7220
0.6 1.0000 -0.6995 -0.7003 -23.2576 0.0000 0.0000 0.0000 0.0000 0.3050 1.7220
0.0000 0.5934 -0.0657
85
Figure 6-15 Graph of Normal Lift across Wingspan (Zoom in)
Figure 6-16 Graph of Normal Lift across Wingspan (Zoom out)
86
Figure 6-17 Graph of Shear Force across Wingspan when the aileron deflects downwards
Figure 6-18 Graph of Shear Force across Wingspan when the aileron deflects upwards
87
Figure 6-19 Graph of Bending Moment across Wingspan when aileron deflects downwards
Figure 6-20 Graph of Bending Moment across Wingspan when aileron deflects upwards
88
Figure 6-21 Graph of Torsion across Wingspan
Based on the all shear force diagrams and bending moment diagrams, the maximum shear
force and maximum bending moment when the aileron deflects upwards and deflects downwards
can be summarised in Table 6-12..
Table 6-12 Maximum Shear force and Bending moment in each position
Position Deflection of Aileron downwards Deflection of Aileron upwards
V (N) M (Nm) T (Nm) V (N) M (Nm) T (Nm)
A 13.8754 2.1912 -3.1419 -14.2892 2.0901 1.9783
D 14.3394 2.2786 -6.4716 13.3231 2.0643 -2.3969
E 7.7098 -0.9903 -5.9923 -7.4977 -1.2044 -2.3969
G 7.3655 -1.0367 -2.8512 -7.1555 -1.1361 1.7220
-8
-6
-4
-2
0
2
4
0 0.2 0.4 0.6 0.8 1 1.2
T (N
m)
2y/b
T vs y
Position A (Down)
Position D (Down)
Position E (Down)
Position G (Down)
Position A (Up)
Position D (Up)
Position E (Up)
Position G (Up)
89
6.2.6 Discussion
After taking into account the lift coefficient for the part without aileron and for the part
with the deflected aileron, the normal lift force is calculated, tabulated and presented in graph
shown in Figure 6-15 when it was zoomed in to a smaller range of normal lift force axis, and in
Figure 6-16 when it was zoomed out into a larger range of normal lift force axis. The sudden spike
of the graph is due to the reaction force due to the wing strut, which we assume to be a concentrated
load force at a small area. In Figure 6-15, the normal lift force is a constant straight line, since the
wing is assumed to have no twist angle due to its small size, and the angle of attack is constant
across the wingspan without aileron. However, there is a slight change in the normal lift force at a
distance of 0.4m away from the wing root. That is where the aileron is located on the wing. The
change of the normal lift force due to the aileron, which is resulted from the change in lift
coefficient, has been discussed in previous section. As the aileron deflects downwards, the lift
coefficient increases, hence the increase in normal lift force, and vice-versa.
Figure 6-17 and Figure 6-18 show the shear force across the wingspan when the aileron
deflects downwards and deflects upwards respectively. The graphs are almost similar due to the
reaction force of the wing strut, that supports the wing. The sudden change in the shear force in
the graph indicates where the wing strut is located. As a result, the shear force diagram differs
from the normal shear force diagram that shows a graph with a decreasing slope along the
wingspan. The difference between Figure 6-17 and Figure 6-18 is that the maximum shear force
is relatively smaller, when the aileron deflects upwards in Figure 6-18. This is because the change
in lift coefficient when the aileron deflects downwards is larger.
Furthermore, Figure 6-19 and Figure 6-20 show that the bending moment diagrams are
affected by the reaction force of the wing strut too. The bending moment diagram and shear force
diagram are derived from the normal lift force diagram in Figure 6-15 and Figure 6-16. As shown
in Figure 6-19 and Figure 6-20, the bending moment is negative when the load factor is negative
and vice-versa. Also, difference between Figure 6-19 and Figure 6-20 is that when the aileron
deflects downward, the moment of reaction at the wing root is positive and when the aileron
deflects upward, the moment of reaction at the wing root is negative. This is to counter the opposite
direction of the applied moment on the wing.
90
Figure 6-21 shows the graph of torsion across the wingspan. As explained in the previous
section, the value of torsion depends on the value of moment coefficient. The change in the
moment coefficient is due to the deflection of the aileron. The change in moment coefficient
depends on the position of the wing during flight. Position A and G experience smaller angle
deflection of aileron than position D and E due to maneuverability and control of the aircraft at
lower speed of the aircraft, therefore the change of torsion is relatively smaller. When the aileron
deflects downward, the torsion decreases as the moment coefficient decreases. The amount of
change in moment coefficient depends on how much the graph in Figure 6-8 is shifting downwards,
as explained in section 6.2.3.
Based on Table 6-12, results show that the maximum shear force is 14.3394 N at position
D, when the aileron deflects downwards. As the shear force is the highest, the value of both the
bending moment and torsion at position D is also the highest. The highest maximum bending
moment and maximum torsion is 2.2786 Nm and -6.4716 Nm respectively. This is because the
speed and the load factor are the highest at position D, as compared to other positions. The shear
force and bending moment at position A and D are higher, as compared to position E and G because
the magnitude of positive load factor of aircraft at position A and D is larger than the negative load
factor of aircraft at position E and G. The maximum shear force, maximum bending moment and
maximum torsion is used in structural analysis conducted in the next section in order to determine
if the structure is safe.
91
6.3 Structural Analysis
6.3.1 Material Selection
The material used in our aircraft’s wing construction is mainly Expanded Polypropylene
(EPP) Foam, and a minority of the components are made up of plywood.
The reason why Expanded Polypropylene (EPP) is selected because it is a highly versatile
closed-cell bead foam that provides a unique range of properties, including outstanding energy
absorption, multiple impact resistance, thermal insulation, buoyancy, exceptionally high strength
to weight ratio and 100% recyclability. EPP can be made in a wide range of densities. In
manufacturing, individual beads are fused into final product form by the steamchest moulding
process resulting in a strong and lightweight shape.
On the other hand, plywood is chosen for some part of the wing, such as rib and wing strut
because it offers increased stability, high impact resistance, high surface dimensional stability,
high strength to weight ratio, and effective resistance in shear. These properties are beneficial in
the construction of our lightweight wing.
Table 6-13 Material Selection of each Wing Component
Wing Components Material
Rib, Wing strut Plywood
Skin, Spar, Aileron Expanded Polyproplyene (EPP) Foam
92
6.3.2 Rib Analysis
All of the four ribs along the wingspan are of the same dimension and thickness. Only the analysis
of the rib which carries the critical load is carried out, since if the rib which carries the critical load is safe,
then the other ribs will be safe as well. The critical position of the aircraft during flight, which is position
D, is selected because the maximum bending moment and the shear force is the highest, as compared to
other positions. The analysis of the bending moment as well as the shear is performed with the design factor
of 1.5. Also, each of the rib cells is compared with the plate buckling strength to determine the margin of
safety and the safety factor.
Based on the graph of normal lifting force across the wingspan, the area under the graph, which is
the normal lifting force is obtained to determine the load distribution along the rib, which is indicated as
the value of w. Since bay 1 and bay 3 have the same lift distribution force, therefore only one of them is
taken for analysis. The material selected for the rib is plywood and its material properties are listed in Table
6-15.
Figure 6-22 Location of ribs along the wingspan
Table 6-14 The chordwise load distribution for each bay
Bay Normal lifting force (N) w (N/m)
Bay 1 13.8393 46.5187
Bay 2 26.5136 89.1213
93
Table 6-15 Material properties of plywood
Material properties Plywood
Ultimate tensile strength 5000psi
Ultimate shear strength 1000psi
Shear Modulus 110ksi
Modulus of Rupture 10ksi
Modulus of Elasticity 1000ksi
Figure 6-23 The load distribution on rib
Total area of the load distribution based on our rib design as shown in Figure 6-23 can be
obtained as follows,
𝑇𝑜𝑡𝑎𝑙 𝑎𝑟𝑒𝑎, 𝐴 = (0.15𝑐 × 3𝑤) + (0.45𝑐 × 2𝑤) + (0.4𝑐 × 𝑤) = 1.75𝑐𝑤
1.75𝑐𝑤 = 26.5136
𝑤 = 89.1213 𝑁/𝑚
Cell 1 Analysis
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑟𝑖𝑏′𝑠 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛, 𝐴1 = 0.000648𝑚2
𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑟𝑖𝑏, 𝑡1 = 0.003𝑚
ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒, 𝑎 = 0.0415𝑚
𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒, 𝑏 = 0.0202𝑚
94
𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑐𝑒𝑛𝑡𝑟𝑜𝑖𝑑, 𝑧 = 0.012𝑚
𝐹0.7 = 700𝑘𝑠𝑖
𝑛 = 10
Figure 6-24 Cell 1 of Rib 2
The compressive stress of cell 1 can be determined by obtaining the bending moment due
to the leading edge’s air load acting on the rib, as shown below,
𝑀1 = [3𝑤 × 0.15𝑐 × (0.15𝑐
2+ 0.15𝑐)] + [2𝑤 × 0.15𝑐 ×
0.15𝑐
2]
𝑀1 = 0.3187 𝑁𝑚
𝐼 = 𝐴1(𝑧1)2 = 9.3312 × 10−8𝑚4
𝜎 =𝑀𝑧
𝐼= 0.040985 𝑀𝑃𝑎
By assuming the perimeter of the rib web is simply supported, and obtaining the 𝑘𝑏value
from Figure 3 of Appendix III, the web buckling strength is
𝜎𝑐𝑟 =𝜋2𝑘𝑏𝐸
12(1 − 𝑣2)(𝑡
𝑏)2
𝜎𝑐𝑟 =𝜋2(36)(1000 × 103)
12(1 − 0.32)(0.003
0.0202)2
𝜎𝑐𝑟 = 7.177 × 105 𝑝𝑠𝑖 = 4948.0988𝑀𝑃𝑎
𝜎𝑐𝑟
𝐹0.7= 1.0253
Since the 𝜎𝑐𝑟value is larger than the 𝐹0.7 value, correction needs to be done. Based on
Figure 5 of Appendix III, the new ratio will be as follows,
95
𝜎𝑐𝑟
𝐹0.7= 0.845
𝜎𝑐𝑟 = 0.845 × 700000
𝜎𝑐𝑟 = 591500𝑝𝑠𝑖 = 4078.24894 𝑀𝑃𝑎
Thus,
𝑅𝑏 =1.5 × 0.040985
4078.24894= 1.5074 × 10−5
𝑀𝑎𝑟𝑔𝑖𝑛 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 =1
𝑅𝑏− 1 = 66336.26
The shear stress due to the total air load of cell 1 can be determined as follows. Then, the
shear is divided by the area of the cross section to obtain the shear stress.
𝑉1 = (3𝑤)(0.15𝑐) + (2𝑤)(0.15𝑐) = 11.363𝑁
𝜏 =𝑉1
𝐴𝑟𝑒𝑎
𝜏 = 17535.494𝑃𝑎
Assuming that the perimeter of the rib web is simply supported, and obtaining the 𝑘𝑠value
from Figure 2 of Appendix III, the buckling strength is
𝜎𝑐𝑟 =𝜋2𝑘𝑠𝐸
12(1 − 𝑣2)(𝑡
𝑏)2
𝜎𝑐𝑟 =𝜋2(6.4)(1000 × 103)
12(1 − 0.32)(0.003
0.0202)2
𝜎𝑐𝑟 = 1.27584 × 105 𝑝𝑠𝑖 = 879.662 𝑀𝑃𝑎
Since the 𝜎𝑐𝑟value is much smaller than the 𝐹0.7 value, correction is not required.
Thus,
𝑅𝑠 =1.5 × 0.017535494
879.662= 2.9902 × 10−5
𝑀𝑎𝑟𝑔𝑖𝑛 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 =1
𝑅𝑠− 1 = 33442.103
Therefore, the combined bending and shear is
96
𝑀𝑎𝑟𝑔𝑖𝑛 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 = 1
√𝑅𝑏2 + 𝑅𝑠
2
− 1
𝑀𝑎𝑟𝑔𝑖𝑛 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 = 1
√(1.5074 × 10−5)2 + (2.9902 × 10−5)2− 1 = 29861.655
𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 = 29861.655 + 1 = 29862.655
Since the factor of safety for cell 1 is more than one, cell 1 of rib 2, which carries the
critical load is safe.
Cell 2 Analysis
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑟𝑖𝑏′𝑠 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛, 𝐴2 = 0.001729𝑚2
𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑟𝑖𝑏, 𝑡1 = 0.003𝑚
ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒, 𝑎 = 0.1285𝑚
𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒, 𝑏 = 0.0202𝑚
𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑐𝑒𝑛𝑡𝑟𝑜𝑖𝑑, 𝑧 = 0.0099𝑚
𝐹0.7 = 700𝑘𝑠𝑖
𝑛 = 10
Figure 6-25 Cell 2 of Rib 2
The compressive stress of cell 2 can be determined by obtaining the bending moment due
to the leading edge’s air load acting on the rib, as shown below,
𝑀2 = [2𝑤 × 0.3𝑐 × (0.3𝑐
2)] + [𝑤 × 0.4𝑐 × (
0.4𝑐
2+ 0.3𝑐)] = 0.7469 𝑁𝑚
𝐼 = 𝐴1(𝑧1)2 = 1.6946 × 10−7𝑚4
97
𝜎 =𝑀𝑧
𝐼= 0.043635 𝑀𝑃𝑎
Assuming the perimeter of the rib web is simply supported, and obtaining the 𝑘𝑏value from
Figure 3 of Appendix III, the web buckling strength is
𝜎𝑐𝑟 =𝜋2𝑘𝑏𝐸
12(1 − 𝑣2)(𝑡
𝑏)2
𝜎𝑐𝑟 =𝜋2(24)(1000 × 103)
12(1 − 0.32)(0.003
0.0202)2
𝜎𝑐𝑟 = 0.478441 × 106 𝑝𝑠𝑖 = 3298.7325 𝑀𝑃𝑎
Since the 𝜎𝑐𝑟value is much smaller than the 𝐹0.7 value, correction is not required.
Thus,
𝑅𝑏 =1.5 × 0.043635
3298.7325= 1.9842 × 10−5
𝑀𝑎𝑟𝑔𝑖𝑛 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 =1
𝑅𝑏− 1 = 50397.9
The shear stress due to the total air load of cell 2 can be determined as follows. Then, the
shear is divided by the area of the cross section to obtain the shear stress.
𝑉2 = (𝑤)(0.4𝑐) + (2𝑤)(0.3𝑐) = 15.1506𝑁
𝜏 =𝑉1
𝐴𝑟𝑒𝑎
𝜏 = 8762.65𝑃𝑎
Assuming that the perimeter of the rib web is simply supported, and obtaining the 𝑘𝑠value
from Figure 2 of Appendix III, the buckling strength is
𝜎𝑐𝑟 =𝜋2𝑘𝑠𝐸
12(1 − 𝑣2)(𝑡
𝑏)2
𝜎𝑐𝑟 =𝜋2(5.5)(1000 × 103)
12(1 − 0.32)(0.003
0.0202)2
𝜎𝑐𝑟 = 0.109642 × 106 𝑝𝑠𝑖 = 755.9595 𝑀𝑃𝑎
Since the 𝜎𝑐𝑟value is much smaller than the 𝐹0.7 value, correction is not required.
Thus,
98
𝑅𝑠 =1.5 × 8762.65
755959536.5537= 1.73872 × 10−5
𝑀𝑎𝑟𝑔𝑖𝑛 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 =1
𝑅𝑠− 1 = 57512.77
Therefore, the combined bending and shear is
𝑀𝑎𝑟𝑔𝑖𝑛 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 = 1
√𝑅𝑏2 + 𝑅𝑠
2
− 1
𝑀𝑎𝑟𝑔𝑖𝑛 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 = 1
√(1.9842 × 10−5)2 + (1.73872 × 10−5)2− 1 = 37903.366
𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 = 37903.366 + 1 = 37904.366
Since the factor of safety for cell 2 is more than one, cell 2 of rib 2, which carries the
critical load is safe.
Figure 6-26 Shear force across the chord
0
5
10
15
20
25
30
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18
Shea
r fo
rce,
V (
N)
Length, x (m)
Shear force across the chord
99
Figure 6-27 Bending Moment across the chord
Based on the chordwise shear force diagram and chordwise bending moment diagram, we
can conclude that the shear force and bending moment decreases across the chord. The trend of
both diagrams depends on the lift distribution across the chord, which is in reference to FAR Part
23. Stepwise lift distribution is selected due to the type and size of the aircraft.
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18
Ben
din
g M
om
ent
, M (
Nm
)
Length, x (m)
Bending Moment across the chord
100
6.3.3 Spar Analysis
Spar is the wing structure that supports an aircraft on ground and allows it to taxi, take-off,
and land. In fact, Spar design tends to have several interferences with the aircraft structural design.
The strength and the weight of spar have become important factors. Efforts are being made to
reduce the weight of the aircraft and consequently increase the payload. The aerodynamic lift load
acting on the top and bottom surfaces of the wing gets transferred to the rib structures, which in-
turn gets transferred to the spar web as concentrated shear loads. These shear loads cause bending
of the spars which are the major loading on the spars. The spar is made up of Expanded
Polypropylene (EPP) foam, which its material properties are listed in Table 6-16.
Table 6-16 Material Properties of EPP Foam
Material Properties EPP Foam
Modulus of Elasticity, E 1000ksi
Yield Stress 4000psi
Density 1.3 lb/ft2
Figure 6-28 Dimension and location of spar in wing
101
Figure 6-29 Cross section of spar
𝑎 = 0.6𝑚, 𝑏 = 0.01𝑚, 𝑡 = 0.005𝑚
E = 6895MPa
𝑀𝑚𝑎𝑥 = 20.97266𝑁𝑚
𝑉 = 52.71447 𝑁
𝑆𝑎𝑓𝑒𝑡𝑦 𝐹𝑎𝑐𝑡𝑜𝑟 = 1.5
𝐹0.7 = 4826.295 𝑀𝑃𝑎
n = 10
𝐼 =𝑏ℎ3
12=
(0.005)(0.01)3
12= 4.1667 × 10−10 𝑚4
Ultimate Bending Moment
𝜎 =𝑀𝑦
𝐼=
20.97266 × 0.005 × 1.5
4.1667 × 10−10= 377.50558 𝑀𝑃𝑎
By assuming the spar is simply supported, and obtaining kb value from Figure 3 in
Appendix III, the critical buckling strength due to compressive force is
𝜎𝑐𝑟 =𝜋2𝑘𝑏𝐸
12(1 − 𝑣2)(𝑡
𝑏)2
=𝜋2 × 24 × 6895𝑀
12(1 − 0.32)(0.005
0.01)2
= 37390.6𝑀𝑃𝑎
𝜎𝑐𝑟
𝐹0.7=
37390.6
4826.295= 7.7473
102
Based on Figure 4 in Appendix III, as 𝜎𝑐𝑟 value is greater than 𝐹0.7 value, correction must
be done, where the new ratio is given as follows,
𝜎𝑐𝑟
𝐹0.7= 1.05
𝜎𝑐𝑟 = 1.05 × 4826.295𝑀𝑃𝑎 = 5067.6𝑀𝑃𝑎
Therefore,
𝑅𝑐 =377.50558
5067.6= 0.07449
𝑀𝑆 =1
0.07449− 1 = 12.424
Ultimate Shear Force
τ =𝑉
𝐴=
52.71447
0.005 × 0.01 × 2= 527144.7 𝑃𝑎
By assuming the spar is simply supported, and obtaining ks value from the Figure 2 in
Appendix III, the critical buckling strength due to compressive force is
𝜎𝑐𝑟 =𝜋2𝑘𝑠𝐸
12(1 − 𝑣2)(𝑡
𝑏)2
=𝜋2 × 5.5 × 6895𝑀
12(1 − 0.32)(0.005
0.01)2
= 8568.683𝑀𝑃𝑎
𝜎𝑐𝑟
𝐹0.7=
8568.683
4826.295= 1.7754
Based on Figure 5 in Appendix III, as 𝜎𝑐𝑟 value is greater than 𝐹0.7 value, correction must
be done, where the new ratio is given as follows,
𝜎𝑐𝑟
𝐹0.7= 0.55
𝜎𝑐𝑟 = 0.55 × 4826.295𝑀𝑃𝑎 = 2654.462𝑀𝑃𝑎
Therefore,
103
𝑅𝑠 =527144.7
2654462250= 1.9859 × 10−4
𝑀𝑆 =1
1.655 × 10−6− 1 = 604264.7
𝑀𝑆 =1
√(𝑅𝑏)2 + (𝑅𝑠)
2=
1
√(0.07449)2 + (1.9859 × 10−4)2= 13.4246
𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 = 13.4246 + 1 = 14.4246
Since the factor of safety is more than 1, the spar will not buckle and it is under a safe condition.
104
6.3.4 Skin Analysis
According to wing loading calculation, the highest value of shear and bending moment are
when the flight is in position D. Therefore, the maximum bending moment and maximum shear
force can be used to determine the suitable thickness of wing skin to withstand the applied load.
The case will be analysed with the condition of compression force acting on the upper wing and
with the tension force acting on the lower skin. Shear force is exerted on both side of the wing at
the same time.
Figure 6-30 Wing’s skin Figure 3.9:
The calculation is being divided into two parts, which is upper wing and lower wing
to find the factor of safety for the skin. For the upper wing, the buckling strength can be calculated
first as below:
𝐹𝑐𝑟 =𝑘𝜋2𝐸
12(1 − 𝑣2)(𝑡
𝑏)2
Upper skin
Lower skin
105
Shear load
Based on Figure 2 in Appendix III, the value of ks is obtained. Buckling strength due to
shear force can be determined as follows,
𝐹𝑠,𝑐𝑟 =𝑘𝑠𝜋
2𝐸
12(1 − 𝑣2)(𝑡
𝑏)2
=(9.6)𝜋2(6.9 𝑥 106)
12(1 − 0.32)(0.005
0.25)2
= 23.95𝑘𝑃𝑎
𝐹𝑠,𝑐𝑟
𝐹0.7=
23.95𝑘𝑃𝑎
4826.3𝑀𝑃𝑎= 4.96𝑥10−6
The ratio value of Fs/F0.7 is so small that the correction can be neglected.
The applied shear loads have been obtained by taking the value of maximum positive and
maximum negative shear stress:
𝜏1 =𝑉
𝐴 𝜏2 =
𝑉
𝐴
=14.3394
(0.25)(1.2) =
−14.2892
(0.25)(1.2)
= 47.798𝑃𝑎 = −47.631𝑃𝑎
𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑆𝑎𝑓𝑒𝑡𝑦 = |𝐹𝑠,𝑐𝑟
𝜏1| 𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑆𝑎𝑓𝑒𝑡𝑦 = |
𝐹𝑠,𝑐𝑟
𝜏2|
= |23.95𝑥103
47.798| = |
23.95𝑥103
−47.631|
= 501.07 = 502.82
106
We can say that the minimum factor of safety for upper wing due to shear load is 501.07. Hence,
it will not buckle due to the shear load.
Compressive Force
For compression force acting on upper wing, the same step used to determine the factor of
safety. Based on Figure 1 in Appendix III, the value of kc is obtained. Buckling strength due to
compressive force can be determined as follows,
𝐹𝑐,𝑐𝑟 =𝑘𝑐𝜋
2𝐸
12(1 − 𝑣2)(𝑡
𝑏)2
=(4)𝜋2(6.9 𝑥 106)
12(1 − 0.32)(0.005
0.25)2
= 9.98𝑘𝑃𝑎
𝐹𝑐,𝑐𝑟
𝐹0.7=
9.98𝑘𝑃𝑎
4826.3𝑀𝑃𝑎= 2.07𝑥10−6
Again, the ratio of Fc/F0.7 is so small that the correction can be neglected.
The applied compression loads have been obtained by taking the value of maximum positive and
maximum negative axial stress:
𝜎1 =𝑀𝑦
𝐼 𝜎2 =
𝑀𝑦
𝐼
=(2.2786)(0.6)
((0.25)(1.23)
12 ) =
(2.0901)(0.6)
((0.25)(1.23)
12 )
107
= 37.98𝑃𝑎 = 34.84𝑃𝑎
𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑆𝑎𝑓𝑒𝑡𝑦 = |𝐹𝑐,𝑐𝑟
𝜎1| 𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑆𝑎𝑓𝑒𝑡𝑦 = |
𝐹𝑐,𝑐𝑟
𝜎2|
= |9.98𝑥103
37.98| = |
9.98𝑥103
34.84|
= 262.77 = 286.45
The smallest factor of safety due to the compression load is 262.77. Based on the observation, the
lowest factor of safety among both shear and compression loads for upper side of the wing is
262.77, showing that the skin will not buckle due to the forces applied. The same steps have been
used to find the factor of safety for lower wing, and the results are being tabulated as below:
Table 6-17 Factor of safety of upper and lower wing
Wing Part Shear Compression Bending Lowest Factor of Safety
Upper Wing 501.07 262.77 - 262.77
Lower Wing 544.62 - 241.72 241.72
Based on the Table 6-17, the lowest factor of safety at the upper and lower side of the wing is
above 1, showing that it will not fail due to the load applied of the aircraft wing’s skin.
108
6.3.5 Stress Analysis of Multi Cell Sections
The shear force at any chordwise section along the span produce shear stresses that are
resisted by the skin and the spars. The shear force at any section obtained from the shear force
diagram evaluated. The resulting shear flow distribution is due to the combined effects of shear
and torsion. In this analysis, two type of wing structure which are monocoque structure and semi-
monocoque structure is taking into account. Both positive (upward) deflection and negative
(downward) deflection of aileron are considered by taking the maximum shear force and torsion
from each case.
6.3.5.1 Shear stress due to shear force (monocoque)
The RC plane wing is supported by one spar which then make it to have 2 cells of wing.
Figure 6-31 shows the RC plane 2 cells wing section with shear flow diagram.
Figure 6-31 Shear flow diagram of 2 cells wing
To simulates monocoque structure, panel AB were cut in order to analyses them as open cell.
Figure 6-32 indicate cell 1 and cell 2 separately.
109
Figure 6-32 Vertical distance from centroid, z of cell 1 and cell 2
Table 6-18 shows the properties of the wing including the area cell 1 and cell 2, vertical
distance from centroid, z, the thickness of the skin and the thickness of the spar.
Table 6-18 Dimensions of cell 1 and cell 2 wing section
𝐴1 = 0.648 × 10−3 𝑚2 𝐴2 = 1.729 × 10−3 𝑚2
𝑧1 = 0.012 𝑚 𝑧1 = 0.0099 𝑚
𝑡𝑠𝑘 = 0.003 𝑚 𝑡𝑠𝑝 = 0.011 𝑚
Calculation of Moment of Inertia
𝐼𝑥𝑥 = Σ𝐴𝑧2
= 2.628 × 10−7 𝑚4
Data from Material Properties
𝐹𝑠 = 13.785 × 106 𝑃𝑎
𝐸 = 6894.76 × 106 𝑃𝑎
110
Calculating Integral part:
Table 6-19 Integral of cell 1 and cell 2
z A ∫𝑧 𝑑𝐴
BA 0.012 0.648 × 10−3 𝑚2 7.776 × 10−6 𝑚3
AB 0.0099 1.729 × 10−3 𝑚2 1.712 × 10−5 𝑚3
Positive (+ve) deflection analysis
The maximum shear force applied for upward deflection for the aircraft is 14.2892 N. The shear
flow is calculated followed by the shear stress and safety factor.
Maximum Shear, 𝑉𝑆 = 14.2892 𝑁
Calculating Shear Flow
𝑞𝑏 ⇒ open section
𝑞𝑏 =𝑉𝑆
𝐼𝑥𝑥∫𝑧 𝑑𝐴
𝑞𝑏𝐵𝐴= 422.8499 𝑁/𝑚
𝑞𝑏𝐴𝐵= 930.8082 𝑁/𝑚
Calculating Shear Stress
𝜏𝑏 =𝑞𝑏
𝑡
Where 𝑡 = 0.003 𝑚
𝜏𝑏𝐵𝐴= 0.1409 × 106
𝜏𝑏𝐴𝐵= 0.3103 × 106
111
Calculating Safety Factor
𝑆𝐹 =𝐹𝑠
𝜏𝑏𝑚𝑎𝑥
=13.785 × 106
0.3103 × 106
= 44.43
According to the safety factor obtained, the structure of the wing is safe and undergo any failure.
Negative (-ve) deflection analysis
The maximum shear force applied for downward deflection for the aircraft is 14.3394 N. The shear
flow is calculated followed by the shear stress and safety factor.
Maximum Shear, 𝑉𝑆 = 14.3394 𝑁
Calculating Shear Flow
𝑞𝑏 =𝑉𝑆
𝐼𝑥𝑥∫𝑧 𝑑𝐴
𝑞𝑏𝐵𝐴= 424.3355 𝑁/𝑚
𝑞𝑏𝐴𝐵= 934.0782 𝑁/𝑚
Calculating Shear Stress
𝜏𝑏 =𝑞𝑏
𝑡
Where 𝑡 = 0.003 𝑚
𝜏𝑏𝐵𝐴= 0.1415 × 106
𝜏𝑏𝐴𝐵= 0.3114 × 106
112
Calculating Safety Factor
𝑆𝐹 =𝐹𝑠
𝜏𝑏𝑚𝑎𝑥
=13.785 × 106
0.3114 × 106
= 44.27
The value shows a positive value with greater than 1. This means a safe condition for the RC plane
to fly with this applied force.
6.3.5.2 Shear stress due to shear force (semi-monocoque)
Shear flow for each cell
𝑞𝑠 = 𝑞𝑏 + 𝑞𝑠,0
𝑞𝐵𝐴 = 𝑞𝑏𝐵𝐴+ 𝑞1 = 1559.9419 + 𝑞1
𝑞𝐴𝐵 = 𝑞𝑏𝐴𝐵+ 𝑞2 = 3433.8583 + 𝑞2
𝑞𝐴𝐵𝑖𝑛= 𝑞1 + 𝑞2
Figure 6-33 shows the line integral is obtained from the wing section.
Figure 6-33 Line integral of wing section
113
𝑙𝐵𝐴 = 0.0888 𝑚
𝑙𝐴𝐵 = 0.2582 𝑚
𝑙𝐴𝐵𝑖𝑛= 0.0202 𝑚
𝑙𝐵𝐴
𝑡𝑠𝑘= 29.6
𝑙𝐴𝐵
𝑡𝑠𝑘= 86.0667
𝑙𝐴𝐵𝑖𝑛
𝑡𝑠𝑝= 1.8364
Modulus of Rigidity
𝐺 =𝐸
2(1 + 𝑣)
=6894.76 × 106
2(1 + 0.3)
= 2651.8308 × 106
Angular displacement
휃𝑖 =1
2𝐴𝑖𝐺𝑖∫
𝑞
𝑡𝑑𝑠
𝑖
Angular displacement for each cell is the same
Cell 1:
휃1𝐺 =1
2𝐴1[𝑞1 (
𝑙𝐵𝐴
𝑡𝑠𝑘) + 𝑞12 (
𝑙𝐴𝐵𝑖𝑛
𝑡𝑠𝑝)]
29.6𝑞1 + 1.8364𝑞1 − 1.8364𝑞2 − 2𝐴1𝐺휃 = 0
31.4364𝑞1 − 1.8364𝑞2 − 3.4368 × 106휃 = 0
Cell 2:
휃2𝐺 =1
2𝐴2[𝑞2 (
𝑙𝐴𝐵
𝑡𝑠𝑘) − 𝑞12 (
𝑙𝐴𝐵𝑖𝑛
𝑡𝑠𝑝)]
114
86.0667𝑞2 − 1.8364𝑞1 + 1.8364𝑞2 − 2𝐴1𝐺휃 = 0
−1.8364𝑞1 + 87.9031𝑞2 − 9.17 × 106휃 = 0
Positive (+ve) deflection analysis
The moment, M produced by the total shear flow about any convenient moment centre is given
by below equation. Taking the moment at point A.
Equilibrium of Moment
𝑀 = Σ2𝐴𝑖𝑞𝑖
2𝐴1𝑞𝐵𝐴 + 2𝐴2𝑞𝐴𝐵 = 0
2𝐴1(1559.9419 + 𝑞1) + 2𝐴2(3433.8583 + 𝑞2) = 0
1.296 × 10−3𝑞1 + 3.458 × 10−3𝑞2 + 0휃 = −3.7667
Solve simultaneously for Cell 1, Cell 2 and Equilibrium of Torque equation
𝑞1 = −838.9335
𝑞2 = −774.8531
휃 = −0.0073
Calculate Shear Stress
𝜏1 =𝑞1
𝑡𝑠𝑘+
(𝑞1 + 𝑞2)
𝑡𝑠𝑝= −0.4264 × 106
𝜏2 =𝑞2
𝑡𝑠𝑘+
(𝑞2 − 𝑞1)
𝑡𝑠𝑝= −0.2525 × 106
115
Calculate Safety Factor
𝑆𝐹1 =𝐹𝑠
𝜏1
=13.785 × 106
0.4264 × 106
= 32.33
𝑆𝐹2 =𝐹𝑠
𝜏2
=13.785 × 106
0.2525 × 106
= 51.60
Therefore, cell 1 and cell 2 both have a safe structural system.
Negative (-ve) deflection analysis
The calculation of moment for downward deflection is same as upward deflection scenario.
Equilibrium of Moment
𝑀 = Σ2𝐴𝑖𝑞𝑖
2𝐴1𝑞𝐵𝐴 + 2𝐴2𝑞𝐴𝐵 = 0
2𝐴1(−1171.3717 + 𝑞1) + 2𝐴2(−2578.5091 + 𝑞2) = 0
1.296 × 10−3𝑞1 + 3.458 × 10−3𝑞2 + 0휃 = −3.78
Solve simultaneously for Cell 1, Cell 2 and Equilibrium of Torque equation
𝑞1 = −841.8957
𝑞2 = −777.5891
휃 = −0.0073
Calculate Shear Stress
𝜏1 =𝑞1
𝑡𝑠𝑘+
(𝑞1 + 𝑞2)
𝑡𝑠𝑝= −0.4279 × 106
𝜏2 =𝑞2
𝑡𝑠𝑘+
(𝑞2 − 𝑞1)
𝑡𝑠𝑝= −0.2534 × 106
116
Calculate Safety Factor
𝑆𝐹1 =𝐹𝑠
𝜏1
=13.785 × 106
0.4279 × 106
= 32.22
𝑆𝐹2 =𝐹𝑠
𝜏2
=13.785 × 106
0.2534 × 106
= 54.41
This highlighted that the applied shear force does not fail the wing.
6.3.5.3 Shear stress due to torsion (monocoque)
The problems involving torsion are common with aircraft structures. The material covered
wing of the airplane are basically thin walled tubular structures subjected to large torsion moments
under many flight and landing conditions. In this analysis, both upward and downward deflection
of aileron is analyses. The monocoque structure of the wing is being analyse. In monocoque
structure, the spar is not taking into account as the support system. Only one cell is investigated
for the shear stress analysis.
Positive (+ve) deflection analysis
The maximum torsion applied for upward deflection for the aircraft is 2.3969 Nm. The shear
flow is calculated followed by the shear stress and safety factor.
Maximum Torsion, 𝑇 = 2.3969 𝑁𝑚
Area, 𝐴 = 𝐴1 + 𝐴2 = 2.377 × 10−3 𝑚2
117
Calculate Shear Flow
𝑞 =𝑇
2𝐴
= 504.1859 𝑁/𝑚
Calculate Shear Stress
𝜏 =𝑞
𝑡
= 168061.9829
Calculate Safety Factor
𝑆𝐹 =𝐹𝑠
𝜏
=13.785 × 106
168.0620 × 103
= 82.02
The safety factor obtain is greater than 1. Thus, this means that the structure of the wing is safe to
be operate.
Negative (-ve) deflection analysis
The maximum torsion applied for downward deflection for the aircraft is 6.4716 Nm. The shear
flow is calculated, followed by the shear stress and safety factor.
Maximum Torsion, 𝑇 = 6.4716 𝑁𝑚
Area, 𝐴 = 𝐴1 + 𝐴2 = 2.377 × 10−3 𝑚2
118
Calculate Shear Flow
𝑞 =𝑇
2𝐴
= 1361.2958 𝑁/𝑚
Calculate Shear Stress
𝜏 =𝑞
𝑡
= 453765.2503
Calculate Safety Factor
𝑆𝐹 =𝐹𝑠
𝜏
=13.785 × 106
453.7653 × 103
= 30.38
This proved that the structure will not fail when there is downward aileron deflection applied to
the aircraft.
6.3.5.4 Shear stress due to torsion (semi-monocoque)
In this part, the investigation is done for the wing semi-monocoque structure. This type of
structure taking the spar into account in the analysis as the support system. The aircraft wing
consists of only one spar thus the analysis will be done with two cell system.
The line integral for shear stress due to torsion is same as in the shear stress due to shear
flow part. The angular displacement in each Cell 1 and Cell 2 equation also same as in the shear
119
stress due to shear flow part. The equilibrium of torque is different in this part for both positive
and negative deflection.
Cell 1:
31.4364𝑞1 − 1.8364𝑞2 − 3.4368 × 106휃 = 0
Cell 2:
−1.8364𝑞1 + 87.9031𝑞2 − 9.17 × 106휃 = 0
Positive (+ve) deflection analysis
The maximum torsion applied for upward deflection for the aircraft is 2.3969 Nm. The shear
flow is calculated followed by the shear stress and safety factor.
Maximum Torsion, 𝑇 = 2.3969 𝑁𝑚
The wing section comprises 2 cells and carries a torque of 2.3969 Nm which generate individual
but unknown torques in each of the cells. Each cell therefore develops a constant shear flow.
Then the equation of torque is
Equilibrium of Torque
𝑇 = Σ2𝐴𝑖𝑞𝑖
2𝐴1𝑞1 + 2𝐴2𝑞2 = 2.3969
1.296 × 10−3𝑞1 + 3.458 × 10−3𝑞2 + 0휃 = 2.3969
Solve simultaneously for Cell 1, Cell 2 and Equilibrium of Torque equation
𝑞1 = 533.8465
𝑞2 = 493.0697
휃 = 0.0046
120
Calculate Shear Stress
𝜏1 =𝑞1
𝑡𝑠𝑘+
(𝑞1 + 𝑞2)
𝑡𝑠𝑝= 0.2713 × 106
𝜏2 =𝑞2
𝑡𝑠𝑘+
(𝑞2 − 𝑞1)
𝑡𝑠𝑝= 0.1607 × 106
Calculate Safety Factor
𝑆𝐹1 =𝐹𝑠
𝜏1
=13.785 × 106
0.2713 × 106
= 50.81
𝑆𝐹2 =𝐹𝑠
𝜏2
=13.785 × 106
0.1607 × 106
= 85.81
Both value of calculated safety factor shows that the structure is able to support the applied torsion.
Thus, it is considered as a safe structure.
Negative (-ve) deflection analysis
The maximum torsion applied for downward deflection for the aircraft is 6.4716 Nm. The shear
flow is calculated followed by the shear stress and safety factor. The calculation and steps are same
as the upward deflection.
Maximum Torsion, 𝑇 = 6.4716 𝑁𝑚
Equilibrium of Torque
𝑇 = Σ2𝐴𝑖𝑞𝑖
2𝐴1𝑞1 + 2𝐴2𝑞2 = 6.4716
1.296 × 10−3𝑞1 + 3.458 × 10−3𝑞2 + 0휃 = 6.4716
121
Solve simultaneously for Cell 1, Cell 2 and Equilibrium of Torque equation
𝑞1 = 1.4414 × 103
𝑞2 = 1.3313 × 103
휃 = 0
Calculate Shear Stress
𝜏1 =𝑞1
𝑡𝑠𝑘+
(𝑞1 + 𝑞2)
𝑡𝑠𝑝= 0.7325 × 106
𝜏2 =𝑞2
𝑡𝑠𝑘+
(𝑞2 − 𝑞1)
𝑡𝑠𝑝= 0.4338 × 106
Calculate Safety Factor
𝑆𝐹1 =𝐹𝑠
𝜏1
=13.785 × 106
0.7325 × 106
= 18.82
𝑆𝐹2 =𝐹𝑠
𝜏2
=13.785 × 106
0.4338 × 106
= 31.78
The safety factor for both cells shows a value of more than 1. Thus, this prove that there will be
no structural damage or fail during the flight operation.
6.3.5.5 Shear Stress Summary
From the analysis and calculation of the shear stresses due to shear force and torsion, the total
shear flow for monocoque and semi-monocoque structure is obtained. The maximum shear flow
value of each cases is taken into consideration. Table 6-20 shows the total shear flow obtained.
122
Table 6-20 Total shear flow for monocoque and semi-monocoque wing structure
Shear Flow, 𝑞
Monocoque Semi-monocoque
Cell Cell Cell 1 Cell 2
Due to Shear Force Torsion Shear Force Torsion Shear Force Torsion
Maximum q 934.078 1361.296 -838.93 1441.400 -774.853 1331.300
Total 𝑞 2295.374 602.467 556.447
The combine total shear stress due to shear force and due to torsion is calculated and tabulated in
Table 6-21 below.
Table 6-21 Total Shear Stress of Monocoque and Semi-monocoque
Total Shear Stress
Monocoque Semi-monocoque
Cell 1 Cell 1 Cell 2
765124.663 306177.930 181298.700
The safety factor for the combine stresses is shown in Table 6-22. The value of the safety factor
indicate that the structure is safe for operation with the applied maximum shear force and torsion.
Table 6-22 Safety Factor of Monocoque and Semi-monocoque
Safety Factor
Monocoque Semi-monocoque
Cell 1 Cell 1 Cell 2
18.02 45.02 76.03
123
6.3.6 Part Attachment
The part attachment between the wing and the fuselage is the wing strut. The wing strut
supports the wing structure. Structural analysis of the wing strut is carried out to determine if the
it is able to support the load on the wing. Furthermore, the analysis of the joint between the wing’s
wall and the wing strut as well as that between the fuselage’s wall are carried out. Also, the analysis
of the part attachment between the aileron and the aerofoil is executed.
6.3.6.1 Joint of the part attachment
The wing strut is attached to the wing and the fuselage by applying glue and using the hinge
with a hinge pin. The material selected for hinge and the hinge pin are mica plastic and AN Steel
respectively. The moment at the joint is assumed to be zero at the initial design stage, and thus,
the wing strut is a two force-member with pin joints at both ends. Since the wing strut is a two-
force member, the result of the analysis at the joint between the wing and the wing strut is the same
as at the joint between the fuselage and the wing strut. Thus, only the joint between the wing and
the wing strut is performed. The hinge pin is used to resist the vertical reaction force of the wing
strut while the glue joint aims to resist the horizontal shear force between the surfaces of the wing
and the wing strut.
Figure 6-34 Part attachment (wing strut)
124
Hinge pin
Table 6-23 Material Properties of AN Steel (hinge pin)
Material Properties AN Steel
Ultimate Tensile Strength, Ftu 125000 psi
Ultimate Shear Strength. Fsu 75000 psi
Bearing Strength, Fb 180000 psi
Safety factor of fitting 1.15
According to Bruhn (1973), the section properties and the ultimate shear, tension and
bending strengths for AN Standard Steel bolt/pin at room temperature are as follows,
Table 6-24 Properties of AN Standard Steel bolt/pin
Size of pin 0.19
Area of solid section (in2) 0.02835
Moment of inertia of solid (in4) 0.0000640
Ultimate single shear strength at full D, lb 2126
Ultimate tensile strength (in thread), lb 2210
Ultimate bending moment in lbs 121
Figure 6-35 Dimension of the connection
125
Dimensions
Diameter of hole, D = 0.19 in
Thickness, t = 0.08 in
Width of hinge, 2R = 0.4 in
From wing loading calculation,
Vertical reaction force on Wing strut, Rstrut = 28.1707 𝑁 (6.333 lb)
𝐷𝑒𝑠𝑖𝑔𝑛 𝑓𝑖𝑡𝑡𝑖𝑛𝑔 𝑙𝑜𝑎𝑑, 𝑃 = 1.15 × 6.333 = 7.283 𝑙𝑏
Check Shear Strength of Hinge Pin
The bolt is double shear. From Table 6-24, the ultimate single shear strength is 2126 lb.
Hence, 𝑃𝑢 = 2 × 2126 = 4252 𝑙𝑏
𝑆𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 =𝑃𝑢
𝑃=
4252
7.283= 583.83
Check Bending Strength of Hinge Pin
Figure 6-36 The moment arm for bending moment on hinge pin
126
Dimensions
𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑤𝑎𝑙𝑙, 𝑡1 = 0.1969 𝑖𝑛
𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 ℎ𝑖𝑛𝑔𝑒, 𝑡2 = 0.08 𝑖𝑛
𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒, 𝑔 = 0.004 𝑖𝑛
𝑏 =𝑡12
+𝑡24
+ 𝑔
𝑏 = 0.12245
Therefore, bending moment on the pin,
𝑀 = 0.5𝑃𝑏 = 0.5(7.283)(0.12245) = 0.4459 𝑙𝑏 ∙ 𝑖𝑛
𝑓𝑏 =𝑀𝑟
𝐼=
(0.4459) (0.192 )
0.000064= 661.883 𝑝𝑠𝑖
As 𝐹𝑏 = 180000𝑝𝑠𝑖, then
𝑆𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝐹𝑏
𝑓𝑏= 271.9515
Hinge
Table 6-25 Material Properties of Mica Plastic (fixed hinge)
Material Properties Mica Plastic
Ultimate Tensile Strength, Ftu 25000 psi
Ultimate Shear Strength. Fsu 38400 psi
Safety factor of fitting 1.15
Check Tensile Strength of Pin Hole
𝑃𝑢 = 𝑘𝑡𝐹𝑡𝑢𝐴𝑡
𝑘𝑡 is the tension efficiency factor to take into account of the stress concentration due to
the pin hole, and its value can be obtained from the graph shown in Figure 6-37.
127
Figure 6-37 Graph of kt against d/w
𝑃𝑢 = 𝑘𝑡𝐹𝑡𝑢(2𝑅 − 𝐷)𝑡
𝑃𝑢 = 2.19(25000)(0.4 − 0.19)(0.08) = 919.8 𝑙𝑏
Therefore,
𝑆𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 =𝑃𝑢
𝑃=
919.8
7.283= 126.2941
Check Shear Out Strength of Hinge
𝑃𝑢 = 𝐹𝑠𝑢𝐴𝑠
𝑃𝑢 = 38400(0.4 − 0.19)(0.08) = 645.12 lb
The value is over the design fitting load of 7.283 lb
Glue
Check Shear Strength of Glue
Figure 6-38 Shear stress on the contact area of the glue
128
𝜏 =𝑉
𝐴
𝜏 =𝑉
𝐴𝑠𝑡𝑟𝑢𝑡 + 𝐴ℎ𝑖𝑛𝑔𝑒
𝜏 =84.5589cos (19.46)
(0.01)(0.002) + (0.01)(0.02)
𝜏 = 362401.88 𝑃𝑎
The shear strength of the glue is assumed as 6000 psi (41.36854 MPa), therefore
𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 = 41.36854
0.3624= 114.151
Since all the values of safety factor are greater than 1, we can conclude that the all the
joints of the wing strut to the wall of the wing and the wall of the fuselage are safe.
6.3.6.2 Structural Analysis of Wing Strut
The wing strut can buckle due to tension and compression, depending on the direction of
the wing’s deflection. The cross section of the wing strut also experiences shear.
Axial force
The applied stress due to the axial force acted on the wing strut can be calculated as follows,
𝜎 =𝐹
𝐴
𝜎 =84.5589
(0.01)(0.002)= 4.2279 𝑀𝑃𝑎
The critical buckling strength of the wing strut can also be obtained, based on the material
properties of the wing strut (plywood). Assuming that the wing strut is simply supported, the
buckling strength is
𝜎𝑐𝑟 =𝜋2𝑘𝑐𝐸
12(1 − 𝑣2)(𝑡
𝑏)2
129
𝜎𝑐𝑟 =𝜋2(4)(1000 × 103)
12(1 − 0.32)(0.002
0.01)2
𝜎𝑐𝑟 = 1.4461 × 105 𝑝𝑠𝑖 = 997.048 𝑀𝑃𝑎
The 𝐹0.7value is 700ksi. Since the 𝜎𝑐𝑟value is much smaller than the 𝐹0.7 value, correction
is not required. Therefore,
𝑅𝑐 = 𝜎
𝜎𝑐𝑟= 4.2404 × 10−3
Shear force
The applied stress due to the shear force acted on the wing strut can be calculated as follows,
𝜏 =𝑉
𝐴
𝜎 =84.5589 sin(19.46)
(0.01)(0.002)= 1.40853 𝑀𝑃𝑎
Assuming that the wing strut is simply supported, the buckling strength is
𝜎𝑐𝑟 =𝜋2𝑘𝑠𝐸
12(1 − 𝑣2)(𝑡
𝑏)2
𝜎𝑐𝑟 =𝜋2(5.5)(1000 × 103)
12(1 − 0.32)(0.002
0.01)2
𝜎𝑐𝑟 = 1.9884 × 105 𝑝𝑠𝑖 = 1370.9410 𝑀𝑃𝑎
The 𝐹0.7value is 700ksi. Since the 𝜎𝑐𝑟value is much smaller than the 𝐹0.7 value, correction
is not required. Therefore,
𝑅𝑠 = 𝜎
𝜎𝑐𝑟= 1.0274 × 10−3
130
Figure 6-39 Combined compression, bending and shear (Bruhn, 1973)
𝑅𝑠
𝑅𝑐= 0.2423, 𝑅𝑏 = 0
From Figure 6-39, 𝑅𝐶𝐴 = 0.91
𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝑅𝐶𝐴
𝑅𝐶=
0.91
4.2404 × 10−3= 214.602
Since the safety factor for the loading of wing strut due to axial load and shear load is larger
than 1, we can conclude that the wing strut is safe.
6.3.6.3 Part attachment between aerofoil and aileron
According to Federal Aviation Regulations (FAR) Appendix A to Part 23 (Simplified Design Load
Criteria), the simplified limit surface loadings for the aileron is specified as follows.
131
Table 6-26 Average limit control surface loading
Figure 6-40 Average limit control surface loading
132
Based on Figure 6-40,
𝑤𝑎𝑟𝑒𝑎 = 0.466 (𝑛𝑊
𝑆)
Where,
𝑊𝑒𝑖𝑔ℎ𝑡,𝑊 = 14.715 𝑁 (3.30806 𝑙𝑏)
𝑊𝑖𝑛𝑔𝑠𝑝𝑎𝑛 𝑎𝑟𝑒𝑎, 𝑆 = 0.20387 𝑚2 (2.19444 𝑓𝑡2)
𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑎𝑖𝑙𝑒𝑟𝑜𝑛 = 0.235𝑚
𝑤𝑎𝑟𝑒𝑎 = 0.466 (3(3.30806)
2.19444) = 2.10745 𝑙𝑏/𝑓𝑡2 (100.905𝑁/𝑚2)
𝑤 = 100.905 × 0.235 = 23.7127𝑁/𝑚
Figure 6-41 Chordwise lift distribution of the aerofoil with deployed aileron
Figure 6-42 Wing loading on the deployed aileron
𝑉 =1
2(0.04)(23.7127) = 0.4743 𝑁
𝑀 =1
2(0.04)(23.7127) (
1
3) (0.04) = 0.0063234 𝑁𝑚
133
The result shows that the amount of the applied load is very small. A fibre tape is used to
withstand the shear and bending moment of the part attachment between the aileron and the
aerofoil. Fibre tape is chosen due to its strong bonding properties. The fibre tape is applied to one
side of the aileron only with the upper part of the wing so that it can be deflect upward and
downward freely. The tape function like a hinge to the aileron. The area of tape use for the
attachment in each aileron is shown in Figure 6-43. It is assumed that the fibre tape is so strong
that it can withstand 4kg load and is able to withstand shear load of 83.65N.
Figure 6-43 Aileron attachment area
134
6.4 Summary of Safety Factor
Therefore, the safety factor obtained from the structural analysis can be summed up in
Table 6-27. Based on Table 6-27, as all the factor of safety is more than 1, the structure of the
wing components is safe.
Table 6-27 Factor of Safety of Wing Components
Structure Factor of safety
Spar Rectangular spar 14.4
Skin Upper Skin 262.8
Lower Skin 241.7
Rib Cell 1 29862.7
Cell 2 37904.4
Part Attachment
Wing Strut 214.6
Hinge Pin 272
Hinge 126.3
Glue 114.2
142
CHAPTER 7
FUSELAGE AND TAIL
7.1 Introduction
The basic functions of an aircraft’s fuselage structure are to transmit and resist applied
load, to provide an aerodynamics shape and to protect passengers, payload, systems and others
from the environmental conditions encountered in flight. These requirements in most aircraft,
result in thin shell structures where the outer surface or skin of the shell is usually supported by
longitudinal stiffening members and transverse frames to enable it to resist bending, compressive
and tensional loads without buckling. Such structures are known as semi-monocoque, while thin
shells which rely entirely on their skins for their capacity to resist loads are referred to as
monocoque.
Fuselage while of different shape to the aerodynamics surfaces, comprise members which
perform similar functions to their counterparts in the wings and tail plane. However, there are
differences in the generation of the various types of load. Aerodynamic forces on the fuselage skin
are relatively low, on the other hand, the fuselage supports large concentrated loads such as wing
reactions, tail plane reactions, undercarriage reactions and it carries payloads of varying size and
weight, which may cause large inertia forces.
Furthermore, aircraft designed for high altitude flight must withstand internal pressure. The
shape of the fuselage cross-section is determined by operational requirements. For example, the
most efficient sectional shape for a pressurized is circular or a combination of circular elements.
Irrespective of shape, the basic fuselage structure is essentially a single cell thin-walled tube
comprising skin, transverse frames and stringers; transverse frames which extend completely
across the fuselage are known as bulkheads.
143
7.2 Monocoque Structure
Monocoque fuselage design as shown in Figure 7-1 relies on the strength of the skin (also
known as the shell or covering) to carry the various loads. True monocoque construction does not
use formers, frame assemblies, or bulkheads to give shape to the fuselage. Instead, the skin carries
all fuselage stresses. Since no bracing members are present, the skin must be strong enough to
keep the fuselage rigid. Thus, the biggest challenge in monocoque design is maintaining enough
strength while keeping the weight within allowable limit.
Figure 7-1: Fuselage monocoque structure
For our fuselage monocoque design, the construction is done by using the two former and
one bulkhead to give shape to the fuselage. The material used for both bulkhead and formers is
plywood with thickness of 3mm. Then, between the two formers, there is a servo platform (blue
colour) where both servo for rudder and elevator was placed. This servo platform was made of
EPP foam with the thickness of 5mm. There is also the battery tray (red colour), made from EPP
foam where the battery is going to be and it will be hold by the battery holder (black colour) which
made from the Velcro tape.
144
7.3 Material Properties
The Medium Density Fiberboard (MDF) with 3mm thickness use to fabricate both formers and
bulkhead. The material properties for the MDF as shown in Table 7-1 below.
Table 7-1: Material properties
Mechanical Properties
Compressive strength 10 Mpa
Elastic (Young’s Tensile) Modulus 4.0 Gpa
Modulus of rupture 35.85 Mpa
Elongation at break 0.5%
Poisson’s ratio 0.25
Shear modulus 1.6 Gpa
Ultimate tensile strength 18 Mpa
Thermal Properties
Specific heat capacity 1700 J/kg-K
Thermal conductivity 0.3 W/m-K
Thermal expansion 12 µm/m-K
Other Properties
Density 0.75 g/cm3
Dielectric strength 0.5 kV/mm
145
7.4 Shear and Bending Moment Diagram for Fuselage
Analysis of the fuselage is taken when the plane flying in cruising condition with
differences G with -1.5, 1 and 3.
7.4.1 𝑮 = 𝟏
The maximum shear force and bending moment are 6.4 N and -0.48 Nm respectively.
146
7.4.2 𝑮 = 𝟑
The maximum shear force and bending moment are 19.24 N and -1.44 Nm respectively.
147
7.4.3 𝑮 = −𝟏. 𝟓
The maximum shear force and bending moment are -9.61 N and 0.7 Nm respectively.
148
Table 7-2: Maximum values of shear force and bending moment for fuselage
G = -1.5 G = 1 G = 3
Shear Force (N) -9.61 6.4 19.24
Bending moment
(Nm) 0.7 -0.48 -1.44
Material used for fuselage skin is EPP foam with material properties of ultimate stress
27.37 MPa with the thickness of 0.005 m.
𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠, 𝜎𝑏
27.57 𝑀𝑃𝑎 = 𝑀𝑦
𝐼
Where
M bending moment (Nm)
y Vertical distance away from the neutral axis (m)
I Second moment of inertia (m4)
The second moment of inertia, 𝐼 = 𝑏𝑑3
12 is assume as figure below:
Figure 7-2: Cross section of maximum bending moment location for fuselage (cm)
149
𝐼 = (0.12)(0.105)3
12−
(0.11)(0.095)3
12
𝐼 = 3.717 × 10−6 𝑚4
From the bending stress, the maximum bending moment can be stand by the foam is given as:
𝑀𝑚𝑎𝑥 =27.57 × 106 × 3.717 × 10−6
0.0525
𝑀𝑚𝑎𝑥 = 1.951 𝑘𝑁𝑚
For the maximum shear stress is assumed to be the same as ultimate stress:
𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠, 𝑣𝑚𝑎𝑥
27.57 𝑀𝑃𝑎 =𝑉𝑄
𝐼𝑏
Where
Q Calculated statically moment (m3)
V Shear force (N)
I Moment of inertia (m4)
b Width of the fuselage (m)
Calculated statically moment, 𝑄 = 𝐴𝑦
Where
A Cross section area of fuselage
y Vertical distance away from the neutral axis (m)
𝑄 = (0.12 × 0.105 × 0.06 ) − (0.11 × 0.095 × 0.055)
𝑄 = 1.813 × 10−4 𝑚3
From the Shear stress, the maximum load force able to resist from damage:
𝑉 = 27.57 × 106 × 3.717 × 10−6 × 0.12
1.813 × 10−4
𝑉 = 67.83 𝑘𝑁
150
Therefore, the ultimate stress from the foam show that maximum shear force and bending
moment can be accept are 67.83 kN and 1.951 kNm respectively. The result from the analysis of
shear force and bending moment diagram having the maximum shear force with 19.24 and
maximum bending moment -1.44 Nm when G is 3. Both of it is within the range of maximum
shear force and bending moment of the foam. Overall, during the cruising condition the EPP foam
will not damage easily.
7.5 Flow Simulation
In order to see the pressure acting on our rc plane, Solidwork flow simulation was used to
see the pressure especially at the area of formers and bulkhead as shown in Figure 7-3; thus it will
be used as an estimation of pressure applied on the internal structure in order to do the structural
analysis. The simulation was run with the maximum speed of 33.6m/s based on the v-n diagram
that we already plotted.
Figure 7-3: Pressure contour
Figure 7-4 shown the air flow on the rc plane and there is some vorticity flow (blue) occur
on the top surface at the middle of the wing where the attachment between the wing and fuselage
is located.
Area of interest
151
Figure 7-4: Flow visualization
7.6 Stress Analysis
7.6.1 Former structure
For the structural analysis, we used Ansys as a tool to do the analyze the structure by
assuming the load applied on the former as shown in Figure 7-5. There is an additional load from
on top of the former due to the wing load and we assume the bottom surface to be fixed support.
The value of pressure applied is based on the result of flow simulation.
152
Figure 7-5: Load applied on former
The result of the stress analysis for former is shown in Table 7-3. We analyze based on two
difference type of stresses which is equivalent (Von-Mises) stress and shear stress at the condition
when the rc plane at the maximum speed.
Table 7-3: Stress result for former
Velocity
(m/s)
Load (MPa) Equivalent (Von-Mises)
Stress (MPa)
Shear Stress (MPa)
Max Min Max Min
33.6 0.10187 5.0187 0.0141 2.2434 -2.2418
153
To identify the location of maximum and minimum of both type of stresses acting on the
former, refer to Figure 7-6 where it show the pressure contour when the load is applied. For
equivalent (Von-Mises) stress, we can see that most of the area has low stress and very small are
has the high stress. Besides, most area of the former has the approximately zero shear stress area.
Figure 7-6: Pressure contour on former structure for (a) equivalent (Von-Mises) stress (b) shear
stress
7.6.2 Bulkhead structure
Figure 7-7 show the assumption on how the pressure acting on the bulkhead. The pressure
was obtained from the flow simulation and no additional load is acting on the structure. For the
analysis, the Ansys tool was used to identify the stress on the structure.
(a) (b)
154
Figure 7-7: Load applied on bulkhead
The result of the stress analysis for bulkhead is shown in Table 7-4. We analyze based on
two difference type of stresses which is equivalent (Von-Mises) stress and shear stress at the
condition when the rc plane at the maximum speed.
Table 7-4: Stress result for bulkhead
Velocity
(m/s)
Load (MPa) Equivalent (Von-Mises)
Stress (Pa)
Shear Stress (Pa)
Max Min Max Min
33.6 0.10105 0.3923 0.0699 0.1680 -0.1680
To identify the location of maximum and minimum on both type of stresses acting on the
bulkhead, refer to Figure 7-8 where it show the pressure contour when the load is applied. For
equivalent (Von-Mises) stress, we can see that most of the area has low stress and very small area
has the high stress. Besides, most area of the bulkhead has the approximately zero shear stress
area.
155
Figure 7-8: Pressure contour on bulkhead structure for (a) equivalent (Von-Mises) stress (b) shear
stress
7.7 Compressive-Buckling Analysis
Based on Bruhn, the equation for the elastic instability of flat sheet in compression is,
𝜎𝑐𝑟 =𝜋2𝑘𝑐𝐸
12(1 − 𝜐2)(𝑡
𝑏)2
Where;
𝑘𝑐 is buckling coefficient which depends on edge boundary conditions and sheet aspect ratio
(a/b). Refer graph of compressive-buckling coefficient for flat rectangular plate in Bruhn.
𝐸 is modulus of elasticity
𝜐 is elastic Poisson’s ratio
𝑏 is short dimension of plate or loaded edge
𝑡 is sheet thickness
(a) (b)
156
7.7.1 Former structure
The former aspect ratio is,
𝑎
𝑏=
110
107= 1.03
From graph compressive-buckling coefficient for flat rectangular plate
𝑘𝑐 = 6.0
The critical elastic compression buckling stress,
𝜎𝑐𝑟 =𝜋2(6.0)(4 × 109)
12(1 − 0.252)(
3 × 10−3
107 × 10−3)
2
𝜎𝑐𝑟 = 16.5513 𝑀𝑃𝑎
To find safety factor,
𝑆𝐹 = (𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑏𝑢𝑐𝑘𝑙𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑠𝑡𝑟𝑒𝑠𝑠 ) = (
16.5513
5.0187) = 3.30
To find the margin of safety,
𝑀𝑆 = 𝑆𝐹 − 1 = 3.30 − 1 = 2.30
Since 𝜎𝑐𝑟 = 16.5513 𝑀𝑃𝑎 the buckling stress is greater than the maximum stress
experience by the structure, which is 5.0187 MPa, the formers structure will not buckle. Besides,
the positive value of margin of safety show that the structure is safe.
157
7.7.2 Bulkhead structure
The former aspect ratio is,
𝑎
𝑏=
750
550= 1.36
From graph compressive-buckling coefficient for flat rectangular plate
𝑘𝑐 = 5.8
The critical elastic compression buckling stress,
𝜎𝑐𝑟 =𝜋2(5.8)(4 × 109)
12(1 − 0.252)(
3 × 10−3
550 × 10−3)
2
𝜎𝑐𝑟 = 0.6056 𝑀𝑃𝑎
To find safety factor,
𝑆𝐹 = (𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑏𝑢𝑐𝑘𝑙𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑠𝑡𝑟𝑒𝑠𝑠) = (
0.6056
0.3923) = 1.54
To find the margin of safety,
𝑀𝑆 = 𝑆𝐹 − 1 = 1.54 − 1 = 0.54
Since 𝜎𝑐𝑟 = 0.6056 MPa the buckling stress is greater than the maximum stress experience
by the structure, which is 0.3923 MPa the bulkhead structure will not buckle. Besides, the positive
value of margin of safety show that the structure is safe.
158
7.8 Shear-Buckling Analysis
The critical elastic shear buckling stress for flat plates with various boundary conditions is
given by the following equation,
𝜎𝑠𝑐𝑟=
𝜋2𝑘𝑠𝐸
12(1 − 𝜐2)(𝑡
𝑏)2
Where;
𝑘𝑠 is shear buckling coefficient and is plotted as a function of the plate aspect ratio a/b.
Refer graph of shear-buckling-stress coefficient of plates in Bruhn.
𝐸 is modulus of elasticity
𝜐 is elastic Poisson’s ratio
𝑏 is short dimension of plate or loaded edge
𝑡 is sheet thickness
159
7.8.1 Former structure
The former aspect ratio is,
𝑎
𝑏=
110
107= 1.03
From graph shear-buckling-stress coefficient of plates plate
𝑘𝑠 = 9.8
The critical elastic compression buckling stress,
𝜎𝑠𝑐𝑟=
𝜋2(9.8)(4 × 109)
12(1 − 0.252)(
3 × 10−3
107 × 10−3)
2
𝜎𝑠𝑐𝑟= 27.0339 𝑀𝑃𝑎
To find safety factor,
𝑆𝐹 = (𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑏𝑢𝑐𝑘𝑙𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑠𝑡𝑟𝑒𝑠𝑠) = (
27.0339
2.2434) = 12.05
To find the margin of safety,
𝑀𝑆 = 𝑆𝐹 − 1 = 12.05 − 1 = 11.05
Since 𝜎𝑠𝑐𝑟= 27.0339 MPa the shear buckling stress is greater than the maximum shear
stress experience by the structure which is 2.2434 MPa , the structure will not buckle. The positive
value of margin of safety show that the structure is safe.
160
7.8.2 Bulkhead structure
The former aspect ratio is,
𝑎
𝑏=
750
550= 1.36
From graph compressive-buckling coefficient for flat rectangular plate
𝑘𝑠 = 7.8
The critical elastic compression buckling stress,
𝜎𝑠𝑐𝑟=
𝜋2(7.8)(4 × 109)
12(1 − 0.252)(
3 × 10−3
550 × 10−3)
2
𝜎𝑠𝑐𝑟= 0.8144 𝑀𝑃𝑎
To find safety factor,
𝑆𝐹 = (𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑏𝑢𝑐𝑘𝑙𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑠𝑡𝑟𝑒𝑠𝑠) = (
0.8144
0.1680) = 4.85
To find the margin of safety,
𝑀𝑆 = 𝑆𝐹 − 1 = 4.85 − 1 = 3.85
Since 𝜎𝑠𝑐𝑟= 0.8144 MPa the shear buckling stress is greater than the maximum stress
experience by the structure, which is 0.1680 MPa the bulkhead structure will not buckle. Besides,
the positive value of margin of safety show that the structure is safe.
161
7.9 Shear Flow of Fuselage due to Torsion
In this analysis, the semi-monocoque structure of the fuselage is being analyse. This type
of structure have former that will be taking account in the analysis as the support system. This
aircraft fuselage consist of two formers. So, the analysis will be consisting of three cell system.
The maximum torsion applied to the aircraft when in cruising speed is 1.119 Nm.
Figure 7-9: Shear flow of the fuselage
Table 7-5: Fuselage parameter
Wall Length (m) Thickness
(m) G (MPa) Cell Area
120 0.45 0.005 10.6 𝐴1 = 0.01569
121 0.12 0.003 291.7
13 0.19 0.005 10.6 𝐴2 = 0.020412
24 0.19 0.005 10.6
341 0.08 0.003 291.7 𝐴3 = 0.019263
340 0.75 0.005 10.6
162
Using equation:
𝑇 = ∑ 2
𝑛
𝑅=1
𝐴𝑅𝑞𝑅
Then,
𝑇 = 2𝐴1𝑞1 + 2𝐴2𝑞2 + 2𝐴3𝑞3
So,
1.119 = 2(0.01569)𝑞1 + 2(0.020412)𝑞2 + 2(0.019263)𝑞3
Equation 1:
1.119 = 0.03138𝑞1 + 0.040824𝑞2 + 0.038526𝑞3
Known that,
휃 =𝑑휃
𝑑𝑧=
1
2𝐴𝑅∫𝑞
𝑑𝑠
𝐺𝑡
Then,
휃1 =1
2𝐴1[
𝑞1
10.6 × 106(
0.45
0.005) +
𝑞12
291.7 × 106(
0.12
0.003)]
휃2 =1
2𝐴2[
𝑞2
10.6 × 106(0.19 + 0.19
0.005) −
𝑞12
291.7 × 106(
0.12
0.003) +
𝑞23
291.7 × 106(
0.08
0.003)]
휃3 =1
2𝐴3[
𝑞3
10.6 × 106(
0.75
0.005) −
𝑞23
291.7 × 106(
0.08
0.003)]
휃1 = 휃2
휃1 = 휃3
𝑞12 = 𝑞1 − 𝑞2
𝑞23 = 𝑞2 − 𝑞3
So, Equation 2:
2.783 × 10−4𝑞1 − 1.856 × 10−4𝑞2 + 2.239 × 10−6𝑞3 = 0
Equation 3:
163
2.7494 × 10−4𝑞1 − 1.997 × 10−6𝑞2 − 3.6967 × 10−4𝑞3 = 0
Solving Equation (1) to (3) simultaneously gives
𝑞1 = 9.225𝑁
𝑚 𝑞2 = 13.915
𝑁
𝑚 𝑞3 = 6.786
𝑁
𝑚
To calculate shear stress using formula:
𝜏1 =𝑞1
𝑡𝑠𝑘+
(𝑞1 + 𝑞2)
𝑡𝑓𝑜𝑟
𝜏2 =𝑞2
𝑡𝑠𝑘+
(𝑞2 − 𝑞1)
𝑡𝑓𝑜𝑟+
(𝑞2 + 𝑞3)
𝑡𝑓𝑜𝑟
𝜏3 =𝑞3
𝑡𝑠𝑘+
(𝑞3 − 𝑞2)
𝑡𝑓𝑜𝑟
So, shear stress for each cell as follow:
𝜏1 = 9.558 𝑘𝑁 𝜏2 = 11.247 𝑘𝑁 𝜏3 = −1.019 𝑘𝑁
To calculate safety factor using formula:
𝑆𝐹 =𝐹𝑆
𝜏
𝐹𝑆 = 13.785 𝑀𝑃𝑎
𝑆𝐹1 = 1442 𝑆𝐹2 = 1226 𝑆𝐹3 = 13528
164
Based on the calculation of safety factor, all cell obtain safety factor that exceed 1. Thus,
the skin of the aircraft fuselage can withstand the applied torsion.
7.10 Structure Analysis on Horizontal Tail
Figure 7-10: Flow simulation with 33.6m/s and 860RPM
Table 7-6: Stress on surface
maximum minimum
Horizontal tail 101476.71 Pa 101215.83 Pa
Vertical tail 101364.90 Pa 101300.00 Pa
The analysis taken from a small element for the maximum and minimum stresses on the
horizontal tail since the values of vertical tail within the range of maximum and minimum for
horizontal tail.
a
b
fixed
free
a
b
165
Element taken with the dimension of a, b, and t are 0.025m, 0.01m, 0.005 respectively.
Where t is the thickness of the horizontal tail. by taking of 1.5 as the safety factor:
maximum minimum
𝐹𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒𝑚𝑎𝑥= 𝐹𝑚𝑎𝑥 × 1.5
= 101476.71 × 1.5
= 152215.07 𝑃𝑎
𝐹𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒𝑚𝑖𝑛= 𝐹𝑚𝑖𝑛 × 1.5
= 101215.83 × 1.5
= 151823.75 𝑃𝑎
Young modulus of the foam is 27.57 MPa. According to Bruhn, Figure 7-11 𝑎
𝑏 is 2.5. so 𝑘𝑠 = 6
and 𝑣𝑒 = 0.3.
Figure 7-11: Shear-Buckling-Stress Coefficient of Plates as a Function of a/b
Critical stress, 𝐹𝑐𝑟 = 𝜋2𝑘𝑠𝐸
12 (1− 𝑣𝑒2)
(𝑡
𝑏)2
𝐹𝑐𝑟 = 𝜋2(6)(27.57 × 106)
12 (1 − (0.3)2)(0.005
0.01)2
= 37.38 𝑀𝑃𝑎
𝐹0.7 = 0.7 𝐸
166
= 0.7 × 27.57 × 106
= 19.30 𝑀𝑃𝑎
Figure 7-12: Chart of Nondimensional Compressive Buckling Stress
Then, correction method must be done. The element taken is assume as one is simply
supported and another one is free while the minimum number of n is 2.
𝐹𝑐𝑟
𝐹0.7=
37.38 𝑀𝑃𝑎
19.30 𝑀𝑃𝑎= 1.94
As referred in figure 3, 𝐹𝑐𝑟
𝐹0.7= 1.3 after applying the correction method.
𝐹𝑐𝑟𝑝= 1.3 × 27.57 𝑀𝑃𝑎
= 35.84 𝑀𝑃𝑎
The value of safety factor and margin of safety must be positive. If values are less than zero it
means the elements will buckle due to the stresses.
Safety factor of horizontal tail given by:
𝑆𝐹 = 𝐹𝑐𝑟𝑝
𝐹𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒𝑚𝑎𝑥/𝑚𝑖𝑛
Maximum Minimum
35.84 𝑀𝑃𝑎
152215.07 𝑃𝑎= 0.234 × 103
35.84 𝑀𝑃𝑎
151823.75 𝑃𝑎= 0.236 × 103
Margin of safety, 𝑀𝑆 = 𝑆𝐹 − 1
167
𝑀𝑆 = 0.234 × 103 − 1
= 233
𝑀𝑆 = 0.236 × 103 − 1
= 235
Therefore, the stresses on the surface will not damage the tail of the plane
7.11 Shear and Bending Moment Diagram for Tail
Analysis taken when the plane flying in cruising condition with differences G with -1.5, 1,
3.
7.11.1.1 G = 1
The lift generated by the horizontal tail is 1.14 N/m along the tail while the weight of
horizontal tail is 0.04 kg and length of tail, L is 0.35m.
168
Maximum shear force and bending moment are 0.1995 N and -0.0512 Nm respectively.
7.11.2 G = 3
The lift generated by the horizontal tail is 3.42 N/m along the tail while the weight of
horizontal tail is 0.04 kg and length of tail, L is 0.35m.
169
Maximum shear force and bending moment are 0.5985 N and 0.1408 Nm respectively.
7.11.3 G = -1.5
The lift generated by the horizontal tail is -1.71 N/m along the tail while the weight of
horizontal tail is 0.04 kg and length of tail, L is 0.35m.
170
Maximum shear force and bending moment are -0.9909 N and -0.1734 Nm respectively.
Overall the maximum values of shear force and bending moment according to the analysis
are illustrated as below:
Table 7-7: Maximum values of shear force and bending moment
G = -1.5 G = 1 G = 3
Shear Force (N) -0.9909 0.1995 N 0.5985
Bending moment
(Nm) -0.1734 -0.0512 0.1408
The material used is EPP foam with the material properties of ultimate stress of 27.57 MPa
with the thickness of 0.005m.
171
𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠, 𝜎𝑏
27.57 𝑀𝑃𝑎 = 𝑀𝑦
𝐼
Where
M bending moment (Nm)
y Vertical distance away from the neutral axis (m)
I Second moment of inertia (m4)
Second moment of inertia, 𝐼 = 𝑏𝑑3
12
𝐼 = (0.1)(0.005)3
12+
(0.135)(0.005)3
12
𝐼 = 5.052 × 10−9 𝑚4
From the bending stress, the maximum bending moment can be stand by the foam is given as:
𝑀𝑚𝑎𝑥 =27.57 × 106 × 5.052 × 10−9
0.0675
𝑀𝑚𝑎𝑥 = 2.06 𝑁𝑚
For the maximum shear stress is assumed to be the same as ultimate stress:
𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠, 𝑣𝑚𝑎𝑥
27.57 𝑀𝑃𝑎 =𝑉𝑄
𝐼𝑏
Where
Q Calculated statically moment (m3)
V Shear force (N)
I Moment of inertia (m4)
b Width of the tail (m)
Calculated statically moment, 𝑄 = 𝐴𝑦
Where
A Cross section area of tail
y length of the tail
172
𝑄 = (0.35 × 0.005 × 0.1) + (0.135 × 0.005 × 0.1)
𝑄 = 0.2425 × 10−3 𝑚3
From the Shear stress, the maximum load force able to resist from damage:
𝑉 = 27.57 × 106 × 5.052 × 10−9 × 0.35
0.2425 × 10−3
𝑉 = 201.03 𝑁
Therefore, the ultimate stress from the foam show that maximum shear force and bending
moment can be accept are 201.03 N and 2.06 Nm respectively. The result from the analysis of
shear force and bending moment diagram having the maximum shear force of 0.5985N when G is
3 and maximum bending moment 0.1734Nm when G is -1.5. Both of it is within the range of
maximum shear force and bending moment of the foam. Overall, during the cruising condition the
EPP foam will not damage easily.
173
CHAPTER 8
LANDING GEAR
8.1 Introduction
The landing gear part for our RC-plane is tailwheel-type landing gear. It is also known as
conventional landing gear which is an aircraft undercarriage consisting of two main wheels
forward of the center of gravity and a small wheel or skid to support the tail. The term
"conventional" persists for historical reasons, but all modern jet aircraft and most modern propeller
aircraft use tricycle gear with a single nose wheel in the front and two or more main wheels aft.
The tailwheel configuration offers several advantages over the tricycle landing gear
arrangement, which make tailwheel aircraft less expensive to manufacture and maintain. Due to
its position much further from the center of gravity, a tailwheel supports a smaller part of the
aircraft's weight allowing it to be made much smaller and lighter than a nosewheel. As a result, the
smaller wheel weighs less and causes less parasitic drag. Because of the way airframe loads are
distributed while operating on rough ground, tailwheel aircraft are better able to sustain this type
of use over a long period of time, without cumulative airframe damage occurring. If a tailwheel
fails on landing, the damage to the aircraft will be minimal. Tailwheel aircraft are easier to fit into
and maneuver inside some hangars.
8.2 Static Load Analysis
The calculation of nose gear load uses the diagram shown in Figure 8-1 and the following
appropriate formulas based on American Institute of Aeronautics & Astronautics (1988):
𝑀𝑎𝑥 𝑠𝑡𝑎𝑡𝑖𝑐 𝑚𝑎𝑖𝑛 𝑔𝑒𝑎𝑟 𝑙𝑜𝑎𝑑 (𝑝𝑒𝑟 𝑠𝑡𝑟𝑢𝑡) = W(F − M)
2𝐹
174
𝑀𝑎𝑥 𝑠𝑡𝑎𝑡𝑖𝑐 𝑛𝑜𝑠𝑒 𝑔𝑒𝑎𝑟 𝑙𝑜𝑎𝑑 = W(F − L)
𝐹
Where W is the maximum gross weight and the other quantities are defined in Figure 8-1.
Figure 8-1 Diagram for nose landing gear load calculation
Figure 8-2 Centre of gravity for nose landing gear (front landing gear)
175
Figure 8-3 Centre of gravity for main landing gear (rear landing gear)
176
𝑊 = 1.5𝑘𝑔
𝐹 = 452.15𝑚𝑚
𝑁 = 19.16mm
𝑀 = 432.89𝑚𝑚
𝐿 = 442.76𝑚𝑚
𝐽 = 103.75𝑚𝑚
𝑀𝑎𝑥 𝑠𝑡𝑎𝑡𝑖𝑐 𝑚𝑎𝑖𝑛 𝑔𝑒𝑎𝑟 𝑙𝑜𝑎𝑑 (𝑝𝑒𝑟 𝑠𝑡𝑟𝑢𝑡)
= W(F − M)
2𝐹
=1.5(0.45215 − 0.43289)
2(0.45215)
= 0.03195𝑁
Since the front landing gear is consist of two tyres and struts;
𝐿𝑟𝑒𝑎𝑟,𝑒𝑎𝑐ℎ =0.03195
2= 0.01598𝑁
𝑀𝑎𝑥 𝑠𝑡𝑎𝑡𝑖𝑐 𝑛𝑜𝑠𝑒 𝑔𝑒𝑎𝑟 𝑙𝑜𝑎𝑑
= W(F − L)
𝐹
=1.5(0.45215 − 0.44276)
(0.45215)
= 0.03115N
177
8.3 Introduction of FEM
Finite element method (FEM) is a fundamental numerical procedure for the analysis of
structure. It is a numerical method for solving problems of engineering and mathematical physics.
The use of numerical methods to solve mathematical models of complex structures has become
essential criteria in design process. Finite element method is also a popular method used and a
powerful simulation tool to analyse structural problems. It has been widely used to solve
engineering problems in visualizing the stresses and displacement of a model. It provides an
outcome of finite element analysis which is able in turn help in producing a better products to an
industry company. Then, an improvement to the product will reduce the spending on prototypes
by optimizing the design through the finite element analysis.
8.3.1 Software Used
ABAQUS is commercial software to be used in order to conduct the simulation in finite
element analysis. ABAQUS can solve problems ranging from relatively simple linear analyses to
the most challenging nonlinear simulations based on finite element method. It offers a wide range
of capabilities for simulation of linear and nonlinear applications. It also contains a vast library of
elements that can model virtually any geometry. It has an equally extensive list of material models
that can simulate the behaviour of most typical engineering materials.
Figure 8-4 Overview of ABAQUS on Front Landing Gear being modelled
178
8.3.2 Material of Landing Gear
Table 8-1 Material used
Components Material Used
Attachment of Fuselage Plywood
Supports 18/10 Stainless Steel
Wheel EPP Foam
8.3.2.1 Material Property for Landing Gear
Table 8-2 Properties of the material used
Specification Parameter
Material Used 18/10 Stainless Steel Plywood
Mass density 0.0077kg/cm3 0.00063kg/m3
Young Modulus 2e7 N/cm2 860000 N/cm2
Poisson ratio 0.265 0.28
8.3.3 Applied Load on Front Landing Gear
The weight of the designed aircraft is 1.5kg. The weight is considered as the only loading
that will exert on the front landing gear per struts is 0.01598N. The load exerted on the surface is
uniform distributed pressure of 5.811E-4N/cm2 since the surface area for front landing gear
is(5.5cm × 5cm). Figure 8-5 illustrate the load exerted on the surface which attached of the
fuselage.
179
Figure 8-5 Loading on Front Landing Gear
8.3.4 Boundary Conditions for Front Landing Gear
Boundary condition for rear landing gear is the main concern of the finite element analysis.
From Figure 8-6, left bottom and right bottom are the parts that attached with the wheel. In our
case, we just considered the analysis is static deformation, and then both of the parts that attached
with wheels are considered as fixed support. The boundary condition of the bottom parts is fixed
in displacement and rotation in the analysis. Beside, only the y direction of displacement will vary
with the loading for the upper part which attached with fuselage. The other boundary conditions
for the upper part are also fixed in displacement and rotation. Figure 8-6 show all the boundary
conditions of the front landing gear.
Figure 8-6 Boundary Condition of Front Landing Gear
180
8.3.5 Meshing for Front Landing Gear
The meshing of the element is the most significant step in the analysis. Meshing of the
element will show every single element of rear landing gear that being to deform and support all
of the conditions. The approximate global size of the meshing is 2 and the meshing control of the
element shape is tetrahedron as shown in Figure 8-7.
Figure 8-7 Meshing of Front Landing Gear
The incrementation of the step is 100 number of increment which is the process of
increasing in size of the element from 0.0001 to 1. Figure 8-8 show the number of increment of
the analysis.
181
Figure 8-8 Incrementation of the analysis
182
8.4 Results of Finite Element Analysis for Front Landing Gear
Front landing gear in the analysis has 1292 number of element after obtaining the result.
Figure 8-9 show the shape of the rear landing gear before deformation that load exerted and
boundary conditions applied.
Figure 8-9 Plot Undeformed Shape of Rear Landing Gear
Contour plot in ABAQUS is based on mises stress of the structure. Contour plot display
the variation of a variable across the surface of the model. From Figure 8-10, the red colour contour
plot is the most critical stress of the element whereas the blue colour is the lowest stress of the
element that followed by the mises stress. The elements that near the load exerted and boundary
conditions applied will experience the highest stress compared to others. The curved area will also
exert more stress compared to others. Figure 8-10 illustrate the contour plot of the element.
183
Figure 8-10 Contour Plot of Front Landing Gear
Figure 8-11 shows the contour plot for the magnitude displacement for front landing gear,
the red colour contour plot is the most critical displacement of the element whereas the blue colour
is the lowest magnitude displacement of the element that followed by the contour plot. Figure 8-
11 illustrate the magnitude displacement of element.
Figure 8-11 Magnitude displacement of Front Landing Gear
184
8.5 Discussion
The main function of the nose landing gear is to balance the aircraft and provide it with the
ability to steer when plane is grounded. The landing gear must be fuselage mounted, but must also
have necessary track width for the aircraft to operate as required. The major material of use in the
landing gear is 18/10 stainless steel because the characteristics of itself have higher yield stress to
withstand the 90% of load.
From the result obtained by using ABAQUS in Figure 8-9, undeformed shape of front
landing gear was plotted. This undeformed shape is the situation when there is no force acting on
it. When 0.01598N of load was acting on the front landing gear, it began to deflect downward and
deformed shape of front landing gear.
Figure 8-11 shows the highest magnitude displacement acting on the front landing gear is
9.539cm whereas Figure 8-10 shows contour plot of rear landing gear. Based on this figure 8-10,
the biggest stress acting on the front landing gear are about 1.127 N/cm2 which acting at near to
the load applied and at contact surface between front landing gear with tyres since the assumption
of contact surface between front landing gear with tyres are fixed. The result can be more accurate
when the numbers of elements is larger by decreasing the size of elements.
185
CHAPTER 9
FLIGHT STABILITY AND CONTROL
9.1 Center of Gravity
There are four types of cg that we need to find:
1. Empty weight cg
2. Maximum weight cg
3. Forward cg
4. Aft cg
Figure 9-1 CG location of each weight on the aircraft
Based on Figure 9.1, we can proceed to find all the four types of cg.
186
9.1.1 Empty Weight cg
Table 9-1 Empty weight cg calculation
Items Mass (g) Weight,W
(N)
Horizontal Vertical
Arm,x
(mm)
Moment Wx
(Nmm)
Arm,y
(mm)
Moment Wy
(Nmm)
Propeller 20 0.1962 24.05 4.71861 0 0
Motor 72 0.70632 74 52.26768 -9.52 -6.7241664
Wing 210 2.0601 273.74 563.931774 48 98.8848
Horizontal Tail 20 0.1962 771.34 151.336908 12.17 2.387754
Vertical Tail 20 0.1962 753.41 147.819042 58.55 11.48751
Fuselage 140 1.3734 347.51 477.270234 -6.91 -9.490194
front landing gear 20 0.1962 263.84 51.765408 -92.02 -18.054324
back landing fear 10 0.0981 718.58 70.492698 -34.46 -3.380526
ESC 38 0.37278 161.21 60.0958638 -28.04 -10.4527512
LiPo battery 229 2.24649 242.96 545.8072104 -27.14 -60.9697386
Servo (aileron) 18 0.17658 323.57 57.1359906 44.67 7.8878286
Servo (elevator) 9 0.08829 294.17 25.9722693 -40.48 -3.5739792
Servo (rudder) 9 0.08829 294.17 25.9722693 -40.48 -3.5739792
Receiver 6.4 0.062784 192.94 12.11354496 -28.04 -1.76046336
∑W=8.058 ∑Wx=2246.7 ∑Wy=2.6677
Xcg= 278.818
Ycg=0.331
187
9.1.2 Maximum Weight cg
Table 9-2 Maximum weight cg calculation
Items Mass (g) Weight,W
(N)
Horizontal Vertical
Arm,x
(mm)
Moment Wx
(Nmm)
Arm,y
(mm)
Moment Wy
(Nmm)
Propeller 20 0.1962 24.05 4.71861 0 0
Motor 72 0.70632 74 52.26768 -9.52 -6.7241664
Wing 210 2.0601 273.74 563.931774 48 98.8848
Horizontal Tail 20 0.1962 771.34 151.336908 12.17 2.387754
Vertical Tail 20 0.1962 753.41 147.819042 58.55 11.48751
Fuselage 140 1.3734 347.51 477.270234 -6.91 -9.490194
front landing gear 20 0.1962 263.84 51.765408 -92.02 -18.054324
back landing fear 10 0.0981 718.58 70.492698 -34.46 -3.380526
ESC 38 0.37278 161.21 60.0958638 -28.04 -10.4527512
LiPo battery 229 2.24649 242.96 545.8072104 -27.14 -60.9697386
Servo (aileron) 18 0.17658 323.57 57.1359906 44.67 7.8878286
Servo (elevator) 9 0.08829 294.17 25.9722693 -40.48 -3.5739792
Servo (rudder) 9 0.08829 294.17 25.9722693 -40.48 -3.5739792
Receiver 6.4 0.062784 192.94 12.11354496 -28.04 -1.76046336
Payload 500 4.905 290 1422.45 -3.57 -17.51085
∑W=8.058 ∑Wx=3669.15 ∑Wy=-14.843
Xcg= 283.05
Ycg=-1.145
188
9.1.3 Forward cg
Figure 9-2 Location of cg empty and payloads
From Figure 9.2, all payloads are located behind empty weight cg on the x axis, thus position of
forward cg is equal to empty weight cg and aft cg located at same position as maximum weight
cg.
Table 9-3 Forward cg calculation
Items Mass (g) Weight,W
(N)
Horizontal Vertical
Arm,x
(mm)
Moment Wx
(Nmm)
Arm,y
(mm)
Moment Wy
(Nmm)
Propeller 20 0.1962 24.05 4.71861 0 0
Motor 72 0.70632 74 52.26768 -9.52 -6.7241664
Wing 210 2.0601 273.74 563.931774 48 98.8848
Horizontal Tail 20 0.1962 771.34 151.336908 12.17 2.387754
Vertical Tail 20 0.1962 753.41 147.819042 58.55 11.48751
Fuselage 140 1.3734 347.51 477.270234 -6.91 -9.490194
front landing gear 20 0.1962 263.84 51.765408 -92.02 -18.054324
back landing fear 10 0.0981 718.58 70.492698 -34.46 -3.380526
ESC 38 0.37278 161.21 60.0958638 -28.04 -10.4527512
LiPo battery 229 2.24649 242.96 545.8072104 -27.14 -60.9697386
Servo (aileron) 18 0.17658 323.57 57.1359906 44.67 7.8878286
Servo (elevator) 9 0.08829 294.17 25.9722693 -40.48 -3.5739792
Servo (rudder) 9 0.08829 294.17 25.9722693 -40.48 -3.5739792
Receiver 6.4 0.062784 192.94 12.11354496 -28.04 -1.76046336
∑W=8.058 ∑Wx=2246.7 ∑Wy=2.6677
Xcg= 278.818
Ycg=0.331
189
9.1.4 Aft cg
Table 9-4 Aft weight cg calculation
Items Mass (g) Weight,W
(N)
Horizontal Vertical
Arm,x
(mm)
Moment Wx
(Nmm)
Arm,y
(mm)
Moment Wy
(Nmm)
Propeller 20 0.1962 24.05 4.71861 0 0
Motor 72 0.70632 74 52.26768 -9.52 -6.7241664
Wing 210 2.0601 273.74 563.931774 48 98.8848
Horizontal Tail 20 0.1962 771.34 151.336908 12.17 2.387754
Vertical Tail 20 0.1962 753.41 147.819042 58.55 11.48751
Fuselage 140 1.3734 347.51 477.270234 -6.91 -9.490194
front landing gear 20 0.1962 263.84 51.765408 -92.02 -18.054324
back landing fear 10 0.0981 718.58 70.492698 -34.46 -3.380526
ESC 38 0.37278 161.21 60.0958638 -28.04 -10.4527512
LiPo battery 229 2.24649 242.96 545.8072104 -27.14 -60.9697386
Servo (aileron) 18 0.17658 323.57 57.1359906 44.67 7.8878286
Servo (elevator) 9 0.08829 294.17 25.9722693 -40.48 -3.5739792
Servo (rudder) 9 0.08829 294.17 25.9722693 -40.48 -3.5739792
Receiver 6.4 0.062784 192.94 12.11354496 -28.04 -1.76046336
Payload 500 4.905 290 1422.45 -3.57 -17.51085
∑W=8.058 ∑Wx=3669.15 ∑Wy=-14.843
Xcg= 283.05
Ycg=-1.145
190
After locating empty weight cg, maximum weight cg, forward cg and aft cg from all
the tables, then plot the cg location on the aircraft as illustrated in Figure 9.3.
Figure 9-3 Location of all cg on the aircraft
191
9.2 Stability Control System
Stability is the ability of an aircraft to correct for conditions that act on it, like
turbulence or flight control inputs. For aircraft, there are two general types of stability:
static and dynamic. Static stability is the initial tendency of an aircraft to return to its
original position when it's disturbed. There are three kinds of static stability which is
positive, natural and negative. Stability of aircraft is the short- and intermediate-time
response of the attitude and velocity of the vehicle. Stability considers the response of
the vehicle to perturbations in flight conditions from some dynamic equilibrium, while
control considers the response of the vehicle to control inputs.
9.2.1 Contribution of the Wing to Mcg
Figure 9-4 Wing geometry for pitching moment.
The moment about the aerodynamic center of the wing denoted by Mac,w and
the wing lift and drag are, Lw and Dw. Those Mac,w, Lw and Dw all contribute to the
moments about the center of gravity. Consider the forces and moments on the wing
only, as shown in Figure 9.4. The relative wind is inclined at the angle αw with respect
to the zero life line, where αw is the absolute angle of attack of the wing. The c is the
mean zero lift chord of the wing. The center of gravity is located a distance hc behind
the leading edge, and the zc above the zero lift line. Hence, h and z are coordinates of
the center of gravity in fractions of chord length. The aerodynamic center is distance
192
hac,wc from the leading edge. Referring to Figure 9.4, the moments about center of
gravity on the aircraft due to wing contributions only is shown as equation below.
𝑀𝑐𝑔𝑤 = 𝑀𝑎𝑐𝑤 + 𝐿𝑤 𝑐𝑜𝑠 𝛼𝑤(ℎ𝑐 − ℎ𝑎𝑐𝑤𝑐) + 𝐷𝑤 𝑠𝑖𝑛 𝛼𝑤(ℎ𝑐 − ℎ𝑎𝑐𝑤𝑐) + 𝐿𝑤
𝑠𝑖𝑛 𝛼𝑤 𝑧𝑐 − 𝐷𝑤 𝑐𝑜𝑠 𝛼𝑤 𝑧𝑐
For normal flight range of an aircraft, αw is small, thus 𝑠𝑖𝑛 𝛼𝑤 ≈ 𝛼𝑤 and 𝑐𝑜𝑠
𝛼𝑤 ≈ 1.
𝑀𝑐𝑔𝑤 = 𝑀𝑎𝑐𝑤 + (𝐿𝑤 + 𝐷𝑤 𝛼𝑤)(ℎ𝑐 − ℎ𝑎𝑐𝑤𝑐) + (𝐿𝑤𝛼𝑤 − 𝐷𝑤) 𝑧𝑐
Dividing the equation above by q∞Sc, Thus, the moment coefficient about
the center of gravity are as equation below.
𝐶𝑀,𝑐𝑔𝑤 = 𝐶𝑀,𝑎𝑐𝑤 + (𝐶𝐿,𝑤 + 𝐷𝐷,𝑤 𝛼𝑤)(ℎ − ℎ𝑎𝑐𝑤) + (𝐶𝐿,𝑤𝛼𝑤 − 𝐶𝐷,𝑤) 𝑧
𝐶𝑀,𝑐𝑔𝑤 = 𝐶𝑀,𝑎𝑐𝑤 + 𝐶𝐿,𝑤(ℎ − ℎ𝑎𝑐𝑤)
9.2.2 Contribution of the Wing-Body to Mcg
When consideration of fuselage is added into contribution of wing to moment
about center gravity, body of aircraft experiences a moment about its aerodynamic
center, because of some lift force and drag force due to the airflow around it. The
airflow about the wing-body combination is different from that over the wing and body
separately. Aerodynamic interference occurs where the flow over the wing affects the
body flow. Due to the interference, the moment due to the wing-body combination is
not the summation of wing and body moments. Similarly, the lift and drag of the wing-
body combination are affected by aerodynamic interference. By modify the equation
for the moment coefficient for wing only, the contribution of the wing-body
combination to Mcg is:-
𝐶𝑀,𝑐𝑔𝑤𝑏 = 𝐶𝑀,𝑎𝑐𝑤𝑏 + 𝐶𝐿,𝑤𝑏(ℎ − ℎ𝑎𝑐𝑤𝑏)
𝐶𝑀,𝑐𝑔𝑤𝑏 = 𝐶𝑀,𝑎𝑐𝑤𝑏 + 𝑎𝑤𝑏𝛼𝑤𝑏(ℎ − ℎ𝑎𝑐𝑤𝑏)
193
9.2.3 Contribution of the the Tail to Mcg
Figure 9-5 Geometry of wing tail combination.
From the Figure 9.5, the summation of moments about the centre of gravity
due to lift, drag and moment of the tail is
𝑀𝑐𝑔𝑡 = −𝑙[𝐿𝑡 cos(𝛼𝑤𝑏 − 휀) + 𝐷𝑡 sin(𝛼𝑤𝑏 − 휀)]
+ 𝑧𝑡𝐿𝑡 sin(𝛼𝑤𝑏 − 휀) + 𝑧𝑡𝐷𝑡 cos(𝛼𝑤𝑏 − 휀) + 𝑀𝑎𝑐,𝑡
𝑧𝑡 ≪ 𝑙𝑡
𝐷𝑡 ≪ 𝐿𝑡
𝛼𝑤𝑏 − 휀 is small, hence sin(𝛼𝑤𝑏 − 휀) ≈ 0 and cos(𝛼𝑤𝑏 − 휀) = 1
𝑀𝑎𝑐, is small in magnitude
𝑀𝑐𝑔𝑡 = −𝑙𝑡𝐿𝑡
𝑀𝑐𝑔𝑡 = −𝑙𝑡𝑞∞𝑆𝑡𝐶𝐿,𝑡
Tail volume ratio,
𝐶𝑀,𝑐𝑔,𝑡 = −𝑉𝐻𝐶𝐿,𝑡
194
9.2.4 Total Pitching Moment about the centre of gravity
Total pitching moment is use the consideration of the aircraft as whole which
is due to contribution of wing body combination and the tail.
9.2.5 Equations for Longitudinal Static Stability
1. Slope of moment coefficient curve
The criteria necessary for longitudinal balance and static stability which discussed
before are 𝐶𝑀,0 must be positive and must be negative, both conditions with
implicit assumption that αe falls within the practical flight range of angle of attack. By
substitute the αa = 0, the equation is obtained shown as below.
𝐶𝑀,0 = (𝐶𝑀,𝑐𝑔)𝐿=0 = 𝐶𝑀,𝑎𝑐,𝑤𝑏 + 𝑉𝐻𝑎𝑡(𝑖𝑡 + 휀0)
While the equation for the slope of moment coefficient curve are shown as
below.
𝜕𝐶𝑚.𝑐𝑔
𝜕𝛼= 𝛼 [ℎ − ℎ𝑎𝑐𝑤𝑏
− 𝑉𝐻
𝑎𝑡
𝑎(1 −
𝜕휀
𝜕𝛼)]
where
𝑎 = 𝑤𝑖𝑛𝑔 𝑏𝑜𝑑𝑦 𝑙𝑖𝑓𝑡 𝑐𝑢𝑟𝑣𝑒 𝑠𝑙𝑜𝑝𝑒 = 0.1066 𝑝𝑒𝑟 𝑑𝑒𝑔𝑟𝑒𝑒
ℎ = 𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑐𝑒𝑛𝑡𝑟𝑒 𝑜𝑓 𝑔𝑟𝑎𝑣𝑖𝑡𝑦
ℎ𝑎𝑐𝑤𝑏= 𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑤𝑖𝑛𝑔 𝑏𝑜𝑑𝑦 𝑎𝑒𝑟𝑜𝑑𝑦𝑛𝑎𝑚𝑖𝑐 𝑐𝑒𝑛𝑡𝑟𝑒 = 0.274 𝑚
195
𝑉𝐻 = 𝑡𝑎𝑖𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑟𝑎𝑡𝑖𝑜 =𝑙𝑡𝑆𝑡
𝑐𝑆=
(0.59)(0.075)
(0.17)(0.204)= 1.276
𝑎𝑡 = 𝑡𝑎𝑖𝑙 𝑙𝑖𝑓𝑡 𝑐𝑢𝑟𝑣𝑒 𝑠𝑙𝑜𝑝𝑒 = 0.106 𝑝𝑒𝑟 𝑑𝑒𝑔𝑟𝑒𝑒
For subsonic flow,
𝜕휀
𝜕𝛼=
2𝐶𝐿𝛼𝑤
𝜋𝐴𝑅𝑤=
2(0.113)
𝜋(7.059)= 0.0102
Therefore,
Empty Weight Centre of Gravity
ℎ = 0.278 𝑚
𝜕𝐶𝑚.𝑐𝑔
𝜕𝛼= 𝛼 [ℎ − ℎ𝑎𝑐𝑤𝑏
− 𝑉𝐻
𝑎𝑡
𝑎(1 −
𝜕휀
𝜕𝛼)]
𝜕𝐶𝑚.𝑐𝑔
𝜕𝛼= −0.133
Maximum Weight Centre of Gravity
ℎ = 0.283 𝑚
𝜕𝐶𝑚.𝑐𝑔
𝜕𝛼= 𝛼 [ℎ − ℎ𝑎𝑐𝑤𝑏
− 𝑉𝐻
𝑎𝑡
𝑎(1 −
𝜕휀
𝜕𝛼)]
𝜕𝐶𝑚.𝑐𝑔
𝜕𝛼= −0.132
Forward Centre of Gravity
ℎ = 0.279 𝑚
𝜕𝐶𝑚.𝑐𝑔
𝜕𝛼= 𝛼 [ℎ − ℎ𝑎𝑐𝑤𝑏
− 𝑉𝐻
𝑎𝑡
𝑎(1 −
𝜕휀
𝜕𝛼)]
196
𝜕𝐶𝑚.𝑐𝑔
𝜕𝛼= −0.133
Aft Centre of Gravity
ℎ = 0.283 𝑚
𝜕𝐶𝑚.𝑐𝑔
𝜕𝛼= 𝛼 [ℎ − ℎ𝑎𝑐𝑤𝑏
− 𝑉𝐻
𝑎𝑡
𝑎(1 −
𝜕휀
𝜕𝛼)]
𝜕𝐶𝑚.𝑐𝑔
𝜕𝛼= −0.132
2. Moment coefficient at zero angle of attack
𝐶𝑀,0 = 𝐶𝑀,𝑎𝑐𝑤𝑏+ 𝑉𝐻𝑎𝑡(𝑖𝑡 + 휀0)
where in our case,
𝐶𝑀,𝑎𝑐𝑤𝑏= 𝑚𝑜𝑚𝑒𝑛𝑡 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑎𝑡 𝑤𝑖𝑛𝑔 𝑏𝑜𝑑𝑦 𝑎𝑒𝑟𝑜𝑑𝑦𝑛𝑎𝑚𝑖𝑐 𝑐𝑒𝑛𝑡𝑟𝑒 = 0.228
𝑉𝐻 = 𝑡𝑎𝑖𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑟𝑎𝑡𝑖𝑜 =𝑙𝑡𝑆𝑡
𝑐𝑆= 1.276
𝑎𝑡 = 𝑡𝑎𝑖𝑙 𝑙𝑖𝑓𝑡 𝑐𝑢𝑟𝑣𝑒 𝑠𝑙𝑜𝑝𝑒 = 0.106 𝑝𝑒𝑟 𝑑𝑒𝑔𝑟𝑒𝑒
𝑖𝑡 = 𝑡𝑎𝑖𝑙 𝑠𝑒𝑡𝑡𝑖𝑛𝑔 𝑎𝑛𝑔𝑙𝑒 = 7°
휀0 = 𝑑𝑜𝑤𝑛𝑤𝑎𝑠ℎ 𝑎𝑛𝑔𝑙𝑒 𝑎𝑡 𝑧𝑒𝑟𝑜 𝑙𝑖𝑓𝑡 = 0°
Therefore,
𝐶𝑀,0 = 𝐶𝑀,𝑎𝑐𝑤𝑏+ 𝑉𝐻𝑎𝑡(𝑖𝑡 + 휀0)
𝐶𝑀,0 = 1.175
197
From the calculation of last two sections, all of the slope of the pitching moment
coefficient curve is negative and the 𝐶𝑚.𝑐𝑔 is positive. Hence, the aircraft model is
statically stable.
3. Total pitching moment about centre of gravity
Assuming linear curve for the moment coefficient versus alpha graph,
𝐶𝑀,𝑐𝑔 =𝐶𝑀,𝑐𝑔
𝜕𝛼𝛼 + 𝐶𝑀,0
Therefore,
Empty Weight Centre of Gravity
𝐶𝑀,𝑐𝑔 = −0.133𝛼 + 1.175
Table 9-5 Pitching moment at various angle of attack for empty weight
Angle (rad) 𝐶𝑀,𝑐𝑔
0 1.175
2 0.909
4 0.643
6 0.377
8 0.111
10 -0.155
12 -0.421
14 -0.687
16 -0.953
198
Figure 9-6: Graph of moment coefficient vs angle of attack for empty weight
Equilibrium angle of attack was obtained from,
0 = −0.133𝛼 + 1.175
𝛼 = 8.83° (𝑡𝑟𝑖𝑚 𝑎𝑛𝑔𝑙𝑒)
-1.5
-1
-0.5
0
0.5
1
1.5
0 2 4 6 8 10 12 14 16 18
Mom
ent
Coef
fici
ent
Angle of Attack
Moment Coefficient vs Angle of Attack
for Empty Weight
199
Maximum Weight Centre of Gravity
𝐶𝑀,𝑐𝑔 = −0.132𝛼 + 1.175
Table 9-6: Pitching moment at various angle of attack for maximum weight
Angle (rad) 𝐶𝑀,𝑐𝑔
0 1.175
2 0.911
4 0.647
6 0.383
8 0.119
10 -0.145
12 -0.409
14 -0.673
16 -0.937
Figure 9-7: Graph of moment coefficient vs angle of attack for maximum weight
Equilibrium angle of attack was obtained from,
0 = −0.132𝛼 + 1.175
𝛼 = 8.9° (𝑡𝑟𝑖𝑚 𝑎𝑛𝑔𝑙𝑒)
-1.5
-1
-0.5
0
0.5
1
1.5
0 5 10 15 20
Mom
ent
Coef
fici
ent
Angle of Attack
Moment Coefficient vs Angle of Attack
for Maximum Weight
200
Forward Centre of Gravity
𝐶𝑀,𝑐𝑔 = −0.133𝛼 + 1.175
Table 9-7: Pitching moment at various angle of attack for forward cg
Angle (rad) 𝐶𝑀,𝑐𝑔
0 1.175
2 0.909
4 0.643
6 0.377
8 0.111
10 -0.155
12 -0.421
14 -0.687
16 -0.953
Figure 9-8: Graph of moment coefficient vs angle of attack for forward cg
Equilibrium angle of attack was obtained from,
0 = −0.133𝛼 + 1.175
𝛼 = 8.83° (𝑡𝑟𝑖𝑚 𝑎𝑛𝑔𝑙𝑒)
-1.5
-1
-0.5
0
0.5
1
1.5
0 5 10 15 20
Mom
ent
Coef
fici
ent
Angle of Attack
Moment Coefficient vs Angle of Attack
for Forward CG
201
Aft Centre of Gravity
𝐶𝑀,𝑐𝑔 = −0.132𝛼 + 1.175
Table 9-8: Pitching moment at various angle of attack for aft cg
Angle (rad) 𝐶𝑀,𝑐𝑔
0 1.175
2 0.911
4 0.647
6 0.383
8 0.119
10 -0.145
12 -0.409
14 -0.673
16 -0.937
Figure 9-9: Graph of moment coefficient vs angle of attack for aft cg
Equilibrium angle of attack was obtained from,
0 = −0.132𝛼 + 1.175
𝛼 = 8.9° (𝑡𝑟𝑖𝑚 𝑎𝑛𝑔𝑙𝑒)
-1.5
-1
-0.5
0
0.5
1
1.5
0 5 10 15 20
Mom
ent
Coef
fici
ent
Angle of Attack
Moment Coefficient vs Angle of Attack
for Aft CG
202
Apparently, the angle of attack falls within the reasonable flight range, 𝛼𝑚𝑖𝑛 ≤ 𝛼 ≤
𝛼𝑚𝑖𝑛 (12.02°). Therefore, the aircraft is longitudinally balanced as well as statically
stable.
4. Calculation of neutral point
ℎ𝑛 = ℎ𝑎𝑐,𝑤𝑏 + 𝑉𝐻
𝑎𝑡
𝑎(1 −
𝜕휀
𝜕𝛼)
where in our case,
ℎ𝑎𝑐𝑤𝑏= 𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑤𝑖𝑛𝑔 𝑏𝑜𝑑𝑦 𝑎𝑒𝑟𝑜𝑑𝑦𝑛𝑎𝑚𝑖𝑐 𝑐𝑒𝑛𝑡𝑟𝑒 = 0.274 𝑚
𝑉𝐻 = 𝑡𝑎𝑖𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑟𝑎𝑡𝑖𝑜 =𝑙𝑡𝑆𝑡
𝑐𝑆= 1.276
𝑎𝑡 = 𝑡𝑎𝑖𝑙 𝑙𝑖𝑓𝑡 𝑐𝑢𝑟𝑣𝑒 𝑠𝑙𝑜𝑝𝑒 = 0.106 𝑝𝑒𝑟 𝑑𝑒𝑔𝑟𝑒𝑒
𝑎 = 𝑤𝑖𝑛𝑔 𝑏𝑜𝑑𝑦 𝑙𝑖𝑓𝑡 𝑐𝑢𝑟𝑣𝑒 𝑠𝑙𝑜𝑝𝑒 = 0.1066 𝑝𝑒𝑟 𝑑𝑒𝑔𝑟𝑒𝑒
For subsonic flow,
𝜕휀
𝜕𝛼=
2𝐶𝐿𝛼𝑤
𝜋𝐴𝑅𝑤= 0.0102
Therefore,
ℎ𝑛 = ℎ𝑎𝑐,𝑤𝑏 + 𝑉𝐻
𝑎𝑡
𝑎(1 −
𝜕휀
𝜕𝛼)
ℎ𝑛 = 1.5298 𝑚
203
5. Calculation of static margin
𝑠𝑡𝑎𝑡𝑖𝑐 𝑚𝑎𝑟𝑔𝑖𝑛 = ℎ𝑛 − ℎ
Therefore,
Empty Weight Centre of Gravity
ℎ = 0.278 𝑚
𝑠𝑡𝑎𝑡𝑖𝑐 𝑚𝑎𝑟𝑔𝑖𝑛 = 1.5298 − 0.278
𝑠𝑡𝑎𝑡𝑖𝑐 𝑚𝑎𝑟𝑔𝑖𝑛 = 1.252 𝑚
Maximum Weight Centre of Gravity
ℎ = 0.283 𝑚
𝑠𝑡𝑎𝑡𝑖𝑐 𝑚𝑎𝑟𝑔𝑖𝑛 = 1.5298 − 0.283
𝑠𝑡𝑎𝑡𝑖𝑐 𝑚𝑎𝑟𝑔𝑖𝑛 = 1.2468 𝑚
Forward Centre of Gravity
ℎ = 0.279 𝑚
𝑠𝑡𝑎𝑡𝑖𝑐 𝑚𝑎𝑟𝑔𝑖𝑛 = 1.5298 − 0.279
𝑠𝑡𝑎𝑡𝑖𝑐 𝑚𝑎𝑟𝑔𝑖𝑛 = 1.2508 𝑚
204
Aft Centre of Gravity
ℎ = 0.283 𝑚
𝑠𝑡𝑎𝑡𝑖𝑐 𝑚𝑎𝑟𝑔𝑖𝑛 = 1.5298 − 0.278
𝑠𝑡𝑎𝑡𝑖𝑐 𝑚𝑎𝑟𝑔𝑖𝑛 = 1.252 𝑚
From the calculation of static margin, the value is positive in all centre of gravity.
Hence, the aircraft is statically stable. We can also conclude that the degree of static
stability is high due to the large value of static margin.
205
9.3 Static Lateral Roll Stability
An airplane possesses static roll stability if a restoring moment is developed
when it is disturbed from a wings-level attitude. The restoring rolling moment can be
shown to be a function of the sideslip angle 𝛽 as illustrated in Figure 9.10. The
requirement for stability is that 𝐶𝑙𝛽 < 0. The roll moment created on an airplane
Figure 9-10 Static Lateral Roll Stability
The major contributor to 𝐶𝑙𝛽 is the wing dihedral angle Γ. The dihedral angle
is defined as the spanwise inclination of the wing with respect to the horizontal. If the
wing tip is higher than the root section, then the dihedral angle is positive; if the wing
tip is lower than the root section, then the dihedral angle is negative. The equation of
𝐶𝑙𝛽 can be written as follows,
𝐶𝑙𝛽= (
𝐶𝑙𝛽
𝛤)𝛤 + ∆𝐶𝑙𝛽
Where,
𝛤: 𝑑𝑖ℎ𝑒𝑑𝑟𝑎𝑙 𝑎𝑛𝑔𝑙𝑒
𝐶𝑙𝛽: 𝑠𝑖𝑑𝑒𝑠𝑙𝑖𝑝 𝑙𝑖𝑓𝑡 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡
As our aircraft’s dimension follows strictly on the model of Cessna 172, we
assume the 𝐶𝑙𝛽value to be the same as the model, which is -0.089. Based on Figure
206
9.10, the static roll stability is achieved since the slope of the graph, which is 𝐶𝑙𝛽value
is negative.
9.3.1 Influence of roll rate on the rolling moment
In this section we will examine how the roll rate creates a rolling moment.
When an airplane rolls about its longitudinal axis, the roll rate creates a linear velocity
distribution over the vertical, horizontal, and wing surfaces. The velocity distribution
causes a local change in angle of attack over each of these surfaces that results in a
change in the lift distribution and, consequently, the moment about the center of
gravity. Figure 9.11 shows a wing planform rolling with a positive rolling velocity. On
the portion of the wing rolling down, an increase in angle of attack is created by the
rolling motion. This results in an increase in the lift distribution over the downward-
moving wing. If we examine the upward-moving part of the wing we observe that the
rolling velocity causes a decrease in the local angle of attack and the lift distribution
decreases. The change in the lift distribution across the wing produces a rolling
moment that opposes the rolling motion and is proportional to the roll rate, p.
The control derivative 𝐶𝑙𝛿𝑎, is a measure of the power of the aileron control; it
represents the change in moment per unit of aileron deflection. The larger 𝐶𝑙𝛿𝑎the more
effective the control is at producing a roll moment. The equation of derivative 𝐶𝑙𝛿𝑎 is
given by,
𝐶𝑙𝛿𝑎= 2
𝐶𝑙∝𝜏
𝑆𝑏𝑤∫ 𝑐𝑦 𝑑𝑦
𝑦2
𝑦1
Wings of large span or high aspect ratio will have larger roll damping than low
aspect ratio wings of small wing span. An estimate of the rolling damping derivative,
𝐶𝑙𝑝, due to the wing surface can be expressed as follows,
𝐶𝑙𝑝 = −𝐶𝑙∝
(1 + 3𝜆)
12(1 + 𝜆)
207
Figure 9-11 Wing planform undergoing a rolling motion.
𝑤𝑖𝑛𝑔𝑠𝑝𝑎𝑛, 𝑏𝑤 = 1.2𝑚
𝑎𝑖𝑙𝑒𝑟𝑜𝑛 𝑐ℎ𝑜𝑟𝑑, 𝑐𝑎 = 0.17𝑚
𝑤𝑖𝑛𝑔 𝑙𝑖𝑓𝑡 𝑐𝑢𝑟𝑣𝑒 𝑠𝑙𝑜𝑝𝑒, 𝐶𝑙∝ = 4.9 𝑟𝑎𝑑−1
𝐷𝑦𝑛𝑎𝑚𝑖𝑐 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒, 𝑞 = 120.05 𝑁/𝑚2
𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝐼𝑛𝑒𝑟𝑡𝑖𝑎, 𝐼𝑥𝑥 = 0.14𝑘𝑔/𝑚2
𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑎𝑖𝑙𝑒𝑟𝑜𝑛, 𝑆𝑎 = 0.0188𝑚2
𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑤𝑖𝑛𝑔, 𝑆𝑤 = 0204𝑚2
𝑇𝑎𝑝𝑒𝑟 𝑟𝑎𝑡𝑖𝑜, 𝜆 = 1
𝐶𝑙𝑝 = −𝐶𝑙∝
(1 + 3𝜆)
12(1 + 𝜆)= −0.8167
𝐿𝑝 =𝑄𝑆𝑏2𝐶𝑙𝑝
2𝐼𝑥𝑢𝑜
𝐿𝑝 =(120.05)(0.204)(1.2)2(−0.8167)
2(0.14)𝑢𝑜=
−102.863
𝑢𝑜
𝑆𝑎
𝑆𝑤=
0.0188
0.204= 0.0922
𝐺𝑖𝑣𝑒𝑛 𝑡ℎ𝑒 𝑔𝑟𝑎𝑝ℎ 𝑖𝑛 𝑓𝑖𝑔𝑢𝑟𝑒 , 𝜏 = 0.2
208
Figure 9-12 Flap effectiveness parameter
𝐶𝑙𝛿𝑎= 2
𝐶𝑙∝𝜏
𝑆𝑏𝑤∫ 𝑐𝑦 𝑑𝑦
𝑦2
𝑦1
𝐶𝑙𝛿𝑎= 2
𝐶𝑙∝𝜏𝑐
𝑆𝑏𝑤[𝑦2
2]𝑦2
𝑦1
𝐶𝑙𝛿𝑎= 2
4.9(0.2)(0.17)
(0.204)(1.2)[𝑦2
2]
0.590.355
𝐶𝑙𝛿𝑎= 0.1511
𝐿𝛿𝑎=
𝑄𝑆𝑏𝐶𝑙𝛿𝑎
𝐼𝑥=
120.05(0.0204)(1.2)(0.1511)
0.14= 3.1718
𝑅𝑜𝑙𝑙 𝑟𝑎𝑡𝑒 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛,
𝑝
𝛿𝑎=
𝐿𝛿𝑎
𝑠 − 𝐿𝑝=
3.1718
𝑠 +102.863
𝑢𝑜
𝑠𝑡𝑒𝑎𝑑𝑦 𝑠𝑡𝑎𝑡𝑒 𝑣𝑎𝑙𝑢𝑒 𝑖𝑠 𝑤ℎ𝑒𝑛 𝑠 → 0, 𝑡ℎ𝑒𝑛
𝑝
𝛿𝑎=
3.1718𝑢𝑜
102.863
Therefore, the value of roll rate is influenced by the airspeed as well as the deflection
angle.
If the deflection angle, 𝛿𝑎 and the airspeed, 𝑢𝑜 are 30o and 14 m/s, respectively, the
transient response of the roll rate can be obtained as follows,
𝑝 =3.1718(14)
102.863(30𝜋
180) = 0.226
209
The Figure 9.13 shows the transient response of roll rate, at which when the
aircraft reaches steady state, the roll rate is levelled off to a value.
Figure 9-13 Transient reponse of roll rate
9.3.2 Influence of deflection angle of aileron on the roll rate
At cruising speed of 14 m/s, the graph of roll rate against the deflection angle
can be plotted based on the equation of roll rate p as follows,
Figure 9-14 Graph of roll rate vs deflection angle
Figure 9-14 shows that the roll rate varies linearly with the deflection angle.
This is because the change in the lift distribution across the wing, controlled by the
0
0.05
0.1
0.15
0.2
0.25
0 5 10 15 20 25 30 35
Ro
ll ra
te (
rad
/s)
Deflection angle (deg)
Graph of roll rate vs deflection angle
210
deflection angle of aileron produces a rolling moment that opposes the rolling motion
and is proportional to the roll rate, p.
9.3.3 Influence of deflection angle of aileron on the rolling moment
Roll control is achieved by the differential deflection of small flaps called
ailerons which are located outboard on the wings, or by the use of spoilers. Figure 9.15
is a sketch showing both types of roll control devices. The basic principle behind these
devices is to modify the spanwise lift distribution so that a moment is created about
the x axis.
Figure 9-15 Ailerons for roll control
The change of lift coefficient has been discussed in wing loading with aileron
section. The variation of the aileron deflection angle will result in change of lift
coefficient, and hence change in the lift force acting on the wing. The torque due to
the change in lift force about the x-axis is therefore calculated as follows,
𝑇 = ∆𝐿 × 𝑥
where,
∆𝐿 = 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑙𝑖𝑓𝑡 𝑓𝑜𝑟𝑐𝑒
𝑥 = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡ℎ𝑒 𝑙𝑖𝑓𝑡 𝑓𝑜𝑟𝑐𝑒𝑠 𝑜𝑓 𝑎𝑖𝑙𝑒𝑟𝑜𝑛
211
Table 9-9 Torque varies with the aileron deflection angle
Deflection angle
(deg)
n Delta Cl Delta L Torque
(Nm)
-30 0.53 -0.10678 -2.61345 -2.46971
-25 0.58 -0.09738 -2.38334 -2.25226
-20 0.69 -0.09268 -2.26828 -2.14353
-15 0.76 -0.07656 -1.8738 -1.77074
-10 0.8 -0.05373 -1.31495 -1.24262
-5 0.8 -0.02686 -0.65747 -0.62131
0 0.8 0 0 0
5 0.8 0.026863 0.657473 0.621312
10 0.8 0.053727 1.314946 1.242624
15 0.76 0.076561 1.873798 1.770739
20 0.69 0.092679 2.268281 2.143526
25 0.58 0.09738 2.383339 2.252255
30 0.53 0.106782 2.613454 2.469714
Figure 9-16 Graph of torque vs deflection angle
-3
-2
-1
0
1
2
3
-40 -30 -20 -10 0 10 20 30 40
Torq
ue
(Nm
)
Deflection angle (Deg)
Graph of Torque vs Deflection Angle
212
Figure 9-16 shows that the graph varies linearly at small angle but eventually
levels off at high angle of deflection. This is because the flap effectiveness factor 𝜏 of
the aileron limits the continuous rising of the torque.
9.3.4 Influence of airspeed on rolling moment
Table 9-10 Torque varies with the airspeed
V (m/s) q (N/m2) Cl Delta Cl Torque (Nm)
4 9.8 7.365138 0.981062 0.927104
6 22.05 3.273395 0.436028 0.412046
8 39.2 1.841284 0.245266 0.231776
10 61.25 1.178422 0.15697 0.148337
11.42 79.88005 0.903584 0.120361 0.113741
14 120.05 0.601236 0.080087 0.075682
16 156.8 0.460321 0.061316 0.057944
18 198.45 0.363711 0.048448 0.045783
20 245 0.294606 0.039242 0.037084
22 296.45 0.243476 0.032432 0.030648
24 352.8 0.204587 0.027252 0.025753
26 414.05 0.174323 0.02322 0.021943
28 480.2 0.150309 0.020022 0.01892
30 551.25 0.130936 0.017441 0.016482
32 627.2 0.11508 0.015329 0.014486
34 708.05 0.10194 0.013579 0.012832
36 793.8 0.090928 0.012112 0.011446
38 884.45 0.081608 0.01087 0.010273
213
Figure 9-17 Graph of torque vs airspeed
At 30o aileron deflection angle, the torque varies with the velocity as shown in
Figure 9.17. Since the value of lift coefficient varies with the dynamic pressure, which
depends on the velocity, the torque that varies with the change in lift coefficient varies
too. The graph in Figure 9.17 shows that the torque decreases drastically when the
speed approaches the stalling speed. When the airspeed increases further, the torque
will level off and approach to zero.
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 5 10 15 20 25 30 35 40
Torq
ue
(Nm
)
Airspeed (m/s)
Graph of Torque vs Airspeed
214
9.4 Static Lateral Directional Stability
Directional stability is related with the static stability of the airplane about the z axis,
which is the tendency of the airplane return to an equilibrium condition when subjected
to yawing disturbance. In order to achieve positive static directional stability, the slope
of the yawing moment curve must be positive (𝐶𝑛𝛽> 0), as shown in Figure 9.18.
Figure 9-18 Static directional stability
The fuselage and engine nacelles contribute more to destabilize to directional
stability, compared to wing. The wing fuselage contribution can be calculated from the
following empirical expression as below:
𝐶𝑛𝛽𝑤𝑓= −𝑘𝑛𝑘𝑅𝑙
𝑆𝑓𝑠𝑙𝑓
𝑆𝑤𝑏 𝑝𝑒𝑟 𝑑𝑒𝑔
Where
𝑘𝑛= empirical wing-body interference factor. (Determine from Figure 9.19)
𝑘𝑅𝑙 = empirical correction factor (Determine from Figure 9.20)
𝑆𝑓𝑠= the projected side area of the fuselage
𝑙𝑓 = the length of the fuselage
In order to complete the get the data, the following data is obtained beforehand.
215
√ℎ1
ℎ2= √
0.12
0.05= 1.549
𝑙𝑓2
𝑆𝑓𝑠=
0.732
0.055= 9.689
𝑥𝑚
𝑙𝑓=
0.153
0.73= 0.2096
𝑅𝑙𝑓 =𝑉𝑙𝑓
𝛾=
9.314 × 0.73
1.46 × 10−5= 4.657 × 105
ℎ
𝑤𝑓= 1
Figure 9-19: Wing Body Interference Factor
216
Figure 9-20: Reynolds number correction factor
Thus from Figure 9.19 and figure 9.20, we obtained that the value of 𝑘𝑛 and
𝑘𝑅𝑙 are 0.007 and 1 respectively. Put the value obtained back to the equation, we get:
𝐶𝑛𝛽𝑤𝑓= −0.007(1)
0.055(0.73)
(0.204)(1.2)= −1.148 × 10−4
Since the wing-fuselage contribution to directional stability is destabilizing, the
vertical tail must be properly sized to ensure that the airplane has directional stability.
The contribution of the vertical tail to directional stability now can be obtained
𝐶𝑛𝛽𝑡= 𝑉𝜈휂𝜈𝐶𝐿𝛼𝑡
(1 +𝑑𝜎
𝑑𝛽)
Where 𝑉𝜈 is the vertical tail volume ratio and 휂𝜈 = 𝑄𝜈/𝑄𝑤 is the ratio of the
dynamic pressure at the vertical tail to the dynamic pressure at the wing. Also, a simple
algebraic equation for estimating the combined sidewash and tail efficiency factor 휂𝜈
is presented in and reproduced here:
휂𝜈 (1 +𝑑𝜎
𝑑𝛽) = 0.724 + 3.06
𝑆𝜈/𝑆
1 + cos Λ𝑐/4𝑤+ 0.4
𝑧𝑤
𝑑+ 0.009𝐴𝑅𝑤
= 0.724 + 3.06130.66 × 10−4/1.2
1 + cos 0+ 0.4
0.065
0.12+ 0.009(7.0588)
= 1.02085
217
𝑉𝜈 =𝑙𝑣𝑆𝑣
𝑆𝑏=
0.08(130.66 × 10−4)
(0.204)(1.2)= 4.27 × 10−3
𝐶𝑛𝛽𝑡= 4.27 × 10−3(0.9198)(1.02085) = 4.0094 × 10−3
Therefore, summing up both contribution of wing fuselage and vertical tail
𝐶𝑛𝛽= 𝐶𝑛𝛽𝑤𝑓
+ 𝐶𝑛𝛽𝑡
𝐶𝑛𝛽= −1.148 × 10−4 + 4.0094 × 10−3 = 3.89 × 10−3
As the 𝐶𝑛𝛽> 0, thus we can conclude that the airplane is directionally stable
with a positive static directional stability.
9.4.1 Pure Yawing Motion
The aircraft data are given as follows,
Moment of inertia, 𝐼𝑧𝑧 : 0.07 kg𝑚2
Vertical tail side force slope, 𝐶𝑦𝛽(𝑉𝑇)∶ 0.21
Fuselage diameter, X : 0.115m
Cruising speed, 𝑈𝑜 : 14 m/s
Body side force slope, 𝐶𝑦𝛽(𝐵𝑜𝑑𝑦): 0.31
Area of vertical tail (side view), 𝑆𝑉𝑇 : 0.0139 𝑚2
Area of fuselage (side view), 𝑆𝐵𝑜𝑑𝑦 : 0.0556 𝑚2
Distance between the cg and the cg of the vertical tail, 𝐿𝑉𝑇 : 0.499m
**Remark: Body = Fuselage
Estimation 𝑁𝛽 and 𝑁𝑟
Body Contribution, 𝑁𝛽𝐵
𝑁𝛽𝐵𝑜𝑑𝑦= −𝑌𝑏𝑜𝑑𝑦𝑋 = −𝐶𝑦𝛽(𝐵𝑜𝑑𝑦)𝛽𝑄𝑆𝐵𝑜𝑑𝑦𝑋
(𝜕𝑁
𝜕𝛽)
𝐵𝑜𝑑𝑦
= −𝐶𝑦𝛽(𝐵𝑜𝑑𝑦)𝑄𝑆𝐵𝑜𝑑𝑦𝑋
218
𝑁𝛽𝐵𝑜𝑑𝑦 = (𝜕𝑁
𝜕𝛽/𝐼𝑧𝑧 )
𝐵𝑜𝑑𝑦= −𝐶𝑦𝛽(𝐵𝑜𝑑𝑦)𝑄𝑆𝐵𝑜𝑑𝑦𝑋/𝐼𝑧𝑧
𝑁𝛽𝐵𝑜𝑑𝑦 = -
(0.31)(1
2(1.225)(14)2)(0.0556)(0.115)
0.07= −3.399
Horizontal Tail Contribution, 𝑁𝛽𝐻𝑇
𝑁𝛽𝑉𝑇= −𝑌𝑉𝑇𝐿𝑉𝑇 = −𝐶𝑦𝛽(𝑉𝑇)
𝛽𝑄𝑆𝑉𝑇𝐿𝑉𝑇
(𝜕𝑁
𝜕𝛽)𝑉𝑇
= −𝐶𝑦𝛽(𝑉𝑇)𝑄𝑆𝑉𝑇𝐿𝑉𝑇
𝑁𝛽𝑉𝑇= (
𝜕𝑁
𝜕𝛽/𝐼𝑧𝑧 )
𝑉𝑇= −𝐶𝑦𝛽(𝑉𝑇)
𝑄𝑆𝑉𝑇𝐿𝑉𝑇/𝐼𝑧𝑧
𝑁𝛽𝐵 = -
(0.21)(1
2(1.225)(14)2)(0.0139)(0.449)
0.07= -2.2477
Body Contribution, 𝑁𝑟(𝐵𝑜𝑑𝑦)
𝛽 = 𝑟|𝑋|/𝑈𝑜
𝑁𝑟(𝐵𝑜𝑑𝑦) = −𝑟|𝑋|
𝑈𝑜𝐶𝑦𝛽(𝐵𝑜𝑑𝑦) 𝑄𝑆𝐵𝑜𝑑𝑦𝑋
(𝜕𝑁
𝜕𝑟)
𝐵𝑜𝑑𝑦= −𝐶𝑦𝛽(𝐵𝑜𝑑𝑦)
|𝑋|
𝑈𝑜 𝑄𝑆𝐵𝑜𝑑𝑦𝑋
𝑁𝑟𝐵𝑜𝑑𝑦= (
𝜕𝑁
𝜕𝑟/𝐼𝑧𝑧 )
𝐵𝑜𝑑𝑦= −𝐶𝑦𝛽(𝐵𝑜𝑑𝑦)
|𝑋|
𝑈𝑜 𝑄𝑆𝐵𝑜𝑑𝑦𝑋 /𝐼𝑧𝑧
𝑁𝑟𝐵𝑜𝑑𝑦 = -
(−0.31)(0.115
14)(
1
2(1.225)(14)2)(0.0556)(0.115)
0.07= 0.0279
Horizontal Tail Contribution, 𝑁𝑟𝑉𝑇
𝛽 = 𝑟|𝐿𝑉𝑇|/𝑈𝑜
𝑁𝑟𝐻𝑇= −𝐶𝑦𝛽(𝑉𝑇)
𝑟|𝐿𝑉𝑇|
𝑈𝑜 𝑄𝑆𝑉𝑇𝐿𝑉𝑇
(𝜕𝑁
𝜕𝑟)𝑉𝑇
= −𝐶𝑦𝛽(𝑉𝑇)
|𝐿𝑉𝑇|
𝑈𝑜 𝑄𝑆𝑉𝑇𝐿𝑉𝑇
𝑁𝑟𝑉𝑇= (
𝜕𝑁
𝜕𝑟/𝐼𝑧𝑧 )
𝑉𝑇= −𝐶𝑦𝛽(𝑉𝑇)
|𝐿𝑉𝑇|
𝑈𝑜 𝑄𝑆𝑉𝑇𝐿𝑉𝑇 /𝐼𝑧𝑧
𝑁𝑟𝐻𝑇 = -
(0.21)(0.499
14)(
1
2(1.225)(14)2)(0.0139)(0.499)
0.07= −0.089
Characteristic equation
219
𝑠2 − (𝑁𝑟𝐵𝑜𝑑𝑦+ 𝑁𝑟𝑉𝑇
) 𝑠 − (𝑁𝛽𝐵𝑜𝑑𝑦+ 𝑁𝛽𝑉𝑇
) = 0
𝑠2 + 0.0611𝑠 + 5.6467 = 0
Roots 𝑠 = −0.0306 ± 2.3761𝑖
Natural frequency, 𝜔𝑛 = √5.6467 = 2.3763 rad/s
Damping ratio, 휁 = 0.0611
2𝜔𝑛= 0.013
Based on the result of pure yawing motion, the aircraft has good directional stability
𝑁𝛽, but has a very low yaw damping 𝑁𝑟. This makes its behavior close to neutral stable.
220
9.5 Dynamic Longitudinal Stability System
Table 9-11: Longitudinal dimensionless derivatives (Nelson, 1998)
221
Table 9-12: Longitudinal dimensional derivatives (Nelson, 1998)
The values of dimensionless and dimensional derivatives are presented in
Table 9-13 and Table 9-14 as below:
Table 9-13: Longitudinal stability coefficients
X-force derivatives Z-force derivatives Pitching moment
derivatives
𝐶𝑋𝑢 = −0.062 𝐶𝑍𝑢 = −0.440 𝐶𝑀𝑢 = 0
𝐶𝑋𝛼 = −0.168 𝐶𝑍𝛼 = −6.508 𝐶𝑀𝛼 = −2.212
𝐶𝑋�̇� = 0 𝐶𝑍�̇� = −2.650 𝐶𝑀�̇� = −7.426
𝐶𝑋𝑞 = 0 𝐶𝑍𝑞 = −5.885 𝐶𝑀𝑞 = −16.490
𝐶𝑋𝛼𝑒= 0 𝐶𝑍𝛿𝑒
= −0.315 𝐶𝑀𝛿𝑒= −0.883
222
Table 9-14: Longitudinal dimensional derivatives
𝑋𝑢 = −0.073 𝑍𝑢 = −0.525 𝑀𝑢 = 0
𝑋𝛼 = −2.814 𝑍𝛼 = −108.65 𝑀𝛼 = −76.744
𝑋𝑤 = −0.211 𝑍�̇� = −0.268 𝑀�̇� = −0.111
− 𝑍𝑞 = −0.596 𝑀𝑞 = −3.473
− 𝑍𝛿𝑒= 5.260 𝑀𝛿𝑒
= −30.630
− 𝑍𝑤 = −7.723 𝑀𝑤 = −5.481
− − 𝑀�̇� = −0.111
By taking a flight case for our RC plane flying at sea level with the trimmed
velocity at 14 m/s, the state-space matrix for longitudinal motion due to elevator
deflection are presented as below:
[
𝛥�̇�𝛥�̇�𝛥�̇�
𝛥휃̇
] = [
−0.0739 −0.2111 0 −9.8100−0.5251 −7.7234 14.0000 00.0587 −4.6187 −5.0380 0
0 0 1 0
] [
𝛥𝑢𝛥𝑤𝛥𝑞𝛥휃
] + [
05.2605
−31.21830
] [𝛥𝛿𝑒]
The state-space matrix can be written in the form of transfer function:
𝛥𝑢
𝛥𝛿𝑒=
−1.11𝑠2 + 392.9𝑠 + 2604
𝑠4 + 12.84𝑠3 + 104.4𝑠2 + 7.848𝑠 + 28.24
𝛥𝑤
𝛥𝛿𝑒=
5.26𝑠3 − 410.2𝑠2 − 30.35𝑠 − 157.8
𝑠4 + 12.84𝑠3 + 104.4𝑠2 + 7.848𝑠 + 28.24
𝛥𝑞
𝛥𝛿𝑒=
−31.22𝑠3 − 267.7𝑠2 − 16.23𝑠
𝑠4 + 12.84𝑠3 + 104.4𝑠2 + 7.848𝑠 + 28.24
223
𝛥휃
𝛥𝛿𝑒=
−31.22𝑠2 − 267.7𝑠 − 16.23
𝑠4 + 12.84𝑠3 + 104.4𝑠2 + 7.848𝑠 + 28.24
Where the characteristic equation is:
𝑠4 + 12.84𝑠3 + 104.4𝑠2 + 7.848𝑠 + 28.24 = 0
From the characteristic equation, all the sub-value of representing the phugoid and
short period is positive in value, means it still in a stable condition. The time response
of the open loop transfer function of longitudinal motion was plotted by using
MATLAB in time interval between 0 to 50s.
Figure 9-21: Time response open-loop for longitudinal stability
As observed, the forward velocity in phugoid mode, 𝑢 is having a large damping ratio
due to long period oscillation while w and q are having a lower damping ratio due to
short period.
224
State-Space Matrix and transfer function for Short Period Motion
The state-space matrix then reduced to short period motion, the matrix and
transfer functions are shown below:
[𝛥�̇�𝛥�̇�
] = [−7.7604 1−75.8770 −3.5854
] [𝛥𝛼𝛥𝑞
] + [0.3757
−30.6725] [𝛥𝛿𝑒]
𝛥𝛼
𝛥𝛿𝑒=
0.3757𝑠 − 29.33
𝑠2 + 11.35𝑠 + 103.7
𝛥𝑞
𝛥𝛿𝑒=
−30.67𝑠 − 266.5
𝑠2 + 11.35𝑠 + 103.7
The time response of the open loop transfer function for short period are as shown:
Figure 9-22: Time response for short period
225
Pole Damping Frequency
(rad/TimeUnit)
Time Constant
(TimeUnit)
-5.67e+00 +
8.46e+00i
0.557 10.2 0.176
-5.67e+00 -
8.46e+00i
0.557 10.2 0.176
From the open loop transfer function of short period motion, it shown that the system
is having a low stability condition where the damping ratio is 0.557 and natural
frequency of 10.2.
226
9.6 Dynamic Lateral Stability System
Table 9-15 Parameters of Cessna 172
Parameters Values Side Force Coefficient
Derivatives
Density, 𝜌 2.3769 ×
10−3𝑠𝑙𝑢𝑔
𝑓𝑡3
1.225 𝑘𝑔/𝑚3
𝐶𝑦𝛽 -0.31
Mach number, M 0.04 𝐶𝑦𝛿𝑟 0.187
Speed of sound, a 1125.33 ft/s @ 343 m/s 𝐶𝑦𝑝 -0.037
Reference flight
speed, 𝑈𝑜
45.9318 ft/s @ 14 m/s 𝐶𝑦𝑟 0.21
Wing area, S 2.1958 𝑓𝑡2 @ 0.204
𝑚2
𝐶𝑦𝛿𝑎 0
Mean chord, 𝑐̅ 2.2966 ft @ 0.17 m Yawing Moment Coefficient
Derivatives
Wing span, b 3.9370 ft @ 1.2m 𝐶𝑛𝛽 0.065
Aspect ratio, AR 7.06 𝐶𝑛𝛿𝑟 -0.0657
Rolling moment of
inertia, 𝐼𝑥
0.14 kg𝑚2 @ 0.1032
𝑠𝑙𝑢𝑔𝑓𝑡2
𝐶𝑛𝑝 -0.03
Pitching moment
of inertia, 𝐼𝑦
0.12 kg𝑚2 @
0.0885 𝑠𝑙𝑢𝑔𝑓𝑡2
𝐶𝑛𝑟 -0.099
Yawing moment of
inertia, 𝐼𝑧
0.07 kg𝑚2 @
0.0516 𝑠𝑙𝑢𝑔𝑓𝑡2
𝐶𝑛𝛿𝑎 -0.053
Product of inertia
about xz, 𝐼𝑥𝑧
0.04 kg𝑚2 @
0.0295 𝑠𝑙𝑢𝑔𝑓𝑡2
Rolling Moment Coefficient
Derivatives
227
Product of inertia
about xy, 𝐼𝑥𝑦
0.03 kg𝑚2 @
0.0221 𝑠𝑙𝑢𝑔𝑓𝑡2
𝐶𝑙𝛽 -0.089
Gravitational
acceleration, g
32. 171 𝑓𝑡2/𝑠 @ 9.81
𝑚2/𝑠
𝐶𝑙𝛿𝑟 0.0147
Weight, W 104.0474 Ibf @ 14.39
N
𝐶𝑙𝑝 -0.47
Mass, m 3.2342 𝐼𝑏 @ 1.467 kg 𝐶𝑙𝑟 0.096
Dynamic pressure,
Q
2.5073 psi @
17287.225 Pascal
𝐶𝑙𝛿𝑎 -0.178
Oswald efficiency
factor, e
0.7
228
Lateral Directional Derivatives
𝑌𝛽 -0.1582
𝑁𝛽 0.7732
𝐿𝛽 -0.5293
𝑌𝑝 −3.8632 × 10−4
𝑁𝑝 -0.0047
𝐿𝑝 -0.0365
𝑌𝑟 0.0272
𝑁𝑟 -0.0154
𝐿𝑟 0.0075
𝑌𝛿𝑎 0
𝑌𝛿𝑟 0.0296
𝑁𝛿𝑎 -0.6304
𝑁𝛿𝑟 -0.7815
𝐿𝛿𝑎 -1.0587
𝐿𝛿𝑟 0.0874
For lateral state-space equation,
[
∆�̇�∆�̇�∆�̇�∆∅̇
] =
[ 𝑌𝛽
𝑢𝑜
𝑌𝑝
𝑢𝑜−(1 −
𝑌𝑟
𝑢𝑜)
𝐿𝛽 𝐿𝑝 𝐿𝑟
𝑁𝛽
0
𝑁𝑝
1
𝑁𝑟
0
𝑔
𝑢𝑜cos 휃0
000 ]
[
∆𝛽∆𝑝∆𝑟∆∅
] +
[ 0
𝑌𝛿𝑟
𝑢𝑜
𝐿𝛿𝑎𝐿𝛿𝑟
𝑁𝛿𝑎
0
𝑁𝛿𝑎
0 ]
[∆𝛿𝑎
∆𝛿𝑟]
229
From the state-space equation, by using MATLAB program, the transfer function of
the lateral motion is obtained. The stability of the aircraft is able to be determined from
the transfer function.
�̇� = 𝐴𝑥 + 𝐵𝑢
𝑨 = [
−0.0034 0 −0.9994−0.5293 −0.0365 −0.00750.7732
0−0.00471.000
−0.01540
0.7004000
]
𝐵 = [
0 0.0006−1.0587 0.0874−0.6304
0−0.7815
0
]
[
∆�̇�∆�̇�∆�̇�∆∅̇
] =[
−0.0034 0 −0.9994−0.5293 −0.0365 −0.00750.7732
0−0.00471.000
−0.01540
0.7004000
] [
∆𝛽∆𝑝∆𝑟∆∅
]
+[
0 0.0006−1.0587 0.0874−0.6304
0−0.7815
0
] [∆𝛿𝛼
∆𝛿𝑟]
Transfer Function of Lateral Motion
Aileron Input
𝛽
𝛿𝛼
=0.63𝑠2 − 0.7235𝑠 − 0.0081
𝑠4 + 0.0553𝑠3 + 0.7734𝑠2 + 0.4014𝑠 + 0.0098
𝑝
𝛿𝛼
=−1.0587𝑠3 − 0.0152𝑠2 − 1.1516𝑠
𝑠4 + 0.0553𝑠3 + 0.7734𝑠2 + 0.4014𝑠 + 0.0098
𝑟
𝛿𝛼
=−0.6304𝑠3 − 0.0202𝑠2 − 0.0001𝑠 − 0.8070
𝑠4 + 0.0553𝑠3 + 0.7734𝑠2 + 0.4014𝑠 + 0.0098
230
∅
𝛿𝛼
=−1.0587𝑠2 − 0.0152𝑠 − 1.1516
𝑠4 + 0.0553𝑠3 + 0.7734𝑠2 + 0.4014𝑠 + 0.0098
Rudder Input
𝛽
𝛿𝑟
=0.006𝑠3 + 0.7811𝑠2 + 0.0901𝑠 + 0.0050
𝑠4 + 0.0553𝑠3 + 0.7734𝑠2 + 0.4014𝑠 + 0.0098
𝑝
𝛿𝑟
=0.0874𝑠3 + 0.0072𝑠2 − 0.3458𝑠
𝑠4 + 0.0553𝑠3 + 0.7734𝑠2 + 0.4014𝑠 + 0.0098
𝑟
𝛿𝑟
=−0.7815𝑠3 − 0.0311𝑠2 − 0.0001𝑠 − 0.2424
𝑠4 + 0.0553𝑠3 + 0.7734𝑠2 + 0.4014𝑠 + 0.0098
∅
𝛿𝑟
=0.0874𝑠2 + 0.0072𝑠 − 0.3458
𝑠4 + 0.0553𝑠3 + 0.7734𝑠2 + 0.4014𝑠 + 0.0098
Flying Qualities
Cessna 172 is classified as Class I airplanes as it is small, light
airplanes, such as light utility, primary trainer and light observation craft. This
airplane is categorized as non-terminal flight under category C which is a
gradual maneuver, accurate flight path.
Lateral Motion
𝜆1,2(𝐷𝑢𝑡𝑐ℎ 𝑟𝑜𝑙𝑙) = −0.191 ± 0.945𝑖
𝜆1,2(𝑆𝑝𝑟𝑖𝑟𝑎𝑙) = −0.0256
𝜆1,2(𝑅𝑜𝑙𝑙) = −0.411
231
Table 9-16 Flying Qualities
Modes Damping
( ξ )
Natural
Frequency
(𝜔𝑛)
Time Constant
(s)
Flying
Qualities
Dutch Roll 0.198 0.945 5.25 Level 1
Spiral - 0.0256 39.1 Level 1
Roll - 0.411 2.43 Level 2
Based on Table 9-16, both spiral and dutch mode has the same flying
qualities, which is Level 1 where the flying qualities clearly adequate for the
mission flight phase. For roll mode, it is a Level 2 flying qualities where flying
qualities adequate to accomplish the mission flight phase but with some
increase in pilot workload and degradation in mission effectiveness.
232
CHAPTER 10
FLIGHT TESTING PLANNING AND PREPARATION
10.1 Flight test and preparation
1. Events and activities during the day of flight test.
Flight test of RC plane
• Understand the nature
• Calibrate electronics part and sensors
• Check the control surfaces of RC plane
• Set the location for pilot to stand
• Record the meteorological conditions
• Record data of RC plane and pilot
• Record the flight path and flight time of RC plane
2. Approximation of the nature and environment of the testing field
Area Padang Ragbi UTM
Obstacles Goal post, trees, net
Crowd Around 50 people
Ambient Temperature 32°C
Atmospheric Pressure 1 atm
Wind Speed 3.08 m/s = 11.1 km/hr
3. Battery consumption of each cell.
1 𝑐𝑒𝑙𝑙 = 35𝑐
233
4. Flight path at its altitude and the location of pilot to stand.
Flight Path :
Condition Altitude (metre)
Takeoff 0-15
Cruising 15-20
Landing 0
Location of pilot:
• Pilot should stand 3 metre when the RC plane on ground before take-
off
• RC plane should not fly more than 92 metre from pilot since the
frequency is 2.4GHz
5. Strategies when the engine suddenly cut-off
Maneuverability Control
Glide angle 1.25°
Glide speed Stall speed – 9.52 m/s
Movement of control
surface
Adjust manually by pilot (aileron, rudder,
elevator)
0
5
10
15
20
25
0 100 200 300 400 500
Alt
itu
de
(met
re)
Time(sec)
Flight Path
234
10.2 Checklist
1. Electronics and sensors calibration checklist including its condition during
the day of testing
Battery capacity
Motor
Propeller
• Clockwise / anticlockwise
Servo
• Aileron
• Rudder
• Elevator
Push rod
• Aileron
• Rudder
• Elevator
2. Control surfaces checklist (accurate angles, movements and directions)
Accurate angle 0° angle
Movements and directions
Parts Movements
Elevator (left and
right)
Upward and downward
Aileron (left and
right)
Upward and downward
Rudder Left and right
235
3. Record of meteorological conditions onsite.
Area
Obstacles
Crowd
Ambient Temperature
Atmospheric Pressure
Wind Speed
Wind Direction
4. Testing form
Trial number
Type of aircraft
Pilots’s name
Location
Date
Time
Purpose of test
Requirement to be met 5 minutes flight time
Ambient airspeed
Ambient temperature
Wind direction
MTOW
Flight path
Flight time
Operating altitude
Battery voltage in each
cell
Battery voltage of
transmitter
Comments
Name of reporter
Signature
236
REFERENCES
Newcome, L.R. (2004). Unmanned aviation: a brief history of unmanned aerial
vehicles. Reston, VA, USA: American Institute of Aeronautics and
Astronautics, Inc.
RC Airplanes Victor (2017, August 13) .How to choose motor and propeller. version
2 [Video file]. Retrieved from:
https://www.youtube.com/watch?v=ug4zKgNV6zY
SunnySky X2216 Brushless Motors. (n.d.). Retrieved from
https://sunnyskyusa.com/products/sunnysky-x2216-brushless-
motors?variant=45677872079
Starlino. (2017, September 18). How much power is needed to hover ? Retrieved from
http://www.starlino.com/power2thrust.html
Lance W. Traub (2011) ‘Range and Edurance Estimates for Battery-Powered Aircraft’,
Journal Of Aircraft, Vol. 48, No 2, pp. 703-707.
All About Multirotor Drone Radio Transmitters and Receivers. (2018, February 8).
Retrieved from Getfpv: https://www.getfpv.com/learn/new-to-fpv/all-about-
multirotor-fpv-drone-radio-transmitter-and-receiver/
Differences Between NiMH and LiPo Batteries. (n.d.). Retrieved from Euro RC:
https://www.eurorc.com/page/69/differences-between-nimh-and-lipo-
batteries
Electronic speed control. (2020, March 4). Retrieved from Wikipedia:
https://en.wikipedia.org/wiki/Electronic_speed_control
FS-i6 Instruction Manual. (2015). Retrieved from Fly SKy:
https://static1.squarespace.com/static/5bc852d6b9144934c40d499c/t/5c0787e
10e2e721a7f17c998/1543997593953/FS-i6+User+manual+20160819.pdf
Introduction to Servos. (2020). Retrieved from UAV Navigation:
https://www.uavnavigation.com/support/kb/general/general-system-
info/introduction-servos
Liang, O. (2013, November 5). PWM AND PPM DIFFERENCE AND
CONVERSION. Retrieved from OscarLiang: https://oscarliang.com/pwm-
ppm-difference-conversion/
237
Liang, O. (2019, March 12). FLYSKY TRANSMITTER & RECEIVER BUYER’S
GUIDE. Retrieved from Oscarliang: https://oscarliang.com/flysky-tx-rx-
buyers-guide/
Monti, S. (2010, October 20). Is Lithium-ion the Ideal Battery? Retrieved from Battery
University:
https://batteryuniversity.com/learn/archive/is_lithium_ion_the_ideal_battery
mpnxt1. (2019, December 20). Omnidirectional Antenna Radiation Patterns
Explained. Retrieved from MP Antenna:
https://www.mpantenna.com/omnidirectional-antenna-radiation-patterns/
Perlman, A. (2016, September 14). How Do Drones Work? Retrieved from
UAVCoach: https://uavcoach.com/infographic-drones-work/
Salt, J. (2019, June). Understanding RC Antenna Operation & Placement. Retrieved
from RC helicopter fun: https://www.rchelicopterfun.com/rc-antenna.html
Salt, J. (2020, March). 11 Things to Know About LiPo Batteries to Get the Best
Performance, Life, Value & Fun Out of Them, Whatever You Fly. Retrieved
from RC Helicopter Fun: https://www.rchelicopterfun.com/lipo-batteries.html
Servo Motor SG-90. (2017, September 18). Retrieved from Component 101:
https://components101.com/servo-motor-basics-pinout-datasheet
SunnySky X2216 Brushless Motors. (n.d.). Retrieved from SunnySky USA:
https://sunnyskyusa.com/products/sunnysky-x2216-brushless-motors
McCormick, B. W. (1995). Aerodynamics, aeronautics and flight mechanics. New
York: Wiley.
Abbott, I. H., E., V. D. A., & Stivers, L. S. (1945). Summary of airfoil data.
Washington, D.C.: National Advisory Committee for Aeronautics.
Anderson, J. D. (2007). Fundamentals of aerodynamics. London: Mcgraw-hill
Publishing Co.
Bruhn, E. F., & Bollard, R. J. H. (1973). Analysis and design of flight vehicle
structures. Carmel: Jacobs.
Federal Aviation Regulations. (n.d.). Retrieved from
https://www.risingup.com/fars/info/23-index.shtml
D. Raymer, Aircraft Design: A Conceptual Approach. American Institute of
Aeronautics and Astronautics, Inc., 2012.
238
Appendix I Graph to determine aerodynamic coefficients
Figure 1: Graph of Cm and Cl vs angle of attack (left), Graph of Cd and Cm vs Cl (right) (Abbott 1994)
239
Appendix II McCormick’s method on wing aileron analysis with aileron
Figure 1: Flap effectiveness factor (McCormick, 1995)
Figure 2: Correction factor to flap effectiveness factor 𝝉 (McCormick, 1995)
240
Figure 3: 𝑪𝒍𝒎𝒂𝒙increment ratio as a function of flap chord ratio (McCormick, 1995)
241
Appendix III Bruhn’s method on structural analysis
Figure 1: Compressive-buckling coefficients, kc for flat rectangular plates (Bruhn,
1973)
Figure 2: Shear-Buckling-Stress Coefficient ks of Plates as a Function of /b for
Clamped and Hinged Edges (Bruhn, 1973)
242
Figure 3: Bending-Buckling Coefficient kb of Plates as a Function of a/b for Various
Amounts of Edge Rotational Restraint. (Bruhn, 1973)
Figure 4: Chart of Nondimensional Compressive Buckling Stress for Long Clamped
Flanges and for Supported Plates with Edge Rotational Restraint. (Bruhn, 1973)
243
Figure 5: Chart of Nondimensional Shear Buckling Stress for Panels with Edge-
Rotational Restraint (Bruhn, 1973)
244
Appendix IV EPP foam material properties
Figure 1: Static and dynamic loading of 80grams/litre EPP Foam
Figure 2: EPP Foam properties at different densities
245
Appendix V MATLAB coding on dynamic stability and control
%%% Cessna 172 rho = 2.3769e-3; %slug/ft^3 M = 0.6; a = 1125.33; %ft/s Vo = 45.9318 ; %ft/s uo = Vo; %ft/s S = 0.204; %ft^2 c = 0.17; %ft b = 1.2; % ft AR = 7.06; Ix = 0.1032; %Slug.ft^2 Iy = 0.0885; %Slug.ft^2 Iz = 0.0516; %Slug.ft^2 Ixz = 0.0295; %Slug.ft^2 Ixy = 0.0221; Iyz = 0.0516; %Slug.ft^2 g = 32.17095; m = 3.2342; %Ibf Weight = m*g %Ib Q = (1/2)*rho*(Vo^2) e = 0.7;
%Lateral Data %Side Force Coefficient Derivatives CYbeta = -1.0; CYdeltar = 0.187; %due to rudder CYp = -0.187; CYr = 0.21; CYdeltaa = 0; %Yawing Moment Coefficient Derivatives CNbeta = 0.065; CNp = -0.03; CNr = -0.099; CNdeltaa = -0.053; %due to aileron CNdeltar = -0.0657; %due to rudder %Rolling Moment Coefficient Derivatives CLbeta = -0.089; CLp = -0.47; CLr = 0.096; CLdeltaa = -0.178; %due to aileron CLdeltar = 0.0147; %due to rudder %Matrix Lateral Derivatives Ybeta = (Q*S*CYbeta)/m Nbeta = (Q*S*b*CNbeta)/Iz Lbeta = (Q*S*b*CLbeta)/Ix Yp = (Q*S*b*CYp)/(2*m*uo) Np = (Q*S*(b^2)*CNp)/(2*Iz*uo) Lp = (Q*S*(b^2)*CLp)/(2*Ix*uo) Yr = (Q*S*b*CYr)/(2*Iz*uo) Nr = (Q*S*(b^2)*CNr)/(2*Iz*uo) Lr = (Q*S*(b^2)*CLr)/(2*Ix*uo) Ydeltaa = (Q*S*CYdeltaa)/m Ydeltar = (Q*S*CYdeltar)/m Ndeltaa = (Q*S*b*CNdeltaa)/Iz Ndeltar = (Q*S*b*CNdeltar)/Iz Ldeltaa = (Q*S*b*CLdeltaa)/Ix Ldeltar = (Q*S*b*CLdeltar)/Ix thetanot = 0;
246
%Matrix A Lateral Alat = [Ybeta/uo, Yp/uo, -(1-(Yr/uo)),(g*cos(thetanot))/uo; Lbeta, Lp, Lr, 0; Nbeta, Np, Nr, 0; 0, 1, 0, 0]
%Matrix B Lateral Blat = [0, Ydeltar/uo; Ldeltaa, Ldeltar; Ndeltaa, Ndeltar; 0,0]
%% %Aileron Alat_a = [-0.0034 0 -0.9994 0.7004;-0.5293 -0.0365 -0.0075 0;0.7732
-0.0047 -0.0154 0;0 1.00 0 0]; Blat_a = [0;-1.0587;-0.6304;0]; Clat_a = eye(size(Alat_a)); Dlat_a = zeros(size(Blat_a)); [numlat_a, denlat_a]=ss2tf(Alat_a,Blat_a,Clat_a,Dlat_a,1); numlat_a denlat_a
%% %Rudder Alat_r = [-0.0034 0 -0.9994 0.7004;-0.5293 -0.0365 -0.0075 0;0.7732
-0.0047 -0.0154 0;0 1.00 0 0]; Blat_r = [0.0006;0.0874;-0.7815;0]; Clat_r = eye(size(Alat_r)); Dlat_r = zeros(size(Blat_r)); [numlat_r,denlat_r]=ss2tf(Alat_r,Blat_r,Clat_r,Dlat_r,1); numlat_r denlat_r damp(Alat_r)
247
Appendix VI Minutes of meetings
FACULTY OF MECHANICAL ENGINEERING
UNIVERSITI TEKNOLOGI MALAYSIA
12 FEBRUARY 2020 MINUTES OF MEETING NO. 1/2020 AIRCRAFT DESIGN II (SKMA 4523)
Date: 12 February, 2020
Time: 9am-12pm Venue: P20-215
Attendance AHMAD AMIRUL AMIN BIN ABDUL WAHAB (A16KM0017) AHMAD FAKHRURRAZI BIN ABDUL LATIF (A16KM0022) ALEXANDER JONG JIUN WEI (A16KM0033) CHENG WEI QIN (A16KM0062) LEONG JOE YEE (B17KM0011) MOHAMAD ASHRAF BIN AB HAN (A16KM0176) MOHAMAD AZMIL BIN BAHJAM (A16KM0178) MUHAMAD AZAMUDDIN BIN MOHD TAHIR (A16KM0222) MUHAMAD HANIF SHAH BIN KHAIRUL ANUAR (A16KM0224) MUHAMMAD HAFIZ AKMAL BIN HARIS FADZILAH (A16KM0289) MUHAMMAD NASRUL BIN YAZI (A16KM0496) MUHAMMAD SHAFRIEZ BIN MD SHAIFUL (A16KM0330) NURUL ‘AFIFAH BINTI ZULKEFLI (A16KM0390) SAMALADEWI A/P MURUKAPPAN (A16KM0417) 1.0 Opening speech from lecturers
1.1 Introduction to the subject of Aircraft Design 2 1.2 Discussion on project done in Aircraft Design 1 1.3 Students divided into groups 1.4 Task given to present the project done in previous class
Action: All students 2.0 Approval of meeting minutes
2.1 First week of meeting Action: All lecturers and students
3.0 Matters arising 3.1 Formation of sub-group on propulsion, wing, fuselage and tail. Landing
gear, avionics and flight testing. 3.2 Conceptual design of aircraft. 3.3 Details of the aircraft are required, such as the v-n diagram, wing NACA,
Cruise Speed, etc. 4.0 Adjournment of meeting
The meeting was adjourned at 12pm Prepared by: Checked and approved by ____________________ ____________________ ____________________
Alexander Jong Jiun Wei Dr. Wan Zaidi bin Wan Omar
Dr. Ing. Mohd Nazri bin Mohd Nasir
248
FACULTY OF MECHANICAL ENGINEERING UNIVERSITI TEKNOLOGI MALAYSIA
17 FEBRUARY 2020 MINUTES OF MEETING NO. 2/2020 AIRCRAFT DESIGN II (SKMA 4523)
Date: 17 February, 2020
Time: 4-6pm Venue: P20-215
Attendance AHMAD AMIRUL AMIN BIN ABDUL WAHAB (A16KM0017) AHMAD FAKHRURRAZI BIN ABDUL LATIF (A16KM0022) ALEXANDER JONG JIUN WEI (A16KM0033) CHENG WEI QIN (A16KM0062) LEONG JOE YEE (B17KM0011) MOHAMAD ASHRAF BIN AB HAN (A16KM0176) MOHAMAD AZMIL BIN BAHJAM (A16KM0178) MUHAMAD AZAMUDDIN BIN MOHD TAHIR (A16KM0222) MUHAMAD HANIF SHAH BIN KHAIRUL ANUAR (A16KM0224) MUHAMMAD HAFIZ AKMAL BIN HARIS FADZILAH (A16KM0289) MUHAMMAD NASRUL BIN YAZI (A16KM0496) MUHAMMAD SHAFRIEZ BIN MD SHAIFUL (A16KM0330) NURUL ‘AFIFAH BINTI ZULKEFLI (A16KM0390) SAMALADEWI A/P MURUKAPPAN (A16KM0417) 1.0 Opening speech from lecturers
1.1 Lecture on calculation of flight envelope and load factor 1.2 Lecture on wing loading distribution
Action: All students 2.0 Approval of meeting minutes
2.1 Task distribution on each group. Action: All lecturers and students
3.0 Matters arising 3.1 Conversion of scale. 3.2 Need to finalise the components and weight of aircraft and its
components. 3.3 Final Solidworks drawing of aircraft is needed. 3.4 Material selection and material properties must be determined. 3.5 Determine preliminary cost estimation to build our rc plane.
4.0 Adjournment of meeting The meeting was adjourned at 6pm
Prepared by: Checked and approved by ____________________ ____________________ ____________________
Alexander Jong Jiun Wei Dr. Wan Zaidi bin Wan Omar
Dr. Ing. Mohd Nazri bin Mohd Nasir
249
FACULTY OF MECHANICAL ENGINEERING UNIVERSITI TEKNOLOGI MALAYSIA
19 FEBRUARY 2020 MINUTES OF MEETING NO. 3/2020 AIRCRAFT DESIGN II (SKMA 4523)
Date: 19 February, 2020
Time: 9am-12pm Venue: P20-215
Attendance AHMAD AMIRUL AMIN BIN ABDUL WAHAB (A16KM0017) AHMAD FAKHRURRAZI BIN ABDUL LATIF (A16KM0022) ALEXANDER JONG JIUN WEI (A16KM0033) CHENG WEI QIN (A16KM0062) LEONG JOE YEE (B17KM0011) MOHAMAD ASHRAF BIN AB HAN (A16KM0176) MOHAMAD AZMIL BIN BAHJAM (A16KM0178) MUHAMAD AZAMUDDIN BIN MOHD TAHIR (A16KM0222) MUHAMAD HANIF SHAH BIN KHAIRUL ANUAR (A16KM0224) MUHAMMAD HAFIZ AKMAL BIN HARIS FADZILAH (A16KM0289) MUHAMMAD NASRUL BIN YAZI (A16KM0496) MUHAMMAD SHAFRIEZ BIN MD SHAIFUL (A16KM0330) NURUL ‘AFIFAH BINTI ZULKEFLI (A16KM0390) SAMALADEWI A/P MURUKAPPAN (A16KM0417) 1.0 Opening speech from lecturers
1.1 Lecture on calculation of flight envelope and load factor 1.2 Lecture on wing loading distribution
Action: All students 2.0 Approval of meeting minutes
2.1 Mistake made in calculation of the flight envelope were corrected Action: All lecturers and students
3.0 Matters arising 3.1 Discussion on the method used to construct the shear load, bending
moment and torsional diagram. 3.2 Task distribution to complete the construction of the diagrams. 3.3 Suggestion of where to buy materials for rc plane.
4.0 Adjournment of meeting The meeting was adjourned at 12pm
Prepared by: Checked and approved by ____________________ ____________________ ____________________
Alexander Jong Jiun Wei Dr. Wan Zaidi bin Wan Omar
Dr. Ing. Mohd Nazri bin Mohd Nasir
250
FACULTY OF MECHANICAL ENGINEERING
UNIVERSITI TEKNOLOGI MALAYSIA
24 FEBRUARY 2020 MINUTES OF MEETING NO. 4/2020 AIRCRAFT DESIGN II (SKMA 4523)
Date: 24 February, 2020
Time: 4pm-6pm Venue: P20-215
Attendance AHMAD AMIRUL AMIN BIN ABDUL WAHAB (A16KM0017) AHMAD FAKHRURRAZI BIN ABDUL LATIF (A16KM0022) ALEXANDER JONG JIUN WEI (A16KM0033) CHENG WEI QIN (A16KM0062) LEONG JOE YEE (B17KM0011) MOHAMAD ASHRAF BIN AB HAN (A16KM0176) MOHAMAD AZMIL BIN BAHJAM (A16KM0178) MUHAMAD AZAMUDDIN BIN MOHD TAHIR (A16KM0222) MUHAMAD HANIF SHAH BIN KHAIRUL ANUAR (A16KM0224) MUHAMMAD HAFIZ AKMAL BIN HARIS FADZILAH (A16KM0289) MUHAMMAD NASRUL BIN YAZI (A16KM0496) MUHAMMAD SHAFRIEZ BIN MD SHAIFUL (A16KM0330) NURUL ‘AFIFAH BINTI ZULKEFLI (A16KM0390) SAMALADEWI A/P MURUKAPPAN (A16KM0417) 1.0 Opening speech from lecturers
1.1 Lecture on avionics component selection Action: All students
2.0 Approval of meeting minutes 2.1 Dr Wan Zaidi introduced the application of aileron in wing loading
analysis. Action: All lecturers and students
3.0 Matters arising 3.1 Calculate the aerodynamic coefficients with and without aileron. 3.2 Calculate the weight distribution, Cg.
4.0 Adjournment of meeting The meeting was adjourned at 6pm
Prepared by: Checked and approved by ____________________ ____________________ ____________________
Alexander Jong Jiun Wei Dr. Wan Zaidi bin Wan Omar
Dr. Ing. Mohd Nazri bin Mohd Nasir
251
FACULTY OF MECHANICAL ENGINEERING UNIVERSITI TEKNOLOGI MALAYSIA
26 FEBRUARY 2020 MINUTES OF MEETING NO. 5/2020 AIRCRAFT DESIGN II (SKMA 4523)
Date: 26 February, 2020
Time: 9am-12pm Venue: P20-215
Attendance AHMAD AMIRUL AMIN BIN ABDUL WAHAB (A16KM0017) AHMAD FAKHRURRAZI BIN ABDUL LATIF (A16KM0022) ALEXANDER JONG JIUN WEI (A16KM0033) CHENG WEI QIN (A16KM0062) LEONG JOE YEE (B17KM0011) MOHAMAD ASHRAF BIN AB HAN (A16KM0176) MOHAMAD AZMIL BIN BAHJAM (A16KM0178) MUHAMAD AZAMUDDIN BIN MOHD TAHIR (A16KM0222) MUHAMAD HANIF SHAH BIN KHAIRUL ANUAR (A16KM0224) MUHAMMAD HAFIZ AKMAL BIN HARIS FADZILAH (A16KM0289) MUHAMMAD NASRUL BIN YAZI (A16KM0496) MUHAMMAD SHAFRIEZ BIN MD SHAIFUL (A16KM0330) NURUL ‘AFIFAH BINTI ZULKEFLI (A16KM0390) SAMALADEWI A/P MURUKAPPAN (A16KM0417) 1.0 Opening speech from lecturers
1.1 Lecture on material selection and stress calculation Action: All students
2.0 Approval of meeting minutes 2.1 Task distribution in each group
Action: All lecturers and students 3.0 Matters arising
3.1 Preliminary selection of motor 3.2 Engine selection and the power curve must be determined 3.3 Internal sketching of the aircraft structure and the part attachment. 3.4 Continue the construction of shear, bending moment and lift distribution
diagrams. 3.5 Electronics components selection for avionics.
4.0 Adjournment of meeting
The meeting was adjourned at 12pm Prepared by: Checked and approved by ____________________ ____________________ ____________________
Alexander Jong Jiun Wei Dr. Wan Zaidi bin Wan Omar
Dr. Ing. Mohd Nazri bin Mohd Nasir
252
FACULTY OF MECHANICAL ENGINEERING UNIVERSITI TEKNOLOGI MALAYSIA
2 MARCH 2020 MINUTES OF MEETING NO. 6/2020 AIRCRAFT DESIGN II (SKMA 4523)
Date: 2 MARCH, 2020
Time: 4-6pm Venue: P20-215
Attendance AHMAD AMIRUL AMIN BIN ABDUL WAHAB (A16KM0017) AHMAD FAKHRURRAZI BIN ABDUL LATIF (A16KM0022) ALEXANDER JONG JIUN WEI (A16KM0033) CHENG WEI QIN (A16KM0062) LEONG JOE YEE (B17KM0011) MOHAMAD ASHRAF BIN AB HAN (A16KM0176) MOHAMAD AZMIL BIN BAHJAM (A16KM0178) MUHAMAD AZAMUDDIN BIN MOHD TAHIR (A16KM0222) MUHAMAD HANIF SHAH BIN KHAIRUL ANUAR (A16KM0224) MUHAMMAD HAFIZ AKMAL BIN HARIS FADZILAH (A16KM0289) MUHAMMAD NASRUL BIN YAZI (A16KM0496) MUHAMMAD SHAFRIEZ BIN MD SHAIFUL (A16KM0330) NURUL ‘AFIFAH BINTI ZULKEFLI (A16KM0390) SAMALADEWI A/P MURUKAPPAN (A16KM0417) 1.0 Opening speech from lecturers
1.1 Dr Wan Zaidi requested students to sketch the wing structure design of their own aircraft.
1.2 Lecture on the wing loading diagram where greater load needed to be at the root of the wing.
Action: All students 2.0 Approval of meeting minutes
2.1 Continue the work assigned during last meeting. Action: All lecturers and students
3.0 Matters arising 3.1 Calculate aircraft performance. 3.2 Determine flight path and flight power consumption. 3.3 Calculate the factor of safety and the stress in the wing components. 3.4 Deciding the best price and place to buy electronic components.
4.0 Adjournment of meeting The meeting was adjourned at 6pm
Prepared by: Checked and approved by ____________________ ____________________ ____________________
Alexander Jong Jiun Wei Dr. Wan Zaidi bin Wan Omar
Dr. Ing. Mohd Nazri bin Mohd Nasir
253
FACULTY OF MECHANICAL ENGINEERING UNIVERSITI TEKNOLOGI MALAYSIA
3 MARCH 2020 MINUTES OF MEETING NO. 7/2020 AIRCRAFT DESIGN II (SKMA 4523)
Date: 3 MARCH, 2020
Time: 9-11am Venue: E07
Attendance AHMAD AMIRUL AMIN BIN ABDUL WAHAB (A16KM0017) AHMAD FAKHRURRAZI BIN ABDUL LATIF (A16KM0022) ALEXANDER JONG JIUN WEI (A16KM0033) CHENG WEI QIN (A16KM0062) LEONG JOE YEE (B17KM0011) MOHAMAD ASHRAF BIN AB HAN (A16KM0176) MOHAMAD AZMIL BIN BAHJAM (A16KM0178) MUHAMAD AZAMUDDIN BIN MOHD TAHIR (A16KM0222) MUHAMAD HANIF SHAH BIN KHAIRUL ANUAR (A16KM0224) MUHAMMAD HAFIZ AKMAL BIN HARIS FADZILAH (A16KM0289) MUHAMMAD NASRUL BIN YAZI (A16KM0496) MUHAMMAD SHAFRIEZ BIN MD SHAIFUL (A16KM0330) NURUL ‘AFIFAH BINTI ZULKEFLI (A16KM0390) SAMALADEWI A/P MURUKAPPAN (A16KM0417) 1.0 Opening speech from lecturers
1.1 Dr Wan Zaidi explained about the wing load analysis. Action: All students
2.0 Approval of meeting minutes 2.1 Continue the work assigned during last meeting.
Action: All lecturers and students 3.0 Matters arising
3.1 Calculate aircraft performance with the correct data chosen. 3.2 Convinced to use the right maximum and minimum value of forces for the
load analysis. 4.0 Adjournment of meeting
The meeting was adjourned at 11am Prepared by: Checked and approved by ____________________ ____________________ ____________________
Alexander Jong Jiun Wei Dr. Wan Zaidi bin Wan Omar
Dr. Ing. Mohd Nazri bin Mohd Nasir
254
FACULTY OF MECHANICAL ENGINEERING UNIVERSITI TEKNOLOGI MALAYSIA
11 MARCH 2020 MINUTES OF MEETING NO. 8/2020 AIRCRAFT DESIGN II (SKMA 4523)
Date: 11 MARCH, 2020
Time: 9am-12pm Venue: P20-215
Attendance AHMAD AMIRUL AMIN BIN ABDUL WAHAB (A16KM0017) AHMAD FAKHRURRAZI BIN ABDUL LATIF (A16KM0022) ALEXANDER JONG JIUN WEI (A16KM0033) CHENG WEI QIN (A16KM0062) LEONG JOE YEE (B17KM0011) MOHAMAD ASHRAF BIN AB HAN (A16KM0176) MOHAMAD AZMIL BIN BAHJAM (A16KM0178) MUHAMAD AZAMUDDIN BIN MOHD TAHIR (A16KM0222) MUHAMAD HANIF SHAH BIN KHAIRUL ANUAR (A16KM0224) MUHAMMAD HAFIZ AKMAL BIN HARIS FADZILAH (A16KM0289) MUHAMMAD NASRUL BIN YAZI (A16KM0496) MUHAMMAD SHAFRIEZ BIN MD SHAIFUL (A16KM0330) NURUL ‘AFIFAH BINTI ZULKEFLI (A16KM0390) SAMALADEWI A/P MURUKAPPAN (A16KM0417) 1.0 Opening speech from lecturers
1.1 No lecture Action: All students
2.0 Approval of meeting minutes 2.1 Information of the Presentation 1 was distributed to the students.
Action: All lecturers and students 3.0 Matters arising
3.1 Discuss the task distribution and compiling slides of presentation 1. 3.2 Discussion on static and dynamic stability system. 3.3 Completion on shear and bending moment diagram for fuselage and tail.
4.0 Adjournment of meeting The meeting was adjourned at 12pm
Prepared by: Checked and approved by ____________________ ____________________ ____________________
Alexander Jong Jiun Wei Dr. Wan Zaidi bin Wan Omar
Dr. Ing. Mohd Nazri bin Mohd Nasir
255
FACULTY OF MECHANICAL ENGINEERING UNIVERSITI TEKNOLOGI MALAYSIA
8 APRIL 2020 MINUTES OF MEETING NO. 9/2020 AIRCRAFT DESIGN II (SKMA 4523)
Date: 8 APRIL, 2020
Time: 9am-12pm Online Discussion
Attendance AHMAD AMIRUL AMIN BIN ABDUL WAHAB (A16KM0017) AHMAD FAKHRURRAZI BIN ABDUL LATIF (A16KM0022) ALEXANDER JONG JIUN WEI (A16KM0033) CHENG WEI QIN (A16KM0062) LEONG JOE YEE (B17KM0011) MOHAMAD ASHRAF BIN AB HAN (A16KM0176) MOHAMAD AZMIL BIN BAHJAM (A16KM0178) MUHAMAD AZAMUDDIN BIN MOHD TAHIR (A16KM0222) MUHAMAD HANIF SHAH BIN KHAIRUL ANUAR (A16KM0224) MUHAMMAD HAFIZ AKMAL BIN HARIS FADZILAH (A16KM0289) MUHAMMAD NASRUL BIN YAZI (A16KM0496) MUHAMMAD SHAFRIEZ BIN MD SHAIFUL (A16KM0330) NURUL ‘AFIFAH BINTI ZULKEFLI (A16KM0390) SAMALADEWI A/P MURUKAPPAN (A16KM0417) 1.0 Opening
1.1 Thank the members for coming. Action: Chairperson
2.0 Approval of meeting minutes 2.1 Rectifying mistakes made in Presentation 1 2.2 Discussion on the comments made by Dr. Nazri on the aircraft design. 2.3 Task distribution of each each group on each comment.
Action: All lecturers and students 3.0 Matters arising
3.1 Assign the work to each group for each comment by Dr. Nazri. 3.2 Discussion on FEM analysis of landing gear system.
4.0 Adjournment of meeting The meeting was adjourned at 12pm
Prepared by: Checked and approved by ____________________ ____________________ ____________________
Alexander Jong Jiun Wei Dr. Wan Zaidi bin Wan Omar
Dr. Ing. Mohd Nazri bin Mohd Nasir
256
FACULTY OF MECHANICAL ENGINEERING UNIVERSITI TEKNOLOGI MALAYSIA
19 APRIL 2020 MINUTES OF MEETING NO. 10/2020
AIRCRAFT DESIGN II (SKMA 4523)
Date: 19 APRIL, 2020 Time: 9am-12pm Online Discussion
Attendance AHMAD AMIRUL AMIN BIN ABDUL WAHAB (A16KM0017) AHMAD FAKHRURRAZI BIN ABDUL LATIF (A16KM0022) ALEXANDER JONG JIUN WEI (A16KM0033) CHENG WEI QIN (A16KM0062) LEONG JOE YEE (B17KM0011) MOHAMAD ASHRAF BIN AB HAN (A16KM0176) MOHAMAD AZMIL BIN BAHJAM (A16KM0178) MUHAMAD AZAMUDDIN BIN MOHD TAHIR (A16KM0222) MUHAMAD HANIF SHAH BIN KHAIRUL ANUAR (A16KM0224) MUHAMMAD HAFIZ AKMAL BIN HARIS FADZILAH (A16KM0289) MUHAMMAD NASRUL BIN YAZI (A16KM0496) MUHAMMAD SHAFRIEZ BIN MD SHAIFUL (A16KM0330) NURUL ‘AFIFAH BINTI ZULKEFLI (A16KM0390) SAMALADEWI A/P MURUKAPPAN (A16KM0417) 1.0 Opening
1.1 Thank the members for coming. Action: Chairperson
2.0 Approval of meeting minutes 2.1 Follow-up on the assigned task of each sub-group on Dr Nazri’s comment. 2.2 Task distribution on each group for unassigned task.
Action: All lecturers and students 3.0 Matters arising
3.1 Recheck any missing part for final report. 3.2 Assign the unassigned work to relevant group for the comment made by Dr. Nazri.
4.0 Adjournment of meeting The meeting was adjourned at 12pm
Prepared by: Checked and approved by ____________________ ____________________ ____________________
Alexander Jong Jiun Wei Dr. Wan Zaidi bin Wan Omar
Dr. Ing. Mohd Nazri bin Mohd Nasir
257
FACULTY OF MECHANICAL ENGINEERING UNIVERSITI TEKNOLOGI MALAYSIA
8 JUNE 2020 MINUTES OF MEETING NO. 11/2020
AIRCRAFT DESIGN II (SKMA 4523)
Date: 8 JUNE, 2020 Time: 9am-12pm Online Discussion
Attendance AHMAD AMIRUL AMIN BIN ABDUL WAHAB (A16KM0017) AHMAD FAKHRURRAZI BIN ABDUL LATIF (A16KM0022) ALEXANDER JONG JIUN WEI (A16KM0033) CHENG WEI QIN (A16KM0062) LEONG JOE YEE (B17KM0011) MOHAMAD ASHRAF BIN AB HAN (A16KM0176) MOHAMAD AZMIL BIN BAHJAM (A16KM0178) MUHAMAD AZAMUDDIN BIN MOHD TAHIR (A16KM0222) MUHAMAD HANIF SHAH BIN KHAIRUL ANUAR (A16KM0224) MUHAMMAD HAFIZ AKMAL BIN HARIS FADZILAH (A16KM0289) MUHAMMAD NASRUL BIN YAZI (A16KM0496) MUHAMMAD SHAFRIEZ BIN MD SHAIFUL (A16KM0330) NURUL ‘AFIFAH BINTI ZULKEFLI (A16KM0390) SAMALADEWI A/P MURUKAPPAN (A16KM0417) 1.0 Opening
1.1 Thank the members for coming. Action: Chairperson
2.0 Approval of meeting minutes 2.1 Follow-up on the assigned task of each sub-group on Dr Nazri’s comment. 2.2 Task distribution on Presentation 2 2.3 Discussion on the format and template of video recording and compiling.
Action: All lecturers and students 3.0 Matters arising
3.1 Preparation of flight test. 3.2 Discussion on the challenges, weakness of our design activities and how
confident we are with the aircraft. 3.3 Choose compiler for our presentation video.
4.0 Adjournment of meeting
The meeting was adjourned at 12pm Prepared by: Checked and approved by ____________________ ____________________ ____________________
Alexander Jong Jiun Wei Dr. Wan Zaidi bin Wan Omar
Dr. Ing. Mohd Nazri bin Mohd Nasir