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BONAFIDE CERTIFICATE

Certified that this is a bonafide record of the project work done by

R AVINASH (20053003) R D BHARATHAN (20053005) M G KAMATCHI (20053011) S NARENDRAN (20053020) S NARMADA (20053021) VIMAL RAJ (20053038) during the period December 2007 to December 2008.

Dr. S ELANGOVAN Professor

Department of Aerospace Engineering MIT Campus

Anna University Chennai Chromepet, Chennai‐600044

Ms. M F LEELAVATHY Lecturer



Dr. B T N SRIDHAR Professor and Head



ACKNOWLEDGEMENT

We are grateful to our faculty members Dr. S Elangovan and Ms. M F Leelavathy for guiding us throughout the course of our project work. We are thankful to our Head of the Department, Dr. B T N Sridhar for inspiring us and enlightening our way. We are also greatly indebted for all the support and help given to us by the faculty members of the department at various stages of our project.

INDEX

Part 1: Aerodynamic Analysis, Performance Analysis and Stability and Control Analysis

• Data for Existing Aircraft……………………………………………………. 2 • Graphs………………………………………………………………………………. 12 • Weight Estimation and Engine Selection……………………………... 20 • Calculation of Wing Dimensions…………………………………………. 23 • Estimation of Runway Length…………………………………………….. 26 • Calculation of Empennage Dimensions……………………………….. 27 • Positioning of Landing Gear and Tire Selection……………………. 30 • 3‐View Diagram of the Aircraft…………………………………………… 32 • Seating Configuration of the Aircraft…………………………………… 35 • Balance Diagram………………………………………………………………… 37 • Drag Estimation of the Aircraft……………………………………………. 43 • Performance Analysis of the Aircraft…………………………………… 47 • Analysis of Stability and Control of the Aircraft……………………. 62

Part 2: Structural Analysis

• Estimation of Load Factor Limits and V‐n Diagram……………… 104 • Lift Distribution for the Aircraft‐Schrenck’s Curve………………. 109 • Shear Force and Bending Moment Diagrams of the Wing…….. 113 • Torque Diagram for the Wing…………………………………………….. 120 • Critical Shear Force, Bending Moment and Torque Calculations 123 • Material Selection for the Aircraft………………………………………. 127 • Structural Design of the Aircraft‐Basic Wing Design……………. 131 • Structural Design of the Aircraft‐Fuselage Design……………….. 145

Conclusion………………………………………………………………………………... 157

References………………………………………………………………………………… 158

NAME OF THE AIRCRAFT CRUISING SPEED

RANGE CRUISING MACH NUMBER CRUISING ALTITUDE

UNITS km/hr km

m

AÉROSPATIALE/BRITISH AEROSPACE CONCORDE 2179 6228 15635

AÉROSPATIALE CARAVELLE Mk 12 825 3465 0.74 7620

AIRBUS A319‐100 837 6800 0.78 10670

AIRBUS A320‐200 903 5263 0.78 8530

BOEING B720B (707‐020B) 896 6687 0.81 8190 BOEING B707‐120B 1000 6820 0.93 9900

BOEING B707‐320C 885 9262 0.83 10500

BOEING B727‐200 917 4002 0.81 7530

BOEING B727‐287 915 4818 7530

BOEING B727‐300 950 4800 12192 BOEING B737‐400 796 4262 0.73 9145 BOEING B737‐700 850 6230 0.79 12497

BOEING B737‐800 938 5425 0.79 11491

2

NAME OF THE AIRCRAFT CRUISING SPEED

RANGE CRUISING MACH NUMBER CRUISING ALTITUDE

UNITS km/hr km

m

BOEING B767‐200 852 12235 11887

BOEING B767‐300 852 9667 13137

DASSAULT MERCURE 858 756 0.79 9145 ILYUSHIN Il‐62 900 7800 0.85 11000

McDONNEL DOUGLAS MD‐11 COMBI 876 9270 0.82 10670

McDONNEL DOUGLAS MD‐82 815 3800 0.80 10688 McDONNEL DOUGLAS MD‐83 811 4635 10668 McDONNEL DOUGLAS MD‐88 811 3798 10668

McDONNEL DOUGLAS MD‐90‐30 811 4329 0.76 10670 TUPOLEV Tu‐154B 850 3500 10670 TUPOLEV Tu‐154M 950 6600 11900 TUPOLEV Tu‐204 830 3500 11500

VICKERS VC‐10 1151 885 7596 11580

3

NAME OF THE AIRCRAFT Wp (Payload Weight)

Wo (Takeoff Gross Weight)

We (Empty Weight)

We/Wo

UNITS kg kg kg

AÉROSPATIALE/BRITISH AEROSPACE CONCORDE 13380 185066 78700 0.4253

AÉROSPATIALE CARAVELLE Mk 12 9100 58000 24185 0.4170

AIRBUS A319‐100 18800 75500 40600 0.5377

AIRBUS A320‐200 19220 77000 42400 0.5506

BOEING B720B (707‐020B) 7711 106141 46785 0.4408 BOEING B707‐120B 19280 116570 55589 0.4769

BOEING B707‐320C 20955 151318 55580 0.3673

BOEING B727‐200 14263 95027 45360 0.4773

BOEING B727‐287 12500 95027 45360 0.4773

BOEING B727‐300 16148 56472 58390 1.0340 BOEING B737‐400 19900 68050 34827 0.5118 BOEING B737‐700 17544 70080 38147 0.5443

BOEING B737‐800 21319 79016 41413 0.5241

4

NAME OF THE AIRCRAFT Wp (Payload Weight)

Wo (Takeoff Gross Weight)

We (Empty Weight)

We/Wo

UNITS kg kg kg

BOEING B767‐200 32568 142881 85595 0.5991

BOEING B767‐300 44497 159755 86955 0.5443

DASSAULT MERCURE 56500 31800 0.5628 ILYUSHIN Il‐62 23000 165000 67500 0.4091

McDONNEL DOUGLAS MD‐11 COMBI 55655 273300 132800 0.4859

McDONNEL DOUGLAS MD‐82 19709 67900 26070 0.3839 McDONNEL DOUGLAS MD‐83 18721 72575 36530 0.5033 McDONNEL DOUGLAS MD‐88 19709 67812 37925 0.5593

McDONNEL DOUGLAS MD‐90‐30 15649 70760 39990 0.5651 TUPOLEV Tu‐154B 18000 98000 50700 0.5173 TUPOLEV Tu‐154M 18000 100000 50700 0.5070 TUPOLEV Tu‐204 21500 93500 58300 0.6235

VICKERS VC‐10 1151 22860 151953 63278 0.4164

5

NAME OF THE AIRCRAFT T/W W/S AR FUSELAGE LENGTH

FUSELAGE DIAMETER TAIL HEIGHT

UNITS

kg/m2

m m m

AÉROSPATIALE/BRITISH AEROSPACE CONCORDE 0.373 516.6 1.82

AÉROSPATIALE CARAVELLE Mk 12 0.227 395.4 8.01

AIRBUS A319‐100 0.311 539.0 9.38 33.84 3.95 11.76

AIRBUS A320‐200 0.294 588.0 9.39 37.57 3.95 11.76

BOEING B720B (707‐020B) 0.308 453.2 6.79 41.25 3.76 12.65 BOEING B707‐120B 0.265 515.1 7.03 44.07 3.76 12.93

BOEING B707‐320C 0.228 534.0 6.96 46.61 3.76 12.93

BOEING B727‐200 0.208 601.7 6.86 46.70 4.32 10.30

BOEING B727‐287 0.208 601.7 7.07

BOEING B727‐300 0.321 528.1 7.80 BOEING B737‐400 0.267 645.6 9.16 36.50 3.76 11.10 BOEING B737‐700 0.388 560.6 9.45 33.60 3.76 12.50

BOEING B737‐800 0.344 632.1 9.42 39.50 3.76 12.50

6

NAME OF THE AIRCRAFT T/W W/S AR FUSELAGE LENGTH

FUSELAGE DIAMETER TAIL HEIGHT

UNITS

kg/m2

m m m

BOEING B767‐200 0.305 504.3 7.99

BOEING B767‐300 0.276 563.8 7.99

DASSAULT MERCURE 0.249 487.1 8.05 34.84 11.35 ILYUSHIN Il‐62 0.259 349.0 6.68 53.12 4.10 12.35

McDONNEL DOUGLAS MD‐11 COMBI 0.305 806.4 7.86 61.23 6.02 17.60

McDONNEL DOUGLAS MD‐82 0.267 575.4 9.12 McDONNEL DOUGLAS MD‐83 0.250 615.0 9.62 McDONNEL DOUGLAS MD‐88 0.267 574.7 9.16

McDONNEL DOUGLAS MD‐90‐30 0.320 630.1 9.58 TUPOLEV Tu‐154B 0.321 486.4 7.00 TUPOLEV Tu‐154M 0.315 496.4 7.00 TUPOLEV Tu‐204 0.342 512.6 9.67 46.10 4.18 13.90

VICKERS VC‐10 1151 0.270 534.0 7.49 48.36 3.90 12.04

7

NAME OF THE AIRCRAFT l/dmax WINGSPAN SWEEPBACK ANGLE No. OF ENGINES

UNITS

m deg

AÉROSPATIALE/BRITISH AEROSPACE CONCORDE 13.650 25.55 70 4 X Olympus 593 Mk 610

turbojets (RR/SNECMA) AÉROSPATIALE CARAVELLE Mk 12 34.29 2 X JT8D‐9 (P&W)

AIRBUS A319‐100 8.567 33.91

2 X CFM56‐5‐A1 (CFM Intl) or

2 X V2500‐A1 (IAE)

AIRBUS A320‐200 9.511 33.91 30 2 X CFM56‐5‐A1 (CFM Intl)

or 2 X V2500‐A1 (IAE)

BOEING B720B (707‐020B) 10.971 39.90 35 4 X JT3D‐3 (P&W) BOEING B707‐120B 11.721 39.90 35 4 X JT3D‐1 (P&W)

BOEING B707‐320C 12.396 44.42 35 4 X JT3D‐3 (P&W)

or 4 X JT3D‐7 (P&W)

BOEING B727‐200 10.800 32.92 32

3 X JT8D‐8 (P&W) or


3 X JT8D‐17R (P&W)

BOEING B727‐287 32.92 32

3 X JT8D‐9A (P&W) or


3 X JT8D‐17R (P&W) BOEING B727‐300 28.88 32 2 X CFM56‐3‐B‐1 (CFM Intl) BOEING B737‐400 8.881 28.90 25 2 X CFM56‐3‐B‐2 (CFM Intl) BOEING B737‐700 8.175 35.70 25.02 2 X CFM56‐7‐B26 (CFM Intl)

BOEING B737‐800 10.497 34.31 25.02 2 X CFM56‐7‐B (CFM Intl)

8

NAME OF THE AIRCRAFT l/dmax WINGSPAN SWEEPBACK ANGLE No. OF ENGINES

UNITS

m deg

BOEING B767‐200 9.644 47.57

2 X JT9D‐7R‐4D (P&W) or

2 X PW4052 (P&W) or

2 X CF6‐80A (GE) or

2 X CF6‐80C2‐B2 (GE)

BOEING B767‐300 10.922 47.57

2 X JT9D‐7R‐4D (P&W) or

2 X CF6‐80A2 (GE) or

2 X RB211‐524H (RR) DASSAULT MERCURE 30.56 2 X JT8D‐15 (P&W)

ILYUSHIN Il‐62 12.956 43.20 4 X D‐30KU (Soloviev)

McDONNEL DOUGLAS MD‐11 COMBI 10.170 51.60

3 X CF6‐80C2D1F (GE) or

3 X PW4460 (P&W) McDONNEL DOUGLAS MD‐82 10.010 32.80 2 X JT8D‐217 (P&W) McDONNEL DOUGLAS MD‐83 10.010 32.87 2 X JT8D‐219 (P&W) McDONNEL DOUGLAS MD‐88 10.010 32.87 2 X JT8D‐217 (P&W)

McDONNEL DOUGLAS MD‐90‐30 10.177 32.80 2 X V2525‐D5 (IAE) TUPOLEV Tu‐154B 9.776 37.55 35 3 X D‐30KU‐154‐D (Soloviev)TUPOLEV Tu‐154M 12.605 37.55 35 3 X D‐30KU‐154‐D (Soloviev)TUPOLEV Tu‐204 11.029 42.00 2 X D‐90A (Soloviev)

VICKERS VC‐10 1151 12.405 44.55 4 X Mk. 301 (RR Conway)

9

NAME OF THE AIRCRAFT No. OF SEATS No. OF CLASSES

UNITS

AÉROSPATIALE/BRITISH AEROSPACE CONCORDE 144

AÉROSPATIALE CARAVELLE Mk 12 140

AIRBUS A319‐100 142 1

AIRBUS A320‐200 180 150

1 2

BOEING B720B (707‐020B) 140 BOEING B707‐120B 179 1

BOEING B707‐320C 147 2

BOEING B727‐200 149

BOEING B727‐287 163 2

BOEING B727‐300 149 1 BOEING B737‐400 168 1 BOEING B737‐700 149 1

BOEING B737‐800 162 146

1 2

10

NAME OF THE AIRCRAFT No. OF SEATS No. OF CLASSES

UNITS

BOEING B767‐200 174 3

BOEING B767‐300 174 3

DASSAULT MERCURE 156 ILYUSHIN Il‐62 176

McDONNEL DOUGLAS MD‐11 COMBI 176

McDONNEL DOUGLAS MD‐82 146 McDONNEL DOUGLAS MD‐83 172 McDONNEL DOUGLAS MD‐88 172

McDONNEL DOUGLAS MD‐90‐30 158 TUPOLEV Tu‐154B 180 TUPOLEV Tu‐154M 164 TUPOLEV Tu‐204 170

VICKERS VC‐10 1151 151

11

0

2000

4000

6000

8000

10000

12000

14000

0 500 1000 1500 2000 2500

RANGE (km)

CRUISING SPEED (km/hr)

RANGE vs CRUISING SPEED

0

2000

4000

6000

8000

10000

12000

14000

16000

18000

0 500 1000 1500 2000 2500

CRUISING ALTITUDE (m

)


CRUISING ALTITUDE vs CRUISING SPEED

12

0

50000

100000

150000

200000

250000

300000

0 500 1000 1500 2000 2500

TAKE

‐OFF GRO

SS W

EIGHT (kg)


TAKE‐OFF GROSS WEIGHT vs CRUISING SPEED

0

10000

20000

30000

40000

50000

60000

0 500 1000 1500 2000 2500

PAYLOAD W

EIGHT (kg)


PAYLOAD WEIGHT vs CRUISING SPEED

13

0

20000

40000

60000

80000

100000

120000

140000

0 500 1000 1500 2000 2500

EMPT

Y WEIGHT (kg)


EMPTY WEIGHT vs CRUISING SPEED

0.0000

0.2000

0.4000

0.6000

0.8000

1.0000

1.2000

0 500 1000 1500 2000 2500

We/Wo


EMPTY WEIGHT RATIO vs CRUISING SPEED

14

0.000

0.050

0.100

0.150

0.200

0.250

0.300

0.350

0.400

0.450

0 500 1000 1500 2000 2500

T/W


T/W vs CRUISING SPEED

0.0

100.0

200.0

300.0

400.0

500.0

600.0

700.0

800.0

900.0

0 500 1000 1500 2000 2500

WING LOADING (kg/m

2 )


WING LOADING vs CRUISING SPEED

15

0.00

10.00

20.00

30.00

40.00

50.00

60.00

0 500 1000 1500 2000 2500

WINGSPAN (m

)


WINGSPAN vs CRUISING SPEED

0

10

20

30

40

50

60

70

80

0 500 1000 1500 2000 2500

SWEEPB

ACK

ANGLE (d

eg)


SWEEPBACK ANGLE vs CRUISING SPEED

16

0.00

2.00

4.00

6.00

8.00

10.00

12.00

0 500 1000 1500 2000 2500

ASPECT RA

TIO


ASPECT RATIO vs CRUISING SPEED

0.000

2.000

4.000

6.000

8.000

10.000

12.000

14.000

16.000

0 500 1000 1500 2000 2500

FINEN

ESS RA

TIO


FINENESS RATIO vs CRUISING SPEED

17

0

10

20

30

40

50

60

70

0 500 1000 1500 2000 2500

FUSELA

GE LENGTH

(m)


FUSELAGE LENGTH vs CRUISING SPEED

0

1

2

3

4

5

6

7

0 500 1000 1500 2000 2500

FUSELA

GE WIDTH

(m)


FUSELAGE WIDTH vs CRUISING SPEED

18

0

2

4

6

8

10

12

14

16

18

20

0 500 1000 1500 2000 2500

TAIL HEIGHT (m

)


TAIL HEIGHT vs CRUISING SPEED

19

WEIGHT ESTIMATION AND ENGINE SELECTION

No. of crew members : 3 (cockpit crew) 4 (cabin crew)

Seating configuration : Economy class: 162 passengers in 27 rows Single aisle configuration (3 + 3)

1st Weight estimation of the aircraft:

Weight of crew, Wc = 7000 N

Weight of passengers, Wp = 162000 N

From the mean value taken from graph, empty weight ratio

0.6

Assuming a fuel fraction of 0.155, we have

0.155

W W W

1 WWWW

Using the above equation,

Wo = 689795.9184 N

Engine selected : CFM56‐7B (CFM International)

No. of Engines : 2

20

Engine Specifications:

Sea level Thrust : 23500 lbs Specific Fuel Consumption : 0.38 lb/lb‐hr Engine Dry Weight : 5205 lb Engine Length : 98.7 in Engine Diameter : 65 in

W No. of Engines x Thrust at altitude x Range x SFC

Cruising Speed x 1.2

where Wf (fuel weight) is in lb Range is in km Thrust is in lb SFC is in lb/lb‐hr Cruising speed is in km/hr

Thrust at altitude T x σ .

where To : Sea level thrust in lb σ : Density ratio at altitude

Considering mean cruising altitude from graph,

Cruising Altitude = 11000 m

At this altitude, σ = 0.2971

Considering mean cruising speed from graph,

Cruising Speed = 915 km/hr

Assuming Range = 4450 km

For CFM56‐3B‐2, To = 23500 lbs

21

Using the above data and equations,

Wf = 24304.63115 lb

= 108131.304 N

Check:

Wf / Wo = 108131.304/689795.9184

= 0.156

Assuming specific weight of fuel (aviation gasoline) to be 0.0235 ft3/lb,

Volume of fuel, Vf = 571.064 ft3

= 16.1707 m3

22

CALCULATION OF WING DIMENSIONS

Length of fuselage, lfus = 38 m

Wo = 689795.9184 N

From the mean value taken from graph,

Wing loading, W/S = 4500 N/m2

Sweepback angle, Ω = 32˚

Using the above values,

Wing area, S = 153.288 m2

From the mean value taken from graph,

Wingspan, b = 36 m

Using the above values,

Aspect ratio, AR = 8.454

c2S

b 1 λ

where S : Wing area in m2

b : Wingspan in m : Wing taper ratio

λ cc

where ct : Wing tip chord cr : Wing root chord

23

Using the above formulae and data, assuming a taper ratio = 0.3,

cr = 6.5508 m

ct = 1.9652 m

Mean chord, ĉ = 4.67 m

V 0.5ĉ tc b ĉ 0.5 0.75 2

where Vf : Volume of fuel in m3

ĉ : Mean chord in m

: Thickness to chord ratio

Using the above formula and data,

0.1098 10.98 %

From the graph plotted between cosΩ and Mcr (critical Mach number),

Mcr = 0.89

From NACA airfoil data considering the required performance parameters, the selected airfoil is NACA 23011‐63 (5 – series).

For this airfoil,

Zero‐lift angle of attack, αo = ‐1.175˚

From leading edge, Xcop = 28.3 %

Xac = 26.5 %

From lift curve, Cl(max) = 1.495 at α = 14.5˚

From drag polar, Cd(min)= 0.0065

From drag bucket, operational Cl = 0.38

24

ESTIMATION OF RUNWAY LENGTH

W/S = 4500 N/m2

Cl(max) = 1.495

Sea level density, ρo = 1.2256 kg/m3

Using above data,

Vstall = 70.0852 m/s

= 252.31 km/hr

VT.O = 1.3 Vstall

= 91.1m/s

= 328 km/hr

VT.O 2 a ST.O

where a : deceleration value (= 0.25g with thrust reversers)

ST.O : take off distance


ST.O = 1700 m = 5580 ft

Runway length = ST.O/0.6 = 2833 m = 9300 ft

From the data on double slotted flaps,

Assuming flap width = 0.2c,

For a maximum flap deflection, (δflap)max = 40˚,

For t/c = 0.11,

ΔCl (max) = 0.635

Flap length = 0.4b

= 4.5 m on each wing

26

CALCULATION OF EMPENNAGE DIMENSIONS

HORIZONTAL TAIL:

HTSHT lHTS c

where HT : Horizontal tail volume ratio

SHT : Horizontal tail area in m2

lHT : Horizontal tail arm in m

S : Wing area in m2

c : Wing chord in m

Using the above formula, and assuming

SHT/S = 0.12

lHT/c = 4,

HT 0.48

SHT = 19.161 m2

lHT = 18.68 m

ARHT = 0.6 * ARwing

= 5.0724

ARHTbHTSHT

Using the above formula,

Horizontal tail span, bHT = 9.859 m

27

The symmetrical airfoil NACA 0012 is selected for the horizontal tail.

Assuming HT = 0.3,

cr (HT) = 2.99 m

ct (HT) = 0.897 m

ĉHT23

c HT1 λ λ1 λ

YHTbHT6

1 2λ1 λ

where YHT : Span‐wise location of mean chord

Using the above formulae,

ĉHT = 2.131 m

YHT = 2.022 m

VERTICAL TAIL:

VTSVT lVTS c

where VT : Vertical tail volume ratio

SVT : Vertical tail area in m2

lVT : Vertical tail arm in m

S : Wing area in m2

C : Wing chord in m

Using the above formula, and assuming

VT 0.05

lVT = 0.95 * lHT = 17.75 m

SVT = 16.15 m2

28

Assuming

ARVT = 1.5

ARVThVTSVT


Vertical tail height hVT = 4.951 m

The symmetrical airfoil NACA 0012 is selected for the vertical tail.

Assuming VT = 0.3,

cr (VT) = 5.007 m

ct (VT) = 1.523 m

ĉVT23 c VT

1 λ λ1 λ

ZVT2 hVT6

1 2λ1 λ

where ZVT : Height‐wise location of mean chord


ĉVT = 3.619 m

ZVT = 2.031 m

29

POSITIONING OF LANDING GEAR AND TIRE SELECTION

For the aircraft, the commonly‐used tricycle landing gear configuration is used.

The nose gear is of double‐bogey type with two wheels. The main gear consists of two sets of wheels (wing‐retracted) each of multi‐bogey type with 4 wheels each.

NOSE GEAR:

Load on nose gear = 0.1Wo

= 15501.93257 lb

Hence, load per tire, Lt = 7750.966287 lb

Wheel diameter 1.51 L . Wheel width 0.75 L .

Using the above empirical relations,

Wheel diameter = 33.7741 in

Wheel width = 12.26141 in

Selecting DUNLOP© tire 33 X 9.75 – 11,

Tire diameter, d = 33.4 in

Tire width, w = 9.65 in

Rolling radius, Rf = 14.3 in

Footprint area, A 2.3√w d 0.5d R


Ap = 99.1in2

Tire pressure = 78.2132 psi

30

MAIN GEAR:

Load on nose gear = 0.9Wo

= 139517.3931 lb

Hence, load per tire, Lt = 8719.837073 lb

Wheel diameter 1.51 L . Wheel width 0.75 L .

Using the above empirical relations,

Wheel diameter = 35.183 in

Wheel width = 12.7204 in

Selecting DUNLOP© tire 36 X 10.75 – 16.5,

Tire diameter, d = 35.6 in

Tire width, w = 10.6 in

Rolling radius, Rf = 15.25 in

Footprint area A 2.3√w d 0.5d R


Ap = 113.932in2

Tire pressure = 76.535 psi

From the mean value taken from graph and extrapolated,

Wheel base = 14.263 m

Wheel track = 0.25 * b

= 9 m

31

BALANCE DIAGRAM

The balance diagram is used to estimate the weight distribution in the aircraft and to approximately locate the centre of gravity of the aircraft. The weights of the individual components in the fuselage and their centers of gravity are tabulated below.

S.No COMPONENT WEIGHT Wi (Xcg)i (from nose)

Wi(Xcg)i

Unit N m N‐m 1 Fixed Equipment

(0.04 Wo) 27591.840 2 55183.68

2 Pilots (3 Wp)

2400.000 2.8 6720.0000

3 Nose Landing Gear (0.01 Wo)

6897.960 4.466 30806.2894

4 Cabin Crew (Front) (2 Wp)

1600.000 14.694 23510.4000

5 Passengers (162 Wp)

129600.000 (distributed load)

1944000.0000

6 Cargo 32400.000 31 1004400.00007 Fuselage Structure

(0.1 Wo) 68979.592 18.338 1264947.7580

8 Cabin Crew (Rear) (2 Wp)

1600.000 24.8115 39698.4000

9 Lavatories (0.08Wo)

55183.670 (distributed load)

977965.0600

10 Vertical Tail (0.02Wo)

13795.920 34.768 479656.5500

11 Horizontal Tail (0.03Wo)

20693.880 36.2 749118.4560

Total 360742.862 6576006.5900

Using the above tabulated values, we obtain the location of the center of gravity of the fuselage section from the nose of the airplane.

X

ΣW XΣW

6576006.59360742.862 18.22907m

37

Similarly, the weights of the individual components in one wing and their centers of gravity are tabulated below.

S.No COMPONENT WEIGHT Wi (Xcg)i (from leading edge)

Wi(Xcg)i

Unit N m N‐m 1 Wing Structure

(0.05 Wo) 34489.796 6.746 232668.1638

2 Fuel (Wf)

54065.652 5.657 305849.3934

3 Main Landing Gear (0.02 Wo)

13795.918 7.109 98095.1811

4 Engine (Weng)

23370.450 4.101 95842.2155

Total 125721.816 732454.9537

Using the above tabulated values, we obtain the location of the center of gravity of the fuselage section from the nose of the airplane.

X

ΣW XΣW

732454.9538125721.816

5.825838m

For the other configuration of the airplane based on weight, the following locations are determined.

X

18.25276m

X

18.23749m

X

5.914388m

39

Using the equation

where

X : Distance of wing root leading edge from fuselage nose

X3 : Distance of aircraft C.G from wing root leading edge

We obtain

.

Hence, . .

The different locations of the aircraft C.G for different configurations of weight are tabulated below.

CONFIGURATION Wfuselage Xfuselage Wwing Xwing X(c.g) aircraft

Full payload + Full fuel

360742.862 18.0494 251443.632 5.8258 18.2115

Full payload + Reserve fuel

360742.862 18.0494 164938.589 5.9144 18.2581

Zero payload + Reserve fuel

198742.862 18.2528 164938.589 5.9144 18.3054

Zero payload + Full fuel

198742.862 18.2528 251443.632 5.8258 18.2209

Half payload + Full fuel

279742.862 18.1217 251443.632 5.8258 18.2149

Half payload + Reserve fuel

279742.862 18.1217 164938.589 5.9144 18.2749

The movement of the aircraft C.G between extreme cases is evaluated.

X . X .

ĉ0.02009418

41

Hence, the aircraft C.G is found to move by 2.01% of the wing mean aerodynamic chord between extreme weight configurations.

42

DRAG ESTIMATION OF THE AIRCRAFT

The drag coefficients at zero‐lift condition of the different components of the aircraft are estimated using different formulae characteristic of them.

An example of the calculation involved is furnished below:

Wing:

Using the formulae

CDC

4 2ct

120tc

C0.427

ln R 0.407 .

Rρ V ĉ

µ

and the values

cr = 0.363924 kg/m3

Vcr 915 km/hr

ĉ = 4.67 m

μcr = 1.4216 * 10‐5 N‐s/m2

t/c = 0.1097

we obtain the following values –

Re = 30384509

Cf = 2.47788 * 10‐4

5.548 10

The various formulae used for calculating drag coefficients are the following:

43

Fuselage : CDC

3 4.5.

21

where ineness ratio

Horizontal and Vertical Tail : same as wing

Retractable Landing Gear : from the graph plotted between CD and e/d, where ‘e’ is the length of the exposed portion of the landing gear, and ‘d’ is the diameter of the wheel.

The values of the individual component drag coefficients are tabulated as follows:

S.No COMPONENT Sπ CDπ CDπ * Sπ Unit m2 m2

1 Canopy Max. Cross Sectional Area

0.073 0.8044

2 Fuselage Max. Cross Sectional Area

0.007531 0.09229

3 Wing Planform Area (Sw) 0.005548 0.85044 4 Horizontal Tail Planform Area (Sh) 0.005172 0.10012 5 Vertical Tail Planform Area (Sv) 0.005172 0.09211 6 Powerplant Max. Cross Sectional

Area 0.046854 0.20477

Total X 2.06797957(a) Undercarriage–

Nose Gear Vertical Cross Section of Tire

0.9 a=0.37429

7(b) Undercarriage‐ Main Gear

Vertical Cross Section of Tire

1 b=0.97383

8(a) Flaps – Take‐off (45o)

Planform Area 0.045 c=0.4221

8(b) Flaps – Landing (60o)

Planform Area 0.045 d=0.29853

The net drag coefficients of the airplane in various phases of flight are calculated:

44

Take‐off : Σ CDπ * Sπ = X + a + b + c = 3.8382042

Cruise : Σ CDπ * Sπ = X = 2.0679795

Landing : Σ CDπ * Sπ = X + a + b + d = 3.7146342

Accounting for the additional component for drag contributed by interference, using the formulae

CDCD SS

CDo net = CDo * 1.05

we have

Take‐off : CDo = 0.02629116

Cruise : CDo = 0.01416555

Landing : CDo = 0.02544465

From the plot of (M∞L/D)max vs M∞ typical of a 160‐seater aircraft, using the cruising mach number of 0.86, we obtain

(L/D)max = 15.4023

Using the formulae

LD CD

kR where e – Ostwald planform efficiency factor

we have

k = 0.071093

e = 0.53

45

Using these values we obtain the drag polar equation.

CD CDO KCL

.

For various phases in flight the drag polar equation reduces to the following:

Take‐off : CD 0.02629116 0.071093 CL

Cruise : CD 0.01416555 0.071093 CL

Landing : CD 0.02544465 0.071093 CL

Using these equations the drag polar is plotted for various flight phases.

0

0.5

1

1.5

2

2.5

0 0.1 0.2 0.3 0.4 0.5

Lift Coe

fficient C

L

Drag Coefficient CD

DRAG POLAR

Cruise Flight

Take‐Off

Landing

46

PERFORMANCE ANALYSIS OF THE AIRCRAFT

Performance analysis of the aircraft is carried out by plotting the various performance curves for the aircraft at various altitudes for different phases of flight, i.e. cruise flight, take‐off, landing etc.

STEADY FLIGHT AIRCRAFT PERFORMANCE:

In this analysis, the aircraft velocity is constant, i.e. this analysis deals with unaccelerated flight of the airplane. The various phases analysed are climb and level cruise flight.

STEADY CLIMB PERFORMANCE:

In the analysis of steady climb, curves are obtained showing the variation of performance parameters like thrust and power required against velocity and Mach number at different altitudes.

The various formulae used in the analysis are the following:

V M a

CL2 W

ρ S V

CLCL1 M

CD CD kCL

DW

CLCD

T D

P T V

47

PP S.L

σ

σρρS.L

Rate of climb, RC

P PW

Using the above formulae, a sample calculation is furnished below:

At altitude 4000m,

4000 = 0.819133 kg/m3

a4000 = 324.6 m/s

At Mach 0.6,

V = 194.76 m/s

CLcomp = 0.289660019

CDtot = 0.020130461

CL/CD = 14.38914005

D =47938.64791 N =Treqd

Preqd = 9336531.067 W

For most turbofan engines, the value of ‘m’ is very close to 1. We therefore make a reasonable assumption that m=1.

Pav = 27223130.86 W

R/C = 25.9302778 m/s

In this way, the climb performance parameters are determined for Mach numbers 0.1 to 1.5 at sea level, as well as altitudes 4000m, 8000m and 11000m. The variation of the parameters with Mach number and altitude are plotted as Climb Performance Curves.

48

0

20000000

40000000

60000000

80000000

100000000

120000000

140000000

160000000

180000000

200000000

0 0.5 1 1.5 2

Power Req

uired (W

)

Mach Number

Power vs Mach Number

Power Required‐Sea Level

Power Required‐4000m



Power Available‐Sea Level

Power Available‐4000m



0

20000000

40000000

60000000

80000000

100000000

120000000

140000000

160000000

180000000

200000000

0 200 400 600

Power Req

uired (W

)

Velocity (m/s)

Power vs Velocity

Power Required‐Sea Level




Power Available‐Sea Level




49

0

200000

400000

600000

800000

1000000

1200000

1400000

1600000

0 0.5 1 1.5 2

Thrust Req

uired (N)

Mach Number

Thrust vs Mach Number

Thrust Required‐Sea Level

Thrust Required‐4000m



Trust Available‐Sea Level

Thrust Available‐4000m



0

200000

400000

600000

800000

1000000

1200000

1400000

1600000

0 200 400 600

Trust R

equired (N)

Velocity (m/s)

Thrust vs Velocity

Thrust Required‐Sea Level




Thrust Available‐Sea Level




50

From the plot of max. climb rate vs altitude,

Absolute ceiling = 13610 m Service Ceiling (R/C=100ft/min) = 13395 m

‐120

‐100

‐80

‐60

‐40

‐20

0

20

40

60

0 100 200 300 400 500 600

Rate of C

limb (m

/s)

Velocity (m/s)

Rate of Climb vs

Velocity

Sea Level

4000 m

8000 m

11000 m

0

2000

4000

6000

8000

10000

12000

14000

16000

‐10 0 10 20 30 40 50

Altitud

e (m

)

Maximum Rate of Climb (m/s)

Maximum Rate of Climb vs

Altitude

51

CRUISE FLIGHT PERFORMANCE:

Cruise flight performance of the aircraft includes determining the critical speeds of the airplane – the minimum power and drag speeds, as well as the maximum range and endurance of the airplane. The performance of the airplane optimized for maximum range and endurance, under varying payload conditions is also analysed.

CD KCL

V 2 WS ρ

KCD

LD

14 K CD


Vminimum drag = 235.3771261 m/s

(L/D)max = 15.75577837

3CD KCL

V 2 WS ρ

K3CD


Vminimum power= 135.8950471 m/s

Since the performance of the turbofan engine can be closely approximated to a jet engine, we have the following relations

R 22

S ρ1c

CLCD

WT.O W

52

E1c

CLCD

lnWT.O

W

For maximum range,

CD 3KCL

For maximum endurance,

CD KCL


= 0.2577167659

Rmax = 6544.588358 km

(L/D)maximum range = 13.64490432

= 0.4463785326

Emaximum range = 6.122 hrs

Emax = 7.069 hrs

(L/D)maximum endurance = 15.75577837

Rmaximum endurance = 5742.108443 km

V 2 W

S ρ CL

V 2 W

S ρ CL

The variation of velocity, range and endurance of the airplane optimized for maximum range and maximum endurance, for different weight configurations of the airplane are plotted.

53

500000

550000

600000

650000

700000

5000 6000 7000 8000

Airplan

e Weight (N)

Range (km)

Variation of Range with Airplane Weight

Maximum Range

Maximum Endurance

500000

550000

600000

650000

700000

5.5 6.5 7.5 8.5 9.5 10.5

Airplan

e Weight (N)

Endurance (hrs)

Variation of Endurance with Airplane Weight

Maximum Range

Maximum Endurance

500000

550000

600000

650000

700000

100 150 200 250 300 350

Airplan

e Weight (N)

Airspeed (m/s)

Variation of Airspeed with Airplane Weight

Maximum Range

Maximum Endurance

54

ACCELERATED FLIGHT AIRCRAFT PERFORMANCE:

This analysis deals with the accelerated phases of aircraft flight. The various phases analysed are take‐off, landing and turn.

TAKEOFF AND LANDING PERFORMANCE:

This analysis mainly focuses on determining the distance required for take‐off and landing.

TakeOff Performance:

V2 W

S ρS.L CL

Here, CLmax is evaluated with flaps deployed. In this condition,

2.13


Vstall = 58.71608315 m/s

VT.O V 1.1

CLT.O2 W

S ρ VT.O

CDT.O CD T.OKCLT.O

DT.OCDT.O ρ VT.O S

2

Using the above formulae and data,

VT.O = 64.58769147 m/s

. = 1.760330579

. = 0.246501571

55

DT.O = 96628.37274 N

From the data on the engine used,

TT.O = 171233.7013 N

The angle of climb is given by

sin γTT.O DT.O

W


sin = 0.1081556538

= 0.1083676327 rad

= 6.196°

V 0.7 VT.O

LCLT.O ρ V S

2

DCDT.O ρ V S

2

S1.21 W

g ρ S CL TT.O D µ W L


Vavg = 45.21138403 m/s

Lavg = 338000.0401 N

Davg = 46347.90265 N

Assuming μr=0.2,

Sgroundroll = 1120.235678 m

RVT.O0.2 g

S R sin γ

56

h R 1 cos γ

SH h

tan γ


Rtransition = 2238.086748 m

Stransition = 242.0617355 m

htransition = 13.12867925 m

Assuming screen height, Hobst =15 m,

Sclimb = 17.20061289 m

ST O S S S

Using the above formula, we obtain

STakeOff = 1379.498028 m

Landing Performance:

V2 W

S ρS.L CL

Using the values,

Wlanding = 581664.6144 N

=2.13

We obtain,

= 53.9179117 m/s

V V 1.3

57

CD CD KCL

V 0.7 V

LCL ρ V S

2

DCD ρ V S

2

T 0.125 TT.O

RV0.2 g


Vapproach = 70.09328521 m/s

Vavg = 49.06529965 m/s

Lavg = 481676.4672 N

= 0.3479864817

Davg = 78693.37988 N

Treverser = 21404.21266 N

Rflare = 2504.112452 m

SH htan γ

S R sin γ

S1.69 W

g ρ S CL T D µ W L

Using the above formulae and data, and assuming

approach 2.5°

58

brakes 0.25

We obtain

Sapproach 288.9685603 m

Sflare 109.2278511 m

Sground roll 1164.361421 m

S S S S


Slanding = 1562.557832 m

59

TURNING PERFORMANCE:

Turning performance of the aircraft is analysed by observing the variation of parameters like velocity, turn rate, turn radius, etc.

Since turn is an accelerated phase of flight, load factor during turn is of vital importance.

Load Factor,

The values of the turn parameters are analysed in the three most critical turn maneuvers explained below.

Maximum Sustained Turn Rate:

It is the turn executed with maximum angular velocity, with the additional constraint that the airplane does not lose any altitude. In other words, it is the turn executed in a horizontal plane with maximum angular velocity.

Sharpest Sustained Turn:

It is the turn executed with minimum turning radius. Here, there is no restriction on the altitude of the aircraft.

Maximum Load Factor Turn:

It is the turn executed while simultaneously sustaining the maximum load factor permissible on the airplane.

Turn performance is analysed with the help of certain non‐dimensional parameters, defined below:

ECLCD

ZTW

E

nLW

60

uV

V

Here, we use

V V 2 WS ρ

KCD


Vreference = 235.3771261 m/s

Em =15.75577837

The various values of velocity, turn radius and rate, and load factor are calculated using the formulae described below:

CASE u n rad/s R (m)

MSTR 1 √2Z 1 g√2Z 2V

Vg√2Z 2

SST 1√Z √2Z 1

Z

gV

Z 1Z

Vg√Z 1

nmax √Z Z gV

Z 1Z

Z Vg√Z 1

Using the above formulae, the various values are obtained.

CASE u n rad/s R (m) MSTR 1 2.92471 0.114549 2054.81696 SST 0.457535 1.398634 0.089074 1209.03353 nmax 2.185626 4.776963 0.089074 5775.50845

61

ANALYSIS OF STABILITY AND CONTROL OF THE AIRCRAFT

STATIC STABILITY AND CONTROL OF THE AIRCRAFT:

Longitudinal StickFixed Stability:

For the aircraft configuration adopted,

xC.G = 0.3049805ĉ

ĉ = 4.67 m

,

= 0.042

= 0.4

H.T = 0.48

t = 0.9

For the wing, NACA 23011‐63 airfoil is selected. For this airfoil,

= 0.125 deg ‐1

xa.c = 0.265ĉ

For the tail, NACA 0012 airfoil is selected. For this airfoil,

= 0.1 deg ‐1

For the longitudinal stick‐fixed static stability of the aircraft, we have the expression

∂C∂CL

xC.G x .

ĉ∂C∂CL ,

a H.T ηH.Ta 1

∂ε∂α

62


= ‐0.1253795

The negative value of CCL

indicates that the airplane has

longitudinal stick‐fixed static stability.

Determination of Stick‐Fixed Neutral Point:

When the value of CCL

reduces to zero, the location of the

C.G is called the Neutral Point. From the above formula,

Nx .

ĉ∂C∂CL ,

a H.T ηH.Ta

1∂ε∂α

Static margin NxC.Gĉ

Using the above formula and data, we obtain

No = 0.43036

(xC.G)neutral = 2.0097812 m (from leading edge of wing at MAC section)

Static margin = 0.1253195

The locations of the aircraft C.G as well as static margins for various configurations of aircraft weight are tabulated below.

S.No WEIGHT CONFIGURATION C.G LOCATION

(%OF ĉ) STATIC MARGIN

1. Full payload + Full fuel 30 0.130360 2. Full payload + Reserve fuel 30.9961 0.120399 3. Zero payload + Reserve fuel 32.0094 0.1102664. Zero payload + Full fuel 30.2020 0.128340 5. Half payload + Full fuel 30.0715 0.129645 6. Half payload + Reserve fuel 31.3561 0.116799

63

Determination of the Zero‐Lift Pitching Moment:

The variation of Cm with CL is a linear variation, given by the relation

C C∂C∂CL

CL

For cruise fight conditions,

(CL)cruise = 0.38282

= ‐0.1253795

At trim condition, Cm=0


= 0.04799778

Determination of Tail Setting Angle:

C a H.T ηH.T i ε i

For the aircraft,

at = 0.1 deg ‐1

H.T = 0.48

H.T = 0.9

iw = 1°

ε2 CLπ AR


o = 0.6299°

it = 0.51884° upward

64

Longitudinal StickFixed Control:

For the aircraft, we assume the following values‐

τe = 0.4

H.T = 0.48

H.T = 0.9

The rate at which the pitching moment coefficient of the aircraft changes with change in elevator deflection is called the ELEVATOR CONTROL POWER.

∂C∂δ

a H.T ηH.T τ


=‐0.01728 deg‐1

Determination of the Extreme Elevator Deflections:

For calculating the extreme elevator deflections required for maintaining trim, we assume that when the airplane is flying level at design conditions, zero elevator deflection is required.

= 0

Elevator deflection has a linear variation, given by

δ δ

∂C∂CL

∂C∂δ

CL

In cruise flight, the minimum value of CL is encountered when the geometric angle of attack is zero.

= 0.146

65

= ‐0.13036

= 0.01903256

C C∂C∂δ

δ

Using the trim conditions (Cm=0) for both CL and cruise conditions,

δC C

∂C∂δ


= 1.676222801°

Hence,

Maximum DownElevator = 2.5°

In cruise flight, the maximum value of CL is encountered during landing.

= 2.13

= ‐0.110266

= 0.2348666

C C∂C∂δ

δ

Using the trim conditions (Cm=0) for both CL and cruise conditions,

δC C

∂C∂δ


= ‐10.8141678°

66

Hence,

Maximum UpElevator = 15°

Determination of Stick‐Fixed Stability Characteristics:

C a H.T ηH.T i ε i τ δ

Using the above formula, the variation of CCL

and C with elevator

deflection is tabulated below.

°

Location of C.GDesign Most Forward Most Rearward

‐15 ‐0.125 0.3071978 ‐0.130 0.3071978 ‐0.110 0.3071978‐12.5 ‐0.125 0.2639978 ‐0.130 0.2639978 ‐0.110 0.2639978‐10 ‐0.125 0.2207978 ‐0.130 0.2207978 ‐0.110 0.2207978‐7.5 ‐0.125 0.1775978 ‐0.130 0.1775978 ‐0.110 0.1775978‐5 ‐0.125 0.1343978 ‐0.130 0.1343978 ‐0.110 0.1343978‐2.5 ‐0.125 0.0911978 ‐0.130 0.0911978 ‐0.110 0.09119780 ‐0.125 0.0479978 ‐0.130 0.0479978 ‐0.110 0.04799782.5 ‐0.125 0.0047978 ‐0.130 0.0047978 ‐0.110 0.0047978

The variation of Cm with CL for various elevator deflections is tabulated and plotted below.

CL Elevator Deflection (°)

‐15 ‐12.5 ‐10 ‐7.5 ‐5 ‐2.5 0 2.5Cm

‐0.25 0.3385 0.2953 0.2521 0.2089 0.1657 0.1225 0.0793 0.03610 0.3072 0.2640 0.2208 0.1776 0.1344 0.0912 0.0480 0.0048

0.25 0.2759 0.2327 0.1895 0.1463 0.1031 0.0599 0.0167 ‐0.02650.5 0.2445 0.2013 0.1581 0.1149 0.0717 0.0285 ‐0.0147 ‐0.05790.75 0.2132 0.1700 0.1268 0.0836 0.0404 ‐0.0028 ‐0.0460 ‐0.08921 0.1818 0.1386 0.0954 0.0522 0.0090 ‐0.0342 ‐0.0774 ‐0.1206

1.25 0.1505 0.1073 0.0641 0.0209 ‐0.0223 ‐0.0655 ‐0.1087 ‐0.15191.5 0.1191 0.0759 0.0327 ‐0.0105 ‐0.0537 ‐0.0969 ‐0.1401 ‐0.18331.75 0.0878 0.0446 0.0014 ‐0.0418 ‐0.0850 ‐0.1282 ‐0.1714 ‐0.21462 0.0564 0.0132 ‐0.0300 ‐0.0732 ‐0.1164 ‐0.1596 ‐0.2028 ‐0.2460

2.25 0.0251 ‐0.0181 ‐0.0613 ‐0.1045 ‐0.1477 ‐0.1909 ‐0.2341 ‐0.2773

67

On varying the C.G location for design conditions, the value of C remains

constant but the slope of the curve CCL changes. The variation of Cm with

CL for different C.G locations is tabulated and plotted below.

CL

.

. .

.

.

0.3 ‐0.1304 0.305 ‐0.1254 0.31 ‐0.1204 0.315 ‐0.1154 0.3201 ‐0.1103

Cm‐0.25 0.080599778 0.079349778 0.078099778 0.076849778 0.075574778 0 0.047999778 0.047999778 0.047999778 0.047999778 0.047999778

0.25 0.015399778 0.016649778 0.017899778 0.019149778 0.0204247780.5 ‐0.01720022 ‐0.014700222 ‐0.012200222 ‐0.009700222 ‐0.007150222 0.75 ‐0.04980022 ‐0.046050222 ‐0.042300222 ‐0.038550222 ‐0.034725222 1 ‐0.08240022 ‐0.077400222 ‐0.072400222 ‐0.067400222 ‐0.062300222

1.25 ‐0.11500022 ‐0.108750222 ‐0.102500222 ‐0.096250222 ‐0.089875222 1.5 ‐0.14760022 ‐0.140100222 ‐0.132600222 ‐0.125100222 ‐0.117450222 1.75 ‐0.18020022 ‐0.171450222 ‐0.162700222 ‐0.153950222 ‐0.145025222 2 ‐0.21280022 ‐0.202800222 ‐0.192800222 ‐0.182800222 ‐0.172600222

2.25 ‐0.24540022 ‐0.234150222 ‐0.222900222 ‐0.211650222 ‐0.200175222

‐0.4

‐0.3

‐0.2

‐0.1

0.0

0.1

0.2

0.3

0.4

‐0.5 0 0.5 1 1.5 2 2.5

C m

CL

Stick‐Fixed Stability Characteristics at Design Conditions

‐15 deg

‐12.5 deg

‐10 deg

‐7.5 deg

‐5 deg

‐2.5 deg

0 deg

2.5 deg

Elevator Deflection

68

In order to find the variation of elevator deflection with the trim value of lift coefficient, we first evaluate the elevator deflection required at zero trim lift coefficient.

δC . a H.T ηH.T α i i

∂C∂δ

For the NACA 23011‐63 airfoil,

o = ‐1.2°

. = ‐0.009


= 3.682°

δ δ

∂C∂CL

∂C∂δ

CL

‐0.3

‐0.25

‐0.2

‐0.15

‐0.1

‐0.05

0

0.05

0.1

‐0.5 0 0.5 1 1.5 2 2.5

C mCL

Stability Characteristics with respect to Location of C.G

0.3c

0.305c

0.31c

0.315c

0.32009c

Location of C.G

69

Using the above formula, the variation of elevator deflection with the trim value of lift coefficient is tabulated and plotted below.

CL

.

.

.

.

.

0.3 ‐0.1304 0.305 ‐0.1254 0.31 ‐0.1204 0.315 ‐0.1154 0.32 ‐0.1103

Elevator Deflection (°)‐0.25 5.568574074 5.496236111 5.423898148 5.351560185 5.277775463 0 3.682 3.682 3.682 3.682 3.682

0.25 1.795425926 1.867763889 1.940101852 2.012439815 2.086224537 0.5 ‐0.09114814 0.053527778 0.198203704 0.34287963 0.4904490740.75 ‐1.97772222 ‐1.760708333 ‐1.543694444 ‐1.326680556 ‐1.105326389 1 ‐3.86429629 ‐3.574944444 ‐3.285592593 ‐2.996240741 ‐2.701101852

1.25 ‐5.75087037 ‐5.389180556 ‐5.027490741 ‐4.665800926 ‐4.2968773151.5 ‐7.63744444 ‐7.203416667 ‐6.769388889 ‐6.335361111 ‐5.892652778 1.75 ‐9.52401851 ‐9.017652778 ‐8.511287037 ‐8.004921296 ‐7.488428241 2 ‐11.4105925 ‐10.83188889 ‐10.25318519 ‐9.674481481 ‐9.084203704

2.25 ‐13.2971666 ‐12.646125 ‐11.99508333 ‐11.34404167 ‐10.67997917

‐15

‐10

‐5

0

5

10

‐0.5 0 0.5 1 1.5 2 2.5

e (°)

CL

Variation of Elevator Deflection Angle with Trim CL

0.3c

0.305c

0.31c

0.315c

0.32009c

Location of C.G

70

Determination of Stick‐Fixed Maneuver Point:

The stick‐fixed maneuver point is of vital importance in accelerated maneuvers of the aircraft. Here, the load factor plays a major role.

Elevator Angle Required per g:

The elevator angle required per unit load factor is used in determining the stick‐fixed maneuver point.

δ δ

∂C∂CL

∂C∂δ

CL63.03gl n 1

τ V

∂δ∂n

∂C∂CL

∂C∂δ

2WSρV

63.03glτ V

For the aircraft,

lt = 18.68 m

Using the above formula and data, the variation of elevator deflection required per g with the location of the C.G is plotted below.

‐40

‐30

‐20

‐10

0

10

20

30

0 20 40 60 80

Elevator Angle Req

uired pe

r g (°)

XC.G (% of chord)

Determination of Stick‐Fixed Maneuver Point

100 m/s

200 m/s

300 m/s

Velocity

71

The location of the C.G at which elevator deflection per g is zero is called the StickFixed Maneuver Point. From the graph,

Nm = 0.45058

(xC.G)maneuver = 2.1042086 m

Longitudinal StickFree Stability:

In this section, the control column of the aircraft is free. Hence, the effect of elevator hinge moments C in different situations is to be considered.

The additional factor that comes into picture when the stick is free is the Free Elevator Factor.

F 11a

∂CL∂δ

∂C∂α

∂C∂δ

For the aircraft, we assume

= ‐0.002 deg‐1

= ‐0.005 deg‐1


F = 0.84

For the stick free case, the stability criterion is the following equation.

∂C∂CL

xC.G x .

ĉ∂C∂CL ,

a H.T ηH.Ta 1

∂∂α F


= ‐0.0922019

72

The negative value of CCL

indicates that the airplane has

stick‐free longitudinal static stability.

Determination of Stick‐Free Neutral Point:

When the value of CCL

reduces to zero, the location of the

C.G is called the Neutral Point. From the above formula,

Nx .

ĉ∂C∂CL ,

a H.T ηH.Ta

1∂ε∂α

F


No’ = 0.3971824

No – No’ = 0.0331776

Longitudinal StickFree Control:

Determination of Stick‐Free Stability Characteristics:

The variation of CCL

and C with elevator deflection is tabulated

below.

Location of C.GDesign Most Forward Most Rearward

‐15 ‐0.092 0.3071978 ‐0.097 0.3071978 ‐0.077 0.3071978‐12.5 ‐0.092 0.2639978 ‐0.097 0.2639978 ‐0.077 0.2639978‐10 ‐0.092 0.2207978 ‐0.097 0.2207978 ‐0.077 0.2207978‐7.5 ‐0.092 0.1775978 ‐0.097 0.1775978 ‐0.077 0.1775978‐5 ‐0.092 0.1343978 ‐0.097 0.1343978 ‐0.077 0.1343978‐2.5 ‐0.092 0.0911978 ‐0.097 0.0911978 ‐0.077 0.09119780 ‐0.092 0.0479978 ‐0.097 0.0479978 ‐0.077 0.04799782.5 ‐0.092 0.0047978 ‐0.097 0.0047978 ‐0.077 0.0047978

73

The variation of Cm with CL for various elevator deflections is tabulated and plotted below.

CL Elevator Deflection (°)

‐15 ‐12.5 ‐10 ‐7.5 ‐5 ‐2.5 0 2.5

Cm‐0.25 0.3302 0.2870 0.2438 0.2006 0.1574 0.1142 0.0710 0.02780 0.3072 0.2640 0.2208 0.1776 0.1344 0.0912 0.0480 0.0048

0.25 0.2841 0.2409 0.1977 0.1545 0.1113 0.0681 0.0249 ‐0.0183 0.5 0.2611 0.2179 0.1747 0.1315 0.0883 0.0451 0.0019 ‐0.04130.75 0.2380 0.1948 0.1516 0.1084 0.0652 0.0220 ‐0.0212 ‐0.0644 1 0.2150 0.1718 0.1286 0.0854 0.0422 ‐0.0010 ‐0.0442 ‐0.0874

1.25 0.1919 0.1487 0.1055 0.0623 0.0191 ‐0.0241 ‐0.0673 ‐0.1105 1.5 0.1689 0.1257 0.0825 0.0393 ‐0.0039 ‐0.0471 ‐0.0903 ‐0.13351.75 0.1458 0.1026 0.0594 0.0162 ‐0.0270 ‐0.0702 ‐0.1134 ‐0.1566 2 0.1228 0.0796 0.0364 ‐0.0068 ‐0.0500 ‐0.0932 ‐0.1364 ‐0.1796

2.25 0.0997 0.0565 0.0133 ‐0.0299 ‐0.0731 ‐0.1163 ‐0.1595 ‐0.2027

‐0.3000

‐0.2000

‐0.1000

0.0000

0.1000

0.2000

0.3000

0.4000

‐0.5 0 0.5 1 1.5 2 2.5

C m

CL

Stick‐Free Stability Characteristics at Design Conditions

‐15 deg

‐12.5 deg

‐10 deg

‐7.5 deg

‐5 deg

‐2.5 deg

0 deg

2.5 deg

Elevator deflection

74

Stick Force Gradients:

In this section, an analysis is done to determine the stick force gradients in non‐accelerated flight.

For non‐accelerated flight,

FSKρV2

A∂C∂δ

δ KWS

∂C∂δ∂C∂δ

∂C∂CL

∂FS∂V

KρV A∂C∂δ

δ

A C∂C∂α

α i i∂C∂δ

δ

K GS c η

For the aircraft, we assume

= 0

= ‐0.012 deg‐1


A = ‐0.01504768

K = ‐215.547552 m2

∂C∂δ

δ2WS

ρV

∂C∂δ∂C∂δ

∂C∂CL

A

Assuming that the airplane needs to be trimmed out at cruise velocity,

= 4.834505289*10‐3

Using the above data,

75

( t)tri m = ‐0.402875°

= 203.6259609 Ns/m

Using the above data, the variation of stick force with velocity at cruise condition is plotted below.

Determination of Stick‐Free Maneuver Point:

Stick Force Required per g:

The stick force required per unit load factor is used in determining the stick‐free maneuver point.

∂F∂n

Gη S cWS

∂C∂δ

∂C∂CL∂C∂δ

63.03ρgl2

∂C∂α

∂C∂δτ

For the aircraft,

G = 2 rad/m

‐30000

‐25000

‐20000

‐15000

‐10000

‐5000

0

5000

10000

0 50 100 150 200 250 300

Stick Force (N)

Velocity (m/s)

Variation of Stick Force with Velocity

76

= 114.592°/m

Se = 4.18 m2

ce = 0.5 m

Using the above formula and data, the variation of stick force required per g with the location of the C.G is plotted below.

The location of the C.G at which stick force required per g is zero is called the StickFree Maneuver Point. From the graph,

Nm’ = 0.41413

‐80000

‐60000

‐40000

‐20000

0

20000

40000

60000

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

F s/n

Location of C.G (% of chord)

Determination of Stick‐Free Maneuver Point

77

Directional Stability:

Directional stability of the airplane is a measure of its tendency to produce restoring moments when disturbed from an equilibrium angle of sideslip – usually taken as zero. It is measured quantitatively by the variation of yawing moment coefficient with sideslip angle.

CN

q S b

In general, C should be negative for the airplane to have static

directional stability. All the components of the aircraft contribute to the

stability coefficient C .

Contribution from Wing:

The wing contribution to directional stability is quite small, as the cross wind effects on the wing are very small. The critical factor is the sweepback of the wing.

∂C∂ψ

0.00006 √Ω

For the aircraft,

= 32°


= ‐0.000339411 deg‐1

Contribution from Fuselage and Nacelle:

The contribution from the fuselage and nacelle is generally destabilizing.

∂C∂ψ ,

0.96k57.3

SS

Lb

hh

ww

78

where k : Empirical constant from graph

Sf : Projected area of fuselage in m2

h1, w1 : Height and width of fuselage at Lf/4 in m

h2, w2 : Height and width of fuselage at 3Lf/4 in m

From the graph of k vs (d/Lf) for varying fuselage fineness ratio,

k = 0.125

For the aircraft,

Sf = 135.149 m2

h1=h2 = 3.95 m

w1=w2 = 3.95 m


,

= 0.00202715 deg‐1

Since the wing configuration is low‐wing,

= 0

Contribution of Vertical Tail:

The vertical tail is the stabilizing component in the aircraft as far as directional stability is concerned.

∂C∂ψ

a η 1∂σ∂ψ

∂C∂ψ

∂C∂ψ V.T

Δ∂C∂ψ

For the vertical tail,

79

av = 0.1 deg‐1

v = 0.05

v = 0.9


. = ‐0.0045

= ‐0.0003 (from graph)

= ‐0.0048 deg‐1

∂C∂ψ

∂C∂ψ

∂C∂ψ ,

∂C∂ψ

∂C∂ψ


= ‐0.002212261 deg‐1

Directional Control:

Determination of Directional Stability Characteristics:

C C∂C∂ψ ψ

C a η τ δ

80

For the ruder, we assume

τr = 0.4

Using the above formulae and data, the variation of C with rudder deflection is tabulated below.

r ±30° 0.054±25° 0.045±20° 0.036±15° 0.027±10° 0.018±5° 0.0090° 0

The variation of Cn with ψ for various rudder deflections is tabulated and plotted below.

Ψ (°) Rudder Deflection (°)

‐30 ‐20 ‐10 0 10 20 30

Cn‐25 0.1093 0.0913 0.0733 0.0553 0.0373 0.0193 0.0013 ‐20 0.0982 0.0802 0.0622 0.0442 0.0262 0.0082 ‐0.0098 ‐15 0.0872 0.0692 0.0512 0.0332 0.0152 ‐0.0028 ‐0.0208‐10 0.0761 0.0581 0.0401 0.0221 0.0041 ‐0.0139 ‐0.0319 ‐5 0.0651 0.0471 0.0291 0.0111 ‐0.0069 ‐0.0249 ‐0.0429 0 0.0540 0.0360 0.0180 0.0000 ‐0.0180 ‐0.0360 ‐0.05405 0.0429 0.0249 0.0069 ‐0.0111 ‐0.0291 ‐0.0471 ‐0.0651 10 0.0319 0.0139 ‐0.0041 ‐0.0221 ‐0.0401 ‐0.0581 ‐0.0761 15 0.0208 0.0028 ‐0.0152 ‐0.0332 ‐0.0512 ‐0.0692 ‐0.0872 20 0.0098 ‐0.0082 ‐0.0262 ‐0.0442 ‐0.0622 ‐0.0802 ‐0.098225 ‐0.0013 ‐0.0193 ‐0.0373 ‐0.0553 ‐0.0733 ‐0.0913 ‐0.1093

81

The rate at which the yawing moment coefficient of the aircraft changes with change in rudder deflection is called the RUDDER CONTROL POWER.

∂C∂δ

a η τ


=‐0.0018 deg‐1

Determination of Asymmetric Power Condition Control Characteristics:

The basic configuration of the aircraft is twin tractor engine. When one engine is inoperative (OEI), we obtain a case of asymmetric thrust or power. The control of the aircraft under this condition with the help of the rudder is investigated in this section.

The equation of moment equilibrium in yaw in case of OEI is as follows‐

‐0.1500

‐0.1000

‐0.0500

0.0000

0.0500

0.1000

0.1500

‐30 ‐20 ‐10 0 10 20 30

Cn

ψ(°)

Directional Stability Characterictics

‐30 deg

‐20 deg

‐10 deg

0 deg

10 deg

20 deg

30 deg

Rudder deflection

82

Ty2

∂C∂δ

δ12ρV Sb 0

The value of C δ in this expression is called the asymmetric‐thrust

yaw coefficient C T.

From the above equation,

C T

T y2

12 ρV Sb

For the aircraft,

ye = 15 m

For take‐off conditions,

VT.O = 64.58769147 m/s

TT.O = 85616.85605 N


( )takeoff = 0.0475390381

r = ‐26.411°

For cruise conditions,

Vcruise = 254.167 m/s

Tcruise = 85616.85605 N

cruise = 0.363924 kg/m3


( )cruise = 0.01033839708

r = ‐5.743554°

For full rudder deflection,

( )full rudder = 0.054

83

The variation of asymmetric‐thrust yawing moment coefficient with velocity at sea level is tabulated and plotted below.

Velocity (m/s) (Cn)T (Cn)Full Rudder 40 0.123945955 0.054 80 0.030986489 0.054 120 0.013771773 0.054 160 0.007746622 0.054 200 0.004957838 0.054 240 0.003442943 0.054 280 0.002529509 0.054

The velocity at which the yaw coefficient due to full rudder deflection equals the yaw coefficient due to asymmetric thrust is called the critical velocity for rudder control, or the unstick speed. From the graph,

Vcritical = 60.601 m/s

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0 100 200 300

C n

Velocity (m/s)

Variation of Cn with Velocity

Yaw Moment Coefficient Required

Full Rudder Yaw Moment Coefficient

84

Determination of Cross Wind Condition Control Characteristics:

When the airplane is in take‐off phase, a cross wind can induce a yawing moment. The control of the aircraft under this condition with the help of the rudder is investigated in this section.

According to FAR standards, for design we assume

Vcrosswind = 7.5 m/s

The sideslip due to cross wind is given by

β sinV

VT.O


= 6.6683°

To maintain sideslip, we have

δ

∂C∂β∂C∂δ

β


r = ‐8.1956°

85

Lateral Stability:

When a small vertical disturbance causes the aircraft to roll to one side, as such, the airplane will continue to roll at the same constant velocity. As such, the airplane is neutrally stable in roll. However, due to the development of sideslip, the lift distribution over the wings is altered, tending to produce restoring moments which restore the aircraft to its original state. This effect is generally called the Dihedral Effect. Lateral stability of the airplane is a measure of this tendency to produce restoring moments when disturbed in roll. It is measured quantitatively by the variation of rolling moment coefficient with sideslip angle.

CL

q S b

In general, C should be positive for the airplane to have static lateral

stability. All the components of the aircraft contribute to the stability

coefficient C .

Contribution from Wing:

The wing dihedral angle has a linear variation with the stability coefficient. There is also an additional component due to the tip shape. The stability coefficient is given by the following equation.

∂C∂ψ

0.0002 Γ Δ∂C∂ψ

Δ∂C∂ψ

where Γ : Dihedral angle in degrees

For the aircraft,

Γ = 3.5°

The tip shape used is shown below.

86

For this tip shape,

= 0.0002 deg‐1

Δ∂C∂ψ

0.5∂C∂ψ


= 0.0001697055 deg‐1


= 0.001069705 deg‐1


= ‐0.0008 deg‐1

Contribution of Vertical Tail:

The vertical tail is stabilizing as far as directional stability is concerned.

∂C∂ψ

a ηZVlV

For the vertical tail,

av = 0.1 deg‐1

v = 0.05

v = 0.9

ZV = 4.78 m

lV = 17.75 m

87


= 0.00121183


= 0.00016 deg‐1

∂C∂ψ

∂C∂ψ

∂C∂ψ

∂C∂ψ


= 0.001641535

Effective dihedral

∂C∂ψ0.0002


Effective dihedral = 8.207675°

Lateral Control:

The angular velocity, with which the airplane rolls under the combined effect of aileron deployment and dihedral effect, is obtained using the strip integration technique.

p τ Vcy dy⁄

⁄

cy dy⁄

The span wise variation for the wing chord is given by

c c λ2yb λ 1

where cr : root chord in m

88

: wing taper ratio cr/ct

y : span wise location of strip

Using the above relation, the aileron rolling power is given by the relation

pb2V 2τ

2 1 λ k k 3λ k kλ 3 δ

For the airplane,

τa = 0.5

k1 = 0.78

k2 = 0.84

= 3.3333333


= 0.06641431579

p 0.06641431579 δ2Vb

Using the above formula, the variation of rolling angular velocity with velocity for different aileron deflections is plotted and tabulated below.

89

Velocity (m/s) Aileron Deflection (°)

5 7.5 10

Rolling Angular Velocity (rad/s) 100 0.033666816 0.050500224 0.067333632 110 0.037033498 0.055550247 0.074066996 120 0.040400179 0.060600269 0.080800359 130 0.043766861 0.065650292 0.087533722 140 0.047133543 0.070700314 0.094267085 150 0.050500224 0.075750336 0.101000449 160 0.053866906 0.080800359 0.107733812 170 0.057233588 0.085850381 0.114467175 180 0.060600269 0.090900404 0.121200538 190 0.063966951 0.095950426 0.127933901 200 0.067333632 0.101000449 0.134667265 210 0.070700314 0.106050471 0.141400628 220 0.074066996 0.111100493 0.148133991 230 0.077433677 0.116150516 0.154867354 240 0.080800359 0.121200538 0.161600718 250 0.08416704 0.126250561 0.168334081 260 0.087533722 0.131300583 0.175067444

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

0 50 100 150 200 250 300

Rolling

Angular Velocity (rad

/s)

Velocity (m/s)

Variation of Rolling Rate with Velocity

5 deg

7.5 deg

10 deg

AileronDeflection

90

DYNAMIC STABILITY AND CONTROL OF THE AIRCRAFT:

Estimation of Moments of Inertia of the Aircraft:

Axis Definition:

For the aircraft,

XC.G = 18.21153 m (from nose)

YC.G = 0 (Plane of symmetry)

ZC.G = 1.692 m (gear down)

= 1.971 m (gear up) (from the lowest point on the fuselage)

S.No Component Weight

W MassM X Y Z

IXX= MX2

IYY= MY2

IZZ=MZ2

Unit (*103) N

(*103) kg m m m

(*106) kg‐m2

(*106) kg‐m2

(*106) kg‐m2

1 Fixed Equipment 27.5918 2.8126 ‐16.21 0.00 1.24 0.0043 0.7435 0.73922 Pilots 2.4000 0.2446 ‐15.41 0.00 1.60 0.0006 0.0587 0.05813 Crew + Passengers 132.8000 13.5372 ‐3.10 0.00 1.60 0.0347 0.1649 0.13024 Cargo 32.4000 3.3028 12.79 0.00 0.20 0.0001 0.5403 0.54015 Fuselage Structure 68.9796 7.0316 0.13 0.00 0.88 0.0055 0.0056 0.00016 Lavatories 55.1837 5.6252 ‐0.49 0.00 1.78 0.0178 0.0192 0.00147 Nose Gear 6.8980 0.7032 ‐13.74 0.00 ‐3.11 0.0068 0.1396 0.13288 Horizontal Tail 20.6939 2.1095 17.99 0.00 2.03 0.0087 0.6913 0.68269 Vertical Tail 13.7959 1.4063 16.56 0.00 4.78 0.0321 0.4177 0.385610 Wing Structure 68.9796 7.0316 0.90 7.70 ‐0.80 0.4216 0.0102 0.422811 Fuel 108.1313 11.0226 ‐0.19 6.23 ‐0.80 0.4349 0.0074 0.428212 Engine 46.7409 4.7646 ‐1.75 7.50 ‐2.52 0.2982 0.0448 0.282613 Main Landing Gear 27.5918 2.8126 1.26 4.08 ‐3.11 0.0742 0.0317 0.0514

TOTAL 1.3395 2.8749 3.8551

91

Multiplying by a factor of 2 to account for the moments about the respective centroidal axes, we have

IXX = 2.6790*106 kg‐m2

IYY = 5.7498*106 kg‐m2

IZZ = 7.7102*106 kg‐m2

92

Longitudinal Dynamic Stability:

In this section, we analyze the response of the aircraft at design conditions after a disturbance in pitch, with all controls locked.

We define some parameters which are used frequently in the dynamic stability analysis.

Time Constant τ =

= 4.959229016 s

Relative Density Factor μ =

= 269.9080746

Radius of Gyration KY =

= 9.04275 m

h =

= 0.02778314996

Evaluation of Stability Derivatives:

= 7.161972439 rad‐1

=

= 0.3895554465 rad‐1

=

= ‐0.8979645234 rad‐1

=

93

= ‐0.01467271509 rad‐1

= .

= ‐0.04034996651 rad‐1

= at tτ

= ‐0.99007107 rad‐1

=

=

Here, A, B, C and D are evaluated from a graph given in NACA TR 709.

For a chordwise hinge location at 80% from leading edge,

A = 0.46

B = 0.18

C = 0.82

D = 0.04


= ‐6.051094702*10‐4 rad‐1

= ‐8.868945584*10‐4 rad‐1

=

= ‐1.2*10‐3 rad‐1

= ‐0.005 rad‐1

94

=

= ‐2.9639721*10‐5 rad‐1

Analysis of Longitudinal Dynamic Stability:

The longitudinal stability quartic is given by

Aλ Bλ Cλ Dλ E 0

A = 1

B =

= 5.586004056

C =

=37.67979774

D =

= 0.9598973397

E =

= 2.369153982

Using the above values, the stability quartic is given as follows‐

λ 5.586004056λ 37.67979774λ 0.9598973397λ 2.3691539820

95

Routh’s Discriminant R = BCDAD2B2E

= 127.1865152

The following inferences are made from the above equation and data‐

• Since all the coefficients in the stability quartic are positive, there is no possibility for pure divergence.

• Since the Routh’s discriminant is positive, all oscillations are convergent.

The roots of the stability quartic are:

1,2 = ‐2.7849 ± 5.4563i

3,4 = ‐0.0081 ± 0.2511i

Modes of Dynamic Response:

1. Phugoid or Long Period Mode:

This is an oscillatory mode with long period and weak damping. The root for this mode is ‐0.0081 ± 0.2511i

Time period T =

= 124.093 s

Damping Factor ζ = 0.032241294

From the above data, the time period of oscillations is very high, and the damping is very low. This indicates that no extra controls are required for dynamic stability.

2. Short Period Mode:

This is an oscillatory mode with short period and heavy damping. The root for this mode is ‐2.7849 ± 5.4563i

Time period T =

96

= 5.71078476 s


From the above data, the time period is very low, and the damping is high. This indicates that the pilot is unaware of any effects of this motion. Hence no extra damping needs to be provided.

97

DirectionalLateral Dynamic Stability:

In this section, we analyze the response of the aircraft at design conditions after a disturbance in coupled yaw and roll, with all controls locked.

Radius of Gyration KX =

= 6.17249 m

Radius of Gyration KZ =

= 10.47145 m

Time Constant τ =

= 4.959229016 s

Relative Density Factor μ =

= 36.60966333

JX =2

=0.06428012909

JZ =2

=0.1849990921

Evaluation of Stability Derivatives:

= ‐0.28 rad‐1

= 0.1267532185 rad‐1

98

= ‐0.09405302742 rad‐1

= ‐0.48 rad‐1 (from graph)

=

= ‐0.313374328 rad‐1

=

= 0.09570500145 rad‐1

=

= ‐0.04785250073 rad‐1

Analysis of Directional‐Lateral Dynamic Stability:

The directional‐lateral stability quartic is given by

Aλ Bλ Cλ Dλ E 0

A = 1

B =

= 4.689515363

C =

=28.98315814

99

D =

= 105.3046303

E = = 2.084736967

Using the above values, the stability quartic is given as follows‐

λ 4.689515363λ 28.98315814λ 105.3046303λ2.084736967 0

Routh’s Discriminant R = BCDAD2B2E

= 3177.77402

The following inferences are made from the above equation and data‐

• Since all the coefficients in the stability quartic are positive, there is no possibility for pure divergence.

• Since the Routh’s discriminant is positive, all oscillations are convergent.

The roots of the stability quartic are:

1 = ‐0.0199

2 = ‐3.997

3,4 = ‐0.3363 ± 5.1077i

Modes of Dynamic Response:

1. Convergent Modes:

These are aperiodic modes which are purely convergent. The roots for these modes are ‐0.0199 and ‐3.997. The first mode is mildly convergent while the second mode is heavily convergent.

100

2. Dutch Roll Mode:

This is an oscillatory mode with short period and moderate damping. The root for this mode is ‐0.3363 ± 5.1077i

Time period T =

= 6.10054 s


From the above data, the time period is low, and the damping is moderate. In the stick‐fixed case, the case is tolerable as the oscillation damps out quite quickly, but in the stick‐free case, the damping is very weak, resulting in prolonged oscillations. This indicates that aileron and rudder balancing is required. Balancing is provided using tabs.

Under certain cases, the first real root can be positive indicating mild divergence. This is called the Spiral Divergence mode.

The stability boundaries for pure divergence (spiral) and divergent oscillations are obtained using the following conditions‐

• E=0 • R=0

Using the first condition, we have

∂C∂β

∂C∂r

∂C∂β

∂C∂r

0

Using the condition C =

CL

we have

∂C∂β

4 ∂C∂β∂C∂r

CL

From the above condition, we see that spiral divergence occurs for the following condition

101

CL 0.80274

Using the second condition, we have

BCD AD B E 0

Using the above formulae, the directional‐lateral stability boundaries are obtained and plotted below.

‐10

0

10

20

30

40

50

60

70

‐20‐15‐10‐50

µCn β

µCl β

Stick‐Fixed Lateral Stability Boundaries

Divergence Boundary

Oscillatory Boundary

102

ESTIMATION OF LOAD FACTOR LIMITS AND Vn DIAGRAM

In accelerated flight, the lift becomes much more compared to the weight of the aircraft. This implies a net force contributing to the acceleration. This force causes stresses on the aircraft structure. The ratio of the lift experienced to the weight at any instant is defined as the Load Factor.

nρ V CL2 W

S

In this section, we estimate the aerodynamic limits on load factor, and attempt to draw the variation of load factor with velocity, commonly known as the Vn Diagram. The V‐n diagram is drawn for Sea Level Standard conditions.

Using the above formula, we infer that load factor has a quadratic variation with velocity. However, this is true only up to a certain velocity. This velocity is determined by simultaneously imposing limiting conditions aerodynamically ((CL)max) as well as structurally (nmax). This velocity is called the Corner Velocity, and is determined using the following formula.

V2 n WS ρ CL

(CL)max is a property of the airfoil selected. For the NACA 23011‐63,

1.495

nmax is determined using maximum conditions of (L/D) and (T/W).

nTW

LD


nmax = 4.776963

104

Vcorner = 153.1800549 m/s

In the V‐n diagram, a horizontal line is drawn at this velocity. This load factor is a limit load factor, beyond which structural damage occurs to aircraft components if load factor is exceeded. The plot is extended upto the maximum cruise velocity possible for the thrust and wing loadings of the aircraft. Both these speeds are greater than the design cruise speed of the aircraft.

V ,

TAW

WS

WS

TAW 4KCD

ρ CD


Vmaximum, cruise = 394 m/s

We also define a new speed called the dive speed, defined as 1.5 times the maximum cruise velocity. Using the above standard,

Vdive = 591 m/s

At this speed, the positive load factor is assumed to be 3.5.

The V‐n diagram is also extended for negative load factors, i.e. when the aircraft is in dives. For this case,

= ‐0.9

nmax = ‐1.910785201

Vcorner = 124.8623655

In order to calculate the gust load factors, the following FAR standards are used.

For velocities up to Vmaximum, cruise, a gust velocity of 50 ft/s at sea level is assumed. For Vdive, a gust velocity of 66 ft/s is assumed.

105

n 1K U VE a

498 WS

Gust Alleviation Factor K0.88µ5.3 µ

Airplane Mass Ratio µ2 W

Sρ ĉ a g

where Uge : equivalent gust velocity (ft/s)

ρ : mean geometric chord (ft)

g : acceleration due to gravity (ft/s2)

VE : airplane equivalent speed (knots)

a : wing lift curve slope (rad‐1)

For the aircraft at sea level,

= 0.002377 slug/ft3

W/S = 93.9845 lb/ft2

ĉ = 15.3215 ft

g = 32.185 ft/s2

a = 7.162 rad‐1

μ = 22.3431

Kg = 0.711278

Normal gust equivalent velocity is 50ft/s from sea level up to 20000 ft, reducing linearly to 25 ft/s at 50000 ft. The rough gust velocity is 66 ft/s from sea level up to 20000 ft, reducing linearly to 38 ft/s at 50000 ft.

Since the dive velocity is rarely achieved, we assume the gust velocity to be half the normal value at this speed.

The variation of gust velocity with altitude is plotted below.

106

Using the above obtained values, the coordinates of various points in the V‐n diagram maneuvering envelope and gust envelope for the aircraft at sea level are computed and tabulated, and the V‐n diagrams are drawn.

Point Load Factor E.A.S (m/s) Maneuvering Envelope

A 4.776963 153.1800549 C 4.776963 394 D 3.5 591 E 0 591 F ‐1.910785201 394 G ‐1.910785201 124.86235655

Gust Envelope B’ 3.64736 112.13632 C’ 5.10225 394 D’ 4.07669 591 E’ ‐2.07669 591 F’ ‐3.10225 394 G’ ‐1.64736 112.13632

0

5

10

15

20

25

0 5000 10000 15000 20000

Gust V

elocity (m

/s)

Altitude (m)

Variation of Gust Velocity with Altitude

Normal Gust

Rough Air Gust

107

‐3

‐2

‐1

0

1

2

3

4

5

6

0 100 200 300 400 500 600

Load

Factor

Velocity (m/s)

V ‐ n Diagram ‐Maneuvering Envelope

VS VA

VC

A

D

E

G

+ nmax

‐ nmax

C

S

F

‐4

‐3

‐2

‐1

0

1

2

3

4

5

6

0 100 200 300 400 500 600

Load

Factor

Velocity (m/s)

V ‐ n Diagram ‐ Gust Envelope

B'

C'D'

E'

F'

G'

108

LIFT DISTRIBUTION FOR THE AIRCRAFT – SCHRENCK’S CURVE

While performing a structural analysis of the aircraft, it is of essence that we investigate the effects of various loads acting on different parts of the aircraft. As such, the lift force is the most important force acting on the aircraft – specifically the wing – and the lift distribution on the wing is critical in determining several factors such as the shear force and bending moment distribution along the span of the wing, the torque distribution on the wing and so on.

The Schrenck’s curve is used to approximate the lift distribution along the span of the wing. In obtaining the Schrenck’s curve, we assume that the actual lift distribution can be approximated by considering the average of two types of distributions – one is a trapezoidal lift distribution, in which, the lift per unit span at any section of the wing is assumed to be proportional to the chord of the wing at that section; the other is an elliptic lift distribution, which represents an ideal case of maximum wing planform efficiency (Ostwald planform efficiency, e=1).

Trapezoidal Lift Distribution:

In this distribution, we assume section lift per unit span to be proportional to the section wing chord. As such,

Lift per unit span,

2

This, in turn, reduces to

2

At cruise conditions,

CL = 0.3825

ρ = 0.363924 kg/m3

109

V = 915 km/hr

The distribution assumes the shape of a trapezium with the ends proportional to the root and tip chords of the wing. At wing tip,

ct = 1.9652 m

L’t = 8836.00765 N/m

Similarly,

cr = 6.5508 m

L’r = 29453.9583 N/m

Knowing the values of L’ at the wing tip and root, we obtain the equation of the trapezoidal lift distribution of the wing:

4496.23837 1.9652 0.25475555556 0 ≤ x ≤ 18

4496.23837 1.9652 0.25475555556 36

18 ≤ x ≤ 36

The trapezoidal lift distribution thus obtained is shown below:

0

5000

10000

15000

20000

25000

30000

35000

0 5 10 15 20 25 30 35 40

Lift per Unit S

pan (N/m

)

Distance along span from wingtip (m)

Lift Distribution ‐ Trapezoidal

110

Elliptic Lift Distribution:

This lift distribution assumes an ideal and efficient use of the wing area for lift generation. The wing area required to generate a required lift is found to be minimum when the planform is elliptic in shape.

In this distribution, we assume the lift per unit span to vary in a semi‐elliptic shape, with the major axis of the semi ellipse coinciding with the span of the wing. The area enclosed between the lift distribution curve and the span axis is equivalent to the total lift, which in turn is equal to the weight of the aircraft. Using this, we obtain the length of the semi‐minor axis of the semi‐ellipse. Thus, the elliptic lift distribution is obtained.

For the aircraft,

W = 689795.9184 N

Span, b = 36 m = 2A

From 0.5*π*AB = W, we have

B = 24396.54003

Using the above values, we have the equation of the elliptic lift distribution as follows:

24396.54003 1

The elliptic lift distribution thus obtained is shown below:

0

5000

10000

15000

20000

25000

30000

0 5 10 15 20 25 30 35 40Lift per Unit S

pan (N/m

)


Lift Distribution ‐ Elliptic

111

Schrenck’s Curve:

Schrenck’s curve is an approximation for the lift distribution along the span for the wing. The equation of the curve is obtained by taking the average of the trapezoidal and elliptic lift distributions. Thus, the equation for Schrenck’s curve is given as follows—

0.5 4496.23837 1.9652 0.25475555556 24396.54003

1 0 ≤ x ≤ 18

0.5 4496.23837 1.9652 0.25475555556 36

24396.54003 1 18 ≤ x ≤ 36

In all the above expressions, the origin (x=0) is assumed to lie at the tip of the port wing.

The Schrenck’s curve for the wing is shown below, along with a comparison with the trapezoidal and elliptic distributions.

0

5000

10000

15000

20000

25000

30000

35000

0 10 20 30 40

Lift per Unit S

pan (N/m

)


Lift Distribution ‐ Schrenck's Curve

Trapezoidal Lift Distribution

Elliptic Lift Distribution

Screnck's Curve

112

SHEAR FORCE AND BENDING MOMENT DIAGRAMS FOR THE WING

Generally, the shear force and bending moment variation along the span of the wing are of significance. The wing has the spars and ribs as major structural elements, along with stringers. Since the wing is subjected to bending as a result of all the forces acting on it, the shear force and bending moment variations help us to predict structurally critical locations on the wing, and accordingly provide sufficient factor of safety.

Shear Force and Bending Moment Normal to Chord:

In order to determine the shear force and bending moment distributions normal to the chord along the wingspan, we first determine the normal load distribution along the span. The major loads acting on the wing are as follows:

Lift:

Lift contributes the most to the net loading on the wing. Lift force acts in a direction perpendicular to the relative wind direction. For small angles of attack, we can approximate the lift to act perpendicular to the wing chord. The lift distribution over the wingspan is approximated by the Schrenck’s curve over the wing. For the aircraft, the wingspan is 18 m. Assuming the origin to lie at one of the wingtips (for this entire stage of estimation, the equation for the Schrenck’s curve is given below:

0.5 4496.23837 1.9652 0.25475555556 24396.54003

1 0 ≤ x ≤ 18

0.5 4496.23837 1.9652 0.25475555556 36

24396.54003 1 18 ≤ x ≤ 36

where x : distance along the wingspan

113

Wing Structural Weight:

For this preliminary analysis, the structural weight of the wing is assumed to vary parabolically along the span, with zero weight per unit span at the wingtips.

Again, the area enclosed between the weight distribution curve and the span axis should be equal to wing structural weight. With this consideration, we assume the following weight distribution:

where x : distance along wingspan

Satisfying the above condition, we have

3

For the aircraft, Ww = 34489.796 N

Using the above data, we get

k = 17.741664609

17.741664609

Engine Weight:

The engine is mounted on the bottom of the wing through pylons. Hence, the engine weight can be assumed to be a point load acting on the wing. For the aircraft, the engine weight is 5205 lbs (23157.045 N) and engine span is 15 m. Hence, the above point loads act at 10.5 m from the wingtips.

Landing Gear Weight:

In the aircraft, the main landing gear is retracted into the wing. Hence, at all times, there exists a point load corresponding to the gear. However, the weight on each wing due to landing gear is only 6897.95 N, and moreover, this load is a relieving load as far as normal shear force and bending moment are concerned. Due to these factors, the contribution of the landing gear is neglected in this section.

114

Fuel Weight:

Fuel is accommodated in wing‐mounted fuel tanks. In the fuel‐occupied region, the fuel weight can be assumed to have an elliptic distribution. Once again, the fuel weight on each wing is only 54065.652 N, and this is a relieving load. Hence, this load is also neglected in this section.

Net Load Distribution:

The net load distribution is found by considering the contribution of both lift and structural weight. Thus we obtain the expression for normal load distribution along the span for one wing as follows:

4418.003826 572.7208526 24396.54003 1

17.7417665 0 ≤ x ≤ 18

Shear Force:

The shear force variation is obtained using the following expression:

The contribution of engine weight is considered in the required range. Using the above expression, the normal shear force distribution is obtained as follows:

4418.003826 286.3604263 677.115298 √36

162 sin 81 5.913888203 0 ≤ x < 10.5

4418.003826 286.3604263 677.115298 √36

162 sin 81 5.913888203 23157.045 10.5 ≤ x ≤ 18

115

Bending Moment:

The bending moment variation is obtained using the following expression:

Using the above expression, the normal bending moment variation is obtained as follows:

2209.001913 95.45327543 677.115298 81

√ 2916 sin cos sin

1.478472051 0 ≤ x < 10.5

2209.001913 95.45327543 677.115298 81

√ 2916 sin cos sin

1.478472051 23157.045 10.5 10.5 ≤ x ≤ 18

Using the above expressions, the variation of shear force and bending moment normal to the chord along the wingspan is obtained and plotted below:

116

0

50000

100000

150000

200000

250000

300000

350000

0 5 10 15 20 25 30 35 40

Normal She

ar Force (N

)

Distance along wingspan from wingtip (m)

Normal Shear Force variation along wingspan

0

500000

1000000

1500000

2000000

2500000

0 5 10 15 20 25 30 35 40

Normal Ben

ding

Mom

emnt (N

‐m)


Normal Bending Moment variation along wingspan

117

Shear Force and Bending Moment Parallel to Chord:

In order to determine the variation of shear force and bending moment in the chordwise direction, we first resolve the aerodynamic loads on the wing into components normal to and in the direction of the chord.

From the above diagram, we see that the normal force coefficient is given by

cos sin

where α : angle of attack

From the above expression, we see that since the angle of attack is small for steady, level flight and CL >> CD, it is a justifiable approximation to take only the variation of CL in the trapezoidal distribution in the previous section.

However, in the chordwise direction, the chordwise force component needs to be used, given by

cos sin

This component is small, yet significant, and must be computed and used in determining the trapezoidal force distribution in the chordwise direction.

Using the above equation,

Cc = 0.1086853395

Hence the trapezoidal variation of force per unit span in the chordwise direction is given by

2

Again assuming cruise conditions for the aircraft,

ρ = 0.363924 kg/m3

V = 915 km/hr

Assuming the origin to lie at one of the wingtips, we get the following variation:

118

1277.582207 1.9652 0.25475555556 0 ≤ x ≤ 18

1277.582207 1.9652 0.25475555556 36

18 ≤ x ≤ 36

Shear Force:

Once again we use the expression

and obtain the following chordwise shear force variation along the span

1277.582207 1.9652 0.127377778 0 ≤ x ≤ 18

1277.582207 1.9652 0.127377778 72

18 ≤ x ≤ 36

Bending Moment:

The bending moment variation is obtained using the following expression:

Using the above expression, the following chordwise bending moment variation is obtained

1277.582207 0.9826 0.04125926 0 ≤ x ≤ 18

1277.582207 0.9826 0.127377778 36

18 ≤ x ≤ 36

Using the above expressions, the variation of shear force and bending moment parallel to the chord along the wingspan is obtained and plotted below:

119

0

20000

40000

60000

80000

100000

120000

0 5 10 15 20 25 30 35 40

Chordw

ise Shear Force (N)


Chordwise Shear Force variation along wingspan

0

50000

100000

150000

200000

250000

300000

350000

400000

0 5 10 15 20 25 30 35 40Chordw

ise Be

nding Mom

ent (N‐m

)


Chordwise Bending Moment variation along wingspan

120

TORQUE DIAGRAM FOR THE WING

In the aircraft wing, at any section, the net aerodynamic force always acts at the centre of pressure. For convenience, we assume the aerodynamic force to act through the aerodynamic centre since this point is fixed for a given airfoil section and does not depend on factors like angle of attack. Along with the force, we also have a moment about the aerodynamic centre Ma.c which has a constant value. Since the centre of pressure is always located behind the aerodynamic centre, Ma.c is always a nose‐down moment.

In this section, we analyze the torque or twist distribution along the wingspan. It is known that when a force acts through the shear centre of a section, no rotation is produced in the section. Applying this concept, we shift the aerodynamic forces to the shear centre. In performing this shift, in addition to the existing Ma.c, an additional moment due to the shifting is also to be considered. The variation of net moment per unit span about the sectionwise shear centre is investigated in this section.

We resolve the aerodynamic forces acting on each section of the wing into components normal and parallel to the chord. Since the parallel component passes through the shear centre, it does not contribute to any torque. Only the normal force component contributes to section torque. For the preliminary analysis, we assume the shear centre to lie at 0.35c from the leading edge. Since this location is very close to the C.G location (0.32c), and because weight is a relieving force as far as section twist is concerned, we neglect the contribution of weight towards torque distribution.

For the aircraft, at cruise

CL = 0.38282

CD = 0.02458431108

ρ = 0.363924 kg/m3

V = 915 km/hr

α = 2.05°

For the wing, airfoil section is NACA 23011‐63. For this,

121

xa.c = 0.265c

CMa.c = ‐0.009

The normal force coefficient is given by

cos sin

where α : angle of attack

Using the above expression,

CN = 0.3834544123

We have the expression for torque per unit span, given by

.

2. .

2

For the wing, we have the variation of chord given by

1.9652 0.25475555556

Using the above data and integrating the above equation, we have the span‐wise torque variation for one wing, given by

362.8838522 1.9652 0.25475555556 0 ≤ x ≤ 18

Using the above expression, the torque diagram for the wing is obtained and plotted below:

0

20000

40000

60000

80000

100000

120000

0 5 10 15 20 25 30 35 40

Torque

(N‐m

)


Torque variation along wingspan

122

CRITICAL SHEAR FORCE, BENDING MOMENT AND TORQUE CALCULATIONS

In this section, we attempt to calculate the maximum or limit values of shear force, bending moment and torque per unit span developed in the wing. This enables us to provide sufficient factor of safety in the form of material selection at required areas.

Initially, from the V‐n diagram described for the aircraft at sea level condition, we determine the critical loading points for the aircraft by comparing corresponding points from the maneuvering and gust envelopes. Referring to the V‐n diagram, the following corner points are identified and tabulated below:

Point Load Factor E.A.S (m/s) CL CD Maneuvering Envelope

A 4.776963 153.1800549 1.495 0.17306 C 4.776963 394 0.22597 0.01779 D 3.5 591 0.0735845 0.01455 E 0 591 0 0 F ‐1.910785201 394 ‐0.090388 0.014746G ‐1.910785201 124.86235655 ‐0.9 0.07175

Gust Envelope B’ 3.64736 112.13632 2.13 0.33671 C’ 5.10225 394 0.241358 0.0183 D’ 4.07669 591 0.085709 0.01468 E’ ‐2.07669 591 ‐0.04366067 0.02771 F’ ‐3.10225 394 ‐0.146749 0.01569G’ ‐1.64736 112.13632 ‐0.962032 0.07996

From the above table, the critical flight conditions are identified and tabulated below:

123

Critical Flight Condition

Point (n, E.A.S) Description

nmax C’ (5.10225, 394)

+ve high AOA A (4.776963, 153.18) Compressive stress is additive in upper flange

of front spar

+ve low AOA D’ (4.07669, 591) Compressive stress is additive in upper flange

of rear spar

‐ve high AOA G (‐1.91, 124.86) Compressive stress is additive in lower flange

of front spar

‐ve low AOA E’ (‐2.07669, 591) Compressive stress is additive in lower flange

of rear spar

In the following section, we attempt to estimate the maximum shear force, bending moment and torque per unit span for one of the above critical flight conditions mentioned above, namely, the positive high angle of attack condition. This point is represented by A (4.776963, 153.18).

To determine the load intensity variation normal to the chord in this condition, we once again assume that the normal force is approximately equal to the lift (for small angle of attack). To determine the lift per unit span variation, we determine the Schrenck’s curve for the point A. If L” represents the total lift under +ve high AOA condition, and L represents the lift under cruise condition, we have

L” = kL

We know

L” = nA W where nA : Load factor at A

L = W

From the above relations we obtain

k = nA

Thus, to obtain the Schrenck’s curve for the +ve high AOA condition, it is sufficient to multiply the expression representing lift intensity variation by nA. Thus, for +ve high angle of attack condition, lift per unit span is given by the following expressions:

124

2.3884815 4496.23837 1.9652 0.25475555556

24396.54003 1 0 ≤ x ≤ 18

2.3884815 4496.23837 1.9652 0.25475555556 36

24396.54003 1 18 ≤ x ≤ 36

A comparison between the Schrenck’s curves for cruise and +ve high AOA condition is shown below:

From the previous section on estimation of variation of shear force, bending moment and torque intensity along the wingspan, it is seen that the maximum values for all the above occur at the wing root (represented by x=18 in the appropriate expressions). Multiplying the appropriate terms corresponding to lift in the shear force, bending moment and torque intensity expressions by nA, and evaluating them at the wing root, we obtain the critical values listed below:

0

20000

40000

60000

80000

100000

120000

140000

0 10 20 30 40

Lift per Unit S

pan (N/m

)


Lift Distribution ‐ Schrenck's Curve

Cruise Condition

+ve High AOA Condition

125

(Normal Shear Force)max = 1588540.96 N

(Normal Bending Moment)max = 12037306.18 N‐m

(Chordwise Shear Force)max = 186287.512 N

(Chordwise Bending Moment)max = 1480209.089 N‐m

(Torque)max = 102011.86 N‐m

126

MATERIAL SELECTION FOR THE AIRCRAFT

The aircraft industry is widely acknowledged as one the branches of engineering which makes the most stringent demands as far as materials are concerned. The very essence of aircraft construction lies in constructing an efficient flying machine with the least weight, which makes an in‐depth knowledge of materials necessary. In addition, aircraft components are subject to a vast variety of structural and thermal loading conditions which necessitates the use of several classes of materials, including metals, non‐metals, plastics and composites. However, unlike most other mechanical engineering fields like civil engineering, the factor of safety involved in aerospace construction is very rarely more than 2, so the limiting properties required of the components closely match the properties of the materials commonly used.

Design Criteria:

There are 3 major criteria which govern material selection and airframe design: strength, stiffness and fatigue resistance.

Members such as wing spars are designed primarily on the basis of strength. In the engineer’s viewpoint, there are 2 important factors:

• It does not matter if the spar partially deforms in the process of carrying load

• There is a certain amount of stress occurring in all flight conditions; it is necessary that the material selected is capable of withstanding all values of stress likely to be experienced by the member without total failure

Stiffness‐based design is customary in components which need not necessarily fail to wreak havoc. A slight deformation can be catastrophic as in turbine blades, whose clearance with the casing is very minute. Even the slightest dimensional changes brought on by either aerodynamic or centripetal loads can cause the engine to come to a sudden halt.

Fatigue is a phenomenon which occurs due to repeated loading of a member, which causes gradual failure. It is common in members subjected

127

to cyclic or vibratory motion. Fatigue failure is one of the most unpredictable forms since in‐service analysis of a member for fatigue damage is very difficult.

Types of Loads:

Loads on aircraft members can be broadly classified as static and dynamic. Static loads are those which are applied gradually such that the time effects of the application of loads are negligible. Dynamic or energy loads are those which are applied impulsively such that the energy with which the structure is loaded also needs to be considered. The weights of different components in the aircraft are static loads, while gust and landing loads on the aircraft are dynamic loads.

Material Properties:

In aircraft construction, weight is a parameter that the engineer tries to limit to a minimum, while strength is attempted to be maximized. Thus, a higher strength to weight ratio is one of the most important properties preferred by aerospace engineers. Strength is commonly determined by tensile testing of a specimen and plotting a graph between the stress applied and the strain developed in it. A common stress‐strain curve obtained for metals is shown below:

In this graph, the initial linear part corresponds to the elastic region, where loading and subsequent unloading causes the material to return to its original state. When the curve fails to be linear by a strain of 0.001, the corresponding stress marks the proportional limit. Immediately after this, the yield stress is achieved. Loading of a member beyond its material elastic limit will result in a certain amount of permanent deformation. Finally, at the ultimate tensile stress, the material fails for good. The yield strength and ultimate strength of the material are crucial in strength‐based designs, and the stresses developed in the members should always be within these limits.

The slope of the linear portion of the stress‐strain curve is called the Young’s Modulus of the material. Stiffness is defined as the ratio between the load applied and the corresponding elongation. Stiffness is related to

128

Young’s modulus through volume. Hence, if it is necessary to have a stiff member, it is necessary to select a material with a high Young’s modulus.

Every time a material is loaded beyond its yield strength but below its ultimate strength, it undergoes a certain amount of plastic deformation and its properties including yield strength itself get modified. Plasticity coupled with repeated loading results in internal stress concentration which builds up with each loading. After certain number of loading cycles, cracks start propagating causing the material to eventually fail. Hence, the plastic region characteristics are critical in members prone to fatigue failure. Fatigue is commonly analyzed using the stress intensity factor ‘k’.

Creep is a highly temperature‐dependent phenomenon. It is one of the most common modes of failure in thermally highly‐stressed members, such as turbine blades and combustor linings. In order to analyze creep, knowledge of the thermal properties like thermal expansivity is required, in addition to carrying out experiments simulating simultaneous thermal and structural loading. Titanium alloys are very well known for their excellent creep‐resistant thermal properties.

In addition to resisting static loads, fatigue and creep, energy loads must also be resisted. The area under the linear portion of the stress‐strain curve is called the resilience modulus, while the area under the whole of the curve is called the toughness modulus. These moduli are indicators of the energy the material can absorb before failure, and are determined using standard tests like the Izod and Charpy tests.

Compression failure is likely in members that are short and thick. The compression characteristic curves are similar to the tensile characteristics, and have corresponding proportional and ultimate limits. Compressive limits have significance only related to bending stresses. All other compression related failures are due to phenomena such as buckling or crippling.

Common Aircraft Materials:

Aluminum‐based alloys such as duralumin are conventionally preferred for the skin and most other structural components of the aircraft. This is because of its relatively high strength to weight ratio coupled with its

129

extreme lightness. Nowadays, composites have come to replace aluminum largely due to their higher strength to weight ratio as well as other attractive properties such as lower overall weight and easier component tailoring. However, steel remains the standard material for crucial components like spars owing to its high strength and stiffness, predictable behavior under loading and reliability. Steel is avoided for other components due to its excessive weight.

Another useful class of metallic alloys is the titanium‐based ones, known for their superior thermal properties. High Mach number aircraft make extensive use of these alloys due to the high temperatures associated with aerodynamic heating.

Glass and clear Perspex are also important in aircraft for windows and viewing shields. The two common types of glass are electrical and structural glass. Though structural glass is more favorable for aircraft construction purposes, it is highly expensive compared to electrical glass. Hence the ideal choice is a compromise between the two types.

Composites like epoxy resin reinforced with glass, carbon or aramid fibers are commonly used for their lightness. Particulate composites and ceramics are used for parts required to resist high temperatures. The major advantages of using composites are their excellent strength to weight ratio, easy tailoring as per requirements and low overall weight.

Some commonly used materials and their properties are listed below:

MATERIAL 5Cr‐Mo‐V STEEL Al 2024 Al 7075 Ti‐6Al‐4V GLASSULTIMATE TENSILE STRESS

(MPa) 1654.7 530.896 228 900 89.6

YIELD TENSILE STRESS (MPa) 1378.9 455.05 103 830 75.8 YIELD COMPRESSIVE STRESS

(MPa) 1516.8 525.1 – – –

YOUNG’S MODULUS (GPa) 206.8 73.1 71.4 114 70 SHEAR MODULUS (GPa) 75.84 21.48 26.95 44 26.9

ULTIMATE SHEAR STRENGTH (MPa)

999.7 206.84 – – –

DENSITY (kg/m3) 7790 2800 2810 4430 2500

130

STRUCTURAL DESIGN OF THE AIRCRAFT – BASIC WING DESIGN

In this section, we perform the structural design of the aircraft wing. This includes design of the various structural members in the wing like spars and skin. For the wing, we adopt a 2‐spar configuration for the following reasons:

• The single cell construction, which is otherwise called the monocoque wing construction, is generally not preferred as since the skin is required to take all structural loads, the skin thickness required is very great, which increases the weight greatly.

• The 2 cell construction making use of a single main spar is generally weak since the rear portion of the wing, which contains control surfaces, is unsupported. The rearward COP case becomes highly critical in this kind of construction.

• The multiple cell construction involving more than 2 spars also is disadvantageous in terms of weight, since higher number of spars unnecessarily increases the weight.

In the 2‐spar construction, the front spar is generally located between 0.15c and 0.25c, and the rear spar is separated by a distance of approximately 0.5c. It is required that the front spar be located close to the position of maximum thickness in the airfoil in order to resist wing loads efficiently. The functions of the front and rear spars are the following:

Front Spar:

• The flange takes bending loads in the forward COP case. • The web takes shear loads transferred to it by the skin. • It transfers wing loads effectively to the fuselage.

Rear Spar:

• The flange takes bending loads in the rearward COP case. • The web takes shear loads transferred to it by the skin.

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• It supports the various control surfaces located on the rear portion of the wing, serving as a hinge.

• Along with the front spar, it forms the boundaries for the fuel tanks in the wings.

Design of the wing spar caps:

The loads taken by the wing spars include bending loads mainly, as well as part of the shear force. The spar is responsible for resisting nearly 90% of the bending. The common section for the spar is the I‐section, in which the flanges resist bending moments while the webs resist shear forces. For the front and rear spars in the wings, the material used is 5Cr‐Mo‐V steel. This material is selected for its high strength, relatively high strength to weight ratio, as well as smooth elastic properties. For this material, the properties are listed below:

Ultimate tensile strength : 1654.7 MPa Yield tensile strength : 1378.9 MPa Yield compressive strength : 1516.8 MPa Ultimate shear strength : 999.7 MPa

Young’s modulus : 206.8 GPa Shear modulus : 75.84 GPa Density : 7790 kg/m3

In the aircraft wing, we assume the front and rear spars to be located at 0.22c and 0.72c from the leading edge respectively. This location satisfies the condition of locating the front spar close to the maximum thickness position in the wing as well as the fuel space requirement. In this section, we design the spars for the critical bending moment developed in the wing. As seen in the previous section, the critical bending moment is developed at the wing root. Hence, we take a section close to the wing root and design the spar based on it.

The airfoil selected for the aircraft is the NACA 23011‐63. The section selected for this analysis has a chord of 6.55 m, and the critical bending moment developed at this section is approximately 12037300 N‐m. At the spar locations, the distance between the top and bottom surfaces of the wing are:

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h1 = 0.768427 m h2 = 0.472026 m

The bending moment is assumed to be shared between the front and rear

spars in the ratio . Using this relation, we have

M1 = 8764236.404 N‐m M2 = 3273063.596 N‐m

We use the Euler bending theory to find the areas of the spar caps. The design is carried out assuming that when this moment is resisted, the stress in the material reaches the yield stress value. Also, we use the lumped mass analysis with the assumption that the areas of the top and bottom caps are the same. With this assumption, the location of the neutral surface of each spar lies midway between the top and bottom surfaces. Using the above assumptions, we have the following formulae:

Front Spar:

8764236.404 0.38421352 A 0.3842135

1.3789514 10

Rear Spar:

3273063.596 0.2360132 A 0.236013

1.3789514 10

Using the above equations, we obtain the following:

A1 = 0.008271085445 m2 A2 = 0.00528512832 m2

For both front and rear spars, each spar cap is designed in two halves, each of the following shape:

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In this shape, assume the ratio l:b = 3:1. With this assumption, we have the following relations:

Front spar:

b7π16

214

0.0082710854452

Rear spar:

b7π16

214

0.005285128322

Using the above relations, we obtain the following:

b1 = 0.024985695 m b2 = 0.019972754 m

Using the above values, the spar caps are sized. The spar webs take the shear force acting on the wing section, as well as part of the torsional loads (along with the skin). Since the shear flow causes compression in the webs, they have to be designed considering strength as well as stability. The thickness which satisfies both these criteria is then taken as the web thickness.

Design of spar webs and wing skin:

The major load taken by the spar webs as well as the wing skin is shear. The shear forces are exerted as a result of the aerodynamic forces acting on the wing, including lift, drag and torque. Due to a higher amount of material used for the skin and webs, we use Al 2024 as the material. This variety of aluminum has fairly good strength to weight ratio, is very lightweight, and has high shear strength. For this material, the properties are listed below:

Ultimate tensile strength : 530.896 MPa Yield tensile strength : 455.05 MPa Yield compressive strength : 525.1 MPa Young’s modulus : 73.1 GPa Shear modulus : 21.48 GPa Ultimate shear strength : 206.84 MPa

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Density : 2800 kg/m3

Shear flow calculations for the wing and Stringer design:

The wing section is the NACA 23011‐63 airfoil. The shear flow in this section needs to be evaluated in order to refine the spar web and wing skin thicknesses. In the aircraft wing, we assume the inter‐rib spacing as 0.6 m, which is a standard value for transport aircraft. In the preliminary analysis, the section is studied without any stringers and the spar flanges are idealized as concentrated area booms. The skin is assumed to be ineffective in bending. Initially, a cut is made in the 2 box sections, using which the open section shear flow is computed. The cut is closed introducing a constant torque, with which the net shear flow in the section is computed.

The shear flow due to bending is calculated using the formula

qV I V II I I

ΣAxV I V II I I

ΣAy

For the section considered, the values of the above parameters are as follows:

Vx = 186287.512 N Vy = 1588540.96 N Ixx = 0.003185351 m4 Iyy = 0.069172304 m4 Ixy = ‐0.007417964 m4

With these values, we have the equation for open section shear flow through the sections as

q 74871400.09 ΣAx 673060662.5 ΣAy

The above equation is modified by adding a constant term while closing the cuts. Apart from the shear flow due to bending, we have an additional constant shear flow due to torque, given by

T 2 A q

The shear flow from the two cases is superimposed, using which the net shear flow through the wing section is computed. Using this distribution, we have the maximum value of shear flow in the skin without any stringers

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to be in the segment between the front and rear spar on the top surface of the wing.

qmax = 903730.7957 N/m

With this value of shear flow, we obtain the skin thickness required from the strength as well as buckling aspects. For strength requirements, we have the relation

τqt

Using the above relation, we have

t = 4.36923 mm

For the buckling analysis, we consider the wing panel with the maximum shear flow. This panel has dimensions 3.28485 m x 0.6 m, and is assumed to be simply supported on all 4 edges by the front and rear spars as well as the ribs. The relation used to determine the thickness required is

qt

π K E12 1 υ

tb

Kb : constant based on the panel supports and aspect ratio

ν : Poisson’s ratio (0.35 for the material considered

b : shorter edge of the panel

For the panel considered,

Kb = 24.5


t = 5.787116817 mm

Selecting the higher of the 2 values computed we have the required thickness of the wing skin without stiffening stringers as 5.787 mm.

For the detailed analysis of the wing, we assume that stringers are used to stiffen the skin panels. The introduction of stringers significantly alters the

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shear flow pattern in the wing. Since the skin is stiffened, the thickness required is also reduced. The purpose of adding stringers is to reduce the skin thickness to a standard value below 2.5 mm. In this section, we first assume a fixed number of stringers based on the size of the cross section. Then, we iterate for the cross section area of the stringers analyzing the shear flow pattern in the wing section for each area, and select the area which gives the minimum skin thickness.

The section selected for the stringers is the Z‐section with end tabs. This section is selected as it gives the maximum area moment of inertia for the minimum cross section area. The properties of the stringer section selected are given below:

A = 0.58382 in2 Ixx = 0.1215 in4 Iyy = 0.09558 in4

For the wing root chord of 6.55 m, we select the number of stringers to be 40. There are no stringers in cell 3 as the region has other components like control surfaces, and the space restriction makes it difficult to insert stringers in this portion.

The exercise performed in the previous section is repeated to compute the new shear flow pattern. It needs to be noted that the shear flow due to the torque will suffer no change due to the addition of the stringers. For the wing section with stringers, the values of area moments of inertia are modified as follows:

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Ixx = 0.01062291 m4 Iyy = 0.222703738 m4 Ixy = ‐0.021768467 m4

The modified shear flow equation due to bending is given below:

q 19324015.23 ΣAx 189137932.8 ΣAy

Once again superimposing the shear flows due to bending and torsion, we evaluate the shear flow and identify the wing skin panel with the maximum value of shear flow. This panel is located between the rear spar and the adjacent stringer on the top surface, and is of size 0.2376 m x 0.6 m. The shear flow in this panel is calculated as

qmax = 357240.7698 N/m

From the strength point of analysis, we have

τqt


t = 1.727136 mm

For the buckling analysis, using the relation

qt

π K E12 1 υ

tb

we have

t = 2.290338558 mm

Again selecting the higher of the 2 values computed we have the required thickness of the wing skin without stiffening stringers as 2.29034 mm.

It is seen that there is a 60.4% reduction in skin thickness due to the addition of stiffening stringers. The values of stringer number and dimensions are arrived at after several iterations, changing the area and section of each stringer. An important factor to be considered simultaneously while designing the stringers is whether the stringers will buckle due to the compressive stress which is exerted on them as a result

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of wing bending. To calculate the bending stress exerted at different locations on the wing section, we use the relation

σM I M II I I

yM I M II I I

x

Using the above relation, the stringer subjected to maximum stress is identified as the stringer one away from the top flange of the rear spar. The stress value acting on this stringer is calculated as

σexerted = 679.7562107 MPa.

Since the stringer is assumed to be simply supported between ribs, the buckling stress for each stringer segment is calculated using the formula

σπ E IA L

Using the data for stringer dimensions and the above equation, we have critical stress

σcr = 686.5569627 MPa

Since the exerted stress is less than the bucking stress, the chosen stringer section is ideal. The stringer properties are tabulated below:

STRINGER AREA x y A*x A*y

m2 m m m3 m3

1 0.000376257 0.2 0.243666 7.52515E‐05 9.16811E‐05 2 0.000376257 0.4 0.332168 0.000150503 0.000124981 3 0.000376257 0.6 0.388728 0.000225754 0.000146262 4 0.000376257 0.8 0.425013 0.000301006 0.000159914 5 0.000376257 1 0.446917 0.000376257 0.000168156 6 0.000376257 1.2 0.458742 0.000451509 0.000172605 7 0.000376257 1.8 0.464498 0.000677263 0.000174771 8 0.000376257 2 0.460796 0.000752515 0.000173378 9 0.000376257 2.2 0.455255 0.000827766 0.000171293 10 0.000376257 2.4 0.448003 0.000903018 0.000168564 11 0.000376257 2.6 0.439071 0.000978269 0.000165204 12 0.000376257 2.8 0.428535 0.00105352 0.000161239 13 0.000376257 3 0.416448 0.001128772 0.000156692 14 0.000376257 3.2 0.402865 0.001204023 0.000151581 15 0.000376257 3.4 0.387847 0.001279275 0.00014593

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16 0.000376257 3.6 0.371451 0.001354526 0.000139761 17 0.000376257 3.8 0.353734 0.001429778 0.000133095 18 0.000376257 4 0.334755 0.001505029 0.000125954 19 0.000376257 4.2 0.314572 0.001580281 0.00011836 20 0.000376257 4.4 0.293242 0.001655532 0.000110334 21 0.000376257 0.2 ‐0.111761 7.52515E‐05 ‐4.20509E‐05 22 0.000376257 0.4 ‐0.143583 0.000150503 ‐5.40242E‐05 23 0.000376257 0.6 ‐0.166394 0.000225754 ‐6.2607E‐05 24 0.000376257 0.8 ‐0.18697 0.000301006 ‐7.03488E‐05 25 0.000376257 1 ‐0.206225 0.000376257 ‐7.75937E‐05 26 0.000376257 1.2 ‐0.223486 0.000451509 ‐8.40882E‐05 27 0.000376257 1.8 ‐0.25482 0.000677263 ‐9.58779E‐05 28 0.000376257 2 ‐0.259801 0.000752515 ‐9.7752E‐05 29 0.000376257 2.2 ‐0.262958 0.000827766 ‐9.89399E‐05 30 0.000376257 2.4 ‐0.264403 0.000903018 ‐9.94836E‐05 31 0.000376257 2.6 ‐0.264184 0.000978269 ‐9.94012E‐05 32 0.000376257 2.8 ‐0.262369 0.00105352 ‐9.87183E‐05 33 0.000376257 3 ‐0.259014 0.001128772 ‐9.74559E‐05 34 0.000376257 3.2 ‐0.254176 0.001204023 ‐9.56356E‐05 35 0.000376257 3.4 ‐0.257917 0.001279275 ‐9.70432E‐05 36 0.000376257 3.6 ‐0.240293 0.001354526 ‐9.0412E‐05 37 0.000376257 3.8 ‐0.231365 0.001429778 ‐8.70528E‐05 38 0.000376257 4 ‐0.221189 0.001505029 ‐8.3224E‐05 39 0.000376257 4.2 ‐0.209824 0.001580281 ‐7.89478E‐05 40 0.000376257 4.4 ‐0.197329 0.001655532 ‐7.42465E‐05

The values of stresses developed in each stringer are tabulated below:

STRINGER AREA x‐Xc y‐Yc A*(x‐Xc)2 A*(y‐Yc)

2 A*(x‐Xc)*(y‐Yc) STRESS

m2 m m m4 m4 m4 N/m2

1 0.000376 ‐2.4624 0.2457 0.002281 0.000023 ‐0.000228 9181174.9772 0.000376 ‐2.2624 0.3342 0.001926 0.000042 ‐0.000284 ‐147093459.43 0.000376 ‐2.0624 0.3908 0.001600 0.000057 ‐0.000303 ‐257563319.54 0.000376 ‐1.8624 0.4270 0.001305 0.000069 ‐0.000299 ‐338958863.65 0.000376 ‐1.6624 0.4490 0.001040 0.000076 ‐0.000281 ‐399732077.86 0.000376 ‐1.4624 0.4608 0.000805 0.000080 ‐0.000254 ‐446052023 7 0.000376 ‐0.8624 0.4665 0.000280 0.000082 ‐0.000151 ‐542394862.48 0.000376 ‐0.6624 0.4628 0.000165 0.000081 ‐0.000115 ‐566449115.79 0.000376 ‐0.4624 0.4573 0.000080 0.000079 ‐0.000080 ‐587866246 10 0.000376 ‐0.2624 0.4500 0.000026 0.000076 ‐0.000044 ‐606829805.311 0.000376 ‐0.0624 0.4411 0.000001 0.000073 ‐0.000010 ‐623384247.312 0.000376 0.1376 0.4306 0.000007 0.000070 0.000022 ‐637638556 13 0.000376 0.3376 0.4185 0.000043 0.000066 0.000053 ‐649668733.4

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14 0.000376 0.5376 0.4049 0.000109 0.000062 0.000082 ‐659553649.215 0.000376 0.7376 0.3899 0.000205 0.000057 0.000108 ‐667380777.516 0.000376 0.9376 0.3735 0.000331 0.000052 0.000132 ‐673231856.117 0.000376 1.1376 0.3558 0.000487 0.000048 0.000152 ‐677188622.918 0.000376 1.3376 0.3368 0.000673 0.000043 0.000170 ‐679335683.819 0.000376 1.5376 0.3166 0.000890 0.000038 0.000183 ‐679756210.720 0.000376 1.7376 0.2953 0.001136 0.000033 0.000193 ‐678531941.621 0.000376 ‐2.4624 ‐0.1097 0.002281 0.000005 0.000102 51886289622 0.000376 ‐2.2624 ‐0.1415 0.001926 0.000008 0.000120 535132675.5 23 0.000376 ‐2.0624 ‐0.1644 0.001600 0.000010 0.000128 538480696.2 24 0.000376 ‐1.8624 ‐0.1849 0.001305 0.000013 0.000130 538623730.625 0.000376 ‐1.6624 ‐0.2042 0.001040 0.000016 0.000128 536872453.2 26 0.000376 ‐1.4624 ‐0.2215 0.000805 0.000018 0.000122 532261783.2 27 0.000376 ‐0.8624 ‐0.2528 0.000280 0.000024 0.000082 489105941.5 28 0.000376 ‐0.6624 ‐0.2578 0.000165 0.000025 0.000064 466885772.129 0.000376 ‐0.4624 ‐0.2609 0.000080 0.000026 0.000045 442049989.7 30 0.000376 ‐0.2624 ‐0.2624 0.000026 0.000026 0.000026 414759202.2 31 0.000376 ‐0.0624 ‐0.2621 0.000001 0.000026 0.000006 385082241.632 0.000376 0.1376 ‐0.2603 0.000007 0.000026 ‐0.000013 353116619.6 33 0.000376 0.3376 ‐0.2570 0.000043 0.000025 ‐0.000033 318942640.2 34 0.000376 0.5376 ‐0.2521 0.000109 0.000024 ‐0.000051 282642041.3 35 0.000376 0.7376 ‐0.2559 0.000205 0.000025 ‐0.000071 258643713.936 0.000376 0.9376 ‐0.2383 0.000331 0.000021 ‐0.000084 204008012.5 37 0.000376 1.1376 ‐0.2293 0.000487 0.000020 ‐0.000098 161842360.5 38 0.000376 1.3376 ‐0.2192 0.000673 0.000018 ‐0.000110 117887078.539 0.000376 1.5376 ‐0.2078 0.000890 0.000016 ‐0.000120 72226772.56 40 0.000376 1.7376 ‐0.1953 0.001136 0.000014 ‐0.000128 24946048.51

The web thickness is calculated based on 2 criteria:

• Strength to withstand the shear flow due to bending and torsion • Stability under compression

From the shear flow pattern obtained the shear flow in the front and rear spar webs are found to be 1748418.192 N/m and 92764.69801 N/m respectively. From the strength point of analysis, using these values and the relation

τqt

we have

t1 = 8.452998414 mm

141

t2 = 0.448485293 mm

For the buckling analysis, we use the relation

qt

π K E12 1 υ

tb

Ks : constant based on the panel supports and aspect ratio

Assuming the web panels to be simply supported at the ribs and clamped at the spar caps, we have the values of Ks for the front and rear spar webs as 9 and 6.6 respectively. Using the above relation, we have

t1 = 9.713582552 mm t2 = 3.766396698 mm

Taking the higher of the above 2 sets of values, we have the web thicknesses as

t1 = 9.714 mm t2 = 3.766 mm

A calculation is also made to determine the percentage of bending stress that can be taken by the stringers. This is done to evaluate the potential reduction in the spar cap areas possible.

The Euler bending theory is used again to compute the bending moment taken by the stringers.

σM yI

Assuming the stress to be the critical yield stress of the stringer material, the above formula reduces to

M σ 2 A Σh2

Using the above relation, we obtain the moment taken by the stringers as

Mstringers = 2113550.199 N‐m

142

Since this is the moment reduction from the load to be taken by the spar caps, the ratio in which this reduction is shared between the two spars is

given by . Using this relation, we have

M1’ = 1534521.986 N‐m M2’ = 579028.2128 N‐m

Using the Euler bending theory again to compute the reduction in areas of the spar caps, we have

ΔA1 = 0.001448176644 m2

ΔA2 = 0.000889579659 m2

These represent possible reductions of approximately 17.5% and 16.8% in the front and rear spar cap areas respectively. Using the dimensions obtained above, the spars as well as the wing section with stringers are shown to scale below:

Front Spar

143

Rear Spar

Wing Cross Section

144

STRUCTURAL DESIGN OF THE AIRCRAFT – FUSELAGE DESIGN

Similar to the previous section, here we perform the structural design of the aircraft fuselage. There are two types of fuselage constructions using metals:

• Monocoque construction • Semi‐monocoque construction

In the monocoque construction, the fuselage is a pure metal shell, with no stiffening members in the axial direction. In this type of construction, all loads on the fuselage are taken by the skin alone. This implies a very thick skin, which potentially increases the aircraft weight greatly.

In the semi‐monocoque construction, the skin is relatively thin. It is stiffened using long, thin members running along the length of the fuselage called stringers (longerons if fewer in number). These members take, apart from stiffening the skin, also directly take some of the bending loads acting on the fuselage.

Design of the fuselage:

Since the demerits of a monocoque construction have discouraged engineers from using this method for a long time, we design a semi‐monocoque fuselage for the aircraft. For carrying out this structural design, the forces acting on the fuselage are required. The weights of various components in the fuselage are already known from the balance diagram. The force acting on the wing and horizontal tail are transmitted to the fuselage through their respective mounting points, which are assumed to be located at 1/4th the root chord. For simplicity, we only analyze the critical symmetrical loading condition (aircraft is pitching), where the maximum load factor is involved.

For the aircraft,

nmax = 4.776963

145

When the aircraft achieves this flight condition, it is in some type of maneuver. The force acting on all components in the fuselage in nmax times their respective weight. We impose moment equilibrium to compute the wing and tail reactions. Using this procedure, the force distribution for the critical case is obtained as below:

Using this force distribution, the shear force and moment variation along the length of the fuselage is obtained and plotted below:

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Structural Design of the Fuselage:

Using the shear force and bending moment diagrams plotted above, we obtain the maximum shear force and bending moment acting on the fuselage. Using these values, the shear flow in the fuselage is computed. Unlike the analysis for the wing, since only symmetric loading case is considered, the shear flow at any section of the fuselage is developed only due to bending; there is no contribution from torque.

‐1200000

‐1000000

‐800000

‐600000

‐400000

‐200000

0

200000

400000

600000

800000

0 5 10 15 20 25 30 35 40

Shear F

orce (N

)

Distance along fuselage from nose (m)

Shear Force

‐6000000

‐5000000

‐4000000

‐3000000

‐2000000

‐1000000

0

1000000

0 5 10 15 20 25 30 35 40

Bend

ing Mom

ent (N‐m

)

Distance along fuselage from nose (m)

Bending Moment

147

For the aircraft, fuselage diameter

D = 3.95 m

We assume the number of stringers as 96. Once again, the section assumed for the stringers is the Z‐section with end tabs. The properties of the stringer section assumed are given below:

A = 0.202248 in2 Ixx = 0.047342886 in4 Iyy = 0.03724307 in4

The above properties are arrived at after an iterative procedure to obtain the minimum skin thickness with the minimum stringer cross section area. The stringer material is Al 2024. The shear flow is determined in a similar manner, by first making a cut in the fuselage section, determining the open shear flow pattern, and then closing the cut by adding a constant shear flow and obtaining the net shear flow pattern.

For the fuselage section with the stringers, the section properties are given below:

Ixx = 0.024430205 m4 Iyy = 0.024430205 m4 Ixy = 0

The equation used to determine the open shear flow distribution is

148

qV I V II I I

ΣAxV I V II I I

ΣAy

Here, for the symmetric loading case,

Vx = 0 Vy = ‐1066569.105 N (maximum shear force in the fuselage)

Using the above values, the above equation is reduced as follows:

q 43657804.14 ΣAy

Using the equation, the open section shear flow is computed. The cut is then closed by adding a constant shear flow. It can be seen that since both the section as well as the loading is symmetric with respect to the normal axis, the shear flow pattern is also symmetric about this pattern. The shear flow variation around the fuselage is tabulated below:

From the table, it is seen that the maximum shear flow is at the bottom, and has a value

qmax = 171837.1086 N/m

Using this shear flow value, similar to the wing structural analysis, we determine the skin thickness required to sustain this shear flow based on strength as well as stability points of analysis.

From the strength point of analysis, we have

τqt 1.5

The factor of safety is included here since the shear strength considered is the ultimate shear strength of the material.


t = 1.24616 mm

For the buckling analysis, we use the relation

qt

π K E12 1 υ

tb

149

For transport aircraft, the bulkhead spacing is standard, given as 18 inches (0.4572 m). The panel with critical shear flow is bounded by a pair of bulkheads and a pair of stringers. Thus, the dimensions for the panel are 0.12926 m x 0.4572 m. For this aspect ration, the value of Kb is 34. Using this value in the above equation, we have

t = 1.072078 mm

Selecting the higher of the 2 values computed we have the required thickness of the fuselage skin without stiffening stringers as 1.24616 mm.

It needs to be noted that unlike the wing, in the fuselage, the strength criterion becomes critical. Once again, we need to consider simultaneously while designing the stringers whether the stringers will buckle due to the compressive stress which is exerted on them as a result of fuselage bending. To calculate the bending stress exerted at different locations on the fuselage section, we use the relation

σM I M II I I

yM I M II I I

x

Here, Mx is the only moment acting on the section, and the value of MX is the maximum value of the bending moment in the fuselage.

Mx = 0 My = ‐4976325.132

Using the above values, the maximum stress acting on a stringer is found to be

σexerted = 401.4374513 MPa

Since the stringer is assumed to be simply supported between bulkheads, the buckling stress for each stringer segment is calculated using the formula

σπ E IA L

150

Using the data for stringer dimensions and the above equation, we have critical stress

σcr = 410.0467306 MPa

Since the exerted stress is less than the critical stress, the chosen stringer section is ideal. The stringer properties are tabulated below:

151

STRINGER AREA x y Ax Ay

m2 m m m3 m3

1 0.000130482 ‐0.12917118 1.970771373 ‐1.68546E‐05 0.000257151 2 0.000130482 ‐0.25778923 1.958103601 ‐3.36369E‐05 0.0002554983 0.000130482 ‐0.385303386 1.937050929 ‐5.02753E‐05 0.000252751 4 0.000130482 ‐0.511167614 1.907703507 ‐6.66983E‐05 0.000248922 5 0.000130482 ‐0.634842944 1.870187006 ‐8.28358E‐05 0.000244026 6 0.000130482 ‐0.755799779 1.824662077 ‐9.86185E‐05 0.0002380867 0.000130482 ‐0.873520163 1.771323665 ‐0.000113979 0.000231126 8 0.000130482 ‐0.9875 1.710400172 ‐0.000128851 0.000223177 9 0.000130482 ‐1.09725121 1.642152484 ‐0.000143172 0.00021427210 0.000130482 ‐1.202303822 1.566872847 ‐0.000156879 0.000204449 11 0.000130482 ‐1.302207985 1.48488362 ‐0.000169915 0.000193751 12 0.000130482 ‐1.396535893 1.396535893 ‐0.000182223 0.000182223 13 0.000130482 ‐1.48488362 1.302207985 ‐0.000193751 0.00016991514 0.000130482 ‐1.566872847 1.202303822 ‐0.000204449 0.000156879 15 0.000130482 ‐1.642152484 1.09725121 ‐0.000214272 0.000143172 16 0.000130482 ‐1.710400172 0.9875 ‐0.000223177 0.00012885117 0.000130482 ‐1.771323665 0.873520163 ‐0.000231126 0.00011397918 0.000130482 ‐1.824662077 0.755799779 ‐0.000238086 9.86185E‐05 19 0.000130482 ‐1.870187006 0.634842944 ‐0.000244026 8.28358E‐05 20 0.000130482 ‐1.907703507 0.511167614 ‐0.000248922 6.66983E‐0521 0.000130482 ‐1.937050929 0.385303386 ‐0.000252751 5.02753E‐05 22 0.000130482 ‐1.958103601 0.25778923 ‐0.000255498 3.36369E‐05 23 0.000130482 ‐1.970771373 0.12917118 ‐0.000257151 1.68546E‐05 24 0.000130482 ‐1.975 0 ‐0.000257703 0 25 0.000130482 ‐1.970771373 ‐0.12917118 ‐0.000257151 ‐1.68546E‐05 26 0.000130482 ‐1.958103601 ‐0.25778923 ‐0.000255498 ‐3.36369E‐05 27 0.000130482 ‐1.937050929 ‐0.385303386 ‐0.000252751 ‐5.02753E‐0528 0.000130482 ‐1.907703507 ‐0.511167614 ‐0.000248922 ‐6.66983E‐05 29 0.000130482 ‐1.870187006 ‐0.634842944 ‐0.000244026 ‐8.28358E‐05 30 0.000130482 ‐1.824662077 ‐0.755799779 ‐0.000238086 ‐9.86185E‐0531 0.000130482 ‐1.771323665 ‐0.873520163 ‐0.000231126 ‐0.00011397932 0.000130482 ‐1.710400172 ‐0.9875 ‐0.000223177 ‐0.000128851 33 0.000130482 ‐1.642152484 ‐1.09725121 ‐0.000214272 ‐0.000143172 34 0.000130482 ‐1.566872847 ‐1.202303822 ‐0.000204449 ‐0.00015687935 0.000130482 ‐1.48488362 ‐1.302207985 ‐0.000193751 ‐0.000169915 36 0.000130482 ‐1.396535893 ‐1.396535893 ‐0.000182223 ‐0.000182223 37 0.000130482 ‐1.302207985 ‐1.48488362 ‐0.000169915 ‐0.000193751 38 0.000130482 ‐1.202303822 ‐1.566872847 ‐0.000156879 ‐0.00020444939 0.000130482 ‐1.09725121 ‐1.642152484 ‐0.000143172 ‐0.000214272 40 0.000130482 ‐0.9875 ‐1.710400172 ‐0.000128851 ‐0.000223177 41 0.000130482 ‐0.873520163 ‐1.771323665 ‐0.000113979 ‐0.00023112642 0.000130482 ‐0.755799779 ‐1.824662077 ‐9.86185E‐05 ‐0.000238086

152

43 0.000130482 ‐0.634842944 ‐1.870187006 ‐8.28358E‐05 ‐0.00024402644 0.000130482 ‐0.511167614 ‐1.907703507 ‐6.66983E‐05 ‐0.000248922 45 0.000130482 ‐0.385303386 ‐1.937050929 ‐5.02753E‐05 ‐0.000252751 46 0.000130482 ‐0.25778923 ‐1.958103601 ‐3.36369E‐05 ‐0.00025549847 0.000130482 ‐0.12917118 ‐1.970771373 ‐1.68546E‐05 ‐0.000257151 48 0.000130482 0 ‐1.975 0 ‐0.000257703 49 0.000130482 0.12917118 ‐1.970771373 1.68546E‐05 ‐0.000257151 50 0.000130482 0.25778923 ‐1.958103601 3.36369E‐05 ‐0.00025549851 0.000130482 0.385303386 ‐1.937050929 5.02753E‐05 ‐0.000252751 52 0.000130482 0.511167614 ‐1.907703507 6.66983E‐05 ‐0.000248922 53 0.000130482 0.634842944 ‐1.870187006 8.28358E‐05 ‐0.00024402654 0.000130482 0.755799779 ‐1.824662077 9.86185E‐05 ‐0.000238086 55 0.000130482 0.873520163 ‐1.771323665 0.000113979 ‐0.000231126 56 0.000130482 0.9875 ‐1.710400172 0.000128851 ‐0.000223177 57 0.000130482 1.09725121 ‐1.642152484 0.000143172 ‐0.00021427258 0.000130482 1.202303822 ‐1.566872847 0.000156879 ‐0.000204449 59 0.000130482 1.302207985 ‐1.48488362 0.000169915 ‐0.000193751 60 0.000130482 1.396535893 ‐1.396535893 0.000182223 ‐0.00018222361 0.000130482 1.48488362 ‐1.302207985 0.000193751 ‐0.000169915 62 0.000130482 1.566872847 ‐1.202303822 0.000204449 ‐0.000156879 63 0.000130482 1.642152484 ‐1.09725121 0.000214272 ‐0.000143172 64 0.000130482 1.710400172 ‐0.9875 0.000223177 ‐0.00012885165 0.000130482 1.771323665 ‐0.873520163 0.000231126 ‐0.000113979 66 0.000130482 1.824662077 ‐0.755799779 0.000238086 ‐9.86185E‐05 67 0.000130482 1.870187006 ‐0.634842944 0.000244026 ‐8.28358E‐0568 0.000130482 1.907703507 ‐0.511167614 0.000248922 ‐6.66983E‐05 69 0.000130482 1.937050929 ‐0.385303386 0.000252751 ‐5.02753E‐05 70 0.000130482 1.958103601 ‐0.25778923 0.000255498 ‐3.36369E‐05 71 0.000130482 1.970771373 ‐0.12917118 0.000257151 ‐1.68546E‐0572 0.000130482 1.975 0 0.000257703 0 73 0.000130482 1.970771373 0.12917118 0.000257151 1.68546E‐05 74 0.000130482 1.958103601 0.25778923 0.000255498 3.36369E‐0575 0.000130482 1.937050929 0.385303386 0.000252751 5.02753E‐05 76 0.000130482 1.907703507 0.511167614 0.000248922 6.66983E‐05 77 0.000130482 1.870187006 0.634842944 0.000244026 8.28358E‐05 78 0.000130482 1.824662077 0.755799779 0.000238086 9.86185E‐0579 0.000130482 1.771323665 0.873520163 0.000231126 0.000113979 80 0.000130482 1.710400172 0.9875 0.000223177 0.000128851 81 0.000130482 1.642152484 1.09725121 0.000214272 0.00014317282 0.000130482 1.566872847 1.202303822 0.000204449 0.000156879 83 0.000130482 1.48488362 1.302207985 0.000193751 0.000169915 84 0.000130482 1.396535893 1.396535893 0.000182223 0.000182223 85 0.000130482 1.302207985 1.48488362 0.000169915 0.00019375186 0.000130482 1.202303822 1.566872847 0.000156879 0.000204449

153

87 0.000130482 1.09725121 1.642152484 0.000143172 0.00021427288 0.000130482 0.9875 1.710400172 0.000128851 0.000223177 89 0.000130482 0.873520163 1.771323665 0.000113979 0.000231126 90 0.000130482 0.755799779 1.824662077 9.86185E‐05 0.00023808691 0.000130482 0.634842944 1.870187006 8.28358E‐05 0.000244026 92 0.000130482 0.511167614 1.907703507 6.66983E‐05 0.000248922 93 0.000130482 0.385303386 1.937050929 5.02753E‐05 0.000252751 94 0.000130482 0.25778923 1.958103601 3.36369E‐05 0.00025549895 0.000130482 0.12917118 1.970771373 1.68546E‐05 0.000257151 96 0.000130482 0 1.975 0 0.000257703

The values of stresses developed in each stringer are tabulated below:

STRINGER AREA x‐Xc y‐Yc A*(x‐Xc)2 A*(y‐Yc)

2 A*(x‐Xc)*(y‐Yc) STRESS

m2 m m m4 m4 m4 N/m2

1 0.00013 ‐0.12917 1.97077 0.00000 0.00051 ‐0.00003 ‐401437451.32 0.00013 ‐0.25779 1.95810 0.00001 0.00050 ‐0.00007 ‐398857081.93 0.00013 ‐0.38530 1.93705 0.00002 0.00049 ‐0.00010 ‐394568745.24 0.00013 ‐0.51117 1.90770 0.00003 0.00047 ‐0.00013 ‐388590804.65 0.00013 ‐0.63484 1.87019 0.00005 0.00046 ‐0.00015 ‐380948858.56 0.00013 ‐0.75580 1.82466 0.00007 0.00043 ‐0.00018 ‐371675630.97 0.00013 ‐0.87352 1.77132 0.00010 0.00041 ‐0.00020 ‐360810831.28 0.00013 ‐0.98750 1.71040 0.00013 0.00038 ‐0.00022 ‐348400984.19 0.00013 ‐1.09725 1.64215 0.00016 0.00035 ‐0.00024 ‐334499230.510 0.00013 ‐1.20230 1.56687 0.00019 0.00032 ‐0.00025 ‐319165099.911 0.00013 ‐1.30221 1.48488 0.00022 0.00029 ‐0.00025 ‐302464255.312 0.00013 ‐1.39654 1.39654 0.00025 0.00025 ‐0.00025 ‐284468212.313 0.00013 ‐1.48488 1.30221 0.00029 0.00022 ‐0.00025 ‐265254032.814 0.00013 ‐1.56687 1.20230 0.00032 0.00019 ‐0.00025 ‐244903994.715 0.00013 ‐1.64215 1.09725 0.00035 0.00016 ‐0.00024 ‐223505240.216 0.00013 ‐1.71040 0.98750 0.00038 0.00013 ‐0.00022 ‐201149401.917 0.00013 ‐1.77132 0.87352 0.00041 0.00010 ‐0.00020 ‐177932211.118 0.00013 ‐1.82466 0.75580 0.00043 0.00007 ‐0.00018 ‐153953087.119 0.00013 ‐1.87019 0.63484 0.00046 0.00005 ‐0.00015 ‐129314712.420 0.00013 ‐1.90770 0.51117 0.00047 0.00003 ‐0.00013 ‐104122592.321 0.00013 ‐1.93705 0.38530 0.00049 0.00002 ‐0.00010 ‐78484603.2 22 0.00013 ‐1.95810 0.25779 0.00050 0.00001 ‐0.00007 ‐52510531.0123 0.00013 ‐1.97077 0.12917 0.00051 0.00000 ‐0.00003 ‐26311600.6624 0.00013 ‐1.97500 0.00000 0.00051 0.00000 0.00000 0 25 0.00013 ‐1.97077 ‐0.12917 0.00051 0.00000 0.00003 26311600.66 26 0.00013 ‐1.95810 ‐0.25779 0.00050 0.00001 0.00007 52510531.0127 0.00013 ‐1.93705 ‐0.38530 0.00049 0.00002 0.00010 78484603.2 28 0.00013 ‐1.90770 ‐0.51117 0.00047 0.00003 0.00013 104122592.3

154

29 0.00013 ‐1.87019 ‐0.63484 0.00046 0.00005 0.00015 129314712.430 0.00013 ‐1.82466 ‐0.75580 0.00043 0.00007 0.00018 153953087.1 31 0.00013 ‐1.77132 ‐0.87352 0.00041 0.00010 0.00020 177932211.1 32 0.00013 ‐1.71040 ‐0.98750 0.00038 0.00013 0.00022 201149401.933 0.00013 ‐1.64215 ‐1.09725 0.00035 0.00016 0.00024 223505240.2 34 0.00013 ‐1.56687 ‐1.20230 0.00032 0.00019 0.00025 244903994.7 35 0.00013 ‐1.48488 ‐1.30221 0.00029 0.00022 0.00025 265254032.8 36 0.00013 ‐1.39654 ‐1.39654 0.00025 0.00025 0.00025 284468212.337 0.00013 ‐1.30221 ‐1.48488 0.00022 0.00029 0.00025 302464255.3 38 0.00013 ‐1.20230 ‐1.56687 0.00019 0.00032 0.00025 319165099.9 39 0.00013 ‐1.09725 ‐1.64215 0.00016 0.00035 0.00024 334499230.540 0.00013 ‐0.98750 ‐1.71040 0.00013 0.00038 0.00022 348400984.1 41 0.00013 ‐0.87352 ‐1.77132 0.00010 0.00041 0.00020 360810831.2 42 0.00013 ‐0.75580 ‐1.82466 0.00007 0.00043 0.00018 371675630.9 43 0.00013 ‐0.63484 ‐1.87019 0.00005 0.00046 0.00015 380948858.544 0.00013 ‐0.51117 ‐1.90770 0.00003 0.00047 0.00013 388590804.6 45 0.00013 ‐0.38530 ‐1.93705 0.00002 0.00049 0.00010 394568745.2 46 0.00013 ‐0.25779 ‐1.95810 0.00001 0.00050 0.00007 398857081.947 0.00013 ‐0.12917 ‐1.97077 0.00000 0.00051 0.00003 401437451.3 48 0.00013 0.00000 ‐1.97500 0.00000 0.00051 0.00000 402298803.9 49 0.00013 0.12917 ‐1.97077 0.00000 0.00051 ‐0.00003 401437451.3 50 0.00013 0.25779 ‐1.95810 0.00001 0.00050 ‐0.00007 398857081.951 0.00013 0.38530 ‐1.93705 0.00002 0.00049 ‐0.00010 394568745.2 52 0.00013 0.51117 ‐1.90770 0.00003 0.00047 ‐0.00013 388590804.6 53 0.00013 0.63484 ‐1.87019 0.00005 0.00046 ‐0.00015 380948858.554 0.00013 0.75580 ‐1.82466 0.00007 0.00043 ‐0.00018 371675630.9 55 0.00013 0.87352 ‐1.77132 0.00010 0.00041 ‐0.00020 360810831.2 56 0.00013 0.98750 ‐1.71040 0.00013 0.00038 ‐0.00022 348400984.1 57 0.00013 1.09725 ‐1.64215 0.00016 0.00035 ‐0.00024 334499230.558 0.00013 1.20230 ‐1.56687 0.00019 0.00032 ‐0.00025 319165099.9 59 0.00013 1.30221 ‐1.48488 0.00022 0.00029 ‐0.00025 302464255.3 60 0.00013 1.39654 ‐1.39654 0.00025 0.00025 ‐0.00025 284468212.361 0.00013 1.48488 ‐1.30221 0.00029 0.00022 ‐0.00025 265254032.8 62 0.00013 1.56687 ‐1.20230 0.00032 0.00019 ‐0.00025 244903994.7 63 0.00013 1.64215 ‐1.09725 0.00035 0.00016 ‐0.00024 223505240.2 64 0.00013 1.71040 ‐0.98750 0.00038 0.00013 ‐0.00022 201149401.965 0.00013 1.77132 ‐0.87352 0.00041 0.00010 ‐0.00020 177932211.1 66 0.00013 1.82466 ‐0.75580 0.00043 0.00007 ‐0.00018 153953087.1 67 0.00013 1.87019 ‐0.63484 0.00046 0.00005 ‐0.00015 129314712.468 0.00013 1.90770 ‐0.51117 0.00047 0.00003 ‐0.00013 104122592.3 69 0.00013 1.93705 ‐0.38530 0.00049 0.00002 ‐0.00010 78484603.2 70 0.00013 1.95810 ‐0.25779 0.00050 0.00001 ‐0.00007 52510531.01 71 0.00013 1.97077 ‐0.12917 0.00051 0.00000 ‐0.00003 26311600.6672 0.00013 1.97500 0.00000 0.00051 0.00000 0.00000 0

155

73 0.00013 1.97077 0.12917 0.00051 0.00000 0.00003 ‐26311600.6674 0.00013 1.95810 0.25779 0.00050 0.00001 0.00007 ‐52510531.0175 0.00013 1.93705 0.38530 0.00049 0.00002 0.00010 ‐78484603.2 76 0.00013 1.90770 0.51117 0.00047 0.00003 0.00013 ‐104122592.377 0.00013 1.87019 0.63484 0.00046 0.00005 0.00015 ‐129314712.478 0.00013 1.82466 0.75580 0.00043 0.00007 0.00018 ‐153953087.179 0.00013 1.77132 0.87352 0.00041 0.00010 0.00020 ‐177932211.180 0.00013 1.71040 0.98750 0.00038 0.00013 0.00022 ‐201149401.981 0.00013 1.64215 1.09725 0.00035 0.00016 0.00024 ‐223505240.282 0.00013 1.56687 1.20230 0.00032 0.00019 0.00025 ‐244903994.783 0.00013 1.48488 1.30221 0.00029 0.00022 0.00025 ‐265254032.884 0.00013 1.39654 1.39654 0.00025 0.00025 0.00025 ‐284468212.385 0.00013 1.30221 1.48488 0.00022 0.00029 0.00025 ‐302464255.386 0.00013 1.20230 1.56687 0.00019 0.00032 0.00025 ‐319165099.987 0.00013 1.09725 1.64215 0.00016 0.00035 0.00024 ‐334499230.588 0.00013 0.98750 1.71040 0.00013 0.00038 0.00022 ‐348400984.189 0.00013 0.87352 1.77132 0.00010 0.00041 0.00020 ‐360810831.290 0.00013 0.75580 1.82466 0.00007 0.00043 0.00018 ‐371675630.991 0.00013 0.63484 1.87019 0.00005 0.00046 0.00015 ‐380948858.592 0.00013 0.51117 1.90770 0.00003 0.00047 0.00013 ‐388590804.693 0.00013 0.38530 1.93705 0.00002 0.00049 0.00010 ‐394568745.294 0.00013 0.25779 1.95810 0.00001 0.00050 0.00007 ‐398857081.995 0.00013 0.12917 1.97077 0.00000 0.00051 0.00003 ‐401437451.396 0.00013 0.00000 1.97500 0.00000 0.00051 0.00000 ‐402298803.9

156

CONCLUSION

Thus, we come to the end of aerodynamic and structural conceptual design of the allocated 160‐seater passenger aircraft. Briefly, the design commenced with the analysis of various aircraft the same class as ours, in passenger capacity. Using this data, a rudimentary design was arrived upon as an average of their characteristics.

In the first section, this design was refined using exhaustive analysis and iterative methods, to finally satisfy the requirements pertaining to aerodynamics and flight mechanics. An analysis was also made into the performance as well as the stability and control of the aircraft.

In the second section the structural design of the aircraft was performed, paying attention to the loads acting on the aircraft, and an attempt was made to ensure structural integrity of the aircraft for all flight conditions.

It can be safely said that a sincere effort in this subject is more than sufficient to understand the procedure involved in the actual design of an aircraft. The challenges we faced at various phases of the project made clear the fact that experience plays a vital role in successful design of any aircraft or aircraft component. With the design project as a base, our understanding of various concepts in the field of aerospace design has been fortified. We would like to express our heartfelt gratitude to all those who have been instrumental in the successful completion of our Aircraft Design Project.

157

REFERENCES

Aerodynamic design:

• Jane’s All the world’s aircraft

• Aircraft design – A Conceptual Approach – Daniel P. Raymer

• Design of Aircraft – Thomas Corke

• Aircraft Performance and Design – J.D. Anderson

• Aircraft Performance, Stability and Control – Perkins and Hage

• Fluid Dynamic Drag ‐ Hoerner

• Summary of Airfoil Data – Abbott, Doenhoff and Stivers

• www.airliners.net

• www.wikipedia.org

• www.aerospaceweb.org

Structural design:

• Analysis of Aircraft Structures – E.F. Bruhn

• Aircraft Structures for Engineering Students – T.H.G Megson

• Aircraft Structures – Peery and Azar

• Airplane Design – Jan Roshkam

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aircraft design project group 5

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