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- 1. Page |1vigneshaeronautical@gmail.com EIGHT SEATER SHORT RANGE BUSINESS JET AIRCRAFT AN AIRCRAFT DESIGN PROJECT-II REPORT Submitted by S.VIGNESH (30609101062) P.VASANTHA PRABHU (30609101059) J.SELVA KUMAR (30609101051) N.VIGNESH (30609101061) in partial fulfillm for the award of the degree fulfillment of BACHELOR OF ENGINEERING in AERONAUTICAL JEPPIAAR ENGINEERING COLLEGE, CHENNAI 600 119 ANNA UNIVERSITY: CHENNAI 600 025 NOV/DEC 2012
- 2. Page |2vigneshaeronautical@gmail.com DEPARTMENT OF AERONAUTICAL ENGINEERING JEPPIAAR ENGINEERING COLLEGE ANNA UNIVERSITY, CHENNAI BONAFIDE CERTIFICATECertified that this project report ‘EIGHT SEATER SHORT RANGEBUSINESS JET AIRCRAFT’ is a bonafide work of__S.VIGNESH___________who carried out project under my supervision.Submitted for the examination held on ____6.11.2012________PROJECT GUIDE HEAD OF THE DEPARTMENTINTERNAL EXAMINER EXTERNAL EXAMINER
- 3. Page |3vigneshaeronautical@gmail.com ACKNOWLEDGEMENT It gives us immense pleasure in expressing our sincere gratitude toHonourable Dr.Jeppiaar, M.A., B.L., Ph.D., founder and Chairman of JeppiaarEngineering College for bestowing us with an opportunity to bring out thisproject as a successful one. We are very much grateful to our principal Dr.Susil Lal Das, M.Sc., Ph.D.,for their encouragement and moral support. We are very much indebted to Mr.G.Prabakaran (HOD) AeronauticalDepartment for giving me his able support and encouragement. At this juncture I must emphasis the point that this AIRCRAFT DESIGNPROJECT-II would not have been possible without the highly informative andvaluable guidance by our respected preceptor (Ms. Puja Sunil and Ms. UshaBharathi), Mr. Balaraman whose vast knowledge and experience has must us goabout this project with great ease. We have great pleasure in expressing our sincere& whole hearted gratitude to them. It is worth mentioning about my team mates,friends and colleagues of the Aeronautical department, for extending their kindhelp whenever the necessity arose. I thank one and all who have directly orindirectly helped me in making this design project a great success.
- 4. Page |4vigneshaeronautical@gmail.com ABSTRACT The aim of this design project is to analysis an 8 Seater Short Range Executive Aircraftby a structural analysis of Shear force and Bending moment. Have to design a more strengthaircraft by give the support of stringer, ribs, spar in Wing section and to give the support ofstinger, bulkhead, longer in Fuselage. The flying strength of aircraft is analysis by Vn diagram.Then the Design of Miscellaneous Members of Wing Fuel Tank, Rib location and direction,Empennage Design, Auxiliary Surfaces, Wing –Fuselage Intersection, Flutter, Aileron Buzz andBuffeting. Then the necessary graphs have to be plotted for further performance calculation.Required diagrams are also drawn.
- 5. Page |5vigneshaeronautical@gmail.com TABLE OF CONTENTS TITLE PAGE Abstract i List of symbols iv List of figures v List of tables vi List of Graphs vii 1 Introduction 1.1 Brief review of ADP - 1 1 1.2 Structural Design - Overview 2 2 V-n diagram 4 3 Structural Design of wing 3.1 Introduction 10 3.2 Air –Inertia Load Estimation 11 3.3 Shear force & Bending moment Distribution 13 3.4 Material Selection 14 3.5 Wing Spar and Stringer Design 18 3.6 Shear flow Distribution 33 4 Structural Design of Fuselage 4.1 Design of fuselage 39 4.2 Stringer Design 39 4.3 Shear flow Distribution 43 4.4 Bulkhead design 46
- 6. Page |6vigneshaeronautical@gmail.com 5 Design of Miscellaneous Members 5.1 Wing Fuel Tank 47 5.2 Rib location and direction 48 5.3 Empennage Design 49 5.4 Auxiliary Surfaces 50 5.5 Wing –Fuselage Intersection 52 5.6 Flutter 53 5.7 Aileron Buzz 54 5.8 Buffeting 55 6 Final design- three view diagram 56 7 Conclusion 57 Bibliography 58 Website reference 59
- 7. Page |7vigneshaeronautical@gmail.com LIST OF SYMBOLSA Total cross sectional area ft2ASPAR Cross sectional area of spar ft2Afuselage Cross sectional area of fuselage ft2at Slope of the CL vs. α curve for a horizontal tail. Deg-1a Distance of the front spar from the nose of the aircraft ftb Distance of the rear spar from the nose of the aircraft ftb Wing span ftbw Width of the web Ftbf Width of the flange FtCwing Chord of the actual wing FtCelliptic Chord of the elliptic wing FtC.G Centre of gravity Ftg Acceleration due to gravity ft/s2D Drag LbE Youngĵs modulus lb/ft2FOS Factor of safetyFtu Tensile ultimate strength lb/ft2H Height of the C.G from the ground level Ftiw Orientation of wing on fuselage DegIxx Second moment of area about X axis ft4Lw Local lift LbWT.O Takeoff weight LbW/S Wing loading lb/ft2y span location Ftα Angle of attack Degβ Turnover angle Degρ Density lb/ft3ρo Density of air at sea level lb/ft3σ Bending stress lb/ft2
- 8. Page |8vigneshaeronautical@gmail.com LIST OF FIGURES FIGURE TITLE PAGE NO. 1 Load on the aircraft 8 2 Typical V-n diagram 9 3 Final V-n diagram 14 4 Linear lift distribution 16 5 Elliptic Lift Distribution 17 6 Wing Separation Of Finite Section 22 7 Torque Distribution Over an Airfoil 26 8 Shear Center On The Chord 26 9 Wing Spar Arrangement 30 10 Different Spar Selection 31 11 Cross Section of Rear Spar 33 12 Cross Section of Middle Spar 35 13 Semimonocoque And Monocoque 39 14 Cross section Of Z-section 42
- 9. Page |9vigneshaeronautical@gmail.com LIST OF TABLES TABLE TITLE PAGE NO. 1 Specification of the Aircraft Design Project I 7 2 Load Factor 10 3 Velocity VS Load Factor 11 4 Load on Factor of safety 13 5 Span VS Linear Lift Distribution 17 6 Span VS Elliptic Lift Distribution 18 7 Span Vs Schrenks value 18 8 Span Vs Load acting on wing 20 9 Centroid table 22 10 Span VS Shear Force 23 11 Span VS Bending Moment 23 12 Span wise VS Shear Force 25 13 Span wise VS Bending Moment0 25 14 Span wise VS Torque at Normal force 27 Spanwise VS Mean Aerodynamic Chord 28 15 16 Frontspar Centroid calculation table 33 17 Front Spar Bending Stress 34 18 Middle Spar Centroid Calculations 35 19 Middle Spar Bending Stress 36 20 Rear Spar Centroid Calculations 37 21 Rear Spar Bending Calculation 38 22 Fuselage Structure Analysis 41 23 Stringer Stress Tabulation 44 22 Weight, Moment, Shear Force, Bending Moment 45
- 10. P a g e | 10vigneshaeronautical@gmail.com LIST OF GRAPHSGRAPHS TITLE PAGE NO. 1 Span VS Linear Lift Distribution 16 2 Span vs Elliptic lift Distribution 19 3 Span Vs Load acting on wing 20 4 Span Vs Fuel weight distribution 21 5 NET SHEAR FORCE vs BENDING MOMENT 25 6 Span wise VS Bending Moment 28 7 Span vs Net Torque Force 32 8 Stringer location in Fuselage 46
- 11. P a g e | 11vigneshaeronautical@gmail.com 1 INTRODUCTION Aircraft Design Project-II is a continuation of Aircraft Design Project-I. As mentioned in ourearlier project, Business jet, private jet or, colloquially, bizjet is a term describing a jet aircraft, usuallyof smaller size, designed for transporting groups of up to 19 business people or wealthy individuals.Business jets may be adapted for other roles, such as the evacuation of casualties or express parceldeliveries, and a few may be used by public bodies, governments or the armed forces. The more formalterms of corporate jet, executive jet, VIP transport or business jet tend to be used by the firms that build,sell, buy and charter these aircraft. In our Aircraft Design Project-I, we have performed a rudimentaryanalysis. We have carried out a preliminary weight estimation, power plant selection, aerofoil selection,wing selection and aerodynamic parameter selection and analysis. Apart from the above mentioned, wehave also determined performance parameters such lift, drag, range, endurance, thrust and powerrequirements. Aircraft Design Project-II deals with a more in-depth study and analysis of aircraft performanceand structural characteristics. In the following pages we have carried out structural analysis of fuselageand wings and the appropriate materials have been chosen to give our aircraft adequate structuralintegrity. The flight envelope of our aircraft has also been established by constructing the V-n diagram.We have also determined the landing gear position, retraction and other accompanying systems andmechanisms. The study of all the above mentioned characteristics, has given us insight into thecomplexity of designing a subsonic multi-role 8 seater business jet.The specifications of the Aircraft design project-I as follows: S.No DESIGN PARAMETER MAGNITUDE UNIT 1. Cruising speed 236.11 m/s 2. Wing span 12.84 m 3. Aircraft Length 19.5 m 4. Wing area 21.84 m2 5. Height 4.8 m
- 12. P a g e | 12vigneshaeronautical@gmail.com 6. Aspect ratio 7.55 (No unit) 7. Wing loading 585.89 Kg/m2 8. Empty weight 7,296 Kg 9. Maximum take-off weight 1,25,568 N 10. Pay load 1280 kg 11. No. of engines 2 (No unit) 12. Thrust power 32.00 Kn 13. Range 5200 Km 14. Service ceiling 1,3700 M 15. Mach no. 0.715 (No unit) 16. Thrust/weight ratio 0.25 (No unit) o 17. Gliding angle 4.23 18. Seating capacity 8 (No unit) 19. Fuselage 10.54 M 20. Take-off distance 1,007.46 M 21. Landing distance 710.3 M 22. Rate of climb 1.298 × 10-3 m/s
- 13. P a g e | 13vigneshaeronautical@gmail.comLOADS ON THE AIRCRAFT: The structure of an aircraft is required to support two classes of loads, first termed groundloads, includes all loads encountered by the aircraft during movement or transportation on theground such as taxing, landing loads, towing etc, while the second is the air loads, comprisesloads imposed on the structure. The two classes of loads of loads may be still classified assurface forces acting on the surface of the structure and body forces acting over the volume ofthe structure. Basically all air loads are the resultant of the pressure distribution over the surfacesof the skin produced by steady flight, maneuver or gust conditions. Generally these causesbending, shear, torsion in all parts of the structure in addition to local normal pressure loadsimposed on the skin. Ground loads encountered in landing and taxing subject the aircraft to concentrated shockloads through the undercarriage system. The majority of the aircraft have their mainundercarriage located in the wings with nose wheel or tail wheel in the vertical plane ofsymmetry. Clearly the position of the undercarriage should be in such a position so as to produceminimum loads on the wing structure.
- 14. P a g e | 14vigneshaeronautical@gmail.com 2 ESTIMATION OF V-N DIAGRAM The control of weight in aircraft design is of extreme importance. Increase in weightrequires stronger structures to support them, which in turn lead to further increase in weight & soon. Excess of structural weight means lesser amounts of payload, affecting the economicviability of the aircraft. Therefore there is need to reduce aircraft’s weight to the minimum compatible withsafety. Thus to ensure general minimum standards of strength & safety, airworthiness regulationslay down several factors which the primary structures of the aircraft must satisfy.These are 1. LIMIT LOAD: the maximum load that the aircraft is expected to experience in normal operation. 2. PROOF LOAD: product of the limit load and proof factor(1.0-1.25) 3. ULTIMATE LOAD : product of limit load and ultimate factor(1.0-1.5) The aircraft’s structure must withstand the proof load without detrimental distortion &should not fail until the ultimate load has been achieved.V-n Diagram: A chart of Velocity versus load factor (V-n diagram) is another way of showing limits of aircraft performance. It shows how much load factor can be safely achieved at different airspeeds.
- 15. P a g e | 15vigneshaeronautical@gmail.com The maneuverability of the aircraft is also dictated by the loads falling on the structuresduring the maneuvers. Both the aerodynamic and structural limitations for a given airplane areillustrated in the V-n diagram, a plot of load factor versus flight velocity.A V-n diagram is type of flight envelope for the aircraft establishing the maneuver boundaries. The BCAR (British civil airworthiness requirements) has given the basic strength andflight performance limits of various categories of the aircraft. They are listed below Category Positive load factor (n+) Negative load factor(n-) Normal 3.8 -1.5 Semi aerobatic 4.5 -2 Fully aerobatic 6 -3 Tabular column 2: LOAD FACTOR The 8 seater executive aircraft comes under the normal category. Therefore the loadfactor limits for the aircraft is 3.8 & -1.5.The V-n diagram for the aircraft is drawn for the two cases namely, 1. Intentional maneuver( pilot induced maneuver ) 2. Unintentional maneuver( gusts) INTENTIONAL MANEUVER: Intentional maneuvers are induced by the pilot during climb, pull up or pull down,banking the plane etc... The load factor is function of velocity. The expression relating the load factor and thevelocity is given by nmax = { {$ Where nmax is the maximum load factor, V is the speed of the aircraft, Vs is the stallingspeed of the aircraft. $ 9 9 The stalling speed of the aircraft Vs 2 = /0?9 Vs= 59.197 m/s
- 16. P a g e | 16vigneshaeronautical@gmail.comFor various values of V, nmax is calculated and tabulated below, Nmax={ {$ V Nmax={ {$ V 4 118.394 -0.25 29.598 3.8 115.396 -0.5 41.858 3.5 110.74 -0.75 51.256 3 102.53 -1 59.197 2.5 93.599 2 83.71 1.5 72.501 Tab3. Velocity VS Load Factor 1 59.197The cruising speed of the aircraft is 236.11 m/s.The dive speed of the aircraft is the maximum speed of the aircraft. The dive speed is equal tothe sum of the cruising speed and 60 knots. VD = 236.11 + 60 knots = 236.11 +30.56 m/s = 266.67 m/sGUSTS: The movement of air in turbulence is known as gusts. It produces changes in wingincidence, thereby subjecting the aircraft to sudden or gradual increases or decreases in lift fromwhich normal accelerations result. These may be critical for large, high speed aircraft and may possibly cause higherloads than control initiated maneuvers. Thus in the gust analysis, the change in load factor due to the gust is calculated. TheBCAR has given standard gust velocities for stall, cruise, dive speeds as 66, 50, 25 ft/s
- 17. P a g e | 17vigneshaeronautical@gmail.comrespectively. The small change in load factor ∆n due to the gust is calculated by assuming asharp gust. ˅ʿˀ The change in load factor ∆n = {ˁÈʽ{Where ρ density at cruising altitude, a lift slope, in radians U gust velocity in m/s V velocity of the aircraft in m/s W/S wing loading in N/m2 In the above formula, gusts are assumed to be sharp but it is usually graded, hencea relief factor called gust alleviation factor K is introduced in the term. The value of the K is obtained from the book “AIRPLANE AERODYNAMICSAND PERFORMANCE” by JAN ROSKAM " Ŷ {È{ Where K = , µ = 0 00 % Where ρ is the density, C is the mean aerodynamic chord, g is the acceleration due to gravity;CLα is the slope lift coefficient.The CLα (corrected for aspect ratio) is 0.0962/deg. ${ È9{ µ= 0/0A0/0Y {$0 { µ= {$ % 0 #0# 0 #0 %{ µ = 50.36 K= % 0 % = % % Κ = 0.796
- 18. P a g e | 18vigneshaeronautical@gmail.com K = Therefore ∆n = H Ŷ{È{For STALL SPEED V= 59.197 m/s, U= 20m/s ∆n = H Ŷ{È{ 0$ % 0 #0 %0$0 # = $0 ∆n = 1.062For CRUISE SPEED V=236.11 m/s, U= 15m/s ∆n = H Ŷ{È{ 0$ % 0 #0 %0#0$% ## = $0 ∆n = 2.725For DIVE SPEED V= 266.67 m/s, U= 7.5 m/s ∆n = H Ŷ{È{ 0$ % 0 #0 %0 0$ = $0 ∆n ∆ = 1.794
- 19. P a g e | 19vigneshaeronautical@gmail.com V 1+∆n ∆ 1-∆n ∆ 59.197 2.062 -0.062 236.11 3.725 -1.725 266.67 2.794 -1.794 Tab 4.load Factor of safety
- 20. P a g e | 20vigneshaeronautical@gmail.com 3 STRUCTURAL ANALYSIS AND DESIGN OF WINGSWING STRUCTURAL ANALYSIS: The structural design of the wing requires a complete quantitative knowledge of thedifferent loads it will be subjected to during its flight regime. These loads can be brieflyclassified as 1. Distributed loads - Loads such as aerodynamic loads, weight of the wing and weight of fuel. 2. Concentrated loads – Loads such as thrust, engine weight, landing gear weight and armament weight.LOADS ACTING ON WING: As both the wings are symmetric, let us consider the starboard wing at first. There are three primary loadsacting on a wing structure in transverse direction which can cause considerable shear forces and bending momentson it. They are as follows: Lift force (given by Schrenk’s curve) Self-weight of the wing Weight of the power plant Weight of the fuel in the wingSCHRENK’S CURVE: Lift is a component of the resultant aerodynamic force acting at the centre of pressure ofan aerodynamic chord, along a direction perpendicular to the direction of the relative wind. At aparticular altitude and at a specific angle of attack, Lift varies along the wing span due to thevariation in chord length along the span. Schrenks curve defines this lift distribution over thewing span of an aircraft. Since the wings of an aircraft are symmetrical about the longitudinalaxis, the Schrenks curve for the starboard wing alone can be obtained at first. This is given by Y Y y= { { where y1 linear variation of lift along the wing semi-span y2 equivalent elliptic lift distribution along the wing semi-span
- 21. P a g e | 21vigneshaeronautical@gmail.comTO FIND y1: Lift force is found along the line joining the aerodynamic centers of chords along thewing span. Hence, the wing is rotated about the wing root so that the line joining theaerodynamic centers becomes the horizontal line. a={ ) = G #$ a={ ) = G## #%É =6.55Lift per unit length at wing root = CL×0.5×ρ×V2×CR = 0.23884×0.5×1.4×236.1112×2.55 = 23766.98 N/mLift per unit length at wing tip = CL×0.5×ρ×V2× Ct =0.23884×0.5×1.4×236.1112×0.84 = 7829.12 N/m 6.55 m Fig.4 Linear lift distributionArea under trapezoid life distribution = 155673.719Equation of linear lift distribution for starboard wing Y1 = -1195.289x + 23766.98Equation of linear lift distribution for port wing we have to replace x by –x in general, Y1 = 1195.289x + 23766.98
- 22. P a g e | 22vigneshaeronautical@gmail.com X Y1 0 23766.98 1 22571.7 2 21376.42 3 20181.14 4 18985.86 5 17790.58 6 16595.3 6.55 15937.89 Tab5. Span VS Linear Lift Distribution 25000 LINEAR LIFT DISTRIBUTION 20000 15000 Series1 10000 Series2 5000 0 0 1 2 3 4 5 6 6.55 SPAN (a) Graph.4 Span VS Linear Lift DistributionElliptic Lift Distribution: Twice the area under the curve or line will give the lift which will be required toovercome weight Considering an elliptic lift distribution we get L/2 = W/2 = Where b1 is Actual lift at root A= And a is wing semi span
- 23. P a g e | 23vigneshaeronautical@gmail.com Lift at Tip b = 15138.35 N/m15138.35 N/m 6.55 m Fig 5. Elliptic Lift DistributionY2 = 1155.60 .Y x Y2 0 15138.317 1 14960.852 2 14415.354 3 13457.142 4 11987.651 5 9779.05 6 6072.292 6.55 0 Tab 6: Span VS Elliptic Lift Distribution
- 24. P a g e | 24vigneshaeronautical@gmail.comConstruction of Schrenk”s Curve:Schrenk”s Curve is given by ː Y Y1+Y2 = Y1+Y2 = -597.64 x + 45489.019 + 1155.60 .Y x Y1+Y2 0 19452.65 1 18766.28 2 17895.89 3 16819.14 4 15486.76 5 13784.82 6 11333.796 6.55 7968.95 Tab 7:SpanVsSchrenks value
- 25. P a g e | 25vigneshaeronautical@gmail.com Graph 2: Span vs Elliptic lift DistributionLoad Estimation on wings Description: The solution methods which follow Euler’s beam bending theory (σ/y=M/I=E/R) use thebending moment values to determine the stresses developed at a particular section of the beamdue to the combination of aerodynamic and structural loads in the transverse direction. Mostengineering solution methods for structural mechanics problems (both exact and approximatemethods) use the shear force and bending moment equations to determine the deflection andslope at a particular section of the beam. Therefore, these equations are to be obtained asanalytical expressions in terms of span wise location. The bending moment produced here isabout the longitudinal (x) axis.Loads acting on wing: As both the wings are symmetric, let us consider the starboard wing at first. There arethree primary loads acting on a wing structure in transverse direction which can causeconsiderable shear forces and bending moments on it. They are as follows: Lift force (given by Schrenk’s curve) Self-weight of the wing Weight of the power plant Weight of the fuel in the wingSelf-Weight (y3): Self-weight of the wing, Wwing = 5548.06 kg ×9.81 = 54426.46 N
- 26. P a g e | 26vigneshaeronautical@gmail.com Wport wing =-27213.23 N Wstar board wing = -27213.23 N { { {{ ˳% = % y3= -290.52 (x-6.55)2 x y 0 -12464.07 1 -8948.74 2 -6014.49 3 -3661.28 4 -1889.11 5 -697.97 6 87.88 6.55 0 Tab 8: SpanVs Load acting on wing Graph 3: Span Vs Load acting on wing
- 27. P a g e | 27vigneshaeronautical@gmail.comPower Plant: According to our design data, Our Aircraft power plant is attach to rear fuselage. So,power plant calculation won’t be calculated.Fuel Weight Distribution: Wf = 23215.55Consider as equation, yf = 1902.91x-12464.07 2 -8658.25 3 -6755.35 4 -4852.43 5 -2949.52 Fuel weight Distribution 0 -1000 2 3 4 5 6 6.55 -2000 -3000 -4000 -5000 -6000 -7000 -8000 -9000 -10000 Graph 4. Span Vs Fuel weight distribution
- 28. P a g e | 28vigneshaeronautical@gmail.com SHEAR FORCE BENDING MOMENTSHEAR FORCE AND BENDING MOMENT DIAGRAMS OF A WING DUETO LOADS IN TRANSVERSE DIRECTION AT CRUISE CONDITION: The solution methods which follow Eulers beam bending theory (σ/y=M/I=E/R) use thebending moment values to determine the stresses developed at a particular section of the beamdue to the combination of aerodynamic and structural loads in the transverse direction. Mostengineering solution methods for structural mechanics problems (both exact and approximatemethods) use the shear force and bending moment equations to determine the deflection andslope at a particular section of the beam. Therefore, these equations are to be obtained asanalytical expressions in terms of span wise location. The bending moment produced here isabout the longitudinal (x) axis. As both the wings are symmetric, let us consider the starboard wing at first. There arethree primary loads acting on a wing structure in transverse direction which can causeconsiderable shear forces and bending moments on it. They are as follows: → Lift force (given by Schrenks curve) → Self-weight of the wing → Weight of the powerplantS.No. Curve/Component Area/Structural weight (N) Centroid 1 Y1 155673.719 5.458 2 Y2 77896.859 2.781 3 Wing 54426.46 1.637 4 Fuel 23215.55 1.31 Tab 9: Centroid table
- 29. P a g e | 29vigneshaeronautical@gmail.com∑V=0 77836.859-54426.46-23215.55-VA = 0 VA = 194.85∑MA = 0 MA+ (54426.465×1.637)+(23215.55×1.31)-(155673.719×5.458)-(77836.859×2.781) MA – 946622.97 = 0 MA= 946622.97
- 30. P a g e | 30vigneshaeronautical@gmail.comSHEAR FORCE S.F1 = -298.84 x2 + 45489.019 x + 577.8 . ʹ - 42.903 Sin- Y YӘ1 ә -290.52 Ә - Y- Y ә -194.85 SPAN Shear Force 0 -19625.79 1 -8113.17 2 1496.55 3 9203.37 4 15007.29 5 18908.31 6 20906.43 6.55 21194.28 Tab 10: Span VS Shear ForceBENDING MOMENT: YB.M = -199.21 x3 + 15163.006 x2 + 288.9 [x (x . ʹ ) + 42.903 Sin -1 ] + 385.2 Y(42.903- x2)1.5-290.52 ( + 21.45 x2- 2.18 x3) + 94662.97
- 31. P a g e | 31vigneshaeronautical@gmail.com SPAN BENDING MOMENT 0 754870.44 1 1043525.353 2 1062831.914 3 1084364.515 4 1108829.591 5 1318721.935 6 1381469.212 6.55 943585.515 Tab 11 : Span VS Bending MomentNET SHEAR FORCE AND BENDING MOMENT DIAGRAM: 1600000 1400000 BENDING MOMENT 1200000 1000000 800000 600000 400000 200000 SHEAR FORCE 0 -8 -6 -4 -2 0 2 4 6 8 WING SPAN Graph.5 NET SHEAR FORCE vs BENDING MOMENT
- 32. P a g e | 32vigneshaeronautical@gmail.comShear force and bending moment diagrams due to loads along chordwisedirection at cruise condition:Aerodynamic center- This is a point on the chord of an airfoil section where the bendingmoment due to the components of resultant aerodynamic force (Lift and Drag) is constantirrespective of the angle of attack. Hence the forces are transferred to this point for obtainingconstant Ma.cShear center- This is a point on the airfoil section where if a force acts, it produces only bendingand no twisting. Hence the force is transferred to this point and the torque is found.Angle of Attack (max) = 15.00, Angle of Attack (Zero lift) = -1.0Cruise CL = 1.40 V = 236.11 m/sρ = 0.23884 kg/m3 CD = 0.0025SHEAR FORCE BENDING MOMENT:Co-efficient of force at normal direction Cn = CL cos ά + CD sin ά = 1.398Co-efficient of force at Chordwise direction CC = CL sin ά + CD cos ά = 0.026Therefore, Force per unit length= Cc×0.5×ρ×V2×C Force at Cr = 441.39 N/m Force at Ct = 145.39 N/mFor Linear, y = 23.05 x + 145.39 1Shear Force:Integrate Eqn. 1 Y - Y
- 33. P a g e | 33vigneshaeronautical@gmail.com Span wise Shear force 0 0 1 156.915 2 336.88 3 539.895 4 765.96 5 1015.07 6 1287.24 7 1582.45 8 1900.72 9 2242.03 10 2606.40 11 2993.81 12 3404.28 12.84 3766.88 Tab 12: Span wise VS Shear ForceBENDING MOMENT: #$ 3.841 x3+72.69x2 Span wise Bending moment 0 0 1 76.531 2 176.108 3 321.77
- 34. P a g e | 34vigneshaeronautical@gmail.com 4 536.584 5 843.575 6 1265.796 7 1826.293 8 2548.112 9 3454.299 10 4567.9 11 5911.961 12 7509.528 12.84 9064.425 Tab 13: Span wise VS Bending Moment 14000 BENDING MOMENT SHEAR FORCE 12000 BENDING MOMENT 10000 8000 6000 SHEAR FORCE 4000 2000 0 CHORDWISE Graph 6. Span wise VS Bending Moment
- 35. P a g e | 35vigneshaeronautical@gmail.comTORQUE DUE TO NORMAL FORCES AND CONSTANT PITCHINGMOMENT AT CRUISE CONDITION:Aerodynamic center- This is a point on the chord of an airfoil section where the bendingmoment due to the components of resultant aerodynamic force (Lift and Drag) is constantirrespective of the angle of attack. Hence the forces are transferred to this point for obtainingconstant Ma.cShear center- This is a point on the airfoil section where if a force acts, it produces only bendingand no twisting. Hence the force is transferred to this point and the torque is found The lift and drag forces produce a moment on the surface of cross-section of the wing,otherwise called a torque, about the shear center. Moment about the aerodynamic center getstransferred to the shear center. The shear center on the chord under which it is locates.Cruise condition (Normal Force) T= ½ Cn ρ V2 c × 0.034 C
- 36. P a g e | 36vigneshaeronautical@gmail.com = 1325.51 C2Where,C chordthe equation for chord can also be represented in terms of x by taking C = mx +k, C = 0.264 x +2.55Therefore, Torque = 30.751x3 + 8418.46 x + 892.33 x2 Span wise Torque at Normal force 0 0 1 9541.54 2 21052.24 3 34716.62 4 50719.18 5 69244.42 6 90476.85 7 114600.98 8 141801.31 9 172262.34 10 206168.6 11 243704.57 12 285054.76 12.84 322871.54
- 37. P a g e | 37vigneshaeronautical@gmail.com Tab 14:Span wise VS Torque at Normal forceTORQUE DUE TO CHORDWISE FORCE:Torque per unit length T2 = FC×0 T2 = 0TORQUE DUE TO MEAN AERODYNAMIC CHORD: Torque due to Ma.c = ˕ ×0.5×ρ×V2×C×C T3 = -3347.9 ×C2 T3 = -77.671 x3- 21768.04 x- 2253.806 X2 Span wise MEAN AERODYNAMIC CHORD 0 0 1 -24099.51 2 -53172.67 3 -87685.49 4 -128104.0 5 -174894.22 6 -228522.19 7 -289453.92 8 -358155.45 9 -435092.80 10 -520732.01 11 -615539.06 12 -719980.03 12.84 -815496.45
- 38. P a g e | 38vigneshaeronautical@gmail.com Tab 15: Spanwise VS Mean Aerodynamic ChordNET TORQUE: Then the different torque components are brought together in a same graph to make acomparisonThe net torque will be sum of all the above torques (i.e.) torques due to normal force, chordwiseforce, powerplant and aerodynamic moment. 400000 200000 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 -200000 Series1 Series2 -400000 Series3 -600000 -800000 -1000000 Graph.7 Span vs Net Torque Force
- 39. P a g e | 39vigneshaeronautical@gmail.com LOAD ESTIMATION OF WINGSWING STRUCTURAL LAYOUTSpecific Roles of Wing (Mainwing) Structure:The specified structural roles of the wing (or main plane) are: To transmit: wing lift to the root via the main span wise beam Inertia loads from the power plants, undercarriage, etc., to the main beam. Aerodynamic loads generated on the aerofoil, control surfaces flaps to the main beam. To react against: Landing loads at attachment points Loads from pylons/stores Wing drag and thrust loads To provide: Fuel tank age space Torsional rigidity to satisfy stiffness and aero-elastic requirements. To fulfill these specific roles, a wing layout will conventionally compromise: Span wise members (known as spars or booms) Chord wise members(ribs) A covering skin Stringers
- 40. P a g e | 40vigneshaeronautical@gmail.comBasic Functions of Wing Structural MembersThe structural functions of each of these types of members may be considered independently as:SPARS Form the main span wise beam Transmit bending and torsional loads Produce a closed-cell structure to provide resistance to torsion, shear and tension loads.In particular: Webs – resist shear and torsional loads and help to stabilize the skin. Flanges - resist the compressive loads caused by wing bending.SKIN To form impermeable aerodynamics surface Transmit aerodynamic forces to ribs stringers Resist shear torsion loads (with spar webs). React axial bending loads (with stringers).STRINGERS Increase skin panel buckling strength by dividing into smaller length sections. React axial bending loadsRIBS Maintain the aerodynamic shape Act along with the skin to resist the distributed aerodynamic pressure loads Distribute concentrated loads into the structure redistribute stress around any discontinuities Increase the column buckling strength of the stringers through end restraint Increase the skin panel buckling strength.SPAR DEFINITION: The maximum bending moment from previous section was found to be as 2897784.51Nm. Therefore we define 3 Spars with front spar at 15% of chord, middle spar at 45% of chordand rear spar at 70% of chord. The position of the three spars from the leading edge of the rootchord is given below as follows:
- 41. P a g e | 41vigneshaeronautical@gmail.comFront spar - 15% of chord = 2.442 mMiddle spar - 45% of chord = 7.326 mRear spar - 70% of chord = 11.396 m Bending moment M = Max BM * FOS * n = 2897784.51 × 1.5 × 3.8 = 16517371.71Nm The Structural load bearing members in the wing are the Spars and Stringers. Thebending moment carried by the Spars is 70% and that of Stringers is 30% of the total BendingMoment.Bending Moment taken by Spars is = 0.7 x 16517371.71 = 11562160.19 NmThe cross section of the spar chosen here is an I-sectionFor each spar we are determining the following parameters: A. Centroid B. Moment of Inertia C. Bending Moment D. Bending Stress
- 42. P a g e | 42vigneshaeronautical@gmail.comFRONT SPAR: Height of the spar = 38 cm Breadth of the spar = 16 cm Thickness of the spar = 4.5 cm Fig 11: Cross Section of Rear SparTo find out the centroid, the following calculations are made: Area(A) x y Ax Ay Ax2 Ay2 Icx IcyElement (cm2 ) (cm) (cm) (cm3 ) (cm3 ) (cm4 ) (cm4 ) (cm4 ) (cm4 ) 1 72 8 2.25 576 162 4608 364.5 121.5 1536 2 130.5 8 19 1044 2479.5 8352 47110.5 9145.8 220.22 3 72 8 35.75 576 2574 4608 92020.5 121.5 1536 Total 274.5 2196 5215.5 17568 139495.5 9388.87 3292.22
- 43. P a g e | 43vigneshaeronautical@gmail.comFront Spar Calculations: -L -MCentroid = X = = 8 cm; Y= = 19 cm - - I xx = Σ Icx + ΣAy2 – ΣAY2 I xx = (9388.87) + (139495.5) – (274.5)(19)2 I xx = 49789.88 cm4 I yy = Σ Icy+ ΣAx2 – ΣAX2 I yy = (3292.22) + (17568) – (274.5) (8)2 I yy = 3292.22 cm4The FRONT SPAR carries 35 % of the BM carried by the Spars. Thus, Front spar BM = 0.35 x 1156216019 N-cm = 404675606.7 N cmFront Spar Bending Stress: Bending Stress, σ z = { {y COORDINATES (y) BENDING STRESS POINTS (cm) (N/cm2) A 19 154425.68 B 14.5 117851.18 C 14.5 117851.18 D -14.5 -117851.18 E -14.5 -117851.18 F -19 -154425.68 The bending stress at various points whose co-ordinates are determined with centroid asthe origin are calculated from above formula and tabulated.
- 44. P a g e | 44vigneshaeronautical@gmail.comMIDDLE SPAR: Height of the spar = 41.6 cm Breadth of the spar = 18 cm Thickness of the spar = 5 cm Fig 12: Cross Section of Middle Spar
- 45. P a g e | 45vigneshaeronautical@gmail.comTo find out the centroid, the following calculations are made: Area(A) x y Ax Ay Ax2 Ay2 Icx IcyElement (cm2 ) (cm) (cm) (cm3 ) (cm3 ) (cm4 ) (cm4 ) (cm4 ) (cm4 ) 1 90 9 2.5 810 225 7290 562.5 187.5 2430 2 158 9 20.8 1422 3286.4 12798 68357.12 13147.7 329.17 3 90 9 39.1 810 3519 7290 137592.9 187.5 2430 Total 338 3042 7030.4 27378 206512.5 13522.7 5189.17Middle Spar Calculations -L -MCentroid = X = - = 9 cm; Y= = 20.8 cm - I xx = Σ Icx + ΣAy2 – ΣAY2 I xx = (13522.7) + (206512.5) – (338) (20.8)2 I xx = 60467.7 cm4 I yy = Σ Icy+ ΣAx2 – ΣAX2 I yy = (5189.17) + (27378) – (338) (9)2 I yy = 5189.17 cm4 The bending moment carried by the middle spar is 40% of the total bending momentcarried by the spars. Middle Spar BM = 462486407.6 N-cm 3 Bending Stress, σ z = { {y
- 46. P a g e | 46vigneshaeronautical@gmail.com COORDINATES (y) BENDING STRESS POINTS (cm) (N/cm2) A 20.8 159088.52 B 15.8 120846.09 C 15.8 120846.09 D -15.8 -120846.09 E -15.8 -120846.09 F -20.8 -159088.52 The bending stress at various points whose co ordinates are determined with centroid as co-ordinatesthe origin are calculated from above formula and tabulated.REAR SPAR Height of the spar = 17.72 cm Breadth of the spar = 7.6 cm Thickness of the spar = 2.5 cm
- 47. P a g e | 47vigneshaeronautical@gmail.comTo find out the centroid, the following calculations are made: Area(A) x y Ax Ay Ax2 Ay2 Icx IcyElement (cm2 ) (cm) (cm) (cm3 ) (cm3 ) (cm4 ) (cm4 ) (cm4 ) (cm4 ) 1 19 3.8 1.25 72.2 23.75 274.36 29.6875 9.896 91.45 2 31.8 3.8 8.86 120.84 281.748 459.19 2496.287 428.76 16.56 3 19 3.8 16.47 72.2 312.93 274.36 5153.957 9.896 91.45 Total 69.8 265.24 618.428 1007.9 7679.932 448.552 199.46Rear Spar Calculations -L -M Centroid = X = = 3.8 cm; Y= = 8.86 cm - - I xx = Σ Icx + ΣAy2 – ΣAY2 I xx = (448.552) + (7679.932) – (69.8) (8.86)2 I xx = 2649.184 cm4 I yy = Σ Icy+ ΣAx2 – ΣAX2 I yy = (199.46) + (1007.9) – (69.8) (3.8)2 I yy = 199.46 cm4Rear Spar carries 25 % of the spar Bending Moments.Bending Moment = 289054004.8 N-cm 3 Bending Stress, σ z = { {y
- 48. P a g e | 48vigneshaeronautical@gmail.comThe bending stresses at various points are obtained as: Rear Spar Bending COORDINATES (y) BENDING STRESS Stress POINTS (cm) (N/cm2) A 8.86 966719.74 B 6.36 693943.29 C 6.36 693943.29 D -6.36 -693943.29 E -6.36 -693943.29 F -8.86 -966719.74 The bending stress at various points whose co-ordinates are determined with Centroid asthe origin are calculated from above formula and tabulated.
- 49. P a g e | 49vigneshaeronautical@gmail.com 4 STRUCTURAL ANALYSIS AND DESIGN OF FUSELAGEFUSELAGE STRUCTURAL LAYOUT: The fuselage is the main structure, or body, of the aircraft. It provides space forpersonnel, cargo, controls, and most of the accessories. The power plant, wings, stabilizers, andlanding gear are attached to it. There are two general types of fuselage construction welded steel truss and monocoque construction—weldeddesigns. The welded steel truss was used in smaller Navy aircraft, and it is still being used insome helicopters. The monocoque design relies largely on the strength of the skin, or covering, to carry onocoquevarious loads. The monocoque design may be divided into three classes - monocoque,semimonocoque and reinforced shell. Fig 13:Semimonocoque And Monocoque
- 50. P a g e | 50vigneshaeronautical@gmail.com The true monocoque construction uses formers, frame assemblies, and bulkheads to give shape to the fuselage. However, the skin carries the primary stresses. Since no bracing members are present, the skin must be strong enough to keep the fuselage rigid. The biggest problem in monocoque construction is maintaining enough strength while keeping the weight within limits. Semimonocoque design overcomes the strength-to-weight problem of monocoque construction. In addition to having formers, frame assemblies, and bulkheads, the semimonocoque construction has the skin reinforced by longitudinal members. The reinforced shell has the skin reinforced by a complete framework of structural members.Different portions of the same fuselage may belong to any one of the three classes. Most areconsidered to be of semimonocoque-type construction. The semimonocoque fuselage is constructed primarily of aluminum alloy, although steeland titanium are found in high-temperature areas. Primary bending loads are taken by thelongerons, which usually extend across several points of support. The longerons aresupplemented by other longitudinal members known as stringers. Stringers are more numerousand lightweight than longerons. The vertical structural members are referred to as bulkheads, frames, and formers. Theheavier vertical members are located at intervals to allow for concentrated loads. These membersare also found at points where fittings are used to attach other units, such as the wings andstabilizers. The stringers are smaller and lighter than longerons and serve as fill-ins. They have somerigidity but are chiefly used for giving shape and for attachment of skin. The strong, heavylongerons hold the bulkheads and formers. The bulkheads and formers hold the stringers. All ofthese join together to form a rigid fuselage framework. Stringers and longerons prevent tensionand compression stresses from bending the fuselage. The skin is attached to the longerons, bulkheads, and other structural members andcarries part of the load. The fuselage skin thickness varies with the load carried and the stressessustained at particular location.There are a number of advantages in using the semimonocoque fuselage. The bulkhead, frames, stringers, and longerons aid in the design and construction of a streamlined fuselage. They add to the strength and rigidity of the structure. The main advantage of the semimonocoque construction is that it depends on many structural members for strength and rigidity. Because of its stressed skin construction, a semimonocoque fuselage can withstand damage and still be strong enough to hold together.
- 51. P a g e | 51vigneshaeronautical@gmail.comLoads and its distribution: To find out the loads and their distribution, consider the different cases. The maincomponents of the fuselage loading diagram are: Weight of the fuselage Engine weight Weight of the horizontal and vertical stabilizers Tail lift Weight of crew, payload and landing gear Systems, equipment, accessoriesSymmetric flight condition, steady and level flight: (Downward forces negative) Values for thedifferent component weights are obtained from aerodynamic design calculations. Length Equipment and Shear BendingS.No. from Ref. Weight (N) Moment Component Force Moment point 1 Nose 0.58 3933.81 2281.609 3933.81 2281.809 2 Pilot(2) 2.02 1962.00 3963.24 5895.81 6244.84 3 Cockpit 2.62 4944.24 12953.908 10840.05 19198.757 4 Wing 6.62 54426.46 360303.16 65266.51 379501.91 5 Passenger(3) 6.63 3237.30 21463.29 68503.81 400965.2 6 Passenger(3) 7.92 3237.30 25639.41 71741.11 426604.61 7 Passenger(2) 9.85 2060.10 20291.98 73801.21 446896.59 8 Crew(2) 10.32 2060.10 21260.23 75861.31 468156.82 Weight of 9 10.50 2000.68 21007.14 77861.99 489163.96 Fuselage Sheet 10 Power Plant 14.23 3937.83 56036.74 81799.92 545200.72 11 Horizontal Tail 17.10 13232.807 226280.9 95032.72 771481.607 12 Vertical tail 17.60 8971.34 157895.58 104004.06 929377.18SHEAR FORCE AND BENDING MOMENT DIAGRAM
- 52. P a g e | 52vigneshaeronautical@gmail.comTo determine the shear force and bending moment diagram for the wing we assume that the wingis a cantilever beam with the root end fixed while the tail end is free.For a cantilever beam the shear force is a given by, Shear Force = Rx Bending Moment = $Tabulation for the values of shear force and bending moment at various positions along the spanis as follows. Graph 9. Length from Ref. point Vs Shear Force 120000 100000 80000 Shear force 60000 40000 20000 0 0.58 2.02 2.62 6.62 6.63 7.92 9.85 10.32 10.5 14.23 17.1 17.6 Length from Ref. point Graph 10. Lengtth from Ref. point Vs Bending Moment 1000000 900000 800000 Bending moment 700000 600000 500000 400000 300000 200000 100000 0 0.58 2.02 2.62 6.62 6.63 7.92 9.85 10.32 10.5 14.23 17.1 17.6 Leght from Ref. point FUSELAGE STRUCTURAL ANALYSIS
- 53. P a g e | 53vigneshaeronautical@gmail.com Structural analysis of fuselage like that of wing is of prime importance while designingan aircraft. As the fuselage is the one which houses the pilot, the power plant and also part of thepayload its structural integrity is a matter of concern. While analyzing the fuselage structure thesection must be idealized. Idealization involves the conversion of a stringer and itsaccompanying skin thickness into a concentrated mass known as a boom. The shear flowanalysis of the fuselage simulating flight conditions is shown below. X Y (m) (m) 1.005 0 0.985 0.26 0.88 0.48 0.72 0.72 0.48 0.88 0.26 0.985 0 1.005 -0.26 0.985 -0.48 0.88 -0.72 0.72 -0.88 0.48 -0.985 0.26 -1.005 0 -0.985 -0.26 -0.88 -0.48 -0.72 -0.72 -0.48 -0.88 -0.26 -0.985 0 -1.005 0.26 -0.985 0.48 -0.88 0.72 -0.72 0.88 -0.48 0.985 -0.26 1.005 0
- 54. P a g e | 54vigneshaeronautical@gmail.com Stringer location in Fuselage 1.5 1 0.5 0 -1.5 -1 -0.5 0 0.5 1 1.5 -0.5 -1 -1.5The stringer used is of Z type. The following are its dimensionsCross sectional area of each stringer is 100mm2 Fig 14: Cross section of Z Z-section
- 55. P a g e | 55vigneshaeronautical@gmail.com The above stringer section is uniformly used throughout the fuselage as shown above inorder to provide the fuselage the required load carrying capacity. The diagram showed adjacentis of the idealized fuselage structure. The idealization process is carried out in the following way.STRESS ANALYSIS:IDEALIZATION:The boom 1 is given by where, B1 Area of Boom 1 tD Thickness of skin panel b Circumferential distance between 2 stringersBy Symmetry, B1 = B9, B2 = B8, B10 = B16, B3 = B7 , B11 = B15, B4 = B6 = Bl2 = B14 ,B5 = B13 # ## # ## B1=100+ (0.65×1.37× ) [2+{ {] + (0.65×1.37× )[2+{ {] =815582.12Similarly for boom 2 , B2 = 815582.12 mm2Similarly B3 = 815582.12 mm2, B4 =815582.12 mm2. We note that stringers 5 and 13 lie on theneutral axis of the section and are therefore unstressed; the calculation of boom areas B5 and B13does not then arise.Thus, we have B1:B16 = 815582.12 mm2
- 56. P a g e | 56vigneshaeronautical@gmail.comWe know that, Ixx = By2 LL = 24.67 m4; LL = 13.77 m4; LL = 6.12 m4; LL = 1.11 m4Maximum bending moment = 2897784.51 NmHence the Bending moment acting on the fuselage M = Max.B.M × n× FOS =2897784.51 × 3.8×1.5 =16517371.71 Nm Ixx = 24.67 m4The value of stress acting is given by the expression: # # % # # 0M ={ { $ STRINGER/BOOM Y (m) STRESS x 10 6 (Nm-2) 1 5.5 3.68 2, 16 4.11 2.75 3, 15 2.74 1.83 4, 14 1.37 0.9 5, 13 0 0 6, 12 -1.37 -0.9 7, 11 -2.74 -1.83 8, 10 -4.11 -2.75 9 -5.5 -3.68
- 57. P a g e | 57vigneshaeronautical@gmail.com 5 Design of Miscellaneous MembersWing fuselage intersection The 8 seater business jet aircraft has low wing configuration, thus the entire wingstructure continues in the way of airplane body.Four pin design concept: This concept is adopted as it is the most simple and straight forward method used inHorizon 4000 transport, during 1950s. The lift and moment loads can be carried between thewing and fuselage by simple shear on the four pins. The drag and thrust is taken by breather web.This design allows the wing spar and fuselage bulkheads to deflect independently of each othersuch that no spar moment is directly transferred to the bulkheads. The wing-body juncture produces aerodynamic interference which in turn promotes flowseparation with its attendant higher drag and unsteady buffeting. This adverse pressure gradientand consequent flow separation can be minimized using contoured surface called fillet.Engine mount An engine mount is a frame that supports the engine and holds it to the fuselage ornacelle. Usually it is made of built up sheet metal, welded steel tubing. The turbofan engine,“HONEY WELL TFE731-20” is wing mounted. A typical turbofan engine installation for alow wing aircraft configuration similar to that of this aircraft is shown below,
- 58. P a g e | 58vigneshaeronautical@gmail.com The pylon has three spars (longerons) – Upper, middle and lower- and three majorbulkheads, and is attached to the wing at four primary points. These are two mid-spar fittings, an mid-upper link and a diagonal brace (drag strut). The attachment pins are secured with “fuse” boltswhich are hollow carbon steel devices that have been heat treated to shear fail at a definite load.In the landing break way condition (wheels (wheels-up landing); the sequence is designed to fail the anding);upper and lower links so that the pylon rotates around the mid spar and upward. The wing pylon mid-spardesign provides considerable load path redundancy such that an upper link can fail, partially orcompletely, and there is an alternate path lower diagonal brace. The below figure shows the ere path-engine mounts.
- 59. P a g e | 59vigneshaeronautical@gmail.comEmpennage DesignHorizontal Stabilizer: The horizontal tail of the aircraft is conventional and consists of a fixed tail box. Thehorizontal stabilizer is usually a two spar structure consisting of a Centre structural box sectionand two outer sections. The stabilizer assembly is interchangeable (symmetrical airfoil section)as a unit at the fuselage attach points and the outer sections are interchangeable at the attachmentto the center box.The two basic horizontal stabilizer box constructions for modern transports are 1. Box constructions with spars, closer light rib spacing (usually less than 10 inches) and surface (may be tapered skins) without stringer reinforcement. The feature of this design is the low manufacturing cost and high torsional stiffness require by the flutter analysis. 2. Box construction with spar stronger ribs and surface skins with stringer reinforcements (skin-stringer or integrally stiffened panels) is a lighter weight structure.
- 60. P a g e | 60vigneshaeronautical@gmail.comVertical Stabilizer: The structural design of the vertical stabilizer is essentially the same as for the horizontalstabilizer is essentially the same as for horizontal stabilizers. The vertical stabilizer box is a twoor multi spar structure (general aviation airplanes usually use single spar design) with coverpanels (with or without ribs). The root of the box is terminated at the aft fuselage conjuncturewith fittings or splices.
- 61. P a g e | 61vigneshaeronautical@gmail.comWING FUEL TANKS: In addition to providing the required strength and stiffness, the structural box almostalways has to provide fuel space. Integral tanks, as opposed to separate internally supportedtypes, are preferred since their use enables the maximum advantage to be taken of the availablevolume. Integrally machined or moulded constructions, which use a small amount of largecomponents, are obviously an advantage since sealing is reduced to a minimum. The majorproblem occurs at tank end ribs, particularly in the corners of the spar web and skins, and atlower surface access panels. The corner difficulty is overcome by using special “suitcase” cornerfittings. Access panels should be large enough for a person to get through so that the inside can beinspected and resealed if necessary. On shallow section wings, the access has to be in the lowersurface so that the operator can work in an acceptable way even if the depth is insufficient toclimb in completely. Apart from the sealing problems, lower surface access panels are in what isprimarily a tension skin and so introduce stress concentrations in an area where crackpropagation is a major consideration. The access panels are arranged in a span-wise line so theedge reinforcing can be continuous and minimum stress concentration due to the cut-outs.Access panels are often designed to carry only shear and pressure loads, the wing bending beingreacted by the edge reinforcing members. A deep wing can avoid these problems by using uppersurface access panels but this is not a preferred aerodynamic solution.AUXILIARY SURFACES The structural layout of the auxiliary lifting surfaces is generally similar to that of thewing but there are differences, in part due to the smaller size and in part due to the need toprovide hinges or supports. The latter implies that each auxiliary surface is a well-defined.HINGED CONTROL SURFACES: Conventional training edge control surfaces are almost invariably supported by a numberof discrete hinges, although continuous, piano type, hinges may be used for secondary tabs. Tosome degree the number and location of the discrete hinges depends upon the length of thecontrol. The major points to be considered are: The bending distortion of the control relative to the fixed surface must be limited so that the nose of the control does mot fouls the fixed shroud. The control hinge loads and the resulting shear forces and bending moments should be equalized as far as is possible. Structural failure of a single hinge should be tolerated unless each hinge is of fail-safe design and can tolerate cracking one load path.
- 62. P a g e | 62vigneshaeronautical@gmail.com These points suggest the use of a relatively large number of discrete hinges but there aredifficulties associated with this solution there are the obvious loads likely to be induced in thecontrol by the distortion under load of the main surface to which it is attached may be significant.These problems do not arise if only two hinge points are used as any span-wise distortion ormisalignment can be accommodated by designing one of the hinges so that it can rotate about avertical axis. When more than two hinges are used the „floating hinge concept cannot fullyovercome the problems. However, it is possible to design the control surface so that it is flexiblein bending and indeed the more hinges there are the easier this is to accomplish. One hinge mustalways be capable of reacting side loads in the plane of the control surface. The hinges aresupported near to the aft extremities of the main surface ribs.PIVOTED CONTROL SURFACES In certain high-performance aircraft, the whole of a stabilizing or control surface on oneside of the aircraft may be pivot about a point on its root chord. Clearly in this case, the structuralconsiderations are dominated by the need to react all the forces and moments at the pivot andoperating points. Thus the structural layout may consist of an integral root rib or pivot or stubspar arrangement to which is attached a number of shear webs fanning out towards theextremities of the surface, possibly in conjunction with full depth honeycomb. High skin shearloading is inevitable due to the need to bring the loads to the two concentrated points. Shearloads due to torsion may be limited by locating the operating point on the root rib some distanceaway from the pivot. Some designs incorporate the pivot into the moving surface with the support bearings onthe fuselage, while on others the pivot is attached to the fuselage and the bearings are in thesurface. The bearings should be as far apart as local geometry allows minimizing loads resultingfrom the reaction of the surface bending moment.HIGH LIFT SYSTEMS There is a wide variety of leading and trailing edge high-lift systems. Some types aresimply hinged to the wing, but many require some degree of chord-wise extension. This can beachieved by utilizing a linkage, a mechanism, a pivot located outside the aerofoil contour or,perhaps most commonly, by some form of track. Trailing edge flaps may consist of two or moreseparate chord-wise segments, or slats, to give a slotted surface and these often move on tractsattached to the main wing structure. The majority of flaps and slats are split into span wise segments of no greater lengthsthan can be supported at two or three locations. As with control surfaces, the locations of thesupport points are established so as to minimize local deformations since the various slots arecritical in determining the aerodynamic performance. Sometimes the actuation may be located at
- 63. P a g e | 63vigneshaeronautical@gmail.coma different pan wise position from the support points. This is often a matter of convenience,layout clearances, and the like. The structural design of flaps is similar to that of control surfaces but it s simpler as thereis no requirement for mass balance, the operating mechanisms normally being irreversible. Onlarge trailing edge flap components, there is often more than one spar member. Especially whenthis assists in reacting the support or operating loading. There may be a bending stiffnessproblem in the case of relatively small chord slat segments and full depth honey combs can beused to deal with this. Figure shows a cross section of a typical slotted flap of metal constructionbut the same layout applies if composite materials are used. In many cases the slipstream or afflux from power plants impinges upon a flap and this islikely to require special consideration in the design. Additional stiffness is not necessarily theanswer because acoustic fatigue characteristics are often worse at higher panel frequencies.However the extensive local support offered by sandwich construction, either in panel or fulldepth configuration, is usually beneficial. This leads naturally to the application of reinforcedplastic materials. Trailing edge flaps tends to be prone to damage by debris thrown up by thelanding gear and it may be desirable to use Kevlar or glass rather than carbon fibers for the lowersurface, but material compatibility needs to be considered.ATTACHMENT OF LIFTING SURFACES The joint of the fuselage with the wing is subjected to heavy load inputs and there is apotential for considerable relative distortion. This distortion is usually accepted and the wingcentre box is built completely into the fuselage, the resulting constraint stresses being allowedfor. It is usual for the wing structure of large aircraft to include a production joint at the side ofthe fuselage and this is virtual essential for swept wings. It is sometimes possible to arrange the wing pick-ups as pivots on the neutral axis or setthem on swinging links. In this case, the relative motion is allowed to take place and there are noinduced stresses. Structural assembly of the wing to the fuselage is relatively simple. Similar remarks also apply to the attachment of the horizontal stabilizer when theincidence setting is fixed. If the surface is also used for trimming or control, some specialconsideration is necessary in the location of the pivot and actuation fittings. These usuallyrequire a relatively heavily loaded rib or a pair of ribs, and where possible at least one of theattachment points should be close to the rib or spar intersection. It is desirable to arrange for thelateral distance between the pivots to be as great as possible to minimize pivot loads resultingfrom asymmetric span-wise loading. When the controls are manually operated, it is simplest ifthe elevator-hinge line and pivot coincide. Fins are usually built integrally with the rear fuselage. This is mainly due to the differentform of loading associated with the geometric asymmetry.
- 64. P a g e | 64vigneshaeronautical@gmail.comFlutter: Flutter as the dynamic instability of an elastic body in an airstream. It is found mostfrequently in aircraft structures subjected to large aerodynamic loads such as wings, tail units andcontrol surfaces. Flutter occurs at a critical or flutters speed Vf which in turn is defined as the lowestairspeed at which a given structure will oscillate with sustained simple harmonic motion. Flightat speeds below and above the flutter speed represents conditions of stable and unstable (that isdivergent) structural oscillation, respectively. Generally, an elastic system having just one degree of freedom cannot be unstable unlesssome peculiar mechanical characteristic exists such as a negative spring force or a negativedamping force. However, it is possible for systems with two or more degrees of freedom to beunstable without possessing unusual characteristics. The forces associated with each individualdegree of freedom can interact, causing divergent oscillations for certain phase differences. The flutter of a wing in which the flexural and torsional modes are coupled is animportant example of this type of instability. Some indication of the physical nature of wingbending–torsion-flutter may be had from an examination of aerodynamic and inertia forcesduring a combined bending and torsional oscillation in which the individual motions are 90 outof phase. In a pure bending or pure torsional oscillation the aerodynamic forces produced by theeffective wing incidence oppose the motion; the geometric incidence in pure bending remainsconstant and therefore does not affect the aerodynamic damping force, while in pure torsion thegeometric incidence produces aerodynamic forces which oppose the motion during one-half ofthe cycle but assist it during the other half so that the overall effect is nil. Thus, pure bending orpure torsional oscillations are quickly damped out. This is not the case in the combinedoscillation when the maximum twist occurs at zero bending and vice versa; i.e. a 90 phasedifference. The type of flutter described above, in which two distinctly different types of oscillatingmotion interact such that the resultant motion is divergent, is known as classical flutter. Othertypes of flutter, non-classical flutter, may involve only one type of motion. For example, stallingflutter of a wing occurs at a high incidence where, for particular positions of the span wise axisof twist, self-excited twisting oscillations occur which, above a critical speed, diverge.Aileron Buzz: Another non-classical form of flutter, aileron buzz, occurs at high subsonic speeds and isassociated with the shock wave on the wing forward of the aileron. If the aileron oscillatesdownwards the flow over the upper surface of the wing accelerates, intensifying the shock and
- 65. P a g e | 65vigneshaeronautical@gmail.comresulting in a reduction in pressure in the boundary layer behind the shock. The aileron,therefore, tends to be sucked back to its neutral position. When the aileron raises the shockintensity reduces and the pressure in the boundary layer increases, tending to push the aileronback to its neutral position. At low frequencies these pressure changes are approximately 180 out of phase with theaileron deflection and therefore become aerodynamic damping forces. At higher frequencies acomponent of pressure appears in phase with the aileron velocity which excites the oscillation. Ifthis is greater than all other damping actions on the aileron a high frequency oscillation results inwhich only one type of motion, rotation of the aileron about its hinge, is present, i.e. aileronbuzz. Aileron buzz may be prevented by employing control jacks of sufficient stiffness to ensurethat the natural frequency of aileron rotation is high.Buffeting: Buffeting is produced most commonly in a tail plane by eddies caused by poor airflow Inthe wing wake striking the tail plane at a frequency equal to its natural frequency; a resonantoscillation having one degree of freedom could then occur. The problem may be alleviated byproper positioning of the tail plane and clean aerodynamic design.
- 66. P a g e | 66vigneshaeronautical@gmail.com
- 67. P a g e | 67vigneshaeronautical@gmail.com 7 CONCLUSION In conclusion, the series of short range aircrafts incorporated many unique design offuture that was never seen on an operational aircraft. The design of these aircrafts points the wayfor the design of future of very high mach airplanes. The airplane has gone through many design modifications since its early conceptualdesigns expected, among these was a growth in weight. The document to provide information onthe trends in various aircraft characteristics that may influence general long-term airport planningand design. These are strong indications that future trends could see the coexistence of very highcapacity aircraft modules of similar capacities for the long range/very long range operations.Cargo payloads, which include mail, express and freight, are increasing in size and weight aslarger aircraft service with the airlines, To ensure continued growth in payload and the profitability of cargo operations,improvements in methods, equipment and terminal facilities will be required in order to reducecargo handling costs and aircraft ground time and to provide improved service for the shippers. We have enough hard work for this design project. A design never gets completed in aflutter sense but it is one step further towards ideal system. But during the design of this aircraft,we learnt a lot about aeronautics and its implications when applied to an aircraft design.

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