Kirubagaran mazhalai Project final report
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Kirubagaran mazhalai Project final report Kirubagaran mazhalai Project final report Document Transcript

  • УДК 629.7Інв. № МІНІСТЕРСТВО ОСВІТИ І НАУКИ, МОЛОДІ ТА СПОРТУ УКРАЇНИ Національний аерокосмічний університет ім. М. Є. Жуковського «Харківський авіаційний інститут» Кафедра проектування літаків i вертольотів ДО ЗАХИСТУ ДОПУСКАЮ Завідувач кафедри д–р техн. наук, проф. _________О. Г. Гребеніков (підпис)PASSENGER AIRCRAFT INTEGRATED DESIGNING AND MODEL ANALYSIS Пояснювальна записка до випускної роботи магістра, напрямок 8.100101 — «Авіація та космонавтика» Фах — «Літаки і вертольоти» (номер зал. книжки без позначки «№»)Виконавець студент гр. 16Е-2 KIRUBAGARAN MAZHALAI PRIYAN (№ групи) (І.Б.П.) (підпис, дата) Керівник–консультант з основного розділу к. техн. наук, доц. S. Trubaev (науковий ступінь, вчене звання) (І.Б.П.) (підпис, дата) Нормоконтролер к. техн. наук, доц. S. Trubaev (науковий ступінь, вчене звання кєрівника)(І.Б.П.) (підпис, дата)
  • [1] Ministry of Science and Education, Youth and Sports of Ukraine National Aerospace University,named by N.E. Zhukovskiyj «Kharkov Aviation Institute» Faculty of aircraft and helicopter construction Aircraft and helicopter design department «Approved by» Head of department №103, prof.___________ Grebenikov А. G. «___»_______________ 201__ ASSIGNMENT FOR A FINAL WORK OF AN APPLICANT for a masters degree 8.05110101 «Aircraft and helicopter»group 16 E- 2 students name KIRUBAGARAN MAZHALAI (Name) SUBJECT OF GRADUATION PROJECT «PASSENGER AIRCRAFT INTEGRATED DESIGNING AND MODEL ANALYSIS»Initial data for design: Vmax – _835____ km/h; Vkr – __760__ km/h; Vу – __14___ m/s; Hmax– _11____ km; Hcr – __10___ km; L – ___2000__ km; Lp – __1970___ m; npas. – _ 47____men; ncrew – ___4__ men.; mp/l – _39___ t;Т – __4____h; Кcr – __0.10___. Graduation Project. Table of Contents Abstract Design section 1. Computer–aided general design of aircraft Introduction, design goal – setting and tasking 1.1. Purpose, aircraft performance requirements, conditions of production and operation, limitations imposed by aviation regulations in design of an
  • [2] aircraft. 1.2. Statistical data collection, processing and analysis. Selection of aircraft main relative initial parameters (Characteristics). 1.3. Selection and grounds of aircraft configuration, type of its power plant. 1.4. Selection of engines and examination of take off run. 1.5. Determination optimization of aircraft components and design parameters. 1.6. Development of design–structural configuration, aircraft center of gravity. 1.7. Standard specification of designed aircraft. Realization of calculations, models and drawings: master–geometry of aircraft surface, outline drawing (format А1); Design–structural layout of aircraft (format А1). 2. Impact analysis in changes of aircraft component design parameters under their optimization in aerodynamic and weight characteristics of aircraft 2.1. Determination of designed aircraft drag. 2.2. Lift force, induced drag, aircraft polar curve, aircraft lift–drag ratio, aircraft polar. 2.3. Longitudinal moment and location of aerodynamic center of aircraft. 2.4. Influence of aircraft design parameters on its aerodynamic and weight characteristics. ______________________________________________________________ ___3. Integrated designing and computer–aided modeling __________SURFACE MODEL__________ of designed aircraft 3.1. Development of unit master–geometry. 3.2. Determination of loads acting on unit. Realization of calculations, models and drawings: unit master–geometry;
  • [3] 4. Integrated designing and computer–aided modeling of aircraft systems 4.1. Hydraulic system designing and modeling. 4.2. Maintenance Manual of designed system. ________________________________________________________________ _ Realization of calculations, models and drawings: system schematic diagram (format А2); Technological Section 5. Development of aircraft unit manufacturing technique 5.1. Development of enlarged production (manufacturing) methods in assembly of units: selection of tools and equipment, specifications for delivery of parts and assembly units, development of production charts for assembly procedure, standardization, assembly cycle schedule._________________________________________________________________ Realization of calculations, models and drawings: Economical section 6. Calculation of economic efficiency characteristics 6.1. Business plan: companys history, aircraft characteristic, product market, marketing, personnel and management, risk analysis and their prevention. 6.2. Project financing: sources of financing, receipts and expenditures – calculation of expenditures for designing and manufacturing, calculation of cost value, price income, calculation of companys minimal internal funds, determination of point of make out, calculation of direct and indirect costs. 6.3. Total transportation cost value and companys revenue. 6.4. Income from project. 6.5. Influence in change of aircraft and its units design parameters on aircraft efficiency criteria. 7. Special assignment Cabin layout and interiors design of the aircraft. Seating arrangement with high comfort level ______________________________________________________
  • [4]2. Explanatory note contents (list of questions subjected to development): in compliance with assignment. Design-explanatory note with Figures, Tables involved in text – up to 120 pages.3. List of Graph materials (with obligatory drawings clearly specified): graph material and presentation in strict correspondence to the assignment Information on CD–R or DVD+/–R medium installed in department computer network prior to defense4. Date of assignment issue:5. Date of final project presentation: Project supervisor (Date, signature) Assignment accepted to fulfillment « » 200 (Date, students signature)
  • [5] 2012 CONTENTSABSTRACT ………………………………………………………………………… 4INTRODUCTION………………………………………………………………… 5AIRCRAFT DESIGN PROCESS………………………………………………… 7 GENERAL DESIGNING OF AIRCRAFTPURPOSE OF THE AIRCRAFT………………………………………………….11REQUIREMENTS FOR FLIGHT PERFORMANCES………………………….14DESIGN CHART OF THE DESIGNED AIRCRAFT……………………………15PROTOTYPE DATAS………………………………………………………………17SELECTION OF AIRCRAFT MAIN RELATIVE INITIAL PARAMETERS…..24CALCULATION OF AIRCRAFT MASSES THROUGH THE SOFTWARE AND ITSRESULTS……………………………………………………………………………..25ZERO APPROXIMATION…………………………………………………………31STATISTICAL COMPUTATION OF MASSES OF AIRCRAFT………………..32AIRCRAFT OPTIMIZATION AND DESIGN PARAMETERS…………………33SELECTION AND GROUNDS OF AIRCRAFT CONFIGURATION…………..43SELECTION OF ENGINE…………………………………………………………..55AVERAGE BETWEEN GRAPHICAL, SEMI-EMPIRICAL & STATISTICALMETHOD……………………………………………………………………………….63MAXIMUM TAKE-OFF MASS………………………………………………………66CENTER OF GRAVITY………………………………………………………………68DESIGN STRUCTURAL CONFIGURATION …………………………………….83 AERODYNAMICSAIRCRAFT DESIGN PARAMETERS ON AERODYNAMIC CHARACTERISTICS………………………………………..83CALCULATION OF AERODYNAMIC PARAMETERS USING THE SOFTWARE…………………………………………92CALCULATION OF ZERO DRAG COEFFICIENT FOR TAKE-OFF ANDLANDING…………………………….109
  • [6] INTEGERATED DESIGN OF AIRCRAFT AND LOAD CALCULATIONAIRCRAFT MASTER GEOMETRY USING UNIGRAPHICS………………….114WING LOAD CALCULATION……………………………………………………..118CALCULATION OF THE DISTRIBUTED FUEL LOAD ON A PLANE WING…119THE WING STRUCTURE MASS LOAD ALLOCATION…………………………125SHEAR FORCE, BENDING MOMENT AND REDUCED MOMENT……………127CALCULATION SCHEME OF REDUCED MOMENT FROM CONCENTRATEDLOADS AND FROM ALL LOADS……………………………………………………132 AIRCRAFT SYSTEMS DESIGN AND SCHEMATIC LAYOUTAIRCRAFT HYDRAULIC SYSTEM……………………………………….134HYDRAULIC FLUID…………………………………………………………135COMPONENTS INVOLVED IN HYDRAULIC SYSTEM…………………137HYDRAULIC DESCRIPTION OF THE DESIGNED AIRCRAFT……….151HYDRAULIC SYSTEM PANEL……………………………………………..154HYDRAULIC SYSTEM MAINTENANCE………………………………….157 MANUFACTURING TECHNOLOGY OF VERTICAL RIBAIRCRAFT RIB……………………………………………………………………..159RIB CONSTRUCTION……………………………………….……………………..163PRODUCTION METHOD OF PARTS OF THE RIB……………………………166ASSEMBLY PROCEDURE OF THE RIB…………………………………………168STAGES OF FORMATION OF RIB DIMENSIONS USING TEMPLATES……171AIRCRAFT VERTICAL ASSEMBLY JIG DESIGN LAYOUT DIAGRAM…….173 ECONOMICAL SECTIONECONOMIC EFFICIENCY CHARACTERISTICS CALCULATION………..174
  • [7] SPECIAL ASSIGNMENTINTERIOR CABIN LAYOUT AND SEATING ARRANGEMENT……..…………..179FULFILLING REQUIREMENTS OF THREE ABREAST SEATING LAYOUT….179.CABIN DIMENSIONING FOR 3- ABREAST SEATING…………………………..…179INTERIOR ARRANGEMENT – CROS SSECTION (TYPICAL)……………………180DETERMINATION OF DESIGNED AIRCRAFT CABIN CROSS-SECTION…......181DETERMINATION OF CABIN LENGTH FOR HIGH COMFORT LEVEL………183LIST OF DIAGRAMS3-VIEW DRAWING OF THE AIRCRAFT …………………………………189DESIGN STRUCTURAL LAYOUT………………………………………….188CENTRE OF GRAVITY LOCATION OF THE AIRCRAFT……………..188HYDRAULIC SYSTEM SCHEMATIC OF THE AIRCRAFT…………….190CABIN SEATING LAYOUT OF THE AIRCRAFT………………………..189CONCLUSION …………………………………………………………………. 191REFERENCE ………………………………………………………………………. 192
  • [8] ABSTRACT The General design of the aircraft is carried out on basis of collection of aircraftstatistical data and in accordance with the pilot project development task and finally thegeneral view of an aircraft is presented. The main purpose of the aircraft design requirementis fulfilled according to the aviation rules and regulations. The main project is categorized into five part, in the first part aircraft take-off mass inzero approximation is determined and followed by designing the weight of main units, fuel,equipment, control system, geometrical dimensions of a wing, tail units, fuselage, landinggear, location of their center of masses, calculating the aircraft’s center-of-gravity. Finallythe design specifications for the aircraft are presented. The second part is focused on determining aerodynamic forces acting upon thedesigned aircraft. The third part is the development of aircraft unit structure using Computer aideddesigning. The loads acting on the designed aircrafts unit structure is calculated and thematerials for unit structure are selected. The fourth part is concerned about the systems developed for designed aircraft. Theschematic layout of the hydraulic system and its purpose are briefed following the operations& maintenance manual of the designed system. The last part is the technological section where the development of production chartsfor assembly of designed aircraft vertical rib structure is done. In special assignment the seating arrangement of the designed aircraft is sketched andinteriors of aircraft components are briefed with the current industry techniques.
  • [9] INTRODUCTIONThe purpose of designing a new aircraft is the creation of a structure with uniquecharacteristics, which should be reliable, economical fulfilling the conditions of operation,performance requirements and its primary goal should be attained. To perform thepreliminary design structure of the aircraft it is necessary to be knowledgeable in the field ofgeneral arrangement of aircraft and helicopters, design of power units and systems,construction of elements of assembly structures and units of the aircraft, aerohydrodynamics, durability, technologies, material science, and economics. The purpose ofdesign is to develop a project, realization of which, being limited to a certain extent, wouldensure the most efficient reaching of the defined goals of the design.In designing a new aircraft the following should be considered,  fulfillment of targeted tasks  stability and controllability of an aircraft on a specified trajectory  control and navigation in various flight conditions  life support  Performance characteristics  Characteristics of technological level of the serial aircraft and its economic efficiency  The special equipment  Standardization and unification level  Requirements to reliability and maintenance system  Power plant and its systems  Perspective of development of the aircraft and its basic systems An aircraft is an element of the aviation complex, which seamlessly unites human andmaterial resources and carries out certain useful functions. The functional-structural diagramof the aviation complex is shown on Figure. The aviation complex is an element of statetransport or defense system. All this defines necessity to use systematic approach to aircraftdesign.To implement the process of aircraft design, there was necessity to create specializeddevelopment design offices, which include complicated laboratory and manufacturingresearch. The activities of development design officers are based on work of branch-wiseresearch institutes, which research the prospects of aviation development in variousdirections, and on experience of aircraft production and operation.
  • [10]FOUR STAGES OF DESIGNING 1) External designing: At this stage the research of complicated organization-technical systems including an aircraft or aircraft family as an element is carried out. 2) The second stage — the development of a technical proposal: At this stage, the scheme is selected and optimal combination of basic aircraft parameters, composition and structure of systems ensuring fulfillment of required functions is determined. 3) Third stage – front end engineering: In the process of design arrangement the aircraft center-of-gravity is specified. The calculation of center-of-gravity is followed by making weight reports on the basis of strength and weight calculations of airframe and power unit, lists of equipment, outfit, cargo etc. 4) The fourth stage – working draft: The purpose of this stage is issuing all technical documentation required for production, assembly, mounting of separate units and systems and the whole aircraft as well. At this stage, on the basis of design- technological elaboration the drawings with general view of aircraft units, assembly and working-out drawings of separate parts of the aircraft. MAIN STAGES OF AIRCRAFT PROJECT DEVELOPMENTAIRCRAFT DESIGN PROCESSThe aircraft design process is the steps by which aircraft are designed. These depend onmany factors such as customer and manufacturer demand, safety protocols, physical andeconomic constraints etc. For some types of aircraft the design process is regulated bynational airworthiness authorities. This article deals with powered aircraft such as airplanesand helicopter designs.
  • [11]Aircraft design is a compromise between many competing factors and constraints andaccounts for existing designs and market requirements to produce the best aircraft.DESIGN CONSTRAINTS IN DESIGNING PROCESSA. Aircraft regulationsAnother important factor that influences the design of the aircraft are the regulations putforth by national aviation airworthiness authorities.Airworthiness Certificates-An airworthiness certificate is an FAA document which grantsauthorization to operate an aircraft in flight.Standard Airworthiness Certificate-A standard airworthiness certificate (FAA form 8100-2displayed in the aircraft) is the FAAs official authorization allowing for the operation of typecertificated aircraft in the following categories:  Normal  Utility  Acrobatic  Commuter  Transport  Manned free balloons  Special classes FUNCTIONAL-STRUCTURAL CHART OF THE AVIATION COMPLEXA standard airworthiness certificate remains valid as long as the aircraft meets its approvedtype design, is in a condition for safe operation and maintenance, preventative maintenance,and alterations are performed in accordance with 14 CFR parts 21, 43, and 91.
  • [12]Airworthiness Certification Process-The FAA requires several basic steps to obtain anairworthiness certificate in either the Standard or Special class.The FAA may issue an applicant an airworthiness certificate when: o Registered owner or operator/agent registers aircraft, o Applicant submits application (PDF) to the local FAA office, and o FAA determines the aircraft is eligible and in a condition for safe operationA. Environmental factorsAn increase in the number of aircraft also means greater carbon emissions. Environmentalscientists have voiced concern over the main kinds of pollution associated with aircraft,mainly noise and emissions. Aircraft engines have been historically notorious for creatingnoise pollution and the expansion of airways over already congested and polluted cities havedrawn heavy criticism, making it necessary to have environmental policies for aircraft noise.Noise also arises from the airframe, where the airflow directions are changed. Improvednoise regulations have forced designers to create quieter engines and airframes. Emissionsfrom aircraft include particulates, carbon dioxide (CO2), Sulphur dioxide (SO2), Carbonmonoxide (CO), various oxides of nitrates and unburnt hydrocarbons. To combat thepollution, ICAO set recommendations in 1981 to control aircraft emissions.[11] Newer,environmentally friendly fuels have been developed and the use of recyclable materials inmanufacturing have helped reduce the ecological impact due to aircraft. Environmentallimitations also affect airfield compatibility. Airports around the world have been built to suitthe topography of the particular region. Space limitations, pavement design, runway endsafety areas and the unique location of airport are some of the airport factors that influenceaircraft design.B. SafetyThe high speeds, fuel tanks, atmospheric conditions at cruise altitudes, natural hazards(thunderstorms, hail and bird strikes) and human error are some of the many hazards thatpose a threat to air travel. Airworthiness is the standard by which aircraft are determined fitto fly.[19] The responsibility for airworthiness lies with national aviation regulatory bodies,manufacturers, as well as owners and operators.The International Civil Aviation Organization sets international standards and recommendedpractices for national authorities to base their regulations on The national regulatoryauthorities set standards for airworthiness, issue certificates to manufacturers and operatorsand the standards of personnel training. Every country has its own regulatory body such asthe Federal Aviation Authority in USA, DGCA (Directorate General of Civil Aviation) inIndia, etc.
  • [13]C. Design optimizationAircraft designers normally rough-out the initial design with consideration of all theconstraints on their design. Historically design teams used to be small, usually headed by aChief Designer who knew all the design requirements and objectives and coordinated theteam accordingly. As time progressed, the complexity of military and airline aircraft alsogrew.D. Design aspectsThe main aspects of aircraft design are: 1. Aerodynamics 2. Propulsion 3. Controls 4. Mass 5. StructureAll aircraft designs involve compromises of these factors to achieve the design mission.E. Computer-aided design of aircraftIn the early years of aircraft design, designers generally used analytical theory to do thevarious engineering calculations that go into the design process along with a lot ofexperimentation. These calculations were labor intensive and time consuming. In the 1940s,several engineers started looking for ways to automate and simplify the calculation processand many relations and semi-empirical formulas were developed. Even after simplification,the calculations continued to be extensive. With the invention of the computer, engineersrealized that a majority of the calculations could be done by computers, but the lack ofdesign visualization and the huge amount of experimentation involved kept the field ofaircraft design relatively stagnant in its progress.F. Financial factors and marketBudget limitations, market requirements and competition set constraints on the designprocess and comprise the non-technical influences on aircraft design along withenvironmental factors. Competition leads to companies striving for better efficiency in thedesign without compromising performance and incorporating new techniques andtechnology.
  • [14] DESIGN SECTION PART-1 COMPUTER AIDED GENERAL DESIGNING OF AIRCRAFT1. PURPOSE OF THE AIRCRAFT2. REQUIREMENTS FOR FLIGHT PERFORMANCES3. DESIGN CHART OF THE DESIGNED AIRCRAFT4. PROTOTYPE DATAS5. SELECTION OF AIRCRAFT MAIN RELATIVE INITIAL PARAMETERS6. CALCULATION OF AIRCRAFT MASSES THROUGH THE SOFTWARE AND ITS RESULTS7. ZERO APPROXIMATION8. STATISTICAL COMPUTATION OF MASSES OF AIRCRAFT9. AIRCRAFT OPTIMIZATION AND DESIGN PARAMETERS10. SELECTION AND GROUNDS OF AIRCRAFT CONFIGURATION11. SELECTION OF ENGINE12. DETERMINATION OF CENTER OF GRAVITY OF THE AIRCRAFT13. AVERAGE BETWEEN GRAPHICAL, SEMI-EMPIRICAL & STATISTICAL METHOD14. MAXIMUM TAKE-OFF MASS15. CENTER OF GRAVITY16. DESIGN STRUCTURAL CONFIGURATION17. DESIGN SPECIFICATION
  • [15] PURPOSE OF THE DESIGNING AIRCRAFT A. Aircrafts Intended Purpose - Commercial usageCommercial usage denotes using the aircraft for a business purpose or gettingdirectly/indirectly financial gain from it. B. Payload category - PassengersAircraft adapted for carrying passengers. C. Type - Regional jetThe term regional jet describes a range of short to medium-haul turbofan powered aircraft,whose use throughout the world expanded after the advent of airline deregulation in theUnited States in 1978.ExamplePRIMARY USERS MANUFACTURER ROLEAeroflot Yakolev Yak-40 regional sized mini-jet airlinersSkyWest Airlines Bombardier CRJ100 Regional jet/Business jetPinnacle AirlinesExpressJetComairAerosvit Airlines Antonov An-148 regional jetRossiya D. Range - Short-rangeshort range refers to distance travelled is between 2500.2 km (less than 1350nm) and Timetaken to travel is less than 5 hoursExampleFROM & TO DISTANCE in km TIMENew York-Miami 2051.914 2 hours 49 minsTokyo-Seoul 1,159.04 1.5 to 2 hoursDenver-Boston 2800 3hrs 42 minsG. Special Requirements - Cargo Carrying capabilityCan be used to carry cargos and can be used as a cargo variant
  • [16]H. Mode of Class - Economy classEconomy class refers to the seating arrangement of the aircraft which is usually reclined andinclude a fold-down table. The seats pitch range from 29 to 36 inches (74 to 91 cm), usually30–32 in (76–81 cm), and 30 to 36 in (76 to 91 cm) for international economy class seats.Domestic economy classes range from 17 to 18.25 in (43 to 46.4 cm). GENERAL REQUIREMENTS 1. The aircraft, its engines, equipment and other parts, and operational publications shall meet the following requirements:  aviation requirements АП-25 and additional requirements for airworthiness of "AIRCRAFT NAME" aircraft, in consideration of its design and operational features, forming the "Certification basis of aircraft of "AIRCRAFT NAME" type" together with mentioned requirements;  engine - aviation requirements АП-33;  APU - aviation requirements АП-ВД. 2. As for engine emission the aircraft shall meet the requirements of Appendix 16 to International Aviation Convention (Volume II «Aviation engines emission», Edition 1981, Revisions 1 to 4) and requirements of Aviation Regulations АП34. 3. As for protection against hijacking the aircraft shall meet the requirements ICAO Appendix 6,8,17 (with Revisions 97 and 98)Ukrainian Air Law (Section 8). 4. Processing and analysis of flight data using the ground personal computer shall be provided to control the correctness of maintaining of preset flight modes and the pilot technique, to evaluate the pilots professional level, technical state of the aircraft, its equipment and functional systems in monitoring of operation conditions within life time limits.The system shall include:  aircraft removable data carrier, receiving the information from corresponding aircraft signal transmitters;  personal computer with printer, input and reproducing device and specific software. 5.Ground facilities and repair equipment shall correspond to this performancespecification. 6.Simulators and training devices should be designed for aircraft according to individual
  • [17]performance specifications. The programs for training of flight and technical staff should bedeveloped up to completion of certification tests.SPECIFIC AIRCRAFT STRUCTURE REQUIREMENTS  . The airplane should be designed and manufactured by a principle of ―fail-safe structure‖.  Weight layout and airplane center-of-gravity should ensure a capability of operational both with total and short number of passengers at all possible operational versions of loading and fuelling according to the instruction of loading and centre-of-gravity not using ballast. Limit of on-ground tail-heavy center of gravity be no less than 5 % of MAC.  The capability of creation of convertible and transport versions should be provided on the basis of this airplane according to special performance specification. REQUIREMENTS FOR FLIGHT PERFORMANCESMaximum passenger capacity withdistance between the seats 750 person 55(762) mm,Maximum payload kg 5000Cruise speed:at long range cruise km/h 835maximumCruise altitude, km 10.5Required length of RWY (SA, Н =0, dry concrete), m 1950for takeoff: 2250for landing: Applied flight range (emergencyfuel reserve for 0.75 hour of flight; km 2500takeoff in SA conditions; Н = 0)with maximum payloadFuel consumption for 1 pass/km g 340while flying for technical rangewith maximum payloadMaintenance and overhaul, 8.8
  • [18]manhourREQUIREMENTS FOR ENVIRONMENTAL PROTECTION  . As for perceived noise the aircraft should meet the requirements of Chapter 4 of "Environmental protection" International Standards, Appendix 16 to the International Civil Aviation Convention (Volume I «Air noise», 2001) and to requirements of Part 36 of Aviation Regulations АП-36.  To decrease atmospheric pollution and reduce fuel flow at ground operation the capability of fulfillment of taxiing before take-off and after landing with one operating engine should be worked out on airplane. DESIGN CHART OF THE DESIGNED AIRCRAFT • Collection and process of statistical data General • Design specification and three view diagram Design Aerodyna mic • Designed aircraft drag Characteri stics • Calculation of loads acting on unit Design structural unit • Modelling of designed unit Systems • Schematic layout of the hydraulic system Design Technological • Design of assembly jigs for developed unit Activity • Cabin layoutSpecial activity
  • [19] STATISTICAL DATA COLLECTIONStatistical data collection is the process of collecting flight, mass, power plant andgeometrical data’s of required prototypes for the design project. In this project I havecollected four different aircraft data’s and their features are explained and tabulated. Theseaircrafts are selected based upon the design requirements and design specification mentionedbelow,TACTICAL TECHNICAL REQUIREMENTS OF THE DESIGNING AIRCRAFTMaximum speed , Vmax 835 km/hCruising speed, Vcruise 760 km/hCruising height, Нcruise 11 kmNumber of passengers, npass 47Number of crew members, ncrew 4Range, L 2000 kmTake-off distance, Lр 1970 mVertical speed, Vy 14 m/sMaximum take-off weight, Ммах 40 tonsDESIGN SPECIFICATION OF THE AIRCRAFTType of the aircraft - Transport category with capacity to carry 47 to 55 passengers including crewAerodynamic configuration Normal configuration with horizontal stabilizer on tail sectionWing Low wing with Dihedral and wing sweepTail T-tail configurationFuselage Cylindrical shapePower plant type Turbofan located at aft part of the fuselageLanding gear Tricycle configuration with nose wheelBased on the tactical technical requirements and the general design specification of thedesigning aircraft we are gathering the similar aircrafts and their detailed specification istabulated. From the critical parameters of the aircraft are listed. With the obtained results wenow ready to input all parameters in the software which would give all the relative massesand some important parameters for further calculation.Aircrafts data are gathered from various sources which include books, magazines, websites,etc., Some of the missing parameters are found manually by calculations or it can be found
  • [20]by scaling the three view picture of the collected aircraft. In obtaining the details it isimportant to have the three view pictures of each aircraft for simplification further in drawingthe designed aircraft three view it is very helpful .A short brief of the aircraft is provided foreach of the aircraft with its variant and their three view picture.Upon the four aircrafts selected we can take any one from that as a main prototype for furthersimplification. I have selected the following aircrafts for comparison, 1.EMBRAER ERJ 1452. BOMBARDIER CRJ100 3. TUPOLEV 134-A 4. BOEING 717-200My main prototype is EMBRAER ERJ 145AIRCRAFTS SELECTED FOR STATISTICAL DATA COLLECTION AND THEIRPARAMETERSEMBRAER ERJ 145The Embraer ERJ 145 family is a series of regional jets produced by Embraer, a Brazilianaerospace company. Family members include the ERJ 135 (37 passengers), ERJ 1(44passengers), and ERJ 145 (50 passengers). The key features of the production designincluded: 1. Rear fuselage-mounted engines 2. Swept wings (no winglets) 3. "T"-tail configuration 4. Range of 2500 km
  • [21]Civilian models  ERJ 135ER - Extended range, although this is the Baseline 135 model. Simple shrink of the ERJ 145, seating thirteen fewer passengers, for a total of 37 passengers.  ERJ 135LR - Long Range - increased fuel capacity and upgraded engines.  ERJ 140ER - Simple shrink of the ERJ 145, seating six fewer passengers, for a total of 44 passengers.  ERJ 140LR - Long Range (increased fuel capacity (5187 kg) and upgraded engines.  ERJ 145STD - The baseline original, seating for a total of 50 passengerMilitary models  C-99A - Transport model  EMB 145SA (R-99A) - Airborne Early Warning model  EMB 145RS (R-99B) - Remote sensing modelBOMBARDIER CRJ100The Bombardier CRJ100 and CRJ200 are a family of regional airliners manufactured byBombardier, and based on the Canadair Challenger business jet.The CRJ100 was stretched 5.92 meters (19 feet 5 inches), with fuselage plugs fore and aft ofthe wing, two more emergency exit doors, plus a reinforced and modified wing. Typicalseating was 50 passengers, the maximum load being 52 passengers. The CRJ100 featured aCollins ProLine 4 avionics suite, Collins weather radar, GE CF34-3A1 turbofans with41.0 kN (4,180 kgp / 9,220 lbf), new wings with extended span, more fuel capacity, and
  • [22]improved landing gear to handle the higher weights. It was followed by the CRJ100 ERsubvariant with 20% more range, and the CRJ100 LR subvariant with 40% more range thanthe standard CRJ100. The CRJ 100 SE sub-variant was produced to more closely meet theneeds of corporate and executive operators.VariantsSeveral models of the CRJ have been produced, ranging in capacity from 40 to 50passengers. The Regional Jet designations are marketing names and the official designationis CL-600-2B19.CRJ100 -The CRJ100 is the original 50-seat version. It is equipped with General ElectricCF34-3A1 engines. Operators include Jazz Aviation, Comair and more.CRJ200 -The CRJ200 is identical to the CRJ100 except for its engines, which were upgradedto the CF34-3B1 model, offering improved efficiency.CRJ440 -Certified up to 44-seat, this version was designed with fewer seats in order to meetthe needs of some major United States airlines.Challenger 800/850 - A business jet variant of the CRJ200TUPOLEV 134-A
  • [23]The Tupolev Tu-134 (NATO reporting name: Crusty) is a twin-engined airliner, similar tothe French Sud Aviation Caravelle and the later-designed American Douglas DC-9, and builtin the Soviet Union from 1966–1984. The original version featured a glazed-nose design and,like certain other Russian airliners (including its sister model the Tu-154), it can operatefrom unpaved airfields.Design and developmentFollowing the introduction of engines mounted on pylons on the rear fuselage by the FrenchSud Aviation Caravelle, airliner manufacturers around the world rushed to adopt the newlayout. Its advantages included clean wing airflow without disruption by nacelles or pylonsand decreased cabin noise. At the same time, placing heavy engines that far back createdchallenges with the location of the center of gravity in relation to the center of lift, which wasat the wings. To make room for the engines, the tailplanes had to be relocated to the tail fin,which had to be stronger and therefore heavier, further compounding the tail-heavyarrangement.VariantsTu-134 The glass nosed version. The first series could seat up to 64 passengers, and this was later increased to 72 passengers. The original designation was Tu- 124A.Tu-134A Second series, with upgraded engines, improved avionics, seating up to 84 passengers. All Tu-134A variants have been built with the distinct glass nose and chin radar dome, but some were modified to the B standard with the radar moved to the nose radome.Tu-134B Second series, 80 seats, radar moved to the nose radome, eliminating the glazed nose. Some Tu-134B models have long-range fuel tanks fitted under the fuselage; these are visible as a sizeable bulge.Tu- Bomber aircrew training version.134UBLTu134UBK Naval version of Tu-134UBL. Only one was ever built.BOEING 717-200Boeing 717 was specifically designed for the short-haul, high frequency 100-passengerairline market. The highly efficient 717 concluded its production run in May 2006, thoughthe airplane will remain in service for years to come.Final assembly of the 717 took place at the Boeing plant in Long Beach, Calif. The airplanewas originally part of the McDonnell Douglas airplane family and designated the MD-95
  • [24]prior to merger with The Boeing Co. in 1997. The program produced 156 717s and pioneeredbreakthrough business and manufacturing process for Boeing Commercial Airplanes The.The standard 717 has a two-class configuration with 106 seats. Its passenger-pleasing interiorfeatures a five-across-seating arrangement in economy class, with illuminated handrails andlarge overhead stow bins.The two-crew flight deck incorporates six interchangeable liquid-crystal-display units andadvanced Honeywell VIA 2000 computers.Flight deck features include an Electronic Instrument System, a dual Flight ManagementSystem, a Central Fault Display System, and Global Positioning System. Category IIIbautomatic landing capability for bad-weather operations and Future Air Navigation Systemsare available.Two advanced Rolls-Royce 715 high-bypass-ratio engines power the 717. The engine israted at 18,500 to 21,000 pounds of takeoff thrust, with lower fuel consumption andsignificantly lower noise and emission levels than the power plants on comparable airplanes.DESIGNThe 717 features a two-crew cockpit that incorporates six interchangeable liquid-crystal-display units and advanced Honeywell VIA 2000 computers. The cockpit design is calledAdvanced Common Flight deck (ACF) and is shared with the MD-11.
  • [25] STATISTICAL TABLEFLIGHT DATA:Flight includes Vmax – the maximum speed of flight; HV max – flight altitude with themaximum speed; Vcruise –cruise speed; Нcruise –cruise altitude; Vland – landing speed; Vto –take-off speed; VY– rate of climb; Hclg– static ceiling; L– flight range; Ltor – distance of thetake-off run; Lto – take-off distance; Lroll–landing roll distance; Lland– landing distance; 1 No 1 2 4 3 2 Name of the EMBRAER Bombardier TUPOLEV BOEING aircraft 145 CRJ200 134-A 717-200 Producer Embraer Bombardier Boeing Country Brazil Canada Tupolev United stated Year of 1989-present 1992 Soviet union 1998–2006 production 1966–1984 3 Source Janes all the world aircraft and WikipediaFLIGHT DATA 4 Vcruise, km/h 833 850 850 811 5 Vmax, km/h 679 785 950 629 6 Нcruise, km 11.277 11 11 10.400 7 HV max, km 9.753 11 11.5 11.280 8 Vto, km/h 170 155 - 150 9 Vland, km/h 233 250 - 244 10 VY, km/h 6.5 6 - 6 11 Hclg, km 11.27 12.49 12.1 11 12 L(mf Max ) , km 3037 2500 - 2645 13 L(mcargo max) , 2963 1800 1020 3800 km 14 Lto, km 1.97 1527 2.4 1.7 15 Lland, km 1.3 1423 2.2 1.52MASS DATA : This includes take-off mass(m0), maximum take-off mass(m0max), payloadmass(mpld), number of passengers(npass), landing mass(mland), empty mass(mempty), mass ofcrew(mc), mass of fuel(mf), empty equipped mass(mempt.eqpd) and total mass(mtotal ).
  • [26] S.NO MASS EMBRAER Bombardier TUPOLEV BOEING 717- 145 CRJ200 134-A 200 16 m0 (mto), kg 19200 21636 47000 49895 17 m0max, kg 20000 22000 47200 22000MASS DATA 18 mpld, kg 5640 6240 8200 12000 19 npass 47 52 84 100 20 mland, kg 18700 20000 43000 43359 21 mempty, kg 11585 19,958 27,960 30000 22 mc, kg 5,284 6,124 8,200 12400 23 mf, kg 2865 4300 - 8500 24 mempt.eqpd, kg 17,100 13730 29050 43545 25 mtotal, kg 20,100 24,041 47,600 49,900POWERPLANT DATA: This includes engine thrust (P0), mass of engine (meng), number ofengines and its type, specific fuel consumption (Cp) and bypass ratio(Y). S.NO ENGINE EMBRAER Bombardier TUPOLEV BOEING 717- SPECS 145 CRJ200 134-A 200 26 P0 (N0), 31.3 31 103 97.9 POWERPLANT DATA daN (kN) 27 meng, kg 1438 2305 4640 28 No of 2 2 2 2 engines Twin-spool Type of non- engine afterburning turbofan 29 Cp, lb/lbf·hr 0.39 - 0.498 - 30 Y, Bypass 3:1 5:1 - - ratioGEOMETRICAL DATA: This includes wing area(S), wing span(L), sweep angle(),aspect ratio of wing(), thickness ratio at chord( c 0 ) and at tip( c tip ),taper ratio(), length of
  • [27] fuselage(Lf), diameter of fuselage(df), area of aileron( S ail ), relative fuselage mid section area(  S mcs ), wing loading(P0) and thrust to weight ratio(t0). S.NO GEOMETRICAL EMBRAER Bombardier TUPOLEV BOEING PARAMETERS 145 CRJ200 134-A 717-200 31 S, m2 51.18 54.54 127.3 92.97 32 L, m 20.04 20.52 29.00 28.45 33  22.73 24.75 35.00 24.50 GEOMETRICAL DATA 34  7.85 7.72 6.61 8.7 35 c0 4.09 5.13 - - c tip 1.04 1.27 - - 36  4 3.4 0.255 5.10 37 Lf, m 29.87 24.38 37.10 33 38 df, m 2.28 2.69 2.9 3.34 39 f 12.25 9.06 11.45 4.30 40 S ail 1.70 1.93 - - , m2 2 41  S mcs , m 7.56 8.38 10.5416 18.84 42 P0=m0g/10S, 375.15 394.3 369.21 556 daN/m2 43 t0=10P0/m0g 0.3326 0.3884 0.289 0.3806 DERIVATIVE VALUE: This includes specific fuel weight (eng), effective load factor ( K eff .load ), relative aileron area ( S ail ), relative horizontal ( S HT ) and vertical stabilizer area ( S VT ). S.NO DERIVATIVE EMBRAER Bombardier TUPOLEV BOEING 717- PARAMETERS 145 CRJ200 134-A 200DERIVATIVE VALUES 44 eng, kg/daN2 306.51 263 - 294 45 m c arg o 0.2752 0.288 0.1744 0.2405 K eff .load  m0 46 K mcs  m 0  S mcs , 2539 2600 4459 2627
  • [28] daN/m2 47 S ail  S ail S 0.0332 0.0353 - - 48 S HT  S HT S 0.219 0.173 0.241 0.205 49 S VT  S VT S 0.141 0.168 0.167 0.210 SELECTION OF AIRCRAFT MAIN RELATIVE INITIAL PARAMETERSThus finally tabulating all the required values it is necessary to find the main relative initialparameters of wing, fuselage and the tail unit. The obtained result is used in the software. WING PARAMETERaspect ratio,  7.85sweep angle,  22.73taper ratio,  4relative width of airfoil, c 18 or 0.18relative chord of flap, b f  b f / b wing 0.25deflection angles of flap, f 18relative area of ailerons, S ail  S ail / S 0.06FUSELAGE PARAMETERfineness ratio f 12.25fuselage diameter Df 2.50TAIL UNIT PARAMETERrelative area of horizontal stabilizer, S HT  S HT S 0.219relative area of vertical stabilizer, S VT  S VT S 0.141aspect ratio of horizontal surface, HS 4.077aspect ratio of vertical surface, VS 1.36Sweep angle of horizontal surface, HS 20Sweep angle of horizontal surface, HS 32Relative thickness of horizontal surface, c HS 12 or 0 .12Relative thickness of vertical surface, c VS 10 or 0.10
  • [29]CALCULATION OF AIRCRAFT MASSES THROUGH THE SOFTWARE AND ITSRESULTSAircraft masses in zero approximation are calculated using software by entering necessaryparameters taken from statistical data and the initial parameters. First the relative masses forfuselage, wing, power plant, tail unit, fuel and landing gear are found with respect to theaspect ratio of wing taken as 4 and aircrafts wing loading attained from graphical result as600 N/m2. The graphs obtained from this result help us to select the desired wing loading andfrom the wing the lowest value of takeoff mass is taken as the final one. The relative massesare changed to direct masses by multiplying it with the finally obtained take-off mass, for meit is 39.11tons. Therefore my relative masses are multiplied with 39.11 ton to get direct massGraphs for each lab are plotted versus each parameter and from that the final wing loading isobtained followed by the takeoff mass. Other parameters obtained include engineperformance data like thrust to weight ratio at take-off, climbing and cruise. My maincomparative parameter is Aspect ratio which I took in three variations asASPECT RATIO 2 , 4 AND 6AIM OF THE LAB AND ITS RESULT TO FIND THE RELATIVE MASS OFAIRCRAFTLAB 5: In this part we are finding the relative mass of power plant respect to the aspectratio of wing taken as 4 and aircrafts wing loading 600 N/m2.
  • [30]RESULT: In table P,denotes wing loading Tk,aspect ratio and SU is the RELATIVE MASSOF POWER PLANT which is 0.086LAB 7A:In this part we are finding the relative mass of wing respect to the aspect ratio ofwing taken as 4 and aircrafts wing loading 600 N/m2.RESULT: In table p, denotes wing loading and Tk, aspect ratio and Mkp is the RELATIVEMASS OF wing which is 0.048.
  • [31]LAB 7B:In this part we are finding the relative mass of fuselage respect to the aspect ratioof wing taken as 4 and aircrafts wing loading 600 N/m2.RESULT: In table DF, refers to diameter of the fuselage (3.84m) and Lf refers to aspect ratioof the fuselage (12.25). By comparing both the values we get the relative value of fuselageequals to 0.350.LAB 7G:In this part we are finding the relative mass of tail unit with respect to the aspectratio of wing taken as 4 and aircrafts wing loading 600 N/m2.RESULT: In table P, denotes wing loading and MOP, denotes RELATIVE MASS OF THETAIL UNIT which is 0.0182
  • [32]LAB 7V:In this part we are finding the relative mass of landing gear with respect to theaspect ratio of wing taken as 4 and aircrafts wing loading p,600 N/m2.RESULT: RELATIVE MASS OF THE LANGING GEAR is 0.062LAB 8: In this part we are finding the mass of equipment, crew and payload.RESULT: MASS OF THE Equipment, crew and payload is 9502.98kgLAB 9: In this part we are finding the take off mass of the aircraft which is equal to39.11tons.
  • [33]RESULT: Take off MASS OF is 39110 kg or 39.11 tonsRESULTS FROM GRAPH WITH RESPECT TO THE RELATIVE PARAMETERLAB PARAMETERS RESULTSNO3 Lift to drag ratio 12.204 Thrust to weight ratio at Take-off 0.2514 Thrust to weight ratio at Landing 0.2714 Thrust to weight ratio at Cruising 0.1795 Mean Thrust to weight 0.2715 Relative mass of powerplant 0.0867a Relative mass of wing 0.0487b Relative mass of fuselage 0.3507g Relative mass of Tail unit 0.01827B Relative mass of Landing gear 0.0626 Relative mass of Fuel 0.2248 MEQ = MCREW+MPAYLOAD+MEQUIPMENT 9502.98 kg9 Take-off mass relative to wing loading 39110 kg
  • [34]DIRECT MASSMass of fuselage 13688.5 kgMass of wings 1877.28 kgMass of tail unit 711.802 kgMass of powerplant 3363.46 kgMass of landing gear 2424.82 kgMass of fuel 760.64 kgCOMPUTATION OF AIRPLANE TAKE-OFF MASS IN ZERO APPROXIMATIONDETERMINATION OF MASS FROM LAB RESULTSMass of fuselage = Relative mass of fuselage * Take off mass = 0.350 * 39110 kg = 13688.5 kgMass of wings = Relative mass of wing * Take off mass = 0.048* 39110 kg = 1877.28 kgMass of tail unit = Relative mass of tail unit * Take off mass = 0.0182* 39110 kg = 711.802 kgMass of power plant = Relative mass of power plant * Take off mass = 0.086* 39110 kg = 3363.46 kgMass of landing gear = Relative mass of landing gear * Take off mass = 0.062* 39110 kg = 2424.82 kgMass of Fuel = Relative mass of fuel * Take off mass = 0.224* 39110 kg = 8760.64 kgMass of crew = 4 * 80 kg = 320kgMass of payload = 47 * 90 kg = 4230kgMass of Equipment and control systems = 4952kg ZERO APPROXIMATIONTake-off mass of the airplane for zero approximation is determined by the formula receivedfrom the equation of mass ratio with statistical data.
  • [35]m 0  m st  m p . p  m f  m pl  m crew  m eq ;Here, m 0 = Take-off mass, m st = Structural mass of the aircraft, m p. p = Power plantmass,m f = Fuel mass, m pl = Payload mass, m crew = Crew mass, m eq = Equipment mass m pl  m crewMass Ratio (dimensionless) equation is, 1  m st  m p. p m f  m eq  m0Re-arranging we get final takeoff mass as, m pl  m crew m0  1  ( m st  m p . p  m f  m eq )m st - Relative airframe mass = Relative mass of fuselage+ Relative mass of wing+ Relative mass of tail unit+landing gear = 0.350+0.048+0.0182+0.062 = 0.4782m p. p - Relative mass of power plant = 0.086m f - Relative mass of fuel = 0.224m eq - Relative mass of Equipment = 0.1266 4230  320m0  1  ( 0 . 4782  0 . 086  0 . 224  0 . 1244 ) m 0 = 53403.755 kg STATISTICAL COMPUTATION OF MASSES OF AIRCRAFT When the airplane takeoff mass in zero approximation is determined it is necessary to m airfr m wing m fuscalculate airframe mass and its components (mass of the wing , fuselage , m tail unit m fuel m pow . pltail unit , landing gears), and also mass of fuel , power plant and mengines eng . Relative masses of airframe, power plant, equipment and control system, andalso of the aircraft performing normal take-off and landing are given in Table below Plane Purpose m airfr m pow . pl m ctl . sys m fuel Subsonic light 0.30…0.32 0.12…0.14 0.12…0.14 0,18…0,22 passenger long- medium 0.28…0.30 0.10…0.12 0.10…0.14 0,26…0,30 distance heavy 0.25…0.27 0.08…0.10 0.09…0.11 0,35…0,40 Multipurpose for local airlines 0.29…0.31 0.14…0.16 0.12…0.14 0.12…0.18
  • [36]Take off mass from lab 9 = 39110 kgRelative mass of Airframe = 0.30Mass of Airframe = Relative mass of airframe * Take off mass = 0.30 * 39110 kg = 12515.2 kgRelative mass of power plant = 0.14Mass of power plant = Relative mass of power plant * Take off mass = 0.14* 39110 kg = 5475.4 kgRelative mass of control systems and equipments = 0.11209Mass of control sys = Relative mass of control systems and equipments * Take off mass = 0.11209* 39110 kg = 4383.839 kgRelative mass of fuel = 0.18Mass of fuel = Relative mass of fuel * Take off mass = 0.18* 39110 kg = 7039.8 kg AIRCRAFT OPTIMIZATION AND DESIGN PARAMETERSGeometrical parameters for designed aircraft are calculated by formulas taken from pilotproject book and rest is determined statistically by comparing with the prototypes. Aftercalculating the geometrical parameters we are drawing the theoretical drawing. Thegeometrical parameters are calculated and obtained satisfying the general requirements of theaircraft.The stages of aircraft optimization include the following:  Determination of Wing Parameters  Determination of Fuselage Parameters  Determination of Tail Unit Parameters  Determination of Position of Center of Mass of the Airplane  Determination of Landing gear parametersDetermination of Wing Parameters:In determining Wing parameters its plan form shape is very important in obtaining number ofuseful relations that apply to a trapezoidal shape. These are based on knowing the wing area,aspect ratio, taper ratio, and leading-edge sweep angle.Before finding the wing area it is necessary to determine the wing loading corresponding totake off mass of 39110kg, which is found in LAB 9 as given below,
  • [37]RESULT: The wing loading is found to be 600 daN/m2Wing statistical parameter aspect ratio,  sweep angle,  taper ratio,  7.85 22.73 4WING AREA m0  gS  Where m 0  39110 ( kg ) , g  9 . 8 ( m / s 2 ) , p 0  600 ( dN / m ) 2 10  p 0 39110  9 . 8S   63 . 87 ( m ) 2 10  600Wing Span ( l )L  S Where λ= 7.85 (choosing from table)L 7 . 85  63 . 87 = 22.39m
  • [38]Wing Chords (b) S  2 b root  b 0      Where  = 4 L  1 63 . 87  2  4 b root  b 0     4 .5 ( m ) 22 . 39  4  1  b 4 .5b tip  0   1 . 1( m )  4Quarter chord line sweep angle of the Wing (  0 .25 ) 0 .25  22 . 73 0 (Choosing from the statistic’s table)Leading edge sweep angle of the Wing (  0 )  1 4 1tg  0  tg  0 . 25   tg 22 . 73   0 . 4953 0     1  7 . 85   4  1  0  arctg 0 . 4953   26 . 349 0Mean Aerodynamic Chord of the Wing (MAC = b Aw )   1 2 2bA   b0  3     1  4  4 1 2 2bA   4 .5   3 . 15 ( m ) 3 4  4  1Vertical distance between horizontal central line to MAC ( z A ) L  2 22 . 39 4  2zA      4 . 478 ( m ) 6  1 6 4 1Horizontal distance between wing root tips to MAC root l  2 22 . 39 4  2xA    tg  0  . 0 . 4953  2 . 2179 ( m ) 6  1 6 4 1Determination of Fuselage Parameters The size and shape of subsonic commercial aircraft are generally determined by thenumber of passengers, seating arrangements and cargo requirements. Seating arrangementson commercial passenger aircraft vary depending on the size and range.Fuselage fineness ratio f fuselage diameter Df Nose section Rear section fineness ratio ,  N fineness ratio,  T 12.25 2.5 m 1.5 2.5
  • [39]Overall Fuselage length, LfL f   f  D f  12 . 25  2 . 5  30 . 62 ( m )Fuselage nose length, L f .nL f . n   N  D f  1 . 5  2 . 5  3 . 75 ( m )Fuselage rear length, L f .rL f . r   T  D f  2 . 5  2 . 5  6 . 25 ( m )Fuselage middle section length, L f .mLfm = Lf – Lf.n – Lf.r = 30.62 - 3.75 - 6.25 = 20.62 mDetermination of Tail Unit ParametersTail unit parameters include Horizontal and Vertical stabilizer. Their geometrical parametersare determined by the same formulae which were used for the wing.Horizontal stabilizer parameterHorizontal stabilizer statistical parameteraspect ratio, sweep angle,  taper ratio,  Relative horizontal  stabilizer area, S h .t 4.077 20 2 0.219Horizontal stabilizer areaS h .t  S h .t  S  0 . 219  63 . 87 ( m )  13 . 98 m 2 2 Where S =63.89, wing areaLength of the Horizontal stabilizerL h .t   h .t  S h .t  4 . 077  13 . 98  7 . 55 ( m )Chords of the Horizontal tail Unit S ht 2  13 . 98 2  2.b root  b 0 . ht      2 . 48 ( m ) L ht   1 7 . 55 2 1 b 0 . ht 2 . 48b tip . ht    1 . 24 ( m )  ht 2 .0
  • [40]Quarter chord line sweep angle of the Horizontal tail unit (  0 .25 . ht ) 0 . 25 . ht  20 (Choosing from the static’s table 1.1) 0 Mean Aerodynamic Cord of the Horizontal Tail ( b A . ht )    ht  1 2.  2.  1 2 2 2 ht 2b A . ht   b 0 . ht    2 . 48   1 . 93 ( m ) 3  ht   ht  1  3 2 2  1Vertical Distance between horizontal central line to MAC ( z A . ht ) L ht  ht  2 7 . 55 2  2z A . ht      1 . 68 ( m ) 6  ht  1 6 2 1Leading edge sweep angle of the Horizontal stabilizer (  0 )  1 2 1tg  0  tg  0 . 25   tg 20   0 . 44572 0     1  4 . 077   2  1  0  arctg 0 . 44572   24 . 023 0Horizontal distance between wing root tip to MAC root ( x A . h .t ) L ht  ht  2 7 . 55 2  2x A . ht    tg  0  . . 0 . 4452  0 . 7488 ( m ) 6  ht  1 6 2 1x A . ht  0 . 30 ( m )Vertical stabilizer parameterVertical stabilizer statistical parameteraspect ratio, sweep angle,  taper ratio, vs Relative horizontal  stabilizer area, S h .t 1.36 32 1 0.141Vertical stabilizer areaS v .t  S v .t  S  0 . 141  63 . 87 ( m )  9 . 01 m 2 2 Where S =63.89, wing area
  • [41]Length of the Vertical stabilizerL v .t   v .t  S v .t  1 . 36  9 . 01  3 . 50 ( m )Chords of the Vertical tail Unit S vt 2  9 . 01 1 2.b root  b 0 . vt      2 . 57 ( m ) L vt   1 3 .5 11 b 0 .vt 2 . 57b tip .vt    2 . 57 ( m )  vt 1Quarter chord line sweep angle of the Vertical tail unit (  0 .25 .vt ) 0 . 25 . vt  32 0 (Choosing from the static’s table 1.1)Mean Aerodynamic Cord of the Vertical Tail ( b A .vt )    vt  1 1 11 2 2 2 vt 2b A . vt   b 0 . vt    2 . 57   2 . 57 ( m ) 3  vt   vt  1  3 11  1 Vertical Distance between horizontal central line to MAC ( y A .v .t ) L v .t  v .t  2 3 . 50 1  2y A . v .t      1 . 76 ( m ) 3  v .t  1 3 11Leading edge sweep angle of the Vertical stabilizer (  0 )  1 11tg  0  tg  0 . 25   tg 32   0 . 62 0     1  1 . 36  1  1  0  arctg 0 . 62   31 . 79 0Horizontal distance between wing root tip to MAC root ( x A .v .t )x A .v .t  Lvs  tg  0  3 . 50  tg  0  3 . 50  0 . 62  2 . 17 ( m )DETERMINATION OF LANDING GEAR PARAMETERSFor nose-wheel tri-cycle landing gear the following parameters are consideredWheel base (b ) : distance between axels of nose wheel and main landing gear wheels inside view. It depends on fuselage length.b  (0.30.5)l fus,b = 0.48 * 30.62 = 14.6976mNose wheel offset (a ) : It is distance between vertical line passing through the airplane centerof gravity and nose wheel axis (or axis of several wheels whenever);a = 0.968 * 14.697 = 14.235m
  • [42]Main Landing Gear offset (e) : It is distance (on side view) between vertical line passingthrough the airplane center of gravity and axis (or center line of several wheels, bogie) ofMLG;e = 0.031 * 14.697 = 0.462mStatic ground angle() : It is the angle between fuselage construction plane and runwaysurface. It is generally between2 to +2. So it is taken as1.Angle of wing setting sett : It is the angle formed between wing construction plane to thefuselage axis. It is generally between sett  (04). So it is taken be 2.Angle of overturning(): It is the angle appearing when fuselage tail part or its tail bumptouches the runway surface;Ψ=2 parking angleamax = 12 maximum angle of attackaw = -1 angle between a wing chord and longitudinal axis of fuselage.Ф = amax –(-1) – 2 = 12+1-2 = 11As a rule  = 1018, smaller values are accepted for non-maneuverable subsonic aircraft.Offset angle(): It is the angle of offset for wheels of MLG relatively to airplane CG.Itprevents airplane overturning backward during landing. =  + (12). =11+ 2 = 13.Wheel track(В ) : It is the distance (on front view) between planes of symmetry of MLGwheels. This can be found byB  ( 0 . 15 .... 0 . 35 ) LWB  0 . 21  22 . 39  4 . 7 mHeight of airplane center of gravity(Н): It is the distance from airplane CG to the ground.H=H= 0.462 / tan (13)H= 2.1 mHeight of the landing gear (h): Distance from leg attachment fittings to the runway surfacewhen shock absorber and tiers compression is of parking state (at take-off mass).DF = 2.5 m h =H – DF/2h = 2.1 – (2.5/2)
  • [43]h = 0.85 mDetermination of Position of Center of Mass of the AirplanePosition of the airplane center of mass is determined relative to nose part of the wing meanaerodynamic chord (MAC).The recommended distance for the center of mass from the nose part of mean aerodynamicchord x m as follows:For airplanes with swept wing:x m  0 . 23  b A  0 . 23  3 . 15  0 . 7245 ( m )Determination of tail armsVertical tail unit arm:It is the distance measured from the aircraft center of massup to the vertical tail unit centre ofpressure. It is selected statistically , Tvs = 12m. For T-Tail Tvs is not equal to Ths.Horizontal tail unit arm :It is the distance measured from the aircraft center of mass up to the horizontal tail unitcentre of pressure. Horizontal tail unit arm is obtained after drawing theoretical diagram.Calculation of high lifting devices parameters:Flap configuration:We will consider relative value of flap span from statistical data L flap  0 . 5 to 0 . 8I considered relative length of flap = 0.6The length of the flap is calculated by the formula L span  D fuselage  2  L flap L flap  L flap 2ΔLflap- is the gap between flap and fuselage = 100mm 22 . 73  2 . 5  2 * 0 . 100 L flap   0 .6  6 m 2Flap root chord is calculated by the following formula    1 D fuselage  2  flap  b 0 flap  b flap  b 0   1        L Here b flap is relative chord of flap is 0.2 to 0.4  4  1 2 . 5  2  0 . 100 There for b 0 flap  0 . 2  4 . 5   1     0 . 818 m  4 22 . 73 Flap tip chord length is calculated by fallowing formula
  • [44]    1 D fuselage  2  flap  2 l flap  b кflap  b flap  b 0   1        L   4  1 2 . 5  2  0 . 100  2  6  b кflap  0 . 2   4 . 5   1     0 . 46 m  4 22 . 73 Slats configuration:We will consider relative value of slat span from statistical data L slat  0 .6 to 0 .85I considered relative length of slat = 0.65The length of the slat span is calculated by below formula L span  D fuselage  2  L slat L slat   L slat 2ΔLslat- is the gap between slat and fuselage = 300mmThere for 22 . 73  2 . 5  2 * 0 . 3 L slat   0 . 65  6 . 3 m 2slat root chord is calculated by the fallowing formula    1 D fuselage  2  slat  b 0 slat  b slat  b 0   1        L Here b slat is relative chord of slat is 0.08 to 0.15 = 0.09  4  1 2 .5  2  0 .3 There for b 0 slat  0 . 09  4 . 5   1     0 . 363 m  4 22 . 73 Slat tip chord length is calculated by fallowing formula    1 D fuselage  2  slat  2 l slat  b кslat  b slat  b 0   1        L   4  1 2 .5  2  0 .3  2  6 .3  b кslat  0 . 09   4 . 5   1     0 . 19 m  4 22 . 73 Calculation of control surfaces parameters:Aileron configuration:AREA ( SAIL) : It is found by formula S AIL  S AIL  S , from statistical data Relative are ofaileron is 0.06, S AIL  0 .06  63 .87  3 .83 m2Length of the aileron is calculated by the following formula
  • [45] L  D fuselage l aileron   l flap   flap   AF   Aw , м, 2here  AF – the gap between flap and aileron= 0.02  Aw – the gap between aileron and wing tip = 1.2 22 . 73  2 . 5 l aileron   6  0 . 10  0 . 02  1 . 2  2 . 7 m 2   1  Aileron root chord b оaileron  b aileron  b 0   1    Z оaileron ,  м,   Here b aileron 0.25…0.3 – from the statistical data = 0.25 D fuselage  2  l flap   Af   Aw  Here Z оaileron  . L 2 . 5  2  6  0 . 02  1 . 2  Z оaileron   0 . 74 22 . 73  4 1 There for b оaileron  0 . 25  4 . 5   1   0 . 74   0 . 50 m  4    1  Aileron tip chord b кaileron  b aileron  b 0   1    Z кaileron ,  м,    D fuselage  2  l flap  l aileron   af   Aw Here Z кaileron  . L 2 . 5  2  6  2 . 7  0 . 02  1 . 2  Z кaileron   0 . 98 22 . 73  4 1 There for b кaileron  0 . 25  4 . 5   1   0 . 98   0 . 29 m  4 Elevator configuration: Length of the elevator l H .S l elevator    HT   Ht , м, 2Here  HT – is the operating gap of elevator =0.02m  Ht – is the gap between elevator stabilizer to horizontal stabilizer tip=0.6m 7 . 55 l elevator   0 . 02  0 . 6  3 . 155 m 2Elevator root chord    1 2  HT  b 0 er  b er  b 0 H . S   1  H . S  , м,   H .S   l H .S 
  • [46]Here b er  0.25…0.35 – relative length of elevator chord = 0.25  2  1 2  0 . 02  b 0 er  0 . 25  2 . 48   1     0 . 617 m  2 7 . 55 Elevator tip chord:   H , S  1 l H . S  2  Ht  b к er  b er  b 0 H . s   1    ,  m.   H .S l H .S   2  1 7 . 55  2  0 . 6  b к er  0 . 25  2 . 48   1     0 . 359 m  2 7 . 55 ELEVATOR AREAS EL  S EL / S HS  0 , 2  0 , 4 (lower values – for supersonic aircraft);SEL= S EL × SHS = 0.3 × 13.98 m2 = 4.194 m2Rudder configuration:rudder areaS RUD  0 ,20  0 ,45 (lower values – for supersonic aircraft);SRD = S RD × SVS = 0.40 × 9.01m2 = 3.64 m2Length of the rudder l rudder  lV . S   Rf   RT , м,Here  RF – the gap between fuselage to rudder root chord= 0.015m  RT – the gap between vertical tail tip to rudder tip chord.=0.6m l rudder  3 . 5  0 . 015  0 . 6  2 . 885 mRudder root chord    1  RF  b 0 rudder  b rudder  b 0 V . S   1  V . S  , м,   V .S   lV . S Here b rudder  0.25…0.4 – relative chord of rudder = 0.25  1  1 0 . 02  b 0 rudder  0 . 25  2 . 57   1     0 . 642 m  1 9 . 56 For T-tail tip and root chord of rudder is same there for bkrudder= 0.642 m
  • [47] SELECTION AND GROUNDS OF AIRCRAFT CONFIGURATIONAERODYNAMIC CONFIGURATIONThe ―normal" classical configuration applies to my aircraft. The advantages of thisconfiguration are:  wing is in the pure, undisturbed airflow and is not shadowed by stabilizers;  nose section of a fuselage is short and does not create destabilizing moment relatively to the vertical axis; this allows to reduce area and mass of vertical stabilizer;  Crew has better observation of the front semi-sphere.Selection of wing position relatively to the fuselageLow-wing aircraft. Advantages:  due to ground shield effect (aerodrome surface) Ycr increases, Vto, Vland; decreases  height of the landing gear struts and their mass is less, their retraction becomes simpler;  high-lift devices can also be located on ventral wing parts;  Safety of passengers and crew increases during emergency landing – the wing provides additional protection;  Floating capacities during emergency landing on water are higher, that allows to evacuate passengers and crew.
  • [48]SELECTION OF WING EXTERIOR SHAPESwept wings are applied at M = 0,82.With increase of sweep angle:  the shockwave drag on moderate subsonic and supersonic speeds (Fig. 2.6) is drastically decreased. 2 M cr  Ì cr   0  1  cos  is increased.  Critical values of flutter speed Vfl increase, divergence speed Vdiv (at swept wings), lateral stability increases.WING SHAPE ON FRONT VIEWIt is characterized by wing dihedral angle Dihedral angle is defined by  angle between wing chords plane and planeperpendicular to aircraft of symmetry plane passing through the inboard chord. Thefollowing types are distinguished:At = 0+7 dihedral angle (for straight wings)
  • [49]SHAPES OF WING CROSS-SECTIONS AND TAIL UNIT STABILIZER CROSSSECTIONDouble convex asymmetrical - high Cy max, smaller Cxр, stable position of the centre ofpressure. These airfoils find wide application in various subsonic aircraft;The airfoil should have low profile drag in a range of Cy factors characteristic for cruiseflight;  It is necessary that the airfoil with the extended flap has small Cxp at Cmax, especially during climb;  Tip wing cross-sections at Cmax should have smooth performances of shock stall;  Internal wing cross-sections should have high values of Cmax with extended flaps;  It is necessary to ensure a high value of Mcr above 0,65;Airfoils for my aircraft are selected as mentioned belowThe NACA four-digit wing sections define the profile by 1. One digit describing maximum camber as percentage of the chord. 2. One digit describing the distance of maximum camber from the airfoil leading edge in tens of percents of the chord. 3. Two digits describing maximum thickness of the airfoil as percent of the chord.WING AIRFOIL - NACA 2415 airfoil has a maximum camber of 2% located 40% (0.4chords) from the leading edge with a maximum thickness of 15% of the chord. Four-digitseries airfoils by default have maximum thickness at 30% of the chord (0.3 chords) from theleading edge.
  • [50]HORIZONTAL STABILIZER AIRFOIL - NACA 2412 airfoil has a maximum camber of2% located 40% (0.4 chords) from the leading edge with a maximum thickness of 12% of thechord. Four-digit series airfoils by default have maximum thickness at 30% of the chord (0.3chords) from the leading edge.Selection of the scheme of ailerons b ail  b ail / b  0 ,25  0 ,3 ;  ail .upw  20  25  ; S ail  S ail / S  0 ,03  0 ,08 ; L ail  L ail / L  0 ,2  0 ,4 ;  ail .downw  10  15  .Requirements to ailerons: Minimum yawing moment (an aircraft yaw motion relatively to OY axis) in bank, with aircraft turning in the side of bank angle. Full weight balancing with least weight of balance weights. Provision of efficiency on all flight phases.
  • [51] Critical speed of reverse thrust should be sufficient.SELECTION OF FUSELAGE STRUCTUREThe basic geometrical sizes of the fuselage are selected statistically by comparing theprototypes and are listed belowLf – length of a fuselage; 30.62mDf – diameter of the greatest mid-section, 2.5mSmcs – the area of fuselage mid-section;Lns, length of nose section of a fuselage. 3.75mLts – tail section of a fuselage. 6.25mn.f – Fitness ratio of fuselage nose section – 1.5t.f – Fitness ratio of tail section – 2.5 NOSE SECTION TAIL SECTION Shapes of nose and tail sections are also determined generally from conditions ofaerodynamics, layout, technology, purpose. For the nose section, an important condition isensuring the demanded observation from the cockpit that leads to smaller fineness andsmaller sharpness. It is reasonable to deflect the tail section of a fuselage upwards to ensure
  • [52]the landing angle  during take-off. Many cargo aircraft have a large door in the tail sectionwith the cargo ramp lowered on ground for loading and an unloading of cargos. For modernaircraft, for aerodynamic drag decrease considerations, the whole tail section is lengthenedand bended. Some cargo aircraft have the cargo door in the fuselage nose (An-124, C-5А,Boeing 747F). The lower part is capable to open back and upwards that simplifies loadingand unloading of transported vehicles and cargos.CONTOURS OF FUSELAGE NOSE & TAIL SECTION Yn.f =  a (Хn.fdn.f/4n.f) m .Хn.f – Length of nose section – 3.75mdn.f – Diameter of fuselage nose section – 2.5mn.f – Fitness ratio of fuselage nose section – 1.5m – 0.5 from tablea – 1 from tableYn.f =  1 (3.75* 2.5 / 4 * 1.5) 0.5 =  1.25Factors m and a are presented in Tab m 0,35 0,4 0,45 0,5 0,55 0,6 0,65 а 1,1293 1,08845 1,04136 1 0,96026 0,9221 0,88546 n  n    Y   в  Х d  Х  d / 4  f .ts   f .ts f   f f .ts   .в – 1 from tableХt.f – Length of tail section – 6.25 mdf. – Diameter of the fuselage – 2.5mt.f – Fitness ratio of tail section – 2.5n – 0.50 from table
  • [53]x – Total length of the fuselage – 30.62mY   1  6 . 25 * 2 . 5  30 . 62  0 . 50 2 . 5 / 4 * 2 . 5  0 . 50 = 0.9839 f .ts    Factors n and в are presented in Table n 0,30 0,35 0,40 0,45 0,50 0,55 0,60 0,65 0,70 в 18,5396 8,9707 4,3174 2,0728 1 0,48122 0,23162 0,11147 0,053649The point of reference of the tail section is located at distance of l f.ns+l f.ts from a fuselagenose. Coordinates of Yf.ns and Yf.ts axis are turned counter-clockwise on n.f and t.f angles. Ifm and n are selected equal to 0,5 of the contours are usual quadratic parabolas. If smallervalues of m and n are selected, shapes of nose and tail sections will be fuller, if greater valuesare selected, the shapes will be more concentrated.SELECTION OF STABILIZERS STRUCTURET-shaped stabilizers: HS is placed off the zone of wing wash on all flight phases,VS isloaded additionally, its mass is increased . DORSAL FIN 1. Dorsal fin installation (fig. 2.35). The dorsal fin improves VS washing, activatesat high speeds, favorably influences side-slip, increases the effective area of VS. Allows toreduce SVS, mVS, VS.
  • [54]Selection of Landing Gear StructureThe landing gear must ensure:  Absorption and dispersion of kinetic energy of impact at landing and taxiing on aerodrome roughness’s protecting the airframe construction from destruction;  Stability and controllability of the aircraft at take-off run, landing roll, taxiing, maneuvering, pushback;The three-strut with nose strut configuration:  Absence of danger of nose-over (overturning onto the canopy);  Capability of effective braking of wheels after landing;  Simplification of piloting, preference of landing on two struts with lowering on the nose strut (the angle of attack decreases and bouncing is excluded);  Good directional stability at take-off run and landing roll: the turn force is compensated by the stabilizing moment of main struts friction forces;  Convenience and comfort for passengers, good observation for crew;  Simplicity of loading and unloading of cargo aircraft owing to horizontal or nearly horizontal position of the fuselage axis.Selection of Power Plant Type It is necessary to select the type of engine by comparing our prototype, two of theengines are mentioned below. Embraer ERJ 145 - Rolls-Royce AE 3007-A1
  • [55] Bombardier CRJ200 - GE CF34-3B1 Then we must select the type of Engine followed by its units and systems and then theintegral components of the power plantENGINE TYPE - Low Bypass turbojet engines with various bypass ratio without theafterburner.Integral components of the power plants 1. Engine nacelles with air intakes; The intake length in front of the engine face = 1.12 × Engine face diameter = 1.12 × 0.825 = 0.931 m
  • [56] The maximum nacelle diameter = 1.5 × = 1.5× 0.825 = 1.2375 m Engine face diameter The exhaust jet-pipe length aft of the last stage turbine disc = 1.5 × engine face = 1.5 ×0.825 = 1.05 m diameter The total length of the Nacelle = Engine = 2.78 + (1.5 + 0.825) = 4.018 m length + (k × Engine face diameter) 2. Systems of engines attachment; 3. Fuel and lubrication systems; 4. Fire protection and anti-icing systems; 5. Fuel and oil tanks;POSITION OR LOCATION OF ENGINE Position of engines on fuselage tail section allows ensuring the following features:"clean" wing, selection of anhedral/dihedral angle for wing from conditions of optimal lateralstability and controllability of the aircraft, noise level decrease in passenger cabins, firesafety improvement, and engine protection against getting of foreign objects. However, themass of fuselage tail section increases, the wing mass is increased because of the absence ofits unloading, the aircraft center of masses is displaced back, the efficiency of stabilizersdecreases, etc. Therefore use of such structure is limited.
  • [57]SELECTION OF THE SCHEME OF ACCOMMODATION OF CREW ANDPAYLOADThe cockpit or service compartment is usually placed in the forward nose section of thefuselage. Its dimensions depend on crew.Number of crew members is established bycalculation:  ncrew = 1 2 people for regional airlines;  ncrew= 2 people for short-range airways;Range of my aircraft ShortNo of Recommended crew members Pilot and A Co-pilotThe number of flight attendants is determined by the following regulations:  For I class cabin – one flight attendant for 15-16 passengers;  For II and III class cabins– one flight attendant for 30-35 passengers.Number of cabin attendant = 2
  • [58] SELECTION OF ENGINEIt is necessary to determine starting thrust of the engine. It is determined on the basis of thegraphical values of starting thrust-to-weight ratio which is determined in lab 4 . For this, it isnecessary to establish the value t 0 for the projected airplane.LAB 4 : In this part we are finding the Thrust to weight ratio with respect to the aspectratio of wing taken as 4 and aircrafts wing loading 600 N/m2.RESULT: In table P, denotes wing loading, Tk, denotes aspect ratio
  • [59]TOW- is the Thrust to weight ratio at take-off which is 0.251TOB- is the Thrust to weight ratio at landing which is 0.271TOK- is the Thrust to weight ratio at cruising which is 0.179Engine Total Thrust Determination P0  t 0  m 0  g Where t 0  0 . 251 , m0 = 39110 kg P0  0 . 251  39110  9 . 8  96202 ( N ) P0  96 . 2 ( kN )Thrust of one engine Further starting thrust of one engine can be determined on the basis of enginesnumbers ― n ‖. Number of engines is taken to be 2 P0 96202 Peng    48101 ( N )  48 . 1( kN ) n eng 2After determining the total thrust the specific type of engine is selected according to thethrust value. From engines catalogue based on thrust 48 kN the following engine is selectedROLLS-ROYCE [Spey RSP 4 Mk 510-14]Type Low bypass Number of spools : 2 turbofanLength 2.8 m Compressor : Axial flowFan 0.8255 m High pressure compressor 12Diameter stages :Dry weight 1048 kg Combustors : 10 can annular combustion chamberWidth 0.92 m High pressure turbine stages : 2Max 0.942 m Low pressure turbine stages : 2diameter
  • [60]PERFORMANCE DATAMaximum thrust 48.9 kN Thrust at cruising 30.70 kNSpecific fuel 0.6 Specific fuel consumption at 0.770consumption cruisingBy pass ratio 0.71:1 Cruising speed 0.77Thrust to weight ratio Cruising altitude 32000Fan pressure ratio 2.70 Airflow 206 lb/s
  • [61] DETERMINATION OF CENTER OF GRAVITY OF THE AIRCRAFTThe centre of gravity for the aircraft in preliminary stage is found statistically but beforefinding it, we need to know the masses of various components. Therefore the mass of thedifferent groups are found by three methods, the first one is the graphical determinationusing the software which is done already, The second one is the statistical method which isalso already done, the last method is the semi-empirical methodsSEMI EMPIRICAL MASS DETERMINATION-FUSELAGE MASSMFcivil – Cfus × ke ×kp×kuc×kdoor×( 2× L × D× VD0.5)1.5×(MTOM×Nut)xWhere, cfus is a generalized constant to fit the regression, as follows:  0.038 for small unpressurized aircraft (leaving the engine bulkhead forward)  0.041 for a small transport aircraft (≤19 passengers)  0.04 for 20 to 100 passengers  0.039 for a midsized aircraft  0.0385 for a large aircraft  0.04 for a double-decked fuselage  0.037 for an unpressurized, rectangular-section fuselageAll k-values are 1 unless otherwise specified for the configuration, as follows:ke = for fuselage-mounted engines = 1.05 to 1.07kp = for pressurization = 1.08 up to 40,000-ft operational altitude1.09 above 40,000-ft operational altitudekuc = 1.04 for a fixed undercarriage on the fuselage 1.06 for wheels in the fuselage recess 1.08 for a fuselage-mounted undercarriage without a bulge 1.1 for a fuselage-mounted undercarriage with a bulgekVD = 1.0 for low-speed aircraft below Mach 0.31.02 for aircraft speed 0.3 < Mach < 0.61.03 to 1.05 for all other high-subsonic aircraft
  • [62]kdoor = 1.1 for a rear-loading doorFor the designed aircraft,VD - Velocity 750km/h = 208 m/s(MTOM×Nut)x 1 for civil aircraftCfus 0.04 for 20 to 100 passenger aircraftke 1.05kp 1.08kuc 1.1kdoor 1.1L - length of the aircraft 30.62mD diameter of the fuselage 2.5mMTOM- maximum takeoff mass 39.11tonsMFcivil = 0.04 × 1.05 ×1.08×1.1×1.1×( 2 30.62× 2.5 × 2080.5)1.5×1 = 5693.969 kg5% reduction in mass due to composite usage therefore 5409.270kgWING MASSwhere Cw = 0.0215 and flaps are a standard fitment to the wing.kuc = 1.002 for a wing-mounted undercarriage; otherwise, 1.0ksl = 1.004 for the use of a slatksp = 1.001 for a spoilerkwl = 1.002 for a winglet (a generalized approach for a standard size)kre = 1 for no engine, 0.98 for two engines, and 0.95 for four engines (generalized)CW 0.0215kuc 1.002kse 1.004ksp 1.001
  • [63]kwl 1.002kre 0.98Sw - Wing area 63.87m2AR – Aspect ratio 7.35λ 0.375Fuel mass 8760kg - Sweep angle 22.73 - Thickness ratio 0.18MWing = 2682.738kgHORIZONTAL TAIL MASSMHT =Kconfg 1.1 for T-tailSh – Area of horizontal stabilizer 13.98 m2AR – Aspect ratio of the horizontal stabilizer 4Λ – Sweep angle of the of the horizontal 20stabilizer - Thickness ratio of the horizontal 0.12stabilizerλ 0.3Kmat 0.98MHT = ; MHT = 806 kg
  • [64]VERTICAL TAIL MASS MVT =Kconfg 1.1 for T-tailSv - Area of vertical stabilizer 9.01 m2AR- Aspect ratio of the vertical stabilizer 1.36Λ - Sweep angle of the of the vertical 32stabilizer - Thickness ratio of the vertical stabilizer 0.10λ 0.3Kmat 0.98MVT = ; MVT =274.212 kgTOTAL TAIL UNIT MASSMTAIL = Horizontal tail group mass + Vertical tail group mass = 806+ 274.212 =1080.212 kgNACELLE MASSFor by-pass ratio less than 4,MNAC = 6.2 × thrust per nacelleMNAC = 6.2 × 48 kN = 297.6 kgFor two nacelles, MNAC = (6.2 × 48) 2 = 595.2 kgUNDERCARRIAGE MASSFor Low wing aircraft,MUG = 0.04× MTOMMUG = 0.04× 39110 kg = 1564.4 kg
  • [65]POWERPLANT MASSFor turbofan ( without thrust reverser )Meng = 1.4 ×Mdry engine (Mdry engine = 1048kg)Meng = 1.4 ×1048 kg = 1467.2 kgFor two engines Meng = 2934.4 kgSYSTEMS MASSMSYS = 0.12 × MTOMMSYS = 0.12 × 39110 kg = 4693.2 kgFURNISHING MASSMFURN = 0.057 × MTOMMFURN = 0.057 × 39110 kg = 2262.56 kgCONTINGENCY & MISCELLANOUSMCONT = 0.015 × MTOMMCONT= 0.015 × 39110 kg = 586.65 kgMMISC = 0.010 × MTOMMMISC= 0.010 × 39110 kg = 391.1 kgCREW MASSCrew members = 4MCREW = 4 × 80 = 320 kgPAYLOAD MASSNo of passengers = 47Mass of one passenger = 90 kgMPAYLOAD = 47 ×90 = 4230 kg
  • [66]AVERAGE BETWEEN GRAPHICAL, SEMI-EMPIRICAL & STATISTICALMETHODMass of Airframe = Mfuselage + Mwing + Mtail + MlgFUSELAGEMfuselage Graphical method = 13688.5 kgMfuselage Statistical method = 4392.83 kgMfuselage Semi-empirical method = 5409.270 kgMfuselage Average =WINGMwing Graphical method = 1877.28 kgM wing Statistical method = 4956.01 kgM wing Semi-empirical method = 2682.738 kgM wing Average =TAILMtail Graphical method = 711.802 kgM tail Statistical method = 863.54 kgM tail Semi-empirical method = 1080.212 kgM tail Average =LANDING GEARMlg Graphical method = 2424.82 kgMlg Statistical method = 2302.79 kgM lg Semi-empirical method = 1564.4 kgMlg Average =POWERPLANTMpowerplant Graphical method = 3363.46 kgM powerplant Statistical method = 5475.4 kgM powerplant Semi-empirical method = 2934.4 kgMlg Average =POWERPLANTMfuel Graphical method = 8760.64 kg
  • [67]M fuel Statistical method = 7039.8 kgM fuel Average = kgEQUIPMENT & CONTROL SYSTEMSMeq & cont Graphical method = 4952 kgM eq & cont Statistical method = 4383.839 kgM eq & cont Semi-empirical method = 4693.2 kgM eq & cont Average = kgFURNISHINGMfurnishing Graphical method = 610.116kgM furnishing Statistical method = 2346.6 kgM furnishing Average = kg MASS BREAKDOWNSTRUCTURAL GROUP MASSStructural group mass = M fuselage + M wing + M tail + M lgM fuselage = kgM wing = 3172.009 kgM tail = M horizontal + M verticalM tail = 480+ 375.18 = 885.1846 kgM lg = M main lg + M nose lgM lg = 1400 + 697.33 = 2097.33kgStructural group mass = + 3172.009 + 885.1846 + 2097.33Structural group mass = 13984.723 kgPOWERPLANT GROUP MASSPowerplant group mass = M dry engine + M nacelle + M fuel system+ M oil + M control systemM dry engine = 2096 kgM nacelle = 595.2 kgM fuel system = 650 kgM oil = 383 kgM control system = 200.22 kgPowerplant group mass = 2096+ 595.2 + 650+ 383+ 200.22Powerplant group mass = 3924.42 kgFUEL MASS GROUPM fuel =
  • [68]M fuel Group 1 = 3950.11 kgM fuel Group 2 = 3950.11 kgAUXILIARY POWER UNITMapu = 7 × Number of seatsMapu = 7 × 52 = 364 kgEQUIPMENT & CONTROL SYSTEMS GROUP MASSEquipment & control systems = M ecs + M fcs + M hyd & pneum+ M electrical+M inst & navi + M avionics &electronics + M operating instM ecs = 15 × Number of seatsM ecs = 15 × 52 = 780 kgMfcs = 550 kgM hyd & pneum = 1400 kgM electrical = 13 × Number of seatsM electrical = 13 × 52 = 672 kgM inst & navi = 365 kgM avionics & electronics = 405 kgM operating inst = 504.34 kgEquipment & control systems = 672+ 405 + 780+ 365+ 550+ 1400 + 504.3463Equipment & control systems = 4676.3463 kgFURNISHING GROUP MASSFurnishing group mass = M seat + M paint + M oxygen systemM seat = 15 × Number of seats = 15× 54 = 810 kgMpaint = 338.358 kgM oxygen system = 300 kgFurnishing group mass = 810+ 338.358+ 300 = 1449 kgCONTINGENCY & MISCELLANEOUS GROUP MASSContingency & miscellaneous = M contingency + MmiscellaneousM contingency = 586.65 kgMmiscellaneous = 391.1 kgContingency & miscellaneous = 586.65 + 391.1 = 977.75 kgPAYLOAD, CREW, LUGGAGE, CONSUMABLES & WATERPAYLOAD MASSNo of passengers = 47
  • [69]Mass of one passenger = 75 kgMpayload = 47 ×75 = 3525 kgCREW MASSCrew members =4Mcrew = 4 × 80 = 320 kgLUGGAGE MASSNo of passengers = 47Mass of luggage for one passenger = 15 kgMluggage = 47 ×15 = 705 kgCONSUMABLES & WATERMConsumables = number of passengers ×10kgMConsumables = 47 ×10kg = 470 kg MAXIMUM TAKE-OFF MASSMTOM(maximum takeoff mass) = OEM + Payload + FuelOperating Empty mass (OEM) is the mass of structure, powerplant, furnishing systems,unusable fuel and other unusable propulsion agents, and other items of equipment that areconsidered anintegral part of a particular airplane configuration. Also included are certainstandard items,personnel, equipment, and supplies necessary for full operations, excludingusable fuel and payload.OEM(operator’s empty mass) = MEM + Crew + ConsumableWhere,MEM (manufacturer’s empty mass) = M fuselage + M wing + M tail + M lg) + (M dry engine + M nacelle+ M fuel system+ M oil + M control system + Mapu) +( M ecs + M fcs + M hyd & pneum+ M electrical+M inst & navi+ M avionics & electronics + M operating inst )+ (M seat + M paint + M oxygen system )+ (M contingency +Mmiscellaneous)MEM (manufacturer’s empty mass) = + 3172.009 + 885.1846 + 2097.33 + 2096+595.2 + 650+ 383+ 200.22+ 364 + 672+ 405 + 780+ 365+ 550+ 1400 + 504.3463+ 810+338.358+ 300 + 586.65 + 391.1MEM (manufacturer’s empty mass) = 25375 kgOEM(operator’s empty mass) = M fuselage + M wing + M tail + M lg) + (M dry engine + M nacelle + Mfuel system+ M oil + M control system + Mapu) +( M ecs + M fcs + M hyd & pneum+ M electrical+M inst & navi + M
  • [70]avionics & electronics + M operating inst )+ (M seat + M paint + M oxygen system )+ (M contingency + Mmiscellaneous) +(special equipments )+ Mcrew + MConsumablesOEM(operator’s empty mass) = + 3172.009 + 885.1846 + 2097.33 + 2096+ 595.2 +650+ 383+ 200.22+ 364 + 672+ 405 + 780+ 365+ 550+ 1400 + 504.3463+ 810+ 338.358+300 + 586.65 + 391.1+320 + 470OEM (operator’s empty mass) = 26165.58 kgMaximum Design Takeoff Weight (MTOM). Maximum weight for takeoff as limited byaircraftStrength and airworthiness requirements. (This is the maximum weight at start of the takeoffrun.)MTOM = (M fuselage + M wing + M tail + M lg) + (M dry engine + M nacelle + M fuel system+ M oil + M controlsystem + M fuel + Mapu) +( M ecs + M fcs + M hyd & pneum+ M electrical+M inst & navi + M avionics & electronics+ M operating inst )+ (M seat + M paint + M oxygen system )+ (M contingency + Mmiscellaneous) + ( specialequipments )+ (Mpayload + Mcrew + Mluggage) + MConsumablesMTOM =( + 3172.009 + 885.1846 + 2097.33) + (2096+ 595.2 + 650+ 383+ 200.22)+ + 364 +(672+ 405 + 780+ 365+ 550+ 1400 + 504.3463)+ (810+ 338.358+ 300) +(586.65 + 391.1 )+(814)+ (3525 + 320 + 705) + 470MTOM = 13984.723+3924.42+7900.22+364+4676.3463+1449+977.75+814 +4550+470MTOM = 39110 kg CENTER OF GRAVITYFrom the required mass data the CG of the aircraft is determined by tabulating all the unitcargo with their position located on the aircraft along with its X and Y co-ordinates. Some ofthe coordinates are assumed based on the statistical data and others can be located on threeview diagram. At last the average of the CG is found and presented in the A1 format.STRUCTUREUNIT CARGO POSTION OF X Y LOCATION MASS CG LOCATIONWing 0.40 × Bmac 18.551m 1.431m 3172 kg 0.40 × 3.15m = 1.26m
  • [71]Fuselage 0.6×Lf 18.372m 2.34m 7830 kg 0.6×30.62 = 18.37mHorizontal 0.45×Bahs 30.533m 5.601m 480 kgstabilizer 0.45×1.93 = 0.8685mVertical tail 0.45 × Bavs 28.919m 4.459m 405 kg 0.45 ×2.57 = 1.1565mMain LG Located aft of CG 17.679m 3.75m 1400 kg of aircraftNose LG From 3 view 4.265m 1.322m 697 kgPOWERPLANTMiddle engine Center section 23.54m 2.887m 1400 kgEnd engine Center section 25.772m 2.837m 696 kgEngine nacelles Center section 24.028m 2.887m 595 kgFuel systems Center of fuel 18m 1.5m 650+ tanks +Oil System 383+ +Engine Control 200system = 1233 kgEQUIPMENT & CONTROL SYSTEMSEnvironmental At tail section 25.772m 2.837m 780Control System ++ 364Auxilliary power =unit 1144 kgNavigation 0.40 × nose length 1.5 2 365+550+
  • [72]+ 0.40 × 3.75 405Flight Control =1.5m =System+ 1320 kgAvionics &Electronics 0.75×Fuselage 22.96 2.1 672 kgElectrical length 0.75 ×30.62 =22.965mHydraulics Near main LG 16.773 1.3 1400+ +Operating unit 504 = 1904 kgFurnishings 0.51×Fuselage 15.611 2.3 1449+ length +Special equip 0.51×30.62 814+ =15.611m +Contingency 977 = 3240 kgOPERATIONAL ITEMSCrew 0.45 × nose length 1.6875 2.499 160 kg 0.45 ×3.75 =1.6875Cabin attendant Near cabin 5 2 160 +470 + =Consumables 630 kgFUELGroup- 1 Center of fuel 19 1.414 3950 kg tanksGroup- 2 Center of fuel 19 1.414 3950 kg tanks
  • [73]PAYLOADPassenger 0.51×Fuselage 15.611m 2.3 3525 kg length 0.51×30.62 =15.611mLuggage 0.51×Fuselage 15.611m 2.3 705 kg length 0.51×30.62 =15.611mDETERMINATION OF Mg X and Mg YSTRUCTURE Mg Mg X Mg YUNIT CARGO MASS X LOCATION Y LOCATIONWing 3172 31117.32 577257.4033 44528.88492Fuselage 7830 768123.3 14111961.27 1797408.522Horizontal 480 4708.8 143773.7904 26373.988stabilizerVertical tail 405 3973.05 114896.633 11715.82995Main LG 1400 13734 242803.386 51502.5Nose LG 697 6837.57 29162.236 25640.887 POWERPLANT Mg Mg X Mg YMiddle engine 1400 13734 323298.36 39650.058End engine 696 6827.76 175965.0307 19370.35512Engine nacelles 595 5836.95 140250.2346 16851.27465Fuel systems 650+ + 12095.73 217723.14 18143.595Oil System 383+ +Engine Control 200system =
  • [74] 1233EQUIPMENT & CONTROL SYSTEMS Mg Mg X Mg YEnvironmental 780Control System + 11222.64 289229.8781 31838.62968+ 364Auxilliary power =unit 1144Navigation 365+550++ 405 12949.2 19423.8 25898.4Flight Control =System+ 1320Avionics &Electronics 672 6592.32 151359.6672 13843.872ElectricalHydraulics 1400+ + 18678.24 313290.1195 24281.712Operating unit 504 = 1904Furnishings 1449+ +Special equip 814 31784.4 496186.2684 73104.12+ +Contingency 977 = 3240OPERATIONAL ITEMSCrew 160 1569.6 2648.7 3922.4304Cabin attendant 160 +470 + = 6180.3 30901.5 12360.6Consumables 630
  • [75]FUELGroup- 1 3950 38749.5 736240.5 54791.793Group- 2 3950 38749.5 736240.5 54791.793PAYLOADPassenger 3525 34580.25 539832.2828 79534.575Luggage 705 6916.05 107966.4566 15906.915 n ( mg ) = 1074960.48 i i 1 n ( mgx ) = 19500411.16 i i 1 n ( mgy ) = 2440660.734 i i 1CG coordinates. n n xm   ( mgx ) /  ( mg ) i i i 1 i 1Xm = n n ym   ( mgy ) i /  ( mg ) i . i 1 i 1Ym =
  • [76] DESIGN STRUCTURAL CONFIGURATIONDESIGN STRUCTURAL CONFIGURATION OF WINGThe choice of load-carrying structures for wing is determined by M  criterion of bending moment intensity acting upon wing, , MPa; 3 H  criterion of wing specific load, p0 = m0g/s, N/m2;  wing lay-out (arrangement of fuel tanks or fuel sections, wells for LG retraction, hatches for instrument and equipment maintenance and so on inside the wing);  fuselage lay-out and possibility to arrange the wing center section inside the fuselage;  Requirements of sufficient strength, rigidity, aeroelasticity, service life, reliability under condition of minimal mass, costs. Wing LCS is selected on the basis of the following factors: To define approximately the wing LCS let us use concept of conditional spar, whichcaps have width bcond. about 0.6b, here b – wing chord in design section
  • [77] Conditional spar bcond = 0.6 × Chord of wing at design section bcond = 0.6 × 3.82 = 2.292 Nbr. cap = bcond. cond. br. Nbr. cap = 2.292 ×2×10-03.×330×106 Nbr. cap = 1.512 Mpa CAP THICKNESS OF CONDITIONAL SPAR Depending upon the Cap thickness of spar of the wing the distance between Ribs and Stringers are selected, from the formula the cap thickness is determined as follows, p 0 Sz A  2 m i gz i  m w ing gz A  n des .  cond .  0.96 cb0  br . 2р0 – wing specific load, N/m2; 600×9.81 = 5886 N/m2S – wing area, м2; 63.87 m2g – gravity acceleration, 9.8 m/s2; 9.8zA – MAC coordinate from longitudinal airplane 4.4maxis spanwise, m;mi, zi – mass of cargo, placed on wing, kg, and its [ mass of fuel×Zfuel+ mass of LG×ZLgCG coordinate, m; ]9.81 [7900 × 5.5 +1400 ×1.8]mwing – wing mass, kg; 3172.009 kgndes. – design g-force, for design case А it equals 31213.5 for maneuverable aircraft, it equals 69for limited maneuverable aircraft and 2.53.8 fornon-maneuverable passenger and transport;c – wing thickness ratio; 0.15 [NACA 2415]b0 – wing root chord, m; 4.5 mbr. – breaking stress of spar capbr = 330 МPа for cap made of aluminum alloyД16Т, 880 МPа for cap made of alloyed steel 330 MPa30ХГСА, 800 МPа for titanium alloy Вт6; [168616 . 8  2 ( 425810  24696 )  136777 . 0281 ]3 = 2mm ln   0 . 00204 ( m )  0 . 96  0 . 15  4 . 5 2   330  10 6
  • [78] cond. = 2 mm defines class of spar-type wings with partially working skin. Bending moment is taken by spar; cross-cut force Q is taken by spar webs, torsion torque is taken by contour shaped by skin and spar webs. To increase skin critical stress in shear it is reinforced by weak stringers and thickly arranged ribs. Recommended Rib pitch 220 mm Stringer pitch 200 mm CRITERION OF LOAD MOMENT INTENSITY AND CROSS-CUT FORCE LOAD MOMENT INTENSITY After finding the rib and stringer pitch, we need to find the type of wing load carrying structure. This can be found using load moment intensity formula and from its result it is desirable to select appropriate LCS. M ( p 0 S  m w  g )  z A  2 m i  g .z i   n p  1 . 03  ( c  b 0 ) 3 3 HM ( 5886  63 . 87  3172 . 009  9 . 8 )  4 . 4  2  24696  1 . 8  9 . 8  2  7900 . 22  5 . 5  9 . 8   3 3 H 0 . 31677M 3 =5.8369 MPaH Since the intensity of bending moment does not exceed 15 МPа, it is better to use spar-type wing. SPAR-TYPE WING CONSTRUCTION: - we are considering two cross-sections of the wing, which are, the root chord and the second one is wing tip chord section. Root chord 4.5 m Tip chord 1.1 m Spar position in the wing:- Front spar Rear spar At root chord = 0.25 × 4.5 = 1.125 m At root chord = 0.74 × 4.5 = 3.33 m At tip chord = 0.25 × 1.1 = 0.275 m At tip chord = 0.74 × 1.1 = 0.814 m
  • [79]CROSS-CUT FORCETo calculate intensity of the CROSS-CUT FORCE the concept of equivalent rectangularwing is used go by turning swept wing so to provide perpendicularity of line of elasticcenters to airplane longitudinal axis. For straight wing such turn is not accomplished.Position of line of elastic centers may be accepted at 40% of chord.The formula for cross cut force is, n  m 0  m w ing  g   m i g S cut Q S i 1 n f  op .   2 2 H 1.28 cb0
  • [80]Q – cross-cut force;Н – design height of wing section airfoil, m;zcut. – distance from design section to point of L sec .  b d es .  2 beq u iv . application of resultat force of aerodynamic z cu t .    = 3  b d es .  beq u iv .   and mass loads, m; = 3.96mScut. – area of wing cut part; 23.866 m2mwing – mass of wing; 3172 kgLcut – length of wing cut part; 9.72 mm0 – airplane take-off mass; 39110mi – mass of concentrated cargo (mass of fuel + mass of LG) 9.81nop × f 3c – wing thickness ratio; 0.15 [NACA 2415]b0 – wing root chord, m; 4.5 m = 0.9493 MPaPosition and Purpose of Strong Ribs in the wingNumber of the Strong Purpose of the Strong RibRib Fuselage rib Front Spar Joint in the Wing Rear Spar Joint in the Wing Main landing gear mechanism support Flap Joint Main landing gear mechanism support Main Landing Gear, Flap Joint Engine Connection, Flap Joint Engine Connection, Flap Joint Flap Joint Flap Joint Flap Joint Aileron Joint Aileron Joint Aileron Joint Aileron Joint
  • [81] Aileron Joint Aileron JointLOAD-CARRYING STRUCTURE OF THE FUSELAGESEMI-MONOCOQUE-Advantages 1) The semi-monocoque structure of a fuselage ensures taking up and transmission ofall loads at high stiffness, strength and low mass. The reason for this is that the material,which carries bending and torsion stresses (skin and stringers), is as much possible distancefrom the central axis. 2) Semi-monocoque has large internal free volumes and allows them to be effectivelyused.The distance between frames depends on thickness of a fuselage skin, configuration andmass of the airplane. On real structures frame pitch is accepted is some limits, for Lighttransport airplanes it is between 300 to 400mm.The distance between stringers in a fuselage is chosen for the same reasons as in a wing.Depending on thickness of a skin distance between stringers is accepted as 80…250mm formedium and light airplanes.Distance between frames = 400mm.Distance between stringers = 150mm.Position and Purpose of Strong Ribs and Frames in Aircraft StructureOn Fuselage:Number of Purpose of the Framethe Frame Nose Radom Support Connection with the Cock-Pit Main Door support beam connection i) Landing gear cutout support connection ii) Main Door Connection i) Connection between Nose and mid fuselage sections. ii) Landing gear cutout support connection Wing front spar connection Emergency door support beam connection Emergency door support beam connection Second emergency exit support beam connection i) Wing rear spar Connection
  • [82] ii) Emergency door support beam connection Connection between fuselage mid section and rear section Rear Door Connection Rear Door Connection Vertical Tail front Spar Connection to the Fuselage Vertical Tail rear Spar Connection to the FuselageLoad-Carrying Structures for the Tail Unit Section Structural members of horizontal and vertical must be coordinated with each otherand with structural members of the fuselage also. In the design of tail unit, the two-sparstructure is usually applied. The lightest structure of tail unit can be in that case if theprimary structural members of the tail unit carrying bending stresses (spar and torsion-box)may be passed through the fuselage. To reduce structure mass whenever possible verticalunit and stabilizer should be fixed to the same strong frames of the fuselage.Load-Carrying Structures of Vertical Tail UnitSpar-type vertical tail unit construction: - we are considering two cross-sections of the wing,which are, the root chord and the second one is wing tip chord section.Root chord 2.57 mTip chord 2.57 mSPAR POSITIONFront spar Rear sparAt root chord = 0.18 × 2.57 = 0.462 m At root chord = 0.63 × 2.57 = 1.6191 mAt tip chord = 0.18 × 2.57 = 0.462 m At tip chord = 0.63 × 2.57 = 1.6191 mPurpose of Strong RibsNumber of the Purpose of the Strong RibStrong Rib Fuselage rib Nose Spar Joint Starting Connection Rear Spar Joint Rudder Joint Rudder Joint Rudder JointLoad-Carrying Structures of Horizontal Tail Unit
  • [83]Spar-type Horizontal Tail Unit construction: - we are considering two cross-sections of thewing, which are, the root chord and the second one is wing tip chord section.Root chord 2.48 mTip chord 1.24 mSpar positionFront spar Rear sparAt root chord = 0.20 × 2.48 = 0.496m At root chord = 0.65 × 2.48 = 1.612 mAt tip chord = 0.20 × 1.24 = 0.248 m At tip chord = 0.65 × 1.24 = 0.806 mPurpose of Strong RibsNumber of the Purpose of the Strong RibStrong Rib Fuselage Starting Connection of the Horizontal Tail Nose spar connection Rear spar connection Elevator Joint Elevator Joint Elevator JointLANDING GEARIn aviation, the under carriage or landing gear is the structure (usually wheels, butsometimes skids, floats or other elements) that supports an aircraft on the ground and allowsit to taxi, takeoff and land. Landing gear usually includes wheels equipped with shockabsorbers for solid ground, but some aircraft are equipped with skis for snow or floats forwater. To decrease drag in flight some undercarriages retract into the wings and/or fuselagewith wheels flush against the surface or concealed behind doors; this is called retractablegear. In this project work, the aircraft is equipped with such type of landing gearconfiguration. Here tricycle type landing gear with nose /front support is chosen, as in mostof the passenger pla
  • [84] STANDARD SPECIFICATION OF THE DESIGNED AIRCAFT MAERO – DA12 The designed aircraft is the all-metal low-wing cantilever monoplane with Т-tail withfixed fuselage-mounted stabilizer. The aircraft is a regional passenger aircraft, designed to be operated on domesticairlines for transporting passenger’s patients and injured man at short- and medium-haulservice airlines. Two turbofan Rolls Royce [ spey RSP 4 Mk 510-15] (2х10790 kgf) and RE220 gasturbine engine of the auxiliary power unit are installed in the aircraft.Aircraft crew consists of tree members: - captain (pilot); - co-pilot;PERFORMANCESThe maximum allowed operational enroute speed and Mach number for enroute aircraftconfiguration are the following: VmaxOp 835 km/h МmaxOp 0.65DESIGN MASSES- Maximum takeoff mass 39110 kg- Empty mass 25375 kgCENTER OF GRAVITY DATAAllowable range of center of gravity at takeoff, in flight and at landing: -- extreme forward 15.5 % MAC -- extreme aft 31.5 % MACDIMENSIONS AMD AREASWing: - area 63.87 m 2 - span 22.39m - inner wing sweep along line of 0.25 chord 26° 34 - ailerons area 3.83 m 2Tail unit: - horizontal tail area 13.98 m2 - elevator area 4.194 m2 - vertical tail area 9.01 m2 - rudder area 3.64 m2Fuselage: - length 30.62 m - maximum diameter 2.5 m
  • [85]Inner dimensions of Passenger compartment: - length 18.03 m - width along floor 2.08 m - maximum width 2.27 m - height 1.86 mEntrance door dimensions 0.91x1.78 mUpper emergency hatch dimensions 0.480.510 mGalley service door dimensions 0.61  1.22mOver wing emergency exit 0.51  0.97mLanding gear: - track 4 .7 m - wheelbase 14.6976 mWheels parameters:а) main landing gear: - tire dimensions 800х310 mm - pressure 0.86 МPаb) nose landing gear: - tire dimensions 500х310 mm - pressure 0.86 МPаConclusion: In the general designing part we calculated the required geometrical parametersof the designed aircraft, the components mass and the three views is sketched finallyaccording to the results.
  • [86] DESIGN SECTION PART 22. CALCULATION OF AERODYNAMIC CHARACTERISTICS OF THE AIRCRAFTPARAMETERS OF AEROPLANEPARAMETER ABBREVATION DIMENSION UNITtake off mass 39110 kgnumber of passengers 47maximum speed 760 km/hcruising speed 835 km/hrange 2500 kmengine thrust 48 knGEOMETRICAL CHARECTERISTICS OF FUSELAGEPARAMETERS ABBREVATION DIMENSION UNITfuselage length 30.62length of fuselage nose 3.75length of fuselage tail 6.25length of mid fuselage 20.62diameter of fuselage 2.5nose deflected angle 3 degtail deflected angle 4 degaspect ratio of fuselage 12.25 -aspect ratio of fuselage nose 1.5 -aspect ratio of fuselage of tail 2.5 -
  • [87] CHARACTERISTICS OF WING WITH VENTRAL SECTIONPARAMETERS ABBREVATION DIMENSION UNITwing span 22.39gross area 63.87root chord 4.5tip chord 1.1mean aerodynamic chord 3.15swept angle 22.73 DEGaspect ratio 7.85 -taper ratio 4 -fuselage diameter D 2.5thickness ratio 0.18 -camber ratio 0.02 -position of chamber relative 0.35 -AERODYNAMIC CHARECTERISTICS OF WINGDetermination of angle of Zero lifts = -60 [1+10 ( ) 2] = -60 [1+10 ( ) 2] = -1.47Determination of derivative of lift coefficient = 2 π ( 1- 0.274 ) = 2 × 3.14(1- 0.274 ) = 5.1781m=m= = 1.64 [ ] [ ] = 0.03292 = = = 3.9149
  • [88]Determination of maximum coefficient of liftRe == = 500.42 × 39.3 . tg (Re . 10-6) 39.3 . tg (500.42 × . 10-6) = 1.6760 = (1- ) = 1- ] = 1.5327 = 57.3 + + = 57.3 + (-1.47) + = 22.463Drag coefficientA=A= = 0.04178 friction = 2 = = = 2.338 = = = 1.0621 = = = 1.6516 friction = 2 =2 1.0621 =8.2024
  • [89]Aerodynamic quality = = = 27.0094 = = = 0.44308 = ( ) = 3.9149( ) = 0.714860 = +A = 8.2024 + 0.04178 = 0.026K= = = 27.4946Determination of Aerodynamic Center -2.8 (1- ) -2.8 x 0.021 ( 1- ) = -0.02532 = -0.25(1-1.6 ) = -0.25(1-1.6x ) = -0.2370 = [cya ( )+ ] = (0.23)-0.0253 = [0.71486(-0.23) + (-0.02532)] = -0.194741CHARECTERISTICS OF HORIZONTAL STABILIZER WITH VENTRALSECTIONPARAMETERS ABBREVATION DIMENSION UNITspan of horizontal stabilizer 7.55gross area 13.98root chord 2.48mean aerodynamic chord 1.93sweep angle 20 DEGaspect ratio 4.077 -taper ratio 2 -tip chord 1.24
  • [90]thickness ratio 0.12 -Determination of angle of Zero lifts = -60 [1+10 ( )2 ] = -60 [1+10 ( )2] = -1.68Determination of derivative of lift coefficient = 2 π ( 1- 0.274 ) = 2 × 3.14( 1- 0.274 ) = 5.54m=m= = 0.78188 [ ] [ ] = 0.041080 = = = 3.452Determination of maximum coefficient of liftRe == = 306.6114 × 39.3 . tg (Re . 10-6) 39.3 . tg (306.6114× . 10-6) = 1.6760 = (1- ) = 1- ] = 1.79 = 57.3 + + = 29.68
  • [91]Drag coefficientA=A= = 0.0804 friction = 2 = = = 2.5106 = = = 1.0621 = = =1.3696 friction = 2 =2 1.0621 =7.3040Aerodynamic quality = = = 20.63296 = = = 0.30140 = ( ) = 3.4526( ) = 0.63059 = +A = 7.3040 + 0.0804 = 0.03927K= = = 16.057Determination of Aerodynamic center -2.8 (1- ) -2.8 x 0.021 ( 1- ) = -0.02532] = -0.25(1-1.6 ) = -0.25(1-1.6x ) = -0.2442 = [cya( )+ ] = 0.630 (-0.24)-0.0253 = -0.17929
  • [92] CHARECTERISTICS OF VERTICAL STABILIZER WITH OUTER PANELPARAMETERS ABBREVATION DIMENSION UNITspan of vertical stabilizer 3.50cross area 9.01root chord 2.57tip chord 2.57mean aerodynamic chord 2.57swept angle 32 DEGaspect ratio 1.36 -tapper ratio 1 -thickness ratio 0.10 -Re =Re= = 408.28576 ×A=A= = 0.24119 friction = 2 = = = 2.4078 = = = 1.0622 = 1+2. +9( = 1+2. +9( = 1.29 friction = 2 =2 2.4078 1.0622 1.29=6.5985
  • [93] DETERMINATION OF AERODYNAMIC CHARECTERISTICS OF FUSELAGELift coefficient : Fuselage = nose+ cylinder+ rearLift coefficient for nose: nose = 2(1- ) =2(1-0)=2Lift coefficient for cylindrical part: cylinder = 0Lift coefficient for rear part: rear= -0.4(1- ) = -0.4 (1-0)= -0.4Total Lift coefficient of fuselage : Fuselage = nose+ cylinder+ rear Fuselage = 2+0+(-0.4)=1.6Zero Lift Angle: =1.25 [ ] = = = = 4.9062 , = , =1.25 [ ]= = (8+ ) .Smf = (8+0 ) . 4.9062 = 9.8124 = = 1- = =0.5333 = 1-0.5333 = 0.4666For cylindrical part: =0For rear part: = -0.5 = 27.495
  • [94]Coordinate of AD centre of fuselage: = = = -6.2905Drag coefficient of fuselage body = +Σ = 0.10984+0.16307 = 0.272911 = ( ) = 44.8007 0.95661 0.002348 1.09162 = 0.10984 = = = 0.0023484 = 1+ + = 1+ + = 1.09126 = = = 0.95661( ) 4 [1-0.2 ]( ) 4.12.25 [1-0.2 ] = 44.8007 = = = 0.0190645 = Defletion of rear ( ) = = 2.2206 = 0.038 Drag coefficient of cockpit = 0.038 = 0.133 = 0.08 Drag coefficient of landing gear = 0.08 = 8.7872
  • [95]PROCEDURE INVOLVED IN AERODYNAMICS CALCULATION USING THESOFTWAREThe calculation aerodynamic centre, wave drag, profile drag, lift coefficient and dragcoefficient of the fuselage, wing, engines, horizontal stabilizer and vertical stabilizer aredone through the software by entering the required parameters under each section. This issimplified by referring our three view diagram for certain parameters like,  distance between the wing to nose of the fuselage,  distance between horizontal stabilizer to wing and  the offset angle between wing and fuselage,etc.,are found easily. Hence before proceeding to enter the values all dimensions in three viewsare mentioned clearly. From the results of the software graphs for various parameters ofaircraft is drawn and compared.The software has seven section the details involved in calculation is as followsMAIN PARAMETERS
  • [96]FUSELAGE PARAMETERSENGINE PARAMETERS
  • [97]WINGHORIZONTAL STABILIZERVERTICAL STABILIZER
  • [98]SECONDARY SECTIONIn The secondary section the additional drag is to 0.10Finally we obtain the general view diagram from the software for the given data’s and it ischecked with our general view diagram.
  • [99]After obtaining the three view diagram the aerodynamic characteristics are obtainedfollowing the longitudinal moment of the designed aircraft. From the centre of gravitycalculation we know that the cg is 18.1 so in calculation longitudinal moment this is enteredand checked with values obtained.
  • [100] RESULTS OF THE SOFTWARECalculation of aircraft aerodynamic characteristics==================================================================================Student: 16E02- kirubagaran Mazhalai PriyanAircraft type: subsonic non-manoeuvre. Configuration of aircraft: normal. Area Sch=63.870GEOMETRICAL PARAMETERS AND AERODYNAMIC CHARACTERISTICSOF THE ISOLATED FUSELAGE==================================================================================Lf=30.620 Df= 2.500 Lmf=12.248 Smf= 4.909 Srlf=0.0769 Fs/Sm=44.292 M*=0.934Ln= 3.750 Dn= 0.000 Lmn= 1.500 Etn= 0.000 Betn= 3.000 Shape: ellipsoidalLr= 6.250 Dr= 0.000 Lmr= 2.500 Etr= 0.000 Betr= 5.000 Shape: curvilinearAir intake - absent Sai= 0.000 Scb= 0.000 Srlcb=0.0000Cockpit Lcpt= 0.000 Scpt= 0.000Presence of the landing gear fairing----------------------------------------------------------------------------------Profile drag of the isolated fuselage M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Point Xt 0.14019 0.13142 0.12710 0.12463 0.12311 0.12213 0.12148 0.12050H= 0.0 0.09032 0.08654 0.08405 0.08218 0.08063 0.07926 0.07800 0.07334H= 4.0 0.09464 0.09065 0.08801 0.08602 0.08437 0.08292 0.08158 0.07663H= 8.0 0.10002 0.09582 0.09302 0.09089 0.08913 0.08756 0.08612 0.08083H=11.0 0.10498 0.10063 0.09770 0.09546 0.09360 0.09194 0.09041 0.08480----------------------------------------------------------------------------------Additional profile drag of the isolated fuselage M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.04255 0.04278 0.04295 0.04308 0.04321 0.04332 0.04343 0.08638H= 4.0 0.04233 0.04253 0.04269 0.04281 0.04292 0.04303 0.04313 0.08528H= 8.0 0.04211 0.04228 0.04241 0.04252 0.04262 0.04271 0.04281 0.08398H=11.0 0.04195 0.04209 0.04220 0.04230 0.04238 0.04246 0.04255 0.08285----------------------------------------------------------------------------------Wave drag of the isolated fuselage and its parts M*=0.934 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Nose part 0.25860Rear part 0.12471Fuselage 0.38332----------------------------------------------------------------------------------Additional wave drag of the isolated fuselage M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 0.00000----------------------------------------------------------------------------------Derivative dCy/dAl of the isolated fuselage and its parts Alfa0= 0.59 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Nose part 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.1632Cylindric 0.2005Rear part -0.4000 -0.4000 -0.4000 -0.4000 -0.4000 -0.4000 -0.4000 -0.4000Fuselage 1.6000 1.6000 1.6000 1.6000 1.6000 1.6000 1.6000 1.9637----------------------------------------------------------------------------------Location aerodynamic center Xac of the isolated fuselage and its parts from nose M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Nose part 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000Cylindric 0.1840Rear part 0.8979 0.8979 0.8979 0.8979 0.8979 0.8979 0.8979 0.8979Fuselage -0.2245 -0.2245 -0.2245 -0.2245 -0.2245 -0.2245 -0.2245 -0.1641
  • [101]===================================================GEOMETRICAL PARAMETERS AND AERODYNAMICCHARACTERISTICS OF THE ISOLATED WING================================================================================== B0= 4.500 Bs= 4.500 Bt= 1.100 Ba= 2.800 L= 22.730 Lp=11.365 Fiwt= 0.00 Bmac= 3.144 Xmac= 2.245 (w/o Extent.)Sp= 63.644 Srl=0.9965 Lm= 8.118 Et= 4.091 (w/o Extent.)Xi00= 26.3 Xi05= 19.1 Xi10= 11.1 Xic= 22.1 Xi25= 22.8 (w/o Extent.)----------------------------------------------------------------------------------Type of airfoil - classical Kp= 2.1 m= 0.350 Ñs= 0.150 Ñt= 0.080 Ñ= 0.136 Xc= 0.300 f= 0.020 Xf= 0.418 M*=0.776----------------------------------------------------------------------------------XB0=15.074 XBs=15.074 Fi= 2.000 distance from fuselage nose and setting angleX14=16.199 D14= 0.000 X12=17.324 D14= 0.000----------------------------------------------------------------------------------Position - wing + circular fuselage Kint= 0.644 Df= 2.500 H=-1.090 Sig=0.110Tip elements - absent----------------------------------------------------------------------------------Deceleration flow coefficient before a wing M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000H= 4.0 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000H= 8.0 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000H=11.0 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000----------------------------------------------------------------------------------Profile drag of the isolated wing M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Xt 0.29436 0.27113 0.25743 0.24863 0.24270 0.23857 0.23567 0.23071H= 0.0 0.00577 0.00549 0.00528 0.00511 0.00497 0.00484 0.00472 0.00428H= 4.0 0.00604 0.00575 0.00554 0.00537 0.00522 0.00508 0.00496 0.00450H= 8.0 0.00636 0.00606 0.00585 0.00568 0.00552 0.00538 0.00525 0.00477H=11.0 0.00665 0.00635 0.00613 0.00595 0.00580 0.00565 0.00552 0.00502----------------------------------------------------------------------------------Wave drag of the isolated wing M*=0.776 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 0.17041----------------------------------------------------------------------------------Derivative dCy/dAl of the isolated wing Alfa0= -1.78 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 4.0314 4.1076 4.2227 4.3875 4.6210 4.9590 5.4772 5.3445----------------------------------------------------------------------------------Location aerodynamic center Xac isolated wing from nose of side chord M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 0.6637 0.6637 0.6637 0.6637 0.6637 0.6637 0.6637 0.8478==================================================================================
  • [102]GEOMETRICAL PARAMETERS AND A/D CHARACTERISTICS OF THEISOLATED HORIZONTAL TAIL================================================================================== B0= 2.480 Bs= 2.480 Bt= 1.240 Ba= 1.860 L= 7.550 Lp= 3.775 Bmac= 1.929 Xmac= 0.748 (w/o Extent.)Sp= 14.043 Srl=0.2199 Lm= 4.059 Et= 2.000 (w/o Extent.)Xi00= 24.0 Xi05= 15.7 Xi10= 6.7 Xic= 19.1 Xi25= 20.0 (w/o Extent.)Type of airfoil - classical Kp= 2.1 m= 0.350 Ñs= 0.120 Ñt= 0.120 Ñ= 0.120 Xc= 0.300 f= 0.020 Xf= 0.417 M*=0.786----------------------------------------------------------------------------------XB0=28.900 XBs=28.900 Fi= 0.000 distance from fuselage nose and setting angleX14=29.520 D14= 0.000 X12=30.140 D14= 0.000 X1= 9.797 B1= 3.935 Xht=13.321 Yht= 4.500 S*/Sw= 0.808----------------------------------------------------------------------------------Position - Tee-tail horizontal tail Sig=0.000 Kint= 0.000Tip elements - absent----------------------------------------------------------------------------------Deceleration flow coefficient before a horizontal tail M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 0.93842H= 4.0 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 0.93842H= 8.0 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 0.93842H=11.0 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 0.93842----------------------------------------------------------------------------------Profile drag of the isolated horizontal tail M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Xt 0.33896 0.31160 0.29423 0.28243 0.27406 0.26794 0.26339 0.00000H= 0.0 0.00583 0.00555 0.00534 0.00516 0.00501 0.00487 0.00474 0.00603H= 4.0 0.00609 0.00580 0.00559 0.00541 0.00525 0.00511 0.00498 0.00635H= 8.0 0.00641 0.00610 0.00589 0.00571 0.00555 0.00540 0.00526 0.00677H=11.0 0.00672 0.00638 0.00616 0.00597 0.00581 0.00566 0.00552 0.00718----------------------------------------------------------------------------------Wave drag of the isolated horizontal tail M*=0.786 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 0.09897----------------------------------------------------------------------------------Derivative dCy/dAl of the isolated horizontal tail Alfa0= -1.77 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 3.2819 3.3328 3.4090 3.5170 3.6678 3.8818 4.2007 4.3515----------------------------------------------------------------------------------Location aerodynamic center Xac isolated horizontal tail from nose of side chord M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 0.4472 0.4472 0.4472 0.4473 0.4474 0.4478 0.4489 0.6794==================================================================================
  • [103]GEOMETRICAL PARAMETERS AND A/D CHARACTERISTICS OF THEISOLATED VERTICAL TAIL==================================================================================Vertical tail - central (1pcs.) B0= 2.570 Bs= 2.570 Bt= 2.570 Ba= 2.570 L= 3.500 Lp= 3.500 Bmac= 2.570 Xmac= 1.085 (w/o Extent.)Sp= 8.995 Srl=0.1408 Lm= 1.362 Et= 1.000 (w/o Extent.)Xi00= 31.8 Xi05= 31.8 Xi10= 31.8 Xic= 31.8 Xi25= 31.8 (w/o Extent.)----------------------------------------------------------------------------------Type of airfoil - classical Kp= 2.1 m= 0.350 Ñs= 0.100 Ñt= 0.100 Ñ= 0.100 Xc= 0.300 f= 0.000 Xf= 0.000 M*=0.837----------------------------------------------------------------------------------XB0=26.730 XBs=26.730 Fi= 0.000 distance from fuselage nose and setting angleX14=27.372 D14= 0.000 X12=28.015 D14= 0.000----------------------------------------------------------------------------------Position a= 1.250 b= 0.000 Lm= 1.362----------------------------------------------------------------------------------Deceleration flow coefficient before a vertical tail M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 0.99809H= 4.0 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 0.99809H= 8.0 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 0.99809H=11.0 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 0.99809----------------------------------------------------------------------------------Profile drag of the isolated vertical tail M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Xt 0.24813 0.22775 0.21552 0.20756 0.20211 0.19827 0.19551 0.00000H= 0.0 0.00566 0.00534 0.00511 0.00492 0.00476 0.00462 0.00449 0.00502H= 4.0 0.00595 0.00561 0.00537 0.00518 0.00501 0.00486 0.00472 0.00529H= 8.0 0.00631 0.00595 0.00570 0.00550 0.00532 0.00517 0.00502 0.00562H=11.0 0.00665 0.00627 0.00601 0.00580 0.00561 0.00545 0.00529 0.00595----------------------------------------------------------------------------------Wave drag of the isolated vertical tail M*=0.837 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 0.02900GEOMETRICAL PARAMETERS AND AERODYNAMICCHARACTERISTICS OF ISOLATED ENGINE NACELLE==================================================================================Engine nacelles in a tail part of a fuselage Nen= 2 Kint= 1.600Len= 4.018 Den= 1.031 Dcb= 1.031 Lm= 3.896 S= 0.835 Srl=0.0131 Fs/Sm=15.583Distance from fuselage nose Lfen=22.182 M*= 0.743Type of engine - Turbo-jet engine----------------------------------------------------------------------------------Profile drag of the isolated engine nacelle M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Xt 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000H= 0.0 0.04601 0.04496 0.04461 0.04455 0.04463 0.04477 0.04492 0.04530H= 4.0 0.04867 0.04748 0.04705 0.04695 0.04700 0.04712 0.04725 0.04759H= 8.0 0.05214 0.05076 0.05023 0.05007 0.05008 0.05017 0.05028 0.05055H=11.0 0.05554 0.05395 0.05331 0.05309 0.05306 0.05312 0.05321 0.05340----------------------------------------------------------------------------------
  • [104]Additional profile drag of the isolated engine nacelle M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000H= 4.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000H= 8.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000H=11.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000----------------------------------------------------------------------------------Wave drag of the isolated engine nacelle M*=0.743 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 0.00861----------------------------------------------------------------------------------Derivative dCy/dAl of the isolated engine nacelle dCy/dAl= 0.0000==================================================================================Critical Mach number of aircraft M*=0.95 min{0.776,0.786,0.837,0.934,0.743}= 0.738DRAG OF THE AIRCRAFT PATRS AND ADDITIONAL ELEMENTS INTOAIRCRAFT SYSTEM==================================================================================Profile drag of the fuselage into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.01021 0.00994 0.00976 0.00963 0.00952 0.00942 0.00933 0.01228H= 4.0 0.01053 0.01024 0.01004 0.00990 0.00978 0.00968 0.00958 0.01244H= 8.0 0.01092 0.01061 0.01041 0.01025 0.01013 0.01001 0.00991 0.01267H=11.0 0.01129 0.01097 0.01075 0.01059 0.01045 0.01033 0.01022 0.01289----------------------------------------------------------------------------------Wave drag of the fuselage into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 0.02946----------------------------------------------------------------------------------Profile drag of the wing into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.00575 0.00547 0.00526 0.00509 0.00495 0.00482 0.00470 0.00426H= 4.0 0.00602 0.00573 0.00552 0.00535 0.00520 0.00506 0.00494 0.00448H= 8.0 0.00633 0.00604 0.00583 0.00566 0.00550 0.00536 0.00523 0.00476H=11.0 0.00663 0.00632 0.00611 0.00593 0.00578 0.00563 0.00550 0.00501----------------------------------------------------------------------------------Interference profile drag of (wing + fuselage) system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000H= 4.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000H= 8.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000H=11.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000----------------------------------------------------------------------------------Wave drag of the wing into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.16980H= 4.0 0.16980H= 8.0 0.16980H=11.0 0.16980----------------------------------------------------------------------------------Interference wave drag of (wing + fuselage) system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.00000 0.00000H= 4.0 0.00000 0.00000H= 8.0 0.00000 0.00000H=11.0 0.00000 0.00000----------------------------------------------------------------------------------
  • [105]Profile drag of the horizontal tail into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.00128 0.00122 0.00117 0.00113 0.00110 0.00107 0.00104 0.00124H= 4.0 0.00134 0.00128 0.00123 0.00119 0.00116 0.00112 0.00109 0.00131H= 8.0 0.00141 0.00134 0.00129 0.00125 0.00122 0.00119 0.00116 0.00140H=11.0 0.00148 0.00140 0.00135 0.00131 0.00128 0.00124 0.00121 0.00148----------------------------------------------------------------------------------Interference profile drag of (horizontal tail + fuselage) system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000H= 4.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000H= 8.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000H=11.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000----------------------------------------------------------------------------------Wave drag of the horizontal tail into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.02042H= 4.0 0.02042H= 8.0 0.02042H=11.0 0.02042----------------------------------------------------------------------------------Interference wave drag of (horizontal tail + fuselage) system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.00000 0.00000H= 4.0 0.00000 0.00000H= 8.0 0.00000 0.00000H=11.0 0.00000 0.00000----------------------------------------------------------------------------------Profile drag of the vertical tail into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.00080 0.00075 0.00072 0.00069 0.00067 0.00065 0.00063 0.00071H= 4.0 0.00084 0.00079 0.00076 0.00073 0.00071 0.00068 0.00066 0.00074H= 8.0 0.00089 0.00084 0.00080 0.00077 0.00075 0.00073 0.00071 0.00079H=11.0 0.00094 0.00088 0.00085 0.00082 0.00079 0.00077 0.00075 0.00084----------------------------------------------------------------------------------Wave drag of the vertical tail into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.00408H= 4.0 0.00408H= 8.0 0.00408H=11.0 0.00408----------------------------------------------------------------------------------Profile drag of the engine nacelle into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.00193 0.00188 0.00187 0.00186 0.00187 0.00187 0.00188 0.00190H= 4.0 0.00204 0.00199 0.00197 0.00196 0.00197 0.00197 0.00198 0.00199H= 8.0 0.00218 0.00212 0.00210 0.00210 0.00210 0.00210 0.00210 0.00212H=11.0 0.00232 0.00226 0.00223 0.00222 0.00222 0.00222 0.00223 0.00223----------------------------------------------------------------------------------Wave drag of the engine nacelle into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 0.00023----------------------------------------------------------------------------------
  • [106]Profile drag of the aircraft (with out additional elements) M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.01997 0.01926 0.01878 0.01841 0.01811 0.01784 0.01759 0.02039H= 4.0 0.02076 0.02002 0.01952 0.01913 0.01881 0.01852 0.01826 0.02097H= 8.0 0.02174 0.02096 0.02044 0.02003 0.01969 0.01939 0.01911 0.02173H=11.0 0.02266 0.02184 0.02129 0.02087 0.02052 0.02020 0.01990 0.02244----------------------------------------------------------------------------------Wave drag of the aircraft (with out additional elements) M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.22398H= 4.0 0.22398H= 8.0 0.22398H=11.0 0.22398----------------------------------------------------------------------------------Summarized drag of the aircrafts additional elemetns KdCx= 0.100 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.00200 0.00193 0.00188 0.00184 0.00181 0.00178 0.00176 0.02444H= 4.0 0.00208 0.00200 0.00195 0.00191 0.00188 0.00185 0.00183 0.02450H= 8.0 0.00217 0.00210 0.00204 0.00200 0.00197 0.00194 0.00191 0.02457H=11.0 0.00227 0.00218 0.00213 0.00209 0.00205 0.00202 0.00199 0.02464----------------------------------------------------------------------------------Drag coefficient of aircraft at zero lift Cya=0 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.02196 0.02118 0.02066 0.02026 0.01992 0.01962 0.01934 0.26881H= 4.0 0.02283 0.02202 0.02147 0.02105 0.02069 0.02038 0.02008 0.26945H= 8.0 0.02391 0.02306 0.02248 0.02204 0.02166 0.02133 0.02102 0.27028H=11.0 0.02492 0.02402 0.02342 0.02296 0.02257 0.02222 0.02189 0.27107----------------------------------------------------------------------------------Wave drag coefficient of aircraft from M*=0.738 to M=1.2 M=0.5 M=0.6 M=0.7 M=0.8 M=0.9 M=1.0 M=1.1 M=1.2H= 0.0 0.00198 0.02860 0.09406 0.17779 0.22398H= 4.0 0.00198 0.02860 0.09406 0.17779 0.22398H= 8.0 0.00198 0.02860 0.09406 0.17779 0.22398H=11.0 0.00198 0.02860 0.09406 0.17779 0.22398----------------------------------------------------------------------------------Drag coefficient of aircraft from M*=0.738 to M=1.2 M=0.5 M=0.6 M=0.7 M=0.8 M=0.9 M=1.0 M=1.1 M=1.2H= 0.0 0.02664 0.05831 0.12880 0.21758 0.26881H= 4.0 0.02738 0.05902 0.12949 0.21824 0.26945H= 8.0 0.02831 0.05992 0.13037 0.21910 0.27028H=11.0 0.02918 0.06077 0.13120 0.21991 0.27107==================================================================================LIFT OF THE AIRCRAFT PATRS INTO AIRCRAFT SYSTEM==================================================================================Average value of deceleration flow coefficient before a wing M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000----------------------------------------------------------------------------------Interference coefficient: wing + fuselage kAl0=1.04423 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 kAl 1.04423 1.04423 1.04423 1.04423 1.04423 1.04423 1.04423 1.04423DkAl 0.04619 0.04619 0.04619 0.04619 0.04619 0.04619 0.04619 0.04619 kFi 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000DkFi 0.04423 0.04423 0.04423 0.04423 0.04423 0.04423 0.04423 0.04423----------------------------------------------------------------------------------
  • [107]Average value of deceleration flow coefficient before a horizontal tail M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 0.93842----------------------------------------------------------------------------------Interference coefficient: horizontal tail + fuselage kAl0=1.00000 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 kAl 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000DkAl 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 kFi 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000DkFi 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000----------------------------------------------------------------------------------Downwash before horizontal tail M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20eps0 0.497° 0.504° 0.515° 0.531° 0.554° 0.587° 0.640° 0.546°epsAl 0.1357 0.1377 0.1407 0.1451 0.1513 0.1605 0.1748 0.1492----------------------------------------------------------------------------------Derivative dCy/dAl & Alfa0 of the wing into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20dCy/dAl 4.38036 4.46318 4.58822 4.76728 5.02104 5.38826 5.95129 5.80711Alfa0 -3.62° -3.62° -3.62° -3.62° -3.62° -3.62° -3.62° -3.62°----------------------------------------------------------------------------------Derivative dCy/dAl & Alfa0 of the horizontal tail into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20dCy/dAl 0.62366 0.63187 0.64406 0.66109 0.68441 0.71651 0.76219 0.76384Alfa0 -1.47° -1.47° -1.46° -1.45° -1.43° -1.41° -1.37° -1.44°----------------------------------------------------------------------------------Derivative dCy/dAl & Alfa0 of the fuselage into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20dCy/dAl 0.12297 0.12297 0.12297 0.12297 0.12297 0.12297 0.12297 0.15092Alfa0 0.59° 0.59° 0.59° 0.59° 0.59° 0.59° 0.59° 0.59°----------------------------------------------------------------------------------Derivative dCy/dAl engine nacelle into aircraft system dCy/dAl= 0.00000----------------------------------------------------------------------------------Derivative dCy/dAl & Alfa0 of the aircraft with out horizontal tail M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20dCy/dAl 4.50333 4.58615 4.71119 4.89025 5.14401 5.51123 6.07425 5.95803Alfa0 -3.50° -3.50° -3.51° -3.51° -3.51° -3.52° -3.53° -3.51°----------------------------------------------------------------------------------Derivative dCy/dAl & Alfa0 of the aircraft with horizontal tail M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20dCy/dAl 5.12699 5.21802 5.35525 5.55134 5.82842 6.22773 6.83644 6.72188Alfa0 -3.12° -3.12° -3.12° -3.13° -3.14° -3.15° -3.17° -3.15°----------------------------------------------------------------------------------Lift coefficient of the aircraft Cya=dCy/dAl(Al-Al0) (line section) M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Al= 0.0° 0.27875 0.28399 0.29190 0.30323 0.31927 0.34247 0.37800 0.36911Al=10.0° 1.17358 1.19471 1.22657 1.27212 1.33652 1.42941 1.57118 1.54229Al=20.0° 2.06841 2.10542 2.16124 2.24101 2.35377 2.51636 2.76437 2.71548----------------------------------------------------------------------------------Maximum lift coefficient of aircraft and stalling angle M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Cya max 1.23511 1.19831 1.15952 1.11875 1.07599 1.03126 0.98453Alfa st 12.19° 11.54° 10.78° 9.92° 8.94° 7.84° 6.58°----------------------------------------------------------------------------------Polar coefficient A M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 0.04750 0.04748 0.04746 0.04744 0.04740 0.04734 0.04726 0.14879----------------------------------------------------------------------------------
  • [108]Maximum lift to drag ratio Kmax M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 15.480 15.765 15.967 16.131 16.273 16.406 16.537 2.500H= 4.0 15.184 15.463 15.663 15.824 15.966 16.098 16.229 2.497H= 8.0 14.837 15.111 15.306 15.464 15.604 15.734 15.863 2.493H=11.0 14.532 14.805 14.996 15.151 15.288 15.416 15.544 2.490----------------------------------------------------------------------------------Most efficiently lift coefficient Cya m.e. M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.68004 0.66794 0.65974 0.65345 0.64826 0.64376 0.63977 1.34412H= 4.0 0.69331 0.68097 0.67257 0.66610 0.66073 0.65606 0.65191 1.34573H= 8.0 0.70951 0.69685 0.68825 0.68161 0.67608 0.67124 0.66693 1.34780H=11.0 0.72440 0.71126 0.70246 0.69568 0.69003 0.68508 0.68066 1.34976----------------------------------------------------------------------------------Polar of aircraft Cxa = Cxo + A·Cya^2 at H= 0.0 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Cya= 0.0 0.02196 0.02118 0.02066 0.02026 0.01992 0.01962 0.01934 0.26881Cya= 0.2 0.02386 0.02308 0.02256 0.02215 0.02181 0.02151 0.02123 0.27476Cya= 0.4 0.02956 0.02878 0.02825 0.02784 0.02750 0.02719 0.02691 0.29261Cya= 0.6 0.03906 0.03828 0.03775 0.03733 0.03698 0.03666 0.03636 0.32237Cya= 0.8 0.05236 0.05157 0.05104 0.05061 0.05025 0.04992 0.04959 0.36403Cya= 1.0 0.06946 0.06867 0.06812 0.06769 0.06731 0.06696 0.41760Cya= 1.2 0.09036 0.48306----------------------------------------------------------------------------------Additional induced drag of aircraft dCxi at H= 0.0 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Cya= 0.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000Cya= 0.2 0.00004 0.00005 0.00005 0.00006 0.00006 0.00007 0.00008Cya= 0.4 0.00034 0.00038 0.00042 0.00047 0.00052 0.00060 0.00425Cya= 0.6 0.00120 0.00132 0.00146 0.00164 0.00186 0.00318 0.02538Cya= 0.8 0.00305 0.00338 0.00379 0.00430 0.00583 0.03290 0.06093Cya= 1.0 0.00689 0.00782 0.00905 0.01537 0.04875 0.08417Cya= 1.2 0.01854Cya= Max 1.43894 1.29362 1.13961 0.97436 0.79364 0.58919 0.33912----------------------------------------------------------------------------------Polar of aircraft Cxa = Cxo + A·Cya^2 + dCxi at H= 0.0 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Cya= 0.0 0.02196 0.02118 0.02066 0.02026 0.01992 0.01962 0.01934 0.26881Cya= 0.2 0.02391 0.02313 0.02261 0.02221 0.02188 0.02159 0.02132 0.27476Cya= 0.4 0.02991 0.02916 0.02867 0.02831 0.02803 0.02779 0.03116 0.29261Cya= 0.6 0.04026 0.03960 0.03921 0.03897 0.03884 0.03984 0.06174 0.32237Cya= 0.8 0.05541 0.05495 0.05483 0.05492 0.05608 0.08282 0.11052 0.36403Cya= 1.0 0.07636 0.07649 0.07717 0.08306 0.11606 0.15113 0.41760Cya= 1.2 0.10890 0.48306Cya= Max 0.13442 0.12937 0.12878 0.14872 0.16692 0.18317 0.19810----------------------------------------------------------------------------------K max 14.924 15.159 15.303 15.396 15.455 15.486 13.460 2.500Ñya m.e. 0.63109 0.61672 0.60513 0.59448 0.58391 0.57286 0.33914 1.34413Alf m.e. 3.94° 3.65° 3.35° 3.01° 2.60° 2.12° -0.33° 8.31°==================================================================================
  • [109]LONGITUDINAL MOMENT AND AERODYNAMIC CENTERLOCATION OF AIRCRAFT PATS INTO AIRCRAFT SYSTEM==================================================================================Derivative dMz/dAl & aerodynamic center Xac/Lf of the wing into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20dMz/dAl -2.5837 -2.6325 -2.7063 -2.8119 -2.9616 -3.1782 -3.5103 -3.5823Xac/Lf 0.5898 0.5898 0.5898 0.5898 0.5898 0.5898 0.5898 0.6169----------------------------------------------------------------------------------Derivative dMz/dAl & aerodynamic center Xac/Lf of the horizontal tail into aircraftsystem M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20dMz/dAl -0.6112 -0.6193 -0.6312 -0.6479 -0.6708 -0.7022 -0.7471 -0.7630Xac/Lf 0.9800 0.9800 0.9800 0.9801 0.9801 0.9801 0.9802 0.9989----------------------------------------------------------------------------------Derivative dMz/dAl & aerodynamic center Xac/Lf of the fuselage into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20dMz/dAl 0.0276 0.0276 0.0276 0.0276 0.0276 0.0276 0.0276 0.0248Xac/Lf -0.2245 -0.2245 -0.2245 -0.2245 -0.2245 -0.2245 -0.2245 -0.1641----------------------------------------------------------------------------------Derivative dMz/dAl engine nacelle into aircraft system dMz/dAl= 0.0000Position of the engine nacelle from fuselage nose Xìãä/Lf= 0.7244----------------------------------------------------------------------------------Derivative dMz/dAl & aerodynamic center Xac/Lf of aircraft w/o horizontal tail M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20dMz/dAl -2.5561 -2.6049 -2.6787 -2.7843 -2.9340 -3.1506 -3.4827 -3.5576Xac/Lf 0.5676 0.5680 0.5686 0.5694 0.5704 0.5717 0.5733 0.5971----------------------------------------------------------------------------------Derivative dMz/dAl & aerodynamic center Xac/Lf of aircraft with horizontal tail M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20dMz/dAl -3.1673 -3.2242 -3.3099 -3.4322 -3.6047 -3.8528 -4.2298 -4.3205Xac/Lf 0.6178 0.6179 0.6181 0.6183 0.6185 0.6187 0.6187 0.6428----------------------------------------------------------------------------------Aerodynamic center Xac/Bmac of aircraft w/o horizontal tail relative to MAC L.E. M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Xac/Bmac 0.0194 0.0233 0.0289 0.0365 0.0463 0.0590 0.0754 0.3067----------------------------------------------------------------------------------Aerodynamic center Xac/Bmac of aircraft with horizontal tail relative to MAC L.E. M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Xac/Bmac 0.5080 0.5092 0.5109 0.5128 0.5149 0.5166 0.5171 0.7513----------------------------------------------------------------------------------Shift of the aerodynamic center of aircraft dXac/Bmac M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20dXac/Bmac 0.4886 0.4859 0.4819 0.4763 0.4685 0.4576 0.4417 0.4446----------------------------------------------------------------------------------Longitudinal moment mz0 of wing and (fuselage+wing) system (relative to MAC) M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20mz0iz.w -0.0333 -0.0333 -0.0333 -0.0333 -0.0333 -0.0333 -0.0333 -0.0333mz0f(w) 0.0932 0.0933 0.0933 0.0934 0.0935 0.0937 0.0939 0.0934mz0* 0.0366 0.0366 0.0366 0.0367 0.0368 0.0369 0.0371 0.0367----------------------------------------------------------------------------------Longitudinal moment mz0 of the parts of the aircraft w/o horizontal tail M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20mz0w’ -0.0599 -0.0599 -0.0599 -0.0600 -0.0601 -0.0602 -0.0603 -0.0706mz0f’ -0.0097 -0.0097 -0.0097 -0.0097 -0.0097 -0.0098 -0.0098 -0.0115----------------------------------------------------------------------------------
  • [110]Longitudinal moment mz0 of the aircraft w/o horizontal tail (relative to MAC) M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 -0.0330 -0.0330 -0.0330 -0.0330 -0.0330 -0.0330 -0.0330 -0.0454----------------------------------------------------------------------------------Longitudinal moment mz0 of the parts of the aircraft with horizontal tail M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20mz0w 0.0752 0.0770 0.0796 0.0833 0.0884 0.0954 0.1056 0.0923mz0h.t. -0.0566 -0.0571 -0.0579 -0.0589 -0.0603 -0.0620 -0.0641 -0.0665mz0f -0.0103 -0.0103 -0.0103 -0.0103 -0.0103 -0.0103 -0.0103 -0.0121----------------------------------------------------------------------------------Longitudinal moment mz0 of the aircraft with horizontal tail (relative to MAC) M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 0.0449 0.0461 0.0480 0.0508 0.0546 0.0600 0.0682 0.0504================================================================================== © Holjavko V.I. Aerodynamic characteristics of aircraft, 1991 - 1998 © Chmovzh V.V. 1991 - 2011 Âåðñèÿ 4.11b îò 01.11.2011LONGITUDINAL MOMEMT OF THE AIRCRAFT RELATIVE TO MASSGRAVITY XMG=18.100 XMG/LF=0.591==================================================================================Longitudinal momemt of the aircraft with out horizontal tail (relative to MAC) M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Cya= 0.0 -0.0330 -0.0330 -0.0330 -0.0330 -0.0330 -0.0330 -0.0330 -0.0454Cya= 0.2 0.0128 0.0120 0.0109 0.0094 0.0074 0.0049 0.0016 -0.0570Cya= 0.4 0.0586 0.0571 0.0548 0.0518 0.0478 0.0428 0.0362 -0.0687Cya= 0.6 0.1044 0.1021 0.0987 0.0942 0.0883 0.0807 0.0708 -0.0803Cya= 0.8 0.1502 0.1471 0.1426 0.1365 0.1287 0.1186 0.1054 -0.0920Cya= 1.0 0.1961 0.1921 0.1865 0.1789 0.1691 0.1565 -0.1036Cya= 1.2 0.2419 -0.1153Longitudinal momemt of the aircraft with horizontal tail (relative to MAC) M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Cya= 0.0 0.0449 0.0461 0.0480 0.0508 0.0546 0.0600 0.0682 0.0504Cya= 0.2 -0.0071 -0.0060 -0.0044 -0.0021 0.0013 0.0064 0.0144 -0.0502Cya= 0.4 -0.0590 -0.0582 -0.0569 -0.0550 -0.0520 -0.0473 -0.0393 -0.1508Cya= 0.6 -0.1109 -0.1103 -0.1094 -0.1079 -0.1053 -0.1009 -0.0930 -0.2514Cya= 0.8 -0.1628 -0.1625 -0.1619 -0.1608 -0.1586 -0.1545 -0.1468 -0.3520Cya= 1.0 -0.2147 -0.2147 -0.2144 -0.2136 -0.2119 -0.2082 -0.4525Cya= 1.2 -0.2666 -0.5531==================================================================================Longitudinal momemt of the aircraft relative to mass gravity Xmg=18.100 Xmg/Lf=0.591==================================================================================Longitudinal momemt of the aircraft with out horizontal tail (relative to MAC) M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Cya= 0.0 -0.0330 -0.0330 -0.0330 -0.0330 -0.0330 -0.0330 -0.0330 -0.0454Cya= 0.2 0.0128 0.0120 0.0109 0.0094 0.0074 0.0049 0.0016 -0.0570Cya= 0.4 0.0586 0.0571 0.0548 0.0518 0.0478 0.0428 0.0362 -0.0687Cya= 0.6 0.1044 0.1021 0.0987 0.0942 0.0883 0.0807 0.0708 -0.0803Cya= 0.8 0.1502 0.1471 0.1426 0.1365 0.1287 0.1186 0.1054 -0.0920Cya= 1.0 0.1961 0.1921 0.1865 0.1789 0.1691 0.1565 -0.1036Cya= 1.2 0.2419 -0.1153----------------------------------------------------------------------------------
  • [111]Longitudinal momemt of the aircraft with horizontal tail (relative to MAC) M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Cya= 0.0 0.0449 0.0461 0.0480 0.0508 0.0546 0.0600 0.0682 0.0504Cya= 0.2 -0.0071 -0.0060 -0.0044 -0.0021 0.0013 0.0064 0.0144 -0.0502Cya= 0.4 -0.0590 -0.0582 -0.0569 -0.0550 -0.0520 -0.0473 -0.0393 -0.1508Cya= 0.6 -0.1109 -0.1103 -0.1094 -0.1079 -0.1053 -0.1009 -0.0930 -0.2514Cya= 0.8 -0.1628 -0.1625 -0.1619 -0.1608 -0.1586 -0.1545 -0.1468 -0.3520Cya= 1.0 -0.2147 -0.2147 -0.2144 -0.2136 -0.2119 -0.2082 -0.4525Cya= 1.2 -0.2666 -0.5531==================================================================================The results obtained from this calculation are further used to draw graphs ofaircraft polar and other aerodynamic characteristics. AIRCRAFT DESIGN PARAMETERS ON AERODYNAMIC CHARACTERISTICSdCy/dAl Cya = 4.86167Alfa0 = -3.16Cya max = 1.23511A = 0.04772Cxa = 0.02196Calculation of lift coefficient for the whole airplaneCya = Cyaf . f Cyaw . w CyaHS . HSCya = 4.86167 ( from software ) o = .( of. Cyaf . f + ow. Cyaw . w) +( oHS.CyaHS . HS) o 3.16 (from software )Using equation, Cya = Cya .( o) we get, S.No α Cyaα α0 Cya 1 0 4.86167 -3.16 0.26799 2 2 4.86167 -3.16 0.437604 3 4 4.86167 -3.16 0.607218 4 6 4.86167 -3.16 0.776832 5 8 4.86167 -3.16 0.946446
  • [112] 6 10 4.86167 -3.16 1.11606 7 12 4.86167 -3.16 1.285674 8 14 4.86167 -3.16 1.455288 9 16 4.86167 -3.16 1.624902 10 18 4.86167 -3.16 1.794516 11 20 4.86167 -3.16 1.964129CALCULATION OF LIFT COEFFICIENT WHILE TAKE-OFF AND LANDINGCya = Cya .( o)Cya = 4.86167 ( from software ) , o= -3.16( from software )FOR TAKE-OFF, o take-off = -8 ( o take-off = 6 ...... 10 ) o = 3.16– 8 = -11.16Alpha zero = -3.16 -8 = -11.16Using equation, Cya take-off = Cya .( o) we get, α Cyaα α0 Cya -6 4.86167 -11.16 0.437604 -4 4.86167 -11.16 0.607218 -2 4.86167 -11.16 0.776832 0 4.86167 -11.16 0.946446 2 4.86167 -11.16 1.11606 4 4.86167 -11.16 1.285674 6 4.86167 -11.16 1.455288 8 4.86167 -11.16 1.624902 10 4.86167 -11.16 1.794516 12 4.86167 -11.16 1.964129 14 4.86167 -11.16 2.133743
  • [113] 16 4.86167 -11.16 2.303357 18 4.86167 -11.16 2.472971 20 4.86167 -11.16 2.642585FOR LANDINGFor landing, o landing = ( o landing = 10 ...... 15 )Alpha zero = -3.16 -12 = -15.16Cya = 4.86167 ( from software ) α Cyaα α0 Cya -12 4.86167 -15.16 0.26799 -10 4.86167 -15.16 0.437604 -8 4.86167 -15.16 0.607218 -6 4.86167 -15.16 0.776832 -4 4.86167 -15.16 0.946446 -2 4.86167 -15.16 1.11606 0 4.86167 -15.16 1.285674 2 4.86167 -15.16 1.455288 4 4.86167 -15.16 1.624902 6 4.86167 -15.16 1.794516 8 4.86167 -15.16 1.964129 10 4.86167 -15.16 2.133743 12 4.86167 -15.16 2.303357 14 4.86167 -15.16 2.472971 16 4.86167 -15.16 2.642585 18 4.86167 -15.16 2.812199 20 4.86167 -15.16 2.981813
  • [114]CALCULATION OF ZERO DRAG COEFFICIENT FOR TAKE-OFF ANDLANDINGAs we know, Cxa = Cxao + A.Cya2So, Cxa = (Cxao flight + Cxao Flaps + Cxao Landing gear) + A.Cya2Cxa = 0.02196A = 0.04772Using the given coefficients,FOR TAKE – OFF, = 0.4.......0.5 = 0.45 = 0.5.....0.6 = 0.55Cxa = Cxao. + A.Cya2 S.No Cya Cxa0 A Cxa 1 0 0.02196 0.04772 0.04392 2 0.2 0.02196 0.04772 0.045829 3 0.4 0.02196 0.04772 0.051555 4 0.6 0.02196 0.04772 0.061099 5 0.8 0.02196 0.04772 0.074461 6 1 0.02196 0.04772 0.09164 7 1.2 0.02196 0.04772 0.112637FOR LANDING,Cxa = 0.02196A = 0.04772 = 1.0.........1.5 = 1.25 = 0.5.........0.6 = 0.55Cxa = Cxao. + A.Cya2 S.No Cya Cxa0 A Cxa
  • [115] 1 0 0.02196 0.04772 0.061488 2 0.2 0.02196 0.04772 0.063397 3 0.4 0.02196 0.04772 0.069123 4 0.6 0.02196 0.04772 0.078667 5 0.8 0.02196 0.04772 0.092029 6 1 0.02196 0.04772 0.109208 7 1.2 0.02196 0.04772 0.130205 0.14 3.2 3 0.12 2.8 2.6 2.4 0.1 2.2 2 For flight 0.08 1.8 For take - off 1.6 For take- 0.06 1.4 For landing 1.2 off 0.04 1 For landing 0.8 0.6 0.02 0.4 0.2 0 0 0 0.5 1 1.5 -20 0 20 40Conclusion: The aerodynamic characteristics of the designed aircraft are calculated bothmanually and through the software.
  • [116] DESIGN SECTION PART 33.1. INTEGRATED DESIGNING AND COMPUTER AIDED MODELING OFDESIGNED AIRCRAFT3.2. WING LOAD CALCULATION
  • [117]3.1.INTEGRATED DESIGNING AND COMPUTER AIDED MODELING WING OFDESIGNED AIRCRAFTFuselage modeling:Fuselage parameters are calculated in PART 1 AND the design Parameters of fuselage areused for the surface modeling in Unigraphics NX6 the surface model of the fuselagedeveloped in NX 6 is shown below Fuselage surface modelWing designing:Wing is the main lifting surface in the aircraft so it should maintain aerodynamic shape forthat we need to consider the airfoil shapes for the wing the Wing airfoil are generated byusing profile 2 software and I used NACA 2415 Mod 40-30 airfoil for wing root and wingtips.The airfoils are imported in to Unigraphics as shown below:
  • [118] Airfoils imported from profili to NX6The airfoil are moved from one another according to wing parameters and by using meshsurface option created the wing surface under rules mesh and also high lifting devices likeflaps and control surfaces like ailerons are also represented in the design as shown in bellowfigure. Wing surfaceDesigning of tail section:Horizontal and vertical stabilizers are generated as same procedure used in wing designing.My configuration of tail is T tail the horizontal stabilizer is located on vertical stabilizer .tailunit control surfaces are calculated in PART-1 from those values elevator and ruddersurfaces are created. The surface design of tail part is shown below.
  • [119] Tail surfaceAssembling the unitsAfter creating the aircraft fuselage, wing and tail unit the individual units are assembledtogether. The assembling process depends up on location of center of gravity of aircraft .sothe center of gravity of the aircraft which is calculated already in first section is used here.As like wings tail unit is also mounted on the fuselage in same mannerThe corresponding surface model of my aircraft is done and represented. Surface model of aircraft
  • [120]Top viewFront view
  • [121] Side view WING LOAD CALCULATIONWING’S GENERAL DATAWING’S GEOMETRICAL DATAbrs = 4.5mbts = 1.1mLw = 22.73mLength of half section of wing = =11.365 mSw = 0.5 (br + bt) Lw= 0.5(4.5 + 1.1) 22.73=63.64 m 2Technical DataTake- off Weight Gt = 383278KgWing Span Lw = 22.73 mWing Area Sw = 63.64 m2Cruising Speed VC = 835 Km/hCruising Altitude HC = 10000 m
  • [122]Wing sweep till 0.25 of chord χ0.25 = 22.730 CALCULATION OF THE DISTRIBUTED FUEL LOAD ON A PLANE WINGApproximately it is possible to consider, that linear load from fuel qf changes linearly onlength of each tank separately. For definition of this function it is necessary to find sizes qfon borders of tanks. With this purpose it is necessary to construct cross section of a wingwith spars and to find the area of cross section, which is occupied by fuel –Sf. It is possible to make manually or with the help of the program "Compass".As a first approximation Sf. is equal: H1  H 2Sf    a 2where μ=0.9 – is factor, which takes into account, what not all cross section is engaged infuel (part is engaged in a structure, fixture, air for overpressure of fuel), Н1, Н2 - heightaccordingly front and rear spars, and a- distance between spars. The linear load qf is under the formula: q f  S f p g [N/m],Where р=800kg/m3 – is density of fuel, g=9.81 m/с2 - is acceleration of free fall.*Location of Front Spar = 0.25 x chord of the Wing.
  • [123]*Location of Rear Spar = 0.74 x chord*Wing’s centre of gravity centre cross-section = 0.40 x chord from leading edge*Landing Gear base = Distance between two landing gears fitted on each side of wing = (0.20…..0.25) x Length of Wing (LW = 22.73) = 0.20 x 22.73 = 4.546 m*Distance b/w Wing Leading edge to LG centre of gravity (bl) = 0.3 x Wing chord (b) = 0.3 x 4.5 = 1.35 m*Weight of primary structure of main LG = 0.45 x Weight of LG = 0.45 x 1400 kg = 630 kg*Weight of primary structure of Nose LG = 0.45 x Weight of Nose LG = 0.45 x 697.33 kg = 311.0915 kg. DISTRIBUTED FUEL LOAD ON WINGSf= Area of cross-section of wing with spars.qf = Linear Load.qfu= Ultimate Linear Load.Gwf = Weight of fuel in tank.Xf= Position of CG for fuel in cross-section.Sf =µ.(H1+ H2/2).µ = 0.9H1 = Height of front sparsH2 = Height of Rear sparsa = Distance b/w two spars.[Total no of fuel tanks = 2], one at 0.2 & another one at 0.5.Equivalent wing is drawn, Airfoilcharacteristics are plotted, from that the value of H1& H2 is found.
  • [124]At 0.2, chord length = 3.82, NACA 2415H1 = 0.5680 m ; H2 = 0.3247 ma = Location of rear spar – Location of front spar.Location of Front spar = 0.25 x 3.82 = 0.955 m.Location of Rear spar = 0.74 x 3.82 = 2.826 m.*a = 2.826 – 0.955 = 1.871 m. H1  H 2Sf    a , 2*Sf fuel tank 1 at 0.2 = 0.9 x (0.3247 + 0.5680/2) x 1.871 = 0.9 x 0.44 x 1.871 = 0.7412 m2.At point 0.2*(Fuel Density (p) = 800 kg / m3; g = 9.8)qf =(0.7412 x 800 x 9.8) = 5811.2Sf fuel tank 1 at 0.8 , chord = 1.78H1 = 0.2646 m ; H2 = 0.1513 ma = Rear position spar – Front position sparFront position spar = 0.25 x 1.78 = 0.445 m.Rear position spar = 0.74 x 1.78 = 1.317 m*a = 1.317 – 0.445 = 0.872 m.*Sf fuel tank 2 at 0.8 = 0.9 x (0.2646 + 0.1513/2) x 0.8722 = 0.9 x 0.20 x 0.8722 = 0.1632 m2.At point 0.8*qf =Sf.p.g(Fuel Density(p) = 800 kg / m3 ; g = 9.8)*qf =(0.1632 x 800 x 9.8)= 1280.7*qfu = qf.nl.fnl = limit load factor,is calculated according to AR-25,FAR-25,JAR-25n ≤ 2.1 +G = Take of weight = 39110 kg: takeThen f = 1.5
  • [125]li= Length b/w two tanks, calculated belowLENGTH OF FUEL TANK CALCULATION:-Total length of wing = 22.73 m.Semi-span = 11.363 m.Equivalent wing of 10.115 m is divided in 11 sectionsNo of section b/w 0.2 & 0.8 = 6i.e.qfi +qfi+1 = 7097.9664 = 19738.674
  • [126] In a wing the fuel can be placed in space between front and rear spar in cross section andon length from root rid up to tip rib. In the transport plane from conditions of safety the fuelis forbidden for placing in a fuselage. They can be placed in a wing and tail unit. On semispan of wing they are placed from two up to five tanks. One tank is forbidden for putting,that at damage of a tank all fuel was not poured out and for observance of positioning. Theheavier plane has the more tanks. If the engines are established on a wing, between theengine and tank there should be a fire-prevention interval not less 200mm. If the engines areestablished on engine pylon, the fuel can be placed on all semi span of wing. For simplicityof border of fuel tanks it is possible to combine with points of the task of aerodynamic load.It is possible to place fuel at once on an equivalent straight wing. a front spar a fuel tank c.g. of fuel H1 H2 a rear spar хf b а cIn The position of a centre of gravity for fuel in cross section хf can be estimated as thecentre of gravity for a trapeze under the formulaAt point 0.2 a H1  2 H 2xf   = 3 H1  H 2= 1.02mWhere хf - is distance from front spar up to a centre of gravity of fuel in the given crosssection. On this size it is possible to find relative size for a position of a centre of gravity inpercentage of chord from leading edge of a wing xf  b xf   100 c
  • [127] = = 51.57 % It is possible to consider this size as constant size on the wing span. AIR LOAD CALCULATION The Y air load is allocated according to the relative circulation low, i.e. n  f gM а zqy (z)  L t Г (z) , z  w 0 ,5 L wwhere Г ( z ) - is relative circulation, Mt- is the take-off weight of the plane (it is given in theinitial data), n, f – limit load factor and factor of safety, Lw – wingspan.Distribution relative circulation on wingspan straight trapezoidal center-section-less wingГf (5    10)2z/l = 40.0 1,38590.1 1,37010,2 1,32450.3 1,25240.4 1,16010.5 1,05530.6 0,94190.7 0,82710.8 0,70510.9 0,54340,95 0,40921 01. Wing has not center-section (2 lc = 0).2. Wing is plane.3. Wing aspect ratio   L w 2 S w = 7.854. Wing taper is   b 0 / b t = 4
  • [128]5. For low-wing monoplane Гf is given from board rib, for midwing and high-wing Гp isgiven from axial rib.6. If wing taper  differentiates from table data, valises Гf are calculated by linearinterpolation.The function Г ( z ) depends from many factors, from which in the given work we takeinto account only the dependence from wing taper and sweepback.The Г ( z ) = Г f ( z ) function values for plane straight trapezoidal center-section-less wingare reduced in table # 2.  = 45°.2z/l ΔΓx(450) ΔΓx(x)0.0 -0.235 -0.11860.1 -0.175 -0.088310,2 -0.123 -0.06210.3 -0.072 -0.03630.4 -0.025 -0.01260.5 0.025 0.01260.6 0.073 0.03680.7 0.111 0.05600.8 0.135 0.06810.9 0.140 0.07070,95 0.125 0.06311 0 0 Essential influence on distribution of air loading renders a wing sweep. Relativecirculation in this case is determined under the formula:Г(z ) = Г f ( z ) +   (z)where  s ( z ) is amendment on the wing sweep. This amendment is calculated upon formulas: о   (  )   ( 45 ) о 45where the  is the designing wing sweep on the chord’s fourth, angle in degree. THE WING STRUCTURE MASS LOAD ALLOCATION.
  • [129] In approximate calculations it is possible to consider, that load per unit of length ofwing mass forces is proportional to chords. Then the next formula is used: w n f G qy ( z ) w  b( z ) Swwhere the b(z) is the wing chord. After the component calculations it is possible to compute the total distributed wingload, acting in the direction of the axis Y in the speed coordinate system. Calculations are putinto the tab.# 4. At this action the coordinates origin is put into the wing root and crosssections are enumerated from the wing root in the wing tip direction, beginning from the i   = 0. The letter Z accentuate relative coordinate Z  2  z / L w . Since on the Z = 1 − 0.95cross sections the q diagram are moved away from a straight line, it is necessary to introduce the cross section with the Z = 0.95 coordinate (see tab. # #2, 4). The total distributed wing load is calculated under the formula: a w fq  qy  qy  qyWhere q yf – is distributed ultimate fuel load from formula (2), if you consider this loaddistributed and it is equal to: fq y  q f nf . a w f It is also necessary to plot the qy , qy, q y and q functions in the samecoordinate system and in the same scale.i ΔΓx Γ b(z)1 2 3 4 5 6 7 8 90 0 1.3731 151.856 -9.073 4.5 142.7831 0.1 1.3580 153.594 -8.387 4.16 145.2072 0.2 1.3143 151.269 -7.702 3.82 143.5673 0.3 1.2435 145.721 -7.016 3.48 138.7054 0.4 1.1565 137.501 -6.331 3.14 131.17
  • [130]5 0.5 1.0551 127.843 -5.645 2.80 122.1986 0.6 0.9464 117.274 -4.839 2.40 112.4357 0.7 0.8351 105.819 -4.274 2.12 101.5458 0.8 0.7164 92.6502 -3.589 1.78 89.0619 0.9 0.5630 73.585 -2.903 1.44 70.68210 0.95 0.4242 56.594 -2.560 1.27 54.03411 1.0 0 0 0 0 -2.217 1.1 -2.217 SHEAR FORCE, BENDING MOMENT AND REDUCED MOMENT At calculation of the allocation law of shear forces and bending moments on the winglength in the beginning functions Q d ( z ) and M d ( z ) affected by the distributed load q (z)are found. For this purpose integrals are calculated by a tabulated way with trapezoids andmean-value methods. z z Q   q ( z )dz , M   Q ( z )dz Lw Lw 2 2We shall use the following formulae for the determination of distributer load q(z), shear forceQ(z) and Bending Moment Md(z).∆Zi = ( - ×∆Qi = ( + ×Qi =∆Mi = ( + ×Mi =i m m KN KN1 2 3 4 5 6 7 8
  • [131]0 0 0 142.78 161.182 1270.162 1331.605 6055.3381 0.1 1.1195 145.20 161.618 1108.98 1150.934 4723.7392 0.2 1.1195 143.56 157.98 947.362 972.055 3572.7983 0.3 1.1195 138.70 151.04 789.382 799.077 2600.754 0.4 1.1195 131.17 141.80 638.342 635.194 1801.6735 0.5 1.1195 122.19 131.316 496.542 482.331 1166.4796 0.6 1.1195 112.43 119.75 365.226 341.809 684.1487 0.7 1.1195 101.54 106.678 245.476 215.078 342.3398 0.8 1.1195 89.06 89.406 138.798 105.329 127.2619 0.9 1.1195 70.68 34.893 49.392 17.876 21.93210 0.95 1.1195 54.03 14.499 14.449 4.056 4.05611 1.0 0 -2.21 0 0 0 0 When we consider the shear forces and bending moments resulting from concentratedmass forces, from engine, we shall note the following results in the table below. z i  Z i  Z i1   Lw / 2 ; z0 = 0, (i = 11, 10, 9... 1),Q i   Q i  1  Q i  1 , Q11 = 0; (i = 10, 9... 1)  M iс  Q i  Q i  1    z i , M0с =0, (i = 11, 10, 9... 1) 2M iс   M i  1с  M i  1с , M11с = 0; (i = 10, 9... 1)I m m KN KN1 2 3 4 5 6 70 0 0 0 165.48 185.238 -643.59841 0.1 1.1195 0 165.48 185.238 -458.36042 0.2 1.1195 111.73 165.48 122.7030 273.1221
  • [132]3 0.3 1.1195 0 53.75 60.1677 150.41914 0.4 1.1195 0 53.75 60.1677 90.25145 0.5 1.1195 53.65 53.75 30.0837 30.08376 0.6 1.1195 0 0 0 07 0.7 1.1195 0 0 0 08 0.8 1.1195 0 0 0 09 0.9 1.1195 0 0 0 010 0.95 1.1195 0 0 0 011 1.0 1.1195 0 0 0 0The calculation scheme is given in, which includes the following values:Qid - is shear force from distributed loadsQic - is shear force from concentrated loadsQtot = Qid+ Qic with account signs;Mid - is bending moment from distributed loadsMic - is bending moment from concentrated loadsMtot = Mid + Mic with account signs. As a wing is calculated on strength in connected coordinate system for design crosssection determination of shear forces and bending moments are carried out in this coordinatesystem. In connected coordinate system the t axis is directed on a chord of a wing, an axis n -is perpendicular to it.The total Qtot (z) shear forces and the total Мtot (z) bending moment affected by all forcescalculation scheme.Summation of total shear forces and bending moments acting on the wing due to distributedand concentrated loads shall be calculated as follows. = and =The total shear force and total bending moment affected by all forces
  • [133]I m m KN KN1 2 3 4 5 6 70 0 0 0 165.48 185.238 -643.59841 0.1 1.1195 0 165.48 185.238 -458.36042 0.2 1.1195 111.73 165.48 122.7030 273.12213 0.3 1.1195 0 53.75 60.1677 150.41914 0.4 1.1195 0 53.75 60.1677 90.25145 0.5 1.1195 53.65 53.75 30.0837 30.08376 0.6 1.1195 0 0 0 07 0.7 1.1195 0 0 0 08 0.8 1.1195 0 0 0 09 0.9 1.1195 0 0 0 010 0.95 1.1195 0 0 0 011 1.0 1.1195 0 0 0 0 DETERMINATION OF REDUCED SHEAR FORCE AND BENDING MOMENT = = qa Reduced axis e = qw If we take angle of attack, α 20
  • [134]ThenTaking from aerodynamic properties of the airfoil, = 3.557 oPosition of cross-section for reduced moments; we shall use the following formulae = = = cos(    )Q n  Q tot , = (1.001) Qn cos  sin(    )Qt  Q , tot cos  sin(    ) ,M n  M  tot cos  cos(    )M t  M tot  cos  Reduced moment’s calculation for distributed loadsI m m KN KN1 2 3 4 5 6 70 0 0 0 165.48 185.238 -643.59841 0.1 1.1195 0 165.48 185.238 -458.36042 0.2 1.1195 111.73 165.48 122.7030 273.12213 0.3 1.1195 0 53.75 60.1677 150.41914 0.4 1.1195 0 53.75 60.1677 90.25145 0.5 1.1195 53.65 53.75 30.0837 30.0837
  • [135] 6 0.6 1.1195 0 0 0 0 7 0.7 1.1195 0 0 0 0 8 0.8 1.1195 0 0 0 0 9 0.9 1.1195 0 0 0 0 10 0.95 1.1195 0 0 0 0 11 1.0 1.1195 0 0 0 0 Reduced Bending Moments as a result of concentrated forces, are calculated as follows, = CALCULATION SCHEME OF REDUCED MOMENT MOMENT FROM CONCENTRATED LOADS AND FROM ALL LOADSi r KN KNM KNM KNM KNM m1 2 3 4 5 6 70 0 0 0 308.2315 -1673.36 -1365.1291 0 0 0 308.2315 -1486.026 -1177.7942 111.73 1.88 210.052 308.2315 -1289.06 -980.828 98.17953 0 0 0 -1087.616 -989.436 98.17954 0 0 0 -887.078 -788.898 98.17955 53.65 1.83 98.1795 -691.125 -592.9456 0 0 0 0 -536.269 -536.2697 0 3 0 0 -391.588 -391.588
  • [136]8 0 0 0 0 -227.238 -227.2389 0 0 0 0 -84.417 -84.41710 0 0 0 0 -26.734 -26.73411 0 0 0 0 0 0 Conclusion: Thus the designed aircrafts 3-D model is done. The wings mass load and air load are found. DESIGN SECTION PART- 4 AIRCRAFT SYSTEMS 1. AIRCRAFT HYDRAULIC SYSTEM 2. COMPONENTS INVOLVED IN HYDRAULIC SYSTEM 3. HYDRAULIC DESCRIPTION OF THE DESIGNED AIRCRAFT AIRCRAFT HYDRAULIC SYSTEM
  • [137]INTROUCTIONAircraft Hydraulics DefinitionHydraulic systems provide a means of remotely controlling a wide range of components bytransmitting a force through a confined fluid. It is a system where liquid under pressure isused to transmit this energy. Hydraulic systems take engine power and convert it to hydraulicpower by means of a hydraulic pump. This power can be distributed throughout the airplaneby means of tubing that runs through the aircraft. Hydraulic power may be reconverted tomechanical power by means of an actuating cylinder, or turbine.NEED FOR THE USE OF HYDRAULIC SYSTEMBecause hydraulics can transmit high forces rapidly and accurately along lightweight pipesof any size, shape and length, they are the prime source of power in aircraft systems such asflying controls, flaps, retractable undercarriages and wheel brakes.Modern aircrafts include many different types of subsystems. These subsystems are veryclosely interlinked to each other. One of these subsystems is hydraulic subsystem, which isusually used for actuating most of the mechanical subsystems, such as landing gear, flightcontrol surfaces, weapons system etc. Thus the hydraulic system is a very essential part ofthe aircraft and its reliability and functionality are very essential to the flight worthiness ofthe whole aircraft.PRINICIPLE OF HYDRAULICSHydraulics is based on a very simple fact of nature - you cannot compress a liquid. You cancompress a gas (think about putting more and more air into a tire, the more you put in, thehigher the pressure). If youre really strong you can compress a solid mass as well. But nomatter how much pressure you apply onto a liquid, it isnt possible to compress it. Now ifyou put that liquid into a sealed system and push on it at one end, that pressure is transmittedthrough the liquid to the other end of the system. The pressure is not diminished.Advantages of Hydraulic Systems 1. It is lighter in weight than alternate existing systems. 2. It is dead beat, that is, there is an absence of sloppiness in its response to demands placed on the system. 3. It is reliable; either it works or doesnt. 4. It can be easily maintained. 5. It is not a shock hazard; it is not much of a fire hazard. .
  • [138]Devices Operated by Hydraulic Systems in Aircraft 1. Primary control boosters 2. Retraction and extension of landing gear 3. Sweep back and forth of wings 4. Opening and closing doors and hatchways 5. Automatic pilot and gun turrets 6. Shock absorption systems and valve lifter systems 7. Dive, landing, speed and flap brakes 8. Pitch changing mechanism, spoilers on flapsHYDRAULIC FLUIDThe primary purpose of hydraulic fluid is to transmit force from one place, throughnonmoving hydraulic tubes, to another location. The advantage of using a liquid is itsincompressibility. Except for minor friction losses as the fluid passes through the tubing, allapplied force is transmitted throughout the system, according to Pascal. Thus, if a number ofpassages exist in a system, pressure can be distributed through all of them by means of theliquid.Hydraulic fluids are different and generally speaking they cannot be mixed. Manufacturers ofhydraulic devices usually specify the type of liquid best suited for use with their equipment,in view of the working conditions, the service required, temperatures expected inside andoutside the systems, pressures the liquid must with stand, the possibilities of corrosion, andother conditions that must be considered.A good hydraulic fluid must have the following properties: 1. Incompressible 2. It flows with minimum friction 3. Strong lubricating properties 4. Maintain its property at high temperature 5. Resistant to foaming Some of the properties and characteristics that must be considered when selecting a satisfactory liquid for a particular system are discussed based upon four primary considerations:• Viscosity• Chemical stability• Flash point• Fire point
  • [139]ViscosityOne of the most important properties of any hydraulic fluid is its viscosity. Viscosity isinternal resistance to flow. A liquid such as gasoline flows easily (has a low viscosity) whilea liquid such as tar flows slowly (has a high viscosity). Viscosity increases with temperaturedecrease. A satisfactory liquid for given hydraulic system must have enough body to give agood seal at pumps, valves, and pistons; but it must not be so thick that it offers resistance toflow, leading to power loss and higher operating temperatures. These factors will add to theload and to excessive wear of parts. A fluid that is too thin will also lead to rapid wear ofmoving parts, or of parts which have heavy loads.There are several types of viscometers that are used as an instrument to measure theviscosity of a hydraulic fluid used on aircraft systems. Such instruments measure the numberof seconds it takes for a fixed quantity of liquid to flows through a small orifice of standardlength and diameter at a specific temperature.Chemical StabilityChemical stability is another property which is exceedingly important in selecting ahydraulic liquid. It is the liquids ability to resist oxidation and deterioration for long periods.All liquids tend to undergo unfavorable chemical changes under severe operating conditions.This is the case, for example, when a system operates for a considerable period of time athigh temperatures.Excessive temperatures have a great effect on the life of a liquid. It should be noted that thetemperature of the liquid in the reservoir of an operating hydraulic system does not alwaysrepresent a true state of operating conditions. Localized hot spots occur on bearings, gearteeth, or at the point where liquid under pressure is forced through a small orifice.Continuous passage of a liquid through these points may produce local temperatures highenough to carbonize or sludge the liquid, yet the liquid in the reservoir may not indicate anexcessively high temperature. Liquids with a high viscosity have a greater resistance to heatthan light or low viscosity liquids which have been derived from the same source. Theaverage hydraulic liquid has a low viscosity. Fortunately, there is a wide choice of liquidsavailable for use within the viscosity range required of hydraulic liquids.Flash pointFlash point is the temperature at which the fluid produces enough combustible vapor that itwill ignite momentarily or flash when a flame is applied. A fluid with a high flash point canget very hot before it becomes susceptible to flashing. Looking at it from anotherperspective, a fluid with a high flash point has minimal evaporation under normal operatingconditions; therefore, a high flash point is a desirable characteristic of hydraulic fluids.
  • [140]Fire PointThis is the next step up the temperature spectrum from flash point. A hydraulic fluid’s firepoint is the temperature at which a substance gives off vapor in sufficient quantity to igniteand continue to burn when exposed to a spark or flame. Like flash point, a high fire point isrequired for desirable hydraulic liquid’s operating condition.Precautions while using hydraulic fluid: 1. Hydraulic fluid should not be mixed 2. Hydraulic fluids are flammable 3. Hydraulic fluids are supposed to have adverse effects like nerve damage, hence avoid contact with skin.Examples of Hydraulic fluid used in Aviation: 1. MIL H-5606 (red petroleum based)- first introduced over fifty years ago and still used on many aircraft. Used on business jets and many U.S. Air Force aircraft, it is highly flammable and considered responsible for the loss of at least one military aircraft, due to the fire created. 2. MIL-H-83282 - first used by the Air Force in 1982 and the U.S. Navy in 1997, it is less flammable than 5606, but much more viscous at low temperature. The lower temperature limit of MIL-H-83282 is considered 40° F, and it is used in virtually all Navy aircraft. 3. Skydrol (purple synthetic)- these alkyl phosphate ester based fluids are used on commercial aircraft, and are less flammable than the military fluids described above. Maximum temperature limit is 160° F. These fluids have been around at least since the 1960s. 4. MIL-H-87257 - this newest fluid is used in C135, E3, and U2 aircraft; it is less flammable than 5606 (similar to 83282) but its viscosity at low temperatures allows use down to 65° F.COMPONENTS INVOLVED IN HYDRAULIC SYSTEMA. hydraulic pump converts mechanical power to hydraulic powerWhen a hydraulic pump operates, it performs two functions. First, its mechanical actioncreates a vacuum at the pump inlet which allows atmospheric pressure to force liquid fromthe reservoir into the inlet line to the pump. Second, its mechanical action delivers thisliquid to the pump outlet and forces it into the hydraulic system.
  • [141]Pump Cross SectionA pump produces liquid movement or flow: it does not generate pressure. It produces theflow necessary for the development of pressure which is a function of resistance to fluid flowin the system. For example, the pressure of the fluid at the pump outlet is zero for a pump notconnected to a system (load). Further, for a pump delivering into a system, the pressure willrise only to the level necessary to overcome the resistance of the load.Types of Power PumpsThere are two types of power pumps, a gear pump and a piston pump. A. Gear pumps have efficiencies that average about 70-80% overall efficiency, where overall efficiency is defined by (mechanical efficiency)*(volumetric efficiency)Gear pumps move fluid based upon the number of gear teeth and the volume spacingbetween gear teeth.External-gear pumpsGear pumps can be divided into external and internal-gear types. These pumps come with astraight spur, helical, or herringbone gears. Straight spur gears are easiest to cut and are themost widely used. Helical and herringbone gears run more quietly, but cost more.
  • [142]A gear pump produces flow by carrying fluid in between the teeth of two meshing gears. Onegear is driven by the drive shaft and turns the idler gear. The chambers formed betweenadjacent gear teeth are enclosed by the pump housing and side plates (also called wear orpressure plates).A partial vacuum is created at the pump inlet as the gear teeth unmesh. Fluid flows in to fillthe space and is carried around the outside of the gears. As the teeth mesh again at the outletend, the fluid is forced out.Internal-gear pumpsInternal-gear pumps, Figure 4, have an internal gear and an external gear. Because thesepumps have one or two less teeth in the inner gear than the outer, relative speeds of the innerand outer gears in these designs are low. For example, if the number of teeth in the inner andouter gears were 10 and 11 respectively, the inner gear would turn 11 revolutions, while theouter would turn 10. This low relative speed means a low wear rate. These pumps are small,compact units. B. Piston pumpsThe piston pump is a rotary unit which uses the principle of the reciprocating pump toproduce fluid flow. Instead of using a single piston, these pumps have many piston-cylindercombinations. Part of the pump mechanism rotates about a drive shaft to generate thereciprocating motions, which draw fluid into each cylinder and then expels it, producingflow. There are two basic types, axial and radial piston; both area available as fixed andvariable displacement pumps. The second variety often is capable of variable reversible (overcenter) displacement.Most axial and radial piston pumps lend themselves to variable as well as fixed displacementdesigns. Variable displacement pumps tend to be somewhat larger and heavier, because they
  • [143]have added internal controls, such as hand wheel, electric motor, hydraulic cylinder, servo,and mechanical stem.Axial piston pumpsThe pistons in an axial piston pump reciprocate parallel to the centerline of the drive shaft ofthe piston block. That is, rotary shaft motion is converted into axial reciprocating motion.Most axial piston pumps are multi-piston and use check valves or port plates to direct liquidflow from inlet to discharge.Axial-piston pump varies displacement by changing angle of swash plate.Inline piston pumpsThe simplest type of axial piston pump is the swash plate design in which a cylinder block isturned by the drive shaft. Pistons fitted to bores in the cylinder block are connected throughpiston shoes and a retracting ring, so that the shoes bear against an angled swashplate.As the block turns, Figure 8, the piston shoes follow the swash plate, causing the pistons toreciprocate. The ports are arranged in the valve plate so that the pistons pass the inlet as theyare pulled out and the outlet as they are forced back in. In these pumps, displacement isdetermined by the size and number of pistons as well as their stroke length, which varieswith the swash plate angle.In variable displacement models of the inline pump, the swash plate swings in a movableyoke. Pivoting the yoke on a pintle changes the swash plate angle to increase or decrease thepiston stroke. The yoke can be positioned with a variety of controls,i.e., manual, servo,compensator, hand wheel, etc.Bent axis pumpsThis pump consists of a drive shaft which rotates the pistons, a cylinder block, and astationary valving surface facing the cylinder block bores which ports the inlet and outlet
  • [144]flow. The drive shaft axis is angular in relation to the cylinder block axis. Rotation of thedrive shaft causes rotation of the pistons and the cylinder block.Because the plane of rotation of the pistons is at an angle to the valving surface plane, thedistance between any one of the pistons and the valving surface continually changes duringrotation. Each individual piston moves away from the valving surface during one-half of theshaft revolution and toward the valving surface during the other half.Radial-piston pumpsIn these pumps, the pistons are arranged radially in a cylinder block; they moveperpendicularly to the shaft centerline. Two basic types are available: one uses cylindricallyshaped pistons, the other ball pistons. They may also be classified according to the portingarrangement: check valve or pintle valve. They are available in fixed and variabledisplacement, and variable reversible (over-center) displacement.In pintle-ported radial piston pump, Figure 9, the cylinder block rotates on a stationary pintleand inside a circular reacting ring or rotor. As the block rotates, centrifugal force, chargingpressure, or some form of mechanical action causes the pistons to follow the inner surface ofthe ring, which is offset from the centerline of the cylinder block. As the pistons reciprocatein their bores, porting in the pintle permits them to take in fluid as they move outward anddischarge it as they move in.Plunger pumpsThese reciprocating pumps are somewhat similar to rotary piston types, in that pumping isthe result of pistons reciprocating in cylinder bores. However, the cylinders are fixed in thesepumps; they do not rotate around the drive shaft. Pistons may be reciprocated by acrankshaft, by eccentrics on a shaft, or by a wobble plate. When eccentrics are used, returnstroke is by springs. Because valving cannot be supplied by covering and uncovering ports asrotation occurs, inlet and outlet check valves may be used in these pumps.Because of their construction, these pumps offer two features other pumps do not have: onehas a more positive sealing between inlet and outlet, permitting higher pressures withoutexcessive leakage of slip. The other is that in many pumps, lubrication of moving parts otherthan the piston and cylindrical bore may be independent of the liquid being pumped.Therefore, liquids with poor lubricating properties can be pumped. Volumetric and overallefficiencies are close to those of axial and radial piston pumps.
  • [145]B. An actuating cylinder (Hydraulic Actuators) converts hydraulic power to mechanicalpowerHydraulic actuators are devices for converting hydraulic pressure to mechanical motion(work). The most commonly utilized actuator is actuating cylinder; however, servo actuatorsand hydraulic motors are also employed for special applications where modified motion isrequired. Actuating cylinders are used for direct and positive movement such asretracting and extending of landing gear, wing flaps, spoilers and slats.Balanced Actuator SchematicDesign of actuating cylinders is determined by the functions that they are to perform.Actuating cylinders can be a single-acting or a double-acting cylinder types. In case of asingle-acting cylinder, hydraulic pressure is applied to one side of the piston to provide forcein one direction only. When hydraulic pressure is removed from the piston, a return springmoves the piston to its start position. Where as a double acting cylinder is designed so thathydraulic pressure can be applied to both sides of the piston. Thus, the cylinder can provideforce in either direction. So, double acting cylinders are widely used for the operation ofretractable landing gear, wing flaps, spoilers, bus doors and other similar applications.
  • [146]Unbalanced ActuatorServo actuators are designed to provide hydraulic power to aid the pilot in the movement ofvarious aircraft controls. Such actuators usually include an actuating cylinder, a multi-portflow control valve, check valves, and relief valves together with connecting linkages. Servoactuators are employed in situations where accurately controlled intermediate positions ofunits are required. The servo unit feeds back position information to the pilot’s control, thusmaking it possible for the pilot to select any control position required. Such actuators areused to move large control surfaces such as aircraft rudder, elevator, and ailerons. Forexample, servo units are used to aid the pilot in the operation of collective pitch and throttlecontrol lever in Mi-24/35 helicopter’s friction clutch.C. Hydraulic System AccumulatorsPurpose of Accumulators in the System: 1. Absorbs the shock due to rapid pressure variations in a hydraulic system 2. Helps to maintain a constant pressure within the hydraulic system 3. Helps the hydraulic pump under peak pressure loads 4. It is an emergency source of power (the braking system has its own accumulator)Piston Accumulator Schematic Metal Bellows Accumulator DiaphragmAccumulatorPrinciple of OperationAt the bottom of the accumulator is a gas valve. Compressed gas at about one half the systempressure is let into the accumulator through the gas valve. This forces the diaphragm thatseparates the oil side from the gas side to "pop" up towards the oil side.
  • [147]Then oil is sent through the system. When the system pressure reaches a point when it isgreater than the pressure of the accumulator, the diaphragm will deploy (inflate). UsingBoyle’s Law, the compressed gas will increase in pressure as its volume decreases. Thediaphragm will move up or down, depending on system pressure.When the diaphragm is at half way, the gas volume will be ½ as much as it was initially,while the accumulator pressure will be twice as much as its pre-load pressure (i.e., 1/2system pressure). Therefore when the accumulator is at half volume of gas, it will be chargedat full system pressure.Accumulator ShapesAccumulator can have different shapes according to its purpose. The shape can be; sphericalor cylindrical or bottle type. The spherical shape is the strongest and effective single shellbody used to withstand high pressure before failing. The bottle type accumulator is notwidely used on most aircrafts because of the effect on bladder used to expand and contract.Also the cylindrical type is not used very often because the friction will cause wearing of thebody and piston, thereby allowing the gas pressure to escape.D.Distribution Devices and ReducersDistribution devices (valves) are the main part of hydraulic system to control a flow of fluidto the actuating mechanisms in aircraft control system. These hydraulic valves may beclassified according to the following features: Drive type: manual or electrically (solenoid) controlled Number of fixed positions of the wing: two or three-position valves Type of distribution device: slide valve or plug type Control method: direct control or servo control methodHydraulic reducers are intended to decrease the fluid pressure on separate portions of thesystem to the required value. They are necessitated owing to the fact that aircraft hydraulicsystems use a number of actuating devices which are operable according to the specificationson lower pressure than the working pressure in the supply line. The hydraulic reducers can beof a constant-pressure reducer which maintains a definite fluid pressure or variable-pressurereducers in which the pressure is set in the course of control.E. Pressure Boosters and De-boostersPressure BoostersPressure boosters are rarely used in aircraft (almost all planes use de-boosters). If we needhigher pressure, we must change the entire power and actuating system. This adds weightthat is not needed. In general, we cannot put-in larger pistons and piston cylinders to increasethe power, because, normally we don’t have the room for it. A simple solution is to raise thepressure in a localized area.
  • [148]The function of a pressure booster is to act like a transformer; i.e, it raises the pressure of asmall circuit connected to the power system. The booster is a cylinder made up of twopistons of different surface areas that are connected. The larger surface area (A1) isconnected to the inlet side of the hydraulic system, and, the smaller surface area (A2) isconnected to the outlet side of the hydraulic system as shown in the figure below. Pressure booster’s operationDisadvantages of a Pressure Booster  Weight of the pressure booster is high if it is put into the aircraft, thereby reducing payload that the aircraft can carry.  It requires a very large booster stroke to meet the requirement  Pressure boosters must be built into the aircraft during the aircrafts construction, if high pressure is needed in the hydraulic system.  Leakage from the pressure booster is an important factor and will increase possibility of fire hazards.Pressure De-BoostersPressure de-boosters are used to reduce the pressure in the system to a level that can be usedby certain devices. Pressure de-boosters are pressure boosters turned upside down (that is,the inlet side of the booster has the smaller area piston and the outlet side has the larger areapiston). They are employed in power brake systems, using engine power to help apply thebrakes. Since aircraft wheels are made of magnesium, any high pressure on the wheels willcause them to split. That is why they must de-boost the pressure gotten from engine.The inlet line to the de-booster comes from the power brake control valve. The outlet linegoes to the brake system. This valve meters hydraulic fluid and hydraulic pressure directly.The force applied to the wheels to make them stop is proportional to how hard you push onthe rudder pedal. It is normally used for a 1 or 2 cubic inch application.
  • [149] Pressure de-booster’s operationBasic Operations in Hydraulic ControlsPressure Control (Pressure limiting devices/relief valves)FunctionPressure limiting device or relief valves are required to limit the pressure in some section ofthe hydraulic system to a predetermined level. That pressure level may be considereddangerous and, therefore, must be limited in such a case.Principle of OperationThe adjustment screw at the top of the pressure relief valve is set for a certain pressure value,let us call it P2. In general, even with a pressure of P1, the poppet would lift up, except thatthe spring is strong and has downward force forcing the poppet closed. Poppet will not moveuntil a pressure greater than that required is felt by the system (i.e., P1>P2).Pressure limiting device operating principleWhen the pressure increases, the poppet will move up, forcing the excess liquid to movethrough opening at high velocity. On other side of seat, pressure is zero because the back sideof the relief valve is connected to the return line. When the pressure in the system decreases
  • [150]below maximum, poppet will return to its seated position, sealing the orifice and allowing thefluid to follow its normal path. These type of pressure relief valves are only made to be usedintermittently.Flow ControlSelector ValvesSelector valves are used as (1) directional control devices to insure the movement ofhydraulic fluid flow in the proper direction, and (2) as stop-locks to lock the selector switchin a certain position.TypesThere are three types of selector valves used in aircraft hydraulic system namely; rotary type,piston type and poppet type.Rotary typesRotary type selector valves are plugs within which are passage ways for the fluid to movethrough. Tubing from the hydraulic pump or return line are connected to the rest of thehydraulic system by movement of the plug. You cannot use high pressure oil because ofleakage around the plug. To reduce leakage, you might want to make the plug fit more tightlyinto the selector valve body. However, the better you make the fit, the more friction will existbetween the plug and the selector valve body, making it difficult to operate.Rotary type selector valvePiston TypePositions 1, 2 and 3 (shown in fig.--) are representative positions for the piston-type selectorvalve. Position (1) is the position of the selector valve, for example, upon the extension oflanding gear or the lowering of flaps. Position (2) is the position of the selector valve uponretraction of the landing gear or the raising of the flaps. Position (3) is the stop-lockingposition of this type of valve. This piston type valve uses the Vickers spool mechanism inwhich the piston "lands" isolate the high pressure oil (red area) from the low pressure oil(blue area).
  • [151]Piston type selector valves operating principlePoppet Type – Stacked PoppetIn this type of valve, any movement of the handle (at the lower right of the diagram) changesthe camshaft and cam settings, thereby opening and closing the poppet valves and lettinghigh and low pressure oil to the proper sides of the actuating cylinder and return line,respectively.These mechanical type selector valves require a fair amount of tubing. In order to reduce theamount of tubing, electric switches have been used to operate solenoids which operate theselector valves. This has the added advantage of reducing the wasteful motions of pilot. Thistype of combined electronic circuits and hydraulic system is called electro-hydraulics. Stacked poppet type selector valves operating principle
  • [152]Flow RestrictorsSince the speed of the actuating cylinder is determined by the rate of flow of the hydraulicfluid, we may need a device to control the rate of flow. This device is called a flow restrictor.Since none of the selector valves meter the flow, we must use the restrictor.There are four types of restrictors used in aircrafts hydraulic system: 1. One way fixed restrictor 2. One way variable restrictor 3. Two way fixed restrictor 4. Two way variable restrictorOne Way Fixed RestrictorThe One Way Fixed Restrictor is not used all the time, but, it is being used more than theother types of restrictors. It is a check valve type restrictor with a drilled hole through theseat to the other side of the check valve. When the flow pressure seats the check valve ball(i.e., flow moving from right to left), some of the fluid can still reach the other side throughthe drilled hole in the seat. However, since the hole size is fixed, the amount of fluid passingthrough the passage to the other side is also fixed. One way fixed restrictorTwo Way Fixed RestrictorThe Two Way Fixed Restrictor is not used because it restricts the flow on the side of therestrictor where we want the flow to occur normally. Because the passage size is fixed, theamount of fluid moving from right to left, or vice versa, is fixed, as well. Two way fixed restrictorOne Way Adjustable RestrictorThe One Way Adjustable Restrictor is being used nowadays. It is the same as the One WayFixed Restrictor but the amount of fluid passed through the drilled opening in the seat isregulated by means of an adjustment screw.
  • [153] One way adjustable restrictorTwo-way Adjustable RestrictorThe Two Way Adjustable Restrictor is the same as the Two Way Fixed Restrictor, but it alsohas an adjustable screw that can be used to further restrict the amount of hydraulic fluidpassing through the opening. Two way adjustable restrictorIf a system relief valve (SRV) were used to regulate pressure, it would have to be replacedin a very short time. This would be due to the overuse of the SRV and the failure of thesprings elasticity. If the SRV were used, the oil pushing on the spring-ball combinationwould cause tremendous vibrations and heat would be dissipated by the oil under highpressure attempting to push the ball away from the seat to get to the low pressure side. Therange of operation of pressure regulator is defined by the difference in force required forbypass and the force required at actuation. And the dual purpose of this pressure regulator isto reduce the load on the hydraulic pump when not needed and to keep the hydraulic pressurewithin the operating range of the hydraulic system.Douglass Pressure RegulatorWhen an actuating cylinder finishes its motion and stops, a high pressure will be felt throughthe system. If so, this high pressure oil coming from the power pump (right side of diagram)will keep check valve C open and also act on piston A. In its movement, piston A pushesBall B off seat D. The oil, taking the passage of least resistance, goes through passage D intothe center chamber (colored blue) back to the reservoir. The pressure on the right side ofcheck valve C will drop and will be less than the pressure on the left side of C, therefore,
  • [154]causing the ball to seat itself in check valve C. When the hydraulic system pressure drops,the pressure on piston A decreases, causing a decrease in pressure on B as well. The path ofleast resistance through D will close and the oil will move in the direction towards checkvalve C. Now, because the pressure on the right side of C is greater than on the left of C, thecheck valve will be forced to open and the oil will move toward the selector valve side of thesystem (left side of diagram). HYDRAULIC DESCRIPTION OF THE DESIGNED AIRCRAFTThe airplane is equipped with two independent hydraulic systems, each powered by oneengine driven-pump and one electric motor-driven pump.Hydraulic power is provided by three independent systems designated No.1, No.2 and No.3.All systems operate at a nominal pressure of 3000 psi (20,685 kPa)Hydraulic fluid ―Skydrol‖- Skydrol 500B-4 (Type IV class 2): The Skydrol 500 series offluids has the longest service history among phosphate ester products. The first version,Skydrol 500, was introduced in 1952. Steady improvements to the formulation led in 1978 tothe current version, Skydrol 500B-4 which contains the same breakthrough anti erosionadditive and acid scavenger found in Skydrol LD-4 . Skydrol 500B-4 is the most workerfriendly of the aviation phosphate esters; it is least irritating to skin and less prone to formmists which can be irritating to the respiratory tract. This has given the product enormouspopularity for use in workshops and indoor test stands. Specification of SkydrolProperty Skydrol® 500B-4Appearance Clear, purple liquidNeutralization number 0.10 max.Moisture content (wt.%) 20 maxSpecific gravity, 77 °F/77 °F 1.050-1.062Viscosity in cSt at 210 °F/99 °C 3.68-4.00at 100 °F/38 °C 11.4-12.4at -65 °F/-54 °C 4200 max.Flash point, COC, 350 °F/177 °CRefractive Index @ 25 °C 466 to 1.474Skydrol is highly corrosive and can produce severe skin and eye irritation.
  • [155]SYSTEM DESCRIPTIONEach hydraulic system consists of a hydraulic fluid reservoir, a manifold, one engine-drivenpump, one electric motor-driven pump, one shutoff valve, one accumulator and a priorityvalve installed in the hydraulic system 1.RESERVOIRThe hydraulic fluid stored in the reservoir is pressurized, to avoid pump cavitations. Thispressurization function is performed by fluid drained from the pressure line. The reservoir isequipped with a quantityIndicator which transmits information to the MFD and EICAS displays for indication andwarning purposes. A thermal switch is responsible for the high temperature message, if thefluid temperature increases above 90°C.SHUTOFF VALVEA shutoff valve is installed between the reservoir and the engine-driven pump. It cuts thehydraulic fluid supply to the engine-driven pump, if there is a fire on the related engine or incase of hydraulic fluid overheats. This valve may be closed either through the engine fireextinguishing handle or through a dedicated button on the overhead panel.ENGINE-DRIVEN PUMPThe engine-driven pump provides continuous fluid flow at 3000 psi for operation of thevarious airplane hydraulically-powered systems. The pump is connected to the engineaccessory gearbox and, as long as engine is running, it generates hydraulic pressure. Duringengine start the fluid remaining in the suction line is sufficient to avoid pump cavitation andprovide reservoir pressurization.ELECTRIC MOTOR-DRIVEN PUMPThe electric motor-driven pump has the same connections as the engine-driven pump, but hasa lower flow capacity. The pump normally operates in the automatic setting mode, turning onwhen the associated hydraulic pressure drops below 1600 psi or the associated engine N2drops below 56.4%. If the pump starts operating in the automatic mode, it will be turned offafter the pressure or N2 are reestablished to normal values. The electric pump may be turnedon at pilot command, through the selector knob on the overhead panel, furnishing continuousfluid flow at 2900 psi.MANIFOLDThe manifold provides the following functions:-Fluid filtering (pressure and return lines).
  • [156]-Overpressure relief (main and electrical pumps).-Pressure indications (main and electrical pumps).Fluid leaving the pump flows to the manifold, where it is filtered and then routed to theairplane systems. Inside the manifold, a check valve prevents the fluid from returning to thepump, while a relief valvediverts the excess fluid to the return line. The return line is supplied by the fluid coming fromthe airplane systems, fluid drained from the pump, fluid from the relief valve, and fluidrefilled by the maintenance personnel. Under any situation the fluid is filtered and returned tothe reservoir. The manifold incorporates two pressure switches to detect low hydraulicpressure, and a pressure transducer to indicate system pressure. Signals from the pressureswitches and pressure transducer are sent to the MFD and EICAS displays.PRIORITY VALVEThe hydraulic system 1 incorporates a priority valve. If the system is powered by the electricmotor-driven pump and the landing gear is commanded to retract, the valve will provideminimum flow to theLanding gear system and give priority to the flight control services. In this case, the landinggear will operate through the accumulator pressure.ACCUMULATOREach hydraulic system has one accumulator. The function of the accumulator is to keep thesurges of the hydraulic pumps at a minimum, and to keep a 3000 psi pressure available foroperation of the landing gear and main door (system 1) or operation of the emergencyparking brake (system 2).EICAS MESSAGESTYPE MESSAGE MEANINGCAUTION HYD SYS 1 (2) FAIL Associated hydraulic system is not pressurized (inhibited when the airplane is on the ground, engine is shut down and parking brake is applied HYD SYS 1 (2) OVHT Associated hydraulic system fluid temperature is above 90C.ADVISORY E1 (2) HYD PUMP FAIL Engine-driven pump is not generating pressure with
  • [157] associated engine running. Associated hydraulic shutoff E1 (2) HYDSOV CLSD valve is closed. Fluid level in the associated reservoir is below one liter. HYD1 (2) LO QTY Report to the maintenance personnel if the hydraulic reservoir operates empty. Associated electric pump HYD PUMP SELEC OFF selected OFF with the parking brake released.CONTROLS AND INDICATORSHYDRAULIC SYSTEM PANEL1- ENGINE PUMP SHUTOFF BUTTON (guarded)  Closes (pressed) or opens (released) the associated engine pump shutoff valve.  A striped bar illuminates in the button to indicate that it is pressed.2- ELECTRIC HYDRAULIC PUMP CONTROL KNOBOFF - Associated pump is turned off.AUTO - Associated pump is kept in standby mode, ready to operate if the engine-drivenpump outlet pressure drops below 1600 psi or the associated engine N2 drops below 56.4%.ON - Associated pump is turned on.HYDRAULIC PAGE ON MFD 1- FLUID QUANTITY INDICATION  Ranges from zero to maximum hydraulic fluid quantity.
  • [158]  Scale (horizontal line) and pointer:  green when greater than 1 liter.  amber when equal to or less than 1 liter.  Pointer disappears if data is invalid. 2- PRESSURE INDICATION  Ranges from 0 to 4000 psi, with a resolution of 100 psi.  Digits:  green from 1300 to 3300 psi.  amber and boxed below 1300 and above 3300 psi.  Digits are replaced by amber dashes if data is invalid.3- ELECTRIC PUMP STATUS  Indicated by the green label ON or OFF. HYDRAULIC SYSTEM MAINTENANCEMAINTAINENCE The set of action including inspection, servicing, and determination of conditionrequired to achieve a derived outcome which restore an A/C part and equipment inserviceable condition.AIRWORTHINESS The continuing capability of the A/C to perform in satisfactory manner, the flightoperation for which it is designed.INSPECTION It is the most important form of function of aviation maintenance. As the A/C givescomplexity, it becomes more important to detect any possible trouble before it becomesserious. To assist this, aero engineers are provided with detail special check list and themaintenance manual for each type of A/C. The engineer has to go through maintenancemanual thoroughly before attempting any kind of activity in aircraft and its components. Theoperations may be carried out on A/C on daily flying hours and/or cycle basis.OVERHAUL
  • [159] Overhaul means stripping a unit and restoring it to its design performance level afterreplacing, reworking of parts to a given standard.SERVICING It means preparing the A/C for flight, includes providing the A/C with fuel and otherfluid and gases but do not include any work that is maintenance.TROUBLE SHOOT It means to analyses and identify the malfunction.REPAIR It means to correct the defective condition.MODIFICATION It is a continuous process to improve its reliability and performance.SERVICING SCHEDULESServicing on Hours/Calendar/Cycle basis, which are to be carried out on aircraft at setHours/Calendar/Cycle basis are mentioned in the manual or A/C servicing schedule. Theservicing includes examination, inspection, lubrication and removal of major componentssuch as landing gear jacks, door locks, air-conditioning equipments; aircraft brake unitswheels etc. landing gear functional test, flying control range and moment check., A/C riggingprocedure, hydraulic fluid contamination test, fuel contamination test & some activitiesrequires replacement of components. Aircraft maintenance checks are periodical checks that have to be done on all aircraftafter a certain amount of time usage. Aircrafts usually refer to as one of the following checks.A CHECK This is performed approximately every month. This is usually done over night. Theactual occurrence of this check varies by the type, cycle or number of hours flown since thelast check. The occurrence can be delayed by the aircraft if certain predetermine conditionsare met.B CHECK This is performed in approximately 3 months.C CHECK
  • [160] This is performed every 12 to 18 months. This check puts aircraft out of service andrequires plenty of space usually at the hanger and maintenance base. Schedule andoccurrence has many factors. The component is described and thus varies with the A/Ccategory and type.D CHECK This is the heaviest check of an A/C. This check is done approximately every 4 to 5year. This is the check that takes the entire A/C apart for inspection. A comprehensive check,analysis Non Destructive Testing (NDT) check and complete health monitoring of the enginehas to be recorded. Complete overhauling of the A/C and its components even A/C paintingis also required in this process.AIRCRAFT GENERAL MAINTENANCEBefore caring out any work on the A/C, the respective maintenance manual is to be referredfor further instructions. The necessary safety precautions are to be strictly followed.  Before switching on the master battery switch ensure that the under carriage selector lever is in down position and latched and all the armament store door switches are in safe condition.  Ensure that the wheel chocks are engaged.  Before operating the control surface, ensure that the control locks are removed.  Before starting the engine.  Chocks are to be kept in front of the wheel.  A/C brake system in serviceable.  A serviceable fire extinguisher is available.  Never tow an aircraft without a person inside the cockpit before towing the A/C, check the brake pressure.  While towing the A/C never exceed the walking speed.  Never drop any tool while working.  While working inside the A/C, collect all the tools and space on completion of the job and ensure no items are left behind.DESIGNED AIRCRAFT HYDRAULIC SYSTEM MAINTENANCE  Always release the system pressure before removing a component from the A/C  Never does any maintenance work on airplane with any other specified oil other than the recommended one.  Carry out the patch test on the system to prevent the contamination of oil. This can be carried out using Millipore patch test kit.
  • [161]  Never mix different grade of hydraulic oil to service the A/C.  Blank all the ports of the removed components and the A/C pipe ends to avoid the entry of dust, dirt and foreign particles.  Follow the necessary precautions to dismantle the hydraulic components.  Avoid spilling of hydraulic fluid on the A/C and in and around from the A/C. If spilled it should be cleaned immediately to avoid slipping.  Before fitting a new hydraulic component, it should be unblanked, degreased, washed and flushed.  While fitting the non return valve and restrictors, ensure that the marked arrows are in the desired direction.Conclusion: The general hydraulic components are briefed and the designed aircrafthydraulic system descriptions are explained with the schematic diagram. TECHNOLOGICAL SECTION INTRODUCTION TO RIB MANUFACTURE AND ASSEMBLYThe main objective of this report is to study and understand the manufacturing process andassembly scheme of the designed unit. Also the assembly base and installation base of eachpart is briefed which is very useful during the assembly of the rib. Some basics concepts ofjig and assembly jigs are explained in this report. Basics of aircraft rib its type, rib loads,parts of rib, rib construction is discussedThe assembly process is the process by which an item or product is put together. Theassembly process may include soldering, wiring, press fitting, brazing, shrink fitting,welding, adhesive bonding, riveting, and mechanical fastening. Within each of theseassembly processes a series of sequences is required to accomplish the process, withoutregard to the product configuration, material, or quantity to be produced, or the rate ofproduction. For example, many of the steps in creating a circuit card assembly, a wireharness, the frame of a truck, or in the installation of fittings on sailboat are essentially the
  • [162]same. The detailed instructions for the sequence should spell out the differences peculiar tothe product at hand.An extremely important part of the assembly process documentation is visual aids. Visualaids can be anything from an actual mockup of the product to a black and whiter colorphotograph, to a three-dimensional isometric or exploded-view drawing, to simple sketch, orto a tracing lifted directly from the engineering drawingIn this report we are going to see about the aircraft vertical fin ribs manufacturing process, itsassembly process, and jigs used to assemble the rib, assembly diagram, exploded viewdiagram, basics of rib construction and rib parts. AIRCRAFT RIBIn the framework of a wing, ribs are the crosspieces running from the leading edge to thetrailing edge of the wing. The ribs give the wing its contour and shape and transmit the loadfrom the skin to the spars. Ribs are also used in ailerons, elevators, fins, and stabilizers.Ribs typically form part of a structure that supports and defines the shape of an aero foilsurface. The rib comprises a generally planar web structure disposed between a series of ribfeet associated with the upper wing skin and a series of rib feet associated with the lowerwing skin.TYPES OF RIB:Two main classification of ribs are explained below,COMPRESSION RIBS:Carry main load in the direction of the flight, it has firm from leading edge to the trailingedge. In some aircrafts the compression ribs is a structural piece of tubing separating twomain spars. The main function of this rib is to absorb the force applied to the spar whenaircraft is in flight.FORMER RIBS:It is made from light metal which is attached to the stringers and wing skins to give The wingaerodynamic shape.  Nose ribs  Trailing edge ribs  Mid ribs running fore and aft, between front and rear spar.The rib is mainly subjected to three kinds of loading.a) Aerodynamic loads transmitted from the skin-stringer wing panels.
  • [163]b) Concentrated forces transmitted to the rib due to landing gear connections, power plant’snacelle connections, etc. RIB PARTS, ITS USES AND LOADS ACTING ON RIB PARTSThe Rib consists of following partsRIB CAPS (RIGHT AND LEFT CAP)USES – Rib caps are nothing but the flange of a beam – Caps enable the rib to hold tightly from top and bottom and it is also a load carrying element. There are different types of cap profiles used in rib.Some of the types of cap are given below.LOADS ACTING ON CAPS • Rib caps carry the bending load as axial load. • Rib caps are designed to have maximum radius of gyration. • High local crippling stress.RIB CAP MATERIAL  Aluminum alloy 7075 (Aluminum & zinc) - High mechanical properties and improved stress corrosion cracking resistance.RIB WEBUSES • Web of the rib is used to define and produce the air foil shape.  It is very long, thin and flexible. Since it gives main shape to the rib, the contour of web should be very exact and to be produced very carefully.
  • [164]LOADS • To Carry inertial loads (fuel, equipment, missiles, rockets). • To Support skin-stringer panel in compression and tension. • Crushing loads due to wing bending.WEB MATERIAL  Aluminum alloy 7075 (Aluminum & zinc) - High mechanical properties and improved stress corrosion cracking resistance.WEB STIFFNERStiffeners are mainly substituted for the corrugated aluminum sheets. They are responsible totransfer the loads or even provide supports for the web and to the caps.STIFFENER LOADS  Prevent the overall instability of the web.  Increase the buckling strength of web.MATERIALS Stiffener - Same material (Aluminum alloy 7075) used for web of ribs. Ductile materials are best suited for the pressing like aluminum, mild steelLIGHTENING HOLESLightening holes are found in web of ribs. They’re purposely made in different shapes. Butmost common shape is circular and elliptical in some cases.
  • [165]USES  It is provided to stiffen the rib web and reinforce the hole edges.LIGHTENING HOLES LOADS • Lightening holes may be introduced to the web of the rib for mass reduction. • It is also an accessibility and to form a passage for wiring and fuel pipes.WEB FASTNERSThe fasteners we use in our rib construction includes rivets and boltsSome of the aluminums alloy rivets used in rib construction is 2017, 2024 and 2117.ATTACHEMENT PLATEUSES  Provides special support to the whole structure.  Increases stiffness of the web at leading edge.  Also provides rigidity to the web.MATERIAL  Attachment plate - Aluminum alloy 7075RIVET HOLESUSES  Rivet holes are provided for installation of the rivets.  Rivets are permanent mechanical fasteners to join the web along with the caps and stiffener.MATERIALS  Rivets - Rivets of pure aluminum are used for riveting nonstructural parts fabricated using the softer aluminum alloys. RIB CONSTRUCTIONFormer ribs, located at frequent intervals throughout the wing, are made of formed sheetmetal and are very lightweight. The bent-up portion of a former rib is the flange and thevertical portion is the web. The latter is generally made with beads pressed between thelightening holes. These holes lessen the ribs weight without decreasing its strength.
  • [166]Lightening hole area rigidity is ensured by flanging the edges of the holes. Basic ribconstruction is given in a figure belowThe reinforced rib is similar in construction to the spar, consisting of upper and lower capstrips joined by a web plate. Vertical and diagonal angles between the cap strips reinforce theweb plate. The reinforced rib is used more frequently than the truss rib.Vertical and diagonal cross members only are used to reinforce and join the cap strips inconstructing truss ribs. These and reinforced ribs are heavier than former ribs and are usedonly at points where the greatest stresses are imposed. ASSEMBLY METHOD BY ASSEMBLING HOLESIt is used for flat form and simple curvature units and panels. In this method mutual positionof the assembled parts is determined by coincidence of the coordinated Assembly hole.Which in advance are done with the help of coordinated template. The basing on the Aircraftis possible at assembly of Aircraft and Helicopter framework and skin part as well as of theproduct inner equipment objects, when a demanded accuracy of the assembly not exceeds by-1.5 to +1.5mm
  • [167]LOFT TEMPLATE METHODDistinctive feature of the A and H manufacture is the LTM of AT manufacturing of manyparts and objects of assembly by a way of dependent on sizes and forms by carrying them onstandard rigging, then on a working rigging and further on products by various ways ofcopying. Three main scheme of coordination process includes, 1) The scheme using flat special rigid carriers as basic means- templates received by theoretical and constructive lofts. 2) The scheme of special volumetric carriers of forms and size- of standards, of mock-up made by construction method. 3) The scheme based on standard assembly unit and technological machine as a whole, received as a result of monitoring assemblies and coordination development of nodal complete set of parts.At LTM realization as the form and sizes standard take theoretical loft, representing the fullscale drawing of the A and H aggregate, made in three projections on the rigid basis i.e.,  The lateral projection Loft  The plane projection  The combined cross-section Loft.On TL the basic templates are made, which bear all necessary information for industrialtemplates manufacturing, and on their basis devices for parts manufacturing and AUassembling are made. CO-ORDINATE TEMPLATE METHODIn Coordinate template method the task of accuracy is increased and labor input is reduced atassembly devices coordination and coordination of their clampers was solved in LTM systemby its addition of special flat and spatial coordinate benches. The Loft Conductors and
  • [168]Instrument Stands which are universal means for exact construction of the sizes coordinate atmounting assembly devices. So coordinate template method of coordination was originated. Figure 1 MANUFACTURING METHODThe rib is machined from a solid cuboidal one-piece block of metal material, known as abillet, in order to provide strength and to remove the problems associated with joiningcomponents that are made separately.Some of the manufacturing process is given below,The forming processes modify metal or work piece by deforming the object, that is, withoutremoving any material. Forming is done with heat and pressure, or with mechanical force, orboth.Stamping includes a variety of sheet-metal forming manufacturing processes, such aspunching using a machine press or stamping press, blanking, embossing, bending, flanging,and coining. This could be a single stage operation where every stroke of the press producethe desired form on the sheet metal part, or could occur through a series of stages. Theprocess is usually carried out on sheet metal, but can also be used on other materials, such aspolystyrene.A punch press is a type of machine press used to cut holes in material. It can be small andmanually operated and hold one simple die set, or be very large, CNC operated, with a multi-station turret and hold a much larger and complex die set.Blanking and piercing both are shearing process in which a punch and die are used tomodify webs. The tooling and processes are the same for two, only the terminology isdifferent. In blanking the punched out piece is called blank. In piercing punched out piece iscalled scrap.Bending is a manufacturing process that produces a V-shape, U-shape, or channel shapealong a straight axis in ductile materials, most commonly metal. Commonly used equipment
  • [169]includes box and pan brakes, brake presses, and other specialized machine presses. Typicalproducts that are made like this are boxes such as electrical enclosures and rectangularductwork. PRODUCTION METHOD OF PARTS OF THE RIBTemplate of the web of rib is made first with perfect contour and then original sheet is mademanually.RIB WEBWeb of the rib is produced by forming with hydro press, punch press with special millingdevice. Since the contour of our web is very complex first the template is made with markingeach dimension clearly and then the manufacturing is processed if we are making itmanually.The rib web is formed from metal that lies near the surface on a first side of the billet, andthe ends of the rib feet are formed from metal near the surface on the opposite side. Thus arelatively small amount of machining is conducted on the first side of the billet to form theweb. Such a rib may suffer from distortion problems, which have to be accounted for duringthe rib machining process because the metal at and close to the surfaces of the billet issubject to residual stresses (resulting from the process used to manufacture the billet) that
  • [170]may cause deformation of, or undesirable internal stresses in, the web produced there frombe damaged.Finally completing the web plate make riveting holes using drill bits for exact dimensions.STIFFENERStiffeners are made by bending process. We know that our stiffener is a L-typeSuitable length of the aluminum stiffener is taken .Dimensions are marked and therectangular shape is made by cutting unnecessary parts and then it bend using the bendingtool.We can also use press brake forming, a work piece is positioned over the die block and thedie and pressed.The L-punch forms a L-shape with a single punch.Finally completing the five stiffener make riveting holes using drill bits for exact dimensions.MAKING THE LIGHTENING HOLESLightening holes are done by drill-bits and flange is made by form blocks. The dimensionsof four holes are marked out clearly and then it is processed.The first and last hole has different dimensions, the second and third are similar indimensions.Drill-bit is a cutting tool used to make cylindrical holes. Bits are held in a tool called grill,which rotates them and provide torque and axil force to create holes.Flanging the Lightening HolesMost of the rib lightening holes has a 3/8" wide flange at 30 degrees around each hole. Thisis to stiffen the rib web and reinforce the hole edges. There are several ways to accomplishthis.Another method is using form blocks. Two large holes were cut in each of the two formblocks that were 3/4" larger than the lightening holes.Rib was clamped between the two form blocks and bolted together through the jig holes. Itwas also clamped in the center between the holes to assure a nice tight squeeze. The piecesare then placed into the bench top press and the plug is then pressed down against the rib(bevel side down).
  • [171]RIGHT, LEFTAND SIDE CAPSThe three caps are manufactured using milling process, with the help of milling cutters. Theright and left caps runs along the length of the rib and it has the same contour of the rib. Sowhile making it the dimensions are marked out clearly first. The side cap which is a T-profiletype attaches to the trailing edge of the rib and special holes are made there using exactdimensions.Finally after completing the caps make riveting holes.RIVETTING HOLESRivet holes for each of the part like web, top and bottom caps, and stiffener are made using adrill bits. Before riveting make sure to mark the dimensions for all rivets.ATTACHMENT PLATEMark out the dimensions clearly first, then cut out the remaining parts using cutting tool andusing milling tool the curved surface is made and required holes are placed using drill bit.ASSEMBLY PROCEDURE OF THE RIBSCHEME OF ASSEMBLY OF THE FIN RIBDepending on a degree of partition of the airframe and airborne systems in the assembly unitand a degree of differentiation of the assembly mounting and attachment works on the object,the assembly mounting works can be classified into three typical schemes of assembly theyare 1. The sequential scheme 2. The parallel scheme 3. The parallel sequential schemeThe Sequential Scheme: This scheme is used for construction of the aircrafts aggregates incondition of the small series or individual production, when the partition scheme doesntdistinguish the panels. The parts and small assembly units are based sequentially on the basicpart or on the basic unit. After that the sections, compartments, aggregates are assembled in asequence which are joined into united airframe.
  • [172]The assembly scheme of the rib we use here is a sequential scheme, where each and everypart of the rib structure is installed separately one by one to complete the whole structure. RIB MOUNTING ATTACHMENT PATE ASSEMBLY DEVICEUPPER CAP LOWER CAP STIFFENERS FASTENERS SIDE CAP RIB WEBFor the assembly of the given rib section of the aircraft, it is recommended to use the verticaljig. It is very easy to use with and we can able to work out from both sides of the jig.Before stepping to assembly process ensure that necessary procedures of manufacturing ofall part is done and kept for assembling. Refer assembly diagramFor the assembly of the given rib section of the aircraft, it is recommended to use the verticaljig. It is very easy to use with and we can able to work out from both sides of the jig.Before stepping to assembly process ensure that necessary procedures of manufacturing ofall part is done and kept for assembling. Refer assembly diagram while assembling.  Prepare the jig  Install first the right cap (T-profile) using clamps and fix it rigidly.  Install the web of rib after installing right cap.  Then install the left cap (T-profile) and fix it firmly.
  • [173]  Install the caps using elements of special clamps.  Assembly holes are made to fix all the caps to the web of rib.  Now install the side cap of section M using special Holes.  The caps and web are clamped together tightly.  Now the two stiffeners are glued to one side of the web and other three stiffeners are glued to another side of the web.  The special attachment plate is placed to the other side of the rib and it is glued initially and then it is clamped.  If necessary use guide holes at required places.  The attachment plates are fixed using special bolts  Then rivet the whole assembled structure using the rivet guns.  Trailing edges of the rib is installed very carefully using assembly holes.  Avoid drilling after assembled because it would damage the whole structure.  Make sure that the contour shape of the web of the rib is not affected throughout the manufacturing and assembling process. STAGES OF FORMATION OF RIB DIMENSIONS USING TEMPLATESThe scheme with application of mathematical model of a surface for manufacturing of ribusing manual template method is shown below.
  • [174]THE SCHEME OF A RIB MANUFACTURING USING SURFACEMATHEMATICAL MODELScheme with application of mathematical model of a surface for manufacturing of rib usingNPC method is shown below.JIGA jig forms a guide for the tool used in machining operation. It is a type of tool used tocontrol the location or motion of another tool. It is primarily used to provide repeatability,
  • [175]accuracy and interchangeability in the manufacturing of products. It does both functions thatis holding the work and guiding a tool.MAIN TYPES OF JIGOPEN JIGIt is also called as plate jig, attaches to work piece without enclosing completely. It must besquare or doughnut shaped and affixed over the work piece with clamps.DIAMETER JIGIt is used for cylindrical work pieces .It encloses a work piece in v-shaped groove.LEAF JIGS AND BOX JIGThese jigs completely enclose the work piece.Some other types of jig include,Trunion, Template, Guide, Sandwich, Angle plate, UniversalASSEMBLY JIGA device or mechanism used in machine building for the attachment, securement, and correctalignment of parts and modules. In individual and small-lot production, general-purposeassembly jigs are used, including plates, assembly beams, knife-edged supports, try squares,clamps, and jacks. A set of such jigs serves as a basis for constructing stands used inassembling modules and complete machines. In mass and large-lot production, specializedassembly jigs are used for securing major components and modules (rotating and multi-position jigs) and for accurate and rapid assembly of parts and modules (single-position,multiposition, stationary, and mobile jigs). Assembly jigs are also used for the preliminarydeformation of elastic elements, such as coil and leaf springs and split rings, and for makingconnections with interference fit. In progressive assembly, jigs are used for varying thepositions of the objects being assembled. The use of assembly jigs improves the quality ofthe items produced, facilitates the work of assemblers, and increases productivity.
  • [176] AIRCRAFT VERTICAL ASSEMBLY JIG DESIGN LAYOUT DIAGRAM THE DESIGNED VERTICAL FIN IN CATIAConclusion: The manufacturing and assembly process of vertical rib of the aircraft isbriefed.
  • [177] ECONOMICAL SECTION 1. ECONOMIC EFFICIENCY CHARACTERISTICS CALCULATIONThe economic calculations of the aircraft are done through required formulae or it can alsobe done on math cad which is pre-programmed already.REQUIRED PARAMETERS:Take-off mass – 39 .11 tonesEmpty mass = takeoff – fuel mass – payload mass = 26.98 tonesThe items of expenses are:Cost of materials and raw materials 4 0 . 93 3 . 32 log( N )M o  1 . 95  10 m  0 .9 dmd-- Mass of the designed airplane in tonesN – Annual quantity of airplane (N=8)md =39.11 tones M o  1 . 95  10  28 . 65  0 .9 4 0 . 93 3 . 32 log( 8 )Thus =391017.802834 $Cost of devices onboard equipmentPurchased devices cost (PD) = 1.95(-1280+2.37 Vmax+14.15 Me.a) N-0.09Me.a – mass of empty airplane in tonesMe.a=26.980tonsVmax–Maximum speed of airplane in km/hrVmax- 835 km/hrPD = 1651.575307$Expenses for manufacturing special technological expensesGeneral expenses of labor manufacturing for special technological equipment  Tli = 0.87*1.03n*Me.a*106 hours  n = total number of engines=2 = 24902081.34 dollars  Expenses in dollar  Texp = Tli* C– price of one hour- (8)$
  • [178]= 199216650.72 $  Expense for one airplane  Toa = Texp/3(N1+N2) N1 – production in the first year=3 N2 - production in the second year=5Toa = 11067591.706667 WagesExpenses on wages for work on direct manufacturing of airplane include a wage fund ofindustrial piece workers and workers creating the airplane by payment per hour.  W = 45200*Me.a0.903*M0.42*N-0.32*Kp Me.a – mass of empty airplane W =204074.65775$M - Maximum speed of airplane=0.7M , N– annual quantity of production Kp– coefficientwhich consider increase of labour charges of industrial workers during the production ofairplane=0.64.Indirect expenses  Indirect shop expenses (IS) IS = 5.837*W*N-0.129 = 839968.49613$Indirect manufacturing expenses (IM) IM = 7.1*W*N-0.359=578606.52$Taxes  T = 0.375*0.7*W = 548063.227042Total manufacturing expense  Tme= Mo+PD+ToA+W+IS+IM+Tax= 13105937.063395$Outside Manufacturing expense  Ome= 5% (Tme)= 524237.482536$Total Manufacturing cost price  Tmcp= Tme+Ome=13630174.545931
  • [179]Planned Profit 15% Tmcp= 2044526.18189Wholesale Price Tmcp+ profit= 15674700.727821Value Added Tax 20% of wholesale price= 3134940.145564$Total Wholesale price of enterprise with VAT  Wholesale price + VAT= 18809640.873385$ Variable expenses:- Expenses which are changed (with manufacturing of airplanes) proportionally to the quantity of airplanes =Mo+PD+W+Tax= 650313.6333551$  Fixed expense :- expenses which are not changed with the manufacturing of airplane  = IS+IM+Toa+Om= 12979860.91238$  Analytical Calculation of break Even Point  NBP=( Fixed expense)/(price – Variable expense)= 0.714777 The whole calculation is concluded in the table ExpensesS.No Articles (in million US $) 0. 391017 1 Materials and Raw Materials 1651thousands 2 Purchased devices , onboard equipments
  • [180] 11.067591 3 Manufacturing Repair of special technological equipment 0. 204074 4 Wage 0. 839968 5 Indirect Shop expenses 0. 578606 6 Indirect Manufacturing expenses 0.548063 7 Taxes 13.105937 8 Total ( Manufacturing expenses) 0. 524237 9 Outside Manufacturing expenses 13.630174 10 Total Manufacturing Cost Price 2.044526 11 Profit (Planned) 15.67470 12 Wholesale Price 3.13494 13 Value Added Tax 18.809640 14 Wholesale Price with VATConclusion:I have efficiently calculated the economic efficiency characteristics for my aircraft and foundthe cost of the aircraft and the amount of profit it that will be realized for the companyannually.
  • [181] SPECIAL ASSIGNMENT INTERIOR CABIN LAYOUT AND SEATING ARRANGEMENTThe design process of the cabin is mainly characterized by the number of passenger’saccommodation and availability of volume, so it is necessary to find some of the baseparameters for interior design of the cabinINITIAL DATA FOR CABIN DESIGN CONSIDERATION  Number of passengers - 47 and can be increased up to 55  Payload mass which includes also mass of luggage - 4230 kg  Diameter of fuselage – 2.5mFactors influencing Fuselage seat dimensions are  The seating layout or arrangement this can be selected based upon the prototype statistical data which is briefed below,  Cross section of the fuselage  Diameter and length of the fuselageSEATING TYPE – 3 ABREAST3 Abreast SEATING refers to the 2 × 1 seat arrangement, each row has two seats on righthand side and one seat to the left hand side of aircraft. This is selected based upon theprototype and diameter of the fuselage.A typical 3-abreast seating arrangement accommodates 24 to 45 passengers, but variantdesigns change that from 20 to 50passengers (e.g., ERJ145). Full standing headroom ispossible; for smaller designs, a floorboard recess may be required. A floorboard recess couldtrip passengers when they are getting to their seat. Space below the floorboards is still notadequate for accommodating any type of payload. Generally, space for luggage in thefuselage is located in a separate compartment at the rear but in front of the aft pressurebulkhead (the luggage-compartment door is sealed).FULFILLING REQUIREMENTS OF THREE ABREAST SEATING LAYOUT  Full standing headroom is possible  A floorboard recess may be required  Space for luggage in the fuselage is located in a separate compartment at the rear but before aft pressure bulk head  Luggage compartment door is sealed if present  Two cabin crews for 50 passengersNUMBER OF ROWSIt is determined by the formula = = 15.6615.66 can be rounded to either 16 , or
  • [182]For 45 passengers = = 15 rowsFor two passengers = = 1 row, it may be either 2 abreast or 3 abreast according todesigners choice.Therefore no of Rows = 16COMFORT LEVEL DETERMINATIONThe comfort level of the aircraft setting layout is the process of distributing the mass of thepayload and its luggage capacity if the aircraft is designed with varying passengersnumbers. In general there are two types of comfort level one is the low level and another ishigh level comfort.LOW LEVEL COMFORTIn low level comfort the number of passenger is increased by decreasing the luggage massand compacting the cabin length. The low level comfort scheme of the designed aircraft isgiven below,The mass of luggage in low level comfort – 7 kg per passengerNumber of passengers – 55Mass of the single passenger – 70 kgTotal mass of passengers – 70 x 55 = 3850 kgTotal mass of luggage – 7 x 55 = 385 kgTotal Mass of payload = 3850 + 385 = 4230 kgHIGH LEVEL COMFORTIn high level comfort the number of passenger is decreased or set to the designed goalstandard by increasing the luggage mass and providing more convenient seating arrangementinside the cabin. The high level comfort scheme of the designed aircraft is given below,The mass of luggage in high level comfort – 15 kg per passengerNumber of passenger – 47The mass of single the passenger - 75 kgTotal mass of the passenger – 75 x 47 = 3525 kgMass of luggage – 15 x 47 = 705 kgHigh comfort level = Number of passengers x mass of the passenger + mass of luggage = 75 x 47 + 705 = 4230kgSTANDARD SEAT AND AISLE PITCH AND WIDTH SEAT PITCH cm SEAT WIDTH cm AISLE WDTH cmECONOMY CLASS 71 to 81 46 to 51 43 to 41BUISNESS CLASS 84 to 91.5 53 to 56 56 to 63.5The dimensions given above are general recommendation it varies according to operators andchanges for different abreast seating of the aircraft.
  • [183]DIMENSIONSSeat width, B (Left hand side) - 48.5 cmAisle width, A - 45.72 cmSeat width – B (Right Hand side) - 2 x 48.5 cmTotal elbow room - 20 cmGap between wall & seat, G - 5 cmCABIN DIMENSIONING FOR 3- ABREAST SEATINGThe cabin dimensioning depends on two main factors,  Fuselage width  Fuselage height  Cargo compartment locationSo the selection of cabin width and floor height may vary according to the designed aircraft.But the normal cabin dimensioning for 3 abreast aircraft is given below we can use more orless value based on it and also we can refer our prototypes data for this selection.Total cabin width Wcabin - 215 cmTotal wall thickness T - 20 cmTotal fuselage width Wfuselage - 236 cmTypical fuselage height, Hfuselage – 215 cm INTERIOR ARRANGEMENT – CROS SSECTION (TYPICAL)
  • [184]The above picture depicts an example of 3 Abreast fuselage cross-section of Embraer 145by comparing with these data’s my aircraft cabin is selected relative to the fuselage diameter,followed by the seat dimensions. Thus the whole aircraft seating layout with its cross sectionare drawn with all the necessary dimensions.DETERMINATION OF DESIGNED AIRCRAFT CABIN CROSS-SECTIONThe cabin cross-section of the designed aircraft is constructed based on the collectedinformation and compared with our aircraft fuselage diameter, and then the cross section isdrawn as given below,Total fuselage width – 2.5 mTotal fuselage width – 2.5 mSeat width – 0.45 mSeat pitch – 0.8277mCabin diameter – 2.27 mMaximum floor height – 1.85 mFloor width – 0.44 mGap between wall and seat - 0.09 mWall thickness = (fuselage diameter – cabin diameter) = 2.5-2.27 = 0.23m
  • [185]DETERMINATION OF CABIN LENGTH FOR HIGH COMFORT LEVELThe length of both the passenger and cockpit compartment is found statistically by selectingfirst selecting the cabin accessories for designed aircraft. The accessories placed insideaircraft varies on the right and left side so it is necessary to determine list of accessories andtheir quantities to place. Then the length of each accessory should be mentioned, along withthe interior spacing and the lavatory dimension.The right hand side of the aircraft cabin includes the following,Closet - one closet = 0.80 mGalley – two galleys = 0.70 m each (2 X 0.70 = 1.4 m)Galley doors – one galley door (width) = 0.60 mSpace between the cockpit compartment and galley = 0.53mSpace between the second galley and first seat (RHS) = 0.30mTotal length of the occupied seat = Number of rows x seat pitch = 16 x 0.8293 = 13.21mSpace between last seat and lavatory = 0.30mLavatory length in side cross-section = 1mTOTAL LENGTH = 0.53 + 0.80+ 0.70+ 0.60 +0.70 +0.30 +13.21 + 0.30 + 1Total length of the passenger cabin = 18.14m (RIGHT HAND SIDE)The Left hand side of the aircraft cabin includes the following,Passenger main door – 0.80 mSpace between cockpit compartment and main door – 0.46 mMain door length – 0.80 mSpace between main door and cabin attendant seat – 1.2879mSpace between cabin attendant seat to 1st row main seat – 0.82mCabin attendant seat length – 0.46Total length of the occupied seat = Number of rows x seat pitch = 16 x 0.8293 = 13.21mSpace between last seat and lavatory = 0.30mLavatory length in side cross-section = 1mTOTAL LENGTH = 0.80 + 0.46 + 0.80 + 1.287 + 0.82 + 0.46 +13.21 + 0.30 + 1 = 18.14mTotal length of the passenger cabin = 18.14m (LEFT HAND SIDE)Therefore length on RHS = length on LHSNOW adding the cockpit length which is equal to 1.90m, we getTotal length of cabin including cockpit = 18.14 + 1.90 = 20.04 m
  • [186]Then the final length is sketched for the foe calculated values by arranging the accessories inthe cabin at appropriate places. The selection of those accessories may be from differentmanufacturers but it is compulsory to follow the rules and regulation of aviation federation.Some of the design parts used in this type of aircraft is given below ,For the designed aircraft Galley no 1 is chosen as below Galley used in embraer 120For galley no 2:
  • [187]Closet of the aircraftService cart for providing food
  • [188]Beverages cart:SEAT DESIGNA two dimensional model of the chair is shown below for the reference to design a chair ifneeded but in practice the seats for an aircraft are selected based on the nominal values setpreviously so it is better to choose a good manufacturer for the selection of aircraft seat someexamples are provided below
  • [189]SEAT SELECTION- ECONOMY RANGE SEATS
  • [190]CONCLUSION: Thus the cabin-cross section, length, seat dimensions are found and it issketched briefly.
  • [191] DIAGRAMSLOAD CARRYING STRUCTURE CENTRE OF GRAVITY
  • [192]GENERAL VIEW DIAGRAM SEATING ARRANGEMENT
  • [193]HYDRAULIC SYSTEM OF THE AIRCRAFT
  • [194]CONCLUSION OF THE PROJECTThus the selected aircraft is processed starting from the initial geometrical parametersfollowing its aerodynamic characteristics, the loads acting on unit structure, integrateddesigning , manufacturing techniques, the hydraulic system, and even its economicalvaluation s are done each step by step according to the valuable lectures given by myprofessors and with the help of their textbooks. At last the designed aircrafts seating layout isconsidered and cabin length are found and sketched. The aircraft which is designed herefollows all the rules and regulations of corresponding aviation federation with necessarylimitations and has its own special requirements which are solely fulfilled according to theprojects calculation and results.
  • [195]REFERENCES1.DEVELOPMENT OF A PILOT PROJECT OF AN AIRCRAFT-Training guide BY A.K. Myalitsa, L.A.Malashenko, A.G. Grebenikov, E.T. Vasilevskiy, V.N. Klimenko,A.A. Serdyukov – Kharkiv: National Aerospace University Kharkov Aviation Institute2. Aircraft Design - A Conceptual Approach 2nd ed. - D. Raymer (1992)3. Torenbeek - Synthesis of Subsonic Airplane Design4. Airplane design volume by Jhon Roskam5. Aircraft Design Ajoy Kumar Kundu Queen’s University Belfast6. Understanding aircraft structures By John Cutler- Jeremy Liber7. Anderson Jr (2001) - Fundamentals of Aerodynamics8. Mechnologies of aircraft manufacturing by Yu. M. Bukin, Vorobyov9. http://triomatic.boico.com/Examples.htm.10. www. Scribd.com