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Mini Project - STRUCTURAL-ANALYSIS-AND-MATERIAL-SELECTION

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Mini Project - STRUCTURAL-ANALYSIS-AND-MATERIAL-SELECTION

  1. 1. STRUCTURAL ANALYSIS AND MATERIAL SELECTION This mini-project report was submitted to the Department of Aeronautical Engineering of Kotelawala Defence University in a partial fulfillment of the requirement for the Semester-5 in Degree of Bachelor of Science By 3888 TUO HMHHS BANDARA 3907 TUO HD MILLEWA 3938 C/SGT DMCD DISSANAYAKE 3930 O/C HUGT PIYARATHNA ENG/AE/12/011 AMDN ATAPATTU Supervised by SQN LDR JI ABEYGOONEWARDENA Mr. S.L.M.D. RANGAJEEVA Department of Aeronautical Engineering Kotelawala Defence University Intake 29 Group 2
  2. 2. CHAPTER ONE 1.1 INTRODUCTION 1. The UAV was designed with the maximum possible simplicity. The structure of the designed UAV is analyzed and the material was selected by dividing the whole task in to four phases as mentioned below. 2. In material selection phase, the front wing-box was mainly taken in to consideration because it is the portion where the highest buckling stress is occurred. It is reasonable to mathematically determine the properties of material which are required to withstand for the highest buckling stress, and apply the same material for the total skin. For the spars and other internal structures, the material with highest strength to weight ratio was considered. 3. Considering the performance requirements of the UAV, the β€œv-n diagram” was designed analyzing the flight loads. The total weight was taken into discussion and calculated according to the data and assumptions considered at the discussions made with other groups. Designing the landing gear was done in the simplest way using two β€œLeaf-spring” shock absorbers instead of landing gear legs either side of the fuselage, and supporting the two β€œFoam filed” tires.
  3. 3. CHAPTER TWO 2.1 MATERIAL SELECTION 4. For the spar we recommend high strength material because the spar strength is given the first priority. From the table we can recognize 2 high strength materials but with different densities. 2.1.1 Aluminum Alloy 7075-T6 0.101 lb/in3 7178-T8 0.102 lb/in3 So, we chose 7075-T6 because of its lower density; hence the lightness of the aircraft is ensured. 2.1.2 Sheet Material 5. The leading edges are the most vulnerable areas for the fatigue loads. Therefore we consider the leading edge area to choose the materials for the sheet by considering the buckling stress.
  4. 4. 2.1.3 Calculating the buckling stress h = 12.14% c π‘Ž = 40π‘š h = 0.1214 m β„Ž 2 = 0.0607 π‘š l =20% =0.2m 𝑏 = 0.06072 + 0.22 𝑏 = 0.209 π‘š Therefore buckling stress = 5 + 6𝑏 π‘Ž 𝐸( 𝑒 𝑏 )2 𝜏C =2.948Γ— 10-4 E Therefore critical buckling stress = 5 + 6π‘Ž 𝑏 𝐸( 𝑒 π‘Ž )2 𝜏C1=1.84532Γ— 10-6 E 𝜏C1< 𝜏C Therefore this is not capable. By putting ribs a is reduced to 0.3125m Therefore a = 0.15625m 𝑏 = 0.209 π‘š Therefore new buckling shear stress = 5 + 6𝑏 π‘Ž 𝐸( 𝑒 𝑏 )2 𝜏C =7.63387Γ— 10-4 E Therefore critical buckling stress = 5 + 6π‘Ž 𝑏 𝐸( 𝑒 π‘Ž )2 𝜏C1 = 9.94642Γ— 10-4 E 𝜏C < 𝜏C1 Therefore this can be stand with loads Number of ribs for a wing = 40 0.15625 = 256 6. Now we have to choose the material which has a less buckling stress and less density from the table. Then we are choosing the β€œHM21A” magnesium alloy because it is fulfilling the above requirements and especially it can stand with the temperature up to 700o F. The material should be stand to high temperature because it fly around the world l h
  5. 5. 2.1.4 Calculating the shear stress t-shear flow 𝜌 - 1.225 kg m3 𝑣 - 38 ms-2 (cruising speed ) 𝑠 - 40 m2 CL max - 2.4 max. lift force per wing = 1 2 πœŒπ‘£2 𝑠𝑐 𝐿 = 1 2 Γ— 1.225 Γ— 382 Γ— 40 Γ— 2.4 = 84 907.2 N But, 𝑑 = 𝑇×𝐴 β„ŽΓ—βˆ , ∝= 𝑑π‘₯ 𝑒 , 𝐴 = 1 ∝1 + 1 ∝2 + 1 ∝3 Length of front box sheet = 461.366 mm Height of the bar = 121.4mm Length of rear box sheet = 1633.1208mm T = 84 907.2 N e1 = 1.6 Γ— 10βˆ’3 m ∝1 = 288.353 e2 = 2 Γ— 10βˆ’3 π‘š ∝2 = 60.7 e3 = 0.3 Γ— 10βˆ’3 π‘š ∝3 = 5443.736 A = 49.6866 Therefore, t1= 120.514 kNm-1 t2 = 572.501 kNm-1 t3 =6.383 kNm-1
  6. 6. 2.1.5 Shear stress (𝜏 ) 𝜏 = 𝑑 𝑙 𝑙 = 0.15625 m Thus, 𝜏1 = 771.2896kN m-2 𝜏2 = 3 664.0064kN m-2 𝜏3 = 40.8512kN m-2 2.1.6 Comparison with the chosen materials Material Identical 𝝉 (kN m-2 ) Critical 𝝉 (kN m-2 ) Comment HM 21A 771.2896 206842.7 Critical shear stress is higher. Therefore shear stress is bearable for the sheet. 7075-T6 3 664.0064 524001.6 Critical shear stress is higher. Therefore shear stress is bearable for the spar.
  7. 7. 2.2 FLIGHT LOADS ANALYSIS 7. The design of the structure is based on a load limit, which is the largest expected load. For aerodynamic forces, this is related to the aerodynamic load factor, n. Load factors were designated for some of the flight phases, such as intercept, and with maximum and sustained turn rates. These will be considered here in the design of the structure. 8. In addition to the loads that occurs at different flight phases, the following are also considered, 1. The loads produced when flying at the highest possible angle of attack without stalling. 2. The loads that occur at a dive speed equal to the 1.5 cruising velocity. (Vc) 3. The loads produced by wing gusts, such as those that can occur in thunder storms or from clear air turbulence. 9. The largest load factor from any of those in this group will be considered to be the β€œdesign load factor”, which will be the basis for the design of the internal structure. The design of the internal structure and the material selection clearly go hand in hand. The use of higher strength materials can reduce the size or number of structural elements. However the structure weight is an important factor that also needs to be considered. Therefore, the structure design and material selection should be done together. 2.2.1 V-N Diagram 10. A v-n diagram shows the flight load factors that are used for the structural design as a function of the air speed. These represent the maximum expected loads that the aircraft will experience. These load factors are referred to as the β€œlimit load” factors. Below shows the table of some limit load factors of some aircrafts. 11. As we are going to design a solar powered UAV and it will not perform high maneuvering part during its flight time, we are going to take the limit loads as General aviation (normal) (-1.25 ≀ n ≀ 3.10). 12. As the group 02 decided the cruising speed will be 38ms-1 VCruise = 38 ms-1 VDive = 1.5 VC= 57ms-1 VStall = 2𝑀 πœŒπ‘π‘  =14.83ms-1 (ρ=0.3025kgm-3 , CL max= 2.3954, S=80m2 , W=650 kg, g=9.81 ms-2 )
  8. 8. For our UAV during the cruising time it will face only a load factor n= 1. n = 𝐿 π‘Š 13. And also during the flight if the UAV come across a load factor more than the limit loads for a long time period the aircraft will subject to structural damage. So during the manufacturing of the aircraft the structure should be able to withstand the limit loads for a minimum of 3 seconds as for the airworthiness requirements. 14. Gust loads are unsteady aerodynamic loads that are produced by atmospheric turbulence. They represent a load factor that is added to the aerodynamic loads, which were presented in the previous sections. 15. The effect of a turbulent gust is to produce a short-time change in the effective angle of attack. This change can be either positive or negative, thereby producing an increase or decrease in the wing lift and a change in the load factor, βˆ†π‘› = Β± βˆ†πΏ π‘Š 16. By considering all above the design load factor can be implemented. Because the design limit factor will shows the allowable strength of the structure for the external loads. As for the airworthiness requirements the UAV should be design with the design load factor be1.5(limit load factor).
  9. 9. Aircraft weight Weight of airframe Thickness of the Front sheet = 1.6x10-3 m Thickness of the Rear sheet = 0.3x10-3 m Length of front box sheet = 461.366 mm Height of the bar = 121.4mm Length of rear box sheet = 1633.1208mm Weight of Skin of a wing = 0.4544 0.0253 Γ— 0.064 Γ— 40 Γ— 0.16331208 Γ— 0.3 Γ— 10βˆ’3 + 0.461366 Γ— 1.6 Γ— 10βˆ’3 = 58.605 kg Thickness of the spar = 0.002m Weight of a spar = 0.454 0.0253 Γ— 0.101 Γ— 40 Γ— 0.002 Γ—0.081 = 28.526 kg Total weight of the wings = (58.605+28.526) Γ—2 = 174.26 kg Assume that the fuselage weight is 75kg
  10. 10. 2.3 Landing gear design The aircraft will employ a single bicycle-type nose landing gear and, two main landing gears by which are supported by two leaf-springs separately for minimal complexity. This configuration is similar to that of Cessna 337 F Sky Master aircraft. To bypass complications with pneumatic tires exploding from the greatly decreased pressure at cruise altitudes, the tires will be filled with lightweight foam instead of compressed air (Ex: - Polyurethane Foam) Landing gear attachment β€’ Assuming the weight of the aircraft during landing as the MTOW Thus, the energy absorbed per main landing gear (one out of two): WABΓ— 1 2 MV2 Γ— 1 2 = 1 4 Γ—650Γ—(6Γ—0.3048)2 = 543.4827 J Energy absorbed by landing gear = Energy absorbed by leaf spring + Energy absorbed by the tire Wab = Wleaf + Wtire Hence assuming Wtire is negligible, Wab = Wleaf Wab = Wleaf = 1 2 Kx2 ; K= Spring Constant. Assuming the max compression of the leaf spring as 100mm (x = 100mm): 1 2 K(0.1)2 = 543.4827 K=108.696 N mm-1 Therefore the Leaf-spring is designed in such manner that the value of K is 108.696 N mm-1 Landing gear location 2T = Track a = Horizontal displacement between nose landing gear and CoG b = Horizontal displacement between center line of main landing gear and CoG a+b = Wheelbase h = Height between CoG and wheel plane ΞΈ = Turnover angle
  11. 11. The turnover angle ΞΈ should be less than 63Β°

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