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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v-n diagramβ 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Leaf-springβ shock absorbers instead of landing gear legs either side of the fuselage, and supporting the two βFoam filedβ 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 = 40π h = 0.1214 m β 2 = 0.0607 π l =20% =0.2m π = 0.06072 + 0.22 π = 0.209 π Therefore buckling stress = 5 + 6π π πΈ( π π )2 πC =2.948Γ 10-4 E Therefore critical buckling stress = 5 + 6π π πΈ( π π )2 πC1=1.84532Γ 10-6 E πC1< πC Therefore this is not capable. By putting ribs a is reduced to 0.3125m Therefore a = 0.15625m π = 0.209 π Therefore new buckling shear stress = 5 + 6π π πΈ( π π )2 πC =7.63387Γ 10-4 E Therefore critical buckling stress = 5 + 6π π πΈ( π π )2 πC1 = 9.94642Γ 10-4 E πC < πC1 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HM21Aβ 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 π - 1.225 kg m3 π£ - 38 ms-2 (cruising speed ) π  - 40 m2 CL max - 2.4 max. lift force per wing = 1 2 ππ£2 π π πΏ = 1 2 Γ 1.225 Γ 382 Γ 40 Γ 2.4 = 84 907.2 N But, π‘ = πΓπ΄ βΓβ , β= ππ₯ π , π΄ = 1 β1 + 1 β2 + 1 β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 10β3 m β1 = 288.353 e2 = 2 Γ 10β3 π β2 = 60.7 e3 = 0.3 Γ 10β3 π β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 (π ) π = π‘ π π = 0.15625 m Thus, π1 = 771.2896kN m-2 π2 = 3 664.0064kN m-2 π3 = 40.8512kN m-2 2.1.6 Comparison with the chosen materials Material Identical π (kN m-2 ) Critical π (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.