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STRUCTURES & WEIGHTS PDR 1
TEAM 4 Jared Hutter, Andrew Faust, Matt Bagg, Tony Bradford, Arun Padmanabhan, Gerald Lo, Kelvin Seah October 28, 2003
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OVERVIEW Materials Wing Analysis Tail Boom Sizing C-G Determination
Landing Gear
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Modulus of Elasticity (ksi)
Material Properties Material Density (lb/ft3) Modulus of Elasticity (ksi) Al 2024-T6 178.2 10500 Balsa 5.1 490 Basswood 24.9 1500 Spruce 24.5 1230 Sources: - - US Dept. of Agriculture
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Wing Analysis Procedure Calculated sectional lift coefficient
Evaluated sectional wing bending moment Sized I-beam to desired proportions Trade Study Minimize material weight Maximize stress loading capacity Selected most suitable material and thickness
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Wing Analysis Actual bending moment at each point along spar
Root Bending Moment = ft-lbf Actual bending moment at each point along spar Based on lifting line theory
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Wing Analysis
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Wing Analysis
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Wing Analysis 508.5 ft-lbf
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Wing Analysis Single spar wing structure selection I-beam
Material: BALSA (Ochroma Pyramidale) 12% Height = ft = 4.28 in Base = ft = 2.59 in Thickness = ft = 0.61 in Weight = 11.0 lbf Saftey Factor = 2 (for Sigma) t factor = 2.6 times increase for required thickness
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Tail Boom Sizing Cylindrical tubes
Availability More efficient than solid rods Used twist and deflection constraints Appropriately sized inner diameters Found corresponding outer diameters
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Tail Boom Sizing Equation for Deflection I: moment of inertia (in4)
P: estimated maximum aerodynamic load applied to end of boom (lbf) E: modulus of elasticity (ksi) L: length of tail boom (in) d: deflection of end of boom (in)
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Tail Boom Sizing Equation for Twist Torsion Constant J
f: angle of twist (rad) T: applied torque (ft-lbf) L: length of tail boom G: shear modulus (ksi) J: torsion constant (in4) Torsion Constant J For circular tube: t: thickness (in) r: radius of tube (in)
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Tail Boom Sizing Known Constants Twist Deflection T = 15 ft-lbf
P = lbf L = 5 ft E = ksi set d = 2 in Twist T = 15 ft-lbf L = 5 ft G = 3920 ksi set f = 5 deg = rad
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Tail Boom Sizing Set inner diameter to be 1.6 in
Solve for the outer diameter that satisfies both constraints Outer diameter = 1.7 in Thickness = 0.05 in Weight for both booms = 5.04 lbf
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Tail Boom Sizing
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C.G. LOCATION ESTIMATION
This figure shows the approximate weights and C.G. locations of the main components: z x Wing W = lb x = 1.55 ft z = 0 ft Avionics Pod W = 20 lb x = ft z = ft Tail Section W = 2.3 lb x = 8.23 ft z = ft Main Gear W = 3 lb x = 0 ft z = ft Tail Booms W = 5.94 lb x = 4.05 ft z = 0 ft Engines, Fuel, Casings W = lb x = -0.3 ft z = -0.5 ft Tail Gear W = 0.5 lb x = 8 ft z = -0.21ft NOT TO SCALE
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C.G. LOCATION ESTIMATION
LIFT z x Total Weight: W = 54.5 lb C.G. Location: x = 0.47 ft, z = ft Wing M.A.C.: x = ft Static Margin: SM = 10.0% WEIGHT SM = – (xCG – xMAC) / c NOT TO SCALE
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TAILDRAGGER LANDING GEAR CONSTRAINTS
RAYMER 11.2 3.1 ft Represents C.G. location 0.47 ft Z 0.38 ft X 18.80 deg. ( deg) 1.35 ft 1.42 ft 10.04 deg. ( deg) 8 ft NOT TO SCALE
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WEIGHT DISTRIBUTION ∑MB = 0 ∑MA = 0 FA W = 54.5 lbf FB y = 7.43 ft
Center of Gravity FA W = 54.5 lbf FB Main Gear Tail Gear y = 7.43 ft x = 0.70 ft Wy ∑MB = 0 FA = = 49.81 lbf x + y 91% of weight carried by main gear 9% by tail gear FB = Wx x + y ∑MA = 0 = 4.68 lbf NOT TO SCALE
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FOLLOW-UP ACTIONS Torsion constraint on spar Geometry of wing ribs
Geometric layout of tail Moments and products of inertia
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QUESTIONS?
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