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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. OVERVIEW. Materials Wing Analysis Tail Boom Sizing C-G Determination Landing Gear. Material Properties. Sources: - www.matweb.com
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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
OVERVIEW • Materials • Wing Analysis • Tail Boom Sizing • C-G Determination • Landing Gear
Material Properties Sources: - www.matweb.com - US Dept. of Agriculture
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
Wing Analysis Root Bending Moment = 508.5 ft-lbf Actual bending moment at each point along spar Based on lifting line theory
Wing Analysis 508.5 ft-lbf
Wing Analysis • Single spar wing structure selection • I-beam • Material: BALSA (Ochroma Pyramidale) 12% • Height = 0.357 ft = 4.28 in • Base = 0.216 ft = 2.59 in • Thickness = 0.051 ft = 0.61 in • Weight = 11.0 lbf
Tail Boom Sizing • Cylindrical tubes • Availability • More efficient than solid rods • Used twist and deflection constraints • Appropriately sized inner diameters • Found corresponding outer diameters
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)
Tail Boom Sizing • Equation for Twist • 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)
Tail Boom Sizing • Twist • T = 15 ft-lbf • L = 5 ft • G = 3920 ksi • set f = 5 deg = 0.0873 rad • Known Constants • Deflection • P = 26.73 lbf • L = 5 ft • E = 10500 ksi • set d = 2 in
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
C.G. LOCATION ESTIMATION • This figure shows the approximate weights and C.G. locations of the main components: z x Wing W = 12.04 lb x = 1.55 ft z = 0 ft Avionics Pod W = 20 lb x = -1.44 ft z = - 0.58 ft Tail Section W = 2.3 lb x = 8.23 ft z = 0.075 ft Main Gear W = 3 lb x = 0 ft z = -1.25 ft Tail Booms W = 5.94 lb x = 4.05 ft z = 0 ft Engines, Fuel, Casings W = 12.72 lb x = -0.3 ft z = -0.5 ft Tail Gear W = 0.5 lb x = 8 ft z = -0.21ft NOT TO SCALE
C.G. LOCATION ESTIMATION LIFT • Total Weight: W = 54.5 lb • C.G. Location: x = 0.47 ft, z = -0.38 ft • Wing M.A.C.: x = 0.775 ft • Static Margin: SM = 10.0% z x WEIGHT SM = – (xCG – xMAC) / c NOT TO SCALE
Represents C.G. location TAILDRAGGER LANDING GEAR CONSTRAINTS RAYMER 11.2 3.1 ft 0.47 ft Z 0.38 ft X 18.80 deg. (16 - 25 deg) 1.35 ft 1.42 ft 10.04 deg. (10 - 15 deg) 8 ft NOT TO SCALE
Wx FB = x + y WEIGHT DISTRIBUTION Center of Gravity FA W = 54.5 lbf FB Main Gear Tail Gear y = 7.43 ft x = 0.70 ft Wy 49.81 lbf FA ∑MB = 0 = = 91% of weight carried by main gear 9% by tail gear x + y ∑MA = 0 4.68 lbf = NOT TO SCALE
FOLLOW-UP ACTIONS • Torsion constraint on spar • Geometry of wing ribs • Geometric layout of tail • Moments and products of inertia