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

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: - 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 Actual bending moment at each point along spar Root Bending Moment = 508.5 ft-lbf Actual bending moment at each point along spar Based on lifting line theory

Wing Analysis

Wing Analysis

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 Saftey Factor = 2 (for Sigma) t factor = 2.6 times increase for required thickness

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 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)

Tail Boom Sizing Known Constants Twist Deflection T = 15 ft-lbf P = 26.73 lbf L = 5 ft E = 10500 ksi set d = 2 in Twist T = 15 ft-lbf L = 5 ft G = 3920 ksi set f = 5 deg = 0.0873 rad

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

Tail Boom Sizing

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 z x 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% WEIGHT SM = – (xCG – xMAC) / c NOT TO SCALE

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. (16 - 25 deg) 1.35 ft 1.42 ft 10.04 deg. (10 - 15 deg) 8 ft NOT TO SCALE

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

FOLLOW-UP ACTIONS Torsion constraint on spar Geometry of wing ribs Geometric layout of tail Moments and products of inertia

QUESTIONS?