Dane BatemaBenoit Blier Drew Capps Patricia Roman Kyle Ryan Audrey Serra John TapeeCarlos Vergara Team 1: Structures 1 PDR Team “Canard” October 12th, 2006

AAE 451 Team 1 2 Wing Sizing Aerodynamics gives the geometry Load case: Resist to 10g (74ft radius at 100mph) Materials 0.4 ft ft 9.3 deg S wing = 4.16 ft 2 MH 43 Thickness:8.5% With a weight of 5 lb Wing should support 50 lb

AAE 451 Team 1 3 Proof of n value

AAE 451 Team 1 4 Sizing Method Discretization of the wing Determination of the loads Quarter chord MAC: application of the lift For each part, we can figure out: The bending moment due to the lift The torsion torque due to the aerodynamic moment

AAE 451 Team 1 5 Distribution of Lift The lift is assumed to be linear: Lift = W loading x Surface

AAE 451 Team 1 6 Bending Moment L1 L2 d1 d2 M=L1.d1 + L2.d2 + ….. MAC

AAE 451 Team 1 7 Minimal thickness Assumptions: –Only bending loading –Foam doesn’t carry the load The balsa should resist the load We assume the shape of the airfoil is an ellipse t is figured out from I G a b polar inertia

AAE 451 Team 1 8 Skin Thickness Easy to built, but 70% heavier than discretized thickness Optimal thickness distribution 1.53 in

AAE 451 Team 1 9 Twist and Deflection Twist –Assumption: Only the aerodynamic twist (twist due to the swept angle is neglected) Deflection Lift at MAC yy’ y’ with Thales theorem =100 mph

AAE 451 Team 1 10 Twist max. twist: 0.7°

AAE 451 Team 1 11 Deflection 0.21 in

AAE 451 Team 1 12 Carbon Spar Advantages The minimum skin thickness is too thick to bend around the airfoil shape (from experience) Method: Assume iso-rigidity for the carbon tube and the skin (EI) skin =(EI) spar 50% of the load in the spar 50% of the load in the skin Redo the calculations for the skin with 50% of load

AAE 451 Team 1 13 Skin Thickness With Spar From this point, the structure resist without spar Cut spar at 1.50ft

AAE 451 Team 1 14 Skin Thickness 4.8 in Balsa sheet thickness t = 0.06 in ($22 for 253”x36”x1/16”) ID ≈ 0.32 in OD ≈ 0.45 in Thickness ≈ 0.13 in Final size will depend on market availability

AAE 451 Team 1 15 CG Location: 33% of Mean Aerodynamic Chord Mean Aero. Chord (MAC) CG Estimation/Spar Location Spar Length = 3.0 ft (≈1/2 of wingspan)

AAE 451 Team 1 16 Geometry Structure layout: –Full sheet of balsa sanded –Foam + balsa skin Horizontal tail structure 0.38 ft 0.07 ft 1.84 ft NACA 0006, 6% thickness ratio

AAE 451 Team 1 17 Horizontal tail calculations Load case: High speed dash + 20° deflection L=0.5ρSV 2 C L From control team we have: V is the most influent parameter for the Lift force Wing downwash = Wing AOA = Incidence of the Htail =0 Control surface deflection 20° =

AAE 451 Team 1 18 Bending moment roottip MAC location Lift a b x y DATAS Sref (ft²)0.56 lift force (lbf)1.98 Ult. Comp. Stress (psi)725.2 Vmax (ft/s)150.0 young modulus (ksi)185.6 shear modulus (psi) balsa density (lb/ft3)6.2 wingspan (ft)1.84 root chord (ft)0.38 wingtip chord (ft)0.23 airfoil thickness (%)6 Results: Min. thickness: 2.18e-2 in Total deflection: 5.3e-1 in Very thin, impossible to find in the market

AAE 451 Team 1 19 Final tail structure layout Horizontal tail: Foam core in balsa sheet (similar to the wing) Vertical tail: The final geometry is not define yet We plan to make it in a full sheet of balsa sanded.

AAE 451 Team 1 20 Components easy to vary Components more difficult to vary Components that cannot be varied Component CG Table Green Components need to be moved until this value = 0

AAE 451 Team 1 21 Graphical Representation of Weight Distribution

AAE 451 Team 1 22 Working Model

AAE 451 Team 1 23 Components Layout Payload Battery S.C. Motor GearBox

AAE 451 Team 1 24 V-n Diagram V Cruise = 115 ft/s V Dive = 1.3*V Cruise -5 g’s (from Raymer Table) V Stall ≈ 35 ft/s C LMAX = 0.9 S = 4.15 ft 2 m = 5 lbs G max = 10 Trade Study on n-loading, turn radius, and speed will provide more technical justification

AAE 451 Team 1 25 Landing Gear Roskam method for landing gear sizing: 1. Landing gear system: Fixed 2. Landing gear configuration: taildragger 3. Locate c.g.: 1.23 ft from the nose 4. Decide preliminary landing gear Goldberg Landing Gear: Glass filled Length: 14-3/8” (from axle to axle) Mount area length: 3-7/16” Height: 4-1/2”

AAE 451 Team 1 26 Landing Gear 5. Longitudinal tip-over analysis 15 deg 12 deg 6. Lateral tip-over analysis Ψ≤ 55 deg Main gear Tail gear

AAE 451 Team 1 27 Landing Gear 7. Ground clearance criteria 8. Max. static load per strut > 5 deg W TO nSPmnSPm PtPt ltlt lmlm l m +l t W TO = Total weight = 5.0 lbs n S = Number of main gear struts = 2 P m = Main gear load P t = Tail gear load l m = distance from main gear to c.g. = 1.6” l t = distance from tail gear from c.g. = 2.55” P t = (W TO l m )/(l m +l t ) = 1.93 lbs P m = (W TO l t )/n S (l m +l t ) = 1.54 lbs

AAE 451 Team 1 28 Landing Gear 9. Number of wheels: 2 for main gear 1 for tail gear Glassfilled Nylon Lightweight Width:.177’’ Diameter: 2.25” OD Hayes Racing Wheels: Ohio Tail Wheel Tiny 4-8 lbs

AAE 451 Team 1 29 Wing-Fuselage Attachment Fuselage Rib Wing top view Carbon rod Nylon screwL max /4

AAE 451 Team 1 30 Wing-Fuselage Attachment Nylon screws from hobby lobby: D = in Length = in Cross-sectional area: A = π(D/2) 2 = in 2 F = n*W/4 = 12.5 lbs n= number of g σ = F/A= 12.5 lbs/ in 2 = psi Ultimate Tensile Strength for Nylon = psi Margin = 34 L max /4 Margin=σ max / σ-1

AAE 451 Team 1 31 Wing-Fuselage Attachment Carbon rod: D= in t = thickness of rib = in σ = F/(D*t) = 12.5 lbs/(( in)( in)) = psi Ultimate Compressive Strength of balsa = psi Margin = 2.2 L max /4 t Carbon rod Front view Top view

AAE 451 Team 1 32 Questions