Team “Canard” September 28th, 2006 Structures 1 QDR Team “Canard” September 28th, 2006

Wing Sizing Aerodynamics gives the geometry Load case: Resist to 10g (640ft radius at 100mph) Materials 0.3 0.89ft 9.3 deg Swing = 4.16 ft2 0.4 ft MH 43 Thickness:8.5% With a weight of 5 lb Wing should support 50 lb balsa Expanded polystyrene Easy to built Good surface AAE 451 Team 1

Sizing Method Discretization of the wing Determination of the loads 1 2 3 4 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

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

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

Minimal thickness Assumptions: The balsa should resist the load 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 IG a b polar inertia AAE 451 Team 1

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

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

Twist max. twist: 0.7° AAE 451 Team 1

Deflection 0.21 in AAE 451 Team 1

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

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

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

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

Tip Over Analysis d2 d3 -(Waircraft-Wwing) + FLG – Wwing – Ftip = 0 z d1 d3 Ftip FLanding Gear Wwing Waircraft-Wwing -(Waircraft-Wwing) + FLG – Wwing – Ftip = 0 ∑MLG = 0 = -(Waircraft)d1 + (FLG)d3 – (Ftip)d2 Ftip ≈ 0.5 lb AAE 451 Team 1

Landing Gear Calcs MLG fwd Angle 20° RLG Angle 5° Rolling AOA 12° CG to Wing TE 0.43 ft Racer wheels MLG min Spread 12° MLG dist from CG 5.5 in Fuselage Height 5.4 in Propeller Radius 5.04 in Prop Clearance carbon or aluminum strut depending on cost AAE 451 Team 1

Questions AAE 451 Team 1