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

2. 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

3. Sizing Method Discretization of the wing Determination of the loads 12 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

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

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

6. 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

7. 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

8. 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

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

10. Deflection
0.21 in
AAE 451 Team 1

11. 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

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

13. 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

14. 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

15. Tip Over Analysis d2 d3 -(Waircraft-Wwing) + FLG – Wwing – Ftip = 0z 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

16. 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

17. Questions
AAE 451 Team 1