Dane Batema Benoit Blier

Drew Capps Patricia Roman

Kyle Ryan Audrey Serra

John Tapee Carlos Vergara

Team 1:

Structures 1 QDRTeam “Canard”

September 28th, 2006

AAE 451 Team 1 2

Wing Sizing• Aerodynamics gives the geometry

• Load case: Resist to 10g (640ft radius at 100mph)

• Materials

0.4 ft

0.3

0.89ft9.3 deg

Swing = 4.16 ft2

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

1 3 42

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 Liftoutline and discretization of the wing

0.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

1.60

1.80

0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50

x axis (ft)

y ax

is (

ft)

0.00

1.00

2.00

3.00

4.00

5.00

6.00

7.00

8.00

9.00

10.00

Lif

t (l

bf)

outline

lif t

The lift is assumed to be linear:

Lift = Wloading x Surface

AAE 451 Team 1 5

Bending Momentbending moment relative to the x position on the wing

0.00

5.00

10.00

15.00

20.00

25.00

30.00

35.00

40.00

-0.30 0.20 0.70 1.20 1.70 2.20 2.70 3.20

distance from the root (ft)

ben

din

g m

om

en

t (l

bf.

ft)

bending moment

L1 L2

d1d2

M=L1.d1 + L2.d2 + …..MAC

AAE 451 Team 1 6

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

maxb

compG

My cste

I

t is figured out from IG

a

b2

2

ratiochord tb

chorda

( , , )

( , , )GI f a b t

J g a b t

polar inertia

psifailurestressncompressiocomp 5.72

AAE 451 Team 1 7

Skin Thickness

Thicknesses

0.00E+00

2.00E-02

4.00E-02

6.00E-02

8.00E-02

1.00E-01

1.20E-01

1.40E-01

1.60E-01

1.80E-01

0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50

root distance (ft)

skin

th

ickn

ess

(in

)

discretizedthicknesseslinearized thickness

constant thickness

Easy to built, but 70% heavier than discretized thickness

Optimal thickness distribution

1.53 in

AAE 451 Team 1 8

Twist and Deflection• Twist

– Assumption: Only the aerodynamic twist (twist due to the swept angle is neglected)

• Deflection

2max _ max

maxmax

1max

2 mT SV C

T L

GJ

_ max

max

0.0318

150 /

23060

m

balsa

C

V ft s

G psi

Lift at MAC

y y’3lift

G

F Ly

EI

y’ with Thales theorempsiEbalsa 6.185

=100 mph

AAE 451 Team 1 9

Twist

twist angle

-1.6E-02

-1.4E-02

-1.2E-02

-1.0E-02

-8.0E-03

-6.0E-03

-4.0E-03

-2.0E-03

0.0E+00

0.00 0.50 1.00 1.50 2.00 2.50

distance from the root (ft)

twis

t ang

le (r

ad)

twist angle

max. twist: 0.7°

AAE 451 Team 1 10

Deflection

deflection

0

0.05

0.1

0.15

0.2

0.25

0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50

distance from the root (ft)

def

lect

ion

(in

)

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 spar50% 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

Thicknesses

0.00E+00

2.00E-02

4.00E-02

6.00E-02

8.00E-02

1.00E-01

1.20E-01

1.40E-01

1.60E-01

1.80E-01

0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50

root distance (ft)

skin

th

ickn

ess

(in

) with sparwithout sparconstant thickness

From this point, the structure resist without spar

Cut spar at 1.50ft

AAE 451 Team 1 13

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 14

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 15

Tip Over Analysis

-(Waircraft-Wwing) + FLG – Wwing – Ftip = 0

∑MLG = 0 = -(Waircraft)d1 + (FLG)d3 – (Ftip)d2

Ftip ≈ 0.5 lb

FLanding Gear

Ftip

Wwing

Waircraft-Wwing

y

z

d1 d3

d2

AAE 451 Team 1 16

Landing Gear Calcs

MLG min Spread 12°

MLG dist from CG 5.5 in

Fuselage Height 5.4 in

Propeller Radius 5.04 in

Prop Clearance 5.04 in

MLG fwd Angle 20°

RLG Angle 5°

Rolling AOA 12°

CG to Wing TE 0.43 ft

Racer wheels

carbon or aluminum strut depending on cost

AAE 451 Team 1 17

Questions