1. STRUCTURES & WEIGHTS PDR 1
TEAM 4
Jared Hutter, Andrew Faust, Matt Bagg, Tony Bradford,
Arun Padmanabhan, Gerald Lo, Kelvin Seah
October 28, 2003

2. OVERVIEW Materials Wing Analysis Tail Boom Sizing C-G Determination
Landing Gear

3. 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: -
- US Dept. of Agriculture

4. 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

5. Wing Analysis Actual bending moment at each point along spar
Root Bending Moment = ft-lbf
Actual bending moment at each point along spar
Based on lifting line theory

6. Wing Analysis

7. Wing Analysis

8. Wing Analysis
508.5 ft-lbf

9. Wing Analysis Single spar wing structure selection I-beam
Material: BALSA (Ochroma Pyramidale) 12%
Height = ft = 4.28 in
Base = ft = 2.59 in
Thickness = ft = 0.61 in
Weight = 11.0 lbf
Saftey Factor = 2 (for Sigma)
t factor = 2.6 times increase for required thickness

10. Tail Boom Sizing Cylindrical tubes
Availability
More efficient than solid rods
Used twist and deflection constraints
Appropriately sized inner diameters
Found corresponding outer diameters

11. 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)

12. 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)

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

14. 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

15. Tail Boom Sizing

16. C.G. LOCATION ESTIMATION
This figure shows the approximate weights and C.G. locations of the main components: z x
Wing
W = lb
x = 1.55 ft
z = 0 ft
Avionics Pod
W = 20 lb
x = ft
z = ft
Tail Section
W = 2.3 lb
x = 8.23 ft
z = ft
Main Gear
W = 3 lb
x = 0 ft
z = ft
Tail Booms
W = 5.94 lb
x = 4.05 ft
z = 0 ft
Engines, Fuel,
Casings
W = lb
x = -0.3 ft
z = -0.5 ft
Tail Gear
W = 0.5 lb
x = 8 ft
z = -0.21ft
NOT TO SCALE

17. C.G. LOCATION ESTIMATION
LIFT z x
Total Weight: W = 54.5 lb
C.G. Location: x = 0.47 ft, z = ft
Wing M.A.C.: x = ft
Static Margin: SM = 10.0%
WEIGHT
SM = – (xCG – xMAC) / c
NOT TO SCALE

18. 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. ( deg)
1.35 ft
1.42 ft
10.04 deg. ( deg)
8 ft
NOT TO SCALE

19. 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

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

21. QUESTIONS?