structures & weights pdr 1
DESCRIPTION
STRUCTURES & WEIGHTS PDR 1. TEAM 4 Jared Hutter, Andrew Faust, Matt Bagg, Tony Bradford, Arun Padmanabhan, Gerald Lo, Kelvin Seah October 28, 2003. OVERVIEW. Materials Wing Analysis Tail Boom Sizing C-G Determination Landing Gear. Material Properties. Sources:- www.matweb.com - PowerPoint PPT PresentationTRANSCRIPT
AAE 451
STRUCTURES & WEIGHTS PDR 1
TEAM 4Jared Hutter, Andrew Faust, Matt Bagg, Tony Bradford,Arun Padmanabhan, Gerald Lo, Kelvin Seah
October 28, 2003
TEAM4OVERVIEW
Materials
Wing Analysis
Tail Boom Sizing
C-G Determination
Landing Gear
TEAM4Material 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: - www.matweb.com- US Dept. of Agriculture
TEAM4
Wing Analysis
ProcedureCalculated sectional lift coefficientEvaluated sectional wing bending momentSized I-beam to desired proportionsTrade Study
Minimize material weight Maximize stress loading capacity
Selected most suitable material and thickness
TEAM4
Wing Analysis
0 1 2 3 4 5 6 7 80
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6Lift Coefficient Distribution, PT40
Half-Spanwise Position (ft)
Se
cti
on
al
Lif
t C
oe
ffic
ien
t, C l
0 1 2 3 4 5 6 7 80
100
200
300
400
500
600Bending Moment Distribution, PT40
Half-Spanwise Position (ft)
Se
cti
on
al
Be
nd
ing
Mo
me
nt
(lb
f)
Root Bending Moment = 508.5 ft-lbf
Based on lifting line theory Actual bending moment at each point along spar
TEAM4
Wing Analysis
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5
Airfoil & I-Beam Spar Profile
Unit Spanwise Position
Un
it H
eig
htw
ise
Po
sit
ion
Airfoil Profilet = 0.0618 ftt = 0.0927 ftt = 0.1236 ftt = 0.1545 ft
TEAM4
Wing Analysis
0 0.005 0.01 0.015 0.02 0.0250
500
1000
1500
2000
2500
3000Maximum Root Bending Moment versus Thickness
Web Thickness (ft)
Ro
ot
Be
nd
ing
Mo
me
nt
(ft.
lbf)
BasswoodSpruceBalsaAl-2024T6
0 0.005 0.01 0.015 0.02 0.0250
10
20
30
40
50
60
70
80
90Spar Weight versus Thickness
Web Thickness (ft)
Sp
ar
We
igh
t (l
bf)
BasswoodSpruceBalsaAl-2024T6
TEAM4
Wing Analysis
0 100 200 300 400 5000
5
10
15Spar Weight versus Maximum Root Bending Moment
Maximum Allowable Root Bending Moment (ft.lbf)
Sp
ar
We
igh
t (l
bf)
BasswoodSpruceBalsaAl-2024T6Design Point
508.5 ft-lbf
TEAM4
Wing Analysis
Single spar wing structure selection I-beam
Material: BALSA (Ochroma Pyramidale) 12% Height = 0.357 ft = 4.28 in Base = 0.216 ft = 2.59 in Thickness = 0.051 ft = 0.61 in Weight = 11.0 lbf
TEAM4
Tail Boom Sizing
Cylindrical tubesAvailability More efficient than solid rods
Used twist and deflection constraints Appropriately sized inner diameters Found corresponding outer diameters
TEAM4
Tail Boom Sizing
BE
PLI
3
3
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) : deflection of end of boom (in)
4 4
64 o iI d d
TEAM4
Tail Boom Sizing Equation for Twist
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)
GJ
TL
trJ 32
2o id d
r
2
o id dt
TEAM4
Tail Boom Sizing
Known Constants
Deflection P = 26.73 lbf
L = 5 ft E = 10500 ksi set = 2 in
Twist T = 15 ft-lbf
L = 5 ft G = 3920 ksi set = 5 deg
= 0.0873 rad
TEAM4
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
TEAM4
Tail Boom Sizing
0 0.5 1 1.5 2 2.5 3 3.5 40
0.2
0.4
0.6
0.8
1
1.2
1.4Tail Boom Thickness vs. Inner Diameter
thic
kn
es
s (
in)
inner diameter (in)
Al 2024-T6SprussBasswoodBalsa
0 0.5 1 1.5 2 2.5 3 3.5 40
2
4
6
8
10
12Tail Boom Weight vs. Inner Diameter
inner diameter (in)
we
igh
t (l
bs
)
Al 2024-T6SprussBasswoodBalsa
34 4 64
3o iB
PLd d
E
2 2
4o id d
W L
TEAM4C.G. LOCATION ESTIMATION
Avionics PodW = 20 lbx = -1.44 ftz = - 0.58 ft
Engines, Fuel,Casings
W = 12.72 lbx = -0.3 ftz = -0.5 ft
WingW = 12.04 lb
x = 1.55 ftz = 0 ft
Tail BoomsW = 5.94 lbx = 4.05 ft
z = 0 ft
Tail SectionW = 2.3 lbx = 8.23 ft
z = 0.075 ft
This figure shows the approximate weights and C.G. locations of the main components:
NOT TO SCALE
Main GearW = 3 lbx = 0 ft
z = -1.25 ft
Tail GearW = 0.5 lb
x = 8 ftz = -0.21ft
x
z
TEAM4C.G. LOCATION ESTIMATION
Total Weight: W = 54.5 lb C.G. Location: x = 0.47 ft, z = -0.38 ft Wing M.A.C.: x = 0.775 ft Static Margin: SM = 10.0%
LIFT
WEIGHT
SM = – (xCG – xMAC) / cNOT TO SCALE
x
z
TEAM4TAILDRAGGER LANDING GEAR CONSTRAINTS
NOT TO SCALE
RAYMER 11.2
18.80 deg. (16 - 25 deg)
10.04 deg. (10 - 15 deg)
3.1 ft
0.47 ft
0.38 ft
8 ft
1.42 ft1.35 ft
Represents C.G. location
ZX
TEAM4WEIGHT DISTRIBUTION
NOT TO SCALE
W = 54.5 lbf FB
FA
y = 7.43 ftx = 0.70 ft
FA =Wy
x + y
FB =Wx
x + y
∑MB = 0
∑MA = 0
=
=
49.81 lbf
4.68 lbf
Center of Gravity
Tail Gear
MainGear
91% of weight carried by main gear
9% by tail gear
TEAM4FOLLOW-UP ACTIONS
Torsion constraint on spar
Geometry of wing ribs
Geometric layout of tail
Moments and products of inertia
AAE 451
QUESTIONS?