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
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AAE 451
STRUCTURES & WEIGHTS PDR 1
TEAM 4Jared Hutter, Andrew Faust, Matt Bagg, Tony Bradford,Arun Padmanabhan, Gerald Lo, Kelvin Seah
October 28, 2003
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TEAM4OVERVIEW
Materials
Wing Analysis
Tail Boom Sizing
C-G Determination
Landing Gear
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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
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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
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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
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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
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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
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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
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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
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TEAM4
Tail Boom Sizing
Cylindrical tubesAvailability More efficient than solid rods
Used twist and deflection constraints Appropriately sized inner diameters Found corresponding outer diameters
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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
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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
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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
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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
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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
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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
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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
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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
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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
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TEAM4FOLLOW-UP ACTIONS
Torsion constraint on spar
Geometry of wing ribs
Geometric layout of tail
Moments and products of inertia
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AAE 451
QUESTIONS?