STRUCTURES & WEIGHTS PDR 1

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?