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1 CADES Proprietary Confidential Design of Front and Rear Spars for The Trainer Aircraft Wing.

44848151 WING SPAR Presentation

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Page 1: 44848151 WING SPAR Presentation

1CADES Proprietary Confidential

Design of Front and Rear Spars for The

Trainer Aircraft Wing.

Page 2: 44848151 WING SPAR Presentation

2CADES Proprietary Confidential

TEAM

Team Members :Akshay A. Pradeep S. Shet

Pavan Kumar N. R.Raghunandan M.Lakshmana H. B.Chetan A. V.

Guide : Mr. H. N. Athavale

Co-ordinator : Mr. Umanath Nayak

CAE

CAD

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3CADES Proprietary Confidential

OBJECTIVE

CAD

To generate the CAD model of wing using the available data and

prepare the assembly of all components

CAE

Determine the Spar locations with respect to chord length.

Determine the dimensions for flange and web of the spars.

Estimate the number of ribs and their positioning

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4CADES Proprietary Confidential

SCOPE OF THE PROJECT

Estimation of spar position.

Dimension calculations of front and rear spars.

Calculations for number of ribs and their positions.

CAD

Profile creation of the wing using the given NACA standards.

Creation of the wing geometry

Use available data to develop CAD models for each individual component

Prepare an assembly of all components using CATIA

CAE

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5CADES Proprietary Confidential

Root chord : 2400 mmTip chord : 700 mmSemi Span length : 5500 mmExposed Span : 4750 mmAirfoil (root) : NACA 64A

1215

(tip) : NACA 64A1210

Aircraft weight : 14000 NLift Load : 6gDesign Factor : 1.5Given Spar Position(in % of chord length)

Front Spar : 18-25Rear Spar : 62-70

INPUT

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DERIVED INPUT

●Limit load : 14000 * 6= 84000 N

●Design Load : 84000 * 1.5= 126000 N

●Load on semi-span : 126000 / 2= 63000 N

●Exposed wing area : 7.3625 E6 mm2

●Pressure load on wing : 63000 / 7.3625 E6 = 8556.87 E-6 N/mm2

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WING GEOMETRY

ALL DIMENSIONS ARE IN mm

RO

OT

CH

OR

D SWEEP AT ¼ CHORD

4750

700

2400

TIP

CH

OR

D

LEADING EDGE

TRAILING EDGE

Top View [RH]

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8CADES Proprietary Confidential

AIRFOIL

Generate the aerofoil section using the Coordinates of NACA 64A

1215 and NACA 64A

1210.

[source : http://www.pdas.com/sections6a.htm]

Generate the CAD model of the wing using CATIA- V5.

Aerofoil at Root NACA 64A1215Aerofoil at Tip NACA 64A

1210

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DESIGN PROCEDURE

Calculation of the Shear force, Bending moment & Torsion for the

given load.

Calculation of load distribution between the front and rear spar.

Estimation of spar positions.

Generation of CAD Model and Drafting.

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10CADES Proprietary Confidential

DESIGN PROCEDURE

Divide the wing area into number of divisions.

Calculate the chord length at each section.

Determine the C.G of each area.

Calculate the shear force, bending moment and Torque at the respective

sections.

Shear force =pressure*area.

Bending moment=shear force*CG distance.

Torque = Shear force*Distance b/w CG and CP.

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METHODS AND METHODOLOGY

A10

A9

A8

A7

A6

A5

A4

A3

A2

A1

475

2400

700

ALL DIMENSIONS IN mm

L1

L2

L9

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12CADES Proprietary Confidential

Chord Length, L1= L

root-((L

root-L

tip) / S) * x

At section 2, L1 = 2400-((2400-700)/4750)*4275

L1 = 870 mm

Area of Trapezium, A1 = 0.5*(L

1+L

tip)*h

A1 = 0.5*(870+700)*475

A1 = 373 E3 mm2

CG of Trapezoid Section = h/3*((Ltip

+2L1)/(L

tip+L

1))

CG=475/3*((700+2*870)/(700+870)) CG = 246 mm from L

tip

DESIGN PROCEDURE

S

xLroot

Ltip

L1

A1

h

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DESIGN PROCEDURE

Limit load = 84000 NDesign Load = Limit Load*Design factorDesign load on wing, = 84000*1.5 = 1,26,000 NDesign load on semi-span wing, = 63000 Npressure load on wing [P] = 8556.87 E-6 N/mm2

Load At Section 2, = P2+P

1 = P*A

2+P

1

= 8557 E-6 * 453625 + 3190.65 = 3881.6 + 3190.65 = 7072.25 N

Bending Moment At Section 2, M2 = P

2 * CG

2 + P

1 * (CG

1 + L

2)

M2 = 3881.6 * 230 + 3190.65 * (229 + 475)

M2 = 3248260 N-mm

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SHEAR FORCE

Root 0 475 950 1425 1900 2375 2850 3325 3800 4275 TIP 4750

0

10000

20000

30000

40000

50000

60000

7000063000

53590.65

44872.25

36844.85

29508.4

22862.9

16908.39

11644.857072.25

3190.653190.65

Shear force diagram for the wing

Wing span [Root to tip] [mm]

She

ar f

orce

[N

]

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BENDING MOMENT

ROOT 0

475 950 1425 1900 2375 2850 3325 3800 4275 TIP 4750

0

20000000

40000000

60000000

80000000

100000000

120000000

140000000123020000

95259000

71809000

52341000

36527000

2403900014548000

77278603248260781700 0

Bending moment diagram for the wing span

Wing span [root to tip] [mm]

Ben

ding

mom

ent [

N-m

m]

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LOAD DISTRIBUTION

Centre of Pressure, CP = 45% of Chord Length (C) from LE [870mm] [1]Front Spar Position = 25% of C from LE [217.5mm]Rear Spar Position = 62% of C from LE [539.4mm]

Chord Length 'C'

45% of C

25% of C a b

c

62% of C

RA R

B

FS RS

ChordCP

a=174mm

b=148mm

c=322mm

C=870mm

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Shear Force Distribution:

Shear Force on Front Spar, = Load * b/cAt Section 1, SF

FS = 3190.65 * (148/322)

SFFS

= 1465.974 N

Shear Force on Rear Spar SFRS

= 3190.65 - 1465.974 SF

RS = 1724.676 N

SF on Front Spar = 45.9% of total loadSF on Rear Spar = 54.1% of total load

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Bending Moment Distribution:

Moment is distributed in same ratio as that of the Shear force.

Bending Moment on Front Spar, M

FS = 0.459 * 781700

MFS

= 359159 N-mm

Bending Moment on Rear Spar, M

RS = 781700 - 359159

MRS

= 422541N-mm

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19CADES Proprietary Confidential

Front Spar

Rear Spar

SHEAR FORCE & BENDING MOMENT

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MATERIAL

Material : AA 2024-T6 Ultimate tensile strength, σ : 427 MPa Shear strength : 283MPa Density : 2.79 E-6 kg/mm3

Young's Modulus, E : 72400 Mpa Poisson's Ratio : 0.33

[Aluminum Association, Inc]. [7]

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Moment of Inertia:

I = M*y/σ Where, I = Moment of Inertia, in mm4

M = Bending Moment, in N-mmy = distance b/w neutral axis to top surface, in mmσ = Tensile strength, in MPa

Moment of Inertia on Front Spar, IFS

= 359159 * 52.8 / 427 I

FS = 44412 mm4

Moment of Inertia on Rear Spar, IRS

= 422541 * 43.44 / 427 I

RS = 42987 mm4

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MOMENT OF INERTIA

Front Spar

Rear Spar

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TORSION

Area of Torque Box, A1 = 30980.3 mm2

CG of Torque Box = 165 mm From Rear sparDistance Between CG & CP = 18.268 mmTorque, T = Load*d = 3190.65 * 18.268

T = 58286 N-mm

Shear flow, q1 = T/(2*A1) [2]

q1

= 58286 / (2 * 30980.3) q

1= 0.941 N/mm

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CG OF TORQUE BOX

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Torque

Shear Flow

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TORQUE DIAGRAM

ROOT 0

475 950 1425 1900 2375 2850 3325 3800 4275 TIP 4750

0

2000000

4000000

6000000

8000000

10000000

1200000011857039.54

8689789.08

6187429.48

4252608.34

2795550.39

1734041.9992888.78

506210.31212789.9958285.91 0

Torque diagram for the wing span

Wing span [root to tip] [mm]

Tor

que

[N-m

m]

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Shear force (SF) on Front SparSF

FS = q * h

FS

SFFS

= 0.941*105.6 = 99.34 NTotal SF on FS = 1465.974+99.34 = 1565.313 N

On Rear SparSF

RS = q*h

RS

SFRS

= 0.941*86.88SF

RS = 81.729 N

Total SF on RS = 1724.676+81.729 = 1806.405 N

SHEAR FORCE DUE TO TORSION

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SHEAR FORCE DUE TO TORSION

Front Spar

Rear Spar

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TOTAL SHEAR FORCE

Front Spar

Rear Spar

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WEB THICKNESS

Thickness of the Web can be calculated from the following formula,

ح shear strength

= SFFS

/ A web

Where,

ح shear strength

= Shear strength of the material AA 2024-T6 in MPa

A web

= Area of the web = (height * thickness) in mm

283 = 1565.313 / (105.602 * t web

)

t web = 0.052 mm

Area of the web = height * thickness = 105.602 * 0.052

A web

= 5.531 mm2

Moment of Inertia of Web:Moment of Inertia of a rectangular section web is given by,

I web

= t web * (hFS

)3 / 12I

web = 0.052 * (105.602)3 / 12I

web = 5140.175 mm4

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Front Spar

Rear Spar

WEB

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FLANGE

MOIflange

= MOIFront Spar

- MOIWeb

I flange

= IFS

- Iweb

= 44411 - 5140.175

I flange

= 39270.825 mm4

Also Moment of Inertia of the flange is given by,

I flange

= Aflange

* (yFS

)2

Where, Iflange

= Moment of Inertia of flange in mm4

yFS

= height from neutral axis to top surface of the flange in mm

Hence, Aflange

= Iflange

/ (yFS

)2

= 39270.825 / (52.801)2

Aflange

= 14.086 mm2

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FLANGE

Front Spar

Rear Spar

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MASS CALCULATIONS

AFS

= Aflange

+ Aweb

AFS

= 14.09 + 5.53 = 19.62 mm2

VFS

= AFS

* 475 = 19.62 * 475

VFS

= 9318.3 mm3

Mass = Density * Total Volume

= 2.78 E-6 * 4218551.12

Mass = 11.73 kg

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62 63 64 65 66 67 68 69 7011.5

12

12.5

13

13.5

14

12.5042471206507

12.6220245652181

12.7465877815849

12.8783982887878

13.0179753996257

13.1658829246678

13.3227251692486

13.4892202653604

13.6661526949271

12.3881208913364

12.4999845046666

12.6186071735573

12.7444287056342

12.8779461219943

13.0197004118034

13.1702724904167

13.3303551420369

13.5007061649515

12.274714489472612.3808256430427

12.4936607138559

12.6136374240938

12.7412301397201

12.876956676848

13.0213742183015

13.1751498097018

13.3390141564684

12.163512014196312.2640129538278

12.3711945626529

12.4854520552158

12.6072367354625

12.7370428555719

12.8754034876527

13.0229595119557

13.1804140972234

12.053900196057912.1489117223174

12.2505530055126

12.3591962678982

12.4752692884713

12.5992423142542

12.7316238829751

12.8730282376668

13.0241305238174

11.945183922182312.0348041336471

12.130996070132712.2341084011342

12.344544850195

12.4627511576641

12.5892108491091

12.7245110055869

12.8719649469388

11.836630462490411.9209354659232

12.011747601955712.1093913423994

12.2142457502486

12.3267314916976

12.4473065436803

12.5945353217656

12.7744354934928

11.727572120147911.806620573670211.8921053901977

11.984326119573712.0836364901705

12.1904314665371

12.339144756082

12.5065546355284

12.6799534484769

1819202122232425

Rear spar position in %

Mas

s [k

g]

MASS CALCULATIONS

Fro

nt

spar

po

siti

on

Hence, from the Calculations it is found that (25% - 62%) combination of Spar Position was found suitable. The Mass of this combination is 11.73 Kg which is least than any other combinations

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BUCKLING

To Check whether the web fails under shear buckling.

Condition: Shear stressinduced

< Buckling stress (safe design)

●The thickness calculation is based on iterations,

Finduced

= q / tweb

Fcritical

= k*E*(tweb

/ b)2

where, q = shear flow, in N/mm

E = Young's Modulus, in MPa

b = height of spar, in mm

tweb

= web thickness, in mm

k = shear buckling coefficient from graph

[4]

[4]

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BUCKLING CALCULATIONSITERATION 1. RIB SPACING FOR EQUAL DISTANCE OF 475mm

Web thickness's of front spar at section 1 is as follows,F

induced = q

1 / t

web------------ (1)

= 0.941 / 0.052 F

induced = 18.09 N/mm2

Fallowable

= K * E * (t web

/ b)2-----------(2)18.09 = 5 * 72400 * (t

web / 105.602)2

The value calculated for tweb

is re substituted in Eqn.(1) and this loop will continue till we get equal consecutive thickness.

Hence, the thickness of the web is 0.30 mm at section 1. Same calculations were repeated for all sections of front spar to optimize the web thickness

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Front Spar

Rear Spar

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MASS CALCULATION

●Web design is safe under buckling.

●From buckling calculation the total mass of the spars is 16.14 kg.

●By this, mass of the spars got increased by 4.41 kg.

●To decrease the mass, one more iteration has been carried out.

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ITERATION-2● For optimum Rib spacing, (a/b) ratio >= 1

Rib no. Rib dist. From root Spar heights (a/b) ratioK from graph

Web thickness Web volumeFS RS FS RS FS RS FS RS

[mm] [mm] [mm] [mm] [mm]0 TIP4750 64.49 54.05 0 0.00 - - - -1 4440 81.97 68.4 3.78 4.53 5.10 5.00 0.22 0.2 5590.56 4240.82 4110 100.58 83.67 3.28 3.94 5.17 5.08 0.34 0.3 11285.3 8283.533 3780 119.19 98.94 2.77 3.34 5.30 5.15 0.44 0.39 17306.53 12734.094 3450 137.8 114.22 2.39 2.89 5.50 5.20 0.53 0.48 24101.05 18091.815 3120 156.41 129.49 2.11 2.55 5.75 5.40 0.64 0.57 33033.37 24356.696 2790 175.02 144.76 1.89 2.28 6.00 5.60 0.74 0.67 42739.15 32006.447 2470 193.06 159.57 1.66 2.01 6.30 5.80 0.84 0.76 51894.8 38807.188 2150 211.11 174.38 1.52 1.84 6.55 6.20 0.94 0.84 63500.68 46872.819 1830 229.15 189.19 1.40 1.69 6.90 6.25 1.03 0.94 75528.17 56907.75

10 1520 246.63 203.53 1.26 1.52 7.25 6.55 1.12 1.02 85630.63 64357.4511 1210 264.11 217.88 1.17 1.42 7.60 6.80 1.2 1.1 98250.04 74297.4212 900 281.59 232.23 1.10 1.33 7.80 7.00 1.31 1.19 114355.32 85668.5413 600 298.51 246.11 1.00 1.22 8.20 7.35 1.39 1.26 124479.09 93029.9614 300 315.43 259.99 1.05 1.15 8.00 7.60 1.51 1.35 142888.88 105297.5715 Root 0 332.35 273.88 1.11 1.10 7.80 7.80 1.63 1.43 162519.15 117494.52

Total volume 1053102.72 782446.56Web volume 1835549.28

[mm3] [mm3]FS RS

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WEIGHT CALCULATION

●Finally mass of the spars reduced by 0.89 kg when compared to 1st iteration.

● These dimensions are taken for modelling

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RESULTS AND DISCUSSION

Root 0 300 600 900 1210 1520 1830 2150 2470 2790 3120 3450 3780 4110 4440 47500.00

0.40

0.80

1.20

1.60

2.00

0.386 0.362 0.34 0.318 0.291 0.27 0.245 0.22 0.198 0.17 0.145 0.12 0.095 0.07 0.042

1.63

1.51

1.391.31

1.21.12

1.030.94

0.840.74

0.64

0.530.44

0.34

0.22

WEB THICKNESS FOR FRONT SPAR

ACTUALFROM BUCKLING

FROM ROOT TO TIP [mm]

TH

ICK

NE

SS

OF

WE

B[m

m]

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Root 0 300 600 900 1210 1520 1830 2150 2470 2790 3120 3450 3780 4110 4440 47500.00

0.40

0.80

1.20

1.60

2.00

0.517 0.488 0.455 0.425 0.39 0.361 0.327 0.295 0.261 0.23 0.195 0.16 0.129 0.095 0.06

1.431.35

1.261.19

1.11.02

0.940.84

0.760.67

0.570.48

0.390.3

0.2

WEB THICKNESS FOR REAR SPAR

ACTUALFROM BUCKLING

FROM ROOT TO TIP

TH

ICK

NE

SS

OF

WE

B[m

m]

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CONCLUSION

● Front Spar positioning is estimated to 25% and Rear Spar to 62% of the

Chord Length.

● Flange and web dimensions are calculated and suitable changes in

dimensions are incorporated from manufacturing point of view.

● Number of Ribs and their positioning for the prevention of bending and

buckling of Spars is calculated.

● Mass of the spars calculated from iterations is 15.25 kg.

● The Detail drawings for the front and rear spars are provided using CATIA V5.

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SCOPE FOR FURTHER WORK

● Spar position can be optimized based on buckling calculations.

● Further optimization of Rib is possible.--Varying number of Ribs and spacing of Ribs.

● Use of other materials for the design of spars can be thought of.

● Detail stress analysis of individual components and its validation with calculations can be carried out.

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CAD

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Taking values from NACA Standards

At Root: Profile: NACA 64A1215.

Leading Edge radius = 1.556% c.

Slope of mean line at leading edge = 0.0842.

At Tip: Profile: NACA 64A1210.

Leading Edge radius = 0.701% c.

Slope of mean line at leading edge = 0.0842.

CAD MODELING OF THE WING SPAR

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1. Generation of the profiles at the root and tip using the NACA profiles.

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● INCORPORATING THE LEADING EDGE RADIUS AS SPECIFIED IN THE PROFILE STANDARD.

1.Giving the slope in the sketcher mode

2.Creating the arc of the required dimension coming out of sketcher.

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● Using the connect curve option to join the leading edge radius and the aerofoil profile.

● Create the surface using multi section surface option.

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INTERSECTION OF THE PROFILES

● Creating the planes at the four sections at ½, ¼, ¾ of the span of the wing.● Intersecting the lofted surface on the planes creating unique sketches on them.

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ANGLE OF ATTACK

● Create a point at the quarter chord and draw a line for reference.

● Rotate the intersected profiles as 0.60 at the quarter, 1.10 at mid span, 1.60 at three

fourths and 20 at the tip.

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By considering the profiles generated with angle of attack at different sections,

the wing surface is created using multi-section surface option.

CREATE THE SURFACE USING MULTI SECTION SURFACE OPTION

Thus the surface is created as per the requirements incorporating all the necessary data.

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CREATION OF REFERENCE AEROFOIL SECTIONS

● 15 planes are created at rib positions along the wing span.● The intersections created are used as the reference for the creation of the spar.

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CONSIDERATIONS MADE DURING THE DESIGN OF SPAR

ELEMENTS

●The maintenance of the nose box is made easy.

● The front spar is I – section.

● The rear spar is C – section.

● Minimum distance required for a single row riveting is kept as 15 mm.

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DESIGNING OF SPAR ON MANUFACTURING BASIS

➢ The front spar is placed at 25% of chord length from leading edge.

➢ The rear spar is placed at 62% of chord length from leading edge.

➢ Thicknesses of the flanges and webs are different.

➢ The flanges are made of T-sections and L- sections.

➢ The webs are made with sheet metal.

➢ The thicknesses are optimized based on the availability of the standard gages of sheet metal.

➢ The final assembly of elements can be fastened with rivets.

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CROSS SECTION SPAR

Skin area, As = (b +2*20*t

s) mm2

Effective flange area = (Af- As)/2

Web thickness is altered as per the availability of sheet metal gages.

where , b= flange width in mm ts =skin thickness in mm Af =designed flange area in mm2

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Rib no. Dist. From root Flange Width Skin Area Available area Flange Thickness Effective Flange area Web thicknessFrom root (mm) (mm) (mm) (mm)

1 Root 0 70 220 266.94 3.81 266.94 1.632 300 70 220 215 3.07 215 1.633 600 70 220 175 2.5 175 1.634 900 65 210 150 2.31 150 1.295 1210 65 155.2 144.9 2.23 144.9 1.296 1520 60 147.2 116.4 2 120 1.297 1830 60 147.2 86.4 2 120 1.298 2150 55 139.2 60.4 2 110 0.919 2470 50 88.8 55.6 2 100 0.91

10 2790 45 82.8 36.1 2 90 0.9111 3120 40 76.8 16.6 2 80 0.6412 3450 35 70.8 -0.4 2 70 0.6413 3780 30 64.8 -12.4 2 60 0.6414 4110 30 64.8 -19.9 2 60 0.6415 4440 30 64.8 -24.9 2 60 0.6416 NO RIB 4750

(mm2) (mm2) (mm2)

FRONT SPAR DIMENSIONS

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REAR SPAR DIMENSIONS

Rib no. Dist. From root Flange Width Skin Area Available area Flange Thickness Effective Flange area Web thicknessFrom root (mm) (mm) (mm) (mm)

1 Root 0 90 260 415.01 4.61 415.01 1.452 300 90 260 340 3.78 340 1.453 600 80 240 295 3.69 295 1.454 900 75 230 257.5 3.43 257.5 1.155 1210 70 204 223 3.19 223 1.156 1520 65 194 180.5 2.78 180.5 1.157 1830 60 184 138 2.3 138 1.158 2150 55 174 100.5 2 110 0.919 2470 50 148 73.5 2 100 0.91

10 2790 45 138 48.5 2 90 0.9111 3120 40 128 21 2 80 0.6412 3450 35 118 4 2 70 0.6413 3780 30 108 -19 2 60 0.6414 4110 30 108 -36.5 2 60 0.6415 4440 30 108 -43 2 60 0.6416 NO RIB 4750

(mm2) (mm2) (mm2)

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CREATION OF THE SPAR SECTIONS

1. Two T sections for the flange, and web section for the front spar.2. Two L sections for the flange, and web section for the rear spar.

FRONT SPAR REAR SPAR FULL PROFILE

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GENERATING SPAR USINGDIFFERENT SECTIONS

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CRIMP HOLES OR LIGHTENING HOLES

The lightening holes are made in the element in order to reduce the weight of the element. the crimp holes are made to the web element of the spar. These holes provided in between the two successive rib locations.

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SPAR WITH LIGHTENING HOLES

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REPRESENTATION OF RIVET HOLES

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FINAL SPAR ASSEMBLY

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BIBLIOGRAPHY

1] Abbot & Albert,'Theory of wing sections',Dover publication,1949.

2] David J. Perry,'Aircraft structures',Mc-Graw Hill publication,1950.

3] E. F. Bruhn,'Analysis and design of flight vehicle structures',1973.

4] Michael C. Y. Niu, 'Airframe Stress Analysis and Sizing', 2001.

5] Michael C. Y. Niu, 'Airframe structural design', Conmilit press Ltd., 1989.

6] Kuethe and Schetzer, 'Foundations of Aerodynamics', 2nd Edition, John Wiley

and Sons, New York, 1959.

7] ASM Material Data Sheet

8] MIL Handbook.

&CADES Library.

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THANK YOU

CADES Digitech Pvt. Ltd.Tel: +91 80 4193 9000Fax: +91 80 4193 9099

URL: www.cadestech.com