1

AIRCRAFT DESIGN PROJECT - II

DESIGN PROJECT OF HUNDRED SEATER TWIN

ENGINE PASSENGER AIRCRAFT

A PROJECT REPORT

Submitted by

ABDU SSALAM.P (32208101001)

GUNASEKARAN.S (32208101019)

LOKESHWARAN.VJ (32208101027)

MIDHUN DAS.AM (32208101032)

PRATHEESH.CP (32208101041)

In partial fulfillment for the awards of the degree

Of

BACHELOR OF ENGINEERING

IN

AERONAUTICAL ENGINEERING

DHANALAKSHMI SRINIVASAN COLLEGE OF ENGINEERING

& TECHNOLOGY, CHENNAI.

ANNA UNIVERSITY: CHENNAI 600 025

APRIL 2011

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BONAFIDE CERTIFICATE

Certified that this report “A DESIGN PROJECT OF HUNDRED SEATER

TWIN ENGINE PASSENGER AIRCRAFT’’ is the bonafide work of

ABDUSSALAM.P (32208101001) who carried out the project work under my

supervision.

Internal Examiner External Examiner

SIGNATURE

G.SARAVANAN, M.E., PhD,

HEAD OF THE DEPARTMENT

AERONAUTICAL ENGINEERING,

DHANALAKSHMI SRINIVASAN

COLLEGE OF ENGINEERING &

TECHNOLOGY,CHENNAI.

SIGNATURE

J.P. RAMESH, M. E.,

LECTURER,

AERONAUTICAL ENGINEERING,

DHANALAKSHMI SRINIVASAN

COLLEGE OF ENGINEERING &

TECHNOLOGY,CHENNAI.

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ACKNOWLEDGEMENT

This report for the Design of HUNDRED SEATER TWIN ENGINE

PASSENGER AIRCRAFT is prepared on the basis of Anna University

Syllabus. This is prepared by references attached in this report.

For getting interested in this subject and nurturing my knowledge base, I would

like to thank my beloved teachers. Mr. Saravanan.G, Head of the Department

and Mr.Ramesh.J.P, lecture who deserve all credit

Last but not least, I am thankful to all of my Department staffs.

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TABLE OF CONTENTS

ABSTRACT 6

SYMBOLS USED 7

INTRODUCTION 8

SPECIFICATIONS 10

1. V N DIAGRAM 11

MANUVERING ENVELOPE

GRAPH

2. GUST ENVELOPE 16

FAA SPECIFICATIONS

GRAPH

3. STRUCTURA L DESIGN APPROACH 19

THEORY

SCHRENKS CURVE

4. LOADING PERFORMANCE & CALCULATIONS 23

SHEAR FORCE & BENDING MOMENT - TRANSVERSE

GRAPHS

SHEAR FORCE & BENDING MOMENT – CHORDWISE

GRAPHS

5

5. CRITICAL FLIGHT CONDITIONS 35

SHEAR FORCE & BENDING MOMENT -NORMAL

GRAPHS

SHEAR FORCE & BENDING MOMENT – CHORDWISE

GRAPHS

6. MATERIAL SELECTION 42

PROPERTIES OF THE MATERIAL

7. DETAILED WING DESIGN 48

SPAR DESIGN

FRONT SPAR

REAR SPAR

8. FUSELAGE DESIGN 51

SYMMETRIC FLIGHT CONDITION

SHEAR FORCE GRAPH

BENDING MOMENT GRAPH

SHEARFLOW GRAPH

9. THREE VIEWS 60

10. SOFTWARE ANALYSIS 51

PRESSURE PLOT

VELOCITY PLOT

11. CONCULSION 64

12. REFERENCES 65

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ABSTRACT

All of the airliners aim at building an aircraft with large

capacity and long range at a higher velocity and with low fuel consumption. Our

project conceptualizes this aim. So in our aircraft design project we have

concentrated on a100seater passenger aircraft with twin engine which can travel

at a cruise Mach number of 0.66 and a minimum range of4500km at an

optimum altitude. For the propulsion system we have chosen an existing engine

for reference. Historic data is being used wherever necessary to make our

project more precise

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LIST OF SYMBOLS

W0 Overall weight

Wf Weight of fuel

S Wing area

b Wing span

T/W Thrust loading

W/S Wing loading

A.R Aspect ratio

cr, ct Chord length of wing root & tip

h1, h2 Thickness of front &rear spar

CL Coefficient of lift

CL,max Maximum coefficient of lift

(L/D)max Maximum lift – drag ratio

Vdiv Dive velocity

Vcruise Cruise velocity

Vs Stall velocity

K Gust alleviation factor

µ Airplane mass ratio

M Lift curve slope

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Umax Maximum gust velocity

k Proportionality factor

Cn Co-eff of forces along normal

direction

Cc Co-eff of forces along chord wise

direction

Mfr, Mr Bending moment taken up by front

& rear spar

Afr, Ar Area of front & rear spar

τult Ultimate shear stress

FOS Factor of safety

E Youngs modulus

υ Poisson ratio

V Shear force at the spar

I Moment of inertia

q Shear flow

α 0 Zero lift angle of attack

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INTRODUCTION

Overview:

The structural design of an airplane actually begins with the flight

envelope or V-n diagram, which clearly limits the maximum load factors that

the airplane can withstand at any particular flight velocity. However in normal

practice the airplane might experience loads that are much higher than the

design loads. Some of the factors that lead to the structural overload of an

airplane are high gust velocities, sudden movements of the controls, fatigue load

in some cases, bird strikes or lightning strikes. So to add some inherent ability

to withstand these rare but large loads, a safety factor of 1.5 is provided during

the structural design.

The two major members that need to be considered for the structural

design of an airplane are wings and the fuselage. As far as the wing design is

concerned, the most significant load is the bending load. So the primary load

carrying member in the wing structure is the spar (the front and rear spars)

whose cross section is an „I‟ section. Apart from the spars to take the bending

loads, suitable stringers need to take the shear loads acting on the wings.

Unlike the wing, which is subjected to mainly unsymmetrical load, the

fuselage is much simpler for structural analysis due to its symmetrical crossing

and symmetrical loading. The main load in the case of fuselage is the shear load

because the load acting on the wing is transferred to the fuselage skin in the

form of shear only. The structural design of both wing and fuselage begin with

shear force and bending moment diagrams for the respective members. The

maximum bending stress produced in each of them is checked to be less than

the yield stress of the material chosen for the respective member.

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SPECIFICATIONS

Take-off gross weight= 548516N

Fuel fraction= 0.264

Payload fraction= 0.186

Empty Weight Ratio=0.55

Structural Weight fraction of

Wings=0.1

Fuselage=0.1

Horizontal Tail=0.04

Vertical Tail=0.2

Cruise Velocity= 225m/s

Cruise altitude= 10000m

Wing area= 99.78 m2

Wing span= 31.59m

Cruise CL= 0.0536; Density at 10000 m=0.412 kg/m3

Chord length at

Wing root= 4.21 m

Wing tip= 2.10 m

Sweepback angle= 25o

CLmax = 1.41; CLmax = -0.91

Cdo = 0.0387; k = 0.864; T/W = 0.34

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1. V-n DIAGRAM

MANEUVERING ENVELOPE:

It is the graph between velocity (v) and load factor (n). It is

mainly used to know the structural damage when it is going at different

angle of attack

Load factor

It is the ratio between lift to weight

ie, nmax = (L/D)max(T/W)max

When the angle of attack is increasing, the lift will increase. The maximum load factor will get when the lift is maximum

Positive limit load factor

It is the limit for the load factor ie, beyond this limits the load factor should

not increase when the angle of attack is increasing positively. If it increases,

structural damage will occur

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Stall velocity

Stalls depend only on angle of attack, not airspeed. Because a

correlation with airspeed exists, however, a "stall speed" is usually used in

practice. It is the speed below which the airplane cannot create enough lift to

sustain its weight in 1g flight. In steady, un accelerated (1g) flight, the faster

an airplane goes, the less angle of attack it needs to hold the airplane up (i.e.,

to produce lift equal to weight). As the airplane slows down, it needs to

increase angle of attack to create the same lift (equal to weight). As the

speed slows further, at some point the angle of attack will be equal to the

critical (stall) angle of attack. This speed is called the "stall speed". The

angle of attack cannot be increased to get more lift at this point and so

slowing below the stall speed will result in a descent. And so, airspeed is

often used as an indirect indicator of approaching stall conditions. The stall

speed will vary depending on the airplane's weight, altitude and

configuration (flap setting, etc.).

From the figure, it is clear that for a particular velocity, it is not possible to

fly at a value of CL higher than the CLmax corresponding to that velocity. If

we wish to increase the lift of the airplane to that value of CLmax, then we

should increase the flying speed of the airplane.

We know that,

The maximum positive limit load factor is given by the formula

nmax = (L/D)max(T/W)max

(L/D)max = 1/[(4*0.0374*0.0387)^0.5] = 13.1

nmax= 4.45

Dive velocity is the highest velocity attainable theoretically given by the

formula

Vdiv = 1.5Vcruise = 1.5*225 = 337.5 m/s

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Consider Curve OA:

Point A: Corner velocity It is the velocity corresponding to the condition of

smallest possible turn radius and largest possible turn rate. It is the corner

point on the V-n diagram where the structural and aerodynamic boundaries

meet

So we can find out the corner velocity as

VA = (2*nmax*(W/S))/ ρCLmax)^0.5 = 92.6m/s

nA= nmax= 4.45

The load factor along the curve OA is given by the expression

n=L/W = CL*0.5*V2 *ρ /(W/S)

n=5.186*10-4

*V2

So this OA curve will get by giving some different velocity to above

equation. From that we can see the Corner velocity

Therefore the OA curve in the V-n diagram is drawn

Velocity (m/s) Load Factor (n)

0 0

10 0.05184

20 0.207

30 0.466

40 0.829

50 1.296

60 1.86

70 2.540

80 3.31

90 4.19

92 4.38

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Consider Curve AB:

The velocity at point B is given by the formula

VB= {{[(T/W)*(W/S)]+[(W/S)*((T/W)2-(4*k*CDo))

0.5]}/ρ*CDo}

0.5

VB= 125.69 m/s

nB = nA

A straight line is drawn to join the points A and B

Consider Curve BC:

The velocity at point C is given by VC=1.5*VB= 188.4 m/s

nC= 0.75*nA= 3.285

A straight line is used to join the points Band C

Consider Line CD:

A vertical line is drawn from the point C to meet the V-axis at point D where

D corresponds to zero load factor

Consider Curve DE:

VE= VB

nE= -1.8 (for a semi-aerobatic aircraft)

A straight line is drawn to join the points D and E

Consider Curve OF:

The negative load factor along the curve OF is given by the expression

n=L/W = CL*0.5*V2 *ρ /(W/S)

n=-3.34*10-4

*V2

Substituting for various values of velocity, the load factor values can be

obtained as follows

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Therefore the OF curve in the V-n diagram is drawn

Now, F is the point on the curve where the load factor is -1.8

Hence, VF= 73.32 m/s

Velocity (m/s) Load Factor (n)

10 -0.0334

20 -0.133

30 -0.30

40 -0.5

50 -0.835

60 -1.202

70 -1.636

73.32 -1.8

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2.GUST ENVELOPE

Gust

Gust is an rough air, when encountered should reduce the aircraft‟s speed.

Gust load factors increase with increasing airspeed, and the strength used

for design purposes usually corresponds to the highest level flight speed.

The effect of turbulence gust is to produce a short time change in the

effective angle of attack. These changes produce a variation in lift and

thereby load factor

The increase in the load factor due to the gust can be calculated by

n+ve = 1+ KUmax ρVm/2(W/S) (for curve above V-axis)

n-ve = 1- KUmax ρVm/2(W/S) (for curve below V-axis)

where , ρ = density at sea level(kg/m3)

K= gust alleviation factor

K can be calculated by K=0.88 µ /(5.3+ µ)

where µ is airplane mass ratio given by µ=2(W/S)gρcm

where

c=mean aerodynamic chord (3.162 m)

m= lift curve slope (0.1)

g= gravitational constant (9.81m/s2)

Umax= max. gust velocity

Using the formulae,

µ = 295.95 and

K= 0.864

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From the manual available,

Points Umax(m/s)

A‟ and F‟ 16

B‟ and E‟ 12

C‟ and D‟ 7

We know that ,

The velocities at points B‟,C‟, D‟, E‟ are the same as those at points B,

C, D, E in the Maneuvering envelope

The load factors at the various points can be found using the formula using

the corresponding values of Umax

n A’ = 2.395

n F’= -0.3950

n B’ = 2.419

n E’ = -0.419

n C’ = 2.241

n D’ = -0.24

CURVE OA’:

The positive load factor along the curve OA‟ is given by the equation

n = L/W = CL*0.5*V2 *ρ /(W/S)

n = 1.535*10-3

* V2

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The velocity at A is that corresponding to nA‟; Hence, VA‟=39 m/s

The curve OA‟ is now drawn

Since the velocities and load factors at B‟, C‟, D‟ and E‟ are known, straight

lines are used to join these points in sequence

LINE E’F’:

VF‟= VA‟; Now a straight line is used to join the points F‟ and E‟

Velocity (m/s) Load Factor

0 0

10 0.153

20 0.612

30 1.377

39 2.3

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3. STRUCTURAL DESIGN

Modern aircraft structures are designed using a semi-monocoque concept-

a basic load-carrying shell reinforced by frames and longerons in the bodies,

and a skin-stringer construction supported by spars and ribs in the surfaces

Proper stress levels, a very complex problem in highly redundant

structures, are calculated using versatile computer matrix methods to solve for

detailed internal loads. Modern finite element models of aircraft components

include tens-of-thousands of degrees-of-freedom and are used to determine the

required skin thicknesses to avoid excessive stress levels, deflections, strains, or

buckling.

The goals of detailed design are to reduce or eliminate stress

concentrations, residual stresses, fretting corrosion, hidden undetectable cracks,

or single failure causing component failure.

The structure of a pressurized fuselage which fulfills this criterion has to

guarantee neither that neither the crack in the skin becomes unstable nor that the

stiffeners perpendicular to the crack (i.e. the frames) fail statically. The two-

bay-crack criterion is the designing criterion for large areas in the upper and

side shells of the pressurized fuselage of medium and long range aircraft. These

aircraft types have lower design service goals in flights compared with short

range aircraft with the result that the fatigue and damage tolerance criteria have

less influence on the design. To limit the implications on the weight due to the

compliance with the two-bay-crack requirement following precautions are

possible:

selection of skin material with high residual strength

selection of frame material with high static strength

limitation of the allowable frame pitch

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SCHRENK’S CURVE:

We know that the Tip Stalling on the wing begins in the

region near the wing tips .This is because the distribution of local lift

coefficient(Cl) is not uniform along the span and as the angle of attack of the

wing increases, the stalling will begin at a location where the local lift

coefficient exceeds the value of maximum lift coefficient (Clmax) there .To

appreciate this phenomenon let us consider an un swept tapered wing. The lift

distribution on such a wing has a maximum at the root and goes to zero at the

tip. This distribution is also known as Γdistribution.

Further, the local lift (ΔL) can be equated to(1/2)ρV∞2 c Cl Δy,

where c is the local chord and Cl is the local lift coefficient over an element

(Δy) of span. Thus Γ distribution is proportional to the product c Cl .The local

lift coefficient (Cl) is proportional to Γ/c and is not uniform along the span .The

Γ distribution along the span can be approximately obtained by Schrenk‟s

method.

Lift is a component of the resultant aerodynamic force

acting at the centre of pressure of an aerodynamic chord, along a direction

perpendicular to the direction of the relative wind. At a particular altitude and at

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a specific angle of attack, Lift varies along the wing span due to the variation in

chord length along the span. Schrenk‟s curve defines this lift distribution over

the wing span of an aircraft. Since the wings of an aircraft are symmetrical

about the longitudinal axis, the Schrenk‟s curve for the starboard wing alone

can be obtained at first. This is given by

y= (y1+y2)/2

where y1= linear variation of lift along the wing semi-span

y2 =equivalent elliptic lift distribution along the wing semi-span

To find y1:

Lift force is found along the line joining the aerodynamic centers of

chords along the wing span. Hence, the wing is rotated about the wing root so

that the line joining the aerodynamic centers becomes the horizontal line.

a= (17.42/cos 25) = 17.42 m

Lift per unit length at wing root= CL*0.5*ρ*V2*CR

= .0536*0.5*0.412*2252*

4.21

= 2357.31 N/m

Lift per unit length at wing tip = CL*0.5*ρ*V2*Ct

= .0536*0.5*0.412*2252*2.10

= 1178.3 N/m

Equation of this line, y1=-67.5x + 2353.1

Area under y1= Total lift= 30795.25 N

2357.31N/m

1178.3N/

m

17.42

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To find y2:

Area of the above quarter ellipse, (π*a1*b1/4) =Area enclosed by

y1=30795.25N

a1= 17.42

Hence, b1=2251

Equation of this curve is (x/a1)2+(y/b1)

2=1

y2= 2248*((1-0.0574x2)^ 0.5)

Equation of Schrenk’s curve is given by

y= (y1+y2)/2

y=-67.5x + 2353.1+2248*((1-0.0574x2)^0.5)

Therefore, for the port wing, replace x by –x

y=67.5x + 2353.1+2248*((1-0.0574x2)^0.5)

Hence, substituting for different values of x, Lift force distribution can be

determined at the required span wise location.

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SCHRENK’S CURVE TABULATION:

Span wise Location Lift Force

-17.42 1177

-15 2485

-13 2974

-11 3356

-9 3673

-7 3942

-5 4172

-3 4368

-1 4533

0 4604

1 4533

3 4368

5 4172

7 3942

9 3673

11 3356

13 2974

15 2485

17.42 1177

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4. LOADING PERFORMANCE & CALCULATIONS

SHEAR FORCE AND BENDING MOMENT DIAGRAMS OF A WING

DUE TO LOADS IN TRANSVERSE DIRECTION AT CRUISE

CONDITION:

The expressions that are derived can be used for load on the

wing to calculate bending moment. The first step is to start by integrating total

load to

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Determine shear force: V(x) = - ∫ q r(x) dx. The bending moment can then be

calculated by integrating shear force: M(x) = ∫ V(x) dx.

The solution methods which follow Euler‟s beam bending theory

(σ/y=M/I=E/R) use the bending moment values to determine the stresses

developed at a particular section of the beam due to the combination of

aerodynamic and structural loads in the transverse direction. Most engineering

solution methods for structural mechanics problems (both exact and

approximate methods) use the shear force and bending moment equations to

determine the deflection and slope at a particular section of the beam.

Therefore, these equations are to be obtained as analytical expressions in terms

of span wise location. The bending moment produced here is about the

longitudinal (x) axis.

As both the wings are symmetric, let us consider the starboard

wing at first. There are three primary loads acting on a wing structure in

transverse direction which can cause considerable shear forces and bending

moments on it. They are as follows:

→ Lift force (given by Schrenk’s curve)

→ Self-weight of the wing

→ Weight of the power plant

LIFT FORCE:

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y= (y1+y2)/2=-67.5x + 2353.1+2248*((1-0.0574x2)^0.5)

where y1=-67.5x + 2353.1 [Trapezium]

y2= 2248*((1-0.0574x2)^0.5) [Ellipse]

This has already been explained under the topic “Schrenk‟s curve”

SELF WEIGHT OF THE WING:

The wing weight varies along the span as the chord length and thickness

decrease on moving from wing root to tip. Hence the spar cross-section should

also decrease from root to tip. This implies weight is higher at the root and it is

assumed to decrease parabolically to zero at the tip.

Weight of the wing= 0.05*W0= 27425.8N

∫k*(x-17.42)2 dx

= 4611.6667 → k=-15.63

y3= -15.63*(x-15.92)2

The negative sign implies downward load; since this will be taken care of

while finding shear force and bending moment, the sign is omitted.

yx=0 = 4743.0

POWERPLANT WEIGHT:

The power plant chosen for 100-seater aircraft is CFM56-5B5

Wpp= 0.03*W0= 23361.4 N

17.42

y0

y3=k*(x-(b/2))2

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The wing is fixed at one end and free at other end.

ΣFV=0

274258-27425.8-23361.4-VA=0; Hence, VA=223471 N

ΣMA=0

MA-(5116.25*5.80)-(10262.9*8.71)+

(23361.4*3.57)+(27425.8*4.355)-(137129*7.393)=0

MA=960068 Nm

CURVE/COMPONENT AREA

ENCLOSED/STRUCTURAL

WEIGHT

CENTROID (from wing

root)

Triangle (under y1/2) 51116.254N 5.80 m (a/3)

Rectangle(under y1/2) 10262.9N 8.71 m (a/2)

y2/2 274258N 7.393 m (4a/3π)

Wing 27425.8N 4.355 m (a/4)

Powerplant 23361.4N 3.57 m

1

1

2

2

y2/2

y1/2

y3

Powerplant

VA

MA

A B

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SHEAR FORCE:

Force acting on a substance in a direction perpendicular to the extension of the

substance, as for example the pressure of air along the front of an airplane wing.

Shear forces often result in shear strain. Resistance to such forces in a fluid is

linked to its viscosity. Also called shearing force.

SHEAR FORCE

S.F= -33027+∫(-67.5x + 2353.1)dx

+∫2248*((1-0.0574x2)^0.5) dx -∫15.63*(x-15.79)

2 dx

BENDING MOMENT

BM= -33027x+∫(-67.5x‟ + 2353.1)(x-x‟)dx‟

+∫2248*((1-0.0574x2)^0.5)(x-x‟)dx+9960068

-∫15.63*(x‟-15.79)2 (x-x‟)dx

These analytical expressions are used to calculate the shear force and

bending moment at different span wise locations of the starboard wing By

symmetry, the shear force and bending moment of the other wing can also be

tabulated.

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SHEAR FORCE AND BENDING MOMENT DIAGRAMS OF A

WING DUE TO LOADS IN TRANSVERSE DIRECTION AT

CRUISE CONDITION TABULATION:

S

NO

SPANWISE DISTANCE

FROM ROOT (M)

SHEAR FORCE (N) BENDING MOMENT

(Nm)

1 17.42 -301178 -3238280

2 15 -304490 -3093293

3 13 -307402 -2636019

4 11 -310205 -2203978

5 9 -312649 -1959324

6 7 -314482 -1425630

7 5 -315455 -846351

8 3 -315316 -233467

9 1 -313815 406016

10 0 -312476 734906

11 -1 -313815 406016

12 -3 -315316 -233467

13 -5 -315455 -846351

14 -7 -314482 -1425630

15 -9 -312649 -1959324

16 -11 -310205 -2636019

17 -13 -307402 -2636019

18 -15 -304490 -3093293

19 -17.42 -301178 -3238280

30

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SHEAR FORCE AND BENDING MOMENT DIAGRAMS DUE

TO LOADS ALONG CHORDWISE DIRECTION AT CRUISE

CONDITION:

SPECIFICATIONS:

Cruise CL=0.0536

Cruise CD= 0.02

Angle of attack= -1.5

Angle of attack at zero lift= 4o

Wing lift curve slope= 0.1

Location of aerodynamic center= 0.25c

Location of shear center= 0.35c

SHEAR FORCE AND BENDING MOMENT:

Lift and Drag are the components of resultant aerodynamic force

acting normal to and along the direction of relative wind respectively. As a

result, components of them act in the chordwise direction also which produce a

bending moment about the normal (z) axis.

Co-efficient of force along the normal direction, Cn= CL cos α + CD sin α

Cn= 0.0536 cos (4) +0.02sin (4)

Cn= -0.185

α

Vrelative

CL

CD

CC

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Co-efficient of force along the chordwise direction, CC= -CL sin α + CD cos α

CC= -0.0536 sin (4) + 0.02cos (4)

CC= 0.02692

Therefore, force per unit length= CC*0.5*ρ*V2*c

Let x represent the spanwise location from the wing tip to root

Hence, force per unit length at the wing root (x=17.42 m) = CC*0.5*ρ*V2*cR

= 1181.9 N/m

Force per unit length at the wing tip (x=0) = CC*0.5*ρ*V2*cT

= 590.96N/m

Equation of force per unit length is linear: in the form y=mx+c

where, slope m= (1181.9-590.96)/17.42 = 33.92

x=0, y=590.9 → c=590.96

Hence, y=33.92x+590.9

On integrating over a length dx, Shear force over the segment can be obtained

Shear force over dx16.95x2+590.9x → „1‟

On integrating again over the length dx, bending moment over the segment can

be obtained

Bending moment over dx= 5.65x3+295.45x

2 → „2‟

These equations can be used to obtain the shear force and bending moment

values at a particular span wise location

33

SHEAR FORCE AND BENDING MOMENT DIAGRAMS DUE

TO LOADS ALONG CHORDWISE DIRECTION AT CRUISE

CONDITION TABULATION:

S.NO SPANWISE

LOCATION(m)

SHEAR

FORCE(N)

BENDING

MOMENT(Nm)

1 17.42 15437 119523

2 15 12677 85545

3 13 10546 62344

4 11 8551 43270

5 9 6691 28050

6 7 4967 16415

7 5 3378 8093

8 3 1925 2812

9 1 608 301

10 0 0 0

34

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5. CRITICAL FLIGHT CONDITION

Optimum Wing structural design consists of determining that stiffness

distribution which is proportional to the local load distribution. The

aerodynamic forces of lift and drag are resolved into components normal and

parallel to the wing chord. The distribution of shear force, bending moment and

torque over the aircraft wing are considered for wing structural analysis.

SHEAR FORCE AND BENDING MOMENT DIAGRAMS OF A

WING DUE TO NORMAL FORCES AT CRITICAL FLIGHT

CONDITION:

In the preliminary stage of structural analysis, the critical flight loading

condition of positive high angle of attack (represented by point A in V-n

diagram) will be investigated.

nA= 4.45; VA = 92.62 m/s; CL = 1.414

Therefore, LA=1.414*0.5*1.2256*92.622*99.78 = 736846.5 N;

In steady flight, L= 0.053*0.5*1.2*2252*99.78 = 160633.3 N

Proportionality factor, k= LA/L = 4.58

The aim is to find the shear forces and bending moments due to

normal forces in critical flight condition. There are three primary loads acting

on a wing structure in transverse direction which can cause considerable shear

forces and bending moments on it. They are as follows:

Lift force (given by Schrenk‟s curve)

Self-weight of the wing

Weight of the power plant

Now, the proportionality constant influences the lift force alone and other

factors remain unaffected.

36

CURVE/COMPONEN

T

AREA

ENCLOSED/STRUCTURAL

WEIGHT

CENTROID (from wing

root)

Triangle (under y1/2) 4.58*51116.254N 5.80 m (a/3)

Rectangle(under y1/2) 4.58*10262.9N 8.71 m (a/2)

y2/2 4.58*274258N 7.393 m (4a/3π)

Wing 27425.8N 4.355 m (a/4)

Power plant 23361.4N 3.57 m

ΣFV = 0

(274258*4.58) – 27425.8 - 0 - VA = 0

VA = 1228675N

ΣMA = 0

MA - (51116*5.80) - (10262*8.71) - (274258*7.39) + (27425.8*4.355) =0

MA = 10934134Nm

SHEAR FORCE

S.F= -1205315+4.58(∫(-67.5x + 2353.1)dx

+∫2248*((1-0.0574x2)^0.5) dx )-∫15.63*(x-15.79)

2 dx

BENDING MOMENT

BM = -1205315x+4.58[∫(-67.5x‟ + 2353.1)(x-x‟)dx‟

+∫2248*((1-0.0574x2)^0.5)(x-x‟)dx]+9960068

-∫15.63*(x‟-15.79)2 (x-x‟)dx

37

SHEAR FORCE AND BENDING MOMENT DIAGRAMS OF A

WING DUE TO NORMAL FORCES AT CRITICAL FLIGHT

CONDITION TABULATION:

S

NO

SPANWISE DISTANCE

FROM ROOT (M)

SHEAR FORCE (N) BENDING MOMENT

(Nm)

1 17.42 -1082727 -17135704

2 15 -1097984 -15275801

3 13 -1111719 -12599960

4 11 -1126206 -10622018

5 9 -1141194 -8520580

6 7 -1156432 -6318162

7 5 -1171670 -4032572

8 3 -1186657 -1676765

9 1 -1201143 -475233.5

10 0 -1208120 741226.63

38

SHEAR FORCE, BENDING MOMENT DIAGRAMS OF A

WING DUE TO CHORDWISE FORCES AT CRITICAL

FLIGHT CONDITION:

In the preliminary stage of structural analysis, the critical flight loading

condition of positive high angle of attack (represented by point A in V-n

diagram) will be investigated.

nA= 4.45; VA=92.26 m/s; CL=1.41

SHEAR FORCE AND BENDING MOMENT:

Lift and Drag are the components of resultant aerodynamic force

acting normal to and along the direction of relative wind respectively. As a

result, components of them act in the chord wise direction also which produce a

bending moment about the normal (z) axis.

α

Vrelative

CL

CD

39

CL= a*(α-α0)

a=0.1; CL=1.41; α0= 4o hence, α= 15

o

Co-efficient of force along the normal direction, Cn= CL cos α + CD sin α

Cn= 1.41 cos (15) + 0.02 sin (15)

Cn= 1.365

Co-efficient of force along the chordwise direction, CC= -CL sin α + CD cos α

CC= -1.41 sin (15) + 0.02 cos (15)

CC= -0.34

Therefore, force per unit length= CC*0.5*ρ*V2*c

Let x represent the spanwise location from the wing tip to root

Hence, force per unit length at the wing root (x=17.42 m) = CC*0.5*ρ*V2*cR

= -14928 N/m

Force per unit length at the wing tip (x=0) = CC*0.5*ρ*V2*cT

= -7464 N/m

Equation of force per unit length is linear: in the form y=mx+c

where, slope m= (-14928+7464)/17.42 = -428.47

x=0, y= -7464 → c= -7464

Hence, y= -428x-7464

On integrating over a length dx, Shear force over the segment can be obtained

Shear force over dx= -214x2-7464x → „1‟

On integrating again over the length dx, bending moment over the segment can

be obtained

Bending moment over dx= -71.3x3-3732x

2 → „2‟

40

These equations can be used to obtain the shear force and bending moment

values at a particular span wise location and plots can therefore be drawn

In the following table and plot, x=0→wing root and x=17.42→wing tip of

starboard wing

SHEAR FORCE, BENDING MOMENT DIAGRAMS OF A

WING DUE TO CHORDWISE FORCES AT CRITICAL

FLIGHT CONDITIONTABULATION

S.No. SPANWISE LOCATION

(m)

SHEAR FORCE (N) BENDING

MOMENT (Nm)

1

17.42 -194963 -1509406.093

2 15 -160110

-1080337.5

3 13 -133198

-787354.1

4 11 -107998

-546472.3

5 9 -84510

-354269.7

6 7 -62734

-207323.9

7 5 -42670

-102212.5

8 3 -24318

-35513.1

9 1 -7678

-3803.3

10 0 0

0

41

42

6. MATERIAL SELECTION

WOOD, ALUMINUM, STEE AND COMPOSITES

PROPERTIES

Aircraft structures are basically unidirectional. This means that one dimension,

the length, is much larger than the others - width or height. For example, the

span of the wing and tail spars is much longer than their width and depth; the

ribs have a much larger chord length than height and/or width; a whole wing has

a span that is larger than its chords or thickness; and the fuselage is much longer

than it is wide or high. Even a propeller has a diameter much larger than its

blade width and thickness, etc.... For this simple reason, a designer chooses to

use unidirectional material when designing for an efficient strength to weight

structure.

Unidirectional materials are basically composed of thin, relatively flexible, long

fibers which are very strong in tension (like a thread, a rope, a stranded steel

wire cable, etc.)

An aircraft structure is also very close to a

symmetrical structure. That means the up and down

loads is almost equal to each other. The tail loads may

be down or up depending on the pilot raising or

dipping the nose of the aircraft by pulling or pushing

the pitch control; the rudder may be deflected to the

right as well as to the left (side loads on the fuselage).

The gusts hitting the wing may be positive or negative,

giving the up or down loads which the occupant

experiences by being pushed down in the seat ... or

hanging in the belt.

Because of these factors, the designer has to use a

structural material that can withstand both tension and

compression. Unidirectional fibers may be excellent in

tension, but due to their small cross section, they have

very little inertia (we will explain inertia another time)

and cannot take much compression. They will escape

the load by bucking away. As in the illustration, you

cannot load a string, or wire, or chain in compression.

In order to make thin fibers strong in compression, they are "glued together"

with some kind of an "embedding". In this way we can take advantage of their

43

tension strength and are no longer penalized by their individual compression

weakness because, as a whole, they become compression resistant as they help

each other to not buckle away. The embedding is usually a lighter, softer "resin"

holding the fibers together and enabling them to take the required compression

loads. This is a very good structural material.

WOOD

Historically, wood has been used as the first unidirectional structural raw

material. They have to be tall and straight and their wood must be strong and

light. The dark bands (late wood) contain many fibers, whereas the light bands

(early wood) contain much more "resin". Thus the wider the dark bands, the

stronger and heavier the wood. If the dark bands are very narrow and the light

bands quite wide, the wood is light but not very

strong. To get the most efficient strength to weight

ratio for wood we need a definite numbers of bands

per inch.

Some of our aircraft structures are two-dimensional

(length and width are large with respect to

thickness). Plywood is often used for such

structures. Several thin boards (foils) are glued

together so that the fibers of the various layers cross

over at different angles (usually 90 degrees today

years back you could get them at 30 and 45 degrees as well). Plywood makes

excellent "shear webs" if the designer knows how to use plywood efficiently.

(We will learn the basis of stress analysis sometime later.)

Today good aircraft wood is very hard to come by. Instead of using one good

board for our spars, we have to use laminations because large pieces of wood

are practically unavailable, and we no longer can trust the wood quality. From

an availability point of view, we simply need a substitute for what nature has

supplied us with until now.

ALUMINUM ALLOYS

So, since wood may not be as available as it was before, we look at another

material which is strong, light and easily available at a reasonable price (there's

no point in discussing Titanium - it's simply too expensive). Aluminum alloys

are certainly one answer. We will discuss the properties of those alloys which

are used in light plane construction in more detail later. For the time being we

will look at aluminum as a construction material.

44

Extruded Aluminum Alloys: Due to the manufacturing process for aluminum

we get a unidirectional material quite a bit stronger in the lengthwise direction

than across. And even better, it is not only strong in tension but also in

compression. Comparing extrusions to wood, the tension and compression

characteristics are practically the same for aluminum alloys so that the linear

stress analysis applies. Wood, on the other hand, has a tensile strength about

twice as great as its compression strength; accordingly, special stress analysis

methods must be used and a good understanding of wood under stress is

essential if stress concentrations are to be avoided!

Aluminum alloys, in thin sheets (.016 to .125 of an inch) provide an excellent

two dimensional material used extensively as shear webs - with or without

stiffeners - and also as tension/compression members when suitably formed

(bent).

It is worthwhile to remember that aluminum is an artificial metal. There is no

aluminum ore in nature. Aluminum is manufactured by applying electric power

to bauxite (aluminum oxide) to obtain the metal, which is then mixed with

various strength-giving additives. (In a later article, we will see which additives

are used, and why and how we can increase aluminum's strength by cold work

hardening or by tempering.) All the commonly used aluminum alloys are

available from the shelf of dealers. When requested with the purchase, you can

obtain a "mill test report" that guarantees the chemical and physical properties

as tested to accepted specifications.

As a rule of thumb, aluminum is three times heavier, but also three times

stronger than wood. Steel is again three times heavier and stronger than

aluminum.

STEEL

The next material to be considered for aircraft structure will thus be steel, which

has the same weight-to-strength ratio of wood or aluminum.

Apart from mild steel which is used for brackets needing little strength, we are

mainly using a chrome-molybdenum alloy called AISI 413ON or 4140. The

common raw materials available are tubes and sheet metal. Steel, due to its high

density, is not used as shear webs like aluminum sheets or plywood. Where we

would need, say.100" plywood, a .032 inch aluminum sheet would be required,

but only a .010 steel sheet would be required, which is just too thin to handle

with any hope of a nice finish. That is why a steel fuselage uses tubes also as

diagonals to carry the shear in compression or tension and the whole structure is

then covered with fabric (light weight) to give it the required aerodynamic

45

shape or desired look. It must be noted that this method involves two

techniques: steel work and fabric covering.

We will be discussing tubes and welded steel structures in more detail later and

go now to "artificial wood" or composite structures.

COMPOSITE MATERIALS

The designer of composite aircraft simply uses fibers in the desired direction

exactly where and in the amount required. The fibers are embedded in resin to

hold them in place and provide the required support against buckling. Instead of

plywood or sheet metal which allows single curvature only, the composite

designer uses cloth where the fibers are laid in two directions .(the woven thread

and weft) also embedded in resin. This has the advantage of freedom of shape in

double curvature as required by optimum aerodynamic shapes and for very

appealing look (importance of esthetics).

Today's fibers (glass, nylon, Kevlar, carbon, whiskers or single crystal fibers of

various chemical compositions) are very strong, thus the structure becomes very

light. The drawback is very little stiffness. The structure needs stiffening which

is achieved either by the usual discreet stiffeners, -or more elegantly with a

sandwich structure: two layers of thin uni- or bi-directional fibers are held apart

by a lightweight core (foam or "honeycomb"). This allows the designer to

achieve the required inertia or stiffness.

From an engineering standpoint, this method is very attractive and supported by

many authorities because it allows new developments which are required in case

of war. But this method also has its drawbacks for homebuilding: A mold is

needed, and very strict quality control is a must for the right amount of fibers

and resin and for good adhesion between both to prevent too "dry" or "wet" a

structure. Also the curing of the resin is quite sensitive to temperature, humidity

and pressure. Finally, the resins are active chemicals which will not only

produce the well-known allergies but also the chemicals that attack our body

(especially the eyes and lungs) and they have the unfortunate property of being

cumulatively damaging and the result (in particular deterioration of the eye)

shows up only years after initial contact.

Another disadvantage of the resins is their limited shelf life, i.e., if the resin is

not used within the specified time lapse after manufacturing, the results may be

unsatisfactory and unsafe.

46

LIGHT AIRCRAFT RAW MATERIALS

The focus of our article is our Table which gives typical values for a variety of

raw materials.

Column 1 lists the standard materials which are easily available at a reasonable

cost. Some of the materials that fall along the borderline between practical and

impractical are:

Magnesium: An expensive material. Castings are the only readily

available forms. Special precaution must be taken when machining

magnesium because this metal burns when hot.

Titanium: A very expensive material. Very tough and difficult to

machine.

Carbon Fibers: Still very expensive materials.

Kevlar Fibers: Very expensive and also critical to work with because it

is hard to "soak" in the resin. When this technique is mastered, the

resulting structure is very strong, but it also lacks in stiffness.

Columns 2 through 6:

Column 2 through 6 list the relevant material properties in metric units.

Column 2 the density (d), is the weight divided by the volume.

Column 3, the yield stress (fy), is the stress (load per area) at which there will

be a permanent deformation after unloading (the material has yielded, given

way ...).

Materials d fy fu e E/103 E/d Root

2 of N/d Root

3 of E/d fu/d

1 2 3 4 5 6 7 8 9 10

Wood Spruce .45 - 3.5/11 - 1.4 2200 70 22.0 (15)

Poplar .43 - 30/12 - 1.0 2200 70 22.0 (15)

Oregon Pine .56 - 4.0/13 - 1.5 2200 70 22.0 (15)

Fiberglass Matte 2.2 - 15 - 1.5 700 17 5.0 7

(70% Glass) Woven 2.2 - 35 - 2.0 900 20 6.0 16

Unidirectional 2.2 - 60 - 3.5 1500 27 7.0 27

Alum.

Alloy 5052-H34 2.7 16 24 4 7.1 2600 30 7.0 11

8086-H34 2.7 22 31 5 7.1 2600 30 7.0 11

6061 -T6 2.7 24 26 9 7.1 2600 30 7.0 11

47

6351 -T6 2.7 25 28 9 7.1 2600 30 7.0 11

6063-T6 2.7 17 21 9 7.1 2600 30 7.0 11

2024-T3 2.8 25 41 12 7.2 2600 30 7.0 14

Steel AISI 1026 7.8 25 38 15 21.0 2700 18 3.5 5

4130 N (4140) 7.8 42 63 10 21.0 2700 18 3.5 7

Lead 11.3 - - - - - - - -

Magnesium Alloy 1.8 20 30 - 4.5 2500 37 9.0 16

Titanium 4.5 50 80 - 11.0 2400 23 5.0 18

Units for above kg/dm3 kg/mm

2 kg/mm2 % kg/mm

2 km kg-m

2 kg2/3

m1/3 km

to obtain: lbs/cu3 KSI KSI % KSI

multiply by: .0357 1420 1420 - 1420

Column 4, the ultimate stress (fu), is the stress (load per area) at which it cannot

carry a further load increase. It is the maximum load before failure.

Column 5, the elongation at ultimate stress (e), in percentage gives an indication

of the 'Toughness" of the material.

Column 6 lists the Youngs Modular or Modulus of Elasticity (E), which is the

steepness of the stress/strain diagram as shown in Figure 1.

Important Note: For wood, the tension is much greater (2 to 3 times) than the

compression. Both values are given in the Table. For fiberglass, the same

applies, but the yield is so dependent on the manufacturing process that we

cannot even give 'typical values'.

48

7. DETAILED WING DESIGN

SPAR DESIGN:

Spars are members which are basically used to carry the bending and

shear loads acting on the wing during flight. There are two spars, one located at

15-25% of the chord known as the front spar, the other located at 60-70% of the

chord known as the rear spar. Some of the functions of the spar include:

They form the boundary to the fuel tank located in the wing.

The spar flange takes up the bending loads whereas the web carries the

shear loads.

The rear spar provides a means of attaching the control surfaces on the

wing.

Considering these functions, the locations of the front and rear spar are

fixed at 0.175c and 0.75c respectively.

The spar design for the wing root has been taken because the maximum

bending moment and shear force are at the root. It is assumed that the flanges

take up all the bending and the web takes all the shear effect. The maximum

bending moment for high angle of attack condition is 734906 Nm. the ratio in

which the spars take up the bending moment is given as

h1=0.686m

h2=0.380m

(Mfr/Mr) = (h12/h2

2) = (0.686

2/0.380

2) = 3.259

Mfr+Mr = 734906Nm

From the above two equations, Mfr = 562312 Nm, Mr = 172594 Nm

The yield tensile stress σy for 6061 T6 Al Alloy is = (24*10^6) MPa. The area

of the flanges is determined using the relation

σy = Mz/(A*z2)

where M is bending moment taken up by each spar,

A is the flange area of each spar,

49

z is the centroid distance of the area = h/2.

Using the available values,

Area of front spar Afr 0.0683m2, Area of rear spar Ar = 0.037m

2

Each flange of the spar is made of two angle sections. For the front spar,

the length of the angle is 6t, angle height is 5t with angle thickness t. Area for

each angle of front spar is found to be 0.017m2 and hence value of t is found to

be 0.412m

Front spar - Dimensions of each angle:

Length = 0.247m

Height 0.206m

Thickness = 0.412m.

For the rear spar, the length of the angle is 8t, angle height is 3.5t with

vertical thickness t and horizontal thickness t/2. Area for each angle of rear spar

is found to be 9.25*10^-3 m2 and hence value of t is found to be 0.0304m.

Rear spar - Dimensions of each angle:

50

Length = 0.2432m

Height = 0.1064m

Thickness = 0.0304m.

Now to determine the thickness of the web portion, the ultimate shear

stress for 6061 T6 Al Alloy is 26Kg/mm2. The maximum shear force at root of

the wing for high angle of attack condition is 1082727 N. The wing chord is

assumed to be a simply supported beam supported at the two spars. The

maximum shear force acts at the center of pressure which is located at 0.2982c.

Vr Vfr

V

0.585c

0.1232c

51

8. FUSELAGE DESIGN

Fuselage contributes very little to lift and produces more drag but it is an

important structural member/component. It is the connecting member to all load

producing components such as wing, horizontal tail, vertical tail, landing gear

etc. and thus redistributes the load. It also serves the purpose of housing or

Accommodating practically all the equipment‟s, accessories and systems in

addition to carrying the payload. Because of large amount of equipment inside

the fuselage, it is necessary to provide sufficient number of cutouts in the

fuselage for access and inspection purposes. These cutouts and discontinuities

result in fuselage design being more complicated, less precise and often less

efficient in design. As a common member to which other components are

attached, thereby transmitting the loads, fuselage can be considered as a long

hollow beam. The reactions produced by the wing, tail or landing gear may be

considered as concentrated loads at the respective attachment points. The

balancing reactions are provided by the inertia forces contributed by the weight

of the fuselage structure and the various components inside the fuselage. These

reaction forces are distributed all along the length of the fuselage, though need

not be uniformly. Unlike the wing, which is subjected to mainly unsymmetrical

load, the fuselage is much simpler for structural analysis due to its symmetrical

cross-section and symmetrical loading. The main load in the case of fuselage

is the shear load because the load acting on the wing is transferred to the

fuselage skin in the form of shear only. The structural design of both wing and

fuselage begin with shear force and bending moment diagrams for the

respective members. The maximum bending stress produced in each of them is

checked to be less than the yield stress of the material chosen for the respective

member.

Loads and its distribution:

To find out the loads and their distribution, consider the different cases.

The main components of the fuselage loading diagram are:

(i) Weight of the fuselage

(ii) Engine weight

(iii) Weight of the horizontal and vertical stabilizers

(iv) Tail lift

(v) Weight of crew, payload and landing gear

(vi) Systems , equipment‟s, accessories

52

Case I: Symmetric flight condition

(i) Steady and level flight: (Downward forces negative)

Values for the different component weights are obtained from aerodynamic

design calculations. Load factor = 4.38 (from v-n diagram)

FUSELAGE Cg TABULATION

Type of parts Distance from the

reference point(M)

weight(N)

Fuselage 18.58 54851.6

Cock pit 3.679 3924

Nose landing gear 5.136 16455.48

Horizontal stabilizer 33.22 21940.6

Vertical stabilizer 30.22 10970.32

Wing 6.9575 54851.6

Pay load (station-1) 10.575 25506

Pay load (station-2) 17.575 47088

Pay load (station-3) 24.575 25506

53

FUSELAGE SHEARFORCE TABULATION

LOCATION(M) SHEAR FORCE(N)

18.58 54851.6

3.679 3924

5.136 16455.48

33.22 21940.6

30.22 10970.32

6.9575 54851.6

10.575 25506

17.575 47088

24.575 25506

54

FUSELAGE BENDING MOMENT TABULATION

LOCATION(m) BENDING MOMENT(Nm)

18.58 -1019143

3.679 -14436.4

5.136 -84515

33.22 -728867

30.22 -331523

6.9575 -381630

10.575 -269726

17.575 -827572

24.575 -626810

55

8. DETAILED DESIGN OF FUSELAGE:

Design of the fuselage can be carried out by considering the maximum bending

moment which is taken as the design bending moment. The cross-sectional area

required to withstand the bending stress is found out by using the formula for

bending stress. This area is divided among several stringers which are spaced

evenly. The stringers spacing is calculated by considering the buckling of the

portion between adjacent stringers which can be modeled as a plate. Now, the

first step is to calculate the required cross-sectional area of the stringers. Use the

following formula for bending stress.

σ= M*y/I

Where,

σ = Tensile strength of the material used (6061 T6 Al Alloy) = 24 MPa

M = Design bending moment = 1019143Nm

I = Second moment of area (m4) = A*(d/2)

2

y = d/2

d =4.9m (diameter of the fuselage)

A = cross-sectional area of the fuselage stringers (m2)

A stringer cross section is chosen satisfying the condition that the actual stress

is less than the yield stress of the material.

The properties of the stringer section chosen are as follows,

56

t = 0.03515m

The total circumference of the fuselage cross section is found to be

5.497775 m. This circumference is distributed with „n‟ number of stringers such

that the total bending moment is taken up by these stringers effectively. Assume

skin is ineffective in bending. Arbitrarily, let us set the number of stringers to be

equal to 40 i.e. 10 stringers in each quadrant. Now, the net IYY is computed

considering these stringers to be lumped masses. As it is a symmetric cross

section,

SHEAR FLOW:

Consider the stringer at Ө = 0° of the first quadrant of the cross section as the

first stringer and number it in anticlockwise direction. Make a cut between

stringers 1 and 2 and determine shear flow using the formula,

No. of stringers = 60

Stringer distribution

57

FUSELAGE SHEARFLOW TABULATION

Angle x y A Ax q*l q fin q abs

1 0 3.5 0 0.017297 0.060541 0 55598896 55598896

2 6 3.480828 0.365839 0.017297 0.060209 -156335.4731 54989400.98 54989400.98

3 12 3.423521 0.72767 0.017297 0.059218 -467293.676 53777088.3 53777088.3

4 18 3.328708 1.081529 0.017297 0.057578 -929467.8864 51975239.52 51975239.52

5 24 3.197427 1.423539 0.017297 0.055307 -1537794.725 49603594.93 49603594.93

6 30 3.031116 1.749953 0.017297 0.05243 -2285609.629 46688137.21 46688137.21

7 36 2.831598 2.057196 0.017297 0.048979 -3164719.865 43260806.86 43260806.86

8 42 2.601058 2.341901 0.017297 0.044991 -4165494.284 39359152.2 39359152.2

9 48 2.342021 2.600949 0.017297 0.040511 -5276968.838 35025918.08 35025918.08

10 54 2.057327 2.831502 0.017297 0.035586 -6486966.697 30308577.49 30308577.49

11 60 1.750094 3.031035 0.017297 0.030272 -7782231.654 25258811.58 25258811.58

12 66 1.423687 3.197361 0.017297 0.024626 -9148573.349 19931943.37 19931943.37

13 72 1.081683 3.328658 0.017297 0.01871 -10571022.74 14386331.71 14386331.71

14 78 0.727828 3.423487 0.017297 0.012589 -12033996.09 8682731.897 8682731.897

15 84 0.366 3.480811 0.017297 0.006331 -13521465.71 2883630.082 2883630.082

16 90 0.000162 3.5 0.017297 2.8E-06 -15017135.52 -2947441.37 2947441.307

17 96 -0.36568 3.480845 0.017297 -0.00633 -16504619.62 -8746599.54 8746599.594

18 102 -0.72751 3.423555 0.017297 -0.01258 -17967621.78 -14450311.3 14450311.73

19 108 -1.08137 3.328758 0.017297 -0.0187 -19390114 -19996090.4 19996090.34

20 114 -1.42339 3.197493 0.017297 -0.02462 -20756512.06 -25323178.2 25323178.29

21 120 -1.74981 3.031197 0.017297 -0.03027 -22051846.3 -30373214.3 30373214.33

22 126 -2.05706 2.831693 0.017297 -0.03558 -23261925.61 -35090872.4 35090872.47

23 132 -2.34178 2.601166 0.017297 -0.04051 -24373492.89 -39424468.1 39424468.1

24 138 -2.60084 2.342142 0.017297 -0.04499 -25374370.29 -43326524.2 43326524.24

25 144 -2.83141 2.057458 0.017297 -0.04898 -26253592.64 -46754291.6 46754291.67

26 150 -3.03095 1.750234 0.017297 -0.05243 -27001527.55 -49670217.2 49670217.27

27 156 -3.19729 1.423835 0.017297 -0.0553 -27609980.99 -52042355.4 52042355.42

28 162 -3.32861 1.081837 0.017297 -0.05758 -28072286.99 -53844718.0 53844718.01

29 168 -3.42345 0.727987 0.017297 -0.05922 -28383380.75 -55057559.1 55057559.15

30 174 -3.48079 0.366161 0.017297 -0.06021 -28539854.04 -5566759.48 55667591.48

31 180 -3.5 0.000324 0.017297 -0.06054 -28539992.62 -55668131.7 55668131.75

32 186 -3.48086 -0.36552 0.017297 -0.06021 -28383794.97 -55059174.0 55059174.04

33 192 -3.42359 -0.72735 0.017297 -0.05922 -28072972.32 -53847389.8 53847389.83

34 198 -3.32881 -1.08122 0.017297 -0.05758 -27610929.9 -52046054.8 52046054.89

35 204 -3.19756 -1.42324 0.017297 -0.05531 -27002729.67 -49674903.8 49674903.87

36 210 -3.03128 -1.74967 0.017297 -0.05243 -26255034.78 -46759914.0 46759914.06

37 216 -2.83179 -2.05693 0.017297 -0.04898 -25376036.66 -43333020.8 43333020.82

38 222 -2.60127 -2.34166 0.017297 -0.045 -24375365.24 -39431767.6 39431767.69

39 228 -2.34226 -2.60073 0.017297 -0.04051 -23263983.42 -35098895.1 35098895.1

40 234 -2.05759 -2.83131 0.017297 -0.03559 -22054067.02 -30381872.1 30381872.11

58

41 240 -1.75037 -3.03087 0.017297 -0.03028 -20758871.36 -25332376.3 25332376.37

42 246 -1.42398 -3.19723 0.017297 -0.02463 -19392586.04 -20005727.9 20005727.95

43 252 -1.08199 -3.32856 0.017297 -0.01872 -17970179.49 -14460283.2 14460283.28

44 258 -0.72815 -3.42342 0.017297 -0.01259 -16507234.96 -8756795.84 8756795.84

45 264 -0.36632 -3.48078 0.017297 -0.00634 -15019779.84 -2957750.54 2957750.546

46 270 -0.00049 -3.5 0.017297 -8.4E-06 -13524110.04 2873320.792 2873320.792

47 276 0.365355 -3.48088 0.017297 0.00632 -12036611.47 8672535.501 8672535.501

48 282 0.727194 -3.42362 0.017297 0.012579 -10573580.51 14376359.92 14376359.92

49 288 1.081066 -3.32886 0.017297 0.0187 -9151045.482 19922305.43 19922305.43

50 294 1.423094 -3.19762 0.017297 0.024616 -7784591.071 25249613.07 25249613.07

51 300 1.749532 -3.03136 0.017297 0.030262 -6489187.551 30299919.19 30299919.19

52 306 2.056802 -2.83188 0.017297 0.035577 -5279026.796 35017894.85 35017894.85

53 312 2.341539 -2.60138 0.017297 0.040502 -4167366.801 39351851.94 39351851.94

54 318 2.600624 -2.34238 0.017297 0.044984 -3166386.426 43254309.54 43254309.54

55 324 2.831216 -2.05772 0.017297 0.048972 -2287051.977 46682514.02 46682514.02

56 330 3.030792 -1.75051 0.017297 0.052425 -1538997.058 49598907.47 49598907.47

57 336 3.197163 -1.42413 0.017297 0.055302 -930417.0309 51971539.15 51971539.15

58 342 3.328507 -1.08215 0.017297 0.057574 -467979.2344 53774415.55 53774415.55

59 348 3.423386 -0.7283 0.017297 0.059215 -156749.9347 54987785.14 54987785.14

60 354 3.48076 -0.36648 0.017297 0.060208 -138.8241828 55598354.78 55598354.78

1 0 3.5 0 0.017297 0.060541 0 55598896 55598896

-856223000.1

59

9.THREE VIEWS OF AIRCRAFT

60

10. SOFTWARE MODELING DESIGN& ANALYSIS

WING MODELING (CATIA)

ANALYSIS (FLUENT)

Meshing the Airfoil

Import the airfoil through IGES format.

Since the imported file contains two faces, delete the airfoil from the

wind tunnel using Boolean operation through the face command.

Select the mesh command and select the face command button.

Select the face to be mesh on Faces, Tri on Element, Pave on Type and

give appropriate size in Interval Size option.

On the Zone command, select Specify Boundary type command.

Using the VELOCITY_INLET, OUTLET_VENT, and WALL type select

the corresponding edge and apply.

Save the file as mesh in Export-Mesh from File command.

61

Computing and Analyzing the Airfoil

Import the mesh file using option File > Read > Case in Fluent.

Grid Check using option Grid > Check.

After the completion of reading by fluent, the report shows "done".

Tick the checkbox on energy equation using option Define > Model >

Energy.

Click Ok. Set the velocity and temperature as required using options

Define > Model > Boundary Conditions > VELOCITY_INLET.

Now open Solution Initialization window using option Solve > Initialize

> Initialize.

On solution Initialization, set Gauge Pressure, x-velocity, and

temperature for required values (At the altitude of aircraft flying) and click

limits and Apply.

Click solve and then Iterate.

In this window give appropriate number of iterations and give iterate.

Wait till the iterations shows "solution is converged".

Save the file as "Case and Data" through File > Save as.

Choose Display > Contours.

Tick the "Filled" Checkbox.

On the drop down list of "Contours of.." select the required conditions

and click display.

62

STATIC PRESSURE (pascal)

VELOCITY VECTORS (m/s)

63

11. CONCLUSION

We have come to a completion of the conceptual design of an aircraft.

Aircraft design involves a variety of faculties of the field of Aerospace

engineering like structures, performance, aerodynamics, stability etc. this

project has enabled us to get a taste of what it is to design a real aircraft. The

fantasies of the flying world seem to be much more than what we thought. With

this design project as the base, we will strive to progress in the field of airplane

design and maintenance. We convey our heartfelt gratitude to all of them who

have provided their helping hand in the completion of this project.

64

12. REFERENCES

Aerodynamic design:

1. Jane‟s All the world‟s aircraft

2. Aircraft design – a conceptual approach – Daniel P. Raymer

3. Design of aircraft – Thomas Corke

4. Aircraft Performance – J.D. Anderson

5. Aircraft performance, Stability and control – Perkins and Hage

6. Fluid dynamic Drag - Hoerner

7. Summary of airfoil data – Abbott, Doenhoff and Stivers

8. www.airliners.net

9. www.wikipedia.org

10. www.aerospaceweb.org

Structural design:

1. Analysis of Aircraft structures – Bruhn

2. Aircraft Structures for engineering students – T.H.G Megson

3. Aircraft structures – Peery and Azar

4. Airplane design – Jan Roskam

5. Airframe Stress Analysis and Sizing – Niu