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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
2
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.
3
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.
4
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
6
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
7
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
8
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
9
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.
10
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
11
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
12
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
13
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
14
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
15
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
16
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
17
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
18
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
19
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
20
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
21
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
22
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.
23
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
24
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
25
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:
26
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
27
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
28
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.
29
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
31
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
32
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
35
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.
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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.
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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.
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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