Post on 01-Nov-2014
CHAPTER 1
CLUSTER DIAGRAM
Cluster diagrams are graphs used to find the necessary data needed to start the preliminary
design of aircraft with historical figures.To plot the graph, data from similar kinds of aircrafts are
collected and fixing the cruise velocity the values are selected.the vales to be selected are Gross
take of Weight (WTO),Aspect Ratio(AR), Range (R), Thrust to Weight ratio (T/W)and Wing
loading (W/S).
Graph 1 Vcruise vs. WTO
Graph 2 Vcruise vs. AR
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Graph 3 Vcruise vs. Range
Graph 4 Vcruise vs. T/W
2
Graph 5 Vcruise vs. W/S
Keeping the Cruise Velocity constant at 890 km/hr, the following values are chosen from the
Cluster Diagram,
Chosen Values from Cluster Diagram
Gross Take Off Weight,WTO 231000 kg (2266.11N)
Aspect Ratio,AR 7.62
Range,R 9620 km
Thrust to Weight ratio,T/W 0.294
Wing Loading,W/S 604.97 kg/m2
Fineness ratio,Lf/Df 10.11
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CHAPTER 2
WEIGHT ESTIMATION
First Weight Estimation:
It is necessary to make the first estimate of the weight as it provides the basis for the preliminary
geometrical sizing of the aircraft. This method provides a quick estimation of the initial gross take off
weight which does not include any combat or payload drops.
The gross take off weight, WTO of the aircraft is broken into
1. weight of the payload, Wp
2. weight of the crew, Wcrew
3. empty weight, We
4. weight of fuel, Wf
Weight of the payload includes the passengers weight in this case.
Empty weight includes structures, landing gear, fixed equipment weight and avionics.
For the ease of calculation empty weight and the weight of fuel are expressed as fraction of gross
take off weight.
In the above equation the unknowns are the empty weight fraction and the fuel weight fraction.
2.1 Determination of Empty Weight Fraction (We / WTO ):
Empty weight fractions are determined from the historical trends plotted using data collected
from already existing aircraft of the particular type.
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Empty weight fraction value varies from 0.3 to 0.7 where the highest weight fraction applies to
the flying boat and the lowest for the military aircrafts and the empty weight fraction diminishes
with increase in aircraft gross take off weight.
Empty Weight Fraction
We /WTO = AWC TO Kus A C
Sailplane- unpowered 0.86 -0.05
Sailplane-powered 0.91 -0.05
Homebuilt 1.19 -0.09
Homebuilt- composite 0.99 -0.09
GA- single engine 2.36 -0.18
GA- twin engine 1.51 -0.10
Agricultural 0.74 -0.03
Twin turboprop 0.96 -0.05
Flying boat 1.09 -0.05
Jet trainer 1.59 -0.10
Jet fighter 2.34 -0.13
Military bomber 0.93 -0.07
Jet transport 1.02 -0.06
K us = 1.04 for variable sweep wing
= 1.00 for fixed swept win
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Fig 1 Empty weight fractions trends
The above graph shows the trend lines of various types of aircraft from which the values of A
and C are computed from the slopes of these trend lines. The negative exponential power is due
to the fact that the empty weight fraction decreases with increase in gross take off weight.
Since composite materials are used for the construction of major parts of the aircraft 0.95 is
multiplied to the obtained empty weight fraction.
2.2 Estimation of Fuel Weight Fraction:
The total weight of the aircraft fuel consists of the fuel used to complete the required mission and
the reserve fuel which also includes the trapped fuel.
Fuel weight fraction is calculated using approximations for specific fuel consumption and
aerodynamics.
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Mission Profile for Transport Jet Aircrafts:
Mission profile is a pictorial representation of different phases the aircraft has to undergo to
complete the required mission
Fig 2 – mission profile without loiter
Fig 3 – mission profile with loiter
1-2 indicates taxi phase ; 2-3 indicates take off phase ;
3-4 indicates climbing phase ; 4-5 indicates cruise phase ;
5-6 indicates loiter phase (in case of fig 2); 6-7 indicates landing phase(in fig 1)
7-8 indicates landing phase( in fig 2)
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In this manual, the gross take off weight is calculated without loiter phase
2.3 Steps for The Calculation of Gross Take Off Weight, WTO:
1. Determination of payload weight and the crew weight
According to the standards followed weight of 1079 N (110 kg) is allowed for each
passenger and the crew in which 196.2N (20 kg) is for the baggage carried
2. Determination of empty weight fraction (We /WTO) using table 1 mentioned according to the
type of the aircraft.
Where WTO is the value chosen from the cluster diagram for the given cruising velocity.
3. Determination of fuel weight fraction using the mission profile which will be explained in
detail in the example to be followed.
4. The gross take off weight ( WTO ) is calculated using equation
5. Using the obtained gross take off weight, empty weight fraction is calculated again until the
value of gross take off weight converges.
Following the above steps,
The problem given here is to design a 250 seater passenger high subsonic aircraft satisfying the
FAR regulations.
Determination of weight of payload and the weight of crew
Weight of the payload, Wp = 242 * 1079.1 = 261.142kN
Weight of the crew, Wcrew = 10 * 1079.1 = 10.791kN
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Wp + Wcrew = 271.933kN
Determination of empty weight fraction
The empty weight fraction, We /WTO = AWC TO Kus * 0.95
For jet transport A = 1.02 , C = - 0.06 , Kus = 1.00
From the cluster diagram ,WTO = 2266.110kN
We /WTO = 0.46185
Determination of fuel weight fraction
The Fuel weight fraction, Wf / WTO = 1.2 ( 1 - mff )
mff is the mission fuel fraction.
The mission fuel fraction is determined as follows
( W1 / Wto ) = 0.990 [ engine start up ]
(W2 / W1 ) = 0.990 [ taxi]
(W3 / W2 ) = 0.995 [ take off ]
Calculation for climb segment, ( W4 / W3 ):
Eclimb = (1/ c) climb (L / D) climb ln (W3 / W4 )……………..(3)
(L / D) max = 1 / [ 2 ( C do * K) 0.5 ]
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C do = 0.02 ( for Jet transport )
AR = 7.62 ( from cluster diagram )
e = 0.8
K = 0.052242
(L / D) max = 15.46838
Assume climb rate to be 1800 ft /min.
Let the cruising altitude be 36000 feet, so the time required to reach 36000 feet is 36000/1800
which equals to 20 mins ( 0.33 hr)
Eclimb = 0.33 hr
Substituting the above values in (3)
( W4 / W3 ) =0.991 (climb phase)
Calculation for Cruise segment:
R = ( V/ sfc)*( L/D) cruise* ln ( W4 / W5)
R is the Range covered during cruise, in km.
V is the cruise velocity, in km / hr.
c for range is 0.33
Range covered during cruise = total range – ( range covered during climb and descent )
Average climb speed is 592.64 km /hr and the time taken is 0.33 hr
Distance covered during climb =592.64 * 0.33 =195.5712 km.
Let the descent rate be 970 ft/min and it descends from 36000ft, so the time taken is 0.618 hr.
Let the average descent speed be 518.56 km/hr and the time taken is 0.618 hr
Distance covered during descent = 320.4700 km
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Range covered during cruise =9620 – 195.5712 – 320.47
= 9103.9588 km
( L/D)cruise = 0.98 ( L/D)max
( L/D)cruise = 15.1590
Substituting the above values in Range formula
( W5/W4) =0.7643
Calculation for descent segment
Edescent = (1/ c) descent (L / D)descent ln (W5/ W6 )…………….(5)
Edescent is the time taken to complete descent, in hours
(L/D)descent = (L/D)max to cover maximum distance
c is 0.35 for descent, in lb/lbf-hr.
Edescent = 0.618 hr ; (L/D)descent = 15.46838
(W5/ W6 ) = 0.988 [ descent ]
(W5/ W6 ) = 0.992 [ shut down]
Mission fuel fraction, mff = (W1/ Wto )*(W2/ W1)*(W3/ W2 )*(W4/ W3 )*(W5/ W4)*
(W6/ W5)*(W7/ W6)
mff = 0.7239
Fuel weight fraction, Wf / WTO = 1.2 ( 1 - mff )
Wf / WTO = 0.33132
From the cluster diagram, WTO is chosen to be 2266.11kN
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We/ WTO = 1.02* (WTO)-0.06*0.95 = 0.4618
Using the equation WTO = 1314.766kN
Using this value the iteration is proceeded until the value of WTO converges as shown below
s.no WTO(initial,kN) We/WTO WTO(final,kN)
1. 2266.11 0.46185 1314.76
2. 1314.766 0.47719 1420.09
3. 1420.090 0.47499 1403.96
4. 1403.960 0.47531 1406.28
5. 1406.284 0.47526 1405.92
6. 1405.920 0.47527 1405.92
WTO = 1405.92kN
2.4 SECOND WEIGHT ESTIMATION:
In second weight estimation we choose an engine based on the WTO value obtained from the
first weight estimation and include the weight of the fuel to calculate the final WTO.
The procedure is illustrated below
Thrust to weight ratio , T/W = 0.294 ( from cluster diagram)
From this T = 413.362kN
Since we are employing two engines , thrust per engine
T = 232.516kN
For safety reasons, T = 1.2T
T = 279.019kN
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So the engine selected to meet the above thrust requirement is GenX- 1B54
Weight of the powerplant , Wpp = 57.054kN
W pp / W TO = 0.081158
Weight of the fuel , Wf = (Ne * r * c * Talt * 1.2)/ V
Where Ne is the no of engines
r is the range, km
c is the specific fuel consumption of engine, lb/lbf-hr
V is the cruising speed, km/hr
T alt is the thrust at the cruising altitude
To is the thrust produced by the engine at sea level , lbf
σ is the density ratio =
W f = 423.718 kN
Wf/ WTO = 0.30136
WTO = (W pl+Wcrew+Wpp) /(1-Wf/ WTO-We/WTO –Wpp/WTO)
WTO =2714.551kN
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s.no WTO
(initial,kN)
Engine selected We/ WTO WTO (final)
1. 1405.99 GenX-1B54 0.4752 0.3014 0.08116 2714.55
2. 2714.55 GE90-110B1 0.4569 0.3275 0.05986 2788.96
3. 2788.96 GE90-115BL 0.4561 0.3188 0.05826 2604.47
4. 2604.47 GE90-110B1 0.4580 0.3188 0.06329 2701.72
5. 2701.72 GE90-110B1 0.4570 0.3291 0.06015 2825.57
6. 2825.568 GE90-110BL 0.4567 0.3294 0.05751 2761.68
Since the error involved is less than 3% between the last two iterations WTO = 2761.68 kN
WTO = 2761.68 kN
So the engine selected is GE90-115BL
Engine specifications:
Type : axial flow, twin shaft bypass turbofan engine.
Compressor : axial, 1 wide chord swept fan, 4 low pressure stage, and 9 high pressure
stages.
Turbine : axial,6 low pressure stages, 2 high pressure stages
Length :7290 m.
Diameter : 3.429m
Dry weight :81.256kN
Pressure ratio : 42:1
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Thrust : 514 kN( at sea level)
c : 0.34 lb/lbf-hr
CHAPTER 3
Placement of Engines
Engines can be placed on the wings, above the wings, suspended on pylons below the wings or
mounted on the aft of the fuselage.
Buried engines:
These engines have the minimum parasite drag and the minimum weight
They pose the threat of damaging the wing structure in the event of blade damage
and in other similar kind of issues
Inlet efficiency is less
Accessibility during maintenance is difficult.
Podded engines:
Acts as vortex generators
When placed outboard this arrangement provides wing bending relief .
Ease of maintenance.
Reduce noise within the cabin.
The placement of pods in front of the wing prevents flutter of wing.
This type of arrangement increases drag.
The engines are prone to foreign object damages.
Aft fuselage engines:
This type of arrangement is preferred for smaller aircrafts to maintain adequate
clearance.
Drag is less due to the elimination of wing pod interference drag
Greater center of gravity travel range which causes balancing problem.
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Vibration and noise reduction are difficult
Wing bending relief is not obtained in this arrangement.
So considering the advantages and disadvantages the engines are suspended by pylons under
the wings.
CHAPTER 4
DETERMINATION OF WING PARAMETERS
Wings are the most important component which produces lift and also produces drag and
pitching moment. The important issue to be kept while determining the necessary wing
parameters is to produce lift at its maximum efficiency and to reduce drag and pitching moment
while satisfying the mission requirements
The parameters to be determined are
1. Wing planform area, S
2. vertical position of the wing i.e. high, mid or low wing position
3. Aspect Ratio, ARw
4. Span, b
5. Taper Ratio,λw
6. Root Chord,
7. Tip chord
8. Mean Aerodynamic Chord,
9. Twist angle
10. Sweep Angle, Λ
11. Wing setting angle, iw
12. High Lift devices
4.1 Wing vertical location:
The vertical loction of wing can be of four types
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1. high wing
2. mid wing
3. low wing
High Wing:
Advantages:
facilitates loading and unloading of loads and cargo in the cargo aircraft
clearance is higher and thus prevents foreign particle damage to the engine.
High wing configuration increases the dihedral effect and lowers the stall speed since Cl
max will be higher
Disadvantages :
The frontal area is more and thus it produces drag.
Since the ground effect is low, it increases the take-off run distance
The landing gear length will be longer and requires more space to be stored while
retracted.
It requires the tail surface to be 20% larger when compared to low wing configuration
The weight of the wing is 20% greater when compared to low wing configuration
This type of configuration produces more induced drag as it produces more lift.
The lateral control of this configuration is weaker as it is more dynamically stable.
Low wing:
The aircraft take off performance is improved due to the influence of ground effect.
Landing gear is shorter and thus reduces weight
The wing has less induced drag
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This type of configuration has more lateral control since the aircraft has less lateral
dynamic stability.
The wing has lesser downwash on the tail thus increasing the effectiveness of tail.
The tail is lighter when compared to high wing configuration
This configuration facilitates the use of wing carry through structure to effectively
transmit the loads and thus this type of configuration is usually selected for passenger
aircraft
Disadvantages:
Since the wing has two separate sections it produces less lift and has higher stall speed.
The take off run is increased due to less lift produced
This type of aircraft requires long landing length.
This type of configuration has negative effect on dihedral effect.
Mid wing:
The aircraft structure is heavier due to the need of reinforcing wing root at the
intersection of the fuselage
The wing is costlier to manufacture.
The mid wing is aerodynamically streamliner and it an attractive option compared to the
other two configurations.
In general high wing configuration is preferred for cargo aircrafts where as low wing
configuration for passenger aircrafts.
4.2 OVERVIEW OF WING PARAMETERS:
Aspect Ratio, AR:
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Aspect ratio is defined as the ratio between the span wing span to the wing Mean Aerodynamic
Chord (MAC).
AR = b/ MAC
For the rectangular or simple tapered wing planform area, S is defined as
S = b* MAC
AR = b*b / (MAC*b) = b2/S
As the Aspect ratio of the wing is increased the aerodynamic characteristics such as Cl,
Cd , Cm of 3-d wings gets reduced to 2-D characteristics.
Fig 4 Effect of increase in AR
As the aspect ratio of wing is increased the stall angle of the wing gets reduced to the
value of the stall angle of the airfoil section. This is the reason to have the tail with low
AR as it has higher stall value and it is more effective in recovery even if the wing stalls.
As the AR is increased the wing becomes heavier as the moment arm gets larger which
results in higher bending stress at the root section which requires the root to be large
enough to withstand the stress. So as the weight of the wing increases the manufacture
cost also increases.
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The increase in AR facilitates an increase in the maximum lift to drag ratio.
As the AR is increased the induced drag is reduced as it follows the relation below
The effect of wing tip vortices on the horizontal tail is reduced as the AR is increased.
As the AR is increased the aileron arm is also increased as they are placed outboard of
the wings which results in high lateral control but because the stiffness is reduced it
results in aileron reversal problem, so an optimum value of AR is to be selected
TYPICAL ASPECT RATIO FOR DIFFERENT AIRCRAFT TYPES
Aircraft type Aspect Ratio , AR
Hang glider 4-8
Sail plane 20-35
Home- built 4-7
General Aviation 5-9
Jet trainer 4-8
Subsonic Aircraft 6-10
Super sonic 3-5
Tactical missile 0.3-1
Hypersonic 1-3
Span, b:
The span is the distance between the tip of one wing to the tip of other.
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Increase in the span results in the reduction of induced drag but it has several constraints
also. There is the concern about wing bending as it affects the stability and results in the
flutter of the wing.
Increase in span results in the increase in the weight of the structure which in turn
increases the manufacturing cost.
Increase in span results in the reduced Reynolds numbers of the section which in turn
reduces the lift produced by the wing.
Taper Ratio, λw:
Taper ratio is defined as the ratio of length of chord at the tip to the length of chord at the root.
The reason for providing taper to the wing is to reduce the lift produced at the tips and to obtain a
lift distribution near to the ellipitical lift distribution. The taper reduces the lift produced at the
tips the tips which in turn reduces the induced drag. The use of taper to the wings also results in
the reduction of the weight. Taper ratio for low swept wing is around 0.4 -0.5 whereas for a
highly swept wing it is around 0.2 – 0.3.This is because the swept wing diverts the air towards
the tip which “results in the production of lift at the tips. So to make the lift distribution nearlt
ellipitical taper ratio has to be reduced.
Fig 5 Effect of taper ratio on CL
Sweep Angle, Λ:
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The angle between the wing leading edge and the longitudinal axis (y –axis) of the aircraft is
called as the leading edge sweep angle, Λ lew. Similarly the angle between the quarter chord line
and the longitudinal axis is the quarter chord sweep, Λ c/4 . In the same way trailing edge sweep is
also defined.
Fig 6 leading edge sweep , Λ lew
Fig 7 quarter chord sweep, Λ c/4
Fig 8 trailing edge sweep, Λ tew
The features of swept wing
Improving aerodynamic features of the wing i.e. lift, drag and pitching moment bby
delaying compressibility effect.
The main reason incorporating this feature is to increase the critical mach number of the
wing.
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Wing maximum lift coefficient decreases with the increase in sweep as followed by the
relation ,
CLmax = Clmax [0.86 – 0.002Λ] where Clmax is the maximum lift coefficient of the
airfoil.
Thus it increases the stall speed.
An increase in the wing sweep tends to reduce the wing curve slope, CLα.
CLα = 2πAR/ ( 2+(AR2(1+tan2Λc/4 – M2+4)0.5)
Swept wing tend to produce negative rolling moment due to the changes in the velocity
vector normal to the leading edge between the left and the right wing sections.
Tip stall is a very serious issue in swept wing. If the outboard section of the wing stalls,
the loss of lift is behind the aerodynamic centre of the wing while the lift maintained at
the inboard section of the wing ahead of the aerodynamic center produces a pitch up
moment moving the aircraft into deeper stall. This when combines with the pitching
moment produced by tail becomes a very serious issue.
The aircraft pitching moment is increased as the aircraft cg is in front the aircraft
aerodynamic center as the aerodynamic center tends to move behind with the increase in
sweep.
For low subsonic aircraft which travels at a speed less than mach of 0.3 as it tends to increase
the cost of manufacture while availing little benefit of sweeping the wing.
Twist Angle, α t:
The main reasons for introducing the twist angle are
To avoid tip stall
To modify the lift distribution near to the elliptical distribution.
The twist can be provided in form of geometric twist or aerodynamic twist. In geometric twist
if the tip is at lower incidence than the root , it is said to be wash out or negative twist. If the tip
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is at higher incidence than the root, it is known positive twist or wash in. If the whole wing
sections is composed of different airfoils each with a different zero lift angle , it is known as
aerodynamic twist.
The disadvantage of incorporation of the twist is that it results in the production of lower lift
since the incidence angle is decreasing towards the tip. The twist angle must not be high enough
such that it produces negative lift towards the tip.
Fig 9 effect of negative twist on lift distribution
The criterion to be followed in selecting the twist angle is
|α t| + i w => |α o |
Where α o is the zero angle of attack.
For aircrafts usually only negative twist is provided. Typical value is between -1 to -4 degree.
Dihedral Angle, Γ:
If the wing tip is higher than the x-y plane , it is said to have positive dihedral or simply dihedral.
If the wing tip is lower than the x-y plane , it is said to have negative dihedral or anhedral
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The reason for incorporating this feature is to improve the lateral stability. Lateral stability is
mainly the tendency of the aircraft to return to the original trim level condition when disturbed
by gust and rolls around the x – axis. The lateral static stability is represented by Clβ which is the
change in lift co efficient with change in aircraft sideslip angle. For lateral stability this
derivative is negative which is due to positive dihedral.
Typical Values of Dihedral Angle for Different Aircraft types
Wing Low wing Mid wing High wing
Unswept 5-10 3-6 -4 to -10
Low subsonic swept 2 to 5 -3 to +3 -3 to -6
High subsonic swept 3to8 -4 to +2 -5 to -10
Supersonic swept 0 to -3 1 to -4 0 to -5
Hypersonic swept 1 to 0 0 to -1 -1 to -2
4.3 DETERMINATION OF WING PARAMETERS:
Following the same example
Planform area, S = WTO / (W/S)
Where W/S and WTO (from 2nd weight estimation) are known
AR = b2 / S (to determine the span, b)
Where AR is known from cluster diagram
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Fig 9 typical trapezoidal wing
The dimensions of the length of root chord, length of the tip chord and MAC aerodynamic chord
is calculating assuming the wing to be a trapezoidal wing.
Root chord,
Tip chord , Ct = λw * Cr
Distance of the MAC from the aircraft centerline
Position for aerodynamic centre is 0.25 MAC for subsonic aircraft
Position of the aerodynamic centre is 0.4 MAC for supersonic aircraft.
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Fig 10 position of the MAC
Geometrical Wing Parameters:
Planform area, S 465.342m2
Span, b 59.55m
AR 7.62 (from cluster diagram)
Taper ratio,λw 0.201
Dihedral angle, Γ 5 deg
Root chord, Cr 13.014m
Tip chord , Ct 2.616 m
8.968 m
Y 11.586m
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4.4 Airfoil Selection:
Fig 11 typical airfoil
Chord of the airfoil is the straight line from the leading edge to the trailing edge of the
airfoil.
Mean camber line is the line equidistant from the upper and lower surfaces of the airfoil.
Airfoil camber is the maximum distance between the mean camber line and the chord line
expressed as percentage of chord.
Thickness to chord ratio is the ratio between the maximum thickness of the airfoil to the
chord of the airfoil
The trailing edge of the airfoil is not perfectly sharp as it is difficult to manufacture, so the
trailing edge has finite thickness.
Requirements for the Selection:
1. highest possible lift co efficient
2. proper design lift co efficient
3. lowest minimum drag co efficient
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4. highest lift to drag ratio
5. highest lift curve slope
6. lowest pitching moment co efficient.
4.4.1 Overview of airfoil parameters:
Thickness to chord Ratio:
t/c affects the maximum lift and stall characteristics by the effect of nose radius
high AR and large nose radius provides higher stall angle and greater Clmax
thickness also affects the structural weight. The airfoil at the root can be thicker than the
tip without causing much drag for accommodating landing gear and to reduce the
structural weight and hence the cost.
Figure 11 effect of t/c on Clmax
For low speed aircraft which requires high lift coefficient like cargo aircraft, (t/c) max is between
15% to18%.
For high subsonic aircraft, (t/c) max is 9% to 12%.
For supersonic aircraft, (t/c) max is 6 % to 9 %.
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Note on NACA airfoils:
Basically there are three groups of airfoils
four - digit NACA airfoils
five – digit NACA airfoils
6 – series NACA airfoils
Four – digit airfoil:
These airfoils are generated by two parabolas. One parabola from the leading edge to the
maximum camber and the other parabola from the maximum camber to the trailing edge. This
type of airfoil generates more drag compared to other airfoils.
The nomenclature is as follows
1st digit gives the maximum camber in % of chord
2nd digit gives the position of maximum camber in tenths of chord.
Last two digits gives the thickness of the airfoil.
For example, NACA 2415 airfoil has a maximum thickness of 15% with a camber of 2% located
at 40% from the airfoil leading edge.
Five – digit airfoil:
These airfoils are generated using one parabola from the leading edge to the maximum camber
and a straight line that connects the end point of parabola to the trailing edge. The maximum
camber has been shifted forward to produce greater lift.
The nomenclature is as follows
1st digit multiplied by 3/2 gives the design lift co efficient in tenths
Next two digits divided by 2 gives the position of maximum camber in tenths of the
chord.
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Last two digits gives the thickness of the airfoil.
For example, NACA 23012 airfoil has a design lift co efficient of 0.3,maximum thickness of
12% and maximum camber located at 15% behind the leading edge of the airfoil.
Six – digit airfoil:
These airfoils were used to maintain laminar flow over a large portion of the airfoil and thus
maintain Cdmin when compared to other two series of the airfoil.
The nomenclature is as follows
6 denotes the series
2nd digit shows the location of minimum pressure in tenths of the chord
The suffix and the digit next to the ‘-‘ denotes the range where low drag is maintained
Final two digits specify the thickness.
For example, NACA 641- 212 airfoil, 6 denotes the series, with the minimum pressure location
of 0.4C,low drag maintained between 0.1 to 0.2 with the thickness of 12%.
Supercritical airfoils:
The key elements in the design of the supercritical airfoil are
A relatively large leading edge radius is used to expand the flow at the upper surface
leading edge, thus obtaining more lift
To maintain supersonic flow along a constant pressure plateau, by slowing the flow going
into the shock , a relatively weak shock compared to the amount of lift generated used to
bring the flow down to subsonic speed
Another means of obtaining lift without shock is to use aft camber
To avoid flow separation, the upper and lower surfaces at the trailing edge is nearly
parallel, resulting in finite thickness trailing edge.
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The major disadvantage is this kind of airfoil produces a large zero lift pitching moment.
The nomenclature is as follows
The first two digits denote the design lift co efficient in tenths
The last two digits denote the thickness in %
For example, SC- 0612 airfoil has a design lift co efficient of 0.6 and thickness of 12%
Wing incidence, iw:
Wing incidence is the angle between the fuselage reference line and the wing chord line of the
airfoil at the root.
The conditions to be satisfied in selecting the incidence angle are
The wing must be able to generate desired lift co efficient during cruise
Minimum drag must be obtained from the wing and fuselage during cruise.
Figure 1 setting of wing incidence angle
The wing incidence angle corresponds to the ideal lift co efficient (Cli) in the Cl vs. α graph.
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4.4.2 Steps involved in selecting the airfoil: (along with example)
The first step in selecting the airfoil is to calculate the design lift co efficient (Cl design)
Cl design = WTO / ( 0.5* ρ *V2*SW )
Cl design = 0.55 ( approx 0.6)
airfoils taken into consideration which produces the desired design lift co efficient are
1. 641- 412
2. SC – 0612
In the figure shown below the center of the drag bucket indicated the ideal lift co efficient of the
airfoil and the right end of drag bucket indicates the design lift co efficient.
Fig 13 location of design lift co efficient
Sweep angle is calculated using the formula given below to satisfy the selected Mach
Number
( 1 – M dd )/ (1 – M dd Λ=0 ) = (1 – Λ / 90 )
Where
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Mdd is the drag divergence Mach number
M dd Λ=0 is the drag divergence Mach number at Λ = 0
For NACA 6 series airfoil, M dd Λ=0 = 0.66
For Super Critical airfoil , M dd Λ=0 = 0.78
For NACA 6 series airfoil
Λc/4w(deg) Mdd
30 0.77
38 0.82
45 0.84
48 0.86
For Super Critical airfoil
It is to be noted that the drag divergence Mach number must be 0.05 greater than the
cruise Mach number.
So with a sweep of 43 deg we obtain the desired Mach number with Supercritical airfoil.
Λc/4w(deg) Mdd
30 0.85
33 0.86
35 0.87
43 0.89
34
So we choose the SC -0612 airfoil.
Specifications :
Design lift coefficient 0.6
Thickness 1.1%
Max Cl 1.364
Stall angle 14 deg
Max L / D 25.94
Zero lift angle -4 deg
zero lift pitching moment co efficient, C mo -0.013
4.5 Selection of high lift devices :
High lift devices are used to produce additional lift by employing the following methods
Increasing the airfoil camber
Boundary layer control from improved pressure distribution and re energizing or
removing low energy boundary layer
Increment in the effective wing area.
Factors involved in the selection are
Incremental drag
Mechanical complexity
Maintenance cost
Development
35
Structural weight
A high lift devices tends to cause the following changes in the airfoil features
Lift coefficient is increased
Maximum lift coefficient is increased
Zero lift angle of attack is changed
Stall angle is changed
Pitching moment coefficient is changed
Drag coefficient is increased
Lift curve slope is increased.
Two main groups of high lift devices
1. leading edge high lift devices such as leading edge flap, leading edge slat and
Kruger flap modify the pressure distribution over the top surface of the main wing
body
2. Trailing edge high lift devices such as plain flap, split flap, slotted flap, double
slotted flap, triple slotted flap and fowler flap when deflected downward increase
the camber of the wing.
In designing the high lift device for a wing, the following items has to be determined
High lift device location along the span
The type of high lift devices
High lift device chord, Cf
High lift device span, bf
High lift device maximum deflection, δ max.
36
Types of high lift devices and the corresponding ΔCL
s.no High lift devices Δ CL
1. Plain flap 0.7
2. Split flap 0.7-0.9
3. Fowler flap 1-1.3
4. Slotted flap 1-1.3
5. Double slotted flap 1-1.6
6. Triple slotted flap 1-1.9
7. Leading edge flap 0.2 – 0.3
8. Leading edge slat 0.3 – 0.4
9. Kruger flap 0.3 – 0.4
Placement of the high lift devices in the wing chord :
Plain flap – 30% of wing chord
Split flap – 25% of wing chord
Slotted flap – 35 % of wing chord
Fowler flap – 40 % of wing chord
Leading edge devices – extends 15% to 20% of wing chord.
The span occupied by the high lift devices is about 70 % of the semi span of the wing.
37
4.5.1 Steps involved in Determination of high lift devices:
The take off speed of the aircraft is assumed and it is usually between 270 km/hr to
345 km/hr.
The Stall velocity of the aircraft is found using Vto = 1.2 Vs
The maximum lift co efficient produced by the aircraft is found using
CLmax = WTO/ (0.5* Vs2*SW)
The landing speed of the aircraft is found using
V ls = 1.3 VS
The landing lift co efficient is found using
CLI = / ( 0.5*ρ* V ls * SW),[ aircraft lands with reserve fuel except for
emergency landing]
The lift produced by the wing is calculated using
CLW = 0.98 (Clmax r + Clmax t)/ 2
Where Clmax r and Clmax t is the maximum lift coefficient produced by the root airfoil
and tip airfoil respectively.
Difference in lift produced by the aircraft and wing helps us to determine the
required high lift devices.
Following the above steps
Lets assume the take off speed, Vto of 300 km/hr i.e. 83.33 m/s
Vs = 69.4441 m/s ; CLmax = 2.009; V ls = 90.2733 m/s ; CLI = 0.9
CLW = 1.2958; Δ Clmax = 0.7132
So we can use split flap to obtain the remaining lift coefficient.
38
C f /C = 0.20
δ f = 45 deg (for take off)
δ f =55 deg (for landing) to reduce the lift of 0.9
4.6 Placement of Aileron:
The aileron is placed at 20% of the chord of the wing.
C aileron / C = 0.20
So the rear spar is placed at 0.80C.
So the font spar is placed at 0.20C.
4.7 Volume of space available in wing for fuel:
Vwf =0.54[(S2 / b) (t / c)root {1+λw ζw0.5 + λw ζw / (1 + λw )2}]
Where
ζw = (t/c)r/ (t/c)t
Density of fuel = 800 kg/m3
Substituting the data in the above equation
Vwf = 315.85m3
Weight of fuel that can be accommodated inside the wing = 2478.754kN.
But the weight of the fuel estimated = 930.995 kN
So the fuel can be accommodated without any external devices.
39
4.8 Winglets:
Winglets are near vertical extension of the wing tips which increase the effective aspect ratioof
wing with less added wing span. This could reduce the strength of the wing tip vortices and
hence lower the induced drag. The winglet converts the wasted energy at the tip into apparent
thrust which in turns helps in less consumption of fuel. But the use of winglets leads to the
increase of parasite drag.
Fig 14 Winglet Shape definition
Where
Cwr is the root chord length of the winglet
Cwt is the tip chord length of the winglet
lw is the length of the winglet
40
Λw25 is the sweep at the quarter chord of the winglet
Prescribed values for the design of winglets:
Root chord length , Cwr / Ct = 0.5 -1.0
Taper ratio, Cwt / Cwr = 0.3 – 1.0
Sweep angle, Λw25 = 24deg – 50deg
Length, lw / Ct = 0.5 – 2.0
Cant angle = 0deg – 84 deg
Toe out (twist) angle = 0 deg – 4 deg
The winglet parameters
Length, lw 2.0928 m
Cant angle 50 deg (assumed)
Sweep angle, Λw25 50 deg (assumed)
Toe out angle 2 deg.
Taper ratio 0.8
Sweep angle chosen is 50 deg as it uses the same airfoil as that of wing, increase in sweep angle
increases the critical Mach number.
As a result of which the span of the wing is increased to 61.8418 m and thus the aspect ratio is
increased to 8.21
41
CHAPTER 5
DETERMINATION OF FUSELAGE PARAMETERS
The fuselage must be strong, rigid and light to avoid fatigue and failure of the pressurized cabins
low stress level must be chosen Drag of the fuselage must be low as it represents 20 to 40% of
the zero lift drag. Larger fuselage results in higher fuselage consumption decreased range and
increased takeoff weight.
Circular cross section of the conventional fuselage has an optimum enclosed volume, minimal
structural weight and minimum wetted area. Circular shell will react to the internal pressure
loads by hoop tension. Usually the cahin altitude is maintained at 8000 feet. Circular cross
section is efficient and lowest in structural weight. Non circular sections impose bending stress
in the shell structure. Sometimes a small radius arc is provided on the upper deck to provide
adequate head room whereas in some design the upper and lower radii arc is blended with a short
straight section.
Fineness ratio which is the length to diameter ratio plays an influential role in the design of
fuselage. Low fineness ratio results in drag penalty but can be stretched in future to meet the
needs. High fineness ratio is a long thin tubular structure which will suffer from dynamic
instability and is not flexible for future stretch.
The geometrical parameters to be determined for the fuselage are
Length of the fuselage , Lf
Diameter of the fuselage , Df
Typically for a 3 class arrangement 8% of seats are allotted for the first class, 13% of seats are
allotted for the business class and remaining for economy class.
42
Typical seat pitch values
Class Seat pitch, mm Seat pitch, inches
First class 950 - 1050 38 – 42
Business class 900 - 950 36 – 38
Economy class 775 - 850 31- 34
Charter class 700 - 775 28 - 31
Fig 15 Seat Pitch
Allocation of flight attendants:
One attendant per 30 to 40 passengers in economy class.
One attendant per 20 to 25 passengers in business class.
One attendant per 10 to 15 passengers in first class.
43
Allocation of facilities:
10 to 60 passengers each galley in which lower value dictates to the business class and
the higher value for the economy class.
15 to 40 passengers each lavatory which follows the same allocation as mentioned above.
Size of Galley = 762*914 mm (30*36 inches)
Size of toilet = 914*914 mm (36*36 inches)
The nose and the tail of the fuselage has to be streamlined to provide a smooth and reduce drag
shape.
Nose fineness ratio = 1.5; Tail cone fineness ratio = 1.8 to 2.0
Dimensions of the seats
First class Economy class
a, cm 50 43.5
b, cm 120 102
l ,cm 7 5.5
h, cm 107 107
m,cm 20 22
n ,cm 81 81
44
Fig 16 Dimensions of seat ( for clear diagram refer Egbert)
class k , cm p / p max
(cm/cm)
α /α max
(deg/deg)
First class 43 71/102 15/45
Economy
class
45 69/95 15/38
45
Fig 17 Dimensions of seat ( for clear diagram refer Egbert)
Emergency exits:
Emergency evacuation of the cabin plays an important role in deciding the fuselage layout. The
manufacturers should demonstrate to the airworthiness officers that all the passengers can
evacuate in 90 seconds or less in case of emergency
Requirements of emergency exits
Seats less than Type 1 Type2 Type 3 Type 4
10 - - - 1
20 - - 1 -
40 - 1 1 -
80 1 - 2 -
110 1 - 2 -
140 2 - 1 -
180 2 - 2 -
46
Dimensions of emergency exits:
Type 1: 24 wide * 48 high inches ( 610*1219 mm)
Type 2: 20* 44 inches (508*1118 mm)
Type 3: 20* 36 inches ( 508* 914 mm)
Type 4: 19* 26 inches ( 483* 600 mm)
Type A: 42*72 inches ( 1067*1829mm)
Type A exit is equivalent to passenger or service loading door. For more than 179 seats a pair of
additional type A exits allows an extra of 110 seats, a pair of type 1 exits will allow a extra of 45
seats. Above 300 seats, all exits must be of type A. Above 600 seats will require six type A exits
on each side of fuselage.
Flight deck:
The length of the flight deck varies from 110 inches (2.75m) for smaller aircrafts to 150
inches(3.75m) for larger aircrafts where the latter type can accommodate a third operator if
needed.
5.1 Steps for determining the geometrical parameters with the example
Total passengers = 242
Cabin attendants = 8 (as per the guidelines mentioned above)
No of passengers in economy class = 80% of 242=198
No of passengers in first class =20% of 242 = 44
Single aisle is chosen for the cabin design.
47
Seating arrangement:
Economy class = 6 abreast [3+3 ] = 33 rows
First class = 4 abreast [2+ 2] = 11 rows.
The total length of cabin = 39.116 m
For better accommodation the economy class is divided in two sections in which the first section
contains 17 rows and the next contains 16 rows
50 ft i.e 15.24 m is accounted for the front and aft of the cabin width
2.54 m for the 2 crew cockpit
Emergency exits
First class 1 type III
Economy class 2 type III
class Seat
pitch(inches)
Aisle
width(inches)
toilets Galley Total
length(m)
Economy
class
34 20.5 5 3 28.2988
First class 38 28.5 2 2 10.6172
48
Total length of fuselage, Lf 56.896 m (initial)
Diameter of the fuselage, D f 5.7 m (initial)
Length of nose, l nose 8.5 m
Tail cone ratio , l tail / D f 1.8 m
Length of tail con e , l tail 10.26 m
Total length of fuselage = 39.116+8.55+10.26+6.096
Total length of fuselage, Lf = 63.962m (final)
Diameter of the fuselage, D f = 6.33m (final)
49
CHAPTER 6
DETERMINATION OF HORIZONTAL PARAMETERS
The empennage is composed of horizontal tail and the vertical tail. The two primary functions
are trim and stability. The first and primary function of horizontal tail is longitudinal trim or
equilibrium or balance. The function of vertical tail is to provide directional stability.
The tail parameters to be determined are
1. Plan form area ,S ht
2. Aspect Ratio , AR ht
3. Span ,b ht
4. airfoil selection
5. Root chord length , C rht
6. Tip chord length , C tht
7. Taper ratio ,λ ht
8. Tail setting angle ,α ht
9. Mean aerodynamic chord ,
10. Sweep angle , Λ ht
The parameters given above have the same characteristics as that of wings. The major
differences which are encountered in the determination of tail parameters will be given below.
The different types of tail configuration are
Aft tail and one vertical tail
Aft tail and twin aft vertical tail
Canard and aft vertical tail
50
Canard and twin wing vertical tail
Triplane i.e aft tail as aft plane, canard as fore plane, wing as fore plane
Tailless i.e delta wing with one vertical tail
No formal tail
Planform area, Sht:
The tail volume co efficient, plays an important role in determining the planform area of the
horizontal tail.
Where
is the tail arm, m
Typical values of
Aircraft type
Glider and motor glider 0.6
Homebuilt 0.5
General aviation- single prop driven 0.7
General aviation- twin prop driven 0.8
General aviation with canard 0.6
agricultural 0.5
Twin turboprop 0.9
Jet trainer 0.7
Fighter 0.4
51
Fighter with canard 0.1
Bomber/ military 1
Jet transport 1.1
Tail arm, lt:
Tail arm is the distance between aircraft centre of gravity and the aerodynamic centre of the
horizontal tail. This serves as the arm for the tail pitching moment about aircraft centre of gravity
to maintain the longitudinal trim. Tail area is responsible in production of the lift and the tail arm
influences this parameter. As the tail arm is increased the tail area is decreased and the vice
versa.
Generally lt / Lf = 0.5
Differences between the ARw and ARt :
1. Elliptical distribution of lift is not required for the tail
2. Lower AR for tail than Wing. Lower AR results in smaller bending moment
3. For a single prop driven engine the tail span should be longer than the propeller
diameter as it avoids being fully immersed in the downwash region making the
tail inefficient.
Typically it follows the relation
AR ht = (3/5) AR w
The value is between 3 to 5.
Taper Ratio, λ ht :
52
Taper Ratio of the tail influences the aircraft lateral stability, control, aircraft performance, tail
efficiency, ŋt and the aircraft weight and the centre of gravity. The main reason for tail taper
ratio is to reduce the weight of the tail.
Typical values for general aviation aircrafts is 0.7 to 1.
Typical values for transport aircraft is 0.4 to 0.7.
Tail incidence ,α ht :
Tail incidence angle is used to nullify the pitching moment about the center gravity at cruising
flight.
α ht = CLt / CLαt
CLαt = Clαt / (1 + Clαt / π* ARht )
Where
Clαt is the lift curve slope of the tail airfoil section.
CLt is the lift coefficient of the tail.
Airfoil selection:
Requirements of empennage airfoil:
1. Since the CG moves during the cruising flight, the airfoil section must be able to create
both negative (-L ht) and positive (+L ht) lift, so the airfoil must be symmetrical.
2. The airfoil should have a low base drag co efficient.
3. Tail lift coefficient must be less than the wing lift co efficient
4. Horizontal tail airfoil section must be thinner than the wing airfoil section. Typically
around 9% thicknesses.
Sweep angle, Λ ht:
53
Sweep angle influences the aircraft longitudinal and lateral stability and control, aircraft
performance and the aircraft center of gravity. As an initial value it is taken 5 deg greater than
the leading edge sweep angle of the wing.
Dihedral angle, Γ ht:
The tail dihedral is same as the wing dihedral. The other dihedral angle which comes into play is
the longitudinal dihedral which has an effect on longitudinal static stability.
Tail vertical location:
The vertical location of the tail has two options
1. At the fuselage aft section
2. At the vertical tail
The wing influences the horizontal tail via downwash, wake and the trailing vortices. The wing
wake degrades the tail efficiency and decreases the tail dynamic pressure.
Three major regions for horizontal tail installation:
1. Out of wake region and downwash which is the safest and the best region.
2. Inside the wake region but out of wing downwash which is not recommended
3. Out of wake region but affected by the downwash which is the safe region from deep
stall and pitch up but tail is not efficient.
54
Fig 15 horizontal tail installation regions
Tail setting configuration:
There are three horizontal tail setting configuration
1. fixed horizontal tail
2. adjustable tail
3. all moving tail
Fixed horizontal tail is attached to the fuselage hinged with a longitudinal control surface
elevator to change the pitching moment of the aircraft. This design is easier and cheaper.
Whereas the in the all moving tail the whole of the horizontal tail plane is moved and it does not
employ an elevator. The trim drag of the fixed horizontal tail is greater than the all moving tail.
The all moving tail is also known as variable incidence tail plane.
Figure 16 various tail setting configurations
55
6.1 Procedure for Determining the Horizontal Tail Parameters:
Following the same example
= l * S ht / MAC * S
Assume = 1.1
Tail arm , lt = 30 m ( as lt / L f = 0.5 approx)
Substituting the values in the formula for V H
S ht = 139.12 m2
Geometrical Parameters of Horizontal Tail
Plan form area ,S ht 139.12 m2
Span ,b ht 23.59m
Root chord length , C rht 7.863m
Tip chord length , C tht 3.932 m
Taper Ratio, λ ht 0.5 (assumed)
Sweep angle Λ ht (leading edge) 50.562 deg
Sweep angle Λ ht (quarter chord ) 48.554 deg
Sweep angle Λ ht (trailing edge) 41.427 deg
Dihedral angle, Γ ht 5 deg
AR ht 4.572
Tail setting angle is determined in the Stability Section
Max C l 0.9
56
Cmo -0.008
Stall angle 13 deg
Cdmin 0.004
Airfoil selected for tail section is NACA 63 – 009
57
CHAPTER 7
DETERMINATION OF VERTICAL TAIL PARAMETERS:
The same parameters determined for the horizontal tail plane must be determined for the vertical
tail also with the suffix vt used in the naming convention.
Planform area, S vt :
The vertical tail volume co efficient ,VV plays an important role in determining the plan form
area of the vertical tail.
VV = S vt*l vt / b*SW
Typical values for VVT
Aircraft type VVT
Homebuilt 0.04
General aviation- single prop driven 0.04
General aviation- twin prop driven 0.07
General aviation with canard 0.05
agricultural 0.04
Twin turprop 0.08
Jet trainer 0.06
Fighter 0.07
Fighter with canard 0.06
Bomber/ military 0.08
Jet transport 0.09
58
Increasing the area will improve the lateral-directional stability, control and trim required. It will
also improve the directional and lateral control. If the vertical is too large, the aircraft will be
lateral directionally too stable but the directional control requirement are not satisfied.
Vertical tail arm, lvt:
The vertical tail arm must be long enough to satisfy the directional stability, control and trim
requirements. An increase in the vertical tail moment arm improves the directional and lateral
control. Since the vertical tail arm decides the position of the vertical plane it should be such that
it is out of wake region of horizontal tail.
Initially it is taken as lvt / lht = 0.95.
Vertical incidence angle, ivt:
The vertical incidence should be zero to maintain the symmetricity about the respective plane to
maintain equilibrium condition. But for propeller driven aircraft, the propeller will create a
torque so to maintain the equilibrium the vertical tail plane must be set at some incidence angle.
Aspect Ratio, ARvt :
If AR vt is higher it weakens aircraft lateral control but it has higher directional control. The
bending moment and the bending stress at the tail root will be higher which in turn makes the
vertical tail plane heavier. If the AR vt is large then the induced drag will be higher. For a T- tail
configuration, high AR vt keeps the horizontal tail from wake. Higher AR vt results in high
aerodynamic efficiency.
Typically it is around 1 to 2.
Taper Ratio, λvt :
The main reasons for giving taper to vertical tail is to reduce the bending stress on the tail root
and to allow the vertical tail to have a sweep angle. But an increase in taper ratio reduces the
lateral stability. Typically the value is around 0.5
59
Sweep angle, Λvt :
If the sweep angle is increased it results in directional control and for T-tail configuration the tail
arm is increased which improves the aircraft longitudinal stability and control. For initial design
assume a same value as that of horizontal tail.
Airfoil selection:
The airfoil must generate lift with minimum drag coefficient. To maintain symmetricity airfoil
must be symmetrical. The airfoil must be thinner so that the Mach number is less on vertical tail
compared to the wing. High value of lift curve slope is preferred as the derivatives of directional
stability depends on the slope.
7.1 Procedure for determining the Vertical tail Parameters:
vertical tail volume co efficient , Vv =(l vt * S vt )/ (b * Sw)
l vt / l ht = 0.95
l vt = 28.5m
The formulae used for the determination of wing parameters are used here.
Vertical Tail Parameters
Plan form area ,S vt 87.51m2
Span ,b vt 13.23 m
Root chord length , C rvt 8.82m
Tip chord length , C tvt 4.41m
Taper Ratio, λ vt 0.5
Sweep angle Λ vt (leading edge) 50.562 deg
Sweep angle Λ vt (quarter chord ) 46.373 deg
Sweep angle Λ vt(trailing edge) 28.772 deg
60
6.86m
AR vt 2.0
The same airfoil used for horizontal tail plane is selected i.e NACA 63 - 009
CHAPTER 8
61
WEIGHT ESTIMATION OF THE COMPONENTS
Till now we have estimated the weight of the aircraft with historical data but in this section the
weights of the individual components are calculated to find the accurate weight of the aircraft.
Basically four methods are followed for the estimation of weight of the components.
1. Cessna method
2. USAF method
3. GD (general dynamics) method
4. Torenbeek method
The overall idea for the designer is to obtain an optimum weight of the aircraft. Over weight of
the aircraft will suffer reductions in range, reduced climbing altitude, reduced maneuverability
and increased takeoff and landing distances. Two methods which are widely used which offer
reduction in weight are the use of composites and relaxation in inherent stability of the aircraft.
Fig 17 various factors which influence the weight of the aircraft
As the payload weight and the fuel weight has been already estimated, in this section the
estimation of the empty weight is done.
62
The empty weight consists of
1. fuselage weight ,Wfuse
2. wing weight ,Wwing
3. horizontal tail weight, Wht
4. vertical tail weight, Wvt
5. engine group weight ,Weg
The estimation of the weight of the aircraft is done following the same example
8.1 Fuselage weight, Wfuse :[all units in fps]
The weight of the fuselage depends upon the size and the aircraft layout. The ratio
of Wfuse / WTO = 7 – 12 %
Where
Sf is the surface area of the fuselage, ft2
Sf = π * Df * Lf
Kf = 1.08 (for pressurized fuselage)
= 1.07 (for main gear attached to the fuselage )
= 1.10 (cargo airplane with cargo floor)
VD is the dive speed., knots
Airworthiness requirement for civil aircraft set a minimum margin of M= 0.05 between the
cruise speed and the maximum dive speed
63
Dive speed is approximately chosen from the graph shown below
Fig 18 selection of dive speed
For the calculation we assume the dive speed, VD = 450 knots
Substituting the values in the above formula
W fuse = 368167.24 N
8.2 Weight of Wing , W wing : [ all units in fps]
Where WMZF = maximum weight of aircraft with out fuel.
tr is the thickness of the airfoil
L1/2 is the sweep angle at half chord, deg
nult is the ultimate load factor = 3.75
formula for calculating chord at half sweep
64
where n = ½ ; m = ¼
L1/2 = 40.17deg
Substituting the values in the above formula
W wing = 350751.056 N
8.3 Weight of Horizontal tail:( all units in fps)
Where K h = 1 (for fixed incidence tail)
= 1.1 ( for variable incidence tail)
= 33111.59 N
The same formula is applied for vertical tail with a suffix v
K v = 1.0 (for conventional tail)
= 17667.83 N
8.4 Weight of the Engine Group, : (all units in fps)
Where is the engine weight
is the thrust reverser weight
is the electrical start system weight
is the fuel system weight
65
= 50.38 [ /1000] 0.459
= 38.93 [ /1000] 0.918
= 80( Ne + Nt -1)+ 15 (Nt)0.5* (Wf /Kfs)
Where Nt is the number of fuel tanks, here it is 6
Kfs = 5.87 lbs/gal ( for aviation gasoline)
= 6.55 lbs/gal (for JP4)
= 39,603.44lb = 176.224kN
8.5 Weight of under carriage: (all units in fps)
Where
is the weight of under carriage, lb
= 1.0 for low wing aircraft
= 1.08 for high wing aircraft
For nose wheel,
= 20.0; = 0.10; Cg = 0; D g = 2*10-6
For main wheel,
= 40.0; = 0.16; Cg = 0.019; D g = 1.5*10-5
Wng = 14279.53 N
Wmg =101027.21N
8.6 Weight of fixed equipment, Wfe: (all units in fps)
66
Where
is the weight of the flight control
is the weight of the electrical system
is the weight of the oxygen system
is the weight of the furnishings
is the weight of the paint
is the weight of cargo and baggage handling equipment , lb
is the weight of the avionics
Where is the passenger cabin volume = volume allotted per passenger* no of passengers
Volume allotted per passenger = 65 ft3 [ transport jets]
Where
= 0.64 (for powered flight controls)
= 0.44( for unpowered flight controls)
=
Where is the number of passengers
= 0.0045*WTO
67
=
Where =0.0646 (with no preload provisions)
= 0.316 ( with preload provisions)
=
Where R is the range, nm
Wfe = 13148.38 N
CHAPTER 9
68
BALANCE DIAGRAM
Balance diagram is drawn to locate the C.G of the aircraft and to verify the shift the C.G is
within 5% to 15%.
Guidelines to construct the balance diagram
C.G locations of the Components:
For wing , horizontal tail and vertical tail are at 40 % of their respective M.A.C
For fuselage at 42% of its length
For the engine at 40 % of its length
Nose landing gear at 13% of fuselage length
Main landing gear at 55 % of the fuselage length
Engine group is about 2 m from the leading edge of the.wing.
All other components have at 42 % of the fuselage length.
For the crew the centre of the nose plane
For the whole aircraft at the quarter chord point of the MAC of wing
The shift has to be verified by finding the C.G for the following cases
Full fuel and full payload
Reserve fuel and full payload
Full fuel and zero payload
Reserve fuel and zero payload
Full fuel and half payload
Reserve fuel and half payload
69
Fig 19 axis system to draw balance diagram
The nose of the aircraft is taken as the reference point. The distance from the nose to the leading
edge of the wing is denoted by Xle and the calculations are preceded for each case.
Components Weight(N) C.G Locations (m)
Wing 350751.056 Xle +12.71
Fuselage 368167.24 26.86
Horizontal Tail 33111.59 Xle + 41.617
Vertical Tail 17667.83 Xle + 42.192
Powerplant 176224.19 Xle + 2
Nose Wheel 14279.53 8.32
Main Landing gear 101027.21 31.81
Fixed Equipment 13148.38 26.86
Fuel 926674.38 Xle + 11.366
Payload 2794091.4 27.5
70
Crew 2158.2 3.2
Gross Weight 2447301.014 Xle + 11.366
The moment equilibrium is applied at the nose and the distance of leading edge of the wing from
the nose tip is calculated.
Xle = 18.16 m
The C.G location for the aircraft is at 28.992m from the leading edge
C.G locations for different cases
Cases C.G Locations(m)
Full Payload + Full Fuel 28.992
Full Payload + Reserve Fuel 29.002
Zero Payload + Full Fuel 29.652
Zero Payload + Reserve Fuel 29.652
Half Payload + Full Fuel 29.26
Half Payload + Reserve Fuel 29.26
% shift of C.G =(most forward point of CG – most Aft point of CG)/ MAC
= 7.3%
CHAPTER 10
71
DETERMINATION OF LANDING GEAR PARAMETERS
The parameters to be determined are
1. wheel base
2. wheel track
3. load on each gear
4. tyre selection
5. stroke length
6. oleo length and the diameter
The basic landing gear configurations are
1. bicycle
2. tail dragger
3. tricycle
4. quadricycle
5. multi-bogey
Bicycle arrangement:
This type of arrangement has two main wheels fore and aft of the C.G and a outrigger
wheels on the wing to prevent the aircraft from tipping sideways. This is the oldest of the design.
Advantages are design simplicity and low weight. Since the wheels are placed at equal distances
about the C.G , they carry similar loads.
Tail dragger:
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This type of arrangement has two wheels forward of the C.G and a auxiliary wheel at the
tail. The main gears in front of the C.G carries most of the load. As the aircraft is always at an
high attitude the tail has to be lifted up during take off. This type of landing gear is directionally
unstable during ground maneuver i.e turn.
Tricycle arrangement:
This type of arrangement has two main wheels aft of C.G and an nose gear at forward of
C.G. this is the most widely used arrangement nowadays. The nose gear carries 5%– 15% of the
load and the main gear carries 85%- 95% of the load. The nose gear configuration is
directionally stable on the ground as well as during taxing.
Quadricycle arrangement:
This type of arrangement is similar to the arrangement in cars with wheels on either sides of thr
fuselage. This type of arrangement requires a flat attitude for take off and landing as it is very
hard to rotate the aircraft during landing and taxiing.
Multi bogey:
As the aircraft becomes heavier it is necessary to employ multiple wheels to carry the loads,
which lead to Multi-bogey arrangement. When multiple wheels are used in tandem, they are
attached to a strut element called a “bogey” or “ truck”. Aircrafts weighing above 50,000 lbs
employ multi bogey arrangement.
73
Fig 20 types of landing gear arrangements
The tip back angle is the maximum aircraft nose up attitude with the tail touching the ground and
the strut fully extended. This angle prevents the aircraft tipping at the tail during take off and this
should be greater than take off rotation angle. Typical take off rotation angle (α to) is 10 to 15 deg
so the tip back angle (α tb )should be around 15 to 20 deg. The tip back angle regulates the most
forward and the most aft position of centre of gravity in tricycle configuration.
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Figure 21 tip back angle and the take off rotation angle
The overturn angle is the aircraft tendency to overturn when taxied around a sharp corner and
this angle should not be greater than 63 deg.
Wheel base:
The wheel base plays an important role in distributing the load between the main gear and the
nose gear. It is the distance between the nose gear and the main gear from the side view. Due to
ground controllability, the nose gear must carry load greater than 5% and not more than 20%
which implies that the landing gear must carry 80% to 90% of total weight.
Wheel track,T :
It is the distance between the left and the right of the main wheels. The minimum value for the
wheel track must satisfy the overturn angle required and the maximum value should satisfy the
structural integrity requirement. The wheel track should be such that it should satisfy the
following requirement that the overturn angle must be greater than or equal to 25 deg.
10.1 Loads acting on the gears: (all units in SI)
Maximum static load:
On nose gear : Fnmax = Bmmax*WTO/ B
On main gear: Fmmax = Bnmax*WTO/ B
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Minimum static load:
On nose gear : Fnmim = Bmmin*WTO/B
On main gear : Fmmin = Bnmin*WTO/B
Figure 22 For the calculation of loads
Dynamic load :
On the nose gear, Fndyn = al* WTO * Hcg /g*B
On the main gear,Fmdyn = at* WTO * Hcg /g*B
Where Hcg is the distance from the ground to the C.G of the aircraft
al is the landing deceleration, 3 m/s2
at is the take off acceleration, 4 m/s2
g is the acceleration due to gravity, 9.81 m/s2
The load carried by the nose gear, Fn = Fnmax + Fndyn
The load carried by the main gear, Fm = Fmmax +Fmdyn
The nose gear must withstand both the maximum static and dynamic loads.
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Following the steps above and as per the balance diagram
Bnmin = 20.672 m
Bnmax= 21.332 m
Bmmax = 2.989m
Bmmin= 2.329m
Fn = 39112.53kg = 383.694kN
Fm =230007.85kg = 2256.377kN.
We are employing two wheels for the nose gear and each wheel carries 191.846kN
The load calculated for the main gear gets equally distributed between two sides and we are
employing six wheels on each side which accounts to a total of 12 wheels. Each wheel carries
188.031kN
10.2 Braking kinetic energy:(all units in fps)
The braking kinetic energy , K.E braking = 0.5*Wlanding*Vstall2
Sometimes there is a need for emergengy landing, so consider Wlanding to be 90% of the WTO
K.E braking = 387*106 ft-lb/s
The western design employs brakes only on the main gear whereas the soviet design employs
brakes in both nose and landing gear.
Here we will employ the brakes in the main gear. So the K.E braking is 32.25*106 ft-lb/s per wheel
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Figure 2 wheel diameter for braking
From the above graph, the wheel diameter should be around 22 inches.
10.3 Tyre selection :
Pressure recommended for different runways
Surface PSI
Aircraft carrier 200+
Major military airfield 200
Major civil airfield 120
Tarmac runway,good foundation 70 – 90
Tarmac runway, poor foundation 50 – 70
Temporary metal runway 50 – 70
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Dry grass on hard soil 45 – 60
Wet grass on soft soil 30 - 45
Soft sand 25 - 35
From the tyre data book,
Dimensions of Wheel Rim
Type of gear Diameter Width
Nose Gear 0.5334m 0.5334m
Main Gear 0.55m 0.34m
Dimensions of Tyre Selected
Type of gear Size(inches) Diameter Width
Nose Gear 48*15.00 – 21 1.22 m 0.38m
Main Gear 50 * 15.75- 22 1.28m 0.4m
10.4 Pressure calculated:
Foot print area , Ap = 2.3 (d*w)*( d/2 – Rr)
Where d is the diameter of the tyre , inches
w is the width of tyre, inches
Rr is the rolling radius = 2/3 of tyre radius
79
Where is the load carried by each wheel, lb
P is the pressure required, PSI
For nose gear , P = 87.6 PSI
For main gear , P = 77.64 PSI
10.5 Stroke :
Stroke is the required deflection of shock absorbing system to absorb the shocks of bad
landing. The most widely used shock absorbing system is oleo pneumatic shock absorbing
system. This type of shock strut absorbs shock by forcing a chamber of oil into the chamber of
compressed air and nitrogen and then compressing both gas and oil together. To improve the
efficiency of the device the orifices change its size as the oleo compresses.
The stroke is calculated using the formula,
E = ŋ LS + ŋt L St
Where
E is the kinetic energy
Ŋ is the efficiency of the shock absorber (for oleo 0.80)
S is the stroke to be calculated
Ŋt is the tyre efficiency (for tyre 0.47)
St is the tyre deflection = d/2 – Rr
E = (0.5*Wlanding*V2)/ g
V is the vertical velocity at touch down, 10 ft/s
Ŋ*S*N + Ŋt*St*N = V2 /2g + (1 – N) ( S +St)
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Where N is the design reaction factor (0.7 – 1.2)
With 1.2 most widely used for transport aircrafts.
From the above formula ,
stroke ,S = 10.77 inches
For safety purpose add 1 inch to the above value
S = 11.77 inches = 0.29 m
Typical value is 8 to 12 inches.
10.6 Oleo sizing :
Figure 24 oleo strut
Static position is approximately 84% of stroke. The total length of the oleo including the stroke
distance and the fixed position of oleo is 2.5 times of stroke.
The static position of oleo = 9.668 inches = 0.25 m
Length of oleo, loleo = 29.45 inches = 0.74 m
Diameter of oleo, Doleo = 0.04*(Loleo)0.5
81
where
Loleo is the load carried by each oleo, lb
For nose gear,
Doleo = 8.31 inches = 0.21m
For main gear
Doleo = 8.22 inches = 0.21m
10.7 Calculation of wheel track:
For a jet aircraft a clearance of 1.5 m is allotted at the inlet of the engine
Hcg = 6.2 m
We are assuming a value of 30 deg in the above fig
From the above fig, tan 30 = (T/2 )/ Hcg
Wheel track, T = 7.14m
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Figure 25 for track calculation
From the right hand side figure,
Tan Φ1 = AC / FC
Φ1= 8.6 deg
Sin Φ1 = DE/AD
DE = 4.334 m
From the left hand side figure,
Tan Φ ot = 4.334/ Hcg
Φ ot = 35 deg
This satisfies the overturn angle criterion that it should be greater than 25 deg
So the minimum track length is 7.14 m but the fuselage diameter is 6.33 m, so the landing gear
cannot be attached to fuselage. But take into consideration the space available while the landing
gear is retracted.
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CHAPTER 11
DRAG POLAR
Drag force is the summation of all forces resisting the motion of the aircraft. Usually drag force
is represented in form of non dimensional parameter,CD which is known as the drag coefficient.
This drag coefficient take sin to account every configuration which contributes to the drag. The
mathematical model of variation of CD vs. CL is given by
The above equation clearly represents that the parabolic variation of CD vs.CL.
The first term CDO is the zero lift drag co efficient
The second term is the induced drag co efficient,CDi = The constant K is the induced drag
correction factor which can be easily calculated using the formula
K = 1/(π*e*AR)
Figure 26 typical variation of CD vs. CL
84
Zero lift drag coefficient, CDO:
This represents the drag associated with the frictional characteristics, shape and protuberances in
the aircraft structure not associated with production of lift. It increases with aircraft velocity and
it is the main factor in determining the maximum speed of the aircraft.
Induced drag coefficient, CDi:
This represents the drag produced by the aircraft wings as a result of induced vortices on a finite
aspect ratio wing. It first decreases with speed with its contribution highest at low velocities and
decreases with increase in speed.
Figure 27 drag classification
Induced drag coefficient can be easily calculated from the given expression but the calculation of
the zero lift drag is a tedious procedure. It has to be kept in mind that this calculation proceeded
85
is applicable for subsonic aircraft and the same smplified expression cannot be used for
supersonic flights.
11.1 CALCULATION OF CDO
CDO is the summation of all contributing components and each and every component of the
aircraft have only positive contribution to CDO
CDO = CDOf +CDOw + CDOht + CDOvt + CDOn+ CDOhld + CDOlg
Where CDOf is the zero lift drag produced by fuselage
CDow is the zero lift drag produced by wing
CDOht is the zero lift drag produced by horizontal tail
CDOvt is the zero lift drag produced by vertical tail
CDon is the zero lift drag produced by nacelle
CDOhld is the zero lift drag produced by high lift devices
CDolg is the zero lift drag produced by landing gear.
11.1.1 Fuselage:
The zero lift drag for the fuselage is given by
Where Cf is the friction coefficient
Cf = 0.455*( log Re ) -2.58 ( for turbulent flows)
Cf = 1.327 / [( Re)0.5] ( for laminar flows)
Reynolds number, Re = ρ*V*l / μ
Where μ is the viscosity coefficient at that altitude
.
86
Mostly aircraft fly under the combined conditions of laminar flow and turbulent flow. But its
better to overestimate the values rather than underestimation in this case.
fld is the function of fuselage fineness ratio (Lf/Df)
fld = 1+ 60 /[ (Lf/Df)3]+ 0.0025(Lf/Df)
fm is the function of mach number
fm = 1- 0.08M1.45
Swetf is the wetted area of the fuselage
S is the wing reference area (mostly the gross area)
Swetf = π*Df*Lf*[1- 2/(Lf/Df)]0.67*[1+ 1/( λ2)]
Following the same example,
For cruise flight
At 11km
μ = 14.17 *10-6 ; ρ = 0.364kg/m3 ; V = 250.92 m/s
Re = 4.12 * 108
Cf = 0.001758
fld = 1.083
fm = 0.9365 ( M = 0.85)
Swetf/S = 2.37
CDOf = 0.004227
87
11.1.2 WING, HORIZONTAL TAIL, VERTICAL TAIL:
The zero lift coefficients of wing is given by
The zero lift coefficient of horizontal tail is given by
The zero lift coefficient of vertical tail is given by
Cfw, Cfht, Cfvt is defined in the same way as that of fuselage. The only difference is instead of lf
we use the MAC of the respective surfaces under consideration.
Cdmin is the minimum drag coefficient of the respective airfoils of the surfaces.
ftc is the function of thickness to chord ratio
ftc = 1+ 2.7 ( t/c ) max + 100 (t/c)max
Swet of lifting surfaces is given by
Swet = 2.0*[1+0.2(t/c)]*S
Following the same example,
For wing , CDOw = 0.00492
For horizontal tail, CDOht = 0.00173
For vertical tail, CDOvt = 0.000961
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11.1.3 Nacelle :
As it is similar to the shape of the fuselage, the same formula can be applied. The diameter of the
nacelle is about 10 % larger than diameter of the engine and the fineness ratio is around 1.5 to 2.
Following the same example
CDOn = 0.00623
For cruise
CDO = CDOf + CDOw + CDOht + CDOvt + CDOn
CDO = 0.0168
11.1.4 Landing gear:
The zero lift drag coefficient due to landing gear comes into the picture only during
take-off and landing because during cruise they are retracted into their respective locations.
CDOlg = Σ CDlg (Slg/S)
Where CDlg is the drag coefficient of each wheel = 0.3 (for landing gear with fairing)
= 0.15 ( for landing gear without fairing)
Slg is the frontal area of each wheel = diameter of tyre * radius of tyre (in meter)
Following the example,
CDOlg = 0.3 [ 12*1.28*0.40 + 2*1.22*0.38]/ 465.342
CDOlg = 0.0045
89
11.1.5 Trailing edge high lift devices:
The zero lift contribution of high lift devices is also obtained during take off and landing. It is
given by
CDOthld = (Cf /C)* A*df B
Where
Cf / C is the ratio of high lift device chord to the wing chord
df is the deflection angle corresponding to take off or landing
Flap type A B
Split flap 0.00014 1.5
Plain flap 0.00016 1.5
Single slotted flap 0.00018 2
Double slotted flap 0.00011 1
Fowler flap 0.00015 1.5
Following the same example
For take – off , df = 45 deg , Cf/C = 0.20 , A= 0.00014,B = 1.5 (as we are using
split flaps)
CDOthld = 0.00845
For landing , df = 55 deg
CDOthld =0.0114
90
11.1.6 Leading edge high lift devices:
The zero lift drag contribution of the leading edge devices is given by
CDOlhld = (Cle/C)*CDOw
Where Cle/C is the ratio between average extended slat chord to the average wing chord
CDow is the zero lift parasite drag of wing
Since we are not employing any leading edge devices, the zero lift drag due to leading edge
devices is not added.
Induced drag coefficient,CDi = 0.05172
For landing , CDO =0.0298
For takeoff, CDO = 0.03275
Drag polar equations
For cruise,
For take off
For landing,
In the same way the drag polar curves are constructed for takeoff and landing configurations.
91
CHAPTER 12
PERFORMANCE OF THE AIRCRAFT
The performance aspects of the aircraft to be studied are under unaccelerated flight
conditions and accelerated flight conditions
Under unaccelerated flight the following are studied
Steady level flight
Steady Climb
Descent and Glide
Range and Endurance
Under accelerated flight the following are studied
Take off performance
Landing performance
12.1 Unaccelerated flight
12.1.1 Steady level flight:
Under the steady level flight performance the maximum and the minimum level of speed for
each altitude is estimated.
The equations of motion for steady level flight are
T – D = 0 ; L – W = 0
W ;
Velocity ,V =
92
Steps involved:
1. Choose the altitude and find out the density of the corresponding altitude.
2. Choose a particular velocity
3. Calculate CL for that particular altitude and velocity chosen using
4. Calculate the CD for the calculated CL using the drag polar equation found for cruise
(project Phase I)
5. Calculate the Thrust required(TR) using
TR =
6. Then choose a different velocity and repeat from step 2 to step 5
7. Thrust available (TA) is calculated using
Where is the thrust produced by the engine at sea level
ρ is the density at altitude chosen
is the density at sea level
Thrust available is constant for the particular altitude.
Following the above steps
CD = 0.0168+ 0.05172 ( at cruise condition)
Velocity (m/s) TR at 11000 m (kN)
100 376.88
120 274.4
140 214.13
160 179.086
93
180 158.934
200 148.24
220 144.288
240 145.372
260 150
280 158.161
300 168.61
320 181.34
TA=152.73kN
Similarly the plot has been carried out for different velocities at different altitudes.
The point of intersection of the thrust Available (straight line) and the thrust required curve gives
the maximum and minimum velocity at that altitude.
The altitude where the thrust required and the thrust available curve becomes tangent to one
other gives the service ceiling of the aircraft.
94
12.1.2 Steady Climb:
Under steady climb performance, the following are evaluated for the aircraft
• Maximum rate of climb
• maximum angle of climb
• maximum attainable ceiling
Maximum angle to climb, :
The maximum angle to climb is given by
θmax = 8.68 º
Velocity Required to attain maximum angle to climb, :
=122.0366 m/s
Service Ceiling and Absolute Ceiling:
Service ceiling:
The altitude at which the climb rate is 100ft/min for the aircraft is known as service ceiling.
Absolute ceiling:
The altitude at which the climb rate is 0 ft/min for the aircraft is known as Absolute ceiling. It is
also the maximum altitude the aircraft can achieve.
Steps involved:
1. Choose the sea level altitude to start with and note the density for the altitude.
95
2. The rate of climb, of the jet aircraft is given by
3. The value of Z is found using
4. All the other values are known from the project phase I
5. Substituting all the values in the formula given in step 2 is calculated for that
altitude.
6. Increase the Altitude and calculate the for each altitude and tabulate the values
using which the graph is plotted to indentify the service and absolute ceiling.
Following the above steps
Altitude (m) R/C (m/s) R/C (ft/min)
0 24.166 4639.98
1000 21.379 4208.413
5000 12.10 2381.8608
10000 2.582 508.436
11000 0.7787 153.2855
11100 0.2797 55.06
11200 0.18405 36.23
11300 0.0789 15.53
11400 -0.05 -9.138
96
From the graph
Service ceiling = 11000 m; Absolute ceiling =11350m
12.1.3 Gliding flight:
The following are determined for the gliding flight performance,
Minimum descent angle
Maximum range
Equilibrium glide velocity
Rate of descent
Minimum descent angle, :
= 3.37º
Maximum range covered over the ground, :
97
R/C max vs. altitude
-1000
0
1000
2000
3000
4000
5000
0 2000 4000 6000 8000 10000 12000
altitude (m)
R/C
max R/C max
=187.074km
Equilibrium glide velocity, :
= 71.76 m/s
Rate of descent, :
= 14.595
= 213.0140
= 3.718 m/s
12.1.4 Range, R:
Range is the total distance traversed by the aircraft on a full tank of fuel.
Steps involved:
1. Calculate (CL0.5/CD)max using
[for maximum Range]
2. The range of the jet aircraft is given by
From the above mentioned steps,
(CL0.5/CD)max = 25.608
R = 9390.87537 miles = 15154.587km
98
12.1.5 Endurance, E:
Endurance is the total time the aircraft can fly on a given amount of fuel.
Endurance of the jet aircraft is given by
For maximum endurance the aircraft must fly at .
= (1/4*K*CDO)0.5
For the aircraft under consideration,
= 16.95
E = 16 hours
12.1.6 Level turn:
Under level turn the following are determined,
Minimum turn radius
Velocity for minimum turn radius
Load Factor corresponding to minimum turn radius
Minimum Radius Speed usually occurs at full flap configuration gives the highest lift harvest for
the lowest speed input
Minimum turn radius, :
= 1451.1651m
99
Velocity corresponding to minimum turn radius, :
= 119.3139 m/s
Load factor corresponding to minimum radius, :
= 1.0522
The two categories in turn are
Maximum sustained turn rate( MSTR)
Sharpest sustained turn (STR)
The conditions to be satisfied for the above mentioned turn to take place are
;
The maximum constant altitude turning rate that can be sustained by the aircraft is known as
Maximum Sustained Turn Rate. This will result in bleeding of speed and eventually reduction in
turn rate.
The aircraft to make sharpest turn or the turn with minimum radius of curvature whilr holding
the altitude constant is the Sharpest Sustained Turn Rate.
Steps Involved:
1. The value of Z and is calculated in common to both turn using
;
2. The load factor(n),turn rate ( ) and the radius of turn (R) are calculated for both the
categories for different altitudes using the formulae given below
100
For maximum sustained turn rate,
For sharpest sustained turn,
Following the above mentioned steps,
At sea level
N ω R
MSTR 2.47 0.1809 678.04
STR 1.4070 0.1447 449.044
At 5000 m
N ω R
MSTR 1.808 0.0934 1682.849
STR 1.33 0.0800 1343.979
At 11000 m
101
n ω R
MSTR 1.0566 0.0149 2083.493
STR 1.0522 0.147 1964.583
12.2 Accelerated Flight:
Under accelerated flight the Takeoff and Landing performance are studied.
Regulations FAR 25 specify that:
Aircraft should lift off 10% above the stalling speed
Aircraft should climb initially at 20% above the stalling speed
Aircraft speed during a regular approach should be 30% above the stalling speed
During take off the aircraft should clear an imaginary 11m obstacle
During landing aircraft should cross the run way threshold 15m above the ground
The above mentioned conditions should be taken into account to estimate the run way length
12.2.1 Take off Performance:
102
R
θ
h
ground run transition length
Steps Involved:
1. The stalling velocity of the aircraft is calculated using
2. The velocity for lift off, is calculated using
3. 70% of the lift off velocity is calculated which is to be used to find the ground roll
distance i.e. 0.7 of
4. The ground roll distance is calculated using
(N
= 1 for small aircrafts; N = 3 for large aircrafts)
5. The radius of take off and the angle for take off is calculated using
;
6. The transition length is calculated
7. The total take off distance is calculated
103
Following the above steps
65.375m/s ; = 78.45m/s; 0.7 of = 54.915m/s
; R= 3032.2404m ; θ = 4.8º; 253.7313m
= 3648.8013m
12.2.2 Landing Performance:
θa
Hf R
Point of touch down
approach flare ground run
Steps Involved:
1. The stalling velocity of the aircraft is calculated using
104
2. The velocity required for landing, is calculated
3. 70% of the lift off velocity is calculated which is to be used to find the ground roll
distance i.e. 0.7 of
4. The velocity for touchdown, , and velocity for flare, is calculated as
; = 1.23
5. The ground roll distance, is calculated using
6. The radius for landing, R and the angle for landing, θ is calculated using
; (for transport aircraft)
7. The height at which the flare phase begins, is calculated using
8. The approach distance, and the flare distance, is found using
;
9. The total landing distance, is
Following the above steps,
65.375m/s; =84.9875m/s; = 75.18m/s; = 80.411m/s
=1483.29 m; R=3295.5805m; = 4.5164m; =204.6184m
=172.4773m
= 1860.38m
12.3 V-n diagram:
105
The restrictions on the speed and the load factor (n) for the aircraft is given in form of V-n
diagram in which the load factor is plotted against the velocity. The operating limits represented
by the V-n diagram are the limits dictated by the airframe.
Steps Involved:
1. The maximum limit load factor on the positive side is found by velocity constraint and
the using which is found for different velocities
For velocity constraint,
=
For velocity constraint,
=
2. The graph is plotted using the tabulated values with velocity along X-axis and along
Y axis. The point of intersection of the two curves gives the positive limit load factor.
3. To find the maximum velocity the aircraft can travel, the thrust required and the thrust
available curves are plotted with velocity and the maximum velocity is found.
4. The negative limit load factor is half of the positive limit load factor .
5. The positive ultimate load factor is 1.5 times of positive limit load factor and the negative
ultimate load factor is 1.5 times of negative limit load factor.
Ultimate limit load factor = limit load factor * 1.5
6. For the construction of V-n diagram on positive side and negative side the following
formula is used
n = 0.5* S/W CLmax ρv2
Where for the positive side + is used
For the negative side- is used which are the function of airfoil chosen
7. The is identified using the + and - on the positive and negative side using
106
8. For the construction of gust V-n diagram, the gust velocities vary linearly from 20000ft.
At 20000ft and below,
66ft / sec at VB (velocity at maneuver point)
50ft / sec at VC (cruise velocity)
25ft / sec at VD (dive velocity or maximum velocity)
At 50000ft and above,
38ft / sec at VB (velocity at maneuver point)
25ft / sec at VC (cruise velocity)
12.5ft / sec at VD (dive velocity or maximum velocity)
So a graph has to be plotted and the gust velocities for that particular altitude must be
identified from the graph.
9. The gust lines are constructed using the formula( all in SI units)
Where K = 0.88μ/ 5.3+μ ;
10. The gust lines are plotted in the graph with velocity along x- axis and load factor (n)
along y axis.
Following the above steps,
constraint
Velocity constraint
V(m/s)
100 1.588
120 2.28
140 3.13
150 3.574
107
Using the above tabulated values, positive limit load factor = 2.5
Negative limit load facto r = -1.25
Ultimate positive load factor = 3.75
Ultimate Negative load factor = 1.875
V(m/s)
100 2.07
150 2.89
200 3.44
210 3.505
220 3.54
240 3.5543
241 3.5507
250 3.51
260 3.44
280 3.16
300 2.62
320 1.49
328 0
108
To find :
From the above graph, = 323.69 m/s
For positive side For negative side
n = 0.5* S/W CLmax ρv2 = 1.5862 * 10-4V2 n = 0.5* S/W CLmax ρv2 = 6.9878* 10-5V2
The equation for gust lines is
V(m/s) n
80 1.01516
100 1.5862
110 1.9193
120 2.2841
130 2.6806
140 3.1089
V(m/s) n
120 1.0062
130 1.18
140 1.36
150 1.57
109
T vs V (at sea level)
0100
200300
400500
600700
800
0 50 100 150 200 250 300 350
V (m/s)
T (
kN) TR
TA
for VD, n =1±0.003129V
For VC , n = 1±0.0069V
For VB n = 1±0.009V
110
CHAPTER 13
STABILITY ANALYSIS OF THE AIRCRAFT
While designing the aircraft it is necessary to prove that the stability of the aircraft is within the
acceptable limits. Basically stability means the ability of the body to return to its original
position if disturbed by some external forces. This stability in the aircraft design is divided into
two major categories
Static stability
Dynamic Stability
Static stability means when the aircraft is disturbed from the original flight path, forces will be
activated in such a way the aircraft returns to its original position
111
Fig 1. Different types of static Stability
Positive static Stability means to return to its original position after the restorative forces act on
the body.
Neutral static Stability means the body will obtain equilibrium in the disturbed position after the
restorative forces act on the body.
Negative static stability means the body continues to be moving away from the original position
as the restorative forces act on the body or in other words the body has no stability.
Dynamic stability means the way in which the restorative forces act on the body with respect to
time. In other ways it is the property which dampens the oscillations set up by the statically
stable aircraft.
112
Fig 2 Different modes of Dynamic stability.
The assumptions in the Analysis of Aircraft Stability and Control:
The flow is incompressible
Airframe is rigid i.e the distortion or deformation of airframe due to aerodynamic forces
and loads not considered.
The relationships between various parameters are generally linear which implies the
disturbances are assumed to be very small
The longitudinal motion is independent of lateral and directional forces.
Under Static stability of the Aircraft we have three different categories
Longitudinal Static stability
Lateral Static stability
Directional Stability.
Longitudinal stability:
Longitudinal stability refers to the tendency of the aircraft to return to trim condition after a
nose up or nose down disturbance.
The criterion for longitudinal stability
113
Cmomust be positive
must be negative
Lateral Stability:
Lateral stability refers to the ability of the aircraft to generate the rolling moment to stabilize
the rolling effect.
=
The criteria for lateral stability is must be greater than zero.
Directional stability:
Stability about the aircraft’s vertical axis i.e. the sideways moment is known as
directional stability or weather cock stability.
=
The criteria for directional stability is must be positive for directional stability.
Stick fixed stability:
Stick fixed stability is concerned with the calculation of the trim angle and the stability of
the aircraft held at a constant location. Here there is no freedom for the control surface to move
and there are fixed at a particular position.
Stick free stability:
In stick free stability the control surface is allowed to float. So when the aircraft
encounters a vertical gust its pitch angle is altered and the elevator which is free to move seeks
114
some momentary equilibrium position from its original position before disturbance. This will
have an effect on the stability characteristics of aircraft.
Neutral point:
Neutral point is the location of C.G where the stability becomes zero and it is usually the
aerodynamic centre where the lift vector acts. It gives the most aft position of C.G of the aircraft
beyond which the C.G moves the aircraft becomes unstable
Maneuver point:
The position of C.G where the stick force required to accelerate the aircraft becomes zero
is the stick free maneuver point.
The position of C.G where the elevator angle required to accelerate the aircraft vanishes is
known as stick fixed maneuver point.
It is always behind the neutral point of the aircraft.
13.1 Longitudinal Stability
13.1.1 Steps involved in determining stick fixed longitudinal stability of the aircraft:
1. must be evaluated which is equivalent to since CL is proportional to α.
= +( ) fuselage + (1- )
2. Xcg can be found out from the balance diagram( from the phase I) and X ac is assumed to
be around 0.25
3. ( ) fuselage is found from the formula given by Gilruth
( ) fuselage =
4. at and aw are the slope of the lift curve found from the equation
115
a=
5. is around 0.90 to 0.80
6. Vht is the tail volume co efficient computed from the formula
Vht =
7. is evaluated from the formulae given below
=
8. Using the above known values the derivative is found to be negative if positive
some changes in the design has to be made.
9. The stick fixed neutral point ( No)is found as it making is zero from the equation
mentioned in step 1
10. From the stick fixed neutral point, the static margin for the aircraft is calculated within
which the movement of C.G of the aircraft should be restricted.
Static margin = No – Xc.g
11. Using the stick fixed neutral point the C.G of the aircraft for different cases is evaluated.
12. Using the newly found C.G of the aircraft for different cases, for each case is
evaluated.
13. The zero lift pitching moment co efficient is evaluated using
where iw is known from phase I,
And it is found using the method given below.
116
14. Steps for finding, it:
During the trim condition, Cm = 0 and substitute CL for cruise
Substitute Cm=0 in the equation
Use the value of found from step 9
After evaluating substitute in the equation given in step 14 and determine the
tail setting angle it
15. Substitute it in the equation of in step 14 and is found
16. Using the equation below the variation of CL vs. Cm is evaluated for different cases
;
Following the above mentioned steps,
Xcg = 0.31 ; X ac = 0.24 ; aw = 4.8433 ;at = 4.09
Vht= 0.6290; =0.4049; ( )fuselage = 0.08589; = 0.90
= -0.13003
So the aircraft has longitudinal static stability.
Substituting = 0, No = 0.42
Static Margin for different cases
Full payload+ Full fuel 0.2295
Full payload+ reserve fuel 0.225
117
Zero payload+ Full fuel 0.228
Zero payload+ Reserve fuel 0.231
Half payload+Full fuel 0.22605
Half payload + Reserve fuel 0.22603
(CL)cruise = 0.2 , = 0.026006 , it = 0.0610 ;
=0.5 using =0.207( Perkins and Hage)
For the design condition i.e are formulated as follows for different elevator deflection,
Elevator deflection, (rad) Equations
-5
-10
-15
-20
-25
25
20
15
10
5
118
Graph for for different elevator deflection
For different cases where C.G varies equations are formulated. For eg. Equations for first three
cases are formulated as follows
119
For different values of ranging from -0.2 to 2.2 is evaluated and graphs are drawn
13.1.2 Determination of Stick- Fixed Stability Characteristics:
For different values for , is evaluated as follows
(deg)
-25 0.76076
-20 0.61826
-15 0.4757
-10 0.3332
-5 0.19072
0 0.0483
5 -0.0942
120
10 -0.2367
15 -0.3792
20 -0.5217
25 -0.6642
13.1.3 Estimation of elevator control power ( ) :
Elevator control power is the rate at which the pitching moment changes with the deflection of
the elevator.
=
= -0.02858
13.1.4 Extreme elevator deflections:
Steps Involved:
1. The value during cruise is identified and is found from the stick fixed
stability analysis.
2. The above values are substituted in assuming trim condition where
= 0 and minimum is obtained.
3. is obtained from the previous section.
4. Substituting the above obtained values the formula below
Maximum down elevator deflection is found.
121
5. For maximum up elevator deflection, during landing and is obtained from
stability analysis.
6. Repeat step 2 assuming trim condition maximum is obtained. Repeat step 3
7. Substituting the above obtained values the formula below
Maximum up elevator deflection is found.
Following the above steps,
δe max = 4.4 deg ( maximum down elevator)
= 19.1deg ( maximum up elevator)
13.1.4 Stick Fixed Maneuver point ( ):
=
=0.52
13.1.5 Steps involved in determining the Stick free longitudinal stability:
1. Free elevator factor, F has to be evaluated using the formula
, where = -0.003 ; = -0.005
2. Evaluate using the formula for stick free longitudinal stability
122
= +( ) fuselage + (1- ) and all the other parameters
are same as in stick fixed longitudinal stability.
3. Determine the stick free neutral point, by keeping = 0 in the formula given in
step 2
4. Evaluate as given by the formulae in stick fixed longitudinal stability
5. Formulate equations for in terms of as given below for different elevator angle
deflections (all angles in degree)
6. Graphs are plotted for the above formulated equations
7. In a similar way for different cases ( )stick fixed is calculated and the graphs are plotted
for vs.
13.1.6 Stick force , :
= (at sea level)
Where
K = ; G = 1.5, =0.90, =0.20
K= -24.929
A=
A= -0.05230
123
=
=0.04637
= -0.3073 (for full payload +full fuel)
= -0.002 ; =-0.005 ; = -0.02858
Substituting the given values in the equation for
=-7048.3914+0.09054*V2
For different values of Velocity , Fs is calculated and graph is plotted.
124
From the graph the trim velocity, = 270 m/s. The slope of the curve at this point is the
measure of stick force to produce a change in speed.
13.1.7 Steps involved in determining the Directional Stability of the aircraft:
1. Contribution of the wing to Directional Stability:
=-0.00006*(˄) 0.5
Where, ˄ is the sweep back angle in degree
2. Contribution of fuselage and nacelle to Directional Stability ( all units in SI):
=
Where, h1 and w1 is the height and width of fuselage at Lf/4
h2 and w1 is the height and weight of fuselage at 3Lf/4
is found from vs. (d/Lf) i.e fineness ratio of fuselage ( Perkins and
Hage)
3. Contribution of Vertical Tail to Directional Stability (all in SI units)
=
Where
; = 0.9
(Perkins and Hage,Pg.71)
4. = + +
125
5. The variation of vs. is found using the relation
6. Evaluate and for various rudder deflection angle , is
evaluated
7. For a particular rudder deflection angle equation is formulated which gives the
variation of with respect to sideslip angle and graph is plotted
Following the above mentioned steps
= + +
=-0.003
Variation for vs.
(deg)
30 -0.06075
-30 0.06075
20 0.0405
-20 -0.0405
15 -0.0303
-15 0.0303
10 -0.02025
-10 0.02025
5 -0.010125
126
-5 0.010125
The equations formulated are
(degree) equation
-30
30
-20
20
15
-15
10
-10
+5
-5
For different values of (25 deg to -25 deg), Cn is evaluated and the graphs are plotted to show
the variation.
127
Under cross-wind conditions,
; = ; where =7.5m/s
=-8.979 deg
13.1.8 Steps involved for determining Lateral Stability of Aircraft:
1. Contribution of the wing fuselage interface to lateral stability
= 0.0006 (high wing)
= 0 (mid wing)
=-0.0008 (low wing)
2. Contribution of vertical tail to lateral stability (all units in S.I)
=
3. Contribution of Wing interference on vertical tail to lateral stability
128
Ψ vs. Cn
-0.15
-0.1
-0.05
0
0.05
0.1
0.15
-30 -20 -10 0 10 20 30
Ψ
Cn
δr = 30 deg
δr =20 deg
δr = 10 deg
δr = 5 deg
δr = 0 deg
δr = - 5 deg
δr = -10 deg
δr = -20 deg
δr = - 30 deg
=-0.00016 (high wing)
=0 (Mid wing)
= 0.00016 ( Low wing)
4. Contribution of wing to lateral Stability
=0.0002(˄) + +
Where
=-0.5* ; is found from Perkins and Hage
5. = + + +
6. Effective Dihedral, = /0.0002
Following the above mentioned steps
=0.00144 deg-1
Effective Dihedral, = 2.14 deg-1
13.1.9 Lateral Control:
129
The common way of determining the lateral control effectiveness is by the use of the non
dimensional parameter . The lateral control deflection is the measure of lateral control power
available.
Aileron Rolling power relation is given by
=
Where is the distance to the point of starting of the aileron on half span
is the distance to the end of the aileron on half span
is the taper ratio of the wing.
=0.006306 deg-1
Substituting for different aileron deflection angle ( ) and for different velocities, P is found
V(m/s) = 5deg =7.5deg =10deg
100 0.1058 0.1588 0.2117
120 0.1269 0.1905 0.2546
140 0.1481 0.2223 0.2963
160 0.1692 0.2541 0.3386
180 0.1904 0.2858 0.3810
200 0.2116 0.3172 0.4233
220 0.2327 0.3493 0.4657
240 0.2539 0.3811 0.5080
260 0.2750 0.4128 0.5504130
V vs. δa
0
0.1
0.2
0.3
0.4
0.5
0.6
0 50 100 150 200 250 300
V(m/s)
P(ra
d / s
)
δa = 5deg
δa = 7.5 deg
δa = 10 deg
13.1.10 Dynamic Stability of the aircraft
Steps involved:
1. Mass moment of inertia about the three principal axis is calculated by taking into account
the c.g position of each component from the reference axis.
I xx = M*(Y2+Z2); I yy = M*(Z2+X2); I zz = M*(X2+Y2)
2. Parameters required for the dynamic Stability analysis like Time constant ( τ), Relative
density factor (μ) and Radius of Gyration about the three axes K x, K y, K z are calculated.
3. The stability derivatives required for the determination of co efficients of the quartic
equation to study the dynamic stability is evaluated.(Refer Perkins and Hage)
4. Using the above determined stability derivatives the constants A , B , C , D and E are
determined using the formulae provided which are the co efficients of the quartic equation which
is given by the form
Aλ4 + Bλ3+Cλ2+Dλ+E = 0.
5. The roots of the quartic equation are found using Quartic Calculator.
6. Based on the roots, the modes of the dynamic response are studied using the time period
and damping factor evaluated.
7. Steps from 2 to 4 are used seperately to study longitudinal dynamic stability and the
lateral directional dynamic stability.
Following the steps mentioned above, keeping C.G of the aircraft as reference
components Mass(kg) Z(m) Y(m) Ixx(kg-m2)
131
nose wheel 1455.6095 2.107 0 6462.104139
main wheel 10298.39042 2.137 11.586 1429438.886
fuselage 37529.79001 2.038 0 155877.8811
wing 35754.43996 -0.809 11.586 4822912.008
H.T 3375.289501 4.929 6.11 208009.442
V.T 1801.002039 11.818 0 251537.1731
payload 28482.07339 2.038 0 118298.6968
F.E 1340.3037 2.038 0 5566.876361
Pilots 220 2.038 0 913.75768
Fuel 94462.22018 0.809 11.586 12741997.26
Engine 17963.72987 1.147 11.586 2435001.637
∑ I xx = 22176015.72kg-m2
components Mass(kg) X(m) Z(m) Iyy(kg-m2)
nose wheel 1455.6095 19.226 2.107 544512.2547
main wheel 10298.39042 4.267 2.137 234536.1407
fuselage 37529.79001 26.86 2.038 27232105.17
wing 35754.43996 27.546 -0.809 27153230.21
H.T 3375.289501 30.381 4.929 3197412.426
V.T 1801.002039 30.708 11.818 1949848.352
132
payload 28482.07339 0.046 2.038 118358.9649
F.E 1340.3037 0.686 2.038 6197.617921
Pilots 220 24.346 2.038 131313.8552
Fuel 94462.22018 0 0.809 61823.72833
Engine 17963.72987 9.366 1.147 1599446.766
∑ I yy = 62228785.49kg-m2
components Mass(kg) X(m) Y(m) Izz(kg-m2)
nose wheel 1455.6095 19.226 0 538050.1506
main wheel 10298.39042 4.267 11.586 1569914.287
fuselage 37529.79001 26.86 0 27076227.29
wing 35754.43996 27.546 11.586 31929341.02
H.T 3375.289501 30.381 6.11 3241416.274
V.T 1801.002039 30.708 0 1698311.179
Payload 28482.07339 0.046 0 60.26806729
F.E 1340.3037 0.686 0 630.74156
Pilots 220 24.346 0 130400.0975
Fuel 94462.22018 0 11.586 12680173.53
Engine 17963.72987 9.366 11.586 3987181.914
133
∑ I yy = 82851706.75 kg-m2
. Time constant,τ = = 5.313 s;
Relative density factor,μ = = 164.228;
Kx = = 9.4282 m; Ky = 15.7937; Kz = = 18.2239 m
Evaluation of Stability Derivatives: ( Refer Perkins and Hage)
= 4.84412 rad-1
= 0.10123 rad-1
= -0.726618
= -0.020449 rad-1
= -0.0555 rad-1
= -1.2397rad-1
= -1.6825*10-3rad-1
= -0.10299 rad-1
= -0.2865 rad-1
= 2.1156*10-3rad-1
134
From the above known values
A =1;
B = = 3.1472;
C=
C = 34.0313;
D = D=
0.58310 ;
E = = 0.65423
The Quartic Equation is given by
λ4 + 3.1472λ3+34.0313λ2+0.58310λ+0.65423 = 0.
, in which real part represents the damping and the imaginary part
represents the frequency
The roots of the above equation are
λ 1,2 = -1.5659 ± i 5.6135 9
λ 3,4 = - 0.0076 ± i 0.1385
Phugoid oscillations are lightly damped and low frequency oscillations. So the Eigen value
chosen should have small real and imaginary part.
Short period oscillations are highly damped high frequency oscillations. So the Eigen value
chosen should have large real and imaginary part
135
For phugoid or long period motion
λ 3,4 = - 0.0076 ± i 0.1385 is selected
Damping factor, .
Damping factor = 0.055
Time period, ; natural frequency,
Time period = 44.957 s
For Short period mode
λ 1,2 = -1.5659 ± i 5.6135
Using the same formulae,
Damping factor = 0.2686
Time period = 1.12 s
2.1.11 Following the same steps for lateral directional dynamic stability
1. τ = 5.313 s; μ = 164.228; K x = 9.4282 m; K y = 15.7937; K z = 18.2239 m
2. Evaluation of damping derivatives
C n r = -0.24409 rad-1
C l r = 0.05 rad-1
C n p = -0.0025 rad-1
C n φ = 0.1719 rad-1
C l β = -0.0825 rad-1
C l p = - 0.45 rad-1
136
3. Based on the above derivatives (Refer Perkins and Hage for the formulae)
A = 1; B = 1.1916; C = 24.94727; D = 108.52602; E = 7.6138
5. The Quartic equation is given by
λ4 + 1.1916λ3+24.9472λ2+108.5260λ+7.6138 = 0
6. The roots of the above equations are
λ 1 = -0.071322
λ 2 = -3.318945
λ 3,4 = 1.09933± i 5.563813
Aperiodic Roll is a very fast motion and very highly damped. The Eigen value is negative.
The Spiral Motion is unstable and the divergence will occur very slowly. The Eigen value is
positive and small.
The Dutch Roll is slightly low damping and slightly high frequency motion. The aircraft
performs alternately yawing and rolling motion.
7. The Aperiodic modes is given by the roots
λ 1 = -0.071322 ( mildly convergent)
λ 2 = -3.318945 ( heavily convergent)
8. The Dutch roll mode is given by the root (using the same formulae as mentioned above
for longitudinal dynamic Stability)
λ 3,4 = 1.09933± i 5.563813
Damping factor = 0.1938
Time period = 1.1285 secs
137
CHAPTER 14
STRUCTURAL DESIGN OF THE AIRCRAFT
In the construction of aircraft two types of structures are used
Monocoque structures which are unstiffened shells and has to be relatively thick to resists
the different kinds of load acting on it
Semi-Monocoque structures which are consist of stiffening members along with the skin.
This is the most efficient type of construction which is widely used now days.
Fig 1.Semi Monocoque Construction of Fuselage
138
Fig 2.Semi Monocoque Construction of Wing
Functions of different structural members:
Skin:
Reacts to applied torsion and shear forces, transmits aerodynamic forces to the
longitudinal and transverse structural members.
Acts with the longidutinal members in resisting the applied bending and axial loads
Acts with transverse members in reacting the hoop or circumferential load when the
structure is pressurized.
Ribs and Frames:
For the structural integration of the wing and fuselage
To maintain the aerodynamic profile of the structure
Spar:
Resists axial and bending loads
Form the wing box for stable torsion resistance.
Stiffener or Stringers:
Resist bending and axial loads along with the skin
Divide the skin into small panels and thereby increase its buckling and failing stresses
Act with the skin in resisting the axial loads caused by pressurization
139
14.1WING DESIGN
The first step towards the Structural Design is to find the lift Distribution of the wing.
The lift distribution of the wing is best approximated using Schrenk’s Curve which is the average
of the Elliptical Distribution of the lift over the wing and the Trapezoidal Distribution of the lift
over the wing. So the elliptical and the trapezoidal distribution of the lift over the wing should be
approximated.
14.1.1 Steps involved in plotting the elliptic lift distribution:
1. The area under the elliptic curve is given by
A = Π*b* w o / 8
This gives the lift produced by a single wing.
2. In steady level flight lift produced is equal to weight
Π*b* w o / 8 = W TO / 2
From the above w o can be calculated
3. Using the equation of ellipse the curve is plotted for different values of y
w y =(4 W TO/ Π*b)/ [ 1- (2y/b)2]0.5
Using the above mentioned Steps the elliptic lift is given by
w y = [52352.32426 ( 1 – 0.00112y2) 0.5]
140
elliptic curve
0
10000
20000
30000
40000
50000
60000
0 5 10 15 20 25 30 35
elliptic curve
14.1.2 Steps involved in plotting trapezoidal lift distribution:
1. The area under the trapezoid is given by
A = b ( w 1 + w 2)/4
This gives the lift Distribution
2. In the steady level flight lift produced is equal to weight
b ( w 1 + w 2)/4 = W TO / 2
3. The lift produced by any section of wing is directly proportional to the chord of the wing.
w 2 / w 1 = c t / c r = λ ( taper ratio)
w 1 = 2 W TO / b (1+λ )
w 2 = 2λ W TO / b (1+λ )
4. Using the equation of trapezoid, the lift distribution is given by
w y = 2 W TO / b (1+λ )*[ 1 + 2y/b*(λ – 1)]
Following the above the steps the trapezoidal lift distribution
w y = 68437.25985 [ 1- 0.02683y]
141
trapezoidal curve
0
10000
20000
30000
40000
50000
60000
70000
80000
0 5 10 15 20 25 30 35
trapezoidal curve
Schrenk’s curve is the average of the elliptical lift distribution and trapezoidal lift distribution
So the lift distribution by Schrenk’s curve is given by
w y = 26176.162 [( 1 – 0.00112y2)0.5]+ 34218.6299 ( 1- 0.02683y)
14.1.3 Steps involved in plotting the Shear force diagram:
1. The different loads acting on the wing are identified and their distribution is evaluated
2. The loads whose distribution is to be known are the lift load, the wing structural weight
and the fuel weight. The point loads acting on the wing are engine weight and the landing
gear and their locations are identified.
142
lift per unit span vs. span
010000
2000030000
4000050000
6000070000
80000
0 5 10 15 20 25 30 35
b ( m )
lift
per
un
it s
pan
(N
/m)
trapezoidal curve
elliptic curve
schrenk's curve
3. The lift load distribution is obtained from the Schrenk’s curve, the fuel and the wing
structural weight is assumed to have trapezoidal distribution and the load acting depends
upon the thickness and the chord of the section of the wing.
4. The shear force diagram is obtained integrating the loads acting on the wing along its half
span taking into account the point loads at their positions with the tip of the wing as
origin.
Following the above steps, for 0 ≤y ≤29.775
The lift distribution is given by
w y = 26176.162 [( 1 – 0.00112y2)0.5]+ 34218.6299 ( 1- 0.02683y)
Fuel weight Distribution:
The fuel load varies linearly as shown below
Y = 6.208m Y = 20.876m Y -2.76m
The weight of the fuel is proportional to the chord of that section.
The chord at any section can be computed using
;
The fuel weight distribution is given by
143
w y = 4854.7392 + 1862.6888 ( y – 2.76)
3.3.2 Structural weight Distribution:
Structural weight varies linearly from the root to tip of the wing and the intensity of the load is
proportional to the chord of the section.
where point 1 and 2 represents the Root and the tip of the wing
=915.015N; =22645.088N
structural weight distribution
w1
w2
Y = 29.775m
The structural weight distribution is given by
w y = 729.80y +915
At y = 6.665 m , the weight of the landing gear is taken into account
At y = 8.968 m , the weight of the engine is taken into account
So the shear force diagram plotted for the wing is
144
Shear force vs. Spanwise distance
0
200000
400000
600000
800000
1000000
1200000
0 5 10 15 20 25 30 35
Spanwise distance(m)
Sh
ear
forc
e(N
)
V
For plotting the bending moment diagram, the obtained shear force is once again integrated.
y (m) Shear force (N)
0 9605004.42
6.665 562308.572
6.665 511794.967(with L.G)
7.165 540671.640
8.968 617816.536
8.968 536560.536 (with eng)
27.015 158845.485
29.775 0
145
14.1.4 Design of the Spar:
Steps involved:
1. Plot the airfoil in the graph and identify the spar location. The front spar is placed near
the maximum (t/c) max of the airfoil and the rear spar is placed at 2/3rd of the chord
location.
y (m) Bending
moment (N-m)
0 16574205.32
6.665 7697715.201
7.165 7486345.164
8.968 6377850.355
27.015 219209.8237
29.775 0
146
Bending moment vs. Spanwise distance
-20000000
-15000000
-10000000
-5000000
0
5000000
0 5 10 15 20 25 30 35
Spanwise distance(m)
Ben
din
g M
om
en
t(N
-m)
Y
2. Using the co ordinates of the airfoil chosen, the height of the front spar and the rear spar
is identified.
To identify the area of Flanges
3. The maximum bending moment is identified from the bending moment diagram
plotted and is given by
M max = M 1 + M 2
Where M 1 is the bending moment at the front spar
M 2 is the bending moment at the rear spar
4. M 1, M 2 is proportional to the height of the front and the rear spar respectively i.e
M 1/ M 2 =( h 1 / h 2 )2
Where h 1 is the height of the front spar
h 2 is the height of the rear spar
5. For the material selected, the σ yield is identified for the material. Based on the σ yield , the
area of the flanges are calculated using the formula
A flange = M / σ yield * h
To identify the thickness of the web:
8. The shear force experienced by the spar is again proportional to the height of the spars
Which is given by V 1 / V 2 = h 1 / h 2
Where V 1 is the shear force acting on the front spar
V 2 is the shear force acting on the rear spar
V 1 + V 2 = V max (identified from the graph plotted for shear force)
147
7. Based on the material selected the shear strength of the material is selected and based on
this shear strength the thickness of the web is calculated using the formula
τ = V/ A = V / h*t
Following the above steps,
h 1 = 1.468 m; h 2 = 1.061 m
Material Selected is 7075 – T6
Area of the rear flange = 7.07 * 10 -3 m 2
Area of the front flange = 9.78 * 10-3 m 2
Thickness of the web = 0.0010317 m
14.1.5 Torque Distribution over the wing:
Before going into the construction of the torque
diagram, the known quantities are C.G of the aircraft, the centre
of pressure of the aircraft and the shear centre which is assumed to
be around 0.35
=integration of lift
distribution (Schrenk’s curve)*(S.C – C.P)
= integration of structural load distribution *(S.C – C.G of wing)
=integration of fuel weight distribution*(S.C – C.G of the fuel)
X (m)
Torque(N-
m)
0 947973.4266
6.665 591379.3779
6.665 840532.6832
7.165 830465.3153
8.968 578187.3949
8.968 759930.8956
27.015 90377.8379
29.775 0
148
14.1.6 Steps involved in finding the thickness of the wing:
1. The torque produced by the various forces acting on the wing is calculated and the torque
diagram is constructed for the wing.
2. The aerofoil is assumed to be of 2 cell structure i.e from the leading edge to the front spar
is considered as one cell and from the front spar to the rear spar is considered as second
cell.
3. The total torque acting on the root airfoil is found from the torque diagram.
149
4. The total torque acting on the root airfoil is divided between the two cells and expressed by
the following equation.
T = ∑ 2Aq = 2 (A 1 *q 1 + A 2 * q 2)
Where A 1 is the area of the first cell
A 2 is the area of the second cellss
q 1 is the shear flow in the first cell
q 2 is the shear flow in the second cell
5. The area of the first cell, A 1= ( 0.23 * C r *h 1)/ 2
The area of the second cell, A 2 = ( 0.55 * C r * { h 1 + h 2})/2
Where the constant 0.23 denotes the position of the front spar from the
leading edge
The constant 0.55 denotes the position of the rear spar from the front spar.
5. The twist equation for the two cells are formulated.
For the first cell it is given by
2Gθ 1 = 1/ A 1 * [ q 1 * a 10 + ( q 1 – q 2) * a 12 ]
For the second cell it is given by
2Gθ 2 = 1/ A 2 * [ ( q 2 – q 1)* a12 + q 2* a 20]
6. The constants a 10 , a 12 and a 20 is found as following
150
From the above figure
S 1 = L 1 + L 2 ; S 2 ; S 3 = L 3 + L 4
In which L 1,L 2,L 3, L 4 are measured using threads on the airfoil co ordinates shown.
Therefore
a 10 = S 1 / t ; a 12 = S 2 / t w ; a 20 = (S 3 / t) + ( h 2 / t w)
7. The values obtained in the above step is substituted in the equations given in the step 5 and
the twist in both cells is assumed to be the same
Gθ 1 = Gθ 2
8. Using the above mentioned condition q 2 is obtained in terms of ‘t’ and ‘q 1’ and using the
torque equation q 1 is obtained in terms of ‘q 2’
9.Using the equation for critical shear buckling stress,
q 2 / t = τ cr = { Π * E * K s / [ 12 * ( 1 – γ 2 ) ] } * ( t / b ) 2 }
q 2 is obtained in terms of t
10. Substituting the q 2 obtained in step 8 in the q 2 obtained in step 9, a quartic equation is
obtained in terms of ‘t’ which is solved using the quartic calculator available online.
151
Following the above procedure
Area of the first cell , A 1 = 2.56705 m2
Area of the second cell , A 2 = 9.23118m2
a 10 = 6.22724/t ; a 12 = 1692.3966 ; a 20 = 13.56546/t + 852.6689
Assuming twists in both cells are same, we obtain
From the critical shear stress equation
q 2 = 1.53756*10 12 * t 3
Equating and solving the above two equations, skin thickness (t) = 0.002834m
Skin thickness (t) = 0.002834m
14.1.7 STRINGER DESIGN:
General Procedure
1. The first step towards to the stringer design is to find the spacing required for the
placements of the stringer in the upper and lower surface.
2. Using the spacing obtained in the above step the number of stringers to be placed on the
upper and lower surfaces are decided.
3. The appropriate material required for the stringer and the stringer size are decided
4. Bending stress developed in the each stringer is calculated and stress obtained should be
less than the ultimate stress of the material. If the stress is more it indicates the failure of
152
the stringer and the area of the stringer has to be increased appropriately to develop stress
less than the ultimate stress of the material.
14.1.7.1 Spacing of Stringer:
Using the expression for critical bending stress of the material, the spacing for the stringer
placement is calculated.
σ cr = {Π 2 * E * K b / [ 12 * ( 1 – γ 2 )]* [ t/b] 2 }
where E is the Young’s Modulus of the material, Pa
K b is the bending buckling co efficient of plates,( 8.5 for aircraft applications)
γ is the Poisson’s ratio of the material.
t is the thickness of the skin, m
b is the spacing of the stiffener, m
Using the above mentioned formula and substituting the required values,
The stringer spacing , b = 0.1408 m
14.1.7.2 Number of Stringers:
The root airfoil is plotted in the design software and the distance are measured between the
points from which the number of stringers are calculated. Usually there are more stringers on
the upper surface than on the lower surface and the placement distance can be modified
slightly according to it.
Following the above mentioned procedure,
For the upper surface 0.12 m stringer spacing is decided and 88 stringers are placed.
For the lower surface 0.14 m stringer spacing is decided and 72 stringers are placed.
153
14.1.7.3 Stinger Size And Bending Stress Calculation Of The Stringers:
Steps Involved:
1. The stringer size is chosen from chapter A3 of “ Analysis and Design of Flight Vehicle
Structures, Bruhn” and the area of the stringer is calculated.
2. The ‘ x ‘ and ‘ y ‘ of the stringers is obtained using the design software and the centroid is
calculated .
X c = ∑ AX / ∑ A
Y c = ∑ AY / ∑ A
3. The moment of inertia about the three axis I xx , I yy , I xy are calculated using the
expressions given below.
I xx = ∑ [ A ( y – Y c ) 2 ]
I yy = ∑ [ A ( x – X c ) 2 ]
I xy = ∑ [ A ( x – X c) ( y – Y c) ]
4. The ‘x ‘ and ‘ y ‘ component of the bending moment is calculated since the wing is placed
at the wing setting angle and the value of the bending moment at the root airfoil is read from the
moment diagram constructed.
M x = M cos α
M y = M sin α
5. The bending stress experienced by the stringers are calculated using the following
expression
6. After calculating the bending stress , it is checked with the ultimate strength of the material.
If the bending stress is greater than the ultimate strength of the material then the area of the
stringer has to be increased proportionately and again the bending stress has to be checked.
154
Following the above mentioned steps,
Tabulation for finding the Centroid:
Upper Surface of the wing:
Stringer X(m) Y(m) Area(m2) Ax(m3) Ay(m3)
1 0.0153 0.11 0.00115 0.000017595 0.0001265
2 0.0614 0.21 0.00115 0.00007061 0.0002415
3 0.14 0.3019 0.00115 0.000161 0.00034719
4 0.23 0.369 0.00115 0.0002645 0.00042435
5 0.33 0.421 0.00115 0.0003795 0.00048415
6 0.43 0.47 0.00115 0.0004945 0.0005405
7 0.541 0.5105 0.00115 0.00062215 0.00058708
8 0.65 0.5439 0.00115 0.0007475 0.00062549
9 0.76 0.5742 0.00115 0.000874 0.00066033
10 0.87 0.6006 0.00115 0.0010005 0.00069069
11 0.98 0.624 0.00115 0.001127 0.0007176
12 1.09 0.6466 0.00115 0.0012535 0.00074359
13 1.2 0.6674 0.00115 0.00138 0.00076751
14 1.32 0.6866 0.00115 0.001518 0.00078959
15 1.4315 0.704 0.00115 0.001646225 0.0008096
16 1.54 0.719 0.00115 0.001771 0.00082685
17 1.65 0.7339 0.00115 0.0018975 0.00084399
18 1.762 0.7478 0.00115 0.0020263 0.00085997
19 1.874 0.7606 0.00115 0.0021551 0.00087469
20 1.987 0.7729 0.00115 0.00228505 0.00088884
21 2.102 0.7855 0.00115 0.0024173 0.00090333
22 2.212 0.796 0.00115 0.0025438 0.0009154
23 2.323 0.8065 0.00115 0.00267145 0.00092748
24 2.4381 0.8158 0.00115 0.002803815 0.00093817
25 2.55 0.8202 0.00115 0.0029325 0.00094323
155
26 2.665 0.8334 0.00115 0.00306475 0.00095841
27 2.7764 0.8412 0.00115 0.00319286 0.00096738
28 2.8865 0.8487 0.00115 0.003319475 0.00097601
29 2.9983 0.8553 0.00115 0.003448045 0.0009836
30F 3.12336 0.862828 0.004233 0.013221183 0.00365235
31 3.2371 0.8685 0.00115 0.003722665 0.00099878
32 3.3516 0.8742 0.00115 0.00385434 0.00100533
33 3.4664 0.8801 0.00115 0.00398636 0.00101212
34 3.5792 0.885 0.00115 0.00411608 0.00101775
35 3.6926 0.8895 0.00115 0.00424649 0.00102293
36 3.8023 0.8937 0.00115 0.004372645 0.00102776
37 3.9133 0.8969 0.00115 0.004500295 0.00103144
38 4.0264 0.9003 0.00115 0.00463036 0.00103535
39 4.1405 0.9084 0.00115 0.004761575 0.00104466
40 4.2546 0.9061 0.00115 0.00489279 0.00104202
41 4.3691 0.9074 0.00115 0.005024465 0.00104351
42 4.4843 0.9081 0.00115 0.005156945 0.00104432
43 4.5952 0.9128 0.00115 0.00528448 0.00104972
44 4.7052 0.919 0.00115 0.00541098 0.00105685
45 4.8181 0.9121 0.00115 0.005540815 0.00104892
46 4.9295 0.9119 0.00115 0.005668925 0.00104869
47 5.0417 0.9125 0.00115 0.005797955 0.00104938
48 5.1562 0.9122 0.00115 0.00592963 0.00104903
49 5.2752 0.9118 0.00115 0.00606648 0.00104857
50 5.3901 0.9096 0.00115 0.006198615 0.00104604
51 5.5011 0.9094 0.00115 0.006326265 0.00104581
52 5.6125 0.9081 0.00115 0.006454375 0.00104432
53 5.7262 0.9057 0.00115 0.00658513 0.00104156
54 5.8414 0.9035 0.00115 0.00671761 0.00103903
55 5.9528 0.9013 0.00115 0.00684572 0.0010365
156
56 6.0631 0.899 0.00115 0.006972565 0.00103385
57 6.1775 0.8962 0.00115 0.007104125 0.00103063
58 6.2932 0.8926 0.00115 0.00723718 0.00102649
59 6.4092 0.8892 0.00115 0.00737058 0.00102258
60 6.5251 0.8856 0.00115 0.007503865 0.00101844
61 6.6371 0.8811 0.00115 0.007632665 0.00101327
62 6.7506 0.8765 0.00115 0.00776319 0.00100798
63 6.8647 0.8721 0.00115 0.007894405 0.00100292
64 6.9809 0.8665 0.00115 0.008028035 0.00099648
65 7.0974 0.8607 0.00115 0.00816201 0.00098981
66 7.2088 0.8546 0.00115 0.00829012 0.00098279
67 7.3233 0.8477 0.00115 0.008421795 0.00097486
68 7.4374 0.8407 0.00115 0.00855301 0.00096681
69 7.5481 0.8329 0.00115 0.008680315 0.00095784
70 7.6628 0.8248 0.00115 0.00881222 0.00094852
71 7.7787 0.8168 0.00115 0.008945505 0.00093932
72 7.8915 0.8082 0.00115 0.009075225 0.00092943
73 8.0047 0.7991 0.00115 0.009205405 0.00091897
74 8.1207 0.7894 0.00115 0.009338805 0.00090781
75 8.2341 0.7787 0.00115 0.009469215 0.00089551
76 8.3461 0.7674 0.00115 0.009598015 0.00088251
77 8.4591 0.7561 0.00115 0.009727965 0.00086952
78 8.5703 0.7438 0.00115 0.009855845 0.00085537
79 8.6804 0.7319 0.00115 0.00998246 0.00084169
80 8.7903 0.7191 0.00115 0.010108845 0.00082697
81 8.8998 0.7058 0.00115 0.01023477 0.00081167
82 9.0125 0.6921 0.00115 0.010364375 0.00079592
83 9.126 0.6772 0.00115 0.0104949 0.00077878
84 9.2399 0.6624 0.00115 0.010625885 0.00076176
85 9.3519 0.6467 0.00115 0.010754685 0.00074371
157
86 9.4621 0.6314 0.00115 0.010881415 0.00072611
87 9.5714 0.6154 0.00115 0.01100711 0.00070771
88 9.6739 0.5998 0.00115 0.011124985 0.00068977
RF 9.7605 0.586931 0.002133 0.020819147 0.00125192
Lower Surface of the wing:
Stringer X(m) Y(m) Area(m2) Ax(m3) Ay(m3)
1 0.02603 -0.1406 0.00115 2.99345E-05 -0.00016169
2 0.09979 -0.2614 0.00115 0.000114759 -0.00030061
3 0.2074 -0.3544 0.00115 0.00023851 -0.00040756
4 0.3319 -0.4266 0.00115 0.000381685 -0.00049059
5 0.4605 -0.4832 0.00115 0.000529575 -0.00055568
6 0.5928 -0.5282 0.00115 0.00068172 -0.00060743
7 0.7238 -0.5656 0.00115 0.00083237 -0.00065044
8 0.8599 -0.5991 0.00115 0.000988885 -0.000688965
9 0.991 -0.6274 0.00115 0.00113965 -0.00072151
10 1.1237 -0.6537 0.00115 0.001292255 -0.000751755
11 1.2599 -0.6777 0.00115 0.001448885 -0.000779355
12 1.3956 -0.6997 0.00115 0.00160494 -0.000804655
13 1.5297 -0.7164 0.00115 0.001759155 -0.00082386
14 1.6447 -0.7536 0.00115 0.001891405 -0.00086664
15 1.8013 -0.7573 0.00115 0.002071495 -0.000870895
16 1.9341 -0.7702 0.00115 0.002224215 -0.00088573
17 2.07 -0.7863 0.00115 0.0023805 -0.000904245
18 2.2043 -0.7996 0.00115 0.002534945 -0.00091954
19 2.3388 -0.8117 0.00115 0.00268962 -0.000933455
20 2.4726 -0.8238 0.00115 0.00284349 -0.00094737
21 2.6085 -0.8346 0.00115 0.002999775 -0.00095979
22 2.7428 -0.8423 0.00115 0.00315422 -0.000968645
23 2.8798 -0.8451 0.00115 0.00331177 -0.000971865
158
F 3.12336 -0.86283 0.004233 0.013221183 -0.003652351
24 3.2607 -0.8752 0.00115 0.003749805 -0.00100648
25 3.3947 -0.8866 0.00115 0.003903905 -0.00101959
26 3.5298 -0.8866 0.00115 0.00405927 -0.00101959
27 3.6657 -0.8924 0.00115 0.004215555 -0.00102626
28 3.7999 -0.896 0.00115 0.004369885 -0.0010304
29 3.9324 -0.9001 0.00115 0.00452226 -0.001035115
30 4.0645 -0.9024 0.00115 0.004674175 -0.00103776
31 4.1999 -0.9053 0.00115 0.004829885 -0.001041095
32 4.3339 -0.9076 0.00115 0.004983985 -0.00104374
33 4.4675 -0.9089 0.00115 0.005137625 -0.001045235
34 4.5999 -0.9099 0.00115 0.005289885 -0.001046385
35 4.7349 -0.9093 0.00115 0.005445135 -0.001045695
36 4.8693 -0.9077 0.00115 0.005599695 -0.001043855
37 5.0029 -0.9066 0.00115 0.005753335 -0.00104259
38 5.1384 -0.9046 0.00115 0.00590916 -0.00104029
39 5.2739 -0.9019 0.00115 0.006064985 -0.001037185
40 5.4071 -0.8985 0.00115 0.006218165 -0.001033275
41 5.5456 -0.8943 0.00115 0.00637744 -0.001028445
42 5.6795 -0.8897 0.00115 0.006531425 -0.001023155
43 5.8162 -0.883 0.00115 0.00668863 -0.00101545
44 5.9543 -0.8762 0.00115 0.006847445 -0.00100763
45 6.0891 -0.8685 0.00115 0.007002465 -0.000998775
46 6.2223 -0.8594 0.00115 0.007155645 -0.00098831
47 6.3569 -0.8488 0.00115 0.007310435 -0.00097612
48 6.4912 -0.837 0.00115 0.00746488 -0.00096255
49 6.6263 -0.8236 0.00115 0.007620245 -0.00094714
50 6.761 -0.8101 0.00115 0.00777515 -0.000931615
51 6.8974 -0.7952 0.00115 0.00793201 -0.00091448
52 7.0316 -0.779 0.00115 0.00808634 -0.00089585
159
53 7.1671 -0.7613 0.00115 0.008242165 -0.000875495
54 7.3015 -0.7424 0.00115 0.008396725 -0.00085376
55 7.438 -0.7217 0.00115 0.0085537 -0.000829955
56 7.5713 -0.7004 0.00115 0.008706995 -0.00080546
57 7.7066 -0.6785 0.00115 0.00886259 -0.000780275
58 7.84 -0.6556 0.00115 0.009016 -0.00075394
59 7.9739 -0.6325 0.00115 0.009169985 -0.000727375
60 8.1105 -0.6082 0.00115 0.009327075 -0.00069943
61 8.2424 -0.5839 0.00115 0.00947876 -0.000671485
62 8.3769 -0.5584 0.00115 0.009633435 -0.00064216
63 8.5103 -0.5325 0.00115 0.009786845 -0.000612375
64 8.6449 -0.5056 0.00115 0.009941635 -0.00058144
65 8.7793 -0.478 0.00115 0.010096195 -0.0005497
66 8.9104 -0.451 0.00115 0.01024696 -0.00051865
67 9.0444 -0.4239 0.00115 0.01040106 -0.000487485
68 9.1782 -0.3953 0.00115 0.01055493 -0.000454595
69 9.3142 -0.3694 0.00115 0.01071133 -0.00042481
70 9.4491 -0.3403 0.00115 0.010866465 -0.000391345
71 9.5823 -0.3125 0.00115 0.011019645 -0.000359375
RF 9.7605 -0.2772 0.002133 0.020819147 -0.000591264
∑A=0.194432m2 ; ∑ Ax = 0.939154737m3 ;∑ Ay = 0.019534231m3
X c = 4.9529 m;Y c = 0.11 m
Tabulation for calculating the moment of inertia:
Upper surface of the wing
stringer Area(m2) X(m) Y(m) Ixx(m3) Iyy(m3) Ixy(m3)
1 0.00115 0.0153 0.11 0 0.028037 0
2 0.00115 0.0614 0.21 1.15E-05 0.027516 -0.00056
160
3 0.00115 0.14 0.3019 4.23E-05 0.026639 -0.00106
4 0.00115 0.23 0.369 7.71E-05 0.025652 -0.00141
5 0.00115 0.33 0.421 0.000111 0.024577 -0.00165
6 0.00115 0.43 0.47 0.000149 0.023525 -0.00187
7 0.00115 0.541 0.5105 0.000184 0.022385 -0.00203
8 0.00115 0.65 0.5439 0.000217 0.021292 -0.00215
9 0.00115 0.76 0.5742 0.000248 0.020217 -0.00224
10 0.00115 0.87 0.6006 0.000277 0.019171 -0.0023
11 0.00115 0.98 0.624 0.000304 0.018152 -0.00235
12 0.00115 1.09 0.6466 0.000331 0.01716 -0.00238
13 0.00115 1.2 0.6674 0.000357 0.016197 -0.00241
14 0.00115 1.32 0.6866 0.000382 0.015178 -0.00241
15 0.00115 1.4315 0.704 0.000406 0.01426 -0.00241
16 0.00115 1.54 0.719 0.000427 0.013395 -0.00239
17 0.00115 1.65 0.7339 0.000448 0.012546 -0.00237
18 0.00115 1.762 0.7478 0.000468 0.011709 -0.00234
19 0.00115 1.874 0.7606 0.000487 0.010902 -0.0023
20 0.00115 1.987 0.7729 0.000505 0.010116 -0.00226
21 0.00115 2.102 0.7855 0.000525 0.009347 -0.00221
22 0.00115 2.212 0.796 0.000541 0.008639 -0.00216
23 0.00115 2.323 0.8065 0.000558 0.007954 -0.00211
24 0.00115 2.4381 0.8158 0.000573 0.007273 -0.00204
25 0.00115 2.55 0.8202 0.00058 0.00664 -0.00196
26 0.00115 2.665 0.8334 0.000602 0.00602 -0.0019
27 0.00115 2.7764 0.8412 0.000615 0.005448 -0.00183
28 0.00115 2.8865 0.8487 0.000628 0.004911 -0.00176
29 0.00115 2.9983 0.8553 0.000639 0.004394 -0.00168
30F 0.004233 3.12336 0.862828 0.002399 0.014169 -0.00583
31 0.00115 3.2371 0.8685 0.000662 0.003386 -0.0015
32 0.00115 3.3516 0.8742 0.000672 0.002949 -0.00141
161
33 0.00115 3.4664 0.8801 0.000682 0.002541 -0.00132
34 0.00115 3.5792 0.885 0.000691 0.00217 -0.00122
35 0.00115 3.6926 0.8895 0.000699 0.001827 -0.00113
36 0.00115 3.8023 0.8937 0.000706 0.001522 -0.00104
37 0.00115 3.9133 0.8969 0.000712 0.001243 -0.00094
38 0.00115 4.0264 0.9003 0.000718 0.000987 -0.00084
39 0.00115 4.1405 0.9084 0.000733 0.000759 -0.00075
40 0.00115 4.2546 0.9061 0.000729 0.000561 -0.00064
41 0.00115 4.3691 0.9074 0.000731 0.000392 -0.00054
42 0.00115 4.4843 0.9081 0.000733 0.000253 -0.00043
43 0.00115 4.5952 0.9128 0.000741 0.000147 -0.00033
44 0.00115 4.7052 0.919 0.000753 7.06E-05 -0.00023
45 0.00115 4.8181 0.9121 0.00074 2.09E-05 -0.00012
46 0.00115 4.9295 0.9119 0.00074 6.3E-07 -2.2E-05
47 0.00115 5.0417 0.9125 0.000741 9.07E-06 8.2E-05
48 0.00115 5.1562 0.9122 0.00074 4.75E-05 0.000188
49 0.00115 5.2752 0.9118 0.000739 0.000119 0.000297
50 0.00115 5.3901 0.9096 0.000735 0.00022 0.000402
51 0.00115 5.5011 0.9094 0.000735 0.000346 0.000504
52 0.00115 5.6125 0.9081 0.000733 0.0005 0.000605
53 0.00115 5.7262 0.9057 0.000728 0.000688 0.000708
54 0.00115 5.8414 0.9035 0.000724 0.000908 0.000811
55 0.00115 5.9528 0.9013 0.00072 0.00115 0.00091
56 0.00115 6.0631 0.899 0.000716 0.001417 0.001007
57 0.00115 6.1775 0.8962 0.000711 0.001725 0.001107
58 0.00115 6.2932 0.8926 0.000704 0.002066 0.001206
59 0.00115 6.4092 0.8892 0.000698 0.002439 0.001305
60 0.00115 6.5251 0.8856 0.000692 0.002843 0.001402
61 0.00115 6.6371 0.8811 0.000684 0.003262 0.001493
62 0.00115 6.7506 0.8765 0.000676 0.003716 0.001585
162
63 0.00115 6.8647 0.8721 0.000668 0.004203 0.001676
64 0.00115 6.9809 0.8665 0.000658 0.00473 0.001764
65 0.00115 7.0974 0.8607 0.000648 0.005289 0.001851
66 0.00115 7.2088 0.8546 0.000638 0.005852 0.001932
67 0.00115 7.3233 0.8477 0.000626 0.006462 0.002011
68 0.00115 7.4374 0.8407 0.000614 0.007099 0.002088
69 0.00115 7.5481 0.8329 0.000601 0.007745 0.002157
70 0.00115 7.6628 0.8248 0.000588 0.008445 0.002228
71 0.00115 7.7787 0.8168 0.000575 0.009183 0.002297
72 0.00115 7.8915 0.8082 0.000561 0.009931 0.002359
73 0.00115 8.0047 0.7991 0.000546 0.010711 0.002418
74 0.00115 8.1207 0.7894 0.000531 0.01154 0.002475
75 0.00115 8.2341 0.7787 0.000514 0.012381 0.002523
76 0.00115 8.3461 0.7674 0.000497 0.013241 0.002565
77 0.00115 8.4591 0.7561 0.00048 0.014137 0.002605
78 0.00115 8.5703 0.7438 0.000462 0.015048 0.002637
79 0.00115 8.6804 0.7319 0.000445 0.015978 0.002666
80 0.00115 8.7903 0.7191 0.000427 0.016934 0.002688
81 0.00115 8.8998 0.7058 0.000408 0.017915 0.002704
82 0.00115 9.0125 0.6921 0.00039 0.018952 0.002718
83 0.00115 9.126 0.6772 0.00037 0.020027 0.002722
84 0.00115 9.2399 0.6624 0.000351 0.021135 0.002723
85 0.00115 9.3519 0.6467 0.000331 0.022254 0.002715
86 0.00115 9.4621 0.6314 0.000313 0.023383 0.002704
87 0.00115 9.5714 0.6154 0.000294 0.02453 0.002684
88 0.00115 9.6739 0.5998 0.000276 0.025631 0.002659
RF 0.002133 9.7605 0.586931 0.000485 0.0493 0.004891
Lower Surface of the wing:
stringer area X Y Ixx Iyy Ixy
163
1 0.00115 0.02603 -0.1406 7.22E-05 0.027915 0.00142
2 0.00115 0.09979 -0.2614 0.000159 0.027086 0.002073
3 0.00115 0.2074 -0.3544 0.000248 0.025898 0.002534
4 0.00115 0.3319 -0.4266 0.000331 0.024557 0.002852
5 0.00115 0.4605 -0.4832 0.000405 0.023209 0.003065
6 0.00115 0.5928 -0.5282 0.000468 0.021862 0.0032
7 0.00115 0.7238 -0.5656 0.000525 0.020568 0.003286
8 0.00115 0.8599 -0.5991 0.000578 0.019266 0.003338
9 0.00115 0.991 -0.6274 0.000625 0.018051 0.00336
10 0.00115 1.1237 -0.6537 0.000671 0.016862 0.003363
11 0.00115 1.2599 -0.6777 0.000714 0.015684 0.003345
12 0.00115 1.3956 -0.6997 0.000754 0.014553 0.003312
13 0.00115 1.5297 -0.7164 0.000785 0.013476 0.003253
14 0.00115 1.6447 -0.7536 0.000858 0.012586 0.003286
15 0.00115 1.8013 -0.7573 0.000865 0.011422 0.003143
16 0.00115 1.9341 -0.7702 0.000891 0.01048 0.003056
17 0.00115 2.07 -0.7863 0.000924 0.009558 0.002972
18 0.00115 2.2043 -0.7996 0.000951 0.008688 0.002875
19 0.00115 2.3388 -0.8117 0.000977 0.007859 0.002771
20 0.00115 2.4726 -0.8238 0.001003 0.007075 0.002664
21 0.00115 2.6085 -0.8346 0.001026 0.006321 0.002547
22 0.00115 2.7428 -0.8423 0.001043 0.005617 0.00242
23 0.00115 2.8798 -0.8451 0.001049 0.004942 0.002277
F 0.004233 3.12336 -0.8628 0.004006 0.014169 0.007534
24 0.00115 3.2607 -0.8752 0.001116 0.003293 0.001917
25 0.00115 3.3947 -0.8866 0.001142 0.002792 0.001786
26 0.00115 3.5298 -0.8866 0.001142 0.002329 0.001631
27 0.00115 3.6657 -0.8924 0.001156 0.001905 0.001484
28 0.00115 3.7999 -0.896 0.001164 0.001529 0.001334
29 0.00115 3.9324 -0.9001 0.001173 0.001198 0.001185
164
30 0.00115 4.0645 -0.9024 0.001179 0.000908 0.001034
31 0.00115 4.1999 -0.9053 0.001185 0.000652 0.000879
32 0.00115 4.3339 -0.9076 0.001191 0.000441 0.000724
33 0.00115 4.4675 -0.9089 0.001194 0.000271 0.000569
34 0.00115 4.5999 -0.9099 0.001196 0.000143 0.000414
35 0.00115 4.7349 -0.9093 0.001195 5.47E-05 0.000256
36 0.00115 4.8693 -0.9077 0.001191 8.04E-06 9.78E-05
37 0.00115 5.0029 -0.9066 0.001188 2.88E-06 -5.8E-05
38 0.00115 5.1384 -0.9046 0.001184 3.96E-05 -0.00022
39 0.00115 5.2739 -0.9019 0.001178 0.000118 -0.00037
40 0.00115 5.4071 -0.8985 0.00117 0.000237 -0.00053
41 0.00115 5.5456 -0.8943 0.00116 0.000404 -0.00068
42 0.00115 5.6795 -0.8897 0.001149 0.000607 -0.00084
43 0.00115 5.8162 -0.883 0.001134 0.000857 -0.00099
44 0.00115 5.9543 -0.8762 0.001118 0.001153 -0.00114
45 0.00115 6.0891 -0.8685 0.001101 0.001485 -0.00128
46 0.00115 6.2223 -0.8594 0.001081 0.001853 -0.00142
47 0.00115 6.3569 -0.8488 0.001057 0.002267 -0.00155
48 0.00115 6.4912 -0.837 0.001031 0.002721 -0.00168
49 0.00115 6.6263 -0.8236 0.001002 0.00322 -0.0018
50 0.00115 6.761 -0.8101 0.000974 0.00376 -0.00191
51 0.00115 6.8974 -0.7952 0.000942 0.004348 -0.00202
52 0.00115 7.0316 -0.779 0.000909 0.004969 -0.00213
53 0.00115 7.1671 -0.7613 0.000873 0.005638 -0.00222
54 0.00115 7.3015 -0.7424 0.000836 0.006343 -0.0023
55 0.00115 7.438 -0.7217 0.000795 0.007102 -0.00238
56 0.00115 7.5713 -0.7004 0.000755 0.007884 -0.00244
57 0.00115 7.7066 -0.6785 0.000715 0.00872 -0.0025
58 0.00115 7.84 -0.6556 0.000674 0.009586 -0.00254
59 0.00115 7.9739 -0.6325 0.000634 0.010495 -0.00258
165
60 0.00115 8.1105 -0.6082 0.000593 0.011466 -0.00261
61 0.00115 8.2424 -0.5839 0.000554 0.012444 -0.00262
62 0.00115 8.3769 -0.5584 0.000514 0.013482 -0.00263
63 0.00115 8.5103 -0.5325 0.000475 0.014553 -0.00263
64 0.00115 8.6449 -0.5056 0.000436 0.015675 -0.00261
65 0.00115 8.7793 -0.478 0.000398 0.016838 -0.00259
66 0.00115 8.9104 -0.451 0.000362 0.018011 -0.00255
67 0.00115 9.0444 -0.4239 0.000328 0.019251 -0.00251
68 0.00115 9.1782 -0.3953 0.000294 0.020531 -0.00246
69 0.00115 9.3142 -0.3694 0.000264 0.021874 -0.0024
70 0.00115 9.4491 -0.3403 0.000233 0.023248 -0.00233
71 0.00115 9.5823 -0.3125 0.000205 0.024646 -0.00225
RF 0.002133 9.7605 -0.2772 0.00032 0.0493 -0.00397
∑ I xx = 0.1121 m 3
∑ I yy = 1.6241 m 3
∑ I xy = 0.0237 m 3
σ zz = 1806359273Y – 30724010.12 X
Tabulation for finding the stress:
Upper surface of the wing:
stringer X(m) Y(m) stresss (Pa)
1 0.0153 0.11 44967669.14
2 0.0614 0.21 85651073.36
3 0.14 0.3019 122818172.1
4 0.23 0.369 149756555.9
5 0.33 0.421 170448319.5
6 0.43 0.47 189911147.7
166
7 0.541 0.5105 205824916
8 0.65 0.5439 218842399.4
9 0.76 0.5742 230583885
10 0.87 0.6006 240727754.5
11 0.98 0.624 249642688.6
12 1.09 0.6466 258229906.5
13 1.2 0.6674 266079763.2
14 1.32 0.6866 273213209.3
15 1.4315 0.704 279661125.7
16 1.54 0.719 285144187.4
17 1.65 0.7339 290577137.7
18 1.762 0.7478 295588247.2
19 1.874 0.7606 300148747.1
20 1.987 0.7729 304498326.5
21 2.102 0.7855 308958603.8
22 2.212 0.796 312589115.5
23 2.323 0.8065 316213529.4
24 2.4381 0.8158 319321367.9
25 2.55 0.8202 320441458.3
26 2.665 0.8334 325147522.7
27 2.7764 0.8412 327663455.5
28 2.8865 0.8487 330064422
29 2.9983 0.8553 332086341.5
30F 3.12336 0.862828 334407554.4
31 3.2371 0.8685 336037493.4
32 3.3516 0.8742 337674268.1
33 3.4664 0.8801 339391142.5
34 3.5792 0.885 340710567.4
35 3.6926 0.8895 341862475.6
36 3.8023 0.8937 342914052.2
167
37 3.9133 0.8969 343548056.5
38 4.0264 0.9003 344251184.4
39 4.1405 0.9084 346873546.6
40 4.2546 0.9061 345235599.2
41 4.3691 0.9074 345069935.3
42 4.4843 0.9081 344654215.8
43 4.5952 0.9128 345903297.6
44 4.7052 0.919 347772335.2
45 4.8181 0.9121 344257337.6
46 4.9295 0.9119 343496109.2
47 5.0417 0.9125 343057718.7
48 5.1562 0.9122 342236622.6
49 5.2752 0.9118 341347121.6
50 5.3901 0.9096 339745260.5
51 5.5011 0.9094 338986471.3
52 5.6125 0.9081 337774633.3
53 5.7262 0.9057 336098160.6
54 5.8414 0.9035 334494470.2
55 5.9528 0.9013 332913951.6
56 6.0631 0.899 331299176
57 6.1775 0.8962 329454576.8
58 6.2932 0.8926 327274334.3
59 6.4092 0.8892 325174191.4
60 6.5251 0.8856 322992729.3
61 6.6371 0.8811 320466368.1
62 6.7506 0.8765 317889895.6
63 6.8647 0.8721 315391693.5
64 6.9809 0.8665 312389111.7
65 7.0974 0.8607 309302771.6
66 7.2088 0.8546 306124636.9
168
67 7.3233 0.8477 302599882.8
68 7.4374 0.8407 299036603.3
69 7.5481 0.8329 295166340.3
70 7.6628 0.8248 291148792.4
71 7.7787 0.8168 287164891.7
72 7.8915 0.8082 282954107.1
73 8.0047 0.7991 278536060.9
74 8.1207 0.7894 273855153.6
75 8.2341 0.7787 268780455.5
76 8.3461 0.7674 263468507.3
77 8.4591 0.7561 258150461.3
78 8.5703 0.7438 252433746.3
79 8.6804 0.7319 246887596.9
80 8.7903 0.7191 240973986.5
81 8.8998 0.7058 234857992.6
82 9.0125 0.6921 228558627.6
83 9.126 0.6772 221762810.2
84 9.2399 0.6624 215005518.1
85 9.3519 0.6467 207891131.3
86 9.4621 0.6314 200951578.6
87 9.5714 0.6154 193730762.4
88 9.6739 0.5998 186715269.6
RF 9.7605 0.586931 180915637.2
Lower surface of the wing:
Stringer X(m) Y(m) stresss (Pa)
1 0.02603 -0.1406 -57754834.32
2 0.09979 -0.2614 -107689744.6
3 0.2074 -0.3544 -146442931.7
4 0.3319 -0.4266 -176778492.3
169
5 0.4605 -0.4832 -200748589.7
6 0.5928 -0.5282 -219989365.5
7 0.7238 -0.5656 -236108910.9
8 0.8599 -0.5991 -250661939.2
9 0.991 -0.6274 -263054323.6
10 1.1237 -0.6537 -274637174.3
11 1.2599 -0.6777 -285299183.5
12 1.3956 -0.6997 -295138853.5
13 1.5297 -0.7164 -302797647.7
14 1.6447 -0.7536 -318737698.7
15 1.8013 -0.7573 -321208307.4
16 1.9341 -0.7702 -327302522.9
17 2.07 -0.7863 -334726506.1
18 2.2043 -0.7996 -340993726.4
19 2.3388 -0.8117 -346770592.1
20 2.4726 -0.8238 -352543189.3
21 2.6085 -0.8346 -357796053.2
22 2.7428 -0.8423 -361769260.7
23 2.8798 -0.8451 -363751671.1
F 3.12336 -0.8628 -372499050
24 3.2607 -0.8752 -378404657.1
25 3.3947 -0.8866 -383891722.2
26 3.5298 -0.8866 -384715540.3
27 3.6657 -0.8924 -387920178.6
28 3.7999 -0.896 -390213231.1
29 3.9324 -0.9001 -392700740
30 4.0645 -0.9024 -394448448.4
31 4.1999 -0.9053 -396462066.8
32 4.3339 -0.9076 -398221361.1
33 4.4675 -0.9089 -399568571.1
170
34 4.5999 -0.9099 -400785570.2
35 4.7349 -0.9093 -401362991.5
36 4.8693 -0.9077 -401527108.9
37 5.0029 -0.9066 -401891170.5
38 5.1384 -0.9046 -401898137.5
39 5.2739 -0.9019 -401618352.8
40 5.4071 -0.8985 -401037791.6
41 5.5456 -0.8943 -400161832.7
42 5.6795 -0.8897 -399093965.7
43 5.8162 -0.883 -397182917.9
44 5.9543 -0.8762 -395239442.6
45 6.0891 -0.8685 -392907163.7
46 6.2223 -0.8594 -389991625.1
47 6.3569 -0.8488 -386470155.7
48 6.4912 -0.837 -382455282.8
49 6.6263 -0.8236 -377789856
50 6.761 -0.8101 -373081025.5
51 6.8974 -0.7952 -367809058.1
52 7.0316 -0.779 -361991136.8
53 7.1671 -0.7613 -355566674.9
54 7.3015 -0.7424 -348643931.3
55 7.438 -0.7217 -340996631.9
56 7.5713 -0.7004 -333084032.2
57 7.7066 -0.6785 -324937841.2
58 7.84 -0.6556 -316370419.2
59 7.9739 -0.6325 -307724117
60 8.1105 -0.6082 -298602704.8
61 8.2424 -0.5839 -289452632.8
62 8.3769 -0.5584 -279826841
63 8.5103 -0.5325 -270030483.4
171
64 8.6449 -0.5056 -259831798.2
65 8.7793 -0.478 -249345141.8
66 8.9104 -0.451 -239084149.6
67 9.0444 -0.4239 -228799876.6
68 9.1782 -0.3953 -217899916.4
69 9.3142 -0.3694 -208119413.2
70 9.4491 -0.3403 -197021338
71 9.5823 -0.3125 -186445435.2
RF 9.7605 #REF! -173070859.1
So from the above tabulation it is found that the stress calculated for each stringer and it is found
to be less than the ultimate strength of the material selected.
So the selected stringer dimensions are shown below.
172
14.1.8 SHEAR FLOW OF THE AIRFOIL: ( using ‘K’ method)
General procedure:
1. The shear force acting on the root airfoil is obtained from the shear force diagram
constructed and the shear force obtained is resolved into ‘x’ and ‘y’ since the wing is
attached at a wing setting angle (α)
V x = V sin α ; V y = V cos α
2. The root airfoil is considered as a 2 cell box and the cut is made at the origin and at
top of the front spar.
173
3.. K 1 , K 2 , K 3 are evaluated using the following relations given
K 1 =
K 2 =
K 3 =
4.. The basic shear flow equation is given by
Where ∑ A(X – X c) and ∑ A(Y- Y c)have been already evaluated.
5. The shear flow due to torque and bending has to be added to the above obtained basic
shear flow for the closed cell. So the twist equation for cell1 and 2 are considered here
again and shear flow obtained in cell 1 and cell 2 are added as constants to the respective
twist equation. There force the equation will be in the form
Constants are found by substituting thickness of the skin in ‘q 1’ and ‘q 2’ which
were evaluated in finding the skin thickness.
6. Assuming twist in both cells are equal one equation is framed and other equation is
framed by taking the moment contribution of the components of shear force.
7. Solving the two equations we get q 1 and q 2 which is added to the basic shear flow and
the shear flow is evaluated in the root airfoil section.
Following the above steps
K 1 = 0.13105 m 3
K 2 = 8.9496 m 3
174
K 3 = 0.6176 m 3
q = -85344.66581∑A ( X – X c ) -6972868.162∑A (Y – Y c)
The shear flow due to torsion and bending
q 1’ = 29436.4041 Nm -1
q 2’ = 34276.4386 Nm -1
Tabulation for Shear Flow of the wing:
Cell 1 of the root airfoil:
stringer area X-Xc Y-Yc ∑A(X-Xc) ∑ A(Y-Yc) basic q q(net)
1 0.00115 -4.9376 0 -0.00567824 0 484.6074951 28951.7966
2 0.00115 -4.8915 0.1 -0.01130346 0.000115 162.8106042 29273.5935
3 0.00115 -4.8129 0.1919 -0.0168383 0.000335685 -903.6281628 30340.03226
4 0.00115 -4.7229 0.259 -0.02226963 0.000633535 -2516.961474 31953.36557
5 0.00115 -4.6229 0.311 -0.02758597 0.000991185 -4557.086939 33993.49104
6 0.00115 -4.5229 0.36 -0.02758597 0.001405185 -7443.854358 36880.25846
7 0.00115 -4.4119 0.4005 -0.03265965 0.00186576 -10222.37116 39658.77526
8 0.00115 -4.3029 0.4339 -0.03760799 0.002364745 -13279.41378 42715.81788
9 0.00115 -4.1929 0.4642 -0.04242982 0.002898575 -16590.2221 46026.6262
10 0.00115 -4.0829 0.4906 -0.04712516 0.003462765 -20123.52279 49559.92689
11 0.00115 -3.9729 0.514 -0.05169399 0.004053865 -23855.25946 53291.66356
12 0.00115 -3.8629 0.5366 -0.05613633 0.004670955 -27779.01708 57215.42118
13 0.00115 -3.7529 0.5574 -0.06045216 0.005311965 -31880.36181 61316.76591
14 0.00115 -3.6329 0.5766 -0.06463 0.005975055 -36147.44503 65583.84913
15 0.00115 -3.5214 0.594 -0.06867961 0.006658155 -40564.99865 70001.40275
16 0.00115 -3.4129 0.609 -0.07260444 0.007358505 -45113.48314 74549.88724
17 0.00115 -3.3029 0.6239 -0.07640278 0.00807599 -49792.24382 79228.64792
175
18 0.00115 -3.1909 0.6378 -0.08007231 0.00880946 -54593.45819 84029.86229
19 0.00115 -3.0789 0.6506 -0.08361305 0.00955765 -59508.30558 88944.70968
20 0.00115 -2.9659 0.6629 -0.08702383 0.010319985 -64532.87472 93969.27882
21 0.00115 -2.8509 0.6755 -0.09030237 0.01109681 -69669.76756 99106.17166
22 0.00115 -2.7409 0.686 -0.09345440 0.01188571 -74901.65388 104338.058
23 0.00115 -2.6299 0.6965 -0.09647879 0.012686685 -80228.63183 109665.0359
24 0.00115 -2.5148 0.7058 -0.09937081 0.013498355 -85641.48125 115077.8853
25 0.00115 -2.4029 0.7102 -0.10213414 0.014315085 -91100.59596 120537.0001
26 0.00115 -2.2879 0.7234 -0.10476523 0.015146995 -96676.84564 126113.2497
27 0.00115 -2.1765 0.7312 -0.10726820 0.015987875 -102326.5755 131762.9796
28 0.00115 -2.0664 0.7387 -0.10964456 0.01683738 -108047.2522 137483.6563
29 0.00115 -1.9546 0.7453 -0.11189235 0.017694475 -113831.8257 143268.2298
30F 0.004233
-
1.82954 0.752828 -0.11963679 0.020881196 -135391.4637 164827.8678
1 0.00115
-
4.92687 -0.2506 -0.12530269 0.020593006 -132898.3985 162334.8026
2 0.00115
-
4.85311 -0.3714 -0.13088377 0.020165896 -129443.9016 158880.3057
3 0.00115 -4.7455 -0.4644 -0.1363411 0.019631836 -125254.2181 154690.6222
4 0.00115 -4.621 -0.5366 -0.14165525 0.019014746 -120497.7965 149934.2006
5 0.00115 -4.4924 -0.5932 -0.14682151 0.018332566 -115300.1326 144736.5367
6 0.00115 -4.3601 -0.6382 -0.15183562 0.017598636 -109754.6075 139191.0116
7 0.00115 -4.2291 -0.6756 -0.15669909 0.016821696 -103922.0365 133358.4406
8 0.00115 -4.093 -0.7091 -0.16140604 0.016006231 -97834.19348 127270.5976
9 0.00115 -3.9619 -0.7374 -0.16596222 0.015158221 -91532.28546 120968.6896
10 0.00115 -3.8292 -0.7637 -0.17036580 0.014279966 -85032.50707 114468.9112
11 0.00115 -3.693 -0.7877 -0.17461275 0.013374111 -78353.64505 107790.0492
176
12 0.00115 -3.5573 -0.8097 -0.17870365 0.012442956 -71511.68793 100948.092
13 0.00115 -3.4232 -0.8264 -0.18264033 0.011492596 -64548.97831 93985.38241
14 0.00115 -3.3082 -0.8636 -0.18644476 0.010499456 -57299.25621 86735.66031
15 0.00115 -3.1516 -0.8673 -0.1900691 0.009502061 -50035.23429 79471.63839
16 0.00115 -3.0188 -0.8802 -0.19354072 0.008489831 -42680.8037 72117.2078
17 0.00115 -2.8829 -0.8963 -0.19685605 0.007459086 -35210.60855 64647.01265
18 0.00115 -2.7486 -0.9096 -0.20001694 0.006413046 -27646.94443 57083.34853
19 0.00115 -2.6141 -0.9217 -0.20302316 0.005353091 -19999.45355 49435.85765
20 0.00115 -2.4803 -0.9338 -0.20587550 0.004279221 -12268.06718 41704.47128
21 0.00115 -2.3444 -0.9446 -0.20857156 0.003192931 -4463.415888 33899.81999
22 0.00115 -2.2101 -0.9523 -0.21111318 0.002097786 3389.799098 26046.605
23 0.00115 -2.0731 -0.9551 -0.21349724 0.000999421 11252.02067 18184.38343
24F 0.004233
-
1.82954
-
0.972828
-
0.221241688 -0.00311856 40627.10563
-
11190.70153
Cell 2 of the root Airfoil:
stringer area X-Xc Y-Yc ∑A(X-Xc) ∑ A(Y-Yc) basic q q(net)
30F 0.004233 -1.8295 0.752828 -0.00774444 0.003186721 -21559.63799 12716.80061
31 0.00115 -1.7158 0.7585 -0.00971761 0.004058996 -27473.49703 6802.94157
32 0.00115 -1.6013 0.7642 -0.01155910 0.004937826 -33444.30098 832.137619
33 0.00115 -1.4865 0.7701 -0.01326858 0.005823441 -39473.68305 -5197.24444
34 0.00115 -1.3737 0.775 -0.01484833 0.006714691 -45553.42813 -11276.9895
35 0.00115 -1.2603 0.7795 -0.01629768 0.007611116 -51680.38761 -17403.9490
36 0.00115 -1.1506 0.7837 -0.01762087 0.008512371 -57851.7927 -23575.3541
37 0.00115 -1.0396 0.7869 -0.01881641 0.009417306 -64059.75219 -29783.3135
38 0.00115 -0.9265 0.7903 -0.01988188 0.010326151 -70306.07594 -36029.6373
39 0.00115 -0.8124 0.7984 -0.02081614 0.011244311 -76628.55047 -42352.1118
177
40 0.00115 -0.6983 0.7961 -0.02161919 0.012159826 -82943.78025 -48667.3416
41 0.00115 -0.5838 0.7974 -0.02229056 0.013076836 -89280.67224 -55004.2336
42 0.00115 -0.4686 0.7981 -0.02282945 0.013994651 -95634.48384 -61358.0452
43 0.00115 -0.3577 0.8028 -0.02324080 0.014917871 -102036.8682 -67760.4296
44 0.00115 -0.2477 0.809 -0.02352566 0.015848221 -108499.7653 -74223.3266
45 0.00115 -0.1348 0.8021 -0.02368068 0.016770636 -114918.4133 -80641.9747
46 0.00115 -0.0234 0.8019 -0.02370759 0.017692821 -121346.3911 -87069.9525
47 0.00115 0.0888 0.8025 -0.02360547 0.018615696 -127790.1922 -93513.7536
48 0.00115 0.2033 0.8022 -0.02337167 0.019538226 -134242.8255 -99966.3868
49 0.00115 0.3223 0.8018 -0.02300103 0.020460296 -140703.9306 -106427.492
50 0.00115 0.4372 0.7996 -0.02249825 0.021379836 -147158.6714 -112882.232
51 0.00115 0.5482 0.7994 -0.02186782 0.022299146 -153622.7026 -119346.264
52 0.00115 0.6596 0.7981 -0.02110928 0.023216961 -160087.243 -125810.804
53 0.00115 0.7733 0.7957 -0.02021998 0.024132016 -166543.6974 -132267.258
54 0.00115 0.8885 0.7935 -0.01919821 0.025044541 -172993.817 -138717.378
55 0.00115 0.9999 0.7913 -0.01804832 0.025954536 -179437.2287 -145160.790
56 0.00115 1.1102 0.789 -0.01677159 0.026861886 -185873.0227 -151596.584
57 0.00115 1.2246 0.7862 -0.01536330 0.027766016 -192297.5921 -158021.153
58 0.00115 1.3403 0.7826 -0.01382196 0.028666006 -198704.6492 -164428.210
59 0.00115 1.4563 0.7792 -0.01214721 0.029562086 -205095.8275 -170819.388
60 0.00115 1.5722 0.7756 -0.01033918 0.030454026 -211469.5132 -177193.074
61 0.00115 1.6842 0.7711 -0.00840235 0.031340791 -217818.1068 -183541.668
62 0.00115 1.7977 0.7665 -0.00633500 0.032222266 -224140.9535 -189864.514
63 0.00115 1.9118 0.7621 -0.00413643 0.033098681 -230439.7159 -196163.277
64 0.00115 2.028 0.7565 -0.00180423 0.033968656 -236704.9778 -202428.539
65 0.00115 2.1445 0.7507 0.000661942 0.034831961 -242935.1646 -208658.726
66 0.00115 2.2559 0.7446 0.003256227 0.035688251 -249127.3702 -214850.931
67 0.00115 2.3704 0.7377 0.005982187 0.036536606 -255275.484 -220999.045
68 0.00115 2.4845 0.7307 0.008839362 0.037376911 -261378.6646 -227102.226
69 0.00115 2.5952 0.7229 0.011823842 0.038208246 -267430.1634 -233153.724
178
70 0.00115 2.7099 0.7148 0.014940227 0.039030266 -273427.9673 -239151.528
71 0.00115 2.8258 0.7068 0.018189897 0.039843086 -279372.996 -245096.557
72 0.00115 2.9386 0.6982 0.021569287 0.040646016 -285260.134 -250983.695
73 0.00115 3.0518 0.6891 0.025078857 0.041438481 -291085.411 -256808.974
74 0.00115 3.1678 0.6794 0.028721827 0.042219791 -296844.2907 -262567.852
75 0.00115 3.2812 0.6687 0.032495207 0.042988796 -302528.499 -268252.060
76 0.00115 3.3932 0.6574 0.036397387 0.043744806 -308133.0873 -273856.647
77 0.00115 3.5062 0.6461 0.040429517 0.044487821 -313658.1538 -279381.715
78 0.00115 3.6174 0.6338 0.044589527 0.045216691 -319095.5028 -284819.064
79 0.00115 3.7275 0.6219 0.048876152 0.045931876 -324448.2341 -290171.795
80 0.00115 3.8374 0.6091 0.053289162 0.046632341 -329709.1111 -295432.675
81 0.00115 3.9469 0.5958 0.057828097 0.047317511 -334874.0851 -300597.646
82 0.00115 4.0596 0.5821 0.062496637 0.047986926 -339940.2626 -305663.824
83 0.00115 4.1731 0.5672 0.067295702 0.048639206 -344898.0996 -310621.661
84 0.00115 4.287 0.5524 0.072225752 0.049274466 -349748.4373 -315471.998
85 0.00115 4.399 0.5367 0.077284602 0.049891671 -354483.8723 -320207.433
86 0.00115 4.5092 0.5214 0.082470182 0.050491281 -359107.4354 -324830.996
87 0.00115 4.6185 0.5054 0.087781457 0.051072491 -363613.425 -329336.986
88 0.00115 4.721 0.4898 0.093210607 0.051635761 -368004.3815 -333727.942
RF 0.002133 4.8076 0.4769314 0.103465218 0.052653056 -375973.0195 -341696.580
F 0.004233
-
1.82954 -0.972828 0.095720775 0.048535075 -346597.9345 -312321.495
24 0.00115 -1.6922 -0.9852 0.093774745 0.047402095 -338531.7311 304255.2925
25 0.00115 -1.5582 -0.9966 0.091982815 0.046256005 -330387.2649 -296110.826
26 0.00115 -1.4231 -0.9966 0.09034625 0.045109915 -322256.0584 -287979.619
27 0.00115 -1.2872 -1.0024 0.08886597 0.043957155 -314091.6809 -279815.242
28 0.00115 -1.153 -1.006 0.08754002 0.042800255 -305911.6069 -271635.168
29 0.00115 -1.0205 -1.0101 0.086366445 0.04163864 -297711.6603 -263435.221
30 0.00115 -0.8884 -1.0124 0.085344785 0.04047438 -289506.2356 -255229.797
31 0.00115 -0.753 -1.0153 0.084478835 0.039306785 -281290.8454 -247014.406
179
32 0.00115 -0.619 -1.0176 0.083766985 0.038136545 -273070.1635 -238793.724
33 0.00115 -0.4854 -1.0189 0.083208775 0.03696481 -264852.1696 -230575.731
34 0.00115 -0.353 -1.0199 0.082802825 0.035791925 -256639.1515 -222362.712
35 0.00115 -0.218 -1.0193 0.082552125 0.03461973 -248444.1944 -214167.755
36 0.00115 -0.0836 -1.0177 0.082455985 0.033449375 -240275.2582 -205998.819
37 0.00115 0.05 -1.0166 0.082513485 0.032280285 -232128.2551 -197851.816
38 0.00115 0.1855 -1.0146 0.08272681 0.031113495 -224010.5884 -189734.149
39 0.00115 0.321 -1.0119 0.08309596 0.02994981 -215927.8713 -181651.432
40 0.00115 0.4542 -1.0085 0.08361829 0.028790035 -207885.4912 -173609.052
41 0.00115 0.5927 -1.0043 0.084299895 0.02763509 -199890.3833 -165613.944
42 0.00115 0.7266 -0.9997 0.085135485 0.026485435 -191945.3037 -157668.865
43 0.00115 0.8633 -0.993 0.08612828 0.025343485 -184067.3667 -149790.928
44 0.00115 1.0014 -0.9862 0.08727989 0.024209355 -176257.5115 -141981.072
45 0.00115 1.1362 -0.9785 0.08858652 0.02308408 -168522.6312 -134246.192
46 0.00115 1.2694 -0.9694 0.09004633 0.02196927 -160873.795 -126597.356
47 0.00115 1.404 -0.9588 0.09166093 0.02086665 -153323.1686 -119046.73
48 0.00115 1.5383 -0.947 0.093429975 0.0197776 -145880.3451 -111603.906
49 0.00115 1.6734 -0.9336 0.095354385 0.01870396 -138558.2331 -104281.794
50 0.00115 1.8081 -0.9201 0.0974337 0.017645845 -131357.5951 -97081.1565
51 0.00115 1.9445 -0.9052 0.099669875 0.016604865 -124289.8244 -90013.3858
52 0.00115 2.0787 -0.889 0.10206038 0.015582515 -117365.1295 -83088.6909
53 0.00115 2.2142 -0.8713 0.10460671 0.01458052 -110595.6662 -76319.2275
54 0.00115 2.3486 -0.8524 0.1073076 0.01360026 -103990.949 -69714.5103
55 0.00115 2.4851 -0.8317 0.110165465 0.012643805 -97565.61788 -63289.1792
56 0.00115 2.6184 -0.8104 0.113176625 0.011711845 -91324.17011 -57047.7311
57 0.00115 2.7537 -0.7885 0.11634338 0.01080507 -85271.61323 -50995.1746
58 0.00115 2.8871 -0.7656 0.119663545 0.00992463 -79415.77956 -45139.3409
59 0.00115 3.021 -0.7425 0.123137695 0.009070755 -73758.32193 -39481.8833
60 0.00115 3.1576 -0.7182 0.126768935 0.008244825 -68309.12789 -34032.6899
61 0.00115 3.2895 -0.6939 0.13055186 0.00744684 -63067.73616 -28791.2976
180
62 0.00115 3.424 -0.6684 0.13448946 0.00667818 -58044.02448 -23767.5858
63 0.00115 3.5574 -0.6425 0.13858047 0.005939305 -53241.09239 -18964.6579
64 0.00115 3.692 -0.6156 0.14282627 0.005231365 -48667.07649 -14390.6379
65 0.00115 3.8264 -0.588 0.14722663 0.004555165 -44327.57029 -10051.1319
66 0.00115 3.9575 -0.561 0.151777755 0.003910015 -40217.43864 -5941.00004
67 0.00115 4.0915 -0.5339 0.15648298 0.00329603 -36337.76804 -2061.32937
68 0.00115 4.2253 -0.5053 0.161342075 0.002714935 -32700.56705 1575.871549
69 0.00115 4.3613 -0.4794 0.16635757 0.002163625 -29284.40085 4992.037751
70 0.00115 4.4962 -0.4503 0.1715282 0.00164578 -26114.82163 8161.616975
71 0.00115 4.6294 -0.4225 0.17685201 0.001159905 -23181.23809 11095.20051
RF 0.002133 4.8076 -0.387198 0.187106621 0.000334011 -18297.56612 15978.87248
14.2 DESIGN OF FUSELAGE:
General Procedure:
1. The load distribution on the fuselage is identified and the shear force, Bending moment
diagrams are constructed.
2. Stringer design for the fuselage is carried out and the bending stress acting on each
stringer is evaluated.
3. Shear flow distribution around the fuselage is found out.
4. Bulkhead design for the fuselage is carried out.
14.2.1 Identification of distributed loads on the fuselage:
The distributed loads on the fuselage is identified are
181
Fuselage structural distribution
Passenger weight distribution
Cargo weight distribution
Structural weight Distribution:
Length of the nose section = 8.5m
Length of the tail section = 10.26m
Length of the cabin = 45.202m
8.5m 45.202m 10.26m
From the diagram,
=-368167.24
= -6745.2134
Load intensity distribution of structural weight of fuselage:
, x varies from 0 to 8.5m
, x varies from 8.5m to 53.702m
, x varies from 53.702m to 63.962m
Passenger Weight Distribution:
182
8.5m 11.219m 1.5m 14.749m 1.5m 14.749m 1.5m 10.26m
From the above figure,
=-218959.2
= -5186.7384
Cargo weight Distribution:
8.5m 11.219m 1.5m 14.749m 1.5m 14.749m 1.5m 10.26m
From the above figure,
=-49050
=-1204.7094
183
Identification of point loads: (phase I balance diagram, side view)
Point load X(m) Load(N) Moment(N-m)
Fixed equipments (P1) 26.86 13148.38 353165.4868
Pilots (P2) 3.2 2158.2 6906.24
Nose landing gear(P3) 8.32 14279.53 118805.6896
Reserve fuel (P4) 19.862 185334.87 3682085.048
Horizontal tail(P5) 55.6172 33111.59 1841573.923
Vertical tail(P6) 56.172 17667.83 992437.346
(Lavatory & Galley)1 (P7) 20.4672 3209 65679.2448
(Lavatory & Galley)2 (P8) 37.4662 3209 120229.0358
(Lavatory&Galley) 3 (P9) 53.7152 3209 172372.0768
P2 P3 P4 P7 P1 P8 P9 P5 P6
Consider the fuselage to be a over-hanging beam with the supports provided by the front and
the rear spar and the reactions provided is and
To find the Reactions provided by the supports:
Step Involved:
184
Calculate the moment provided by the point loads along the length of the fuselage.
Convert the distributed loads on the fuselage into point loads.
Determine the moment provided by the distributed load by integration
Following the steps above,
+ = -903723.7305 (equilibrium by summation of point loads)
17.25 +23.76 =-26905689.43 (equilibrium by summation of moments)
Solving the above equations = -837269.3586N
=1740993.089N
Using the method described in Standard Strength of Materials books the variation of shear force
and the bending moment is calculated and graph is plotted as done for the wing.
185
14.2.2 STRINGER DESIGN:
In the Stringer Design, the following are calculated
Number of stringers
Area of the Stringers
Steps followed in Stringer Design:
1. The maximum bending moment experienced by the fuselage is calculated by allowing a
Factor of Safety of 2
Where is identified from the bending moment diagram of fuselage
2. The material for the Stringer is selected and the Critical Shear stress of the material is
identified.
186
3. The moment of inertia for the fuselage is given by
Where n is the number of stringers
A is the area of the stringer, m2
R is the radius of the fuselage, m
4. The bending stress experienced by the stringer in the fuselage is
;
Substituting the known values nA is found.
5. The skin should withstand the buckling stress so using the formula
Where Kb =8.5 ( for aircraft applications) and’ b’ is the stringer spacing.
6. Number of stringer is calculated as
n=
where d is the diameter of the fuselage.
7. The area of the Stringer is calculated as
A=
8. From the book ,Airframe Stress Analysis and Sizing,Niu the properties of the section is
known choosing the Area to be closer to the Area calculated in Step 7
Following the above steps,
= 18919332.26 N-m
The material Selected is Al 2024 for which the critical stress = 216MPa
nA = 0.0553488m2 ; b = 9.04cm; n = 220, A= 2.5158*10-4 m2(calculated)
Area of the stringer= 2.6*10-4 m2( from Niu)
187
Calculation of bending Stress in the Stringer:
The bending stress of the Stringer is given by
; Z=Rsinθ
θ= 0.02893
= 0.2792m2
Substituting in the bending stress formula,
Tabulation for Bending Stress of the Stringers
stringer angle (rad) bending stress (Mpa)
1 0 0
2 0.02893 6.125159803
3 0.05786 12.24519354
4 0.08679 18.35497944
5 0.11572 24.44940431
6 0.14465 30.5233678
7 0.17358 36.57178669
8 0.20251 42.58959915
9 0.23144 48.57176895
10 0.26037 54.51328969
11 0.2893 60.40918899
12 0.31823 66.25453265
13 0.34716 72.04442879
14 0.37609 77.77403192
15 0.40502 83.438547
16 0.43395 89.03323349
17 0.46288 94.55340927
18 0.49181 99.99445457
188
19 0.52074 105.3518159
20 0.54967 110.6210096
21 0.5786 115.7976262
22 0.60753 120.8773332
23 0.63646 125.8558797
24 0.66539 130.7290991
25 0.69432 135.492913
26 0.72325 140.1433348
27 0.75218 144.6764724
28 0.78111 149.0885323
29 0.81004 153.375822
30 0.83897 157.5347536
31 0.8679 161.5618464
32 0.89683 165.4537303
33 0.92576 169.2071482
34 0.95469 172.818959
35 0.98362 176.2861398
36 1.01255 179.6057892
37 1.04148 182.7751289
38 1.07041 185.7915065
39 1.09934 188.6523978
40 1.12827 191.3554084
41 1.1572 193.8982762
42 1.18613 196.2788732
43 1.21506 198.495207
44 1.24399 200.5454229
45 1.27292 202.427805
46 1.30185 204.140778
47 1.33078 205.6829084
48 1.35971 207.0529054
189
49 1.38864 208.2496227
50 1.41757 209.2720587
51 1.4465 210.1193578
52 1.47543 210.7908107
53 1.50436 211.2858557
54 1.53329 211.6040784
55 1.56222 211.7452125
56 1.59115 211.7091398
57 1.62008 211.4958907
58 1.64901 211.1056435
59 1.67794 210.5387248
60 1.70687 209.7956091
61 1.7358 208.8769182
62 1.76473 207.7834211
63 1.79366 206.5160329
64 1.82259 205.0758141
65 1.85152 203.4639702
66 1.88045 201.68185
67 1.90938 199.730945
68 1.93831 197.6128879
69 1.96724 195.3294512
70 1.99617 192.8825459
71 2.0251 190.2742198
72 2.05403 187.5066558
73 2.08296 184.5821699
74 2.11189 181.5032098
75 2.14082 178.272352
76 2.16975 174.8923006
77 2.19868 171.3658841
78 2.22761 167.6960538
190
79 2.25654 163.8858809
80 2.28547 159.9385542
81 2.3144 155.857377
82 2.34333 151.6457649
83 2.37226 147.3072425
84 2.40119 142.8454407
85 2.43012 138.2640934
86 2.45905 133.5670348
87 2.48798 128.7581957
88 2.51691 123.8416006
89 2.54584 118.8213642
90 2.57477 113.7016878
91 2.6037 108.4868559
92 2.63263 103.1812329
93 2.66156 97.78925883
94 2.69049 92.31544629
95 2.71942 86.76437619
96 2.74835 81.14069416
97 2.77728 75.44910658
98 2.80621 69.69437665
99 2.83514 63.88132044
100 2.86407 58.01480281
101 2.893 52.09973338
102 2.92193 46.14106238
103 2.95086 40.14377654
104 2.97979 34.11289493
105 3.00872 28.05346468
106 3.03765 21.97055688
107 3.06658 15.86926221
108 3.09551 9.754686763
191
109 3.12444 3.631947755
110 3.15337 -2.493830782
111 3.1823 -8.617522265
112 3.21123 -14.73400186
113 3.24016 -20.83815077
114 3.26909 -26.92486051
115 3.29802 -32.98903719
116 3.32695 -39.0256058
117 3.35588 -45.0295144
118 3.38481 -50.99573841
119 3.41374 -56.91928477
120 3.44267 -62.79519615
121 3.4716 -68.61855508
122 3.50053 -74.38448806
123 3.52946 -80.08816966
124 3.55839 -85.72482656
125 3.58732 -91.2897415
126 3.61625 -96.77825729
127 3.64518 -102.1857807
128 3.67411 -107.5077861
129 3.70304 -112.7398198
130 3.73197 -117.877503
131 3.7609 -122.9165361
132 3.78983 -127.852702
133 3.81876 -132.6818697
134 3.84769 -137.3999978
135 3.87662 -142.0031376
136 3.90555 -146.487437
137 3.93448 -150.8491429
138 3.96341 -155.0846053
192
139 3.99234 -159.1902794
140 4.02127 -163.1627293
141 4.0502 -166.9986306
142 4.07913 -170.6947729
143 4.10806 -174.2480631
144 4.13699 -177.6555274
145 4.16592 -180.9143142
146 4.19485 -184.0216963
147 4.22378 -186.9750731
148 4.25271 -189.7719729
149 4.28164 -192.4100552
150 4.31057 -194.8871121
151 4.3395 -197.2010705
152 4.36843 -199.3499941
153 4.39736 -201.3320843
154 4.42629 -203.1456824
155 4.45522 -204.7892706
156 4.48415 -206.2614735
157 4.51308 -207.5610589
158 4.54201 -208.6869392
159 4.57094 -209.6381723
160 4.59987 -210.413962
161 4.6288 -211.0136591
162 4.65773 -211.4367617
163 4.68666 -211.6829157
164 4.71559 -211.7519151
165 4.74452 -211.6437022
166 4.77345 -211.3583676
167 4.80238 -210.89615
168 4.83131 -210.2574362
193
169 4.86024 -209.4427608
170 4.88917 -208.4528056
171 4.9181 -207.2883991
172 4.94703 -205.9505157
173 4.97596 -204.440275
174 5.00489 -202.7589411
175 5.03382 -200.9079209
176 5.06275 -198.8887636
177 5.09168 -196.703159
178 5.12061 -194.3529361
179 5.14954 -191.8400619
180 5.17847 -189.1666393
181 5.2074 -186.3349057
182 5.23633 -183.347231
183 5.26526 -180.2061154
184 5.29419 -176.9141878
185 5.32312 -173.474203
186 5.35205 -169.88904
187 5.38098 -166.1616992
188 5.40991 -162.2952999
189 5.43884 -158.2930778
190 5.46777 -154.1583824
191 5.4967 -149.8946739
192 5.52563 -145.5055206
193 5.55456 -140.9945957
194 5.58349 -136.3656743
195 5.61242 -131.6226303
196 5.64135 -126.7694331
197 5.67028 -121.8101442
198 5.69921 -116.7489141
194
199 5.72814 -111.5899784
200 5.75707 -106.3376546
201 5.786 -100.9963382
202 5.81493 -95.5704993
203 5.84386 -90.06467876
204 5.87279 -84.48348431
205 5.90172 -78.83158676
206 5.93065 -73.11371612
207 5.95958 -67.3346576
208 5.98851 -61.49924762
209 6.01744 -55.61236973
210 6.04637 -49.67895061
211 6.0753 -43.70395584
212 6.10423 -37.69238582
213 6.13316 -31.64927155
214 6.16209 -25.57967043
215 6.19102 -19.48866203
216 6.21995 -13.38134383
217 6.24888 -7.262826968
218 6.27781 -1.13823194
219 6.30674 4.987315659
220 6.33567 11.10868944
14.2.3 Shear Flow for the Fuselage:
Steps involved:
195
1. K 1 , K 2 , K 3 are evaluated using the following relations given below
K 1 =
K 2 =
K 3 =
2. The basic shear flow equation is given by
3. Since the cross section is symmetric,
K1=0; K2= ; K3=
4. Since there is no side force acting on the aircraft, Vy=0
5. The shear flow is given by =
Where is the maximum shear force identified from the shear force diagram
is calculated in the previous section
6. A cut is made at the first stringer and the shear flow due to the cut, q o is calculated using
the equation
Where As is the area of the sector
7. The tabulation is done for calculating the shear flow around the fuselage
Following the above mentioned steps,
196
; As= =0.1412 m ;qo= -1.5569Nm-1
Tabulation for Shear flow:
stringer theta ( rad) basic shear flow net shear flow due to cut
1 0 0 0 -1.5569
2 0.02893 102.7466089 102.7466089 101.1797089
3 0.05786 205.4072305 308.1538393 306.5869393
4 0.08679 307.8959495 513.30318 511.73628
5 0.11572 410.1269946 718.0229442 716.4560442
6 0.14465 512.01481 922.1418046 920.5749046
7 0.17358 613.474127 1125.488937 1123.922037
8 0.20251 714.4200358 1327.894163 1326.327263
9 0.23144 814.7680561 1529.188092 1527.621192
10 0.26037 914.434208 1729.202264 1727.635364
11 0.2893 1013.335082 1927.76929 1926.20239
12 0.31823 1111.38791 2124.722992 2123.156092
13 0.34716 1208.510632 2319.898542 2318.331642
14 0.37609 1304.621968 2513.1326 2511.5657
15 0.40502 1399.641483 2704.263451 2702.696551
16 0.43395 1493.489658 2893.131141 2891.564241
17 0.46288 1586.087951 3079.577608 3078.010708
18 0.49181 1677.358868 3263.446818 3261.879918
19 0.52074 1767.226026 3444.584894 3443.017994
20 0.54967 1855.614217 3622.840243 3621.273343
21 0.5786 1942.449469 3798.063685 3796.496785
22 0.60753 2027.659111 3970.108579 3968.541679
23 0.63646 2111.171832 4138.830943 4137.264043
24 0.66539 2192.917743 4304.089575 4302.522675
25 0.69432 2272.82843 4465.746173 4464.179273
26 0.72325 2350.837017 4623.665447 4622.098547
197
27 0.75218 2426.878221 4777.715238 4776.148338
28 0.78111 2500.888403 4927.766624 4926.199724
29 0.81004 2572.805625 5073.694028 5072.127128
30 0.83897 2642.569701 5215.375326 5213.808426
31 0.8679 2710.122246 5352.691947 5351.125047
32 0.89683 2775.406726 5485.528972 5483.962072
33 0.92576 2838.368506 5613.775232 5612.208332
34 0.95469 2898.954893 5737.323398 5735.756498
35 0.98362 2957.115184 5856.070076 5854.503176
36 1.01255 3012.800705 5969.915888 5968.348988
37 1.04148 3065.964853 6078.765558 6077.198658
38 1.07041 3116.563137 6182.527991 6180.961091
39 1.09934 3164.553212 6281.116349 6279.549449
40 1.12827 3209.894914 6374.448126 6372.881226
41 1.1572 3252.550299 6462.445213 6460.878313
42 1.18613 3292.483668 6545.033966 6543.467066
43 1.21506 3329.661602 6622.145269 6620.578369
44 1.24399 3364.052986 6693.714588 6692.147688
45 1.27292 3395.629041 6759.682027 6758.115127
46 1.30185 3424.363339 6819.992379 6818.425479
47 1.33078 3450.231833 6874.595172 6873.028272
48 1.35971 3473.212875 6923.444708 6921.877808
49 1.38864 3493.287232 6966.500107 6964.933207
50 1.41757 3510.438104 7003.725336 7002.158436
51 1.4465 3524.651137 7035.089241 7033.522341
52 1.47543 3535.914438 7060.565575 7058.998675
53 1.50436 3544.218579 7080.133017 7078.566117
54 1.53329 3549.556612 7093.775191 7092.208291
55 1.56222 3551.924068 7101.48068 7099.91378
56 1.59115 3551.318967 7103.243035 7101.676135
198
57 1.62008 3547.741815 7099.060782 7097.493882
58 1.64901 3541.195606 7088.937421 7087.370521
59 1.67794 3531.685817 7072.881423 7071.314523
60 1.70687 3519.220409 7050.906226 7049.339326
61 1.7358 3503.809812 7023.030221 7021.463321
62 1.76473 3485.466924 6989.276736 6987.709836
63 1.79366 3464.207096 6949.67402 6948.10712
64 1.82259 3440.04812 6904.255216 6902.688316
65 1.85152 3413.010213 6853.058333 6851.491433
66 1.88045 3383.116005 6796.126218 6794.559318
67 1.90938 3350.390512 6733.506516 6731.939616
68 1.93831 3314.861122 6665.251634 6663.684734
69 1.96724 3276.55757 6591.418692 6589.851792
70 1.99617 3235.511911 6512.069481 6510.502581
71 2.0251 3191.758495 6427.270406 6425.703506
72 2.05403 3145.33394 6337.092436 6335.525536
73 2.08296 3096.277098 6241.611038 6240.044138
74 2.11189 3044.629022 6140.90612 6139.33922
75 2.14082 2990.432938 6035.06196 6033.49506
76 2.16975 2933.734201 5924.167139 5922.600239
77 2.19868 2874.580261 5808.314461 5806.747561
78 2.22761 2813.020623 5687.600884 5686.033984
79 2.25654 2749.106807 5562.12743 5560.56053
80 2.28547 2682.8923 5431.999107 5430.432207
81 2.3144 2614.432516 5297.324816 5295.757916
82 2.34333 2543.78475 5158.217266 5156.650366
83 2.37226 2471.008124 5014.792873 5013.225973
84 2.40119 2396.163544 4867.171668 4865.604768
85 2.43012 2319.313648 4715.477192 4713.910292
86 2.45905 2240.522749 4559.836397 4558.269497
199
87 2.48798 2159.856787 4400.379537 4398.812637
88 2.51691 2077.38327 4237.240058 4235.673158
89 2.54584 1993.171219 4070.554489 4068.987589
90 2.57477 1907.29111 3900.462329 3898.895429
91 2.6037 1819.814814 3727.105924 3725.539024
92 2.63263 1730.81554 3550.630355 3549.063455
93 2.66156 1640.36777 3371.183311 3369.616411
94 2.69049 1548.547198 3188.914968 3187.348068
95 2.71942 1455.430668 3003.977866 3002.410966
96 2.74835 1361.096108 2816.526776 2814.959876
97 2.77728 1265.622464 2626.718572 2625.151672
98 2.80621 1169.089638 2434.712102 2433.145202
99 2.83514 1071.578417 2240.668055 2239.101155
100 2.86407 973.1704061 2044.748823 2043.181923
101 2.893 873.9479621 1847.118368 1845.551468
102 2.92193 773.9941228 1647.942085 1646.375185
103 2.95086 673.3925383 1447.386661 1445.819761
104 2.97979 572.2274006 1245.619939 1244.053039
105 3.00872 470.5833735 1042.810774 1041.243874
106 3.03765 368.5455215 839.128895 837.561995
107 3.06658 266.1992388 634.7447603 633.1778603
108 3.09551 163.6301774 429.8294162 428.2625162
109 3.12444 60.92417625 224.5543537 222.9874537
110 3.15337 -41.83281158 19.09136467 17.52446467
111 3.1823 -144.5547901 -186.3876017 -187.9545017
112 3.21123 -247.1557926 -391.7105827 -393.2774827
113 3.24016 -349.5499538 -596.7057465 -598.2726465
114 3.26909 -451.6515814 -801.2015352 -802.7684352
115 3.29802 -553.3752278 -1005.026809 -1006.593709
116 3.32695 -654.635762 -1208.01099 -1209.57789
200
117 3.35588 -755.3484403 -1409.984202 -1411.551102
118 3.38481 -855.4289776 -1610.777418 -1612.344318
119 3.41374 -954.793618 -1810.222596 -1811.789496
120 3.44267 -1053.359204 -2008.152822 -2009.719722
121 3.4716 -1151.043249 -2204.402453 -2205.969353
122 3.50053 -1247.764001 -2398.807249 -2400.374149
123 3.52946 -1343.440515 -2591.204516 -2592.771416
124 3.55839 -1437.992723 -2781.433238 -2783.000138
125 3.58732 -1531.341494 -2969.334217 -2970.901117
126 3.61625 -1623.408706 -3154.7502 -3156.3171
127 3.64518 -1714.117309 -3337.526014 -3339.092914
128 3.67411 -1803.39139 -3517.508698 -3519.075598
129 3.70304 -1891.156237 -3694.547627 -3696.114527
130 3.73197 -1977.338401 -3868.494638 -3870.061538
131 3.7609 -2061.865757 -4039.204159 -4040.771059
132 3.78983 -2144.667566 -4206.533323 -4208.100223
133 3.81876 -2225.674531 -4370.342097 -4371.908997
134 3.84769 -2304.818859 -4530.49339 -4532.06029
135 3.87662 -2382.034315 -4686.853175 -4688.420075
136 3.90555 -2457.256279 -4839.290595 -4840.857495
137 3.93448 -2530.421798 -4987.678078 -4989.244978
138 3.96341 -2601.469642 -5131.89144 -5133.45834
139 3.99234 -2670.34035 -5271.809992 -5273.376892
140 4.02127 -2736.976287 -5407.316637 -5408.883537
141 4.0502 -2801.321684 -5538.297971 -5539.864871
142 4.07913 -2863.322694 -5664.644378 -5666.211278
143 4.10806 -2922.927427 -5786.250121 -5787.817021
144 4.13699 -2980.086002 -5903.013429 -5904.580329
145 4.16592 -3034.750583 -6014.836585 -6016.403485
146 4.19485 -3086.875422 -6121.626005 -6123.192905
201
147 4.22378 -3136.416897 -6223.292319 -6224.859219
148 4.25271 -3183.333546 -6319.750443 -6321.317343
149 4.28164 -3227.586107 -6410.919653 -6412.486553
150 4.31057 -3269.137544 -6496.723651 -6498.290551
151 4.3395 -3307.953084 -6577.090628 -6578.657528
152 4.36843 -3344.000243 -6651.953327 -6653.520227
153 4.39736 -3377.248853 -6721.249095 -6722.815995
154 4.42629 -3407.671088 -6784.919941 -6786.486841
155 4.45522 -3435.24149 -6842.912579 -6844.479479
156 4.48415 -3459.936985 -6895.178475 -6896.745375
157 4.51308 -3481.736905 -6941.673889 -6943.240789
158 4.54201 -3500.623006 -6982.35991 -6983.92681
159 4.57094 -3516.579483 -7017.202489 -7018.769389
160 4.59987 -3529.592982 -7046.172464 -7047.739364
161 4.6288 -3539.652612 -7069.245593 -7070.812493
162 4.65773 -3546.749954 -7086.402566 -7087.969466
163 4.68666 -3550.87907 -7097.629024 -7099.195924
164 4.71559 -3552.036502 -7102.915571 -7104.482471
165 4.74452 -3550.221283 -7102.257785 -7103.824685
166 4.77345 -3545.434931 -7095.656214 -7097.223114
167 4.80238 -3537.681453 -7083.116384 -7084.683284
168 4.83131 -3526.967337 -7064.64879 -7066.21569
169 4.86024 -3513.301549 -7040.268886 -7041.835786
170 4.88917 -3496.695527 -7009.997076 -7011.563976
171 4.9181 -3477.163167 -6973.858694 -6975.425594
172 4.94703 -3454.720817 -6931.883984 -6933.450884
173 4.97596 -3429.387257 -6884.108073 -6885.674973
174 5.00489 -3401.183689 -6830.570945 -6832.137845
175 5.03382 -3370.133716 -6771.317404 -6772.884304
176 5.06275 -3336.263323 -6706.397039 -6707.963939
202
177 5.09168 -3299.600857 -6635.864181 -6637.431081
178 5.12061 -3260.176999 -6559.777857 -6561.344757
179 5.14954 -3218.024744 -6478.201743 -6479.768643
180 5.17847 -3173.179366 -6391.20411 -6392.77101
181 5.2074 -3125.678398 -6298.857764 -6300.424664
182 5.23633 -3075.561591 -6201.239989 -6202.806889
183 5.26526 -3022.870888 -6098.432479 -6099.999379
184 5.29419 -2967.650386 -5990.521274 -5992.088174
185 5.32312 -2909.946296 -5877.596682 -5879.163582
186 5.35205 -2849.806912 -5759.753208 -5761.320108
187 5.38098 -2787.282563 -5637.089475 -5638.656375
188 5.40991 -2722.425575 -5509.708138 -5511.275038
189 5.43884 -2655.290225 -5377.7158 -5379.2827
190 5.46777 -2585.932699 -5241.222924 -5242.789824
191 5.4967 -2514.41104 -5100.343739 -5101.910639
192 5.52563 -2440.785105 -4955.196145 -4956.763045
193 5.55456 -2365.116509 -4805.901614 -4807.468514
194 5.58349 -2287.46858 -4652.585089 -4654.151989
195 5.61242 -2207.906299 -4495.374879 -4496.941779
196 5.64135 -2126.49625 -4334.402549 -4335.969449
197 5.67028 -2043.306566 -4169.802817 -4171.369717
198 5.69921 -1958.406866 -4001.713432 -4003.280332
199 5.72814 -1871.868202 -3830.275068 -3831.841968
200 5.75707 -1783.762996 -3655.631198 -3657.198098
201 5.786 -1694.164983 -3477.927979 -3479.494879
202 5.81493 -1603.149146 -3297.31413 -3298.88103
203 5.84386 -1510.791656 -3113.940802 -3115.507702
204 5.87279 -1417.169804 -2927.96146 -2929.52836
205 5.90172 -1322.361942 -2739.531746 -2741.098646
206 5.93065 -1226.447413 -2548.809355 -2550.376255
203
207 5.95958 -1129.506486 -2355.953899 -2357.520799
208 5.98851 -1031.620291 -2161.126778 -2162.693678
209 6.01744 -932.870747 -1964.491038 -1966.057938
210 6.04637 -833.3404958 -1766.211243 -1767.778143
211 6.0753 -733.1128332 -1566.453329 -1568.020229
212 6.10423 -632.2716383 -1365.384471 -1366.951371
213 6.13316 -530.9013037 -1163.172942 -1164.739842
214 6.16209 -429.0866651 -959.9879688 -961.5548688
215 6.19102 -326.9129295 -755.9995946 -757.5664946
216 6.21995 -224.465605 -551.3785345 -552.9454345
217 6.24888 -121.8304282 -346.2960332 -347.8629332
218 6.27781 -19.09329318 -140.9237214 -142.4906214
219 6.30674 83.65982077 64.56652758 62.99962758
220 6.33567 186.3429209 270.0027417 268.4358417
∑q= 338.116639Nm-1
From this summation of shear flow , qo is calculated
14.2.4 Thickness of the Skin of Fuselage:
Steps:
The change of lift produced due to the deflection of aileron is calculated using
Where = (in radians) ; V is the cruise velocity(m/s);
The change in lift is felt at the root attachment point and the torque produced due to
aileron deflection is
From the torque relation, T =2Aq and from which q is calculated
204
The appropriate material for skin is chosen
The stress in the skin is found as ,from which the skin thickness is found.
Following the above steps,
; T=20725023.32N-m; q=329448.8193Nm-1
For Al2024, the critical shear stress is188MPa
The skin thickness = 0.00176m
14.2.5 Bulkhead Design:
Steps involved:
1. The torque and the radial load(F.O.S of 3 is considered) acting on the fuselage is
identified from the previous sections.
2. The Normal force, Shear force, Bending moment due to the torque and the radial load
acting on the fuselage is found for various angles( all in radians)
Normal force,
Shear force,
Bending moment,
3. For various angular position , Normal force, Shear force and the bending moment is
calculated and plotted for the rear spar, front spar and at the aft bulkhead position(i.e
starting of the tail section of fuselage)
4. Maximum values of Normal force, Shear force and Bending moment for each case is
identified.
5. The appropriate material for the bulkhead is chosen.
6. The design condition for bulkhead is
;
7. The cross section of the bulk head is chosen and the dimensions of the bulkhead are
assumed.
205
8. ; is evaluated and the
design condition checked.
9. If it does not satisfy the design conditions, the dimensions of the bulk head is changed
appropriately.
Following the steps above,
For the front spar,
Q= 194782.0844 N = 584346.25 N; T=20725023.32 N-m
Using the equations mentioned above the variation of shear force, normal force and
bending moment for different angular positions (0 to 360 deg).
206
From the graphs maximum values for Shear force, Normal force and Bending moment is
found.
Since I section is chosen,
Area, A =
207
Section Modulus, Z =
Dimensions of Bulkhead Chosen:
Location A(inches) B(inches) T(inches)
Rear spar 35 35 2.16
Front spar 40 40 3.33
At 57.702 m 34 34 2.615
208
CHAPTER 15
DIRECT OPERATING COSTS FOR THE PASSENGER AIRCRAFT
The costs to be estimated in the direct operating cost are as follows
Flight crew and Cabin crew costs
Airframe maintenance labour cost
Airframe maintenance material cost
Engine maintenance labour cost
Engine maintenance materials cost
Landing fee
Navigation fee
Depreciation
Insurance
Before estimating the above mentioned costs, the block time ( ) of the aircraft has to be
calculated. Block Time is defined as the time from which the plane departs the gate of one
airport to the time the plane arrives at a gate of another airport.
15.1 Calculation of Block Time, ( ):
From phase I project,
Time taken to climb = 0.33 hour
Time taken to descent= 0.618 hour
Time taken for Start up, Taxi out, and Take off is around 8 mins (0.13 hour)
Time taken for landing and Taxi to stop is around 5mins (0.083 hour)
Cruise time = distance to be cruised/ cruise velocity = 10 hours
= 10+0.33+0.618+0.13+0.083=11.61hours
=11.61hours
209
15.2 Estimation of Flight Crew and Cabin Crew costs:
Flight crew Cost =
=$18051.09
Where is $440/hr
Cabin crew Cost =
= $6988.8
Where is $78/hr
15.3 Estimation of Airframe Maintenance Cost:
Airframe Maintenance Labour Cost=
= $1304.3
Where =
Airframe Maintenance Materials Cost =
=$1179
Applied maintenance Burden = 2* Airframe Maintenance Labour Cost
= $2609
15.3 Estimation of Engine Maintenance Cost:
Engine Maintenance Labour Cost =
= $459
210
Where is $25/hr
Engine Maintenance Labour Cost =
=$830
Applied maintenance Burden = 2* Engine Maintenance Labour Cost
= $918
15.5 Landing Fee:
Landing Fee =
=$1119 [for Domestic Airports]
=$3178 [for international Airports]
Where =$2.20 [for Domestic Airports]
=$6.25[for international Airports]
is the landing mass of the Aircraft.
15.6 Navigation Fee:
Navigation Fee =
= $2346
Where is $0.20
211
Trips per Year:
Short Range Aircraft = 2100 trips / year
Medium Range Aircraft = 625 trips/year
Long Range Aircraft= 480 trips/year
15.7 Estimation of the Cost of Aircraft:
Total weight of the aircraft = 2447301.014N
Empty weight of the aircraft = 1239059.294n (with the engines)
= 126305.7384kg
So from the above graph using this empty weight estimated the cost of the aircraft = $100m
212
15.8 Estimation of Engine Cost:
Engine Cost= - 2228
=$8.39M
15.9 Depreciation:
Depreciation/year =
= $33.138M
Where R = 0.1
Depreciation/trip = $33.138M/480 = $0.069M
15.10 Insurance:
Annual Insurance = 0.0035*
= $0.35M
Insurance/ trip = $0.00073M
15.11 Interest:
Total investment = cost of Airframe+ cost of Engine+ cost of spares
= $113.35
Where cost of spares =10% of Cost of Airframe + 30% of Cost of Engine
Interest = 5.4% of Total investment
= $6.1212M
Interest/trip = $0.01275M
213
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