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

1

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

3

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.

4

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

5

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.

6

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)

7

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

8

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 ]

9

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

10

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

11

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

12

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

13

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

14

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.

15

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

16

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

17

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:

18

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.

19

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.

20

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:

21

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.

22

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

23

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

24

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

25

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.

26

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

27

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

28

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 %.

29

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.

30

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.

31

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.

32

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

33

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:

72

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.

74

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

75

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.

76

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

77

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

78

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)

80

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

82

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.

83

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

88

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

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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

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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

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