Aircraft Design Project 2

1. Introduction 1.1 Overview: The structural design of an airplane actually begins with the flight envelope or V-n diagram, which clearly limits the maximum load factors that the airplane can withstand at any particular flight velocity. However in normal practice the airplane might experience loads that are much higher than the design loads. Some of the factors that lead to the structural overload of an airplane are high gust velocities, sudden movements of the controls, fatigue load in some cases, bird strikes or lightning strikes. So to add some inherent ability to withstand these rare but large loads, a safety factor of 1.5 is provided during the structural design. The two major members that need to be considered for the structural design of an airplane are wings and the fuselage. As far as the wing design is concerned, the most significant load is the bending load. So the primary load carrying member in the wing structure is the spar (the front and rear spars) whose cross section is an ‘I’ section. Apart from the spars to take the bending loads, suitable stringers need to take the shear loads acting on the wings. Unlike the wing, which is subjected to mainly unsymmetrical load, the fuselage is much simpler for structural analysis due to its symmetrical crossing and symmetrical loading. The main load in the case of fuselage is the shear load because the load acting on the wing is transferred to the fuselage skin in the form of shear only. The structural design of both wing and fuselage begin with shear force and bending moment diagrams for the respective members. The maximum bending stress produced in each of them is checked to be less than the yield stress of the material chosen for the respective member.

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1.2 Outline: The Structural design involves: Determination of loads acting on aircraft:  V-n diagram for the design study  Gust and maneuverability envelopes  Schrenk’s Curve  Critical loading performance and final V-n graph calculation Determination of loads acting on individual structures  Structural design study – Theory approach  Load estimation of wings  Load estimation of fuselage.  Material Selection for structural members  Detailed structural layouts  Design of some components of wings, fuselage

1.3 Parameters forwarded from ADP – 1 Take off Gross Weight,

Maximum Velocity,

Cruise Velocity, Stall Velocity, 2

Table 1-1: Mass ratio Split up

Components Crew Landing Gear Payload Fixed Equipment Fuselage mass Horizontal Stablizer Vertical Stabilizer Wing Structure Fuel Power plant Total

Mass Fraction 0.00053146 0.042516824 0.190144687 0.002262033 0.085033649 0.012597578 0.006298789 0.125975776 0.4810995 0.053539705 1

Cruise Altitude = 12 km The airfoil used her is NACA 653 - 418 Density at cruise altitude,

Cruise C_L @ Cruise altitude,

@ 16 ˚ aoa @ 14 ˚ since tail angle is 15.56 ˚ @ -14 ˚ aoa

cr = 11.593 m 3

ct = 5.797 m ( )

( )

4

2. V-n Diagram 2.1 Maneuvering Envelope: In accelerated flight, the lift becomes much more compared to the weight of the aircraft. This implies a net force contributing to the acceleration. This force causes stresses on the aircraft structure. The ratio of the lift experienced to the weight at any instant is defined as the Load Factor (n).

Using the above formula, we infer that load factor has a quadratic variation with velocity. However, this is true only up to a certain velocity. This velocity is determined by simultaneously imposing limiting conditions aerodynamically ((CL)max) as well as structurally (nmax). This velocity is called the Corner Velocity, and is determined using the following formula,

√

In this section, we estimate the aerodynamic limits on load factor, and attempt to draw the variation of load factor with velocity, commonly known as the V-n Diagram. The V‐n diagram is drawn for Sea level Standard conditions.

5

Figure 2-1: Typical V-n diagram for a private airliner.

Figure 2-2: V-n diagram nomenclature

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V-n diagram is used primarily in the determination of combination of flight conditions and load factors to which the airplane structure must be designed. Vn diagram precisely gives the structural (maximum load factor) and aerodynamic (maximum CL) boundaries for a particular flight condition.

2.2 Construction of V-n diagram 2.2.1 Curve OA: Maximum Load Factor,

( )

( )

Hence along the curve OA, Using the above equation we get

Table 2-1: Velocity vs. positive load factor (n)

Velocity V (m/s)

Load Factor (n) 0 20 40 60 80 100 120 140 160 180 200 220 240

0 0.08492371 0.339694842 0.764313394 1.358779368 2.123092762 3.057253578 4.161261814 5.435117472 6.87882055 8.49237105 10.27576897 12.22901431

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At A,

2.2.2 Curve AC: AC is a line limiting the maximum amount of load that can be withstood by the weakest structure of the aircraft

√

*

+

√[

( )

]

VC= 408.32 m/s nC=nA

2.2.3 Along CD: The velocity at point D is given by VD=1.5VC= 416.66 m/s nD= 0.75nA= 4.80864 A straight line is used to join the points C and D This VD is the dive velocity or the maximum permissible EAS in which the aircraft is at the verge of structural damage due to high dynamic pressure.

2.2.4 Along DE: E corresponds to zero load factor point i.e

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n= 0 For Bombers the load factor can vary from -3 to +6.5 Hence the negative load factor of aircraft is limited to -2

2.2.5 Along EF The point F corresponds to the velocity VC = VF = 408.32 m/s

2.2.6 Curve OG: nF= -2 (for a typical bomber aircraft)

Hence along the curve OG,

Hence we get, Table 2-2: Velocity vs. negative Load factor (n)

Velocity V (m/s)

Load Factor (n) 0 20 40 60 80 100 120 140 160 180 200 220

0 -0.034356779 -0.137427115 -0.309211009 -0.549708461 -0.85891947 -1.236844037 -1.683482161 -2.198833843 -2.782899083 -3.43567788 -4.157170234

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2.2.7 Along GF: Also nG=nF Finally join GF by using a straight line

2.3 Nomenclature of curves: • • • • • • • •

PHAA – Positive High Angle of Attack PSL – Positive Structural Limit PLAA – Positive Low Angle of Attack HSL –High Speed Limit NHAA – Negative High Angle of Attack NSL – Negative Structural Limit NLAA – Negative Low Angle of Attack LSL – Low Speed Limit

Figure 2-3: Four basic flight conditions showing how location of maximum stresses in wing depends on angle of attack

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2.4 Low Speed Limit: Stall velocity is the maximum speed at which the aircraft can maintain level flight. This implies the intersection of this line at cruise n=1 with OA curve corresponds to stall velocity Vs. Vs = 68.630 m/s

Rough V-n Diagram PHAA

NHAA

PSL

HSL

NSL

PLAA

NLAA

LSL

Load Factor

10

5 0 0

100

200

300

400

500

-5 -10

Velocity (m/s)

Figure 2-4: Rough V-n Diagram

From the V-n diagram, it is observed that the stall curve corresponds to maximum value of CLmax and any point beyond this curve for a particular velocity is not achievable in flight as it enters the stall region there. The upper horizontal line corresponds to limit load factor as well as ultimate load factor. It

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shows that there is outright structural failure when the aircraft is flown beyond this value of load factor. n=-2 gives the negative limit load factor and negative ultimate load factor. From the figure, it is clear that for a particular velocity, it is not possible to fly at a value of CL higher than the CLmax corresponding to that velocity. If we wish to increase the lift of the airplane to that value of C Lmax, then we should increase the flying speed of the airplane.

Maneuvering Envelope PIAA

7 6 5

3

HSL

Load Factor

4

2 1

LSL

0 -1

0

50

100

150

200

250

300

-2 -3

NIAA Velocity

Figure 2-5: Maneuvering Envelope

12

350

400

450

Maneuvering Envelope D

C

7 6

E

5

Load Factor

4 3 2 1

B

0A 0 -1

50

H

100

150

200

250

300

350

400

450

-2

G

-3

F Velocity

Figure 2-6: Maneuvering Envelope

Maneuvering Envelope with coordinates 408.32, 6.41152

173.77, 6.41152

7 6

416.66, 4.80864

5

Load Factor

4 3 2

68.630 2,1

1 0 -1

0

50 68.63, -0.4046

100

150

200

250

300

350

400

416.66 , 0 450

-2 -3

159.5944,-2

Velocity

Figure 2-7: Maneuvering envelope with coordinates

Hence for the strategic bomber aircraft we get, Safety Factor = 1.5

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408.32,-2

Caution Speed = 325 m/s Corner Velocity = 173.77 m/s Stall speed = 59.669 m/s Safety load factor limit i.e., indications given to pilot n = -2/ 1.5 = -1.3333 n = 6.41152/ 1.5 = 4.2743 Dive Velocity = 416.66 m/s

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Figure 2-8: V-n diagram with safety factor or safety limit consideration

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3. Gust Envelope 3.1 Description: Gust is a sudden, brief increase in the speed of the wind. Generally, winds are least gusty over large water surfaces and most gusty over rough land and near high buildings. With respect to aircraft turbulence, a sharp change in wind speed relative to the aircraft; a sudden increase in airspeed due to fluctuations in the airflow, resulting in increased structural stresses upon the aircraft. Sharp-edged gust (u) is a wind gust that results in an instantaneous change in direction or speed. Derived gust velocity (U or Umax) is the maximum velocity of a sharp-edged gust that would produce a given acceleration on a particular airplane flown in level flight at the design cruising speed of the aircraft and at a given air density. As a result a 25% increase is seen in lift for a longitudinally disturbing gust. The effect of turbulence gust is to produce a short time change in the effective angle of attack. These changes produce a variation in lift and thereby load factor For velocities up to Vmax, cruise, a gust velocity of 15 m/s at sea level is assumed. For Vdiv, a gust velocity of 10 m/s is assumed. Effective gust velocity: The vertical component of the velocity of a sharpedged gust that would produce a given acceleration on a particular airplane flown in level flight at the design cruising speed of the aircraft and at a given air density. Reference Gust Velocity (Uref ) —at sea level 15m/s. Design Gust Velocity (Uds) — Uref X K

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Variation in aerodynamic limits

12

Load factor

10 8

Normal Stall curve

6

Gust stall curve

4

Normal neg stall curve

2

Gust neg stall curve

0 0

50

100

150

200

250

Flaps Retracted

-2 -4 Velocity -6 Figure 3-1: Variation in Aerodynamic limits due to gust

3.2 Construction The increase in the load factor due to the gust can be calculated by For curve above V-axis:

Where K  Gust Alleviation Factor U max  Maximum derived Gust Velocity a  Lift Curve Slope for wing For curve below V-axis: 17

Gust Alleviation Factor: Gust Alleviation Factor (K):

Lateral Mass Ratio (µ):

̂ Where g  Acceleration due to Gravity ĉ  Mean Aerodynamic Chord ̂

(

)

ct  Chord at tip cr Chord at root cr = 11.593 m ct = 5.797 m

√ a= 0.1213507 /degree

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a= 6.9528829 /radian for a’ =0.15/ degree where a’ is lift curve slope for the chosen airfoil NACA 65(3) 418 a’  lift curve slope for airfoil  Sweep angle at leading Edge of Wing

( )

Table 3-1: Equivalent air speed and corresponding Derived Gust Velocity

For Velocity at points

Equivalent air speed Derived Gust Velocity V (m/s)

Umax (m/s)

B,G

173.77

15

C,F

408.32

10

D,E

416.66

5

19

By using the equations and for various speeds of Umax we get the following gust lines

Gust Lines U=15m/s U=0m/s U=-15m/s

4

U=10m/s U=-5m/s

U=5m/s U=-10./s

250

Design speed 277.77 m/s 350 300

3

Load factor

2

Level

1

0 0

50

100

150

200

-1

Velocity -2

Figure 3-2: Gust Lines

20

400

450

Overlaped Maneuver envelope and gust lines Gust stall curve PLAA NIAA U=5m/s

7

Gust neg stall curve NLAA U=15m/s U=0m/s

HSL PIAA U=10m/s U=-5m/s

6 5 4

Load factor

3 2 1 0 0

50

100

150

200

250

300

350

400

450

-1 -2 Velocity -3

Figure 3-3: Overlapped maneuvering envelope and gust lines.

The load factors at the various points can be found using the formula using the corresponding values of Umax n B’ = 1.4195966 n G’= 0.846720 nC’ = 2.5617017 nF’ = -0.5617017 n D’ = 1.796799 n E’ = 0.203200 21

The positive load factor along the curve OB’ is given by the equation

Hence along the curve OA,

But also

Equating the above two equations we get an intersecting point B where velocity is VB = 73.1379 m/s Since the velocities and load factors at C, D, E and F are known and straight lines are used to join these points in sequence

3.2.1 Line FG: It is found that negative gust line of U= -15 m/s intersects the positive high angle of attack condition at G.

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Equating the above equation with the OA curve equation we get the point G where VG = 51.52026m/s

Gust Envelope 3

C

2.5 2

D B

Load factor

1.5 1

G 0.5

O

0 0

50

100

150

200

250

300

350

400

E

-0.5 -1 -1.5 -2

F Velocity (m/s)

Figure3-4: Gust Envelope

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450

Gust Envelope with coordinates 3

408.32, 2.561702

2.5 2

Load factor

1 0.5

416.66, 1.7968

73.13798, 1.419596

1.5

51.52026, 0.704426

0 0

50

100

150

200

250

300

350

400

450

-0.5

416.66, 0

-1

410.9975, -1.3791

-1.5 -2

Velocity

Figure 3-5: Gust Envelope with coordinates

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4. Schrenk’s Curve 4.1 Description Lift varies along the wing span due to the variation in chord length, angle of attack and sweep along the span. Schrenk’s curve defines this lift distribution over the wing span of an aircraft, also called simply as Lift Distribution Curve. Schrenk’s Curve is given by

Where y1 is Linear Variation of lift along semi wing span also named as L1 y2 is Elliptic Lift Distribution along the wing span also named as L2

a = 44.8285 m

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Figure 4-1: Wing geometry showing sweep angle and semi span along the root.

4.2 Linear Lift Distribution: Lift at root

Lroot = 90978.038 N/m Lift at tip

Ltip = 45492.942 N/m By representing this lift at sections of root and tip we can get the equation for the wing.

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Figure 4-2: Linear lift distribution

Equation of linear lift distribution for starboard wing

Equation of linear lift distribution for port wing we have to replace x by –x in general,

Twice the area under y1= Total lift= 2491907.5 N ≈ Take off Gross Weight

Thousands

Lift per meter (N/m)

Linear variation of Lift along wing Semi span 100 90 80 70 60 50 L1

40 30 20 10 0 0

10

20

30

Wing Semi Span (m)

Figure 4-3: Linear Variation of lift along wing semi span

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40

50

For the Schrenk’s curve we only consider half of the linear distribution of lift and hence we derive y1/2

4.3 Elliptic Lift Distribution: Twice the area under the curve or line will give the lift which will be required to overcome weight Considering an elliptic lift distribution we get

Where b1 is Actual lift at root And a is wing semi span Lift at tip

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Figure 4-4: Elliptic lift distribution

Equation of elliptic lift distribution √

Thousands

Lift per meter (N/m)

Elliptic variation of Lift along wing Semi span 80 70 60 50 40 L2

30 20 10 0 0

10

20

30

Wing Semi Span (m)

Figure 4-5: Elliptic lift distribution

√

29

40

50

4.4 Construction of Schrenk’s Curve: Schrenk’s Curve is given by √

√ Substituting different values for x we can get the lift distribution for the wing semi span Table 4-1: Lift distribution table along semi span

x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

L1 90978.04 89963.39 88948.75 87934.1 86919.45 85904.81 84890.16 83875.52 82860.87 81846.22 80831.58 79816.93 78802.29 77787.64 76772.99 75758.35 74743.7 73729.06 72714.41 71699.76 70685.12 69670.47

L2 70776.189 70758.577 70705.716 70617.525 70493.872 70334.571 70139.378 69907.993 69640.055 69335.14 68992.759 68612.349 68193.275 67734.819 67236.175 66696.443 66114.615 65489.57 64820.058 64104.686 63341.899 62529.962

x 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 44.8285

L 80877.11 80360.98 79827.23 79275.81 78706.66 78119.69 77514.77 76891.75 76250.46 75590.68 74912.17 74214.64 73497.78 72761.23 72004.58 71227.4 70429.16 69609.31 68767.23 67902.22 67013.51 66100.22

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L1 68655.83 67641.18 66626.53 65611.89 64597.24 63582.6 62567.95 61553.3 60538.66 59524.01 58509.37 57494.72 56480.07 55465.43 54450.78 53436.14 52421.49 51406.84 50392.2 49377.55 48362.91 47348.26 46333.61 45492.98

L2 61666.935 60750.639 59778.626 58748.129 57656.013 56498.706 55272.111 53971.505 52591.398 51125.351 49565.739 47903.426 46127.307 44223.675 42175.265 39959.82 37547.784 34898.417 31952.741 28619.406 24742.227 20007.494 13543.872 0

L 65161.38 64195.91 63202.58 62180.01 61126.63 60040.65 58920.03 57762.4 56565.03 55324.68 54037.55 52699.07 51303.69 49844.55 48313.02 46697.98 44984.64 43152.63 41172.47 38998.48 36552.57 33677.88 29938.74 22746.49

Thousands

Lift per meter span (N/m)

Schrenk's Curve 100 90 80 70 60 50

L

40

L1

30

L2

20 10 0 0

10

20

30

40

50

Wing span loaction (m)

Figure 4-6: Schrenk’s curve with linear and elliptic lift distribution

Replacing x by –x for port wing we can get lift distribution for entire span.

Lift per meter span (N/m)

Thousands

Schrenk's Curve 90 80 70 60 50 40

L

30 20 10 0 -60

-40

-20

0 Wing span loaction (m)

Figure 4-7: Schrenk’s curve

31

20

40

60

5. Load Estimation on wings 5.1 Description: The solution methods which follow Euler’s beam bending theory (σ/y=M/I=E/R) use the bending moment values to determine the stresses developed at a particular section of the beam due to the combination of aerodynamic and structural loads in the transverse direction. Most engineering solution methods for structural mechanics problems (both exact and approximate methods) use the shear force and bending moment equations to determine the deflection and slope at a particular section of the beam. Therefore, these equations are to be obtained as analytical expressions in terms of span wise location. The bending moment produced here is about the longitudinal (x) axis.

5.2 Loads acting on wing: As both the wings are symmetric, let us consider the starboard wing at first. There are three primary loads acting on a wing structure in transverse direction which can cause considerable shear forces and bending moments on it. They are as follows:  Lift force (given by Schrenk’s curve)  Self-weight of the wing  Weight of the power plant  Weight of the fuel in the wing

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5.3 Shear force and bending moment diagrams due to loads along transverse direction at cruise condition: Lift Force given by Schrenk’s Curve:

Linear lift distribution (trapezium):

√

Elliptic lift distribution (quarter ellipse)

√

Linear lift distribution (y1/2) 50000

Lift per unit length (N/m)

45000 40000 35000 30000 25000 20000 15000 10000 5000 0 0

5

10

15

20

25

30

Span wise location (m)

Figure 5-1: Lift distribution (linear)

33

35

40

45

50

Elliptic lift distribution (y2/2) 40000

Lift per unit length (N/m)

35000 30000 25000 20000 y2/2

15000 10000 5000 0 0

10

20

30

Span wise location (m)

Figure 5-2: Lift distribution (Elliptic)

Self-Weight (y3): Self-weight of the wing,

Assuming parabolic weight distribution ( Where b  span 34

( ))

40

50

When we integrate from x=0 (root location) to x=b (tip location) we get the net weight of port wing. ∫

(

∫ (

)

)

Substituting various values of x in the above equation we get the self-weight of the wing.

Self Weight 0

WEight of empty wing (N/m)

0

5

10

15

20

25

30

-5000

-10000

-15000

-20000

-25000

Span wise location (m)

Figure 5-3: Self weight of wing

Power plant weight: Power plant is assumed to be a point load,

35

35

40

45

50

Acting at x= 8 m and x= 14 m from the root. Fuel weight: This design has fuel in the wing so we have to consider the weight of the fuel in the wing.

Again by using general formula for straight line y=mx + c we get,

Fuel distribution 0 -2000

0

5

10

15

20

25

30

Fuel weight (N/m)

-4000 -6000 -8000 -10000 -12000 -14000 -16000 -18000 -20000

Span wise location (m)

Figure 5-4: Fuel Distribution

36

35

40

45

50

Load distribution 60000

Load acting on wing (N/m)

40000 20000 0 0

5

10

15

20

25

30

35

40

45

50

-20000 -40000 -60000 -80000

Span wise location (m)

Figure 5-5: Overall Load distribution Table 5-1: Loads simplified as point loads

Curve / component

Area enclosed / structural Centroid (from wing root) weight (N)

y1/2

1529447.31

19.923 m

y2/2

1245953.75

3.510534 m

Wing

313917

16.8107 m

Fuel

365752.803

16.4606 m

Power plant

66708

14 m, 8 m

37

Figure 5-6: Reaction force and Bending moment calculations

Now we know VA and MA, using this we can find out shear force and Bending moment.

5.3.1 Shear Force: ∫

38

√ (

)

(

)

∫ ∫

By using the corresponding values of x in appropriate equations we get the plot of shear force Note: Shear force is a discrete function along y axis so in order to make it continuous we introduce straight lines.

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Shear Force (N)

Thousands

Shear Force 1000 500

0 -44.8285 -34.8285 -24.8285 -14.8285 -4.8285 -500

5.1715

15.1715 25.1715 35.1715

-1000 -1500 -2000 -2500 -3000 Location in wing (m)

Figure 5-7: Shear force diagram - discrete

Shear Force (N)

Thousands

Shear Force (Actual) 1000 500

0 -44.8285 -34.8285 -24.8285 -14.8285 -4.8285 -500

5.1715

-1000 -1500 -2000 -2500 -3000 Location in wing (m)

Figure 5-8: Shear force diagram- continuous

40

15.1715 25.1715 35.1715

5.3.2 Bending moment: ∫∫

( ( √

)

(

))

(

)

∫∫

By substituting the values of x for the above equations of bending moments obtained we can get a continuous bending moment curve for the port wing. Note: if we replace the x by -x in each term we get the distribution of starboard wing

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Bending Moment (Nm)

Millions

Bending Moment 70 60 50 40 30 20 10 0 -50

-40

-30

-20

-10

0

10

20

30

40

50

Location in wing (m)

Figure 5-9: Bending moment diagram

5.4 Shear force and bending moment diagrams due to loads along chordwise direction at cruise condition: Aerodynamic center- This is a point on the chord of an airfoil section where the bending moment due to the components of resultant aerodynamic force (Lift and Drag) is constant irrespective of the angle of attack. Hence the forces are transferred to this point for obtaining constant Ma.c Shear center- This is a point on the airfoil section where if a force acts, it produces only bending and no twisting. Hence the force is transferred to this point and the torque is found. Cruise CL=0.204908 @ V= 250 m/s Cruise CD= 0.0055 Angle of attack= -0.811439˚ (obtained from the lift curve slope) Angle of attack @ zero lift= -3o 42

Wing lift curve slope (a)= 0.1213507 /degree Co-efficient of moment about aerodynamic centre= -0.0543 Location of aerodynamic centre:

Location of shear centre:

Lift and drag are the components of resultant aerodynamic force acting normal to and along the direction of relative wind respectively. As a result, components of them act in the chordwise direction also which produce a bending moment about the normal (z) axis.

Figure 5-10: Normal and chord wise coefficients

Co-efficient of force along the normal direction,

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Chordwise force at root

Chordwise force at tip

By using y = mx +c again we get the equation as

The above equation gives the profile of load acting chordwise, by integrating this above equation we get a component of Shear force and again by integrating the same we get the component of Bending Moment ∫ ∬

44

Load along Chordwise direction

Load along chord wise direction (N)

1400 1200 1000 800 600 400 200 0 0

5

10

15

20

Spanwise location (m)

Figure 5-11: Load along chordwise direction

To find fixing moment and the reaction force,

5.4.1 Shear Force:

45

25

30

35

Thousands

Shear Force 0 0

5

10

15

20

25

30

35

25

30

35

-5

Shear Force (N)

-10 -15 -20 -25 -30 -35

Spanwise location (m)

Figure 5-12: Shear force

5.4.2 Bending Moment:

Thousands

Bending moment (Nm)

Bending Moment 600 500 400 300 200 100 0 0

5

10

15

20

Spanwise location (m)

Figure 5-13: Bending moment

46

Torque due to normal forces and constant pitching moment at cruise condition:

Figure 5-14: Moment about aerodynamic center

The lift and drag forces produce a moment on the surface of crosssection of the wing, otherwise called a torque, about the shear center. Moment about the aerodynamic center gets transferred to the shear center. The powerplant also produces a torque about the shear center on the chord under which it is located.

Figure 5-15:Torque due to normal force and moment

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5.5 Torque at cruise condition: 5.5.1 Torque due to normal force:

Where c  chord the equation for chord can also be represented in terms of x by taking c= mx +k,

Therefore torque ∫ ∫

48

Thousands

Torque due to normal forces 1000 900 800

Torque (Nm)

700 600 500 400 300 200 100 0 0

5

10

15

20

25

Spanwise location (m)

Figure 5-16: Torque due to normal force

5.5.2 Torque due to chord wise force:

5.5.3 Torque due to moment:

∫

49

30

35

40

Millions

Torque due to Moment 0 0

5

10

15

20

25

30

35

40

-1

Torque (Nm)

-2 -3 -4 -5 -6 -7 -8

Spanwise location (m)

Figure 5-17: Torque due to moment

5.5.4 Torque due to powerplant: The powerplant is situated under a chord (8 m and 14 m from the wing root; chord length 10.7504 m and 10.1184m ) from 0.1c to 0.5c at 10.7504m and from 0.1 c to 0.5249c an Uniformly Distributed Load of 15513.488 N/m is assumed to be present for this 4.3 m since the powerplant weight is 66708N. The centroid of the applied UDL is at 0.3c for first case and at 0.31245c at second location. Torque produced about shear center

50

Thousands

70

Torque due to Powerplant

60

Torque (Nm)

50 40 30 20 10 0 0

5

10

15

20

25

30

35

40

Spanwise location (m)

Figure 5-18: Torque due to powerplant

Then the different torque components are brought together in a same graph to make a comparison

Thousands

Torque comparison 2000 1000 0

Torque (Nm)

-1000

0

10

20

30

40

-2000

Torque due to Normal Forces

-3000

Torque due to moment

-4000

Torque due to powerplant

-5000 -6000 -7000 -8000

Spanwise location (m)

Figure 5-19: Torque comparison

51

The net torque will be sum of all the above torques i.e. torque due to normal force, chordwise force, powerplant and aerodynamic moment

Millions

Torque 0 0

5

10

15

20

25

-1

Torque (Nm)

-2 -3 -4 -5 -6 -7

Spanwise location (m)

Figure 5-20: Net torque

52

30

35

40

5.6 Load at Critical flight condition: Optimum Wing structural design consists of determining that stiffness distribution which is proportional to the local load distribution. The aerodynamic forces of lift and drag are resolved into components normal and parallel to the wing chord. The distribution of shear force, bending moment and torque over the aircraft wing are considered for wing structural analysis. Identification of critical points from the maneuvering and gust envelopes: 1. Maneuvering envelope Table 5-1: Coordinates of V-n diagram

Point

Load factor

E.A.S. (m/s)

A

6.41152

173.77

C

6.41152

408.32

D

4.80864

416.66

E

0

416.66

F

-2

408.16

G

-2

159.5944

2. Gust envelope Table 5-2: Coordinates of gust envelope

Point

Load factor

E.A.S. (m/s)

B’

1.41959

173.137975

C’

2.5617

408.16

D’

1.7968

416.66

E’

0.2032

416.66

F’

-1.3255

408.16

G’

0.5822

51.52026

53

Corner points are representative of critical flight load conditions – a summary is given below. Table 5-3: Coordinates of critical conditions

Critical flight condition

Point (‘n’, E.A.S.)

‘n’ max point

C’ (2.5617, 408.16)

Positive H.A.A.

A (6.41152, 173.77)

Positive L.A.A

D (4.80864, 416.66)

Negative H.A.A

G (-2, 408.16)

Negative L.A.A

E’ (0, 416.66)

Shear force and bending moment diagrams of a wing due to normal forces at critical flight condition: In the preliminary stage of structural analysis, the critical flight loading condition of positive high angle of attack (represented by point A in v-n diagram) will be investigated.

It is seen that lift has increased by 6.41152 times. So we introduce a constant of proportionality for the lift alone 54

Thousands

Lift per meter (N/m)

Linear variation of Lift along wing Semi span (critical condition) 700 600 500 400 300

L1

200 100 0 0

10

20

30

40

50

Wing Semi Span (m)

Figure 5-21: Linear Variation of lift along wing semi span

Thousands

Lift per meter (N/m)

Elliptic variation of Lift along wing Semi span (Critical Condition) 500 450 400 350 300 250 200

L2

150 100 50 0 0

10

20

30

Wing Semi Span (m)

Figure 5-22: Elliptic variation of lift along wing semi span

55

40

50

The aim is to find the shear forces and bending moments due to normal forces in critical flight condition. There are three primary loads acting on a wing structure in transverse direction which can cause considerable shear forces and bending moments on it. They are as follows:  Lift force (given by Schrenk’s curve)  Self-weight of the wing  Weight of the power plant  Weight of the fuel in the wing Now, the proportionality constant influences the lift force alone and other factors remain unaffected. Table 5-4: Loads simplified as point loads at critical flight condition

Curve / component

Area enclosed / structural Centroid (from wing root) weight (N)

y1/2

1529447.31×6.41152

19.923 m

y2/2

1245953.75 ×6.41152

3.510534 m

Wing

313917

16.8107 m

Fuel

365752.803

16.4606 m

Power plant

66708

14 m, 8 m

56

Lift per meter span (N/m)

Thousands

Schrenk's Curve (Critical Condition) 600 500 400 300 L 200 100 0 -60

-40

-20

0 Wing span loaction (m)

Figure 5-23: Critical schrenk’s curve

Figure 5-24: load distribution at critical condition

57

20

40

60

Load distribution (Critical Condition) Millions

0.35 0.3

Load (N)

0.25 0.2 0.15 0.1 0.05 0 -0.05

0

5

10

15

-0.1

20

25

30

35

40

45

50

Location in wing (m)

Figure 5-25: load distribution at critical condition

Now we know VA and MA, using this we can find out shear force and Bending moment,

5.7 Shear force and bending moment diagrams due to loads along transverse direction at critical condition: ∫

58

( ( √

( (

))) )

∫ ∫

By using the corresponding values of x in appropriate equations we get the plot of shear force

59

Thousands

Shear Force (Critical condition) 2000

0 -44.8285 -34.8285 -24.8285 -14.8285 -4.8285 -2000

5.1715

15.1715 25.1715 35.1715

-4000 Shear Force (N)

-6000 -8000 -10000 -12000 -14000 -16000 -18000 -20000 Location in wing (m)

Figure 5-26: Transverse Shear force diagram at critical condition

Thousands

Shear Force (Actual) (Critical Condition) 2000

0 -44.8285 -34.8285 -24.8285 -14.8285 -4.8285 -2000

5.1715

-4000 Shear Force (N)

-6000 -8000 -10000 -12000 -14000 -16000 -18000 -20000 Location in wing (m)

Figure 5-27: Transverse Shear force diagram at critical condition

5.7.1 Bending moment: ∫∫ 60

15.1715 25.1715 35.1715

( ( ( √ (

) )))

(

)

∫∫

By substituting the values of x for the above equations of bending moments obtained we can get a continuous bending moment curve for the port wing.

61

Millions

Bending Moment (Critical Condition) 600 500

Bending Moment (Nm)

400 300 200 100 0 -50

-40

-30

-20

-10

0

10

20

30

40

50

-100 Location in wing (m)

Figure 5-28: Transverse bending moment diagram at critical condition

5.8 Shear force and bending moment diagrams due to loads along chordwise direction at critical condition: Critical CL=2.71925 @ V= 250 m/s Critical CD= 0.0084 Angle of attack= 16˚ (obtained from the lift curve slope) Wing lift curve slope (a) = 0.1213507 /degree Co-efficient of moment about aerodynamic centre= -0.025 Location of aerodynamic centre:

Location of shear centre:

62

Figure 5-29: Determination of chordwise force components at critical condition

Co-efficient of force along the normal direction,

Chordwise force at root

Chordwise force at tip

63

By using y = mx +c again we get the equation as

The above equation gives the profile of load acting chordwise, by integrating this above equation we get a component of Shear force and again by integrating the same we get the component of Bending Moment ∫ ∬

Load along Chordwise direction (critical condition) Load along chord wise direction (N)

400000 350000 300000 250000 200000 150000 100000 50000 0 0

5

10

15

20

Spanwise location (m)

Figure 5-30: Load along chord wise direction at critical condition

64

25

30

35

To find fixing moment and the reaction force,

5.8.1 Shear Force:

Shear Force (N)

Thousands

Shear Force 0 0

5

10

15

20

-2000 -4000 -6000 -8000 -10000 -12000

Spanwise location (m)

Figure 5-31: Chordwise Shear force diagram at critical condition

5.8.2 Bending Moment:

65

25

30

35

Thousands

Bending moment (Nm)

Bending Moment 180000 160000 140000 120000 100000 80000 60000 40000 20000 0 0

5

10

15

20

25

30

35

Spanwise location (m)

Figure 5-32: Chordwise Bending moment diagram at critical condition

Torque due to normal forces and constant pitching moment at cruise condition:

Figure 5-33: Determination of various components of torque

66

Figure 5-34: Determination of various components causing torque

5.9 Torque at critical flight condition: 5.9.1 Torque due to normal force:

Where c  chord the equation for chord can also be represented in terms of x by taking c= mx +k,

Therefore torque ∫

67

Torque (Nm)

Thousands

Torque due to normal forces 12000 10000 8000 6000 4000 2000 0 0

5

10

15

20

25

Spanwise location (m)

Figure 5-35: Torque due to normal force at critical condition

5.9.2 Torque due to chord wise force:

5.9.3 Torque due to moment:

∫

68

30

35

40

Millions

Torque due to Moment 0 -0.2

0

5

10

15

20

25

30

35

40

-0.4

Torque (Nm)

-0.6 -0.8 -1 -1.2 -1.4 -1.6 -1.8 -2

Spanwise location (m)

Figure 5-36: Torque due to moment at critical condition

5.9.4 Torque due to powerplant: The powerplant is situated under a chord (8 m and 14 m from the wing root; chord length 10.7504 m and 10.1184m ) from 0.1c to 0.5c at 10.7504m and from 0.1 c to 0.5249c an Uniformly Distributed Load of 15513.488 N/m is assumed to be present for this 4.3 m since the powerplant weight is 66708N. The centroid of the applied UDL is at 0.3c for first case and at 0.31245c at second location. Torque produced about shear center

Hence tis is weight this will remain same as that of the cruise condition.

69

Thousands

70

Torque due to Powerplant

60

Torque (Nm)

50 40 30 20 10 0 0

5

10

15

20

25

30

35

40

Spanwise location (m)

Figure 5-37: Torque due to powerplant at critical condition unchanged

Then the different torque components are brought together in a same graph to make a comparison The net torque will be sum of all the above torques i.e. torque due to normal force, chordwise force, powerplant and aerodynamic moment

70

Thousands

Torque comparison 12000 10000 8000

Torque (Nm)

6000 Torque due to Normal Forces 4000

Torque due to moment

2000

Torque due to powerplant

0 0

10

20

30

40

-2000 -4000

Spanwise location (m)

Figure 5-38: Torque comparison at critical condition

Millions

Torque 10 9 8

Torque (Nm)

7 6 5 4 3 2 1 0 0

5

10

15

20

25

Spanwise location (m)

Figure 5-39: Net torque at critical condition

71

30

35

40

5.10 Interim Summary: DUE TO NORMAL FORCES: Table 5-5: Determination of maximum values of normal force

Cruise condition Max. Shear force (N) -2461484.863 Max. Bending moment (Nm)

65115382.95

+ve high AOA condition

At

-17480623.2

Wing root

486567918.1

Wing root

DUE TO CHORDWISE FORCES: Table 5-6: Determination of maximum values of chordwise force

Cruise condition

+ve high AOA condition

At

Max. Shear force (N) -31590.839

-10102394.13

Wing root

Max. Bending moment (Nm)

51270.9081

163971959.1

Wing root

Max. Torque (Nm)

-6372183

8781344

Wing root

72

6. Material Selection: 6.1 Description: Aircraft structures are basically unidirectional. This means that one dimension, the length, is much larger than the others - width or height. For example, the span of the wing and tail spars is much longer than their width and depth; the ribs have a much larger chord length than height and/or width; a whole wing has a span that is larger than its chords or thickness; and the fuselage is much longer than it is wide or high. Even a propeller has a diameter much larger than its blade width and thickness, etc.... For this simple reason, a designer chooses to use unidirectional material when designing for an efficient strength to weight structure. Unidirectional materials are basically composed of thin, relatively flexible, long fibers which are very strong in tension (like a thread, a rope, a stranded steel wire cable, etc.) An aircraft structure is also very close to a symmetrical structure. That means the up and down loads are almost equal to each other. The tail loads may be down or up depending on the pilot raising or dipping the nose of the aircraft by pulling or pushing the pitch control; the rudder may be deflected to the right as well as to the left (side loads on the fuselage). The gusts hitting the wing may be positive or negative, giving the up or down loads which the occupant experiences by being pushed down in the seat ... or hanging in the belt. Because of these factors, the designer has to use a 73

structural material that can withstand both tension and compression. Unidirectional fibers may be excellent in tension, but due to their small cross section, they have very little inertia (we will explain inertia another time) and cannot take much compression. They will escape the load by bucking away. As in the illustration, you cannot load a string, or wire, or chain in compression. In order to make thin fibers strong in compression, they are "glued together" with some kind of an "embedding". In this way we can take advantage of their tension strength and are no longer penalized by their individual compression weakness because, as a whole, they become compression resistant as they help each other to not buckle away. The embedding is usually a lighter, softer "resin" holding the fibers together and enabling them to take the required compression loads. This is a very good structural material. WOOD Historically, wood has been used as the first unidirectional structural raw material. They have to be tall and straight and their wood must be strong and light. The dark bands (late wood) contain many fibers, whereas the light bands (early wood) contain much more "resin". Thus the wider the dark bands, the stronger and heavier the wood. If the dark bands are very narrow and the light bands quite wide, the wood is light but not very strong. To get the most efficient strength to weight ratio for wood we need a definite numbers of bands per inch. Some of our aircraft structures are two-dimensional (length and width are large with respect to thickness). Plywood is often used for such structures. Several thin boards (foils) are glued together so that the fibers of the various layers cross over at different angles (usually 90 degrees today years back you could get 74

them at 30 and 45 degrees as well). Plywood makes excellent "shear webs" if the designer knows how to use plywood efficiently. (We will learn the basis of stress analysis sometime later.) Today good aircraft wood is very hard to come by. Instead of using one good board for our spars, we have to use laminations because large pieces of wood are practically unavailable, and we no longer can trust the wood quality. From an availability point of view, we simply need a substitute for what nature has supplied us with until now. ALUMINUM ALLOYS So, since wood may not be as available as it was before, we look at another material which is strong, light and easily available at a reasonable price (there's no point in discussing Titanium - it's simply too expensive). Aluminum alloys are certainly one answer. We will discuss the properties of those alloys which are used in light plane construction in more detail later. For the time being we will look at aluminum as a construction material. Extruded Aluminum Alloys: Due to the manufacturing process for aluminum we get a unidirectional material quite a bit stronger in the lengthwise direction than across. And even better, it is not only strong in tension but also in compression. Comparing extrusions to wood, the tension and compression characteristics are practically the same for aluminum alloys so that the linear stress analysis applies. Wood, on the other hand, has a tensile strength about twice as great as its compression strength; accordingly, special stress analysis methods must be used and a good understanding of wood under stress is essential if stress concentrations are to be avoided! Aluminum alloys, in thin sheets (.016 to .125 of an inch) provide an excellent two dimensional material used extensively as shear webs - with or without

75

stiffeners - and also as tension/compression members when suitably formed (bent). It is worthwhile to remember that aluminum is an artificial metal. There is no aluminum ore in nature. Aluminum is manufactured by applying electric power to bauxite (aluminum oxide) to obtain the metal, which is then mixed with various strength-giving additives. (In a later article, we will see which additives are used, and why and how we can increase aluminum's strength by cold work hardening or by tempering.) All the commonly used aluminum alloys are available from the shelf of dealers. When requested with the purchase, you can obtain a "mill test report" that guarantees the chemical and physical properties as tested to accepted specifications. As a rule of thumb, aluminum is three times heavier, but also three times stronger than wood. Steel is again three times heavier and stronger than aluminum. STEEL The next material to be considered for aircraft structure will thus be steel, which has the same weight-to-strength ratio of wood or aluminum. Apart from mild steel which is used for brackets needing little strength, we are mainly using a chrome-molybdenum alloy called AISI 413ON or 4140. The common raw materials available are tubes and sheet metal. Steel, due to its high density, is not used as shear webs like aluminum sheets or plywood. Where we would need, say.100" plywood, a .032 inch aluminum sheet would be required, but only a .010 steel sheet would be required, which is just too thin to handle with any hope of a nice finish. That is why a steel fuselage uses tubes also as diagonals to carry the shear in compression or tension and the whole structure is then covered with fabric (light weight) to give it the required

76

aerodynamic shape or desired look. It must be noted that this method involves two techniques: steel work and fabric covering. We will be discussing tubes and welded steel structures in more detail later and go now to "artificial wood" or composite structures. COMPOSITE MATERIALS The designer of composite aircraft simply uses fibers in the desired direction exactly where and in the amount required. The fibers are embedded in resin to hold them in place and provide the required support against buckling. Instead of plywood or sheet metal which allows single curvature only, the composite designer uses cloth where the fibers are laid in two directions .(the woven thread and weft) also embedded in resin. This has the advantage of freedom of shape in double curvature as required by optimum aerodynamic shapes and for very appealing look (importance of esthetics). Today's fibers (glass, nylon, Kevlar, carbon, whiskers or single crystal fibers of various chemical compositions) are very strong, thus the structure becomes very light. The drawback is very little stiffness. The structure needs stiffening which is achieved either by the usual discreet stiffeners, -or more elegantly with a sandwich structure: two layers of thin uni- or bi-directional fibers are held apart by a lightweight core (foam or "honeycomb"). This allows the designer to achieve the required inertia or stiffness. From an engineering standpoint, this method is very attractive and supported by many authorities because it allows new developments which are required in case of war. But this method also has its drawbacks for homebuilding: A mold is needed, and very strict quality control is a must for the right amount of fibers and resin and for good adhesion between both to prevent too "dry" or "wet" a structure. Also the curing of the resin is quite sensitive to temperature,

77

humidity and pressure. Finally, the resins are active chemicals which will not only produce the well-known allergies but also the chemicals that attack our body (especially the eyes and lungs) and they have the unfortunate property of being cumulatively damaging and the result (in particular deterioration of the eye) shows up only years after initial contact. Another disadvantage of the resins is their limited shelf life, i.e., if the resin is not used within the specified time lapse after manufacturing, the results may be unsatisfactory and unsafe. HEAVY AIRCRAFT RAW MATERIALS The focus of our article is our Table which gives typical values for a variety of raw materials. Column 1 lists the standard materials which are easily available at a reasonable cost. Some of the materials that fall along the borderline between practical and impractical are: Magnesium: An expensive material. Castings are the only readily available forms. Special precaution must be taken when machining magnesium because this metal burns when hot. Titanium: A very expensive material. Very tough and difficult to machine. Carbon Fibers: Still very expensive materials. Kevlar Fibers: Very expensive and also critical to work with because it is hard to "soak" in the resin. When this technique is mastered, the resulting structure is very strong, but it also lacks in stiffness. Columns

2

through

6:

Columns 2 through 6 list the relevant material properties in metric units. Column 2, the density (d), is the weight divided by the volume. 78

Table 6-1: Material property table Materials

d

fy

fu

e

E/10

1

2

3

4

5

Spruce

.45

-

Poplar

.43

Oregon Pine Fiberglas s (70% Glass)

3

2

3

E/d

Root of N/d

Root of E/d

fu/d

6

7

8

9

10

3.5/11 -

1.4

220 70 0

22.0

(15)

-

30/12

-

1.0

220 70 0

22.0

(15)

.56

-

4.0/13 -

1.5

220 70 0

22.0

(15)

Matte

2.2

-

15

-

1.5

700 17

5.0

7

Woven

2.2

-

35

-

2.0

900 20

6.0

16

Unidirectio nal

2.2

-

60

-

3.5

150 27 0

7.0

27

5052-H34

2.7

16

24

4

7.1

260 30 0

7.0

11

8086-H34

2.7

22

31

5

7.1

260 30 0

7.0

11

6061 -T6

2.7

24

26

9

7.1

260 30 0

7.0

11

6351 -T6

2.7

25

28

9

7.1

260 30 0

7.0

11

6063-T6

2.7

17

21

9

7.1

260 30 0

7.0

11

7075-T3

2.8

25

41

1 2

7.2

260 30 0

7.0

14

7.8

25

38

1 5

21.0

270 18 0

3.5

5

7.8

42

63

1 0

21.0

270 18 0

3.5

7

Lead

11.3

-

-

-

-

-

-

-

Magnesium Alloy

1.8

20

30

-

4.5

250 37 0

9.0

16

Titanium

4.5

50

80

-

11.0

240 23 0

5.0

18

Units for above

kg/d 3 m

kg/m 2 m

kg/m 2 m

% kg/m 2 m

to obtain:

lbs/cu KSI

KSI

% KSI

multiply by:

.0357 1420

1420

-

Wood

Alum. Alloy

Steel AISI 1026 4130 N (4140)

3

79

1420

km

-

-

kg m

2

2/3

kg m

1/3

km

Column 3, the yield stress (fy), is the stress (load per area) at which there will be a permanent deformation after unloading (the material has yielded, given way ... ) Column 4, the ultimate stress (fu), is the stress (load per area) at which it cannot carry a further load increase. It is the maximum load before failure. Column 5, the elongation at ultimate stress (e), in percentage gives an indication of the 'Toughness" of the material. Column 6 lists the Yongs Modular or Modulus of Elasticity (E), which is the steepness of the stress/strain diagram as shown in Figure 1. Important Note: For wood, the tension is much greater (2 to 3 times) than the compression. Both values are given in the Table. For fiberglass, the same applies, but the yield is so dependent on the manufacturing process that we cannot even give 'Iypical values'.

Figure 6-1: Stress strain curves for different materials

Columns 7 to 10: Columns 7 to 10 are values which allow the comparison of materials from a weight standpoint (the above referenced text by Timoshenko will also show you why we use those "funny" looking values). Column 7 gives the stiffness of a sandwich construction. The higher the value, the stiffer the construction. From the Table, we see that metals are high wood comes

80

close, but fiberglass is low: which means fiberglass will be heavier for the same stiffness.

Figure 6-2: Stress strain curve

Column 8 shows the column buckling resistance for the same geometric shapes. This time, wood is better than the light alloys, coming before steel and fiberglass. (Surprisingly, the usual welded steel tube fuselage is not very weight efficient.) Column 9 gives the plate buckling stiffness, which is also a shear strength measure. Here again, wood (plywood) is in a very good position before aluminum and fiberglass, with steel not very good. Column 10 provides a crude way of measuring the strength to weight ratio of materials because it does not take into account the various ways the material is used in "light structures". According to this primitive way of looking, unidirectional fibers are very good, followed by high strength (2024) aluminum and wood, then the more common aluminum alloys and finally steel. From just this simple table, we find there is not one material that provides an overwhelming solution to all the factors that must be considered in designing a light aircraft. Each material has some advantage somewhere. The designer's choice (no preconceived idea) will make a good aircraft structure ... if the choice is good!

81

7. Detailed wing design 7.1 Spar design: Spars are members which are basically used to carry the bending and shear loads acting on the wing during flight. There are two spars, one located at 15-20% of the chord known as the front spar, the other located at 60-70% of the chord known as the rear spar. Some of the functions of the spar include: They form the boundary to the fuel tank located in the wing.  The spar flange takes up the bending loads whereas the web carries the shear loads.  The rear spar provides a means of attaching the control surfaces on the wing. Considering these functions, the locations of the front and rear spar are

fixed

at 0.17c and 0.65c respectively. The NACA 65 (3) 418 airfoil is drawn to scale using any design software and the chord thickness at the front and rear spar locations are found to be 1.9708 m and 7.5354 m respectively.

7.1.1 Geometric dimensions: The spar design for the wing root has been taken because the maximum bending moment and shear force are at the root. It is assumed that the flanges take up all the bending and the web takes all the shear effect. The maximum bending moment for high angle of attack condition is 486567918.4 Nm. the ratio in which the spars take up the bending moment is given as

Where

82

h1  height of front spar h2  height of rear spar

From the above two equations,

The yield tensile stress σy for 7075 Al Alloy is 455.053962 MPa. The area of the flanges is determined using the relation

where M is bending moment taken up by each spar, A is the flange area of each spar, z is the centroid distance of the area = h/2. Using the available values, Area of front spar,

Area of rear spar, 83

Each flange of the spar is made of two angle sections. For the front spar, the length of the angle is 6t, angle height is 5t with angle thickness t. Area for each angle of front spar is found to be 0.1799507 m2 and hence value of t is found to be

Length of the front angle section:

Height of the front angle section:

For the rear spar, the length of the angle is 8t, angle height is 3.5t with vertical thickness t and horizontal thickness t/2. Area for each angle of rear spar is found to be 0.164486 m2 and hence value of t is found to be.

Length of the rear angle section:

Height of the rear angle section:

Now to determine the thickness of the web portion, the ultimate shear stress of 7075 Al Alloy is 317.1588MPa. The maximum shear force at root of the wing 84

for high angle of attack condition is 17480623.2 N. The wing chord is assumed to be a simply supported beam supported at the two spars. The maximum shear force acts at the centre of pressure which can be located by using the formula,

Figure 7-1: Reaction force determination at spars

Considering force and moment equilibriums for the given simply supported configuration, the reactive shear force at the spar supports are found to be

We know that,

V  shear force at the spar

85

t  thickness of the web. Thus,

FOS = 1.5 z is the centroidal distance of the area = h/2 Thus the thicknesses of the web portions are,

0.43806

0.670

0.0763

0.02346

0.13414

1.00128

0.1251

0.8048

Rear spar

Front Spar

All dimensions are in m It becomes necessary to check whether the shear stress due to this thickness is less than the allowable of the material.

86

( ) For the web, the dimensions of a and b will be a = 1.6186 m ( rib spacing) and b = spar height. The value of ks is obtained using a/b from the given plot in figure, ks is obtained and thus the actual shear stress in each web

Figure 7-2: Shear buckling coefficients fro plates as a ratio of a and b for hinged and clamped edges

Both these values are less than 211.4392 GPa. Thus, the web does not fail due to shear buckling. 87

7.1.2 Shear flow: The shear flow can be considered for the two cells in the airfoil cross section. The shear flow will be due to the torque as well due to the bending moments. These are computed separately and summed up to obtain the net shear flow pattern for the wing cross section. Due to torque, A  area of each cell Q  shear flow due to the torque The maximum torque experienced at the root of the wing is -25338 Nm. Using GAMBIT software, the areas and perimeters of the cells in the airfoil formed by the spars and the skin are found to be, Cell1 A1 = 2.1783 m2 Cell2 A2= 10.5862 m2 Cell3 A3= 2.7528 m2 l1 = 4.45527 m, l2 = 1.6186 m, l3 = 5.6214 m, l4 = 5.5827 m, l5 = 1.4795 m, l6 = 4.092 m, l7 = 4.176 m . l1, l2 belong to cell1,l2, l3, l4, l5 belong to cell2, l5, l6 and l7 to cell3. l2 and l5 are the spars. The equations for the two cells involving shear flow of cell1 q1 and shear flow of cell2 q2 are

[

(

) [ [

( ) ( )

(

( )] ( )]

)

( )] 88

The second, third and fourth equations are obtained from the condition that the cell twist is zero.

Solving these equations, we get shear flow values due to torque alone.

The shear flow due to bending is given by the formula, *

+

*

Vx = 1012394.13 N (shear force due to chordwise forces) Vy = 17480623.2 N (shear force due to normal forces)

89

+

90

179935.396 179935.396

179935.396 179935.396 164482.769 164482.769

164482.769 164482.769

Cell 2 F_U_2 F_L_2 R_U_1 R_L_1

Cell 3 R_U_2 R_L_2

Area

Spar Cell 1 F_U_1 F_L_1

145398236.2 -49300499.2 118961011.2 -42045249.8 173013498.4

1280625007 118961011.2 1280625007 -42045249.8 2561250014 76915761.31

402889746.5 402889746.5 1198278353 1198278353 2396556707

306343610.4 145398236.2 306343610.4 -49300499.2 612687220.8 96097737.07

A*x

A*y

7785.77 7785.77

2239.08 2239.08 7285.13 7285.13

1702.52 1702.52

x_bar

x_c

y_c

-1238.82 808.058 -1238.82 -273.99 3807.23 723.243 3807.23 -255.621

723.243 4307.87 723.243 -255.621 4307.87 -255.621

808.058 -273.99 723.243 -255.621

808.058 -1775.38 808.058 -273.99 -1775.38 -273.99

y bar

Ixx

88696143051 13406073270

1.20998E+11 17015324658 88696143051 13406073270

1.20998E+11 17015324658

3.05509E+12 3.05509E+12

2.7965E+11 2.7965E+11 2.38684E+12 2.38684E+12

5.70659E+11 5.70659E+11

Iyy

q

5.12469E+11 549369.11 -233830 -1.81125E+11 315539.12

-1.80122E+11 -1964438 61074444357 555765.26 4.52912E+11 607569.23 -1.60076E+11 -265631.1 -1066735

-2.58137E+11 -1485868 87527120181 421804.49 -1064063

Ixy

q Ixy Iyy Ixx y_c x_c y bar x_bar A*y A*x Area 1.20998E+11 5.70659E+11 -2.58137E+11 -1485868 808.058 -1775.38 808.058 1702.52 179935.396 306343610.4 145398236.2 2.7965E+11 -1.80122E+11 -1964438 1.20998E+11 808.058 -1238.82 808.058 2239.08 179935.396 402889746.5 145398236.2 87527120181 421804.49 17015324658 5.70659E+11 -273.99 -1775.38 -273.99 1702.52 179935.396 306343610.4 -49300499.2 61074444357 555765.26 2.7965E+11 17015324658 -273.99 -1238.82 -273.99 2239.08 179935.396 402889746.5 -49300499.2 4.52912E+11 607569.23 88696143051 2.38684E+12 723.243 3807.23 723.243 7285.13 1198278353 118961011.2 164482.769 5.12469E+11 549369.11 88696143051 3.05509E+12 723.243 4307.87 723.243 7785.77 1280625007 118961011.2 164482.769 13406073270 2.38684E+12 -1.60076E+11 -265631.1 7285.13 -255.621 3807.23 -255.621 1198278353 -42045249.8 164482.769 -233830 13406073270 3.05509E+12 -1.81125E+11 7785.77 -255.621 4307.87 -255.621 1280625007 -42045249.8 164482.769 3.34521E+11 4.8023E+11 1.25845E+13 6376273434 346026996.8 1377672.66

Figure 7-3: Shear flow in spars

Spar F_U_1 F_U_2 F_L_1 F_L_2 R_U_1 R_U_2 R_L_1 R_L_2 Sum

( ) For the web, the dimensions of a and b will be a = 1.6186 m9rib spacing and b =5.6214 (length of the cell containing critical shear flow) KS =32; τcr is found in terms of t2 Using relation, τcr = qcr/t The value of ks is obtained using a/b from the given plot in figure, ks is obtained and thus the thickness of the skin without using stringer will be

7.2 Stringer design: The thickness of the skin determined above is too high for the skin of an aircraft. Therefore in order to reduce skin thickness and redistribute the shear flow in the wing skin, stringers are added. The number of stringers can be determined by evaluating the amount by which the skin thickness should be reduced. Roughly 36 stringers can be added to the wing, 18 on the upper surface of the airfoil and 18 on the lower surface of the airfoil. The stringer cross section is chose from the standard cross sections available in Analysis of Aircraft structures – Bruhn. The Z cross section is chosen and scaled up or down determining the critical stress in each stringer and iterating if it is less than the critical buckling stress of the stringer cross section.

91

7.2.1 Geometric dimensions based on shear flow: The stress of each stringer is found using the formula, *

+

*

+

Thus the section satisfying iterations has the following properties, A = 10000mm2,

The shear flow for each section is determined using the formula, *

+

92

*

+

Table 7-1: Shear flow and Bending stress tabulation

Stringer

A 1 2 3 4 5 6 7

R_U_1 R_U_2 8 9 10 11 12 13 F_U_2 F_U_1 14 15 16 17 18 19 20 21 22 23 F_L_1 F_L_2 24 25 26 27 28 29 R_L_1 R_L_2 30 31 32 33 34

10000 10000 10000 10000 10000 10000 10000 179935 179935 10000 10000 10000 10000 10000 10000 179935 179935 10000 10000 10000 10000 10000 10000 10000 10000 10000 10000 164483 164483 10000 10000 10000 10000 10000 10000 164483 164483 10000 10000 10000 10000 10000

Ax Ay x_c y_c σ q 114498474 134228 6483.1 -238.3 -7.497 9.95E+08 111931155 516227 6226.4 -200.1 -7.809 8.5E+08 108296803 1223363 5862.9 -129.3 -8.298 5.64E+08 103680870 2287145 5401.3 -22.96 -9.009 1.04E+08 98180431 3698022 4851.3 118.13 -10.01 -5.6E+08 91896585 5394407 4222.9 287.77 -11.41 -1.4E+09 84934997 7276236 3526.8 475.95 -13.35 -2.5E+09 1.311E+09 1.3E+08 2318.4 471.57 0.6891 -1521292 1.401E+09 1.3E+08 2819 471.57 -0.492 -1457428 69429422 1.1E+07 1976.2 850.91 -18.73 -4.9E+09 61134483 1.3E+07 1146.7 1000.8 -18.58 -4.4E+09 52674321 1.3E+07 300.69 1094.3 -6.489 51024314 44249964 1.4E+07 -541.7 1119.9 12.091 4.67E+09 36077346 1.3E+07 -1359 1084.6 18.094 4.55E+09 28357220 1.2E+07 -2131 995.83 17.126 3E+09 402889746 1.5E+08 -2728 556.38 78.162 1233401 306343610 1.5E+08 -3264 556.38 48.757 757536.7 14973466 9507529 -3469 699.08 13.242 8.9E+08 9612440.3 7640616 -4005 512.39 11.88 3.98E+08 5310518 5662783 -4436 314.6 10.904 1.37E+08 2177421.6 3695896 -4749 117.92 10.243 24461501 330609.2 1889583 -4934 -62.72 9.856 -4256124 423361.3 -730095 -4924 -324.7 9.7933 954223.2 2501869.1 -2E+06 -4717 -463.8 10.129 -2.1E+07 5958680.3 -3E+06 -4371 -569.6 10.795 -9.6E+07 10487170 -4E+06 -3918 -664.5 11.811 -2.4E+08 15969273 -5E+06 -3370 -749 13.29 -4.7E+08 280035204 -5E+07 -3264 -525.7 2.2024 -159628 368290078 -5E+07 -2728 -525.7 6.9376 -280746 29304388 -6E+06 -2036 -882.3 18.122 -1.2E+09 36882186 -7E+06 -1279 -930.6 20.113 -1.1E+09 44867767 -7E+06 -480 -963.9 12.749 1.59E+09 53102607 -7E+06 343.52 -964.6 -11.2 7.38E+09 61406999 -7E+06 1174 -920.3 -21.32 8.28E+09 69601782 -6E+06 1993.4 -838.7 -19.23 6.5E+09 1.198E+09 -4E+07 2318.4 -507.3 2.6696 542408.6 1.281E+09 -4E+07 2819 -507.3 2.1096 510540.7 85045683 -4E+06 3537.8 -629.2 -13.15 3.62E+09 92011967 -3E+06 4234.5 -524.2 -11.25 2.73E+09 98301217 -2E+06 4863.4 -430.7 -9.905 2.08E+09 103796399 -1E+06 5412.9 -354.8 -8.944 1.62E+09 108391643 -484586 5872.4 -300.1 -8.263 1.31E+09

93

35 36

10000 10000

111993602 114525411

-154167 -13414

6232.6 6485.8

-267.1 -253

-7.793 -7.492

1.14E+09 1.06E+09

NACA 653-418 y (mm)

2000 1000 0

-2000

0

2000

4000

-1000

6000

8000

10000

12000

14000

x (mm)

Stringer design ratio: 5t

6t

t

5t

The critical shear flow is found to be 162206.0558N/m acting between the upper flange of front spar and stringer 5. Using the formula, ( ) where a = 1.6186 m, (rib spacing), b = 0.526 m, kb is obtained from following plot for given a/b as 40

94

Figure 7-4: Shear- buckling coefficient for hinged and clamped plates

Thus skin thickness after using stringers is found to be t = 10.953 mm

95

Shear flow distribution on skin of wing Millions

34

35

33

36 10000

1

2

3

4 5

8000

32

6

6000

31

7

4000

30

R_U_1

2000 R_L_2

R_U_2

0

R_L_1

8

-2000

29

-4000

9

28

-6000

10

27

11

26

12

25

13 24

F_U_2

F_L_2

F_U_1

F_L_1

14 23

15 22

21

20

19

18

17

16

Figure 7-5: Shear flow diagram for wing represented in polar coordinates, 19 th stringeer is leading edge and 1st stringer is at trailing edge

96

8. Fuselage design 8.1 Description Fuselage contributes very little to lift and produces more drag but it is an important structural member/component. It is the connecting member to all load producing components such as wing, horizontal tail, vertical tail, landing gear etc. and thus redistributes the load. It also serves the purpose of housing or accommodating practically all equipment, accessories and systems in addition to carrying the payload. Because of large amount of equipment inside the fuselage, it is necessary to provide sufficient number of cutouts in the fuselage for access and inspection purposes. These cutouts and discontinuities result in fuselage design being more complicated, less precise and often less efficient in design. As a common member to which other components are attached, thereby transmitting the loads, fuselage can be considered as a long hollow beam. The reactions produced by the wing, tail or landing gear may be considered as concentrated loads at the respective attachment points. The balancing reactions are provided by the inertia forces contributed by the weight of the fuselage structure and the various components inside the fuselage. These reaction forces are distributed all along the length of the fuselage, though need not be uniformly. Unlike the wing, which is subjected to mainly unsymmetrical load, the fuselage is much simpler for structural analysis due to its symmetrical cross-section and symmetrical loading. The main load in the case of fuselage is the shear load because the load acting on the wing is transferred to the fuselage skin in the form of shear only. The structural design of both wing and fuselage begin with shear force and bending moment diagrams for the respective members. The maximum bending stress produced in each of them is checked to be less than the yield stress of the material chosen for the respective member.

97

8.2 Loads and its distribution: To find out the loads and their distribution, consider the different cases. The main components of the fuselage loading diagram are:  Weight of the fuselage  Engine weight  Weight of the horizontal and vertical stabilizers  Tail lift  Weight of crew, payload and landing gear  Systems, equipment, accessories Symmetric flight condition, steady and level flight: (Downward forces negative) Values for the different component weights are obtained from aerodynamic design calculations. Table 8-1: Loads acting on Fuselage

Condition 1

S. No 1 2 3 4 5 6 7 8 9 10 11 12

Full Payload and Full Fuel Fuselage alone analysis

Distance from reference Moment Components line (m) Mass (kg) Weight (N) (Nm) Crew 3.043 270 2648.7 8059.9941 Nose landing gear 6.086 3600 35316 214933.176 Payload bay 1 12.172 37500 367875 4477774.5 Fixed equipment 18.641 1149.19 11273.554 210150.318 Excess mass 22.641 21600 211896 4797537.34 Fuselage mass 22.641 43200 423792 9595074.67 Fuel in fuselage 22.641 95280.42 934700.92 21162563.5 Main landing gear assembly 1 22.641 10800 105948 2398768.67 Main landing gear assembly 2 30.3211 7200 70632 2141639.94 Payload bay 2 30.321 37500 367875 11154337.9 Horizontal stabilizer 45.367 6400 62784 2848321.73 Vertical Stabilizer 48.501 3200 31392 1522543.39 Total 267699.6 2626133.2 60531705.1 Cg from Nose 23.04974695

98

Figure 8-1: Balance diagram showing loads acting on fuselage

99

8.3 Shear Force and bending moment calculations: Table 8-2: Shear force and bending moment tabulation

Distance(m)

Load (kg) 0 3.043 6.086 12.172 18.641 22.641 22.641 22.641 22.641 23.0497 30.3211 30.321 45.367 48.501 54.849

SF (N)

BM (Nm)

0 -2648.7 -35316 -367875 -11273.6 -211896 -423792 -934701 -105948 2626133 -70632 -367875 -62784 -31392 0

0 -2648.7 -37964.7 -405840 -417113 -629009 -1052801 -1987502 -2093450 532683 462051 94176 31392 0 0

0 -8059.9941 -222993.17 -4700767.67 -4910917.99 -9708455.32 -19303530 -40466093.5 -42864862.2 17666719.62 15525079.69 4370741.812 1522420.084 0 0

Thousands

Shear Force 1000 500

Shear Force(N)

0 0

10

20

30

40

50

-500 -1000 -1500 -2000 -2500

Distance from nose cone(m)

Figure 8-2: Shear force on the fuselage (free-free beam with one reaction at its centre) at fully loaded condition

100

Shear Force(N)

Thousands

Bending Moment 30000 20000 10000 0 0

10

20

30

40

50

-10000 -20000 -30000 -40000 -50000

Distance from nose cone(m)

Figure 8-3: Bending moment on the fuselage (free-free beam with one reaction at its centre) at fully loaded condition

101

9. Detailed Design of Fuselage: 9.1 Stringer Design: Design of the fuselage can be carried out by considering the maximum bending moment which is taken as the design bending moment. The cross-sectional area required to withstand the bending stress is found out by using the formula for bending stress. This area is divided among several stringers which are spaced evenly. The stringers spacing is calculated by considering the buckling of the portion between adjacent stringers which can be modelled as a plate. Now, the first step is to calculate the required cross-sectional area of the stringers. Use the following formula for bending stress.

Where, σ  Tensile strength of the material used (Aluminium 7075) = 455 MPa M  Design bending moment = -42864862.2 Nm I  Second moment of area (m4) ( ) y  d/2 d  diameter of the fuselage (3.5m ) A  cross-sectional area of the fuselage stringers (m2) A stringer cross section (Z section) is chosen satisfying the condition that the actual stress is less than the yield stress of the material.

102

Stringer location in fuselage 4

3

2

1

0 -4

-3

-2

-1

0

1

2

3

4

-1

-2

-3

-4

Figure 9-1: Location of Z shaped Stringer in the fuselage.

The properties of the stringer section chosen are as follows, Length of stringer: Height of stringer: Where t thickness of stringer The total circumference of the fuselage cross section is found to be 21.9914 m. This circumference is distributed with ‘n’ number of stringers such that the total bending moment is taken up by these stringers effectively. Assume skin is ineffective in bending. Arbitrarily, let us set the number of stringers to be equal to 60 i.e. 15 stringers in each quadrant. Now, the net IYY is computed 103

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

9.2 Shear flow along skin of fuselage: Consider the stringer at Ө = 0° of the first quadrant of the cross section as the first stringer and number it in anticlockwise direction. Make a cut between stringers 1 and 2 and determine shear flow using the formula, *

+

*

+

Since VX = 0 , VY = -2093450N ( Max. Shear Force from shear force diagram)

The shear flow equation gets simplified to [

]

∑q_l = -1.81599E+11 N Now, on closing the cut, and considering cell twist is zero for the fuselage cross section, we obtain the equation, -1.81599E+11+ 21.9914qo = 0 Thus constant shear flow to be added to the cell is qo = 8257728021 N/m

104

Table 9-1: Shear flow along the stringers of fuselage

S.No 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

ʘ 0 6 12 18 24 30 36 42 48 54 60 66 72 78 84 90 96 102 108 114 120 126 132 138 144 150 156 162 168 174 180 186 192 198 204 210 216 222 228 234 240 246

A Ax Ay si Ax Si Ay q open q*l q fin 0.0269 0.0942 0 0.0942 0 0 0 8.26E+09 0.0269 0.0937 0.00984 0.1878 0.00984 -9E+07 -33157611 8.17E+09 0.0269 0.0921 0.01958 0.28 0.02942 -2.7E+08 -99109571 7.99E+09 0.0269 0.0896 0.0291 0.3695 0.05852 -5.4E+08 -197133342 7.72E+09 0.0269 0.086 0.0383 0.4556 0.09683 -8.9E+08 -326155016 7.37E+09 0.0269 0.0816 0.04709 0.5371 0.14391 -1.3E+09 -484761088 6.94E+09 0.0269 0.0762 0.05535 0.6133 0.19927 -1.8E+09 -671213940 6.43E+09 0.0269 0.07 0.06301 0.6833 0.26228 -2.4E+09 -883470876 5.85E+09 0.0269 0.063 0.06998 0.7463 0.33227 -3.1E+09 -1.119E+09 5.2E+09 0.0269 0.0554 0.07619 0.8017 0.40845 -3.8E+09 -1.376E+09 4.5E+09 0.0269 0.0471 0.08156 0.8488 0.49001 -4.5E+09 -1.651E+09 3.75E+09 0.0269 0.0383 0.08603 0.8871 0.57604 -5.3E+09 -1.94E+09 2.96E+09 0.0269 0.0291 0.08956 0.9162 0.66561 -6.1E+09 -2.242E+09 2.14E+09 0.0269 0.0196 0.09212 0.9358 0.75772 -7E+09 -2.552E+09 1.29E+09 0.0269 0.0098 0.09366 0.9456 0.85138 -7.8E+09 -2.868E+09 4.33E+08 0.0269 4E-06 0.09418 0.9456 0.94556 -8.7E+09 -3.185E+09 -4.32E+08 0.0269 -0.0098 0.09366 0.9358 1.03922 -9.6E+09 -3.501E+09 -1.29E+09 0.0269 -0.0196 0.09212 0.9162 1.13134 -1E+10 -3.811E+09 -2.14E+09 0.0269 -0.0291 0.08957 0.8871 1.2209 -1.1E+10 -4.113E+09 -2.96E+09 0.0269 -0.0383 0.08604 0.8488 1.30694 -1.2E+10 -4.402E+09 -3.75E+09 0.0269 -0.0471 0.08156 0.8017 1.3885 -1.3E+10 -4.677E+09 -4.5E+09 0.0269 -0.0553 0.07619 0.7464 1.46469 -1.3E+10 -4.934E+09 -5.2E+09 0.0269 -0.063 0.06999 0.6833 1.53468 -1.4E+10 -5.169E+09 -5.85E+09 0.0269 -0.07 0.06302 0.6134 1.59771 -1.5E+10 -5.382E+09 -6.43E+09 0.0269 -0.0762 0.05536 0.5372 1.65307 -1.5E+10 -5.568E+09 -6.93E+09 0.0269 -0.0816 0.04709 0.4556 1.70016 -1.6E+10 -5.727E+09 -7.37E+09 0.0269 -0.086 0.03831 0.3696 1.73847 -1.6E+10 -5.856E+09 -7.72E+09 0.0269 -0.0896 0.02911 0.28 1.76758 -1.6E+10 -5.954E+09 -7.99E+09 0.0269 -0.0921 0.01959 0.1879 1.78717 -1.6E+10 -6.02E+09 -8.17E+09 0.0269 -0.0937 0.00985 0.0943 1.79702 -1.7E+10 -6.053E+09 -8.26E+09 0.0269 -0.0942 8.7E-06 8E-05 1.79703 -1.7E+10 -6.053E+09 -8.26E+09 0.0269 -0.0937 -0.0098 -0.0936 1.78719 -1.6E+10 -6.02E+09 -8.17E+09 0.0269 -0.0921 -0.0196 -0.1857 1.76762 -1.6E+10 -5.954E+09 -7.99E+09 0.0269 -0.0896 -0.0291 -0.2753 1.73853 -1.6E+10 -5.856E+09 -7.72E+09 0.0269 -0.086 -0.0383 -0.3613 1.70024 -1.6E+10 -5.727E+09 -7.37E+09 0.0269 -0.0816 -0.0471 -0.4429 1.65316 -1.5E+10 -5.569E+09 -6.94E+09 0.0269 -0.0762 -0.0553 -0.5191 1.59781 -1.5E+10 -5.382E+09 -6.43E+09 0.0269 -0.07 -0.063 -0.5891 1.5348 -1.4E+10 -5.17E+09 -5.85E+09 0.0269 -0.063 -0.07 -0.6521 1.46482 -1.3E+10 -4.934E+09 -5.2E+09 0.0269 -0.0554 -0.0762 -0.7074 1.38864 -1.3E+10 -4.678E+09 -4.5E+09 0.0269 -0.0471 -0.0816 -0.7545 1.30709 -1.2E+10 -4.403E+09 -3.75E+09 0.0269 -0.0383 -0.086 -0.7929 1.22106 -1.1E+10 -4.113E+09 -2.96E+09

105

43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

252 258 264 270 276 282 288 294 300 306 312 318 324 330 336 342 348 354

0.0269 0.0269 0.0269 0.0269 0.0269 0.0269 0.0269 0.0269 0.0269 0.0269 0.0269 0.0269 0.0269 0.0269 0.0269 0.0269 0.0269 0.0269

-0.0291 -0.0196 -0.0099 -1E-05 0.0098 0.0196 0.0291 0.0383 0.0471 0.0553 0.063 0.07 0.0762 0.0816 0.086 0.0896 0.0921 0.0937

-0.0896 -0.0921 -0.0937 -0.0942 -0.0937 -0.0921 -0.0896 -0.086 -0.0816 -0.0762 -0.07 -0.063 -0.0554 -0.0471 -0.0383 -0.0291 -0.0196 -0.0099 8.7E-06

-0.822 -0.8416 -0.8514 -0.8514 -0.8416 -0.822 -0.7929 -0.7547 -0.7076 -0.6522 -0.5892 -0.5193 -0.4431 -0.3615 -0.2755 -0.1859 -0.0938 -0.0002

106

1.1315 1.03938 0.94573 0.85155 0.75789 0.66577 0.5762 0.49016 0.40859 0.3324 0.2624 0.19937 0.144 0.0969 0.05858 0.02947 0.00987 8.7E-06

-1E+10 -9.6E+09 -8.7E+09 -7.8E+09 -7E+09 -6.1E+09 -5.3E+09 -4.5E+09 -3.8E+09 -3.1E+09 -2.4E+09 -1.8E+09 -1.3E+09 -8.9E+08 -5.4E+08 -2.7E+08 -9.1E+07 -80333 -5E+11

-3.811E+09 -3.501E+09 -3.186E+09 -2.868E+09 -2.553E+09 -2.243E+09 -1.941E+09 -1.651E+09 -1.376E+09 -1.12E+09 -883868023 -671567406 -485066999 -326410021 -197334648 -99254973 -33245515 -29443.594 -1.816E+11

-2.14E+09 -1.29E+09 -4.34E+08 4.32E+08 1.29E+09 2.14E+09 2.96E+09 3.75E+09 4.5E+09 5.2E+09 5.85E+09 6.43E+09 6.93E+09 7.37E+09 7.72E+09 7.99E+09 8.17E+09 8.26E+09

Shear Flow distribution in the skin of fuselage Millions 54

55

60 58 599000 57 56 8000

1

2

3 4

5

6

7000

7

8

53 52 51 50 49 48

6000

9

2000

10 11 12 13 14

47

1000

15

46

0

16

5000 4000 3000

45

17

44 43 42 41 40 39

18 19 20 21 22 23 38

37

36

35

34 33 32

31

28 30 29

27

26

25

24

Figure 9-2: Shear flow distribution along fuselage, View 16 th stringer is at the bottom and 46 th stringer at the top

The critical shear flow is found to occur in elements between 1 and 60, 30 and 31. The critical shear flow value is 82577258021 N/m. We know that, ( ) Where, E = 7.17e10 N/m2, 107

a = 1.099557 m(bulk head spacing), b = 0.366519 m (Stringer circumferential spacing) a/b = 3 ks = 40

Figure 9-3: Shear buckling coefficient for plates with hinge and clamps

ν = 0.3 Thus we obtain, t=0.00748 m The skin thickness is thus found to be t = 7.4833 mm Using τ = 1.5*(q/t), τMat = 211.4392e6, we get = 0.133 mm. 108

Considering the maximum of the two, we get t = 7.4833 mm The above value of skin thickness is well within the standard limits. Therefore, the above design is acceptable.

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10. Computational Fluid Dynamics CFD analysis at tip: The tip stall is the worst initiation step in all instability problems so the tip at an angle of incidence 5˚ is checked for separation.

Figure 10-1: Velocity vector plot for aoa= 5deg

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Figure 10-2: CP plot for aoa= 5deg

Figure 10-3: Contours of static Pressure

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Figure 10-4: Velocity contours plot for aoa= 5deg

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11. Three view diagram:

Figure 11-1 Front view

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Figure 11-2: Top view

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Figure 11-3: Side view

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Conclusion The Conceptual Design phase of an aircraft is probably the most interesting and intriguing phase of aircraft design. It is a clear indication of the compromise that has to be made between various divisions of an Aircraft design center, and yet satisfy an incredible number of real-world constraints and design specifications. Aircraft design involves a variety of the field of Aerospace engineering like structures, performance, aerodynamics, stability etc. Among this we went through the structure part in this project which has enabled us to get a taste of what it is to design a real aircraft. The fantasies of the flying world seem to be much more than what we thought. With this design project as the base, we will strive to progress in the field of airplane design and maintenance. We convey our heartfelt gratitude to all of them who have provided their helping hand in the completion of this project.

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Reference Books:  Analysis of Aircraft structures – Bruhn  Aircraft Structures for engineering students – T.H.G Megson  Aircraft structures – Peery  Airplane design – Jan Roskam  Fundamentals of Aerodynamics - Anderson J D Websites:  www.wikipedia.org  www.joeclarksblog.com  http://www.docstoc.com/  http://www.flightsimaviation.com/  www.tc.gc.ca  http://www.risingup.com/  http://www.aerospacemetals.com/contact-aerospace-metals.html  http://www.aerospaceweb.org/  www.faa.gov/regulations_policies/faa_regulations/

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