Aircraft Design Report

AEROSPACE ENGG. , IIT KANPUR

Project Report AE-462 Disaster Management & Surveillance Drone

Group 6

Pranav Kumar Singh

Jatin Mitruka

Parveen Kumar

4/25/2014

Table of Contents 1. Introduction: .................................................................................................................................................... 3

2. Wing Box Configuration:.................................................................................................................................... 4

3. V-n Diagram with Gust Envelope: ....................................................................................................................... 5

4. Procedures involved:......................................................................................................................................... 7

4.1. Material Description: .................................................................................................................................. 7

4.2 Calculation of Wing Section properties : ....................................................................................................... 7

4.3 Normal and Chordwise load per unit length Calculation:................................................................................ 8

4.4 Inertia-Gravity load per unit length Calculation: ............................................................................................ 8

4.5 Computation of Shear Force and Moment Distribution along span: ................................................................ 8

4.6 Net Loads at ith Wing Bay : ........................................................................................................................... 8

4.7 Idealization of Forces:.................................................................................................................................. 9

4.8 Shear forces and Moments at a Wing Section: .............................................................................................. 9

4.9 Wing Box Idealization: ............................................................................................................................... 10

4.10 Normal Stress Computation: .................................................................................................................... 12

4.11 Shear Stress Computation: ....................................................................................................................... 12

4.11.1 Shear Stress due to Shear Forces:....................................................................................................... 12

4.11.2 Shear stress due to Torsion: ............................................................................................................... 12

4.12 Factor of Safety calculations:.................................................................................................................... 13

4.13 Crushing Loads: ....................................................................................................................................... 13

4.14 Buckling of stringers and spars: ................................................................................................................ 13

4.15 Buckling of panels:................................................................................................................................... 13

5. Results and Discussion: ................................................................................................................................... 14

6.Iterations for Weight Reduction:....................................................................................................................... 29

7. Summary: ....................................................................................................................................................... 30

Appendix:........................................................................................................................................................... 31

1. Introduction:

Mission of our project is to design a drone aircraft for Disaster Monitoring and Surveillance Purpose. MQ-9

Reaper, Global Hawk are used for base calculations. V-n diagram and gust envelope are drawn using details

given in subsequent sections. Later we calculated Distributed Aerodynamic loads on the points obtained

along the span of the wing. All details have been described in respective sections.

Details of our design:

Range 2000km

Endurance 30 hrs

W(Total) 5400 kg

W(Payload) 1500 kg

Airfoil NACA 653 618

Flaps Plain

Details of wing:

b (in m) 26

Sw(in m2) 45

e 0.9

0.4

CD0 0.058

Details of V-n diagram:

Cza max 1.22

Cza min -0.8

Cz alpha 5.41

Vcruise 360.89 ft/s

Vdiving 451.11 ft/s

Vstall

positive

130.14 ft/s

2. Wing Box Configuration:

A typical wing box of an aircraft consists of spars, ribs, stringers as well as skin covering them.

As an initial guess for half span of wing (13 m) two spars are situated at c/4 and 3c/4 from the leading edge .

There are around 20 ribs installed in the half span. Both ribs and spars are made of aluminium. The cutaway

of Global Hawk was used as baseline for the initial guess of wing box structure. The detailed structure of the

ribs would be decided later. The aerofoil chosen for the drone is 6-series NACA aerofoil named as

NACA-653-618. The plots obtained through the data for this aerofoil are given below:

Figure 1 Variation of Normal force Coefficient with angle of attack

Figure 2 Variation of Pitching moment Coefficient with angle of attack

3. V-n Diagram with Gust Envelope:

Figure 3 V-n Diagram with Gust Envelope

3.1 Steps to complete V-n Diagram :

Different attributes of the aircraft used in the following procedure are as given in introduction.

1) Positive Limit Manoeuvring Load Factor (n1) : Calculated using formula

In diagram line AC represents this limit.

2) Negative Limit Manoeuvring Load Factor (n2): with negative sign

This limit is indicate by line BE in the diagram.

3) Parabolic portion OA where Maximum positive lift coefficient of the airfoil section is the limiting

factor is obtained using the following relation:

where, - is in slug/ft3 at sea level

v Air relative speed in ft/s

W/S in lb/m2

Czmax - is maximum value of normal force coefficient.

Here normal lift coefficient is calculated using following expression:-

where, CL , CD, Cma are obtained from airfoil data.

cw and lt are wing chord and distance of tail a.c. from C.G. of aircraft.

4) Parabolic portion OE where Maximum negative lift coefficient of airfoil is the limiting fac tor is

obtained through following relation:

where Czmin is maximum value of negative normal force coefficient.

5) The upper limit of attainable air relative speed marked (also called Dive Speed ) through line BC is

obtained from the relation:

6) Point D corresponds to cruise velocity of drone. Points D and C are joined via a straight line.

3.2 Steps to complete Gust Envelope:

The normal load factor for drone corresponding to three different kinds of gusts (both positive and negative) are analysed here. All are considered at sea level conditions up to different respective ranges

of speeds.

Rough Air Gust ( Speed 66 ft/s ) : It is considered only up to velocity at point A.

High Speed Gust ( Speed 50 ft/s ) : It is considered only up to cruise velocity.

Dive Speed Gust ( Speed 25 m/s ) : It is considered only up to dive speed.

The gust load factor is calculated from:

Where V is in knots

Ude in ft/s

where a= slope of a/c normal force coefficient

4. Procedures involved:

A body axes coordinate system was chosen for the aircraft to carry on further analysis. The y-

axis is along the locus of AC of the wing. Since there is no sweep for the wing y-axis is perpendicular to

upstream velocity. Z- axis is positive upwards while x is positive forward as shown in the Fig

Figure 4. The body axis representation

To simplify the analysis wing was divided into 27 wing bays. Thus 28 wing sections were formed starting

from the root section to tip section. The partition was done to create bays of different lengths at different

locations along the wing as follows:

0 5 m - 1 m

5 10 m - 0.5 m

10 13 m - 0.25 m

4.1. Material Description:

Aluminium alloy 2024-T3 was chosen for present design of wing box. The mechanical properties of the

material obtained from Ref are as follows:

Mechanical /Physical Properties

Value

Density 2780 kg/m3

Tensile Yield Strength 345 MPa

Ultimate Tensile Strength 483 MPa

Poissons Ratio 0.33

Shear Modulus 28 GPa

Modulus of Elasticity 73.1 GPa

4.2 Calculation of Wing Section properties :

As we know the wing is tapered so the mass per unit length varies accordingly apart from the

discontinuities arising because of ribs. A typical wing box structure was designed with 2 spars, 8 longerons

(up to 9 m along the span) and skin in SolidWorks. General wing box structure for different aircraft designs

was studied from Ref for this purpose. The cross-sectional characteristics of spars, longerons and skin were

assumed for the root section. Continuous reduction in area for all the components (except skin) proportional

to the taper was assumed as observed in the SolidWorks snapshot of the model in Fig. Properties of the wing

section that were extracted using SolidWorks tools are:

Coordinates of centroid of the section ( Since material is uniform it is also the CG for the section).

Area moment of Inertia (Ixx , Iyy, Ixy ).

Area of the section.

4.3 Normal and Chordwise load per unit length Calculation:

These are obtained via resolution of sectional lift and drag force along the two directions one

along the chord and other perpendicular to the chord. Former gives chordwise force per unit lenth while

latter gives normal force per unit length as shown below :

Here is angle of attack of the wing.

4.4 Inertia-Gravity load per unit length Calculation:

An initial design of the wing box was set up by locating spars and longerons at proper places in the

wing section. Ribs were distributed along the span with some prior assumptions. For half span, starting from

the root section, first 4 ribs were placed at distance of 0.25 m and the remaining 16 ribs were placed at a

distance of 0.75 m between them till the tip section. Inertia- gravity load per unit length was calculated using

following expression:

where nz is load factor and mi is the mass per unit length at ith station.

4.5 Computation of Shear Force and Moment Distribution along span:

We know that lift, drag and pitching moment act about AC of the wing section while the inertia-

gravity loads act about CG. These loads at each station were computed in initial steps. All these external

forces give rise to the shear forces ( Vz and Vx ), bending moments ( Mz and Mx ) and torsion ( MT )at each

section.

4.6 Net Loads at ith

Wing Bay :

A linear variation of loads per unit length (along span) is assumed between stations i and i+1 on the ith

wing bay. The total load is computed by taking average of load per unit length at 2 stations and mult iplying

it with width of bay.

Normal Force :

Chordwise Force :

Pitching Moment:

Inertia-Gravity Load :

4.7 Idealization of Forces:

This step includes idealization of normal, chordwise and inertia-gravity loads. Here it is assumed

that net force on ith wing bay acts through a point which of course the centroid of force distribution. Three

different centroidal locations were computed for each of the three different loads. It was done for each bay

as a prior step required for computation of shear forces and moments. Locations of the 3 centroid namely A,

B and C corresponding to normal, chordwise and inertia-gravity loads respectively, were calculated with

respect to i+1 section for any ith section. The expressions used for this purpose are as follows:

),,(

),,(

),,(

IGIGIG

CCC

NNN

ZYXC

ZYXB

ZYXA

ii

IG

i

IG

i

IG

i

IG

IG

i

i

zIG

i

zIG

IG

i

i

w

i

wW

ii

w

i

w

i

w

i

w

N

IGNWzxx

i

i

w

i

wW

ci

w

i

w

i

w

i

w

c

WXzz

WXX

IGWzz

YdY

dF

dY

dF

dY

dF

dY

dFY

Y

dY

Fd

dY

FdF

Y

dY

dN

dY

dNN

YdY

dN

dY

dN

dY

dN

dY

dNY

YFYNYVMM

Y

dY

dC

dY

dCC

YdY

dC

dY

dC

dY

dC

dY

dCY

Where

YCYVMM

CVV

FNVV

ZZ

ZZ

Zi

i

IGZiiiiii

i

iiiiii

iii

Ziiii

3

21

2

2

3

21

2

3

21

,

1

1

1

1

1

1

1

1

1

1

1

1

1

4.8 Shear forces and Moments at a Wing Section:

These were calculated at each of the 28 stations set up earlier. The calculations were done starting

from tip to root i.e. i = 1 denotes station at wing tip while i = 28 refers to wing root. For i= 1 i.e. wing

tip the shear forces and moments are zero. This is free end boundary condition similar to that in case of

cantilever beam. The expressions for loads at i+1 station at positive Y face are :

ii

IG

i

IG

i

IG

i

IG

IG

acac

i

wacac

i

w

i

IG

i

IG

c

i

i

ac

i

ac

ac

cWIGNWactt

YdY

dF

dY

dF

dY

dF

dY

dFY

ZZdY

dCZZ

dY

dC

dY

dF

dY

dFZ

Y

dY

dM

dY

dMM

ZCXFXNMMM

ZZ

ZZ

iiii

ZZ

ii

i

iIGZiiiii

3

21

222

3

1

2

1

1

1

1

1

11

1

4.9 Wing Box Idealization:

Wing box was idealized in such a manner that both area and moments of inertia were conserved using

the procedure given in Ref. Spars were idealized into 6 booms each. It was assumed that whole C.S. area of

a longeron was concentrated into single boom. The skin segments remaining between the longerons and

spars were idealized into 2 booms each. Consequently, the portion of wing from root to 9m along span was

idealized to 48 booms cum webas shown in Fig. The remaining portion extending to tip due to absence of

longerons was idealized to 40 booms cum web as shown in Fig.

Coordinates of all the booms, their areas and thickness of we between them is given in following table. For

the wing section without stringers only corresponding booms are removed from the list while all other

values remain same.

Booms X Y Xcentroid Ycentroid Area Thickness 1 0.066145 0.0918 -1.03733 -0.0196 0.000139 0.00075

2 0.24685 0.17745 -0.85662 0.06605 0.000139 0.00075 3 0.313 0.2 -0.79047 0.0886 0.000196 0.003

4 0.3525 0.21 -0.75097 0.0986 7.24E-05 0.00075

5 0.4605 0.2378 -0.64297 0.1264 7.24E-05 0.00075 6 0.54964 0.25561 -0.55383 0.14421 0.001188 0.01

7 0.68526 0.27639 -0.41821 0.16499 0.001188 0.01 8 0.7738 0.28738 -0.32967 0.17598 0.000148 0.00164

9 0.8801 0.29662 -0.22337 0.18522 0.000148 0.00164 10 0.919 0.3 -0.18447 0.1886 0.000196 0.003

11 0.98197 0.29894 -0.1215 0.18754 0.000238 0.00164

12 1.15402 0.29606 0.05055 0.18466 0.000238 0.00164 13 1.217 0.295 0.11353 0.1836 0.000196 0.003

14 1.27955 0.28627 0.17608 0.17487 0.000239 0.00164 15 1.45045 0.26243 0.34698 0.15103 0.000239 0.00164

16 1.513 0.2537 0.40953 0.1423 0.000196 0.003 17 1.56815 0.24151 0.46468 0.13011 0.000214 0.00164

18 1.71884 0.2082 0.61537 0.0968 0.000214 0.00164

19 1.8059 0.1877 0.70243 0.0763 0.00078 0.01 20 1.8931 0.1652 0.78963 0.0538 0.00078 0.01

21 2.040172 0.15458 0.936702 0.04318 0.000463 0.00164 22 2.35483 0.04142 1.25136 -0.06998 0.000463 0.00164

23 2.35652 -0.00613 1.25305 -0.11753 0.00043 0.00164 24 2.04648 -0.02287 0.94301 -0.13427 0.00043 0.00164

25 1.8994 -0.03424 0.79593 -0.14564 0.000805 0.01 26 1.8076 -0.04856 0.70413 -0.15996 0.000805 0.01

27 1.7195 -0.0624 0.61603 -0.1738 9.8E-05 0.00075

28 1.5705 -0.08598 0.46703 -0.19738 9.8E-05 0.00075 29 1.516 -0.0946 0.41253 -0.206 0.000196 0.003

30 1.4528 -0.1023 0.34933 -0.2137 0.000113 0.00075 31 1.2802 -0.1233 0.17673 -0.2347 0.000113 0.00075

32 1.217 -0.131 0.11353 -0.2424 0.000196 0.003 33 1.1536 -0.1335 0.05013 -0.2449 0.000113 0.00075

34 0.9804 -0.1404 -0.12307 -0.2518 0.000113 0.00075

35 0.917 -0.143 -0.18647 -0.2544 0.000196 0.003 36 0.8785 -0.14215 -0.22497 -0.25355 6.83E-05 0.00075

37 0.7734 -0.13984 -0.33007 -0.25124 6.83E-05 0.00075 38 0.6864 -0.136 -0.41707 -0.2474 0.001152 0.01

39 0.5536 -0.128 -0.54987 -0.2394 0.001152 0.01 40 0.4646 -0.12056 -0.63887 -0.23196 7.21E-05 0.00075

41 0.3544 -0.10844 -0.74907 -0.21984 7.21E-05 0.00075

42 0.314 -0.104 -0.78947 -0.2154 0.000196 0.003 43 0.24764 -0.08202 -0.85583 -0.19342 0.000124 0.00075

44 0.0663 -0.02198 -1.03717 -0.13338 0.000124 0.00075 45 0.61 0.1801 -0.49347 0.0687 0.001985 0.01

46 0.61 -0.0491 -0.49347 -0.1605 0.001985 0.01 47 1.8448 0.1297 0.74133 0.0183 0.001094 0.01

48 1.8448 0.0034 0.74133 -0.108 0.001094 0.01

4.10 Normal Stress Computation:

Normal stresses at any wing section arise predominantly due to bending moments. This leads to a particular

portion of wing under compression while other under tension. It is assumed that there is no normal force

acting on any wing station. Also there are no thermal stresses. The normal stress at any boom is computed

using following expression. Here x and y are with respect to centroid of the section as shown in Fig.

4.11 Shear Stress Computation:

4.11.1 Shear Stress due to Shear Forces:

Shear stress at any wing section was calculated by initially calculating shear flow in the idealized section

(procedure given in Ref ).

Imaginary small cut was made in between booms location (6, 7), (19, 20) and (22, 23).

Shear flow (q(0)) with open section acted by Vz and Vx (wrt to body axis) is calculated using the

formulae.

The remaining part of the shear flow (q(1)) which is constant for a particular cell was calculated by

solving three equations.

o The first equation is obtained from equilibrium condition.

o The remaining equations were obtained from the constraints that the section in rigid in its

own plane so twist is equal for all of them.

iix

iix

YY

YX

XX

xy

ii

AxQ

AyQ

I

QV

I

QVqq

1

2211432

3

2 211221

1 2121

1

21

)0(1)0(111111

cellcellwallCELL wallCELL wall t

dsq

At

dsq

At

ds

Aq

t

ds

At

ds

Aq

t

ds

At

ds

Aq

4.11.2 Shear stress due to Torsion:

Torsion generates constant shear flow in each of the cells. Shear flow due to Torsion is calculated by solving

three equation as in earlier case. The first equation is equilibrium equation while other two are derived

through constraints similar to above step. Since material is uniform throughout there is no requirement of

modulus weighted entities.

4.12 Factor of Safety calculations:

Factor of Safety is a term describing the structural capacity of a system beyond the expected loads or actual

loads. Essentially, how much stronger the system is than it usually needs to be for an intended load. FOS

was computed for all the booms at all stations, for critical cases of loading. von Mises yield criterion was

used for this purpose.

von Mises stress ( v ) is calculated using following formula for plane stress condition:

where 1 and 2 are normal stresses and 12 is shear stress for plane stress loading.

FOS is given by :

4.13 Crushing Loads:

Crushing loads on box beams occurs due to curvature developed from bending, which can be ignored for

solid but not for box beams and hence intermediate ribs are required.

Firstly change in Aero loads (wCw, Mac)and inertial + gravity loads (F) are calculated using

formulae

Linear variation of aerodynamic centre (A.C.) is used between all the stations

Using the bending moment distribution crushing force (Pcrush) is calculated directly using

4.14 Buckling of stringers and spars:

Buckling is characterized by a sudden failure of a structural member subjected to high compressive stress,

where the actual compressive stress at the point of failure is less than the ultimate compressive stresses that

the material is capable of withstanding.

Buckling of stringers due to normal stress considering simply supported at each ribs location is

calculated

It is then compared with the critical stress (cr) for the component.

4.15 Buckling of panels:

Buckling failure can occur at reduced primary critical stress levels if the structure is subjected to orthogonal

compressive stresses or high shear stresses.

Buckling of panels due to force Vy (wrt to body axis) considering simply supported at each ribs

location is calculated

Firstly the wing is divided into many rectangular panels (Aspect ratio 3) and some trapezoidal pane l

Then buckling is calculated on these panels using the maximum normal stress and minimum cross-

section are present in the panel

It is then compared with the critical stress (cr) of the material

5. Results and Discussion:

Wing section properties:

Figure 5 plots shown for C.G.,Mass per unit

length and Area Moment of inertia along the

span location.

Normal and chordwise load per unit length:

0 2 4 6 8 10 12 140

1

2

3

4

5

6

7

8

9

10x 10

5

Span Location (y) (in m )

Are

a M

om

ent

of

Inert

ia (

in c

m4)

Ixx

Iyy

Ixy

0 2 4 6 8 10 12 14-2

-1.5

-1

-0.5

0

0.5

1

Span Location (y) (in m )along AC

Dis

tance (

in m

)

XCG

Leading Edge

Trailing Edge

0 2 4 6 8 10 12 140

500

1000

1500

2000

2500

Span Location (y)

Mass/length

(in

kg/m

)

0 2 4 6 8 10 12 14-2000

-1500

-1000

-500

0

500

1000

Span Location (y)

dC

/dy (

in N

/m)

n = 3.19,v = 63.4 m/s

n = 3.19,v = 137.5 m/s

0 2 4 6 8 10 12 141000

2000

3000

4000

5000

6000

7000

8000

9000

10000

Span Location (y)

dN

/dy (

in N

/m)

n = 3.19,v = 63.4 m/s

n = 3.19,v = 137.5 m/s

0 2 4 6 8 10 12 14-1000

-500

0

500

1000

1500

Span Location (y)

dC

/dy (

in N

/m)

n = 2.13,v = 109.7 m/s

n = 1.2,v = 38.9 m/s

n = 1.7,v = 137.5 m/s

n = 0.62,v = 28 m/s

0 2 4 6 8 10 12 140

1000

2000

3000

4000

5000

6000

7000

Span Location (y)

dN

/dy (

in N

/m)

n = 2.13,v = 109.7 m/s

n = 1.2,v = 38.9 m/s

n = 1.7,v = 137.5 m/s

n = 0.62,v = 28 m/s

0 2 4 6 8 10 12 14-1000

-900

-800

-700

-600

-500

-400

-300

-200

-100

0

Span Location (y)

dC

/dy (

in N

/m)

n = 1.6,v = 45 m/s

n = 1.86,v = 63 m/s

n = 1.44,v = 42 m/s

0 2 4 6 8 10 12 14-3500

-3000

-2500

-2000

-1500

-1000

-500

0

Span Location (y)

dN

/dy (

in N

/m)

n = -1.27,v = 110 m/s

n = -1.27,v = 55.1 m/s

n = -0.125,v = 109.7 m/s

0 2 4 6 8 10 12 140

200

400

600

800

1000

1200

1400

Span Location (y)

dC

/dy (

in N

/m)

n = -1.27,v = 110 m/s

n = -1.27,v = 55.1 m/s

n = -0.125,v = 109.7 m/s

0 2 4 6 8 10 12 14500

1000

1500

2000

2500

3000

3500

4000

4500

5000

5500

Span Location (y)

dN

/dy (

in N

/m)

n = 1.6,v = 45 m/s

n = 1.86,v = 63 m/s

n = 1.44,v = 42 m/s

Figure 6 : Normal and chord force variation along the span location

0 2 4 6 8 10 12 140

500

1000

1500

2000

2500

Span Location (y)

dN

/dy (

in N

/m)

n = 0.14,v = 63.9 m/s

n = 0.7,v = 29.6 m/s

n = 0.83,v = 32.4 m/s

n = 0.29,v = 137.5 m/s

0 2 4 6 8 10 12 14-500

0

500

1000

1500

2000

Span Location (y)

dC

/dy (

in N

/m)

n = 0.14,v = 63.9 m/s

n = 0.7,v = 29.6 m/s

n = 0.83,v = 32.4 m/s

n = 0.29,v = 137.5 m/s

Inertia-Gravity load per unit length:

Figure 7: Plot shows the variation of Inertia

gravity load per unit length with the span locat ion.

0 2 4 6 8 10 12 140

100

200

300

400

500

600

700

Span Location (y)

dF

IGz/

dy (

in N

/m)

n = -1.27,v = 110 m/s

n = -1.27,v = 55.1 m/s

n = -0.125,v = 109.7 m/s

0 2 4 6 8 10 12 14-1800

-1600

-1400

-1200

-1000

-800

-600

-400

-200

0

Span Location (y)

dF

IGz/

dy (

in N

/m)

n = 3.19,v = 63.42 m/s

n = 3.19,v = 137.5 m/s

0 2 4 6 8 10 12 14-450

-400

-350

-300

-250

-200

-150

-100

-50

0

Span Location (y)

dF

IGz/

dy (

in N

/m)

n = 0.14,v = 63.9 m/s

n = 0.7,v = 29.6 m/s

n = 0.83,v = 32.4 m/s

n = 0.29,v = 137.5 m/s

0 2 4 6 8 10 12 14-1200

-1000

-800

-600

-400

-200

0

Span Location (y)

dF

IGz/

dy (

in N

/m)

n = 2.13,v = 109.7 m/s

n = 1.2,v = 38.9 m/s

n = 1.7,v = 137.5 m/s

n = 0.62,v = 28 m/s

0 2 4 6 8 10 12 14-1000

-900

-800

-700

-600

-500

-400

-300

-200

-100

0

Span Location (y)

dF

IGz/

dy (

in N

/m)

n = 1.6,v = 45 m/s

n = 1.86,v = 63 m/s

n = 1.44,v = 42 m/s

Shear Force and Moment

distribution:

Figure 8: Shear force and Moment distribution

shown along the span location

0 2 4 6 8 10 12 14

-12000

-10000

-8000

-6000

-4000

-2000

0

Span Location (y)

MT (

in N

m)

n = 2.13,v = 109.7 m/s

n = 1.2,v = 38.9 m/s

n = 1.7,v = 137.5 m/s

n = 0.62,v = 28 m/s

0 2 4 6 8 10 12 14-8

-6

-4

-2

0

2

4x 10

4

Span Location (y)

Mz

(in N

m)

n = 2.13,v = 109.7 m/s

n = 1.2,v = 38.9 m/s

n = 1.7,v = 137.5 m/s

n = 0.62,v = 28 m/s

0 2 4 6 8 10 12 140

0.5

1

1.5

2

2.5

3x 10

5

Span Location (y)

Mx

(in N

m)

n = 2.13,v = 109.7 m/s

n = 1.2,v = 38.9 m/s

n = 1.7,v = 137.5 m/s

n = 0.62,v = 28 m/s

0 2 4 6 8 10 12 140

1

2

3

4

5

6x 10

4

Span Location (y)

Vz

(in N

)

n = 2.13,v = 109.7 m/s

n = 1.2,v = 38.9 m/s

n = 1.7,v = 137.5 m/s

n = 0.62,v = 28 m/s

0 2 4 6 8 10 12 14-8000

-6000

-4000

-2000

0

2000

4000

6000

8000

10000

12000

Span Location (y)

Vx

(in N

)

n = 2.13,v = 109.7 m/s

n = 1.2,v = 38.9 m/s

n = 1.7,v = 137.5 m/s

n = 0.62,v = 28 m/s



0 2 4 6 8 10 12 140

1

2

3

4

5

6x 10

4

Span Location (y)

Mz

(in N

m)

n = 1.6,v = 45 m/s

n = 1.86,v = 63 m/s

n = 1.44,v = 42 m/s

0 2 4 6 8 10 12 140

0.5

1

1.5

2

2.5

3x 10

5

Span Location (y)

Mx

(in N

m)

n = 1.6,v = 45 m/s

n = 1.86,v = 63 m/s

n = 1.44,v = 42 m/s

0 2 4 6 8 10 12 14-8000

-7000

-6000

-5000

-4000

-3000

-2000

-1000

0

Span Location (y)M

T (

in N

m)

n = 1.6,v = 45 m/s

n = 1.86,v = 63 m/s

n = 1.44,v = 42 m/s

0 2 4 6 8 10 12 140

0.5

1

1.5

2

2.5

3

3.5

4

4.5x 10

4

Span Location (y)

Vz

(in N

)

n = 1.6,v = 45 m/s

n = 1.86,v = 63 m/s

n = 1.44,v = 42 m/s

0 2 4 6 8 10 12 14-10000

-9000

-8000

-7000

-6000

-5000

-4000

-3000

-2000

-1000

0

Span Location (y)

Vx

(in N

)

n = 1.6,v = 45 m/s

n = 1.86,v = 63 m/s

n = 1.44,v = 42 m/s



0 2 4 6 8 10 12 14-2

-1.5

-1

-0.5

0

0.5

1x 10

4

Span Location (y)

Vx

(in N

)

n = 3.19,v = 63.42 m/s

n = 3.19,v = 137.5 m/s

0 2 4 6 8 10 12 14-4

-2

0

2

4

6

8

10

12x 10

4

Span Location (y)

Mz

(in N

m)

n = 3.19,v = 63.42 m/s

n = 3.19,v = 137.5 m/s

0 2 4 6 8 10 12 14-18000

-16000

-14000

-12000

-10000

-8000

-6000

-4000

-2000

0

Span Location (y)

MT (

in N

m)

n = 3.19,v = 63.42 m/s

n = 3.19,v = 137.5 m/s

0 2 4 6 8 10 12 140

0.5

1

1.5

2

2.5

3

3.5

4

4.5x 10

5

Span Location (y)

Mx

(in N

m)

n = 3.19,v = 63.42 m/s

n = 3.19,v = 137.5 m/s

0 2 4 6 8 10 12 140

1

2

3

4

5

6

7

8x 10

4

Span Location (y)

Vz

(in N

)

n = 3.19,v = 63.42 m/s

n = 3.19,v = 137.5 m/s

Figure 11: Shear force and Moment

distribution shown along the span location

0 2 4 6 8 10 12 14-3

-2.5

-2

-1.5

-1

-0.5

0x 10

4

Span Location (y)

Vz

(in N

)

n = -1.27,v = 110 m/s

n = -1.27,v = 55.1 m/s

n = -0.125,v = 109.7 m/s

0 2 4 6 8 10 12 14-7

-6

-5

-4

-3

-2

-1

0x 10

4

Span Location (y)

Mz

(in N

m)

n = -1.27,v = 110 m/s

n = -1.27,v = 55.1 m/s

n = -0.125,v = 109.7 m/s

0 2 4 6 8 10 12 14-18

-16

-14

-12

-10

-8

-6

-4

-2

0x 10

4

Span Location (y)

Mx

(in N

m)

n = -1.27,v = 110 m/s

n = -1.27,v = 55.1 m/s

n = -0.125,v = 109.7 m/s

0 2 4 6 8 10 12 140

2000

4000

6000

8000

10000

12000

Span Location (y)

Vx

(in N

)

n = -1.27,v = 110 m/s

n = -1.27,v = 55.1 m/s

n = -0.125,v = 109.7 m/s

0 2 4 6 8 10 12 14-1600

-1400

-1200

-1000

-800

-600

-400

-200

0

Span Location (y)

MT (

in N

m)

n = -1.27,v = 110 m/s

n = -1.27,v = 55.1 m/s

n = -0.125,v = 109.7 m/s



0 2 4 6 8 10 12 14-10

-8

-6

-4

-2

0

2

4x 10

4

Span Location (y)

Mz

(in N

m)

n = 0.14,v = 63.9 m/s

n = 0.7,v = 29.6 m/s

n = 0.83,v = 32.4 m/s

n = 0.29,v = 137.5 m/s

0 2 4 6 8 10 12 140

2

4

6

8

10

12x 10

4

Span Location (y)

Mx

(in N

m)

n = 0.14,v = 63.9 m/s

n = 0.7,v = 29.6 m/s

n = 0.83,v = 32.4 m/s

n = 0.29,v = 137.5 m/s

0 2 4 6 8 10 12 14-3000

-2500

-2000

-1500

-1000

-500

0

Span Location (y)

MT (

in N

m)

n = 0.14,v = 63.9 m/s

n = 0.7,v = 29.6 m/s

n = 0.83,v = 32.4 m/s

n = 0.29,v = 137.5 m/s

0 2 4 6 8 10 12 140

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2x 10

4

Span Location (y)

Vz

(in N

)

n = 0.14,v = 63.9 m/s

n = 0.7,v = 29.6 m/s

n = 0.83,v = 32.4 m/s

n = 0.29,v = 137.5 m/s

0 2 4 6 8 10 12 14-5000

0

5000

10000

15000

Span Location (y)

Vx

(in N

)

n = 0.14,v = 63.9 m/s

n = 0.7,v = 29.6 m/s

n = 0.83,v = 32.4 m/s

n = 0.29,v = 137.5 m/s

Normal stress distribution along span:

Figure 13: Normal stress distribution shown along the span location for booms 26 and 35

Shear stress distribution along span:

0 2 4 6 8 10 12 140

0.5

1

1.5

2

2.5x 10

8

Span Location (y)

Norm

al S

tress (

in P

a)

Boom no. 26

n = 3.19,v = 63.42 m/s

n = 3.19,v = 137.5 m/s

n = 0.29,v = 137.5 m/s

0 2 4 6 8 10 12 140

0.5

1

1.5

2

2.5

3x 10

8

Span Location (y)

Norm

al S

tress (

in P

a)

Boom no. 35

n = 3.19,v = 63.42 m/s

n = 3.19,v = 137.5 m/s

n = 0.29,v = 137.5 m/s

0 2 4 6 8 10 12 14-5

-4

-3

-2

-1

0

1

2

3

4x 10

7

Span Location (y)

Shear

Str

ess d

ue t

o T

ors

ion (

in P

a)

Boom no. 23

n = 3.19,v = 63.42 m/s

n = 3.19,v = 137.5 m/s

n = 0.29,v = 137.5 m/s

0 2 4 6 8 10 12 14-12

-10

-8

-6

-4

-2

0

2

4

6

8x 10

7

Span Location (y)

Shear

Str

ess d

ue t

o s

hear

forc

e(in P

a)

Boom no. 23

n = 3.19,v = 63.42 m/s

n = 3.19,v = 137.5 m/s

n = 0.29,v = 137.5 m/s

Figure 14: Shear stress distribution due to Torsion ,

shear force and Total are shown along the span

location for booms 23 and 9

0 2 4 6 8 10 12 14-3

-2

-1

0

1

2

3x 10

7

Span Location (y)

Shear

Str

ess d

ue t

o T

ors

ion (

in P

a)

Boom no. 9

n = 3.19,v = 63.42 m/s

n = 3.19,v = 137.5 m/s

n = 0.29,v = 137.5 m/s

0 2 4 6 8 10 12 14-2

-1.5

-1

-0.5

0

0.5

1

1.5x 10

8

Span Location (y)

Shear

Str

ess T

ota

l (in P

a)

Boom no. 23

n = 3.19,v = 63.42 m/s

n = 3.19,v = 137.5 m/s

n = 0.29,v = 137.5 m/s

0 2 4 6 8 10 12 14-1.5

-1

-0.5

0

0.5

1x 10

8

Span Location (y)

Shear

Str

ess d

ue t

o s

hear

forc

e(in P

a)

Boom no. 9

n = 3.19,v = 63.42 m/s

n = 3.19,v = 137.5 m/s

n = 0.29,v = 137.5 m/s

0 2 4 6 8 10 12 14-2

-1.5

-1

-0.5

0

0.5

1

1.5x 10

8

Span Location (y)

Shear

Str

ess T

ota

l (in P

a)

Boom no. 9

n = 3.19,v = 63.42 m/s

n = 3.19,v = 137.5 m/s

n = 0.29,v = 137.5 m/s

Crushing load variation:

Figure 15 : Variation of crushing load with span location is shown for 4 different booms 6,7,19,20

0 2 4 6 8 10 12 140

50

100

150

200

250

300

350

Span Location (y)

Cru

shin

g load (

in N

)Boom no. 20

n = 3.19,v = 63.42 m/s

n = 3.19,v = 137.5 m/s

n = 0.29,v = 137.5 m/s

0 2 4 6 8 10 12 140

100

200

300

400

500

600

700

800

Span Location (y)

Cru

shin

g load (

in N

)

Boom no. 19

n = 3.19,v = 63.42 m/s

n = 3.19,v = 137.5 m/s

n = 0.29,v = 137.5 m/s

0 2 4 6 8 10 12 140

500

1000

1500

2000

2500

3000

3500

4000

Span Location (y)

Cru

shin

g load (

in N

)

Boom no. 7

n = 3.19,v = 63.42 m/s

n = 3.19,v = 137.5 m/s

n = 0.29,v = 137.5 m/s

0 2 4 6 8 10 12 140

500

1000

1500

2000

2500

3000

3500

Span Location (y)

Cru

shin

g load (

in N

)

Boom no. 6

n = 3.19,v = 63.42 m/s

n = 3.19,v = 137.5 m/s

n = 0.29,v = 137.5 m/s

Buckling of Panels:

Figure 16 : Buckling of Panels with station

location is shown

Buckling of stringers:

Figure 17 : Buckling of stringers with station

location is shown

0 1 2 3 4 5 6 7 8 90

0.5

1

1.5

2

2.5

3

3.5

Span Location (y)

(Critical str

ess/N

orm

al str

ess)

ratio

n = 3.19,v = 63.42 m/s

0 1 2 3 4 5 6 7 8 90

1

2

3

4

5

6

7

8

9

Span Location (y)

(Critical str

ess/N

orm

al str

ess)

ratio

n = 3.19,v = 137.5 m/s

0 1 2 3 4 5 6 7 8 90

2

4

6

8

10

12

Span Location (y)

(Critical str

ess/N

orm

al str

ess)

ratio

n = 0.29,v = 137.5 m/s

0 1 2 3 4 5 6 7 8 91

1.5

2

2.5

3

3.5

4

4.5

5

5.5

6

Span Location (y)

(Critical str

ess/N

orm

al str

ess)

ratio

n = 3.19,v = 63.5 m/s

0 1 2 3 4 5 6 7 8 95

10

15

20

25

30

35

40

Span Location (y)

(Critical str

ess/N

orm

al str

ess)

ratio

n = 0.295,v = 137.5 m/s

0 1 2 3 4 5 6 7 8 90

0.5

1

1.5

2

2.5

3

3.5

Span Location (y)

(Critical str

ess/N

orm

al str

ess)

ratio

n = 3.19,v = 63.42 m/s

0 1 2 3 4 5 6 7 8 90

1

2

3

4

5

6

7

8

9

Span Location (y)

(Critical str

ess/N

orm

al str

ess)

ratio

n = 3.19,v = 137.5 m/s

Factor of safety:

Figure 18: Factor

of safety is plotted

for n= 0.29,

v=137.5

Figure 19: Factor

of safety is plotted

for n=3.19, v=63.5

Figure 20: Factor of

safety is plotted for

n=3.19,v=137.5

Wing Section Properties:

Plot for mass distribution for wing box along span is shown in Fig5. The kinks at regular distances show

position of ribs along span. Location of CG for any wing section along span is plotted with respect to locus

AC ( y = 0 ) in Fig5. Variation of CS moments of inertia for wing box (without ribs) is shown in Fig5. It can

be observed that Iyy is significantly higher than other two.

Normal and Chordwise Loads for Critical Loading Conditions:

The normal and chordwise load per unit length variation along span is plotted in Fig6. Both decrease from

root to tip. The normal force per unit length is positive or negative depending upon whether load factor is

positive or negative respectively. While the chord force per unit length doesnt follow such behaviour as it

can be seen in Fig6. the negative values correspond to the PHAA flight condition in V-n diagram where

contribution due to lift force dominates over drag force. The reason for this difference is that lift force

contributes predominantly to normal force in all cases while in case of chord force it depends on the angle of

attack whether lift or drag force is dominant. High non- linear behaviour towards the tip basically for normal

force is because of similar behaviour of lift. It is also important to notice that in many case shown in Fig6.

for chord force, the variation is nearly linear. The reason can be attributed to the dominance of drag force

contribution and also within that predominance of parasite drag.

Variation of Inertia-Gravity Loads:

Inertia gravity loads depend on the load factor and mass per unit length of a wing section. Thus for a

particular load condition it decreases from root to tip proportionally to the reducing mass per unit length of

tapered wing. Overlap of the plots can be seen in case of PHAA and PLAA flight conditions from Fig.7

Variation of Shear Forces and Moments:

Shear forces and moments increase from tip to root. As a result root section handles highest magnitude of

internal forces and moments. Shear force along normal (Vz) is positive or negative depending on the

behaviour of normal force which dominates or the sign of load factor. While for chordwise shear force

shows different behaviours depending on variation of chord force. In case of PHAA this is negative because

of dominance of lift component over that of drag. The highest shear forces and moments are obtained at root

section corresponding to PHAA and PLAA points of the flight envelope. One more point at n = 0.29,

V=137.5 m/s shows peak load resultants at root. These 3 points were used to further analyse the different

aspects of the design and are used for plotting as shown in Fig 8, 9, 10 , 11, 12.

Variation of Normal and Shear stress:

It is observed that at the root section the magnitude of Normal as well as shear stress is highest. The plots as

shown in Fig. 13, 14 are made for three different values of n and v namely n= 3.19, v=63.5, n= 3.19,

v=137.5, n= 0.29, v=137.5 for two booms having no. 23 and 9 in case of shear stress for all the three cases only due to shear force, only due to Torsion and due to both, while in case of Normal stress the plots

are made for booms having no. 26 and 52, these booms number for both shear as well as normal stress case

have the most critical values and are selected out of all the 48 booms.

Variation of Crushing load:

Crushing load is plotted for 4 critical booms namely 6, 7, 19, 20 for the three critical n and v values namely

n= 3.19, v=63.5, n= 3.19, v=137.5, n= 0.29, v=137.5 as shown in Fig 15. There are 4 plots

corresponding for each boom with all three n values together in each plot. As expected the crushing load

values decreases from the root to tip and having the maximum value at the root for all the cases. Because of

the low n value crushing load values for the case n=0.29 is very less compared to the remaining cases.

Variation of Buckling of Panels:

It is observed from the graph that the panels for which the value of critical stress/Normal stress is less than 1

buckles but other panels do not buckle having the ratio greater than 1. The analysis is done for all the three

critical n values namely n= 3.19, v=63.5, n= 3.19, v=137.5, n= 0.29, v=137.5 as shown in Fig 16.

Variation of Buckling of stringers:

It is observed from the graph that the ratio of critical stress/Normal stress is greater than 1 for all the cases

and hence concludes that none of the stringer buckles. The analysis is done for all the three critical n cases

as shown in Fig 17.

Variation of Factor of safety:

Factor of safety for all the booms at each station was calculated using von Mises criterion. It can be

observed through the scatter plots given in Fig.18, 19, 20. The two sets of plot show first and last stages of

iterations done to reduce the weight of wing box structure. Initially FOS was very high for almost all the

booms along the span. After iterations the weight was significantly reduced basically due to faster tapering

of spars along the span.

6.Iterations for Weight Reduction:

A number of iteration were done to reduce the weight of current wing box design. Some of the steps

pertaining to the procedure include reduction of skin thickness for some portion of the wing while increasing

at those locations where we encountered shear failure. Another step included faster tapering of spars along

the span. This helped to reduce wing weight considerably while maintaining the strength of spars to requisite

level. Comparisons of wing weight estimated through current design procedure is done with that obtained

through empirical relation given in literature is as follows.

Empirical relation estimate for wing weight = 428.89 kg

Initial Design:

Wing weight = 701.72 kg

Percentage difference w.r.t theoretical = 63.6 %

Improved Design after few Iterations:

Wing weight = 640 kg

Percentage difference w.r.t theoretical = 49.2 %

7. Summary:

Present work is dedicated to understand important techniques and procedures involved in design of aircraft

wing box. Here the design was specifically done for a UAV aircraft (Drone). The whole work can be

summarized into following steps:

Initial Take Off weight estimate, wing design features and aerofoil were directly taken from earlier

design work.

V-n Diagram and Gust Load Factor Diagram were drawn according to current mission requirements.

16 critical points were chosen specifically for further analysis.

Wing was partitioned into 27 wing bays and hence 28 stations. 20 ribs were placed in a half span of

the wing.

Lift and drag per unit length was calculated along the span for the 16 points. Normal and chordwise

forces were further obtained.

Wing section properties were obtained using both SolidWorks and MatLab.

Inertia gravity loads were estimated.

Wing was idealized with 48 booms for 1st part and 40 for second.

Shear forces and moments were obtained at the stations.

Resultant Normal and Shear stresses were computed.

Critical booms were analysed.

FOS was plotted for all the booms.

Wing weight was estimated and compared with the theoretical estimate.

Iterations were done to reduce the wing weight and finalize the design.

Appendix:

%% Name of each item is mentioned above its part in the code.

y =

[0,1,2,3,4,5,5.5,6,6.5,7,7.5,8,8.5,9,9.5,10,10.25,10.5,10.75,11,11.25,11.5,11.75,12,12.

25,12.5,12.75,13]; ynew =

[13,12.75,12.5,12.25,12,11.75,11.5,11.25,11,10.75,10.5,10.25,10,9.5,9,8.5,8,7.5,7,6.5,6

,5.5,5,4,3,2,1,0]; S = 45; b = 26; t = 0.4; d = 1.225; %densit W = 5400; ar = 15; e = 0.9; Cd1 = 0.05815; points3D = zeros(48,3,28);

n = [3.19,3.19,-1.27,-1.27,-

0.1256,1.61,1.865,1.436,2.1256,1.199,1.705,0.6213,0.135,0.696,0.8338,0.2947]; v =

[63.425832,137.498328,109.999272,55.065168,109.7219,45.01408,63.8885,42.52112,109.7215,

38.8559,137.5029,27.96631,63.8885,29.6058,32.3984,137.4989]; CL = [1.55675,0.329,-0.19675,-0.79,-

0.02425,1.55975,0.8915,1.559125,0.343875,1.559,0.174125,1.55938,0.06138,1.5588,1.55938,

0.02725]; Cm = [-0.0224,-0.0294,-0.0098,-0.0271, -0.00087,-0.0221,-0.0511,-0.0221,-0.0296,-

0.0222,-0.017,-0.0221, -0.0075, -0.0222, -0.0221, -0.0053]; alpha = [16.51975806,6.618548387,2.378629032,-2.405645161,

3.7697,16.54395,11.15484,16.53891,6.7385,16.5379,5.36955,16.5409, 4.4603, 16.5359,

16.5409, 4.1851];

Cd = 0;

den = 2720; c = zeros(1,28); m1 = zeros(1,28); m2 = zeros(1,28);

x = zeros(1,28);

L = zeros(1,28); D = zeros(1,28); M = zeros(1,28);

N = zeros(1,28); C = zeros(1,28);

dN = zeros(1,27); dC = zeros(1,27);

dM = zeros(1,27); dF = zeros(1,27);

F = zeros(1,28); yC = zeros(1,27); Fz = zeros(1,27); Mass = zeros(1,27); dy = zeros(1,27);

Vx = zeros(1,28); Vz = zeros(1,28);

Mz = zeros(1,28); Mx = zeros(1,28); MT = zeros(1,28);

X = zeros(3,27);

Y = zeros(3,27); Z = zeros(3,27);

Ixx = zeros(1,28); Iyy = zeros(1,28); Ixy = zeros(1,28);

sigmaxx = zeros(48,28);

for i = 1:27 dy(i) = y(i+1)-y(i); end

for j = 1

Vz(1) = 0; Vx(1) = 0; Mz(1) = 0; Mx(1) = 0; MT(1) = 0;

for i = 1:28

c(1,i) = (2*S/((1+t)*b))*(1-(2*(1-t)*y(i)/b)); if i < 15 m1(1,i) = (5.1015*0.001175*((c(1,i)/2.47)^2))*den; %skin

Thickness - 0.02" m2(1,i) = (( 0.008363+0.005134)*((c(1,i)/2.47)^3)+(0.00003*(c(1,i)^2)))*den; else m1(1,i) = (5.1015*0.001175*((c(1,i)/2.47)^2))*den; m2(1,i) = ( ( 0.008363+0.005134)*((c(1,i)/2.47)^3))*den; end F(1,i) = -n(j)*9.8*(m1(1,i)+m2(1,i));

L(1,i) = (1/2)*(c(i) + (4*S/(pi()*b))*(1-

(2*y(i)/b)^2)^(1/2))*(CL(j)*(0.5*d*v(j)^2)) ; Cl = L(1,i)/(0.5*d*c(i)*(v(j)^2));

%Section lift coeff Cd = Cd1 + Cl*CL(j)/(pi()*ar*e); D(1,i) = Cd*(0.5*d*c(i)*(v(j)^2)); M(1,i) = Cm(j)*(0.5*d*(c(i)^2)*(v(j)^2));

N(i) = L(i)*cosd(alpha(j))+D(i)*sind(alpha(j)); C(i) = D(i)*cosd(alpha(j))-L(i)*sind(alpha(j));

x(i) = -(1.10347*c(i)/2.47-(c(i)/4));

if ( c(i) < 9 ) Ixx(i) =

(((4312.505*1.6)*((c(i)/2.47)^4))+((28798.823+4882.13)*((c(i)/2.47)^5)))*10^(-8); Iyy(i) =

(((284293.1384*1.6)*((c(i)/2.47)^4))+((434706.159+18393.025)*((c(i)/2.47)^5)))*10^(-8); Ixy(i) = (((19714.5*1.6)*((c(i)/2.47)^4))+((-4123.445-

350)*((c(i)/2.47)^5)))*10^(-8);

else Ixx(i) = (((4312.505*1.6)*((c(i)/2.47)^4))+((28798.823)*(((c(i)/2.47)^5))))*10^(-

8); Iyy(i) =

(((284293.1384*1.6)*((c(i)/2.47)^4))+((434706.159)*((c(i)/2.47)^5)))*10^(-8); Ixy(i) = (((19714.5*1.6)*((c(i)/2.47)^4))+((-4123.445)*((c(i)/2.47)^5)))*10^(-8);

end

end

for i = 1:27

dN(i) = ((N(i)+N(i+1))*dy(i))/2; dC(i) = ((C(i)+C(i+1))*dy(i))/2; dM(i) = ((M(i)+M(i+1))*dy(i))/2; dF(i) = ((F(i)+F(i+1))*dy(i))/2; end % % Mt = 0; % for j = 1:27 % yC(1,j) = (1/2)*(y(j) + y(j+1)); % Fz(1,j) = (1/2)*(y(j+1) - y(j))*(F(1,j)+F(1,j+1)); % Mass(1,j) = (1/2)*(m(j) + m(j+1)); % Mt = Mt + Mass(1,j)*(y(j+1) - y(j)); % end % for i=1:27 Y(1,i) = (1/(N(28-i+1)+N(28-i)))*(N(28-i+1)+2*N(28-i))*(dy(28-i)/3);

% tip to root Y(2,i) = (1/(C(28-i+1)+C(28-i)))*(C(28-i+1)+2*C(28-i))*(dy(28-i)/3); Y(3,i) = (1/(F(28-i+1)+F(28-i)))*(F(28-i+1)+2*F(28-i))*(dy(28-i)/3);

X(3,i) = (1/(3*(F(28-i+1)+F(28-i))))*((F(28-i+1)*(2*x(28-i+1)+x(28-i)))+(F(28-

i)*(2*x(28-i)+x(28-i+1)))); X(1,i) = 0; X(2,i) = 0;

Z(2,i) = 0; Z(1,i) = 0; end

for i=1:27 Vz(i+1) = Vz(i)+ dN(28-i)+dF(28-i); % i here

is from tip to root but earlier i is from root to tip Vx(i+1) = (Vx(i) + dC(28-i)); Mz(i+1) = Mz(i) +Vx(i)*dy(28-i)+dC(28-i)*Y(2,i); Mx(i+1) = (Mx(i) - Vz(i)*dy(28-i)-dN(28-i)*Y(1,i)-dF(28-i)*Y(3,i)); MT(i+1) = MT(i) + dM(28-i)-dF(28-i)*X(3,i); % here

two terms are zero end

Point1 = zeros(48,6); Crushload = zeros(4,28);

%% Calculation of Normal stress and crush load

for k = 1:48 for i = 1:28 Point1(:,3) = Point(:,3)*c(i)/2.47; Point1(:,4) = Point(:,4)*c(i)/2.47; points3D(:,1,i) = Point1(:,3); points3D(:,2,i) = Point1(:,4); points3D(:,3,i) = y(i); sigmaxx(k, i) = (- ( Mz(28-i+1)*Ixx(i)+ Mx(28-i+1)*Ixy(i) )*(Point1(k,3)) +(

Mx(28-i+1)*Iyy(i)+ Mz(28-i+1)*Ixy(i))*(Point1(k,4)))/(Ixx(i)*Iyy(i)-Ixy(i)*Ixy(i));

Crushload(1,i) = (2*0.001*(10^(-9))*sigmaxx(6,i)^2)/(2*Point1(6,4)*73.1); Crushload(2,i) = (2*0.001*(10^(-9))*sigmaxx(7,i)^2)/(2*Point1(7,4)*73.1); Crushload(3,i) = (2*0.001*(10^(-9))*sigmaxx(19,i)^2)/(2*Point1(19,4)*73.1); Crushload(4,i) = (2*0.001*(10^(-9))*sigmaxx(20,i)^2)/(2*Point1(20,4)*73.1);

end

end

%% Buckling of stringers

Ixxst = zeros(1,15); Areast = zeros(1,15); lengthst = zeros(1,14);

for i=1:14 Ixxst(i) = (4.9*((c(i))^4)/((2.47)^4))*10^(-8); Areast(i) = (1.96*((c(i))^2)/((2.47)^2))*10^(-4); lengthst(i) = y(i+1)-y(i);

end

lengthst(1) = 0.6998*lengthst(1); lengthst(14) = 1.5*lengthst(14); ratiost = zeros(1,14); sigmacr = zeros(1,14);

for l= 1:14 sigmacr(l) = (pi()^2*73.1*(10)^9)*Ixxst(l+1)/(Areast(l+1)*lengthst(l)^2); ratiost(l) = sqrt((sigmacr(l)/sigmaxx(18,l))^2); end

xr = zeros(1,14); xr = y(1:14); figure(1) plot(xr,ratiost(:),'ko-'); XLABEL('Span Location (y)','fontsize',15) YLABEL('(Critical stress/Normal stress) ratio','fontsize',15) h_legend = legend('n = 3.19,v = 63.5 m/s'); set(h_legend,'FontSize',13); set(gca,'fontsize',13)

%% Calculation of Shear stress separately due to shear force and Torsion

for r = 1:28 for p=1:44 if (p ~= 3)&&(p ~= 6)&&(p ~= 7)&&(p ~= 10)&&(p ~= 13)&&(p ~= 16)&&(p ~= 19)&&(p

~= 20)&&(p ~= 21)&&(p ~= 25)&&(p ~= 26)&&(p ~= 29)&&(p ~= 32)&&(p ~= 35)&&(p ~= 38)&&(p

~= 39)&&(p ~= 42) sigmaxx(p,r) = sigmaxx(p,r)/10; end end end

qo = zeros(52,28); qs1o = zeros(3,28); % down to up qs2o = zeros(3,28);

ds1 = zeros(14,28); ds2 = zeros(30,28); ds3 = zeros(8,28);

Aij = zeros(52,28);

sumcell1 = zeros(14,28); sumcell2 = zeros(30,28); sumcell3 = zeros(8,28);

sumcellq1 = zeros(14,28);

sumcellq2 = zeros(30,28); sumcellq3 = zeros(8,28);

sumqA = zeros(51,28);

wall1 = zeros(1,28); wall2 = zeros(1,28); A1 = zeros(1,28); A2 = zeros(1,28); A3 = zeros(1,28);

Totalq = zeros(52,28); shear = zeros(52,28); Totalq1 = zeros(52,28);

shear1 = zeros(52,28);

shearT = zeros(52,28);

sigma1 = zeros(52,28); sigma2 = zeros(52,28); Tau = zeros(52,28);

generalstress = zeros(48,28); Factorofsafety = zeros(48,28);

Q = zeros(3,1); Q1 = zeros(3,1);

for k=1:28

Point1(:,3) = Point(:,3)*c(k)/2.47; Point1(:,4) = Point(:,4)*c(k)/2.47;

for ii = 1:48 if (p ~= 6)&&(p ~= 7)&&(p ~= 19)&&(p ~= 20)&&(p ~= 25)&&(p ~= 26)&&(p ~= 38)&&(p

~= 39)&&(p ~= 45)&&(p ~= 46)&&(p ~= 47)&&(p ~= 48) Point1(ii,5) = Point(ii,5)*(c(k)/2.47)^2; else Point1(ii,5) = Point(ii,5)*(c(k)/2.47)^(3); end end Point1(:,1) = Point(:,1)*c(k)/2.47; Point1(:,2) = Point(:,2)*c(k)/2.47; Point1(:,6) = Point(:,6)*c(k)/2.47;

if k>14 Point1(3,5) = 0; Point1(10,5) = 0; Point1(13,5) = 0; Point1(16,5) = 0; Point1(29,5) = 0; Point1(32,5) = 0; Point1(35,5) = 0; Point1(42,5) = 0; end for i=0:51 if i==0 qo(i+1,k) = 0 + ((Vx(29-

k)*Point1(Pointer(i+1),3)*Point1(Pointer(i+1),5))/Ixx(k)) + ((Vz(29-

k)*Point1(Pointer(i+1),4)*Point1(Pointer(i+1),5))/Iyy(k)); elseif i>0 && i

qo(i+1,k) = 0 + ((Vx(29-


k)*Point1(Pointer(i+1),4)*Point1(Pointer(i+1),5))/Iyy(k)); elseif i>14 && i44 qo(i+1,k) = qo(i,k) + ((Vx(29-


k)*Point1(Pointer(i+1),4)*Point1(Pointer(i+1),5))/Iyy(k));

end end

qs1o(1,k) = qo(12,k) - qo(29,k); qs1o(2,k) = qo(13,k) - qo(28,k) - ((Vx(29-

k)*Point1(Pointer(13),3)*Point1(Pointer(13),5))/Ixx(k)) - ((Vz(29-

k)*Point1(Pointer(13),4)*Point1(Pointer(13),5))/Iyy(k)); qs1o(3,k) = qo(14,k) - qo(27,k);

qs2o(1,k) = qo(42,k) - qo(49,k); qs2o(2,k) = qo(43,k) - qo(48,k) - ((Vx(29-

k)*Point1(Pointer(43),3)*Point1(Pointer(43),5))/Ixx(k)) - ((Vz(29-

k)*Point1(Pointer(43),4)*Point1(Pointer(43),5))/Iyy(k)); qs2o(3,k) = qo(44,k) - qo(47,k);

qo(12,k) = qs1o(1,k); qo(13,k) = qs1o(2,k); qo(14,k) = qs1o(3,k);




for i=1:52 if i14 && i

ds2(30,k) = sqrt((Point1(Pointer(15),2)-Point1(Pointer(i),2))^2 +

(Point1(Pointer(15),1)- Point1(Pointer(i),1))^2); Aij(44,k) = abs((Point1(Pointer(i),1)*Point1(Pointer(15),2)-

Point1(Pointer(i),2)*Point1(Pointer(15),1))/2); elseif i>44 && i0 && i15 && i44 sumcell3(i+1-44,k) = sumcell3(i-44,k) + ds3(i-44,k)/Point1(Pointer(i),6); % use 52

for cell 3, 8 sumcellq3(i+1-44,k) = sumcellq3(i-44,k) + (ds3(i-

44,k)*qo(i,k))/Point1(Pointer(i),6); sumqA(i+1,k) = sumqA(i,k) + 2*Aij(i,k)*qo(i,k); end end

A1(k) = 0.173495596*(c(k)^2)/(2.47^2); A2(k) = 0.473302955*(c(k)^2)/(2.47^2); A3(k) = 0.057928*(c(k)^2)/(2.47^2);

wall1(k) = sumcell1(14,k)-sumcell1(11,k); wall2(k) = sumcell2(30,k)-sumcell2(27,k);

A11 = sumcell1(14,k)/A1(k) + wall1(k)/A2(k);

A12 = -(sumcell2(30,k)/A2(k) + wall1(k)/A1(k)); A13 = wall2(k)/A2(k);

A21 = sumcell1(14,k)/A1(k); A22 = wall2(k)/A3(k) - wall1(k)/A1(k);

A23 = -sumcell3(8,k)/A3(k);

A31 = 2*A1(k); A32 = 2*A2(k); A33 = 2*A3(k);

C1 = -sumcellq1(14,k)/A1(k) + sumcellq2(30,k)/A2(k); C2 = -sumcellq1(14,k)/A1(k) + sumcellq3(8,k)/A3(k); C3 = -Vz(29-k)*(2.47)/4 - sumqA(52,k);

D3 = -MT(29-k);

A = [A11 A12 A13 ; A21 A22 A23 ; A31 A32 A33]; C = [C1 C2 C3]; D = [C1 C2 D3];

Q = A\transpose(C); Q1 = A\transpose(D);

for i = 1:52 if i=15 && i=45 Totalq(i,k) = qo(i,k)+ Q(3,1); shear(i,k) = Totalq(i,k)/Point1(Pointer(i),6);

Totalq1(i,k) = qo(i,k)+ Q1(3,1); shear1(i,k) = Totalq1(i,k)/Point1(Pointer(i),6);

shearT(i,k) = shear(i,k) + shear1(i,k); end end

%% Calculation of Factor of safety

for i=1:52 sigma1(i,k) = sigmaxx(Pointer(i),k)/2 + sqrt((sigmaxx(Pointer(i),k)/2)^2 +

shear(i,k)^2); sigma2(i,k) = sigmaxx(Pointer(i),k)/2 - sqrt((sigmaxx(Pointer(i),k)/2)^2 +

shear(i,k)^2); Tau(i,k) = sqrt((sigmaxx(Pointer(i),k)/2)^2 + shear(i,k)^2); end

for m = 1:48 generalstress(m,k) = sqrt(sigmaxx(m,k)^2 + 3*(shearT(PointerI(m),k)^2)); Factorofsafety(m,k) = (345*10^6)/generalstress(m,k); end

end

%% Plots

% if j==1 % figure(1) % plot(y,Crushload(1,:),'o-', 'LineWidth',2); % XLABEL('Span Location (y)','fontsize',15) % YLABEL('Crushing load (in N)','fontsize',15)

% hold on % figure(2) % plot(y,Crushload(2,:),'o-', 'LineWidth',2); % XLABEL('Span Location (y)','fontsize',15) % YLABEL('Crushing load (in N)','fontsize',15) % hold on % figure(3) % plot(y,Crushload(3,:),'o-', 'LineWidth',2); % XLABEL('Span Location (y)','fontsize',15) % YLABEL('Crushing load (in N)','fontsize',15) % hold on % figure(4) % plot(y,Crushload(4,:),'o-', 'LineWidth',2); % XLABEL('Span Location (y)','fontsize',15) % YLABEL('Crushing load (in N)','fontsize',15) % hold on % figure(5) % plot(y,shear1(25,:),'o-', 'LineWidth',2); % XLABEL('Span Location (y)','fontsize',15) % YLABEL('Shear Stress due to Torsion (in Pa)','fontsize',15) % hold on % figure(6) % plot(y,shear1(52,:),'o-', 'LineWidth',2); % XLABEL('Span Location (y)','fontsize',15) % YLABEL('Shear Stress due to Torsion (in Pa)','fontsize',15) % hold on % figure(7) % plot(y,shearT(25,:),'o-', 'LineWidth',2); % XLABEL('Span Location (y)','fontsize',15) % YLABEL('Shear Stress Total (in Pa)','fontsize',15) % hold on % figure(8) % plot(y,shearT(52,:),'o-', 'LineWidth',2); % XLABEL('Span Location (y)','fontsize',15) % YLABEL('Shear Stress Total (in Pa)','fontsize',15) % hold on

% Aluminium alloy 2024-T3, young modulus 73.1 GPa

%% Buckling of Panels

y =

[0,1,2,3,4,5,5.5,6,6.5,7,7.5,8,8.5,9,9.5,10,10.25,10.5,10.75,11,11.25,11.5,11.75,12,12.

25,12.5,12.75,13]; ynew =

[13,12.75,12.5,12.25,12,11.75,11.5,11.25,11,10.75,10.5,10.25,10,9.5,9,8.5,8,7.5,7,6.5,6

,5.5,5,4,3,2,1,0]; S = 45; b = 26; t = 0.4; d = 1.225; %densit W = 5400;

ar = 15; e = 0.9; Cd1 = 0.05815; points3D = zeros(48,3,28);

n = [3.19,3.19,-1.27,-1.27,-

0.1256,1.61,1.865,1.436,2.1256,1.199,1.705,0.6213,0.135,0.696,0.8338,0.2947]; v =

[63.425832,137.498328,109.999272,55.065168,109.7219,45.01408,63.8885,42.52112,109.7215,

38.8559,137.5029,27.96631,63.8885,29.6058,32.3984,137.4989]; CL = [1.55675,0.329,-0.19675,-0.79,-

0.02425,1.55975,0.8915,1.559125,0.343875,1.559,0.174125,1.55938,0.06138,1.5588,1.55938,

0.02725]; Cm = [-0.0224,-0.0294,-0.0098,-0.0271, -0.00087,-0.0221,-0.0511,-0.0221,-0.0296,-

0.0222,-0.017,-0.0221, -0.0075, -0.0222, -0.0221, -0.0053]; alpha = [16.51975806,6.618548387,2.378629032,-2.405645161,

3.7697,16.54395,11.15484,16.53891,6.7385,16.5379,5.36955,16.5409, 4.4603, 16.5359,

16.5409, 4.1851]; Cd = 0;

den = 2720; c = zeros(1,28); m = zeros(1,28);

x = zeros(1,28);

L = zeros(1,28);

D = zeros(1,28); M = zeros(1,28);

N = zeros(1,28); C = zeros(1,28);

dN = zeros(1,27); dC = zeros(1,27); dM = zeros(1,27);

dF = zeros(1,27);

F = zeros(1,28); yC = zeros(1,27); Fz = zeros(1,27); Mass = zeros(1,27); dy = zeros(1,27);

Vx = zeros(1,28);

Vz = zeros(1,28);

Mz = zeros(1,28); Mx = zeros(1,28); MT = zeros(1,28);

X = zeros(3,27);

Y = zeros(3,27); Z = zeros(3,27);

Ixx = zeros(1,28);

Iyy = zeros(1,28); Ixy = zeros(1,28);

sigmaxx = zeros(48,28);

for i = 1:27 dy(i) = y(i+1)-y(i); end

for j = 1:16

Vz(1) = 0; Vx(1) = 0; Mz(1) = 0; Mx(1) = 0; MT(1) = 0;

for i = 1:28

c(1,i) = (2*S/((1+t)*b))*(1-(2*(1-t)*y(i)/b)); m(1,i) = (2.0654*0.001*c(1,i)+ ( 0.00139833+0.00044256)*(c(1,i)^2))*den;

%skin Thickness - 0.02" F(1,i) = -n(j)*9.8*m(1,i);

L(1,i) = (1/2)*(c(i) + (4*S/(pi()*b))*(1-

(2*y(i)/b)^2)^(1/2))*(CL(j)*(0.5*d*v(j)^2)) ; Cl = L(1,i)/(0.5*d*c(i)*(v(j)^2));

%Section lift coeff Cd = Cd1 + Cl*CL(j)/(pi()*ar*e);

D(1,i) = Cd*(0.5*d*c(i)*(v(j)^2)); M(1,i) = Cm(j)*(0.5*d*(c(i)^2)*(v(j)^2));

N(i) = L(i)*cosd(alpha(j))+D(i)*sind(alpha(j)); C(i) = D(i)*cosd(alpha(j))-L(i)*sind(alpha(j));

x(i) = -(1.10347*c(i)/2.47-(c(i)/4));

if ( c(i) < 9 )

Ixx(i) = (78097.37183*((c(i)*100)^4)/((2.47*100)^4))*10^(-8); Iyy(i) = (759626.49564*((c(i)*100)^4)/((2.47*100)^4))*10^(-8); Ixy(i) = (22151.37743*((c(i)*100)^4)/((2.47*100)^4))*10^(-8); else Ixx(i) = (73215.24*((c(i)*100)^4)/((2.47*100)^4))*10^(-8); Iyy(i) = (741233.47*((c(i)*100)^4)/((2.47*100)^4))*10^(-8); Ixy(i) = (22401.02*((c(i)*100)^4)/((2.47*100)^4))*10^(-8); end

end

for i = 1:27 dN(i) = ((N(i)+N(i+1))*dy(i))/2; dC(i) = ((C(i)+C(i+1))*dy(i))/2; dM(i) = ((M(i)+M(i+1))*dy(i))/2;

dF(i) = ((F(i)+F(i+1))*dy(i))/2; end % % Mt = 0; % for j = 1:27

% yC(1,j) = (1/2)*(y(j) + y(j+1)); % Fz(1,j) = (1/2)*(y(j+1) - y(j))*(F(1,j)+F(1,j+1)); % Mass(1,j) = (1/2)*(m(j) + m(j+1)); % Mt = Mt + Mass(1,j)*(y(j+1) - y(j)); % end % for i=1:27 Y(1,i) = (1/(N(28-i+1)+N(28-i)))*(N(28-i+1)+2*N(28-i))*(dy(28-i)/3);

% tip to root Y(2,i) = (1/(C(28-i+1)+C(28-i)))*(C(28-i+1)+2*C(28-i))*(dy(28-i)/3);

Y(3,i) = (1/(F(28-i+1)+F(28-i)))*(F(28-i+1)+2*F(28-i))*(dy(28-i)/3);

X(3,i) = (1/(3*(F(28-i+1)+F(28-i))))*((F(28-i+1)*(2*x(28-i+1)+x(28-i)))+(F(28-

i)*(2*x(28-i)+x(28-i+1)))); X(1,i) = 0; X(2,i) = 0;

Z(2,i) = 0; Z(1,i) = 0;

end

for i=1:27 Vz(i+1) = Vz(i)+ dN(28-i)+dF(28-i); % i here

is from tip to root but earlier i is from root to tip Vx(i+1) = (Vx(i) + dC(28-i)); Mz(i+1) = Mz(i) +Vx(i)*dy(28-i)+dC(28-i)*Y(2,i); Mx(i+1) = (Mx(i) - Vz(i)*dy(28-i)-dN(28-i)*Y(1,i)-dF(28-i)*Y(3,i)); MT(i+1) = MT(i) + dM(28-i)-dF(28-i)*X(3,i); % here

two terms are zero end

Point1 = zeros(48,6);

for k = 1:48 for i = 1:28 Point1(:,3) = Point(:,3)*c(i)/2.47; Point1(:,4) = Point(:,4)*c(i)/2.47; points3D(:,1,i) = Point1(:,3); points3D(:,2,i) = Point1(:,4); points3D(:,3,i) = y(i);

sigmaxx(k, i) = (- ( Mz(28-i+1)*Ixx(i)+ Mx(28-i+1)*Ixy(i)

)*(Point1(k,3)) +( Mx(28-i+1)*Iyy(i)+ Mz(28-i+1)*Ixy(i))*(Point1(k,4)))/(Ixx(i)*Iyy(i)-

Ixy(i)*Ixy(i)); end

end

end

Ixxst = zeros(1,15); Areast = zeros(1,15);

lengthst = zeros(1,14);

for i=1:14 Ixxst(i) = (4.9*((c(i))^4)/((2.47)^4))*10^(-8); Areast(i) = (1.96*((c(i))^2)/((2.47)^2))*10^(-4); lengthst(i) = y(i+1)-y(i);

end

lengthst(1) = 0.6998*lengthst(1); lengthst(14) = 1.5*lengthst(14); ratiost = zeros(1,14);

sigmacr = zeros(1,14);

for l= 1:14 sigmacr(l) = (pi()^2*73.1*(10)^9)*Ixxst(l+1)/(Areast(l+1)*lengthst(l)^2); ratiost(l) = sqrt((sigmacr(l)/sigmaxx(18,l))^2); end

xr = zeros(1,14); xr = y(1:14);

figure(1) plot(xr,ratiost(:),'ko-'); XLABEL('Span Location (y)','fontsize',15) YLABEL('(Critical stress/Normal stress) ratio','fontsize',15) h_legend = legend('n = 0.295,v = 137.5 m/s'); set(h_legend,'FontSize',13); set(gca,'fontsize',13)

%% Calculation of mass distribution and C.G. location

y =

[0,1,2,3,4,5,5.5,6,6.5,7,7.5,8,8.5,9,9.5,10,10.25,10.5,10.75,11,11.25,11.5,11.75,12,12.

25,12.5,12.75,13];

ylong = 0:0.05:13; S = 45; b = 26; t = 0.4; n = -1.28; den = 2720; c = zeros(1,261); m1 = zeros(1,261); m2 = zeros(1,261); mtt = zeros(1,261); F = zeros(1,261); yC = zeros(1,260); Fz = zeros(1,260); M1 = zeros(1,260); M2 = zeros(1,260); loc = 1; for i = 1:261 c(1,i) = (2*S/((1+t)*b))*(1-(2*(1-t)*ylong(i)/b)); if i < 15 m1(1,i) = (5.1015*0.001175*((c(1,i)/2.47)^2))*den; %skin

Thickness - 0.02" m2(1,i) = (( 0.008363+0.005134)*((c(1,i)/2.47)^3)+(0.00003*(c(1,i)^2)))*den; else m1(1,i) = (5.1015*0.001175*((c(1,i)/2.47)^2))*den; m2(1,i) = ( ( 0.008363+0.005134)*((c(1,i)/2.47)^3))*den; end F(1,i) = -n*9.8*(m1(1,i)+m2(1,i)); if ( ylong(i) == yrib(loc)) mtt(i) = (m1(1,i)+m2(1,i)+mlrib(loc)); loc = loc + 1; else mtt(i) = (m1(1,i)+m2(1,i)); end end Mt1 = 0; Mt2 = 0; for j = 1:260

yC(1,j) = (1/2)*(ylong(j) + ylong(j+1)); Fz(1,j) = (1/2)*(ylong(j+1) - ylong(j))*(F(1,j)+F(1,j+1)); M1(1,j) = (1/2)*(m1(j) + m1(j+1)); M2(1,j) = (1/2)*(m2(j) + m2(j+1)); Mt1 = Mt1 + M1(1,j)*(ylong(j+1) - ylong(j)); Mt2 = Mt2 + M2(1,j)*(ylong(j+1) - ylong(j)); end % plot(y,c); % hold on

plot(ylong,mtt,'r','LineWidth',2); XLABEL('Span Location (y)','fontsize',15) YLABEL('Mass/length (in kg/m)','fontsize',15) % YLABEL('d(F)IGz/dy (in N/m)','fontsize',15)

Aircraft Design Report

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