15

3. Turbo-jet aircraft: The fuel is split into trip fuel and reserve fuel. Trip fuel is given by figure (6-3). Reserve fuel is estimated from the following formula:

)aircrafttransport(

A/C*18.0

MTOWW Treserve,f θ

=

TC : s.f.c. in 1/s. θ : Relative atmospheric

temperature = OT/T . A : Aspect ratio.

Figure (6-3): estimation of trip fuel weight fraction for jet airliners an executive aircraft.

7-Engine Selection

The choice of engine lies between turbo-props, turbo-jets and turbo-fans engines while piston engines have not been considered since the most piston engines in production were designed three to four decades ago. Turbo-jets are generally inefficient at low altitude and also noisy. Turbo-props are generally inefficient at high altitude, besides propellers can be dangerous on ground. Also slipstream from an idealing propeller can be uncomfortable and nuisance. Turbo-fans are the best choice. The sizing of propulsion units depends on the amount of thrust required during take-off stage. This thrust is called static thrust ( ) where take-off velocity is zero ( ). Take-off thrust affects the acceleration during this stage.

oT0v .o.t =

Total take-off distance can be sub divided into, see figure (7-1):- 1. Total ground distance:-

- Total ground distance, in m, (GS s.o.t v15.1v = ):- = total ground distance… (For propeller a/c). ×8.0= total ground distance… (For jet a/c). ×9.0

- Rotation distance, in m, (RS s.o.t v15.1v = ):- = (for light a/c, it takes three seconds to take-off at this stage). .o.tv0.3 ×= (for large a/c, it takes three seconds to take-off at this stage). .o.tv0.1 ×

2. Transition distance, in m, (TRS s.o.t v15.1v = ).

m60ft200 ≈≈ , It depends on the radius of rotation.

3. Air distance, or in m, (AS CS s.o.t v20.1v = ). It is the distance needed to climb from ground level to a height of ( m25.15ft50 ≈ ). Recently the height of obstacle is ( f50 ) for military a/c and ( ) for civil a/c. t ft35

m575.86)1510(tan

25.15tan

25.15SorS CA ÷=÷

=γ= …7-1

Where ( ) is the angle of climb ( ). γ oo 15to10=γ

1

Figure (7-1a): Take-off distance.

Figure (7-1b): Landing distance.

2

Arithmetic method:- Kinetic energy for the a/c during take-off is:-

Go2av S)RT(v

gw

21

×−= …7-2

G

2av

o S2v

gWRT += …7-3

WD)LW(DFrDR μ+≈−μ+=+= …7-4

.o.t,Dw.o.t,Dw2av CqSCSv

21D =ρ= ..7-5

For minimum thrust required, the resistant force, R, should be minimum. Where :-

e.A.C

CKCCC2

.o.t,LDo

2.o.t,LDo.o.t,D π

+=+= …7-6

)LW(DFrDR −μ+=+=

Lwav

2L

wavDowav CSqWA.

C.kSqCSqR μ−μ+π

+= …7-7

To evaluate the value of ( ) for minimum resistance force, equation (7-7) is differentiating with respect to ( ).

LCLC

0SqCA.k2Sq

CdRd

wavLwavL

=μ−π

= …7-8

k2.A.C Rimummin,Lμπ

=∴ …7-9

Substitute equation (7-9) into equation (7-7) gives:-

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ μπμ−⎟⎟

⎠

⎞⎜⎜⎝

⎛ μππ

++μ=k2.A.

k2.A.

.A.kCSqWR

2

Dowav.min

⎟⎟⎠

⎞⎜⎜⎝

⎛ μπ−+μ=

k4.A.CSqWR Dowav.min …7-10

Where ( e1k = ), (e) is called Oswald span efficiency factor, ( ). 85.07.0e ÷=

64.0)A045.01(78.1e 68.0 −−= For straight wing where ( o30<Λ ) 1.3)(cos)A045.01(61.4e 15.068.0 −Λ×−= For swept wing where ( ) o30>Λ

At take-off ( ) to account for flap deflection and U.C extended. Where ( ) is the average value from zero to take-off velocity. For linear relation ( ). Equations (7-10) and (7-3) are used to evaluate thrust at take-off ( ).

40.0to20.0CD =

avqo.tav q5.0q ≈

oT

3

Graphical method:-

If the take off distance is known the following graph can be used to evaluate thrust at take-off ( ), see figure (7-2). oT

Rapid method:-

A rapid method depends on thrust/weight ratio which is used to estimate required thrust roughly. This ratio was take at sea level, static, standard day condition at design take-off weight and maximum throttle setting.

,0v =

WWTTo ×=

4.0WT

= Jet trainer.

Jet fighter (dog fighter).

Figure (7-2): Take-off chart W & T are in lbf, where 1 lbf = 0.45359237 kg = 4.44822 N. S is inf 2t , where 1 ft = 0.3048 m

96

.0= Jet fighter (others). .0= Military cargo/bomber. 25.0= Jet transporter. 25.0=From the required thrust at take-off, a suitable engine (one or more ) is chosen to

account for plus 10% as a save margin. See table 7.1 4

5

Table (7-1a): Principles characteristics of engines

6

Table (7-1b): Principles characteristics of engines

From the selected engine all useful information about weight, sizing and cost are become known. The following relations are useful for subsonic no afterburning engine which it is used usually on commercial aircraft and covers by-pass ratio from zero to 6:-

B045.01.1

e eT084.0W −××= LB 2.04.0 MT22.2L ××= in

B04.05.0 eT393.0D −××= in B02.09.0

cruise eT60.0T ××= LB B12.0

T.max e67.0sfc −×= l/hr B05.0

cruise e88.0sfc −×= l/hr Where W: weight. L : Length. D : Diameter. sfc: specific fuel consumption. B : by-pass ratio.

7

8-Airplane Center of Gravity

Center of gravity is the point at which a/c would balance if suspended. Variation in the (c.g.) has an effect on:

1. Stability and control characteristics. 2. Tail maneuver loads. 3. Ground loads acts on the nose u/c. Acceptable (c.g.) limit, which is the extreme locations of the (c.g.) within which the

a/c must be operated to a given weight, must be established taking into account: 1. Fore and aft position of the wing relative to the fuselage. 2. Provision of suitable locations for payload and fuel. 3. Design of the horizontal tail plane, the elevator and the longitudinal

flight control system. 4. Location of u/c. The (c.g.) must be established in both, longitudinal and vertical direction, Side view

and suitable system of coordinate axes should be chosen, the position of (c.g.) for each part of the a/c. the data must be tabulated in a table similar to table (1). Then:

∑∑=

i

ii.g.c W

WYY

∑∑=

i

ii.g.c W

WXX ----------(8-1): ---------- (8-2)

Load And Balance Diagram, Loading Loop. The loading loop is a diagram

showing the relation between a/c different weights and the position of (c.g.) as percentage of AMC.

As an example, for passenger transport, short range, with a cabin with 4-sets a breast, one aisle (your own project may differ in sets number a breast).

Let us use “Window Seating Rule” A :c.g. position at OEW. B : maximum c.g. aft position. D1 : maximum c.g. fore position.

B1D :c.g. position limit. ABC : sets nearest to the window are

occupied starting from a/c rear side. AB1C: sets nearest to the window are

occupied starting from a/c rear side. CDE : other seats starting from a/c rear side. Figure (8-1) typical loading loop

1CD1E : other seats starting from a/c front side.

BF2: c.g. position improvement due to addition of fuel weight at B. FF1: c.g. position improvement due to addition of fuel weight at F.

There are many ways that enable the designer to make the c.g. limit within the

specification: 1. For empty weight, the longitudinal location of the wing is rearranged to

ensure that ( ) which is point A. ==χ c%25.0OEW

2. Rearrange cabin layout, engine location, cargo compartments, fuel tanks, systems, etc.

3. Suitable tail plane and control system design. And u/c position should provide an acceptable fore and aft c.g. limit.

After computing ( ) from equation (1), see table (1), one should add the weight of each two passenger starting from rear and front side, taking into consideration all possible ways of seating. Window seating rule is an example, see figure (1).

OEWX

A simple procedure to determine c.g. limit and to choose the wing location

accordingly. Step 1:

Subdivided the a/c into the following: 1. The fuselage group, containing parts whose location is fixed relative to

the fuselage such as. - Fully furnished and

equipped fuselage. - Several airframe services. - Vertical tail plane. - Fuselage mounted engine. - Nose wheel u/c. 2. The wing group: - Wing structure. - Fuel system. - Main u/c. - Wing mounted engine. 4. The variable payload. Figure (8-2) Wing group & fuselage group 5. The variable fuel load.

Step 2: Draw fuselage group with x-axis parallel to cabin floor or propeller axis, determine

c.g. for the complete group in both longitudinal and vertical direction, see table (1). Step 3:

The empty wing group is drawn on a separate transparent sheet. Root chord, tip chord and AMC are indicated. And c.g. is computed as relative to the mean aerodynamic chord leading edge ( ).

=c%LEAMC

Step 4: Assume a value for ( ), say ( ) == c25.0xOEWOEWx

2

Step 5: Calculate the coordinate of the wing leading edge relative to the fuselage coordinate system.

)(WW

XX .E.O.g.c.g.w.g.f

.g.w.E.O.g.c,g.f.C.M.A.E.L χ−χ+χ−= ---------- (8-3)

Step 6: Compute a load and balance diagram, figure (1), considering various possible combinations of payload and fuel loading. A window seat rule is applied to civil transport. Step 7: Estimate the fore and aft limits that are acceptable use table (3) for comparison. Step 8: In case of unacceptable c.g. limit, a revise choice of ( ) or other revisions are recommended. Repeat the procedure until the result is considered satisfactory.

OEWX

: Distance from operational empty weight center of gravity (wing, tail body, u/c main, u/c nose, surfaces controls, nacelles, power plant, etc --- and crew) to the mean aerodynamic mean chord leading edge. The numeric value is usually (% ).

.E.Oχ

=c : Distance from operational empty weight center of gravity (wing, tail body, u/c

main, u/c nose, surfaces controls, nacelles, power plant, etc --- and crew) to the fuselage nose. The numeric value is usually (% length of fuselage).

.E.OX

.: Distance from fuselage group center of gravity (body, tail, u/c if attached, power plant if attached, etc…and crew) to the mean aerodynamic mean chord leading edge. The numeric value is usually (% ).

g.c.g.fχ

=cg.c.g.fX . : Distance from fuselage group center of gravity (body, tail, u/c if

attached, power plant if attached, etc…and crew) to the fuselage nose. The numeric value is usually (% length of fuselage).

. : Distance from wing group center of gravity (wing, u/c if attached, power plant if attached) to the mean aerodynamic mean chord leading edge. The numeric value is usually (% ).

g.c.g.wχ

=cg.c.g.wX . : Distance from wing group center of gravity (wing, u/c if attached,

power plant if attached) to the fuselage nose. The numeric value is usually (% length of fuselage).

3

.g.cX

.c.aX

nW

tl

.g.c,WingX

AMC.E.LX

.g.c,FuselageX

wL tL

Aircraft c.g.

Fuselage grp, c.g.

Wing grp, c.g.

Aerodynamic center

Figure (8-3a): Notation and main dimension, with reference to aircraft nose

oM qB&

.g.cχ

.g.c,fuselageχ

.g.c,wingχ

.c.aχ

tl

Figure (8-3b): notation and main dimension, with reference to aerodynamic mean chord.

4

Table (8-1) weight breakdown

5

Table (8-2) typical c.g. position for a/c main different parts

6

Table (8-3) typical c.g. limits

7

9. Payload-Range Diagram

It is a diagram that shows an interrelation ship between various airplane payload that can be carried and flight range taking into consideration other limitations.

The actual take off weight, landing weight and payload for an aircraft particular flight should never exceed the limiting weight define bellow:

1. Operational Landing Weight (OLW).

It is the maximum weight authorized for landing, and it is the lowest value of the following:

a. Maximum landing weight ( ). MTOW*95.0MLW =

b. Permissible landing weight based on available performance.

c. Maximum zero fuel weight ( ) plus the fuel load on landing.

MZFW

2. Operational Take off Weight ( ). OTOW

Is the maximum weight authorized for take off, and it is the lowest value of the following:

a. Maximum take off weight. b. Maximum take off weight based on available

performance. c. Operational landing weight plus trip fuel. d. Maximum zero fuel weight plus fuel on take

off. e. Take off weight restricted by operation weight

(due to useful weight).

3. Payload. It is the weight of passengers and their baggage, cargo and /or mail that can be loaded in the aircraft without exceeding the MFZW.

4. Operation Empty Weight (OEW).

It is the weight of the airplane without payload and fuel.

1

5. Maximum Zero Fuel Weight ( ). MZFWIt is the maximum weight load of an aircraft less the weight of total fuel load (and other consumable propulsion agents).

6. Total fuel.

It is all usable fuel, engine injection fluid and other consumable propulsion agents. And it is:

a. Fuel consumed during run up and taxing prior to take off..

b. Trip fuel, the fuel consumed during flight up to the moment of touch down in landing.

c. Additional fuel for holding, diversion. d. Reserve fuel, according to the relevant

operation rules.

The range is usually evaluated from Breguet range formula. For propeller aircraft:

2

1maxp

2

1maxp

2

1maxp

WWln*)D/L**

c1

WWln*)D/L**

c1*367

WWln*)D/L**

c1*3600

η=

η=

η=

hr.kWN0.4c.,c.f.s ≈

Range, propeller a/c

hr.kW

kg4.0c.,c.f.s ≈

s.Wkg001.0c.,c.f.s ≈

For maximum range aerodynamic efficiency, (L/D), should be maximum.

( )

( ) 5.0DoDo

5.0Do

Do

d.min,L

drag.min,D

drag.min,L

max kC21

C2k/C

C2C

CC

DL

====⎟⎠⎞

For jet aircraft:

2

1maxt W

Wln*)D/L*U*c1

= Range, jet a/c hr.N

N9.0c.,c.f.s ≈ For maximum range the, term , should be maximum. The true air speed is in (km/hr)

(L/D)Ut)(Ut

( )( )( ) range.maxD

range.maxLrange.maxt

maxt C

C*U

DLU =⎟⎠⎞

⎜⎝⎛

2

( )3

Ck3

CC drag.min,LDorange.maxL ==

( ) 3C.4C Do

range.maxD =

Dorange.max,D

range.max,L

kC1

43

CC

=

range.maxLoLodrag.minrange.max C

)S/W(2C

)S/W(2316.1U*316.1Udrag.min

ρ=

ρ== .

Atypical payload-range diagram is shown.

Reserve fuel

Maximum structural payload

A`

B

C

C`

D`

TOW limited by fuel capacity

D1D

A

MTOW

OEW

MLW

MZFW

Total fuel

Trip fuel

Payload

TOW limited by MLW

Weight kg

DR CR

A1

1DR

Range, km

3

`AR : Minimum range due to MLW restriction.

BR : Maximum payload range.

CR : Maximum range based on maximum available fuel based on fuel tank capacity.

DR : Maximum range based on maximum available fuel based on fuel tank capacity.

1DR : Maximum range based on maximum possible fuel.

`CC : Maximum useful fuel weight restricted by fuel tanks capacity.

`DD : Maximum useful fuel weight restricted by fuel tanks capacity.

`C1D : Maximum possible fuel weight.

For ranges : BR≤1. Maximum payload is maximum structural payload,

which is limited by allowable floor loading. 2. OTOW is limited by MLW. 3. MZFW plus reserve fuel ≤MLW. 4. Point B corresponds to the maximum flight range

with maximum payload and reserve fuel, with relevant cruising condition.

BR

For ranges : CR≤Usable fuel load limited by the fuel tank capacity and the operating

weight reaches its limit point C.

For ranges : CR> A considerable reduction in take-off weight is noticed, which results in a further payload reduction.

For point D: No payload and DR is maximum range due to useful fuel load.

For normal commercial aircraft, region CD is of miner importance. And is frequently referred to as maximum range. Both and

may be increased by adding additional fuel tanks internally or externally.

CR CRDR

4

Ex. Draw payload range diagram for a turboprop aircraft having the following data:

MTOW = 12750 kg. Volume of fuel tanks = 4.958 3m OEW = 7459 kg. Payload = 4140 kg.

Designed range =1000 km. Specific fuel consumption, c = 0.3 hr/kW/kg

max)D/L =15.127 Propeller efficiency, pη = 0.8 = 0.8 fuelγ

Solution. Maximum possible fuel weight = MTOW - OEW = 12750 - 7459 = 5291 kg. Maximum available fuel weight = volume of fuel tanks * fuel density. = 4.958 * 800 = 3966.4 kg. MZFW = OEW + payload = 7459 + 4140 11599 kg. From Breguet formula for range

2

1maxp W

Wln*)D/L**c1*367 η= Range, propeller a/c

1. Maximum range with full payload:

kg12750MTOWW1 == kg11599MZFWW2 ==

km14001159912750ln*127.15*8.0*

3.01*367R B ==

2. Maximum possible range: kg12750MTOWW1 ==

kg7459OEWW2 ==

km79367459

12750ln*127.15*8.0*3.0

1*367R 1D ==

3. Maximum available range:

kg12750MTOWW1 == , −= MTOWW2 Fuel tank capacity kg6.87834.396612750 =−=

5

km55156.8783

12750ln*127.15*8.0*3.0

1*367R C ==

4. Maximum available range: += OEWW1 Fuel tank capacity kg4.114254.39667459 =+=

kg7459OEWW2 ==

km8.63127459

4.11425ln*127.15*8.0*3.0

1*367R D ==

5. Minimum allowable range:

kg12750MTOWW1 == , MTOW95.0MLWW2 ∗==

km76095.01ln*127.15*8.0*

3.01*367R B ==

6. What will be the landing weight if the aircraft flew about 1000 kg?

kg12750MTOWW1 == , ?WeightLandingW2 ==

2W12750ln*127.15*8.0*

3.01*3671000 =

Landing weight = kg11917.2

7. Reserve fuel = landing weight - MZFW = 11917.2 – 11599 = 318.2 kg.

This mount of fuel should be sufficient to account for: - a. Descent and climb stages. b. 200 km diversion. c. 0.75 hr. holding.

6

11917.2 kg

3966.4 kg

12750 kg

11599 kg

12112.5 kg

8783.6 kg

7459 kg

11425.4 kg

5291 kg

1000 km

760 km

5515 km6312.8 km

7936 km

Range km

Weight kg

7

1

10. Air-Inertia load Distribution 10.1. Spanwise air load distribution:

This subject concerns both the aerodynamicist and the stress analyst. The aerodynamicist is usually concerned with properties, which affect the performance, stability and control of the airplane.

The stress analyst is concerned with the load distribution which will represent the most sever conditions for various parts of the internal structure of the airplane.

Exact equations for span-wise load distribution which can be found in many aerodynamic books, can be solved for many wing planform. Numerical methods to solve such system of equation are available but the calculation is not simple (quit tedious).

Approximation solutions for span-wise lift distribution are available, such as: - Fourier series method. - Diederich method. - Schrenk method.

10.2. Schrenk method:

A simple approximate solution for lift distribution has been proposed by Dr. Ing Oster Schrenk and has been accepted by the Civil Aeronautics Administration (CAA) as a satisfactory method for civil a/c.

Schrenk method relies on the fact that the lift distribution does not differ much from elliptical planform if:

- The wing is upswept. - The wing has no aerodynamic twist, i.e. zero lift lines for all wing sections lie in

the same plane (constant airfoil section).

LL2 qSCSCV5.0L =ρ= ------------1

L2 CcbV5.0L ρ= ------------2

LCcqbL = ------------3 LCcq/L = Per unit span length ------------4

SSCq/L L == For unit lift coefficient ( 1CL = ) ------------5 WL = For level flight ------------6

S/WqS/L == For unit lift coefficient ( 1CL = ) ------------8

For rectangular wing (for example): take lift coefficient equal to one for entire wing planform ( 1CL = ).

gulartanrecL cCc = Which is constant at any section along the wing. ---10

For elliptical wing (like British spitfire of word II war)

bS4croot π

= Which is wing chord at wing symmetry ---11

2

22rootellipes )

by2(1

bS41cc −π

=η−= ---------12

This is elliptical wig chord at section (y). The lift distribution for elliptical wing is exactly look like the chordal distribution for such wing. Lift distribution at section (y) for elliptical wing is: see figure (10-1).

2L )

by2(1

bS4Cc −π

= at ( 1CL = ) ---------13

Wing lift distribution “Schrenk distribution”:

⎟⎟⎠

⎞⎜⎜⎝

⎛−

π+=+= 2

recLellipesrecL )by2(1

bS4c

21C)cc(

21Cc at ( 1CL = ) ---14a

Wing lift distribution “Schrenk load distribution”: see figure (10-2).

)SW(*)Cc()Cc(qb/L LL == at ( 1CL = ) ---14b

Local lift coefficient at section (y) at ( 1CL = ) is:

wing2

wingwing

L c)by2(1

bS4c

21

c)Cc(C ⎟⎟

⎠

⎞⎜⎜⎝

⎛−

π+==l --------15a

Local lift load at section (y) at ( 1CL = ) is:

y*SW*)Cc(LW Δ= l --------15b

wingc is wing chord at any section, whether the wing is rectangular or trapezoidal.

2)by2(1

bS4

−π

bS4croot π

=⎟⎟⎠

⎞⎜⎜⎝

⎛−

π+= 2

recL )by2(1

bS4c

21Cc

Spanwise, m

Lift distribution, N/mLoad N/m

b/2 0

Figure (10-1) Schrenk load distribution

3

The result should be tabulated in the following table:

y 2y/b 2)b/y2(1bS4

−π

m

wingc m

LCc )43(5.0 +

lC 5÷4

LCc).S/W( m/N

1 2 3 4 5 6 7 8 1. 2. 3.

10.

0 . . .

b/2

0 . . . 1

c : Standard mean chord (m). wingc : Wing chord at any section (m).

LC : Wing lift coefficient. lC : Local lift coefficient at each section.

b : Wing span (m). S : Wing area ( 2m ).

S/W : Wing loading ( 2m/N ). η : Non-dimensional parameter.

WL : Local lift. Column (5) gives Schrenk distribution while column (7) gives Schrenk air-load

distribution. Column (8) gives local air-load value. This lift distribution is obviously inaccurate at the wing tips and empirical corrections

are often applied. For wings with aerodynamic twist, the distribution is evaluated in two parts. Firstly lift

distribution at zero wing lift (i.e. 0CL = ) is evaluated but with wing twist. Then lift distribution for the wing with no twist is evaluated as it was explained previously where ( 1CL = ). The details are left for the student that is interest. Diederich method seems simpler and more general.

10.3. S.F & B.M distribution:

Schrenk distribution can be used to evaluated shear force and bending moment

distribution along wing.

y*2

wwdy.w.F.S 21

2/b

0

Δ⎟⎠

⎞⎜⎝

⎛ +== ∑∫ ----------16

y*2

.F.S.F.Sdy.wdy..F.S.M.B 21

2/b

0

Δ⎟⎠⎞

⎜⎝⎛ +

=== ∑∫∫∫ ----------17

4

Solution:

2wing m2.47127.270/12750SareawingwieghtS/W ==⇒=

m73.212.47*10bS/bc/bratioAspect w2 ==⇒==

m173.210/73.21AR/bc === 716.2)6.01/2(*173.2c2/)1(c2/)cc(c rootroottiproot =+=⇒λ+=+=

m63.16.0*716.2cc/c tiproott ==⇒=λ

m218.26.01

36.06.01*716.2*32

11c

32c

2

root =+++

=λ+λ+λ+

==

m979.4)6.01(36.0*21*863.10

)1(321*

2by =

++

=λ+λ+

==

y

m

2y/b 2)b/y2(1bS4

−π

m

wingc m

LCc 43(5.0 +

lC 5÷4

LCc).S/W( N/m

1 2 3 4 5 6 7 8 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

0

1

2

3

4

5

6

7

8

9

10

10.863

0

0.092

0.184

0.276

0.368

0.460

0.552

0.644

0.736

0.828

0.920

1.000

2.763 2.753 2.718 2.658 2.571 2.455 2.306 2.112 1.872 1.550 1.084 0.000

2.716 2.616 2.516 2.416 2.316 2.216 2.116 2.016 1.916 1.816 1.716 1.63

2.741 2.685 2.617 2.537 2.444 2.336 2.211 2.066 1.894 1.683 1.400 0.815

1.009 1.026 1.040 1.050 1.055 1.054 1.049 1.025 0.989 0.927 0.816 0.500

740.419 725.292 706.923 685.313 660.191 631.017 597.251 558.083 511.621 454.624 378.178 220.154

Example: Find air-load spanwise, S.F. and B.M. distributions over a straight taper wing for

an aircraft has the following data: Aircraft weight: 12750 Kg : Wing loading: 270.127 2m/Kg Aspect ratio: 10.0 m : Taper ratio: 0.6 m Load factor: 1.0

5

y

Load intensity

w

Interval yΔ

Shear increment

2y)ww( 21 Δ+

= FΔ

Shear force∑F

Shear increment

2y)FF( 21 Δ+

= MΔ

Bending moment =∑M

m m/N m N N m.N m.N 1 2 3 4 5 6 7 8

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

0 1 2 3 4 5 6 7 8 9

10

10.863

740.419

725.292

706.923

685.313

660.191

631.017

597.251

558.083

511.621

454.624

378.178

220.154

1

1

1

1

1

1

1

1

1

1

0.863

732.856

716.108

696.118

672.752

645.604

614.134

577.667

534.852

483.123

416.401

299.166

6388.781

5655.925

4939.817

4243.699

3570.947

2925.343

2311.209

1733.542

1198.690

0715.567

0299.166

0000.000

6022.353

5297.871

4591.758

3907.323

3248.145

2618.276

2022.376

1466.116

957.129

0507.367

0129.090

30767.804

24745.451

19447.580

14855.822

10948.499

07700.354

05082.078

03059.702

01593.586

00636.457

00129.090

00000.000

Check: half weight = shear force at root 0.5(12750) = 6375 Kg

Error = (6375-6388.781)/ 6375= -0.216% Half weight * =y = Bending moment at root 0.5(12750)*4.979=31741.125 m.Kg

Error = (31741.125-30767.804)/31741.125= +3.07%

6

7

8

10.4. Diederich method:

The lift may be divided into additional lift ( aL ) and basic lift ( bL ), then: ba CCC lll += ------------18 In terms of non-dimensional parameter ( ba L&L ) used by used by Anderson R .F. (NACA report 572, 1936)

b0t

La LEaCL

ccC ∈

+=l ------------19

cCcaL

L

0a = ,

0t

0b a

EccaL∈

= ------------20

Dederich F. W. (NACA TR 2751, 1952) proposed the following semi-empirical method, which yields a satisfactory result for pre-design purpose. It is valid for wing with arbitrary planform and lift distribution, provided that the quarter chord line of a wing half is approximately straight. This method can be used for straight and swept wings incompressible, compressible and sub-critical flow.

a- Additional lift distribution:

fC14CccCL 3

221a +η−π

+= -----------21

For coefficients ( 1C , 2C and 3C ) see fig. (10-4), for lift distribution function (f) see fig. (10-5). for straight wings (f) is elliptical and the equation can be simplified to:

2321a 14)CC(CL η−π

++= for o25.0 0=Λ -----------22

If ( 5.0)CC(C 321 =+= ), the distribution becomes Schrenk distribution. b- Basic lift distribution:

)](cosCL[EL 01t

4ab α+∈∈

Λβ= β -----------23

βΛ=Λβ 25.0 -----------24

∫ η∈∈

−=α1

0 at

01 dL -----------25

The factor ( 4C ) is evaluated from fig. (4). The factor ( 01α ) is equal to the local aerodynamic twist at the spanwise station for which ( 0C b =l ), assuming a wingtip twist angle of one degree relative to the root. For the case of a linear twist distribution where ( t∈η∈= ) and elliptic distribution for ( aL ), ( π=α 3401 ).

For straight-taper unswept wings with linear twist distribution and substitution for ( aL ) from equation (19), ( 01α ) is evaluated as:

9

π++

λ+λ+

=α−34)CC(

)1(321C 32101 ----------26

For straight wings with linear lofted (geometric) twist, figure (6) can be used to determine ( 01α ). For swept wings the value of ( 01α ) should be reduced by approximately (0.006) per degree of ( βΛ ).

A linear lofted (geometric) twisted is obtained on a wing where the intermediate sections are formed by linear lofting between the root and tip sections, i.e.

ηλ−−λη

∈=∈)1(1

)( tgg ----------27

( g∈ ) is the geometric twist angle, which is the angle between root chord and the section chord).

Figure (3) definition of twist

10

Figure (6) evaluation the factor ( 01α ) for linear lofted (geometric) twist

Figure (4) evaluation of coefficients ( 1C , 2C , 3C and 4C )

Figure (5) evaluation of lift distribution function (f)

11

10.5. Inertia Loads The maximum load on any part of the airplane structure is at the stage where it is accelerated. The loads produced by landing impaction, maneuvering or encountering gust in flight case are always greater than steady state or equilibrium conditions. Therefore various loading factors should be considered. During stress analysis different inertia loads for different airplane parts should be considered. Since the sever conditions occurs at wing due to many different dynamic loads during flying, our attention will be focused on wing group comp. The wing, from structural analysis point of view, can be regarded as a simple cantilever beam supported at root and free to deflect at tip. Loads at wing are due to:

- Wing structural weight distribution. - Fuel weight distribution. - Concentrated loads due to power unit. - Concentrated loads due to undercarriage. - Other loads due to different parts accommodated in the wing.

S.F. and B.M. diagrams for inertia loads are evaluated by many methods, such as: - By considering forces to the left of each section. - By integration of equations defining loads and shear curves. - By obtaining areas under curves geometrically.

Example:

Find S.F. and B.M. diagram for the beam shown.

Solution: Since load intensity increases linearly from ( cmKg10 ) at ( 0x = ) to

( cmKg20 ) at ( 100x = ), then load at any section is:

x1.010w += ----------28 A- Divide the load distribution into two regions rectangular and triangular, then:

2x1.0x10.F.S

2

+= ----------29

cm100

cmKg20

cmKg10

12

3x1.0

2x10

2x*

2x1.0

2x*x10.M.B

322

+=+= ----------30

B- by direct integration of load distribution (w).

2x1.0x10dx).x1.010(dx.w.F.S

2x

0

x

0

+=+== ∫∫ ----------31

6x1.0

2x10dx).

2x1.0x10(dx.F.S.M.B

32x

0

2x

0

+=+== ∫∫ ----------32

C- by dividing load distribution into strips, and then: And so on for all sections. After that S.F. diagram is drawn and also is divided into

many strips to find B.M. diagram. Although the method lengthy, it is quite beneficial for irregular distribution.

That means that (S.F.) at any section is equal to the area under load curve positioned to the left of the section. (For the above example)

S.F. = rectangular area ( x*10 ) + triangular area ( )2/x(*x10 ) and that (B.M.) at any section is equal to the area under shear force curve positioned

to the left of the section. B.M. = triangular area ( )2/x(*x10 ) + parabola area ( )3/x(*)2/x(*1.0 2

Area of parabola = (maximum ordinate * third of the base)

100 x

x10 2/x

2x1.0

2

3/x

x1.0

10

F

M

Area of strip 1 Area of strip 1+2 Area of strip 1+2+3 Area of strip 1+2+3+4

=S.F. at x1 =S.F. at x2 =S.F. at x3 =S.F. at x4

13

10.6. Wing group load distribution For precise calculation of fuel tank volume it is necessary to account for the actual section shape a wing structural layout. But a first guess for total fuel volume tank is needed. Fuel tank cross-section area =

2c.c.

tc

Volume of fuel tank: *Truncated pyramid

( )2A1A2A1A3

++=l

----------33 *Obelisk

⎟⎠⎞

⎜⎝⎛ +

++=2

baba2A1A3

1221l

----------34

Volume of wing shape:

2root

2

)1(1)c

t(b

SBVλ+

λτ+τλ+=

----------35

B = Constant from statistical data = 0.54. S = Gross wing area. b = Wingspan.

r)c/t( = Thickness/chord ratio at wing root. λ = Taper ratio. τ = Ratio roottip )c/t()c/t(

2/c

c

c.ct

l

2/b

rC

tC 1A 2A

14

The distribution is evaluated as follow: For homogenous distribution, and cross section areas ( 1A and 2A ) are known, then:

2w1w

2A1A=

2A1A2w1w = ----------36

Fuel load distribution for half wing is:

wf*)2wf1wf(21

=+ l (Fuel weight) ----------37a

81.9*Mf*)2wf1wf(21

=+ l ----------37b

where ( 1wf ) and( 2wf ) are fuel load per unit length ( m/N ) at end cross section areas. From equations (36) and (37), one can find ( 1wf ) and ( 2wf ). Similar procedure is used to evaluate wing structure load distribution. Notes:

- In order to estimate area of the airfoil section at root or tip or other sections, Simpson’s rule with graphical paper or computer aided design software are recommended.

( ) ( )[ ]n642321OSimpson fetc...fff2etc...fff4f3eA +++++++++= , where

( n ) is number of sub-divisions, and must be even, ( )n/chord(e = .

- Take tip.maxtip

root.maxroot

tip

root

)c/t(*c)c/t(*c

AA

≈

- For maximum S.F. and B.M., all weight should be multiply by maximum load

factor.

- Although the type of units used in previous example was ( Kg ) and ( cm.Kg ) for moment, it is necessary to use S.I. units.

Finally:

inertiaairtotal .)F.S.)F.S.)F.S += ----------38 inertiaairtotal .)M.B.)M.B.)M.B += ----------39

15

.distWfuel

c/uW engineW

.distWwing

Load distribution

B.M. distribution

S.F. distribution

16

10.7. Fuselage group load distribution Load distribution includes the mass of fuselage and any parts attached to it. Pressure loads are so small, except for fighters with integrated fuselage, compared with inertia loads. The load distribution and integrated from front and rear airplane edges to amid point which is usually lies on (y-axis) passing through (1/4) the chord (usually aerodynamic mean chord).

Fuselage

Engine Main u/c

Baggage Nose u/c

V.tail

H.tail

Cockpit

Pilot

Load distribution

S.F. distribution

B.M. distribution

11. Gust and Flight Envelope 11.1. Flight envelope:

- The various loading conditions are ploted against aircraft speed. For a particular aircraft to indicate the flight performance limits. This inter relation ship diagram is often referred to as flight envelope or (v-n) diagram. These limits are selected by air worthiness authorities.

Figure (11-1): Flight envelope Limited load: maximum load that the a/c is expected to experience in normal

operation. It also called “applied load”. Proof load: the maximum load that a/c structure can withstand without

distortion.

Ultimate load: maximum design load which should be taken into account for

various uncertainties.

SW

CvSW

CvWLn LeqoLt

22max,

2max,

2

maxρρ

=== … (11-1)

:tv True airspeed. v Equivalent airspeed. :eq

1

Using ( ) makes ( ) diagram to be drawn for a range of altitudes from sea level to the operation ceiling of the a/c, while using ( ) makes the ( ) diagram universal.

tv vn −

eqv vn −

Line OA: Limiting condition by stalling characteristics for positive value of

. max,LCLine AC: Maximum load factor ( ) for which a/c is designed. n

Point A: Maximum ( ) for highest angle of attack, positive value of . n max,LCPoint C: Maximum ( ) for lowest angle of attack, positive value of . n max,LCLine OF: Limiting condition by stalling characteristics for negative value of

. max,LCLine FE: Maximum load factor (n ) for negative maneuvers. Point F: Maximum ( ) for highest angle of attack, negative value of . n max,LCPoint E: Maximum ( ) for lowest angle of attack, negative value of . n max,LC

EC vv &

2

The envelope (OACD1D2EFO) is called flight envelope or ( ) diagram for particular a/c at steady flight.

vn −

Load factors laid down by (BCAR):-

: Design cruise speed. 21 & DD vv : Maximum diving speed.

Load factor Normal

Semi-aerobatics aerobatics 1n 4.5 6.0

10000240001.2+

+W

2n 0.275.0 1 >butn 3.5 4.5 3n -1.0 1.8 3.0 According to (FAR) part 32:-

8.310000

240001.2 ≤+

+=+ butW

n

14.0 nn −=− (For normal and utility categories) Note: Weight of a/c should be in pound (1 kg = 2.202 lb.).

A D1

D2

EFFigure (11-2): Flight envelope

According to (FAR) part 32