aircraft design3
TRANSCRIPT
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