V – n DIAGRAM INTRODUCTION Airplanes may be subjected to a variety of loading conditions in flight. The structural design of the aircraft involves the estimation of the various loads on the aircraft structure and designing the airframe to carry all these loads, providing enough safety factors, considering the fact that the aircraft under design is a multirole fighter aircraft. It is obviously impossible to investigate every loading condition that the aircraft may encounter; it becomes necessary to select a few conditions such that each one of these conditions will be critical for some structural member of the airplane. Using the V-n diagram two important load factor values can be plotted, which are 1) Limit load factor- Value of load factor corresponding to which there is Permanent structural deformation 2) Ultimate Load factor – Value of load factor corresponding to which there is outright structural failure. 33
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Transcript
V – n DIAGRAM
INTRODUCTION
Airplanes may be subjected to a variety of loading conditions in flight. The structural design of the aircraft involves the estimation of the various loads on the aircraft structure and designing the airframe to carry all these loads, providing enough safety factors, considering the fact that the aircraft under design is a multirole fighter aircraft. It is obviously impossible to investigate every loading condition that the aircraft may encounter; it becomes necessary to select a few conditions such that each one of these conditions will be critical for some structural member of the airplane.
Using the V-n diagram two important load factor values can be plotted, which are
1) Limit load factor- Value of load factor corresponding to which there is
Permanent structural deformation
2) Ultimate Load factor – Value of load factor corresponding to which there is
outright structural failure.
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VELOCITY – LOAD FACTOR (V – n) DIAGRAM
The various external loads on the airplane are usually represented on a graph
of the limit load factor n plotted against the indicated airspeed (IAS). This diagram is known
better as the V-n diagram. The indicated airspeeds are used, since all air loads are
proportional to q or . The value of q is the same for the air density ρ and the actual
airspeed at altitude, as it is for the standard sea level density and the IAS. The V-n diagram
is therefore the same for all altitudes if indicated airspeeds are used. However, in this design
case, corrections involving compressibility have to be taken into consideration while
calculating the True airspeeds from Indicated airspeeds. Therefore, calculations involving
high speeds have been performed with respect to sea-level conditions only.
212 lV SC
nW
ρ ∞=
The load factor n is basically the ratio of the wing lift produced to the weight of the
aircraft and hence represents the amount of acceleration produced along the z-axis of the
plane. For supersonic fighter aircraft, the ultimate positive load factor ranges from 7.75 to
8.67 and negative load factor between -3 and -4.5. The positive and negative load factors are
arbitrarily chosen as 7.75 and -3 respectively.
For level flight at unit load factor, the value of V corresponding to CLmax would be the
stalling speed of the airplane. In accelerated flight, the maximum lift coefficient can be
achieved at higher speeds. The wing is usually analyzed for a coefficient of 1.25CLmax, and
various values of n are obtained by varying the velocity, until the ultimate positive load factor
is reached. It can be made out from this boundary that it is impossible to maneuver at speeds
and load factors corresponding to points above or to the left of line because this would
represent positive high angles of attack (+HAA). This load factor is usually arrived at by
considering both aerodynamic and structural design capabilities.
The structural design diving speed is usually specified as 1.2 times the cruise velocity,
or is limited by Compressibility effects. Here, a never exceed Mach no. (Mne) of 2.5 and the
design diving speed were considered and were found to be of the same order. The line of the
34
V-n diagram represents this speed and is also known as the buffeting boundary. The velocity
at this boundary is 3172.5 km/hr.
In a similar manner, the maneuver boundary can be carried to the negative
load factor region which is indicative of inverted flight. The negative maneuver boundary is
seldom made use of in transport aircraft. However, the gust loads in the negative region are
indispensable and can be more severe than the manoeuvre load factor itself.
Thus in order to establish the safe flight envelope of our aircraft, we have plotted as per FAR
25 norms,
There are two types of V – n diagram for military airplanes:
V – n manoeuvre diagram and
V – n gust diagram
V – n MANOEUVRE DIAGRAM
There are four important speeds used in the V – n diagram
1 – g stall speed VS
Design maneuvering speed VA
Design cruise speed VC
Design diving speed VD
Rules for determining these speeds are given below:
For purposes of constructing a V – n diagram, the 1-g stall speed VS, may be
computed from:
35
Where: WFDGW is the flight design gross weight. For most aircraft: WFDGW equals the
maximum design take-off weight.
The maximum normal force coefficient follows from:
The midpoint of the difference between the elliptic curve and the planform shape are marked and joined together. This is required Schrenk’s curve.
51
The load intensity at each grid point on the wing plan-form is calculated as follows.
Load intensity at root =
Where is the lift distribution at the root
Load intensity at root = (7595.517 X 3.511)/19.68 = 1355.07N/m
Area of Schrenks curve = 19.68607 m2
Load at any location ‘n’ = Load Intensity at root ×
Where is the lift distribution at the corresponding grid point
Lift on each element is calculated using the following formula and a graph is plotted
between lift on element and wing span.
Lift on element = Load intensity at grid point × Distance between two grid points
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Next, the shear force diagram, the bending moment diagram and the torque diagram
are to be drawn. It can be drawn using the following procedure. Here, the wing is assumed to
be a cantilever beam.
Structural load of the wing =
= A + Bx where is the chord at each station
At x = 0, = = 4.5871 m
At x = , = = 1.3761 m
Using the above conditions, we get,
A = 4.5871; B = -0.45
= 4.5871 – 0.45x
To find the value of K, first the total structural weight of the wing is taken as the wing load.
=
On solving the above equation, we get K = 23.504N/m2
Using the above value of K, the wing structural loads at other locations are calculated
and tabulated.
The resultant can be found as the difference between the lift on element and the
structural load at each station.
Resultant R = Lift on element – Structural load
span YiLIFT LOAD INTENSITY
STRUCTURAL LOAD INTENSITY
RESULTANT INTENSITY
0 3.511 1355.07 3528.679 2173.609
0.25 3.5088411 1354.236 3357.717 2003.481
0.5 3.5023684 1351.738 3191.001 1839.263
0.75 3.4915490 1347.562 3028.529 1680.967
1 3.4763453 1341.695 2870.303 1528.608
53
1.25 3.4566994 1334.112 2716.321 1382.209
1.5 3.4325350 1324.786 2566.584 1241.798
1.75 3.4037560 1313.679 2421.093 1107.414
2 3.3702440 1300.745 2279.846 979.101
2.25 3.3318563 1285.929 2142.844 856.915
2.5 3.2884222 1269.166 2010.087 740.921
2.75 3.2397387 1250.376 1881.575 631.199
3 3.1855651 1229.468 1757.308 527.84
3.25 3.1256159 1206.331 1637.286 430.955
3.5 3.0595517 1180.833 1521.509 340.676
3.75 2.9869668 1152.819 1409.976 257.157
4 2.9073728 1122.100 1302.689 180.589
4.25 2.8201764 1088.446 1199.646 111.2
4.5 2.7246477 1051.577 1100.849 49.272
4.75 2.6198753 1011.140 1006.296 4.844
5 2.5046998 966.688 915.989 50.699
5.25 2.3776095 917.638 829.926 87.712
5.5 2.2365744 863.205 748.108 115.097
5.75 2.0787580 802.296 670.535 131.761
6 1.8999834 733.298 597.207 136.091
6.25 1.6936266 653.654 528.124 125.53
6.5 1.4479424 558.833 463.286 95.547
6.75 1.1377324 531.578 402.693 128.885
7 0.6797537 262.350 346.345 83.995
Plotting the variation of the lift, structural load and resultant load intensity along the spanwise direction.
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55
56
SHEAR FORCE AND BENDING MOMENT DIAGRAM
To determine the shear force and bending moment diagram for the wing we assume that the wing is a cantilever beam with the root end fixed while the tail end is free.
For a cantilever beam the shear force is a given by,
ShearForce Rx =
Tabulation for the values of shear force and bending moment at various position along the span is as follows.
R 72 177.2 -10.49 31399.84 110.2311 2260788.48 7936.64 -133952
S 20.88 118.11 -9.49 13948.51 90.21048 291338.2932
1884.20 -23429.5
T 20.88 59.01 -7.39 3481.897 54.73188 72725.3351 1143.17 -9117.97
TOTAL
654.4138
104857445.6
310393.7691
ΣAx2= 104857445.6 cm^4
ΣAy2 = 310393.77 cm^4
Solving we get,
Ixx = 230588.0449 cm^4
Iyy = 22460575.7 cm^4
Ixy = 0
Sx = Drag × FOS × n = 17756280 N where drag = (1/2) ρV2S Cd = 1.3128 x 10^6 N
Sy = Lift × FOS × n = 43114680 N where lift = (1/2) ρV2S Cl = 3.19368 x 10^6 N
SX = = 11756280 N
SY = = 43114680 N
Section ds(sectional length) (m)
Stringer X (from centroid)
Y(from centroid)
BC 59.0962 A -259.829 -3.920
CR 29 B -236.733 3.9385
RS 59.0962 C -177.637 8.678
ST 59.0962 D -111.187 11.646
TA 70.36 E -44.737 14.747
AB 59.0962 F 21.712 14.730
CD 66.45 G 88.162 13.910
DE 66.45 H 154.612 11.380
EF 66.45 I 221.062 8.368
FG 66.45 J 287.512 4.824
GO 31.6 K 353.962 -1.643
OP 99.675 L 353.962 -19.806
PQ 62.02 M 254.287 -22.995
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QR 104.105 N 187.837 -24.767
RC 29 O 88.162 -27.727
GH 66.45 P -11.512 -29.596
HI 66.45 Q -73.532 -29.880
IJ 66.45 R -177.637 -29.419
JK 66.45 S -236.733 -28.418
KL 12.72 T -295.829 -26.318
LM 99.675
MN 66.45
MO 99.675
OG 31.6
Applying the condition
qs = - ∫yds - ∫xds
qs = - ΣydA - ΣxdA
SECTION SHEAR FLOW in N-cm FINAL SHEAR FLOW N-cm
Qbc -42682.25581 -661502.857
Qcr -386458.627 -1005279.228
Qrs 823246.4897 204425.8885
Qst 1163305.887 544485.2856
Qta 1479502.018 860681.4169
Qab 0 -618820.6012
Qcd 0 -189299.5739
Qde -135932.7013 -325232.2752
Qef -309643.0155 -498942.5894
Qfg -484240.1809 -673539.7548
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Qgo -1199524.96 -1388824.534
Qop 207453.7053 18154.13136
Qpq 557735.6805 368436.1066
Qqr 912394.7219 723095.148
Qrc 2122099.839 1932800.265
Qgh 0 -108474.1111
Qhi -137174.4334 -245648.5445
Qij -239812.2023 -348286.3134
Qjk -301625.1089 -410099.22
Qkl -289256.288 -397730.3991
Qlm -81447.85475 189921.9658
Qmn 186369.8307 77895.71956
Qno 476244.6339 367770.5228
Qog 1883223.3 1774749.189
Shear Flow of Wing
Now applying the condition ∫qds=0
And solving we get
For cell 1:
(qbc+qs0) X 59.0962 + (qcr+qs0) X 29 +(qrs+qs0) X 59.0962 +(qst+qs0) X 59.0962 +(qta+qs0) X
70.36 +(qs0) X 59.0962 = 0
Hence qo for cell 1 is qs0 = -618820.6012
Similarly solving for cells 2 and 3 we get
qs0 = -189299.5739
qs0= -108474.1111 respectively and the corrected shear flow is as shown above.
EX NO:3C
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WING STRUCTURAL LAYOUT
SPECIFIC ROLES OF WING (MAINPLANE) STRUCTURE:
The specified structural roles of the wing (or main plane) are:
• To transmit: wing lift to the root via the main span wise beam
Inertia loads from the power plants, undercarriage, etc, to the main beam.
Aerodynamic loads generated on the aerofoil, control surfaces & flaps to the main beam.
• To react against:
Landing loads at attachment points
Loads from pylons/stores
Wing drag and thrust loads
• To provide:
Fuel tank age space
Torsional rigidity to satisfy stiffness and aeroelastic requirements.
To fulfill these specific roles, a wing layout will conventionally compromise:
• Span wise members (known as spars or booms)
• Chord wise members(ribs)
• A covering skin
• Stringers
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Basic Functions of wing Structural Members
The structural functions of each of these types of members may be considered independently as:
SPARS
• Form the main span wise beam
• Transmit bending and torsional loads
• Produce a closed-cell structure to provide resistance to torsion, shear and tension
loads.
In particular:
• Webs – resist shear and torsional loads and help to stabilize the skin.
• Flanges - resist the compressive loads caused by wing bending.
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SKIN
• To form impermeable aerodynamics surface
• Transmit aerodynamic forces to ribs & stringers
• Resist shear torsion loads (with spar webs).
• React axial bending loads (with stringers).
STRINGERS
• Increase skin panel buckling strength by dividing into smaller length sections.
• React axial bending loads
RIBS
• Maintain the aerodynamic shape
• Act along with the skin to resist the distributed aerodynamic pressure loads
• Distribute concentrated loads into the structure & redistribute stress around any discontinuities
• Increase the column buckling strength of the stringers through end restraint
• Increase the skin panel buckling strength.
WING BOX CONFIGURATIONS
Several basic configurations are in use now-a-days:
• Mass boom concept
• Box Beam(distributed flange) concept-built-up or integral construction
• Multi-Spar
• Single spar D-nose wing layout
Mass Boom Layout
In this design, all of the span wise bending loads are reacted against by substantial booms or flanges. A two-boom configuration is usually adopted but a single spar “D-nose” configuration is sometimes used on very lightly loaded structures. The outer skins only react
74
against the shear loads. They form a closed-cell structure between the spars. These skins need to be stabilized against buckling due to the applied shear loads; this is done using ribs and a small number of span wise stiffeners.
BOX BEAM OR DISTRIBUTED FLANGE LAYOUT
This method is more suitable for aircraft wings with medium to high load intensities and differs from the mass boom concept in that the upper and lower skins also contribute to the span wise bending resistance
Another difference is that the concept incorporates span wise stringers (usually “z” section) to support the highly –stressed skin panel area. The resultant use of a large number of end-load carrying members improves the overall structural damage tolerance.
Design Difficulties Include:
• Interactions between the ribs and stringers so that each rib either has to pass below the stringers or the load path must be broken. Some examples of common design solutions are shown in figure
• Many joints are present, leading to high structural weight, assembly times, complexity, costs & stress concentration areas.
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The concept described above is commonly known as built-up construction method. An alternative is to use a so-called integral construction method. This was initially developed for metal wings, to overcome the inherent drawbacks of separately assembled skin-stringer built-up construction and is very popular now-a-days. The concept is simple in that the skin-stringer panels are manufactured singly from large billets of metal. Advantages of the integral construction method over the traditional built-up method include:
• Simpler construction & assembly
• Reduced sealing/jointing problems
• Reduced overall assembly time/costs
• Improved possibility to use optimized panel tapering
Disadvantages include:
• Reduced damage tolerance so that planks are used
• Difficult to use on large aircraft panels.
TYPES OF SPARS
In the case of a two or three spar box beam layout, the front spar should be located as far forward as possible to maximize the wing box size, though this is subject to there being:
• Adequate wing depth for reacting vertical shear loads.
• Adequate nose space for LE devices, de-icing equipment, etc.
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This generally results in the front spar being located at 12 to 18% of the chord length. For a single spar D-nose layout, the spar will usually be located at the maximum thickness position of the aerofoil section. For the standard box beam layout, the rear spar will be located as far as aft as possible, once again to maximize the wing box size but positioning will be limited by various space requirements for flaps control surfaces spoilers etc.
This usually results in a location somewhere between about 55 and 70% of the chord length. If any intermediate spars are used they would tend to be spaced uniformly unless there are specific pick-up point requirements
RIBS
For a typical two spar layout, the ribs are usually formed in three parts from sheet metal by the use of presses and dies. Flanges are incorporated around the edges so that they can be riveted to the skin and the spar webs Cut-outs are necessary around the edges to allow for the stringers to pass through Lightening holes are usually cut into the rib bodies to reduce the rib weight and also allow for passage of control runs fuel electrics etc.
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RIB CONSTRUCTION AND CONFIGURATION
The ribs should be ideally spaced to ensure adequate overall buckling support to spar flanges .In reality however their positioning is also influenced by
• Facilitating attachment points for control surfaces, flaps, slats, spoiler hinges, power plants, stores, undercarriage attachments etc
• Positions of fuel tank ends, requiring closing ribs
• A structural need to avoid local shear or compression buckling.
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RIB ALIGNMENT POSSIBILITIES
There are several different possibilities regarding the alignment of the ribs on swept-wing aircraft
(a) Is a hybrid design in which one or more inner ribs are aligned with the main axis while the remainder is aligned perpendicularly to the rear spar
(b) Is usually the preferred option but presents several structural problems in the root region
(c) Gives good torsonal stiffness characteristics but results in heavy ribs and complex connection
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EX NO:4
FUSELAGE:
fuselage (from the French fuselé "spindle-shaped") is an aircraft's main body section that
holds crew and passengers or cargo. In single-engine aircraft it will usually contain an engine,
although in some amphibious aircraft the single engine is mounted on a pylon attached to the
fuselage which in turn is used as a floating hull. The fuselage also serves to position control
and stabilization surfaces in specific relationships to lifting surfaces, required for aircraft
stability and maneuverability.
EX NO:4A
FUSELAGE STRUCTURAL ANALYSIS
Structural analysis of fuselage like that of wing is of prime importance while
designing an aircraft. As the fuselage is the one which houses the pilot, the power plant and
also part of the payload its structural integrity is a matter of concern. While analysing the
fuselage structure the section must be idealized. Idealization involves the conversion of a
stringer and its accompanying skin thickness into a concentrated mass known as a boom. The
shear flow analysis of the fuselage simulating flight conditions is shown below.
The fundamental purpose of the fuselage structure is to provide an envelope to support the payload, crew, equipment, systems and (possibly) the power plant. Furthermore, it must react against the in-flight maneuver, pressurization and gust loads; also the landing
88
gear and possibly any power plant loads. Finally, it must be able to transmit control and trimming loads from the stability and control surfaces throughout the rest of the structure.
FUSELAGE LAYOUT CONCEPTS
There are two main categories of layout concept in common use:
• Mass boom and longeron
• Semi-monocoque layout
MASS BOOM & LONGERON
This is fundamentally very similar to the mass-boom wing-box concept. It is used
when the overall structural loading is relatively low or when there are extensive cut-outs in
the shell. The concept comprises four or more continuous heavy booms (longerons), reacting
against any direct stresses caused by applied vertical and lateral bending loads. Frames or
solid section bulkheads are used at positions where there is distinct direction changes and
possibly elsewhere along the lengths of the longeron members. The outer shell helps to
support the longerons against the applied compression loads and also helps in the shear
carrying. Floors are needed where there are substantial cut-outs and the skin is stabilized
against buckling by the use of frames and bulkheads.
SEMI MONOCOQUE LAYOUT
This is the most common layout, especially for transport types of aircraft, with a
relatively small number and size of cut-outs in use. The skin carries most of the loading with
the skin thickness determined by pressurization, shear loading & fatigue considerations.
89
Semi Monocoque fuselage layout
Longitudinal stringers provide skin stabilization and also contribute to the overall load
carrying capacity. Increased stringer cross-section sizes and skin thicknesses are often used
around edges of cut-outs. Less integral machining is possible than on an equivalent wing
structure. Frames are used to stabilize the resultant skin-stringer elements and also to transmit
shear loads into the structure. They may also help to react against any pressurization loads
present. They are usually manufactured as pressings with reinforced edges. Their spacing
(pitch) is usually determined by damage tolerance considerations, i.e. crack-stopping
requirements. The frames are usually in direct contact with the skin; stringers pass through
them and are seated into place.
Bulkhead designs
90
PRESSURE CALCULATION OF FUSELAGE
( / 4 )Pd tσ =
Where,
P=pressure difference between sea level and
optimum altitute
σ =stress at (Kn/m^2)
d = diameter of fuselage
t = thickness of the sheet(skin)
P =Pressure at sea level – pressure at operating level
= (1.2256-0.82)
= 0.3556 2/kg m
Hoop stress:
((0.355*2) / (2*0.13)σ =
20.1846 /kg m=
Longitudinal stress:
( / 4 )Pd tσ =
P=pressure difference between sea level and
optimum altitute
91
σ =stress at (kN/m^2)
d = diameter of fuselage
t = thickness of the sheet(skin)
P =Pressure at sea level – pressure at operating level
((0.355*2) / (4*0.13)σ =
EX NO:9
SHEAR FORCE UNDER W/S:
92
20.3692 /kg m=
Shear force
0
/ 2
(1.030*14.3) / 2
7.36
B
A
F
F wl
=
===
BENDING MOMENT DIAGRAM:
2
0
(14.3) *1.03 / 6 35.2 /
B
A
M
M kN m
=
= =
93
EXNO:10
94
SHEAR FORCE UNDER 3W/S
(MANUEVERABILITY) CONDITION:
0
(3.04*(14.33 / 2)) 12.87
B
A
SHEARFORCE
F
F kN=
=− =
95
BENDING MOMENT AT 3W/S:
(MANUVERABILITY)
2
0
(3.04)*(14.3) / 6 34.15
B
A
M
M kN
== =
96
STABILITY AND CONTROL
A general treatment of the stability and control of the aircraft requires a study of the dynamic of flight. In which we consider not the motion of the aircraft but only its equilibrium states. It is commonly called stability and control analysis
Mean aerodynamic cord:
108.1284c in=
Location of center of gravity percent of chord:
0.5cgcg
XX
c= =
Location of aerodynamic center in percent of chord:
0.24acac
XX
c= =
For wing:
97
2
2
2
12*3.14
1.0.25
L
L
CM
C
α
α
π=−
=
6.477LCα
= per radian
0MCα
= , because of symmetric airfoil.
Side slip, β
2 21 Mβ = −
2 21 0.25
0.968
ββ
= −=
Airfoil efficiency, η
2
6.477*0.968
2*3.140.998
LCαη π
β
η
η
=
=
=
Airfoil efficiency is approximately assumed to be 0.95
Fuselage lift factor, F
98
2
2
1.07(1 )
3.561.07(1 )
11.31.176
dF
b
F
F
= +
= +
=
20.058effA m=
3wet
ref
S
S= from the historical data.
For the airfoil t
c<0.5,
Swet=2.003Sexposed
exp 31.5
2.003osed
ref
S
S= =
max 0t
Λ = , because of no sweep
exp
22 2max
2 2
2*
tan2 4 (1 )t
osedL
wet
SAC F
SAα
π
βη β
=Λ
+ + +
2 2
2
2*3.14*10.3*1.497*1.176
10.3 *0.9682 4 *1
0.998
LCα
=+ +
5.588LCα
= rad
99
Downwash:
216.05*2
42.50.755
tlrb
r
r
=
=
=
( )hL L n oLhC C i
αα ε λ= + − −
1nα εα α
∂ ∂= −∂ ∂
1.62
1.62*5.588
3.14*10.3
0.279
LC
Aαε
α πεαεα
∂ =∂∂ =∂∂ =∂
1 0.279 0.72nαα
∂ = − =∂
100
fus h
h
ph ph nach pL m n L
W Wnp
phh nL n L
W W
FSC C C X X
S qSX
FSC C
S qS
α α
α
ααηα α
αηα
∂∂− + +∂ ∂=
∂+ +∂
0.712npX =
*
0.712*108.1284
76.98
npnp
np
np
X X c
X
X
=
=
=
i.e. 23% of chord
Static margin:
The difference between the CG position and the NP is sometimes called static margin
static margin=( np cgX X− )
76.98 71.02
108.12840.0551
−=
=
i.e. 5.51% stable
If mC were given by the curve b, the moment acting when disturbed would be positive, or nose-up, and would tend to rotate the airplane still farther from its equilibrium attitude. We see that the pitch stiffness is determined by the sign and
magnitude of the slope mC
α∂
∂ .If the pitch stiffness is to be
positive at the equilibrium α , mC must be zero,and mC
α∂
∂ must be negative. It will be appreciated from the figure that an
alternative statement is “ 0mC must be positive,and mC
α∂
∂ negative if the airplane is to meet this (limited) condition for
101
stable equilibrium.”. the various possibilities corresponding to
the possible signs of 0mC and mC
α∂
∂ are shown in figures.
( )
6.47(0.0551)
0.356
np cgM L
M
M
C C X X
C
C
α α
α
α
= − −
= −
= −
Stick free:
Elevator area ; 40% of tail area
So, 0.7n hL LC C
α α; (fixed)
fus h
h
ph ph nach pL m n L
W Wnp
phh nL n L
W W
FSC C C X X
S qSX
FSC C
S qS
α α
α
ααηα α
αηα
∂∂− + +∂ ∂=
∂+ +∂
0.68
*
0.68*108.1284
73.527
np
npnp
np
np
X
X X c
X
X
=
=
=
=
static margin=( np cgX X− )
102
73.527 71.02
108.12840.0231
−=
=
i.e. 2.31% stable
THE FORWARD LIMIT:
As the CG moves forward, the stability of the airplane increases, and larger control movements and forces are required to maneuver or change the trim. The forward CG limit is therefore based on control considerations and may be determined by any one of the following requirements:
The control force per g shall not exceed a specified value. The control-force gradient at trim, P
V∂
∂ , shall not exceed a specified value.
The control force required to land, from trim at the approach speed shall not exceed a specified value.
The elevator angle required to land shall not exceed maximum up elevator.
The elevator angle required to raise the nose wheel off the ground at take off speed shall not exceed maximum up elevator
( )
6.47*0.0231
0.15
np cgM L
M
M
C C X X
C
C
α α
α
α
= − −
= −
= −
( ) ( ) ( )cg N ff fus h
hcg acW ach cg cg pM L M M p m n L
W W
S TC C X X C C f C C X X X X
S qSη= − + + + − − − −
246.47 (0.5 0.24) 0 0 0.21 [0.7* * (3.21 0.5)
108.1284cgM eC α α δ= − − + + + − −
0.5065 0.391cgM eC α δ= −
103
Trim plot:
α \ eδ 0o 4o 8o 12o
0o 0 1.404 2.808
4.212
4o 4.976 6.3 7.784
9.188
8o 9.952 11.356
12.76
14.164
12o 14.928
16.332
17736
19.14
Since the criterion to be satisfied is 0mC < 0, that is, positive
pitch stiffness , then we see that we must have h< hn, or nk > 0 .In other words the CG must be forward of the NP. The farther
forward the CG the greater is nk ,and the sense of “static stability” the more stable the vehicle.
The neutral point has sometimes defined as the CG location
at which the derivative 0m
L
dCdC = .
104
0
0
[( )(1 ) ( ) ]
[0.7292 (0 0) ]
h h
h h
L L w h w Lh
L L Lh
C C i i i
C C
α
α
εα αα
α
∂= + − + − −∂
= + − − ∆
0
0
0
1* *
0.7*6.47*0.5*
7.910.286
LLh f
L f
Lh f e
Lh e
C
C
k
α
α δδ
α δ
α δ
∂−∆ =∂
−∆ =
∆ = −
0.9 * * *cosflappedLf HL
F f ref
SC Ck
Sλ
δ δ∂ ∂=∂ ∂
0.5065 0.391*791(0.6 0.286 )
1.3302 0.9734cg
cg
M e
M e
C
C
α α δ
α δ
= − +
= − −
Lift:
( )
240.5065 0.7 7.91*(0.6 0.286 )
108.12841.224 0.351
h
hTotal L W n L
W
Total e
Total e
SC C i C
S
C
C
αα η
α α δ
α δ
= + +
= + +
= +
Trim plot:
105
α \ eδ 0o 4o 8o 12o
0o 0 -3.8936 -7.7872 -11.6808
4o -5.3208 -9.2144 -13.108 -17.0016
8o -10.6416
-14.5352
-18.4288
-22.3224
12o -15.9624
-19.856 -23.7496
-27.6432
106
DESIGN OF LANDING GEAR
We have designed the landing gear characteristics by following a step by step
method.
1) Landing gear System
We have chosen a Retractable system landing gear which will be retracted in to
the fuselage after the take off.
2) Landing Gear Configuration
The landing gear configuration we have adapted is the Conventional type or Tri-
cycle type with a nose wheel in front. From an ease of ground manoeuvring
viewpoint as well as ground looping the nose wheel configuration is preferred.
3) Preliminary landing gear strut disposition
There are two geometric criteria which are required to be considered on deciding
the disposition of landing gear struts are:
A) Tip-over criteria
B) Ground clearance criteria
A) Tip-over Criteria :
107
a) Longitudinal Tip-over Criterion :
For tricycle gears the main landing gear must be behind the aft CG location. The 15 deg angle
as shown in the Fig. represents the usual relation between main gear and the aft CG.
Longitudinal tip over criterion
b) Lateral Tip-over Criterion :
The lateral tip-over is dictated by the angle ψ in the Fig.
108
Lateral Tip-over Criterion
B) Ground Clearance Criterion :
a) Longitudinal Ground Clearance Criterion :
Longitudinal Ground Clearance Criterion
b) Lateral Ground Clearance Criterion :
109
Lateral Ground Clearance Criterion
4) Number of Wheels :
Nose landing gear - 1
Main landing gear - 2
TYRE SIZING;
Nearly 90% of the load is carried by the main landing gear.
• Only of 10% of aircraft is carried by nose wheel. But it experience dynamic loads.
• Nose wheel size could be 60-100% of size of main wheel.• But in the bicycle and quarter cycle configuration the size
same.
110
WW → Weight of the wheel
rR → Rolling radius
pA → Area of the foot print
d → Diameter
W → width
PERFORMANCE PARAMETER;
*Operating a tyre at the lower pressure will greatly improve the tyre life.
*Largest tyre cause drag, weights the space occupied etc.
1. VERTICAL LOAD FACTOR 2. SPIN UP 3. SPIN BACK 4. BRAKIN G 5. ON E WHEEL ARRESTED6. TURNING LOADS 7. TAXYING LOADS
• when the aircraft touches the ground wheels are not rotating .Then after fraction of second it will spin up. it is called spin load.
• spin up loads nearly 50% of the actual load acting on the landing gear. Once it starts to rotate, the rearward force is released and gear strut springs back forward.
• Breaking load can be estimated by braking co-efficient i.e. 0.8 Spinback load ≥ spinup load test
• The aircraft is subjected to find out the vertical load factor(from 23 to 48 cm).
• For entire layout• Shock absorbers• Skid controls• Steering systems• Retracting mechanisms• Cockpit requirements and strength
PRIMARY LANDINGGEAR DESIGN:
• Selection of tyres• Construction method• Temperature effects• Tyre friction
Kinetic analysis of the brakes, skids, controls and wheels also performed in the design.
DETAILED LANDINGGEAR DESIGN:
• Material selection• Laws• Pushing seals• Lubrication
TYRE SIZING:
Main wheel diameter or width = BwAW
For bomber aircraft the A = 1.63, B = 0.315
d = 0.3151.63[19335.0]
d =36.50 in
wW - weight on wheel
114
10% - nose wheel
90% - main wheel
Width A = 0.1043, B = 0.480
Width of the wheel = BwAW
w = 0.480.1043[19335.0]
w = 11.90 in
Maximum pressure, P =350 2
lb
in
Type Plys
Outside
Dia
in
Normal
Width
in
Rated
Load
lb
Max.rated
Speed
Km
hr
Normal loaded
Radius
in
Tyre
Weight
lb
32*11.5
6 36 13 6350 257 14.1 53.2
STATIC MARGIN:
It is the distance between C.G and neutral point. When the static margin is minimum, the lateral stability of aircraft is poor and when it is maximum, more power is required to maneuver. Hence optimization of static margin in aircraft is required.
Foot print area, PA :
*W PW P A=
PA = 19335
350
PA = 55.242 2in
115
Rolling radius, rR :
2.3* * [ ]2p r
dA w d R= −
55.242 2.3* 11.9*36.5[18.25 ]rR= −
rR =17.098 in
Shock absorber:
Shock absorber is generally based on the compression ratios. For larger aircraft compression ratio from the static condition to the fully extended condition i.e.4:1 and compressed to static is 3:1.
:
116
Stoke of the shock absorber:
2
*2* * *
1*
2
vertical tT
gear
T r
VS S
g N
S d R
ηη η
= −
= −
TS =1.152 in
2283.3 0.47
*2.927*102*9.81*0.8*3*60*60 0.82.37
S
S cm
−= −
=
tη → Tyre efficiency(0.45-0.47)
N → Gear load factor(3-4)
η → Shock absorber efficiency(0.75-0.9)
OLEO SIZING:
The size of oleo strut will be approximate that of stroke length.
117
0.04oleo oleoD L=
oleoD =2.5*23.7
oleoD =0.5925m
The landing gear will be placed in the fuselage.
SHOCK ABSORBER DESIGN:
Static load = 0.3* 0W
= 5846.808
Load to extended = 0.25 * SP
= 0.25*5846.808
=1461.702 kg
Load to compress = 0.3* SP
= 0.3*5846.808
=1754.0424 kg.
Load (kg)
Stroke (cm)
1461.702
0
5846.808
15.8
8770.212
23.7
Static pressure in strut = 350psi
118
Piston area =5846.808*2.204585
350
= 36.82 2inch
(1 kg = 2.204585 lb),
3V =10 % displacement
= 0.1*23.7*0.3937008*36.82
3V = 34.35 3inch
3P = maximum strut pressure = 1050psi (3*350)
1V = 3V +0 =34.35 3inch
Calculation of static volume:
It is confirmed that the piston and cylinder volume are enough for the static load.
Oleo sizing:
For Oleo –pneumatic metered orifice
η =0.8,
The size of oleo strut (D) =1.34 oleol
pπ
L = Load on oleo
119
D = 1.33
4*9.81*19489.36
7239.491*3.14*10
D = 0.2384m
VERTICAL KINETIC ENERGY:
2
2
1. * *
2
. .5*83.33 *8770.21
vertical vertical
vertical
wK E V
g
K E
=
=
. verticalK E =30.44 MJ
120
. * *
. 0.8* *0.02927*9.81absorbed T
absorbed
K E L S
K E L
η==
For nose wheel, L= 0.1* 0W
. absorbedK E = 0.477 KJ
For main wheel, L=0.45* 0W
. absorbedK E = 2.014 KJ
Therefore the landinggear for bomber aircraft is tricycle type arrangement with 4 tyres in a single strut for main and 1 tyre for nose wheel.
121
VIEW DIAGRAM
AIMTo draw the 3-view diagram of the aircraft that has been designed.