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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
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:
2 2max max( ) ( ) ( max( )atCN contolable CL controlable CD CL con= +
1.1*1.16
1.278contrable
contrable
N
N
C
C
=
=
16.08sV = m/s
DESIGN LIMIT LOAD FACTORS nlimpos and nlimneg
The positive design limit load factor, nlimpos, must be selected by the designer, but must
meet the following condition:
For fighter aircraft, nlimpos = 3
nlimneg = -1
3 ≥ 2.1 + (24000 / (19485.2+10000))
≥ 2.24
Hence our design satisfied the above condition.
36
The negative design limit load factor nlimneg, must be selected by the designer, but must
meet the following condition:
nlimneg ≥ 0.4 nlimpos for normal and for utility category airplanes
nlimneg ≥ 0.5 nlimpos for acrobatic category airplanes
therefore,
3 ≥ 0.4 × 1
3 ≥ 0.4
Hence our design satisfied the above condition.
DESIGN MANEUVERING SPEED VA
The design maneuvering speed VA , must be selected by the designer, but must satisfy the
following relationship:
VA ≥ 70.76m/s
DESIGN CRUISE SPEED VC
The design cruise speed VC, must be selected by the designer, but must satisfy the following
relationship:
kc = 33 for utility category
17540.2433
1.224*86.7424.25
C
C
V
V
≥
≥
37
For our designed aircraft the cruise speed is VC = 424.25m/s which satisfy the above
condition.
DESIGN DIVING SPEED VD
The design diving speed must satisfy the following relationship:
VD ≥ 1.5 VC
VD ≥ 710.99m/s
For our designed aircraft the diving speed is VD = 710.99m/s which satisfy the above
condition.
Thus the manoeuvring V-n diagram is obtained as shown below
38
V – n MANOEUVRE DIAGRAM
Cl=1.16
212 lV SC
nW
ρ ∞=
AT ALTITUDE = 2000m
n = 5.1409 × 10-5 V2
(-n) V (+n)0 0 0
-0.12 50 0.122-0.22 100 0.78-0.36 150 1.2-0.48 200 1.8-0.66 250 2.2-0.78 300 2.6
-1 350 3
39
GUST V – n DIAGRAM
The gust V-n diagram is given by the following formulae
which gives positive limit as 9.7 and negative limit -3.5
Where
40
Substituting the values in the given equations we get,
2( / )
2*17540.24
1.224*4.08*9.81*0.1*6.782.59
FDGWg
L
g
g
W S
CgCα
µρ
µ
µ
=
=
=
.88
5.3g
g
kgµµ
=+ (subsonic aircraft)
41
0.826kg =
We know,
At high angle of attack, u = kû = 11.47m/s
At level flight, u = 7.548m/s
At dive condition, u = 3.774m/s
42
43
EX NO:3 WING
INTRODUCTION TO COMPONENTS OF THE
WING:
FUEL TANKS
The study of the fuel tanks and its corresponding feed systems is of importance to us
because, as the fuel is stored in the wings and also externally. Thus structural analysis results
are incorporated here to ensure that the fuel tanks are positioned suitably and are adequate in
capacity so that our fighter aircraft has an endurance of 1.5 hours. The C.G analysis carried
out in ADP-I let us know that our aircraft’s C.G remains within acceptable limits even on
consumption of fuel consumption.
Internally, the fuel is carried in the wings. The wing tanks are capable of withstanding
up to 17200lb of fuel. Externally however we had two choices, conformal tanks or drop
tanks. Drop tanks are heavy and are an aerodynamic liability (drag). Drop tanks severely
hamper agility, and our aircraft being a fighter requires excellent manoeuvrability. Taking
into consideration the above mentioned points, we have adopted conformal fuel tanks.
Conformal Fuel Tanks (CFTs) are additional fuel tanks fitted closely to the profile of
an aircraft which extend either the range or time on station of the aircraft, with little
aerodynamic penalty compared to the same fuel capacity carried in external drop tanks.
Conformal tanks are plumbed into the aircraft, and can only be removed on the ground. CFTs
also have the advantages of not significantly increasing an aircraft's radar cross-section, and
allowing a higher maximum speed than with drop tanks. We have also incorporated in-flight
refuelling using a probe and drogue system.
44
Fig.33 Fuel Tank Layout Fig. 34 Conformal Fuel Tanks
RIB LOCATION AND DIRECTION
The span-wise location of ribs is of some consequence. Ideally, the
rib spacing should be determined to ensure adequate overall buckling support to the
distributed flanges. This requirement may be considered to give a maximum pitch of
the ribs. In practice other considerations are likely to determine the actual rib
locations such as:
a) Hinge positions for control surfaces and attachment/operating points for flaps,
slats, and spoilers.
b) Attachment locations of power plants, stores and landing gear structure.
c) A need to prevent or postpone skin local shear or compression buckling, as
opposed to overall buckling. This is especially true in a mass boom form of
construction.
d) Ends of integral fuel tanks where a closing rib is required. When the wing is
unswept, it is usual for the ribs to be arranged in the flight direction and thereby
define the aerofoil section. While the unswept wing does give torsional stiffness,
the ribs are heavier, connections are more complex and in general the
disadvantages overweigh the gains.
Ribs placed at right angles to the rear spar are usually he most satisfactory in
facilitating hinge pick-ups, but they do cause layout problems in the root regions. Some
designs overcome this by fanning the ribs so that the inclination changes from
perpendicular to the spars outboard to stream-wise over the inboard portion of the wing.
45
Tank 1
Tank 2
Tank 3
There is always the possibility of special exceptions, such as power plant or store
mounting ribs, where it may be preferable to locate them in the flight direction.
FIXED SECONDARY STRUCTURE
A fixed leading edge is often stiffened by a large number of closely
pitched ribs, span-wise members being absent. Providing care is taken in the detail design
of the skin attachment it is possible to arrange for little span-wise end load to be diffused
into the leading edge and buckling of the relatively light structure is avoided. This may
imply short spam-wise sections. The presence of thermal de-icing, high-lift devices or
other installations in the leading edge also has a considerable influence upon the detail
design. Bird strike considerations are likely to be important.
Installations also affect the trailing edge structure where much depends
upon the type of flaps, flap gear, controls and systems. It is always aerodynamically
advantageous to keep the upper surfaces as complete and smooth as is possible. Often
spoilers can be incorporated in the region above flaps or hinged doors provided for ease
of access. The hinges should be flexible and frequently use a continuous, ‘piano’, type.
AUXILIARY SURFACES
The structural layout of the auxiliary lifting surfaces is generally similar to that
of the wing but there are differences, in part due to the smaller size and in part due to the
need to provide hinges or supports. The latter implies that each auxiliary surface is a well-
defined.
HINGED CONTROL SURFACES
Conventional training edge control surfaces are almost invariably supported by a
number of discrete hinges, although continuous, piano type, hinges may be used for
secondary tabs. To some degree the number and location of the discrete hinges depends upon
the length of the control. The major points to be considered are:
a) The bending distortion of the control relative to the fixed surface must be
limited so that the nose of the control does mot fouls the fixed shroud.
b) The control hinge loads and the resulting shear forces and bending moments
should be equalized as far as is possible.
46
c) Structural failure of a single hinge should be tolerated unless each hinge is of
fail-safe design and can tolerate cracking one load path.
These points suggest the use of a relatively large number of discrete hinges but there
are difficulties associated with this solution there are the obvious loads likely to be induced in
the control by the distortion under load of the main surface to which it is attached may be
significant. These problems do not arise if only two hinge points are used as any span-wise
distortion or misalignment can be accommodated by designing one of the hinges so that it can
rotate about a vertical axis. When more than two hinges are used the ‘floating’ hinge concept
cannot fully overcome the problems. However, it is possible to design the control surface so
that it is flexible in bending and indeed the more hinges there are the easier this is to
accomplish. One hinge must always be capable of reacting side loads in the plane of the
control surface. The hinges are supported near to the aft extremities of the main surface ribs.
PIVOTED CONTROL SURFACES
In certain high-performance aircraft, the whole of a stabilizing or control surface
on one side of the aircraft may be pivot about a point on its root chord. Clearly in this case,
the structural considerations are dominated by the need to react all the forces and moments at
the pivot and operating points. Thus the structural layout may consist of an integral root rib or
pivot or stub spar arrangement to which is attached a number of shear webs fanning out
towards the extremities of the surface, possibly in conjunction with full depth honeycomb.
High skin shear loading is inevitable due to the need to bring the loads to the two
concentrated points. Shear loads due to torsion may be limited by locating the operating point
on the root rib some distance away from the pivot.
Some designs incorporate the pivot into the moving surface with the support
bearings on the fuselage, while on others the pivot is attached to the fuselage and the bearings
are in the surface. The bearings should be as far apart as local geometry allows to minimize
loads resulting from the reaction of the surface bending moment.
HIGH LIFT SYSTEMS
There is a wide variety of leading and trailing edge high-lift systems. Some types
are simply hinged to the wing, but many require some degree of chord-wise extension. This
can be achieved by utilizing a linkage, a mechanism, a pivot located outside the aerofoil
47
contour or, perhaps most commonly, by some from of track. Trailing edge flaps may consist
of two or more separate chord-wise segments, or slats, to give a slotted surface and these
often move on tracts attached to the main wing structure.
The majority of flaps and slats are split into span wise segments of no greater lengths
than can be supported at two or three locations. As with control surfaces, the locations of the
support points are established so as to minimize local deformations since the various slots are
critical in determining the aerodynamic performance. Sometimes the actuation may be
located at a different pan wise position from the support points. This is often a matter of
convenience, layout clearances, and the like.
The structural design of flaps is similar to that of control surfaces but it s simpler as
there is no requirement for mass balance, the operating mechanisms normally being
irreversible. On large trailing edge flap components, there is often more than one spar
member. Especially when this assists in reacting the support or operating loading. There may
be a bending stiffness problem in the case of relatively small chord slat segments and full
depth honey combs can be used to deal with this. Figure shows a cross section of a typical
slotted flap of metal construction but the same layout applies if composite materials are used.
In many cases the slipstream or afflux from power plants impinges upon a flap and
this is likely to require special consideration in the design. Additional stiffness is not
necessarily the answer because acoustic fatigue characteristics are often worse at higher panel
frequencies. However the extensive local support offered by sandwich construction, either in
panel or full depth configuration, is usually beneficial. This leads naturally to the application
of reinforced plastic materials. Trailing edge flaps tends to be prone to damage by debris
thrown up by the landing gear and it may be desirable to use Kevlar or glass rather than
carbon fibers for the lower surface, but material compatibility needs to be considered.
ATTACHMENT OF LIFTING SURFACES
The joint of the fuselage with the wing is subjected to heavy load inputs and there is
a potential for considerable relative distortion. This distortion is usually accepted and the
wing centre box is built completely into the fuselage, the resulting constraint stresses being
allowed for. It is usual for the wing structure of large aircraft to include a production joint at
the side of the fuselage and this is virtual essential for swept wings.
48
It is sometimes possible to arrange the wing pick-ups as pivots on the neutral axis or
set them on swinging links. In this case, the relative motion is allowed to take place and there
are no induced stresses. Structural assembly of the wing to the fuselage is relatively simple.
Similar remarks also apply to the attachment of the horizontal stabilizer when the
incidence setting is fixed. If the surface is also used for trimming or control, some special
consideration is necessary in the location of the pivot and actuation fittings. These usually
require a relatively heavily loaded rib or a pair of ribs, and where possible at least one of the
attachment points should be close to the rib or spar intersection. It is desirable to arrange for
the lateral distance between the pivots to be as great as possible to minimize pivot loads
resulting from asymmetric span-wise loading. When the controls are manually operated, it is
simplest if the elevator-hinge line and pivot coincide.
Fins are usually built integrally with the rear fuselage. This is mainly due to the
different form of loading associated with the geometric asymmetry
EX NO:3A
AIR LOADS ON WING
With the Vn-diagram complete, the actual loads and load distribution on the wing can
be determined. Before the actual structural members can be sized and analyse, the loads they
will sustain must be determined. Aircraft loads estimation, a separate discipline of aerospace
engineering, combines aerodynamics, structures and weights.
Initially we have to calculate the lift produced by the wings. Once the lift on the
wings is known, the spanwise and chordwise load distributions can be determined.
According to classical wing theory, the span wise lift or load distribution is
proportional to the circulation at each station. A vortex lifting –line calculation will yield the
spanwise lift distribution. For an elliptical plan form wing, the lift and load distributions is of
elliptical shape.
For a nonelliptical wing, a good semi-empirical method for spanwise load estimation
is known as Schrenk’s Approximation. This method assumes that the load distribution on an
49
untwisted wing has a shape that is the average of the actual planform shape and an elliptical
shape of same and area. The total area under the lift load curve must sum to the required total
lift.
Fig. 3 Schrenk’s approximation for various Wing Planforms
To find the lift distribution in aircraft wing, the following procedure is followed:
1) Plan-form shape wing is plotted.
2) Ellipitic distribution is drawn using the formula
CALCULATIONS:
We know,
4 2
ab Planform Areaπ =
Where; a =
a = 14.27/2 = 7.135m
π x 7.135 x b / 4 = 39.35/2
b = 3.511m
To construct the ellipse,
2 2
2 21
x y
a b+ =
2
2 22
1x
y ba
⇒ = −
50
22
21
xy b
a
∴ = −
Using the above equation, for various values of x, the values of y are found and the
ellipse is drawn
X (m) Y (m)0 3.511
0.25 3.50884110.5 3.50236840.75 3.4915490
1 3.47634531.25 3.45669941.5 3.43253501.75 3.4037560
2 3.37024402.25 3.33185632.5 3.28842222.75 3.2397387
3 3.18556513.25 3.12561593.5 3.05955173.75 2.9869668
4 2.90737284.25 2.82017644.5 2.72464774.75 2.6198753
5 2.50469985.25 2.37760955.5 2.23657445.75 2.0787580
6 1.89998346.25 1.69362666.5 1.44794246.75 1.1377324
7 0.67975377.135 0
.
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
52
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.
54
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.
SPANRESULTANT
INTENSITY(Kg/m)SHEAR FORCE(Kg) BENDING MONENT(Kg-m)
0 2173.609 0 0
0.25 2003.481 500.8703 62.60878
0.5 1839.263 919.6315 229.9079
0.75 1680.967 1260.725 472.772
1 1528.608 1528.608 764.304
1.25 1382.209 1727.761 1079.851
1.5 1241.798 1862.697 1397.023
1.75 1107.414 1937.975 1695.728
2 979.101 1958.202 1958.202
2.25 856.915 1928.059 2169.066
2.5 740.921 1852.303 2315.378
2.75 631.199 1735.797 2386.721
3 527.84 1583.52 2375.28
3.25 430.955 1400.604 2275.981
3.5 340.676 1192.366 2086.641
3.75 257.157 964.3388 1808.135
4 180.589 722.356 1444.712
57
2
2
RxBending Moment =
4.25 111.2 472.6 1004.275
4.5 49.272 221.724 498.879
4.75 4.844 23.009 54.64638
5 50.699 253.495 633.7375
5.25 87.712 460.488 1208.781
5.5 115.097 633.0335 1740.842
5.75 131.761 757.6258 2178.174
6 136.091 816.546 2449.638
6.25 125.53 784.5625 2451.758
6.5 95.547 621.0555 2018.43
6.75 128.885 869.9738 2936.161
7 83.995 587.965 2057.878
58
EX NO:3A
STRUCTURAL ANALYSIS OF WING
Wing is the major lift producing surface. Therefore, the analysis has to be very
accurate. The structural analysis of the wing consists of the following steps.
1) Spar Definition
2) Shear Flow Calculation
SPAR DEFINITION
The maximum bending moment from previous section was found to be as 9706.27
Nm. Therefore we define 3 Spars with front spar at 20% of chord, middle spar at 50% of
chord and rear spar at 80% of chord. The position of the three spars from the leading edge of
the root chord is given below as follows:
Front spar - 20% of chord = 1.77 m
Middle spar - 50% of chord = 4.43 m
Rear spar - 80% of chord = 7.088 m
59
Bending moment M = Max BM × FOS × n = 9706.27 × 1.5 × 9 = 131034.645 Nm
The Structural load bearing members in the wing are the Spars and Stringers. The
bending moment carried by the spars is 70% and that of stringers is 30% of the total Bending
Moment.
The bending moment carried by Spars is = 0.7 x 131034.645 = 91724.25 Nm
The cross section of the spar chosen here is an I-section.
For each spar we are determining the following parameters:
1) Centriod
2) Moment of Inertia
3) Bending Moment
4) Bending Stress
FRONT SPAR
Height of the spar = 38 cm
Breadth of the spar = 16 cm
Thickness of the spar = 4.5 cm
Cross Section of Front Spar
To find out the centre of gravity, the following calculations are made.
60
Element
Area(A)
(c )
x
(cm)
y
(cm)
Ax
(c )
Ay
(c )
Ax2
(c )
Ay2
(c )
Icx
(c )
Icy
(c )
1 72 8 2.25 576 162 4608 364.5 121.5 1536
2 130.5 8 19 1044 2479.5 8352 47110.5 9145.8 220.22
3 72 8 35.75 576 2574 4608 92020.5 121.5 1536
Total274.5 2196 5215.5 17568
139495.5
9388.87
3292.22
Front Spar Calculations
Centroid = X = = 8 cm ; Y= = 19 cm
Ixx= ΣIcx + ΣAy2 – ΣAY2
Ixx = (9388.87) + (139495.5) – (274.5)(19)2
Ixx = 49789.88 cm4
Iyy = ΣIcy+ ΣAx2 – ΣA X2
Iyy = (3292.22) + (17568) – (274.5)(8)2
Iyy = 3292.22 cm4
The bending moment carried by the front spar is 35% of the total bending moment carried by
the spars.
Bending moment carried by front spar = 0.35 x 91724.25 = 3210350 N cm
Bending Stress σz = (Mx/Ixx)y
The bending stress at various points whose co-ordinates are determined with centriod as the
origin are calculated from above formula and tabulated.
POINTS COORDINATES (y) (cm)BENDING STRESS
(N/cm2)A 19 1225.081405
61
B 14.5 934.9305456C 14.5 934.9305456D -14.5 -934.9305456E -14.5 -934.9305456F -19 -1225.081405
Table.4 Front Spar Bending Stress
Fig .10 Bending Stress diagram for I-Section
MIDDLE SPAR
Height of the spar = 41.6 cm
Breadth of the spar = 18 cm
Thickness of the spar = 5 cm
62
Cross Section of Middle Spar
To find out the centre of gravity, the following calculations are made.
Element
Area(A)
(c )
x
(cm)
y
(cm)
Ax
(c )
Ay
(c )
Ax2
(c )
Ay2
(c )
Icx
(c )
Icy
(c )
1 90 9 2.5 810 225 7290 562.5 187.5 2430
2158 9 20.8 1422 3286.4 12798
68357.12
13147.7 329.17
390 9 39.1 810 3519 7290
137592.9 187.5 2430
Total338 3042 7030.4 27378
206512.5
13522.7
5189.17
Middle Spar Calculations
Centroid = X = = 9 cm ; Y= = 20.8 cm
Ixx= ΣIcx + ΣAy2 – ΣAY2
Ixx = (13522.7) + (206512.5) – (338)(20.8)2
Ixx = 60467.7 cm4
Iyy = ΣIcy+ ΣAx2 – ΣA X2
Iyy = (5189.17) + (27378) – (338)(9)2
Iyy = 5189.17 cm4
The bending moment carried by the front spar is 40% of the total bending moment carried by
the spars.
63
Bending moment carried by front spar = 0.40 x 91724.25 = 3668970 N cm
Bending Stress σz = (Mx/Ixx)y
The bending stress at various points whose co-ordinates are determined with centroid
as the origin are calculated from above formula and tabulated.
Middle Spar Bending Stress
POINTS COORDINATES (y) (cm)BENDING STRESS
(N/cm2)A 20.8 1262.071751B 15.8 958.6891183C 15.8 958.6891183D -15.8 -958.6891183E -15.8 -958.6891183F -20.8 -1262.071751
REAR SPAR
Height of the spar = 17.72 cm
Breadth of the spar = 7.6 cm
Thickness of the spar = 2.5 cm
Cross Section of Rear Spar
To find out the centre of gravity, the following calculations are made.
Area(A) x y Ax
(c )
Ay Ax2 Ay2 Icx
64
Element (c ) (cm) (cm) (c ) (c ) (c ) (c )
119 3.8 1.25 72.2 23.75
274.36 29.6875 9.896 91.45
231.8 3.8 8.86 120.84
281.748
459.19
2496.287 428.76 16.56
319 3.8 16.47 72.2 312.93
274.36
5153.957 9.896 91.45
Total69.8 265.24
618.428
1007.9
7679.932
448.552 199.46
Rear Spar Calculations
Centroid = X = = 3.8 cm ; Y= = 8.86 cm
Ixx= ΣIcx + ΣAy2 – ΣAY2
Ixx = (448.552) + (7679.932) – (69.8)(8.86)2
Ixx = 2649.184 cm4
Iyy = ΣIcy+ ΣAx2 – ΣA X2
Iyy = (199.46) + (1007.9) – (69.8)(3.8)2
Iyy = 199.46 cm4
The bending moment carried by the front spar is 25% of the total bending moment carried by
the spars.
Bending moment carried by front spar = 0.25 x 91724.25 = 2293105 N cm
Bending Stress σz = (Mx/Ixx)y
The bending stress at various points whose co-ordinates are determined with centroid
as the origin are calculated from above formula and tabulated.
POINTS COORDINATES (y) (cm)BENDING STRESS
(N/cm2)
65
A 8.86 7669.120461B 6.36 5505.147419C 6.36 5505.147419D -6.36 -5505.147419E -6.36 -5505.147419F -8.86 -7669.120461
Rear Spar Bending Stress
SHEAR FLOW ANALYSIS OF WING
INTRODUCTION
Shear flow is in a solid body, the gradient of a shear stress force through the body;
in a fluid, it is the flow induced by such a force gradient In solid mechanics, shear flow is
given in dimensions of force per length. This corresponds to units of Newton per meter in the
SI system and pound-force per foot in the English Engineering and British Gravitational
systems.
The following pages give us a detailed account of the shear flow calculations.
Initially we must define the area of each stringer in the following way.
Area of front spar = 274.5 cm^2
Area of middle spar = 338 cm^2
Area of rear spar = 69.8 cm^2
Hence the total area = Area of front spar + Area of middle spar + Area of rear spar
= 274.5 + 338 + 69.8 = 682.8 cm^2
A = Total area of stringers = Area of the spar × (100/70)
= 974.714 cm^2
As our wing section has 14 stringers running along its length, the area of each stringer is
given by
Area of each stringer = (total area of stringers) / 14
Hence area of each stringer = 20.8867 cm^2
66
67
Stringer Calculations
68
Stringer Area(A) (cm2)
X (cm) Y (cm) x2 y2 Ax2 Ay2 Axy
A 20.88 59.0076 14.99 3481.897 224.9994 72725.335 4699.49 18487.09
B 20.88 118.11 22.85 13948.51 522.5247 291338.29 10913.82
56388.06
C 72 177.2 27.6 31399.84 761.6993 2260788.48 54842.35
352117.8
D 20.88 243.65 30.56 59365.32 934.3415 1239945.681
19515.31
155556.8
E 20.88 310.1 33.67 96162.01 1133.534 2008507.054
23675.79
218066.5
F 20.88 376.55 33.65 141789.9 1132.341 2961523.157
23650.87
264655.6
G 90 443 32.83 196249 1077.857 17662410 97007.11
1308961
H 20.88 509.45 30.30 259539.3 918.1627 5420919.55 19177.39
322426.9
I 20.88 575.9 27.29 331660.8 744.6786 6927299.84 15553.88
328247.4
J 20.88 642.35 23.74 412613.5 563.8155 8618134.86 11776.25
318573.8
K 19 708.8 17.27 502397.4 298.4947 9545551.36 5671.40 232672.8
L 19 708.8 -0.886 502397.4 0.784996 9545551.36 14.91 -11931.9
M 20.88 609.125 -4.075 371033.3 16.61052 7749660.509
346.94 -51852.3
N 20.88 542.675 -5.84 294496.2 34.19443 6151052.854
714.21 -66280.7
O 90 443 -8.81 196249 77.56043 17662410 6980.44 -351129
P 20.88 343.325 -10.67 117872.1 113.9834 2461958.264
2380.74 -76559
Q 20.88 281.305 -10.96 79132.5 120.1177 1652816.851
2508.86 -64394.8
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
69
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
70
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
71
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
72
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.
73
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.
75
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.
76
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.
77
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.
78
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
79
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.
80
5m0.75m
0.693m
0.53m
0.287m
0.2945m
Stringer Position on Fuselage
The stringer used is of Z type. The following are its dimensions
Cross sectional area of each stringer is 100mm^2
Cross section of Z-section
The above stringer section is uniformly used throughout the fuselage as shown above
in order to provide the fuselage the required load carrying capacity. The diagram showed
adjacent is of the idealized fuselage structure. The idealization process is carried out in the
following way.
EX NO:4B
STRESS ANALYSIS:
IDEALIZATION:
The boom 1 is given by
Where
B1 = Area of Boom 1
81
0.2c
20m
50m
tD = Thickness of skin panel
b = Circumferential distance between 2 stringers
B1 = 100+(5X(29.45/6))[ 2+ (693/750)]+ +(5X(29.45/6))[ 2+ (693/750)]
= 1535.196 mm^2
Similarly for boom 2 we get
B2 = 100+(5X(29.45/6))[ 2+ (530/693)]+ +(5X(29.45/6))[ 2+ (650/693)]
= 1535.196 mm^2
Thus solving we B1:B16 = 1535.196 mm^2. But B5=B13 = 0
We know that Ixx=ΣAy^2
Ixx =[(2 X 750^2)+(4 X 693^2)+(4 X 530^2)+(4 X 287^2)] X 1535.196
= 6.90695 X 10^9 mm^4
We know that the maximum bending moment is B.M = 28803.73 N-m
Hence the bending moment acting on the fuselage M = B.M X n X FOS
= 28803.73 X 7.75 X 3
= 669686.722 N-m
The value of stress acting is given by the expression :
= 0.0969 X y
The stress results are tabulated as shown below
82
STRINGER/BOOM Y in mm STRESS N/mm^2
1 750 72.675
2, 16 693 67.151
3, 15 530 51.357
4, 14 287 27.8103
5, 13 0 0
6, 12 -287 -27.8103
7, 11 -530 -51.357
8, 10 -693 -67.151
9 -750 -72.675
Stress in Stringers
SHEAR FORCE DIAGRAM
EX NO:4B
BENDING MOMENT;
It is defined as the sum of moments, about a section, of all external forces acting to one side of that section.
83
( )
( )
( )
0
(0.5*974.468*3.21*1.07)
1673.5
G
F
F
BM
BM
BM kg m
=
= −
= − −
( ) (0.5*1948.936*6.42*2.14) (487.234*3.21)EBM = − −
( ) 14952.04EBM kg m= − −
( ) (0.5*1948.936*6.42*12.065) (487.234*13.125) (1948.936*9.925)DBM = − − −
(104.08*9.925*4.9625) (342.68*6.715*3.357) (487.2*2.915*1.4575)− − −
( ) 116144.748DBM kg m= − −
( ) (487.234*25.68) (0.5*1948.936*6.42*24.61) (487.234* 22.34) (500*12.545)CBM = − − − −
(104.08*22.47*11.235) (342.68*16.05*11.235) (487.2*5.83*12.545)− − −
(16670.5*12.545)+
( ) 98263.8175CBM kg m= − −
( ) (487.234*28.89) (0.5*1948.936*6.42*25.68) (487.234*23.54)BBM = − − −
(500*13.615) (104.08*22.47*12.305) (342.68*16.05*12.305)− − −
(487.23*5.83*13.615) (16670.5*13.615) (194.89*1.07)− + − (2816.91*1.07)+
( ) 97319.11BBM kg m= − −
( )(487.234*28.89) (0.5*1948.936*6.42*27.32) (487.23*25.68)
ABM = − − −
(500*15.75) (104.08*22.47*14.445) (1342.68*16.05*14.445)− − −
(487.2*5.83*15.775) (16670.5*15.775) (194.89*3.21)− + −
(2816.91*3.21) (180*2.14)+ −
84
( ) 0ABM =
BENDING MOMENT DIAGRAM
CALCULATION FOR FUSELAGE SHEAR FLOW DISTRIBUTION;
The equation of shear flow at some point
q= 0yy
vq a
Iτ− ∑
Max bending moment = 80374Nm
62
9.28*10b
N
mσ =
*0.67* *(0.75* )y e bM A Hσ=
Fuselage dia,d=3.56
85
H=3.56
2=1.78
680734 *0.67*9.28*10 *0.75*1.78eA=
3 29.68*10eA m−=
39.68*10 *4*1.5effA −=
20.058effA m=
No .of stringes = 4
0.058
20*10− 28=
stringer area z A* 2 4*10z −
1 20 1.78 63.36
2 20 1.735 60.2
3 20 1.6 51.2
4 20 1.39 38.64
5 20 1.11 24.64
6 20 0.77 11.85
7 20 0.39 3.042
4 4252.932*10zzI m−=
Total, zzI =4*252.932 4 4*10 m−
Maximum shear force =117571.21N
A=20 4 2*10 m−
86
4
4
117571.21*20*10
1011.72*102324.18
Vq ydA
I
q y
q y
−
−
=
=
=
∫
stringer
distance
Shear flow
1 1.78 4137.04
2 1.735 4032.45
3 1.6 3718.688
4 1.39 3230.61
5 1.11 2579.847
6 0.77 1789.61
7 0.39 906.43
FUSELAGE SHEAR FLOW DIAGRAM
87
EX NO:4C
FUSELAGE STRUCTURAL LAYOUT
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.
ww (wt. of the wheel)=P* PA
PA (area of foot print )=2.3* *w d (2
d- rR )
Braking kinetic energy= 21* ( )
2landing
stall
WV
g
(horizontal)
Braking kinetic energy= 21* ( )
2landing
stall
WV
g
(vertical)
111
GEAR RETRACTION GEOMENTRY:
112
1. INSIDE THE WING.2. WING PADDED.3. INSIDE FUSELAGE.4. FUSELAGE PADDED5. WING FUSELAGE JUNCTION.
LOADS ON LANDING GEAR:
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).
THREE STAGES OF LANDINGGEAR DESIGN:
1. Preliminary landinggear design.2. Primary landinggear design.3. Detailed landinggear design.
PRELIMINARY LANDINGGEAR DESIGN REQUIREMENTS:
113
• 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.
FRONT VIEW
122
TOP VIEW
123
SIDE VIEW
124
125
126
