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1
1. Introduction
1.1 Overview:
The structural design of an airplane actually begins with the flight
envelope or V-n diagram, which clearly limits the maximum load factors that
the airplane can withstand at any particular flight velocity. However in normal
practice the airplane might experience loads that are much higher than the
design loads. Some of the factors that lead to the structural overload of an
airplane are high gust velocities, sudden movements of the controls, fatigue load
in some cases, bird strikes or lightning strikes. So to add some inherent ability
to withstand these rare but large loads, a safety factor of 1.5 is provided during
the structural design.
The two major members that need to be considered for the structural
design of an airplane are wings and the fuselage. As far as the wing design is
concerned, the most significant load is the bending load. So the primary load
carrying member in the wing structure is the spar (the front and rear spars)
whose cross section is an ‘I’ section. Apart from the spars to take the bending
loads, suitable stringers need to take the shear loads acting on the wings.
Unlike the wing, which is subjected to mainly unsymmetrical load, the
fuselage is much simpler for structural analysis due to its symmetrical crossing
and symmetrical loading. The main load in the case of fuselage is the shear load
because the load acting on the wing is transferred to the fuselage skin in the
form of shear only. The structural design of both wing and fuselage begin with
shear force and bending moment diagrams for the respective members. The
maximum bending stress produced in each of them is checked to be less than
the yield stress of the material chosen for the respective member.
2
1.2 Outline:
The Structural design involves:
Determination of loads acting on aircraft:
V-n diagram for the design study
Gust and maneuverability envelopes
Schrenk’s Curve
Critical loading performance and final V-n graph calculation
Determination of loads acting on individual structures
Structural design study – Theory approach
Load estimation of wings
Load estimation of fuselage.
Material Selection for structural members
Detailed structural layouts
Design of some components of wings, fuselage
1.3 Parameters forwarded from ADP – 1
Take off Gross Weight,
Maximum Velocity,
Cruise Velocity,
Stall Velocity,
3
Table 1-1: Mass ratio Split up
Components Mass Fraction
Crew 0.00053146
Landing Gear 0.042516824
Payload 0.190144687
Fixed Equipment 0.002262033
Fuselage mass 0.085033649
Horizontal Stablizer 0.012597578
Vertical Stabilizer 0.006298789
Wing Structure 0.125975776
Fuel 0.4810995
Power plant 0.053539705
Total 1
Cruise Altitude = 12 km
The airfoil used her is NACA 653 - 418
Density at cruise altitude,
Cruise C_L @ Cruise altitude,
@ 16 ˚ aoa
@ 14 ˚ since tail angle is 15.56 ˚
@ -14 ˚ aoa
cr = 11.593 m
4
ct = 5.797 m
(
)
(
)
5
2. V-n Diagram
2.1 Maneuvering Envelope:
In accelerated flight, the lift becomes much more compared to the weight
of the aircraft. This implies a net force contributing to the acceleration. This
force causes stresses on the aircraft structure. The ratio of the lift experienced to
the weight at any instant is defined as the Load Factor (n).
Using the above formula, we infer that load factor has a quadratic variation
with velocity. However, this is true only up to a certain velocity.
This velocity is determined by simultaneously imposing limiting
conditions aerodynamically ((CL)max) as well as structurally (nmax). This velocity
is called the Corner Velocity, and is determined using the following formula,
√
In this section, we estimate the aerodynamic limits on load factor, and
attempt to draw the variation of load factor with velocity, commonly known as
the V-n Diagram. The V‐n diagram is drawn for Sea level Standard conditions.
6
Figure 2-1: Typical V-n diagram for a private airliner.
Figure 2-2: V-n diagram nomenclature
7
V-n diagram is used primarily in the determination of combination of flight
conditions and load factors to which the airplane structure must be designed. V-
n diagram precisely gives the structural (maximum load factor) and
aerodynamic (maximum CL) boundaries for a particular flight condition.
2.2 Construction of V-n diagram
2.2.1 Curve OA:
Maximum Load Factor, (
)
(
)
Hence along the curve OA,
Using the above equation we get
Table 2-1: Velocity vs. positive load factor (n)
Velocity V (m/s) Load Factor (n)
0 0
20 0.08492371
40 0.339694842
60 0.764313394
80 1.358779368
100 2.123092762
120 3.057253578
140 4.161261814
160 5.435117472
180 6.87882055
200 8.49237105
220 10.27576897
240 12.22901431
8
At A,
2.2.2 Curve AC:
AC is a line limiting the maximum amount of load that can be withstood by the weakest structure of the aircraft
√*
+ √[
(
)
]
VC= 408.32 m/s
nC=n
A
2.2.3 Along CD:
The velocity at point D is given by
VD=1.5V
C= 416.66 m/s
nD= 0.75n
A= 4.80864
A straight line is used to join the points C and D
This VD
is the dive velocity or the maximum permissible EAS in which the
aircraft is at the verge of structural damage due to high dynamic pressure.
2.2.4 Along DE:
E corresponds to zero load factor point i.e
9
n= 0
For Bombers the load factor can vary from -3 to +6.5
Hence the negative load factor of aircraft is limited to -2
2.2.5 Along EF
The point F corresponds to the velocity
VC = VF = 408.32 m/s
2.2.6 Curve OG:
nF= -2 (for a typical bomber aircraft)
Hence along the curve OG,
Hence we get,
Table 2-2: Velocity vs. negative Load factor (n)
Velocity V (m/s) Load Factor (n)
0 0
20 -0.034356779
40 -0.137427115
60 -0.309211009
80 -0.549708461
100 -0.85891947
120 -1.236844037
140 -1.683482161
160 -2.198833843
180 -2.782899083
200 -3.43567788
220 -4.157170234
10
2.2.7 Along GF:
Also nG=n
F Finally join GF by using a straight line
2.3 Nomenclature of curves:
• PHAA – Positive High Angle of Attack • PSL – Positive Structural Limit • PLAA – Positive Low Angle of Attack • HSL –High Speed Limit • NHAA – Negative High Angle of Attack • NSL – Negative Structural Limit • NLAA – Negative Low Angle of Attack • LSL – Low Speed Limit
Figure 2-3: Four basic flight conditions showing how location of maximum stresses in wing depends on
angle of attack
11
2.4 Low Speed Limit:
Stall velocity is the maximum speed at which the aircraft can maintain level
flight. This implies the intersection of this line at cruise n=1 with OA curve
corresponds to stall velocity Vs.
Vs = 68.630 m/s
Figure 2-4: Rough V-n Diagram
From the V-n diagram, it is observed that the stall curve corresponds to
maximum value of CLmax and any point beyond this curve for a particular
velocity is not achievable in flight as it enters the stall region there. The upper
horizontal line corresponds to limit load factor as well as ultimate load factor. It
-10
-5
0
5
10
0 100 200 300 400 500
Load
Fac
tor
Velocity (m/s)
Rough V-n Diagram
PHAA NHAA PSL HSL
NSL PLAA NLAA LSL
12
shows that there is outright structural failure when the aircraft is flown beyond
this value of load factor.
n=-2 gives the negative limit load factor and negative ultimate load
factor.
From the figure, it is clear that for a particular velocity, it is not possible
to fly at a value of CL higher than the CLmax corresponding to that velocity. If we
wish to increase the lift of the airplane to that value of CLmax, then we should
increase the flying speed of the airplane.
Figure 2-5: Maneuvering Envelope
-3
-2
-1
0
1
2
3
4
5
6
7
0 50 100 150 200 250 300 350 400 450
Load
Fac
tor
Velocity
Maneuvering Envelope
LSL
PIAA
NIAA
HSL
13
Figure 2-6: Maneuvering Envelope
Figure 2-7: Maneuvering envelope with coordinates
Hence for the strategic bomber aircraft we get,
Safety Factor = 1.5
-3
-2
-1
0
1
2
3
4
5
6
7
0 50 100 150 200 250 300 350 400 450
Load
Fac
tor
Velocity
Maneuvering Envelope
A
H
C
E
D
B
G F
-3
-2
-1
0
1
2
3
4
5
6
7
0 50 100 150 200 250 300 350 400 450
Load
Fac
tor
Velocity
Maneuvering Envelope with coordinates
68.63, -0.4046
68.6302,1
408.32, 6.41152
173.77, 6.41152
159.5944,-2
416.66, 0
416.66, 4.80864
408.32,-2
14
Caution Speed = 325 m/s
Corner Velocity = 173.77 m/s
Stall speed = 59.669 m/s
Safety load factor limit i.e., indications given to pilot
n = -2/ 1.5 = -1.3333
n = 6.41152/ 1.5 = 4.2743
Dive Velocity = 416.66 m/s
15
Figure 2-8: V-n diagram with safety factor or safety limit consideration
16
3. Gust Envelope
3.1 Description:
Gust is a sudden, brief increase in the speed of the wind. Generally, winds are
least gusty over large water surfaces and most gusty over rough land and near
high buildings. With respect to aircraft turbulence, a sharp change in wind speed
relative to the aircraft; a sudden increase in airspeed due to fluctuations in the
airflow, resulting in increased structural stresses upon the aircraft.
Sharp-edged gust (u) is a wind gust that results in an instantaneous change in
direction or speed.
Derived gust velocity (U or Umax) is the maximum velocity of a sharp-edged
gust that would produce a given acceleration on a particular airplane flown in
level flight at the design cruising speed of the aircraft and at a given air density.
As a result a 25% increase is seen in lift for a longitudinally disturbing gust.
The effect of turbulence gust is to produce a short time change in the effective
angle of attack. These changes produce a variation in lift and thereby load factor
For velocities up to Vmax, cruise, a gust velocity of 15 m/s at sea level is
assumed. For Vdiv, a gust velocity of 10 m/s is assumed.
Effective gust velocity: The vertical component of the velocity of a sharp-
edged gust that would produce a given acceleration on a particular airplane
flown in level flight at the design cruising speed of the aircraft and at a given air
density.
Reference Gust Velocity (Uref ) —at sea level 15m/s.
Design Gust Velocity (Uds) — Uref X K
17
Figure 3-1: Variation in Aerodynamic limits due to gust
3.2 Construction
The increase in the load factor due to the gust can be calculated by
For curve above V-axis:
Where
K Gust Alleviation Factor
U max Maximum derived Gust Velocity
a Lift Curve Slope for wing
For curve below V-axis:
-6
-4
-2
0
2
4
6
8
10
12
0 50 100 150 200 250
Load
fac
tor
Velocity
Variation in aerodynamic limits
Normal Stall curve
Gust stall curve
Normal neg stallcurve
Gust neg stall curve
Flaps Retracted
18
Gust Alleviation Factor:
Gust Alleviation Factor (K):
Lateral Mass Ratio (µ):
Where
g Acceleration due to Gravity
ĉ Mean Aerodynamic Chord
(
)
ct Chord at tip
cr Chord at root
cr = 11.593 m
ct = 5.797 m
√
a= 0.1213507 /degree
19
a= 6.9528829 /radian
for a’ =0.15/ degree where a’ is lift curve slope for the chosen airfoil NACA 65-
(3) 418
a’ lift curve slope for airfoil
Sweep angle at leading Edge of Wing
( )
Table 3-1: Equivalent air speed and corresponding Derived Gust Velocity
For Velocity at points Equivalent air speed
V (m/s)
Derived Gust Velocity
Umax (m/s)
B,G 173.77 15
C,F 408.32 10
D,E 416.66 5
20
By using the equations and for various speeds of Umax we get the following gust
lines
Figure 3-2: Gust Lines
-2
-1
0
1
2
3
4
0 50 100 150 200 250 300 350 400 450
Load
fac
tor
Velocity
Gust Lines U=15m/s U=10m/s U=5m/sU=0m/s U=-5m/s U=-10./sU=-15m/s
Level
Design speed 277.77 m/s
21
Figure 3-3: Overlapped maneuvering envelope and gust lines.
The load factors at the various points can be found using the formula using the
corresponding values of Umax
n B’ = 1.4195966
n G’= 0.846720
nC’ = 2.5617017
nF’ = -0.5617017
n D’ = 1.796799
n E’ = 0.203200
-3
-2
-1
0
1
2
3
4
5
6
7
0 50 100 150 200 250 300 350 400 450
Load
fac
tor
Velocity
Overlaped Maneuver envelope and gust lines Gust stall curve Gust neg stall curve HSL
PLAA NLAA PIAANIAA U=15m/s U=10m/s
U=5m/s U=0m/s U=-5m/s
22
The positive load factor along the curve OB’ is given by the equation
Hence along the curve OA,
But also
Equating the above two equations we get an intersecting point B where velocity
is
VB = 73.1379 m/s
Since the velocities and load factors at C, D, E and F are known and straight
lines are used to join these points in sequence
3.2.1 Line FG:
It is found that negative gust line of U= -15 m/s intersects the positive high
angle of attack condition at G.
23
Equating the above equation with the OA curve equation we get the point G
where
VG = 51.52026m/s
Figure3-4: Gust Envelope
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
3
0 50 100 150 200 250 300 350 400 450
Load
fac
tor
Velocity (m/s)
Gust Envelope
B
C
D
G
O
E
F
24
Figure 3-5: Gust Envelope with coordinates
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
3
0 50 100 150 200 250 300 350 400 450
Load
fac
tor
Velocity
Gust Envelope with coordinates
416.66, 1.7968
410.9975, -1.3791
416.66, 0
416.66, 1.7968
408.32, 2.561702
73.13798, 1.419596
51.52026, 0.704426
25
4. Schrenk’s Curve
4.1 Description
Lift varies along the wing span due to the variation in chord length, angle of
attack and sweep along the span. Schrenk’s curve defines this lift distribution
over the wing span of an aircraft, also called simply as Lift Distribution Curve.
Schrenk’s Curve is given by
Where
y1 is Linear Variation of lift along semi wing span also named as L1
y2 is Elliptic Lift Distribution along the wing span also named as L2
a = 44.8285 m
26
Figure 4-1: Wing geometry showing sweep angle and semi span along the root.
4.2 Linear Lift Distribution:
Lift at root
Lroot = 90978.038 N/m
Lift at tip
Ltip = 45492.942 N/m
By representing this lift at sections of root and tip we can get the equation for
the wing.
27
Figure 4-2: Linear lift distribution
Equation of linear lift distribution for starboard wing
Equation of linear lift distribution for port wing we have to replace x by –x in
general,
Twice the area under y1= Total lift= 2491907.5 N ≈ Take off Gross Weight
Figure 4-3: Linear Variation of lift along wing semi span
0
10
20
30
40
50
60
70
80
90
100
0 10 20 30 40 50
Lift
pe
r m
ete
r (N
/m)
Tho
usa
nd
s
Wing Semi Span (m)
Linear variation of Lift along wing Semi span
L1
28
For the Schrenk’s curve we only consider half of the linear distribution of lift
and hence we derive y1/2
4.3 Elliptic Lift Distribution:
Twice the area under the curve or line will give the lift which will be required to
overcome weight
Considering an elliptic lift distribution we get
Where b1 is Actual lift at root
And a is wing semi span
Lift at tip
29
Figure 4-4: Elliptic lift distribution
Equation of elliptic lift distribution
√
Figure 4-5: Elliptic lift distribution
√
0
10
20
30
40
50
60
70
80
0 10 20 30 40 50
Lift
pe
r m
ete
r (N
/m) Th
ou
san
ds
Wing Semi Span (m)
Elliptic variation of Lift along wing Semi span
L2
30
4.4 Construction of Schrenk’s Curve:
Schrenk’s Curve is given by
√
√
Substituting different values for x we can get the lift distribution for the wing
semi span
Table 4-1: Lift distribution table along semi
span
x L1 L2 L
0 90978.04 70776.189 80877.11
1 89963.39 70758.577 80360.98
2 88948.75 70705.716 79827.23
3 87934.1 70617.525 79275.81
4 86919.45 70493.872 78706.66
5 85904.81 70334.571 78119.69
6 84890.16 70139.378 77514.77
7 83875.52 69907.993 76891.75
8 82860.87 69640.055 76250.46
9 81846.22 69335.14 75590.68
10 80831.58 68992.759 74912.17
11 79816.93 68612.349 74214.64
12 78802.29 68193.275 73497.78
13 77787.64 67734.819 72761.23
14 76772.99 67236.175 72004.58
15 75758.35 66696.443 71227.4
16 74743.7 66114.615 70429.16
17 73729.06 65489.57 69609.31
18 72714.41 64820.058 68767.23
19 71699.76 64104.686 67902.22
20 70685.12 63341.899 67013.51
21 69670.47 62529.962 66100.22
x L1 L2 L
22 68655.83 61666.935 65161.38
23 67641.18 60750.639 64195.91
24 66626.53 59778.626 63202.58
25 65611.89 58748.129 62180.01
26 64597.24 57656.013 61126.63
27 63582.6 56498.706 60040.65
28 62567.95 55272.111 58920.03
29 61553.3 53971.505 57762.4
30 60538.66 52591.398 56565.03
31 59524.01 51125.351 55324.68
32 58509.37 49565.739 54037.55
33 57494.72 47903.426 52699.07
34 56480.07 46127.307 51303.69
35 55465.43 44223.675 49844.55
36 54450.78 42175.265 48313.02
37 53436.14 39959.82 46697.98
38 52421.49 37547.784 44984.64
39 51406.84 34898.417 43152.63
40 50392.2 31952.741 41172.47
41 49377.55 28619.406 38998.48
42 48362.91 24742.227 36552.57
43 47348.26 20007.494 33677.88
44 46333.61 13543.872 29938.74
44.8285 45492.98 0 22746.49
31
Figure 4-6: Schrenk’s curve with linear and elliptic lift distribution
Replacing x by –x for port wing we can get lift distribution for entire span.
Figure 4-7: Schrenk’s curve
0
10
20
30
40
50
60
70
80
90
100
0 10 20 30 40 50
Lift
pe
r m
ete
r sp
an (
N/m
)
Tho
usa
nd
s
Wing span loaction (m)
Schrenk's Curve
L
L1
L2
0
10
20
30
40
50
60
70
80
90
-60 -40 -20 0 20 40 60
Lift
pe
r m
ete
r sp
an (
N/m
)
Tho
usa
nd
s
Wing span loaction (m)
Schrenk's Curve
L
32
5. Load Estimation on wings
5.1 Description:
The solution methods which follow Euler’s beam bending theory
(σ/y=M/I=E/R) use the bending moment values to determine the stresses
developed at a particular section of the beam due to the combination of
aerodynamic and structural loads in the transverse direction. Most engineering
solution methods for structural mechanics problems (both exact and
approximate methods) use the shear force and bending moment equations to
determine the deflection and slope at a particular section of the beam.
Therefore, these equations are to be obtained as analytical expressions in terms
of span wise location. The bending moment produced here is about the
longitudinal (x) axis.
5.2 Loads acting on wing:
As both the wings are symmetric, let us consider the starboard wing at first.
There are three primary loads acting on a wing structure in transverse direction
which can cause considerable shear forces and bending moments on it. They
are as follows:
Lift force (given by Schrenk’s curve)
Self-weight of the wing
Weight of the power plant
Weight of the fuel in the wing
33
5.3 Shear force and bending moment diagrams due to
loads along transverse direction at cruise condition:
Lift Force given by Schrenk’s Curve:
Linear lift distribution (trapezium):
Elliptic lift distribution (quarter ellipse)
√
√
Figure 5-1: Lift distribution (linear)
0
5000
10000
15000
20000
25000
30000
35000
40000
45000
50000
0 5 10 15 20 25 30 35 40 45 50
Lift
pe
r u
nit
len
gth
(N
/m)
Span wise location (m)
Linear lift distribution (y1/2)
34
Figure 5-2: Lift distribution (Elliptic)
Self-Weight (y3): Self-weight of the wing,
Assuming parabolic weight distribution
( (
))
Where b span
0
5000
10000
15000
20000
25000
30000
35000
40000
0 10 20 30 40 50
Lift
pe
r u
nit
len
gth
(N
/m)
Span wise location (m)
Elliptic lift distribution (y2/2)
y2/2
35
When we integrate from x=0 (root location) to x=b (tip location) we get the net
weight of port wing.
∫
∫ ( )
( )
Substituting various values of x in the above equation we get the self-weight of
the wing.
Figure 5-3: Self weight of wing
Power plant weight:
Power plant is assumed to be a point load,
-25000
-20000
-15000
-10000
-5000
0
0 5 10 15 20 25 30 35 40 45 50
WEi
ght
of
em
pty
win
g (N
/m)
Span wise location (m)
Self Weight
36
Acting at x= 8 m and x= 14 m from the root.
Fuel weight:
This design has fuel in the wing so we have to consider the weight of the fuel
in the wing.
Again by using general formula for straight line y=mx + c we get,
Figure 5-4: Fuel Distribution
-20000
-18000
-16000
-14000
-12000
-10000
-8000
-6000
-4000
-2000
0
0 5 10 15 20 25 30 35 40 45 50
Fue
l we
igh
t (N
/m)
Span wise location (m)
Fuel distribution
37
Figure 5-5: Overall Load distribution
Table 5-1: Loads simplified as point loads
-80000
-60000
-40000
-20000
0
20000
40000
60000
0 5 10 15 20 25 30 35 40 45 50
Load
act
ing
on
win
g (N
/m)
Span wise location (m)
Load distribution
Curve / component Area enclosed / structural
weight (N)
Centroid (from wing root)
y1/2 1529447.31 19.923 m
y2/2 1245953.75 3.510534 m
Wing 313917 16.8107 m
Fuel 365752.803 16.4606 m
Power plant 66708 14 m, 8 m
38
Figure 5-6: Reaction force and Bending moment calculations
Now we know VA and MA, using this we can find out shear force and Bending
moment.
5.3.1 Shear Force:
∫
39
√
(
)
(
)
∫
∫
By using the corresponding values of x in appropriate equations we get the plot
of shear force
Note: Shear force is a discrete function along y axis so in order to make it
continuous we introduce straight lines.
40
Figure 5-7: Shear force diagram - discrete
Figure 5-8: Shear force diagram- continuous
-3000
-2500
-2000
-1500
-1000
-500
0
500
1000
-44.8285 -34.8285 -24.8285 -14.8285 -4.8285 5.1715 15.1715 25.1715 35.1715
She
ar F
orc
e (
N)
Tho
usa
nd
s
Location in wing (m)
Shear Force
-3000
-2500
-2000
-1500
-1000
-500
0
500
1000
-44.8285 -34.8285 -24.8285 -14.8285 -4.8285 5.1715 15.1715 25.1715 35.1715
She
ar F
orc
e (
N)
Tho
usa
nd
s
Location in wing (m)
Shear Force (Actual)
41
5.3.2 Bending moment:
∫∫
( ( √ )
(
))
(
)
∫∫
By substituting the values of x for the above equations of bending moments
obtained we can get a continuous bending moment curve for the port wing.
Note: if we replace the x by -x in each term we get the distribution of starboard
wing
42
Figure 5-9: Bending moment diagram
5.4 Shear force and bending moment diagrams due to
loads along chordwise direction at cruise condition:
Aerodynamic center- This is a point on the chord of an airfoil section where the
bending moment due to the components of resultant aerodynamic force (Lift
and Drag) is constant irrespective of the angle of attack. Hence the forces are
transferred to this point for obtaining constant Ma.c
Shear center- This is a point on the airfoil section where if a force acts, it
produces only bending and no twisting. Hence the force is transferred to this
point and the torque is found.
Cruise CL=0.204908 @ V= 250 m/s
Cruise CD= 0.0055
Angle of attack= -0.811439˚ (obtained from the lift curve slope)
Angle of attack @ zero lift= -3o
0
10
20
30
40
50
60
70
-50 -40 -30 -20 -10 0 10 20 30 40 50
Be
nd
ing
Mo
me
nt
(Nm
)
Mill
ion
s
Location in wing (m)
Bending Moment
43
Wing lift curve slope (a)= 0.1213507 /degree
Co-efficient of moment about aerodynamic centre= -0.0543
Location of aerodynamic centre:
Location of shear centre:
Lift and drag are the components of resultant aerodynamic force
acting normal to and along the direction of relative wind respectively. As a
result, components of them act in the chordwise direction also which produce a
bending moment about the normal (z) axis.
Figure 5-10: Normal and chord wise coefficients
Co-efficient of force along the normal direction,
44
Chordwise force at root
Chordwise force at tip
By using y = mx +c again we get the equation as
The above equation gives the profile of load acting chordwise, by integrating
this above equation we get a component of Shear force and again by integrating
the same we get the component of Bending Moment
∫
∬
45
Figure 5-11: Load along chordwise direction
To find fixing moment and the reaction force,
5.4.1 Shear Force:
0
200
400
600
800
1000
1200
1400
0 5 10 15 20 25 30 35
Load
alo
ng
cho
rd w
ise
dir
ect
ion
(N
)
Spanwise location (m)
Load along Chordwise direction
46
Figure 5-12: Shear force
5.4.2 Bending Moment:
Figure 5-13: Bending moment
-35
-30
-25
-20
-15
-10
-5
0
0 5 10 15 20 25 30 35
She
ar F
orc
e (
N)
Tho
usa
nd
s
Spanwise location (m)
Shear Force
0
100
200
300
400
500
600
0 5 10 15 20 25 30 35
Be
nd
ing
mo
me
nt
(Nm
) Tho
usa
nd
s
Spanwise location (m)
Bending Moment
47
Torque due to normal forces and constant pitching moment at cruise condition:
Figure 5-14: Moment about aerodynamic center
The lift and drag forces produce a moment on the surface of cross-
section of the wing, otherwise called a torque, about the shear center. Moment
about the aerodynamic center gets transferred to the shear center. The
powerplant also produces a torque about the shear center on the chord under
which it is located.
Figure 5-15:Torque due to normal force and moment
48
5.5 Torque at cruise condition:
5.5.1 Torque due to normal force:
Where
c chord
the equation for chord can also be represented in terms of x by taking c= mx
+k,
Therefore torque
∫
∫
49
Figure 5-16: Torque due to normal force
5.5.2 Torque due to chord wise force:
5.5.3 Torque due to moment:
∫
0
100
200
300
400
500
600
700
800
900
1000
0 5 10 15 20 25 30 35 40
Torq
ue
(N
m)
Tho
usa
nd
s
Spanwise location (m)
Torque due to normal forces
50
Figure 5-17: Torque due to moment
5.5.4 Torque due to powerplant:
The powerplant is situated under a chord (8 m and 14 m from the
wing root; chord length 10.7504 m and 10.1184m ) from 0.1c to 0.5c at
10.7504m and from 0.1 c to 0.5249c an Uniformly Distributed Load of
15513.488 N/m is assumed to be present for this 4.3 m since the powerplant
weight is 66708N. The centroid of the applied UDL is at 0.3c for first case and
at 0.31245c at second location.
Torque produced about shear center
-8
-7
-6
-5
-4
-3
-2
-1
0
0 5 10 15 20 25 30 35 40
Torq
ue
(N
m)
Mill
ion
s
Spanwise location (m)
Torque due to Moment
51
Figure 5-18: Torque due to powerplant
Then the different torque components are brought together in a same graph to
make a comparison
Figure 5-19: Torque comparison
0
10
20
30
40
50
60
70
0 5 10 15 20 25 30 35 40
Torq
ue
(N
m)
Tho
usa
nd
s
Spanwise location (m)
Torque due to Powerplant
-8000
-7000
-6000
-5000
-4000
-3000
-2000
-1000
0
1000
2000
0 10 20 30 40
Torq
ue
(N
m)
Tho
usa
nd
s
Spanwise location (m)
Torque comparison
Torque due to Normal Forces
Torque due to moment
Torque due to powerplant
52
The net torque will be sum of all the above torques i.e. torque due to normal
force, chordwise force, powerplant and aerodynamic moment
Figure 5-20: Net torque
-7
-6
-5
-4
-3
-2
-1
0
0 5 10 15 20 25 30 35 40
Torq
ue
(N
m)
Mill
ion
s
Spanwise location (m)
Torque
53
5.6 Load at Critical flight condition:
Optimum Wing structural design consists of determining that stiffness
distribution which is proportional to the local load distribution. The
aerodynamic forces of lift and drag are resolved into components normal and
parallel to the wing chord. The distribution of shear force, bending moment and
torque over the aircraft wing are considered for wing structural analysis.
Identification of critical points from the maneuvering and gust envelopes:
1. Maneuvering envelope
Table 5-1: Coordinates of V-n diagram
Point Load factor E.A.S. (m/s)
A 6.41152 173.77
C 6.41152 408.32
D 4.80864 416.66
E 0 416.66
F -2 408.16
G -2 159.5944
2. Gust envelope
Table 5-2: Coordinates of gust envelope
Point Load factor E.A.S. (m/s)
B’ 1.41959 173.137975
C’ 2.5617 408.16
D’ 1.7968 416.66
E’ 0.2032 416.66
F’ -1.3255 408.16
G’ 0.5822 51.52026
54
Corner points are representative of critical flight load conditions – a summary
is given below.
Table 5-3: Coordinates of critical conditions
Critical flight condition Point (‘n’, E.A.S.)
‘n’ max point C’ (2.5617, 408.16)
Positive H.A.A. A (6.41152, 173.77)
Positive L.A.A D (4.80864, 416.66)
Negative H.A.A G (-2, 408.16)
Negative L.A.A E’ (0, 416.66)
Shear force and bending moment diagrams of a wing due to normal forces at
critical flight condition:
In the preliminary stage of structural analysis, the critical
flight loading condition of positive high angle of attack (represented by point A
in v-n diagram) will be investigated.
It is seen that lift has increased by 6.41152 times.
So we introduce a constant of proportionality for the lift alone
55
Figure 5-21: Linear Variation of lift along wing semi span
Figure 5-22: Elliptic variation of lift along wing semi span
0
100
200
300
400
500
600
700
0 10 20 30 40 50
Lift
pe
r m
ete
r (N
/m)
Tho
usa
nd
s
Wing Semi Span (m)
Linear variation of Lift along wing Semi span (critical condition)
L1
0
50
100
150
200
250
300
350
400
450
500
0 10 20 30 40 50
Lift
pe
r m
ete
r (N
/m)
Tho
usa
nd
s
Wing Semi Span (m)
Elliptic variation of Lift along wing Semi span (Critical Condition)
L2
56
The aim is to find the shear forces and bending moments due to
normal forces in critical flight condition. There are three primary loads acting
on a wing structure in transverse direction which can cause considerable shear
forces and bending moments on it. They are as follows:
Lift force (given by Schrenk’s curve)
Self-weight of the wing
Weight of the power plant
Weight of the fuel in the wing
Now, the proportionality constant influences the lift force alone and other
factors remain unaffected.
Table 5-4: Loads simplified as point loads at critical flight condition
Curve / component Area enclosed / structural
weight (N)
Centroid (from wing root)
y1/2 1529447.31×6.41152 19.923 m
y2/2 1245953.75 ×6.41152 3.510534 m
Wing 313917 16.8107 m
Fuel 365752.803 16.4606 m
Power plant 66708 14 m, 8 m
57
Figure 5-23: Critical schrenk’s curve
Figure 5-24: load distribution at critical condition
0
100
200
300
400
500
600
-60 -40 -20 0 20 40 60
Lift
pe
r m
ete
r sp
an (
N/m
)
Tho
usa
nd
s
Wing span loaction (m)
Schrenk's Curve (Critical Condition)
L
58
Figure 5-25: load distribution at critical condition
Now we know VA and MA, using this we can find out shear force and Bending
moment,
5.7 Shear force and bending moment diagrams due to
loads along transverse direction at critical condition:
∫
-0.1
-0.05
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0 5 10 15 20 25 30 35 40 45 50
Load
(N
)
Mill
ion
s
Location in wing (m)
Load distribution (Critical Condition)
59
(
( √ (
)))
(
)
∫
∫
By using the corresponding values of x in appropriate equations we get the plot
of shear force
60
Figure 5-26: Transverse Shear force diagram at critical condition
Figure 5-27: Transverse Shear force diagram at critical condition
5.7.1 Bending moment:
∫∫
-20000
-18000
-16000
-14000
-12000
-10000
-8000
-6000
-4000
-2000
0
2000
-44.8285 -34.8285 -24.8285 -14.8285 -4.8285 5.1715 15.1715 25.1715 35.1715
She
ar F
orc
e (
N)
Tho
usa
nd
s
Location in wing (m)
Shear Force (Critical condition)
-20000
-18000
-16000
-14000
-12000
-10000
-8000
-6000
-4000
-2000
0
2000
-44.8285 -34.8285 -24.8285 -14.8285 -4.8285 5.1715 15.1715 25.1715 35.1715
She
ar F
orc
e (
N)
Tho
usa
nd
s
Location in wing (m)
Shear Force (Actual) (Critical Condition)
61
(
( ( √ )
(
)))
(
)
∫∫
By substituting the values of x for the above equations of bending moments
obtained we can get a continuous bending moment curve for the port wing.
62
Figure 5-28: Transverse bending moment diagram at critical condition
5.8 Shear force and bending moment diagrams due to
loads along chordwise direction at critical condition:
Critical CL=2.71925 @ V= 250 m/s
Critical CD= 0.0084
Angle of attack= 16˚ (obtained from the lift curve slope)
Wing lift curve slope (a) = 0.1213507 /degree
Co-efficient of moment about aerodynamic centre= -0.025
Location of aerodynamic centre:
Location of shear centre:
-100
0
100
200
300
400
500
600
-50 -40 -30 -20 -10 0 10 20 30 40 50
Be
nd
ing
Mo
me
nt
(Nm
)
Mill
ion
s
Location in wing (m)
Bending Moment (Critical Condition)
63
Figure 5-29: Determination of chordwise force components at critical condition
Co-efficient of force along the normal direction,
Chordwise force at root
Chordwise force at tip
64
By using y = mx +c again we get the equation as
The above equation gives the profile of load acting chordwise, by integrating
this above equation we get a component of Shear force and again by integrating
the same we get the component of Bending Moment
∫
∬
Figure 5-30: Load along chord wise direction at critical condition
0
50000
100000
150000
200000
250000
300000
350000
400000
0 5 10 15 20 25 30 35
Load
alo
ng
cho
rd w
ise
dir
ect
ion
(N
)
Spanwise location (m)
Load along Chordwise direction (critical condition)
65
To find fixing moment and the reaction force,
5.8.1 Shear Force:
Figure 5-31: Chordwise Shear force diagram at critical condition
5.8.2 Bending Moment:
-12000
-10000
-8000
-6000
-4000
-2000
0
0 5 10 15 20 25 30 35
She
ar F
orc
e (
N)
Tho
usa
nd
s
Spanwise location (m)
Shear Force
66
Figure 5-32: Chordwise Bending moment diagram at critical condition
Torque due to normal forces and constant pitching moment at cruise condition:
Figure 5-33: Determination of various components of torque
0
20000
40000
60000
80000
100000
120000
140000
160000
180000
0 5 10 15 20 25 30 35
Be
nd
ing
mo
me
nt
(Nm
) Tho
usa
nd
s
Spanwise location (m)
Bending Moment
67
Figure 5-34: Determination of various components causing torque
5.9 Torque at critical flight condition:
5.9.1 Torque due to normal force:
Where
c chord
the equation for chord can also be represented in terms of x by taking c= mx
+k,
Therefore torque
∫
68
Figure 5-35: Torque due to normal force at critical condition
5.9.2 Torque due to chord wise force:
5.9.3 Torque due to moment:
∫
0
2000
4000
6000
8000
10000
12000
0 5 10 15 20 25 30 35 40
Torq
ue
(N
m)
Tho
usa
nd
s
Spanwise location (m)
Torque due to normal forces
69
Figure 5-36: Torque due to moment at critical condition
5.9.4 Torque due to powerplant:
The powerplant is situated under a chord (8 m and 14 m from the
wing root; chord length 10.7504 m and 10.1184m ) from 0.1c to 0.5c at
10.7504m and from 0.1 c to 0.5249c an Uniformly Distributed Load of
15513.488 N/m is assumed to be present for this 4.3 m since the powerplant
weight is 66708N. The centroid of the applied UDL is at 0.3c for first case and
at 0.31245c at second location.
Torque produced about shear center
Hence tis is weight this will remain same as that of the cruise condition.
-2
-1.8
-1.6
-1.4
-1.2
-1
-0.8
-0.6
-0.4
-0.2
0
0 5 10 15 20 25 30 35 40
Torq
ue
(N
m)
Mill
ion
s
Spanwise location (m)
Torque due to Moment
70
Figure 5-37: Torque due to powerplant at critical condition unchanged
Then the different torque components are brought together in a same graph to
make a comparison
The net torque will be sum of all the above torques i.e. torque due to normal
force, chordwise force, powerplant and aerodynamic moment
0
10
20
30
40
50
60
70
0 5 10 15 20 25 30 35 40
Torq
ue
(N
m)
Tho
usa
nd
s
Spanwise location (m)
Torque due to Powerplant
71
Figure 5-38: Torque comparison at critical condition
Figure 5-39: Net torque at critical condition
-4000
-2000
0
2000
4000
6000
8000
10000
12000
0 10 20 30 40
Torq
ue
(N
m)
Tho
usa
nd
s
Spanwise location (m)
Torque comparison
Torque due to Normal Forces
Torque due to moment
Torque due to powerplant
0
1
2
3
4
5
6
7
8
9
10
0 5 10 15 20 25 30 35 40
Torq
ue
(N
m)
Mill
ion
s
Spanwise location (m)
Torque
72
5.10 Interim Summary:
DUE TO NORMAL FORCES:
Table 5-5: Determination of maximum values of normal force
Cruise
condition
+ve high AOA
condition
At
Max. Shear force (N) -2461484.863 -17480623.2
Wing root
Max. Bending
moment (Nm)
65115382.95 486567918.1 Wing root
DUE TO CHORDWISE FORCES:
Table 5-6: Determination of maximum values of chordwise force
Cruise
condition
+ve high AOA
condition
At
Max. Shear force (N) -31590.839 -10102394.13 Wing root
Max. Bending
moment (Nm)
51270.9081 163971959.1 Wing root
Max. Torque (Nm) -6372183 8781344 Wing root
73
6. Material Selection:
6.1 Description:
Aircraft structures are basically unidirectional. This means that one dimension,
the length, is much larger than the others - width or height. For example, the
span of the wing and tail spars is much longer than their width and depth; the
ribs have a much larger chord length than height and/or width; a whole wing
has a span that is larger than its chords or thickness; and the fuselage is much
longer than it is wide or high. Even a propeller has a diameter much larger than
its blade width and thickness, etc.... For this simple reason, a designer chooses
to use unidirectional material when designing for an efficient strength to weight
structure.
Unidirectional materials are basically composed of thin, relatively flexible,
long fibers which are very strong in tension (like a thread, a rope, a stranded
steel wire cable, etc.)
An aircraft structure is also very close to a
symmetrical structure. That means the up and down
loads are almost equal to each other. The tail loads
may be down or up depending on the pilot raising or
dipping the nose of the aircraft by pulling or pushing
the pitch control; the rudder may be deflected to the
right as well as to the left (side loads on the fuselage).
The gusts hitting the wing may be positive or
negative, giving the up or down loads which the
occupant experiences by being pushed down in the
seat ... or hanging in the belt.
Because of these factors, the designer has to use a
74
structural material that can withstand both tension and compression.
Unidirectional fibers may be excellent in tension, but due to their small cross
section, they have very little inertia (we will explain inertia another time) and
cannot take much compression. They will escape the load by bucking away. As
in the illustration, you cannot load a string, or wire, or chain in compression.
In order to make thin fibers strong in compression, they are "glued together"
with some kind of an "embedding". In this way we can take advantage of their
tension strength and are no longer penalized by their individual compression
weakness because, as a whole, they become compression resistant as they help
each other to not buckle away. The embedding is usually a lighter, softer
"resin" holding the fibers together and enabling them to take the required
compression loads. This is a very good structural material.
WOOD
Historically, wood has been used as the first unidirectional structural raw
material. They have to be tall and straight and their wood must be strong and
light. The dark bands (late wood) contain many fibers, whereas the light bands
(early wood) contain much more "resin". Thus the wider the dark bands, the
stronger and heavier the wood. If the dark bands are very narrow and the light
bands quite wide, the wood is light but not very strong. To get the most
efficient strength to weight ratio for wood we need
a definite numbers of bands per inch.
Some of our aircraft structures are two-dimensional
(length and width are large with respect to
thickness). Plywood is often used for such
structures. Several thin boards (foils) are glued
together so that the fibers of the various layers cross
over at different angles (usually 90 degrees today years back you could get
75
them at 30 and 45 degrees as well). Plywood makes excellent "shear webs" if
the designer knows how to use plywood efficiently. (We will learn the basis of
stress analysis sometime later.)
Today good aircraft wood is very hard to come by. Instead of using one good
board for our spars, we have to use laminations because large pieces of wood
are practically unavailable, and we no longer can trust the wood quality. From
an availability point of view, we simply need a substitute for what nature has
supplied us with until now.
ALUMINUM ALLOYS
So, since wood may not be as available as it was before, we look at another
material which is strong, light and easily available at a reasonable price
(there's no point in discussing Titanium - it's simply too expensive). Aluminum
alloys are certainly one answer. We will discuss the properties of those alloys
which are used in light plane construction in more detail later. For the time
being we will look at aluminum as a construction material.
Extruded Aluminum Alloys: Due to the manufacturing process for aluminum
we get a unidirectional material quite a bit stronger in the lengthwise direction
than across. And even better, it is not only strong in tension but also in
compression. Comparing extrusions to wood, the tension and compression
characteristics are practically the same for aluminum alloys so that the linear
stress analysis applies. Wood, on the other hand, has a tensile strength about
twice as great as its compression strength; accordingly, special stress analysis
methods must be used and a good understanding of wood under stress is
essential if stress concentrations are to be avoided!
Aluminum alloys, in thin sheets (.016 to .125 of an inch) provide an excellent
two dimensional material used extensively as shear webs - with or without
76
stiffeners - and also as tension/compression members when suitably formed
(bent).
It is worthwhile to remember that aluminum is an artificial metal. There is no
aluminum ore in nature. Aluminum is manufactured by applying electric power
to bauxite (aluminum oxide) to obtain the metal, which is then mixed with
various strength-giving additives. (In a later article, we will see which additives
are used, and why and how we can increase aluminum's strength by cold work
hardening or by tempering.) All the commonly used aluminum alloys are
available from the shelf of dealers. When requested with the purchase, you can
obtain a "mill test report" that guarantees the chemical and physical properties
as tested to accepted specifications.
As a rule of thumb, aluminum is three times heavier, but also three times
stronger than wood. Steel is again three times heavier and stronger than
aluminum.
STEEL
The next material to be considered for aircraft structure will thus be steel,
which has the same weight-to-strength ratio of wood or aluminum.
Apart from mild steel which is used for brackets needing little strength, we are
mainly using a chrome-molybdenum alloy called AISI 413ON or 4140. The
common raw materials available are tubes and sheet metal. Steel, due to its
high density, is not used as shear webs like aluminum sheets or plywood.
Where we would need, say.100" plywood, a .032 inch aluminum sheet would
be required, but only a .010 steel sheet would be required, which is just too thin
to handle with any hope of a nice finish. That is why a steel fuselage uses tubes
also as diagonals to carry the shear in compression or tension and the whole
structure is then covered with fabric (light weight) to give it the required
77
aerodynamic shape or desired look. It must be noted that this method involves
two techniques: steel work and fabric covering.
We will be discussing tubes and welded steel structures in more detail later and
go now to "artificial wood" or composite structures.
COMPOSITE MATERIALS
The designer of composite aircraft simply uses fibers in the desired direction
exactly where and in the amount required. The fibers are embedded in resin to
hold them in place and provide the required support against buckling. Instead
of plywood or sheet metal which allows single curvature only, the composite
designer uses cloth where the fibers are laid in two directions .(the woven
thread and weft) also embedded in resin. This has the advantage of freedom of
shape in double curvature as required by optimum aerodynamic shapes and for
very appealing look (importance of esthetics).
Today's fibers (glass, nylon, Kevlar, carbon, whiskers or single crystal fibers of
various chemical compositions) are very strong, thus the structure becomes
very light. The drawback is very little stiffness. The structure needs stiffening
which is achieved either by the usual discreet stiffeners, -or more elegantly
with a sandwich structure: two layers of thin uni- or bi-directional fibers are
held apart by a lightweight core (foam or "honeycomb"). This allows the
designer to achieve the required inertia or stiffness.
From an engineering standpoint, this method is very attractive and supported
by many authorities because it allows new developments which are required in
case of war. But this method also has its drawbacks for homebuilding: A mold
is needed, and very strict quality control is a must for the right amount of fibers
and resin and for good adhesion between both to prevent too "dry" or "wet" a
structure. Also the curing of the resin is quite sensitive to temperature,
78
humidity and pressure. Finally, the resins are active chemicals which will not
only produce the well-known allergies but also the chemicals that attack our
body (especially the eyes and lungs) and they have the unfortunate property of
being cumulatively damaging and the result (in particular deterioration of the
eye) shows up only years after initial contact.
Another disadvantage of the resins is their limited shelf life, i.e., if the resin is
not used within the specified time lapse after manufacturing, the results may be
unsatisfactory and unsafe.
HEAVY AIRCRAFT RAW MATERIALS
The focus of our article is our Table which gives typical values for a variety of
raw materials.
Column 1 lists the standard materials which are easily available at a reasonable
cost. Some of the materials that fall along the borderline between practical and
impractical are:
Magnesium: An expensive material. Castings are the only readily available
forms. Special precaution must be taken when machining magnesium because
this metal burns when hot.
Titanium: A very expensive material. Very tough and difficult to machine.
Carbon Fibers: Still very expensive materials.
Kevlar Fibers: Very expensive and also critical to work with because it is hard
to "soak" in the resin. When this technique is mastered, the resulting structure
is very strong, but it also lacks in stiffness.
Columns 2 through 6:
Columns 2 through 6 list the relevant material properties in metric units.
Column 2, the density (d), is the weight divided by the volume.
79
Table 6-1: Material property table
Materials d fy fu e E/103 E/d Root
2 of
N/d Root
3 of
E/d fu/d
1 2 3 4 5 6 7 8 9 10
Wood Spruce .45 - 3.5/11 - 1.4 2200
70 22.0 (15)
Poplar .43 - 30/12 - 1.0 2200
70 22.0 (15)
Oregon Pine
.56 - 4.0/13 - 1.5 2200
70 22.0 (15)
Fiberglass
Matte 2.2 - 15 - 1.5 700 17 5.0 7
(70% Glass)
Woven 2.2 - 35 - 2.0 900 20 6.0 16
Unidirectional
2.2 - 60 - 3.5 1500
27 7.0 27
Alum. Alloy
5052-H34 2.7 16 24 4 7.1 2600
30 7.0 11
8086-H34 2.7 22 31 5 7.1 2600
30 7.0 11
6061 -T6 2.7 24 26 9 7.1 2600
30 7.0 11
6351 -T6 2.7 25 28 9 7.1 2600
30 7.0 11
6063-T6 2.7 17 21 9 7.1 2600
30 7.0 11
7075-T3 2.8 25 41 12
7.2 2600
30 7.0 14
Steel AISI 1026 7.8 25 38 15
21.0 2700
18 3.5 5
4130 N (4140)
7.8 42 63 10
21.0 2700
18 3.5 7
Lead 11.3 - - - - - - - -
Magnesium Alloy 1.8 20 30 - 4.5 2500
37 9.0 16
Titanium 4.5 50 80 - 11.0 2400
23 5.0 18
Units for above kg/dm
3 kg/mm
2 kg/mm
2 % kg/m
m2
km kg-m
2 kg2/3
m1/3 km
to obtain: lbs/cu3
KSI KSI % KSI
multiply by: .0357 1420 1420 - 1420
80
Column 3, the yield stress (fy), is the stress (load per area) at which there will
be a permanent deformation after unloading (the material has yielded, given
way ... )
Column 4, the ultimate stress (fu), is the stress (load per area) at which it cannot
carry a further load increase. It is the maximum load before failure.
Column 5, the elongation at ultimate stress (e), in percentage gives an indication
of the 'Toughness" of the material.
Column 6 lists the Yongs Modular or Modulus of Elasticity (E), which is the
steepness of the stress/strain diagram as shown in Figure 1.
Important Note: For wood, the tension is much greater (2 to 3 times) than the
compression. Both values are given in the Table. For fiberglass, the same applies,
but the yield is so dependent on the manufacturing process that we cannot even
give 'Iypical values'.
Figure 6-1: Stress strain curves for different materials
Columns 7 to 10: Columns 7 to 10 are values which allow the comparison of
materials from a weight standpoint (the above referenced text by Timoshenko will
also show you why we use those "funny" looking values).
Column 7 gives the stiffness of a sandwich construction. The higher the value, the
stiffer the construction. From the Table, we see that metals are high wood comes
81
close, but fiberglass is low: which means fiberglass will be heavier for the same
stiffness.
Figure 6-2: Stress strain curve
Column 8 shows the column buckling resistance for the same geometric shapes.
This time, wood is better than the light alloys, coming before steel and fiberglass.
(Surprisingly, the usual welded steel tube fuselage is not very weight efficient.)
Column 9 gives the plate buckling stiffness, which is also a shear strength
measure. Here again, wood (plywood) is in a very good position before aluminum
and fiberglass, with steel not very good.
Column 10 provides a crude way of measuring the strength to weight ratio of
materials because it does not take into account the various ways the material is
used in "light structures". According to this primitive way of looking,
unidirectional fibers are very good, followed by high strength (2024) aluminum
and wood, then the more common aluminum alloys and finally steel.
From just this simple table, we find there is not one material that provides an
overwhelming solution to all the factors that must be considered in designing a
light aircraft. Each material has some advantage somewhere. The designer's
choice (no preconceived idea) will make a good aircraft structure ... if the choice
is good!
82
7. Detailed wing design
7.1 Spar design:
Spars are members which are basically used to carry the bending and
shear loads acting on the wing during flight. There are two spars, one located at
15-20% of the chord known as the front spar, the other located at 60-70% of
the chord known as the rear spar. Some of the functions of the spar include:
They form the boundary to the fuel tank located in the wing.
The spar flange takes up the bending loads whereas the web carries the
shear loads.
The rear spar provides a means of attaching the control surfaces on the
wing.
Considering these functions, the locations of the front and rear spar are fixed
at 0.17c and 0.65c respectively. The NACA 65 (3) 418 airfoil is drawn to scale
using any design software and the chord thickness at the front and rear spar
locations are found to be 1.9708 m and 7.5354 m respectively.
7.1.1 Geometric dimensions:
The spar design for the wing root has been taken because the maximum
bending moment and shear force are at the root. It is assumed that the flanges
take up all the bending and the web takes all the shear effect. The maximum
bending moment for high angle of attack condition is 486567918.4 Nm. the
ratio in which the spars take up the bending moment is given as
Where
83
h1 height of front spar
h2 height of rear spar
From the above two equations,
The yield tensile stress σy for 7075 Al Alloy is 455.053962 MPa. The area of
the flanges is determined using the relation
where M is bending moment taken up by each spar,
A is the flange area of each spar,
z is the centroid distance of the area = h/2.
Using the available values,
Area of front spar,
Area of rear spar,
84
Each flange of the spar is made of two angle sections. For the front spar, the
length of the angle is 6t, angle height is 5t with angle thickness t. Area for each
angle of front spar is found to be 0.1799507 m2 and hence value of t is found to
be
Length of the front angle section:
Height of the front angle section:
For the rear spar, the length of the angle is 8t, angle height is 3.5t with vertical
thickness t and horizontal thickness t/2. Area for each angle of rear spar is
found to be 0.164486 m2 and hence value of t is found to be.
Length of the rear angle section:
Height of the rear angle section:
Now to determine the thickness of the web portion, the ultimate shear stress of
7075 Al Alloy is 317.1588MPa. The maximum shear force at root of the wing
85
for high angle of attack condition is 17480623.2 N. The wing chord is assumed
to be a simply supported beam supported at the two spars. The maximum shear
force acts at the centre of pressure which can be located by using the formula,
Figure 7-1: Reaction force determination at spars
Considering force and moment equilibriums for the given simply supported
configuration, the reactive shear force at the spar supports are found to be
We know that,
V shear force at the spar
86
t thickness of the web.
Thus,
FOS = 1.5
z is the centroidal distance of the area = h/2
Thus the thicknesses of the web portions are,
All dimensions are in m
It becomes necessary to check whether the shear stress due to this thickness is
less than the allowable of the material.
1.00128
0.02346
0.1251
Rear spar
0.43806
0.0763
0.13414
0.8048
0.670
Front Spar
87
(
)
For the web, the dimensions of a and b will be
a = 1.6186 m ( rib spacing) and
b = spar height.
The value of ks is obtained using a/b from the given plot in figure, ks is
obtained and thus the actual shear stress in each web
Figure 7-2: Shear buckling coefficients fro plates as a ratio of a and b for hinged and clamped edges
Both these values are less than 211.4392 GPa. Thus, the web does not fail due
to shear buckling.
88
7.1.2 Shear flow:
The shear flow can be considered for the two cells in the airfoil cross
section. The shear flow will be due to the torque as well due to the bending
moments. These are computed separately and summed up to obtain the net
shear flow pattern for the wing cross section.
Due to torque,
A area of each cell Q shear flow due to the torque The maximum torque experienced at the root of the wing is -25338 Nm.
Using GAMBIT software, the areas and perimeters of the cells in the airfoil
formed by the spars and the skin are found to be,
Cell1 A1 = 2.1783 m2
Cell2 A2= 10.5862 m2
Cell3 A3= 2.7528 m2
l1 = 4.45527 m, l2 = 1.6186 m, l3 = 5.6214 m, l4 = 5.5827 m, l5 = 1.4795 m, l6 =
4.092 m, l7 = 4.176 m .
l1, l2 belong to cell1,l2, l3, l4, l5 belong to cell2, l5, l6 and l7 to cell3. l2 and l5
are the spars.
The equations for the two cells involving shear flow of cell1 q1 and shear flow
of cell2 q2 are
[ (
) (
) (
)]
[ (
) (
)]
[ (
) (
)]
89
The second, third and fourth equations are obtained from the condition that
the cell twist is zero.
Solving these equations, we get shear flow values due to torque alone.
The shear flow due to bending is given by the formula,
*
+ *
+
Vx = 1012394.13 N (shear force due to chordwise forces)
Vy = 17480623.2 N (shear force due to normal forces)
90
Spar
Are
aA
*xA
*yx_
bar
y b
arx_
cy_
cIx
xIy
yIx
yq
F_U
_117
9935
.396
3063
4361
0.4
1453
9823
6.2
1702
.52
808.
058
-177
5.38
808.
058
1.20
998E
+11
5.70
659E
+11
-2.5
8137
E+11
-148
5868
F_U
_217
9935
.396
4028
8974
6.5
1453
9823
6.2
2239
.08
808.
058
-123
8.82
808.
058
1.20
998E
+11
2.79
65E+
11-1
.801
22E+
11-1
9644
38
F_L_
117
9935
.396
3063
4361
0.4
-493
0049
9.2
1702
.52
-273
.99
-177
5.38
-273
.99
1701
5324
658
5.70
659E
+11
8752
7120
181
4218
04.4
9
F_L_
217
9935
.396
4028
8974
6.5
-493
0049
9.2
2239
.08
-273
.99
-123
8.82
-273
.99
1701
5324
658
2.79
65E+
1161
0744
4435
755
5765
.26
R_U
_116
4482
.769
1198
2783
5311
8961
011.
272
85.1
372
3.24
338
07.2
372
3.24
388
6961
4305
12.
3868
4E+1
24.
5291
2E+1
160
7569
.23
R_U
_216
4482
.769
1280
6250
0711
8961
011.
277
85.7
772
3.24
343
07.8
772
3.24
388
6961
4305
13.
0550
9E+1
25.
1246
9E+1
154
9369
.11
R_L
_116
4482
.769
1198
2783
53-4
2045
249.
872
85.1
3-2
55.6
2138
07.2
3-2
55.6
2113
4060
7327
02.
3868
4E+1
2-1
.600
76E+
11-2
6563
1.1
R_L
_216
4482
.769
1280
6250
07-4
2045
249.
877
85.7
7-2
55.6
2143
07.8
7-2
55.6
2113
4060
7327
03.
0550
9E+1
2-1
.811
25E+
11-2
3383
0
Sum
1377
672.
6663
7627
3434
3460
2699
6.8
4.80
23E+
111.
2584
5E+1
33.
3452
1E+1
1
Spar
Are
aA
*xA
*yx_
bar
y b
arx_
cy_
cIx
xIy
yIx
yq
Ce
ll 1
F_U
_117
9935
.396
3063
4361
0.4
1453
9823
6.2
1702
.52
808.
058
-177
5.38
808.
058
1.20
998E
+11
5.70
659E
+11
-2.5
8137
E+11
-148
5868
F_L_
117
9935
.396
3063
4361
0.4
-493
0049
9.2
1702
.52
-273
.99
-177
5.38
-273
.99
1701
5324
658
5.70
659E
+11
8752
7120
181
4218
04.4
9
6126
8722
0.8
9609
7737
.07
-106
4063
Ce
ll 2
F_U
_217
9935
.396
4028
8974
6.5
1453
9823
6.2
2239
.08
808.
058
-123
8.82
808.
058
1.20
998E
+11
2.79
65E+
11-1
.801
22E+
11-1
9644
38
F_L_
217
9935
.396
4028
8974
6.5
-493
0049
9.2
2239
.08
-273
.99
-123
8.82
-273
.99
1701
5324
658
2.79
65E+
1161
0744
4435
755
5765
.26
R_U
_116
4482
.769
1198
2783
5311
8961
011.
272
85.1
372
3.24
338
07.2
372
3.24
388
6961
4305
12.
3868
4E+1
24.
5291
2E+1
160
7569
.23
R_L
_116
4482
.769
1198
2783
53-4
2045
249.
872
85.1
3-2
55.6
2138
07.2
3-2
55.6
2113
4060
7327
02.
3868
4E+1
2-1
.600
76E+
11-2
6563
1.1
2396
5567
0717
3013
498.
4-1
0667
35
Ce
ll 3
R_U
_216
4482
.769
1280
6250
0711
8961
011.
277
85.7
772
3.24
343
07.8
772
3.24
388
6961
4305
13.
0550
9E+1
25.
1246
9E+1
154
9369
.11
R_L
_216
4482
.769
1280
6250
07-4
2045
249.
877
85.7
7-2
55.6
2143
07.8
7-2
55.6
2113
4060
7327
03.
0550
9E+1
2-1
.811
25E+
11-2
3383
0
2561
2500
1476
9157
61.3
131
5539
.12
Fig
ure
7-3
: S
hea
r fl
ow
in
sp
ars
91
(
)
For the web, the dimensions of a and b will be
a = 1.6186 m9rib spacing and
b =5.6214 (length of the cell containing critical shear flow)
KS =32;
τcr is found in terms of t2
Using relation, τcr = qcr/t
The value of ks is obtained using a/b from the given plot in figure, ks is
obtained and thus the thickness of the skin without using stringer will be
7.2 Stringer design:
The thickness of the skin determined above is too high for the skin of an
aircraft. Therefore in order to reduce skin thickness and redistribute the shear
flow in the wing skin, stringers are added. The number of stringers can be
determined by evaluating the amount by which the skin thickness should be
reduced. Roughly 36 stringers can be added to the wing, 18 on the upper
surface of the airfoil and 18 on the lower surface of the airfoil. The stringer
cross section is chose from the standard cross sections available in Analysis of
Aircraft structures – Bruhn. The Z cross section is chosen and scaled up or
down determining the critical stress in each stringer and iterating if it is less
than the critical buckling stress of the stringer cross section.
92
7.2.1 Geometric dimensions based on shear flow:
The stress of each stringer is found using the formula,
*
+ *
+
Thus the section satisfying iterations has the following properties, A =
10000mm2,
The shear flow for each section is determined using the formula,
*
+ *
+
93
Table 7-1: Shear flow and Bending stress tabulation
Stringer A Ax Ay x_c y_c σ q
1 10000 114498474 134228 6483.1 -238.3 -7.497 9.95E+08
2 10000 111931155 516227 6226.4 -200.1 -7.809 8.5E+08
3 10000 108296803 1223363 5862.9 -129.3 -8.298 5.64E+08
4 10000 103680870 2287145 5401.3 -22.96 -9.009 1.04E+08
5 10000 98180431 3698022 4851.3 118.13 -10.01 -5.6E+08
6 10000 91896585 5394407 4222.9 287.77 -11.41 -1.4E+09
7 10000 84934997 7276236 3526.8 475.95 -13.35 -2.5E+09
R_U_1 179935 1.311E+09 1.3E+08 2318.4 471.57 0.6891 -1521292
R_U_2 179935 1.401E+09 1.3E+08 2819 471.57 -0.492 -1457428
8 10000 69429422 1.1E+07 1976.2 850.91 -18.73 -4.9E+09
9 10000 61134483 1.3E+07 1146.7 1000.8 -18.58 -4.4E+09
10 10000 52674321 1.3E+07 300.69 1094.3 -6.489 51024314
11 10000 44249964 1.4E+07 -541.7 1119.9 12.091 4.67E+09
12 10000 36077346 1.3E+07 -1359 1084.6 18.094 4.55E+09
13 10000 28357220 1.2E+07 -2131 995.83 17.126 3E+09
F_U_2 179935 402889746 1.5E+08 -2728 556.38 78.162 1233401
F_U_1 179935 306343610 1.5E+08 -3264 556.38 48.757 757536.7
14 10000 14973466 9507529 -3469 699.08 13.242 8.9E+08
15 10000 9612440.3 7640616 -4005 512.39 11.88 3.98E+08
16 10000 5310518 5662783 -4436 314.6 10.904 1.37E+08
17 10000 2177421.6 3695896 -4749 117.92 10.243 24461501
18 10000 330609.2 1889583 -4934 -62.72 9.856 -4256124
19 10000 423361.3 -730095 -4924 -324.7 9.7933 954223.2
20 10000 2501869.1 -2E+06 -4717 -463.8 10.129 -2.1E+07
21 10000 5958680.3 -3E+06 -4371 -569.6 10.795 -9.6E+07
22 10000 10487170 -4E+06 -3918 -664.5 11.811 -2.4E+08
23 10000 15969273 -5E+06 -3370 -749 13.29 -4.7E+08
F_L_1 164483 280035204 -5E+07 -3264 -525.7 2.2024 -159628
F_L_2 164483 368290078 -5E+07 -2728 -525.7 6.9376 -280746
24 10000 29304388 -6E+06 -2036 -882.3 18.122 -1.2E+09
25 10000 36882186 -7E+06 -1279 -930.6 20.113 -1.1E+09
26 10000 44867767 -7E+06 -480 -963.9 12.749 1.59E+09
27 10000 53102607 -7E+06 343.52 -964.6 -11.2 7.38E+09
28 10000 61406999 -7E+06 1174 -920.3 -21.32 8.28E+09
29 10000 69601782 -6E+06 1993.4 -838.7 -19.23 6.5E+09
R_L_1 164483 1.198E+09 -4E+07 2318.4 -507.3 2.6696 542408.6
R_L_2 164483 1.281E+09 -4E+07 2819 -507.3 2.1096 510540.7
30 10000 85045683 -4E+06 3537.8 -629.2 -13.15 3.62E+09
31 10000 92011967 -3E+06 4234.5 -524.2 -11.25 2.73E+09
32 10000 98301217 -2E+06 4863.4 -430.7 -9.905 2.08E+09
33 10000 103796399 -1E+06 5412.9 -354.8 -8.944 1.62E+09
34 10000 108391643 -484586 5872.4 -300.1 -8.263 1.31E+09
94
35 10000 111993602 -154167 6232.6 -267.1 -7.793 1.14E+09
36 10000 114525411 -13414 6485.8 -253 -7.492 1.06E+09
Stringer design ratio:
The critical shear flow is found to be 162206.0558N/m acting between the
upper flange of front spar and stringer 5. Using the formula,
(
)
where a = 1.6186 m, (rib spacing),
b = 0.526 m,
kb is obtained from following plot for given a/b as 40
-1000
0
1000
2000
-2000 0 2000 4000 6000 8000 10000 12000 14000
y (m
m)
x (mm)
NACA 653-418
5t
5t
6t
t
95
Figure 7-4: Shear- buckling coefficient for hinged and clamped plates
Thus skin thickness after using stringers is found to be t = 10.953 mm
96
Figure 7-5: Shear flow diagram for wing represented in polar coordinates, 19th
stringeer is leading edge
and 1st stringer is at trailing edge
-6000
-4000
-2000
0
2000
4000
6000
8000
10000
12
34
5
6
7
R_U_1
R_U_2
8
9
10
11
12
13
F_U_2
F_U_1
14
1516
1718
1920
2122
23
F_L_1
F_L_2
24
25
26
27
28
29
R_L_1
R_L_2
30
31
32
3334
3536Millions
Shear flow distribution on skin of wing
97
8. Fuselage design
8.1 Description
Fuselage contributes very little to lift and produces more drag but it is an
important structural member/component. It is the connecting member to all
load producing components such as wing, horizontal tail, vertical tail, landing
gear etc. and thus redistributes the load. It also serves the purpose of housing or
accommodating practically all equipment, accessories and systems in addition
to carrying the payload. Because of large amount of equipment inside the
fuselage, it is necessary to provide sufficient number of cutouts in the fuselage
for access and inspection purposes. These cutouts and discontinuities result in
fuselage design being more complicated, less precise and often less efficient in
design. As a common member to which other components are attached, thereby
transmitting the loads, fuselage can be considered as a long hollow beam. The
reactions produced by the wing, tail or landing gear may be considered as
concentrated loads at the respective attachment points. The balancing reactions
are provided by the inertia forces contributed by the weight of the fuselage
structure and the various components inside the fuselage. These reaction forces
are distributed all along the length of the fuselage, though need not be
uniformly. Unlike the wing, which is subjected to mainly unsymmetrical load,
the fuselage is much simpler for structural analysis due to its symmetrical
cross-section and symmetrical loading. The main load in the case of fuselage is
the shear load because the load acting on the wing is transferred to the fuselage
skin in the form of shear only. The structural design of both wing and fuselage
begin with shear force and bending moment diagrams for the respective
members. The maximum bending stress produced in each of them is checked to
be less than the yield stress of the material chosen for the respective member.
98
8.2 Loads and its distribution:
To find out the loads and their distribution, consider the different cases. The
main components of the fuselage loading diagram are:
Weight of the fuselage
Engine weight
Weight of the horizontal and vertical stabilizers
Tail lift
Weight of crew, payload and landing gear
Systems, equipment, accessories
Symmetric flight condition, steady and level flight: (Downward forces
negative) Values for the different component weights are obtained from
aerodynamic design calculations.
Table 8-1: Loads acting on Fuselage
Condition 1 Full Payload and Full Fuel
Fuselage alone analysis
S. No Components
Distance from reference line (m) Mass (kg) Weight (N)
Moment (Nm)
1 Crew 3.043 270 2648.7 8059.9941
2 Nose landing gear 6.086 3600 35316 214933.176
3 Payload bay 1 12.172 37500 367875 4477774.5
4 Fixed equipment 18.641 1149.19 11273.554 210150.318
5 Excess mass 22.641 21600 211896 4797537.34
6 Fuselage mass 22.641 43200 423792 9595074.67
7 Fuel in fuselage 22.641 95280.42 934700.92 21162563.5
8 Main landing gear assembly 1 22.641 10800 105948 2398768.67
9 Main landing gear assembly 2 30.3211 7200 70632 2141639.94
10 Payload bay 2 30.321 37500 367875 11154337.9
11 Horizontal stabilizer 45.367 6400 62784 2848321.73
12 Vertical Stabilizer 48.501 3200 31392 1522543.39
Total 267699.6 2626133.2 60531705.1
Cg from Nose 23.04974695
99
Figure 8-1: Balance diagram showing loads acting on fuselage
100
8.3 Shear Force and bending moment calculations:
Table 8-2: Shear force and bending moment tabulation
Distance(m) Load (kg) SF (N) BM (Nm)
0 0 0 0
3.043 -2648.7 -2648.7 -8059.9941
6.086 -35316 -37964.7 -222993.17
12.172 -367875 -405840 -4700767.67
18.641 -11273.6 -417113 -4910917.99
22.641 -211896 -629009 -9708455.32
22.641 -423792 -1052801 -19303530
22.641 -934701 -1987502 -40466093.5
22.641 -105948 -2093450 -42864862.2
23.0497 2626133 532683 17666719.62
30.3211 -70632 462051 15525079.69
30.321 -367875 94176 4370741.812
45.367 -62784 31392 1522420.084
48.501 -31392 0 0
54.849 0 0 0
Figure 8-2: Shear force on the fuselage (free-free beam with one reaction at its centre) at fully loaded
condition
-2500
-2000
-1500
-1000
-500
0
500
1000
0 10 20 30 40 50
She
ar F
orc
e(N
)
Tho
usa
nd
s
Distance from nose cone(m)
Shear Force
101
Figure 8-3: Bending moment on the fuselage (free-free beam with one reaction at its centre) at fully
loaded condition
-50000
-40000
-30000
-20000
-10000
0
10000
20000
30000
0 10 20 30 40 50
She
ar F
orc
e(N
)
Tho
usa
nd
s
Distance from nose cone(m)
Bending Moment
102
9. Detailed Design of Fuselage:
9.1 Stringer Design:
Design of the fuselage can be carried out by considering the maximum bending
moment which is taken as the design bending moment. The cross-sectional area
required to withstand the bending stress is found out by using the formula for
bending stress. This area is divided among several stringers which are spaced
evenly. The stringers spacing is calculated by considering the buckling of the
portion between adjacent stringers which can be modelled as a plate. Now, the
first step is to calculate the required cross-sectional area of the stringers. Use
the following formula for bending stress.
Where,
σ Tensile strength of the material used (Aluminium 7075) = 455 MPa
M Design bending moment = -42864862.2 Nm
I Second moment of area (m4)
(
)
y d/2
d diameter of the fuselage (3.5m )
A cross-sectional area of the fuselage stringers (m2)
A stringer cross section (Z section) is chosen satisfying the condition that the
actual stress is less than the yield stress of the material.
103
Figure 9-1: Location of Z shaped Stringer in the fuselage.
The properties of the stringer section chosen are as follows,
Length of stringer:
Height of stringer:
Where
t thickness of stringer
The total circumference of the fuselage cross section is found to be 21.9914 m.
This circumference is distributed with ‘n’ number of stringers such that the
total bending moment is taken up by these stringers effectively. Assume skin is
ineffective in bending. Arbitrarily, let us set the number of stringers to be equal
to 60 i.e. 15 stringers in each quadrant. Now, the net IYY is computed
-4
-3
-2
-1
0
1
2
3
4
-4 -3 -2 -1 0 1 2 3 4
Stringer location in fuselage
104
considering these stringers to be lumped masses. As it is a symmetric cross
section,
9.2 Shear flow along skin of fuselage:
Consider the stringer at Ө = 0° of the first quadrant of the cross section as the
first stringer and number it in anticlockwise direction. Make a cut between
stringers 1 and 2 and determine shear flow using the formula,
*
+ *
+
Since
VX = 0 , VY = -2093450N ( Max. Shear Force from shear force diagram)
The shear flow equation gets simplified to
[
]
∑q_l = -1.81599E+11 N
Now, on closing the cut, and considering cell twist is zero for the fuselage
cross section, we obtain the equation, -1.81599E+11+ 21.9914qo = 0
Thus constant shear flow to be added to the cell is qo = 8257728021 N/m
105
Table 9-1: Shear flow along the stringers of fuselage
S.No ʘ A Ax Ay si Ax Si Ay q open q*l q fin
1 0 0.0269 0.0942 0 0.0942 0 0 0 8.26E+09
2 6 0.0269 0.0937 0.00984 0.1878 0.00984 -9E+07 -33157611 8.17E+09
3 12 0.0269 0.0921 0.01958 0.28 0.02942 -2.7E+08 -99109571 7.99E+09
4 18 0.0269 0.0896 0.0291 0.3695 0.05852 -5.4E+08 -197133342 7.72E+09
5 24 0.0269 0.086 0.0383 0.4556 0.09683 -8.9E+08 -326155016 7.37E+09
6 30 0.0269 0.0816 0.04709 0.5371 0.14391 -1.3E+09 -484761088 6.94E+09
7 36 0.0269 0.0762 0.05535 0.6133 0.19927 -1.8E+09 -671213940 6.43E+09
8 42 0.0269 0.07 0.06301 0.6833 0.26228 -2.4E+09 -883470876 5.85E+09
9 48 0.0269 0.063 0.06998 0.7463 0.33227 -3.1E+09 -1.119E+09 5.2E+09
10 54 0.0269 0.0554 0.07619 0.8017 0.40845 -3.8E+09 -1.376E+09 4.5E+09
11 60 0.0269 0.0471 0.08156 0.8488 0.49001 -4.5E+09 -1.651E+09 3.75E+09
12 66 0.0269 0.0383 0.08603 0.8871 0.57604 -5.3E+09 -1.94E+09 2.96E+09
13 72 0.0269 0.0291 0.08956 0.9162 0.66561 -6.1E+09 -2.242E+09 2.14E+09
14 78 0.0269 0.0196 0.09212 0.9358 0.75772 -7E+09 -2.552E+09 1.29E+09
15 84 0.0269 0.0098 0.09366 0.9456 0.85138 -7.8E+09 -2.868E+09 4.33E+08
16 90 0.0269 4E-06 0.09418 0.9456 0.94556 -8.7E+09 -3.185E+09 -4.32E+08
17 96 0.0269 -0.0098 0.09366 0.9358 1.03922 -9.6E+09 -3.501E+09 -1.29E+09
18 102 0.0269 -0.0196 0.09212 0.9162 1.13134 -1E+10 -3.811E+09 -2.14E+09
19 108 0.0269 -0.0291 0.08957 0.8871 1.2209 -1.1E+10 -4.113E+09 -2.96E+09
20 114 0.0269 -0.0383 0.08604 0.8488 1.30694 -1.2E+10 -4.402E+09 -3.75E+09
21 120 0.0269 -0.0471 0.08156 0.8017 1.3885 -1.3E+10 -4.677E+09 -4.5E+09
22 126 0.0269 -0.0553 0.07619 0.7464 1.46469 -1.3E+10 -4.934E+09 -5.2E+09
23 132 0.0269 -0.063 0.06999 0.6833 1.53468 -1.4E+10 -5.169E+09 -5.85E+09
24 138 0.0269 -0.07 0.06302 0.6134 1.59771 -1.5E+10 -5.382E+09 -6.43E+09
25 144 0.0269 -0.0762 0.05536 0.5372 1.65307 -1.5E+10 -5.568E+09 -6.93E+09
26 150 0.0269 -0.0816 0.04709 0.4556 1.70016 -1.6E+10 -5.727E+09 -7.37E+09
27 156 0.0269 -0.086 0.03831 0.3696 1.73847 -1.6E+10 -5.856E+09 -7.72E+09
28 162 0.0269 -0.0896 0.02911 0.28 1.76758 -1.6E+10 -5.954E+09 -7.99E+09
29 168 0.0269 -0.0921 0.01959 0.1879 1.78717 -1.6E+10 -6.02E+09 -8.17E+09
30 174 0.0269 -0.0937 0.00985 0.0943 1.79702 -1.7E+10 -6.053E+09 -8.26E+09
31 180 0.0269 -0.0942 8.7E-06 8E-05 1.79703 -1.7E+10 -6.053E+09 -8.26E+09
32 186 0.0269 -0.0937 -0.0098 -0.0936 1.78719 -1.6E+10 -6.02E+09 -8.17E+09
33 192 0.0269 -0.0921 -0.0196 -0.1857 1.76762 -1.6E+10 -5.954E+09 -7.99E+09
34 198 0.0269 -0.0896 -0.0291 -0.2753 1.73853 -1.6E+10 -5.856E+09 -7.72E+09
35 204 0.0269 -0.086 -0.0383 -0.3613 1.70024 -1.6E+10 -5.727E+09 -7.37E+09
36 210 0.0269 -0.0816 -0.0471 -0.4429 1.65316 -1.5E+10 -5.569E+09 -6.94E+09
37 216 0.0269 -0.0762 -0.0553 -0.5191 1.59781 -1.5E+10 -5.382E+09 -6.43E+09
38 222 0.0269 -0.07 -0.063 -0.5891 1.5348 -1.4E+10 -5.17E+09 -5.85E+09
39 228 0.0269 -0.063 -0.07 -0.6521 1.46482 -1.3E+10 -4.934E+09 -5.2E+09
40 234 0.0269 -0.0554 -0.0762 -0.7074 1.38864 -1.3E+10 -4.678E+09 -4.5E+09
41 240 0.0269 -0.0471 -0.0816 -0.7545 1.30709 -1.2E+10 -4.403E+09 -3.75E+09
42 246 0.0269 -0.0383 -0.086 -0.7929 1.22106 -1.1E+10 -4.113E+09 -2.96E+09
106
43 252 0.0269 -0.0291 -0.0896 -0.822 1.1315 -1E+10 -3.811E+09 -2.14E+09
44 258 0.0269 -0.0196 -0.0921 -0.8416 1.03938 -9.6E+09 -3.501E+09 -1.29E+09
45 264 0.0269 -0.0099 -0.0937 -0.8514 0.94573 -8.7E+09 -3.186E+09 -4.34E+08
46 270 0.0269 -1E-05 -0.0942 -0.8514 0.85155 -7.8E+09 -2.868E+09 4.32E+08
47 276 0.0269 0.0098 -0.0937 -0.8416 0.75789 -7E+09 -2.553E+09 1.29E+09
48 282 0.0269 0.0196 -0.0921 -0.822 0.66577 -6.1E+09 -2.243E+09 2.14E+09
49 288 0.0269 0.0291 -0.0896 -0.7929 0.5762 -5.3E+09 -1.941E+09 2.96E+09
50 294 0.0269 0.0383 -0.086 -0.7547 0.49016 -4.5E+09 -1.651E+09 3.75E+09
51 300 0.0269 0.0471 -0.0816 -0.7076 0.40859 -3.8E+09 -1.376E+09 4.5E+09
52 306 0.0269 0.0553 -0.0762 -0.6522 0.3324 -3.1E+09 -1.12E+09 5.2E+09
53 312 0.0269 0.063 -0.07 -0.5892 0.2624 -2.4E+09 -883868023 5.85E+09
54 318 0.0269 0.07 -0.063 -0.5193 0.19937 -1.8E+09 -671567406 6.43E+09
55 324 0.0269 0.0762 -0.0554 -0.4431 0.144 -1.3E+09 -485066999 6.93E+09
56 330 0.0269 0.0816 -0.0471 -0.3615 0.0969 -8.9E+08 -326410021 7.37E+09
57 336 0.0269 0.086 -0.0383 -0.2755 0.05858 -5.4E+08 -197334648 7.72E+09
58 342 0.0269 0.0896 -0.0291 -0.1859 0.02947 -2.7E+08 -99254973 7.99E+09
59 348 0.0269 0.0921 -0.0196 -0.0938 0.00987 -9.1E+07 -33245515 8.17E+09
60 354 0.0269 0.0937 -0.0099 -0.0002 8.7E-06 -80333 -29443.594 8.26E+09
8.7E-06
-5E+11 -1.816E+11
107
Figure 9-2: Shear flow distribution along fuselage, View 16th
stringer is at the bottom and 46th
stringer
at the top
The critical shear flow is found to occur in elements between 1 and 60, 30 and
31. The critical shear flow value is 82577258021 N/m.
We know that,
(
)
Where,
E = 7.17e10 N/m2,
0
1000
2000
3000
4000
5000
6000
7000
8000
9000
12 3 4
56
78
910
11
12
13
14
15
16
17
18
19
20
21
2223
2425
2627
28293031
32333435
3637
3839
40
41
42
43
44
45
46
47
48
49
50
51
5253
5455
5657
58 59 60
Millions
Shear Flow distribution in the skin of fuselage
108
a = 1.099557 m(bulk head spacing),
b = 0.366519 m (Stringer circumferential spacing)
a/b = 3
ks = 40
Figure 9-3: Shear buckling coefficient for plates with hinge and clamps
ν = 0.3
Thus we obtain, t=0.00748 m
The skin thickness is thus found to be t = 7.4833 mm
Using τ = 1.5*(q/t), τMat = 211.4392e6, we get = 0.133 mm.
109
Considering the maximum of the two, we get t = 7.4833 mm
The above value of skin thickness is well within the standard limits. Therefore,
the above design is acceptable.
110
10. Computational Fluid Dynamics
CFD analysis at tip:
The tip stall is the worst initiation step in all instability problems so the tip at
an angle of incidence 5˚ is checked for separation.
Figure 10-1: Velocity vector plot for aoa= 5deg
111
Figure 10-2: CP plot for aoa= 5deg
Figure 10-3: Contours of static Pressure
112
Figure 10-4: Velocity contours plot for aoa= 5deg
113
11. Three view diagram:
Figure 11-1 Front view
114
Figure 11-2: Top view
115
Figure 11-3: Side view
116
Conclusion
The Conceptual Design phase of an aircraft is probably the most
interesting and intriguing phase of aircraft design. It is a clear indication of the
compromise that has to be made between various divisions of an Aircraft
design center, and yet satisfy an incredible number of real-world constraints
and design specifications. Aircraft design involves a variety of the field of
Aerospace engineering like structures, performance, aerodynamics, stability
etc. Among this we went through the structure part in this project which has
enabled us to get a taste of what it is to design a real aircraft. The fantasies of
the flying world seem to be much more than what we thought. With this design
project as the base, we will strive to progress in the field of airplane design and
maintenance. We convey our heartfelt gratitude to all of them who have
provided their helping hand in the completion of this project.
117
Reference
Books:
Analysis of Aircraft structures – Bruhn
Aircraft Structures for engineering students – T.H.G Megson
Aircraft structures – Peery
Airplane design – Jan Roskam
Fundamentals of Aerodynamics - Anderson J D
Websites:
www.wikipedia.org
www.joeclarksblog.com
http://www.docstoc.com/
http://www.flightsimaviation.com/
www.tc.gc.ca
http://www.risingup.com/
http://www.aerospacemetals.com/contact-aerospace-metals.html
http://www.aerospaceweb.org/
www.faa.gov/regulations_policies/faa_regulations/