Maximum bending moment at the root
=1768011.725Nm
On considering the safety factor (1.5) and the load factor (2.5) into account, we
have Maximum design bending moment
= 1.5*n*Mmax
=6630043.969 Nm

Area required for the stringers and spars are calculated from the bending formula
Y=0.3808m
M=6630043.969
Nm
σb = 476*10^6 Nm^-2
From this we can get the area required
Hence A=0.03658 m^2
the area required to support the maximum bending moment
coming on the wing is 0.03658 m^2
Total area required is divided between spar and stringers as follows;
Total area required is summation of 60 % of this area is allotted for spars and 40 % for
stringers.
Area of the spar =0.02195 m^2
Area of the stringers
=0.01

463m
2

Front spar:
60% of the spar area is taken for the front spar.
A=0.35C-
0.15C
B=0.65C-0.35C
Area of the top flange
= ( a/a+b)*A
fs
=8.78*10^-3
Area of the bottom flange
= (b/a+b)*A
fs
=4.39*10^-3
Rear spar:
40% of the spar area is taken for the front spar.
Area of the top flange
= ( c/c+d)* A
rs
=6.585*10^-3 m
2
Area of the bottom flange
= (d/c+d)* A
rs
=2.195*10^-3 m
2
Ssitable angle section are selected with respect to area for front spar and rear spar.

The properties of the angle section selected are tabulated.
Component
Area
m
2
Width
m
Thickness
m
Radius of
Curvature
M
Front Spar
Top
Flange
94
17.98
2.81
2.25
Bottom
Flange
47
12.75
12.75
1.59
Rear Spar
Top
Flange
70.5
15.60
2.44
1.95
Bottom
Flange
23.5
8.99
1.41
1.21
T
W
Stringer Selection:
For a double riveted joint we have K
c
= 10.2
R

We choose aluminium Al – 2024 material whose yield stress is 340 MPa, E = 70 MPa and
n
= 0.3375
Let us assume the skin thickness be 2.5 mm
On substituting these values in the above formula we get
Then the stringer spacing b =10, 89 Cm
No of stringers at top surface, Top=((perimeter)/2)/10.5
= 40 stringers
No of stringers at bottom surface, Bottom=((perimeter)/2)/14.0
= 30 stringers
Area of one stringers, Astr = 4.4857 cm^2
the stringers are usually Z SECTIONS
The specification of the Z –section chosn for the stringersare given below;
B
T
R
A

The various dimensions are as follows:
A=2.765 cm
B=7.374 cm
T=0.376 cm
T1=0.346
cm R=0.356
cm
Area = 4.4857 cm^2
Redrived data from tabulation
is given by
Ixy
=425647.8612 cm^4
Iyy
=15930092.35 cm^4
I
xx
=640587.3284 cm^4
Xcg=22.9420cm
Ycg=18.3754cm
ΣA=827.19cm^2
ΣAx=184415.3561cm^2
ΣAy=15200.2041cm^2
ΣAxy=3814409.798 cm^4
ΣIGX==43176.9810
cm^4 ΣIGY=12916,521
cm^4
For the airfoil selected, cruise angle of attack
=5
0.
M
x
=M cos

= 7103065.13*cos 5
0
=7076035.823Nm
M
y
=M sin
=713065.13*sin 5
0
=619072.9172Nm
Substituting M
x
, M
z
, I
xx
,I
zz
and I
xz
in the formula, we get,
b
=11.0199*y -0.2563*x
For the top most stringers, (15th)
X=227.5596-222.9420
=54.6176cm
Y=53.1250-18.3757
= 34.7493 cm
b=368.9353MPa, this is less than the allowable stress,
Marigen of safety = (
b
allowable/
b
max
)-1=0.2902
=29.02%

FUSELAGE DESIGN:
The fuselage is the main structure or body of the fixed-wing aircraft. It provides space for
cargo, controls, accessories, passengers, and other equipment. In single-engine aircraft, the
fuselage houses the power plant. In multiengine aircraft, the engines may be either in the
fuselage, attached to the fuselage, or suspended from the wing structure. There are two
general types of fuselage construction: truss and monocoque. Truss Type A truss is a rigid
framework made up of members, such as beams, struts, and bars to resist deformation by
applied loads.

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- Winter '12
- Aerodynamics, Fixed-wing aircraft, Airfoil, Wing design, aircraft wing