Design, Analysis and Testing of Wing Spar for Optimum Weight

International Journal of Research and Scientific Innovation (IJRSI) | Volume IV, Issue VII, July 2017 | ISSN 2321–2705
www.rsisinternational.org Page 104
Design, Analysis and Testing of Wing Spar for
Optimum Weight
Mutturaj Girennavar#1
, Soumya H V*2
, Subodh H M #3
, Tanvi J Heraje#4
, Deepak Raj P.Y#5
1,2,3,4
Student, 5
Associate Professor,
Department of Aeronautical Engineering, Srinivas Institute of Technology, Mangaluru, India.
Abstract— Aircraft is a complex mechanical structure with flying
capability. The structure of an airframe represents one of the
finest examples of a minimum weight design in the field of
structural engineering. Surprisingly such an efficient design is
achieved by the use of simple “strength-of-material” approach.
Aircraft has two major components, which are fuselage and
wing. For a wing of an aircraft the primary load carrying ability
is required in bending. A typical aluminium material 6082-T6 is
chosen for the design. A four-Seater aircraft wing spar design is
considered in the current study. Wings of the aircraft are
normally attached to the fuselage at the root of the wing. This
makes the wing spar beam to behave almost like a cantilever
beam. Minimum two spars are considered in the wing design. In
a conventional beam design approach one will end up in heavy
weight for the spar of the wing. In the current project the spar is
considered as a beam with discrete loads at different stations.
The design is carried out as per the external bending moment at
each station. A finite element approach is used to calculate the
stresses developed at each station for a given bending moment.
Several stress analysis iterations are carried out for design
optimization of the spar beam. Linear static analysis is used for
the stress analysis. The spar beam is designed to yield at the
design limit load. Weight optimization of the spar will be carried
out by introducing lightening cut-outs in the web region. The
results from the conventional design approach and the optimized
design are compared. Weight saving through the design
optimization is calculated. Spar will be a built-up structure. A
scale-down model of the spar will be fabricated using aluminium
alloy 6082-T6 material. Static testing of the spar will be carried
out to validate the design and stress analysis results.
Keywords— Design, Aircraft wing, Design optimization, Finite
Element Analysis, Static testing.
______________________________________________
I. INTRODUCTION
n a fixed-wing aircraft, the spar is usually the most support
of the wing, running span wise at right angles to the body.
The spar carries flight loads and also the weight of the wings
whereas on the bottom so it is important to make it to
withstand the twisting load because that causes the failure if
the material soon. Alternative structural and forming members
like ribs are also connected to the spar or spars, with stressed
skin construction conjointly sharing the loads wherever it is
used have to withstand almost all types of loading action like
bending, torsion, tensile and compression. There is also quite
one spar during a wing or none in any respect. The wing spar
provides the bulk of the load support and dynamic load
integrity of cantilever monoplanes, usually mention the
strength of the wing 'D' box itself. Together, these two
structural elements or components put together offer the wing
rigidity required to alter the aircraft to fly safely. The spar
carries flight loads and the weight of the wings while on the
ground. Other structures such as ribs may also be attached to
spars. There may be more than one spar in a wing. However,
where a single spar carries the majority of forces on it, is
known as main spar. As a rule, a wing has two spars. One spar
is sometimes settled close to the front of the wing, and
therefore the alternative regarding common fraction of the gap
toward the wing’s edge. No matter kind, the spar is that the
most significant a part of the wing.
II. DESIGN, ANALYSIS AND FABRICATION
2.1 Design of Wing Spar
This chapter focuses on the detailed design of Spar. The spar
may be considered as the important component of an aircraft
wing, since it carries 80% of the total load on the wing. Since
the Spar geometry and its features are influencing all other
wing components, we begin the detailed design process by
Spar design. The primary function of a Spar is to carry the
bending load acting on the wing. A Spar is a beam which
extends from wing root to tip carrying the compressive, shear
and tensile loads. In the current project, the spar is considered
as a beam with discrete loads at different stations. The design
is carried out as per the external bending moment at each
station. The design calculations includes selection of
materials; estimation of geometrical characteristics. Spar is
designed for the existing aircraft and its configuration. The
reference aircraft has the following specifications:
 Aircraft: PIPER PA28-161 WARRIOR II
 Cantilever low wing monoplane
 Wing span: 10.67m
 Aerofoil series: NACA 65-415 at root and NACA
65- 415 at the tip
 Wing chord at tip: 1.07m
 Wing chord at root: 1.60m
 All of weight of aircraft: 1106 kg
2.2 Material Selection
A Spar generally consists of an aluminium sheet Spar webs
I

and caps which is welded or riveted to the top or bottom of the
sheet to prevent buckling on application of loads.
Most commonly used materials are aluminium alloys. In the
current project, the material selected is aluminium 6082-T6
because of its following properties:
 Axial yield strength = 26 kg/mm2
 Shear yield strength= 18kg/mm2
 Density = 2700 kg/m3
 Modulus of elasticity = 70 GPa
 Poisons ratio = 0.33
 Excellent corrosion resistance
 Good welding property
Fig. 1 Aluminium 2024- T3
2.3 Cross sectional shape of spar
The shape of the Spar is decided in the further level of the
design process. I-section shape of spar is considered after all
the case study of different shapes. Stress and deflection for I-
section is less when compared to other sections because
moment of inertia for the section is more. Since it is bending,
shape of the cross section will have vital role in calculation of
stress and deflection of beam.
2.4 Load Estimation
Any aircraft structure will be designed for the particular
load given by the aerodynamic experts. In this project the
steady level flight is the load case considered to design the
spar.
To carry on the design a typical load distribution was obtained
by using XFLR software. The inputs given to the software are:
 Aero foil series: NACA 65-415 at root and tip
 Wing area : 15.8m2
 Wing span: 9.62m (excluding fuselage diameter)
 Maximum speed of the aircraft : 282 km/h
 Wing chord : 1.07m at the tip and 1.60m at the root
New wing was defined by the considering the rib stations to
be placed at particular distance according to the standard
thumb rule and by giving inputs to the software. The analysis
definition properties are free stream speed is 282 km/h and
density is 1.225 kg/ . By running the analysis for different
angle of attack of the wing the load distribution satisfying the
load case i.e. steady level flight where lift is equal to the
weight of the aircraft (L=W) was obtained at 2.3 degree angle
of attack.
Lift distribution obtained at each rib station were considered
for the worst case of 3g condition by multiplying lift three
times. Lift was assumed to be acting at the centre of pressure
(Cp) which is at 20% of the chord length from leading edge.
The front spar is considered to be placed at 15% of chord
length from the leading edge and the rear spar is considered to
be placed at 50% of chord length from the leading edge. By
considering the simply supported beam condition i.e. load
acting at Cp and the reaction forces acting at front and rear
spar. Solving the simply supported beam conditions the loads
acting the front and rear spar was obtained. Loads at the front
spar at each station were carried for designing of the spar.
Loads on the spar are acting on the ribs. 17 such rib stations
are considered due to wing buckling. The first load acting is at
100 mm from the fixed end and the second load acting is at a
spacing of 300 mm from the first rib.
P1 P2 P3 …………………. P16 P17
Fig. 2 Loads acting on Cantilever beam
Magnitude of load at each rib sections:
Table 1 Loads acting on each rib station
Sections Loads(kg)
1 19.38
2 39.27
3 46.35
4 51.30
5 58.18
6 58.34
7 61.06
8 63.46
9 65.61
10 67.54
11 69.28
12 70.89
13 72.32
14 73.62
15 74.84
16 75.86
17 25.14

P tensile
D
B
Thickness (T)
b/2 b/2
t D
2.5 Design process
Considering the bending moment acting in that cross section
we find out the normal forces acting in the flange of I section.
After finding the normal force acting on the flange the areas of
flanges will be determined as below calculations shown.
BM at rib 1:
BM=
The value of loads P1 is 25.144 kg, P2 is 75.863 kg, P3 is
74.847 kg and so on. And distance with respect to the first rib
L1 is 4710 mm, L2 is 4500 mm, L3 is 4200 mm and so on.
Therefore, BM = 2589594.24 kg-mm
Area of flange:
BM= (D)………..(2)
Substituting the values for BM = 2589594.24 kg-mm and
perpendicular distance (D) is 175.47mm.
We get,
Ptensile = 14758.045 kg
Since the forces are acting normal to the flange,
We consider stress= P/A to find out area of flange.
…………………………….….(3)
Substituting = 35 kg/mm2
, we get
A =421.658
We know that,
Area= …………………………………………… (4)
Considering the width (B) of the flange to be 60 mm,
Width (B)
We get thickness as:
t = 7.027 mm thickness (t)
Bending stress calculation only for flanges:
………………………(5)
y =distance from neutral axis to the point where stress is
calculated = 87.7355 mm
M= Bending moment = 2589594.24 kg-mm
I= Moment of Inertia.
Moment of inertia is given as:
………………………………………... (6)
Substituting the values of B = 60mm, D=175.471mm,
d , we get
=5984862.321
Therefore, bending stress is:
= 37.96 kg/m
Area of the web:
Considering the Shear Force (SF) acting in the same cross
section, we find out the web cross section area.
nPPPPPSF ........4321   …(7)
Substituting the values of the loads, we get shear force.
SF = 992.5089874 kg
We know that, Shear yield strength for 2024 is 27 kg/mm2
.
Now substituting for shear yield strength and shear force, we
get the area of the web.
A=52.23731513
We know the depth of the web is d = 161.417 mm. Therefore,
from equation 4.4, we get the thickness of the web as
t = 0.32361 mm
Bending stress with flange and web:
Substituting the values of B, D and d and b = 59.676 mm in
equation
I = - ………………………………………… (9)
We get, I = 609 41 .
Now, substituting the values of M, I and y in equation 5,
bending stress obtained is
= 37.255 kg/mm2
d
Here it was observed that bending stress in spar of I-section
with web is less than that of the bending stress without web
due to the increase in moment of inertia. From the bending
stress equation, moment of inertia is inversely proportional to
the bending stress.

60 mm
Since the yield strength of the considered aluminium 2024-T3
is 35 kg/mm2
, the bending stress is further reduced by
increasing the moment of inertia. This is done by changing the
dimensions of flanges or web of the I-section. Since the depth
of I-section is fixed, thickness and width of the flanges and
thickness of the web can be changed.
One such iteration is carried out to make the stress to 35
kg/mm2
by changing the thickness of web. For web thickness
= 0.3236 mm stress obtained is 37.255 kg/mm2
,if the load is
continuously applied on a structure, web buckles before it
reaches the value of bending stress. So to overcome the
buckling phenomenon, Shear buckling concept is introduced.
Shear buckling concept plays an important role.
If buckling factor =1 the section buckles or may not buckle.
>1 the section does not buckle.
< 1 the section buckles.
The buckling factor at thickness 0.3236 mm is 0.009. Some
iterations are carried out to increase the buckling factor
greater than 1. At thickness 1.6 mm, the buckling factor is
1.12. So the web does not buckle and is safe.
Determining buckling factor is as follows:
( )
2
…………………. (10)
Where,
Ks =a/b2 = 6.6 (Bruhns book) a= distance between ribs
b2=smallest length of web
E =Young’s Modulus = 000 kg/mm2
t = thickness of the web = 1.6 mm
Substituting these equations in the above equation, we get
= 4.314
To find critical load Pcr:
= ……………………………………..……(11)
A = t x b1…………………………………..…… (12)
Where,b1= depth of longest web = 161.417 mm, t = 1.6 mm
Therefore, area A = 258.267 mm2
Substituting in equation 11, we get
Pcr = 1114.164 kg
Papplied in the fixed section obtained is 992.508 kg.
Buckling factor = ……………………………. (13)
Therefore, buckling factor = 1.12
Bending stress and final dimension of 1st
section:
Now, we have M = 2589594.24 kg-mm, y = 87.7355 mm,
Since the thickness of web is changed due to buckling factor,
the moment of inertia at a given section will alter.
Thickness (t) = 1.6 mm and b = 58.4 mm
Substituting these values in equation 9, we get moment of
inertia,
I = 6545634.619 mm4
Therefore, from equation 5, Bending stress
= 34.71 kg/ mm2
The final dimension of the I-section at rib station 1 is shown
below:
161.417mm
Fig. 3 Final dimension of 1st
I – section (Al 2024)
The dimension of the I-section at the 1st
rib section was
finalised and similarly same procedure was carried out to find
the dimensions at remaining sections.
But due to the non-availability of aluminium 2024-T3,
forcefully the material had to be changed to aluminium 6082-
T6. The thickness of flange and web is 3mm and 2mm
respectively in all sections because to reduce the machining
cost of sheet metal. The design process of 6082-T6 is carried
out as similar to previous procedure.
Area of the flange:
Substituting the values of the bending moment and
perpendicular distance in equation 2, we get
Ptensile =14757.79kg
We have Yield stress of aluminium 6082 is 2
Therefore from equation 3, A = 567.607 mm2
Considering thickness of the flange to be 3 mm, the width of
the flange is calculated.
Hence from equation 4, Width of the flange B = 190 mm.
Bending stress of the flanges:
We have, B = 190 mm, D = 175.471 mm and d = 169.471 mm
Substituting these in equation 6, we get moment of inertia.
175.471mm
7.027 mm
1.6mm

I =8478535.065mm4
Therefore, bending stress is obtained from equation 5,
26.79kg/mm2
Area of web:
Considering the Shear Force (SF) acting in the same cross
section, we find out the web cross section area. Thickness of
web is 2 mm as mentioned above.
As considered previously, shear force is 992.5089874 kg.
We have Shear yield strength for aluminum 6082 =18 kg/mm2
Substituting the above values in equation 8, the area of web
obtained is A= 55.1393 mm2
We know the depth of the web is d =169.471 mm, t =
thickness of web.
Equation 4 gives the thickness of web t,
t = 0.325 mm
Bending stress is calculated further for the flanges including
the web.
Substituting the values B = 190 mm, D = 175.471 mm, b =
189.675 mm and d = 169.471 mm in equation 9 gives moment
of inertia of flanges including web. i.e., I = 8610357.184 mm4
Bending stress is obtained from equation 5,
=26.38kg/mm2
Now we are considering shear buckling concept as mentioned
previously.
Substituting t = 0.325 mm, b2 = 165.82 mm, from equation 10
we get,
= 0.163
To find Critical load Pcr:
From equation 12, we get area of the I-section to be
A = 55.078 mm2
.
And equation 11 gives, Pcr= 8.977 kg
Papplied on rib section 1 obtained is 992.508 kg.
Substituting the values of Pcr and Papp in equation 13, gives
buckling factor = 0.009
We know that, if buckling factor is less than 1, the section
buckles. If buckling factor is greater than 1, the section will
not buckle. Here, for thickness = 0.325 mm, BF=0.009.
Hence, to increase the buckling factor the thickness of the web
should be increased. After several iterations, we got to know
that at t = 1.6 mm, the buckling factor is equal to 1. The
available thickness of material in market is 2 mm, due to high
expense in machining, 2 mm thickness is considered and
further calculation is continued.
Moment of inertia after considering web thickness = 2 mm, b
= 188 mm. Substitute in equation 9,
Hence, I = 9289748.11 mm4
Therefore, bending stress from eq.5, we get, =
24.45kg/mm2
Maximum deflection of spar:
Deflection of tapered section( x x L x …..(14)
Length of the spar considered is 1500 mm.
Substituting the values in above equation, we get the
maximum deflection, = 29.99 mm
The obtained total bending stress of I-section is less than 26
kg/mm2
. The finalized dimensions of I-section at rib 1 are as
shown below.
190 mm
3 mm
169.47mm 2mm 175.47 mm
Fig. 4 Dimension of 1st
I – section (Al 6082)
Similar iterations are carried out to all the 17 sections over
the length of spar and finalized the dimensions of the I-section
spar. The finalized dimensions of 17 rib sections are as shown
below.
Table 2 Dimension at each rib section
Rib stations Load (kg)
Total I-section dimension (mm)
Width (B) Depth (D)
R1 19.38 190 175.47
R2 39.27 172 171.82
R3 46.35 154 168.17
R4 51.30 137 164.52
R5 58.18 120 160.86
R6 58.34 104 157.21
R7 61.06 88 153.56
R8 63.46 73 149.91
R9 65.61 60 146.26
R10 67.54 47 142.61
R11 69.28 35 138.96
R12 70.89 23 135.31
R13 72.32 16 131.66
R14 73.62 9 128.01
R15 74.84 4 124.36
R16 75.86 1 120.71

From the above table, it is observed that, the flange width is
being reduced from rib location 0 to 17. Since it is difficult to
manufacture the flange of smaller dimensions and considering
damage tolerance, riveting process, the lower dimension are
not considered. Hence a scale-down model of 1.5m is
considered for analysis and testing in current study.
The table below shows the dimensions of considered scale
down model and according to that dimensions over the length
of spar is modelled using CATIA V5 R20.
Table 3 Dimension at rib sections (scale down model)
Rib stations Load (kg)
Total I-section dimension (mm)
Width (B) Depth (D)
R1 19.38 190 175.47
R2 39.27 172 171.82
R3 46.35 154 168.17
R4 51.30 137 164.52
R5 58.18 120 160.86
R6 58.34 104 157.21
Fig. 5 Spar model
2.6 Stress analysis of the spar
Stress analysis of scale down model of length 1.5m is
carried out using Nastran- Patran.3d model has done using
CATIA V5 as shown below.
Fig. 6 Load acting on the spar
The length of the build-up model is 1500 mm and extra
200 mm is for test fixture. Load is applied at the free end of
the Spar.
As calculated previously, bending moment obtained at rib
1 is 2589594.24 kg-mm
Bending moment = load x perpendicular distance
Perpendicular distance = 1500 (build up for testing)
Load = = 1726.39 kg
Therefore the load applied is 1727 kg.
The CATIA model is then imported to the PATRAN
software. The model is set for default dimensions and
extraction of mid surface has been carried out for meshing. A
fine and good quality meshing is generated on each part of the
structure using MSC PATRAN. Figure 5.2 represents the
meshed model of spar. Fine meshing is done for getting better
results.
Fig.7 Meshed spar model
The boundary conditions are now applied to the spar
model. Since the model considered is a cantilever beam, one
end is fixed, ie., all the six degrees of freedom are constrained
and at the free end load is applied.
Fig. 8 Boundary conditions
The loads are applied. Due to the application of load, the
deflection is observed and also the stress at each section is
obtained.
Maximum Deflection = 35.5 mm
Fig. 9 Deflection of the spar

Maximum stress = 23.8 kg/ mm2
Fig. 10 Maximum stress of the spar
2.7 Fabrication of the wing spar
Metal fabrication is the process of manufacturing metal
structures by cutting, bending and assembling processes.
Initially, aluminium 6082-T6 metal sheet of 3mm and 2
mm thickness are purchased. To obtain a structure of I-
section, the sheet is first marked to cut according to the
required dimensions. Sheet of 3mm thickness is marked
according to the dimensions of flange and 2mm thickness
sheet is marked according to the dimensions of web.
Fig. 11 Fabricated spar model
2.8 Testing Setup
After riveting the spar model, the structure is then introduced
for testing. Test rig is required to test the fabricated spar. Spar
length of 1700 mm out of which 200 mm is for fixing the
model on the setup. The determined load is applied on the free
end of the spar, correspondingly the deflection is noted for
each interval of load application. Finally, test results which
are obtained validates the results of calculated and analysis.
Fig. 12 Test rig with spar
In the scale down model, the calculated bending moment
magnitude should be at the fixed end, so the load acting on the
spar is calculated by the formula.
Load applied on the spar at free end is obtained as follows:
Bending moment = load x length of spar
2589594.24 = P x 1500
Therefore, P = 1727 kg
The deflection at a tip of structure is determined while
testing with the application of 1727kg load at the free end.
The yield stress value of the material is 26kg/mm2
, which is
expected at the load of 1727kg.
Since, the spar needs to be considered for further studies, a
conscious decision was taken not to load the spar up to
yielding level. Therefore, the testing was stopped at 1000kg
peak value and the corresponding results are tabulated below.
Total Weight of spar =7.38kg
III. RESULTS AND DISCUSSION
3.1 Test Results
Table 4 Tabulation
Load(kg) Deflection(mm)
0 0
100 2.4
200 4.7
300 6.9
400 9.2
500 11.4
600 13.7
700 16
800 18.4
900 20.5
1000 22.35
3.2 validation
Deflection of a spar for 1000kg by considering linear
interpolation, the below graphs which shows the interpolation
method of deflection versus load we have considered and
compared with calculated, analysis and tested results.
3.3 Calculated result comparison
Fig. 13 Calculated results

3.4 Analysis result comparison
Fig. 14 Analysis results
3.5 Tested result comparison
Fig. 15 Tested results
Maximum deflection of a spar:
Calcula
ted
result(mm)
Analysis
result(mm)
Tested
result(mm)
17.37 20.55 22.35
Maximum stress of a spar:
Calculated
result(kg/mm2
)
Analysis
result(kg/mm2
)
24.45 23.8
IV. CONCLUSION
Wing spar design was carried out by using strength of
material approach. Spar is designed for minimum weight.
Finite Element Analysis approach was used for stress analysis
of the structure. Iterative analysis was carried out to achieve
minimum weight for the structure. The maximum stress
obtained from the finite element analysis is 23.8 kg/mm2
. The
maximum deflection at the tip is 35.5 mm. Steady level lift
load condition was considered for the design. Transport
category aircraft are generally designed for ―3g‖ condition.
The design limit load magnitude corresponding to 3g
condition was considered. The spar was fabricated based on
the design configuration. Aircraft standard material
Aluminium alloy 6082-T6 material is used for the fabrication
of the spar. It is a build-up construction. Spar web and flanges
are connected by using L- angles which are in turn connected
using rivets.
Static testing of the spar is carried out. A test rig was designed
and developed. Vertical deflection of the spar was measured
at several locations. A good co-relation between stress
analysis and test results was observed.
In current study the weight optimization is done through
design process, few iteration are carried out by varying the
dimensions of spar and model is finalized with the dimensions
which has less weight. The finalised structure which is in
tapered form.
In further studies, the following can be considered.
 The weight can also be reduced by introducing cut-
outs in the web of the spar model
 Fatigue damage calculation for crack initiation
 Damage tolerance design can be carried out.
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