Four cross-section shapes and seven materials were analyzed to determine the best design for our wing spar. Mathematica code was written to analyze the drag, lift and static loads.

1. AOE 3024 FALL 2012
AERO PROJECT
TEAM 18

2.  A new wing spar must be created given the data and
parameters provided.
 The design of the spar will be formulated by aspects of
the engine placement, shape of the cross section and the
material used.
 The engine must be placed close enough to the plane to
minimize the bending load but far enough to eliminate
aerodynamic interference on the fuselage.
 A well-chosen material for this wing spar will be low in
cost, have a low density and have a high yield stress and
Young’s modulus.
INTRODUCTION

3. Engine
PlacementPART I

4.  For our first design we went without the engine attached just
to give us a base to start from.
 Next we added to the code to determine the different effects
from the placement of the engine, from 20ft to 90ft, would
have on the beam in terms of meeting the given requirements
for safety.
 After doing this with an initial solid beam, we decided as a
group that the best place to put the engine was as close to
the fuselage as possible to keep the beam from bending too
much and possibly even breaking.
ENGINE PLACEMENT

5.  This was chosen mainly to be able to meet the
requirements of Case 3’s static loading conditions
because the further out you put the engine, the more
the beam deflects at the tip (see Figures 1, 2, 3).
Another reason we chose this location is because in
the event of an engine failure it would produce a
smaller moment about the center of the aircraft,
which in turn would put less force on the rudder.
 Given the simplicity of our beam we have concluded
that 20ft would the best location for the engine on
our design.
ENGINE PLACEMENT

6. Figure 1: Static displacement with
engine at 20ft for solid beam (uy=-
2.47664ft)
Figure 2: Static displacement with
engine at 55ft for solid beam (uy=-
2.68195ft)
Figure 3: Static displacement with
engine at 90ft for solid beam (uy=-
2.98233ft)
ENGINE DISPLACEMENT GRAPHS

7. Cross
Section
Analysis
PART II

8.  During this design iteration we modified the initial box beam
with the maximum breadth and width, and came up with
different designs that changed the properties that we want
the most, in particular the Moment of Inertia about the z-axis.
 The reason we want Izz maximized is to resist bending due to
the static load and the lifting force, which will reduce the
stress and deflection on the z-axis.
CROSS SECTION ANALYSIS

9.  The reason we decided to do a T beam initially is to test if
small modifications will be enough to satisfy all cases for this
problem
 . The T beam will modify the cross section so that more area
is away from the centroid, and thus better resist bending
across the axis perpendicular.
 The T beam failed in the static case, where it deflected more
than the allowed 1.2 feet downward with all possible
dimensions.
T-BEAM

10. T-BEAM
20 40 60 80
Position on bar in ft
3
2
1
1
2
3
uyo x in ft
Displacement Resultant

11.  Before we went into the next iteration, we decided to choose
an arbitrary area that is reasonable and less than the
maximum possible area for the beam.
 This area will be the area that is used for all future
comparisons to identify which design is best. The reason is
that if the area is not normalized for each cross section the
cross sections will not be evenly compared.
 We chose the area to be 4.9 ft2. This number does not matter;
the number could be any number as long as the dimensions
do not exceed the given limits in the problem statement.
CROSS SECTIONAL ANALYSIS

12.  The I-beam with the same area as the T beam, will give higher
possible moments of inertia due to more area located away
from the center.
 We found the approximate max Izz without breaking the beam
in stress by going through trial and error.
 The I-beam was not successful in meeting all the critical
conditions in lift and static loading with the given the
following dimension and properties:
I-BEAM

13. I-BEAM
20 40 60 80
Position on bar in ft
2.0
1.5
1.0
0.5
uyo x in ft
Displacement Resultant

14. HOLLOWED SQUARE BEAM
 The hollow square beam follows the same principles as the I-
beam by putting mass away from the centroid but this time it
is to resist bending about the y-axis.
 The hollowed box beam did not meet the criteria for
deflection in static loading with the following dimensions:

15. HOLLOWED SQUARE BEAM

16.  We considered this beam because it can resist moments in
the z and y axis well and resist stress.
 This is because the beam is similar to the hollow beam and in
addition to more local centroids that are away from the beam.
 This final beam design also took consideration of carrying
stress better than other cross sections since there are
supports in the middle of the beam to help carry it.
DOUBLE I-BEAM

17. DOUBLE I-BEAM

18.  With all of the designs analyzed we created a decision matrix
to pick the most efficient cross section.
 Here we ranked Izz as number one because we wanted to make
sure the beam was able satisfy the conditions for static
loading.
 We ranked the ability to manufacture as number two in order
to account for the price it will cost to manufacture the beam.
 The third ranking was for performance on the Iyy because it
deals with the deflection due to the drag.
CROSS SECTION DECISION MATRIX

19. Choice of Beam Ability to Manufacture Izz Iyy Total Total Rank
Double I Beam 4 1 1 12 1
T beam 1 4 3 15 4
I beam 2 2 4 18 2
Hollow Beam 3 3 2 15 2
CROSS SECTION DECISION MATRIX

20. Torsional
Loads for
Spar DesignPART III

21.  Torsion would be created by the engine because it is placed
below the wing spar. The torsion from the engine would make
the wing spar want to rotate about the centroid of the chosen
cross-section.
 The weight of the fuel would also create a torsional force on
the wing spar that would want to cause the spar to further
rotate.
 The aerodynamics of the airfoil also causes torsional forces to
be applied on the wing spar.
 Given more time on this spar design these torsional forces
can be investigated and the overall design of the spar can be
maximized to resist all force the spar encounters.
TORSIONAL LOADS

22.  We are confident that the current wing spar design we are
using now is worth investigating further because of the basic
thought process on torsion.
 When looking at wing spar designs, it is best to not have a
cross section that has openings such as the cross section of I-
Beams.
 The closed cross section of the double I-Beam that we are
currently proposing would be resistant to the torsional
moment experienced inside the wing of an airplane.
 Our wing spar did not come close to the yield stress of our
material when undergoing bending stresses in the lifting, drag
and static cases.
TORSIONAL LOADS

23. Choosing
the Right
Material
PART IV

24.  A decision matrix was created to choose among six different
materials that were determined to be potential materials for
the spar.
 The six materials are 4140 Steel, AISI 1010 Steel, 6061
Aluminum Alloy, 7075 Aluminum Alloy, Nickel, Titanium, and
Epoxy/Carbon Fiber Composite
 The decision matrix is given below by Table 2 and the way it
was utilized is as follows:
 Each category (density, elastic modulus, cost, etc.) is ranked on
importance from 1-5.
 Each material is scored from 1 through 7 (7 materials) with 7 being
the worst ranking for the category and 1 being the best in that
category.
 The totals for each category are then summed and the material with
the lowest score is the material of choice.
CHOOSING A MATERIAL

25. MATERIAL CANDIDATES
Table 1
Material Modulus Cost $/lb Yield Strength Density lb/ft^3
4140 Steel 29.7100E6 psi 0.450 94.98 ksi 490.752
Aluminum Alloy 6061 10.0000E6 psi 10.25 21.03 ksi 168.480
Aluminum Alloy 7075 10.4350E6 psi 19.06 72.94 ksi 176.250
Nickel 30.0000E6 psi 39.02 8.550 ksi 554.688
Titanium 16.8120E6 psi 57.77 20.30 ksi 281.664
Epoxy/Carbon Fiber 32.6090E6 psi 34.00 29.00 ksi 95.5230
AISI 1010 STEEL 29.0075E6 psi 0.325 26.10 ksi 490.752

26. DECISION MATRIX
Table 2
Choice of the material Cost Yield Strength Density
Modulus of
Elasticity
Cost to
Manufacture
Total
Importance of Each Category (1 most
important)
1 5 2 4 3
Aluminum Alloy 6061 3 5 2 7 3 80
Aluminum Alloy 7075 4 3 3 6 2 74
Titanium 7 6 4 5 5 94
4140 Steel 1 1 5 3 1 64
Nickel 6 7 6 2 4 90
Carbon Fiber 5 2 1 1 5 38
AISI 1010 Steel 2 4 5 4 1 76

27.  From the results given by Table 1, it can be inferred that
carbon fiber composite should in fact be the material
selected.
 However, that material was included in the decision matrix
only as a comparison to see how it stacks up against steel.
 The problem with carbon fiber is that it is very expensive and
the manufacture of carbon fiber needs a lot of time and
specific conditions in order to get a quality product
 Since 4140 steel is a much easier material to work with, it led
to 4140 Steel being chosen as the material to be used in the
spar.
CHOOSING A MATERIAL

28.  Developing the cost function for this spar is a simple
calculation of the volume of the spar multiplied with the
density of the material. When this value is multiplied with the
cost of the material a final cost for the spar is obtained. The
cost function is given as follows:
 𝑻𝒐𝒕𝒂𝒍 𝑪𝒐𝒔𝒕 = 𝑨 𝑪𝒓𝒐𝒔𝒔 𝑺𝒆𝒄𝒕𝒊𝒐𝒏 ∗ 𝑳 ∗ 𝝆 𝒎𝒂𝒕𝒆𝒓𝒊𝒂𝒍 ∗ (𝑴𝒂𝒕𝒆𝒓𝒊𝒂𝒍 𝑪𝒐𝒔𝒕)
 The area of the cross section for the beam is 4.9 ft2, length is
90 ft, the density and cost of 4140 Steel is 490.752 lb/ft3
and $0.45 respectively from Table 1.
 𝑻𝒐𝒕𝒂𝒍 𝑪𝒐𝒔𝒕 = 𝟐. 𝟏 𝒇𝒕 𝟐
∗ 𝟗𝟎 𝒇𝒕 ∗ 𝟒𝟗𝟎. 𝟕𝟓𝟐
𝒍𝒃
𝒇𝒕 𝟑 ∗ 𝟎. 𝟒𝟓
$
𝒍𝒃
=
$𝟒𝟏𝟕𝟑𝟖. 𝟕𝟑
CHOOSING A MATERIAL

29.  The total cost of the spar as $41,738.73 is a very low cost
compared to the other materials.
 For comparison, a carbon fiber composite spar would cost
$7,358,340.00.
 The 4140 steel spar would weigh 97389 lbs while the carbon
fiber composite spar would weigh 42125.6 lbs.
 However, the difference in cost is what makes the 4140 spar
win over the carbon fiber composite as the 4140 steel is
$7,260,945.76 less.
 The 4140 Steel passes all cases set for it, and therefore is
deemed both a safe and cheap material to use in this wing
spar.
CHOOSING A MATERIAL

30.  The third and fourth best scoring materials according to Table
2 were 7075 Aluminum and AISI 1010 Steel respectively.
 The aluminum is ruled out because it is far more expensive
than the steel and would cost a total of $1,481,462.33 per
spar whereas 1010 steel would have a price comparable to
that of the 4140 Steel.
 However, because the 1010 Steel did not score as well as
4140 Steel in the decision matrix it was ruled out as a
possible material to be used and we made the decision to use
4140 Steel as our material for the spar.
CHOOSING A MATERIAL

31. V-N DIAGRAM
The V-N diagram was created using the given load and coefficient of lift maxes and the take-off
weight of the aircraft. The stall velocity on the upper side of the flight envelope is 83.35 feet per
second. The maximum velocity of the aircraft was taken if the airplane were to do a nose dive
and fall under the aid of gravity only.

32.  The wing spar we designed is safe, cost efficient and reliable.
 The final design of the wing spar is the one with a “double-I”
cross-section, 4140 Steel as the material, engine placement
at 20 feet from the fuselage, total weight of 97389 lbs, and
an estimated material cost of $100000 per wing spar.
 According to the calculations and the decision matrix, the
design achieved a good balance of safety, performance and
cost. The wing spar meets all the requirements set forth, with
a factor of safety of 1.15.
 The deflection of the wing tip and the stress in the entire wing
spar does not exceed the yield stress of the material in
multiple situations, including taking off, one engine
malfunctioning during cruise and static conditions.
CONCLUSION