2. 2
Swept forward wing has significant aerodynamic advantages over straight wings when
Mach number approaching or exceeding 1. However, different from Straight wing which
experience only bending deformation, swept forward wing is easily affected by torsional
deformation which cause the wing to twist. The main focus of the project is to come out with a
design with reduced stress and twist for a swept forward wing with Gottingen 398 as the airfoil
shape by changing the internal structure of the wing. The airfoil coordinates are shown in
Appendix A. Throughout our study, we will use the property of balsa wood for material, which
will be shown in Table 2.
Part I – Comparison between straight and swept forward wing
The first step in designing the wing, is comparing the displacement difference between
straight wing and swept forward wing. In this part both of the wings are solid, and 1 newton
force is applied along the ¼ cord with the wing ends fixed. We will keep using 1 newton force
during the first part and when we analyze the wing section geometry to make sure the force is
constant and not affecting the result.
From the study we found displacement is very even along the wing section in the straight
wing and that the difference is very small compared to what we get from a swept forward wing.
The displacement plot of both wings will be shown in Figure 1. By adjusting the position of the
lifting force along the wing section on the straight wing, we are able to get analysis results with
angles of twist close to zero (shown in figure 2). This means that we have applied the lifting
force on the shear center. However, for the swept forward wing, the leading edge will always be
experiencing higher displacement than the trailing edge so we believe the shear center is outside
the wing and that twisting is so serious in a swept forward wing. From the stress analysis which
will be shown in figure 3, we also found that the stress is concentrated around the center part of
the wing section at the wing end, while for the swept forward wing the maximum stress will be
found around front part at the wing end.
Table 1: comparison of displacement
lift at 1/4 cord
leading(m) trailing(m) difference(m)
straight 6.99E-05 6.76E-05 2.27E-06
swept forward 8.71E-05 7.36E-05 1.35E-05
lift at 4.8cm from leading edge
straight 6.91E-05 6.89E-05 1.97E-07
Table 2: material properties
property of the material
Elastic Modulus: 3.71Gpa
Density: 0.13g/cm^3
3. 3
a)
b)
Figure 1: displacement plot :
a) the straight wing, 6.764e-5m at the trailing edge,6.9909e-5m at leading edge
b) the swept wing: 9.259039157e-5 m at trailing edge,8.041581446e-5 mat leading edge
4. 4
Figure 2: displacement of the straight wing when lifting force is applied along the 4.8cm edge from the leading edge:
6.9097e-5m at leading edge, 6.89e-5m at trailing edge
a)
b)
Figure 3: stress plot with lift along ¼ chord: a)straight wing,b)swept forward wing
5. 5
Part II -- Calculation
While hollowing, the first challenge was to determine how many spars we would use and
the location of the spars. We did some online research on how people calculated the shear center
of a multicell wing section to find any approximations in order to make the calculations easier1.
We noticed that the calculations for the shear flow on the spars were the largest and that the ribs
did not have that much effect on the deflection. We decided to focus on two or three spars
because we did not want to add more weight to the wing and it appeared that two to three spars
were typically used in previous designs.
To find the location of the spars we used MATLAB write a function code to calculate the
rigidity of the straight wing if we used two or three spars. The function code had the normalized
coordinates of the Gottingen 398 airfoil and was able to calculate new y coordinates if we
changed the spacing of each x coordinate by linear interpolation. Then the code calculated the
height of the cell walls that would be the height of the spars. After that, the code calculated the
cell areas and the top and bottom wall lengths. With these values the code created a matrix of the
linear equations of the equation theta = 1/(2*Aof cell)*(line integral of q/t ds) for each cell and at
the end of the matrix it would have the equation for torque T = 2*(sum of Aof each cell times
shear flow of each cell). The shear flow for each cell was solved and the end result had the
torsional stiffness of G*theta/T (called “Stiffness”) for the cross section. This torsional stiffness
was compared to the single cell, i.e. no spars case and gave us a ratio called “StiffScaled”. For
the two spar code, the largest increase in torsional rigidity(rigidity ratio) was 16% compared to
the single cell. The code shows the results of the locations of the spars that had over 16%
increase in torsional rigidity. For the three spar code, the largest increase in torsional rigidity was
20% and again our code shows the results of the locations of the sars that had over 20% increase
in torsional rigidity. The MATLAB codes can be referenced in Appendix B.
With the result we found in MATLAB, we made different SolidWorks models with two
or three spars and did a lot of trials in ANSYS, then decided to use the normalized locations of
0.1, 0.45 and 0.7 from the leading edge with normalized thicknesses of 0.025, 0.01, and 0.01 for
the spars. This combination of location and thicknesses had the lowest maximum principal stress
on the wing and fairly low angle of twist. At this point we realized we should do our ANSYS
simulation with a more realistic lift force, so we used 27 Newtons assuming that was the lift
force in the wing at 75 mph wind speed (33.5 m/s). From the plot of cl vs. angle of attack, shown
in graph 1, we found the cl to 0.8 at 10-degree angle attack.The calculation used the following
formula, assuming an air density of 1.225 kg/m3:
FLift = 0.5 ∗ 𝑐𝑐 ∗ 𝑐 ∗ 𝑐2
∗ ∫𝑐
𝑐𝑐
𝑐𝑐𝑐𝑐
FLift = (0.5)*0.8*(1.225 kg/m3)*[(75 mph)(0.447 m/s/m1ph)]2*[12cm * 40cm]= 27N
1
Website for multicell beams in torsion: http://www.aeromech.usyd.edu.au/structures/acs1-p83.html
6. 6
graph1:lift coefficient vs angle of attack for Gottingen 398 airfoil
Part III – Analysis ofthe Wing Section
To find out which wing section design gives us the least twist and maximum stress, we
will keep the lifting force constant as 1 Newton and the designs will have no rib adding in this
part of study. During the analysis, we need to keep in mind that twist is unavoidable for a swept
forward wing because its shear center is at the back of the wing and we will be trying our best to
move the shear center forward. From the result of the MatLab calculation, we have decided the
designs we want to look into. Then we change the thicknesses of the section walls, which are the
spars in the wing, and analyze to decide the best wing section geometry. The results and analysis
of the wing section geometry will be shown in the table 3 and table 4, the thickness of the wing
is designed to be 0.1cm, the scales are based on 1cm cord-length for wing section study.
7. 7
Table 3: Analysis of two spar (three cells) wing sections: 1st
wall and 2nd
wall indicate the position walls are placed
from the leading edge,under the thicknesscolumn, the first number is the thickness of the 1st
wall,and the second
number is the thicknessof the 2nd
wall.
From the table 3, we see the calculation for the straight wing matches with our analysis
for a swept forward wing in the following way: With two equally thick cell walls, when the 2nd
wall is place at 0.7cm, it gives us lower max stress. [Note that all locations and thicknesses
presented for the matlab code are in terms of a cord length of 1cm.] Since the twisting is very
similar in both cases, we decided to keep studying the case of first one. When we increased the
thickness of the first wall to 0.025cm, the stress, displacement, and twist decreases. We found
that past this point, increasing the thickness will not yield better results.
Table 4: Analysis of three spar (four cells) wing sections:1st
wall, 2nd
wall,and 3rd
wall indicate the position walls
are placed from the leading edge,under the thickness column, the first number, second number, third number,
represent the thicknessof 1st
wall, 2nd
wall,3rd
wall,and one number means same thickness of the three walls. The
design has refined mesh only when it is indicated.
For the table 4, we did same thing similar to what we did in table 3. We compared the
best stiffness wing section designs, and choose the better one to continue study. Since the 3-wall
design gives us better results, we give up the 2-wall design, so at this time, we have decide the
wall position to at (0.1cm, 0.4cm, 0.7cm), and the next step is to decide the wall thicknesses. We
get very similar result between using thickness of (0.025, 0.01, 0.01) and thickness of (0.035,
0.01, 0.01), so we made refined mesh for both to compare and made our decision of choosing
(0.025, 0.01, 0.01). We can have a look at the thickness (0.025, 0.02, 0.01), in which we increase
the thickness of the center spar resulting in higher stress on the wing. This might be because of
higher twist (displacement difference is larger comparing to other two designs). Adding more
materials backward might have moved the shear center backward and caused higher twist.
In order to move the shear center of the wing forward, it is apparent that the front spar
should be bulkier than the rear spars. But when the thickness of the front spar is increased to
0.035cm, we see in the refined mesh part the stress is increased, but displacement difference is
8. 8
smaller compared to the design with 0.025cm-thickness (smaller twist, forward shear center.
This may occur because of the increased weight causing larger bending moment which leads to
higher stress, stress equals to Mc/I. So we need to keep in mind the side effect of weight when
we move forward.
Part IV --- Analysis of Wing with Ribs
Since the locations of the spars are decided, we calculate the lifting force to be 27N using
the speed of 75mph, 10-deg angle attack with the lift equation. For the following study we will
keep using 27N as applied force and analysis will be based on refined mesh at the wing end since
that is place the maximum stress will be. Throughout this part of analysis, the wing tip is closed
with one solid rib. The analysis result will be shown in table 5. The thickness of the rib is 0.5cm,
and the scales are based on 40cm wingspan.
Table 5: rib0, rib1, rib 2, and rib3 each represent different rib designs, the numbers under the 1rib,2 rib, 3 rib are
the location ofthe ribs from the wing end, there is one with on rib at the wing end as it indicated,the last two
analysishas four ribs in total with one rib at the wing end.
Along the study, we see that displacement difference doesn’t differ a lot from each other
comparing to the output of the max stress,so we want to focus on stress analysis on this part. From the
analysis, we see that adding three solid ribs does help to decrease the stress a lot, but if we take away the
rib at the wing end which is the most critical place, the stress goes up a lot even compared to the design
with no ribs. This indicates that ribs at wrong positions only increase weight which will cause higher
bending moment leading to higher max stress. To verify the effect of bending moment and to compare
designs with three ribs to those with four ribs, we produced two four-rib designs, which are shown at the
bottom part of the table 5. When the ribs are moved closer to the wing tip, the stress goes up a lot because
it moves the center of the weight further away from the wing end. Moment = force*distance, so with the
same amount of weight, larger the distance gives us higher bending moment. So we made our decision of
not putting ribs too far away from the wing end.
Before we think about the position of ribs, we should consider the design of the ribs themselves.
The last two studies they are done using the best rib design we have found, rib3. In order to determine the
final rib locations, we compared all of our designs. By comparing the results of the last three designs
9. 9
which all used rib3 (shown in table 5), we decided to use only two ribs at the location of 9cm, 16cm from
the wing end which gives us the lowest stress. For the rib design, the hollow parts are in shape of square
to maximize the empty area with fillets on the corner to avoid stress concentration. The rib designs are
shown in the Figure 4, and final design on the Figure 5.
a)
b)
Figure 4: rib design: a)rib 0, b) rib2
Figure 5: rib3, final rib design for laser cut, with the leading and trailing part trimmed to add in supporting
material during the assembly process.
The rib 1 is very similar to rib 0, just increasing the area of each hole, and we found that
increasing area does help to reduce the stress a lot. This might be because of the reducing weight leading
to less bending moment. So for the rib2, we increase the hole-areas even more by merging all the small
holes into one big hole, but the result doesn’t go as well as we thought they would. Even though the
bending moment decreased,the inertia of the rib is decreased in rib2. Since stress equals to
Moment*c/inertia, the smaller the inertia, the larger the stress will be. We need to find the design gives us
a lower moment, but a larger inertia. Next, we separated the hole on the center into three holes we
designed in rib1, shown in the Figure 5, as our final design, which gives us the lowest stress and twist.
Part V --- Final design
The pictures in this section were taken after we decided the final design. The results may
be a little different, from the data shown in table 5, and the material properties of the design will
be shown in Appendix C.
a)
11. 11
c)
Figure 6: final design results: a) displacement plot,b) stress plot, c) stress concentration
Figure 7: wing structure, the first spar (closest spar to the stress concentration) is 1cm longer to
insert into the block
Part VI --- Improvement
12. 12
There are a lot of place that we need to spend more time looking into it, we found that
there are a lot of parts that we should have look more into but didn’t give enough analysis.
1) No analysis done based on varying the thickness of the rib. (Since there is no analysis on
this part, we can’t give strong comments about it)
2) No calculation for the more walls analysis. We have calculation based on 1-wall (which
is not very good comparing to the other two so we didn’t talk about it), 2-wall, 3-wall,
and we have found that as the wall number increased, the stiffness ratio increase which
gives us better result. From the analysis, we also see that 3-wall gives less twist than 2-
wall that increasing the number of walls more might help to more the shear center
forward.
3) On the Part IV, when we found that using three ribs gives us better result than using four
ribs, we chose to use the design of (0cm, 9cm, 16cm) from the wing end because that is
the design we keep using and forgot to study the design of (0cm, 4cm, 16cm), or (0cm,
4cm, 9cm) from the wing end. As we have said that Moment = force*distance, then the
moment will be reduced more by get the ribs closer to the end we have realize this, but
didn’t spend enough time into it.
4) The lift force is calculated based on 75mph, but during the completion, we found the
speed to be larger than 75mph.
Appendix A: Gottingen 398 Cross-Section
15. 15
xnew=[0:xspacing:1];
ytopnew=zeros(1,1/xspacing+1);
ybottomnew=zeros(1,1/xspacing+1);
d=0;
%Create new x and y coords spaced according to xspacing values
for i=1:1/xspacing+1
for j=1:length(x)
if x(j)==xnew(i)
ytopnew(i)=ytop(j);
ybottomnew(i)=ybottom(j);
break
elseif xnew(i)>x(j) && xnew(i)<x(j+1) %linearly interpolate to
find new y coords
ytopnew(i)=(ytop(j+1)-ytop(j))*(xnew(i)-x(j))/(x(j+1)-
x(j))+ytop(j);
ybottomnew(i)=(ybottom(j+1)-ybottom(j))*(xnew(i)-
x(j))/(x(j+1)-x(j))+ybottom(j);
break
end
end
d(i)=ytopnew(i)-ybottomnew(i);
end
%xloc=round(xloc,3);
xnum=xloc/xspacing;
D=zeros(1,length(xloc));
%Calculate height of cell walls
for i=1:length(xloc)
D(i)=d(round(xnum(i)+1));
end
n=1;
Atot=0;
Stot=0;
A=zeros(1,length(xloc)+1);
S=zeros(1,length(xloc)+1);
%Calculate cell areas and top+bottom wall lengths
for i=1:round(1/xspacing)
if n<=length(xloc)
if i==round(xnum(n)+1)
n=n+1;
end
end
a(i)=((ytopnew(i)-ybottomnew(i))+(ytopnew(i+1)-
ybottomnew(i+1)))/2*xspacing;
s(i)=sqrt((ytopnew(i+1)-
ytopnew(i))^2+xspacing^2)+sqrt((ybottomnew(i+1)-ybottomnew(i))^2+xspacing^2);
Atot=Atot+a(i);
Stot=Stot+s(i);
A(n)=A(n)+a(i);
S(n)=S(n)+s(i);
end
C=zeros(length(A),length(A));
%Formulate q matrix
for i=1:length(A)-1
C(i,i)=S(i)/t0+D(i)/t(i);
C(i,i+1)=-D(i)/t(i);
C(i+1,i)=-D(i)/t(i);
17. 17
end
end
Ymax=max(max(Y))
[r,c]=ind2sub(size(Y), find(Y==max(Y(:))));
rc=rc*.05
%maximum improvement over single cell design is 16 percent
%Matrix rc gives list of all wall locations for which rigidity improves by
%at least 16 percent over single cell design
Results: the first two numbers are the locations of the wall and the last number is the rigidity
ratio
rc =
0.0500 0.6500 1.1631
0.0500 0.7000 1.1657
0.0500 0.7500 1.1625
0.1000 0.6000 1.1611
0.1000 0.6500 1.1675
0.1000 0.7000 1.1689
0.1000 0.7500 1.1644
0.1500 0.6500 1.1632
0.1500 0.7000 1.1640
3. Three Spar Locations and Ratios:
clc;
clear all;
xloc1=[.05:.05:.95];
xloc2=[.05:.05:.95];
xloc3=[.05:.05:.95];
n=1;
for i=1:19
for j=1:19
for k=1:19
if xloc1(i)>=xloc2(j) || xloc2(j)>=xloc3(k) || xloc1(i)>=xloc3(k)
Y(i,j,k)=1;
else
Y(i,j,k)=TRigidityFunc(.025,[xloc1(i),xloc2(j),xloc3(k)],[.075 .075 .075]);
if Y(i,j,k)>1.213
rch(n,1)=i; %first spar location
rch(n,2)=j; %second spar location
rch(n,3)=k; %third spar location
rch(n,4)=Y(i,j,k)/0.05; %corresponding Stiffscaled ratio
of locations
n=n+1;
end
18. 18
end
end
end
end
Ymax=max(max(max(Y)));
[r,c,h]=ind2sub(size(Y), find(Y==max(Y(:))));
rch=rch*.05
%maximum improvement over single cell design is 20 percent
%Matrix rch gives list of all wall locations for which rigidity improves by
%at least 21.3 percent over single cell design. 20% had too many results.
Results: the first three numbers are the locations of the wall and the last number is the rigidity
ratio
rch =
0.1000 0.3000 0.7000 1.2130
0.1000 0.3500 0.7000 1.2138
0.1000 0.4000 0.7000 1.2145
0.1000 0.4500 0.7000 1.2141
0.1000 0.4500 0.7500 1.2137
0.1000 0.5000 0.7500 1.2138
0.1500 0.3500 0.7000 1.2130
0.1500 0.4000 0.7000 1.2135
