User:Eas4200c.f08.nine.s
Homework 6
Group nine - Homework 3
Introduction
[edit]Reasons to use open thin wall cross section for stringers
1) manufacturing: stamping of thin flat sheets
2) construction of aircraft: riveting
Why do the stringer have angled walls compared to vertical walls?
One possible reason is for storage purposes. Having angled walls allows the stringers to be stacked while saving space.
Shear Panels
[edit]![](http://upload.wikimedia.org/wikipedia/commons/thumb/4/42/Fig1aero4200c.jpg/220px-Fig1aero4200c.jpg)
= change in the right angle due to shear deformation.
= torsional shear strain =
Curved Panels
[edit]![](http://upload.wikimedia.org/wikipedia/commons/thumb/f/fa/Fig2asd231.jpg/220px-Fig2asd231.jpg)
(Shear flow)
where = and =
![](http://upload.wikimedia.org/wikipedia/commons/thumb/e/e7/Fig3asd22.jpg/220px-Fig3asd22.jpg)
Resultant shear force vector
![](http://upload.wikimedia.org/wikipedia/commons/thumb/4/40/Fig4are0sd.jpg/220px-Fig4are0sd.jpg)
Resultant Magnitude
where = length of straight line .
We can relate to (3.48)
![](http://upload.wikimedia.org/wikipedia/commons/thumb/7/7c/Fig6asdqw2.jpg/220px-Fig6asdqw2.jpg)
Closed thin wall cross section:
where
![](http://upload.wikimedia.org/wikipedia/commons/thumb/4/4b/Fig5asdew1.jpg/220px-Fig5asdew1.jpg)
Open thin wall cross section:
NACA 4-digit air foil series
[edit]![](http://upload.wikimedia.org/wikipedia/commons/thumb/1/13/Fig7asdew1.jpg/220px-Fig7asdew1.jpg)
ns = number of segments to the y-axis
![](http://upload.wikimedia.org/wikipedia/commons/thumb/3/36/Fig8asdq1.jpg/220px-Fig8asdq1.jpg)
Twist and Warping
[edit]Torsion of uniform, non circular bars (leads to warping of cross section)
Warping = axial displacement along x-axis of a part on the deformed cross section.
rate of twist.
(3.11)
(3.12)
Warping displacement along x-axis
Road Map for torsional Analysis of aircraft wing
[edit]![](http://upload.wikimedia.org/wikipedia/commons/thumb/e/e6/Fig9asd1.jpg/220px-Fig9asd1.jpg)
A) Kinematic assumptions (Section 3.2)
B) Strain - displacement relationship (Section 3.2)
C) Equilibrium Equations ( Stresses) (Ch.2; Section 3.2)
D) Pranal Stress Function (Section 3.2)
E) Strain compatibility Eqn.(3.15)(3.17)
F) Eqn. for (3.19)
G) B conditions for (3.24)
H) (3.25)
I) Thin walled cross section
Formal Derivation ( Section 3.5)
(3.48)
J) Twist angle : Method 1
(3.56)
K) Multicell Section (Section 3.6)
Multicell Thin-Walled Sections
[edit]Multicell Thin-Walled Sections refers to wing sections that are composed of airfoil skin supported by thin vertical supports called spar webs that form multicell constructions. There are often stiffeners incorporated in the construction of multicell sections which are usually located above and below the spar webs themselfs. These stiffeners are very effective when used to carry bending loads, but they do little to help counteract torsional loads and are often neglected in the consideration of torsional rigidity.
For a two-cell section there are three boundary contours which we will label S0, S1, and S2. This will give us
(S0)=C0
(S1)=C1
(s2)=C2
Where C0, C1, and C2 are all constants and is the stress function. The shear flow between two boundary contours is equal to the difference between the values of along these contours. The shear flow for each cell is considered positive if it forms a counterclockwise torque about the cell and its value is equal to the value of on the inside contour minus that on the outside contour.
q1=C1-C0
q2=C2-C0
q12=C1-C2
The Torque contributed by each cell can be calculated by using T=2Aq. The total torque of a two-cell section is
T=2A1q1+2A2q2
where A1 and A2 are the areas enclosed by the shear flows q1 and q2 respectively.
The equations for the twist angles 1 and 2 are
1=
2=
==[1]
Homework Problems & Matlab Code
[edit]Homework Problems |
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Homework Problem 1 Prove:
Homework Problem 2 ![]() Show that:
__________________________________________________________________________________________________________________________________________________________ Homework Problem 3 Show: and we know that: since: we can neglect this term, leaving the same holds true for b since: since: we can neglect this term, leaving ___________________________________________________________________________________________________________________________________________________________ Homework Problem 4 ![]() Given: Find:
since:
___________________________________________________________________________________________________________________________________________________________ Homework Problem 5 Compare solid circular cross section to hollow thin wall cross section: a) solid circular cross section: b) hollow circular cross section:
Matlab Code[edit] fprintf('\n NACA Airfoil calculation program \n \n')
m = input('Enter first digit of airfoil: ');
p = input('Enter second digit: ');
t = input('Enter the third and fourth digits: ');
Py = input('Enter Py: ');
Pz = input('Enter Pz: ');
segment = input ('Enter number of segments: ');
y = 0;
n = 1;
c=1;
m = (m/100)*c;
p = (p/10)*c;
t = (t/100)*c;
zc = size(segment);
dzdy = size(segment);
yu = size(segment);
zu = size(segment);
yl = size(segment);
zl = size(segment);
j=1;
while y<=p
zc(n) = (m/p^2)*(2*p*y-y^2);
dzdy(n) = (m/p^2)*(2*p-2*y);
y = y + c/segment;
zcc(n)=zc(n)*c;
n = n+1;
end
while y<=c
zc(n) = (m/((1-p)^2))*((1-2*p)+2*p*y-y^2);
dzdy(n) = (m/((1-p)^2))*(2*p-2*y);
y = y + c/segment;
zcc(n)=zc(n)*c;
n = n+1;
end
y=0;
n=1;
figure(2)
plot(zcc,y,'-k')
while y<=c
theta = size(segment);
zt = size(segment);
zt(n) = 5*t*(0.2969.*sqrt(y)-0.1260.*y-0.3516.*y.^2+0.2843.*y.^3-0.1015.*y.^4);
theta(n)= atan(dzdy(n));
yu(n)= y-zt(n)*sin(theta(n));
zu(n) = zc(n) + zt(n)*cos(theta(n));
yl(n) = y+zt(n)*sin(theta(n));
zl(n) = zc(n)- zt(n)*cos(theta(n));
y = y+c/segment;
n = n+1;
end
y=0;
n = 1;
a1 = 0;
a2 = 0;
while n<segment
line = [0 yu(segment-(n))-yu(segment-(n-1)) zu(segment-(n))-zu(segment-(n-1))];
r = [0 yu(segment-(n))-c zu(segment-(n))-Pz];
a = 0.5*cross(r,line);
a1 = a1 + a;
n = n+1;
end
n =1;
line =0;
while n<segment
line = [0 yl(segment-(n))-yl(segment-(n-1)) zl(segment-n)-zl(segment-(n-1))];
r = [0 yl(segment-(n))-c zl(segment-(n))-Pz];
b = 0.5*cross(r,line);
a2 = a2 + b;
n = n+1;
end
avg = abs(a1-a2);
area(j) = avg(1);
segment_max = segment;
fprintf('\n')
fprintf(1,'The average area is: %4.3f\n',avg(1))
figure(1)
plot(yl,zl,'-k',yu,zu,'-b')
axis([0 1 -0.3 0.3])
R=1;
percentage = 2;
segment = 1;
j=1;
while percentage>1
y = 0:2/segment:2;
z = sqrt(R^2-(y-1).^2);
zl = -z;
Py = 0.25;
Pz = -1;
n=1;
areau = 0;
areal = 0;
c=2;
line = 0;
while n<segment
line = [0 y(segment-(n))-y(segment-(n-1)) z(segment-(n))-z(segment-(n-1))];
r = [0 y(segment-(n))-c z(segment-(n))-Pz];
a = 0.5*cross(r,line);
areau = areau + a;
n = n+1;
end
n =1;
line =0;
while n<segment
line = [0 y(segment-(n))-y(segment-(n-1)) zl(segment-n)-zl(segment-(n-1))];
r = [0 y(segment-(n))-c zl(segment-(n))-Pz];
b = 0.5*cross(r,line);
areal = areal + b;
n = n+1;
end
avg = abs(areau - areal);
area(j) = avg(1);
percentage = ((pi-avg(1))/pi)*100;
segment = segment+1;
j = j+1;
end
fprintf('The minumum number segments required to have the average area accurate within 1 percent is: %4.3f\n',segment)
fprintf('\n Figure 1 shows the cross-section of the NACA airfoil and Figure 2 shows the centroid line \n')
![]() Sample Run[edit]NACA Airfoil calculation program Enter first digit of airfoil: 2 The average area is: 0.103 Figure 1 shows the cross-section of the NACA airfoil and Figure 2 shows the centroid line |
References
[edit]- ^ Sun, C.T. Mechanics of Aircraft Structures. New York: Wiley, 2006.
Contributing Team Members
[edit]The following students contributed to this report:
David Phillips Eas4200C.f08.nine.d 3:10, 8 October 2008 (UTC)
Oliver Watmough Eas4200c.f08.nine.o 16:07, 8 October 2008 (UTC)
Stephen Featherman Eas4200c.f08.nine.s 1:07, 8 October 2008 (UTC)
Ricardo Albuquerque Eas4200c.f08.nine.r 4:30, 8 October 2008 (UTC)
Felix Izquierdo Eas4200c.f08.nine.F 18:34, 8 October 2008 (UTC)