User:Eas4200c.f08.nine.s/Lecture 4
Group nine - Homework 4
Ad-Hoc
[edit]"Ad hoc is a Latin phrase which means "for this [purpose]". It generally signifies a solution designed for a specific problem or task, non-generalizable, and which cannot be adapted to other purposes."[1]
"Common examples are organizations, committees, and commissions created at the national or international level for a specific task; in other fields the term may refer, for example, to a tailor-made suit, a handcrafted network protocol, or a purpose-specific equation. Ad hoc can also have connotations of a makeshift solution, inadequate planning, or improvised events. Other derivates of the Latin include AdHoc, adhoc and ad-hoc."[2]
Angle of Twist Derivation
[edit]Derive
Consider we have a uniform non circular cross section subject to twist.
Going through the geometry we can see that,
where
Now we define the following:
and
And now we take a look at the strain.
Hooke's Law:
and now integrating along the contour C
where and
Integrating and rearranging terms yields the final result
Question
What is ad hoc about the derivation of the angle of twist shown above?
Answer
1) The strain must be obtained using the displacement of in the direction that is tangent to the contour curve at
2) It was assumed that the shear force was uniform across the wall the thickness.
3) To get line assume is small; to get assume is finite and ; and then reintroduced small after that.
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Hooke's law: (Ricardo Albuquerque)
Normal strains:
Shear strains:
Strain tensor:
Note: It is important to note here that the above yields six independent strain terms because .
Hooke's law (isotropic material):
or equivaletly:
Question
What material has a Poisson's ration of 0?
Answer
Cork
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Homework & MATLAB Code
[edit]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')
![](http://upload.wikimedia.org/wikipedia/commons/thumb/2/23/AirfoilNINE.png/400px-AirfoilNINE.png)
Sample Run of Code
[edit]NACA Airfoil calculation program
Enter first digit of airfoil: 2
Enter second digit: 4
Enter the third and fourth digits: 15
Enter Py: 0
Enter Pz: 0
Enter number of segments: 60
The average area is: 0.103
The minumum number segments required to have the average area accurate within 1 percent is: 24.000
Figure 1 shows the cross-section of the NACA airfoil and Figure 2 shows the centroid line
Matlab Code Certification
[edit]I, the undersigned, certify that I can read, understand, and write matlab codes, and can thus contribute effectively to my team.
Ricardo Albuquerque Eas4200c.f08.nine.r 22:46, 21 October 2008 (UTC)
Felix Izquierdo Eas4200c.f08.nine.F 3:46, 22 October 2008 (UTC)
David Phillips Eas4200C.f08.nine.d (talk) 18:34, 23 October 2008 (UTC)
Oliver Watmough Eas4200c.f08.nine.o 10:07, 23 October 2008 (UTC)
Stephen Featherman Eas4200c.f08.nine.s 11:48, 23 October 2008 (UTC)
References
[edit]- ^ "Symonds, Peter." Wikipedia.2008. Wikipedia. 21 Oct. 2008 <http://en.wikipedia.org/wiki/Ad_hoc>
- ^ "Symonds, Peter." Wikipedia.2008. Wikipedia. 21 Oct. 2008 <http://en.wikipedia.org/wiki/Ad_hoc>
Contributing Team Members
[edit]The following students contributed to this report:
David Phillips Eas4200C.f08.nine.d (talk) 18:34, 23 October 2008 (UTC)
Oliver Watmough Eas4200c.f08.nine.o 10:07, 23 October 2008 (UTC)
Stephen Featherman Eas4200c.f08.nine.s 11:48, 23 October 2008 (UTC)
Ricardo Albuquerque Eas4200c.f08.nine.r 4:30, 23 October 2008 (UTC)
Felix Izquierdo Eas4200c.f08.nine.F 4:34, 23 October 2008 (UTC)