User:Eas4200c.f08.nine.s/Lecture 4

Homework 1

Homework 2

Homework 3

Homework 4

Homework 5

Homework 6

Group nine - Homework 4

Ad-Hoc

"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."

"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."

Angle of Twist Derivation

Derive $\theta ={\frac {1}{2G{\bar {A}}}}\oint ^{}{\frac {q}{t}}ds$

Consider we have a uniform non circular cross section subject to twist.
Going through the geometry we can see that,

${\frac {P{\dot {P}}}{OP}}=tan\alpha$

$P{\ddot {P}}=P{\dot {P}}cos\alpha$

$\rightarrow =(OPtan\alpha )cos\alpha$ where $OPcos\alpha =O{\ddot {P}}$

Now we define the following:
$OP=r$ and $O{\ddot {P}}=\rho$

$P{\ddot {P}}=rcos\alpha tan\alpha =\rho \alpha$

And now we take a look at the strain.

$\gamma ={\frac {P{\ddot {P}}}{ds}}={\frac {\rho \alpha }{ds}}=\rho \theta =\rho {\frac {\alpha }{dx}}$

Hooke's Law: $\tau =G\gamma =G\rho \theta$

$\rightarrow \tau (s)=G\rho (s)\theta (x)$

and now integrating along the contour C $\rightarrow$

$\oint _{C}\tau (s)ds=G\theta (x)\oint _{C}^{}\rho (s)ds$

where $\tau (s)={\frac {q(s)}{t(s)}}$ and $\oint _{C}\rho (s)ds=2{\bar {A}}$

Integrating and rearranging terms yields the final result$rightarrow$

$\theta ={\frac {1}{2G{\bar {A}}}}\oint ^{}{\frac {q}{t}}ds$

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 $P$ in the direction that is tangent to the contour curve at $P{\ddot {P}}$

2) It was assumed that the shear force was uniform across the wall the thickness.

3) To get line $P{\dot {P}}$ assume $\alpha$ is small; to get $P{\ddot {P}}$ assume $\alpha$ is finite and $\rho =rcos\alpha$  ; and then reintroduced small $\alpha$ after that.

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Hooke's law: (Ricardo Albuquerque)

$\tau =G\gamma$

Normal strains:

$\epsilon _{xx}={\frac {\sigma _{xx}}{E}}-{\frac {\nu \sigma _{yy}}{E}}-{\frac {\nu \sigma _{zz}}{E}}$

$\epsilon _{yy}=-{\frac {\nu \sigma _{xx}}{E}}+{\frac {\sigma _{yy}}{E}}-{\frac {\nu \sigma _{zz}}{E}}$

$\epsilon _{zz}=-{\frac {\nu \sigma _{xx}}{E}}-{\frac {\nu \sigma _{yy}}{E}}+{\frac {\sigma _{zz}}{E}}$

Shear strains:

$\gamma _{xy}=2\epsilon _{xy}={\frac {\tau _{xy}}{G}}$

$\gamma _{yz}=2\epsilon _{yz}={\frac {\tau _{yz}}{G}}$

$\gamma _{zx}=2\epsilon _{zx}={\frac {\tau _{zx}}{G}}$

Strain tensor:

$\epsilon ={\begin{bmatrix}\epsilon _{11}&\epsilon _{12}&\epsilon _{13}\\\epsilon _{21}&\epsilon _{22}&\epsilon _{23}\\\epsilon _{31}&\epsilon _{32}&\epsilon _{33}\end{bmatrix}}$

Note: It is important to note here that the above yields six independent strain terms because $\epsilon _{ij}=\epsilon _{ji}$ .

Hooke's law (isotropic material):

${\begin{bmatrix}\epsilon _{11}\\\epsilon _{22}\\\epsilon _{33}\\\epsilon _{23}\\\epsilon _{31}\\\epsilon _{12}\\\end{bmatrix}}~=$ ${\begin{bmatrix}{\frac {1}{E}}&{\frac {-\nu }{E}}&{\frac {-\nu }{E}}&0&0&0\\{\frac {-\nu }{E}}&{\frac {1}{E}}&{\frac {-\nu }{E}}&0&0&0\\{\frac {-\nu }{E}}&{\frac {-\nu }{E}}&{\frac {1}{E}}&0&0&0\\0&0&0&{\frac {1}{2G}}&0&0\\0&0&0&0&{\frac {1}{2G}}&0\\0&0&0&0&0&{\frac {1}{2G}}\\\end{bmatrix}}~$ ${\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{33}\\\sigma _{23}\\\sigma _{31}\\\sigma _{12}\\\end{bmatrix}}$

or equivaletly:

${\begin{bmatrix}\epsilon _{xx}\\\epsilon _{yy}\\\epsilon _{zz}\\\gamma _{yz}\\\gamma _{zx}\\\gamma _{xy}\\\end{bmatrix}}~=$ ${\begin{bmatrix}{\frac {1}{E}}&{\frac {-\nu }{E}}&{\frac {-\nu }{E}}&0&0&0\\{\frac {-\nu }{E}}&{\frac {1}{E}}&{\frac {-\nu }{E}}&0&0&0\\{\frac {-\nu }{E}}&{\frac {-\nu }{E}}&{\frac {1}{E}}&0&0&0\\0&0&0&{\frac {1}{G}}&0&0\\0&0&0&0&{\frac {1}{G}}&0\\0&0&0&0&0&{\frac {1}{G}}\\\end{bmatrix}}$ ${\begin{bmatrix}\sigma _{xx}\\\sigma _{yy}\\\sigma _{zz}\\\tau _{yz}\\\tau _{zx}\\\tau _{xy}\\\end{bmatrix}}$


Question
What material has a Poisson's ration of 0?

Answer
Cork

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Homework & MATLAB Code

Homework Problems

Multicell Problem: Multicell Problem Question: Find $\theta$ as a function of T and J. Givens: $\left[t_{1}=0.3cm\right]\left[t_{2}=0.5cm\right]\left[t_{12}=0.4cm\right]\left[a=30cm\right]\left[b=60cm\right]\left[c=40cm\right]$ Equations: $\left[T=T_{1}+T_{2}=2q_{1}{\bar {A_{1}}}+2q_{2}{\bar {A_{2}}}\right]\left[\theta =\theta _{1}=\theta _{2}={\frac {T}{GJ}}\right]\left[\theta ={\frac {1}{2G{\bar {A}}}}\oint _{}^{}{{\frac {q_{i}}{t_{i}}}ds}\right]$ First thing we need to do is compute the areas, ${\bar {A_{1}}}=ac=0.3\times 0.4[m^{2}]\rightarrow {\bar {A_{2}}}=bc=0.6\times 0.4[m^{2}]$ we know that: $\theta _{1}={\frac {1}{2G{\bar {A_{1}}}}}[{\frac {2q_{1}a}{t_{1}}}+{\frac {q_{1}c}{t_{1}}}+{\frac {(q_{1}-q_{2})c}{t_{12}}}]\rightarrow \theta _{2}={\frac {1}{2G{\bar {A_{2}}}}}[{\frac {2q_{2}b}{t_{2}}}+{\frac {q_{2}c}{t_{2}}}+{\frac {(q_{2}-q_{1})c}{t_{12}}}]$ plugging in the known values for a, b, c, and the areas, we obtain the following equations: $\theta _{1}={\frac {1}{G}}[1444.{\bar {4}}q_{1}-333.{\bar {3}}q_{2}]\rightarrow \theta _{2}={\frac {1}{G}}[875q_{2}-208.{\bar {3}}q_{1}]$ Setting $\theta _{1}=\theta _{2}={\frac {T}{GJ}}$ we have: $1652.{\bar {7}}q_{1}=1208.{\bar {3}}q_{2}={\frac {T}{GJ}}$ we now have the followig equations: $q_{1}={\frac {T}{1652.{\bar {7}}GJ}}$ and since $T=2q_{1}{\bar {A_{1}}}+2q_{2}{\bar {A_{2}}}$ and $q_{2}={\frac {1652.{\bar {7}}q_{1}}{1208.{\bar {3}}}}$ we can solve for GJ! $q_{1}={\frac {2q_{1}{\bar {A_{1}}}+2({\frac {1652.{\bar {7}}q_{1}}{1208.{\bar {3}}}}){\bar {A_{2}}}}{GJ1652.{\bar {7}}}}\rightarrow GJ\approx 0.000578754$ and with this result we can solve for T: $T\approx 0.95655q_{1}\approx 0.69933q_{2}$ And now, $q_{1}\approx 1.0454T\rightarrow q_{2}\approx 1.4299T$

MATLAB Code

 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')

Figure 1

Sample Run of Code

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

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)

Contributing Team Members

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)