User:Eas4200c.f08.nine.s/Lecture 5

Group nine - Homework 5

Stress-Strain Relation

$\left\{\epsilon _{ij}\right\}_{6x1}={\begin{bmatrix}A_{3x3}&O_{3x3}\\O_{3x3}&B_{3x3}\end{bmatrix}}_{6x6}\left\{\sigma _{ij}\right\}_{6x1}$ where the {6x6} determinant is denoted by $C$

$\sigma -\epsilon$ relation:

$\left\{\epsilon _{ij}\right\}_{6x1}={\begin{bmatrix}A_{3x3}^{-1}&O_{3x3}\\O_{3x3}&B_{3x3}^{-1}\end{bmatrix}}_{6x6}\left\{\sigma _{ij}\right\}_{6x1}$ where the {6x6} determinant is denoted by $C^{-1}$

Verification of the Identity Matrix: $C_{6x6}^{-1}C_{6x6}=I_{6x6}$

$I={\begin{bmatrix}A^{-1}A&0\\0&B^{-1}{B}\end{bmatrix}}$

$\left\{\sigma _{ij}\right\}={\begin{bmatrix}{\bar {A}}^{-1}&0\\0&{\bar {B}}^{-1}\end{bmatrix}}{\begin{Bmatrix}0\\0\\0\\0\\\epsilon _{31}\\\epsilon _{12}\end{Bmatrix}}$

Stress Tensors

four zero stress components using this equation:

$\sum _{i=1,j=1}^{3}{\frac {\delta \sigma _{ij}}{\delta x_{i}}}=0$

gives:

${\frac {\delta \sigma _{11}}{\delta x_{1}}}+{\frac {\delta \sigma _{12}}{\delta x_{2}}}+{\frac {\delta \sigma _{13}}{\delta x_{3}}}=0$

${\frac {\delta \sigma _{21}}{\delta x_{1}}}+{\frac {\delta \sigma _{22}}{\delta x_{2}}}+{\frac {\delta \sigma _{23}}{\delta x_{3}}}=0$

${\frac {\delta \sigma _{31}}{\delta x_{1}}}+{\frac {\delta \sigma _{32}}{\delta x_{2}}}+{\frac {\delta \sigma _{33}}{\delta x_{3}}}=0$

Bidirectional Bending

$M_{y}=\int _{aA}^{}{z\sigma _{xx}dA}$

$M_{z}=\int _{aA}^{}{y\sigma _{xx}dA}$

Moment of Inertia Tensors:

$I_{yy}=\int _{A}^{}{z^{2}dA}$

$I_{zz}=\int _{A}^{}{y^{2}dA}$

$I_{yz}=\int _{A}^{}{zydA}$

Hooke's Law

$\sigma _{xx}=E\varepsilon _{xx}$

$\sigma ={\frac {IyMz-IyzMy}{IyIz-(Iyz)^{2}}}(y)+{\frac {IzMy-IyzMz}{IyIz-(Iyz)^{2}}}(z)$

$I={\begin{bmatrix}I_{11}&I_{12}&I_{13}\\|&I_{22}&I_{23}\\symm&-&I_{33}\end{bmatrix}}$

Nonuniform Stress Field

Nonuniform Stress Field in 3-D.

$\sum Fx=0=-\sigma (x)A+\sigma (x+dx)A+f(x)dx$

0=A[ $\sigma (x+dx)-\sigma (x)$ ]+f(x)dx

${\frac {d\sigma }{dx}}+{\frac {f(x)}{A}}=0$ were f(x) is the force per unit length and A is the applied load.

$\sum {Fx}=0=dydz[-\sigma _{xx}(x,y,z)+\sigma _{xx}(x+dx,y,z)]+dzdx[-\sigma _{yx}(x,y,z)+\sigma _{yx}(x,y+dy,z)]+dxdy[-\sigma _{zx}(x,y,z)+\sigma _{zx}(x,y,z+dz)]$

$0=(dxdydz)\left[{\frac {\delta \sigma _{xx}}{\delta x}}+{\frac {\delta \sigma _{yx}}{\delta x}}+{\frac {\delta \sigma _{zx}}{\delta z}}\right]$

Similarly for the forces in the y direction:

$\sum Fy=0=dydz[-\sigma _{yx}(x,y,z)+\sigma _{yx}(x+dx),y,z)]+dzdx[-\sigma _{yy}(x,y,z)+\sigma _{yy}(x,y+dy,z)]+dxdy[\sigma _{yz}(x,y,z)+\sigma _{yz}(x,y,z+dz)]$

which becomes:

$0=(dxdydz)\left[{\frac {\delta \sigma _{yx}}{\delta _{x}}}+{\frac {\delta \sigma _{yy}}{\delta _{y}}}+{\frac {\delta \sigma _{yz}}{\delta _{z}}}\right]$

Z direction:

$\sum Fy=0=dydz[-\sigma _{zx}(x,y,z)+\sigma _{zx}(x+dx),y,z)]+dzdx[-\sigma _{zy}(x,y,z)+\sigma _{zy}(x,y+dy,z)]+dxdy[\sigma _{zz}(x,y,z)+\sigma _{zz}(x,y,z+dz)]$

which becomes:

$0=(dxdydz)\left[{\frac {\delta \sigma _{zx}}{\delta _{x}}}+{\frac {\delta \sigma _{zy}}{\delta _{y}}}+{\frac {\delta \sigma _{zz}}{\delta _{z}}}\right]$

Dimensional Analysis

$[F]={\frac {F}{L}}={\frac {force}{lenght}}$

$[{\frac {F}{A}}]={\frac {F}{L^{3}}}$

$[A]=L^{2}$

$[\sigma ]={\frac {f}{L^{2}}}\rightarrow [{\frac {d\sigma }{dx}}]={\frac {[F/L^{2}]}{[L]}}=MLT^{-2}L^{-3}=ML^{-2}T^{-2}$

$[dx]=L$

$[EI]={\frac {F}{L^{3}}}\times L^{4}=FL^{2}$

$[D]=[E][h^{3}]={\frac {F}{L^{2}}}\times L^{3}=FL$

$\epsilon ={\frac {du}{dx}}={\frac {\Delta L}{L}}\rightarrow [\epsilon ]={\frac {[du]}{[dx]}}={\frac {L}{L}}=1$

$\nu ={\frac {\epsilon _{yy}}{\epsilon _{xx}}}\rightarrow [\nu ]={\frac {[\epsilon _{yy}]}{[\epsilon _{xx}]}}=1$

MATLAB Code

 fprintf('\n NACA Airfoil calculation program \n \n')
m = 2/100;
p = 4/100;
t = 15/100;
segment = 50;
y = 0;
n = 1;
c=1;
 zc = size(segment);
 dzdy = size(segment);
 yu = size(segment);
 zu = size(segment);
 yl = size(segment);
 zl = size(segment);
Dy = 1/4;
Dz = 0;
Gy = 3/4;
Gz = 0;
a1a= 0;
a1b= 0;
a2a = 0;
a2b = 0;
a3a = 0;
a3b = 0;
 
 
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

figure(1)
plot(yl,zl,'-k',yu,zu,'-b')
axis([0 1 -0.3 0.3])
 

j=1;
while yu(j)<c/4                                
    j=j+1;
end
Ey = yl(j);
Ez = zl(j);
j=2;
while j<j                                       
    segment = [yu(j-1)+Dy zu(j-1)+Dz 0];                    
    r = [yu(j)+Dy zu(j)+Dz 0];
    a = 0.5*cross(r,segment);
    a1a = a1a + a;
    j = j+1;
end
j =2;
while j<j
    segment = [yl(j-1)+Dy zl(j-1)+Dz 0];                    
    r = [yl(j)+Dy zl(j)+Dz 0];
    a = 0.5*cross(r,segment);
    a1b = a1b + a;
    j = j+1;
end
k=j;
j=1;
while yu(j)<3*c/4                               
    j=j+1;
end
Fy = yu(j);
Fz = zu(j);
while j<j                                      
    segment = [yu(j-1)+Ey zu(j-1)+Ez 0];                    
    r = [yu(j)+Ey zu(j)+Ez 0];
    a = 0.5*cross(r,segment);
    a2a = a2a + a;
    j = j+1;
end
j =k;
while j<j
    segment = [yl(j-1)+Fy zl(j-1)+Fz 0];                    
    r = [yl(j)+Fy zl(j)+Fz 0];
    a = 0.5*cross(r,segment);
    a2b = a2b + a;
    j = j+1;
end
j=j+1;
while j<ns                                      
 
   segment = [yu(j-1)+Gy zu(j-1)+Gz 0];                    
   r = [yu(j)+Gy zu(j)+Gz 0];
   a = 0.5*cross(r,segment);
   a3a = a3a + a;
   j = j+1;
 
end
j=j+1;
while j<ns                                      
   segment = [yl(j-1)+Gy zl(j-1)+Gz 0];                    
   r = [yl(j)+Gy zl(j)+Gz 0];
   b = 0.5*cross(r,segment);
   area3b = area3b + b;
   j = j+1;
end

atotal = 0;
cy = 0;
cz = 0;
for j = 1:ns
    hu = yu(j+1)-yu(j);
    hl = yl(j+1)-yl(j);
    bsu= zu(j+1)+zu(j);
    bsl= abs(zl(j+1)+zl(j));
    au = (hu*bsu)/2;
    al = (hl*bsl)/2;
    atotal = atotal + au + al;
 
    cenuy = hu*(2*zu(j+1)+zu(j))/(3*(zu(j+1)+zu(j))) + yu(j);cy
    cenly = hl*(2*zl(j+1)+zl(j))/(3*(zl(j+1)+zl(j))) + yl(j);
    cenuz = (zu(j)^2 + zu(j)*zu(j+1) + zu(j+1)^2)/(3*(zu(j+1)+zu(j)));
    cenlz = (zl(j)^2 + zl(j)*zl(j+1) + zl(j+1)^2)/(3*(zl(j+1)+zl(j)));
    cy = cy+cenuy*au+cenly*al;
    cz = cz+cenuz*au+cenlz*al;
end
 
ceny = cy/atotal;
cenz = cz/atotal;
 

fprintf('Abar of cell 1 is: %5.4f\i',a2a+a2b)
fprintf('Abar of of cell 2 is: %5.4f\i',a3a+a3b)
fprintf('Abar of cell 3 is: %5.4f\i',a1a+a1b)
fprintf('Abar of the airfoil is: %5.4f\i',abar(1))
fprintf('Top length of cell 1 is: %5.4f\i',yu(k))
fprintf('Bottom length of cell 1 is: %5.4f\i',yl(k))
fprintf('Top length of cell l 2 is: %5.4f\i',yu(j)-yu(k))
fprintf('Bottom length of cell 2 is: %5.4f\i',yl(j)-yl(k))
fprintf('Top length of cell 3 is: %5.4f\i',yu(ns)-yu(j))
fprintf('Bottom length of cell 3 is: %5.4f\i',yl(ns)-yl(j))
fprintf('END OF PROGRAM')

Figure (2)
plot(3*c/4,zl(j):0.001:zu(j),'-k')
axis([0 0.5 -0.2 0.2])
 
end

Sample Run of Code (NACA Plot)

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 the centroid line

Contributing Team Members

The following students contributed to this report:

Felix Izquierdo Eas4200c.f08.nine.F 18:34, 6 November 2008 (UTC)
Ricardo Albuquerque Eas4200c.f08.nine.R 18:39, 6 November 2008 (UTC)
Dave Phillips Eas4200c.f08.nine.D 18:46, 6 November 2008 (UTC)
Stephen Featherman Eas4200c.f08.nine.S 18:49, 6 November 2008 (UTC)