Calculate principal stresses - matlab

Question

Code result fails to appear here in the table produced by it.
In essence only the fifth row fills up for some reason. Aim of this code is to calculate principal stresses with MATLAB. No clue what specifically went wrong.
syms
sigma_max sigma_min tau_max alpha primary_direction_x_x_value primary_direction_x_y_value primary_direction_y_x_value primary_direction_y_y_value
sigma_max=0.5*((sigma_xx)+(sigma_yy))+sqrt(0.25*(((sigma_xx)-(sigma_yy))^2)+(sigma_xy)^2); %physics
sigma_min=0.5*((sigma_xx)+(sigma_yy))-sqrt(0.25*(((sigma_xx)-(sigma_yy))^2)+(sigma_xy)^2);
tau_max=0.5*((sigma_max)-(sigma_min));
%physics
alpha=(180/pi)*0.5*(atan((2*sigma_xy)/(sigma_xx-sigma_yy)));
primary_direction_x_x_value=cosd(alpha);
primary_direction_x_y_value=sind(alpha);
%
primary_direction_y_x_value=-sind(alpha);
primary_direction_y_y_value=cosd(alpha);
%physics
xx=[60 200 900 400 0];
yy=[0 -200 100 800 240];
xy=[40 -200 200 600 120];
for n = 1:length(xx)
sigma_xx=xx(n);
sigma_yy=yy(n);
sigma_xy=xy(n);
A=zeros(1,5);
A(n)=sigma_max;
B=zeros(1,5);
B(n)=sigma_min;
H=zeros(1,5);
H(n)=tau_max;
C=zeros(1,5);
C(n)=alpha;
D=zeros(1,5);
D(n)=primary_direction_x_x_value;
E=zeros(1,5);
E(n)=primary_direction_x_y_value;
F=zeros(1,5);
F(n)=primary_direction_y_x_value;
G=zeros(1,5);
G(n)=primary_direction_y_y_value;
end
varNames = {'σ max [Mpa]';'σ min [Mpa]';'tau max [Mpa]';'alpha [°]';'primary direction x x value';'primary direction x y value';'primary direction y x value';'primary direction y y value'};
T=table(A',B',H',C',D',E',F',G','VariableNames',varNames)

Answer 1

Shouldn't the initialization statements like A=zeros(1,5); be outside of the loop?
Right now for each n the vectors A, B, C, D, E, F and G are cleared overwriting them with all zeros.

Answer 2

syms n
format bank
A=zeros(1,5);
B=zeros(1,5);
H=zeros(1,5);
C=zeros(1,5);
D=zeros(1,5);
E=zeros(1,5);
F=zeros(1,5);
G=zeros(1,5);
I=zeros(1,5);
J=zeros(1,5);
K=zeros(1,5);
Input=[60 200 900 400 0
0 -200 100 800 240
40 -200 200 600 120]';
for n = 1:5
sigma_xx=Input(n,1);
sigma_yy=Input(n,2);
sigma_xy=Input(n,3);
sigma_max=0.5*((sigma_xx)+(sigma_yy))+sqrt(0.25*(((sigma_xx)-(sigma_yy))^2)+(sigma_xy)^2);%physics
sigma_min=0.5*((sigma_xx)+(sigma_yy))-sqrt(0.25*(((sigma_xx)-(sigma_yy))^2)+(sigma_xy)^2);%physics
tau_max=0.5*((sigma_max)-(sigma_min)); %physics
alpha=(180/pi)*0.5*(atan((2*sigma_xy)/(sigma_xx-sigma_yy)));%physics
primary_direction_x_x_value=cosd(alpha);%physics
primary_direction_x_y_value=sind(alpha);%physics
primary_direction_y_x_value=-sind(alpha);%physics
primary_direction_y_y_value=cosd(alpha);%physics
sigma_xxRe= 0.5*((sigma_xx)+(sigma_yy))+0.5*(((sigma_xx)-(sigma_yy))*cosd(2*alpha))+(sigma_xy)*sind(2*alpha);%physics - Rotation of stress component.
sigma_yyRe= 0.5*((sigma_xx)+(sigma_yy))-0.5*(((sigma_xx)-(sigma_yy))*cosd(2*alpha))-(sigma_xy)*sind(2*alpha);%physics
sigma_xyRe=0.5*((-(sigma_xx)+(sigma_yy))*sind(2*alpha))+(sigma_xy)*cosd(2*alpha);%physics
A(n)=sigma_max;
B(n)=sigma_min;
H(n)=tau_max;
C(n)=alpha;
D(n)=primary_direction_x_x_value;
E(n)=primary_direction_x_y_value;
F(n)=primary_direction_y_x_value;
G(n)=primary_direction_y_y_value;
I(n)=sigma_xxRe;
J(n)=sigma_yyRe;
K(n)=sigma_xyRe;
end
varNames = {'σ max [Mpa]';'σ min [Mpa]';'tau max [Mpa]';'alpha [°]';'primary direction x x value';'primary direction x y value';'primary direction y x value';'primary direction y y value';'σ xx revolved';'σ yy revolved';'σ xy revolved'};
T=table(A',B',H',C',D',E',F',G',I',J',K','VariableNames',varNames)
This code solves the problem since
1.A=zeros(1,5) no longer annihlates all that was put into it when it was in the loop, but rather it SAVES it in proper cells.
2.The input was changed to one single matrix since for some reason it is much easier for the loop to deal with a whole coherent matrix rather than several vectors, each of which contains some part of input data necessary for calculation.
3. May this be useful to whomever might need to calculate principal stresses...
Example of Output