Generates for loop for Stress and Strain

I am given a problem where I have to generate data point to plot. This values will show how a biomaterial is reacting.

Here are the values given to me.

Question : Model the elastic-plastic behavior of the sample polymer given above. Create a Matlab

program to model the above equations to plot a stress-strain curve.

This what I done so far yet my for loop is not working; thus I can generate the nessary data points to plot. Please help I am not very verse in Matlab.

E1=4.0; %4 is in GPa
E2=5.0; %5 is in GPa
Mu= 1.2*10^5; %this measures creep
stressO = 800*10^6; %this is initial stress

for i=0:stressO
    data.less(i)=i/(E1); % Yielding when applies stress is less than initial stress  
  end

for i=stressO:(1.5*10^9)
    data.great(i)=(i/(E1))+((i-stressO)/E2); % Yielding when applies stress is less than initial stress  
  end

  
subplot (2,1,1)
plot (data.less);

subplot (2,1,2)
plot (data.great);

Answer 1

[darova]

Here is the problem

The solution:

sp = stress0.plus10:1500
for i = 1:length(sp);
    strain.stressplus10(i)=(sp(i)/E1)+((sp(i)-stress0.plus10)/E2); 
end

Answer 2

[darova]

here are modifications needed to be done. No for loop is needed

[Pablo Tejada Jr.]

I Thank you for your answer and aiding me. Yet the reason I opt to a "For Loop" becuase I wanted to have a singular data variable that is chaning over stresses. I was able to modify my code to anwers question A through D; However, I am facing another problem with part E. I get an error message when I add/subtract .1 from stress0 in the "for Loop"

a) Model the elastic-plastic behavior of the sample polymer given above. Create a Matlab, Simulink (or other appropriate platform) program to model the above equations to plot a stress-strain curve.

b) Use the given constants and yield point above to plot the stress-strain curve. Vary stress from 0 to 1.5 GPa for the graphs. Note this plot has two ranges and two equations that go into the same graph.

c) Change E1 by + and – 10% and plot the result on the graphs above. Describe how this element changes the behavior of the material.

d) Change E2 by + and – 10% and plot the result on a new set of graphs. Describe how this element changes the behavior of the material.

e) Change the yield stress0 by + and – 10% and plot the result on a new set of graphs. Describe how this element changes the behavior of the material.

Here is my code

%% Assigning Values to the Variables

clc
clear all
close all

E1=4000; %4GPa written in 4000 MPa
E2=5000; %5GPa written in 5000 MPa
Mu= 1.2*10^5; %this measures creep
stress0.norm = 800; %this is initial stress value is in MPa

%The Following values are used to answer part E 
stress0.plus10  = 800+.1;  
stress0.minus10  = 800-.1;


%% Generating Data Points Using "For Loop"


% The following loop generates points with E1 and E2. Answers for part
% B,C&D

for i=1:stress0.norm 
    strain.normal(i)=i/E1; % This is storing the values before yielding
    
    %Answer for part C
    strain.E1plus10(i)=i/(E1+.1); % This is storing the values before yielding with E1 + 10%
    strain.E1minus10(i)=i/(E1-.1); % This is storing the values before yielding with E1 - 10%
    
    %Answer for part D
    strain.E2plus10(i)=i/E1; % This is storing the values before yielding
    strain.E2minus10(i)=i/E1; % This is storing the values before yielding
end

for i=stress0.norm:1500  %this is the finising stress in MPa (after yielding 
    strain.normal(i)=(i/E1)+((i-stress0.norm)/E2); % This is storing the values after yielding
    
    %Answer for part C
    strain.E1plus10(i)=(i/E1+.1)+((i-stress0.norm)/E2); % This is storing the values after yielding with E1 + 10%
    strain.E1minus10(i)=(i/E1-.1)+((i-stress0.norm)/E2); % This is storing the values after yielding with E1 + 10%
    
    %Answer for part D
    strain.E2plus10(i)=(i/E1)+((i-stress0.norm)/(E2+.1));
    strain.E2minus10(i)=(i/E1)+((i-stress0.norm)/(E2-.1));
end


% Creating a another seperated loop for changes in stress0

% for stress0+10%
for i=1:stress0.plus10
    strain.stress0plus10(i)=i/E1; 
end

for i=stress0.plus10;1500
    strain.stressplus10(i)=(i/E1)+((i-stress0.plus10)/E2); 
end

% for stress0-10%

for i=1:stress0.minus10
    strain.stressMinus10(i)=i/E1; 
end

for i=stress0.minus10;1500
    strain.stressMinus10(i)=(i/E1)+((i-stress0.minus10)/E2); 
end



%% The following lines plots the data before stress0 is modified