Is there a way for reducing my code lines in a code for ploting

So I have this code for plotting and I wanna know that if there is a quicker way or a code with lesser lines like with a for loop or something can do the same thing for me.

And also there is a warning message that comes up and it doesnt show all my legend entries and I dont know why. (Warning: Ignoring extra legend entries.)

Es = 200*10^9; % Young's modulus of the solid of the steel(Pa)
Beta = 0;
h = 0.02; % Thickness(m)
e1 = 1-((1-Beta).^2);
z = -h/2:0.001:h/2;
Ez = (Es*(1-(e1.*(cos((pi*z)/h)))));
plot(z,Ez)
xlabel('z')
ylabel('E(z)')
title('Porousity factor: Beta')
legend('0','0.1','0.2','0.3','0.4','0.5','0.6','0.7','0.8','0.9','1')

Warning: Ignoring extra legend entries.

hold on
Beta = 0.1;
e1 = 1-((1-Beta).^2);
z = -h/2:0.001:h/2;
Ez = (Es*(1-(e1.*(cos((pi*z)/h)))));
plot(z,Ez)
Beta = 0.2;
e1 = 1-((1-Beta).^2);
z = -h/2:0.001:h/2;
Ez = (Es*(1-(e1.*(cos((pi*z)/h)))));
plot(z,Ez)
Beta = 0.3;
e1 = 1-((1-Beta).^2);
z = -h/2:0.001:h/2;
Ez = (Es*(1-(e1.*(cos((pi*z)/h)))));
plot(z,Ez)
Beta = 0.4;
e1 = 1-((1-Beta).^2);
z = -h/2:0.001:h/2;
Ez = (Es*(1-(e1.*(cos((pi*z)/h)))));
plot(z,Ez)
Beta = 0.5;
e1 = 1-((1-Beta).^2);
z = -h/2:0.001:h/2;
Ez = (Es*(1-(e1.*(cos((pi*z)/h)))));
plot(z,Ez)
Beta = 0.6;
e1 = 1-((1-Beta).^2);
z = -h/2:0.001:h/2;
Ez = (Es*(1-(e1.*(cos((pi*z)/h)))));
plot(z,Ez)
Beta = 0.7;
e1 = 1-((1-Beta).^2);
z = -h/2:0.001:h/2;
Ez = (Es*(1-(e1.*(cos((pi*z)/h)))));
plot(z,Ez)
Beta = 0.8;
e1 = 1-((1-Beta).^2);
z = -h/2:0.001:h/2;
Ez = (Es*(1-(e1.*(cos((pi*z)/h)))));
plot(z,Ez)
Beta = 0.9;
e1 = 1-((1-Beta).^2);
z = -h/2:0.001:h/2;
Ez = (Es*(1-(e1.*(cos((pi*z)/h)))));
plot(z,Ez)
Beta = 1;
e1 = 1-((1-Beta).^2);
z = -h/2:0.001:h/2;
Ez = (Es*(1-(e1.*(cos((pi*z)/h)))));
plot(z,Ez)

Answer 1

[Dyuman Joshi]

Store your data in array and use indices to access the data -

Es = 200*10^9; % Young's modulus of the solid of the steel(Pa)

h = 0.02; % Thickness(m)

%Define Beta as an array
Beta = (0:10)/10;

%Corresponding value of e
e = 1-((1-Beta).^2);

%As z does not depend on the loop index, bring it out of the loop
%This avoids calculating z over and over again.
z = -h/2:0.001:h/2;

figure()
for k=1:numel(Beta)
    Ez = (Es*(1-(e(k).*(cos((pi*z)/h)))));
    plot(z,Ez)
    hold on
end

xlabel('z')
ylabel('E(z)')
title('Porousity factor: Beta')
legend('0','0.1','0.2','0.3','0.4','0.5','0.6','0.7','0.8','0.9','1','location','best')

As you learn things about MATLAB, you will find that vectorization is a powerful tool, as MATLAB is optimized for operations involving arrays. The above for loop can be vectorized as well -

e0 = e';
%What you will get below is an array, where each column will be
%Ez value calculated for corresonding value in beta
Ez0 = (Es*(1-(e0*(cos((pi*z)/h)))));

%Make a new figure to plot and observe the results
figure()
%When first input is a vector, and second input is a non-vector array
%plot() will plot each curve corresponding to each column of the second input
plot(z,Ez0)

As for the warning - You have plotted only 1 curve before using legend(), but you have included more than 1 names, so the legend() ignores the rest of the names, which is what the warning says as well.

[mahdi]

Nice, thanks a lot.

[mahdi]

@Dyuman Joshi

How about something like this

P = [-1000,-5000,-10000,-50000,-100000];
b = [0;500;1000;1500;2000];
c = [0,0.5,1,1.5,2];
b(4) = P;

Unable to perform assignment because the left and right sides have a different number of elements.

plot(c,b)

[Dyuman Joshi]

Please explain what you are trying to do with the above code.

[mahdi]

I have matrix equations like Ax = b and b is a 21*1 zeros but I must change the value in b(7) to be -1000,-5000,-10000,-50000,-100000 and solve the equations, so I kinda want to have all the solutions to be plotted in one plot.

If you did not understand what I mean I can copy the full code here.

[Dyuman Joshi]

Yes, please copy and paste the full code here.

[mahdi]

@Dyuman Joshi

I figured my last comment's answer too and here it is:

P = [-1000,-5000,-10000,-50000,-100000];
b = [0;500;1000;1500;2000];
c = [0,0.5,1,1.5,2];
figure(1)
for l=1:numel(P)
    b(4)=P(l);
    plot(c,b)
    hold on
end

Again thanks for your help.

[Dyuman Joshi]

That's good to know.

You are welcome!