Borrar filtros
Borrar filtros

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

2 visualizaciones (últimos 30 días)
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)

Respuesta aceptada

Dyuman Joshi
Dyuman Joshi el 9 de Ag. de 2023
Editada: Dyuman Joshi el 9 de Ag. de 2023
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.
  7 comentarios
mahdi
mahdi el 10 de Ag. de 2023
Editada: mahdi el 10 de Ag. de 2023
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
Dyuman Joshi el 11 de Ag. de 2023
That's good to know.
You are welcome!

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre 2-D and 3-D Plots en Help Center y File Exchange.

Productos


Versión

R2020b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by