Borrar filtros
Borrar filtros

My plot won't show any graph

2 visualizaciones (últimos 30 días)
Steve
Steve el 14 de Dic. de 2020
Editada: VBBV el 14 de Dic. de 2020
I try to plot the data i get from step below. But My plot doesn't have any data. Just a clean sheet. Can anyone tell which part I'm wrong
clear
%%Given :
T= 5;
del_t=0.5;
t=0:del_t:T;
E=5;
v=0.3;
Sya=25;
Sy=15;
d=0.4;
vis=0;
gamma=0;
%Assume :
epn=0;
alfa=0;
DevS=0;
M= E/(2*(1+v));
L= (E*v)/[(1+v)*(1-2*v)];
G = norm(DevS)-sqrt(2/3)*Sy+(Sya-Sy)*(1-exp(-d*alfa))-[M*gamma*2];
Gd=-2*M*(1+(Sya-Sy)*d*exp(-d*gamma));
for n = 1:length(t);
strain=6*n*[1 0 0; 0 1 0; 0 0 1]+6*sin(n)*[0 2 0; 2 1 0; 0 0 3/2];
el=strain-(1/3)*trace(strain)*eye(3,3);%
DevStrial=DevS+2*M*(el-epn); %Devatoric Stress Trial
etrial=DevStrial;
y=Sy+(Sya-Sy)*(1-exp(-d*alfa));
ftrial=norm(DevS)-sqrt(2/3)*y;
if ftrial<=0;
DevS=DevStrial;
else
while and((abs((G/Gd)/gamma)) >= 0.001, gamma==0);
%g[dγ(k)]
G = ftrial-[M*gamma*2];
%Dg[dγ(k)]
Gd=-2*M*(1+(Sya-Sy)*d*exp(-d*gamma));
gamma= gamma - G/Gd;
% Derivative of yield limit function: K'(α)
end
del_g= gamma;
normal=etrial/norm(etrial);
alfa=alfa+sqrt(2/3)*del_g;
%Update back stress, plastic strain, and stress
epn=epn+del_g*normal;
k = L+(2/3)*M;
S=k*trace(strain)*eye(3,3)+DevStrial-2*M*del_g;
% Compute consistent elastoplastic tangent moduli
A=eye(3,3);
B=eye(6,6);
Cp=1/3*kron(A,A)-kron(normal,normal);
end
end
figure
hold on
plot(t,S(1:1,1:1),'linewidth',5,'color','r')
plot(t,S(2:2,2:2),'linewidth',5,'color','r')
plot(t,S(3:3,3:3),'linewidth',5,'color','r')
legend('s11','s22','s33')
plot(t, alfa ,'linewidth',5,'color','r')
xlabel( 'Time', 'FontSize', 16)
ylabel( 'alfa', 'FontSize', 18)
grid on
figure
hold on
plot(t,Cp(1:1,1:1),'linewidth',5,'color','r')
plot(t,Cp(4:4,4:4),'linewidth',5,'color','r')
plot(t,Cp(5:5,5:5),'linewidth',5,'color','r')
legend('s1','s2','s3')

Respuesta aceptada

VBBV
VBBV el 14 de Dic. de 2020
Editada: VBBV el 14 de Dic. de 2020
clear
%%Given :
T= 5;
del_t=3;
t=linspace(0,T,del_T);
E=5;
v=0.3;
Sya=25;
Sy=15;
d=0.4;
vis=0;
gamma=0;
%Assume :
epn=0;
alfa=zeros(size(t));
DevS=0;
M= E/(2*(1+v));
L= (E*v)/[(1+v)*(1-2*v)];
G = norm(DevS)-sqrt(2/3)*Sy+(Sya-Sy)*(1-exp(-d*alfa))-[M*gamma*2];
Gd=-2*M*(1+(Sya-Sy)*d*exp(-d*gamma));
for n = 1:length(t);
strain=6*n*[1 0 0; 0 1 0; 0 0 1]+6*sin(n)*[0 2 0; 2 1 0; 0 0 3/2];
el=strain-(1/3)*trace(strain)*eye(3,3);%
DevStrial=DevS+2*M*(el-epn); %Devatoric Stress Trial
etrial=DevStrial;
y=Sy+(Sya-Sy)*(1-exp(-d*alfa));
ftrial=norm(DevS)-sqrt(2/3)*y;
if ftrial<=0;
DevS=DevStrial;
else
while and((abs((G/Gd)/gamma)) >= 0.001, gamma==0);
%g[dγ(k)]
G = ftrial-[M*gamma*2];
%Dg[dγ(k)]
Gd=-2*M*(1+(Sya-Sy)*d*exp(-d*gamma));
gamma= gamma - G/Gd;
% Derivative of yield limit function: K'(α)
end
del_g= gamma;
normal=etrial/norm(etrial);
alfa(n)=alfa(n)+sqrt(2/3)*del_g;
%Update back stress, plastic strain, and stress
epn=epn+del_g*normal;
k = L+(2/3)*M;
S=k*trace(strain)*eye(3,3)+DevStrial-2*M*del_g;
% Compute consistent elastoplastic tangent moduli
A=eye(3,3);
B=eye(6,6);
Cp=1/3*kron(A,A)-kron(normal,normal);
end
end
% figure plots
figure(1)
plot(t,S(1:3,1:3),'linewidth',5,'color','r')
legend('s11','s22','s33')
hold on
plot(t, alfa ,'linewidth',5,'color','r')
xlabel( 'Time', 'FontSize', 16)
ylabel( 'alfa', 'FontSize', 18)
grid on
figure(2)
plot(t,Cp(1:3,1:3),'linewidth',5,'color','r')
legend('s1','s2','s3')

Más respuestas (1)

Jan
Jan el 14 de Dic. de 2020
I assume, this is an effect or drawing a line with just 1 point, which produces no line.
Try
plot(t, S(1,1), 'linewidth', 5, 'color', 'r', 'Marker', 'o')
  2 comentarios
Steve
Steve el 14 de Dic. de 2020
It still doesn't work, the S function is a matrix 3x3, and I want to plot the value of the first row and first column when t=0 till t=5
Jan
Jan el 14 de Dic. de 2020
Editada: Jan el 14 de Dic. de 2020
The first row is S(1, :), not S(1:1, 1:1).
What about:
plot(t,S,'linewidth',5,'color','r')

Iniciar sesión para comentar.

Categorías

Más información sobre Stress and Strain en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

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

Start Hunting!

Translated by