HYSTERISIS LOOPS USING DATA

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Parth Kulkarni
Parth Kulkarni el 20 de Jul. de 2021
Comentada: Parth Kulkarni el 20 de Jul. de 2021
I prepared this code to randomise values of stress(Sigma in a range), Kp and Np to get new strains added to the previous iteration( Operation ) using for loop system. I want to create hysterisis loop of stress vs strain ( Sigma vs epsilon ) for all the iterations but i am not able to figure out the way.
clc;
clear all;
close all;
sgna=150;
sgnb=650;
sgnN=200; %%%%%%
sgnrand=sgna+(sgnb-sgna).*rand(sgnN,1);
sgn=sgnrand;
kpa=950;
kpb=1370;
kpN=1; %%%%
kprand=kpa+(kpb-kpa).*rand(kpN,1);
kp=kprand;
npa=0;
npb=1;
npN=1; %%%%
nprand=npa+(npb-npa).*rand(npN,1);
np=nprand;
E=21000;
ee1=(sgn(1)/(2*E))+((sgn(1)/(2*kp(1)))^(1/np(1)))
ll=0;
mm=0;
nn=0;
rr=0;
for ll=1:sgnN
for mm=1:kpN
for nn=1:npN
rr=rr+1;
if rr==1
ee(rr)=ee1;
else
ee(rr)=(sgn(ll)/(2*E))+((sgn(ll)/(2*kp(mm)))^(1/np(nn)))+ee(rr-1);
end
end
end
end
plot(ee',sgn)
hold on;
scatter(ee',sgn,'.')
hold off
  3 comentarios
Parth Kulkarni
Parth Kulkarni el 20 de Jul. de 2021
It is not forming a curve. I also wanted to know if my code is correct for my application. Basically the equation is ee1=(sgn(1)/(2*E))+((sgn(1)/(2*kp(1)))^(1/np(1))) . where ee1 is the intial strain, sgn1 is intial stress, E is the elastic modulus, kp and np are components of hardening. I wanted to randomise Kp, np and stress ( Sgn) for every iteration which would gove a value of ee(Strain) this genrated value will be added to the next generation. I just hope that I am on the right path.
Parth Kulkarni
Parth Kulkarni el 20 de Jul. de 2021
I also wanted to know if my code is correct for my application. Basically the equation is ee1=(sgn(1)/(2*E))+((sgn(1)/(2*kp(1)))^(1/np(1))) . where ee1 is the intial strain, sgn1 is intial stress, E is the elastic modulus, kp and np are components of hardening. I wanted to randomise Kp, np and stress ( Sgn) for every iteration which would gove a value of ee(Strain) this genrated value will be added to the next generation. I just hope that I am on the right path.

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Respuestas (1)

Tesfaye Girma
Tesfaye Girma el 20 de Jul. de 2021
you can try this code
x = rand(10,1);
y = rand(10,1);
quiver(x(1:end-1),y(1:end-1),diff(x),diff(y),0);

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