Hello!
I have the follwoing code and the result i want is in the end. So, im coming in the for loop and ask if a certain value exist. If then that exist i want to assign crit(:,i)=crit_1 but here is the thing; I want also crit(:,i+1)=crit_1 but the problem now is when im coming to the top of the for loop again and i=2, my code replaces column number two with my new condition which i dont want. Now i bascially want to start at i=3 so it replaces the two next columns. Hope this makes sense, thank you!
% Testing
clc
Lam_1=[40 9.8 2.8 0.3]; Lam_2=[210 20 6 0.3]; Lam_3=[136 10 5.2 0.3]; Lam_4=[151 9.4 4.8 0.31]; % Materials [E1 E2 G12 v12] Lam_5=[75 6 2 0.34]; Lam_6=[147 9 3.3 0.31]; Lam_7=[181 10.3 7.17 0.28];
crit_1=[1100 20 600 140 70]'; crit_2=[1400 80 2800 280 120]'; crit_3=[1800 40 1200 220 80]'; crit_4=[2260 50 1200 190 100]'; crit_5=[1400 30 280 140 60]'; crit_6=[2260 50 1200 190 100]'; crit_7=[1500 40 1500 246 68]';
Layup=[Lam_1 0 0.127; Lam_2 pi/2 0.127; Lam_3 0 0.127]; % Can change rowsizes
sigma_local =ones(3,6); % This one contains a bunch of other values usaually
crit=zeros(5,size(sigma_local,2));
for i=1:size(Layup,1)
if Layup(i,1) == 40
crit(1:5,i)=crit_1;
end
if Layup(i,1) == 210
crit(1:5,i)=crit_2;
end
if Layup(i,1) == 136
crit(1:5,i)=crit_3;
end
if Layup(i,1) == 151
crit(1:5,i)=crit_4;
end
if Layup(i,1) == 75
crit(1:5,i)=crit_5;
end
if Layup(i,1) == 147
crit(1:5,i)=crit_6;
end
if Layup(i,1) == 181
crit(1:5,i)=crit_7;
end
end
% Below is the desired output matrix i want, hence this one
crit=[crit_1 crit_1 crit_2 crit_2 crit_3 crit_3];
