# Need to Produce a different matrix after each loop

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DARLINGTON ETAJE on 25 Jun 2021
Hello Friends,
I need to produce a different set of deli after each loop but it remains the same. Help me out please...see code below.
deli is suppose to be different set of 12 by 1 after each loop
L1=[1 33 34 2 35 36 3 37 38 4 39 40];
L2=[3 37 38 4 39 40 5 41 42 6 43 44];
L3=[5 41 42 6 43 44 7 45 46 8 47 48];
L4=[7 45 46 8 47 48 9 49 50 10 51 52];
L5=[9 49 50 10 51 52 11 53 54 12 55 56];
L6=[11 53 54 12 55 56 13 57 58 14 59 60];
L7=[13 57 58 14 59 60 15 61 62 16 63 64];
L8=[15 61 62 16 63 64 17 65 66 18 67 68];
L9=[17 65 66 18 67 68 19 69 70 20 71 72];
L10=[19 69 70 20 71 72 21 22 23 24 25 26];
L11=[21 22 23 24 25 26 27 28 29 30 31 32];
L=[L1;L2;L3;L4;L5;L6;L7;L8;L9;L10;L11];
delu=[-107.9277784;-473.7489066;-456.4464814;-298.1506906;-38.38102414;169.6616599;218.8447121;105.0260144;-78.47051372;-205.707;-1.1961463;701.7499816;-2084.55571;-1229.851404;-0.390010302;39.67805573;-68.88297022;2231.216584;-5473.430599;-3189.887351;-0.390010302;41.85037402;-72.64553587];
delr=zeros(49,1);
del=[delu;delr];
deli=zeros(12,1);
delbar=([]);
mbar=repmat({zeros(12,1)},n,1);
for i=1:n
for p=1:12
deli(p,1)=del((L(i,p)),1);
end
mbar{i}=(kg{i}*deli)+fembar{i};
end
##### 3 CommentsShow 1 older commentHide 1 older comment
Walter Roberson on 26 Jun 2021
It looks to me as if deli is indeed different after each for i loop. You do not ask to record the value for each i value though.
L1=[1 33 34 2 35 36 3 37 38 4 39 40];
L2=[3 37 38 4 39 40 5 41 42 6 43 44];
L3=[5 41 42 6 43 44 7 45 46 8 47 48];
L4=[7 45 46 8 47 48 9 49 50 10 51 52];
L5=[9 49 50 10 51 52 11 53 54 12 55 56];
L6=[11 53 54 12 55 56 13 57 58 14 59 60];
L7=[13 57 58 14 59 60 15 61 62 16 63 64];
L8=[15 61 62 16 63 64 17 65 66 18 67 68];
L9=[17 65 66 18 67 68 19 69 70 20 71 72];
L10=[19 69 70 20 71 72 21 22 23 24 25 26];
L11=[21 22 23 24 25 26 27 28 29 30 31 32];
L=[L1;L2;L3;L4;L5;L6;L7;L8;L9;L10;L11];
delu=[-107.9277784;-473.7489066;-456.4464814;-298.1506906;-38.38102414;169.6616599;218.8447121;105.0260144;-78.47051372;-205.707;-1.1961463;701.7499816;-2084.55571;-1229.851404;-0.390010302;39.67805573;-68.88297022;2231.216584;-5473.430599;-3189.887351;-0.390010302;41.85037402;-72.64553587];
delr=zeros(49,1);
del=[delu;delr];
deli=zeros(12,1);
delbar=([]);
n = 2; kg = {3;5}; fembar = {2; 103};
mbar=repmat({zeros(12,1)},n,1);
for i=1:n
for p=1:12
deli(p,1)=del((L(i,p)),1);
end
deli
mbar{i}=(kg{i}*deli)+fembar{i};
end
deli = 12×1
-107.9278 0 0 -473.7489 0 0 -456.4465 0 0 -298.1507
deli = 12×1
-456.4465 0 0 -298.1507 0 0 -38.3810 0 0 169.6617
DARLINGTON ETAJE on 28 Jun 2021
Thank you so much.

Jayant Gangwar on 7 Jul 2021
It is my understanding that you want to produce and store the set of deli after each iteration. In your current code the deli produced after each set is correct you just need to store it as well after each iteration. In the code below each column of delstore represents the set of deli after the corresponding iteration.
L1=[1 33 34 2 35 36 3 37 38 4 39 40];
L2=[3 37 38 4 39 40 5 41 42 6 43 44];
L3=[5 41 42 6 43 44 7 45 46 8 47 48];
L4=[7 45 46 8 47 48 9 49 50 10 51 52];
L5=[9 49 50 10 51 52 11 53 54 12 55 56];
L6=[11 53 54 12 55 56 13 57 58 14 59 60];
L7=[13 57 58 14 59 60 15 61 62 16 63 64];
L8=[15 61 62 16 63 64 17 65 66 18 67 68];
L9=[17 65 66 18 67 68 19 69 70 20 71 72];
L10=[19 69 70 20 71 72 21 22 23 24 25 26];
L11=[21 22 23 24 25 26 27 28 29 30 31 32];
L=[L1;L2;L3;L4;L5;L6;L7;L8;L9;L10;L11];
delu=[-107.9277784;-473.7489066;-456.4464814;-298.1506906;-38.38102414;169.6616599;218.8447121;105.0260144;-78.47051372;-205.707;-1.1961463;701.7499816;-2084.55571;-1229.851404;-0.390010302;39.67805573;-68.88297022;2231.216584;-5473.430599;-3189.887351;-0.390010302;41.85037402;-72.64553587];
delr=zeros(49,1);
del=[delu;delr];
deli=zeros(12,1);
delbar=([]);
delstore=[];
n = 2; kg = {3;5}; fembar = {2; 103};
mbar=repmat({zeros(12,1)},n,1);
for i=1:n
for p=1:12
deli(p,1)=del((L(i,p)),1);
end
delstore=[delstore,deli];
mbar{i}=(kg{i}*deli)+fembar{i};
end
DARLINGTON ETAJE on 7 Jul 2021
Thank you