What is wrong with my for loop?

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%check for stitching global nxn matrix
elements = 4;
nodes = 5; %nodes
s=3; %size of matrix nxn
k=[1 1 1; 1 1 1; 1 1 1];
size = (s* elements) - (elements-1);
K = zeros(size,size);
K(1:s,1:s) = k(1:s,1:s);
%this fills the matrix with values
for i =2:elements-1
% K((s:((i*s)-(i-1))), (s:((i*s)-(i-1)))) = k(1:s,1:s);
for j = 1:elements-1
K((j*s-j-1):(i*s-i-1), (j*s-j-1):(i*s-i-1)) = k(1:s,1:s);
end
% K((s:((2*s)-(2-1))), (s:((2*s)-(2-1)))) = k(1:s,1:s);
% K((s:((3*s)-(3-1))), (s:((3*s)-(3-1)))) = k(1:s,1:s);
%K(s:((j*s)-(j-1)),s:((j*s)-(j-1))) = k(1:s,1:s);
% K(s:5,s:5) = k(1:s,1:s);
% K(5:7,5:7) = k(1:s,1:s);
% K(7:9,7:9) = k(1:s,1:s);
end
%this adds matrix elements at stitching location
for i = 1:elements-1
K(((i*s)-(i-1)), ((i*s)-(i-1))) = k(1,1)+k(s,s);
end
K
the lines including K(s:5,s:5) to K(7:9,7:9) was my verification to see if the matrix was right and it is. the code above in the for loops should give me the same answers as these but I keep getting an error saying unable to perform assignment because the size of the left matrix doesnt match the right
this is the error I recieve:
Unable to perform assignment because the size of the left side is 1-by-1 and the size of the right side is
3-by-3.
  4 Comments
Kyle McLaughlin
Kyle McLaughlin on 29 Oct 2020
Edited: Kyle McLaughlin on 29 Oct 2020
I was able to come up with the following code which is closer to what I want but doesnt quite fit:
elements = 4;
nodes = 5; %nodes
s=3; %size of matrix nxn
k=[1 1 1; 1 1 1; 1 1 1];
size = (s* elements) - (elements-1);
K = zeros(size,size);
%K(1:s,1:s) = k(1:s,1:s);
for j = 2:elements
for ii = 0: elements -2
%K(j*s-s+ii*(s-1):j*s-1+ii*(s-1), j*s-s+ii*(s-1):j*s-1+ii*(s-1)) = k(1:s,1:s);
%K(j*s-s-1:j*s-1, j*s-s-1:j*s-1) = k(1:s,1:s);
K(j*s-s-ii:j*s-1-ii,j*s-s-ii:j*s-1-ii)= k(1:s,1:s);
end
end
%K(size-s+1:size,size-s+1:size) = k(1:s,1:s);
for i = 1:elements-1
K(((i*s)-(i-1)), ((i*s)-(i-1))) = k(1,1)+k(s,s);
end
K
K =
1 1 1 0 0 0 0 0 0 0 0
1 1 1 1 0 0 0 0 0 0 0
1 1 2 1 1 0 0 0 0 0 0
0 1 1 1 1 1 0 0 0 0 0
0 0 1 1 2 1 1 0 0 0 0
0 0 0 1 1 1 1 1 0 0 0
0 0 0 0 1 1 2 1 1 0 0
0 0 0 0 0 1 1 1 1 1 0
0 0 0 0 0 0 1 1 1 1 1
0 0 0 0 0 0 0 1 1 1 1
0 0 0 0 0 0 0 0 1 1 1

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Accepted Answer

Kyle McLaughlin
Kyle McLaughlin on 29 Oct 2020
elements = 4;
s=3; %size of matrix nxn
k=[1 1 1; 1 1 1; 1 1 1];
length = (s* elements) - (elements-1);
K = zeros(length,length);
for j = 0:elements-1
index =(1:s);
n = index + j*(s-1);
K(n,n) = k(1:s,1:s);
end
for i = 1:elements-1
K(((i*s)-(i-1)), ((i*s)-(i-1))) = k(1,1)+k(s,s);
end
K
This does exactly what I want I answered my own question nevermind. Here it is for reference.
  3 Comments

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More Answers (1)

Matt J
Matt J on 28 Oct 2020
Edited: Matt J on 28 Oct 2020
It seems clear that when i=j, the left hand side of
K((j*s-j-1):(i*s-i-1), (j*s-j-1):(i*s-i-1)) = k(1:s,1:s);
will be 1x1 whereas the right hand side will always be 3x3. Generally speaking, the size of the left hand side is changing in a highly i,j-dependent way whereas the right hand side is not.
  3 Comments
Kyle McLaughlin
Kyle McLaughlin on 29 Oct 2020
Edited: Kyle McLaughlin on 29 Oct 2020
This is the begining of an FEA script, the code that I am writing here assembles the global stiffness matrix from the element matricies. I need to apply this to the mass matrix and damping matrix for nxn size so that it is modular in the event of different sized element matrixies for given element types. The goal is to add the first and lass matrix values which creates the nodal connection between the two elements and apply that to the lot of them. The first comment I made in the post, directly under my question, shows the desired results of the matrix.

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