%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 件のコメント

Kyle McLaughlin
Kyle McLaughlin 2020 年 10 月 28 日
using those verification lines of code and deleting the first for loop gives me what I want and looks like this:
K =
1 1 1 0 0 0 0 0 0
1 1 1 0 0 0 0 0 0
1 1 2 1 1 0 0 0 0
0 0 1 1 1 0 0 0 0
0 0 1 1 2 1 1 0 0
0 0 0 0 1 1 1 0 0
0 0 0 0 1 1 2 1 1
0 0 0 0 0 0 1 1 1
0 0 0 0 0 0 1 1 1
madhan ravi
madhan ravi 2020 年 10 月 28 日
Why did you delete your question?
Kyle McLaughlin
Kyle McLaughlin 2020 年 10 月 28 日
I deleted it to post it correctly so there would be no further confusion
Kyle McLaughlin
Kyle McLaughlin 2020 年 10 月 29 日
編集済み: Kyle McLaughlin 2020 年 10 月 29 日
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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 採用された回答

Kyle McLaughlin
Kyle McLaughlin 2020 年 10 月 29 日

0 投票

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 件のコメント

madhan ravi
madhan ravi 2020 年 10 月 29 日
編集済み: madhan ravi 2020 年 10 月 29 日
Please use the formatting tools. Last time when I edited your code you deleted your question, so edit it yourself appropriately.
Kyle McLaughlin
Kyle McLaughlin 2020 年 10 月 29 日
Excuse me, clearly i didnt see any answer you posted which is why it was deleted. Sorry you didnt answer my question and get an accepted answer. I am aware of how it works now, dont beat a dead horse.
madhan ravi
madhan ravi 2020 年 10 月 29 日
Lol, good luck

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その他の回答 (1 件)

Matt J
Matt J 2020 年 10 月 28 日
編集済み: Matt J 2020 年 10 月 28 日

1 投票

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 件のコメント

Kyle McLaughlin
Kyle McLaughlin 2020 年 10 月 28 日
I am trying to avoid them being equal with the nested for loop in the larger one, additionally I begin the larger for loop with i =2 and begin the nested loop with j =1, so shouldnt they never be the same? how do I make it such that they are not the same value at the same time if this is not the case
Matt J
Matt J 2020 年 10 月 29 日
I'm not sure what you are trying to achieve. If the idea is just to make tiled copies of k, then youcould just use repmat
K=repmat(k,elements,elements)
Kyle McLaughlin
Kyle McLaughlin 2020 年 10 月 29 日
編集済み: Kyle McLaughlin 2020 年 10 月 29 日
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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