How to combine a matrix that has elements up to n?

So, I have this pre calculate matrixes of A that are 4x4 and they are all positioned to be along the diagonal I then have matrix r that are 4x1, set up in a similar pattern but begin with 2. Now, I have set this matrix to go up to 4 calculated values but I need it to completed this pattern for a number of n=1:ntotal, where ntotal is any number. The BA matrix will always be something like (ntotal*4)X(ntotal*5) while by BR will be something like (ntotal*4)X(ntotal-1).

 clear; clc;
nspan=4;
EI=[200,600,600,200];
L=[10,20,20,10];
%%constants setup
a=zeros(1,nspan);
b=a;c=a; d=a;
for n=1:nspan
a(n)=L(n);
b(n)=L(n)/EI(n);
c(n)=L(n)^2/(2*EI(n));
d(n)=L(n)^3/(6*EI(n));
end
%%Compute A to the n
for n = 1:nspan
    A{n} = [1,0,0,0;a(n),1,0,0;c(n),b(n),1,0;-d(n),-c(n),-a(n),1];
end
for n = 1:nspan
    r{n} = [1;a(n);c(n);-d(n)];
end
I=eye(4);
Z = zeros(4,4);
Z1 = zeros(4,1);
BA=[A{1},-I,Z,Z,Z;...
    Z,A{2},-I,Z,Z;...
    Z,Z,A{3},-I,Z;...
    Z,Z,Z,A{4},-I];
BR=[Z1,Z1,Z1;...
    -r{2},Z1,Z1;...
    Z1,-r{3},Z1;...
    Z1,Z1,-r{4}];
B=[BA,BR];

[Rik]

This code runs without errors for me. What is your actual question?

[N/A]

Well, it runs fine but this is set up for the total of nspans set to 4, it needs to set up to automatically to account for nspans of 3, 4, 5, 6, 7, or up to 20 and more without having to change the matrix BA and BR every time. if I wanted to set up a system for 10 spans, I only would need to change the array L and EI to have 10 values within. The matrix of BR and BA are hard coded for 4 spans as in needs to account for n spans. Thanks for any help.

[N/A]

EI and L and initial conditions. Everything else is based on these values. If EI and L looked like this:

 nspan=5;
EI=[200,600,600,200,200];
L=[10,20,20,10,10];

Then the matrix BA and BR would be like this:

 BA=[A{1},-I,Z,Z,Z,Z;...
    Z,A{2},-I,Z,Z,Z;...
    Z,Z,A{3},-I,Z,Z;...
    Z,Z,Z,A{4},-I,Z;...
    Z,Z,Z,Z,A{5},-I];

Same for something like BR

 BR=[Z1,Z1,Z1,Z1;...
    -r{2},Z1,Z1,Z1;...
    Z1,-r{3},Z1,Z1;...
    Z1,Z1,-r{4},Z1;...
    Z1,Z1,Z1,-r{5}];

Answer 1

[Rik]

This code should do the trick. I also slightly modified some earlier code.

nspan=4;
EI=[200,600,600,200];
L=[10,20,20,10];
%%constants setup
L=L(:)';%force direction
EI=EI(:)';%force direction
a=L;
b=L./(n);
c=L.^2./(2*EI);
d=L.^3./(6*EI);
%%Compute A to the n
A=cell(1,nspan);r=cell(1,nspan);
for n = 1:nspan
    A{n} = [1,0,0,0;a(n),1,0,0;c(n),b(n),1,0;-d(n),-c(n),-a(n),1];
    r{n} = [1;a(n);c(n);-d(n)];
end
I=eye(4);
Z = zeros(4,4);
Z1 = zeros(4,1);

%pre-allocate BA as a cell to make indexing easier
BA=repmat({Z},nspan,nspan+1);
BA((nspan+1)*(1:nspan))={-I};
BA(((nspan+1)*(1:nspan))-nspan)=A;
BA=cell2mat(BA);
% BA=[A{1},-I,Z,Z,Z;...
%     Z,A{2},-I,Z,Z;...
%     Z,Z,A{3},-I,Z;...
%     Z,Z,Z,A{4},-I];

%pre-allocate BA as a cell to make indexing easier
BR=repmat({Z1},nspan,nspan-1);
BR(2:(nspan+1):end)=cellfun(@(x) -x,r(2:end),'UniformOutput',false);
BR=cell2mat(BR);
% BR=[Z1,Z1,Z1;...
%     -r{2},Z1,Z1;...
%     Z1,-r{3},Z1;...
%     Z1,Z1,-r{4}];

B=[BA,BR];