I am trying to add a symbolic matrix to cell array matrix using for loop but this error is coming: DOUBLE cannot convert the input expression into a double array.Can you please help me to resolve this issue?

Q=sym('Q',[8,1]);
prompt4='specify global external member joint force in [;] form'
Q=[0; 0; Q(3); Q(4); -2400; Q(6); 0 ;9600]

%specify global external member joint force%

Fall{1}=   [-1200;1200;Q(3) - 1200;Q(4) - 1200];
Fall{2}=[ -Q(3);-Q(4);-2400;Q(6)]
Fall{3}=[2400;-Q(6); 0; 9600]

F=zeros(2*jo,1);

for i=1:1:8
    for m=1:1:4
        for t=1:3
            p=c{t};
            if p(m)==i
                F(i)=F(i)+Fall{t}(m)
            end
        end
    end
end

Answer 1

[Walter Roberson]

You have

Q=sym('Q',[2*jo,1]);
Q=[0; 0; Q(3); Q(4); -2400; Q(6); 0 ;9600]

so Q has unresolved symbolic entries (not an error)

Then you have

 qall{i}=T{i}*[Q(m);Q(m+1);Q(m+2);Q(m+3)]

so qall makes reference to those symbolic variables.

Then

      Fall{i}=((f(i)*h)/12)*[6; -h; 6; h]+ qall{i};

so Fall{i} makes reference to those symbolic variables.

You never give a value to those symbolic entries, so by the time of

                 F(i)=F(i)+Fall{t}(m);    %HERE

then those Q3 and so on are still unresolved. That prevents them from being converted into double precision values such as would be needed to add to F(i) which you initialized to double precision zeros.

If you use

F=zeros(2*jo,1,'sym');

then it happens that by the end of the loop that none of those variables will remain in F, but there are intermediate entries in F that use them. I do not know if it can be guaranteed that they will all cancel out.

[Prateek jain]

Thanks Walter, I wanted my final F in symbolic form and using 'sym' works for it.

F=zeros(2*jo,1,'sym');