Dear Murat,

I don't know if you expect to always have N(i)N(i+1), then at the iteration i+1 the variables in your inner loop will have elements in the last positions that are left over from the previous iteration. I would suggest you try to clear yi, Lcr2, and Lcr3 at the end of the outer loop. Costanzo

Dear Murat,

I agree with the above comments. Not good to use I and j as variable names, and arrays used in inner loop, Lcr2 and Lcr3, are  not cleared or emptied before using them, and they may be longer then N(I) in which case the old data at index N(I)+1, ... persists. Thay will included in e.g. sum(Lcr2).

Also, yi does no have to be an array, since yi(j) is only used in the inner loop, so could be just yi (a scalar). Not causing problems, though. And, since only the sums of Lcr2 and Lcr3 are used, it can be simplified to

sumLcr2 = 0;

sumLcr3 = 0;

for k=1:N(i)  %%%%% problem is here %%%%%

    sumLcr2=sumLcr2 + sin(pi*k*b(I)/Btotal(i))^2;

    sumLcr3=sumLcr3 + cos(pi*k*b(I)/Btotal(i))^2;

end

A next step could be to change to a faster vectorized expression, but that is besides the point of the question.

Thak you so much all of you for your kind answers,

The problem has been solved.

Dear Costanzo,

You are rigth the problem is about vector dimensions,  N(i) > N(i+1) in some cases, so new iteration writes  new vector over the previous one. I had missed that point. Thanks again.

Best regards.

Murat 

Dear Murat,

I am glad to hear that was helpful. Good luck with the rest of your work!

Kind Regards, 

Costanzo

Dear Costanzo,

Thank you for your nice wishes. Good luck you too.

Best Regards.

Murat

In addition, the best advantage of matlab is vectorization, which means that you can avoid "for", for example you can calculate m and Lcr1 just by:

m = a./b

Lcr1 = (pi^2*E./(12*(1-poisson^2)))*((tp./Btotal)^2)*(Btotal./a+a./Btotal)^2+((pi^2*E.*e^2)/(a^2))*(be/b);

Be careful with / operator (matrix division) and ./ (element by element division). The same for * and ^ operators.

When the vectors are bigger, the performance is completely different.

Dear Mr. Sepulveda,

Thank you for your nice support. I know that advantage but I have not used that technique yet. If the dimension of vectors is very large, I guess, I have to use it.

Best Regards.

Murat