Symbolic Linear system returns wrong solution.
Afficher commentaires plus anciens
I would like to solve the symbolic defined integrals U and V in terms of the variable X, how can I do that?
syms C0 C1 C2 C3 C4 X L EI q;
%__________________________________________________
% POLYNOMIAL 4TH
Y3(X) = C0 + C1*X + C2*X^2 + C3*X^3 + C4*X^4;
% Y3(X) DERIVATIVES:
dY3(X) = diff(Y3(X),X);
d2Y3(X) = diff(dY3(X),X);
% RAYLEIGH RITZ METHOD
U = + (EI/2) * int((d2Y3(X))^2,[0 L]);
V = - q * int(Y3(X),[0 L]);
PI = U + V;
% BOUNDARY CONDITIONS IN Y3(X)
cc1 = Y3(0) == 0;
cc2 = Y3(L) == 0;
cc3 = -EI*d2Y3(L) == 0;
cc4 = -EI*d2Y3 (0) == 0;
% NEW BOUNDARY CONDITION FROM THE RR METHOD
dPIdC4 = diff (PI,C4);
cc5 = dPIdC4 == 0;
% SYSTEM LINEAR SOLVING
R = solve([cc1,cc2,cc3,cc4,cc5],[C0,C1,C2,C3,C4]);
C0 = R.C0;
C1 = R.C1;
C2 = R.C2;
C3 = R.C3;
C4 = R.C4;
disp 'CONSTANTES C'
disp ([C0;C1;C2;C3;C4]);
Y3(X) = C0 + C1*X + C2*X^2 + C3*X^3 + C4*X^4;
disp 'EQUAÇÃO DA LINHA ELÁSTICA';
disp (Y3(X));
disp 'EQUAÇÃO DO MOMENTO';
M3(X) = -EI * diff(diff(Y3(X),X),X);
disp(M3(X));
As it is, the the integral solution for U and V is in terms of X, what is wrong. 'X' should disappear .
2 commentaires
Walter Roberson
le 19 Avr 2021
specify the variable of integration for int()
Tiago Araujo
le 19 Avr 2021
Réponse acceptée
Plus de réponses (0)
Catégories
En savoir plus sur Calculus dans Centre d'aide et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!