Symbolic Linear system returns wrong solution.

Question

Tiago Araujo 2021-4-19

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/806091-symbolic-linear-system-returns-wrong-solution

回答： Divija Aleti 2021-4-21

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 .

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Walter Roberson 2021-4-19

specify the variable of integration for int()

Tiago Araujo 2021-4-19

在 MATLAB Online 中打开

rr.jpg

Even doing this, I am not getting correct values... I dont know why.

I did:

clear all;clc;
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 = + int((EI/2)*(d2Y3(X))^2,X,[0 L]);
V = - int(q*Y3(X),X,[0 L]);
PI = U + V;
% BOUNDARY CONDITION FROM THE RR METHOD
%dPIdC2 = diff (PI,C2);
%dPIdC3 = diff (PI,C3);
dPIdC4 = diff (PI,C4);
%cc3 = dPIdC2 == 0;
%cc4 = dPIdC3 == 0;
cc5 = dPIdC4 == 0;
% ESSENTIAL BOUNDARY CONDITIONS IN Y3(X)
cc1 = Y3(0) == 0;
cc2 = Y3(L) == 0;
cc3 = -EI*d2Y3(L) == 0;
cc4 = -EI*d2Y3 (0) == 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 'CONSTANTS C'
disp ([C0;C1;C2;C3;C4]);
disp 'PI';
disp(PI);
disp 'METHOD RR - ELASTIC CURVE EQUATION';
Y3(X) = C0 + C1*X + C2*X^2 + C3*X^3 + C4*X^4;
disp (Y3(X));
disp 'ANALYTICAL - ELASTIC CURVE EQUATION';
Yr(X) = - (q*X)/(24*EI) * (X^3 - 2*L*X^2 + L^3);
disp (Yr(X));
disp 'METHOD RR - MOMENT EQUATION';
M3(X)  = -EI * diff(diff(Y3(X),X),X);
disp(M3(X));
disp 'ANALYTICAL - MOMENT EQUATION';
Mr(X) = -EI* diff((diff (Yr,X)),X);
disp(Mr(X));

I should get the results of the attached image, or the same as analytical equations...

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Answer 1

Divija Aleti 2021-4-21

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/806091-symbolic-linear-system-returns-wrong-solution#answer_681095

Hi Tiago,

I understand that you are getting wrong solutions but there is nothing wrong with the working of the code. I suggest you re-check your initial assumption of Y3(X) as a polynomial (try taking a combination of trigonometric functions and later use Taylor series expansion to expand them) or maybe try using different boundary conditions.

Hope this helps!

Regards,

Divija