error empty sym: 0-by-1

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Mattia Penasa
Mattia Penasa 2021-12-28
编辑： Walter Roberson 2024-2-29
%it seems that the system is ultra-defined, but it is a system with 17 equation and 17 variables
syms ux1 ux2 ux3 ux4 ux5 ux6 uy1 uy2 uy3 uy4 uy5 uy6 fx1 fx2 fx3 fx4 fx5 fx6 fy1 fy2 fy3 fy4 fy5 fy6 eta xi Ep x U(x) Eps
E=1000;
ni=0.3;
H=2;
P=1000;
L=6;
Xe1=[0,0, L/2,0, L/2,H];
Xe2=[0,0, L/2,H, 0,H];
Xe3=[L/2,0, L,0, L,H];
Xe4=[L/2,0, L,H, L/2,H];
Dlte1=[0,0,ux2,uy2,ux5,uy5];
Fe1=[fx1,fy1,fx2,fy2,fx5,fy5];
Dlte2=[0,0,ux5,uy5,0,uy6];
Fe2=[fx1,fy1,fx5,fy5,fx6,fy6];
Dlte3=[ux2,uy2,ux3,uy3,ux4,uy4];
Fe3=[fx2, fy2, P/2, 0, P/2, 0];
Dlte4=[ux2,uy2,ux4,uy4,ux5,uy5];
Fe4=[fx2, fy2,P/2,0,fx5,fy5];
[Ke1,Ne1,Be1,D1]=TL1funz(E,ni,h,Xe1); %TL1funz is a function which is correct and gives 4 matrix
[Ke2,Ne2,Be2,D2]=TL1funz(E,ni,h,Xe2);
[Ke3,Ne3,Be3,D3]=TL1funz(E,ni,h,Xe3);
[Ke4,Ne4,Be4,D4]=TL1funz(E,ni,h,Xe4);
A1=Ke1*Dlte1';
A2=Ke2*Dlte2';
A3=Ke3*Dlte3';
A4=Ke4*Dlte4';
A11=A1(1);
A12=A1(2);
A13=A1(3);
A14=A1(4);
A15=A1(5);
A16=A1(6);
A21=A2(1);
A22=A2(2);
A23=A2(3);
A24=A2(4);
A25=A2(5);
A26=A2(6);
A31=A3(1);
A32=A3(2);
A33=A3(3);
A34=A3(4);
A35=A3(5);
A36=A3(6);
A41=A4(1);
A42=A4(2);
A43=A4(3);
A44=A4(4);
A45=A4(5);
A46=A4(6);
[Ux2,Ux3,Ux4,Ux5,Uy2,Uy3,Uy4,Uy5,Uy6,Fx1,Fx2,Fx5,Fx6,Fy1,Fy2,Fy5,Fy6]=solve(A13==Fe1(3),A14==Fe1(4),A15==Fe1(5),A16==Fe1(6),A21==Fe2(1),A22==Fe2(2),A24==Fe2(4),A25==Fe2(5),A26==Fe2(6),A33==Fe3(3),A34==Fe3(4),A35==Fe3(5),A36==Fe3(6),A41==Fe4(1),A42==Fe4(2),A43==Fe4(3),A44==Fe4(4), ux2, ux3, ux4, ux5, uy2, uy3, uy4, uy5, uy6, fx1, fx2, fx5, fx6, fy1, fy2, fy5, fy6)

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回答（1 个）

sai charan sampara
sai charan sampara 2024-2-29
Hello Mattia,
I understand that you are getting an “empty sym” when trying to solve a system of equations.
For a system of equations of “n” variables to have a solution there should be “n” linearly independent equations. In your case there are 17 variables, so you need 17 independent equations. Even though there are 17 equations corresponding to 17 variables, the absence of a solution suggests the presence of two or more dependent equations within the generated set. It can be thought of as solving for point of intersection of two parallel lines in 2D space. There are 2 variables, 2 equations but there exists no solution.
If the matrices “Ke1”, “Ke2”, “Ke3”, “Ke4” are replaced with random valued 6x6 matrices, it can be observed that there exists a solution for the system of equations. This suggest that the matrices “Ke1”, “Ke2”, “Ke3”, “Ke4” are generating some linearly dependent equations. Ensuring that this does not happen by correcting these matrices might give a solution to the system of equations.

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