Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 1.919170e-17.

Question

Wyatt Gagnon 2023-4-30

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1955314-warning-matrix-is-close-to-singular-or-badly-scaled-results-may-be-inaccurate-rcond-1-919170e

回答： Neeraj 2023-5-16

I need help, when I run I get the error message in the title.

E=210e9; %in Pascal
t=5;     %in mm
v=0.3;   %poissons dimensionless
C=(E/(1-(v.^2)))*[1 v 0; v 1 0; 0 0 ((1-v)/2)];
ix1=0;
iy1=0;
jx1=500;
jy1=0;
mx1=250;
my1=125;
%solve beta and gamma values
bi1=jy1-my1;
bj1=my1-iy1;
bm1=iy1-jy1;
gi1=mx1-jx1;
gj1=ix1-mx1;
gm1=jx1-ix1;
%find area in mm^2 using (1/2)b*h
A1= 0.5*125*500;
B1=(1/(2*A1))*[bi1, 0, bj1, 0, bm1, 0;
              0, gi1, 0, gj1, 0, gm1;
              gi1 bi1 gj1 bj1 gm1 bm1];
Bt1=B1.'; %transpose of B
k1=(t*A1)*(Bt1)*(C)*(B1) %stiffness matrix
F=[(40000*sind(30)); -(40000*cosd(30)); 0; 0] %force on the node
k1_d=k1(3:6,3:6) %adjust matrix for boundary conditions and correct size
d=k1_d\F %solve displacement

2 个评论
显示无隐藏无

Rik 2023-4-30

It is not an error, it is a warning that you might not be able to trust the result. If the condition is very small, there is a high chance of variability in the result. That is not a Matlab issue but is caused by the underlying mathematics.

John D'Errico 2023-4-30

The matrix k_1d has rank 3. It was formed as a product of rank 3 matrices, but it has size 4x4.

You CANNOT solve for the displacement. What you are doing, or exactly what the equations mean is impossible to know, so I cannot diagnose what you have done wrong. If I had to guess, it is that you are mssing a boundary condition in some form.

Essentially, there is a fundamental problem in the equations you have formulated, so that no solution will be possible as you have written it. What you did wrong, we cannot know at this point.

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Neeraj 2023-5-16

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1955314-warning-matrix-is-close-to-singular-or-badly-scaled-results-may-be-inaccurate-rcond-1-919170e#answer_1236494

在 MATLAB Online 中打开

Hello,

I understand that you are getting the warning: Matrix is close to singular or badly scaled. Results may be inaccurate.

The warning message is clear and it occurs for the line

d=k1_d\F %solve displacement

This occurs because k1_d is ill-conditioned and there is no inverse of it. This is equivalent to division by zero. There is no ‘help’ for this case, but the inputs do not allow this operation.