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Warning: Matrix is singular to working precision. HELP!

Asked by Charlie
on 18 Mar 2013

Hi all, I'm new to the forums. I'm having issues with my code and I have no idea where to go to. My instructor advised I tried the helpdesk and this was the only place where it seemed right to ask. This is my code:

function [Kstruct] = HomeworkCode(E,A,L1,L3,h)
%establish geometry
L2 = sqrt((L3-L1)^2+h^2);
L4 = sqrt(L1^2+h^2);
%Local, trans, global element stiffness matrices
Kprime0 = (E*A/L1)*[1,-1;-1,1];
Trans0 = [cos(0),sin(0),0,0;0,0,cos(0),sin(0)];
K0 = Trans0.'*Kprime0*Trans0
Kprime2 = (E*A/L2)*[1,-1;-1,1];
Trans2 = [(L3-L1)/L2,h/L2,0,0;0,0,(L3-L1)/L2,h/L2];
K2 = Trans2.'*Kprime2*Trans2
Kprime3 = (E*A/L3)*[1,-1;-1,1];
Trans3 = [cos(0),sin(0),0,0;0,0,cos(0),sin(0)];
K3 = Trans3.'*Kprime3*Trans3
Kprime4 = (E*A/L4)*[1,-1;-1,1];
Trans4 = [L1/L4,h/L4,0,0;0,0,L1/L4,h/L4];
K4 = Trans4.'*Kprime4*Trans4
%Local structure stiffness
Kstruct = [K0(3,3)+K2(1,1)+K4(3,3),K0(3,4)+K2(1,2)+K4(3,4),K2(1,3),K2(1,4),K0(3,1),K0(3,2),K4(3,1),K4(3,2);

Everything seems fine for me so far up until I try to take the inverse of the variable Kstruct. I've looked over the lines 5 times and even rewrote it incase it was something I'm just ignoring by sight, but I'm still getting a singular matrix for my Kstruct. Any help on this issue would be greatly appreciated.

  1 Comment

on 19 Mar 2013

the structure stiffness matrix will be singular until you impose boundry conditions


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1 Answer

Answer by Jan Simon
on 19 Mar 2013
Edited by Jan Simon
on 19 Mar 2013

Please note, that it is not possible in theorie to give a suggestion for an improvement based on failing code only. How could we fix an error, when all we have is the code, which produces the error? We even do not know, if there is an error at all, because it could be a valid answer, that the matrix is (near to) singular. Does anything disallow this result?

I'm convinced, that terms like "K2(3,4)+K3(3,4)" attract typos as much as possible. Perhaps something like this would simplifiy the program optically, such that finally the problem becomes visible automatically:



K4(1:2,3:4), K3(1:2,3:4), zeros(2), K4(1:2,1:2)+K3(1:2,1:2)


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