How can solve x=K\F in MATLAB using stiffness matrix from APDL?

- May 1, 2024 at 11:07 am
  
  dhjang4522
  Subscriber
  
  I have exported a stiffness matrix from modal analysis using the SMAT command and then carried out a static analysis by applying a load on a specific node. Knowing the node where the force was applied and having the symmetric stiffness matrix, I attempted a simple static analysis in MATLAB using X=K\F to calculate displacement. However, the values turned out to be very small across the board.
  
  I addressed the issue of the difference in node IDs between the stiffness matrix and those displayed in Workbench by using MapForward vectors and NOD2BCS to ensure the load is applied to the correct node. Despite this, the results are still discrepant.
  
  I would be grateful if you could let me know where I might have gone wrong!
- May 3, 2024 at 11:17 am
  
  Ashish Khemka
  Forum Moderator
  
  Hi,
  Please see if the following post helps: Update of the stiffness matrix by the Newton Raphson method (ansys.com)
  Regards,
  Ashish Khemka
  - May 4, 2024 at 8:35 am
    
    dhjang4522
    Subscriber
    
    Hi,
    Thank you for your reply. I understand that the Newton-Raphson method is used in nonlinear analysis, but I am currently working on linear static analysis. I solved x=k\F in MATLAB, and while the tendency seems to somewhat match, the values differ significantly. Could you tell me why such discrepancies might occur?