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An efficient convergence test for the fixed point method

The fixed point method consists to find the solution of F(X)=X.

One can not get fixed with the convergence condition |F'(X)|<1 because if the function has an optimum then |F'(X)|=0 even if the solution is not yet reached.


We introduce an efficient convergence test with the condition:

|Xn+1 - Xn| ≤ epsilon1 And |F(Xn+1)-Xn+1| ≤ epsilon2

Converged, but to the right solution??

Hello eveybody,

I am confused in using numerical schemes: What can convergence of a numerical scheme tell us and what not??? lets assume that a numerical scheme does converge to "A result" or "A number", but how could we be sure that the scheme has actually converged to the "RIGHT solution" and not a WRONG one??

Any input is appreciated!


ABAQUS Reinforced Concrete Beam Model

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Hi All,

I am trying to model a simply supported reinforced concrete beam with longitudinal reinforcement. I am using 8-node solid elements for the concrete, and 2-node truss elements for the longitudinal reinforcement bars. I have embedded the reinforcement bars in the concrete using an Embedded Region constraint. I am then applying a displacement at the midspan (deflection control). I am using the Concrete Damaged Plasticity model for the concrete. My model begins to run but will not converge. Some of the error messages that I get are:

Reg. Convergence and Mesh Density

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I am using a FORTRAN FE code for 4 quad elements with a viscoelastic material and with contact analysis. The routine follows a Newton Raphson iteration scheme and which makes the residual zero for convergence. I am using direct sparse matrix solver from the FORTRAN math library.

Numerical phase modeling of BTO nanostructure using Landis' model with FEAP

  The difference to the common model used is the f,g terms. The 3D FEM formulation has been derived and a new fortran program is written. The derivative matrix is symmetric. But the calculation result does not converge. I have tried every means to get a converging result. But it doesn't work, the residual norm always increases.

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