Percentage of x & y - convergence

[acanfield]

Hello

I found this graph thanks to google (& whoever created it!)

If I'm aiming for 3% convergence ie 3% change in the tip displacement how many extra nodes should of been added (3%? m- surely not)?

Is it a case of adding 50% more nodes or a 100% more?

Thanks in advance

Andrew

[AstroJohnPE]

If I am reading the graph correctly, the hexahedron linear elements have practically converged to within 3%:

change in displacement between last two iterations approximately = 0.001 units/4 = 0.00025 units

0.00025 units/0.008 units displacement * 100 = 3.1% change.

The tetra linear elements are still too far from converging to guess how many nodes are required to reach convergence. It could only require 50 additional nodes -- you just need to add them at the correct location.

But the graph seems to be indicating to use quad elements, which I'm guessing would be the "with midside nodes" in Simulation Mechanical. The quad elements in the graph give accurate results regardless of how many nodes are in the model.

[acanfield]

Hello John

After sleeping I can now answer my own question (I think!).

I didn't create the graph, just borrowed it to illustrate the issue.

My question was - how do know when you have reached a convergence of 3%?

I guess the answer is shown by the flatness of the curve as the curve is alway progressing to a flatline.

I was guessing there was a % change between the result & number of nodes used, there is but it's the curve that shows this.

Thanks

Andrew