Hello
I made a 10 mm thick drilled sheet metal on solidworks and I wanted to test it with Simulation Xpress.
So I entered my values and started my calculation.
My safety factor was then 1.26, so my part was OK.
However, I tried to change the mesh from normal (fine and coarse) to fine. Now, my coefficient goes to 0.73... So it's no longer good.
My question then: what should be entered as a mesh value?!
I was 2 fingers away from validating my part but finally, I have a big doubt...
Thank you!
The mesh affects the accuracy of the calculation.
So if you want something specific, set your mesh to the finest
Yes, okay, but I didn't think it played so much!
So basically, if I stay on a "middle" mesh, my calculation is wrong?
And if I only want correct calculations, I should always set the mesh to the finest?
If so, who would take the risk of having false results?!
The coarse mesh makes your calculation lighter.
In an assembly for example, where you have parts under force constraints or others, and other pieces that do not play or can in the simulation.
You adjust the parts that will be subjected to a bending force or other in a fine mesh, and the rest that will not move or almost not in a coarse mesh.
It's true that it's quite a troubling difference in value.
In addition, I would have tended to think the opposite: the bigger the mesh, the more the stresses are concentrated on the nodes and therefore the part resists less.
Hello
We would need some additional information to be really sure (a view of the result for example) but it is not illogical to have this type of phenomenon.
It may be counterintuitive, but in areas where deformations are large, the finer the mesh, the higher the calculated stresses will be.
In fact, it must be understood that finite element calculus determines displacements and then deduces the stresses (via Young's modulus). However, if an element is smaller, its deformation can be greater than if it is "smoothed" over a larger area. The calculated stress will then be greater locally.
In areas where the deformations are regular, the difference in mesh will change almost nothing in the result because the average deformations will be equivalent. A coarser mesh is therefore sufficient.
What we ask of the calculation engineers is precisely to know how to say whether this phenomenon is significant or can be neglected.
To get an idea, you can try changing your display (provided you can do it with SW Xpress) so that the areas where the stress exceeds the result of the first calculation are displayed in red. If the extent of the area does not vary too much, you should be able to consider that it is not significant. If not, it may be that your initial mesh was really too coarse.
A colleague from the "calculus" had explained to me that in fact there are at least 3 meshes in the smallest dimension but, in the case of a sheet metal, having 10 in the thickness did not bring anything significant in precision but on the other hand the calculation time increases exponentially.
Since then, when I evaluate a sheet metal part, I start the calculation with the default parameters and eventually (often the mesh crashes for sheet metal), I modify by taking a minimum value of mesh corresponding to 1/3 or 1/4 of the thickness and I keep the maximum default value.
You also have to pay attention to the layout of the maximum stress, sometimes it's just a mesh that shows a large value (often near a hole, a cut) that is not significant (in absolute value). This is where the "calculus" experiment comes in.
If you can, send an image of your room with the constraints.
Side note: personally, I would not validate a critical part with Xpress. I only use it to have an order of magnitude of the stresses/strains and to test the impact of a design modification.