bending/shift of neutral axis/nonlinear simulation 2

I assume you're doing a finite element analysis. In that case, if modeled properly the FEA program handles this neutral axis location internally and you don't need to explicitly calculate it.

You mentioned different "strengths" in tension and compression and also different "stress-strain behavior", which means different stiffness. The strength in many applications is modeled using a material failure condition. Having different stiffness in tension and compression is easy for 1D elements, harder but doable in 2D and 3D.

Nagi Elabbasi
Veryst Engineering

have you defined a customised stress/strain curve for your material ? this is implied (in my mind) by being concerned about the NA.

maybe you mean that fty and fcy are different ?

you are right elabbasi, it is internally handled and the results (fem) are very well compared to experiments.
rb1957, I have defined a stress strain curve (NA).

Best Regards,
Sartor

sorry, but what's the question ? ... if the FE is matching the experiment ??

Actually,
When I simulate bending on an element(different geometry than a normal bending coupon), I have maximum principal at bottom (due to tension) and minimum principal at top (due to compression), which is also the case for normal bending coupon.
However, I have different values for Max (60 MPa) and Min (-100MPa) principal. I am trying to understand why it is like this.

At first, I thought that this might be due to the displacement of neutral axis. I have different tension and compression behavior but I do not have any input for my compression!

When I simulate bending with a bending bending coupon, as elabasi suggested, fe-code is capable of arranging it, and simulation results are the same as my experiments.

Do you have any idea why min. principal is higher than max. principal?

Best Regards,
Sartor

are the tension and compression areas different ?

a sketch would help.

is there any axial load, in the FEA or in the real world ??

I did not understand what you mean by tension and compression areas.
Imagine a normal bending case, bottom surface is under tension, and top surface (where I apply the dsplacement)is under compression.
Normal X is almost same as Max Principal.

I am just applying a displacement!

I attached an excel graph, here you can see for the distance 0, i have almost two times higher minimum principal stress than maximum principal.
I am trying to understand why that is like this.

again: Max principal is located at the bottom
Min principla is located at the top (where displacement is introduced during the test)


Best Regards,
Sartor

stress acts on an area to create a load. for a beam in bending, the stresses at different places on the X-section are +ve and -ve to create a couple, that reacts the internal moment with no net (axial) force.

in a rectangular beam in pure bending the upper and lower fibers have the same magnitude stress 'cause the areas they act on are the same.

if your beam has a tapering x-section (like a trapezoid) then the NA will be lower on the section (closer to the fat end) and the stresses on the fat end will be lower than on the narrow end (bigger area, lower stress = the same load as smaller area, bigger stress, therefore only a couple is produced by the internal stresses).

That makes great sense.
Thank you very much for your patience rb1957.

Best Regards,
Sartor