## Numerical solution of simple model doesn't match the analytical solution

## Numerical solution of simple model doesn't match the analytical solution

(OP)

Hello everyone,

Yesterday I experimented a little bit with abaqus in order to understand different constraints. I noticed, that the numerical solutions for a simple model are very different to the results that I expected.

I created a VERY short column with an I-Shape cross-section. Now, since the column is so short that buckling doesn't occur, I expect the axial resistance to be N=A*fy, where A is the area of the cross-section and fy is the strength of the material. In this case I used a simple ideally plastic behaviour of a S235 steel (fy=23.5kN/cm²).

The cross-sectional area of my attached model (if the area is duplicated, since only one half is modelled) is around A=744mm²=7.44cm². Therefore the analytical plastic axial force is N=7.44*23.5kN/cm²=175kN. Creating the model with S4 elements leads to the right solution. With the model in the attached input-file I get around 120kN (duplicated Reaction force). So it is far too low. On the other hand I created the whole cross-section with C3D8R elements without using the symmetry plane. Then the axial force was around 200kN. So it was too high. I am honestly very confused. I'm not sure if its because of the boundary conditions or constraints or something else... I just can't figure it out...

It would be great if someone could help me out with this one.

Yesterday I experimented a little bit with abaqus in order to understand different constraints. I noticed, that the numerical solutions for a simple model are very different to the results that I expected.

I created a VERY short column with an I-Shape cross-section. Now, since the column is so short that buckling doesn't occur, I expect the axial resistance to be N=A*fy, where A is the area of the cross-section and fy is the strength of the material. In this case I used a simple ideally plastic behaviour of a S235 steel (fy=23.5kN/cm²).

The cross-sectional area of my attached model (if the area is duplicated, since only one half is modelled) is around A=744mm²=7.44cm². Therefore the analytical plastic axial force is N=7.44*23.5kN/cm²=175kN. Creating the model with S4 elements leads to the right solution. With the model in the attached input-file I get around 120kN (duplicated Reaction force). So it is far too low. On the other hand I created the whole cross-section with C3D8R elements without using the symmetry plane. Then the axial force was around 200kN. So it was too high. I am honestly very confused. I'm not sure if its because of the boundary conditions or constraints or something else... I just can't figure it out...

It would be great if someone could help me out with this one.

## RE: Numerical solution of simple model doesn't match the analytical solution

## RE: Numerical solution of simple model doesn't match the analytical solution

## RE: Numerical solution of simple model doesn't match the analytical solution

## RE: Numerical solution of simple model doesn't match the analytical solution

## RE: Numerical solution of simple model doesn't match the analytical solution

As seen below the total axial reaction (using displacement control, thus enforced displacements on one face, pink arrows in the image), is exactly that and the whole section has yielded at this load with 200 MPa all over the section at yield. Now all general purpose NL FEA packages can predict this using HEX8 or HEX20 elements (no need for reduced integration) in pure axial loading so very easy.

## RE: Numerical solution of simple model doesn't match the analytical solution

## RE: Numerical solution of simple model doesn't match the analytical solution

## RE: Numerical solution of simple model doesn't match the analytical solution

So the boundary condition to use (skip and RBE2 or RBE3 constraints) are (1/4 symmetry): bottom face at mid web height in the image use Y displ. = 0, for the large face at the symmetry of the web fix X=0, and finally the two ends (z= 0 and 100 mm) use on one enforced displacement of Z=-0.15 mm, and Z=0 mm on the other. At the common edges for these faces it is important that the BC are added (not sure exactly what abaqus does), I enforce this since I am applying BC straight on the mesh nodes in Strand7 so I know what the solver sees, and so I know what I am doing exactly. So just follow these steps and it should be fine (mesh is all conencted so not using any tie-like constraints, evenhough that should not cause any problems, and not using RBE2 and RBE3 constraints, just simple/standard BC as explained above).

## RE: Numerical solution of simple model doesn't match the analytical solution

please excuse the late reply. Thank you for your input.

It seems, that the overconstrained nodes were the problem in this case. I didn't expect them to have such a huge impact on the results.

Thanks again!