The example I was tought to show stiffening after buckling was the example that you use, but with the addition of a higher order term to rotational spring:

Mspring=C*phi + C_NL * phi^3.

I did not look in detail at your mathematical derivation, but the example was studied before, so should be able to be found on the internet.

Thank You Stephan, may be the key to the problem is in including the non-linear term in my example?

It is also easier to leave the shortining out in the beginning, in that case you have only one D.o.F.: phi.

Your equilibrium equation becoms P*sin(phi)*L=C*phi+C_NL*phi^3, which can be used to describe your post buckling path. Depending on C_NL/C. you have stable post buckling solutions or only unstable solutions. (phi

very true

After re-looking at old work I have to reiterate my previous remark,

You do not need a non linear spring stiffness. The origianal equation gives a stable post buckling curve. With a negative C_NL it is possible to create an unstabe post buckling curve.

I have applied the load aligned with the axis of the bar, do u suppose its better if I consider the load to be acting at an angle with respect to the bar?

Honestly I do not want to remove the axial deformation response from my analysis, that way I think we will miss out the reduction in load when bending starts to dominate, what do u say?

I tried, by using F as force which keeps its original direction:

F*sin(phi)*(L-delta)=c*phi

F*cos(phi)=k*delta

Which can be solved for each -pi/4

First of all I thank you very much Stephan for your efforts to make me understand this. I have the following observations:

1. Your delta is actually the axial contraction of the bar. My delta is the vertically downward displacement of the dead load P (your F). Hence in the beginning when bending do not start, I expect them to be equal.

2.Please show the plots with grids on, or only 2d curves for a few theta's (your phi)

This would be them.

Thanks again.

we use a beam under bending load to proof the euler classic buckling fomula for a slender culomn

If you consider the boundary conditions &

column slimming limit

ofcourse you can do that for other problems.