Most often when a column is under axial load, we say that it has bucked when we observe a decrease in the applied force (assuming a displacement controlled loading). On the other hand we know that if we consider large deflection analysis of beam-columns under axial load (elastica,Euler-Bernoulli), we see an increase in load as the displacement increases. I want to simulate this phenomenon with the help of a single simple example to enhance my understanding. I have tried with a case attached here but have been unsuccessful. Please note that I may be wrong at some points in seeking such an explanation and I will be genuinely grateful to the who comes forward to explain me where I am wrong.
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.
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
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)
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.
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