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What is the critical buckling force needed to be applied on a system of made out of two parts?

The parts of the system are as depicted in the picture:

  1. incompressible elastic beam - on top
  2. compressible support beam - on the bottom

Upon applying enough compression force the top beam will buckle and the bottom will compress. I managed to follow worked examples for calculating the Euler buckling of the top beam leading to $P_{cr} = \frac{π^2\kappa}{l^2}$ where $\kappa$ is the bending rigidity and $l$

is the compressed length of the bottom beam and the projection of the top beam on the x-axis. The bottom beam is modeled as a spring using Hooke's law $F = -k(l_0-l)$

I am having problem connecting all the pieces of the system together to be able to find the two behaviors where under the critical load the system is flat and above it it undergoes compression and buckling. I would like to also know the energy of the system but that should be relatively easy to obtain from the force by integration.

enter image description here

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    $\begingroup$ I suspect you mean the two bars are connected somehow and the force is commonly applied to both bars? If yes then the question is how are they connected together? What is the distance between them. Imho more details are required $\endgroup$
    – NMech
    Commented Jan 31, 2023 at 15:08
  • $\begingroup$ Yes, they are connected. I am a chemist, but I understood the way the top part is pinned. The pins are connected to both beams. Shouldn't there be a general solution with the distance between the beams as a parameter? $\endgroup$ Commented Jan 31, 2023 at 15:23
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    $\begingroup$ You need to work out the stiffness of the buckling member. As a column approaches elastic buckling it becomes less stiff, and so will take a smaller proportion of the total force applied. Sadly I cannot remember how to do this, 40 years ago it was the topic of my final year project. $\endgroup$ Commented Jan 31, 2023 at 20:27

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Assuming that the support to which they are attached cannot rotate or deflect other than laterally, the buckling limit of the system equals the buckling limit of the system when the weaker bar buckles.

There is no incompressible elastic material. Actually, there is no rigid material, only a convenient way of solving problems!

Therefore when the first bar has buckled the entire load $F$ has to be supported by the remaining member and that member already could not stop the deflection under force $F$ even with the assistance of the pre-buckling resistance of the failed member.

But if the support can deflect or bend, the system has to be assessed as a whole. The share of lower bar in carrying $F $ is

$$F\_l=F*A_l*E_l/ (A_lE_l+A_{top}E{top})$$ So we can calculate back $F$ from this.

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