# Buckling of Parallel rods

I am trying to calculate critical force $$P$$, if two rods (same diameter and length - rigid fiberglass sandwiched on two steel plates) are parallel to each other.

$$P = \frac{\pi^2 EI}{(KL)^2}$$

I believe the configuration of both rods pinned would be $$K=1$$ (as shown in figure below - B). I am a bit confused if this configuration would behave as springs in this scenario? How would you calculate the the critical force besides using FEA?

If we consider the pins on the top and bottom brackets free to rotate the system will buckle randomly either way, to right or left. With the first buckling column changing the geometry of the system and sparing the other one from buckling.

Unless the width, $$\theta \geq L/10$$, or whatever short column index for this material, the $$K=1$$.

However, The effective force will be $$1/2P$$ at the start of bifurcation of the column buckling.

EDIT

After OP's comment.

Long slender columns behave even more clearly the way I explained; as we increase load P the columns are sharing equally half of the load, then at exactly the point where P, reaches the critical buckling load, randomly and explosively one of the columns fails and becomes the pathway for the force P. And due to the freedom of the pin connection the top or bottom header or both rotate and the system becomes a mechanism, collapsing in a nonrecoverable buckle.

It is notable that even for a system of 3 or more columns the collapse always starts from an end column and then sometimes progresses to the next one down the line. I have observed similar situations in damage to the soft-story buildings in the Northridge earthquake of 1994 when I was preparing Seismic damage estimate reports for the owners of the buildings.

• What if these were really long rods with the width less than L/10? Oct 21 '20 at 17:58
• @kamran would it be possible for you to share a reference with the short column index for a material? Oct 22 '20 at 15:45
• @NMech, this is from one of my excel sheets from a long time ago. -"Short columns: Steel (A36) columns with slenderness ratios ℓ/r ≤ 40 are defined as “short columns”. • The mode of failure is crushing." -And I am kind of positive for Doug Fur columns it used to be L/r<10. Oct 22 '20 at 17:29
• @NMech, but later they changed it and fused short and intermediate together at kl/r<121, I think. Oct 22 '20 at 17:39
• I happen to be doing some work on buckling optimisation, and I realised that the ratio between material and moment of area, has something to do. However, its not in most literature. And I wonder how much knowledge has been forgotten or lost... Oct 22 '20 at 18:02

I believe that the columns will buckle in the direction out of plane.

The total value of the moment of inertia will be $$I_{total}= 2\frac{\pi d^4}{64}= \frac{\pi d^4}{32}$$

Regarding the K value, it depends how the rods are fixed. Theoretically, if they are welded/bonded then $$K = 0.5$$, while if they are free to rotate it should be $$K=1$$.

The behavior of such a system isn't necessarily well-described by simple buckling.

As other answers have mentioned, the most important thing is how the rods are connected to the horizontal plates:

If the connections allow for small rotations, then buckling theory is valid and $$K=1$$, as per your image.

If not, then it depends on the materials used. This is because this case will mean the rods won't be simply resisting an axial force of $$P/2$$. They'll also have to resist a bending moment due to the force being eccentric to their axes. The magnitude of this bending moment will be a function of the stiffness of the vertical rods vs. the horizontal plates: if the plates are orders of magnitude stiffer, then the bending moment will be insignificant; if not, it'll need to be taken into consideration and you're out of the realm of simple buckling.

In this particular case, you state the rods are fiberglass and the plates are steel. This will likely fall into the category of "trivial bending moment", in which case you can still use simple buckling, but adopting $$K=0.5$$ (as per your image) since we're talking about the scenario where the connections are highly fixed.