# Can a thicker model have a lower critical buckling load?

I perform a linear buckling analysis for two similar geometries, where one model has thickness t = 0.01 m, and the other has thickness t = 0.035 m. Otherwise everything is exactly the same. The thinner model (t = 0.01m) however gives me a higher critical buckling load compared to thicker model (t = 0.035m)- could this be right? The slope in the linear region or Young's Modulus for both models seem to correspond well with previous studies, it is just that I'm getting lower buckling load as I increase the thickness.  Edit: I had to increase the Dimensions of the Krylov space from 0 to 300 to capture any buckling modes for the t = 0.035 m model.

• How is the Critical Load Factor defined? It seems to me that the thinner has smaller values (0.04), while the thicker is about 2.4.
– NMech
Jun 25, 2021 at 18:55
• What is the number below the critical load factor represents? It is a negative number for the thicker case, and positive for the thiner case. And how the program select the factor from the results of the 3 modes?
– r13
Jun 26, 2021 at 0:25
• In Comsol critical load factor is the number by which you have to multiply your loads to reach buckling. Jun 26, 2021 at 1:02
• @Materialscientist actually, itseems that the critical load factor is the inverse of that. see the excerpt of the documentation in my updated answer.
– NMech
Jun 26, 2021 at 14:00

Based on what you are presenting and the definition of critical load factor in the COMSOL documentation I don't know why you think that the model with 0.01[m] exhibits higher critical buckling load.

In the documentation it is stated that:

COMSOL reports a critical load factor, which is the value of λ at which the structure becomes unstable. The level of the initial load used is immaterial since a linear problem is solved. If the initial load actually was larger than the buckling load, then the critical value of λ is smaller than 1.

I am assumming that in both analysis you've used the same load parameters (e.g. 1000 N). So (my revised) interpretation of the value $$\lambda$$ is the load multiplier that you need to divide the applied load in order to obtain the critical buckling load.

• in the case of t=0.01m, where $$\lambda = 2.4$$ the load already exceeds the critical load. I.e. critical load = 1000N/2.4 ~ 400N
• in the case of t=0.035m, where $$\lambda \approx 0.04$$ the load is lower than the critical load. I.e. critical load = 1000N/0.04= 25000N
• The load applied is the same for both cases unfortunately, seems like something must be wrong Jun 25, 2021 at 19:05
• It is as I expected. I've made an update to make clear my point.
– NMech
Jun 25, 2021 at 19:18
• Youre right in your explanation, but t=0.01m gives λ=2.4, and t=0.035m gives λ=0.04 in the pictures above, which is opposite of what one would expect. I think this might be a solver issue. Jun 26, 2021 at 1:00
• @Materialscientist "The load applied is the same for both cases". Ok, as the applied force is constant, F = Fcr1*CLF1 = Fcr2*CLF2 (CLF = critical load factor), so for t=0.035, Fcr1 = F/0.04, and for t=0.01, Fcr2 = F/2.46, thus the critical buckling force Fcr1 (t=0.035) > Fcr2 (t = 0.01). Bingo, if the intepretation of "critical load factor" is correct.
– r13
Jun 26, 2021 at 1:44
• @Materialscientist I read the documentation hastily yesterday. The following line in the documentation stated just the opposite of what I initially assumed (I've updated the answer and included the lambda interpretation from the documentation).
– NMech
Jun 26, 2021 at 3:44

No. Something is wrong.

The Euler critical buckling load equation is $$Pcr = \pi^2 EI/(KL)^2$$. For everything else holding the same of the two models, the only variable is the second moment of inertia ($$I$$).

$$I = th^3/12$$

$$I_{t=0.01}/I_{t=0.035} = 0.01/0.035 \approx 0.29$$, or $$I_{t=0.01} \approx 0.29I_{t=0.035}$$

Look back on the Euler equation, a smaller "$$I$$" will yield a smaller critical buckling load. Since $$I_{t=0.01} < I_{t=0.035}$$, the resulting buckling load for the thinner material (t = 0.01) is less than that for the thicker material (t = 0.035).

• For a complex 3D structure like this, the Euler formula is irrelevant. The second picture shows that the buckling deflection is completely different from the assumption in Euler's approximation - it shows two buckled ribs with one edge approximately fixed along the whole length of the "column". Jun 25, 2021 at 19:33
• @alephzero I disagree. We can simplify the entire structure as a cantilever, or a simply supported column that consisting of multi-cells, then the Euler equation/estimate still applies. Of course, there are advanced buckling analysis formulations, but the concept remains the same. You shall provide your reference to invalidate the concept otherwise.
– r13
Jun 25, 2021 at 19:56
• The Euler buckling formula only applies to slender columns. In this case, the structs can be approximated to be slender but not the entire structure. It is impossible to guess what the effective behavior of the structure will be without some sort of renormalization over individual cells and then groups of cells. Jun 26, 2021 at 11:47
• @BiswajitBanerjee Please provide examples or references that invalidate the concept.
– r13
Jun 26, 2021 at 11:52
• The Euler formula applies to the buckling of beams. Since the definition of what makes a beam seems to be the point of contention, let me point you to Section 4.3 of research-collection.ethz.ch/bitstream/handle/20.500.11850/90522/…, a relatively recent work. Jun 26, 2021 at 12:21