In Euler's bending theory, it's stated that the critical stress of a beam is always larger than the yield stress of the beam. I don't really undertstand the difference between them. Critical stress is defined as the stress that the maximum stress applied before the beam starts to buckle.

Does the yield stress here mean the stress which the beam start to deform plastically? Why shouldn't the yield stress be smaller than the critical stress?

When the object is subjected to stress, the beam will deform plastically before it starts to bend and break. So I think that yield stress is smaller than the critical stress.


I was going to put this as a comment but it felt like more of an answer.

Buckling is a fairly unique failure mode. When buckling, the compression force doesn't have to be as high as the yield strength. If a beam is slender enough, the compressive force can cause substantial bending in the beam.

This bending goes beyond the elastic range and causes the beam to fail, even though the applied force usually wouldn't make it fail according to direct equations.

The biggest reason that critical stress is lower than yield stress is that it is an assumption for Euler buckling. The equation might not work in the plastic range.

  • $\begingroup$ do you mean The equation might not work outside the plastic range ?. $\endgroup$ – kelvinmacks Nov 29 '16 at 12:03
  • $\begingroup$ under what circumstances critical stress is lower than yield stress? can you provide example and explain ? $\endgroup$ – kelvinmacks Nov 29 '16 at 12:07
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    $\begingroup$ No, I mean the equation might not work outside of the elastic range. The equation was formulated for a beam that is loaded in compression below its yielding stress. If you use it for a beam that is beyond its yielding stress, there's no indication that it would be accurate. Critical stress is lower than yield stress any time you want to use this equation. If critical stress is higher than yield stress, the beam you are using isn't as slender, therefore buckling is not the only concern, since your load already goes outside of the elastic region. $\endgroup$ – JMac Nov 29 '16 at 12:24
  • $\begingroup$ if the load already goes outside of the elastic region, the member will break by crushing instead of buckling? $\endgroup$ – kelvinmacks Nov 29 '16 at 12:53
  • $\begingroup$ I'm not sure if crushing would be the right word. It would introduce permanent deformations in the beam, which would effect its strength and possibly orientation depending on the application. It could still buckle, but it would fail regardless of buckling. $\endgroup$ – JMac Nov 29 '16 at 13:14

I think you mean Euler Buckling Theory right? Buckling is a different mode of failure to yielding. The plastic strength of a member will usually be constant as it depends on the material. However the "buckling strength or Euler Strength" of a member depends on its slenderness as well as the material.

The image below illustrates this very well.

Euler Strength and Buckling Strength


Your statement

In Euler's bending theory, it's stated that the critical stress of a beam is always larger than the yield stress of the beam

seems just wrong, or you misunderstood the source you got it from.

Euler buckling has nothing to do with the yield stress. The theory assumes the material behavior is linear elastic, at least up to the load at which the bucking failure starts. (Euler's theory of buckling says nothing about what happens after it has started).

You can easily demonstrate that beam buckling can be purely elastic. Get a steel engineer's ruler (the longer the better, preferably 0.5 or 1 meter), or a thin wooden dowel or a sheet of thin plywood at least a meter long. Put one end on a flat surface and apply a load to the other end with your hands until it starts to buckle (i.e. bow outwards). Remove the load, and it will return to its initial straight configuration.

Of course if you continue to apply the load after it starts to buckle, it will continue to bow outwards until the material yields or fractures, but that post-buckling behaviour is not considered by Euler's theory.

  • $\begingroup$ Or an even easier thing to buckle: spaghetti! $\endgroup$ – Wasabi Nov 30 '16 at 11:48
  • $\begingroup$ so , after only the object buckle , then the object will yield and break ? $\endgroup$ – kelvinmacks Nov 30 '16 at 13:34

Buckling and yielding are completely different concepts.

Buckling occurs when an elastic structure does not move back to its initial position as long as the load is applied. Buckling is a purely elastic phenomenon. When elastic material is subjected to a small load, the deformation is usually small as well. This is different with buckling. Buckling requires a significant load in one direction, but then a very, very small load in a perpendicular direction can lead to destructive deformation (depending on the amount of deformation allowed by the system). Buckling is a stability problem, and the sample geometry is essential.

Yielding occurs when the behavior of the material itself changes (due to the high load). When a material yields, the relative position of the atoms change.

Edit: To realise the 'very, very small load' in perpendicular direction it is usually sufficient to apply the large load not perfectly in line with the stability axis. Since buckling is a stability problem, the large load leads to an instable system in the first place. In order to actually deform, some initial deformation in perpendicular direction is required which is then amplified.

I also slightly changed the first sentence of the buckling paragraph to account for Wasabi's comment.

  • $\begingroup$ Buckling require 2 loads applied in different direction ? $\endgroup$ – kelvinmacks Nov 30 '16 at 12:00
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    $\begingroup$ The first sentence of this answer is incorrect. Buckling is elastic: so long as it is restricted so as not to lead to excessively large deformations, a buckled element will return to its initial position once the load is released. Also, buckling does not require two loads. A single axial load that is applied an infinitesimal distance away from the element's centroid will cause it to buckle due to the infinitesimal bending moment created. This bending moment will cause an infinitesimal lateral deflection of the element, which due to second-order effects will be arbitrarily increased. $\endgroup$ – Wasabi Dec 1 '16 at 12:46
  • $\begingroup$ @Wasabi, I agree that my first sentence on buckling was badly formulated. However, since buckling is a stability problem, some force that causes the instable system to move away from the equilibrium point is required. You can call that as you like, but usually this movement will be caused by the same force that causes instability, because usually that force will not be applied exactly along the axis where the system is strong. $\endgroup$ – Robin Dec 5 '16 at 7:38
  • $\begingroup$ @Robin There's no extra force that causes an unstable system to move away from equilibrium. In Euler's buckling theory the force is applied to what should be a "stable" system. It's driven unstable by the fact that you can't realistically continuously apply a load to the exact centre of an object. There's always some uncertainty and variation over time (external factors mixed with the fact you'll never load something exactly on centre). Eulers Buckling theory is useful because you can always assume there is a non-ideal loading. $\endgroup$ – JMac Dec 5 '16 at 11:10
  • $\begingroup$ @JMac It's driven unstable by the fact that you can't realistically continuously apply a load to the exact centre of an object. This is exactly what I wrote. $\endgroup$ – Robin Dec 6 '16 at 12:07

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