# Why is local yielding considered the limiting case in context of buckling in Eurocode?

In Eurocode, when use buckling curves to calculate the maximum load that a compressive member can be subjected to before it risks buckling. Buckling curves are based on the Perry-Robertson formula. I'm reading a derivation of the formula here.

The formula is derived basically like this: we assume some initial eccentricity of the column in shape of a sinusoid, and then arrive at this expression for the deflection of the column at the mid-point:

$$y(\frac{L}{2})=\frac{\sigma_E}{\sigma_E-\sigma}a$$

where $$\sigma$$ is the stress compressing the column, $$\sigma_E$$ the Euler critical load of the column and $$a$$ the assumed eccentricity of the column at mid-point.

Then we use an expression for the maximum stress inside the column, summing the direct compressive stress as well as the stress that occurs from the bending moment:

$$\sigma_{max}=\sigma+\frac{Mc}{Ar^2}$$

where $$\sigma_{max}$$ is the stress on the extreme fiber on the mid-point of the column, $$M$$ the moment of the column due to load, $$c$$ the distance from the neutral axis to the extreme fiber, $$A$$ the cross sectional area and $$r$$ the radius of gyration.

Knowing that $$M = Py(\frac{L}{2})$$, we get:

$$\sigma_{max}=\sigma+\sigma\frac{c}{r^2}\frac{\sigma_E}{\sigma_E-\sigma}a$$

It is then said:

At failure the maximum stress is the yield stress, $$\sigma_y$$ .

and we proceed to set $$\sigma_{max}=\sigma_y$$ and solve for the compressive stress $$\sigma$$.

This last step puzzles me. Why do we set the stress to yield stress, if we are interested specifically in designing against buckling? If we design a column in this way, then yes we can limit the compressive load so that the extreme fiber (and thus the entire column) never reaches the yield limit anywhere, but how does this prevent a failure by buckling? We know buckling is a failure occuring from sudden sideways deflection of a member. But buckling can very well occur entirely elastically. This formula seems to only ensure that the member does not yield, but does not seem to say anything about how much the member deflects.

To me, it would make more sense to take the very first equation, the deflection of the midpoint, and set some limits on that, based on the slenderness of the column. So don't we really care about how much the member bends sideways as long as it does not yield, or have I misunderstood something?

One of the most fundamental principles in the eurocodes is the distinction between the serviceability limit state (SLS) and the ultimate limit state (ULS). An element that has large, visible deformations but still is able to support the load is ugly and potentially scary, but not dangerous. Therefore, initial buckling is considered an SLS failure, but not necessarily a ULS failure. The column will not fail completely before the deformation becomes great enough for the maximum stress to reach the yield stress, and that means you need different equations for the SLS check and for the ULS check. In the ULS check, we generally don't care how large the deformations are as long as nothing fails.

• I see. What kind of a SLS check could possibly be performed here? I mean, if the SLS check makes sure the column doesn't suffer any large, visible deformation, isn't it safe enough to pass ULS checks too? – S. Rotos Apr 12 at 20:44
• It's the other way around. The safety factors in ULS are much larger than in SLS (and much larger than the small reserve capacity you find by doing allowing these deformations in most columns), so in practical terms only the ULS check is relevant for typical cases of column buckling. For plate buckling, things can get more complex, though. – ingenørd Apr 13 at 4:48
• But doesn't that contradict what you said in your answer? If you are now saying ULS check is more strict, then it would be ok to allow possibly large, visible deformations on columns (well, as long as they dont yield). – S. Rotos Apr 13 at 8:08
• The ULS check is usually more strict because the safety factors are much larger, especially on the loads. You need a greater safety against collapse than against ugliness. The ratio will depend heavily on load type and national legislations but in ULS the loads are often around 1.5 times the largest SLS loads. – ingenørd Apr 13 at 8:44
• I'm thinking about it this way: high load -> high delflection -> ugly but no collapse. Small load -> small deflection -> pretty structure, just how we like it. We need to satisfy both limit states. Let's say it takes 50kN to produce the high deflection state, almost to the yield point. Then let's say it takes 5kN to buckle it only slightly, just to the point it still satisfies the SLS limit (no large visible deflections). But now the 5kN is the limiting force, and the load needs to be lower than that. So why concern us with the ULS in this case at all, as the SLS limit is much lower? – S. Rotos Apr 13 at 11:48

Euler buckling is a perfectly elastic behavior: if a beam buckles, it takes a sinusoidal shape of arbitrary (possibly infinite) amplitude. If you then remove the applied force, the beam will return to its original, perfectly-straight shape.

Real-world beams, however, don't behave this way: if you buckle a beam, it gets destroyed. After all, Euler buckling's infinite amplitude implies infinite strain (and therefore stress) throughout the column. So a real-world column buckles until it reaches its yield strain/stress, at which point it collapses.

That's why the Perry-Robertson considers the yield stress: while buckling is an elastic behavior, the column's eventual destruction happens at its yield stress (on the extreme fiber).