Why is local yielding considered the limiting case in context of buckling in Eurocode? - Engineering Stack Exchange most recent 30 from engineering.stackexchange.com 2022-01-20T17:30:50Z https://engineering.stackexchange.com/feeds/question/35128 https://creativecommons.org/licenses/by-sa/4.0/rdf https://engineering.stackexchange.com/q/35128 4 Why is local yielding considered the limiting case in context of buckling in Eurocode? S. Rotos https://engineering.stackexchange.com/users/13372 2020-04-12T11:10:26Z 2020-04-13T21:55:58Z <p>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 <a href="http://www.colincaprani.com/files/notes/Column%20Notes.pdf" rel="noreferrer">here.</a></p> <p>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:</p> <p><span class="math-container">$$y(\frac{L}{2})=\frac{\sigma_E}{\sigma_E-\sigma}a$$</span></p> <p>where <span class="math-container">$\sigma$</span> is the stress compressing the column, <span class="math-container">$\sigma_E$</span> the Euler critical load of the column and <span class="math-container">$a$</span> the assumed eccentricity of the column at mid-point.</p> <p>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:</p> <p><span class="math-container">$$\sigma_{max}=\sigma+\frac{Mc}{Ar^2}$$</span></p> <p>where <span class="math-container">$\sigma_{max}$</span> is the stress on the extreme fiber on the mid-point of the column, <span class="math-container">$M$</span> the moment of the column due to load, <span class="math-container">$c$</span> the distance from the neutral axis to the extreme fiber, <span class="math-container">$A$</span> the cross sectional area and <span class="math-container">$r$</span> the radius of gyration.</p> <p>Knowing that <span class="math-container">$M = Py(\frac{L}{2})$</span>, we get:</p> <p><span class="math-container">$$\sigma_{max}=\sigma+\sigma\frac{c}{r^2}\frac{\sigma_E}{\sigma_E-\sigma}a$$</span></p> <p>It is then said:</p> <blockquote> <p>At failure the maximum stress is the yield stress, <span class="math-container">$\sigma_y$</span> . </p> </blockquote> <p>and we proceed to set <span class="math-container">$\sigma_{max}=\sigma_y$</span> and solve for the compressive stress <span class="math-container">$\sigma$</span>.</p> <p><strong>This last step puzzles me.</strong> 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 <em>yield</em>, but does not seem to say anything about how much the member deflects. </p> <p>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. <strong>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?</strong></p> https://engineering.stackexchange.com/questions/35128/-/35131#35131 2 Answer by ingenørd for Why is local yielding considered the limiting case in context of buckling in Eurocode? ingenørd https://engineering.stackexchange.com/users/15064 2020-04-12T16:14:48Z 2020-04-12T16:14:48Z <p>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.</p> https://engineering.stackexchange.com/questions/35128/-/35146#35146 2 Answer by Wasabi for Why is local yielding considered the limiting case in context of buckling in Eurocode? Wasabi https://engineering.stackexchange.com/users/1832 2020-04-13T21:55:58Z 2020-04-13T21:55:58Z <p>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.</p> <p>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.</p> <p>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).</p>