1
$\begingroup$

I'm reading here on buckling resistance of members according to Eurocode.

Non-dimensional slenderness is defined as:

$$\bar{\lambda}=\frac{\lambda}{\lambda_1}$$

where

$$\lambda_1=\pi \sqrt{\frac{E}{f_y}}$$ $$\lambda = \frac{L_{cr}}{i}$$

The plain $\lambda$ is by definition the slenderness of the column, the ratio of its effective length to its radius of gyration. But what is the meaning of $\lambda_1$? No explanation besides its definition is given in the text. It is some factor by which we can divide the plain lambda or slenderness to make it "non-dimensional", but why does it take the form it does? What meaning does the square root of the ratio of elastic modulus to yield stress multiplied by pi have?

$\endgroup$
1
$\begingroup$

$\lambda_1$ is the minimum slenderness ratio at which buckling will be the dominant condition for a purely axially loaded element. Interestingly, it is an intrinsic property of the material, not the geometry.

It's derivation is simple:

$$\begin{align} P_{crit} &= \dfrac{\pi^2 EI}{L^2} \\ \dfrac{P_{crit}}{A} &= \pi^2 E \cdot \dfrac{I}{L^2A} \\ \sigma_{crit} &= \pi^2 E \cdot \dfrac{i^2}{L^2} \\ \sigma_{crit} \equiv f_y &= \pi^2 E \cdot \dfrac{1}{\lambda_1^2} \\ f_y &= \dfrac{\pi^2 E}{\lambda_1^2} \\ \therefore \lambda_1 &= \pi\sqrt{\dfrac{E}{f_y}} \end{align}$$

From this we can calculate a slenderness ratio for steel and another for aluminium, above which any element made of each material will fail by buckling, and below which they will fail by simple compression.

And that's all that's described by $\bar\lambda$: if it's greater than 1 (the element's slenderness ratio is greater than $\lambda_1$), the element will fail by buckling; if lower, by compression. If exactly equal to 1, it'll do both simultaneously.

In reality, codes usually have "fuzzier" rules for elements with $\bar\lambda \approx 1$, since there can be interactions between both failure states which lead to ugly math.

$\endgroup$
3
  • $\begingroup$ Great answer, thank you! I'm just curious though, why do we use the ideal Euler buckling formula? Looking at buckling curves, we see that the ideal Euler buckling curve gives the value of yield stress at $\bar{\lambda}=1$ as expected, but the non ideal curves a-d do not. So why do we normalize by this Euler critical slenderness when this idea works only for the ideal case, which is one that we are not usually interested in? $\endgroup$
    – S. Rotos
    Dec 18 '20 at 12:46
  • $\begingroup$ You'd have to ask whoever wrote the Eurocode. I assume it is merely a useful conceptual reference point. $\endgroup$
    – Wasabi
    Dec 18 '20 at 14:09
  • 1
    $\begingroup$ The most important parameters in real-world buckling are outside the scope of any theory - for example the tolerances and imperfections in assembling the real-world structure. A theory that is simple, wrong, but (sometimes) useful is better than one that is complicated and equally wrong. $\endgroup$
    – alephzero
    Dec 18 '20 at 14:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.