The AISC 360-10 Specification for Structural Steel Buildings gives provisions for calculating the maximum unbraced length of a compression flange that separates yielding moment from lateral torsional buckling (LTB). This formula is (AISC 360-10, Eqn. F2-5):

$$ L_p = 1.76r_y\sqrt{\frac{E}{F_y}} $$


$L_p =$ limiting length that separates full yield moment and LTB
$r_y =$ radius of gyration about the $y$-axis
$E =$ Young's modulus
$F_y =$ yield strength of material

Assuming that one is using regular structural steel, the Young's modulus of the material is assumed to be the same regardless of steel grade.

This equation works out such that a steel with a lower yield strength may actually be braced at a lesser interval than one with a higher yield strength. In other words, given the same beam size, the material with the higher yield strength buckles first.

I have also found this to be applicable to design using the ASME Boiler & Pressure Vessel code, specifically Division III, Subsection NF for supports. With temperature effects on yield strength and Young's modulus taken into account, it's possible that a member at an elevated temperature may buckle at a longer length than one at room temperature.

This seems counter-intuitive to me. Why would a weaker material exhibit less LTB action with the same given length?


2 Answers 2


As discussed in the earlier answer, if we look at the curve of moment capacity versus unbraced length we see three regions of behavior -- yielding, inelastic LTB, and elastic LTB (see Fig. C-F1.1 in the AISC Steel Construction Manual). It is important to note that we only have inelastic LTB due to residual stresses. This is where that $0.7F_yS_x$ term is coming from (residual stresses are assumed to be $0.3F_y$). It is also important to note that the equation for critical stress at elastic LTB is in the form $\alpha + \sqrt{1+\beta}$ and is not a function of yield stress. Alpha is a term for out-of-plane buckling of the compression flange and beta is a term for torsional stiffness.

Moment Capacity

So, conceptually, we could look at a curve that disregards residual stresses - meaning we have just yielding and elastic LTB. When we increase $F_y$, the elastic LTB curve remains the same while $M_p$ increases. The result is that we transition to elastic LTB at a smaller unbraced length. One way to think about it is that with an increased $F_y$, it takes more force to yield the member, making it more likely that it will buckle before yielding.

Moment Comparison

  • $\begingroup$ This is a good explanation - I like the hand-drawn figures! I'll give this one the check mark since you brought in discussion on inelastic LTB, which I totally forgot about. Thanks for the Answer. $\endgroup$
    – grfrazee
    Aug 25, 2015 at 14:42
  • $\begingroup$ I left out inelastic LTB from my answer because i thought it would only make the discussion more complex than it needed to be. This question needs only be answered with one sentence which was stated at the end: with an increased yield strength, it takes more force to yield the member, making it more likely that it will buckle before yielding (and I thought I addressed that in my answer haha). $\endgroup$
    – pauloz1890
    Aug 25, 2015 at 14:55

Slenderness ($\lambda = L/r$) is the ratio of a member's length to its smallest radius of gyration. It should make sense that:

  • The less slender a member, the more its plastic strength needs to be considered rather than its Euler (buckling) strength.
  • The more slender a member, the more its Euler (buckling) strength needs to be considered rather than its plastic strength.

In other words as slenderness increases there becomes a point where critical buckling stress becomes the limiting factor rather than plastic yield strength ($F_y$). The maximum allowed compressive strength is the minimum of yield strength and buckling strength. This is illustrated in the diagram below:

Euler Strength and Buckling Strength

$$ \lambda = L_p/r_y = 1.76\sqrt{\frac{E}{F_y}} $$

The formula you have provided has separated yielding moment from lateral torsional buckling (LTB) as you stated. This would be the slenderness point where the critical strength changes from the plastic strength to the Euler strength. If $F_y$ increases, then this point on the x-axis would move to the left. This means slenderness $\lambda$ would be smaller and hence the length of the member (or length between bracing points), $L$ should be smaller.

Slenderness decreases if yield strength increases

Looking at the formula it does seem counter-intuitive. But what you have to remember is that it's either going to fail due to plastic yielding or LTB. And so at higher yield strengths, the buckling strength falls below the yield strength at lower slenderness (smaller member length) than lower yield strengths.

Hope that helps.

  • 2
    $\begingroup$ Just to add to the point, what the original equation is stating is not that the maximum load of a higher-yield section is lower than that of a weaker section. It is merely defining the point where the failure mode changes. And since buckling is not affected by yield-strength (since by definition the section never reaches that stress level), the $\lambda$ where buckling is the controlling factor is inversely proportional to the yield strength. A higher-yield section will, however, always support a greater or equal load than a lower-yield one. $\endgroup$
    – Wasabi
    Aug 18, 2015 at 11:19
  • $\begingroup$ While I understand that the point $L_p$ is really just the point where the equations for $F_y$ and the line of the Euler buckling meet, I don't really think that it explains why a stronger material initiates buckling sooner. Looks like I need to read about the phenomenon a little further. $\endgroup$
    – grfrazee
    Aug 18, 2015 at 13:20
  • $\begingroup$ Like I said, I understand the math, just not why it works out the way it does. $\endgroup$
    – grfrazee
    Aug 18, 2015 at 14:04
  • $\begingroup$ Yeah it seemed counter-intuitive to me too. But if you think of it in terms of what the limiting factor is, it makes sense that for a higher yield strength $F_y$, it won't fail due to plastic yielding, rather it will fail due to buckling instead - which is why $L_p$ would get smaller. It is difficult to put into words. Sorry I removed my last comment - couldn't edit it and it wasn't quite what I was trying to say ;P $\endgroup$
    – pauloz1890
    Aug 18, 2015 at 14:09
  • 1
    $\begingroup$ @grfrazee - You're thinking about it the wrong way (or you were, I think you might understand better from cablestay's answer). The stronger material does not initiate buckling sooner. It initiates buckling at the same load. But it initiates yielding at a higher load. Or try thinking about it this way: let's say you've designed a beam for yielding with 100% utilisation, ignoring buckling. You then remember that you need to check it for buckling. This formula tells you the maximum unbraced length, and the higher your yield is, the bigger the moment was, and hence the shorter unbraced length. $\endgroup$
    – AndyT
    Aug 25, 2015 at 14:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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