Roof Beam with Double Cantilever

[BAretired]

Here is an example of a roof beam similar to one I encountered in the last year of my practice. Assuming the top and bottom flanges are laterally supported at points b and c and nowhere else, what is the buckling length of the beam?

BA

 

[nutte]

If the beam is still braced at b and c, where the point loads are located, yes, 16'.

 

[BAretired]

nutte,

I agree that we are talking about lateral torsional buckling but I have always considered that to be similar to Euler buckling, i.e. the compression flange buckles in a sine wave or similar curve from end to end. It is not clear to me why the sine wave cannot fit smoothly when only two points along its length are fixed against translation.

The beam is fixed against rotation about its axis at b and c, but the ends of the cantilevers are free to rotate in a direction opposite to that of the span.

In the case of moment at a and d, no reactions are required at b and c. If these supports are removed, the beam buckles over its entire length. How does its buckled shape differ from that of the supported beam?

BA

 

[nutte]

If you remove the supports at b and c, where is the beam braced against twist and lateral displacement? If it's braced at a and d, then your unbraced length is 28'. This is different than the case originally presented, though. In the original case, the twist at b and c is zero. In this new case, the twist at a and d is zero, and something greater than zero at b and c. That's why these two are different.

 

[BAretired]

nutte,

The hypothetical beam is floating in the air or is weightless and suspended by the top flange at any two points, say b and c. Both flanges are laterally unbraced throughout the entire 28' length.

If equal and opposite moments are applied at a and d, what is the effective length of the beam to be considered for lateral torsional buckling? How does the buckled shape differ from that of the beam laterally supported top and bottom at b and c?

BA

 

[paddingtongreen]

BA, you threw me a curve when you put the axial force (rope) into the mix. I assume in the discussion that followed, Euler buckling means elastic buckling. For me, that is when the beam tries to rotate through about 90 degrees and allow the tension flange to shrink back and the compression flange to stretch back to their original lengths. It is a low energy position.

Given the above, and going back to the beam as diagrammed, the effective length of the span is 16' and the cantilevers get 2x6'= 12'

I may be showing my age here, but 2 was once the multiplier for an unrestrained cantilever.

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[nutte]

We keep talking about different things.

This last scenario, with the beam floating in air, supported at the top flange at b and c, is just like a lifting beam an erector might use. The appropriate buckling length for this scenario would entail a much longer discussion on its own.

I maintain that for the original scenario presented, with a beam braced against twist and lateral displacement at b and c, loaded with either vertical loads or applied moments at the ends, the unbraced length is 16', with little room for discussion.

 

[ishvaaag]

I agree with nutte, within the terms of his last paragraph.

 

[BAretired]

Seems like I am a minority of one.

paddington,

The effective buckling length of a 6' cantilever is 2 x 6' = 12' when the root is fixed. If the root is a rotational spring, it is longer. The greater the flexibility, the greater the length. In the present case, the root's flexibility is determined by the stiffness of the span.

nutte,

It seems to me that the lifting beam analogy is a good one. That is the way I believe the beam as sketched would behave.

There should always be room for discussion but I guess there is no point in beating a dead horse.

BA

 

[BAretired]

If we make the member a truss instead of a beam and consider only the case of equal and opposite end moments, the top chord is in pure tension. The bottom chord is in pure compression from a to d and the web members are unstressed.

If the moments are gradually increased until Euler buckling occurs, the bottom chord will buckle laterally in a half sine wave extending from a to d. The buckling length will be 28', not 16'.

BA

 

[hokie66]

BA,
Change the central span to 2' rather than 16'. Same moment. What will the buckling length be then?

 

[civeng80]

BA

Your replacing the end cantilever with a moment at b and c. Now at b and c the body is pinned laterally so that the effective length of the bottom chord is 16'i.e the distance between the restraints.

If you wish to check the cantiver portion for buckling then the effective length is 12'. So obviously the critical part of the member is between b and c.

 

[ishvaaag]

To make my view even more clear, the lateral and rotational fixity called for nutte are definitive to give the unbraced length as 16 feet. Unbraced. Normally we don't think full rotational restraint proper is a general requirement of bracing, but maybe we must understand so is the case if we ask the compressed flange be restricted against LTB, i.e., tumbling and hence we may surmise for many cases, rotation.

If by whatever the means, stiffness of the column, maybe helped by orthogonal beams at the support points, we may assume the conditions above are met, we are talking of 16' unbraced length. The buckling length may be even less (K=0.5 for full bending fixity on weak axis)

But contrarily, if in spite of good vertical support and lateral restraint, there is no rotational restraint at all, as in the hanged beam later called for by BAretired, the buckling length would become the total length, for as he is assuming, a single curvature LTB shape can form.

Again it is shown the convenience of the modern methods of design that trough proper introduction of material and geometrical nonlinearities, and maybe initial imperfections, free the designer of these issues. Once this is accounted, K is 1 or less for every segment member.

 

[ishvaaag]

The hanged beam wouldn't even have "good" lateral restraint except if we intently so provide. If lacking the lateral restraint it might require a bit of thought to assert definitively that K=1 with the length tip top tip is enough for the buckling length, even if the likely answer is yes because of the likely single curvature buckling shape between tips. But there maybe unsymmetrical tip loads' cases -specially accounting the horizontal action of the spandrels bringing the loads- that require further thought.

 

[BAretired]

hokie,

That's easy. The buckling length would be 6 + 2 + 6 = 14'. The location of the supports are not important because they don't do anything. The buckling length would be the full length of the beam.

civeng80,

Your replacing the end cantilever with a moment at b and c. Now at b and c the body is pinned laterally so that the effective length of the bottom chord is 16'i.e the distance between the restraints.

You are mistaken. A column of length L with axial load applied at each end and laterally supported at two intermediate points has an effective length of L no matter where the intermediate points are.

If you wish to check the cantiver portion for buckling then the effective length is 12'. So obviously the critical part of the member is between b and c.

Again, you are mistaken. As I told paddington earlier, the effective length of a 6' long cantilever is 12' ONLY IF the support is fully fixed. If the support is a rotational hinge, the effective length is greater than 12'.

ishvaag,

I don't understand your argument, so I won't attempt to refute it, but you can rest assured that I believe you to be mistaken.

BA

 

[ishvaaag]

BAretired, in my last posts I point to the fact of KL be the relevant magnitude for the calculation of the buckling stress (LTB as well). I won't say there may not be room for discussion about many things, particularly as to the relevance of the quality of the restraints at bracing points, that maybe not always is stated with clarity enough in the codes, a proof of which is the present discussion.

For example, at AISC 360-05

Lb = length between points that are either braced against lateral displacement of compression flange or braced against twist of the cross section, in.(mm)

So it is clear that it wants the compression flange at bracing points not move outside the web midplane. OK, then take your same brought cantilever in your last post, even fixed at root; what is Lb? at the other end there's no such fixity in all cases. I mean, codes give guide but not always. And so you resource to literature or maybe commentary etc (I haven't looked at) and say, well twice or more the length of the cantilever. But that, even if it is not named so in the code, is a K buckling factor approach, akin to the column fixed at the base and free to displace atop. And it is this Lb (actually an implicit KL) that the code wants as input for the check of the fixed at root cantilever.

Examining the LTB of some compressed flange for which Lb is equal to the length between the physical restraints, the implicit K equals 1 because relative displacement outside the web midplane as per the definition in the code is prevented and then K=1 is applicable. But, if the restriction was enough to provide full fixity at the bracing point in Mx, My and Mz, an implicit favourable K=0.5 could be considered, that, well, the code may not acknowledge by its calibration, method, consistency or intent, but as you can see by my explanation here has as much basis as doubling the actual length of the fixed cantilever for the checks as Lb. It is exactly the same thing, if you can find proper to double the length of the cantilever for Lb, you might be tempted to halve it for full moment fixity for inner segments.

So I maybe may have mislead someone about the straightforward application of some code, but hope not about the intricacies of LTB.

On the other hand I feel not at all wrong in reminding that the method making resource of material and geometrical nonlinearities of the current code (and since LRFD 93 on at least) relieves the designer of much of these considerations, by automatically producing segment solicitations to be always checked at K=1 or less. After the relaxation brought by the P-Deltas and deltas, the nodes are bracing points.

 

[BAretired]

ishvaaag,

To make my view even more clear, the lateral and rotational fixity called for nutte are definitive to give the unbraced length as 16 feet.

I do not agree that the lateral and rotational fixity specified give an unbraced length of 16'. Moreover, I think it is a dangerous assumption.

If by whatever the means, stiffness of the column, maybe helped by orthogonal beams at the support points, we may assume the conditions above are met, we are talking of 16' unbraced length. The buckling length may be even less (K=0.5 for full bending fixity on weak axis)

If joints b and c are fixed against rotation about a vertical axis, then I agree with effective length of 12' for each cantilever and 16' for the span. However, these points are not fixed against rotation in the example given.

But contrarily, if in spite of good vertical support and lateral restraint, there is no rotational restraint at all, as in the hanged beam later called for by BAretired, the buckling length would become the total length, for as he is assuming, a single curvature LTB shape can form.

In this paragraph, you seem to be agreeing with me. Please confirm if I am interpreting this correctly.

BA

 

[BAretired]

ishvaaag,

Sorry, I was busy typing when your latest message came through.

If lateral bracing is located at all four points, a, b, c and d then I agree that 16' would be a conservative estimate of the unbraced length (or the buckling length). It is probably more like 0.75*16 = 12' because of the stiffness of the two cantilever sections.

If the columns at b and c are infinitely stiff in torsion, the K value for the span would be 0.5, and the buckling length would be only 8' for the span and 12' for the cantilevers.

However, if points b and c are the only braced points and the beam is free to rotate about a vertical axis at those points, then for constant moment from a to d, the buckling length should be taken as the full length of the beam.

Perhaps you and I concur on this point.

BA

 

[ishvaaag]

Yes, I concurr that for symmetrical loading the full length would be a good assumption of the length to port to the checks, if the sections free to rotate about vertical axes within the web. But I took (tanslated to spanish) twist in nutte's statement as fixity precisely about such rotation (whereas it refers to torsional restraint, as the code, was wrong in that) and so there was no doubt that in such amovility Lb 16'.

I must point however that even if this seems to me rational, the code seems not particularly well suited to the consideration of braced lengths other than those physical between members (and this includes from the 2 factor for cantilever to any reduction for inner segments of beams between braces). For as you see for the definition given, nutte's statement is exactly what demanded for the code: it has lateral restraint and torsional restraint, in between 16', and hence by literal code, Lb=16'.

When further applying Cb factor for more precise critical moment or moment strength, also, I doubt would be reasonably consistent with the kind of solicitations present in the whole length of the beam plus cantilevers ... and application to subsegments might be adequate but of necessary investigation, for the commentary says "moment diagrams within the unbraced segment", for our interpretation the whole beam tip to tip if free to rotate at column points.

The P-Delta and deltas plus material nonlinearities pursue the deformation to stable limit position if available, and then all issues respect to the restriction to rotation in the axis of the column are cleared. plus the facto of that we are liberated of any estimations of Ks if so we want, (become 1 for every segment modeled) for hand estimation of the adequate K corresponding to the restriction to rotation on its own axis provided by the column may become from risky to adventurous, or require substructure analyses that support any K to factor unto our buckling length.

 

[BAretired]

I am not familiar with other codes, but the CAN/CSA S16-01 (the Canadian standard) seems to assume that the ends of members will be laterally braced and that there may be 1, 2, 3, 4 or more equally spaced braces in between (Article 9.2.6.2). The code does not consider the possibility of a cantilever at one or both ends of the member.

If the member to be braced is in compliance with the code assumption, I do not believe we would be arguing the issue now. I am merely attempting to point out that, in the event of a cantilever at one or both ends of the beam, the code does not apply.

BA

 

[BAretired]

Let me modify that last statement to:

In the event of a cantilever, unbraced at the tip, at one or both ends of a beam, S16-01 bracing requirements do not apply.

BA