Laterally unsupported length of compression flange Lb 1

[GraemeL]

In Section F1 of the Allowable Stress Design manual, Lb is specified as the laterally unsupported length of the compression flange.

One of the weaknesses I see in the AISC is that I cannot find any guidance on what this length actually is in situations more complicated than a simply supported beam.

For instance, in the beam diagrams and formulas for various static loads, pages 2-296ff, using Example 15, "Beam Fixed at Both Ends-Uniformly Distributed Loads" as an example and assuming that there are not intermediate flange braces between the two supports:

For the bottom flange:
Is Lb = 0.2113*L (i.e. the length away from the support and the middle part of the flange that is tension provides the restraint to the part that is in compression) or is Lb = L (the length between actual physical restraints)?

Similarly for the top flange:
Is Lb = (1-2*0.2113)*L (i.e. the length of the top flange that is in compression) or is Lb = L?

I would appreciate hearing what others believe to be the correct interpretation and also any reference within AISC or elsewhere that justifies their belief.

 

[krd]

Unbraced length means distance between centers of gravity of bracing members. If your Lb is greater than Lc then go to section F3 where you can use the effects of reverse curvature if necessary.

 

[GraemeL]

Thank you for your reply. I have some clarifying questions regarding it.

The only location I can find that gives a definition of braced length that resembles your definition is in Appendix F for equation A-F7-1 which is to do with the design of tapered members. Lb is not defined in the Symbols section, p5-201ff. Therefore have you taken this A-F7-1 definition to be the AISC's intention for Lb?

If your Lb > Lc, the AISC leads you to Section F1.3 rather than Section F3 as you state, is Section F1.3 what you meant? Section F3 deals with the bending of box members, rectangular tubes and circular tubes.

Using Section F1.3, the effects of reversed curvature are used in the derivation of Cb. In the example I gave above, the M1 = M2 and there are two points of contraflexure along the beam. M1 and M2 are not of the same sign, one is anti-clockwise and the other clockwise. Is M1/M2 positive or negative in a case where there are two points of contraflexure between M1 and M2? I would have thought it positive.

In practice, the end moments on "fixed" beams are usually not greater than the sagging moments in the middle when the analysis work is done. The supporting columns or beams rotate and cannot provide wl^2/12. Hence Cb = 1.0 if there are no flange restraints between the supports and one gets no benefit from the reverse curvature.

I was trying to see whether it is common practice to equate Lb as the distance between physical restraints to a flange or whether there is a justification for saying that if between two physical restraints to a flange, the flange is not always in compression, one could take Lb as being less than the distance between the physical restraints.

If I understand your reply, you believe you cannot and that Lb can only = the distance between the physical restraints and use Cb to try and increase the allowable Fb using equations F1-6, F1-7 & F1-8. Have I got you correct?

 

[JAE]

GraemeL:

For years there has been a lot of confusion in AISC (in my opinion) concerning the definition of "unbraced length" in beams where there is more complexity than a simple span where the compression flange is not always on top along the length. If the beam in question is supporting perpendicular beams, then the Lb is simply/always, the distance between the connecting beams.

If the beam supports steel joists on the top chord only, then you have a more complex problem.

In one office where I worked some years ago, we had a policy of designing continuous steel beams by taking the unbraced length of the top compression flange as the distance between intersecting joists - which is the expected concept.

For the bottom flange where it is in compression near the columns, we would take the distance between the column and the inflection point as the Lb and use Cb = 1.0 just to be conservative. Sometimes we would even add 30% to the Lb just for good measure.

At an AISC seminar in Texas in the late 80's, Dr. Joe Yura (Univ. of Texas) stated that as long as Cb=1.0 this method was probably OK. However, since that time, AISC has altered the Cb equation and I have heard through various sources that the inflection point should not be used as a brace point.

The correct method, I believe, is to take the full span of the beam as Lb, and compute the Cb from the newer equation, ignoring any inflection point. If you have a problem with the design at this point, then simply adding an angle brace to the bottom chords can establish a shorter Lb and a smaller beam.

 

[krd]

JAE, interesting reply - feel free to comment on mine:
Graemel, yes I meant F1.3. The only way I make sense of F1.3 saying "moments with opposite signs = single curvature bending" is clockwise +ve, counterclockwise -ve at each end of a freebody of the beam section.

I would say that for your symetrical fixed end beam with no bracing of the top compression flange at midspan, M1/M2 doesn't apply and Cb = 1.0 because the mid span moment is larger than the moments at either end of the section of beam that has top flange compression. I wouldn't compare to the the fixed end moments. If the stress at mid span is greater than allowable then you need to brace.

If you brace the top compression flange at midspan, now there is no place in the unbraced length where the moment is larger than at the braced ends so M1/M2 figures. On the freebody of the beam bewteen the fixed end and midspan brace points the moments at each end are either both clockwise or both counterclockwise so they "have the same sign" and M1/M2 is +ve (reverse curavature bending). +ve M1/M2 gives a larger Cb and thus larger allowable stress than -ve M1/M2. So here's where you seem to get the benefit of additional support as the flange transitions into tension.

 

[atklinger]

Graemel,

The basic in Beam Diagrams tell you 0.2113*L (Case 15) is the place when M=0. Nothing about Lc or Lb.

You have to know about the compact section criteria.

Table B5.1 would be very helpful.
Chapter F is next.
Look at equation (F1-2)
and according with the compact section criteria you can go ahead.
Compact Section? F1.1 Fb=0.66*Fy
Non Compact Section? F1.2 or F1.3 Alberto