## bending coefficient "Cb" - W-beams with unbraced length > Lu

## bending coefficient "Cb" - W-beams with unbraced length > Lu

(OP)

does anyone understand this coefficient very well?

I have put lots of time into trying to figure out what the "AISC ASD Manual" means in its description of the coefficient.

In a simply supported beam with a point load in the middle, I need to know if I should use Cb=1.75 or 1.32 or 1.0. I checked three different sources and they all gave me different values. These different values are linearly proportional to the allowable stress in the beam.

The explination in section F1-3 says that when the bending moment at any point within an unbraced length is larger than the moments at the ends of this length, Cb=1.0

this would lead one to believe that all simply supported beams should use Cb=1.0, however I found sources that say otherwise.

Please help me with this

I have put lots of time into trying to figure out what the "AISC ASD Manual" means in its description of the coefficient.

In a simply supported beam with a point load in the middle, I need to know if I should use Cb=1.75 or 1.32 or 1.0. I checked three different sources and they all gave me different values. These different values are linearly proportional to the allowable stress in the beam.

The explination in section F1-3 says that when the bending moment at any point within an unbraced length is larger than the moments at the ends of this length, Cb=1.0

this would lead one to believe that all simply supported beams should use Cb=1.0, however I found sources that say otherwise.

Please help me with this

## RE: bending coefficient "Cb" - W-beams with unbraced length > Lu

However, if you have SOME kind of intermittent lateral bracing, you have portions of the compression flange along the beam that is unbraced. Each of these portions have a length, L that is used in the calculation of unbraced length.

So let's say you have a point load in the middle of the span and that point load has lateral restraint capability. You then have two half-span lengths of beam that are unbraced. Assuming a symmetrical arrangement, you would look at only half the span with zero moment at the end and you maximum moment at the other end (at the midspan). You then design your beam for the maximum M with a Cb calculated based on M1 = 0 and M2 = M(max). This leaves Cb = 1.75 in the ASD.

If you have other braces creating various and different lengths of beam unbraced between, you have to check each and every section of unbraced length for the maximum moment within that unbraced length.

Hope this helps.

## RE: bending coefficient "Cb" - W-beams with unbraced length > Lu

I also just talked to an expert structural engineer, and he informed me that when analizing a trolley beam used for lifting, Lu can be assumed almost double due to the bottom load not generating the beams tendancy to laterally buckle.

## RE: bending coefficient "Cb" - W-beams with unbraced length > Lu

## RE: bending coefficient "Cb" - W-beams with unbraced length > Lu

Have a W12x16 beam, 50'-0" lg. The ends are supported by top plated columns. There is no load on the beam except the weight from the beam itself(assume no lateral force for example). Question is this: Will the beam fail in bending due to the unbraced length of 50'-0"?

bjm