AISC ASD L3.1 beam depth limit

[tonyuk]

I would appreciate a little help on the interpretation of L3.1.1 (ASD Green book p.5-181). In simple terms for floors it states that the depth of fully stressed beams should be a minimum of L.Fy/800 and if the actual depth is less than this the "unit stress in bending should be decreased in the same ratio as the depth is decreased from that recommended above".

No problems with a short span beam: if (say) Fb is 24, the formula gives 15" and your section is 10" then you must not stress it more than 24 x 10/15 =16ksi.

If though the beam is unrestrained with a longer span the various formulae already reduce Fb to take account of this. So it would appear to be double counting to get a stress of (say) 12ksi and then have to reduce this by a further 10/15. Logically it would seem to me that in my hypothetical example if you get an Fb of 18 you have to chop it down to 16, but not if Fb is already 15ksi.

Crawley & Dillon example 1 p.117 has a 30ft beam with intermediate restraints at 10ft so Lb (10ft) is more than Lc and less than Lu so Fb is 0.6Fy and it is this figure they reduce (by 14/16.2). If the restraints were every 6ft then Fb would be 0.66Fy and this figure would then be being reduced. To me it seems more logical to limit the stress to the lesser of 0.66Fy(d1/d2) or (in this case) 0.6Fy, not 0.6Fy(d1/d2)

Comments gratefully received

TonyB UK

 

[par060]

I'm not so sure you use Fb in place of Fy...they are 2 different things

 

[Grizzman]

I think the "fully stressed" part is what needs your attention.
Rarely do I find that the allowable bending moment is the same as the actual bending moment. If it is not, then the beam is not fully stressed.

 

[tonyuk]

Well yes, it is the meaning of "fully stressed" that I am wrestling with. My initial reading of it was that it meant that fb was more or less equal (for a compact section) to 0.66Fy, i.e. any greater load would (notionally) cause a failure in bending. Where Fb is much lower so as to prevent lateral buckling failure (e.g. Fb=16ksi, fb=16ksi) is the member "fully stressed" within the meaning of this clause?

 

[JAE]

I don't think you are "double counting" it at all. The limits given in ASD for maximum allowable stress in the beam, Fb, is dealing with the level of safety in the beam regarding STRENGTH - responding to the possibility of lateral torsional buckling. This is NOT a servicability issue here, but the Fb formulae are only dealing with LTB and strength.

Then....if you go beyond their suggested Fy/800 limit for depth, they are suggesting that, in terms of SERVICABILITY, you will have deflection issues (not strength issues). So they suggest further limiting your fb to a lower limit of Fb (true depth/recommended depth).

They are dealing with two different things here so its not double counting at all.

 

[Grizzman]

I think if fb is equal to Fb, then it is considered fully stressed. Since the paragraph doesn't clarify fully stressed as being the maximum possible stress(Fb=0.66Fy), then when fb=Fb, the beam is fully stressed at that state.
I think the idea is to keep you from getting too "springy" a beam.

 

[JAE]

Yes, I'd agree with Grizzman on that point.