faromic
Structural
- Aug 28, 2007
- 52
Hi,
I have a question about lateral bracing for a beam I'm designing. It's an existing 2 story structure, built 2 years ago. The setup is repetitive. Anyway, they want to remove the 2nd story mezzanine and and 2 interior columns from the interior of the building and hang the roof from a new beam stubbed up from the roof. The span is 75', with 2 point loads (interior columns)acting symmetrically. I sized it as continuously braced, but have questions about the lateral bracing. I am going to attach another W section horizontally to the top flange of the beam. This horizontal section will resist the lateral forces due to Lateral torsional buckling. I'm sizing it for lateral load of 2% * half the reaction of the beam.
The point loads are equal to 35 kips. The reaction of the beam is 35k + .262k*75'/2 = 44.9k. 2% of this is .8985 kips. This is applied per foot so the moment is .8985*75^2/8 = 632 k*ft. I also designed it for L/360 -> Ix required =8823 in^4. I get a W24x250. Does this sound reasonable for this kind of span? Or is it 2% of the reaction from self-weight only? I don't think that's the case, buy it yields a much smaller section, W21x93. I know it's a 75' span so the lateral loads are large but I just want to justify my calc.
Thanks
I have a question about lateral bracing for a beam I'm designing. It's an existing 2 story structure, built 2 years ago. The setup is repetitive. Anyway, they want to remove the 2nd story mezzanine and and 2 interior columns from the interior of the building and hang the roof from a new beam stubbed up from the roof. The span is 75', with 2 point loads (interior columns)acting symmetrically. I sized it as continuously braced, but have questions about the lateral bracing. I am going to attach another W section horizontally to the top flange of the beam. This horizontal section will resist the lateral forces due to Lateral torsional buckling. I'm sizing it for lateral load of 2% * half the reaction of the beam.
The point loads are equal to 35 kips. The reaction of the beam is 35k + .262k*75'/2 = 44.9k. 2% of this is .8985 kips. This is applied per foot so the moment is .8985*75^2/8 = 632 k*ft. I also designed it for L/360 -> Ix required =8823 in^4. I get a W24x250. Does this sound reasonable for this kind of span? Or is it 2% of the reaction from self-weight only? I don't think that's the case, buy it yields a much smaller section, W21x93. I know it's a 75' span so the lateral loads are large but I just want to justify my calc.
Thanks