1. LRFD – Floor beam Unbraced top flange

2. Lateral Torsion Buckling  We have to check if there is plastic failure (yielding) or lateral-torsion buckling.  This depends on the length between the lateral braces, related to the limiting lengths.  L p is the limiting length for plastic failure  L r is the limit length for torsional buckling.  If L b < L p it is plastic failure  If L p < L b < L r we have a different failure criteria  If L b > L r we use the lateral buckling stress criteria

3. Plastic Failure  If L b < L p  M n = M p =  y Z x  Z x is the plastic section modulus about the x axis

4. L p < L b < L r

5. L b > L r  M n =  cr S x ≤ M p

6. The following definitions apply

7. c  For a doubly symmetric I-shape  c=1  For a channel,  Where h 0 = distance between flange centroids

8. Conservative simplifications

9.  A beam of A992 steel with a span of 20 feet supports a stub pipe column with a factored load combination of 55 kips A992 Steel: structural steel, used in US for I-beams. Density = 7.85 g/cm 3. Yield strength = 50 ksi.

10. No flooring – no lateral bracing on top flange Find max moment. Assume beam weighs 50 lbs/ft From distributed load, M max = w L 2 /8 From point load, M max = P L / 4 M max = 55,000 * (20/4) + 50 * (20^2)/8 = 277.5 kip-ft

11. Use trial method Find a beam that has a  M p of at least 277.5 kip-ft Need to check if it will fail in plastic mode (M p ) or from flange rotation (M r ) Tables will show limiting unbraced lengths. L p is full plastic capacity L r is inelastic torsional buckling. If our length is less than L p, use M p. If greater than L r, use M r

12. Selected W Shape Properties – Grade 50 PropW18x35W18x40W21x50W21x62  M p (kip-ft) 249294416540 L p (ft)4.314.494.596.25 L r (ft)11.512.012.516.7  M r (kip-ft) 173205285381 S x (in 3 )57.668.494.5127 I y (in 4 )15.319.124.957.5 h o (in)17.2817.3820.2820.39 r y (in)1.221.271.301.77 J (in 4 )0.5060.811.1457.5 CwCw 1140144025605970

13. W18 x 40 looks promising 294 > 277.5 But, L p = 4.49. Our span is 20 feet. And, L r = 12.0 again, less than 20’  M r = 205, which is too small. W21x50 has L r = 12.5, and  M r = 285. That could work!

14. Nominal flexural design stress  M n =  cr S x  The buckling stress,  cr, is given as

15. Terms in the equation  r ts = effective radius of gyration  h 0 = distance between flange centroids  J = torsional constant (torsional moment of inertia)  C w = warping constant  c = 1.0 for doubly symmetric I-shape

16. Effective radius of gyration

17. So the critical stress is

18. Then the nominal moment is  M n =  cr S x  = 18.77 94.5 = 1,774 kip-in = 147.9 kip-ft  We need 277.5!!  If we had the AISC design manual, they show unbraced moment capabilities of beams.  We would have selected W21x62, which turns out to handle 315.2 kip-ft unbraced.