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LRFD – Floor beam

LRFD – Floor beam. Unbraced top flange. 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

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LRFD – Floor beam

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  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. • Lp is the limiting length for plastic failure • Lr is the limit length for torsional buckling. • If Lb < Lp it is plastic failure • If Lp < Lb < Lr we have a different failure criteria • If Lb > Lr we use the lateral buckling stress criteria

  3. Plastic Failure • If Lb < Lp • Mn = Mp = syZx • Zx is the plastic section modulus about the x axis

  4. Lp < Lb < Lr

  5. Lb > Lr • Mn = scrSx ≤ Mp

  6. The following definitions apply

  7. c • For a doubly symmetric I-shape • c=1 • For a channel, • Where h0 = 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/cm3. 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, Mmax = w L2/8 From point load, Mmax = P L / 4 Mmax = 55,000 * (20/4) + 50 * (20^2)/8 = 277.5 kip-ft

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

  12. Selected W Shape Properties – Grade 50

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

  14. Nominal flexural design stress • Mn = scrSx • The buckling stress, scr , is given as

  15. Terms in the equation • rts = effective radius of gyration • h0 = distance between flange centroids • J = torsional constant (torsional moment of inertia) • Cw = 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 • Mn = scrSx • = 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.

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