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  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

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

L p < L b < L r

L b > L r  M n =  cr S x ≤ M p

The following definitions apply

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

Conservative simplifications

 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.

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 = kip-ft

Use trial method Find a beam that has a  M p of at least 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

Selected W Shape Properties – Grade 50 PropW18x35W18x40W21x50W21x62  M p (kip-ft) L p (ft) L r (ft)  M r (kip-ft) S x (in 3 ) I y (in 4 ) h o (in) r y (in) J (in 4 ) CwCw

W18 x 40 looks promising 294 > But, L p = 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!

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

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

Effective radius of gyration

So the critical stress is

Then the nominal moment is  M n =  cr S x  = = 1,774 kip-in = 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 kip-ft unbraced.