ee web compression 9asd0

huangyhg

tee web compression (asd)
for your case use qs equal to 1, and fb equal to 0.6*fy
visitor (visitor)27 mar 02 16:54
even with, say, a span of 58' (hypothetically)!?  fb=.6fy*(qs=1)?  this doesn't seem right unless stringent lateral bracing criteria are met for the stem (76bf/(square root of fy), maybe?).  and, well, if that's the case, why even bother looking at it as an inverted tee, just examine it as a plate.  seems awfully conservative to design as a plate bent about its strong axis, though...  anyone???  why give guidelines for inverted tees that exceed noncompact limits, but be entirely too vague for scenarios in which noncompact limits are satisfied.
visitor (visitor)27 mar 02 16:57
"even with, say, a span of 58' (hypothetically)!?  fb=.6fy*(qs=1)?  this doesn't seem right unless stringent lateral bracing criteria are met for the stem (76bf/(square root of fy), maybe?).  and, well, if that's the case, why even bother looking at it as an inverted tee, just examine it as a plate.  seems awfully conservative to design as a plate bent about its strong axis, though...  anyone???  why give guidelines for tee stems in compression that exceed noncompact limits, but be entirely too vague for scenarios in which noncompact limits are satisfied."
the above statement refers to a simple span inverted tee beam; not exactly the original problem, but definitely related...
tum (visitor)27 mar 02 19:47
visitor,
    i think you are probably confused with concept of local buckling and global stability (lateral-torsional buckling, torsional buckling, etc.) the criteria you mentioned (b/t < (76bf/(square root of fy)) is only for local buckling of member.
     you may want to check max. allowable bending based on global stability criteria (lateral bracing will play important role on this criteria, see dr.galambos 's book for reference) and compare with max. allowable bending stress based on yielding criteria (0.6fy)...use that smaller one..say fb.
     then check the local buckling criteria to calculate q. the max. allowable stress that you can use is q*fb