deflections

huangyhg

deflections
i'm fairly new out of school and have taught myself asd, since we only learned lrfd in school.  my question is this:
for lrfd, when calculating deflections, you use a factored load.  so you use 1.2d + 1.6l to get your loading, then plug in the simple beam formula to get your deflection.
for asd, when calculating deflections, you use an unfactored load.  so you use d + l to get your loading, then plug in the simple beam formula to get your deflection.
am i missing something, or is lrfd more conservative from a deflection standpoint?  thanks
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[from the load factors, i presume you are designing with steel.] under asd, the allowable material stresses are always less than unity. this is equivalent to the resistance side of the lrfd equation. these 'factors' are much lower than the lrfd phi values. hence :  
l+d = 0.67m  --> 1.5(l+d) = m (fb= 0.67fy; m oc fb)[asd]
1.2d + 1.6l = 0.9m --> 1.33d + 1.78l = m  [lrfd]
lrfd design methods reflect that the dead load is fairly certain (factor closer to 1), but that the ll is more variable (larger factor). asd lumps the factor of safety together, but reduces the resistance instead (  
tparaf:  thanks, but i don't exactly agree with you.  the strength factors are basic enough, but we aren't on the same page on deflections.  i think my question still remains.
aisc chapter l:  the closest thing i could find to saying that "while the lrfd method yields larger deflections (since w & p have been factored), it is not usually used as a governing criteria (see aisc chapter l)." is this:
"limiting values of structural behavior to ensure serviceability (max deflections, accelerations, etc) shall be chosen with due regard to the intended function of the structure.  
essentially, the ibc (or whatever code you are going by) controls deflections.  you always check deflections for the appropriate limit, whether it is l/240, 360, 600 etc.
errr, make that his question.  i guess i spent too long typing the response.
actually, the factored loads are only used for the strength calculations -- unfactored loads should be used for deflection calculations.
lrfd designs yeild larger real world deflections because you can usually use smaller sections from a strength standpoint.  ie., if the asd design says a w14x30 works, maybe the lrfd says a w14x22 will work (both purely from a capacity standpoint).  then you need to check servicability using unfactored loads -- the w14x22 will obviously deflect more, so it may or may not be acceptable depending on your servicabilty criteria.
if serviceability (deflection, vibration, bldg drift) is governing your design, then lrfd vs. asd is immaterial -- you'll get the same answer when you're done.  only when strength is governing your design does the potential exist for lrfd to come out ahead.
don't use factored loads for deflections, even with lrfd.  deflection is a serviceability issue, so factored loads are not used.
lrfd 3rd edition, page 16.1-79, section l3.1, "deformations in structural   
now i've had to get out my brand spanking new (looking -- this is probably only the 2nd or 3rd time i've cracked the binding on the 2nd ed -- didn't even bother with the 3rd ed) lrfd book to read chapter l.
dasman and broekie are very correct - do not use factored loads for deflections.  
didn't the lrfd bridge code remove deflection limits altogether?
hg
dasman,
i agree with you that unfactored loads should be used for deflection calculations.
however, every time i use lrfd for strength calculations, i typically end up with bigger required member sizes.
maybe this is only because i commonly work on industrial projects where live loads exceed dead loads by four or five times. the increased load factor for live loads ends up increasing the   
deflections are computed based on service loads in both lrfd and asd, as stated above.