huangyhg

live load reduction for stud walls
i want to apply live load reduction to a wood stud wall. the problem is (for me at least) that i don't know how to determine what the tributary area should be for the wall.
if i consider the wall on a "per foot" basis then the tributary area is only a 1 ft strip of all of the floors that this wall carries. doing it like that, it'll be almost imposible to get the area of influence over 400 ft^2.
obiously, how much trib area i use will depend on the width of wall that i consider in my analysis. i think that the width i choose should somehow reflect the ability of the wall to distribute the load out. for example, i would choose a smaller width for a wood stud wall than i would for a masonry wall.
either way, i need some algorithm do determine how much tributary area i should be assigning to the wall for the purpose of live load reductions. what is everyone else doing??
my feeling on this is that in a wood stud wall, each stud acts pretty much like its own individual column; therefore, this column could be overloaded by a load concentration in one particular spot, even though the uniform load over a 400 sf area might be less than the allowable.
af&pa's wood frame construction manual for one- and two-family dwellings, table 2.11 calculates loadbearing wall loads (for wall studs and headers) from snow or live loads using the following multiple level reduction approach based on lrfd procedures:
(1.6l + 0.5s)/1.6 or
(0.5l + 1.6s)/1.6
adamp
for live loads, i wouldn't use a live load reduction for wood stud walls as there is no mechanism to calculate the spread of the load across the sheathing to adjacent studs. this is because the usual specified nailing for the sheathing does not take into account this kind of shear transfer.
also, the floor or roof framing is also usually of wood and these place the gravity loads directly onto the top plate which directly transfers it to the stud. statistically, you still have a good probability of loading one individual joist to 100% of the live load which in turn will load the stud 100%.
your trib area then is the spacing of your studs times half the span of the supported
thanks guys!
buddy,
can you give me more information about the wfcm? specifically, is the document actually referenced by any of the model codes? i was under the impression that the wfcm was more of a "recommendation" kind of thing -- albeit a very usefull one. is there an extension of this method for 3 story wood frame buildings? thanks again.
adam
thanks jae. just got your message as i was typing in my last response. do you have any recommendations for effective width on masonry/concrete walls?
well, the codes typically let you spread a concentrated load over a distance of 4t, where t is the nominal thickness. another way to look at it would be to view the elevation of your wall and draw a 45 degree line down from either side of a fictitious point at the top of your wall. then take the mid-height width between the lines (which is simply 1/2 of the height) as your tributary width. masonry/concrete work together in a much more unified way than individual wood studs.
unfortunately, i don't know of any code or spec formulae that would give a standard way to look at it.
adam,
the 1995 wfcm is referenced in all major model building codes in the u.s. including:
» 2000 ibc international building code: 2308.2.1 for winds > 100 mph (3 sec. gust)
» 2000 irc international residential code: 301.2.1.1
» currently adopted in 1997 sbcci standard building code
» 1999 boca national building code
» approved by state of florida board of building codes and standards
» approved by north carolina - department of insurance
» approved by georgia - department of community affairs
the new 2001 wfcm is an ansi approved document and is being adopted in the 2003 ibc, 2003 irc, and new 2002 nfpa building code.
here's the webpage to see code approvals for other wood standards, like nds and lrfd:
have you tried using repetitive
i use the repetitive member increase whenever i'm working with exterior walls (ie. wind = bending). my interpretation was that i could apply the repetitive bending factor and ll reduction whenever i meet the criteria for their respective applications. thanks for your input mreng!

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