huangyhg

calculation of preliminary area of steel for slab
could someone please provide me a quick, if preliminary, formula for finding the area of steel required for a foundation, for instance, if i have the factored moment, effective depth and width. please also provide the reference for the formula. my fc' is 4000 psi. i am
thinking that
as = mu/b*d^2
but the units do not seem consistent.


i would use mu/(0.81*as*d)
the 0.81 comes from approximating d-a/2 to 0.9d and multiplying that by the phi factor of 0.9
there is no need for b in the equation since that only factors into the compressive stress block depth, a. a is accounted for in the 0.9d as noted above.
hope this helps.

sorry, the as in my equation above should be fy
it should be mu/(0.81*fy*d)
sounds like your trying to do something you're not qualified to do. i could give you a formula, but that doesn't account for the various code minimums required for flexure, and temperature and shrinkage reinforcement. if you are an engineer, you need to get aci 318 out, and get familiar with it.
thanks for all the responses, guys.
structuraleit, could you please provide a reference for the formula you provided?
spats, your comments are noted and a thorough understanding of the formulas and how they are derived will be done before being put to use. your commens here have provided a head start and that is much appreciated.
there is no reference, really. the reference is the phi mn formula with the assumption that (d-a/2)=0.9d
this is a conservative assumption.
just plug in 0.9d for d-a/2 in the phimn formula and solve for as, nothing more complicated than that.
ok, you're being so nice, so i'll return the favor. the exact solution formula you are looking for is
ρ = 0.85 f'c/fy (1 – √(1-(28.32mu/(?bd2f'c)))
as = ρbd
where mu = ft-k, f'c & fy = ksi, and b & d = inches.
? = 0.9 for bending.
where tension steel required by analysis, aci 318 requires:
as≥(200/fy)bd or as 1/3 greater than req'd. by analysis, otherwise
as≥0.0018ag, where ag is gross area of the footing.
hope this helps.
the reference is aci318 - derive the approximate formula from there yourself.
as others have stated, there are a number of other things that can dictate the required steel in a slab. i would suggest you do some reading on the matter so that you can come up with an accurate prediction.
i think he was looking for a rule of thumb for preliminary analysis.
i use as req= mu/4d for a ball park check (note: mu in k-ft, and d in inches)
strleit is also applicable.
thanks, spats, structuraleit and everyone else for your support and understanding. i am just getting a head start and a reference from this forum and work from there.
jkstruct, you hit the head on the nail! i was exactly looking for a rule of thumb, and wanted confirmation that it was as req = mu/4d. could you point out a reference for this, or show how this was derived?
as = mu / phi fy jd
where jd = d - a/2
like structuraleit said, assume jd = 0.9d
then as = mu / phi fy 0.9d
= mu / [0.9 (60 ksi) 0.9 d / 12 ]
note 12 for unit conversion
= mu / 4.05 d
much math and derivation omitted here but understand the concept of whitney stress block and the anticipated depth. long hand derivation involves solving a quadratic equation for a.
i use as,est = mu / 4.0d all the time.
for masonry, as,est is m/1.8d for 60 ksi steel and m/1.48d for 40 ksi reinforcing. m is the service level load not ultimate. try to derive these. very similar.

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