lateral soil bearing pr, unconstrained soils and foudations

huangyhg

lateral soil bearing pr, unconstrained soils and foudations
has anyone tried to use equation 18-1 out the ibc 2006 for unconstrained soils and foundations to determine lateral bearing strength?
i tried using it and found it to be very conversative.  i have an 8' cantilever post with a concentrated load of 375 lbs at the end. and according to the equations, it concludes that a 1' diameter footing 3' deep will not restrain the lateral load (i believe it suggest a ~12' deep footing). did i miss something here?  
please let me know if you have used this in a design along with any comments or concerns you have about it.
thanks.
this equation is nothing in the codes (ibc, aitc, ibc) in this form and has been used for many years.  
the 3 to 4 foot embedment with a 1 foot diameter footing seems reasonable to me.  if you are getting a 12 foot embedment, i would look to your units to start.  otherwise, look to the lateral soil bearing you are using and the computation  of the "s" value.
mike mccann
mmc engineering
msquared48, did you mean "nothing new in the codes" ?
enginerrred,
three references you may find useful are:
pole building design by donald patterson, 1957. and
archon engineering's winpost ( about $40 )available at sales at archonengineering.com
and chart for embedment of posts, outdoor advertising association of america ( circa 1960 ) which has been reprinted in many structural engineering handbooks.
good luck
rwf7437:
yep...thanks.
mike mccann
mmc engineering
i get a 5 foot embedment considering allowable passive pressure of 250 psf.  a 12 ft embedment results if allowable passive pressure is reduced to only 25-30 psf, about a factor of 10 difference.
thanks guys for the advice.
lobstaeata,
can you confirm that we are using the same variables? in my orginal calcs i used s1 = 100 psf (conservative due to non-testing), but in my calcs below i tried to match your results using s1 = 250 psf.
i have:
   b = 1 ft
   d = 3 ft (assumed)
   h = 4 ft
   p = 375 lbs
   s1 = 250 psf
resulting in:
   a = 3.51 ft
   d = 6.04 ft (required)
how did you get a 5 foot embedment? can you let me know what we did differently?