embedded pole footing

huangyhg

embedded pole footing
i'm working on some billboards which are using embedded pole footings.  they're in the range of 3'to 4' diameter and anywhere from 12' to 30' deep.  some of the signs are pretty high.  the columns are fixed at the base and free at the top.  sometimes the column is offset from the center which adds additional moment to the length of the column.  while working on this a question arose in my head.
i'm designing these footings per section 1806.8.2.1 of the ubc, nonconstrained.  there's a similar section in the ibc.  
when the sign is center mounted, the moment on the column is from the wind load only.  when it's offset, you get additional moment as stated above.  say that, for some odd reason, there was no wind load and only the moment from the offset dead load on the column.  how would you go about solving the foundation depth equation?  you have no 'p'.  you do know the moment and you could find an arbitrary 'p' and 'h' that give you this resultant moment.  but as you vary 'p' and 'h' to give you the same moment at the base, you get different values for 'd'.  
what is the correct way to analyze this situation?
thanks for any help.
john
i've run across this same problem before... pole foundation with moment and no lateral load.  there's a book called "pole building design" by patterson (which slideruleera has a pdf of on his personal site...
oops, thats an old link for slideruleera's page.
you know, i was just thinking to myself that i need to work on my algebra and statics skills...  ;)
i appreciate the links.  i'm going to try to work out some equations as well.  i'll post them here later and if you would compare, i'd greatly appreciate it.
when you did your analysis, did you simply call the pole a beam supported my q2 and p with a point load q1 in the span?  well, plus that moment also...  i'm having a hard time in my head getting rid of the fixity at the ground level and reducing the moment along the beam with a non-rotationally constrained reaction.  
here goes nothing...i'll post more questions or some equations later.
thanks again for your help.
john
ok...here i go.
using patterson's assumptions for a pole with moment but without lateral load.
i came up with the following:
d = (m/(.381*b*s1))^(1/2)
hopefully that looks familiar to you.
the number's i'm getting appear reasonable when solving with s1 based on 1/3*d.
now to spend the rest of the afternoon trying to derive a formula for loading with moment and lateral load...this should be fun.
azcats:
exactly what i got.  
as far as the "moment + lateral load"... couldn't you just adjust your lateral load's height so that p*hadjusted = m + (p*hactual)?
that's why i started this whole thing because i didn't think that simply adjusting the height would yeild an accurate answer.
i got something worked out yesterday but my algebra skills didn't allow me to simplify the equation so it's still in the form of the quadratic formula.  that's fine for a spreadsheet, but impossible to post here.
when i did get it into a spreadsheet to compare, it's different by about 5% with the adjusting height method being more liberal - less depth.  however, i think i'm being a bit conservative because my moment load comes from two axes where the lateral load is in one.  without getting into a ton of trigonometry, these are difficult to combine.
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so i decided to check them in one axis.
it yeilded a difference of only 1% which is way too accurate for the assumptions anyway.  looks like just adjusting the height is a perfectly acceptable way to look at the issue.  although there could be larger differences if the moment is large compared to the moment from the lateral load.
thanks for your help in this one.  the link you provided was exceedingly helpful.
john
one last thing that i just noticed.  
in my latest equation including moment and lateral load, the only place that m occurs is in one term: (m+p*h).  so obviously it's not only acceptable to simply adjust h, it's perfectly accurate.  
to check your work, use an equivelant p and height for your moment, but use a very light p and very high h.  this should make the effect of the shear minimal and accurately model a moment without the effects of the shear.  
i know enercalc can perform these calcs in the anaylsis tab.