What qualifies "Restrained" for flagpole footings? 2

[Hunter B]

Hello,

I am working on a HSS T-Post design that has a 3k lateral point load at the top 10' in the air. Using IBC Table 1806.2 and a lateral bearing pressure of 150psf/ft w/ the 1.33 & 2.0 factors included, I am getting an unconstrained pole size of ~2.75' round x 7' deep.

I would like to be able to justify this as constrained by pouring a slab (t/slab = t/adjacent grade) around the pole. However, I am unsure how big a pad is really needed to justify this. Plan dimension is not really a concern but I need the pad to be ~8" thick or less.

My first thought was to do a quick check using the coefficient of friction 0.35 between the slab and soil (i.e. 0.35 x 6' wide x 9' long x 6" deep x 150pcf = 1420# < 3000# NOT GOOD). However, this seems excessively conservative to me as it basically assumes the lateral restraint a) needs to take out all the lateral from the point load and b) the slab is sitting on top of grade instead of embedded.

I'd appreciate any tips on how to approach this justification and if the code actually gives clarification which I have been unable to find.

Thanks,

Hunter B.

 

[Celt83]

Grab the pole building design book from slideruleera's site: Link
If I recall correctly I believe it goes into the restrained condition.

have you considered utilizing 1806.3.4, which allows you to use 2x qh if the design can tolerate 1/2" displacement at the ground surface and the loading type is short term?

I'm making a thing:

http://www.thestructuraltoolbox.com

(It's no Kootware and it will probably break but it's alive!)

 

[phamENG]

needs to take out all the lateral from the point load

It does. Draw a free body diagram and prove those equations to yourself and you'll see what kind of restraint you need to provide.

the slab is sitting on top of grade instead of embedded

It might as well be. Freezing, erosion, general unreliability of organic top soils. There are several reasons to discount <8" of soil at the surface.

Friction is the only way if it's going to be flat. You could turn it down and embed it to get below frost and into a competent soil that you can estimate a lateral capacity for.

 

[Celt83]

I get a bit worse that 7':
w/1806.3.4:

w/o 1806.3.4:

I'm making a thing:

http://www.thestructuraltoolbox.com

(It's no Kootware and it will probably break but it's alive!)

 

[phamENG]

Celt - slick as usual. I really like the the shear and moment diagrams for the foundation.

 

[Hunter B]

Apologies, not sure how to reply to individuals but @Celt, you're correct. I rounded slightly in my original post but 7' works. I have run all the same checks. And yes, I used the 2x increase for 1806.3.4 as well as the 1.33x increase from 1806.1.

Pham, fair enough. That was my assumption and what I had been working with so far but was hoping to hear someone with a bit more experience tell me otherwise.

Thanks for the help both of you, looks like I'll just stick w/ my 7' embed and 2.75' diameter unconstrained.

 

[Hunter B]

Also, @Celt fantastic reference/link. Thank you for that. Definitely getting bookmarked!

 

[phamENG]

Happy to help.

not sure how to reply to individuals

You can't. I'm pretty sure this forum is still running on the same basic platform it was built on in the late 90s. If you want to use the nifty quote box, it's the person with a speaking bubble, 6 buttons left of the big Preview button.

 

[Hunter B]

You can't. I'm pretty sure this forum is still running on the same basic platform it was built on in the late 90s. If you want to use the nifty quote box, it's the person with a speaking bubble, 6 buttons left of the big Preview button.

Thanks!

 

[Celt83]

If there are more than one of these see if you can get the client to swing $1500 for a Geotech evaluation that will usually save at least that much on pier material from the reduced embedment.

I'm making a thing:

http://www.thestructuraltoolbox.com

(It's no Kootware and it will probably break but it's alive!)

 

[Aesur]

I wanted to point out that if using "restraint provided at surface" you shouldn't be using "isolated pole factor" as these are opposites, one allows movement, the other doesn't.

Additionally in my experience, I have rarely found that when actually calculating the surface restraint force that you will have enough weight for friction to actually provide surface restraint where these are typically used. In your case, assuming wind force, I come up with around 8500 lbs restraint force required at surface. Assuming friction value of 0.35 and 0.6 Dead load factor, that is around 40,000 pound of concrete.... I am curious to get others comments on this as I understand this 8500 pounds to be due to requiring not only sliding restraint but also rotational restraint.

Is a 33" dia. auger common in your area, typically I go in 6" increments to avoid custom sizes that may not be locally available for rent.