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Foundings 1

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SMiK21

Civil/Environmental
Apr 26, 2011
30
I'm a recently graduated Civil Engineer and I'm a bit clueless about something my boss asked me to do.

I am supposed to check if the foundations of some highway signs are properly or overly designed. The signs are supported by 3 steel columns which are 4.3' embedded in the continuous concrete footing. The concrete footing is 14'x5.9' and 6.6' high.

The axial load is very small, so when I calculate the eccentricity of the load (M/N)it is bigger than the width of the founding. Does this mean that I have to (drastically) broaden my founding?

I also would like to know how to calculate the embedding length of my steel columns. I have seen 1 or 2 answers here in the forum but I am not quite sure about them, and I cannot obtain the articles which they refer to.

Lastly, I would also like to check the bottom reinforcement of the footing. Can I use a strut-and-tie model to do so, with the embedded steel in the footing? Also, where do I place my axial load - top of the footing or bottom of the steel column?

Thanks!
 
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In a case like this, you need to include the axial load from the footing itself (0.6 of it, anyway). This will bring your P up and reduce the eccentricity.

As far as embedding the columns - I, personally, wouldn't do that, but if I were checking it I would take the area of concrete in contact with the column, find the area and section modulus, then do a P/A + M/S to get the max bearing pressure at the top of the footing. Please note that P, in this case, is the lateral shear, not the axial load.

As far as checking the bottom reinforcement. I guess you could use a strut and tie model, but why would you want to. It's too time consuming and won't yield any more useful results compared to treating it like any other footing.
 
Thank you for your kind reply.

I had already included the self-weight of the founding, but the eccentricity problem persisted. The self-weight of the sign is very low (a thin steel panel + 3 steel profiles) but the wind forces on it cause a significant bending moment on its base.

I use strut-and-tie because it's the only one I know to calculate reinforcements in footings =)
 
the terms normally used are Footings or Foundations, never heard the term Foundings but maybe that is a local variation.

Is there soil over the top of the concrete that could contribute to the axial load?

Also, you have given us no indicationof how high this sign is so I cannot tell if lateral resistance might be useful against overturning.





 
How large and tall is your sign? With the footing size you gave, I get a 0.6*DL = 49kips (neglecting the sign). If I make a conservative assumption and say that the overturning is such that you only have the 5.9' dimension working for you then you need a 145k-ft moment to be outside the footprint of the footing. That's a 30' tall sign with 4.8 kips of wind load on it. That's a pretty tall sign with a large wind area (almost 100 sq ft if I assume a 50 psf wind).
 
Does soil pressure provide rotational restraint? A sketch would help.

BA
 
First of all, I'm sorry for my mistake(s) - of course i meant footing, not founding. English isn't my first language, so i mixed footing with foundation and came up with founding :)

I think I have solved it, but nevertheless I have uploaded a rough sketch of the footing + sign.

According to my country's regulation, I should use 42psf for the wind pressure over the surface
 
 http://files.engineering.com/getfile.aspx?folder=c523f921-ee96-417d-b3c4-b35e952a0527&file=sign.pdf
Two more dimensions needed...(1) top of concrete to bottom of sign and (2) top of concrete down to grade.

BA
 
Respectively, 4.9' and 4.3'

I thank you all for your time and patience
 
H(wind) = 42(15.7)21.5 = 14,177#

M(overturning) = 14,177(21.5/2 + 4.9 + 6.6) = 314,440'#

W(concrete) = 14(5.9)6.6*150 = 81,774#

M(stabilizing) = 0.6*81774*5.9/2 = 144,730'#<<M(overturning)

The concrete is buried only 2.3' in the ground, so there won't be much benefit from lateral pressure.

The foundation is not adequate.

BA
 
Now with that sketch I can tell just by looking at it that it doesnt have a chance.

Even if the overturning would work the bearing stress would e so high as to be rediculous.
 
Okay, so the 4.3' dimension on your latest sketch has no meaning. What is the revised width?

BA
 
doesnt look like it'll work.
This probably needs to be a pole type footing or caisson/ drilled pier.
 
...or at least rotate the footing in a direction such that it will actually help you!
 
Changed the foundation's measurements to 14' x 8.9' by 3.3' high.

Here, we should use 0.9 instead of 0.6 for permanent actions that stabilize the foundation.

BAretired, the 4.3' dimension doesn't have to do with the general equilibrium, yes, but I needed it to check the minimum embedding length of my steel columns.

I don't want to change the layout of the footing very much - what I initially had is a design that is being constructed now, my boss just wanted to see if the next few signs of the same highway could take less concrete.
 
Your revised dimensions are still not adequate even when based on the 0.9 factor. The 0.9 factor is not in accordance with any code I know.

You need to expand the width. One way to save on concrete is to use a wide footing with a smaller pedestal. Then the weight of footing plus pedestal plus fill on top of the footing all help to provide a stabilizing moment.

BA
 
It's in Eurocode 7 - Part 1. It will amount to the same 0.6 factor, since the actions are multiplied by 1.5. Resistances x 0.9 / actions x 1.5 = 0.6

Also, we use about 157 pound-force/cubic feet for the concrete's self-weight and the measurementes are rounded (up or down) when converted from meters to feet.

With a 9' width, the overturning moment is just a little bit smaller than the stabilizing one. I don't have any data about the soil, so I didn't take it into account.

Anyway, the final call is up to my boss, I'll give him my calculations and he'll decide on the final dimensions.

Once again, thanks a lot for your help.

 
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