Metal Building Foundation Question 1

[SteelPE]

Unfortunately we are having a little bit of a disagreement in our office about something that should be simple and it is driving me nuts.

Currently, we are working on the design of a metal building foundation. The client decided that he wanted to place the bases of the column 1’-6” above the slab on grade which is causing a bit of a problem. I have attached a sketch.

Normally we would attach hairpins into the slab in order to resist the sliding forces. However, in this case the hairpins are located 1’-6” below the bottom of the footing. In my opinion, we are now getting a prying force acting on the hairpin equal to the horizontal force (H) + (the load developed in the soil to resist the force H). This is all predicated about summing the moments about A and figuring the passive pressure needed resist the horizontal load.

My colleague disagrees with this analysis. He says that you should sum forces about location B or C only because the footing will not rotate about A and therefore my analysis is incorrect. In this instance, the hairpins become virtually useless.

I know I could just not install the hairpins and size the footing appropriately, but I do not believe the client is going to be happy with the results.

How would others approach the problem?

 

[msquared48]

The force in the hairpin will counter the lateral kick of "H", but there will be a moment generated that the concrete plinth will have to take. There is no prying action to the hairpin that I can see.

The footing will just se the "P" plus the moment described above from the plinth.

Mike McCann
MMC Engineering

 

[dhengr]

Seems to me that the 4.5' vert. dim. might represent the exterior grade, but that’s potentially variable, so pick the min. The wall on which the column rests is a retaining wall of sorts, resting on discrete ftgs. More specifically, the wall is a tied back wall, restrained at the ftgs. and at the hairpins, but without fixity at either restraint elev.; and active and passive earth pressures, plus an interior surcharge. I think you both missed the important moment elev., namely “D” at the top of the ftg. For any of your “H’s” the horiz. force at the hairpins “A” will be greater than your H at the col. base pl. because it is applied on a canti. 1.5' above A. I would distribute these horiz. loads over some length of wall as a function of its thickness and its vert. and horiz. reinf’g. above the ftg. and under the stl. col. There should be a big, bold, note on the found. plan saying this design is predicted upon the slab on grade and hairpins and tie rods being in place for all loads except erection DL’s.

 

[ztengguy]

Would you not take the moment around B or C to solve for the force at A? I would ignore any passive pressure or earth pressure on the pedestal and just design the 1'-6" as a cantilever beam from top of footing pinned at A and cantilevered above A?

 

[ron9876]

You can take moments about any point that you like. You have the vertical load, vertical soil pressure, horizontal force at the hairpins and column and horizontal passive soil pressure that offsets column load x 1.5.

 

[SteelPE]

dhenger,

I'm not sure, but we may be saying the same thing (maybe I am confused). I am saying that for case #1 the load at the column base is 8.22 kips.

If we install a hairpin at A we will then end up resiting this loading through passive pressure on the "A" side of the footing (because the load is negative in this instance). Ignoring the top 12" of soil places a resultant at 3.5/3 = 1.167ft from the base of the footing leaving a moment arm of 5.33ft-1.167ft = 4.167ft.

This would mean we would need to develop a passive resistance on the A side equal to (8.22k)*(1.5ft + .25ft)/4.167ft = 3.45 kips. This would make the force in the hairpin 8.22k + 3.45k = 11.67 kips.

This is what we are arguing about.

Zten,

I need to use the passive resistance on the side of the wall to resist the horizontal force H. With the large uplift load I will not be able to use count on the weight of the footing. In your instance the force in my hairpin would be (8.22k)*(6.83ft)/(5.083ft) = 11.045kips

 

[msquared48]

The moment generated in the plinth by the tension in the hairpin and the horizintal kick of the mainframe is taken out by the uneven bearing pressure on the footing, and the moment that generates on the footing - just P +- M. No passive pressures needed here.

Mike McCann
MMC Engineering

 

[hokie66]

I think you have it figured out, SteelPE, although I don't like using hairpins to a nonstructural slab to resist a structural force. Not sure why Mike says passive pressure is not involved.

 

[dcarr82775]

In my mind it rotates about point C.

I would say the force in the hairpin is equal to the frame thrust. the eccentricity is taken out in bearing on the footing base as Mike described. If the eccentricity is large enough to cause uplift at point B and exceed the allowable bearing pressures then you either make the footings bigger or start looking at passive pressures which could change the hairpin force to keep with sum Fx=0

 

[msquared48]

Hokie:

I rarely rely on the lateral bearing pressure to resist any kick as lateral deformation must occur to generate the resistance. It can be tolerated in some structures such as pole buildings, but I just generally do not like to rely on it for other structures. Moreover, the IBC lateral bearing values are predicated on an undisturbed soil matrix, i.e., as in a drilled pier, not dug, formed, poured, and backfilled, which is the case here.

I totally agree that some lateral bearing will play in here, but I rather rely on the differential bearing pressure on the footing to resist the induced moment. Anything else is gravy. That's all.



Mike McCann
MMC Engineering

 

[hokie66]

Mike,
But in this case, SteelPE indicated that the uplift has negated the ability to use friction at the base. He has a propped cantilever, with the prop being the floor slab. Something has to resist the backspan reaction, and I think it has to be passive pressure. Maybe you are thinking of passive pressure to resist the outward force, but in this case it is to resist the inward force below the building.

 

[msquared48]

Ok. I understand now with the uplift problem. I missed that. Agreed.

Mike McCann
MMC Engineering

 

[hokie66]

Although it does raise a question as to the formulation of the problem. SteelPE's force directions would normally be reversed for uplift, both vertical and horizontal, as uplift in a portal frame causes closing rather than opening at the base. Is this a trick question, SPE?

 

[slickdeals]

See document in the link below

http://bse.wisc.edu/bohnhoff/Publications/Copyrighted/EP486_1.pdf

I haven't used it in a while, but looks like the conditions there can be modified or adapted for your situation. It talks about point of rotation for embedded posts.

 

[SteelPE]

FYI, in my previous post, meant to say the C side of the footing develops the passive pressure not the A side (for the 0.6D +1.0W combination).

hokie66

In this particular instance the wind load is rather large and overcome the kick force of the frame in the first load combination (0.6D+1.0W). There are other load combinations that have the kick force pushing out of the structure (two of them are listed on the page), however, with most of these cases, I have a decent amount of downward force. So in essence, the base of the column is pushing into the building with this load combination but there are other cases where the frame is pushing out from the building. I assume this is what you mean by "closing the base".

In case 2, I have H=6.3k

resultant for the passive pressure on the B side of the footing = 5.083/3 = 1.69 ft

Distance from hairpin to resultant = 5.33ft-0.5ft-1.69ft = 3.14ft

Passive force required = 6.3k*(1.5ft + .25ft)/3.14ft = 3.51k

Force in hairpin = 3.51k+6.3k = 9.81k

Of course with using the passive pressure to resist the outward kick, I will need to make sure the system is stable during construction. I do not believe I will have a problem with this though.

I can use friction, I just need to make the footing much larger than what I have now.

 

[msquared48]

On thinking here a little further here, it makes no sense to have a net uplift on the footing as it is an unstable situation.

If there is a net uplift, including the weight of the footing, then yes, the footing does have to be larger, either in mass, or in area, or both, with a net FS - 1.5 against uplift. With the larger footing, there would be a net bearing pressure to resist the moment due to the lateral thrust, but may not be enough. You would have to run the numbers.


Mike McCann
MMC Engineering

 

[hokie66]

My point was that with uplift, the frame force is inwards, thus the slab resistance is compressive, and the resistance at the footing needs to be inward. With talk of hairpins providing the prop force, the frame force is outwards, as is the footing resistance.

 

[SteelPE]

msquared

I am only required to satisfy the load combination 0.6D+1.0W. As long as I take the reduction of 0.6*(footing+soil weight). I have developed enough force to satisfy the code. In this instance I will end up with 1.0/0.6 = 1.67 FS. Is this not correct?

hokie,

I agree with the fact that the hairpins work really well with when the force at the base of the column is pushing outwards. Do you not agree with the analysis when the force is pushing inwards? Or do you not think the hairpins will be required to support any force because of the presence of the slab?

 

[hokie66]

It doesn't matter which way, the slab (in tension or compression) will brace the base of the column. I just don't like the philosophy of depending on a non-structural slab on ground to provide this support. May be all right for a lot of applications, but in many industrial facilities, you can't count on the slab staying where you put it. Trenches, pits, etc. just appear. So my preferred method is a pair of bored piers with a capping beam...probably just my industrial background talking.