L shaped wall footing 1

[LicensedToPEe]

I am designing a strip footing that is, due to a property line restriction, constructed as an L. The wall (stem) is propped at the top with a floor diaphragm and at the bottom with a slab on grade. In other words, the foundation system (wall + footing) cannot rotate and can only move in the vertical direction. There is about 6 kips/ft (5 story bldg) of gravity load at the top of the wall with 7" eccentricity to the center of the footing (click on the link to see the section).

My question is, under these conditions, is the pressure under the footing uniform? If the L-bar is designed for the negative moment at the fixity between the footing and the wall, the system is assumed to act as rigid (i.e. the angle stays 90deg).

I am just so used to be anylyzing footings with the good old p_max = Q/A + (6Q*e)/(B^2L) whic always results in no-uniform pressure distribution when load eccentricities are present. This formulation also assumes a pinned stem at the footing to allow rotation of the footing.

Is my approach correct to assume uniform pressure distribution as long as the stem/footing interface is designed (i.e. has enough rebar) to transfer the negative moment and the floor at the top and slab on grade at the bottom prevet rotation?

Any input/comments are appreciated,

Adam U
G M Structure LLC

 

[kslee1000]

Per your detail, ignore the slab on grade, it contributes very little compared to the loads from floors above.

You may make this system simpler by lowering the slab on grade to the strip footing level, and make it a mat foundation. It may cause you more concrete (I would prefer 8" mat as the minimum to allow for double layer reinforcement), but simplify the excavation/backfill/compaction efforts.

 

[STVU]

AdamU

Diagrams always help a lot. In your formula, how do you account for earth pressure thrust?

It is not a uniform pressure due to two reasons. First, you have 7" eccentricity & the center of the stem does not coincide with the center of the footing. Second, you have earth pressure thrust which will cause higher bearing pressures on the right edge of the footing.

I am getting 1.93 ksf for comparison.

 

[LicensedToPEe]

Making it simpler might be the case but I would also make it SO much more expensive. 2 layers of rebar make residential contractors cringe!!!

If the footing were to becentered, I would only need about a 3 foot wide strip footing (given 2,500 psf allowable bearing). But it is the eccentricity that, given the standard non-uniform pressure distribution approach, creates a bearing problem at the shorter end (the heel?).

However, there are footings like this constructed all the time, I just need to "prove" it on paper that it works...

 

[LicensedToPEe]

STVU,

if the stem/footing is not rotating (fixed at the interface, propped at footing and top of wall) how can there be a pressure variation? the footing can only move up an down, theoretically, so it will always be loaded uniformly? Am I not seeing this right?

 

[hokie66]

The pressure will not be uniform. The soil load will tend to even out the distribution, but the pressure will still be trapezoidal. You just have to sum the forces to get everything in equilibrium.

 

[LicensedToPEe]

Also for STVU,

I would say the pressure is larger on the left edge of the footing (at the shorter side).

 

[LicensedToPEe]

Hokie,

if the footing had no heel, the longer I would make the toe, the higher the pressure I would get since the eccentricity is moving away from the center of the footing. By the trapezoidal approach, you will never get the footing to work since your pressure will always increse linearly at the corner of the L...

 

[hokie66]

That's correct. Just increasing the width of an offset footing doesn't help. The pressure goes to a triangular distribution, with the remote toe carrying no load. When a footing has to be totally offset, no heel, you have to have a rectifying force to keep the toe down. This is commonly achieve by a "strap" to a footing within the building.

 

[kslee1000]

"if the stem/footing is not rotating (fixed at the interface, propped at footing and top of wall) how can there be a pressure variation? the footing can only move up an down, theoretically, so it will always be loaded uniformly? Am I not seeing this right?"

Again, per your detail, the stem (wall) and the footing is connected by a single bar in the middle of the wall (as pinned condition - no moment capacity at the joint), and the slab is simply sitting on the footing, while the lateral movement is prevented by the slab, how could rotation not to occur with the build in eccentricity?

Worst of all, when the footing rotates, the water can seep through, and a gap may form at the slab/footing interface, the gap could be filled by fines, then what?

Double the bars in stem (wall) make it a rigid joint? Now you have moment on footing to deal with, not a solution.

Tie the slab to footing? You are going to have deal with severe cracks, also not a solution.

Cheap has a price! (Sorry, not an attack, just express personal feeling. I didn't do the math, your system colud be worked out just fine)

 

[hokie66]

kslee,

The wall to footing joint shown by AdamU's section is not a pinned joint. The bar has an effective depth of roughly half the wall thickness, 5". It takes bending. Whether or not it is strong enough, I don't know. His solution may work, he just has to do the numbers to find out.

 

[LicensedToPEe]

Exactly what Hokie66 says, the section is designed with d=5" for the negative moment resulting at the interface.

If the joint is rigid, the lateral support provided against the stem by the slab on grade provides a "roller" support, allowing up and down movement only, i.e., no rotation.

I am not trying to convince anyone on the right approach, I am just saying these footing ARE build like this in residential construction and the footing don't experiance a bearing failure nor settlement even though the analysis (trapezoidal pressure) shows high overstress. But I guess this will only lead us into the whole discussion into the accuracy of allowable pressure guess (I don't call that a calculation ha-ha).

 

[STVU]

AdamU

See attached file for my approach. My 1.93 ksf was wrong. Must check to see if Qmax is less than allowable bearing capacity.

 

[LicensedToPEe]

STVU,

if the wall is propped at the top, the condition that develops is the "at rest" and Ko needs to be used in lieu of Ka. That's just a side note.

Also, you are missing the axial load at the top of the wall (Q = 6 kip/ft) contribution to the pressure under the footing given e=7" (per sketch). In other words, the q_max needs an additional term in the form of

+ [(6,000*(7"/12")]/[(4^2*1)/6] = + 1,312 psf

This is waht kills the pressure diagram, it adds 1,312 psf to the tip (if trapazeidal distribution is used).

 

[kslee1000]

hokie:

I agree it has some moment capacity, but not to the extent of fixidity for 10" wall (the tension side wall-footing interface will open up as a triangle wedge). I believe you know very well that finding degree of support fixidity is not a simple task, even for overhead beam-columns, not to mention there is effect from the soil to be considered. Therefore, for practical concerns in the US, single layer bar in the center of wall is most likely to be considered as pinned connection.

Also, will moment be helpful/desirable for his case (he can move the bar to the outer face)? I doubt, and rather to have single vertical load to be dealt with, not both. However, as both of us have pointed out that we didn't do the math, so, his detail may turn out just fine without all these arguments.

 

[STVU]

You are correct-Ko applies instead of Ka. Only exception would be if you backfill before the floor beams are installed. In which case the wall has already rotated and Ka is fine.

The formula I used assumes the axial load is in the center and not 7" off. Let me digest that. I don't know if it is going to add 1312 psf though. I will repost later.

 

[LicensedToPEe]

It's Friday 8pm and I am still at the office...

Thanks for your contributions, I will reply to additianal posts on Monday...

 

[STVU]

I just looked at my reference "Structural Concrete" 2nd edition, by Hassoun. If the eccentricity is less than B/6, then only one formula applies. So, for e=0 to e=B/6 or 8" qmax is the same 2.53 ksf per my previous post. I learned something!

Once you have "e" = B/6 or 8" exact we jump into the triangular distribution, where qmin is zero and q max is still 2.53 ksf.

When "e" > B/6 it is triangular, but the zero portion of the triangle moves to the right.

 

[kslee1000]

Please have a look on the attached sketch.

 

[kslee1000]

One more note:

No matter the fixidity of the lower joint, your wall could have positive moment on the inside face, how do you take care of it, by using the same bar in the middle? Won't there be cracks on the wall? Double check your forces along the wall.