footing with biaxial moment 5

[Lion06]

I've been looking for a reference as to how to determine the max soil pressure for a footing that has moments in both directions, but only has partial bearing. The moment in one direction would the load in the kern (if it were the only moment), and the moment in the other direction would put the load outside the kern. I can't find a reference on this in my foundations book, and I could go through the math of it (but that would take a REALLY long time, and I honestly don't want to spend an entire day to figure it out), but I figured someone else has to have done this before.

My first inclination was to take the max pressure of the moment causing partial bearing and adding that to the max pressure caused by the full bearing (M/S), then I realized that I couldn't us the full S of the footing (for the smaller moment) because the whole footing isn't in bearing anymore. I tried estimating the amount of the footing that would be in bearing and using that S. That would get me close, but I'm really trying to be exact because I'm evaluating a program. The line of zero bearing stress is not perpendicular to either edge of the footing because of the moments in both directions, but again, I don't know how to address this without a day-long geometry session.

 

[WillisV]

1.89ksf

 

[haynewp]

I get about 1.9ksf also.

 

[Lion06]

Interesting. I come up with 1.88ksf and the program is spitting out 1.93 ksf. That seems close enough to me, it's just that my method involves a very "non-engineering" type estimation, which I really hate to use.

 

[BAretired]

I think you are using the wrong shape of footing. Why not elongate it in the direction of the larger moment? That should permit a reduction in footing area. Even better, if possible, use a combined footing so that the loads are balancing each other.

herewegothen,

You've got to be kidding, right?

Best regards,

BA

 

[Lion06]

BA-
I'm just trying to verify what the program is spitting out, nothing more. There may be a time when this comes up in a job and I want to trust the program when it does.

 

[BAretired]

StructuralEIT,

Okay...no offense intended.

Best regards,

BA

 

[Lion06]

none taken.

 

[BAretired]

There still seems to be something wrong with the results you guys are getting. First of all, a 10'x10'x2' footing weighs approximately 30 kips. I don't know what you mean by "an appropriate DL factor" but, for the sake of the problem, let us say that:

P = 20k

Mx = 88k-ft

Assume that My = 0 for now.

The eccentricity in the x direction is 88/20 = 4.4', so for a 10'x10' footing, the effective length of bearing on the soil is (5 - 4.4)3 = 1.8' and the effective width is 10'.

The soil pressure has a triangular distribution. Average pressure = 20/(10*1.8) = 1.11 ksf. Maximum pressure is 2.22 ksf. Minimum pressure is zero. So far, we have said that My = 0.

If My = 4k-ft, surely the maximum pressure will increase slightly, so the maximum pressure by my calculation is approximately 2.22 + 0.13 = 2.35 ksf.

Am I misinterpreting the problem?

Best regards,

BA

 

[Lion06]

If the controlling load combination is 0.6DL-W, then you need to multiply the ftg weight by 0.6 - that's what I meant by "the appropriate DL factor".
I did tweak the original loads I gave a little to be exact, see the later post.

 

[BAretired]

So, are we all in agreement that the maximum soil pressure with the given loading is in the order of 2.35 ksf, not 1.9 ksf? If not, then I still do not understand how you arrive at the maximum pressure.

Best regards,

BA

 

[WillisV]

I agree with myself. =)

 

[chichuck]

The 10' X 10' X 2' footing weighs 30 k
the soil above it weighs about 16.28 k for bearing depth at 3.5' below grade.

If total P = 20 k, then is there an uplift load on the footing?

chichuck

 

[Lion06]

chicuck,
No, there is no uplift on the footing. The axial load from the column (for the controlling load combination of 0.6DL-W is 2.818k. While the weight of the footing is 30k, you can only use 18k (because the load combination is 0.6DL-W), hence the TOTAL P=18k+2.818=20.818K (as noted in an earlier post). I am not considering any soil overburden on the footing (whether that is right or wrong is irrelevant to what I am trying to do), I can specifiy overburden in the program, but all I want to do is verify that any situation that the program is faced with will be done correctly.


BA,
No, the max pressure is around 1.9ksf (for P=20.818, Mx=87.842k-ft, and My=4.226k-ft).

 

[miecz]

For the loads that were specified, the formulas I attached in my earlier post match WillisV's results exactly. Although the PDF is a printout of a Mathcad file, all the formulas are shown, and can easily be followed in a hand calc.

 

[BAretired]

Using the revised values of 20.818k and 87.842k-ft, I would agree that the maximum pressure is in the order of 1.9 ksf.

If the original values of 20k and 88k-ft are used, the maximum pressure is in the order of 2.35 ksf.

This indicates that the footing is extremely sensitive to minor changes in load which suggests to me that this would be a very bad design, no matter what your computer program tells you.

One point to note is that the factor of safety against overturning is only 20.818*5/87.842 = 1.185 which is completely inadequate by any standard.

Best regards,

BA

 

[Lion06]

I don't think that is such a bad safety factor for OT when you consider that it is the 0.6DL-W load combination.
That's actually better than 1.5 for 1.0DL and wind.

 

[BAretired]

Well, I agree with you on that point. The 0.6 factor is certainly not in my code (Alberta Building Code) and from your latest comments, it would appear that you don't take it very seriously either.

Best regards,

BA

 

[haynewp]

I never design a footing to land outside the kern distance. Several geotech books I have seen recommend not doing this either.

 

[Lion06]

I would generally agree with not having a footing with partial bearing, but sometimes you could end up with footing with huge plan dimensions and/or much thicker than they need to be.

 

[csd72]

haynewp, what you say makes sense if it is a permanent load, but it is a bit conservative if you are looking at overturning from wind load.