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# Re: Stack Foundation

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• To: <seaint(--nospam--at)seaint.org>
• Subject: Re: Stack Foundation
• From: "John MacLean" <john_maclean(--nospam--at)pomeroy.bc.ca>
• Date: Fri, 22 Sep 2000 18:47:47 -0700
• Cc: <bryson(--nospam--at)civil.ubc.ca>
```Michael Bryson wrote:

<<You say you use a factor of safety of 1.76 for overturning? But this using
allowable soil pressures or factored soil pressures? If the former, then
you're factor of safety would be higher, yes?>>

I guess I shouldn't really use the term "factor of safety" with a limit
states code. It's probably not correct, so my apologies. You are right that
you often see a factor of safety of 2 or more for overturning in allowable
stress design.

Limit States codes are supposed to be clearly superior to allowable stress
codes for overturning. They are supposed to provide a clear and consistent
safety margin in design. I think the inconsistencies in the allowable stress
approach show up in calculating bending moments and shears in the footing.
The downside of the limit states codes is that they seem to generate very
high factored soil pressures.

Here's an example. Take a 10 ft x 10 ft square footing with a 250 kip-ft
overturning moment on it and a 100 kip concentric dead load. The "factor of
safety against overturning" using allowable stress design would be 100 x 5 /
250 = 2.0. (5 ft being the distance from the centre of the footing to the
outside edge)

For a limit states load case of 1.5 Wind + 0.85 Dead, the factored
overturning moment would be 1.5 x 250 kip-ft = 375 kip-ft. The factored
vertical load would be 0.85 x 100 kip = 85 kip.

The factored resisting moment must be >= the factored overturning moment. In
this case 85 x 5 = 425 kip-ft >= 375 so okay.

To check the soil pressure I revert to specified load levels since the
geotechnical guy will only give me "allowable" soil pressures. The
eccentricity e = M/P = 250/100 = 2.5 ft. Which is outside the kern distance
(10ft/6 = 1.67ft) So my soil pressure is a triangular distribution with a
peak value of (2 x 100) / (3 x (5 - 2.5) x 10) = 2.67 ksf So I'm okay if the
"allowable bearing pressure" is greater than  2.67 ksf.

To check the concrete I go back to my factored loads. The eccentricity e =
M/P = 1.5 x 250 / (0.85 x 100) = 4.41 ft. Which means under factored loads
the resultant is way out near the edge of the footing. The factored soil
pressure is triangular with a peak value of (2 x 85) / (3 x (5 - 4.41) x 10)
= 9.60 ksf. This would make the geotechnical guy go weak at the knees so I
don't tell him about it. It could be worse. The closer I get to having the
factored overturning moment and factored resisting moment equal the closer
the factored soil pressure approaches to infinity. My gut feeling is that
for this "ultimate" condition with soil pressure concentrated over a small
area at the toe of the footing I'm prepared to go 3 or 4 times allowable
soil pressure and even calculate the pressure as a rectangular stress block
instead of triangular to reduce it. Especially if it all works at allowable
stress levels.

If I had say a 2ft x 2ft square pier on the footing the factored bending
moment in the footing at the face of the pier would be 85 x (4.41 - 1) = 290
kip-ft.

If I was designing using allowable stress I might calculate the moment in
the footing at the face of the pier using the triangular stress distribution
at allowable load levels. This works out to 176 kip-ft by my reckoning. If I
applied a 1.5 load factor to this moment I'd get 264 kip-ft. So the moment
calculated this way is about 10% lower than with the limit states code and
the difference can vary quite a bit depending on the eccentricity e. Shear
differences can be even worse. Bottom line....it's not consistent and
doesn't correspond to a uniform safety margin on overturning.

I'd like to hear anyone else's opinion on this.

Cheers,
John MacLean

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