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Although you are checking the factor of safety for overturning across the
flat sections, have you checked the maximum soil pressure across the diagonal?

Neil Moore, S.E.

At 06:47 PM 09/22/2000 -0700, John MacLean wrote:
>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
>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'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.
>John MacLean

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