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Re: Stack Foundation

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In case the eccentricity e is greater than kern distance h/6. Does any code
specify a minimum value for the length of the soil stress triangular
distribution [3 x ( h/2 - e )] with respect to the footing length?

Regards

Hasan Hindawi

> >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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