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# Re: FOS forOvtg clarification

• To: seaoc(--nospam--at)seaoc.org
• Subject: Re: FOS forOvtg clarification
• Date: Fri, 26 Jun 1998 10:29:23 -0700

```Thanks for clarifying the question and its context.

The answer to #1 below depends on acceptance of your premise in #2, that the
axis of rotation is at the right toe. Accepting that premise provisionally,
then every force that does not pass through the axis of rotation at the toe
is an overturning force. (Or an "underturning" force, if it acts in the
other direction and you like playful names.) On this basis, the answer to #1
is "yes".

However I am not aware of any directive that "overturning" is necessarily
about any particular location of axis. In the old-time method of truss
analysis by sections, you would select convenient axes of rotation about
which to sum the moments so as to have unknown truss member forces pass
through those axes and thereby eliminate their effect on the summation.

Here, you can put your axis of rotation under the applied vertical load if
you want and get rid of that load as an "overturning force". As with
trusses, statics should give a consistent, determinate result no matter
where you put the rotational axis.

If there were a single, rigorously "correct" location of the axis of
rotation, it arguably would be where no part of the footing translated or
displaced. To find that location is an exercise in soil mechanics and
strength of materials, and maybe a quest into indeterminacy. This axis might
be at the toe if both footing and soil were essentially rigid and
exceedingly strong, and the overturning motion already underway . More
typically it would be toward the heel, or beyond the heel out in the dirt. I
don't think we are expected to determine that location for FOS purposes. I
don't want to.

I think you should amend the sketch to indicate the chosen axis of rotation
at the toe, and make clear that the FOS is figured on that basis. Then
question #2 should answer itself without controversy.

As for that basis being "correct", I think it is like determining whether
you live your life piously. It depends on what religious conventions you
judge it by. But it isn't a question of science.

Charles O. Greenlaw, SE    Sacramento  CA

___________________________________
At 10:03 PM 6/25/98 EDT, you wrote:
>I am trying to find a consensus on whether the vertical load should be
>considered an overturning force or not.
>The foundation is poured on top of the ground.  Therefor there is no passive
>soil pressure present.
>For now though, I am concerned with a centered column with a 10k upward load
>
>#1.  Is the vertical force an overturning force?..and why/why not
>#2.  What would be the correct calculation for FOS about the right toe as
>
>I have included the original posting below.....Thanks again for the help...I
>am trying to fine tune a footing program that at least models what a consensus
>of experienced engineers beleive.  I was hoping this group would at least
>clear up whether the force is an overturning one or not and the correct
>calculation for the FOS.
>Assume the footing is 15 kips in weight and is 8'x 6' x 2' thick. Pardon the
>phony graphics, I am using AutoBadd.
>
>
>            ^ 10k
>            |
>            |
>     ------> 3k
>     ---------------
>     |             |
>     |             | 2'
>     |             |
>     ---------------
>           8'
>
>
>1.	In calculating the FS for overturning, is it correct to call the 10k load
>an overturning load?  The load is at the center of the foundation and does not
>actually induce any overturning except if you do statics about a toe.  The
>load itself actually does not cause overturning.  Even statics claims that is
>overturns about both sides simultaneously.  I realize that the footing weight
>is always a stabilzing force in both directions of rotation.
>
>2.	Assuming you do think it should be included as an overturning force, would
>the correct calcualtion be Movtg = 3x2 + 10x4 = 46 and Mstab= 15x4 = 60,
>therefore FS = 1.304   OR is the correct one: Movt = 3x2=6 and Mstab=
>(15-10)x4=20  FS = 20/6 = 3.333.  By definition, it would appear that the 1.3