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# Re: FOS for Overturning

• To: seaoc(--nospam--at)seaoc.org, MSSROLLO(--nospam--at)aol.com
• Subject: Re: FOS for Overturning
• From: jerome.tan(--nospam--at)PAREURO.COM
• Date: Fri, 26 Jun 1998 2:21:16 +0000

```     Does the 10 kip uplift the resultant of vertical loads or just an
applied load? IF it is a resultant, then the downward component of it
should be included in the stabilizing moment and the upward component
plus the 3 kips horizontal force should comprise the overturning
moment. If it is an applied load, then this still to be added to the
overturning moment. In which case, your first argument in (2.) is
right.

The lateral pressure on the soil could not be fully develop at such
depth to help resist the overturning.

For stability check, you always take the worst condition , combination
and apparently in this case, the worst side of the footing which is
obviously the right toe to sum up your moments.

Jerome

Subject: FOS for Overturning
Author:  MIME7:MSSROLLO(--nospam--at)aol.com at INTERNET
Date:    6/25/98 2:08 PM

I have a question regarding the correct factor of safety for overturning for a
foundation subjected to an uplift force.  For purposes of this discussion,
assume I have a single footing with a horizontal force or 3 kips (to the
right) and a upward vertical force of 10 kips, both applied at the center of
the top of the footing.

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
permitted.

3.	Assuming you do not think the load is an overturning load, what if I move
it off center to the right 1".  What is the magnitude of moment to attribute
to overturning of the right side?

Thanks for any help on this..

```