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Re: FOS for Overturning

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     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 
     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.

______________________________ Reply Separator _________________________________
Subject: FOS for Overturning
Author:  MIME7:MSSROLLO(--nospam--at) 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'
     |             |
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 
answer is correct and that algebraically adding common forces is not 
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..