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Re: FOS for Overturning[Subject Prev][Subject Next][Thread Prev][Thread Next]
- 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 ______________________________ Reply Separator _________________________________ 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 answer is correct and that algebraically adding common forces is not 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..
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