Need a book? Engineering books recommendations...
Return to index: [Subject] [Thread] [Date] [Author]
Re: FOS for Overturning
[Subject Prev][Subject Next][Thread Prev][Thread Next]- To: seaoc(--nospam--at)seaoc.org
- Subject: Re: FOS for Overturning
- From: Mark Florence <mdf(--nospam--at)daystarsoftware.com>
- Date: Thu, 25 Jun 1998 08:25:42 -0500
I always check overturning about each side of the footing and take the worst case. Depending on which side you are checking, the forces can be overturning or resisting forces. In this case overturning moment about the left side would be 10x4=40 and resisting moment would be 3x2+15x4=66 with FS=66/40=1.65. Overturning moment about the right side would be 10x4+3x2=46 and resisting moment would be 15x4=60 with FS=60/46=1.3. Mark Florence, P.E. At 08:50 AM 6/25/98 EDT, you wrote: >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.. > > >
- References:
- FOS for Overturning
- From: MSSROLLO
- FOS for Overturning
- Prev by Subject: Re: FOS for Overturning
- Next by Subject: Re: FOS for Overturning
- Previous by thread: FOS for Overturning
- Next by thread: Re: FOS for Overturning
- About this archive
- Messages sorted by: [Subject][Thread][Author][Date]