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Re: FOS for Overturning
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- Subject: Re: FOS for Overturning
- From: "ALEX C. NACIONALES" <anacio(--nospam--at)skyinet.net>
- Date: Fri, 26 Jun 1998 01:28:11 +0800
The first one is correct. It is stated in the Structurel code of the Philippines 1992 (based on UBC 1988) that The base overturning moment for the entire structure or for any one of the individual primary lateral resisting elements shall not exceed 2/3 of the dead load resisting momemt. If your structure is less than 20 m tall or has a height to width ratio of .50 ot less in the wind direction, the combination effects of uplift and overturning maybe reduced by 1/3. It is not correct to say that " Even statics claims that is overturns about both sides simultaneously" because you have to choose which side you analyze depending on the wind direction and take moment about the corresponding toe. In your case: overturning Moment = 10 x 4 + 3x2 = 46 K Resisting Moment = 15 x 4(2/3) =40K < 46 NG But it would be conservative not to include the effect of soil pressure at the sides of the footing. Resisting Moment 2 = 1/2(k)(q)(h)(h)*1/3(2) To use this for force you should add a slab at the top of the footing. see the book " Design in Structural Steel" by John Lothers 3rd edition. page 274. Footings for tall isolated commercial signs. My 2 cents worth. I hope this helps Alex C. Nacionales, C.E. Iloilo City, Philippines -----Original Message----- From: MSSROLLO(--nospam--at)aol.com <MSSROLLO(--nospam--at)aol.com> To: seaoc(--nospam--at)seaoc.org <seaoc(--nospam--at)seaoc.org> Date: Friday, June 26, 1998 11:50 AM Subject: FOS for Overturning >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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