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Re: Soil Pressure[Subject Prev][Subject Next][Thread Prev][Thread Next]
- To: seaint(--nospam--at)seaint.org
- Subject: Re: Soil Pressure
- From: EphHirsch(--nospam--at)aol.com
- Date: Tue, 26 Jan 1999 20:44:23 EST
Sorry, last post contained nothing (I hit the wrong key before anything was typed). Based on the data your geotech gave you, the equivalent pressure at the base of the wall (assuming 8 ft. ht. as you stated) due to 35 pcf equivalent fluid pressure is 8 x 35 = 280 psf with a triangular distribution diminishing to zero at the top and therefore an overall horizontal total force = 1/2 x 280 x 8 = 1,140 plf (i.e. pounds per horizontal linear foot of wall). How did you arrive at 75 psf for an answer?? How this force is resisted and what the reaction/shear /moment distribution is of course a function of whether or not the wall is a cantilered one from its base or restrained its top by the floor framing system. Strictly speaking of course this would also affect the soil pressure distribution itself, but the geotech presumably just assumed triangular distribution in either case, based on how the data appears to be stated. Now, in addition to the above there will be added horizontal pressure of 15H (psf), where H is height of surcharge, NOT wall, with a total horizontal force of 8 x 15H (plf). Since you didn't give the surcharge height, the numerical values are not obtainable. The inverted triangular earthquake equivalent fluid pressure of 20 pcf will not increase the horizontal pressure at the base of the wall, but will impose an additional total horizontal force of 1/2 x 8^2 @ 20 pcf = 640 plf applied @ 2/3 the wall height. Does that answer your question?
- RE: Soil Pressure
- From: Alex C. Nacionales
- RE: Soil Pressure
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