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RE: Soil Pressure

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The correct pressure should be p=1/2x(35)x(8)^2=1120 plf.
Maybe you just mistyped it.

I have my own question.

Regarding the seismic pressure, What is the engineering basis for
the upside down triangle pressure distribution. A rectangular
distribution would be safer applied at 1/2 the wall height. 
Assumming soil weight is 150 pcf and g=.40 you get 20 pcf of siesmic
load due to the earth. 


Alex C. Nacionales


> -----Original Message-----
> From: EphHirsch(--nospam--at)aol.com [mailto:EphHirsch(--nospam--at)aol.com]
> Sent: Wednesday, January 27, 1999 9:44 AM
> To: seaint(--nospam--at)seaint.org
> Subject: Re: Soil Pressure
> 
> 
> 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? 
> 
> 
>