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Re: wind load to retaining wall

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Jordan,
 
This would be my approach, too.
 
Steve Gordin SE
Irvine CA
----- Original Message -----
Sent: Monday, October 09, 2006 1:55 PM
Subject: Re: wind load to retaining wall

It is not given because for a wall with equal heights on both sides or
for a sign the positive and negative pressures are not salient to the
analysis - only the net total pressure is.  The only place it is
mentioned is for C&C loads (on a building), as you need to know which
direction the pressure is going on either side of the building.  For a
flat plate in a free stream, the Cp is 2.0. This is in line with the
sign formula topping out at about 1.95 when it's not near the ground
(going by memory here).   Since the force on a unit area is simply the
static pressure (1/2 x density x v^2), or a Cp = 1.0,  it follows that
the remainder of the force is frictional drag, boundary layer effects,
and "suction" on the back side, which happens to be equal to static
pressure (1+1=2).

This case is a little more complicated because we have a boundary with
zero velocity enforced at the bottom, so the Cp = 1.3.  I'd bet a dollar
that the force is distributed approximately parabolically from the top
(at Cp=2) to the bottom (Cp=0), within say 20%. Integrating over the
area would get you about 1.3.  If I had five minutes to come up with an
answer, I'd do a linear max-to-zero on both sides with half the force,
add 20% and call it a day.  If I had one minute, I'd use the full force
and add a couple of extra bars to the wall. ;-)

Jordan



jrgrill(--nospam--at)cableone.net wrote:
>
> Yes, but what is it, it is not specified in ASCE for this application.
>
> Joe
>
>

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