At Rest Earth Pressure on Basement Wall

[Yvet]

Would you recommend using an equivalent fluid pressure for at rest pressure or a rectangular distribution.

If rectangular, what formula do you recommend?

 

[dgillette]

Neither. Depending on the height and rigidity of the wall, and the compaction of the fill on the wall, you can justify a triangular distribution based on Ka or Ko, or else a more complicated shape that comes with compaction of the fill against the wall. For the latter, I recommend some work by done by Mike Duncan and friends in the early '90s. Search around on this forum, because I think I gave the full references a few weeks ago. That work suggests you shouldn't get too carried away compacting the fill against the wall any more than you really need to because it can raise the pressure above Ko in the upper part of the wall.

I'm on a crusade to abolish the term "equivalent fluid pressure," in part because the stress on the wall is not strictly normal to it like fluid pressure is, unless there is greased Teflon between the fill and a smooth wall. (Also because it isn't necessarily a triangle like fluid pressure is.)

Bon chance!

DRG

 

[fattdad]

for residential construction there are tables in the IBC. Use at-rest pressures, use triangular shape, include a surcharge "rectangular" pressure, based the at-rest earth pressure coefficient and what you think to be the surcharge pressure (backhoe next to the house, or such).

f-d

¡papá gordo ain’t no madre flaca!

 

[Stillerz]

I have a question closely related to this if you all dont mind.
I was recently looking over some problems in order to prepare for SE/PE test. I know this is frowned upon in this forum but I hope you'll consider answering.

The problem is a simple one.
Calculate the total lateral FORCE per meter on a retaining wall AT REST given:
H= 4m (wall height)
Gamma = 16.0 kn/m^3
phi= 28º
c=0
q= 10.0 kn/m^2 surcharge.

I calcualte the total force as 89.04 Kn/M
The problem says 78.44...whats wrong here?

 

[CarlB]

I get the same as you.

At rest coef = 1-sin phi=.53
Avg vert press = 10+ 2 x 16=42
Avg lateral pressure = 42 x .53=22.26
Force = 4 x 22.26=89.04

Maybe some other formula was used for Ko, maybe to account for wall friction?

 

[Stillerz]

CarlB-
Thank you.

They simply added (gamma + q) and multiplied by Ko to get a triangular ditribution. Ko= 0.53, like you show.
My thinking was to have a retangular distribution from "q" + triangular from gamma adding up to make a trapezoid.

So I have:

(q * Ko) = 5.3 KN/m
+
(Gamma *h * Ko) = 33.92 KN/m @ base of all and zero at top.

so the total force is (5.3*4) + (33.92*4)/2 = 89.04

I think the problem must be wrong.