Soil Pressure Vs. Hydrostatic Pressure 2

[starcasm]

I am working on retaining wall structures (2 walls forming a containment area) where the load cases are dictated to be saturated soil to a level 3 ft. above the wall OR filled with water. I understand the soil pressures are calculated using Ra = 1/2 * ka * p.soil * g * H^2, where p.soil * g = the saturated soil unit weight. What I don't understand is what equations to use for the water case. I have references that state, R = p.water * g * h, where p.water * g = the water unit weight for uniform pressure at depth, h. And R = 1/2 * p.water * g * h, for resultant force on the vertical surface.

When comparing the load cases, do water depths (h) get squared similarly to the soil depth equations? My goal is to get the governing vertical loads on the heel and toe and lateral loads. Thank you in advance.

 

[Ron]

For water only, the pressure is (unit weight of water)*(depth of water) acting at 2/3 of the depth (triangular pressure distribution)

 

[BigH]

If you have saturated soil, you will have pressures for both soil and water that have to be used concurrently. This is why you try to "drain" the backfill so that you don't have the hydrostatic component.

 

[six06]

i think is better if you dont stuck on "formulas" and study where stuff comes from.

The formula you showed for calculation the soil force, which i have to say is the resultant force comes from.

If its a triangular distribution with cero on the top and max on the bottom. The Lateral pressure is calculated as, P = (gamma of soil) * H * Ka. That gives you max pressure at the bottom. Now to get the soil reaction you calculate the area of the triangle and get Ra = 1/2*P*H and thats how you ended up with
Ra= 1/2 * (gamma of soil) * H^2 * Ka.

I hope that would help you calculate the water force.

 

[fattdad]

Step 1: Know whether you are dealing with active or at-rest earth pressure. Let's say it's active.
Step 2: Know the friction angle of the soil. Let's say it's 30 degrees.
Step 3: Know the moist or saturated unit weight. Let's say it's 125 pcf.
Step 4: Know the position of the water table. Let's say it's three feet below the top of the wall.
Step 5: Know the wall height. Let's say it's 12 ft.
Step 4: Calculate the coefficient of active earth pressure - Tan^2(45-phi/2)= 1/3 (0.333333)
Step 5: Develop the stress profile.

Theoritically, the earth pressure at zero feet is zero. I usually include some measure of surcharge pressure. We'll do that later.

At zero feet - use zero psf

At the depth of 3 ft (i.e. at the ground water table) use 125 psf*1/3*3 ft = 125 psf

From 3 ft to 12 ft (i.e., below the ground water table), use [(125 pcf-62.4 pcf)*1/3*12 ft] + (62.4 pcf*12 ft) = 999.2 psf.

Connect the dots.

To allow for a surcharge pressure (i.e., let's use 200 psf at the ground surface for example), that would be a rectangular earth pressure distribution tempered by the coefficient of active earth pressure (i.e., 666 psf for the whole height of the wall).

To get force vectors, calculate the area of the stress graph and apply the vector at the centroid of the mass (or determine force vectors for each geometric area).

Hope this helps.

f-d

¡papá gordo ain’t no madre flaca!

 

[fattdad]

errata: The total horizontal stress at 12 ft (see post above) would then be the sum of 125.0 psf and 999.4 psf or 1,124.4 psf - not counting the presence of a surcharge pressure, which would make it 1,190.4

f-d

¡papá gordo ain’t no madre flaca!

 

[hokie66]

Another errata is required, fattdad. 3 to 12 ft is only 9 ft for the second part of your summation. I get 874 psf at the bottom.

 

[fattdad]

Yes, this forum needs an edit button. Sorry. I'll blame the last week of vacation and Monday's return.

f-d

¡papá gordo ain’t no madre flaca!

 

[BigH]

We could have just pushed him towards a number of geotechnical texts that handle this problem - with worked examples, too. It seems like no one reads texts anymore . . .