calculation of stress increase at clay for consolidation settlement

[ketanco]

In six minute solution for geaotechnical, problem 51

you place a very large area of fill on a clay layer of 30' thick. water table is at first at ground surface at first.

they ask the consolidation settlement of clay due to the loading by the fill surcharge and also they are saying that the water table is being lowered to 12' below ground at the same time so there is a stress increase for it too of course..

here is my question... for calculating the final effective stress at the center of clay layer, (which is at 15' depth), i took the dry density of clay above water table (since water is lowered by 12' feet now) so 12 feet of dry clay plus 3 feet of bouyant clay plus surcharge. thats what i used to calculate the final effective stress.

but in the solution, when calculating final effective stress, they only took the initial stress (which is buoyant unit weight clay times 15 feet), plus new surcharge from fill, plus the increase of stress due to lowering of water by 12' so they added (15-3)x62.4 to the surcharge and found final effective stress that way.

but why can we not take the dry clay that is now above water table, plus wet clay of 3 feet plus the surcharge when calculating final stress? there is about 10 percent difference in final stress amounts between these two approach. they dont consider the dry density at all. also isnt their approach of just deducting water of 12 feet wrong? because that 12 feet was not entirely water. it contained soil. they treat that 12 feet as if it was entirely water that way. or am i wrong?

 

[civilized_naah]

A soil that exists above the water table is not submerged, but it is not DRY either. The dry density of a soil is calculated as:

Weight of soil solids only divided by total volume.

If the soil is above the water table, it will still have some moisture (as expressed by the water content w) and the density used should be the actual (or moist) density, which is

Dry density times (1+w)

The effect of lowering the water table is simply to replace the submerged density of the affected soil with the actual density for the depth of the lowering, therefore the difference is

unit weight of water x depth of the lowering

 

[djgint]

K,

Good question. The reason why they used the saturated density above water table after lowering it instead of using a dry density or moist density is because clay is a very slow draining soil. So even though the water table has been lowered, the clay soil above it should still be saturated. Regarding your second question, I don't know why they added 12 ft x unit weight of water. But my alternative solution for calculating log (p'o + p'v /p'o) is log (effect stress after lowering w.t. and surch./p'o)

effect. stress after lowering w.t. and surch. = (120.76 pcf)(12 ft) + [(120.76 - 62.4) pcf (3 ft)] + (115 pcf)(10 ft)

= 2774.2 psf

so, log (2774.2 psf / 875 psf) = 0.5011

S = (0.28) (30 ft / 1 + 0.807) (0.5011)

S = 2.329 ft x 12 in / ft

S = 27.95 in or 28 in.

 

[civilized_naah]

"Isn't their approach of just deducting water of 12 feet wrong? because that 12 feet was not entirely water. it contained soil. they treat that 12 feet as if it was entirely water that way."

If you assume that the density of the soil does not change much, the DIFFERENCE between the original effective stress and the new effective stress is still 12x62.4, regardless of what the density was to begin with. Hence, since they already calculated the original effective stress, they added on this DIFFERENCE to calculate the new effective stress

e.g if the clay layer (fully submerged) had a thickness of 20 ft and unit weight of 120 pcf and then the ground water was lowered 12 ft, then the original effective stress at a depth of 20 ft would be:

p1 = 20x(120-62.4)

But after lowering the GWT by 12 ft, the effective stress is

p2 = 12x120 + 8x(120-62.4) = 20x(120-62.4) + 12x62.4

The increase (in bold) is independent of the clay density (120 pcf)