Q4 A soil embankment is to be constructed at a project site

Q4 A soil embankment is to be constructed at a project site, and a Proctor test was conducted on the fill soil and the maximum dry density (d,max) was found to be 1988 kg/m3. The project requires that the in situ dry density of soil be above 95% of d,max by the Proctor test. In order to determine the in situ density of the soil at the embankment, a sand replacement test was performed and the following results recorded:

Mass of soil extracted from the hole 2.26 kg

Mass of soil after oven drying 2.08 kg

Mass of sand pouring cylinder initially 5.32 kg

Mass of sand pouring cylinder finally 3.11 kg

Volume of cone in sand pouring cylinder 0.25x10-3 m3

Density of sand in cylinder 1650 kg/m3

If the specific gravity of the soil particles is assumed to be 2.65, calculate the dry density, the air content and the degree of saturation of the soil. Explain whether the field compaction has satisfied the project specifications.

Solution

SOLUTION-

GIVEN DATA-

Mass of soil extracted from the hole= 2.26 kg= W (Say)

Mass of soil after oven drying=2.08 kg= W_S (Say)

Therefore Weight of Water (W_w)= (2.26-2.08) Kg = 0.18 Kg.

Mass of sand pouring cylinder initially = 5.32 kg = W₁( Say)

Mass of sand pouring cylinder finally = 3.11 kg= W₄ (Say)

Volume of cone in sand pouring cylinder = 0.25x10^-3 m³

Density of sand in cylinder =1650 kg/m³

Therefore Mass of sand in the cylinder= Volume of cone * Density of sand = 0.25x10^-3x 1650 Kg = 0.4125 Kg= W₂(Say).

From this we can determine the Weight of sand in the Pit W_p(say)  = (W₁- W₄-W₂) Kg = (5.32- 3.11-0.4125) Kg= 1.7975 Kg.

Therefore Volume of Sand required to fill the pit (V_p ) = W_p/ Density of sand= 1.7975/ 1650 = 1.0893x10^-3 m³

Bulk Density or Wet Density of Soil= Mass of soil extracted from the hole/ Volume of the pit

=2.26/(1.08* 10^-3) = 2074.5 Kg/m³

and Water Content= (Weight of Water/ Weight of soil after drying) x 100 = (0.18/ 2.08) x 100 = 8.65%.( Note- Here the water content is in percentage but in solving numericals we will take the water content in units i.e 0.0865).

and Dry Density of soil= Bulk Density or Wet Density of Soil/ (1+ water content)= 2074.5/(1+ 0.0865)= 1909.34 Kg/m³

We alsoknow that Dry density = (Specific gravity of the soil * Unit Weight of Water)/ ( 1 + void ratio ).

Now putting the values of Dry density obtained from above and specific gravity and Unit Weight of Water Which is 1000 kg/m³ in the above equation we get the value of void ratio which is= 0.3879.

We also know the relation between void ratio , degree of saturation, water and specific gravity Which is :-

ex S = W x G

where e is void ratio, S is degree of saturation, W is water content and G is specific gravity, putting all the known values, we get Degree of Saturation= ( WxG)/e = 0.0865x 2.65/ 0.3879 = 0.5909 which is equal to 59.09%.

(Note- Air content couldnt be determined because Volume of air was not specified.)

The field compaction has satisfied the project specifications because the in-situ dry density obtained = 1909.34 kg/m³ Which is about 96.04% of the maximum dry density by Procter test i.e 1988 kg/m³.

(1909.34/ 1988)x 100 = 96.04%

(requirement was 95%).