The soil sample could be split up into the following classifications…
Approx 15% clay (Cl)
Approx 10% - fine silt (FSi)
Approx 25% - Medium silt (MSi)
Approx 25% - Course Silt (CSi)
Approx 20% - Fine sand (FSa)
Approx 5% - Fine Sand (FSa)
Soil H - Clay
This soil sample would be classified as Clay as the majority of soil within this sample falls under less than <0.002 – Clay (Cl)
The soil sample could be split up into the following classifications…
Approx 15% of this soil sample is classed as Fine Slit (FSi),
Approx 57% is classed as Clay (Cl)
Approx 15% medium silt (MSi)
The remainder is classed as course silt (CSi), all of this soil sample is classed as fine soil.
Calculating Coefficient
Cc = D230 / (D60D10)
Cc = between 1 and 3 means a well graded soil or gravel.
Soil E
0.552/ 1 x 0.178 = 1.6994
Task 2
Describe how you would assess the nature of a sandy gravel soil sample from the site and describe, with the aid of diagrams the relationship of the three phases. Explain how you would the calculation of the moisture content, bulk density, dry density, saturated density, particle density (specific gravity), degree of saturation, voids ratio, porosity, and air voids.
The relationships of these three phases are,
Void ratio, e = Vv / Vs So when Vs = 1, e = Vv / 1 so Vv = e
Saturation, S = Vw / Vv So when Vs = 1, and Vv = e then S=Vw/e or Vw = Se
Moisture content; Water/moisture content can be calculated by dividing the mass of water by the mass of dry soil, this is then read as a percentage so would be multiplied by 100.
E.g. 50g / 100g = 0.5 x 100 = 50%
to calculate the moisture/water mass in a soil sample you would gather the weight of the wet soil sample e.g. 150g then dry the soil, this would then lower the weight e.g. to 100g.
So mass of water = mass of wet – mass of dry...
… 150g – 100g = 50g
Bulk density Pm, this would be the density of the soil per cm3, this can be calculated by dividing the mass by the volume.
The mass is the total weight of the soil for example would be 150g and the volume 125cm3 so this would give a Bulk density of 1.2g/cm3
Dry density Pd; the dry density would be very similar to calculating the bulk density although in stead of using the mass of the wet sample it would be the mass of the dry sample, Therefore would be calculated by Mass of dry soil / Volume e.g. 100g / 120g = 0.83g/cm3
Saturated density + Degree of saturation;
To calculate the degree of saturation this would be done by dividing the mass of the water (this is determined by drying the soil sample and weighing the difference) by the volume of the voids, this is also generally read as a percentage so would be multiplied by 100 again. This then giving a percentage e.g. 48% to then calculate the density of saturation you would take the 48% off of the total soil sample weight, this then giving an answer e.g. 61.5g.
Particle density; this would be the weight of the soil particles after it has been dried, to calculate this, take the voids volume away from the original soil sample, and the weight of the original soil sample by the weight of the voids. This would then give you a soil weight and a volume.
Divide the soil weight by the soil volume = weight in g / cm3 (Particle density)
Voids ratio; are calculated by adding the volume of air by the volume of water to give a total volume for both (Vv), to work this out as a ratio you would then divide the total volume of the voids by the volume of the soil (Vs) = Vv / Vs, getting a value that’s called “e”.
Porosity (n); can be determined by dividing the total volume of the voids (Vv) by the total Volume of the total soil sample (Vt) then multiplied by 100 to get a percentage.
Air voids; to calculate the air voids you would take the volume of the total soil sample – solid volume – water volume.
Task 3
Describe how you would use the Proctor standard laboratory soil compaction test in BS 1377 to establish the maximum dry density and how the information established from the test would be used on site to confirm that the site compaction had reached the required level of dry density.
The Proctor compaction test and the related modified Proctor compaction test, named for engineer Ralph R. Proctor (1933), are tests to determine the maximum practically-achievable density of soils and aggregates, and are frequently used in geotechnical engineering.
The test consists of compacting the soil or aggregate to be tested into a standard mould using a standardized compaction energy at several different levels of moisture content. The maximum dry density and optimum moisture content is determined from the results of the test.
Soil in place is tested for in-place dry bulk density, and the result is divided by the maximum dry density to obtain a relative compaction for the soil in place.
This test covers the determination of the dry density of soil passing a 20 mm test sieve when it is compacted in a specified manner over a range of moisture contents. The range includes the optimum moisture content at which the maximum dry density for this degree of compaction is obtained. In this test a 2.5 kg rammer falling through a height of 300 mm is used to compact the soil in three layers into a 1 L compaction mould.
This information gained from this test would enable you to on site determine how many times the soil needs to be rolled or compacted, and what moisture the soil will have to hold for it to achieve this.
