Problem 1: The table below present the results of a standard proctor test of a borrow pit soil. Mass of wet soil +Mould ...

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Transcribed: Problem 1: The table below present the results of a standard proctor test of a borrow pit soil. Mass of wet soil +Mould in 3.69 3.75 3.82 3.92 3.98 3.93 (kg) Water content (%) 6.2 8 9.8 11.5 12.4 13.3 The volume and the mass of the mould was 1000cm³ and 2 kg, respectively. The specific gravity of the soil is 2.72, determine the following : a) Plot the compaction curve of this soil. b) Optimum water content c) Maximum dry density. d) Plot the zero air void line. e) Determine the degree of saturation at the maximum dry density. Problem 2: The above borrow pit soil has a natural bulk unit weight of 14.5 kN/m and water content of 4.5%. An embankment is to be constructed using this soil and the specification require the fill to be compacted to 96% of the above proctor compaction. Determine the cost for 50,000 m of the embankment considering the following: a) The distance between the borrow pit and the embankment is 55 km b) The cost of transporting 1 m of the borrow pit soil to the site is 16 OMR/km. c) The cost of compaction is 8 OMR/m. Problem 3: plot the variation of the total, neutral, and effective stresses with the depth under the center of the embankment, refer to problem 2 Elevation (m) Water table above. The embankment has a 0. height of 10 m and supported by a ground having the soil profile shown in the figure below, SAND. e= 0.67, G=2.68 silty CLAY. w=34%, G=2.75 Pervious Rock



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Density of water, ρw= 1 g/cm3

Bulk density of soil, ρb= mass of soilvolume of soilHere, volume of soil = volume of mould =1000 cm3Dry density of soil, ρd=ρb1+w=Gρw1+wGS=(1-na)G ρw1+wGFor Zero air void line, Air void (na)= 0Therefore, Dry density for zero air void line, (ρd)ZAV=Gρw1+wG

Mass of wet soil + Mould Mass of mould

Mass of wet soil


Volume of mould             (cm3)

Bulk Density, ρb


Water content, w      (%)

Dry density,





(kg) (kg)
3.69 2 1690 1000 1.69 6.2 1.04 2.33
3.75 2 1750 1000 1.75 8 1.62 2.23
3.82 2 1820 1000 1.82 9.8 1.65 2.15
3.92 2 1920 1000 1.92 11.5 1.72 2.07
3.98 2 1980 1000 1.98 12.4 1.76 2.03
3.93 2 1930 1000 1.93 13.3 1.703 1.99



Civil Engineering homework question answer, step 2, image 1


Optimum moisture content = 12.4 %


Maximum dry density =1.76 g/cm=1.76 x 10-3 kg/cm3


Zero air void line has been plotted in the above figure.


Let the degree of saturation, water content and the void ratio at maximum dry density be S, w and e respectively.


ρd= G ρw1+wGS1.76=2.72 x 11+12.4 x 2.72SSolving the above equation S = 61.83 %

Therefore, the required degree of saturation at maximum dry density is 61.83%.


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