Volume of the soil excavated from the hole = 0.001337 m³ Weight of soil from the hole when wet = 2220 g Weight of soil when dry = 1890 g What is the in situ water content of the soil?
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- The laboratory compaction test of a certain type of soil gives a maximum dry density of 1486 g/m3 with an optimum moisture content of 12.5%. The following are the results of a field unit weight determination test using sand cone method. Volume of soil excavated from the hole = 0.001337 m3 Weight of soil from the hole when wet = 2220 g Weight of soil from the hole when dried = 1890 g Determine the field unit weight of soil in g/cm3. Determine the relative compaction, then tell whether it is within the allowable range for relative compaction of 95 ± 0.20?The laboratory compaction test of a certain type of soil gives a maximum dry density of 1.486 g/cc3 with an optimum moisture content of 12.5%. Determine the field unit weight of the soil in g/cc3 using the following results of a filed unit weight determination test using sand cone method. Volume of soil excavated from the hole = 0.001337 m3 Weight of moist soil from the hole = 2220 g Weight of oven dried soil = 1890 gThe following results are obtained from standard compaction test for a soil, if Gs = 2.5 Water Content (%) 6.2 8.1 9.8 11.5 12.3 13.2 Unit weight (kN/m3) 16.9 18.7 19.5 20.5 20.4 20.1 Required: 1- The optimum water content (Wc %) 2- The maximum dry unit weight ( dry ) 3- The void ratio (e) 4- Degree of saturation (S %) 5- Find the weight of water needed to be added to 1 m3 to reach 100% saturation.
- The following are results of a field unit weight determination test using sand cone method: Volume of hole = 0.0019 m3 Mass of moist soil from hole = 3.48 kg Water content = 12.43% Max. dry unit weight from a laboratory compaction test = 19.56 kN/m3 Determine the relative compaction, in percent. Hint: Relative compaction = Field dry unit weight / Max dry unit weight (lab)From a compaction test of soil, the following data were obtained in the laboratory:Max. dry unit weight = 18.31 kN/cu.mMin. dry unit weight = 15.25 kN/cu.mRelative density = 64%Find the relative compaction in the field.8 The following are results of a field unit weight determination test using sand cone method: Volume of hole = 0.0019 m3 Mass of moist soil from hole = 3.24 kg Water content = 12.01% Max. dry unit weight from a laboratory compaction test = 19.70 kN/m3 Determine the relative compaction, in percent. Round off to two decimal places. Hint: Relative compaction = Field dry unit weight / Max dry unit weight (lab)
- The ff are results of a field unit weight determination test Volume hole 0.0016 Mass of moist soil from 3.10kg Water content 12.20% Mas dry 19.50 Determine the relative compaction in %The following are results of a field unit weight determination test using sand cone method: Volume of hole= 0.0015 m^3 Mass of moist soil from hole = 3.07 kg Water content = 12.24% Maximum dry unit weight from a laboratory compaction test = 19.28 kN/m^3 Determine the relative compaction, in percent.A proctor compaction test was conducted on a soil sample, and the following observations weremade:Water content, percent 7.7 11.5 14.6 17.5 19.7 21.2Mass of wet soil, g 1739 1919 2081 2033 1986 1948 1. determine the wet density in g/cm32. determine the dry density in g/cm33. determine the dry unit weight in kN/m34. construct the compaction curve 5. determine the optimum moisture content in percent 6. determine the maximum dry density in kN/m37. construct the 80% and 100% saturation lines using w% = 8, 12, 16, 20, and 24 8. calculate the saturation for the maximum dry density (ϒdmax in kN/m3) and the optimummoisture content (wopt %)