Question

A field unit weight determination test for the soil described in Problem 6.6 yielded the following data: moisture content=10.5% and moist density=1705 kg/m³. Determine the relative compaction.

6.6: The results of a standard Proctor test are given in the following table. Determine the maximum dry density (kg/m²) of compaction and the optimum moisture content. Volume of Proctor mold (cm) Moisture Content (%) Moist unit weight (kg/m3) Dry unit weight 943.3 943.3 943.3 943.3 943.3 943.3 943.3 943.3 Mass of Wet soil in the mold (kg) 1.68 1.71 1.77 1.83 1.86 1.88 1.87 1.85 9.9 10.6 12.1 13.8 15.1 17.4 19.4 21.2 1780.98 1812.78 1876.39 1939.99 1971.80 1993.00 1982.40 1961.20 1620.55 1639.04 1673.85 1704.74 1713.12 1697.61 1660.30 1618.15

Answer 1

ANSWER :
Maximum dry density 35 Papmax) = 1718 kg/m2 Moisture content is W = 10.5% Moist density is p= 1785 kg/m3. Relative compaction is determined using the relation (%) = Pa (field Palmar) x 100 where Palfield- field density of the soil of the soil is calculated using Irelationes, field density a Palfield) = (. اما It 100 = 1705 kg/m3 17. 10.5 100 -X100 =1542.99 kg/m3 Relative compaction is 9. REA) = 15 42.99 kg/m3 1718 kg/m3 R = 89.81%. .. the relative compaction B R = 89.81%