Compaction Mould Less Its Collar And Base

The following results were obtained from a compaction test using the 2.5 kg rammer.

Mass of mould + wet soil (g) 2783 3057 3224 3281 3250 3196

Moisture content (per cent) 8.1 9.9 12.0 14.3 16.1 18.2

The weight of the compaction mould, less its collar and base, was 1130 g and the soil had a particle specific gravity of 2.70.

Plot the curve of dry density against moisture content and determine the optimum moisture content. On your diagram plot the lines for 5 per cent and 0 per cent air voids.

Solution

The calculations are best tabulated:

Mass of mould Mass of wet soil Mass of dry soil, Dry density + wet soil = M] M2 = M^l + w) pA = M2/1000

8.1 2783 1653 1529 1.53

9.9 3057 1927 1753 1.75

12.0 3224 2094 1870 1.87 14.3 3281 2151 1882 1.88

16.1 3250 2120 1826 1.83

18.2 3196 2066 1748 1.75

The relationship between pa and Va is given by the expression: 1-Va

Values for the 0 per cent and 5 per cent air voids lines can be obtained by substituting Va = 0 and Va = 0.05 in the above formula along with different values for w.

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