The wet density of a gravel was found to be 2.32 Mg/m^3 and the field water content was 16%. In the laboratory, the density of the solids was found to be 2.70 Mg/m^3, and the maximum and minimum void ratios were 0.59 and 0.28, respectively. Calculate the relative density (Dr) of the gravel in the field.

the relative density of the gravel in the field is 0.7365 or 73.65%

Explanation:

Given that;

wet density of gravel = 2.3 Mg/m³

field water content w = 16%

density of solids in lab = 2.70 Mg/m³

Maximum void ratio = 0.59

Minimum void ratio = 0.28

first we determine the dry density of gravel   =   / 1+ w

  = 2.3 Mg/m³ / 1 + 0.16 = 2.3/1.16 = 1.9827 Mg/m³

we know that;   = 1000 kg/m³ = 1 g/cm³

Specific Gravity of soil G =  /  = 2.70 Mg/m³ / 1 = 2.70 Mg/m³  

eo = ((G×)/) - 1

eo = (( 2.70 × 1) / 1.9827 ) - 1

eo = 1.3617 - 1

Co = 0.3617

so Relative density will be;

=  - eo  / -

we substitute

= 0.59  - 0.3617 / 0.59 - 0.28

= 0.2283 / 0.31

= 0.7365 or 73.65%

Therefore;  the relative density of the gravel in the field is 0.7365 or 73.65%

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explanation: