Question

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.

Answer 1

Answer:

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

Explanation:

Given that;

wet density of gravel $r_{b}$ = 2.3 Mg/m³

field water content w = 16%

density of solids in lab $r_{s}$ = 2.70 Mg/m³

Maximum void ratio $e_{max}$ = 0.59

Minimum void ratio $e_{mini}$ = 0.28

first we determine the dry density of gravel  $r_{d}$ =  $r_{b}$ / 1+ w

 $r_{d}$ = 2.3 Mg/m³ / 1 + 0.16 = 2.3/1.16 = 1.9827 Mg/m³

we know that;  $r_{w}$ = 1000 kg/m³ = 1 g/cm³

Specific Gravity of soil G = $r_{s}$  / $r_{w}$  = 2.70 Mg/m³ / 1 = 2.70 Mg/m³  

eo = ((G×$r_{w}$)/$r_{d}$) - 1

eo = (( 2.70 × 1) / 1.9827 ) - 1

eo = 1.3617 - 1

Co = 0.3617

so Relative density $I_{D}$ will be;

$I_{D}$ = $e_{max}$  - eo  / $e_{max}$ - $e_{mini}$

we substitute

$I_{D}$ = 0.59  - 0.3617 / 0.59 - 0.28

$I_{D}$ = 0.2283 / 0.31

$I_{D}$ = 0.7365 or 73.65%

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