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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 :

Nuhulawal20go

Answer:

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

Explanation:

Given that;

wet density of gravel [tex]r_{b}[/tex] = 2.3 Mg/m³

field water content w = 16%

density of solids in lab [tex]r_{s}[/tex] = 2.70 Mg/m³

Maximum void ratio [tex]e_{max}[/tex] = 0.59

Minimum void ratio [tex]e_{mini}[/tex] = 0.28

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

 [tex]r_{d}[/tex] = 2.3 Mg/m³ / 1 + 0.16 = 2.3/1.16 = 1.9827 Mg/m³

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

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

eo = ((G×[tex]r_{w}[/tex])/[tex]r_{d}[/tex]) - 1

eo = (( 2.70 × 1) / 1.9827 ) - 1

eo = 1.3617 - 1

Co = 0.3617

so Relative density [tex]I_{D}[/tex] will be;

[tex]I_{D}[/tex] = [tex]e_{max}[/tex]  - eo  / [tex]e_{max}[/tex] - [tex]e_{mini}[/tex]

we substitute

[tex]I_{D}[/tex] = 0.59  - 0.3617 / 0.59 - 0.28

[tex]I_{D}[/tex] = 0.2283 / 0.31

[tex]I_{D}[/tex] = 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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