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Sagot :
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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