Answer:
The submerged effective density is 86.93 kN/m³ or 8.693 Mg/m³
Explanation:
Given;
wet density of soil sample = 2.5 Mg/m³ = 25 kN/m³
Specific gravity of solid particle = 2.7
The dry unit weight of soil;
[tex]\gamma _d = \frac{\gamma _t}{1 +w} = \frac{25}{1+0.25} = 20 \ kN/m^3[/tex]
for undisturbed state, the volume of the soil is;
[tex]V_s =\frac{\gamma _d}{G_s \gamma _w} = \frac{20}{2.7*9.81} = 0.76 \ m^3\\\\Void \ volume, V_v = 1-0.76 = 0.24 \ m^3[/tex]
[tex]Void \ ratio, \ e = \frac{0.24}{0.76} = 0.32[/tex]
Submerged effective density is given as;
[tex]\rho _b = \frac{\rho_w(\gamma_s -1)}{1+e}[/tex]
density of water (ρw) = 2.7 x 25 kN/m³ = 67.5 kN/m³, substitute this in the above equation;
[tex]\rho _b = \frac{\rho_w(\gamma_s -1)}{1+e} = \frac{67.5(2.7 -1)}{1+0.32} = 86.93 \ kN/m^3[/tex]
Therefore, the submerged effective density is 86.93 kN/m³ or 8.693 Mg/m³
