Question

5. An undisturbed soil sample has a wet density of 2.5 Mg/m3 when the water content is 25%. The specific gravity of the soil particles is 2.7. Determine the submerged effective density.

Answer 1

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;

$\gamma _d = \frac{\gamma _t}{1 +w} = \frac{25}{1+0.25} = 20 \ kN/m^3$

for undisturbed state, the volume of the soil is;

$V_s =\frac{\gamma _d}{G_s \gamma _w} = \frac{20}{2.7*9.81} = 0.76 \ m^3\\\\Void \ volume, V_v = 1-0.76 = 0.24 \ m^3$

$Void \ ratio, \ e = \frac{0.24}{0.76} = 0.32$

Submerged effective density is given as;

$\rho _b = \frac{\rho_w(\gamma_s -1)}{1+e}$

density of water (ρw) = 2.7 x 25 kN/m³ = 67.5 kN/m³, substitute this in the above equation;

$\rho _b = \frac{\rho_w(\gamma_s -1)}{1+e} = \frac{67.5(2.7 -1)}{1+0.32} = 86.93 \ kN/m^3$

Therefore, the submerged effective density is 86.93 kN/m³ or 8.693 Mg/m³