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

The submerged effective density is 86.93 kN/m³ or 8.693 Mg/m³

Step-by-step 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 = (\gamma _t)/(1 +w) = (25)/(1+0.25) = 20 \ kN/m^3`

for undisturbed state, the volume of the soil is;

`V_s =(\gamma _d)/(G_s \gamma _w) = (20)/(2.7*9.81) = 0.76 \ m^3\\\\Void \ volume, V_v = 1-0.76 = 0.24 \ m^3`

`Void \ ratio, \ e = (0.24)/(0.76) = 0.32`

Submerged effective density is given as;

`\rho _b = (\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 = (\rho_w(\gamma_s -1))/(1+e) = (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