Relative density of sand numerical

Given

$e_{max} = 1.25$, $e_{min} = 0.45$, $D_r = 40\%$, $G = 2.65$, $H = 3$ m, $\gamma_w = 9.81$ kN/m³.

1. Present state

$$D_r = \frac{e_{max} - e}{e_{max} - e_{min}} \Rightarrow e_0 = 1.25 - 0.40(0.80) = 0.93$$ $$\gamma_d = \frac{G\gamma_w}{1+e_0} = \frac{2.65 \times 9.81}{1.93} = 13.47\ \text{kN/m}^3\ (\rho_d \approx 1373\ \text{kg/m}^3)$$

2. After densifying to $D_r = 60\%$

$$e_1 = 1.25 - 0.60(0.80) = 0.77$$

Reduction in thickness (volume of solids constant, area constant):

$$\frac{\Delta H}{H} = \frac{e_0 - e_1}{1+e_0} \Rightarrow \Delta H = 3 \times \frac{0.93 - 0.77}{1.93} = 0.249\ \text{m}$$

New thickness $= 3 - 0.249 = 2.751$ m.

3. New unit weights

$$\gamma_d = \frac{2.65 \times 9.81}{1.77} = 14.69\ \text{kN/m}^3$$ $$\gamma_{sat} = \frac{(G+e_1)\gamma_w}{1+e_1} = \frac{(2.65+0.77) \times 9.81}{1.77} = 18.95\ \text{kN/m}^3$$

Answer: present $\gamma_d = 13.47$ kN/m³; thickness reduces by 0.249 m; new $\gamma_d = 14.69$ kN/m³ and $\gamma_{sat} = 18.95$ kN/m³.