Relative consistency, shrinkage numerical

Relative consistency

The relative consistency (consistency index) $I_c$ is the ratio of the difference between the liquid limit and the natural water content to the plasticity index:

$$I_c = \frac{w_L - w}{I_p}$$

$I_c = 1$ means the soil is at the plastic limit (stiff), and $I_c = 0$ means it is at the liquid limit (very soft).

Given

$w_L = 60\%$, $w_P = 30\%$, $w_S = 20\%$, $V_{LL} = 40$ cc, $V_{SL} = 23.5$ cc, $I_p = 60 - 30 = 30\%$.

Mass of solids

The soil is saturated at the liquid limit and at the shrinkage limit, and the volume of solids is the same. The volume difference is the volume of water lost between $w_L$ and $w_S$ (density of water 1 g/cc):

$$\begin{aligned} V_{LL} - V_{SL} &= \frac{(w_L - w_S) W_s}{\rho_w} \\ 40 - 23.5 &= (0.60 - 0.20) W_s \\ W_s &= \frac{16.5}{0.40} = 41.25\ \text{g} \end{aligned}$$

Specific gravity

At the shrinkage limit the soil is just saturated: $V_{SL} = V_s + w_S W_s$.

$$\begin{aligned} 23.5 &= V_s + 0.20 \times 41.25 = V_s + 8.25 \\ V_s &= 15.25\ \text{cc} \\ G &= \frac{W_s}{V_s \rho_w} = \frac{41.25}{15.25} = 2.70 \end{aligned}$$

Shrinkage ratio

$$R = \frac{W_s}{V_{SL}\,\rho_w} = \frac{41.25}{23.5} = 1.76$$

Answer: $G = 2.70$, $R = 1.76$.

Phase diagrams (per the sample)

 At liquid limit        At shrinkage limit
 V = 40 cc              V = 23.5 cc
 +--------+            +--------+
 | Water  | 24.75 cc   | Water  | 8.25 cc
 | 24.75 g|            | 8.25 g |
 |--------|            |--------|
 | Solids | 15.25 cc   | Solids | 15.25 cc
 | 41.25 g|            | 41.25 g|
 +--------+            +--------+
 W = 66.0 g             W = 49.5 g
 S = 100%, Va = 0       S = 100%, Va = 0

Weights of water: $0.60 \times 41.25 = 24.75$ g at the liquid limit and $0.20 \times 41.25 = 8.25$ g at the shrinkage limit. At both limits there is no air.