Saturated clay shrinkage limit numerical

Given

Initial: $V_1 = 97$ cm³, $M_1 = 202$ g. Dry: $V_d = 87$ cm³, $M_s = 167$ g. $\rho_w = 1$ g/cm³.

(i) Initial water content

$$w_1 = \frac{M_1 - M_s}{M_s} = \frac{202 - 167}{167} = 0.2096 = 20.96\%$$

(ii) Specific gravity

The sample is saturated initially, so $V_1 = V_s + V_{w1}$:

$$V_{w1} = 35\ \text{cm}^3 \Rightarrow V_s = 97 - 35 = 62\ \text{cm}^3$$ $$G = \frac{M_s}{V_s\rho_w} = \frac{167}{62} = 2.69$$

(iii) Shrinkage limit

$$w_s = w_1 - \frac{(V_1 - V_d)\rho_w}{M_s} = 0.2096 - \frac{(97 - 87)}{167} = 0.2096 - 0.0599 = 0.1497$$

So $w_s = 15.0\%$.

Check using the dry volume: $V_d = V_s + w_s M_s = 62 + 0.1497 \times 167 = 87.0$ cm³ (agrees).

Answer: (i) $w = 20.96\%$, (ii) $G = 2.69$, (iii) shrinkage limit $= 15.0\%$.