Saturated clay 20 cc numerical

Given

$V = 20$ cc, $W = 38$ g, dry weight $W_s = 28$ g, $S = 100\%$.

Solution

Weight (and volume) of water: $W_w = 38 - 28 = 10$ g, so $V_w = 10$ cc (taking $\rho_w = 1$ g/cc).

As the clay is saturated, $V_v = V_w = 10$ cc, so:

$$\begin{aligned} V_s &= V - V_v = 20 - 10 = 10\ \text{cc} \\ e &= \frac{V_v}{V_s} = \frac{10}{10} = 1.0 \\ G &= \frac{W_s}{V_s\,\rho_w} = \frac{28}{10 \times 1} = 2.80 \end{aligned}$$

Check: $w = 10/28 = 35.7\%$, and $e = wG = 0.357 \times 2.8 = 1.0$ (agrees).

Answer: void ratio $e = 1.0$, specific gravity $G = 2.80$.