Cylindrical specimen, 95% compaction

Given

$D = 10$ cm, $H = 20$ cm, $G = 2.65$, $w = 15\%$, $\gamma_w = 9.81$ kN/m³.

Assumption: the weight "30 kN" is a misprint for 30 N, since 30 kN in this small mould would give an impossible unit weight of about 19,000 kN/m³.

Volume and unit weights

$$V = \frac{\pi}{4}(0.10)^2(0.20) = 1.571 \times 10^{-3}\ \text{m}^3$$ $$\gamma = \frac{30\ \text{N}}{1.571 \times 10^{-3}} = 19\,099\ \text{N/m}^3 = 19.10\ \text{kN/m}^3$$ $$\gamma_d = \frac{\gamma}{1+w} = \frac{19.10}{1.15} = 16.61\ \text{kN/m}^3$$

Void ratio

$$e = \frac{G\gamma_w}{\gamma_d} - 1 = \frac{2.65 \times 9.81}{16.61} - 1 = 0.565$$

Degree of saturation

$$S = \frac{wG}{e} = \frac{0.15 \times 2.65}{0.565} = 0.703 = 70.3\%$$

(ii) Field requirement

Taking the specimen's dry unit weight as the maximum (laboratory) dry unit weight:

$$\gamma_{d,field} = 0.95 \times \gamma_{d,max} = 0.95 \times 16.61 = 15.78\ \text{kN/m}^3$$

Answer: (i) $\gamma_d = 16.61$ kN/m³, $e = 0.565$, $S = 70.3\%$; (ii) field $\gamma_d \ge 15.78$ kN/m³.