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Answer:
(a) To find the in-situ moist unit weight of the sand, we can use the following equation:
\[
\gamma = (1 + e) \times \frac{\gamma_w \times G}{1 + w}
\]
where:
\(\gamma\) = in-situ moist unit weight of the sand (kN/m³)
\(e\) = void ratio
\(\gamma_w\) = unit weight of water (9.81 kN/m³)
\(G\) = specific gravity of the sand
\(w\) = moisture content
Given:
\(e = 1 - \frac{D_r}{D_{r \text{ max}}} = 1 - \frac{60}{100} = 0.4\)
\(G = 2.65\)
\(w = 0.10\)
Substitute the values into the formula:
\[
\gamma = (1 + 0.4) \times \frac{9.81 \times 2.65}{1 + 0.1} = 1.4 \times \frac{9.81 \times 2.65}{1.1} = 21.09 \, \text{kN/m³}
\]
So, the in-situ moist unit weight of the sand is 21.09 kN/m³.
(b) To calculate the maximum and minimum dry unit weight, we use the formula:
\[
\text{Maximum dry unit weight} = \frac{G}{1 + e_{\text{min}}} \times \gamma_w
\]
\[
\text{Minimum dry unit weight} = \frac{G}{1 + e_{\text{max}}} \times \gamma_w
\]
Given:
\(e_{\text{max}} = 0.94\)
\(e_{\text{min}} = 0.33\)
Calculate:
\[
\text{Maximum dry unit weight} = \frac{2.65}{1 + 0.33} \times 9.81 = \frac{2.65}{1.33} \times 9.81 = 19.72 \, \text{kN/m³}
\]
\[
\text{Minimum dry unit weight} = \frac{2.65}{1 + 0.94} \times 9.81 = \frac{2.65}{1.94} \times 9.81 = 13.36 \, \text{kN/m³}
\]
Therefore, the maximum dry unit weight that the sand can have is 19.72 kN/m³, and the minimum dry unit weight that the sand can have is 13.36 kN/m³.