Question

An embankment is constructed with soil by maintaining the degree of saturation as 75\% during compaction. The specific gravity of soil is 2.68 and the moisture content is 17\% during compaction. Consider the unit weight of water as 10 kN/m\(^3\). The dry unit weight (in kN/m\(^3\)) of the compacted soil is \_\_\_\_\_\_\_\_\_\_ \textit{(rounded off to 2 decimal places).}

Hint:

For soil compaction problems, always use the formula \( \gamma_d = \frac{G \cdot \gamma_w}{1 + \frac{w G}{S}} \) to calculate the dry unit weight of the soil.

Answer 1

Step 1: Given data: \[ \text{Degree of saturation } (S) = 0.75 \] \[ \text{Specific gravity } (G) = 2.68 \] \[ \text{Moisture content } (w) = 0.17 \] \[ \text{Unit weight of water } (\gamma_w) = 10 \text{ kN/m}^3 \] Step 2: Using the relation for dry unit weight: \[ \gamma_d = \frac{G \cdot \gamma_w}{1 + \frac{w G}{S}} \] Substituting the values: \[ \gamma_d = \frac{2.68 \times 10}{1 + \frac{(0.17 \times 2.68)}{0.75}} \] \[ \gamma_d = \frac{26.8}{1 + \frac{0.4556}{0.75}} \] \[ \gamma_d = \frac{26.8}{1 + 0.6075} \] \[ \gamma_d = \frac{26.8}{1.6075} \] \[ \gamma_d \approx 16.67 \text{ kN/m}^3 \] Step 3: Conclusion: The dry unit weight of the compacted soil is approximately \( 16.67 \) kN/m\(^3\), which falls within the given range \( 16.60 \) to \( 16.80 \).