The natural bulk unit weight of a sandy stratum is 18.54 kN/m3 and it has a water content of 8%. For determining the density index, dried sand from the stratum was first filled loosely in a 300 cm3 mould and then vibrated to give a maximum density. The loose dry mass in the mould was 480 g and the dense dry mass at the maximum compaction was 570 g. If the specific gravity of solids is 2.66, find the density index of the sand in the stratum.

Sand Stratum Density Index (GATE Exam Problem)

Problem Statement (GATE Exam)

A sandy stratum has the following properties:

Determine the density index (\( I_D \)) of the sand.

\( \gamma_w = 9.81 \, \text{kN/m}^3 \)
Mould volume = 0.0003 m³
Loose dry density: \( \rho_{d\text{-loose}} = \frac{480}{300} = 1.6 \, \text{g/cm}^3 = 15.696 \, \text{kN/m}^3 \)
Dense dry density: \( \rho_{d\text{-dense}} = \frac{570}{300} = 1.9 \, \text{g/cm}^3 = 18.639 \, \text{kN/m}^3 \)

\( \gamma_d = \frac{\gamma}{1 + w} = \frac{18.54}{1.08} = 17.167 \, \text{kN/m}^3 \)

Natural void ratio: \( e = \frac{G\gamma_w}{\gamma_d} – 1 = \frac{2.66 \times 9.81}{17.167} – 1 = 0.519 \)

Max void ratio (loose): \( e_{\text{max}} = \frac{G\gamma_w}{\gamma_{d\text{-loose}}} – 1 = \frac{2.66 \times 9.81}{15.696} – 1 = 0.662 \)

Min void ratio (dense): \( e_{\text{min}} = \frac{G\gamma_w}{\gamma_{d\text{-dense}}} – 1 = \frac{2.66 \times 9.81}{18.639} – 1 = 0.400 \)

\( I_D = \frac{e_{\text{max}} – e}{e_{\text{max}} – e_{\text{min}}} \times 100 = \frac{0.662 – 0.519}{0.662 – 0.400} \times 100 = 54.58\% \)

Final Result:

Critical Steps:

1. Density Index (54.6%):
Indicates medium-dense sand condition (40-70% range)

2. Engineering Significance:

3. Void Ratio Analysis:
The natural void ratio (0.519) lies between loose (0.662) and dense (0.400) states, showing moderate compaction in situ.