A core cutter 12.6 cm in height and 10.2 cm in diameter weighs 1071 g when empty. It is used to determine the in-situ unit weight of an embankment. The weight of core cutter full of soil of2970 g. If the water content is 6%, what are in situ-dry weight and porosity? (ii) if the embankment gets fully saturated due to heavy rains what will be the increase in water content and bulk unit weight, if nonvolume change occurs? The specific gravity of soil solids is 2.69.

A core cutter 12.6 cm in height and 10.2 cm in diameter weighs 1071 g when empty. It is used to determine the in-situ unit weight of an embankment. The weight of core cutter full of soil of2970 g. If the water content is 6%, what are in situ-dry weight and porosity? (ii) if the embankment gets fully saturated due to heavy rains what will be the increase in water content and bulk unit weight, if nonvolume change occurs? The specific gravity of soil solids is 2.69.

Problem Statement (Civil Services Exam 1997)

A core cutter with:

  • Height = 12.6 cm
  • Diameter = 10.2 cm
  • Empty weight = 1071 g
  • Soil-filled weight = 2970 g
  • Water content (\( w \)) = 6%
  • Specific gravity (\( G \)) = 2.69

Determine:

  1. In-situ dry weight and porosity
  2. If saturated by rain: water content increase and bulk unit weight change

Solution

1. Volume Calculation

\( V = \frac{\pi}{4}D^2H = \frac{\pi}{4}(10.2)^2(12.6) = 1029.58 \, \text{cm}^3 \)

2. Bulk Density Determination

\( \text{Mass of soil} = 2970 – 1071 = 1899 \, \text{g} \)
\( \rho = \frac{M}{V} = \frac{1899}{1029.58} \approx 1.844 \, \text{g/cm}^3 \)
\( \gamma = \rho \cdot g = 1.844 \times 9.81 = 18.094 \, \text{kN/m}^3 \)

3. Dry Density Calculation

\( \rho_d = \frac{\rho}{1 + w} = \frac{1.844}{1 + 0.06} \approx 1.74 \, \text{g/cm}^3 \)
\( \gamma_d = \rho_d \cdot g = 1.74 \times 9.81 \approx 17.07 \, \text{kN/m}^3 \)

4. Void Ratio Determination

\( e = \frac{G \cdot \gamma_w}{\gamma_d} – 1 = \frac{2.69 \times 9.81}{17.07} – 1 \approx 0.546 \)

5. Saturation Analysis

\( w_{\text{sat}} = \frac{e}{G} = \frac{0.546}{2.69} \approx 20.3\% \)
\( \gamma_{\text{sat}} = \gamma_d(1 + w_{\text{sat}}) = 17.07(1 + 0.203) \approx 20.535 \, \text{kN/m}^3 \)
Results:
  • Dry unit weight: \( \gamma_d \approx 17.07 \, \text{kN/m}^3 \)
  • Void ratio: \( e \approx 0.546 \)
  • Water content increase: \( 14.3\% \)
  • Bulk unit weight increase: \( 2.441 \, \text{kN/m}^3 \)

Explanation

Key Relationships:

  • Bulk density relates total mass to volume
  • Dry density accounts for water content reduction
  • Void ratio connects particle arrangement to measurable properties
  • Saturation transforms air voids to water-filled voids

Physical Meaning

1. Density Values:

  • High dry density (17.07 kN/m³) indicates compact soil structure
  • Void ratio 0.546 suggests moderate porosity

2. Saturation Effects:

  • 14.3% water content increase shows significant air void replacement
  • 2.44 kN/m³ weight gain equals ~2.5 tons/m³ extra load on substrate

Engineering Significance:

  • Original state: Good bearing capacity
  • Saturated state: Reduced shear strength, increased pore pressure
  • Highlights need for proper drainage in embankments

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