An embankment, having total volume of 2000 m3 is to be constructed having a bulk density of 1.98 g/cm3 and a placement water content of 18%. The soil is to be obtained either from borrow area A or borrow area B, which have voids ratio of 0.78 and 0.69, respectively and water content of 16% and 12%, respectively. Taking G = 2.66, for both the soils, determine the volume of soil required to be excavated from each of the areas. If the cost of excavation is 35 per m3 in each area, but cost of transportation is 32 and 36 per m3 from areas A and B respectively, which of the borrow areas is more economical?

Soil Mechanics Problems

Problem Statement

An embankment, having a total volume of 2000 m³, is to be constructed with a bulk density of 1.98 g/cm³ and a placement water content of 18%. The soil is to be obtained either from borrow area A or borrow area B, which have voids ratios of 0.78 and 0.69, respectively, and water contents of 16% and 12%, respectively. Taking \( G = 2.66 \) for both soils, determine the volume of soil required to be excavated from each area. If the cost of excavation is \( \text{₹} 35 \) per m³ in each area, but the cost of transportation is \( \text{₹} 32 \) and \( \text{₹} 36 \) per m³ from areas A and B respectively, which borrow area is more economical?

Solution

1. Calculate Dry Density of Embankment

For the embankment, the dry density \( \rho_d \) is given by:

\( \rho_d = \frac{\rho}{1 + w} = \frac{1.98}{1 + 0.18} = \frac{1.98}{1.18} = 1.678 \, \text{g/cm}^3 = 1.678 \, \text{t/m}^3 \)

2. Calculate Dry Mass of Soil

The dry mass \( M_d \) of the embankment is:

\( M_d = \rho_d \cdot V = 1.678 \cdot 2000 = 3356 \, \text{t} \)

3. Borrow Area A

For borrow area A, the dry density \( \rho_d \) is calculated as:

\( \rho_d = \frac{G \cdot \gamma_w}{1 + e} = \frac{2.66 \cdot 1}{1 + 0.78} = \frac{2.66}{1.78} = 1.494 \, \text{t/m}^3 \)

Volume of soil excavated from area A:

\( V_A = \frac{M_d}{\rho_d} = \frac{3356}{1.494} = 2246 \, \text{m}^3 \)

Cost of excavation and transportation from area A:

\( \text{Cost}_A = (35 + 32) \cdot 2246 = \text{₹} 150482 \)

4. Borrow Area B

For borrow area B, the dry density \( \rho_d \) is calculated as:

\( \rho_d = \frac{G \cdot \gamma_w}{1 + e} = \frac{2.66 \cdot 1}{1 + 0.69} = \frac{2.66}{1.69} = 1.574 \, \text{t/m}^3 \)

Volume of soil excavated from area B:

\( V_B = \frac{M_d}{\rho_d} = \frac{3356}{1.574} = 2132 \, \text{m}^3 \)

Cost of excavation and transportation from area B:

\( \text{Cost}_B = (35 + 36) \cdot 2132 = \text{₹} 151372 \)

Result:

Volume of soil excavated from area A: \( V_A = 2246 \, \text{m}^3 \)
Cost from area A: \( \text{₹} 150482 \)
Volume of soil excavated from area B: \( V_B = 2132 \, \text{m}^3 \)
Cost from area B: \( \text{₹} 151372 \)
Conclusion: Borrow area A is more economical.

Explanation

This problem compares the cost-effectiveness of obtaining soil from two different borrow areas with different voids ratios and transportation costs. The calculations involve determining the required volume of soil and the total cost, including excavation and transportation, for each area.

Borrow area A, despite requiring a larger volume of soil, is more economical due to its lower transportation cost, highlighting the significance of both material properties and logistical expenses in construction projects.

Physical Meaning

The selection of a borrow area for soil depends on factors such as voids ratio, water content, and cost of excavation and transportation. A soil with a lower voids ratio provides higher dry density, reducing the volume required but may incur higher transportation costs. Balancing these factors ensures cost-effective construction while meeting design specifications.