4. A soi-layer is fully saturated with a unit weight of \( 20 \mathrm{kN} / \mathrm{m}^{3} \) and a water table located 1 m below the surface. Calculate the total stress ( \( \sigma \) ), pore water pressure (u), and effective stress ( \( \sigma^{\prime} \) ) at depths of: a) 1 m b) 3 m c) 5 m

At a depth of 1 m, which is at the water table, the total stress (\( \sigma \)) is calculated as the unit weight of the soil times the depth, so \( \sigma = 20 \, \mathrm{kN/m^3} \times 1 \, \mathrm{m} = 20 \, \mathrm{kN/m^2} \). The pore water pressure (\( u \)) at this depth is 0, since it's at the water table. The effective stress (\( \sigma' \)) is then \( \sigma - u = 20 \, \mathrm{kN/m^2} - 0 = 20 \, \mathrm{kN/m^2} \). At a depth of 3 m, total stress is \( \sigma = 20 \, \mathrm{kN/m^3} \times 3 \, \mathrm{m} = 60 \, \mathrm{kN/m^2} \). The pore water pressure (\( u \)) can be calculated as the unit weight of water (≈ 10 kN/m³) times the depth below the water table, so at 2 m (3 m - 1 m), \( u = 10 \, \mathrm{kN/m^3} \times 2 \, \mathrm{m} = 20 \, \mathrm{kN/m^2} \). The effective stress is \( \sigma' = 60 \, \mathrm{kN/m^2} - 20 \, \mathrm{kN/m^2} = 40 \, \mathrm{kN/m^2} \). At a depth of 5 m, total stress is \( \sigma = 20 \, \mathrm{kN/m^3} \times 5 \, \mathrm{m} = 100 \, \mathrm{kN/m^2} \). For the pore water pressure, at 4 m below the water table (5 m - 1 m), \( u = 10 \, \mathrm{kN/m^3} \times 4 \, \mathrm{m} = 40 \, \mathrm{kN/m^2} \). Thus, the effective stress is \( \sigma' = 100 \, \mathrm{kN/m^2} - 40 \, \mathrm{kN/m^2} = 60 \, \mathrm{kN/m^2} \). In summary: - At 1 m: \( \sigma = 20 \, \mathrm{kN/m^2}, u = 0, \sigma' = 20 \, \mathrm{kN/m^2} \) - At 3 m: \( \sigma = 60 \, \mathrm{kN/m^2}, u = 20 \, \mathrm{kN/m^2}, \sigma' = 40 \, \mathrm{kN/m^2} \) - At 5 m: \( \sigma = 100 \, \mathrm{kN/m^2}, u = 40 \, \mathrm{kN/m^2}, \sigma' = 60 \, \mathrm{kN/m^2} \)