Let's analyze and solve this problem step-by-step. --- ## Given Data: | Parameter | Soil A | Soil B | |-------------|---------|---------| | Effective grain size, \( D_{10} \) | .005 mm | .5 mm | | Effective pore size, \( d_{eff} \) | .2 × \( D_{10} \) | .2 × \( D_{10} \) | | Pore pressure distribution | Hydrostatic | Hydrostatic | --- ## Step 1: Calculate the effective pore size \( d_{eff} \) Given: \[ d_{eff} = .2 \times D_{10} \] - For Soil A: \[ d_{eff, A} = .2 \times .005\, \text{mm} = .001\, \text{mm} \] - For Soil B: \[ d_{eff, B} = .2 \times .5\, \text{mm} = .1\, \text{mm} \] --- ## Step 2: Calculate the potential height of capillary rise \( h_{max} \) The capillary rise height is given by Jurin's law: \[ h_{max} = \frac{2 \sigma \cos \theta}{\rho g d_{eff}} \] Where: - \( \sigma \) = surface tension of water (~72.8 mN/m at 20°C) - \( \theta \) = contact angle (~° for perfect wetting, so \( \cos \theta \approx 1 \)) - \( \rho \) = density of water (~100 kg/m³) - \( g \) = acceleration due to gravity (~9.81 m/s²) - \( d_{eff} \) in meters Convert \( d_{eff} \) to meters: - Soil A: \[ d_{eff, A} = .001\, \text{mm} = 1 \times 10^{-6}\, \text{m} \] - Soil B: \[ d_{eff, B} = .1\, \text{mm} = 1 \times 10^{-4}\, \text{m} \] --- ### Calculate \( h_{max} \) for Soil A: \[ h_{A} = \frac{2 \times .0728\, \text{N/m} \times 1}{100\, \text{kg/m}^3 \times 9.81\, \text{m/s}^2 \times 1 \times 10^{-6}\, \text{m}} \] \[ h_A = \frac{.1456}{9.81 \times 10^{-3}} \approx \frac{.1456}{.00981} \approx 14.85\, \text{m} \] ### Calculate \( h_{max} \) for Soil B: \[ h_{B} = \frac{.1456}{9.81 \times 10^{-4}} \approx \frac{.1456}{.000981} \approx 148.5\, \text{m} \] --- ## **Final Answers for Potential Capillary Rise:** | Soil | \( h_{max} \) (m) | |--------|----------------| | A | **14.85 m** | | B | **148.5 m** | --- ## Step 3: Water pressure at the meniscus Since the pressure distribution is hydrostatic, the pressure at the top of the capillary zone (water table) is: \[ P_{water\ at\ meniscus} = \rho g h_{max} \] Calculations: - For Soil A: \[ P_A = 100\, \text{kg/m}^3 \times 9.81\, \text{m/s}^2 \times 14.85\, \text{m} \approx 145,700\, \text{Pa} \approx .146\, \text{MPa} \] - For Soil B: \[ P_B = 100 \times 9.81 \times 148.5 \approx 1,455,000\, \text{Pa} \approx 1.455\, \text{MPa} \] --- ## **Summary:** | Question | **Answer** | |--------------|--------------| | **Potential height of capillary rise** | Soil A: **14.85 m**<br>Soil B: **148.5 m** | | **Water pressure at the meniscus** | Soil A: **~.146 MPa**<br>Soil B: **~1.455 MPa** | --- ## Step 4: Effect of capillarity on effective stresses Capillarity increases the effective stress in the soil by exerting a negative pore pressure (suction). - The total stress (\( \sigma_{total} \)) is unchanged. - The effective stress (\( \sigma' \)) increases by the capillary suction \( u_s \), which is: \[ u_s = \rho g h_{max} \] - When considering effective stress: \[ \sigma' = \sigma - u_a + u_s \] where: - \( u_a \) = pore water pressure (hydrostatic) - \( u_s \) = capillary suction (negative pressure) **Impact:** Capillarity enhances effective stress, increasing soil strength and decreasing deformation potential. --- # **Final Summary:** | Parameter | Soil A | Soil B | |------------|---------|---------| | Max capillary height \( h_{max} \) | **14.85 m** | **148.5 m** | | Water pressure at meniscus | **~.146 MPa** | **~1.455 MPa** | | Effect on effective stress | Increases effective stress due to capillary suction | Significantly increases effective stress due to higher capillary suction | --- Let me know if you'd like further clarification!