Let's analyze and solve the problem step-by-step. --- # Given Data: - Soil A: - Effective grain size, \( D_{10A} = .005\, \text{mm} \) Soil B: - Effective grain, \( D_{10} = .5\, \text{mm} \) - Effective pore size \( d_{} = .2 \times D10} \) - Pore pressure distribution is hydrostatic. - Assumption: Uniform soil layers. --- # Step 1: Calculate the effective pore size (\( d_{eff} \)) for each soil \[ d_{eff} = .2 \times D_{10} \] ### For Soil A: \[ d_{effA} = .2 \times .005\, \text{mm} = .001\, \text{mm} \] ### For Soil B: \[ d_{effB} = .2 \times .5\, \text{mm} = .1\, \text{mm} \] --- # Step 2: Determine the potential height of capillary rise (\( h_{max} \)) The capillary rise height is given by the Jurin's Law: \[ h_{max} = \frac{2 \sigma \cos \theta}{\rho g d_{eff}} \] Where: - \( \sigma \): surface tension of water (~.0728 N/m at 20°C), - \( \cos \theta \): contact angle (assumed to be zero for perfect wetting, so \( \cos ^\circ = 1 \)), - \( \rho \): density of water (~100 kg/m³), - \( g \): acceleration due to gravity (~9.81 m/s²), - \( d_{eff} \): effective pore size in meters. ### Convert \( d_{eff} \) to meters: - \( .001\, \text{mm} = 1 \times 10^{-6}\, \text{m} \), - \( .1\, \text{mm} = 1 \times 10^{-4}\, \text{m} \). ### Calculate \( h_{max} \) for each soil: \[ h_{max} = \frac{2 \times .0728 \times 1}{100 \times 9.81 \times d_{eff}} \] --- ### For Soil A: \[ h_{maxA} = \frac{2 \times .0728}{100 \times 9.81 \times 1 \times 10^{-6}} \\ = \frac{.1456}{9.81 \times 10^{-3}} \\ = \frac{.1456}{.00981} \approx 14.85\, \text{meters} \] ### For Soil B: \[ h_{maxB} = \frac{.1456}{100 \times 9.81 \times 1 \times 10^{-4}} \\ = \frac{.1456}{.981} \approx .148\, \text{meters} \] --- # **Answer for Potential Capillary Rise:** | Soil | \( h_{max} \) (meters) | |---------|---------------------| | Soil A | **14.85 m** | | Soil B | **.148 m** | --- # Step 3: Water pressure at the meniscus (at the top of capillary rise) Since the pressure distribution is hydrostatic: \[ P = \rho g h \] - At the water table (reference point): pressure = atmospheric. - At the top of the capillary rise (meniscus): pressure is negative (suction pressure). ### Water pressure at meniscus: \[ P_{meniscus} = - \rho g h \] ### For Soil A: \[ P_{A} = -100 \times 9.81 \times 14.85 \approx -145,713\, \text{Pa} \approx -146\, \text{kPa} \] ### For Soil B: \[ P_{B} = -100 \times 9.81 \times .148 \approx -1,450\, \text{Pa} \approx -1.45\, \text{kPa} \] --- # **Summary:** | Parameter | Soil A | Soil B | |--------------|--------------|--------------| | Max capillary height \( h_{max} \) | **14.85 m** | **.148 m** | | Water pressure at meniscus \( P \) | **-146 kPa** | **-1.45 kPa** | --- # **Effect of Capillarity on Effective Stress:** Capillarity induces a negative pore pressure (suction), which effectively increases the total stress resistance in the soil. The effective stress principle: \[ \sigma' = \sigma - u \] Where: - \( \sigma' \): effective stress, - \( \sigma \): total stress, - \( u \): pore water pressure (negative due to capillarity). **Impact:** - Larger capillary rise (so higher negative pore water pressure) increases the effective stress and stability. - Fine-grained soils with small pores (like Soil A) exhibit higher capillary rise and greater negative pore pressures, enhancing their shear strength. --- # **Final Notes:** - The potential capillary height varies significantly between fine and medium soils. - The hydrostatic pressure at the meniscus indicates the suction that can develop due to capillarity. - Capillarity impacts the effective stress, hence influencing soil stability and strength. --- **Let me know if you'd like further elaboration or specific calculations!**