Let's analyze the problem step-by-step. --- ## **Given Data:** | Parameter | Value | Units/Remarks | |--------------|---------|----------------| | Wall height, \( H \) | 7 | m | | Backfill soil type | Cohesionless, horizontal | | | Relative density / void ratio, \( e \) | 0.6 | | | Specific gravity, \( G \) | 2.7 | | | Internal friction angle, \( \phi \) | 32° | | | Water table depth | 2 m | below top of wall | | Degree of saturation above water table | 60% | | | Wall is vertical and smooth | | --- ## **Part (i): When backfill is dry and wall is restrained against yielding** --- ### **Step 1: Understand the scenario** - The backfill is dry, so no pore water pressure. - The wall is restrained, so no movement; earth pressure can be calculated using Rankine's theory assuming active or at-rest condition. - For a vertical, smooth wall, the earth pressure is generally calculated using Rankine's passive or active earth pressure coefficients. --- ### **Step 2: Calculate soil properties** - **Void ratio (\( e \))**: 0.6 - **Specific gravity (\( G \))**: 2.7 - **Saturation level**: Since the backfill is dry, no water pressure effects apply. - **Density of soil particles:** \[ \rho_s = G \times \rho_w = 2.7 \times 1000 \, \mathrm{kg/m^3} = 2700\, \mathrm{kg/m^3} \] - **Unit weight of dry soil (\( \gamma_{dry} \))**: \[ \text{Using the relation:} \quad \gamma_{dry} = \frac{\rho_s (1+e)}{1} \times g \times \frac{1}{1+e} \] But more straightforwardly, the typical approximation is: \[ \gamma_{dry} \approx \frac{G \times \rho_w (1 + e)}{1+e} \approx G \times 9.81 \times \frac{1+e}{1+e} = G \times 9.81 \approx 2.7 \times 9.81 \approx 26.4\, \mathrm{kN/m^3} \] **Alternatively**, for cohesionless soils, the approximate dry unit weight: \[ \boxed{\gamma_{dry} \approx 17 \text{ to } 20\, \mathrm{kN/m^3}} \] Given the typical values, and since the problem states properties explicitly, let's use: \[ \boxed{\gamma_{dry} \approx 19\, \mathrm{kN/m^3}} \] (This is standard for such soils; more precise calculations would require detailed analysis, but for this problem, an approximate value suffices.) --- ### **Step 3: Calculate the soil unit weight** \[ \boxed{ \gamma_{dry} \approx 19\, \mathrm{kN/m^3} } \] --- ### **Step 4: Calculate earth pressure coefficients** - **Active earth pressure coefficient (\( K_a \))**: \[ K_a = \tan^2 \left(45^\circ - \frac{\phi}{2}\right) \] \[ K_a = \tan^2 \left(45^\circ - 16^\circ\right) = \tan^2 (29^\circ) \approx (0.55)^2 = 0.3025 \] - **At-rest earth pressure coefficient (\( K_0 \))**: For smooth walls and no yielding, the earth pressure is often approximated using \(K_0\): \[ K_0 = 1 - \sin \phi \] \[ K_0 = 1 - \sin 32^\circ \approx 1 - 0.53 = 0.47 \] **Since the wall is restrained against yielding, the earth pressure is at the at-rest condition.** --- ### **Step 5: Calculate the total earth pressure** - **Total horizontal earth thrust per unit length**: \[ P = \frac{1}{2} \gamma_{dry} H^2 K_0 \] \[ P = 0.5 \times 19\, \mathrm{kN/m^3} \times (7\, \mathrm{m})^2 \times 0.47 \] \[ P = 0.5 \times 19 \times 49 \times 0.47 \] \[ P = 0.5 \times 19 \times 49 \times 0.47 \approx 0.5 \times 19 \times 23 \approx 0.5 \times 437 \approx 218.5\, \mathrm{kN/m} \] --- ### **Step 6: Location of the resultant earth thrust** - The earth pressure diagram is triangular with pressure increasing linearly from zero at the top to maximum at the bottom. - The **resultant** acts at a height of \( H/3 \): \[ \boxed{ \text{Location} = \frac{H}{3} = \frac{7}{3} \approx 2.33\, \mathrm{m} } \] --- ## **Summary for Part (i):** | Parameter | Value | |--------------|---------| | Total earth thrust, \( P \) | **~218.5 kN/m** | | Location of resultant | **~2.33 m from the top** | --- ## **Part (ii): When the wall is free to yield, water table at 2 m, and no drainage** --- ### **Step 1: Understand the scenario** - The wall can move, so active earth pressure applies. - Water table at 2 m depth affects pore pressures. - Soil above water table is partially saturated (60%), affecting the effective unit weight. - Below water table, soil is fully saturated, and pore water pressure contributes to reducing the effective stress. --- ### **Step 2: Adjust soil weight considering water table and saturation** - **Above water table (0-2 m):** - Degree of saturation \( S = 60\% \) - **Bulk unit weight (\( \gamma_{bulk} \))**: \[ \gamma_{bulk} = \gamma_{dry} + \text{pore water pressure contribution} \] For partially saturated soil, the effective unit weight: \[ \gamma_{eff} = \gamma_{dry} - u_w \] where \( u_w \) is pore water pressure, but since the soil is only 60% saturated, the water content contributes less. Alternatively, approximate the **unit weight of the soil above water table**: \[ \gamma_{above} \approx \gamma_{dry} \times (1 - S \times \text{degree of saturation adjustment}) \] For simplicity, approximate as: \[ \gamma_{above} \approx 0.6 \times \gamma_{saturated} \] - **Below water table (2-7 m):** - Soil is saturated; unit weight: \[ \gamma_{sat} \approx 10\, \mathrm{kN/m^3} + \text{pore water pressure effects} \] Typically, for saturated fine or coarse sands: \[ \gamma_{sat} \approx 9.81 \times G_{soil} \times \frac{1+e}{1+e} \approx 19\, \mathrm{kN/m^3} \] But considering the pore water pressure, the **effective unit weight**: \[ \gamma' = \gamma_{sat} - u_w \] At 2 m depth, the pore pressure: \[ u_w = \gamma_{w} \times \text{depth} = 9.81 \times 2 \approx 19.6\, \mathrm{kPa} \] The **total** soil weight at this depth is: \[ \gamma_{sat} \times 2\, \mathrm{m} = 19 \times 2 = 38\, \mathrm{kPa} \] The **effective stress**: \[ \sigma' = \text{total stress} - u_w \] - **Note:** For earth pressure calculations, the **effective weight** is used to calculate the pressure distribution. --- ### **Step 3: Calculate the earth pressure** - **Assuming the water table at 2 m:** - **Above water table (0-2 m):** \[ \text{Effective weight} \approx \gamma_{dry} \times (1 - S) = 19 \times 0.4 = 7.6\, \mathrm{kN/m^3} \] - **Below water table (2-7 m):** \[ \gamma_{sat} \approx 19\, \mathrm{kN/m^3} \] Effective weight: \[ \sigma' = \gamma_{sat} \times (H - 2\, \mathrm{m}) = 19 \times 5 = 95\, \mathrm{kPa} \] --- ### **Step 4: Compute earth pressure coefficients** - Since the wall is free to yield, active earth pressure applies: \[ K_a = \tan^2 \left( 45^\circ - \frac{\phi}{2} \right) \approx 0.303 \] - The **average effective unit weight**: \[ \bar{\gamma}_\text{avg} = \frac{ \int_0^{2} \gamma_{effective} dz + \int_2^{7} \gamma_{effective} dz }{H} \] Calculate the approximate **average effective unit weight**: \[ \gamma_{above} \approx 7.6\, \mathrm{kN/m^3} \] \[ \gamma_{below} \approx 19\, \mathrm{kN/m^3} \] Average: \[ \gamma_{avg} = \frac{(7.6 \times 2) + (19 \times 5)}{7} = \frac{15.2 + 95}{7} = \frac{110.2}{7} \approx 15.74\, \mathrm{kN/m^3} \] --- ### **Step 5: Calculate the earth pressure** \[ P_a = \frac{1}{2} \times \gamma_{avg} \times H^2 \times K_a \] \[ P_a = 0.5 \times 15.74 \times 49 \times 0.303 \] \[ P_a \approx 0.5 \times 15.74 \times 49 \times 0.303 \approx 0.5 \times 15.74 \times 14.85 \approx 0.5 \times 233.9 \approx 117\, \mathrm{kN/m} \] --- ### **Step 6: Location of the resultant** - For active earth pressure, the resultant acts at approximately **0.5 H from the bottom** (or at \( \frac{H}{3} \) from the top). \[ \boxed{ \text{Location} \approx \frac{H}{3} = \frac{7}{3} \approx 2.33\, \mathrm{m} } \] --- ## **Summary for Part (ii):** | Parameter | Value | |--------------|---------| | Total earth thrust, \( P_a \) | **~117 kN/m** | | Location of resultant | **~2.33 m from the top** | --- ## **Final Conclusions:** - **Part (i):** The total earth thrust when the wall is restrained is approximately **218.5 kN/m**, acting at about 2.33 m from the top. - **Part (ii):** When the wall is free to yield with a water table at 2 m, the earth thrust reduces to approximately **117 kN/m**, also acting around 2.33 m from the top, considering the effective stress distribution affected by pore pressures. --- **Note:** These calculations involve simplifications and approximations typical in geotechnical engineering, such as assuming linear pressure distribution, using typical unit weights, and approximating the effects of partial saturation. For precise design, detailed analysis and site-specific data are recommended.