CELE Geotechnical Engineering — Lateral Earth Pressure and Retaining StructuresStudy Notes

Lateral earth pressure is the horizontal force exerted by soil against a retaining structure. Unlike vertical loads (which compress soil), lateral pressure tries to move a wall horizontally. The magnitude depends on three factors: (1) the soil's strength and density, (2) the height of the retained soil, and (3) whether the wall moves relative to the soil. In practice, a retaining wall experiences three distinct earth-pressure states: **At-Rest State (K₀):** The wall is rigid and does not move. No strain occurs in the soil parallel to the wall. This condition occurs in basements with stiff bracing or in very old, settled structures. The pressure coefficient K₀ = 1 − sin(φ), where φ is the soil's effective friction angle. **Active State (Kₐ):** The wall moves away from the soil (or allows the soil to expand). Soil shear strength is fully mobilized in tension. This is the most common design condition because it represents the worst-case scenario for wall design. Active pressure is the smallest of the three states. **Passive State (Kₚ):** The wall is pushed into the soil, compressing it. This state occurs at the toe of a retaining wall or in front of a pile or sheet pile. Passive pressure is the largest and is often neglected conservatively in design. The physical reason for these differences: in the active state, the soil is trying to expand (it's in tension), so the horizontal stress is reduced. In the passive state, the soil is compressed, so the horizontal stress is increased. Engineers exploit passive pressure as a resisting force when calculating stability.

Rankine theory (1857) is the most widely used method for calculating earth pressure on smooth, vertical walls with horizontal backfill. It assumes no friction between soil and wall, and uses principal stress concepts from soil mechanics. **Rankine Earth-Pressure Coefficients:** For a cohesionless soil (dry sand or gravel): Active pressure coefficient: Kₐ = (1 − sin φ)/(1 + sin φ) = tan²(45° − φ/2) Passive pressure coefficient: Kₚ = (1 + sin φ)/(1 − sin φ) = tan²(45° + φ/2) = 1/Kₐ At-rest coefficient: K₀ = 1 − sin φ These formulas show that Kₐ and Kₚ are reciprocals, and all three are functions of φ alone (for Rankine). **Physical Interpretation:** The formula tan²(45° ± φ/2) comes from the Mohr circle of stress. When a soil element reaches the active state, its major principal stress becomes horizontal (pointing toward the wall), and the minor principal stress becomes vertical. The angle 45° ± φ/2 represents the orientation of the failure plane within the soil (the slip surface). **Pressure Distribution:** In active state, pressure increases linearly with depth: σₐ(z) = Kₐ γ z where z is the depth below the surface and γ is the unit weight of soil. This linear distribution means the pressure profile is triangular, with zero pressure at the top and maximum pressure at the base. **Passive Pressure:** Similarly, σₚ(z) = Kₚ γ z. However, passive pressure is usually calculated only at the toe of a wall or in front of sheet piles, where the soil is being compressed. **Effect of Friction Angle:** - φ = 25°: Kₐ ≈ 0.406, Kₚ ≈ 2.46 - φ = 30°: Kₐ ≈ 0.333, Kₚ ≈ 3.00 - φ = 35°: Kₐ ≈ 0.271, Kₚ ≈ 3.69 - φ = 40°: Kₐ ≈ 0.217, Kₚ ≈ 4.60 Notice that as φ increases, Kₐ decreases dramatically. A 5° increase in φ can reduce active pressure by 20–30%.

Coulomb theory (1776), developed before Rankine, is more general and accounts for: 1. Wall friction (angle δ between soil and wall) 2. Sloping backfill (angle β to horizontal) 3. Sloping wall face (angle α from vertical) 4. Cohesion in the soil Coulomb assumes a planar failure surface (straight slip line), whereas Rankine assumes a curved surface. For vertical walls with horizontal backfill and no wall friction (δ = 0, β = 0, α = 0), Coulomb reduces to Rankine. **General Coulomb Formula (for cohesionless soil):** Kₐ = sin²(α + φ) / [sin²α × sin(α − δ) × (1 + √[sin(φ + δ) sin(φ − β) / sin(α − δ) sin(α + β)])²] (The full expression is complex, but specialized tables and software are typically used.) **Key Variables:** - δ = wall friction angle (angle between soil-wall interface and horizontal) - β = slope angle of backfill - α = angle of wall from vertical (0° for a vertical wall) - φ = soil friction angle **Physical Meaning of Wall Friction:** Wall friction (δ) reduces the active pressure coefficient but increases the passive one. In active state, friction between soil and wall helps hold the soil back, so Kₐ decreases. In passive state, friction adds to the wall's resistance, so Kₚ increases. For design: - Active: δ typically = 0 to φ/2 (often 0 for smooth concrete; higher for rough surfaces) - Passive: δ typically = φ/2 to φ (higher values mean better resistance) **Sloping Backfill (β > 0):** When the backfill slopes upward at angle β, the active pressure increases significantly. This is why retaining walls with level backfill are preferred from a design standpoint. For a 1-in-2 slope (β ≈ 27°), active pressure can increase by 50% or more compared to horizontal backfill. **Practical Application:** In many Philippine projects, especially hillside construction in Metro Manila, Batangas, and Cebu, sloping backfill is unavoidable. Coulomb theory must be used to account for this. Conservative design often uses a vertical imaginary wall behind the actual wall to create horizontal backfill, which is then analyzed using Rankine.

The total horizontal force (thrust) exerted by soil on a wall is the resultant of the pressure distribution. For linear pressure distribution (cohesionless soil), this resultant is easy to calculate. **For Cohesionless Soil (Active State):** With pressure distribution σₐ(z) = Kₐ γ z (linear from 0 to Kₐ γ H), the resultant thrust is: Pₐ = (1/2) Kₐ γ H² This is the area of the triangular pressure diagram. The resultant acts at the centroid of the triangle, which is at height H/3 above the base (or 2H/3 below the top surface). **Key Insight:** Notice the H² term. If wall height doubles, the thrust quadruples. This is why tall walls are disproportionately expensive to design and build. **Passive Pressure:** Similarly, Pₚ = (1/2) Kₚ γ H² Passive pressure typically acts at H/3, but it is often neglected in design (used only if mobilization is guaranteed, such as at the toe with surcharge or deep embedment). **Effect of Surcharge (Additional Surface Load):** If a uniform surcharge q (in kPa or kN/m²) is applied over the backfill surface, it adds a rectangular pressure block to the triangular diagram: Surcharge contribution: Pₐ,ₛᵤᵣ = Kₐ q H This acts at height H/2 (centroid of rectangle). Total active thrust: Pₐ,ₜₒₜₐₗ = (1/2) Kₐ γ H² + Kₐ q H **Effect of Cohesion:** Cohesion reduces active pressure. Near the surface, where depth z is small, the pressure can become negative (tension). This is because cohesion "glues" the soil together. The active pressure with cohesion is: σₐ(z) = Kₐ γ z − 2c√(Kₐ) where c is the cohesion (undrained shear strength for clay, or apparent cohesion for sandy clay). The pressure becomes zero (tension crack forms) at depth: zc = 2c√(Kₐ) / (Kₐ γ) = 2c / (γ√(Kₐ)) Below this depth, the pressure follows the standard formula. The reduced active thrust is: Pₐ,reduced = (1/2) Kₐ γ (H − zc)² and it acts at height (H − zc)/3 above the base. **Effect of Water (Hydrostatic Pressure):** If water is present in the backfill, two things happen: 1. Soil below the water table has reduced effective unit weight: γ' = γsat − γw (typically 8–10 kN/m³ for saturated soil) 2. Water exerts full hydrostatic pressure: Pwater = (1/2) γw hw², where hw is the depth of water The soil contribution is calculated using γ' below the water table, and the water pressure is added separately. In analysis, these can be combined: Total pressure at depth z below water table: σtotal = Kₐ γ' z + γw z For water, the pressure coefficient is effectively 1.0 (no restraint). **Example Pressure Profile (Mixed Conditions):** Consider a wall 8 m high with: - Upper 4 m: dry sand, γ = 18 kN/m³, φ = 30°, Kₐ = 0.333 - Lower 4 m: saturated sand (water table at 4 m depth), γsat = 20 kN/m³, γ' = 10 kN/m³ At z = 4 m (water table): σ = Kₐ γ (4) = 0.333 × 18 × 4 = 24 kPa At z = 8 m (base): Soil contribution: Kₐ γ' (4) = 0.333 × 10 × 4 = 13.3 kPa Water contribution: γw (4) = 9.81 × 4 = 39.2 kPa Total: 24 + 13.3 + 39.2 = 76.5 kPa Notice how water dominates below the water table.

Once the earth pressures are known, a retaining wall must be checked for three external failure modes. These are checked by analyzing the wall as a rigid body in equilibrium. (Internal design—reinforcement and concrete strength—follows separately.) **Check 1: Overturning** The wall must not tip over. The earth pressure creates a moment (clockwise, trying to rotate the wall away from the soil) that must be resisted by the moment from the wall's weight and any other stabilizing forces (like passive pressure at the toe). Resisting moment = ΣMR (created by wall weight, multiplied by its horizontal distance from the pivot point, usually the toe) Overturning moment = ΣMO (created by active pressure, multiplied by its height from the toe) Factor of safety against overturning: FSOT = ΣMR / ΣMO Typical requirement: FSOT ≥ 1.5 to 2.0 (NSCP 2015 typically specifies 1.5 for most cases, 2.0 for critical structures) Calculation Steps: 1. Identify the toe as the pivot point (it's the most likely hinge). 2. Calculate moments of all weights (wall, soil on heel, surcharge) about the toe. Assume they act at their geometric centroids. Weights produce counterclockwise moments (resisting). 3. Calculate the moment of the active thrust about the toe. It produces a clockwise moment (overturning). 4. Divide resisting by overturning. **Check 2: Sliding** The wall must not slide horizontally. Friction between the wall base and the foundation soil resists the horizontal component of the active thrust. Resisting force = μ ΣW + Pₚ (if passive pressure is counted) Sliding force = Pₐ,H (horizontal component of active thrust) μ = coefficient of friction between wall base and foundation soil (typically 0.4–0.6 for concrete on sand/gravel; lower for clay or silt) W = total vertical weight of wall + soil on heel + surcharge Pₚ = passive pressure at the toe (often neglected) Factor of safety against sliding: FSslide = [μ ΣW + Pₚ] / Pₐ,H Typical requirement: FSslide ≥ 1.5 (NSCP 2015) Note: Passive pressure at the toe is sometimes included if the wall can move slightly into the soil (embedment depth is sufficient). However, in conservative design, passive is often neglected entirely. **Check 3: Bearing Capacity** The foundation soil must not fail under the combined stresses from the wall. Additionally, the resultant of all forces (vertical and horizontal) must fall within the middle third of the base to avoid excessive tension at the toe. Resultant position: e = B/2 − ΣM / ΣW where: B = base width of wall e = eccentricity (distance of resultant from center of base) ΣM = net moment about the center of the base (taking all moments including overturning) ΣW = total vertical weight Condition: e ≤ B/6 (ensures the resultant is in the middle third) If e > B/6, tension occurs at the heel, which is problematic for concrete walls. If e is very large, bearing capacity is exceeded. Base pressure (at toe and heel) using linear distribution: qtoe = ΣW / B + ΣM / (B²/6) [neglecting horizontal loads for simplicity] qheel = ΣW / B − ΣM / (B²/6) Both must satisfy: q ≤ qallowable (from site geotechnical investigation) **Typical Wall Geometry (Cantilever Wall):** Cantilever walls are the most common type in moderate-height applications (up to about 8 m). They consist of: - A vertical stem (the part in contact with soil) - A base slab with heel and toe - Sometimes a shear key (a small vertical wall at the toe to resist sliding) The heel is the part where soil rests on top of the base slab (this adds weight, helping stability). The toe is the unsupported part (or supported by passive pressure). **Design Workflow:** 1. Assume initial wall dimensions (height, base width, stem thickness). 2. Calculate earth pressure (active at back, passive at toe if counted). 3. Calculate wall weight and soil weight on heel. 4. Check overturning about the toe. 5. Check sliding along the base. 6. Check bearing capacity and resultant position. 7. If any check fails, increase wall dimensions and repeat. 8. Once external stability is satisfied, design internal reinforcement (ACI 318). **NSCP 2015 Requirements:** The National Structural Code of the Philippines, Section 419 (Design of Reinforced Concrete Cantilever Walls), specifies: - Minimum FSOT = 1.5 - Minimum FSslide = 1.5 - Bearing pressure must not exceed the allowable capacity of the foundation soil - Resultant force should ideally be in the middle third (e ≤ B/6) - Reinforcement must be adequate for both flexure and shear (per ACI 318) **Common Failure Scenarios:** 1. **Overturning:** Wall is too thin or too light. Increasing base width or adding a thicker stem helps. 2. **Sliding:** Base friction is insufficient. Increase base width, improve soil friction, or add a shear key (extends into soil). 3. **Bearing Failure:** Foundation soil is weak. Use a wider base, improve foundation, or deepen embedment. 4. **Eccentric Loading:** Heel not wide enough. Add more heel length or reduce active pressure (e.g., by reducing height or using batter).

Real soils are more complex than the idealized cases. This section covers practical scenarios commonly encountered in Philippine geotechnical design. **6.1 Cohesive Soils (Clays and Sandy Clays)** Cohesion is the shear strength that exists even when normal stress is zero. In the Mohr-Coulomb criterion, τ = c + σ tan(φ), the c term is cohesion. For active earth pressure in cohesive soil: σₐ(z) = Kₐ γ z − 2c√(Kₐ) The negative 2c√(Kₐ) term represents the "pull" that cohesion provides. Near the surface, where z is small, this can make σₐ negative (tension in the soil). **Tension Crack Formation:** The pressure becomes zero (a tension crack forms) at depth: zc = 2c / [γ √(Kₐ)] Below this depth, the wall does not receive pressure from the cracked zone. So, for design purposes, only the pressure below zc contributes to the thrust. Reduced active thrust: Pₐ,reduced = (1/2) Kₐ γ (H − zc)² This acts at height (H − zc)/3 from the base. **Example (Stiff Clay):** A wall 6 m high backs stiff clay: c = 20 kPa, φ = 22°, γ = 18 kN/m³. Kₐ = tan²(45° − 22°/2) = tan²(34°) = 0.475 √(Kₐ) = 0.689 Tension crack depth: zc = 2 × 20 / [18 × 0.689] = 40 / 12.4 = 3.23 m So the upper 3.23 m is in tension and does not push on the wall. The active thrust comes from the lower 6 − 3.23 = 2.77 m: Pₐ,reduced = 0.5 × 0.475 × 18 × (2.77)² = 0.5 × 0.475 × 18 × 7.67 = 32.8 kN/m Compare to the case with no cohesion (c = 0): Pₐ = 0.5 × 0.475 × 18 × 36 = 153.9 kN/m Cohesion reduces the thrust by a factor of 4.7! This is why clay-backed walls are often more stable than sand-backed walls. **Important Caveat:** Cohesion is time-dependent. In the short term (immediately after construction), undrained cohesion cu can be significant. In the long term, especially in humid climates (common in the Philippines), effective cohesion φ' is much lower (often close to zero in heavily weathered clay). Design typically uses the lower of the two (long-term strength) for safety. **6.2 Saturated Soils and Seepage** Water in soil affects lateral pressure in three ways: 1. **Effective Stress:** Saturated soil has reduced effective unit weight γ' = γsat − γw. For typical soil γsat ≈ 19–20 kN/m³ and γw = 9.81 kN/m³, we get γ' ≈ 9–10 kN/m³. 2. **Hydrostatic Pressure:** Water exerts full lateral pressure independent of soil friction: σwater = γw × (depth of water). 3. **Seepage and Uplift:** Water flowing through soil during rainfall or groundwater rise can create seepage forces that reduce effective stress further. **Calculation with Water Table:** If the water table is at depth zw below the surface: - Above the water table: σ = Kₐ γ z (using dry γ) - Below the water table: σ = Kₐ γ' (z − zw) + Kₐ γ zw (or approximately Kₐ γ' × (depth below water table) + soil pressure at water table) - Water pressure: σwater = γw × (depth below water table) - Total: σtotal = Kₐ γ' × depth(below water) + γw × depth(below water) + pressure at water table from above The water pressure can dominate, especially in deep walls or with a high water table. **Example (Saturated Fill, High Water Table):** Wall 8 m high, water table at 2 m below surface: - Upper 2 m: dry sand, γ = 18, φ = 30°, Kₐ = 0.333 - Lower 6 m: saturated sand, γsat = 20, γ' = 10, φ = 30° Pressure at z = 2 m (water table): σ = Kₐ γ (2) = 0.333 × 18 × 2 = 12 kPa At z = 8 m (base): - Soil: σsoil = 12 + Kₐ γ' (6) = 12 + 0.333 × 10 × 6 = 12 + 20 = 32 kPa - Water: σwater = γw (6) = 9.81 × 6 = 58.9 kPa - Total: 32 + 58.9 = 90.9 kPa For comparison, if the entire 8 m were dry (γ = 18): σ = 0.333 × 18 × 8 = 47.9 kPa. Saturated conditions almost double the pressure! This is critical in Philippine projects during the rainy season or in low-lying areas near rivers and coasts. **Drainage and Design:** Many retaining walls in the Philippines fail during or shortly after heavy rainfall because drainage is poor. Best practices include: - Weep holes in the wall to relieve water pressure - Gravel drainage layer behind the wall - Perforated pipe at the base to direct water away - Good surface drainage (slopes away from the wall) **6.3 Sloping Backfill** When the backfill is not horizontal, active pressure increases. This is common in hillside construction. For a backfill at angle β (measured from horizontal) and a vertical wall, Coulomb theory gives approximately: Kₐ increases with β. For example: - β = 0° (horizontal): Kₐ = 0.333 (for φ = 30°) - β = 15°: Kₐ ≈ 0.38 (about 14% increase) - β = 20°: Kₐ ≈ 0.45 (about 35% increase) - β = 30°: Kₐ ≈ 0.60 (about 80% increase) For design, when sloping backfill is unavoidable, engineers often use an equivalent-height method: imagine a vertical imaginary surface behind the wall such that the soil between it and the wall is horizontal. This reduces the design height and uses Rankine coefficients, though it is somewhat conservative. **Example (Hillside Wall):** Wall backs a 1-in-2 slope (β ≈ 27°) with φ = 30°. Using Coulomb theory or tables: Kₐ ≈ 0.50 (compared to 0.333 for horizontal backfill)—a 50% increase! For a 5 m wall with γ = 18: - Horizontal backfill: Pₐ = 0.5 × 0.333 × 18 × 25 = 75 kN/m - Sloping backfill: Pₐ ≈ 0.5 × 0.50 × 18 × 25 = 112.5 kN/m The additional 37.5 kN/m (50% increase) must be accommodated in wall design. **6.4 Surcharges (Live Loads)** A uniform surcharge q (e.g., a building, stockpile, or road near the wall) adds a rectangular pressure block: Pₐ,surcharge = Kₐ × q × H This acts at height H/2 (the centroid of a rectangle). If a point load P (concentrated force) is applied at distance d behind the wall, the pressure distribution is more complex, but for design, it can be approximated as: σ ≈ (P / H) × (angle influence factor), which decreases with distance. In practice, most design codes (including NSCP 2015) use simplified methods or Boussinesq-type formulas. **Example (Road Surcharge):** A wall backs a parking lot with typical vehicle loads equivalent to q = 15 kPa. For a 5 m wall: Pₐ,surcharge = 0.333 × 15 × 5 = 25 kN/m This is added to the soil pressure: Pₐ,total = 75 + 25 = 100 kN/m. **6.5 Seismic Conditions (Pseudo-Static Analysis)** In earthquake-prone regions (most of the Philippines), retaining walls must be designed for lateral inertial forces. The pseudo-static method adds an equivalent horizontal force: Fₑq = kₕ × W (horizontal) Fₑq = kᵥ × W (vertical) where kₕ and kᵥ are seismic coefficients (typically 0.1–0.4 for the Philippines, depending on seismic zone and return period). These act as equivalent surcharges or weight modifications. Detailed seismic analysis is beyond this chapter but is essential for Philippine design. Most modern Philippine guidelines (e.g., NSCP 2015, Section 503) require seismic design for structures with long-term importance and in high-seismicity zones.

This section summarizes practical design procedures and highlights the most common mistakes encountered in PRC exams and engineering practice. **7.1 Step-by-Step Design Procedure** **Step 1: Define the Problem** - Determine wall height H, soil properties (γ, φ, c, drainage conditions) - Identify soil layers, water table depth, surcharges - Obtain or estimate foundation soil allowable bearing capacity - For seismic regions, obtain seismic coefficient kₕ **Step 2: Calculate Earth Pressures** - Choose appropriate coefficient: Rankine (for vertical wall, horizontal backfill) or Coulomb (for sloping backfill or wall friction) - Calculate Kₐ, Kₚ - Determine active thrust Pₐ = (1/2) Kₐ γ H² and its point of application (H/3) - Account for surcharge, cohesion, water, seismic effects if applicable **Step 3: Assume Initial Wall Geometry** - Typically, base width B ≈ (0.4–0.6) × H for moderate conditions - Heel length ≥ (0.4–0.5) × H - Stem thickness ≥ 0.3 m (practical minimum) - These are starting points; adjust as needed **Step 4: Calculate Wall Weight and Resisting Moments** - Calculate volume of concrete (stem + base) and soil on heel - Sum vertical weights: ΣW - Calculate moment of each component about the toe - Sum resisting moments: ΣMR **Step 5: Check Overturning** - Calculate overturning moment: MO = Pₐ × (height of action) - Check: FSOT = ΣMR / MO ≥ 1.5 - If FSOT < 1.5, increase base width or heel length and repeat **Step 6: Check Sliding** - Calculate horizontal component of active thrust: Pₐ,H - Assume coefficient of friction μ (typically 0.4–0.6 for concrete on sand) - Check: FSslide = [μ ΣW + Pₚ] / Pₐ,H ≥ 1.5 - If FSslide < 1.5, increase base width, friction (shear key), or both **Step 7: Check Bearing Capacity and Eccentric Loading** - Calculate resultant position (eccentricity e) - Check: e ≤ B/6 (middle-third rule) - Calculate base pressure: qmax = ΣW/B × (1 + 6e/B) - Check: qmax ≤ qallowable - If either fails, increase base width **Step 8: Design Internal Reinforcement** - Once external stability is satisfied, design reinforcement per ACI 318 - Check flexure: moment at base, above heel, etc. - Check shear: along stem, at base - Provide minimum reinforcement and proper lap lengths **Step 9: Detail and Review** - Check drainage provisions (weep holes, gravel layer) - Confirm concrete cover, bar spacing - Review for constructability and cost optimization **7.2 Common Board-Exam Mistakes** **Mistake 1: Using the Wrong Pressure Coefficient** Students often confuse Kₐ and Kₚ or use Rankine when Coulomb is needed. - **Correct:** Active (Kₐ) is for the driving side; passive (Kₚ) is only where the wall pushes soil (e.g., at toe, usually neglected). - **Exam tip:** Read the problem carefully. If backfill is sloping, use Coulomb or state an assumption. If the problem mentions "worst case," use active and neglect passive. **Mistake 2: Forgetting the H² in Thrust Calculation** Students sometimes write Pₐ = Kₐ γ H (missing the factor of 1/2 and the H²). - **Correct formula:** Pₐ = (1/2) Kₐ γ H² - **Impact:** A wall twice the height has four times the thrust, not twice. - **Exam tip:** Dimensionally, Pₐ should have units of force per unit length (e.g., kN/m). Check: [dimensionless] × [kN/m³] × [m²] = [kN/m] ✓ **Mistake 3: Wrong Point of Application** The active thrust acts at H/3 above the base for triangular distribution, not at H/2 or H/4. - **Correct:** Centroid of a triangle is at 1/3 of the height from the base. - **Exam tip:** For a surcharge (rectangular pressure block), it acts at H/2. The combined resultant of soil + surcharge will be at an intermediate height; calculate carefully. **Mistake 4: Neglecting Water Pressure** In problems with a water table, students forget to add hydrostatic pressure. - **Correct:** Water pressure = (1/2) γw hw², acting at hw/3 above the base (for triangular profile), or calculate separately. - **Impact:** Water can increase total pressure by 50% or more. - **Exam tip:** If any part of the backfill is below the water table, mention water pressure in your answer. **Mistake 5: Incorrect Moments About the Toe** Students make sign errors or forget to include soil on the heel. - **Correct:** Toe is typically the pivot point (hinge). Weights create resisting (positive/counterclockwise) moments; active pressure creates overturning (negative/clockwise) moments. - **Exam tip:** Draw a clear diagram with all forces and their lines of action. Use a consistent sign convention (e.g., counterclockwise = positive). **Mistake 6: Ignoring Cohesion in Clay** Students often use Kₐ for clay as if it were cohesionless sand. - **Correct:** For clay with cohesion c, pressure is reduced: σₐ(z) = Kₐ γ z − 2c√(Kₐ). A tension crack forms near the surface. - **Impact:** Active pressure in clay can be much smaller than in sand, sometimes even negligible for short-term conditions. - **Exam tip:** If the problem specifies clay with c > 0, explicitly calculate the tension crack depth and reduced thrust. **Mistake 7: Using Passive Pressure Without Justification** Students include passive resistance at the toe without checking if it is mobilized. - **Correct:** Passive pressure requires wall movement into the soil. If the toe is not firmly embedded or if the wall is designed for minimal movement, passive should be neglected. - **Conservative approach:** Design without passive; if passive helps, it's a bonus. - **Exam tip:** Unless the problem explicitly states that passive is available (e.g., "wall embedded 1.5 m in dense sand"), neglect it. **Mistake 8: Forgetting to Account for Soil on the Heel** Students calculate wall weight and sometimes forget the weight of soil on the heel slab. - **Correct:** The soil above the heel acts as a downward force (weight), which increases the normal force and thus improves sliding stability. - **Impact:** Omitting this can make the wall appear unstable when it's actually safe. - **Exam tip:** In the problem setup, identify the heel area, multiply by soil unit weight and height, and include in ΣW. **Mistake 9: Wrong Friction Coefficient** Students use μ = tan(φ) without considering the contact surface. - **Correct:** μ between concrete base and soil is typically 0.4–0.6 (for sand/gravel). μ = tan(φ) applies to shear within the soil, not at the wall-soil interface. - **Exam tip:** The problem should state or imply μ. If not, use a reasonable value (0.4–0.5 for concrete on sand is typical). **Mistake 10: Exceeding the Middle-Third Limit** Students don't check eccentricity or don't understand why it matters. - **Correct:** Resultant should fall in the middle third of the base (e ≤ B/6) to avoid excessive tension at the heel. - **For concrete walls:** Some tension is acceptable if reinforced, but e > B/6 is generally avoided. - **Exam tip:** Calculate e = B/2 − ΣM / ΣW and compare to B/6. If e is too large, widen the base or increase the heel. **7.3 Exam Strategy** 1. **Read Carefully:** Identify all given data (H, γ, φ, c, μ, water table, surcharges, seismic effects). 2. **Draw a Diagram:** Sketch the wall, soil layers, water table, and force directions. This helps catch errors. 3. **State Assumptions:** If information is missing, state reasonable assumptions (e.g., "assuming no wall friction" or "using conservative estimates"). 4. **Show Calculations:** Present all formulas and intermediate results. Partial credit is often given for correct methodology even if the final answer is wrong. 5. **Check Units:** Ensure consistency (e.g., all in kN, m, kPa). Dimensional analysis is a quick error detector. 6. **Verify Reasonableness:** Does the answer make sense? For a 5 m wall with active pressure coefficient of 0.33, is a thrust of 75 kN/m reasonable? (Yes, roughly.) Is a friction coefficient of 0.9 realistic? (No.) 7. **Summarize:** State whether the wall is safe or unsafe for each check, and suggest modifications if needed. **7.4 Typical Exam Question Formats** **Format 1: Calculate earth-pressure coefficients** "A retaining wall is designed for soil with friction angle φ = 33°. Calculate the active and passive pressure coefficients using Rankine theory." → Use Kₐ = tan²(45° − φ/2), Kₚ = 1/Kₐ. **Format 2: Find the total active thrust** "A 6 m high wall retains sand with γ = 18 kN/m³ and φ = 30°. Find the total active thrust and its point of application. Assume a uniform surcharge of 12 kPa." → Calculate Kₐ, then Pₐ from soil and surcharge separately, combine, and find resultant height. **Format 3: Check stability** "For the wall in Format 2, with an assumed base width of 2.8 m and heel length of 1.8 m, check overturning, sliding, and bearing capacity. Use μ = 0.55, concrete unit weight = 24 kN/m³, and qallowable = 200 kPa." → Calculate wall and soil weights, moments, and apply three stability criteria. **Format 4: Design a wall** "Design a retaining wall for a cut-slope in sandy clay (φ = 28°, c = 15 kPa, γ = 19 kN/m³) to a height of 5 m. Assume horizontal backfill, no surcharge, and water table at 3 m below the surface. Check overturning (FS ≥ 1.5) and sliding (FS ≥ 1.5). Provide a sketch with dimensions and factors of safety." → This requires iterative design: assume dimensions, check stability, adjust. **Format 5: Comparative analysis** "Compare the active thrusts for (a) dry sand (γ = 18, φ = 30°), (b) saturated sand (γsat = 20, γ' = 10, φ = 30°), and (c) sandy clay (γ = 19, φ = 28°, c = 12 kPa) at a wall height of 5 m." → Calculate Kₐ for each, then Pₐ, accounting for water and cohesion effects.