CELE Geotechnical Engineering — Lateral Earth Pressure and Retaining StructuresDetailed Explanation
The Lateral Earth Pressure and Retaining Structures chapter rewards slow, careful thinking over quick pattern matching, especially on Professional Regulation Commission (PRC) — Board of Civil Engineering's scenario-based CELE items. This detailed explanation walks through the full derivation of every core idea, then links each one to a worked example pulled from recent CELE Geotechnical Engineering papers.
Exam context
Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Geotechnical Engineering section sits under a "Core" weighting, and Lateral Earth Pressure and Retaining Structures is the 8th chapter in the 11-chapter CELE Geotechnical Engineering rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Geotechnical Engineering.
Lateral Earth Pressure and Retaining Structures - Detailed Explanation
Lateral earth pressure is one of the most frequently tested topics in the PRC Civil Engineer Licensure Examination under Geotechnical Engineering. It deals with the horizontal forces that soil exerts on retaining walls, basement walls, sheet piles, and other earth-retaining structures. A civil engineer must be able to determine the magnitude, direction, and point of application of these lateral forces, and then verify that the retaining structure is stable against overturning, sliding, and bearing capacity failure. This chapter systematically develops the Rankine and Coulomb earth-pressure theories, the computation of resultant thrust for various loading conditions, and the three standard stability checks. All formulas are presented in SI units and at the rigor expected in board examinations.
Concepts
States of Lateral Earth Pressure and Pressure Coefficients
The lateral pressure a soil exerts on a wall depends on how much the wall moves relative to the soil. Three distinct states are recognized: **1. At-Rest State (K₀):** The wall does not move at all — this happens in rigid basement walls or walls braced before any excavation. The soil is in its natural in-situ horizontal stress condition. For normally consolidated soils (Jaky's formula): K₀ = 1 − sin φ **2. Active State (Kₐ):** The wall moves away from the backfill by a small amount (about 0.1–0.5% of wall height for dense sand). The soil mass expands laterally, shear planes develop within the backfill, and the soil reaches the active (minimum) Rankine failure condition. For a smooth vertical wall with horizontal cohesionless backfill: Kₐ = (1 − sin φ) / (1 + sin φ) = tan²(45° − φ/2) **3. Passive State (Kₚ):** The wall is pushed INTO the soil (e.g., at the toe of a retaining wall or for sheet piles resisting lateral load). The soil is compressed until it reaches passive (maximum) failure: Kₚ = (1 + sin φ) / (1 − sin φ) = tan²(45° + φ/2) = 1/Kₐ **Relative magnitudes:** Kₐ < K₀ < Kₚ. For φ = 30°: Kₐ = 0.333, K₀ = 0.500, Kₚ = 3.000. Note that Kₚ = 1/Kₐ only for the Rankine smooth-wall case. **Coulomb's Theory** generalizes these coefficients for walls with friction angle δ between wall and soil, and with a sloping backfill surface at angle β and a wall back face inclined at angle α. The Coulomb active coefficient is: Kₐ = sin²(α + φ) / { sin²α · sin(α − δ) · [1 + √(sin(φ+δ)·sin(φ−β) / (sin(α−δ)·sin(α+β)))]² } For the board exam, Coulomb is usually applied to the standard case (α = 90°, δ given) using tabulated values or simplified formulas provided in the problem.
Examples
This is the most common starting computation. Notice that the at-rest coefficient K₀ = 0.500 is exactly between Kₐ = 0.333 and Kₚ = 3.000. In the exam, always verify with the alternative formula tan²(45 ± φ/2).
Scenario
Board-style Problem: For a backfill with φ = 30°, compute Kₐ, K₀, and Kₚ.
Solution
Kₐ = tan²(45° − 30°/2) = tan²(30°) = (0.5774)² = 0.333 K₀ = 1 − sin 30° = 1 − 0.500 = 0.500 Kₚ = tan²(45° + 30°/2) = tan²(60°) = (1.7321)² = 3.000 Check: Kₚ = 1/Kₐ = 1/0.333 = 3.000 ✓
As φ increases, Kₐ decreases (less active pressure) and Kₚ increases (more passive resistance). This is why well-compacted granular backfill with high φ is preferred — it reduces the lateral load on the wall.
Scenario
For φ = 34°, find all three pressure coefficients.
Solution
Kₐ = tan²(45° − 17°) = tan²(28°) = (0.5317)² = 0.2827 ≈ 0.283 K₀ = 1 − sin 34° = 1 − 0.5592 = 0.441 Kₚ = tan²(45° + 17°) = tan²(62°) = (1.8807)² = 3.537 Check: 1/Kₐ = 1/0.283 = 3.534 ≈ Kₚ ✓
Applications
- Design of gravity retaining walls, cantilever walls, and counterfort walls
- Analysis of basement walls and underground structures
- Sheet pile wall design for temporary excavations and permanent quay walls
- Design of mechanically stabilized earth (MSE) walls
- Abutment design for bridges resisting soil thrust
- Anchor plate and deadman anchor design (passive resistance)
Misconceptions
- Kₚ = 1/Kₐ is true only for Rankine (smooth wall, horizontal backfill). For Coulomb with wall friction, this relationship does not hold exactly.
- The active state does NOT mean the wall is actively resisting — it means the soil is in its minimum-pressure (active failure) condition as the wall moves away.
- K₀ is NOT the average of Kₐ and Kₚ. It is computed separately as 1 − sin φ.
- φ used in Rankine formulas must be the drained friction angle of the backfill, not undrained strength.
- For soft clays, Kₐ computed from Rankine may give a negative (tension) pressure near the surface — this tension crack zone must be handled separately.
Related Concepts
- Mohr's Circle for stress (Rankine theory is derived from it)
- Shear strength of soils (φ and c parameters)
- Coulomb's failure criterion
- Wall friction and interface shear strength
- Retaining wall stability (overturning, sliding)
Common Exam Questions
Example
Given φ = 36°, find Kₐ. Answer: tan²(45°−18°) = tan²(27°) = 0.2596
Approach
Apply Kₐ = tan²(45 − φ/2), K₀ = 1 − sin φ, Kₚ = tan²(45 + φ/2) directly. No intermediate steps needed.
Question Type
Direct coefficient computation
Example
A rigid basement wall: use K₀. A typical gravity retaining wall backfill: use Kₐ.
Approach
If the problem says 'retaining wall' or 'active pressure,' use Kₐ. If it says 'wall pushed into soil' or 'toe resistance,' use Kₚ. If 'no movement' or 'rigid wall,' use K₀.
Question Type
Identifying which K to use
Example
Wall with δ = 15° and β = 10° slope — use Coulomb Kₐ formula.
Approach
If wall friction δ is given or backfill is sloping, the problem likely uses Coulomb. Use the formula or provided table. If not mentioned, default to Rankine.
Question Type
Rankine vs. Coulomb scenario
Key Points To Remember
- K₀ = 1 − sin φ (Jaky) — at-rest, no wall movement
- Kₐ = tan²(45° − φ/2) — active, wall moves away from soil (minimum pressure)
- Kₚ = tan²(45° + φ/2) = 1/Kₐ — passive, wall pushes into soil (maximum pressure)
- Always: Kₐ < K₀ < Kₚ for any value of φ
- Rankine assumes smooth vertical wall and horizontal backfill
- Coulomb accounts for wall friction δ and sloping backfill β
- Doubling φ does NOT double Kₐ — use the tan² formula each time
- φ is the effective friction angle of the backfill, not the foundation soil
Resultant Lateral Thrust — Computation and Point of Application
Once the pressure coefficient K is determined, the lateral earth pressure distribution can be drawn as a pressure diagram. The resultant thrust is the area of this diagram, and its point of application is the centroid of the diagram. **Case 1 — Dry cohesionless backfill (triangular diagram):** The active pressure at depth z: σₐ = Kₐ γ z At the base (z = H): σₐ,max = Kₐ γ H Resultant: Pₐ = ½ Kₐ γ H² (per unit length of wall) Point of application: H/3 from the base (centroid of triangle) **Case 2 — With a uniform surcharge q on the backfill surface:** The surcharge adds a uniform lateral pressure = Kₐ q throughout the full height. Pressure diagram = rectangle (from surcharge) + triangle (from soil weight) Pₐ = Kₐ q H + ½ Kₐ γ H² The rectangle part acts at H/2; the triangle part at H/3. Total point of application: take moments to find ȳ from the base. **Case 3 — Cohesive backfill (c–φ soil):** Rankine active pressure: σₐ = Kₐ γ z − 2c√Kₐ At z = 0: σₐ = −2c√Kₐ (tension, ignored in practice) Depth of tension crack: z_c = 2c / (γ√Kₐ) Below z_c, pressure is positive and increases linearly. Resultant thrust acts on the triangle below z_c. **Case 4 — Submerged backfill (water table at some level):** Above water table: use γ (total unit weight), pressure = Kₐ γ z₁ Below water table: use γ' = γₛₐₜ − γ_w for soil, PLUS add full hydrostatic water pressure u = γ_w z₂ Water pressure is NOT multiplied by K — it acts as full hydrostatic on the wall. This significantly increases total thrust; water table management (weepholes, drains) is critical in practice. **Passive thrust:** Pₚ = ½ Kₚ γ H² (acts at H/3 from base, directed toward the wall)
Examples
The pressure diagram is a right triangle with maximum ordinate 30.0 kPa at the base. The resultant of 75.0 kN/m acts at the centroid of this triangle, which is 1/3 of 5 m = 1.667 m from the base. This is the most common board exam setup.
Scenario
Board Problem (2019-type): A 5 m retaining wall supports dry cohesionless backfill with γ = 18 kN/m³ and φ = 30°. Find the active thrust per meter of wall and its point of application.
Solution
Step 1: Kₐ = tan²(45° − 15°) = tan²(30°) = 0.333 Step 2: σₐ at base = Kₐ γ H = 0.333 × 18 × 5 = 30.0 kPa Step 3: Pₐ = ½ × 0.333 × 18 × 5² = ½ × 0.333 × 18 × 25 = 75.0 kN/m Step 4: Point of application = H/3 = 5/3 = 1.667 m from base
The surcharge produces a uniform (rectangular) pressure diagram, while the soil weight produces a triangular one. These are analyzed separately and combined using moment equilibrium. The combined point of application (2.149 m) is between H/3 = 2.0 m and H/2 = 3.0 m, which makes physical sense.
Scenario
A 6 m wall retains backfill (γ = 19 kN/m³, φ = 32°) with a 10 kPa uniform surcharge on top. Find total active thrust and its location from the base.
Solution
Step 1: Kₐ = tan²(45° − 16°) = tan²(29°) = 0.2596 ≈ 0.260 Step 2: Surcharge component: P₁ = Kₐ × q × H = 0.260 × 10 × 6 = 15.6 kN/m (rectangle, acts at H/2 = 3.0 m) Step 3: Soil weight component: P₂ = ½ × Kₐ × γ × H² = ½ × 0.260 × 19 × 36 = 88.9 kN/m (triangle, acts at H/3 = 2.0 m) Step 4: Total Pₐ = 15.6 + 88.9 = 104.5 kN/m Step 5: Location from base: ȳ = (P₁×3.0 + P₂×2.0) / Pₐ = (15.6×3.0 + 88.9×2.0) / 104.5 ȳ = (46.8 + 177.8) / 104.5 = 224.6 / 104.5 = 2.149 m from base
In cohesive soils, the tension zone near the top is ignored because soil cannot sustain tension reliably and cracks fill with water (which increases pressure). The effective thrust acts only on the triangular positive-pressure zone below z_c. With only 1.619 m of effective height and a 4 m wall, the clay cohesion has dramatically reduced the lateral thrust.
Scenario
A 4 m wall retains clay backfill: c = 15 kPa, φ = 20°, γ = 18 kN/m³. Find the tension crack depth and the net active thrust.
Solution
Step 1: Kₐ = tan²(45° − 10°) = tan²(35°) = 0.4903 ≈ 0.490 √Kₐ = 0.700 Step 2: Tension crack depth: z_c = 2c / (γ√Kₐ) = 2×15 / (18×0.700) = 30/12.6 = 2.381 m Step 3: Effective height below crack = H − z_c = 4 − 2.381 = 1.619 m Step 4: Pressure at base: σₐ = Kₐγ(H) − 2c√Kₐ = 0.490×18×4 − 2×15×0.700 = 35.3 − 21.0 = 14.3 kPa Step 5: Net active thrust (triangle below crack): Pₐ = ½ × 14.3 × 1.619 = 11.6 kN/m (acts at 1.619/3 = 0.540 m from base)
Applications
- Computing design loads for retaining walls, including seismic surcharge (Mononobe-Okabe for earthquake zones per NSCP 2015 Section 208)
- Designing weepholes and drainage blankets to prevent hydrostatic pressure buildup
- Checking that the tension crack in clay does not extend to the full wall height (which would mean zero lateral support)
- Sheet pile penetration depth calculations
- Determining the anchor force in anchored sheet pile walls
Misconceptions
- Water pressure behind the wall is NOT multiplied by Kₐ — water is a fluid and exerts full hydrostatic pressure regardless of wall movement.
- Passive pressure (Pₚ = ½ Kₚ γ H²) acts at H/3 from the base, NOT from the top.
- The tension zone in cohesive soil gives negative pressure in the formula, but this is physically ignored (cracks form). Do NOT include negative areas in the thrust calculation.
- Doubling H gives 4× the thrust (H² relationship), NOT twice the thrust.
- The point of application (H/3 for triangular load) is measured from the BASE, not the top of the wall.
Related Concepts
- Pressure coefficient K (Kₐ, K₀, Kₚ)
- Retaining wall stability checks
- Hydrostatic pressure and effective stress principle
- Shear strength parameters c and φ
- Seismic earth pressure (Mononobe-Okabe method)
Common Exam Questions
Example
q = 20 kPa, H = 5 m, γ = 18, φ = 30°: P₁ = 0.333×20×5 = 33.3 kN/m at 2.5 m; P₂ = ½×0.333×18×25 = 75 kN/m at 1.667 m; Total = 108.3 kN/m
Approach
Separate into rectangle (surcharge) and triangle (soil). Compute each force separately, then add. Find location by moment equilibrium about the base.
Question Type
Compute total active thrust with surcharge
Example
c = 20 kPa, φ = 25°, γ = 17: Kₐ = 0.406, √Kₐ = 0.637; z_c = 2×20/(17×0.637) = 3.69 m
Approach
Use z_c = 2c/(γ√Kₐ). Always verify that z_c < H. If z_c ≥ H, the wall has no net active thrust (temporary condition — not reliable for design).
Question Type
Tension crack depth in cohesive soil
Example
Water table at 2 m depth in a 5 m wall: compute three separate pressure diagrams and sum.
Approach
Split into: (1) soil pressure above water table using γ_total, (2) soil pressure below using γ' = γ_sat − 9.81, (3) water pressure below using γ_w = 9.81 kN/m³. Add all three components.
Question Type
Submerged backfill thrust
Key Points To Remember
- Pₐ = ½ Kₐ γ H² for dry cohesionless soil — memorize this
- Active thrust acts at H/3 from the base for triangular pressure
- Surcharge q adds a rectangle of pressure Kₐq over the full height, acting at H/2
- For c–φ soil, tension crack depth z_c = 2c/(γ√Kₐ); ignore tension zone
- Water below the table: use γ' for soil, add γ_w × z_w separately (no K applied to water)
- Passive thrust Pₚ = ½ Kₚ γ H² — much larger than active
- H² relationship means doubling wall height QUADRUPLES the thrust — critical board exam trap
- For combined loads, find the resultant by summing forces and taking moments for location
Retaining Wall Stability Analysis
A retaining wall must be checked for three external failure modes. Philippine practice (and most international codes) require minimum factors of safety as follows: **1. OVERTURNING about the toe:** The active thrust creates an overturning moment (Mₒ = Pₐ × ȳ) about the toe. The weight of the wall, stem, base, and retained soil above the heel creates resisting moments (ΣMᴿ). FSₒᴛ = ΣMᴿ / Mₒ ≥ 1.5 (minimum), 2.0 (recommended) **2. SLIDING along the base:** The horizontal active thrust tries to push the wall forward. Resistance comes from friction between the wall base and foundation soil (and passive resistance at the toe if mobilized): FS_slide = (μ × ΣW + Pₚ) / Pₐ,H ≥ 1.5 where: μ = tan(δ_base) ≈ tan(⅔φ) to tan(φ) for cast-in-place concrete on soil Pₐ,H = horizontal component of active thrust Pₚ = passive resistance at toe (often conservatively neglected) A shear key (concrete projection below the base) increases sliding resistance by mobilizing additional passive resistance. **3. BEARING CAPACITY (Eccentricity/Pressure Distribution):** The resultant of all vertical forces must fall within the middle third of the base (base width B) to prevent tension under the footing: e = B/2 − (ΣMᴿ − Mₒ)/ΣW ≤ B/6 (middle third rule) Toe and heel pressures: q_toe = ΣW/B × (1 + 6e/B) ≤ q_allowable q_heel = ΣW/B × (1 − 6e/B) ≥ 0 (no tension) **Systematic Stability Procedure:** 1. Identify all forces: wall weight, base weight, soil on heel, active thrust (and passive if applicable) 2. Resolve active thrust into horizontal (Pₐ,H) and vertical (Pₐ,V) components if inclined 3. Compute ΣW (sum of vertical forces) and ΣMᴿ (resisting moments about toe) 4. Compute Mₒ (overturning moment) about the toe 5. Compute FSₒᴛ, FS_slide, eccentricity e, and base pressures
Examples
This problem shows that a wall can pass one check (overturning FSₒᴛ = 2.40 ✓) but fail others (sliding FS = 1.33 ✗ and eccentricity ✗). ALL THREE checks must be satisfied simultaneously. In the board exam, be prepared to perform all three checks and state which ones pass or fail.
Scenario
Complete Board-Type Problem: A gravity retaining wall (H = 5 m, base width B = 3 m) retains dry backfill (γ = 18 kN/m³, φ = 30°). ΣW = 200 kN/m (resultant acts at 1.2 m from toe), μ = 0.5. Active thrust Pₐ = 75 kN/m acting at 1.667 m from base. Check overturning, sliding, and eccentricity.
Solution
Step 1: Pₐ = 75.0 kN/m (from earlier example) Mₒ = 75.0 × 1.667 = 125.0 kN·m/m Step 2: Overturning check ΣMᴿ = ΣW × x̄_from_toe = 200 × 1.2 = 240.0 kN·m/m Note: If ΣMᴿ given = 300 kN·m/m (as in reference examples): FSₒᴛ = 300/125.0 = 2.40 > 2.0 ✓ SAFE Step 3: Sliding check FS_slide = μ × ΣW / Pₐ = 0.5 × 200 / 75.0 = 100/75 = 1.33 < 1.5 ✗ UNSAFE → A shear key or increased base width is required. Step 4: Eccentricity Using ΣMᴿ = 300, Mₒ = 125: x̄_resultant = (ΣMᴿ − Mₒ) / ΣW = (300 − 125) / 200 = 175/200 = 0.875 m from toe e = B/2 − x̄ = 3/2 − 0.875 = 1.5 − 0.875 = 0.625 m B/6 = 3/6 = 0.500 m e = 0.625 m > B/6 = 0.500 m → Resultant OUTSIDE middle third ✗ → Tension would develop under heel; redesign needed.
Passive resistance at the toe significantly improves the sliding factor of safety. However, in final design, Pₚ is often conservatively neglected unless the engineer can guarantee that the soil in front of the wall will never be excavated. In this case, even without Pₚ the wall passes sliding, which is a safer design assumption.
Scenario
Sliding check with passive resistance at toe: ΣW = 250 kN/m, μ = 0.5, Pₐ,H = 80 kN/m, passive resistance Pₚ = 30 kN/m
Solution
FS_slide = (μ × ΣW + Pₚ) / Pₐ,H = (0.5 × 250 + 30) / 80 = (125 + 30) / 80 = 155 / 80 = 1.938 > 1.5 ✓ SAFE Without passive: FS_slide = 125/80 = 1.563 > 1.5 ✓ (barely safe) With passive: FS_slide improves to 1.938.
Applications
- Design of highway retaining walls per DPWH standards and NSCP 2015
- Analysis of existing walls for additional loads (road widening, surcharge from construction equipment)
- Forensic analysis of wall failures — identifying which stability check was violated
- Optimization of wall dimensions (base width, stem thickness) to satisfy all three checks economically
- Design of shear keys to increase sliding resistance
- Evaluation of water pressure effects when drainage systems are blocked (typhoon scenario)
Misconceptions
- Overturning is taken about the TOE (outermost point on the soil side), not the heel or center of the base.
- The weight of soil on the heel (above the base slab) is a RESISTING force — it acts downward and creates a resisting moment about the toe.
- ΣW in the sliding formula includes ALL vertical forces: wall stem, base, soil on heel, and the vertical component of active thrust (if the thrust is inclined).
- Passive resistance at the toe is NOT always included — it must be explicitly stated or it should be neglected conservatively.
- The middle third rule is a SERVICE check — it does not replace the bearing capacity check against ultimate bearing failure.
Related Concepts
- Resultant lateral thrust (Pₐ, magnitude and location)
- Bearing capacity of shallow foundations
- Effective stress and drainage conditions
- Foundation eccentricity and combined loading
- Shear key design and passive resistance
Common Exam Questions
Example
ΣMᴿ = 450 kN·m/m, Pₐ = 90 kN/m at ȳ = 2.0 m: FSₒᴛ = 450/(90×2.0) = 2.5 ✓
Approach
List all forces, compute moments about the toe (resisting and overturning). Divide. Remember: active thrust moment is overturning; wall weight and soil-on-heel moments are resisting.
Question Type
Compute FSₒᴛ given moment table
Example
If wall weight = 50 kN/m (fixed), soil on heel adds γ×B×H terms — algebraic solution for B.
Approach
Set FS_slide = μΣW/Pₐ = 1.5. Express ΣW as function of B (include base weight = γ_conc × B × t_base). Solve for B.
Question Type
Find minimum base width for FS_slide ≥ 1.5
Example
B = 4 m, ΣW = 300 kN/m, ΣMᴿ − Mₒ = 700 kN·m/m: x̄ = 700/300 = 2.33 m; e = 2 − 2.33 = −0.33 m (negative means resultant is on heel side of center) → |e| = 0.33 m < B/6 = 0.67 m ✓
Approach
Compute net moment about toe (ΣMᴿ − Mₒ), divide by ΣW to get location of resultant from toe. Then e = B/2 − location. Apply trapezoid formula for q_toe and q_heel.
Question Type
Eccentricity and base pressure
Key Points To Remember
- FSₒᴛ ≥ 1.5 (minimum) to 2.0 (recommended) for overturning
- FS_slide ≥ 1.5 for sliding along base
- Middle third rule: resultant must fall within B/3 from center (e ≤ B/6)
- Passive resistance at toe is conservative to neglect; include only if reliably mobilized (no future excavation in front of wall)
- For a cantilever wall, include the weight of soil on the heel base as a resisting force and moment
- The overturning moment arm = ȳ = point of application of Pₐ from the base
- Shear key below the base increases FS_slide by adding passive resistance
- μ for cast concrete on moist sand ≈ 0.45–0.55; use tan(⅔φ) as a conservative estimate
Practice Problems
This is a two-component active thrust problem — the most common board exam setup. The surcharge creates a rectangular pressure diagram (uniform over full height) while the soil self-weight creates a triangular diagram. These are treated separately and combined using the principle of moments. The combined point of application (2.175 m) lies between H/3 = 2.0 m and H/2 = 3.0 m, closer to the triangle-dominated case since P₂ >> P₁.
Problem
Problem 1 (Coefficient and Thrust): A 6 m gravity retaining wall retains dry cohesionless sand with φ = 32° and γ = 19 kN/m³. A uniform surcharge of 12 kPa is applied to the backfill surface. (a) Find Kₐ. (b) Find the total active thrust per meter of wall. (c) Find the point of application of the total active thrust from the base.
Solution
(a) Kₐ = tan²(45° − 32°/2) = tan²(45° − 16°) = tan²(29°) tan(29°) = 0.5543 Kₐ = (0.5543)² = 0.3072 ≈ 0.307 (b) Two components: Surcharge (rectangle): P₁ = Kₐ × q × H = 0.307 × 12 × 6 = 22.1 kN/m Acts at ȳ₁ = H/2 = 3.0 m from base Soil weight (triangle): P₂ = ½ × Kₐ × γ × H² = ½ × 0.307 × 19 × 36 = 104.8 kN/m Acts at ȳ₂ = H/3 = 2.0 m from base Total Pₐ = 22.1 + 104.8 = 126.9 kN/m (c) Location from base (moment equilibrium): ȳ = (P₁×ȳ₁ + P₂×ȳ₂) / Pₐ = (22.1×3.0 + 104.8×2.0) / 126.9 = (66.3 + 209.6) / 126.9 = 275.9 / 126.9 = 2.175 m from base
The tension crack at 3.05 m depth means that only the bottom 1.95 m of the wall sees positive active pressure. The net thrust of only 15.1 kN/m (compared to 104+ kN/m for sand at the same height) illustrates the dramatic effect of cohesion in reducing lateral pressure. In practice, the tension crack is assumed to fill with rainwater, adding a hydrostatic force of ½ × γ_w × z_c² — but this problem asks for the soil thrust only. Note: if the tension crack fills with water, the total thrust increases significantly.
Problem
Problem 2 (Cohesive Backfill): A 5 m retaining wall retains a c–φ soil with c = 18 kPa, φ = 22°, γ = 17.5 kN/m³. (a) Compute Kₐ and √Kₐ. (b) Find the depth of tension crack z_c. (c) Compute the net active thrust per meter of wall and its point of application above the base.
Solution
(a) Kₐ = tan²(45° − 22°/2) = tan²(45° − 11°) = tan²(34°) tan(34°) = 0.6745 Kₐ = (0.6745)² = 0.4550 ≈ 0.455 √Kₐ = 0.6745 (b) Tension crack depth: z_c = 2c / (γ × √Kₐ) = (2 × 18) / (17.5 × 0.6745) = 36 / 11.803 = 3.050 m Since z_c = 3.05 m < H = 5 m, a tension crack exists. (c) Effective height below crack: h = H − z_c = 5.0 − 3.05 = 1.95 m Pressure at base of wall: σₐ,base = Kₐ × γ × H − 2c × √Kₐ = 0.455 × 17.5 × 5 − 2 × 18 × 0.6745 = 39.8 − 24.3 = 15.5 kPa Net active thrust (triangle from z_c to base): Pₐ = ½ × σₐ,base × h = ½ × 15.5 × 1.95 = 15.1 kN/m Point of application: h/3 = 1.95/3 = 0.650 m from base
This comprehensive problem demonstrates the board exam scenario where multiple checks must be performed. The wall PASSES overturning (FSₒᴛ = 2.17 ✓) but FAILS both sliding (FS = 1.079 ✗) and the eccentricity check (e > B/6 ✗). The negative heel pressure confirms that the base would have to resist tension — impossible for a soil-wall interface. Solutions: (1) increase base width B, (2) add a shear key for sliding, (3) add more backfill weight on the heel slab. In the board exam, be alert to problems where only one or two checks are asked — do not perform all three unless instructed.
Problem
Problem 3 (Full Stability Check): A cantilever retaining wall has the following properties: H = 4.5 m (total height), base width B = 2.8 m, backfill φ = 30°, γ_backfill = 18 kN/m³. Assume the sum of vertical forces ΣW = 180 kN/m acting at x̄ = 1.10 m from the toe. The wall face is vertical and smooth, and the backfill is horizontal. μ = tan(20°). Neglect passive resistance. Check (a) overturning, (b) sliding, and (c) eccentricity.
Solution
Step 1: Compute active thrust Kₐ = tan²(45° − 15°) = tan²(30°) = 0.333 Pₐ = ½ × 0.333 × 18 × (4.5)² = ½ × 0.333 × 18 × 20.25 = 60.7 kN/m ȳ = H/3 = 4.5/3 = 1.50 m from base (a) Overturning: ΣMᴿ = ΣW × x̄ = 180 × 1.10 = 198.0 kN·m/m Mₒ = Pₐ × ȳ = 60.7 × 1.50 = 91.1 kN·m/m FSₒᴛ = 198.0 / 91.1 = 2.17 > 2.0 ✓ SAFE (b) Sliding: μ = tan(20°) = 0.3640 FS_slide = μ × ΣW / Pₐ = 0.3640 × 180 / 60.7 = 65.5 / 60.7 = 1.079 < 1.5 ✗ UNSAFE → A shear key is needed to increase FS_slide to ≥ 1.5. (c) Eccentricity: x̄_resultant = (ΣMᴿ − Mₒ) / ΣW = (198.0 − 91.1) / 180 = 106.9 / 180 = 0.594 m from toe e = B/2 − x̄_resultant = 2.8/2 − 0.594 = 1.4 − 0.594 = 0.806 m B/6 = 2.8/6 = 0.467 m e = 0.806 m > B/6 = 0.467 m ✗ OUTSIDE MIDDLE THIRD → Base must be widened or wall configuration adjusted. Toe pressure: q_toe = (ΣW/B)(1 + 6e/B) = (180/2.8)(1 + 6×0.806/2.8) = 64.3 × (1 + 1.728) = 64.3 × 2.728 = 175.4 kPa Heel pressure: q_heel = 64.3 × (1 − 1.728) = 64.3 × (−0.728) = −46.8 kPa < 0 → TENSION (not allowed for soil)
Rigid basement walls and braced excavations do not allow the wall to deflect sufficiently to mobilize active conditions. Using Kₐ in these cases UNDERESTIMATES the lateral pressure by about 47% for φ = 28°. This is a critical design consideration: using the wrong pressure coefficient can lead to unsafe under-design of structural elements. The board exam may include a comparison question highlighting this distinction.
Problem
Problem 4 (At-Rest Pressure): A rigid basement wall 3.5 m high retains backfill with φ = 28° and γ = 17 kN/m³. Since the wall cannot move, use K₀. Find the at-rest lateral force per meter of wall and its point of application from the base.
Solution
Step 1: K₀ = 1 − sin φ = 1 − sin 28° = 1 − 0.4695 = 0.5305 ≈ 0.531 Step 2: At-rest lateral force (triangular distribution): P₀ = ½ × K₀ × γ × H² = ½ × 0.531 × 17 × (3.5)² = ½ × 0.531 × 17 × 12.25 = ½ × 110.6 = 55.3 kN/m Step 3: Point of application: ȳ = H/3 = 3.5/3 = 1.167 m from base Comparison with active: Kₐ = tan²(45°−14°) = tan²(31°) = 0.3607 Pₐ = ½ × 0.361 × 17 × 12.25 = 37.6 kN/m Ratio: P₀/Pₐ = 55.3/37.6 = 1.47 → at-rest is 47% higher than active
Exam Preparation Tips
- MEMORIZE the three formulas: Kₐ = tan²(45−φ/2), K₀ = 1−sinφ, Kₚ = tan²(45+φ/2). These appear in nearly every board exam in lateral earth pressure.
- Always draw the pressure diagram — a simple sketch of the triangular (or trapezoidal with surcharge) pressure distribution prevents errors in identifying forces and their locations.
- For stability problems, create a systematic table: Force | Magnitude | Moment arm | Moment about toe. This organized approach reduces arithmetic errors under exam pressure.
- The H² relationship is a common board exam trap: if wall height doubles, active thrust quadruples (not doubles). Watch for problems that change H and ask for the new thrust.
- Remember that water pressure is NEVER multiplied by K. If there is a water table, add γ_w × z_w separately as a hydrostatic force below the water table.
- When φ = 30° (very common in board problems): Kₐ = 1/3, Kₚ = 3.0, K₀ = 0.5 — memorize these exact values to save computation time.
- Know the minimum FS values: FSₒᴛ ≥ 1.5 (min.) to 2.0 (recommended), FS_slide ≥ 1.5. Stating the minimum without checking it is a guaranteed partial-credit loss.
- For the eccentricity check, the formula e = B/2 − (ΣMᴿ − Mₒ)/ΣW is derived every time. Practice this derivation so you do not confuse the formula under pressure.
- Passive resistance at the toe — if the problem says 'neglect passive resistance,' do NOT include Pₚ in the sliding formula. Only include it if explicitly stated.
- In Coulomb problems, the wall friction angle δ is usually given as a fraction of φ (e.g., δ = 2φ/3). Read the problem carefully before applying any formula.
- Tension crack depth z_c = 2c/(γ√Kₐ) — practice this for cohesive soils. If z_c ≥ H, the entire wall theoretically has no net active thrust (but assume the crack fills with water for safety).
- Practice converting between degrees and the tan/sin functions quickly on your calculator. Set it to degrees mode before starting the exam and verify with tan(30°) = 0.5774.
- In the middle third check, |e| ≤ B/6 must hold. If e > B/6, the base pressure formula (trapezoid) no longer applies — only the triangular contact zone formula should be used, or the design must be revised.
- Review NSCP 2015 Section 208 for seismic provisions: the Mononobe-Okabe method adds a seismic coefficient component to the active thrust. This appears occasionally in advanced board problems.
- Time management: a typical lateral earth pressure board problem can be solved in 4–6 minutes if you memorize the formulas and use a systematic table approach. If a problem takes longer, skip and return.
In summary
Lateral earth pressure is a foundational topic in geotechnical engineering that directly governs the safe design of retaining walls, basement structures, and earth-support systems. The PRC Civil Engineer Licensure Examination consistently tests three core skills: (1) computing the correct pressure coefficient (Kₐ, K₀, or Kₚ) using the Rankine or Coulomb framework; (2) determining the resultant lateral thrust and its point of application for various backfill conditions — dry cohesionless, with surcharge, cohesive, and submerged; and (3) performing all three stability checks — overturning (FSₒᴛ ≥ 2.0), sliding (FS_slide ≥ 1.5), and eccentricity (e ≤ B/6) — for a retaining wall. The most important formulas to internalize are Kₐ = tan²(45°−φ/2), Pₐ = ½Kₐγ H², and the moment-based stability equations. Common errors include confusing the point of application (H/3 vs. H/2), applying K to water pressure (incorrect — water gives full hydrostatic), and using Kₐ in a situation that requires K₀. Systematic problem-solving — drawing pressure diagrams, building a force table, and checking all three stability criteria — is the most reliable strategy for achieving full marks on these problems. With rigorous practice of the worked examples and board-style problems in this chapter, Filipino engineering reviewees will be well-equipped to answer any lateral earth pressure question on the PRC examination with confidence and precision.
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