CELE Geotechnical Engineering — Lateral Earth Pressure and Retaining StructuresRevision Notes
Quick revision notes for Lateral Earth Pressure and Retaining Structures — the one-page refresher for CELE aspirants. Every item on this page has appeared in recent CELE Geotechnical Engineering papers, so revising these is the shortest path to a confident performance in Professional Regulation Commission (PRC) — Board of Civil Engineering's CELE 2026.
Exam context
The Civil Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Civil Engineering and is scheduled for May and November 2026. The Geotechnical Engineering subtest is marked as "Core" in the official pattern, and Lateral Earth Pressure and Retaining Structures appears in position 8th of 11 in the CELE Geotechnical Engineering review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent CELE 2026 papers have drawn roughly a meaningful share of questions from this subject.
Lateral Earth Pressure and Retaining Structures - Revision Notes
Lateral earth pressure is a fundamental topic in Geotechnical Engineering and consistently appears in the PRC Civil Engineer Licensure Examination. This chapter covers the theoretical basis for computing soil pressure against retaining walls, sheet piles, and basement walls — and how to check whether those structures are stable. Mastery requires understanding three pressure states (at-rest, active, passive), their corresponding coefficients (Rankine and Coulomb theories), the resultant thrust magnitude and location, and the three external stability checks: overturning, sliding, and bearing. Expect 3–6 board exam items per examination cycle drawing directly from these concepts.
Sections
Formulas
Example
φ = 30°: K₀ = 1 − sin 30° = 1 − 0.5 = 0.50
Formula
K₀ = 1 − sin φ
Variables
K₀ = at-rest earth pressure coefficient (dimensionless); φ = effective internal friction angle of the soil (degrees)
Application
Use for walls that are restrained from any lateral movement, such as basement retaining walls tied to floor slabs.
Example
φ = 30°: Kₐ = (1 − 0.5)/(1 + 0.5) = 0.5/1.5 = 0.333; check: tan²(45 − 15) = tan²30° = 0.333 ✓
Formula
Kₐ = (1 − sin φ)/(1 + sin φ) = tan²(45 − φ/2)
Variables
Kₐ = active earth pressure coefficient; φ = friction angle
Application
Used for the driving (backfill) side of gravity walls, cantilever walls, and sheet piles where the wall can yield slightly.
Example
φ = 30°: Kₚ = 1/0.333 = 3.0; check: tan²(45 + 15) = tan²60° = 3.0 ✓
Formula
Kₚ = (1 + sin φ)/(1 − sin φ) = tan²(45 + φ/2) = 1/Kₐ
Variables
Kₚ = passive earth pressure coefficient; φ = friction angle
Application
Used at the toe side of a retaining wall where the embedded portion pushes against the soil (passive resistance).
Exam Tips
- Memorize the trio: Kₐ = tan²(45 − φ/2), Kₚ = tan²(45 + φ/2), K₀ = 1 − sin φ. Verify Kₐ × Kₚ = 1 as a rapid self-check.
- For φ = 30°, the common values Kₐ = 1/3 and Kₚ = 3 appear very frequently in board problems — recognize them instantly.
- For φ = 0 (pure clay, short-term undrained): Kₐ = Kₚ = 1, meaning σₐ = σₚ = γz (pressure is isotropic in all directions before drainage).
- When Coulomb formulas appear in the exam, they are usually given — you need to substitute correctly, not derive from scratch.
Key Points
- Three pressure states exist depending on wall movement: at-rest (K₀), active (Kₐ), and passive (Kₚ).
- Active pressure is the SMALLEST lateral pressure — it develops when the wall moves AWAY from the backfill, allowing the soil to expand (shear failure in one direction).
- Passive pressure is the LARGEST lateral pressure — it develops when the wall is PUSHED INTO the soil, causing it to compress (shear failure in the opposite direction).
- At-rest pressure is intermediate — it acts on walls that cannot move (e.g., basement walls braced by floor slabs).
- Rankine theory assumes a smooth (frictionless) vertical wall and a horizontal backfill surface; it is the standard for most board exam problems.
- Coulomb theory is more general — it accounts for wall friction (δ) and sloping backfill (β), and is used when those conditions are specified.
- For a given φ, the relationship is always: Kₐ < K₀ < Kₚ.
- Note: Kₐ × Kₚ = 1, so Kₚ = 1/Kₐ (Rankine only).
Definitions
Term
Active Earth Pressure
Definition
The minimum lateral pressure exerted by a retained soil mass on a wall that has moved sufficiently away from the backfill for the soil to reach its active (Rankine) failure state.
Importance
Governs the design of most retaining walls and is the dominant horizontal load in overturning and sliding stability checks.
Term
Passive Earth Pressure
Definition
The maximum lateral resistance offered by soil in front of a retaining wall (at the toe) when the wall is displaced toward the soil mass.
Importance
Contributes to sliding resistance at the toe; however, it is often neglected conservatively in design or included only with a reduced factor.
Term
At-Rest Earth Pressure
Definition
The lateral pressure in a soil mass when no lateral strain occurs (zero lateral deformation condition), characterized by the coefficient K₀.
Importance
Critical for designing basement walls, bridge abutments, and any structure where wall movement is prevented.
Term
Rankine Theory
Definition
A classical lateral earth pressure theory assuming a smooth, vertical, frictionless wall with a horizontal, homogeneous, cohesionless backfill. It treats the retaining-wall problem as a limiting stress state in a semi-infinite soil mass.
Importance
The standard theory for PRC board exam problems unless wall friction or sloped backfill is explicitly given.
Term
Coulomb Theory
Definition
An earth-pressure theory based on a planar failure wedge behind the wall. It accounts for wall friction angle δ and backfill slope angle β, making it more realistic than Rankine for practical design.
Importance
Board exams may give a Coulomb formula and ask you to substitute values; understand the variables so you can identify which formula to use.
Section Title
Earth-Pressure States and Coefficients
Common Mistakes
- Using Kₐ on the passive (toe) side and Kₚ on the active (backfill) side — always match the correct coefficient to the correct side.
- Forgetting that Kₚ = 1/Kₐ (Rankine, cohesionless) — you can quickly verify your values with this reciprocal relationship.
- Applying Rankine formulas when the problem specifies wall friction δ ≠ 0 or a sloped backfill — in those cases, Coulomb's formula must be used.
- Confusing K₀ = 1 − sin φ (normally consolidated) with the overconsolidated case K₀ = (1 − sin φ)·OCR^(sin φ) — for standard board problems, use the simpler form unless told otherwise.
Formulas
Example
H = 5 m, γ = 18 kN/m³, φ = 30° (Kₐ = 0.333): Pₐ = ½(0.333)(18)(5²) = ½(0.333)(18)(25) = 75.0 kN/m, acting at 5/3 = 1.667 m above base.
Formula
Pₐ = ½ Kₐ γ H²
Variables
Pₐ = active resultant thrust per unit length of wall (kN/m); Kₐ = active pressure coefficient; γ = unit weight of soil (kN/m³); H = retained height (m)
Application
Dry cohesionless backfill with no surcharge. This is the most common form in board exams.
Example
q = 10 kPa, H = 5 m, Kₐ = 0.333: ΔPₐ = 0.333(10)(5) = 16.65 kN/m, acting at H/2 = 2.5 m above base.
Formula
ΔPₐ(surcharge) = Kₐ · q · H
Variables
q = uniform surcharge pressure on backfill surface (kPa); all other variables as before
Application
Added to the triangular soil pressure when a uniform load (e.g., traffic load, stored material) sits on the backfill.
Example
c = 15 kPa, γ = 18 kN/m³, φ = 20°: Kₐ = tan²(45−10) = tan²35° = 0.490, √Kₐ = 0.700; zc = 2(15)/(18 × 0.700) = 30/12.6 = 2.38 m
Formula
zc = 2c / (γ √Kₐ)
Variables
zc = depth of tension crack (m); c = cohesion (kPa); γ = unit weight (kN/m³); Kₐ = active coefficient
Application
For cohesive soils — the zone from 0 to zc carries negative (tensile) pressure. In practice, this zone is assumed to crack and fill with water, which is then added as a hydrostatic load.
Example
At the base of a 4 m wall with c = 15 kPa, φ = 20°, γ = 18 kN/m³: σₐ = (0.490)(18)(4) − 2(15)(0.700) = 35.28 − 21.0 = 14.28 kPa
Formula
σₐ = Kₐ γ z − 2c√Kₐ
Variables
σₐ = active pressure at depth z (kPa); c = cohesion (kPa); z = depth below the backfill surface (m)
Application
Active pressure in c-φ (cohesive-frictional) soils. Board exams may ask for the total active thrust over the cracked zone or full height.
Example
Wall H = 6 m, WT at 2 m below surface, γsat = 20 kN/m³, γ' = 10 kN/m³, γw = 10 kN/m³. Below WT (4 m depth): soil active pressure triangle adds γ'Kₐ; water pressure triangle = ½(10)(4²) = 80 kN/m.
Formula
Uw = γw · hw
Variables
Uw = hydrostatic water pressure at depth hw below the water table (kPa); γw = 9.81 kN/m³ ≈ 10 kN/m³ in many board problems
Application
Added separately to the effective soil pressure whenever the backfill is saturated. Use γ' = γsat − γw for the soil pressure below the WT.
Exam Tips
- Pressure diagrams: triangular for soil (acts at H/3), rectangular for surcharge (acts at H/2). DRAW the diagrams before computing — it prevents errors.
- When water is present, split the analysis: soil effective stress (γ' below WT) + hydrostatic water pressure (γw·hw). Add both force components to get total lateral thrust.
- For tension crack problems: find zc first, then compute the active thrust only over the zone from zc to H (the positive pressure zone).
- The H² dependence is a favorite exam trick — doubling H quadruples Pₐ. This is why tall walls are disproportionately more critical than short walls.
Key Points
- For a dry, cohesionless, homogeneous backfill of height H, the lateral pressure varies LINEARLY from zero at the top to Kₐγ H at the base — a triangular distribution.
- The resultant thrust Pₐ = ½ Kₐ γ H² acts at H/3 from the base of the wall (centroid of the triangle).
- A uniform surcharge q (kPa) on the backfill surface adds a RECTANGULAR pressure diagram of intensity Kₐ q over the full height H.
- The surcharge thrust ΔPₐ = Kₐ q H acts at H/2 from the base.
- Cohesion c reduces active pressure: σₐ = Kₐ γ z − 2c√Kₐ. The net pressure is negative near the top, indicating a tension zone. A tension crack forms to depth zc = 2c/(γ√Kₐ).
- Below the water table, use the submerged (buoyant) unit weight γ' = γsat − γw for soil stresses, PLUS add the full hydrostatic water pressure u = γw·hw separately.
- The total active thrust with water is the sum of: (1) active soil pressure using γ' below WT, (2) hydrostatic water pressure diagram (triangle below WT). These must be computed and added as separate resultants.
- Passive thrust: Pₚ = ½ Kₚ γ H² at H/3 from the base of the embedded depth.
- For layered soils, compute σᵥ incrementally through each layer and apply the appropriate K coefficient for that layer.
Definitions
Term
Tension Crack Depth (zc)
Definition
The depth from the backfill surface to where the active pressure becomes zero in a cohesive soil. Above this depth, the soil theoretically goes into tension, but since soil cannot sustain tension, a crack forms.
Importance
In stability analyses, the tension crack is typically assumed to fill with water, adding a destabilizing hydrostatic force — a conservative and code-consistent assumption.
Term
Surcharge
Definition
Any additional load applied at the backfill surface (e.g., traffic loads, building foundations, stored materials) that increases lateral pressure by a uniform amount Kₐ q over the full height.
Importance
Surcharges are commonly included in board exam problems to add a rectangular pressure component; its resultant acts at H/2, not H/3.
Section Title
Resultant Thrust and Pressure Distributions
Common Mistakes
- Forgetting to add water pressure separately — a very common board exam trap. Always check if the problem says 'saturated backfill' or gives a water table location.
- Using γ (bulk unit weight) below the water table instead of γ' (buoyant unit weight) for the soil stress — this significantly overestimates soil pressure and underestimates water pressure.
- Computing the location of the resultant incorrectly when both a surcharge (acts at H/2) and soil (acts at H/3) are present — you must take moments about the base to find the combined resultant location.
- In cohesive soils, summing the full pressure diagram (including the tension zone) instead of only the positive pressure zone — the tension zone contributes no thrust.
- Doubling the height quadruples the thrust (H² relationship) — a 50% increase in wall height increases the thrust by 125%, not 50%.
Formulas
Example
ΣMᴿ = 300 kN·m/m, Pₐ = 75 kN/m at H/3 = 1.667 m: Mₒ = 75(1.667) = 125 kN·m/m; FS(OT) = 300/125 = 2.4 > 2.0 ✓ Safe
Formula
FS(OT) = ΣMᴿ / ΣMₒ ≥ 1.5 to 2.0
Variables
ΣMᴿ = sum of resisting moments about the toe (kN·m/m); ΣMₒ = sum of overturning moments about the toe (kN·m/m)
Application
The primary stability check against toppling. Moments from all vertical loads (wall, soil, surcharge) resist overturning; horizontal active thrust drives overturning.
Example
ΣW = 250 kN/m, μ = 0.5, Pₐ,H = 80 kN/m, Pₚ neglected: FS(slide) = (0.5 × 250)/80 = 125/80 = 1.56 > 1.5 ✓ Safe
Formula
FS(slide) = (μ ΣW + Pₚ) / Pₐ,H ≥ 1.5
Variables
μ = coefficient of base friction = tan δ (typically 0.45–0.60 for concrete on soil); ΣW = total vertical load per unit length (kN/m); Pₚ = passive resistance at toe (kN/m), often neglected; Pₐ,H = horizontal component of active thrust (kN/m)
Application
Check against horizontal sliding along the base. The passive resistance at the toe is included only when its mobilization is reliable (key or embedment present).
Example
B = 3 m, ΣMᴿ − ΣMₒ = 175 kN·m/m, ΣW = 120 kN/m: x̄ = 175/120 = 1.458 m from toe; e = 1.5 − 1.458 = 0.042 m; B/6 = 0.5 m; e < B/6 ✓ Resultant in middle third
Formula
e = B/2 − x̄ ≤ B/6
Variables
e = eccentricity of the resultant from the centre of the base (m); B = base width of the wall (m); x̄ = distance of the resultant vertical load from the toe (m) = (ΣMᴿ − ΣMₒ) / ΣW
Application
Middle-third rule check — ensures the resultant falls within the kern so no tension develops under the base.
Example
ΣW = 120 kN/m, B = 3 m, e = 0.042 m: q_toe = (120/3)(1 + 6×0.042/3) = 40(1.084) = 43.4 kPa; q_heel = 40(0.916) = 36.6 kPa
Formula
q_toe = (ΣW/B)(1 + 6e/B), q_heel = (ΣW/B)(1 − 6e/B)
Variables
q_toe = bearing pressure at the toe (kPa); q_heel = bearing pressure at the heel (kPa); ΣW = total vertical force (kN/m); B = base width (m); e = eccentricity (m)
Application
Determines the actual bearing pressure distribution under the base footing. The toe pressure must not exceed the allowable bearing capacity of the foundation soil.
Exam Tips
- Organize your stability calculations in a TABLE: Component | Weight (kN/m) | Moment Arm from Toe (m) | Moment (kN·m/m). This prevents arithmetic errors under exam pressure.
- The three checks have different required FS: OT ≥ 1.5–2.0, slide ≥ 1.5, bearing ≤ qₐ. Remember the minimum FS for sliding is always 1.5.
- When passive resistance is given but you are conservative: FS(slide) without Pₚ < 1.5 but FS(slide) with Pₚ ≥ 1.5 — the board may ask 'is it safe?' Always clarify your assumption.
- If FS(OT) >> 2.0 and FS(slide) is borderline, consider adding a shear key to increase passive resistance — this is a common design recommendation asked in board exams.
Key Points
- Three external stability modes are checked for every retaining wall: overturning, sliding, and bearing capacity (eccentricity).
- All moments in the overturning and sliding checks are taken PER UNIT LENGTH of wall (kN·m/m and kN/m).
- The overturning moment acts about the TOE of the wall; all resisting moments are also taken about the toe.
- Minimum required factors of safety: FS(overturning) ≥ 1.5 to 2.0; FS(sliding) ≥ 1.5; bearing: resultant must fall within the MIDDLE THIRD of the base.
- Resisting forces in overturning include: weight of the wall stem, weight of the base footing, weight of soil above the heel, and any surcharge on the heel side.
- Active thrust provides the overturning moment (horizontal component × its moment arm from the toe).
- For sliding: resisting force = μ × ΣW, where μ = tan δ (base friction). A passive wedge at the toe adds resistance but is often NEGLECTED conservatively.
- For bearing: the resultant of all vertical loads must act within the middle third of the base width B. If it falls outside, tension develops in the base — generally not acceptable for unreinforced masonry or mass concrete walls.
- Middle third rule: eccentricity e = B/2 − x̄ ≤ B/6, where x̄ is the distance of the resultant from the toe.
- Toe and heel bearing pressures: q_toe = (ΣW/B)(1 + 6e/B), q_heel = (ΣW/B)(1 − 6e/B). The toe pressure must not exceed the allowable bearing capacity qₐ.
Definitions
Term
Factor of Safety (FS)
Definition
The ratio of the resisting force or moment to the driving force or moment in a stability check. Values ≥ 1.5 to 2.0 are typically required for retaining structures.
Importance
PRC board exams frequently ask you to COMPUTE FS and then state whether it is SAFE (FS ≥ required value) — always include the comparison in your answer.
Term
Middle Third Rule
Definition
A design criterion that requires the resultant of all vertical forces on the base to fall within the middle third of the base width (eccentricity e ≤ B/6). This ensures only compressive stresses act under the entire base area.
Importance
If the resultant falls outside the middle third, part of the base lifts off the soil (tension), which is unacceptable for unreinforced or mass-concrete gravity walls.
Term
Toe and Heel of a Retaining Wall
Definition
The TOE is the front face (low side, away from backfill) of the base footing; the HEEL is the rear face (high side, under the backfill). Overturning moment is taken about the TOE.
Importance
Many exam errors arise from confusing toe and heel. Remember: the wall TIPS OVER toward the toe, so moments are taken about the toe.
Term
Base Friction Coefficient (μ)
Definition
The coefficient of friction between the base of the retaining wall and the foundation soil, equal to tan δ where δ is the interface friction angle. Typical values: 0.45–0.55 for concrete on dense sand/gravel.
Importance
Used directly in the sliding FS calculation. The problem will either give μ directly or give δ and ask you to compute tan δ.
Section Title
Retaining-Wall Stability Checks
Common Mistakes
- Taking overturning moments about the HEEL instead of the TOE — moments must always be about the toe for conventional stability checks.
- Forgetting to include the weight of soil ABOVE THE HEEL BASE SLAB in the resisting moment calculation — this soil contributes both vertical force and a moment arm about the toe.
- Including passive resistance at the toe without checking if it is reliable — if the problem says 'neglect passive resistance,' do not include Pₚ in FS(slide).
- Using the wrong moment arm for the active thrust — if Pₐ is inclined (Coulomb), separate it into horizontal (Pₐ,H) and vertical (Pₐ,V) components; the horizontal component causes overturning, and the vertical component may increase or decrease the resisting moment.
- Forgetting that eccentricity e is measured from the CENTER of the base, not from the toe — e = B/2 − x̄.
Exam Tips
- For φ = 30°: memorize Kₐ = 1/3, Kₚ = 3, K₀ = 0.5. These appear in at least 60% of lateral earth pressure board problems.
- For φ = 45°: Kₐ = tan²(22.5°) = 0.172, Kₚ = 1/0.172 = 5.83 — less common but has appeared in recent board exams.
- When two conditions are given (e.g., FS for overturning and sliding), compute both before concluding — a wall can pass one check and fail another.
- Time management: lateral earth pressure problems with full stability checks typically take 8–12 minutes. Budget accordingly for the 5-hour board exam.
Key Points
- PITFALL 1 — Wrong K coefficient: Kₐ drives the wall (backfill side); Kₚ resists at the toe (embedded portion). Never interchange them.
- PITFALL 2 — H² effect: Doubling the retained height multiplies the active thrust by 4, and the overturning moment by 8 (since Mₒ = Pₐ × H/3 ∝ H³). Extremely sensitive parameter.
- PITFALL 3 — Water: Always use γ' for soil below the water table AND add a separate hydrostatic pressure. Using γsat throughout is a common and costly error.
- PITFALL 4 — Location of resultant: Triangular soil pressure → H/3 from base. Rectangular surcharge → H/2 from base. Must use correct arm for moment calculations.
- PITFALL 5 — Cohesion: In active state, cohesion REDUCES pressure (subtract 2c√Kₐ). In passive state, cohesion INCREASES resistance. The tension crack depth is zc = 2c/(γ√Kₐ).
- STRATEGY: Always draw the pressure diagram first, then compute resultants, then sum moments about the toe in a table.
- SI units throughout: forces in kN, moments in kN·m, pressures in kPa, lengths in m. Never mix units.
- In Coulomb problems, the formula for Kₐ is given — substitute δ (wall friction) and β (backfill slope) carefully, watching sign conventions.
Section Title
Board-Exam Pitfalls and Quick Strategies
Common Mistakes
- Computing Pₐ = Kₐ γ H (forgetting the ½ factor) — the correct formula is Pₐ = ½ Kₐ γ H². This error underestimates the thrust by half.
- Using φ in radians instead of degrees when computing sin φ and tan φ — always use degrees for Rankine coefficients.
- Forgetting to convert surcharge from kN/m² to the correct units when computing ΔPₐ = Kₐ q H — the units should yield kN/m (force per unit length of wall).
- Not verifying the Kₐ × Kₚ = 1 identity after computing both — if it does not check out, you made an arithmetic error.
Connections
- Shear Strength of Soils (c and φ): Kₐ and Kₚ are derived directly from Mohr-Coulomb failure criterion — you cannot use earth pressure formulas without understanding shear strength parameters.
- Soil Classification and Index Properties: The unit weight γ used in Pₐ = ½KₐγH² comes from the soil's density and void ratio, linking back to basic soil classification.
- Slope Stability: The active failure wedge assumed in Rankine/Coulomb theory is related to the critical failure surface in slope stability analysis (Culmann's method bridges both topics).
- Foundation Engineering (Bearing Capacity): The bearing pressure check (q_toe ≤ qₐ) connects retaining wall design to foundation bearing capacity — Terzaghi/Meyerhof equations give qₐ for the foundation soil beneath the base.
- Shallow Foundations and Settlement: The eccentricity check determines the pressure distribution under the footing, which also governs differential settlement — a connection to consolidation theory.
- Structural Design of the Wall Stem: Once earth pressure is known, the retaining wall stem is designed as a cantilever beam (structural concrete — ACI 318/NSCP 2015 Section 406) subjected to the active pressure as a distributed triangular load.
- Seismic Design (NSCP 2015 / Mononobe-Okabe Method): Under earthquake loading, an additional seismic active thrust ΔPₐE is added using the Mononobe-Okabe equation — an extension of Coulomb theory for dynamic conditions.
- Consolidation and Pore-Water Pressure: The use of effective stress parameters (c', φ', γ') in earth pressure calculations is rooted in the principle of effective stress — a key concept in consolidation theory.
Exam Strategy
For PRC board exam problems on lateral earth pressure: (1) IDENTIFY the wall condition (active/passive/at-rest) and the backfill type (dry, saturated, cohesive, surcharge). (2) COMPUTE Kₐ and/or Kₚ using Rankine formulas — verify Kₐ × Kₚ = 1. (3) DRAW the pressure diagram: triangular for soil, rectangular for surcharge, separate triangle for water below WT. (4) COMPUTE each resultant force and its moment arm from the toe. (5) For stability: set up a moment table (overturning) and force table (sliding), then compute FS for each. (6) STATE whether FS meets the minimum requirements (OT ≥ 1.5–2.0, slide ≥ 1.5). (7) CHECK bearing by computing eccentricity e and verifying e ≤ B/6. Budget 8–12 minutes per full stability problem. Partial-credit problems (asking only for Kₐ or only Pₐ) should take 2–3 minutes. Always label your units and formula steps clearly — examiners check methodology, not just the final number.
Quick Review Questions
For a cohesionless backfill with φ = 34°, compute Kₐ, Kₚ, and K₀.
Use the Rankine formulas: Kₐ = tan²(45 − φ/2), Kₚ = tan²(45 + φ/2) = 1/Kₐ, K₀ = 1 − sin φ. Always verify Kₐ × Kₚ = 0.283 × 3.537 ≈ 1.0 ✓
A 6 m retaining wall has a dry cohesionless backfill with γ = 19 kN/m³, φ = 32°, and a 10 kPa uniform surcharge. Find the total active thrust per meter of wall.
Two pressure components: triangular soil (Pₐ = ½KₐγH²) and rectangular surcharge (ΔPₐ = KₐqH). They are added as separate forces. The resultant of the triangular part acts at H/3 = 2.0 m, and the surcharge part at H/2 = 3.0 m from the base.
A wall has ΣW = 250 kN/m and μ = 0.5. The horizontal active thrust is 80 kN/m. Passive resistance is neglected. Is the wall safe against sliding?
The sliding FS compares the frictional resisting force (μ × total vertical load) to the horizontal driving force (active thrust). A minimum FS of 1.5 is required. This result just barely passes, which is a common board exam scenario.
The resultant of a retaining wall acts at x̄ = 1.2 m from the toe. The base width B = 3.6 m. Is the middle-third requirement satisfied?
The middle-third rule requires e ≤ B/6. When e = B/6 exactly, the heel pressure becomes zero (q_heel = 0) but no tension occurs. This is the limiting case. In practice, a slightly larger base width is preferred.
A 4 m retaining wall holds a clay backfill with c = 15 kPa, φ = 20°, γ = 18 kN/m³. What is the depth of the tension crack?
The tension crack formula is zc = 2c/(γ√Kₐ). Since zc = 2.38 m and the wall is only 4 m tall, the tension zone occupies more than half the wall height — a significant reduction in active thrust. However, the crack is conservatively assumed to fill with water, adding a hydrostatic destabilizing force.
Why is passive resistance at the wall toe often neglected in sliding stability calculations?
The mobilization of passive resistance requires strains 10 to 15 times larger than those needed to mobilize active resistance. For a wall that has moved only slightly, Kₐ is fully mobilized, but Kₚ at the toe may be only 20–30% mobilized. Including full Kₚ would overestimate resistance.
State the relationship between Kₐ and Kₚ (Rankine) and explain physically why Kₚ > Kₐ.
The reciprocal relationship Kₐ × Kₚ = 1 is specific to Rankine theory for horizontal cohesionless backfill. Physically: active state → horizontal stress < vertical stress (soil wants to expand laterally). Passive state → horizontal stress > vertical stress (soil is being compressed horizontally). The ratio between these two states reflects the mobilization of full shear strength in opposite directions.
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