CELE Geotechnical Engineering — Slope Stability and Soil ImprovementExam Answer Templates

Marks

1

Topic

Factor of Safety

Difficulty

easy

Template Id

T1

Examiner Tip

A 1-mark answer must be self-contained in one sentence plus the formula. Do not waste time elaborating; move on.

Model Answer

The factor of safety of a slope is the ratio of the shear strength available along the potential failure surface to the shear stress required to maintain equilibrium: FS = τ_f / τ. A slope is stable when FS > 1.

Question Type

very_short_answer

Answer Structure

• Line 1: State FS as the ratio of shear strength to mobilised shear stress and write the formula FS = τ_f/τ [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining FS as 'safety factor against overturning' — this is imprecise for slope problems.
• Omitting the formula and giving only a worded description — formula is expected at this level.
• Stating FS = driving/resisting (inverted ratio) — this is a fatal error.

Key Phrases To Include

• ratio
• shear strength available (τ_f)
• shear stress required for equilibrium (τ)
• FS = τ_f/τ
• stable when FS > 1

Marks

1

Topic

Infinite Slope — Dry Cohesionless

Difficulty

easy

Template Id

T2

Examiner Tip

The phrase 'independent of depth' is a hallmark of full understanding and is often worth the difference between a full and half mark in borderline grading.

Model Answer

A dry cohesionless infinite slope (c′ = 0) is stable when the slope angle β is less than the friction angle φ′. The factor of safety is FS = tanφ′/tanβ, which is independent of depth.

Question Type

very_short_answer

Answer Structure

• Line 1: State stability condition β < φ′ and write FS = tanφ′/tanβ [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing FS = tanβ/tanφ (inverted) — zero marks.
• Forgetting to state the independence from depth — misses a key property of this formula.

Key Phrases To Include

• FS = tanφ/tanβ
• β < φ for stability
• independent of depth
• c′ = 0

Marks

2

Topic

Taylor's Stability Number

Difficulty

easy

Template Id

T3

Examiner Tip

Examiners specifically check whether you state FS = 1 for Hcr. Without it, the formula is incomplete and marks are deducted.

Model Answer

Taylor's stability number is a dimensionless parameter Ns = c/(γ·H·FS) obtained from published stability charts as a function of slope angle and friction angle φ. It is used to find the critical height (at FS = 1) by rearranging: H_cr = c/(γ·Ns). A higher Ns indicates a less stable configuration.

Question Type

short_answer

Answer Structure

• Line 1: Define Ns as dimensionless ratio Ns = c/(γHFS) from stability charts [1 mark]
• Line 2: Write the critical-height formula H_cr = c/(γNs) and state the FS = 1 condition [1 mark]

Scoring Breakdown

Common Mark Deductions

• Giving Ns units (e.g., kPa) — Ns is dimensionless; this errors costs 1 mark.
• Omitting the FS = 1 qualifier when writing the H_cr formula.
• Confusing Ns with bearing capacity factors Nc or Nq.

Key Phrases To Include

• dimensionless
• stability chart
• Ns = c/(γHFS)
• H_cr = c/(γNs)
• FS = 1 (critical condition)

Marks

2

Topic

Soil Improvement

Difficulty

easy

Template Id

T4

Examiner Tip

Always match the improvement method to the soil type in the question. For soft clay the keywords are consolidation, drainage, and chemical stabilisation — not densification, which is for sands and gravels.

Model Answer

1. Preloading with prefabricated vertical (wick) drains: A surcharge fill is placed over the site to pre-consolidate the clay. Wick drains shorten the drainage path, accelerating settlement and strength gain before construction. 2. Lime or cement stabilisation: Quicklime or Portland cement is mixed in-situ with the clay; chemical reactions reduce plasticity, increase strength, and lower compressibility of the treated layer.

Question Type

short_answer

Answer Structure

• Item 1: Name the method (preloading/wick drains) and explain the mechanism in 1–2 sentences [1 mark]
• Item 2: Name a second method (lime/cement stabilisation) and explain the mechanism in 1–2 sentences [1 mark]

Scoring Breakdown

Common Mark Deductions

• Listing methods only without any explanation — each method must be described to earn its mark.
• Repeating the same method in different words (e.g., 'preloading' and 'surcharge fill') — only one mark awarded.
• Suggesting vibroflotation without noting it is most effective in sands, not soft clay — context mismatch.

Key Phrases To Include

• preloading/surcharge
• prefabricated vertical drains (wick drains)
• drainage path
• consolidation
• lime/cement stabilisation
• strength gain

Marks

2

Topic

Infinite Slope — Dry Cohesionless

Difficulty

easy

Template Id

T5

Examiner Tip

For any numerical answer, always write a one-line interpretation: 'FS = 1.32 > 1.0, therefore the slope is stable.' This costs zero extra time and often secures the final half-mark.

Model Answer

Given: φ′ = 35°, β = 28°, c′ = 0 (dry sand — cohesionless infinite slope). Formula: FS = tanφ′/tanβ FS = tan35°/tan28° = 0.7002/0.5317 = 1.32 ∴ FS = 1.32 > 1.0 — the slope is stable. (Note: Since β = 28° < φ′ = 35°, stability is confirmed.)

Question Type

numerical

Answer Structure

• Line 1: State given data and identify the applicable formula [0.5 mark]
• Line 2: Substitute and compute tan values correctly [1 mark]
• Line 3: State FS and interpret stability (FS > 1) [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using sinφ/sinβ instead of tanφ/tanβ — formula error, loses formula mark.
• Not stating whether the slope is stable or unstable — loses interpretation mark.
• Arithmetic errors in trigonometric values — verify tan values with calculator.

Key Phrases To Include

• FS = tanφ/tanβ
• tan35° = 0.7002
• tan28° = 0.5317
• FS = 1.32
• β < φ′, therefore stable

Marks

2

Topic

Infinite Slope — Seepage

Difficulty

medium

Template Id

T6

Examiner Tip

The γ′/γ_sat ratio is the examiner's favourite trap. Always distinguish between submerged and saturated unit weights in seepage problems.

Model Answer

When seepage occurs parallel to the slope surface, a positive pore-water pressure develops at the potential failure plane, reducing the effective normal stress. For a cohesionless infinite slope with full seepage (phreatic surface at ground level), the FS becomes: FS_seepage = (γ′/γ_sat) × (tanφ′/tanβ) Since γ′ (submerged unit weight ≈ 9–10 kN/m³) < γ_sat (≈ 18–20 kN/m³), the ratio γ′/γ_sat ≈ 0.5, roughly halving the dry-slope FS. Seepage therefore significantly destabilises slopes.

Question Type

short_answer

Answer Structure

• Line 1: State that seepage generates pore-water pressure, reducing effective normal stress [1 mark]
• Line 2: Write the modified FS formula showing γ′/γ_sat factor and note the roughly 50% reduction [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating only that 'seepage reduces FS' without explaining the pore-pressure mechanism.
• Using total unit weight γ instead of submerged unit weight γ′ in the seepage formula.
• Confusing parallel seepage with artesian pressure conditions.

Key Phrases To Include

• pore-water pressure
• effective normal stress
• γ′ (submerged unit weight)
• γ_sat
• FS = (γ′/γ_sat)(tanφ′/tanβ)
• factor of safety approximately halved

Marks

3

Topic

Cohesive Infinite Slope

Difficulty

medium

Template Id

T7

Examiner Tip

Write cos²β explicitly as (cosβ)² = (0.866)² = 0.750 to show the examiner you have not confused it with cosβ. This single calculation step is worth a mark.

Model Answer

Given: c′ = 15 kPa, φ′ = 25°, γ = 19 kN/m³, β = 30°, z = 4 m Step 1 — Compute intermediate values: γz = 19 × 4 = 76 kPa cos²β = cos²30° = (0.8660)² = 0.7500 tanφ′ = tan25° = 0.4663 sinβ = sin30° = 0.5000 cosβ = cos30° = 0.8660 Step 2 — Apply cohesive infinite-slope formula: FS = [c′ + γz·cos²β·tanφ′] / [γz·sinβ·cosβ] FS = [15 + 76(0.7500)(0.4663)] / [76(0.5000)(0.8660)] FS = [15 + 26.58] / [32.91] FS = 41.58 / 32.91 FS = 1.26 ∴ The slope has FS = 1.26. This is below the typical design target of 1.3–1.5, indicating the slope may require improvement or closer investigation.

Question Type

numerical

Answer Structure

• Step 1: Write the formula FS = [c′ + γz·cos²β·tanφ′] / [γz·sinβ·cosβ] [1 mark]
• Step 2: Compute all intermediate values (γz, cos²β, tanφ′, sinβcosβ) correctly [1 mark]
• Step 3: Substitute, compute FS = 1.26, and interpret result relative to design target [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using cosβ instead of cos²β in the numerator — single most common error in this formula.
• Omitting the c′ term entirely — cohesive slope formula loses the numerator's first term.
• Computing sinβ·cosβ incorrectly; note sin30°·cos30° = 0.5 × 0.866 = 0.433, not sin60°.
• Not comparing FS to 1.3–1.5 design range — misses the engineering interpretation mark.

Key Phrases To Include

• FS = [c′ + γz·cos²β·tanφ′] / [γz·sinβ·cosβ]
• cos²30° = 0.7500
• γz = 76 kPa
• FS = 1.26
• below design target of 1.3–1.5

Marks

3

Topic

Taylor's Stability Number

Difficulty

medium

Template Id

T8

Examiner Tip

Adding the bonus calculation of allowable height at FS = 1.5 demonstrates higher-order thinking. Even if it is not asked, it signals engineering judgement and may earn bonus credit from the examiner.

Model Answer

Given: c = 20 kPa, γ = 18 kN/m³, Ns = 0.06 Assumption: Critical height corresponds to FS = 1 (incipient failure). Step 1 — State Taylor's formula: H_cr = c / (γ · Ns) Step 2 — Substitute values: H_cr = 20 / (18 × 0.06) H_cr = 20 / 1.08 H_cr = 18.5 m ∴ The maximum height the slope can stand without support at FS = 1 is H_cr = 18.5 m. For a design with FS = 1.5, the allowable height would be H_allow = H_cr/FS = 18.5/1.5 = 12.3 m.

Question Type

numerical

Answer Structure

• Line 1: State assumption FS = 1 and write formula H_cr = c/(γ·Ns) [1 mark]
• Line 2: Substitute values correctly with units shown [1 mark]
• Line 3: Final answer H_cr = 18.5 m with physical interpretation or design height at FS = 1.5 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not stating FS = 1 assumption — partial credit only.
• Giving Ns units or not stating it is dimensionless.
• Inverting the formula to H_cr = γ·Ns/c — zero marks for formula.

Key Phrases To Include

• H_cr = c/(γ·Ns)
• FS = 1 (critical, incipient failure)
• Ns is dimensionless
• H_cr = 18.5 m
• design height at FS = 1.5

Marks

3

Topic

Method of Slices — Finite Slopes

Difficulty

medium

Template Id

T9

Examiner Tip

A simple sketch of the slope cross-section showing 3–4 slices with W arrows and the angle α on one slice earns the diagram mark and reinforces your written explanation.

Model Answer

The method of slices analyses the stability of a finite slope by dividing the soil mass above a trial circular failure arc into a series of vertical slices. For each slice, the weight W acts vertically; α is the angle of the slice base to the horizontal. Governing formula (Fellenius): FS = Σ(c′ℓ + N′tanφ′) / Σ(W·sinα) where ℓ = arc length of slice base, N′ = W·cosα − u·ℓ (effective normal force), and u = pore-water pressure. The critical failure surface is found by trial: multiple circles with different centres and radii are analysed, and the circle yielding the minimum FS governs design. Computer programs (e.g., slope stability software) automate this search. One trial circle is insufficient — the minimum-FS circle must be located.

Question Type

short_answer

Answer Structure

• Part 1: Describe the slicing procedure (circular arc, vertical slices, forces) [1 mark]
• Part 2: Write the Fellenius FS formula with correct terms [1 mark]
• Part 3: Explain the critical circle search — trial multiple circles, minimum FS governs [1 mark]

Scoring Breakdown

Common Mark Deductions

• Drawing slices without defining α or W — marks for formula set-up are lost.
• Stating that one trial circle gives the answer — this is explicitly wrong; minimum FS governs.
• Confusing the Fellenius method (ignoring interslice forces) with Bishop's simplified method — be specific about which method is being described.

Key Phrases To Include

• circular failure arc
• vertical slices
• FS = Σ(c′ℓ + N′tanφ′)/Σ(W·sinα)
• N′ = W·cosα − uℓ
• critical circle — minimum FS
• multiple trial circles

Marks

3

Topic

Taylor's Stability Number — Reverse Calculation

Difficulty

medium

Template Id

T10

Examiner Tip

The trap in this question is that students memorise H_cr = c/(γ·Ns) for FS = 1 and forget to include FS when a specific factor of safety is required. Always check whether the question specifies FS ≠ 1.

Model Answer

Given: H = 12 m, FS = 1.5, Ns = 0.05, γ = 18 kN/m³ Step 1 — From Taylor's formula: Ns = c / (γ · H · FS) Step 2 — Rearrange for c: c = Ns · γ · H · FS Step 3 — Substitute: c = 0.05 × 18 × 12 × 1.5 c = 0.05 × 324 c = 16.2 kPa ∴ The soil must have a minimum cohesion of c = 16.2 kPa to maintain FS = 1.5 for the 12 m slope.

Question Type

numerical

Answer Structure

• Step 1: Write Ns = c/(γ·H·FS) and identify the unknown as c [1 mark]
• Step 2: Rearrange correctly to c = Ns·γ·H·FS [1 mark]
• Step 3: Substitute all values with units and compute c = 16.2 kPa [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using H_cr = c/(γ·Ns) without incorporating FS = 1.5 — this gives c for FS = 1 only.
• Forgetting to include FS in the formula — most common error for this question type.
• Arithmetic errors in multiplication; verify: 18 × 12 = 216, 216 × 0.05 = 10.8, 10.8 × 1.5 = 16.2.

Key Phrases To Include

• Ns = c/(γ·H·FS)
• c = Ns·γ·H·FS
• c = 0.05 × 18 × 12 × 1.5
• c = 16.2 kPa
• minimum cohesion required

Marks

5

Topic

Comparison of Slope Analysis Methods

Difficulty

hard

Template Id

T11

Examiner Tip

For 5-mark compare-and-contrast questions, use a structured approach: state each point as a heading (Geometry, Formula, Critical Surface, Pore Pressure, Application) to make it easy for the examiner to follow and award marks sequentially.

Model Answer

INFINITE SLOPE METHOD vs. METHOD OF SLICES 1. Geometry and Applicability The infinite slope method applies to long, uniform slopes where the failure plane is shallow and parallel to the slope surface, making it suitable for translational (planar) failures in residual soils or colluvium. The method of slices (Fellenius/Bishop) applies to finite slopes of any geometry, where the failure surface is assumed circular — making it the general-purpose method for embankments, cuts, and natural slopes. 2. Governing Equations Infinite slope (cohesive, no seepage): FS = [c′ + γz·cos²β·tanφ′] / [γz·sinβ·cosβ] Method of slices (Fellenius): FS = Σ(c′ℓ + N′tanφ′) / Σ(W·sinα) 3. Critical Failure Surface In the infinite slope method, the failure plane depth z is specified or assumed. In the method of slices, the critical failure surface is found by trial — numerous circles are analysed and the circle with the minimum FS governs design. 4. Pore Pressure Treatment Both methods accommodate pore pressures. In the infinite slope, u = γ_w·z_w·cos²β for seepage. In the method of slices, u is applied to each slice base as N′ = (W·cosα − uℓ). 5. Computational Effort and Accuracy The infinite slope formula is closed-form and simple to compute. The method of slices requires iteration over multiple trial circles and is more labour-intensive (typically done with software), but is more accurate for finite slopes. Bishop's simplified modification to the slice method improves accuracy by accounting for interslice forces. In summary, use the infinite slope method for long, translational failures in uniform slopes; use the method of slices for finite slopes with circular failure mechanisms.

Question Type

long_answer

Answer Structure

• Part 1: Define applicability/geometry of each method (translational vs. circular) [1 mark]
• Part 2: Write governing equations for both methods with terms defined [1 mark]
• Part 3: Explain how the critical failure surface is identified in each method [1 mark]
• Part 4: Discuss pore-pressure treatment in both methods [1 mark]
• Part 5: Compare computational effort/accuracy and state when to use each [1 mark]

Scoring Breakdown

Common Mark Deductions

• Listing differences without writing either formula — loses 1–2 formula marks.
• Not mentioning the critical circle search for the method of slices.
• Treating the two methods as interchangeable — they apply to different failure geometries.
• Exceeding allocated space with irrelevant content (Taylor's method, reinforced earth) — wastes time.

Key Phrases To Include

• translational failure plane
• circular failure arc
• FS = [c′ + γz·cos²β·tanφ′]/[γz·sinβ·cosβ]
• FS = Σ(c′ℓ + N′tanφ′)/Σ(W·sinα)
• minimum-FS critical circle
• pore-water pressure u
• Bishop's simplified method

Marks

5

Topic

Cohesive Infinite Slope with Seepage

Difficulty

hard

Template Id

T12

Examiner Tip

In seepage problems, always show the pore-pressure calculation as a separate step. Examiners award a dedicated mark for u = γ_w·z·cos²β. If you bury it inside a combined expression, the mark may be missed.

Model Answer

Given: c′ = 10 kPa, φ′ = 28°, γ_sat = 19 kN/m³, β = 22°, z = 5 m Seepage: phreatic surface at ground level (full seepage) Step 1 — Compute γ′ (submerged unit weight): γ′ = γ_sat − γ_w = 19 − 9.81 = 9.19 kN/m³ Step 2 — Compute pore pressure at failure plane: u = γ_w · z · cos²β = 9.81 × 5 × cos²22° cos²22° = (0.9272)² = 0.8597 u = 9.81 × 5 × 0.8597 = 42.17 kPa Step 3 — Compute stresses: Normal stress: σ = γ_sat · z · cos²β = 19 × 5 × 0.8597 = 81.67 kPa Effective normal stress: σ′ = σ − u = 81.67 − 42.17 = 39.50 kPa Shear stress: τ = γ_sat · z · sinβ · cosβ = 19 × 5 × sin22° × cos22° sin22° = 0.3746, cos22° = 0.9272 τ = 19 × 5 × 0.3746 × 0.9272 = 32.98 kPa Step 4 — Apply FS formula: τ_f = c′ + σ′ · tanφ′ = 10 + 39.50 × tan28° = 10 + 39.50 × 0.5317 = 10 + 21.00 = 31.00 kPa FS = τ_f / τ = 31.00 / 32.98 FS = 0.94 ∴ FS = 0.94 < 1.0 — the slope is UNSTABLE under full-seepage conditions. Immediate remedial action (dewatering, slope flattening, or drainage installation) is required.

Question Type

numerical

Answer Structure

• Step 1: Compute γ′ = γ_sat − γ_w correctly [1 mark]
• Step 2: Compute pore pressure u = γ_w·z·cos²β [1 mark]
• Step 3: Compute effective normal stress σ′ and shear stress τ [1 mark]
• Step 4: Compute τ_f = c′ + σ′·tanφ′ and FS = τ_f/τ [1 mark]
• Step 5: State FS = 0.94, declare slope unstable, and recommend remedial action [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using γ_sat throughout without subtracting pore pressure — ignores seepage effect entirely.
• Using cos β instead of cos²β for pore pressure calculation.
• Not computing effective normal stress (using total stress in Coulomb equation).
• Stopping at FS without declaring stability status or recommending action.

Key Phrases To Include

• γ′ = γ_sat − γ_w
• u = γ_w·z·cos²β
• σ′ = σ − u
• τ_f = c′ + σ′·tanφ′
• FS = 0.94 < 1.0
• slope is UNSTABLE
• dewatering or drainage

Marks

3

Topic

Soil Improvement — Densification

Difficulty

medium

Template Id

T13

Examiner Tip

Mentioning liquefaction mitigation as a benefit of vibroflotation is a high-value addition for Philippine board exams given the country's seismic context and NSCP 2015 Section 208 requirements for liquefiable soil treatment.

Model Answer

Vibroflotation (vibratory compaction) uses a torpedo-shaped vibrating probe inserted into the ground using water jetting. As the probe vibrates horizontally, it densifies the surrounding granular soil and the cavity is backfilled with compacted stone or sand. Most effective for: loose to medium-dense sands and gravels (cohesionless soils with less than 15–20% fines). It is NOT effective for soft clays or silts because cohesive soils do not densify through vibration. Mechanism: The vibration temporarily reduces interparticle friction, allowing soil particles to rearrange into a denser packing. The resulting improvement increases relative density (Dr), bearing capacity, and shear strength while reducing settlement and liquefaction potential.

Question Type

short_answer

Answer Structure

• Part 1: Describe the equipment and process (vibrating probe, water jetting, backfill) [1 mark]
• Part 2: State applicable soil types (loose sands and gravels; NOT clays) [1 mark]
• Part 3: Explain the mechanism (particle rearrangement, increased Dr, improved bearing capacity) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating vibroflotation works in soft clay — this is incorrect; clay does not respond to vibration densification.
• Describing a generic 'vibrating roller' — vibroflotation is a deep probe method, not a surface compaction technique.
• Omitting the mechanism and only naming the soil type — mechanism mark is separate.

Key Phrases To Include

• vibrating probe
• horizontal vibration
• loose sands and gravels
• cohesionless soils
• NOT effective in clays
• relative density (Dr)
• densification
• liquefaction mitigation

Marks

2

Topic

Soil Improvement — Consolidation Acceleration

Difficulty

easy

Template Id

T14

Examiner Tip

Draw a small cross-section showing the clay layer with vertical drain strips and horizontal arrows indicating drainage direction. This visual earns the mechanism mark faster than a paragraph of text.

Model Answer

Prefabricated vertical drains (PVDs), also called wick drains, are factory-made drain strips consisting of a plastic core wrapped in a geotextile filter sleeve. They are inserted vertically into soft compressible clay at regular spacing. Mechanism: By introducing vertical drains throughout the clay layer, the drainage path for pore water is shortened from the full layer thickness (H) to the horizontal spacing between drains (s/2 or s for single-drained). Water expelled by consolidation flows horizontally to the nearest drain and discharges upward, dramatically reducing the time required to achieve a target degree of consolidation (typically from years to months).

Question Type

short_answer

Answer Structure

• Line 1: Define PVDs — construction and installation into soft clay [1 mark]
• Line 2: Explain mechanism — shortened drainage path, horizontal drainage, accelerated consolidation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Describing PVDs as 'pipes' — they are filter-wrapped drain strips, not perforated pipes.
• Not explaining why drainage is accelerated (shorter path) — mechanism mark is essential.
• Confusing PVDs with stone columns — stone columns improve shear strength; PVDs improve drainage.

Key Phrases To Include

• plastic core with geotextile filter sleeve
• vertical insertion
• shortened drainage path
• horizontal drainage
• reduced consolidation time
• pore-water dissipation

Marks

5

Topic

Soil Improvement — Geosynthetic Reinforcement

Difficulty

hard

Template Id

T15

Examiner Tip

Use sub-headings for each mechanism in 5-mark answers. Examiners scan answer papers; clearly labelled headings allow them to award marks efficiently and reduce the risk of your answer being misread.

Model Answer

GEOSYNTHETICS IN SLOPE STABILISATION 1. Types and Forms Geogrids are stiff, open-mesh polymer grids that provide tensile reinforcement. Geotextiles are woven or non-woven fabric sheets that provide reinforcement and/or filtration/drainage. Both are classified as geosynthetics — synthetic materials placed in soil to improve its behaviour. 2. Mechanism 1 — Tensile Reinforcement Geosynthetics placed horizontally within a slope or embankment intercept potential failure surfaces. The tensile strength of the reinforcement adds a resisting horizontal force component, effectively increasing the factor of safety. This is analogous to adding a cohesion increment to the reinforced soil mass. The FS formula is modified to include the sum of reinforcement tensile forces in the denominator-moment calculation: FS_reinforced > FS_unreinforced 3. Mechanism 2 — Filtration and Drainage Geotextile separators and filters prevent fine-grained soil particles from migrating into coarse drainage layers (piping and internal erosion). By maintaining permeability of drainage layers behind retaining walls and within slopes, geotextiles prevent pore-pressure build-up — a primary cause of slope failure. 4. Mechanism 3 — Separation In road embankments over soft ground, geotextiles prevent the mixing of subgrade soil with aggregate base courses, preserving the drainage and structural integrity of each layer. 5. Application Example — Reinforced Earth Wall A reinforced earth retaining wall in the Philippines (e.g., along the SLEX or EDSA widening projects) uses horizontal geogrid layers placed at regular vertical intervals within the compacted fill. The geogrids carry the tensile forces that a conventional gravity wall would resist through mass; this reduces the required wall cross-section and allows construction on weaker foundations. In summary, geosynthetics improve slope stability principally through tensile reinforcement (increasing FS) and drainage/filtration (reducing pore pressures), making them among the most versatile soil improvement tools available to the practising engineer.

Question Type

long_answer

Answer Structure

• Part 1: Define geogrids and geotextiles, distinguishing their forms [1 mark]
• Part 2: Mechanism 1 — tensile reinforcement (intercepts failure surface, adds resisting force) [1 mark]
• Part 3: Mechanism 2 — filtration and drainage (prevents pore-pressure build-up) [1 mark]
• Part 4: Mechanism 3 — separation (optional additional mechanism) or further elaboration on Mechanism 1 or 2 [1 mark]
• Part 5: Specific application example with engineering context [1 mark]

Scoring Breakdown

Common Mark Deductions

• Describing only one mechanism when two are required — loses 1–2 marks.
• Not differentiating geogrids from geotextiles — treated as the same product.
• Giving an application without a mechanism explanation — application alone earns at most 1 mark.
• Writing vague statements like 'strengthens the soil' without explaining how — mechanism must be explained.

Key Phrases To Include

• geogrid — open-mesh polymer grid
• geotextile — woven or non-woven fabric
• tensile reinforcement
• intercepts failure surface
• FS_reinforced > FS_unreinforced
• filtration — prevents piping
• pore-pressure reduction
• separation function
• reinforced earth wall