CELE Geotechnical Engineering — Stresses in Soil MassExam Answer Templates

Marks

1

Topic

Effective Stress Principle

Difficulty

easy

Template Id

T1

Examiner Tip

Examiners expect the equation in a 1-mark definition question. One clean sentence with σ' = σ − u and 'soil skeleton' earns the mark every time.

Model Answer

Effective stress (σ') is the portion of total stress in a soil mass that is carried by the soil skeleton (intergranular contact forces), defined by Terzaghi's principle as σ' = σ − u, where σ is the total vertical stress and u is the pore-water pressure.

Question Type

very_short_answer

Answer Structure

• One sentence: name the concept, identify it as intergranular/skeleton stress, and state the equation σ' = σ − u [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining effective stress as 'stress below the water table' without mentioning intergranular stress — incomplete definition
• Omitting the equation — the formula is part of the standard definition at this level
• Confusing effective stress with total stress

Key Phrases To Include

• effective stress
• σ' = σ − u
• soil skeleton
• pore-water pressure
• Terzaghi

Marks

1

Topic

Stress Increase — 2:1 Method

Difficulty

easy

Template Id

T2

Examiner Tip

The formula itself is the answer. Write it clearly, define every symbol, and you earn the mark.

Model Answer

For a rectangular footing B × L carrying total load Q, the vertical stress increase at depth z below the footing is: Δσ = Q / [(B + z)(L + z)], where the load is assumed to spread at a 2-vertical to 1-horizontal slope.

Question Type

very_short_answer

Answer Structure

• State the formula with all variables defined [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing Δσ = Q / (B·L) — forgetting to add z to each dimension, the most common error
• Using B+z in both dimensions for a non-square footing instead of (B+z)(L+z)
• Omitting the physical meaning of the 2:1 slope

Key Phrases To Include

• Δσ = Q / [(B+z)(L+z)]
• 2-vertical : 1-horizontal
• rectangular footing
• depth z

Marks

2

Topic

Total Stress and Pore Pressure

Difficulty

easy

Template Id

T3

Examiner Tip

A 2-mark numerical question awards one mark for the formula/method and one for the correct answer. Always write the formula first.

Model Answer

Total stress at z = 4 m: σ = γ × z = 17 × 4 = 68 kPa Pore-water pressure (water table at 4 m, so z_w = 0): u = γ_w × z_w = 9.81 × 0 = 0 kPa Therefore: σ = 68 kPa, u = 0 kPa

Question Type

numerical

Answer Structure

• Line 1–2: Compute total stress σ = Σγᵢzᵢ with substitution [1 mark]
• Line 3–4: State pore pressure u = γ_w z_w and note z_w = 0 at the water table [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using γ_sat for soil above the water table — wrong unit weight
• Assigning u = 9.81 × 4 = 39.24 kPa when the point is at the water table, not below it
• Failing to state the formula before substituting

Key Phrases To Include

• σ = γz
• u = γ_w z_w
• z_w = 0
• 68 kPa
• 0 kPa

Marks

2

Topic

Effective Stress Principle

Difficulty

easy

Template Id

T4

Examiner Tip

Distinguish questions expect a parallel structure — define both, give both formulas, and state why the distinction matters. Two crisp paragraphs, one for each concept.

Model Answer

Total stress (σ) is the entire stress at a point in the soil mass due to the weight of the soil and water above it; it is computed as σ = Σγᵢzᵢ using the saturated unit weight below the water table. Effective stress (σ') is the stress carried solely by the soil skeleton (grain-to-grain contacts). It governs shear strength and consolidation settlement. It is obtained from Terzaghi's equation: σ' = σ − u, where u = γ_w z_w is the hydrostatic pore-water pressure.

Question Type

short_answer

Answer Structure

• Sentence 1: Define total stress and give its formula σ = Σγᵢzᵢ [1 mark]
• Sentence 2: Define effective stress, state its engineering significance (governs strength and settlement), and give σ' = σ − u [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining them without their respective equations
• Omitting the engineering significance of effective stress (why it matters)
• Not mentioning γ_sat for the saturated zone

Key Phrases To Include

• total stress σ = Σγᵢzᵢ
• effective stress σ' = σ − u
• soil skeleton
• shear strength
• consolidation
• Terzaghi

Marks

2

Topic

Stress Increase — 2:1 Method

Difficulty

easy

Template Id

T5

Examiner Tip

Write the formula before substituting numbers. Even if arithmetic is wrong, the correct formula earns the method mark.

Model Answer

Given: B = 2 m, L = 3 m, Q = 900 kN, z = 4 m Using the 2:1 approximation: Δσ = Q / [(B + z)(L + z)] Δσ = 900 / [(2 + 4)(3 + 4)] Δσ = 900 / [(6)(7)] Δσ = 900 / 42 Δσ = 21.4 kPa

Question Type

numerical

Answer Structure

• Line 1: List given data [no separate mark, but required for method mark]
• Line 2: Write the 2:1 formula [1 mark — method]
• Lines 3–5: Substitute and compute correctly [1 mark — correct answer]

Scoring Breakdown

Common Mark Deductions

• Using Δσ = Q / (B × L) — ignoring the z spread, the most common error
• Mixing up B and L in substitution (does not matter for this formula since multiplication is commutative, but dimensions must both be increased by z)
• Wrong units for final answer (e.g., kN instead of kPa)

Key Phrases To Include

• Δσ = Q / [(B+z)(L+z)]
• 2:1 method
• 21.4 kPa

Marks

3

Topic

Effective Stress Profile — Multi-layer

Difficulty

medium

Template Id

T6

Examiner Tip

Board exams award marks at each step. Show all three steps clearly labeled. A student who correctly finds σ and u but makes an arithmetic error in σ' still earns 2 of 3 marks.

Model Answer

Given: Layer 1 — sand, 3 m, γ = 18 kN/m³ (above WT); Layer 2 — sat. clay, γ_sat = 20 kN/m³; WT at 3 m; z_w = 5 − 3 = 2 m below WT. Step 1 — Total stress at z = 5 m: σ = (18 × 3) + (20 × 2) σ = 54 + 40 = 94 kPa Step 2 — Pore-water pressure (hydrostatic): u = γ_w × z_w = 9.81 × 2 = 19.62 kPa Step 3 — Effective stress: σ' = σ − u = 94 − 19.62 = 74.4 kPa ∴ Effective stress at 5 m depth = 74.4 kPa

Question Type

numerical

Answer Structure

• Step 1: Compute total stress layer by layer using σ = Σγᵢzᵢ (use γ_sat for saturated layer) [1 mark]
• Step 2: Compute pore pressure u = γ_w × z_w where z_w is depth below the water table [1 mark]
• Step 3: Apply σ' = σ − u and state the answer with units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using γ_w z = 9.81 × 5 instead of γ_w × z_w = 9.81 × 2 — measuring z_w from the surface instead of the water table
• Using the moist/dry unit weight (18 kN/m³) for the saturated clay layer below the water table
• Skipping the intermediate steps and writing only the final answer — loses method marks

Key Phrases To Include

• σ = Σγᵢzᵢ
• u = γ_w z_w
• σ' = σ − u
• γ_sat
• z_w = 2 m
• 74.4 kPa

Marks

3

Topic

Boussinesq Stress Distribution

Difficulty

medium

Template Id

T7

Examiner Tip

Mentioning 'when r = 0, [1/(1+(r/z)²)]^(5/2) = 1' shows the examiner you understand the full formula and are not just memorizing the simplified form.

Model Answer

Given: Q = 150 kN, z = 3 m, r = 0 (directly below the load) Step 1 — Boussinesq equation for a point load directly below (r = 0): Δσ = 3Q / (2π z²) Note: When r = 0, the term [1/(1+(r/z)²)]^(5/2) = [1/1]^(5/2) = 1, so the equation reduces to the above. Step 2 — Substitute: Δσ = 3(150) / [2π(3)²] Δσ = 450 / [2π × 9] Δσ = 450 / 56.55 Δσ = 7.96 kPa ∴ Vertical stress increase at z = 3 m directly below the load = 7.96 kPa

Question Type

numerical

Answer Structure

• Step 1: State the full Boussinesq formula and simplify for r = 0 [1 mark]
• Step 2: Substitute Q = 150 kN and z = 3 m correctly [1 mark]
• Step 3: Compute the correct numerical answer with units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using the full Boussinesq formula but forgetting to set r = 0 (failing to simplify)
• Computing 2πz instead of 2πz² in the denominator
• Forgetting to multiply by 3 in the numerator (writing Q/2πz² instead of 3Q/2πz²)

Key Phrases To Include

• Δσ = 3Q/(2πz²)
• Boussinesq
• r = 0
• 7.96 kPa
• point load

Marks

3

Topic

Significance of Effective Stress

Difficulty

medium

Template Id

T8

Examiner Tip

Three marks = three distinct points. Structure your answer as a numbered list. Examiners scan for the Mohr-Coulomb equation — include it and you guarantee 1 mark.

Model Answer

1. Physical basis: Shear strength in soils arises from friction and interlocking at grain contacts. Only the effective stress σ' — carried by the soil skeleton — generates these contact forces. Pore water, which carries the remainder (u), has negligible shear resistance; it can only sustain hydrostatic pressure. 2. Mohr-Coulomb criterion: The shear strength equation is τ_f = c' + σ' tan φ', where c' and φ' are effective cohesion and friction angle respectively. Total stress σ does not appear in this equation, confirming that strength depends on σ'. 3. Settlement (compression): Consolidation settlement occurs because increased effective stress compresses the soil skeleton, expelling pore water over time. A rise in total stress with no change in effective stress (e.g., immediately after load application in a saturated clay) produces no immediate settlement because the skeleton stress has not changed.

Question Type

short_answer

Answer Structure

• Point 1: Physical explanation — grain contacts carry shear, pore water does not [1 mark]
• Point 2: Mohr-Coulomb equation τ_f = c' + σ' tan φ' — only σ' appears [1 mark]
• Point 3: Consolidation — settlement occurs when σ' increases, not merely σ [1 mark]

Scoring Breakdown

Common Mark Deductions

• Giving only a verbal explanation without the Mohr-Coulomb equation — misses the quantitative mark
• Saying 'total stress includes buoyancy' — irrelevant and demonstrates confusion
• Not linking settlement to changes in effective stress

Key Phrases To Include

• grain contacts
• pore water has no shear resistance
• τ_f = c' + σ' tan φ'
• Mohr-Coulomb
• consolidation
• soil skeleton compression

Marks

3

Topic

Effect of Water Table Rise on Effective Stress

Difficulty

medium

Template Id

T9

Examiner Tip

Comparison questions always earn the final mark through a meaningful conclusion. Never leave the last sentence blank. Relating the σ' reduction to slope instability shows engineering judgment.

Model Answer

BEFORE (WT at 4 m depth, assume γ_bulk = 17 kN/m³ above WT): σ = (17 × 4) + (19 × 2) = 68 + 38 = 106 kPa u = 9.81 × 2 = 19.62 kPa σ' = 106 − 19.62 = 86.4 kPa AFTER (WT rises to surface — now z_w = 6 m for the full 6 m depth): σ = 19 × 6 = 114 kPa (γ_sat throughout) u = 9.81 × 6 = 58.86 kPa σ' = 114 − 58.86 = 55.1 kPa Conclusion: Effective stress decreases from 86.4 kPa to 55.1 kPa — a reduction of 31.3 kPa. This reduction in σ' corresponds to a loss of shear strength, illustrating why a rising water table can trigger slope failures.

Question Type

numerical

Answer Structure

• Part A: Compute σ, u, and σ' at 6 m with WT at 4 m [1 mark]
• Part B: Recompute σ, u, and σ' at 6 m with WT at surface [1 mark]
• Conclusion: Quantify the change and note the engineering significance [1 mark]

Scoring Breakdown

Common Mark Deductions

• After the WT rises, still using γ_bulk above 4 m instead of γ_sat for the entire profile
• Using z_w = 6 m before and after (forgetting that z_w changes with the water table position)
• Omitting the engineering conclusion — the comparison question specifically asks you to interpret the result

Key Phrases To Include

• σ' = σ − u
• z_w increases
• effective stress decreases
• shear strength loss
• slope failure

Marks

3

Topic

Boussinesq Stress Distribution

Difficulty

hard

Template Id

T10

Examiner Tip

The exponent 5/2 is the most frequently miswritten part of this formula. Write it out explicitly and evaluate step-by-step to avoid arithmetic errors under pressure.

Model Answer

Given: Q = 200 kN, z = 2 m, r = 1.5 m Step 1 — Full Boussinesq formula: Δσ = [3Q / (2πz²)] × [1 / (1 + (r/z)²)]^(5/2) Step 2 — Compute r/z: r/z = 1.5 / 2 = 0.75 (r/z)² = 0.5625 1 + (r/z)² = 1.5625 [1.5625]^(5/2): first [1.5625]^(1/2) = 1.25, then [1.25]^5 = 3.0518 So the influence term = 1 / 3.0518 = 0.3277 Step 3 — Compute the lead term: 3Q / (2πz²) = 3(200) / [2π(4)] = 600 / 25.133 = 23.87 kPa Step 4 — Final answer: Δσ = 23.87 × 0.3277 = 7.82 kPa ∴ Vertical stress increase at z = 2 m, r = 1.5 m = 7.82 kPa

Question Type

numerical

Answer Structure

• Step 1: Write the full Boussinesq formula [1 mark]
• Step 2–3: Correctly evaluate r/z and the influence factor [1 mark]
• Step 4: Correct final answer with units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Raising the bracket to the power 2 instead of 5/2 — common misremembering of the exponent
• Computing [1/(1+(r/z)²)]^(5/2) as [1/(1+(r/z)^(5/2))] — wrong placement of the exponent
• Not showing the intermediate steps for the influence factor

Key Phrases To Include

• Boussinesq
• Δσ = [3Q/(2πz²)][1/(1+(r/z)²)]^(5/2)
• r/z = 0.75
• influence factor
• 7.82 kPa

Marks

3

Topic

Stress Increase — 2:1 Method Applied to Settlement

Difficulty

hard

Template Id

T11

Examiner Tip

Always anchor z at the footing base. State this assumption explicitly — examiners reward students who recognize the reference datum.

Model Answer

Given: B = 2 m, L = 4 m, Q = 600 kN, foundation base at 1.5 m depth. Clay layer: from 3 m to 7 m below ground surface = 1.5 m to 5.5 m below the footing base. Mid-depth of clay layer below footing base: z = (1.5 + 5.5) / 2 = 3.5 m Step 1 — Formula: Δσ = Q / [(B + z)(L + z)] Step 2 — Substitute: Δσ = 600 / [(2 + 3.5)(4 + 3.5)] Δσ = 600 / [(5.5)(7.5)] Δσ = 600 / 41.25 Δσ = 14.5 kPa ∴ Stress increase at mid-depth of clay layer = 14.5 kPa Note: In settlement analysis, this Δσ at mid-depth is used as representative of the average stress increase throughout the clay layer.

Question Type

numerical

Answer Structure

• Step 1: Establish z as measured from the footing base, compute mid-depth z = 3.5 m [1 mark]
• Step 2: Apply 2:1 formula with correct B, L, and z [1 mark]
• Step 3: Correct answer and note on settlement application [1 mark]

Scoring Breakdown

Common Mark Deductions

• Measuring z from the ground surface instead of the footing base — the depth must be from the foundation level
• Using the layer boundaries instead of the mid-depth
• Forgetting that the 2:1 spread starts at the footing base, not the ground surface

Key Phrases To Include

• z measured from footing base
• mid-depth
• Δσ = Q/[(B+z)(L+z)]
• 14.5 kPa
• settlement analysis

Marks

5

Topic

Effective Stress Profile with WT at Surface

Difficulty

medium

Template Id

T12

Examiner Tip

A 5-mark question requires a combination of computation, diagram, and insight. Allocate your time: 2 min for the table, 1.5 min for the sketch, 1 min for the observation. The diagram alone can earn 1 of the 5 marks with minimal effort.

Model Answer

Given: γ_sat = 19.5 kN/m³, γ_w = 9.81 kN/m³, WT at ground surface (z_w = z at every depth). Stress Table: ┌──────┬──────────────────────┬──────────────────┬────────────────────┐ │ z(m) │ σ = 19.5z (kPa) │ u = 9.81z (kPa) │ σ' = σ − u (kPa) │ ├──────┼──────────────────────┼──────────────────┼────────────────────┤ │ 0 │ 0.0 │ 0.0 │ 0.0 │ │ 2 │ 39.0 │ 19.62 │ 19.4 │ │ 4 │ 78.0 │ 39.24 │ 38.8 │ │ 6 │ 117.0 │ 58.86 │ 58.1 │ └──────┴──────────────────────┴──────────────────┴────────────────────┘ Key observations: 1. σ increases at γ_sat = 19.5 kN/m³ per metre — a steep linear profile. 2. u increases at γ_w = 9.81 kN/m³ per metre — a less steep linear profile. 3. σ' increases at γ' = γ_sat − γ_w = 19.5 − 9.81 = 9.69 kN/m³ per metre — the submerged (buoyant) unit weight. 4. Both σ and σ' are linear with depth; all three profiles start at zero at the surface. [Sketch: Three vertical lines labeled σ, u, and σ' emanating from the origin, all linear, with σ having the steepest slope, u an intermediate slope, and σ' the gentlest slope. The gap between σ and σ' equals u at every depth.]

Question Type

diagram_based

Answer Structure

• Introduction: State γ' = γ_sat − γ_w = 9.69 kN/m³ — this is the slope of the σ' profile [1 mark]
• Table: Correct σ, u, and σ' values at all four depths (z = 0, 2, 4, 6 m) [2 marks — 0.5 each for the three quantities at each non-zero depth, rounded]
• Diagram: Labeled sketch showing three linear profiles with correct relative slopes [1 mark]
• Observation: Identifying σ' increases at γ' = γ_sat − γ_w (buoyant unit weight concept) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using u = γ_w × z = 9.81z but then not subtracting it to get σ' — computing the table without finishing it
• Drawing only σ and σ' without the pore pressure profile u
• Not identifying γ' as the gradient of the σ' profile — missing the conceptual insight
• Plotting non-linear profiles — both σ and σ' are linear for homogeneous soil

Key Phrases To Include

• σ = γ_sat × z
• u = γ_w × z
• σ' = (γ_sat − γ_w) × z
• buoyant unit weight γ'
• linear profile
• 19.4 kPa at 2 m
• 38.8 kPa at 4 m
• 58.1 kPa at 6 m

Marks

5

Topic

Combined Effective Stress and Stress Increase

Difficulty

hard

Template Id

T13

Examiner Tip

Case-study questions are the highest-yield items in the board exam. Organize your solution with clear part labels (a) and (b). Show every step — this is where partial marks make a significant difference to your total score.

Model Answer

PART (a) — Effective overburden stress at z = 7 m: Layer 1: Fill, 0–2 m, γ = 16 kN/m³ (above WT) Layer 2: Silty sand, 2–5 m, γ_sat = 18.5 kN/m³ Layer 3: Soft clay, 5–7 m (mid-depth), γ_sat = 17 kN/m³ Total stress: σ = (16 × 2) + (18.5 × 3) + (17 × 2) σ = 32 + 55.5 + 34 = 121.5 kPa Pore pressure (z_w = 7 − 2 = 5 m below WT): u = 9.81 × 5 = 49.05 kPa Effective overburden stress: σ'_v0 = 121.5 − 49.05 = 72.45 kPa PART (b) — Stress increase from footing (2:1 method): Foundation base at 1 m depth; mid-depth of clay at 7 m from surface. Depth below footing base: z = 7 − 1 = 6 m B = L = 3 m Δσ = Q / [(B + z)(L + z)] Δσ = 1200 / [(3 + 6)(3 + 6)] Δσ = 1200 / [(9)(9)] Δσ = 1200 / 81 = 14.8 kPa Summary: • Effective overburden at mid-clay = 72.45 kPa • Stress increase from footing = 14.8 kPa • Total effective stress after loading = 72.45 + 14.8 = 87.25 kPa

Question Type

case_study

Answer Structure

• Part (a) Step 1: Compute total stress by summing all three layers [1 mark]
• Part (a) Step 2: Compute pore pressure with z_w = 5 m (below WT) [1 mark]
• Part (a) Step 3: σ' = σ − u = 72.45 kPa [1 mark]
• Part (b) Step 1: Determine z = 6 m below the footing base [0.5 mark]
• Part (b) Step 2: Apply 2:1 formula and compute Δσ = 14.8 kPa [1 mark]
• Summary: Combine and state both answers clearly [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using z = 7 m (from ground surface) instead of z = 6 m (from footing base) in the 2:1 formula
• Applying γ_sat = 16 kN/m³ for the fill layer below the WT — the fill is above the WT so use the given γ = 16 kN/m³
• Computing z_w from the ground surface (z_w = 7 m) instead of from the water table (z_w = 5 m)
• Omitting the final summary that combines both results — the question asks for both

Key Phrases To Include

• σ = Σγᵢzᵢ
• z_w = 5 m
• σ' = 72.45 kPa
• z = 6 m below footing base
• Δσ = 14.8 kPa
• 2:1 method

Marks

5

Topic

Seepage and Quick Condition

Difficulty

hard

Template Id

T14

Examiner Tip

A 5-mark long answer requires 5 distinct scorable points. Writing five numbered paragraphs guarantees you don't accidentally merge two points into one line. The critical gradient formula i_cr ≈ 1.0 is a 'golden phrase' — any student who writes it signals mastery of this topic.

Model Answer

EFFECT OF UPWARD SEEPAGE ON EFFECTIVE STRESS 1. Hydrostatic (no seepage) baseline: In a static water condition, pore pressure is hydrostatic: u = γ_w z_w. Effective stress increases with depth at the rate of the buoyant unit weight: σ' = γ' z (where γ' = γ_sat − γ_w). 2. Upward seepage — increased pore pressure: When water seeps upward through a soil (e.g., beneath an excavation, behind a dam), the seepage exerts an upward seepage pressure on the soil skeleton. The pore pressure at depth z below the seepage entry point becomes: u = γ_w (z + h_L) where h_L is the excess head driving the upward flow. Equivalently, the seepage force per unit volume acting upward is: j = i γ_w (upward), where i is the hydraulic gradient. 3. Reduction of effective stress: The upward seepage force reduces the effective stress: σ' = σ − u = γ_sat z − γ_w (z + h_L) = γ' z − i γ_w z As the hydraulic gradient i increases, σ' decreases progressively. 4. The Quick Condition (Boiling / Piping Initiation): The quick condition is reached when the upward seepage force exactly equals the submerged weight of the soil, causing the effective stress to fall to zero: σ' = 0 → i_cr = γ'/γ_w = (G_s − 1)/(1 + e) For typical sandy soils, i_cr ≈ 1.0. When i ≥ i_cr, the soil grains lose all contact stress, the soil behaves like a liquid ('quicksand'), and piping or heaving failure occurs. This is catastrophically sudden in cohesionless soils. 5. Engineering significance: Upward seepage beneath sheet-pile cofferdams, retaining walls, and under dam foundations reduces bearing capacity and can cause catastrophic heave failure. Safety is evaluated using the factor of safety against piping: FS = i_cr / i_actual ≥ 3 to 4 (depending on project risk).

Question Type

long_answer

Answer Structure

• Para 1: Baseline — hydrostatic σ' = γ' z [1 mark]
• Para 2: Mechanism — upward seepage adds to u, formula u = γ_w(z + h_L) or seepage force j = iγ_w [1 mark]
• Para 3: Effect on σ' — write modified σ' = γ' z − i γ_w z showing decrease [1 mark]
• Para 4: Quick condition — define i_cr = (G_s − 1)/(1+e) ≈ 1.0, σ' → 0 [1 mark]
• Para 5: Engineering significance — cofferdam, dam foundation, FS ≥ 3 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing upward seepage (increases u, reduces σ') with downward seepage (decreases u, increases σ')
• Defining quick condition qualitatively without the critical gradient formula
• Not giving the engineering application — a long-answer question always expects a 'so what?' paragraph

Key Phrases To Include

• upward seepage
• seepage force j = iγ_w
• u = γ_w(z + h_L)
• σ' decreases
• quick condition
• i_cr = (G_s − 1)/(1+e)
• i_cr ≈ 1.0
• σ' = 0
• piping
• FS against piping

Marks

5

Topic

Comparison of Stress Distribution Methods

Difficulty

hard

Template Id

T15

Examiner Tip

Comparison questions are answered best in a parallel structure — for each criterion (theory, formula, accuracy, use), address both methods side by side. Examiners scan for coverage of all five scoring dimensions.

Model Answer

COMPARISON: BOUSSINESQ vs. 2:1 METHOD 1. Theoretical basis: • Boussinesq (1885): Based on the theory of elasticity for a semi-infinite, homogeneous, isotropic, linearly elastic half-space. Provides a rigorous solution for point loads; extended to distributed loads using integration (Newmark's chart, stress influence factors). • 2:1 Method: Purely empirical/geometric assumption. The load is assumed to spread laterally at a 2(vertical):1(horizontal) slope from each side of the footing. No elasticity theory involved. 2. Formulas: • Boussinesq (point load, directly below): Δσ = 3Q / (2πz²) • 2:1 (rectangular footing): Δσ = Q / [(B+z)(L+z)] 3. Accuracy and distribution: • Boussinesq gives a more realistic 'bulb' stress distribution — high directly below and decreasing with r and z. It captures the stress concentration near the surface. • The 2:1 method assumes uniform distribution over the spread area, overestimates stress near the edges and underestimates directly below the footing center compared to Boussinesq at shallow depths. 4. Comparison at depth: • At large depths (z >> B), both methods give similar results because the load appears as a point source. • At shallow depths, the 2:1 method underestimates the center stress and is considered non-conservative for shallow foundation design. 5. Appropriate use cases and practical recommendation: • Boussinesq: Use for precise settlement analysis, stress under embankments, and when the footing geometry and load distribution are well-defined. Required for corner/edge effects. • 2:1 Method: Suitable for quick preliminary estimates of settlement-inducing stress in consolidating layers. Widely used in practice because of its simplicity and slightly conservative results at depth. • Both methods assume the soil is homogeneous and elastic, which is a simplification. For layered soils, more advanced methods (Westergaard for stiff layers, finite element analysis) may be warranted.

Question Type

long_answer

Answer Structure

• Point 1: Theoretical basis — elasticity theory vs. empirical geometric rule [1 mark]
• Point 2: State both formulas explicitly [1 mark]
• Point 3: Accuracy — Boussinesq's bulb vs. 2:1 uniform distribution [1 mark]
• Point 4: Comparison at depth — convergence for z >> B [1 mark]
• Point 5: Appropriate use cases and engineering recommendation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing only qualitative comparisons without the formulas — formulas are worth 1 mark
• Claiming the 2:1 method is always more conservative — it is only more conservative at large depths
• Not discussing appropriate use cases — comparison questions require a recommendation

Key Phrases To Include

• semi-infinite elastic half-space
• Boussinesq (1885)
• 2-vertical:1-horizontal
• stress bulb
• uniform distribution
• z >> B convergence
• preliminary estimate
• settlement analysis