CELE Geotechnical Engineering — Bearing Capacity of SoilsExam Answer Templates

Marks

1

Topic

Bearing Capacity Fundamentals

Difficulty

easy

Template Id

T1

Examiner Tip

The word 'shear' must appear — it distinguishes bearing capacity failure from settlement.

Model Answer

The ultimate bearing capacity (qu) is the maximum gross pressure the soil beneath a footing can sustain before shear failure occurs in the supporting soil mass.

Question Type

very_short_answer

Answer Structure

• One complete sentence defining qu as maximum gross pressure at shear failure [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing 'maximum load the footing can carry' — this confuses pressure with force
• Omitting the word 'shear' — settlement or other failure modes are not bearing capacity failure
• Defining allowable bearing capacity instead of ultimate

Key Phrases To Include

• maximum gross pressure
• shear failure
• soil beneath the footing

Marks

2

Topic

Failure Modes

Difficulty

easy

Template Id

T2

Examiner Tip

Pair each mode with its soil condition (dense, medium, loose) — examiners award the mark for the characteristic, not just the name.

Model Answer

The three modes are: (1) general shear failure — occurs in dense/stiff soils with a well-defined slip surface; (2) local shear failure — occurs in medium-density soils with partial slip surface; and (3) punching shear failure — occurs in loose/soft soils with compression and shear along vertical planes. Terzaghi's general bearing-capacity equation applies to general shear failure.

Question Type

very_short_answer

Answer Structure

• Name all three modes with a brief distinguishing characteristic for each [1 mark]
• Correctly state that Terzaghi's equation applies to general shear failure [1 mark]

Scoring Breakdown

Common Mark Deductions

• Listing only two modes
• Stating Terzaghi applies to all three modes without qualification
• Confusing local shear with punching shear

Key Phrases To Include

• general shear failure
• local shear failure
• punching shear failure
• well-defined slip surface
• dense soil

Marks

2

Topic

Terzaghi's Bearing Capacity Equation

Difficulty

easy

Template Id

T3

Examiner Tip

Write the equation first, then define variables — this format alone earns the formula mark even if later identification has a minor error.

Model Answer

For a square footing: qu = 1.3cNc + qNq + 0.4γBNγ where: • c = soil cohesion (kPa) • Nc, Nq, Nγ = dimensionless bearing-capacity factors (functions of φ) • q = γDf = overburden pressure at footing level (kPa) • γ = unit weight of soil (kN/m³) • B = footing width (m) • Shape factors for square: 1.3 on cohesion term; 0.4 on width term

Question Type

short_answer

Answer Structure

• Write the correct equation with shape factors 1.3 and 0.4 [1 mark]
• Identify all terms with correct symbols and units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using 0.5 instead of 0.4 for the square footing (0.5 is for strip footings)
• Writing q = γ without multiplying by Df
• Omitting the Nq term entirely

Key Phrases To Include

• 1.3cNc
• qNq
• 0.4γBNγ
• bearing-capacity factors
• overburden pressure
• q = γDf

Marks

2

Topic

Allowable Bearing Capacity

Difficulty

easy

Template Id

T4

Examiner Tip

Always write the subtraction formula explicitly — the difference between gross and net is a frequently tested concept.

Model Answer

Gross ultimate bearing capacity (qu) is the total pressure at the base of the footing at shear failure, including the weight of overburden. Net ultimate bearing capacity (qu,net) subtracts the original overburden pressure: qu,net = qu − γDf The net allowable bearing capacity is: qa,net = qu,net / FS where FS = 2.5 to 3 for most foundation designs.

Question Type

short_answer

Answer Structure

• Define gross qu and net qu,net with the relationship qu,net = qu − γDf [1 mark]
• Write the net allowable formula qa,net = qu,net / FS and state typical FS range [1 mark]

Scoring Breakdown

Common Mark Deductions

• Subtracting the full overburden for gross capacity (gross already includes it)
• Using FS = 1.5 — too low for bearing capacity design
• Confusing net pressure applied by the structure with net bearing capacity

Key Phrases To Include

• gross bearing capacity
• net bearing capacity
• qu,net = qu − γDf
• factor of safety
• FS = 2.5 to 3

Marks

2

Topic

Water Table Correction

Difficulty

medium

Template Id

T5

Examiner Tip

Identify the water table position relative to (a) the ground surface, (b) the footing base, and (c) depth B below the base — each position affects a different term.

Model Answer

When the water table is at the base of the footing (at depth Df), the surcharge term q = γDf uses the moist/bulk unit weight above the water table (no correction needed for the Nq term). However, the width term ½γBNγ must use the effective (submerged) unit weight γ' = γsat − γw in place of γ, because the soil below the footing base is saturated and buoyancy reduces the effective stress. γ' = γsat − 9.81 kN/m³ No correction is needed if the water table is at or below a depth B below the footing base.

Question Type

short_answer

Answer Structure

• Identify which term is affected (width term ½γBNγ) and state that γ' replaces γ [1 mark]
• State the condition for no correction (water table deeper than B below base) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Applying γ' to the surcharge term q when WT is exactly at the base (incorrect — q uses bulk γ above WT)
• Stating no correction is needed when WT is at the base
• Using γw = 10 kN/m³ without justification (use 9.81 unless told to approximate)

Key Phrases To Include

• submerged unit weight
• γ' = γsat − γw
• width term
• effective stress
• depth B below the footing base

Marks

3

Topic

Terzaghi's Equation — Strip Footing

Difficulty

medium

Template Id

T6

Examiner Tip

For strip footings, write 'qu = cNc + qNq + ½γBNγ' with no shape factors — the coefficient in the width term is exactly 0.5 = ½.

Model Answer

Given: B = 1.5 m, Df = 1.2 m, c = 10 kPa, φ = 30° γ = 17 kN/m³, Nc = 37.16, Nq = 22.46, Nγ = 19.13 Step 1 — Overburden pressure: q = γDf = 17 × 1.2 = 20.4 kPa Step 2 — Terzaghi's equation for strip footing: qu = cNc + qNq + ½γBNγ qu = 10(37.16) + 20.4(22.46) + ½(17)(1.5)(19.13) qu = 371.6 + 458.2 + 243.9 qu = 1,073.7 kPa ≈ 1,074 kPa

Question Type

numerical

Answer Structure

• List all given data with symbols and units [0.5 mark]
• Compute q = γDf = 20.4 kPa [0.5 mark]
• Write Terzaghi's strip equation correctly (no shape factors) [0.5 mark]
• Substitute values correctly [0.5 mark]
• Add all three terms and state final answer with unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using shape factor 1.3 (for square/circular) on a strip footing
• Using 0.4 instead of 0.5 in the width term for strip
• Forgetting to multiply ½ × γ × B in the third term
• Omitting unit (kPa) in the final answer

Key Phrases To Include

• qu = cNc + qNq + ½γBNγ
• q = γDf
• strip footing
• kPa

Marks

3

Topic

Bearing Capacity — Clay (φ = 0)

Difficulty

medium

Template Id

T7

Examiner Tip

For φ = 0 clay, always state explicitly that Nγ = 0 so the third term vanishes — this shows the examiner you know the condition, not just the arithmetic.

Model Answer

Given: B = 2 m (square), Df = 1 m, cu = 50 kPa, φ = 0 γ = 18 kN/m³, Nc = 5.7, Nq = 1, Nγ = 0 Step 1 — Overburden: q = γDf = 18 × 1 = 18 kPa Step 2 — Terzaghi's equation for square footing: qu = 1.3cNc + qNq + 0.4γBNγ qu = 1.3(50)(5.7) + 18(1) + 0.4(18)(2)(0) qu = 370.5 + 18 + 0 (a) qu = 388.5 kPa Step 3 — Net ultimate: (b) qu,net = qu − γDf = 388.5 − 18 = 370.5 kPa

Question Type

numerical

Answer Structure

• State given data, identify square footing, recall shape factors 1.3 and 0.4 [0.5 mark]
• Compute q = 18 kPa [0.5 mark]
• Substitute into square equation correctly, note Nγ = 0 makes third term zero [1 mark]
• State qu = 388.5 kPa [0.5 mark]
• Compute qu,net = qu − γDf = 370.5 kPa [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using strip equation (no 1.3 and 0.4) for a square footing
• Not recognizing Nγ = 0 when φ = 0, and computing 0.4γBNγ as non-zero
• Subtracting γDf from qu to get qu,net but using incorrect qu

Key Phrases To Include

• 1.3cNc
• φ = 0
• Nγ = 0
• qu,net = qu − γDf
• square footing shape factors

Marks

3

Topic

Allowable Column Load

Difficulty

medium

Template Id

T8

Examiner Tip

Two separate steps: (1) convert net bearing capacity to allowable pressure, (2) multiply by area for load — show both steps for full marks.

Model Answer

Given: qu,net = 370.5 kPa, FS = 3, B = 2 m × 2 m Step 1 — Net allowable bearing capacity: qa,net = qu,net / FS = 370.5 / 3 = 123.5 kPa Step 2 — Footing area: A = B × B = 2 × 2 = 4 m² Step 3 — Allowable column load: Qa = qa,net × A = 123.5 × 4 = 494 kN The allowable column load is Qa = 494 kN.

Question Type

numerical

Answer Structure

• Apply qa,net = qu,net / FS = 123.5 kPa [1 mark]
• Compute footing area A = 4 m² [0.5 mark]
• Multiply qa,net × A to get Qa = 494 kN [1 mark]
• State conclusion with correct unit [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using gross qu instead of net qu,net in the FS equation
• Forgetting to multiply by the footing area (reporting only qa,net without converting to load)
• Using FS = 2 or FS = 4 — state FS = 3 as given

Key Phrases To Include

• qa,net = qu,net / FS
• A = B × B
• Qa = qa,net × A
• 494 kN

Marks

3

Topic

Water Table Correction

Difficulty

hard

Template Id

T9

Examiner Tip

The water-table position dictates which unit weight goes into which term — a table in your scratch notes (position / surcharge term / width term) prevents confusion under time pressure.

Model Answer

Given: B = 2 m (strip), Df = 1 m, c = 15 kPa, φ = 25° γsat = 20 kN/m³, γw = 9.81 kN/m³ Water table at footing base. Nc = 25.13, Nq = 12.72, Nγ = 8.34 Water-table correction: • q = γsat × Df = 20 × 1 = 20 kPa (soil above WT is saturated, use γsat for surcharge) Note: If soil above WT is moist with γ = 18 kN/m³, use moist γ; here γsat given for full profile. • For the width term, use γ' = γsat − γw = 20 − 9.81 = 10.19 kN/m³ Terzaghi's strip equation: qu = cNc + qNq + ½γ'BNγ qu = 15(25.13) + 20(12.72) + ½(10.19)(2)(8.34) qu = 376.95 + 254.40 + 85.1 qu = 716.5 kPa

Question Type

numerical

Answer Structure

• State water-table correction rule: γ' used in width term when WT at base [0.5 mark]
• Compute γ' = γsat − γw = 10.19 kN/m³ [0.5 mark]
• Compute q = γsat × Df = 20 kPa [0.5 mark]
• Write strip equation and substitute with γ' in width term [1 mark]
• Sum all terms and state qu = 716.5 kPa [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Applying γ' to the surcharge term q as well when WT is at the base (not at ground surface)
• Using γ' = γsat − γw = 20 − 10 = 10 instead of 9.81 without justification
• Using full γsat = 20 in the width term without any correction

Key Phrases To Include

• γ' = γsat − γw
• water table at footing base
• width term correction
• γ' in ½γBNγ

Marks

3

Topic

Local Shear Failure

Difficulty

medium

Template Id

T10

Examiner Tip

The phrase 'reduce both c and tan φ to two-thirds of their values' is the key technical statement — write it explicitly, not just 'use reduced parameters.'

Model Answer

Local shear failure occurs in medium-density or moderately compressible soils where only a partial failure surface develops below the footing. Unlike general shear failure, there is no sudden collapse — settlement increases progressively without a clear peak bearing pressure. Terzaghi accounts for local shear by using reduced shear-strength parameters: • c* = (2/3)c • tan φ* = (2/3) tan φ → φ* = arctan(2/3 × tan φ) These reduced values are substituted into the standard bearing-capacity factors to obtain modified factors Nc*, Nq*, Nγ*, which are then used in the usual equation: qu = c*Nc* + qNq* + ½γBNγ* (strip, local shear) This reduction reflects the less-efficient mobilization of soil shear resistance in compressible soils.

Question Type

short_answer

Answer Structure

• Define local shear failure and contrast with general shear (progressive settlement, no sudden peak) [1 mark]
• State the reduction rules: c* = 2c/3 and tan φ* = (2/3) tan φ [1 mark]
• Write the modified equation with starred factors and explain the purpose of reduction [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating that local shear uses a lower FS instead of reduced shear parameters
• Applying the reduction to only c but not φ (or vice versa)
• Confusing local shear with punching shear failure

Key Phrases To Include

• partial failure surface
• c* = (2/3)c
• tan φ* = (2/3) tan φ
• modified bearing-capacity factors
• progressive settlement

Marks

5

Topic

Complete Bearing Capacity Analysis — Clay

Difficulty

medium

Template Id

T11

Examiner Tip

A 5-mark question expects a multi-step solution. Use a 'Given / Required / Solution / Answer' structure and label each sub-answer (a), (b), (c) clearly — examiners award partial marks per step.

Model Answer

Given: B = 2.5 m (square), Df = 1.5 m, cu = 75 kPa, φ = 0 γ = 18 kN/m³, Nc = 5.7, Nq = 1, Nγ = 0, FS = 3 Step 1 — Overburden pressure: q = γDf = 18 × 1.5 = 27 kPa Step 2 — Terzaghi's equation for square footing: qu = 1.3cNc + qNq + 0.4γBNγ qu = 1.3(75)(5.7) + 27(1) + 0.4(18)(2.5)(0) qu = 555.75 + 27 + 0 (a) qu = 582.75 kPa ≈ 582.8 kPa Step 3 — Net ultimate bearing capacity: (b) qu,net = qu − γDf = 582.75 − 27 = 555.75 kPa Step 4 — Net allowable bearing capacity: qa,net = qu,net / FS = 555.75 / 3 = 185.25 kPa Step 5 — Footing area: A = 2.5 × 2.5 = 6.25 m² Step 6 — Allowable column load: (c) Qa = qa,net × A = 185.25 × 6.25 = 1,157.8 kN Summary: (a) qu = 582.8 kPa (b) qu,net = 555.75 kPa (c) Qa = 1,157.8 kN

Question Type

numerical

Answer Structure

• List given data and identify square footing [0.5 mark]
• Compute q = γDf = 27 kPa [0.5 mark]
• Write correct square equation with 1.3 and 0.4 shape factors [1 mark]
• Compute qu = 582.8 kPa correctly [1 mark]
• Compute qu,net = 555.75 kPa [0.5 mark]
• Compute qa,net = 185.25 kPa [0.5 mark]
• Compute A = 6.25 m² and Qa = 1,157.8 kN [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using strip equation (no shape factors 1.3 and 0.4)
• Computing a non-zero third term when Nγ = 0
• Using gross qu in the FS formula instead of net qu,net
• Using FS on gross qu to get gross allowable, then forgetting to subtract overburden
• Arithmetic error in 1.3 × 75 × 5.7 (should be 555.75, not 547.5)

Key Phrases To Include

• qu = 1.3cNc + qNq
• Nγ = 0 for φ = 0
• qu,net = qu − γDf
• qa,net = qu,net / FS
• Qa = qa,net × A

Marks

3

Topic

Shape Factors

Difficulty

medium

Template Id

T12

Examiner Tip

A comparison table is the most efficient way to present shape factor data — it earns the mark faster and is visually clear for the examiner.

Model Answer

Terzaghi's original equation was derived for an infinitely long strip footing assuming plane-strain conditions (no end effects). Shape factors correct for the three-dimensional failure geometry of finite footings: | Footing Shape | Cohesion term | Width term | |--------------|--------------|------------| | Strip (L→∞) | 1.0 × cNc | 0.5 × γBNγ | | Square (L = B) | 1.3 × cNc | 0.4 × γBNγ | | Circular (D) | 1.3 × cNc | 0.3 × γBNγ | The surcharge term qNq uses no shape factor in Terzaghi's original formulation. Why shape factors differ: • Square/circular footings mobilize soil resistance along all four sides (3-D failure mechanism), which increases the cohesion contribution — hence the 1.3 multiplier. • However, the Nγ term is reduced (0.4 for square, 0.3 for circular vs 0.5 for strip) because the 3-D geometry also creates additional soil movement at corners, effectively reducing the width-term efficiency. • Empirical calibration of these factors (Terzaghi 1943) was later verified by Hansen and Meyerhof.

Question Type

short_answer

Answer Structure

• State the origin of shape factors (strip = plane-strain, 3-D correction for others) [1 mark]
• Present a clear comparison table or list of shape factors for all three shapes [1 mark]
• Explain physically why the cohesion factor increases (3-D mobilization) while Nγ factor decreases [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating that shape factors are applied to all three terms including qNq (Nq has no shape factor in Terzaghi's original)
• Mixing up 0.4 and 0.3 (0.4 is square, 0.3 is circular)
• No physical explanation — just listing numbers earns only 2 of 3 marks

Key Phrases To Include

• plane-strain
• shape factors
• 1.3cNc
• 0.4γBNγ
• 0.3γBNγ
• three-dimensional failure mechanism

Marks

2

Topic

Settlement vs Shear Failure

Difficulty

easy

Template Id

T13

Examiner Tip

Citing NSCP 2015 settlement limits (25 mm total, 20 mm differential) signals professional-level knowledge and earns bonus credibility with the examiner.

Model Answer

The bearing capacity of a soil is its ability to support the loads applied to the foundation without shear failure or excessive settlement. Specifically, the ultimate bearing capacity qu is the pressure at which the soil shears beneath the footing. Settlement may govern over shear failure when: 1. The soil is compressible (e.g., soft clay, loose fill) and undergoes significant consolidation under sustained load — the footing may settle excessively while the bearing pressure remains below qu. 2. Structural serviceability limits (typically 25 mm total, 20 mm differential for most buildings per NSCP 2015) may be exceeded at bearing pressures well below qu, especially on normally consolidated clay. 3. In such cases, the design bearing pressure is set by the settlement limit, not by qu / FS.

Question Type

short_answer

Answer Structure

• Define bearing capacity including reference to both shear failure and settlement [0.5 mark]
• State at least two specific reasons settlement may govern over shear (compressible soil, serviceability limit) [1 mark]
• Mention NSCP or standard settlement limits to show professional awareness [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Stating only 'settlement is larger' without explaining the mechanism (consolidation, compressibility)
• Not mentioning that settlement can occur at pressures below qu
• No reference to a standard or limit — purely qualitative with no quantitative anchor

Key Phrases To Include

• shear failure
• excessive settlement
• compressible soil
• serviceability limit
• NSCP 2015
• normally consolidated clay

Marks

5

Topic

Net Allowable Bearing Capacity — Strip Footing

Difficulty

medium

Template Id

T14

Examiner Tip

Show every intermediate term on a separate line — 376.95, 228.96, 150.12 — so partial marks can be awarded even if your final arithmetic has an error.

Model Answer

Given: B = 2 m (strip), Df = 1 m c = 15 kPa, φ = 25°, γ = 18 kN/m³ Nc = 25.13, Nq = 12.72, Nγ = 8.34, FS = 3 Step 1 — Overburden: q = γDf = 18 × 1 = 18 kPa Step 2 — Ultimate bearing capacity (strip, no shape factors): qu = cNc + qNq + ½γBNγ qu = 15(25.13) + 18(12.72) + ½(18)(2)(8.34) qu = 376.95 + 228.96 + 150.12 qu = 756.03 kPa Step 3 — Net ultimate bearing capacity: qu,net = qu − γDf = 756.03 − 18 = 738.03 kPa Step 4 — Net allowable bearing capacity: qa,net = qu,net / FS = 738.03 / 3 qa,net = 246.0 kPa Conclusion: The net allowable bearing capacity of the strip footing is qa,net = 246.0 kPa.

Question Type

numerical

Answer Structure

• State given data; identify strip footing (no shape factors) [0.5 mark]
• Compute q = 18 kPa [0.5 mark]
• Write correct strip equation and substitute [1 mark]
• Compute all three terms: 376.95 + 228.96 + 150.12 = 756.03 kPa [1 mark]
• Compute qu,net = 738.03 kPa [0.5 mark]
• Apply FS: qa,net = 246.0 kPa; state conclusion with unit [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Using gross qu (756 kPa) directly in the FS formula instead of net qu,net
• Applying shape factors 1.3 and 0.4 for a strip footing
• Forgetting to subtract the overburden for the net capacity
• Rounding intermediate values causing final error beyond acceptable tolerance (±1%)
• Missing the conclusion statement with unit

Key Phrases To Include

• qu = cNc + qNq + ½γBNγ
• qu,net = qu − γDf
• qa,net = qu,net / FS
• strip footing
• 246.0 kPa

Marks

2

Topic

Bearing-Capacity Factors

Difficulty

easy

Template Id

T15

Examiner Tip

Always attribute Nc = 5.7 to Terzaghi specifically — Prandtl gives 5.14, Hansen gives 5.14 with inclination factors. In board exams, state the source to avoid ambiguity.

Model Answer

The bearing-capacity factors are dimensionless coefficients that quantify the contribution of each component of soil resistance to the ultimate bearing capacity: • Nc — accounts for soil cohesion c; multiplied by the cohesion term • Nq — accounts for the overburden (surcharge) pressure q = γDf at the footing base; reflects depth benefit • Nγ — accounts for the shear resistance contributed by the footing width B and the unit weight of the failure wedge All three are functions of the internal friction angle φ only (from tables or Terzaghi's original charts). For purely cohesive soil (φ = 0°), per Terzaghi: • Nc = 5.7 • Nq = 1 • Nγ = 0 This means that for saturated clay under undrained conditions, the bearing capacity depends only on cohesion and overburden — the width of the footing does not directly contribute to qu.

Question Type

short_answer

Answer Structure

• Define all three factors (Nc = cohesion, Nq = surcharge/depth, Nγ = width/unit weight) [1 mark]
• State the three φ = 0 values: Nc = 5.7, Nq = 1, Nγ = 0, with implication [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using Nc = 5.14 (Prandtl value) instead of Terzaghi's 5.7 — state which author's value you use
• Stating Nγ = 0 without explaining what this implies for design
• Confusing Nc with the shape factor 1.3

Key Phrases To Include

• functions of φ
• Nc = 5.7
• Nq = 1
• Nγ = 0
• φ = 0
• footing width does not contribute