CELE Geotechnical Engineering — Shear Strength of SoilsExam Answer Templates

Marks

1

Topic

Mohr-Coulomb Failure Criterion

Difficulty

easy

Template Id

T1

Examiner Tip

The prime notation (') is non-negotiable when effective stress parameters are specified. Examiners are trained to look for it. One missing prime can cost you the entire mark.

Model Answer

The Mohr-Coulomb failure criterion in terms of effective stress is: τ_f = c' + σ' tan φ', where τ_f is the shear stress at failure, c' is the effective cohesion, σ' is the effective normal stress on the failure plane, and φ' is the effective angle of internal friction.

Question Type

very_short_answer

Answer Structure

• Line 1: Write the equation with all symbols defined [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing τ = c + σ tan φ without the prime (') notation when effective stress is specifically asked — loses the mark.
• Omitting symbol definitions — some rubrics require at least one symbol defined to award the mark.
• Writing the total stress form when the question says 'effective stress'.

Key Phrases To Include

• τ_f = c' + σ' tan φ'
• effective cohesion c'
• effective angle of internal friction φ'
• effective normal stress σ'

Marks

1

Topic

Unconfined Compression Test

Difficulty

easy

Template Id

T2

Examiner Tip

This is a guaranteed 1-mark question in almost every board exam cycle. The formula cᵤ = qᵤ/2 must be written — it signals to the examiner that you know the derivation (σ₃ = 0, so the Mohr circle diameter equals qᵤ and the radius equals cᵤ).

Model Answer

Using the relationship for the unconfined compression test: cᵤ = qᵤ / 2 cᵤ = 84 / 2 cᵤ = 42 kPa

Question Type

numerical

Answer Structure

• Line 1: State the formula cᵤ = qᵤ/2 [formula mark — 1 mark]
• Line 2: Substitute and solve (answer = 42 kPa)

Scoring Breakdown

Common Mark Deductions

• Writing cᵤ = qᵤ (i.e., 84 kPa) instead of half — most common error, zero marks.
• Correct value 42 but no unit (kPa) — may lose the mark depending on rubric.
• Confusing qᵤ with the deviator stress in a triaxial test.

Key Phrases To Include

• cᵤ = qᵤ / 2
• 42 kPa
• unconfined compression test

Marks

2

Topic

Drained vs Undrained Shear Strength

Difficulty

easy

Template Id

T3

Examiner Tip

In a 'differentiate' question, always present both sides in parallel. Examiners allocate one mark per side. Frame each with: condition → parameter used → governing scenario.

Model Answer

Undrained shear strength (cᵤ) applies when loading is rapid and excess pore-water pressures cannot dissipate; for saturated clay, φ = 0 and strength = cᵤ. This governs short-term stability (e.g., immediately after embankment construction). Drained shear strength is defined by effective parameters c' and φ', applicable when pore pressures have fully dissipated; it governs long-term stability. Since sands drain rapidly, they are always analysed using drained (effective stress) parameters.

Question Type

short_answer

Answer Structure

• Sentence 1: Define undrained condition and state φ = 0, give short-term example [1 mark]
• Sentence 2: Define drained condition, state effective parameters c' and φ', give long-term example [1 mark]

Scoring Breakdown

Common Mark Deductions

• Describing only one condition instead of both — loses 1 mark.
• Saying 'undrained means no water' — conceptually wrong, loses the mark.
• Not mentioning pore pressure behavior — examiners specifically look for this.
• Omitting the φ = 0 condition for undrained saturated clay.

Key Phrases To Include

• excess pore-water pressure
• φ = 0 for saturated undrained clay
• effective parameters c' and φ'
• short-term stability
• long-term stability

Marks

2

Topic

Direct Shear Test

Difficulty

medium

Template Id

T4

Examiner Tip

The simultaneous equation approach is the cleanest and most mark-safe method for direct shear data. Always show the subtraction step explicitly — it demonstrates method clearly to the examiner.

Model Answer

Using the Mohr-Coulomb equation τ_f = c + σ tan φ: Test 1: 82 = c + 120 tan φ ... (1) Test 2: 134 = c + 240 tan φ ... (2) Subtracting (1) from (2): 52 = 120 tan φ tan φ = 52/120 = 0.4333 φ = arctan(0.4333) = 23.4° Substituting into (1): c = 82 − 120(0.4333) = 82 − 52 = 30 kPa Therefore: c = 30 kPa, φ = 23.4°

Question Type

numerical

Answer Structure

• Step 1: Write Mohr-Coulomb equation and set up two simultaneous equations [½ mark]
• Step 2: Solve for tan φ by subtraction, find φ [½ mark]
• Step 3: Back-substitute to find c [½ mark]
• Step 4: State final answers with units and degrees [½ mark]

Scoring Breakdown

Common Mark Deductions

• Setting up equations correctly but arithmetic error in subtraction — loses 1 mark.
• Finding tan φ but forgetting to take arctan — reports tan φ as the angle.
• Not writing units for c (kPa) or not writing degree symbol for φ.
• Using only one data point and assuming c = 0 without justification.

Key Phrases To Include

• τ_f = c + σ tan φ
• simultaneous equations
• tan φ = 0.4333
• c = 30 kPa
• φ = 23.4°

Marks

2

Topic

Triaxial Test — Friction Angle

Difficulty

medium

Template Id

T5

Examiner Tip

The step 'σ₁ = σ₃ + deviator stress' is critical and earns its own mark in many rubrics. Never skip it. Write it as the very first computational line.

Model Answer

Given: σ₃ = 150 kPa, deviator stress = 300 kPa σ₁ = σ₃ + deviator = 150 + 300 = 450 kPa For c = 0 sand, using: sin φ = (σ₁ − σ₃) / (σ₁ + σ₃) sin φ = (450 − 150) / (450 + 150) sin φ = 300 / 600 = 0.500 φ = arcsin(0.500) = 30°

Question Type

numerical

Answer Structure

• Step 1: Compute σ₁ = σ₃ + deviator stress [½ mark]
• Step 2: Write the sin φ formula for c = 0 [½ mark]
• Step 3: Substitute and solve sin φ [½ mark]
• Step 4: Compute φ = 30° [½ mark]

Scoring Breakdown

Common Mark Deductions

• Using the deviator stress directly as σ₁ without adding σ₃ — most frequent board exam error.
• Applying the c ≠ 0 formula unnecessarily when the problem states c = 0 sand.
• Reporting sin φ = 0.5 as the final answer instead of taking arcsin.
• Wrong formula: using cos instead of sin.

Key Phrases To Include

• σ₁ = σ₃ + deviator stress
• sin φ = (σ₁ − σ₃)/(σ₁ + σ₃)
• c = 0 (sand)
• φ = 30°

Marks

2

Topic

Mohr-Coulomb Failure Criterion

Difficulty

easy

Template Id

T6

Examiner Tip

Always write tan 28° = 0.5317 explicitly — it shows you computed it correctly. Examiners appreciate this transparency and it protects your method mark even if there is a minor arithmetic slip.

Model Answer

Applying the Mohr-Coulomb failure criterion (effective stress form): τ_f = c' + σ' tan φ' τ_f = 25 + 180 × tan 28° τ_f = 25 + 180 × 0.5317 τ_f = 25 + 95.7 τ_f = 120.7 kPa

Question Type

numerical

Answer Structure

• Line 1: Write τ_f = c' + σ' tan φ' (formula mark) [1 mark]
• Line 2: Substitute values and compute; state final answer with units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using tan⁻¹ (arctan) instead of tan — misreading the formula direction.
• Omitting the unit kPa in the final answer.
• Using total stress parameters when effective stress parameters are given.
• Rounding tan 28° to 0.5 (tan 30°) without acknowledgment — introduces significant error.

Key Phrases To Include

• τ_f = c' + σ' tan φ'
• tan 28° = 0.5317
• 120.7 kPa

Marks

3

Topic

Triaxial Test — UU Condition

Difficulty

medium

Template Id

T7

Examiner Tip

Structure your 3-mark answer in exactly three parts: (1) what happens in the test, (2) what you get from it, (3) when you use it. This maps directly onto the 3-mark rubric every time.

Model Answer

In the Unconsolidated-Undrained (UU) triaxial test, the soil sample is NOT allowed to consolidate under the confining pressure σ₃, and drainage is prevented throughout both the consolidation stage and the shear stage. Because excess pore pressures are not allowed to dissipate, the measured strength is in terms of total stress. For a saturated clay tested under UU conditions, all Mohr circles (at different confining pressures) have the same diameter, yielding a horizontal failure envelope. This gives: • Undrained shear strength: cᵤ (total cohesion) • Friction angle: φ = 0° (φᵤ = 0) The UU test is used to evaluate the short-term, immediate stability of clay foundations and embankments, where rapid loading prevents pore pressure dissipation.

Question Type

short_answer

Answer Structure

• Paragraph 1: Describe the test procedure — no drainage during consolidation and shear, total stress analysis [1 mark]
• Paragraph 2: State parameters yielded — cᵤ and φ = 0; mention horizontal envelope for saturated clay [1 mark]
• Paragraph 3: State the practical application — short-term stability of clays [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing UU with CU — stating that consolidation is allowed loses the first mark.
• Saying φ = 0 but not explaining why (horizontal envelope from equal-diameter Mohr circles) — partial mark only.
• Not mentioning any practical application — loses the third mark.
• Using effective stress notation c' and φ' for UU results — conceptual error.

Key Phrases To Include

• no drainage / drainage valve closed
• total stress parameters
• φ = 0° (or φᵤ = 0)
• undrained shear strength cᵤ
• short-term stability
• horizontal failure envelope

Marks

3

Topic

Direct Shear Test

Difficulty

medium

Template Id

T8

Examiner Tip

When three data points are given, always verify your answer with the unused third point. This earns the verification mark and shows confidence. It typically takes only 10 seconds but signals a thorough exam response.

Model Answer

Using τ_f = c + σ tan φ, set up simultaneous equations from Tests 1 and 3 (widest range for accuracy): Test 1: 52 = c + 50 tan φ ... (1) Test 3: 104 = c + 200 tan φ ... (2) Subtracting (1) from (2): 52 = 150 tan φ tan φ = 52/150 = 0.3467 φ = arctan(0.3467) ≈ 19.1° Substituting into (1): c = 52 − 50(0.3467) = 52 − 17.3 = 34.7 kPa ≈ 35 kPa Verification with Test 2: τ_f = 35 + 100(0.3467) = 35 + 34.7 = 69.7 kPa ≈ 70 kPa ✓ Therefore: c ≈ 35 kPa, φ ≈ 19.1°

Question Type

numerical

Answer Structure

• Step 1: Write Mohr-Coulomb equation and select two test points; set up simultaneous equations [1 mark]
• Step 2: Solve for tan φ → φ [1 mark]
• Step 3: Solve for c; verify with third data point; state answers with units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using only two points and not attempting verification — loses 1 mark if three data points are given.
• Arithmetic error in subtraction — loses up to 2 marks (downstream effect).
• Not writing c in kPa.
• Selecting adjacent tests (1 and 2) with a narrow stress range — introduces larger rounding error, may yield slightly off answers.

Key Phrases To Include

• τ_f = c + σ tan φ
• simultaneous equations
• tan φ = 0.3467
• φ ≈ 19.1°
• c ≈ 35 kPa
• verification

Marks

3

Topic

Mohr-Coulomb Failure Criterion

Difficulty

medium

Template Id

T9

Examiner Tip

For 'explain and derive' questions, always lead with the engineering significance before the mathematics. Examiners who read quickly look for engineering judgment first, then verify the math. The significance mark is the easiest to earn.

Model Answer

The Mohr-Coulomb failure criterion defines the limiting condition of stress in a soil mass: failure occurs on any plane where τ_f = c + σ tan φ. It is fundamental to the design of foundations, retaining walls, and slopes, as it provides the boundary between stable and unstable stress states. For a cohesionless soil (c = 0), the failure envelope passes through the origin: τ_f = σ tan φ. At failure, the Mohr circle is tangent to this envelope. Using Mohr circle geometry, the relationship between principal stresses is derived as: sin φ = (σ₁ − σ₃) / (σ₁ + σ₃) Rearranging to express σ₁ in terms of σ₃: σ₁ = σ₃ tan²(45 + φ/2) This relationship is used directly in triaxial test analysis to determine φ from measured principal stresses at failure.

Question Type

short_answer

Answer Structure

• Paragraph 1: Significance of Mohr-Coulomb in geotechnical design [1 mark]
• Paragraph 2: State the c = 0 simplification; derive or state sin φ formula from Mohr circle geometry [1 mark]
• Paragraph 3: Express σ₁ in terms of σ₃; state practical use [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not mentioning that failure occurs when the Mohr circle is tangent to the failure envelope — loses derivation mark.
• Writing the formula for c ≠ 0 when c = 0 is specified — formula contains extra term, suggesting confusion.
• Omitting practical applications in geotechnical design — loses significance mark.
• Errors in the tan²(45 + φ/2) expression — common if derived from memory without checking.

Key Phrases To Include

• τ_f = c + σ tan φ
• failure envelope
• Mohr circle tangent to failure envelope
• sin φ = (σ₁ − σ₃)/(σ₁ + σ₃)
• σ₁ = σ₃ tan²(45 + φ/2)
• cohesionless (c = 0)

Marks

3

Topic

Unconfined Compression Test

Difficulty

medium

Template Id

T10

Examiner Tip

Unit consistency is the number one source of errors in this type of problem. The safest approach: convert everything to kN and m before any calculation. 180 N = 0.180 kN; A in m²; result in kPa. Do it in one consistent system.

Model Answer

Given: Failure load P = 180 N, diameter d = 38 mm, height H = 76 mm (a) Cross-sectional area: A = π d² / 4 = π (0.038)² / 4 = 1.134 × 10⁻³ m² Unconfined compressive strength: qᵤ = P / A = 180 / (1.134 × 10⁻³) = 158,730 Pa ≈ 158.7 kPa (b) Undrained shear strength: cᵤ = qᵤ / 2 = 158.7 / 2 ≈ 79.4 kPa Note: In the UU condition, φ = 0, so the Mohr circle has diameter = qᵤ and radius = cᵤ.

Question Type

numerical

Answer Structure

• Step 1: Compute cross-sectional area in m² [1 mark]
• Step 2: Compute qᵤ = P/A in kPa with correct unit conversion [1 mark]
• Step 3: Apply cᵤ = qᵤ/2; state φ = 0 condition [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using diameter instead of radius in area formula — A = πr² instead of πd²/4 — very common error.
• Forgetting to convert N to kN before dividing by m², yielding answer in Pa not kPa.
• Using area in mm² and load in N, giving qᵤ in MPa — unit inconsistency.
• Computing cᵤ = qᵤ (not halving) — most penalized conceptual error.

Key Phrases To Include

• A = πd²/4
• qᵤ = P/A
• unit conversion: N/m² = Pa → kPa
• cᵤ = qᵤ/2
• φ = 0 (undrained saturated clay)

Marks

5

Topic

Triaxial Test — CU Condition

Difficulty

hard

Template Id

T11

Examiner Tip

For any problem with two or more Mohr circles, the key formula is sin φ = ΔR/ΔC (change in radius over change in center). This is derived from the geometry of two circles with a common external tangent. Master this — it appears in almost every board exam CU/CD triaxial problem with c ≠ 0.

Model Answer

Step 1 — Identify principal stresses and compute Mohr circle parameters: For each test, center of Mohr circle = (σ₁ + σ₃)/2, radius = (σ₁ − σ₃)/2 Test 1: Center₁ = (220 + 100)/2 = 160 kPa; Radius₁ = (220 − 100)/2 = 60 kPa Test 2: Center₂ = (380 + 200)/2 = 290 kPa; Radius₂ = (380 − 200)/2 = 90 kPa Step 2 — Determine φ from the slope of the common tangent: sin φ = (R₂ − R₁) / (C₂ − C₁) = (90 − 60) / (290 − 160) = 30 / 130 = 0.2308 φ = arcsin(0.2308) = 13.3° Step 3 — Determine c from the failure envelope intercept: The distance from origin to the tangent line along the τ-axis: c cos φ = R₁ − C₁ sin φ c cos φ = 60 − 160 × 0.2308 = 60 − 36.9 = 23.1 c = 23.1 / cos(13.3°) = 23.1 / 0.9730 = 23.7 kPa ≈ 24 kPa Step 4 — Verify with Test 2: c cos φ = 90 − 290 × 0.2308 = 90 − 66.9 = 23.1 ✓ (consistent) Step 5 — State final results: c ≈ 24 kPa (total stress cohesion) φ ≈ 13.3° (total stress friction angle) Note: These are total stress parameters from CU test. For effective stress parameters, pore pressure measurements at failure would be required.

Question Type

numerical

Answer Structure

• Step 1: Compute center and radius of both Mohr circles [1 mark]
• Step 2: Apply sin φ = ΔR/ΔC formula and solve for φ [1 mark]
• Step 3: Apply intercept formula to find c [1 mark]
• Step 4: Verify consistency with both circles [1 mark]
• Step 5: State final c and φ with units; note total vs effective distinction [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using sin φ = (σ₁ − σ₃)/(σ₁ + σ₃) directly — valid only for c = 0 sand; wrong for c ≠ 0 clay.
• Not computing the circle center and radius — attempting to apply formulas without these intermediate values.
• Skipping the verification step — loses 1 mark even if answer is correct.
• Not noting that CU gives total stress parameters (without pore pressure) — loses the final commentary mark.
• Confusing deviator stress with σ₁ (not adding σ₃).

Key Phrases To Include

• center = (σ₁ + σ₃)/2
• radius = (σ₁ − σ₃)/2
• sin φ = ΔR / ΔC
• c cos φ = R − C sin φ
• total stress parameters
• CU test
• c ≈ 24 kPa
• φ ≈ 13.3°

Marks

5

Topic

Drained vs Undrained — Engineering Application

Difficulty

hard

Template Id

T12

Examiner Tip

This is a classic 'design judgment' question. Examiners at PRC level reward students who not only know the formulas but can explain the engineering reasoning behind test selection. Structure your answer in labeled parts (a), (b), (c) — this tells the examiner exactly where to award each mark.

Model Answer

(a) Appropriate Test Type — UU Triaxial Test: For rapid embankment construction on saturated clay, the Unconsolidated-Undrained (UU) triaxial test is the appropriate test. Rapid loading means that excess pore water pressures generated during loading cannot dissipate before the critical stability condition is reached. The drainage condition in the UU test (valves closed throughout) simulates this field condition. (b) Shear Strength Parameters — Undrained Condition: From the UU test on saturated clay: • Failure envelope is horizontal → φ = 0° (φᵤ = 0) • Intercept gives the undrained shear strength: sᵤ = cᵤ • All Mohr circles at different σ₃ values have equal diameters (pore pressures change exactly to neutralize confining pressure change) The undrained stability is checked using: τ_f = cᵤ (φ = 0 analysis) This is also equivalent to: cᵤ = qᵤ/2 if an unconfined compression test is used. (c) Long-Term Stability Reassessment — Drained Condition: Over time, excess pore pressures dissipate and the clay consolidates. Long-term (drained) stability must be assessed using effective stress parameters c' and φ' obtained from either: • Consolidated-Drained (CD) triaxial test, or • Consolidated-Undrained (CU) triaxial test with pore pressure measurement. The long-term analysis uses: τ_f = c' + σ' tan φ' For overconsolidated clays, special attention must be given to the possibility of negative pore pressures (suction) in the short term, which may make the long-term (drained) condition more critical. This is the scenario that has led to delayed failures in clay embankments and cuts worldwide. Conclusion: Short-term → UU test → φ = 0, use cᵤ. Long-term → CD or CU test → use c', φ' with effective stresses.

Question Type

long_answer

Answer Structure

• Part (a): Identify UU test; explain why (rapid loading, no drainage time) — [1 mark for test type + 1 mark for reasoning]
• Part (b): State φ = 0, cᵤ; explain equal-diameter Mohr circles for saturated clay — [1 mark]
• Part (c): Describe drained reassessment using CD or CU; cite effective parameters c' and φ'; mention overconsolidated clay risk — [2 marks]

Scoring Breakdown

Common Mark Deductions

• Recommending CU or CD for rapid loading — incorrect reasoning, loses both test-type marks.
• Stating φ = 0 without explaining WHY (equal pore pressure response for saturated clay) — partial mark.
• Not distinguishing between total and effective stress parameters for short vs long term — loses up to 2 marks.
• No mention of the long-term drained reassessment — loses the final 2 marks entirely.
• Vague conclusion without specifying which parameters to use in each scenario.

Key Phrases To Include

• UU triaxial test
• rapid loading / no time to drain
• φ = 0 analysis
• undrained shear strength cᵤ
• equal Mohr circle diameters
• long-term drained condition
• effective parameters c' and φ'
• CD or CU test with pore pressure
• overconsolidated clay / delayed failure

Marks

1

Topic

Mohr-Coulomb — Principal Stress Relationship

Difficulty

medium

Template Id

T13

Examiner Tip

Memorize the full expression. The 2c tan(45 + φ/2) term is algebraically identical to 2c√Kₚ where Kₚ is the Rankine passive pressure coefficient — useful cross-check between topics.

Model Answer

At failure in a triaxial test for a c-φ soil: σ₁ = σ₃ tan²(45 + φ/2) + 2c tan(45 + φ/2)

Question Type

very_short_answer

Answer Structure

• Line 1: Write the complete formula with both terms [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing only σ₁ = σ₃ tan²(45 + φ/2) — missing the 2c term; loses the mark.
• Writing tan(45 − φ/2) instead of tan(45 + φ/2) — sign error, common confusion with active earth pressure.
• Leaving out the '2' coefficient on the cohesion term.

Key Phrases To Include

• σ₁ = σ₃ tan²(45 + φ/2) + 2c tan(45 + φ/2)
• Rankine active/passive coefficient
• both terms (σ₃ term and cohesion term)

Marks

5

Topic

Triaxial Test — CU vs CD

Difficulty

hard

Template Id

T14

Examiner Tip

For comparison questions at 5 marks, use numbered sections or a clear structure. Examiners allocate marks to specific points; a wall of paragraph text makes it hard for them to find your marks. Use headers or numbered points.

Model Answer

1. Drainage Conditions: In the CU test, the sample is first consolidated (drained) under the confining pressure σ₃ — drainage is allowed until excess pore pressures fully dissipate. During the shear stage, drainage is closed. If pore pressures are measured during shear, effective stress parameters can be calculated. In the CD test, drainage is allowed throughout — both during consolidation and during shearing. Shearing must be done very slowly to prevent pore pressure build-up. This is impractical for clays (may take days) but straightforward for sands. 2. Parameters Obtained: CU test (without pore pressure): Total stress parameters — c and φ (total) CU test (with pore pressure measurement): Both total AND effective parameters — c, φ (total) and c', φ' (effective) CD test: Directly yields effective parameters — c' and φ'; no pore pressure measurement needed since Δu = 0 throughout. 3. Appropriate Application: CU test: Used when it is necessary to model the situation where the soil has been consolidated under one stress state and then subjected to rapid undrained loading — for example, staged embankment construction, or rapid drawdown in an earth dam. Also used when both total and effective parameters are needed from one test. CD test: Used for slowly drained materials (gravels, sands, silts) or for long-term stability analysis of drained clay slopes. The CD test is the most fundamental effective stress test because pore pressure is zero throughout shear. 4. Key Difference: CD gives c' and φ' directly. CU with pore pressure also gives c' and φ', but requires a pore pressure transducer. For most practical purposes, c'(CU) ≈ c'(CD) and φ'(CU) ≈ φ'(CD) for the same soil. 5. Summary Table: • CU: Consolidated → undrained shear → total c,φ (and effective if Δu measured) • CD: Consolidated → drained shear → effective c', φ' directly

Question Type

long_answer

Answer Structure

• Section 1: Drainage conditions for CU and CD during consolidation and shear stages [1 mark]
• Section 2: Parameters from CU (with and without pore pressure) [1 mark]
• Section 3: Parameters from CD [1 mark]
• Section 4: Engineering applications of CU [1 mark]
• Section 5: Engineering applications of CD; comparison conclusion [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating that CU test has no consolidation stage — fundamental error, loses first mark.
• Stating that CD gives total stress parameters — wrong, CD gives effective; loses up to 2 marks.
• Not mentioning pore pressure measurement as the condition for obtaining effective parameters from CU.
• No engineering application examples — loses 2 marks for application sections.
• Confusing rapid drawdown application (CU) with long-term drained analysis (CD).

Key Phrases To Include

• consolidated under σ₃ with drainage
• shear with drainage closed (CU)
• shear with drainage open (CD)
• pore pressure measurement
• total stress parameters c, φ
• effective parameters c', φ'
• staged construction / rapid drawdown
• slow shear rate for CD
• Δu = 0 in CD

Marks

3

Topic

Triaxial Test — c-φ Soil

Difficulty

hard

Template Id

T15

Examiner Tip

This is the most algebraically demanding shear strength problem type. The key insight is that σ₁ = σ₃ Nφ + 2c'√Nφ creates a quadratic in √Nφ. Substitute x = √Nφ to get a standard form quadratic. Show this substitution explicitly — it earns the quadratic setup mark even if arithmetic goes slightly astray.

Model Answer

Step 1 — Determine principal stresses: σ₃ = 100 kPa σ₁ = σ₃ + deviator = 100 + 280 = 380 kPa Step 2 — Apply principal stress failure relationship: σ₁ = σ₃ tan²(45 + φ'/2) + 2c' tan(45 + φ'/2) Let N_φ = tan²(45 + φ'/2) and note 2c'tan(45 + φ'/2) = 2c'√N_φ 380 = 100 N_φ + 2(15)√N_φ 380 = 100 N_φ + 30√N_φ Step 3 — Let x = √N_φ: 100x² + 30x − 380 = 0 Using the quadratic formula: x = [−30 ± √(900 + 4×100×380)] / (2×100) x = [−30 ± √(900 + 152000)] / 200 x = [−30 ± √152900] / 200 x = [−30 ± 391.0] / 200 Taking positive root: x = 361/200 = 1.805 N_φ = x² = (1.805)² = 3.258 tan²(45 + φ'/2) = 3.258 tan(45 + φ'/2) = √3.258 = 1.805 45 + φ'/2 = arctan(1.805) = 61.0° φ'/2 = 16.0° φ' = 32.0° Verification: σ₁ = 100(3.258) + 2(15)(1.805) = 325.8 + 54.2 = 380 kPa ✓

Question Type

numerical

Answer Structure

• Step 1: Compute σ₁ = 380 kPa [½ mark]
• Step 2: Write principal stress failure formula for c-φ soil [1 mark]
• Step 3: Solve the quadratic (substitution x = √N_φ) for N_φ [1 mark]
• Step 4: Compute φ' = 32° and verify [½ mark]

Scoring Breakdown

Common Mark Deductions

• Using the c = 0 formula sin φ = (σ₁ − σ₃)/(σ₁ + σ₃) when c' = 15 kPa is given — wrong formula, may lose all 3 marks.
• Not computing σ₁ first and using 280 kPa as σ₁.
• Setting up the equation correctly but using linear approximation instead of solving the quadratic.
• Correct x = 1.805 but not squaring to get N_φ before extracting the angle.

Key Phrases To Include

• σ₁ = σ₃ + deviator
• σ₁ = σ₃ tan²(45 + φ'/2) + 2c' tan(45 + φ'/2)
• let x = √N_φ (substitution)
• quadratic formula
• φ' = 32°
• verification