CELE Geotechnical Engineering — Consolidation and SettlementExam Answer Templates

Marks

1

Topic

Primary Consolidation Settlement

Difficulty

easy

Template Id

T1

Examiner Tip

The key discriminator for a 1-mark definition is the phrase 'dissipation of excess pore water pressure' — include it and you earn the mark.

Model Answer

Primary consolidation settlement is the time-dependent compression of a saturated clay layer caused by the gradual dissipation of excess pore water pressure and the corresponding transfer of applied stress to the soil skeleton.

Question Type

very_short_answer

Answer Structure

• One sentence: identify it as time-dependent compression of saturated clay due to pore pressure dissipation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing 'settlement due to load' without mentioning pore pressure — too vague for full credit
• Confusing primary consolidation with immediate (elastic) settlement
• Omitting the word 'saturated' or 'clay'

Key Phrases To Include

• saturated clay
• excess pore water pressure
• dissipation
• time-dependent
• compression of soil skeleton

Marks

1

Topic

Time Rate of Consolidation

Difficulty

easy

Template Id

T2

Examiner Tip

Always state 'double drainage' or 'single drainage' explicitly — it shows the examiner you understand why Hdr is H/2 or H.

Model Answer

For double drainage (drainage at both top and bottom), the drainage path length is: Hdr = H/2 = 6/2 = 3 m

Question Type

very_short_answer

Answer Structure

• State drainage condition (double drainage) [0.5 mark]
• Write and evaluate Hdr = H/2 = 3 m [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using Hdr = H = 6 m (single drainage error on a double-drainage problem)
• Omitting the unit 'm'

Key Phrases To Include

• double drainage
• Hdr = H/2
• 3 m

Marks

2

Topic

Time Rate of Consolidation

Difficulty

easy

Template Id

T3

Examiner Tip

Write both formulas in a table or numbered list — examiners scan for both quickly, and clear formatting earns both marks efficiently.

Model Answer

The two approximate expressions are: (1) For U ≤ 60%: Tv = (π/4) U² [where U is expressed as a decimal, e.g., U = 0.50] (2) For U > 60%: Tv = 1.781 − 0.933 log₁₀(100 − U%) [where U is in percent] Mnemonic: the parabolic form applies for early consolidation; the logarithmic form applies for later stages.

Question Type

short_answer

Answer Structure

• Write Formula 1 with its U range (U ≤ 60%) [1 mark]
• Write Formula 2 with its U range (U > 60%) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Interchanging which formula applies at which range
• Using natural log (ln) instead of log₁₀
• Not stating the boundary condition U = 60%

Key Phrases To Include

• Tv = (π/4)U²
• Tv = 1.781 − 0.933 log(100 − U%)
• U ≤ 60%
• U > 60%

Marks

2

Topic

Primary Consolidation Settlement — OC vs NC

Difficulty

medium

Template Id

T4

Examiner Tip

Mention 'preconsolidation pressure σ'c' by name — it signals to the examiner that you understand the physical mechanism, not just the formula.

Model Answer

A normally consolidated (NC) clay is one whose current effective overburden stress equals the maximum stress it has ever experienced (OCR = 1). Its settlement is governed by the compression index Cc, which is the slope of the virgin e–log σ' line — typically ranging from 0.2 to 0.5 for soft clays. An overconsolidated (OC) clay has experienced a higher past stress (OCR > 1). When the final stress σ'f remains below the preconsolidation pressure σ'c, only the much smaller recompression index Cr (≈ Cc/5 to Cc/10) is used. If σ'f exceeds σ'c, settlement is computed in two parts: Cr for the range σ'₀ to σ'c, and Cc for σ'c to σ'f.

Question Type

short_answer

Answer Structure

• Define NC clay and state Cc is used [1 mark]
• Define OC clay, state Cr is used below σ'c, and explain the two-part calculation when σ'f > σ'c [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating Cc for OC clay when the stress increment stays below σ'c
• Not mentioning preconsolidation pressure at all
• Confusing OCR direction (OCR < 1 does not physically exist)

Key Phrases To Include

• OCR = 1
• OCR > 1
• compression index Cc
• recompression index Cr
• preconsolidation pressure σ'c
• virgin e–log σ' line

Marks

3

Topic

Primary Consolidation Settlement — NC Clay

Difficulty

medium

Template Id

T5

Examiner Tip

The log₁₀ vs ln mistake is the single most common error in settlement problems. Write 'log₁₀' explicitly — not 'log' — to show the examiner you know which base to use.

Model Answer

Given: H = 4 m, Cc = 0.30, e₀ = 0.80 σ'₀ = 100 kPa, Δσ = 80 kPa Clay is normally consolidated (NC) → use Cc Step 1 — Identify the formula: Sc = [Cc / (1 + e₀)] × H × log₁₀[(σ'₀ + Δσ) / σ'₀] Step 2 — Substitute: Sc = [0.30 / (1 + 0.80)] × 4 × log₁₀[(100 + 80) / 100] = [0.30 / 1.80] × 4 × log₁₀(1.80) = 0.1667 × 4 × 0.2553 Step 3 — Compute: Sc = 0.6667 × 0.2553 = 0.1702 m ∴ Sc ≈ 170 mm

Question Type

numerical

Answer Structure

• List all given data and state NC condition [0.5 mark]
• Write the settlement formula in symbolic form [0.5 mark]
• Substitute values correctly, including log₁₀ evaluation [1 mark]
• Correct numerical result with unit (mm or m) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using ln instead of log₁₀ — gives Sc ≈ 392 mm (grossly wrong)
• Forgetting to multiply by H (layer thickness)
• Writing (1 + e₀) in the numerator instead of denominator
• Not converting to mm — leaving answer as 0.17 without unit

Key Phrases To Include

• Sc = [Cc/(1+e₀)] × H × log₁₀[(σ'₀+Δσ)/σ'₀]
• normally consolidated
• log₁₀
• 0.1667 × 4 × 0.2553

Marks

3

Topic

Time Rate of Consolidation

Difficulty

medium

Template Id

T6

Examiner Tip

Memorize the two critical Tv–U pairs: (U=50%, Tv=0.197) and (U=90%, Tv=0.848). These appear in nearly every board exam time-rate problem.

Model Answer

Given: H = 4 m, double drainage → Hdr = H/2 = 2 m cv = 3 m²/yr, U = 90% → Tv = 0.848 (standard value) Step 1 — Write time factor formula: Tv = cv × t / Hdr² → t = Tv × Hdr² / cv Step 2 — Substitute: t = 0.848 × (2)² / 3 = 0.848 × 4 / 3 = 3.392 / 3 Step 3 — Compute: t = 1.131 yr ∴ t ≈ 1.13 years (≈ 413 days)

Question Type

numerical

Answer Structure

• State double drainage and compute Hdr = 2 m [0.5 mark]
• Recall Tv = 0.848 for U = 90% [0.5 mark]
• Write and apply formula t = Tv × Hdr² / cv [1 mark]
• Correct answer with unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using Hdr = 4 m (single drainage) instead of 2 m — multiplies t by 4×
• Using Tv = 0.197 (which corresponds to U = 50%, not 90%)
• Forgetting to square Hdr

Key Phrases To Include

• double drainage
• Hdr = H/2 = 2 m
• Tv = 0.848
• t = Tv Hdr² / cv

Marks

3

Topic

Time Rate of Consolidation — U > 60%

Difficulty

hard

Template Id

T7

Examiner Tip

Always verify the 60% boundary by computing Tv at U=60% = 0.2827. If your computed Tv exceeds this, the logarithmic form is mandatory — this boundary check is itself worth marks.

Model Answer

Given: cv = 3 m²/yr, Hdr = 2 m (double drainage), t = 0.5 yr Step 1 — Compute Tv: Tv = cv × t / Hdr² = 3 × 0.5 / (2)² = 1.5 / 4 = 0.375 Step 2 — Identify formula branch: Check if U = 60% applies: at U = 60%, Tv = (π/4)(0.60)² = 0.2827 Since Tv = 0.375 > 0.2827 → use the U > 60% formula: Tv = 1.781 − 0.933 log₁₀(100 − U%) Step 3 — Solve for U: 0.375 = 1.781 − 0.933 log₁₀(100 − U) 0.933 log₁₀(100 − U) = 1.781 − 0.375 = 1.406 log₁₀(100 − U) = 1.406 / 0.933 = 1.5069 100 − U = 10^1.5069 = 32.13 U = 100 − 32.13 ∴ U ≈ 67.9% ≈ 68%

Question Type

numerical

Answer Structure

• Compute Tv correctly [1 mark]
• Check the 60% boundary and select the correct formula branch [0.5 mark]
• Algebraic manipulation to isolate U [1 mark]
• Correct final answer U ≈ 68% [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using the parabolic formula Tv = (π/4)U² even though Tv > 0.2827 (wrong branch)
• Arithmetic error in antilog step — forgetting that 10^x is needed, not e^x
• Not checking the 60% boundary at all

Key Phrases To Include

• Tv = cv t / Hdr²
• Tv = 0.375
• U > 60% formula
• Tv = 1.781 − 0.933 log(100 − U)
• 10^1.5069 = 32.13

Marks

3

Topic

Primary Consolidation Settlement — OC Clay

Difficulty

hard

Template Id

T8

Examiner Tip

The stress comparison step (σ'f vs σ'c) is worth a full mark. Write it explicitly: 'Since σ'f = 180 kPa < σ'c = 200 kPa, the clay remains in the OC range; use Cr only.'

Model Answer

Given: σ'₀ = 120 kPa, σ'c = 200 kPa, σ'f = 180 kPa Cc = 0.40, Cr = 0.05, e₀ = 0.75, H = 5 m Step 1 — Compare σ'f with σ'c: σ'f = 180 kPa < σ'c = 200 kPa → Entire stress increase is within the OC range → Use only Cr Step 2 — Apply settlement formula with Cr: Sc = [Cr / (1 + e₀)] × H × log₁₀(σ'f / σ'₀) = [0.05 / (1 + 0.75)] × 5 × log₁₀(180 / 120) = [0.05 / 1.75] × 5 × log₁₀(1.50) = 0.02857 × 5 × 0.1761 = 0.02518 m ∴ Sc ≈ 25.2 mm

Question Type

numerical

Answer Structure

• Compare σ'f = 180 kPa with σ'c = 200 kPa and conclude OC range only [1 mark]
• Select Cr and write correct formula [0.5 mark]
• Correct substitution and log₁₀ evaluation [1 mark]
• Final answer with unit [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using Cc = 0.40 instead of Cr = 0.05 — overestimates Sc by 8×
• Performing a two-part calculation (Cr + Cc) when σ'f never exceeds σ'c
• Using the stress increase Δσ as the ratio instead of (σ'₀+Δσ)/σ'₀

Key Phrases To Include

• σ'f < σ'c → OC range only
• use Cr
• Sc = [Cr/(1+e₀)] × H × log₁₀(σ'f/σ'₀)
• log₁₀(1.50) = 0.1761

Marks

2

Topic

Time Rate of Consolidation

Difficulty

easy

Template Id

T9

Examiner Tip

Name both fitting methods if you can — examiners often reward the extra detail, and it differentiates you from students who write only the formula.

Model Answer

The coefficient of consolidation cv [units: m²/yr or cm²/s] governs the rate at which excess pore water pressure dissipates in a clay layer — a higher cv means faster consolidation. It is defined as: cv = k / (mv × γw) where k = hydraulic conductivity, mv = coefficient of volume compressibility, and γw = unit weight of water. In the laboratory, cv is obtained from an oedometer (consolidation) test using either: (a) Taylor's square-root-of-time method, or (b) Casagrande's logarithm-of-time method, both of which match observed settlement–time curves to the Terzaghi theory.

Question Type

short_answer

Answer Structure

• State physical meaning of cv (rate of pore pressure dissipation) and give units [1 mark]
• Name the laboratory test and at least one method for determining cv [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing cv with Cc (compression index) — entirely different parameters
• Not mentioning units
• Writing only 'consolidation test' without specifying oedometer or the fitting method

Key Phrases To Include

• coefficient of consolidation
• rate of pore pressure dissipation
• oedometer test
• Taylor's square-root-of-time method
• Casagrande's log-time method

Marks

5

Topic

Primary Settlement + Time Rate — Combined

Difficulty

hard

Template Id

T10

Examiner Tip

For a 5-mark problem, the examiner expects clearly separated parts. Label your work 'Part (a)' and 'Part (b)' with a horizontal line between them — this shows organization and makes partial credit easy to award.

Model Answer

Given: H = 6 m, single drainage → Hdr = H = 6 m Cc = 0.25, e₀ = 0.90, σ'₀ = 120 kPa, Δσ = 100 kPa cv = 2 m²/yr, NC clay ─── Part (a): Primary Consolidation Settlement ─── Formula (NC clay): Sc = [Cc / (1 + e₀)] × H × log₁₀[(σ'₀ + Δσ) / σ'₀] Substitute: Sc = [0.25 / (1 + 0.90)] × 6 × log₁₀[(120 + 100) / 120] = [0.25 / 1.90] × 6 × log₁₀(220 / 120) = 0.13158 × 6 × log₁₀(1.8333) = 0.13158 × 6 × 0.2632 = 0.2078 m ∴ Sc ≈ 208 mm ─── Part (b): Time for U = 50% ─── For U = 50% → Tv = 0.197 (standard value) Single drainage → Hdr = 6 m t = Tv × Hdr² / cv = 0.197 × (6)² / 2 = 0.197 × 36 / 2 = 7.092 / 2 ∴ t = 3.55 yr

Question Type

numerical

Answer Structure

• Identify NC condition and single drainage; state Hdr = 6 m [0.5 mark]
• Write NC settlement formula symbolically [0.5 mark]
• Correct substitution and log₁₀ calculation for Part (a) [1.5 marks]
• Final Sc ≈ 208 mm with unit [0.5 mark]
• Recall Tv = 0.197 for U = 50% and apply t formula for Part (b) [1.5 marks]
• Final t = 3.55 yr with unit [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using Hdr = 3 m (double drainage) for a single-drainage problem — reduces t by 4× (wrong by factor of 4)
• Using Tv = 0.848 (for 90%) instead of 0.197 (for 50%)
• Mixing up the stress ratio: using Δσ/σ'₀ instead of (σ'₀+Δσ)/σ'₀ inside the log
• Not showing the log₁₀ step explicitly
• Failing to answer both parts (a) and (b)

Key Phrases To Include

• single drainage
• Hdr = H = 6 m
• Sc = [Cc/(1+e₀)] × H × log₁₀(σ'f/σ'₀)
• log₁₀(1.8333) = 0.2632
• Tv = 0.197 for U = 50%
• t = Tv Hdr²/cv

Marks

5

Topic

Primary Consolidation Settlement — OC Clay with σ'f > σ'c

Difficulty

hard

Template Id

T11

Examiner Tip

The two-part OC calculation is a classic board exam trap. Show the transition point (σ'c = 200 kPa) as a separate numbered step — it demonstrates conceptual mastery and earns the decision mark even if arithmetic errors follow.

Model Answer

Step 1 — Find preconsolidation pressure σ'c: OCR = σ'c / σ'₀ → σ'c = OCR × σ'₀ = 2.5 × 80 = 200 kPa Step 2 — Compare σ'f with σ'c: σ'f = 250 kPa > σ'c = 200 kPa → Stress path crosses σ'c → Two-part calculation required: Part 1 (OC range): from σ'₀ = 80 to σ'c = 200 kPa → use Cr Part 2 (NC range): from σ'c = 200 to σ'f = 250 kPa → use Cc Step 3 — Settlement from OC range (Sc1): Sc1 = [Cr / (1 + e₀)] × H × log₁₀(σ'c / σ'₀) = [0.04 / (1 + 0.82)] × 4 × log₁₀(200 / 80) = [0.04 / 1.82] × 4 × log₁₀(2.50) = 0.02198 × 4 × 0.3979 = 0.03499 m = 35.0 mm Step 4 — Settlement from NC range (Sc2): Sc2 = [Cc / (1 + e₀)] × H × log₁₀(σ'f / σ'c) = [0.35 / 1.82] × 4 × log₁₀(250 / 200) = 0.19231 × 4 × log₁₀(1.25) = 0.19231 × 4 × 0.09691 = 0.07455 m = 74.6 mm Step 5 — Total settlement: Sc = Sc1 + Sc2 = 35.0 + 74.6 ∴ Sc ≈ 109.6 mm ≈ 110 mm

Question Type

numerical

Answer Structure

• Compute σ'c from OCR [0.5 mark]
• Compare σ'f with σ'c and establish two-part calculation [1 mark]
• Part 1: Correct Sc1 using Cr [1.5 marks]
• Part 2: Correct Sc2 using Cc [1.5 marks]
• Sum correctly for total Sc ≈ 110 mm [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using only Cc for the entire range (ignoring the OC portion) — underestimates Sc1
• Using only Cr for the entire range — grossly underestimates Sc2
• Reversing the stress limits in either part (e.g., log(80/200) = negative)
• Forgetting to compute σ'c from OCR — jumping directly to the formula

Key Phrases To Include

• σ'c = OCR × σ'₀ = 200 kPa
• σ'f > σ'c → two-part calculation
• Cr for OC range (σ'₀ to σ'c)
• Cc for NC range (σ'c to σ'f)
• log₁₀(2.50) = 0.3979
• log₁₀(1.25) = 0.0969

Marks

1

Topic

Time Rate of Consolidation

Difficulty

easy

Template Id

T12

Examiner Tip

Specify Hdr (drainage path) explicitly — it is distinct from H (layer thickness) and using H instead of Hdr is a classic board-exam error.

Model Answer

The time factor Tv is a dimensionless parameter in Terzaghi's consolidation theory that normalizes elapsed time with respect to drainage path and consolidation rate: Tv = cv × t / Hdr² where cv = coefficient of consolidation (m²/yr), t = elapsed time (yr), and Hdr = drainage path length (m).

Question Type

very_short_answer

Answer Structure

• State that Tv is dimensionless; give the formula Tv = cv t / Hdr² [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing Tv = cv t / H² without specifying Hdr (drainage path, not total thickness)
• Inverting the formula as Tv = Hdr² / (cv t)

Key Phrases To Include

• dimensionless
• Tv = cv t / Hdr²
• drainage path length Hdr

Marks

2

Topic

Time Rate of Consolidation — U < 60%

Difficulty

medium

Template Id

T13

Examiner Tip

Always compute the boundary Tv at U=60% (= 0.2827) first and write it down — this one check prevents choosing the wrong branch, which is the most common error in U-from-Tv problems.

Model Answer

Step 1 — Check which formula to use: At U = 60%: Tv = (π/4)(0.60)² = 0.2827 Since Tv = 0.12 < 0.2827 → use the parabolic formula (U ≤ 60% range) Step 2 — Solve for U: Tv = (π/4) U² 0.12 = (π/4) U² U² = 0.12 × (4/π) = 0.1528 U = √0.1528 = 0.3909 ∴ U ≈ 39.1% ≈ 39%

Question Type

numerical

Answer Structure

• Identify Tv < 0.2827 → use parabolic formula [0.5 mark]
• Apply Tv = (π/4)U² and solve for U [1 mark]
• Final answer U ≈ 39% [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using the logarithmic formula (U > 60%) for a small Tv — wrong branch entirely
• Forgetting to take the square root of U²
• Expressing U as a decimal (0.391) without converting to percent

Key Phrases To Include

• Tv = 0.12 < 0.2827
• parabolic formula
• Tv = (π/4)U²
• U ≈ 39%

Marks

3

Topic

Time Rate of Consolidation — Conceptual

Difficulty

medium

Template Id

T14

Examiner Tip

For explain-type questions, show the algebra even when not explicitly asked — a two-line derivation earns more marks than a verbal statement alone.

Model Answer

Terzaghi's time factor is: Tv = cv × t / Hdr² For a fixed degree of consolidation (fixed U, hence fixed Tv) and constant cv, rearranging gives: t = Tv × Hdr² / cv Time t is proportional to Hdr² (the square of the drainage path). If the clay thickness H is doubled (e.g., from H to 2H), the drainage path — for double drainage — also doubles (Hdr becomes 2Hdr). Substituting: t_new = Tv × (2 Hdr)² / cv = 4 × (Tv × Hdr² / cv) = 4 t_old Thus, consolidation time scales with the square of the drainage path. A 2× increase in drainage path length results in a 4× increase in time. This is the reason that thick clay layers consolidate very slowly — the quadratic dependence on Hdr makes layer thickness the dominant design parameter for time rate.

Question Type

short_answer

Answer Structure

• State Terzaghi's time factor formula [0.5 mark]
• Rearrange to show t ∝ Hdr² [1 mark]
• Demonstrate algebraically: doubling Hdr → 4× time [1 mark]
• State the physical implication [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Stating the answer without algebraic proof — earns only partial credit
• Confusing H (layer thickness) with Hdr (drainage path) when drainage is single vs double
• Not mentioning the fixed Tv assumption (U must be specified)

Key Phrases To Include

• Tv = cv t / Hdr²
• t ∝ Hdr²
• (2Hdr)² = 4Hdr²
• quadratic dependence
• drainage path

Marks

5

Topic

Time Rate of Consolidation — Comprehensive Case Study

Difficulty

hard

Template Id

T15

Examiner Tip

Case study questions reward students who read boundary conditions carefully. The phrase 'impermeable rock' = single drainage = Hdr = H. State this explicitly in your answer — it is the foundation on which both parts depend, and it earns half a mark before any calculation.

Model Answer

Given: H = 10 m, single drainage (impermeable rock below, granular fill above) → Hdr = H = 10 m cv = 1.5 m²/yr, Sc(total) = 360 mm ─── Verify U = 50% at t = 2 yr ─── Compute Tv: Tv = cv × t / Hdr² = 1.5 × 2 / (10)² = 3.0 / 100 = 0.030 For U = 50%: Tv (theoretical) = 0.197 Since computed Tv = 0.030 ≠ 0.197, U after 2 yr is NOT 50%. Actual U after t = 2 yr: Tv = 0.030 < 0.2827 → parabolic branch U² = 4 × Tv / π = 4 × 0.030 / π = 0.03820 U = √0.03820 = 0.1954 ≈ 19.5% ∴ After 2 years, only about 19.5% consolidation has occurred — NOT 50%. The statement is incorrect. ─── Time for settlement = 324 mm ─── Required degree of consolidation: U = (324 / 360) × 100% = 90% For U = 90%: Tv = 0.848 t = Tv × Hdr² / cv = 0.848 × (10)² / 1.5 = 0.848 × 100 / 1.5 = 84.8 / 1.5 ∴ t ≈ 56.5 yr

Question Type

case_study

Answer Structure

• Identify single drainage and Hdr = 10 m [0.5 mark]
• Compute Tv at t = 2 yr = 0.030 [0.5 mark]
• Use parabolic formula to find actual U ≈ 19.5% and state that 50% has NOT been reached [1 mark]
• Compute required U = 90% for S = 324 mm [0.5 mark]
• Apply t = Tv Hdr²/cv with Tv = 0.848 [1.5 marks]
• Final answer t ≈ 56.5 yr with unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Assuming double drainage despite impermeable rock — Hdr = 5 m gives a very different answer
• Accepting the 50% claim without computing the actual Tv
• Computing U = 324/360 = 0.90 but using Tv = 0.197 (for 50%) — misreading the problem
• Using Hdr = 5 m in part 2 after correctly using 10 m in part 1

Key Phrases To Include

• single drainage — impermeable boundary
• Hdr = H = 10 m
• Tv = 0.030 at t = 2 yr
• U ≈ 19.5% — NOT 50%
• U = 90% for S = 324 mm
• Tv = 0.848 for U = 90%
• t ≈ 56.5 yr