CELE Geotechnical Engineering — Consolidation and SettlementExam Answer Templates

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.

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

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

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

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.

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%

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.

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

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

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

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)

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

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%

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

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

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

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.

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

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

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

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

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

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).

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

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%

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%

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.

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

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

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

Do

• Include the most discriminating keyword (e.g., 'excess pore water pressure', 'drainage path Hdr')
• Write the formula in symbolic form if it is a formula-type question
• State units for any quantity you name

Don't

• Do not write a paragraph — you waste time and the examiner cannot find your answer
• Do not use vague language like 'compression of soil' without specifying the mechanism
• Do not leave units out

Do

• Label your points clearly: (1) and (2), or Step 1 and Step 2
• For comparison questions (NC vs OC), use a two-column or two-paragraph structure
• Show the boundary condition or selection criterion (e.g., Tv at U=60%)

Don't

• Do not merge two separate points into one dense sentence — the examiner must see two distinct ideas
• Do not skip the formula even for 'explain' questions — show it to anchor your explanation
• Do not omit units in the final numerical answer

Do

• Write 'Given:' with a bulleted list of all data before any formula
• State the soil condition (NC or OC) and drainage condition (single or double) as explicit sentences
• Show intermediate computations (e.g., log value, ratio) as numbered steps

Don't

• Do not skip the symbolic formula step — even if you know the answer, write the formula first
• Do not use a single block of arithmetic — break it into visible steps
• Do not forget to check the 60% boundary for U–Tv formula selection

Do

• Label separate parts with Part (a) / Part (b) and a visible separator line
• Draw and label a soil profile diagram showing H, Hdr, and drainage direction
• State 'NC condition' or 'OC range: Cr applies / NC range: Cc applies' as explicit sentences
• Box or underline each final numerical answer
• Show the antilog step explicitly for U > 60% problems: 100 − U = 10^x → U = …

Don't

• Do not solve Part (b) before establishing Part (a) — examiners follow your logic linearly
• Do not skip the two-part OC calculation when σ'f > σ'c
• Do not write only the final answer — partial marks come from intermediate steps
• Do not leave out the summary sentence: 'Therefore, t ≈ 56.5 yr'