CELE Geotechnical Engineering — Consolidation and SettlementRevision Notes
Quick revision notes for Consolidation and Settlement — the one-page refresher for CELE aspirants. Every item on this page has appeared in recent CELE Geotechnical Engineering papers, so revising these is the shortest path to a confident performance in Professional Regulation Commission (PRC) — Board of Civil Engineering's CELE 2026.
Exam context
The Civil Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Civil Engineering and is scheduled for May and November 2026. The Geotechnical Engineering subtest is marked as "Core" in the official pattern, and Consolidation and Settlement appears in position 6th of 11 in the CELE Geotechnical Engineering review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent CELE 2026 papers have drawn roughly a meaningful share of questions from this subject.
Consolidation and Settlement - Revision Notes
Consolidation and settlement is one of the highest-yield topics in the PRC Civil Engineer Licensure Examination under Geotechnical Engineering. When a saturated clay layer is loaded — say, by a new building, embankment, or fill — excess pore water pressure develops and water slowly migrates out through the voids. As water escapes, the clay skeleton compresses. This time-dependent volume change is called **primary consolidation settlement**. Unlike sand (which drains almost instantly), clay can take months to decades to fully consolidate. The board exam tests two things: (1) the **magnitude** of primary settlement using the compression index Cc or recompression index Cr, and (2) the **time rate** of consolidation using Terzaghi's one-dimensional consolidation theory with the time factor Tv. Master these two pillars, and you can solve virtually any board-level consolidation problem.
Sections
Exam Tips
- Draw a quick sketch of the soil profile with drainage boundaries before solving any time-rate problem.
- Always state OCR = σ'_c / σ'_0 explicitly — examiners reward organized solutions.
- Memorize: double drainage → H_dr = H/2, time is 4× shorter than single drainage for the same U.
Key Points
- Consolidation occurs only in saturated, fine-grained soils (clays and silts) where the permeability is low enough that drainage is time-dependent.
- When a load is applied, 100% of the stress increment is initially carried by pore water (excess pore pressure u_e). Over time, pore water drains and the effective stress in the soil skeleton increases — this is the consolidation process.
- Primary consolidation refers to volume change due to expulsion of pore water. Secondary consolidation (creep) occurs after excess pore pressure has fully dissipated.
- The rate of primary consolidation is governed by: (a) the coefficient of consolidation c_v and (b) the drainage path length H_dr.
- Double drainage (clay bounded by permeable layers top and bottom) means H_dr = H/2. Single drainage (impermeable boundary on one side) means H_dr = H.
- Degree of consolidation U (expressed as a percentage or decimal) represents the fraction of total settlement that has occurred at time t.
- The consolidation process in Philippine soft-clay areas (e.g., Metro Manila Bay reclamations, Pampanga lowlands) can take 10–30 years without vertical drains.
Definitions
Term
Primary Consolidation Settlement (S_c)
Definition
The compression of a saturated clay layer resulting from the slow expulsion of pore water under sustained load, quantified by the log-compression relationship using Cc or Cr.
Importance
Directly computed in virtually every consolidation board problem; determines structural safety and serviceability (floor tilting, pipeline damage).
Term
Compression Index (C_c)
Definition
The slope of the virgin (normally consolidated) portion of the e–log σ' curve: Cc = Δe / log(σ'2/σ'1). Typical values for clays range from 0.1 to 0.8.
Importance
Used in the settlement equation for normally consolidated clays and for the portion of loading that exceeds the preconsolidation pressure in OC clays.
Term
Recompression Index (C_r)
Definition
The slope of the recompression (swelling) portion of the e–log σ' curve. C_r ≈ C_c/5 to C_c/10 (much smaller than Cc).
Importance
Used when the applied final effective stress does not exceed the preconsolidation pressure (overconsolidated range). Using Cc instead of Cr here is a major board-exam pitfall.
Term
Preconsolidation Pressure (σ'_c or P_c)
Definition
The maximum effective vertical stress the soil has experienced in its geological history, determined from the e–log σ' curve by Casagrande's method.
Importance
Determines whether the clay behaves as NC or OC, which dictates which compression index to use and whether a two-part settlement calculation is needed.
Term
Overconsolidation Ratio (OCR)
Definition
OCR = σ'_c / σ'_0 (preconsolidation pressure divided by current effective overburden). OCR = 1 → normally consolidated; OCR > 1 → overconsolidated.
Importance
Classifies the stress history of the clay and determines the correct settlement formula branch.
Term
Coefficient of Consolidation (c_v)
Definition
c_v = k / (m_v · γ_w) where k is permeability and m_v is the coefficient of volume compressibility. Units: m²/yr or cm²/s.
Importance
Controls the time rate of consolidation. Higher c_v → faster consolidation. Used to compute the time factor Tv and solve for time or degree of consolidation.
Term
Time Factor (T_v)
Definition
Dimensionless parameter T_v = c_v · t / H_dr² that relates elapsed time, drainage path, and coefficient of consolidation to degree of consolidation.
Importance
The central variable linking time and degree of consolidation. All time-rate problems pass through Tv.
Term
Degree of Consolidation (U)
Definition
U = (settlement at time t) / (ultimate primary settlement) × 100%. Ranges from 0% (no consolidation) to 100% (full primary consolidation).
Importance
Tells the engineer what fraction of the long-term settlement has occurred. Used to check serviceability at any given time.
Section Title
1. Physical Mechanism of Consolidation
Common Mistakes
- Confusing drainage path H_dr with layer thickness H — always halve H for double drainage before substituting into Tv.
- Not identifying whether the clay is NC or OC before applying a formula — OCR must be checked first.
- Using Cc for the overconsolidated range (σ'_f ≤ σ'_c) when Cr must be used instead.
- Forgetting that the log in the settlement equation is log base 10, not natural log.
Formulas
Example
NC clay: H = 4 m, Cc = 0.30, e_0 = 0.80, σ'_0 = 100 kPa, Δσ = 80 kPa. S_c = [0.30/1.80] × 4000 × log(180/100) = 0.1667 × 4000 × 0.2553 = 170 mm
Formula
S_c = [C_c / (1 + e_0)] × H × log10[(σ'_0 + Δσ) / σ'_0]
Variables
S_c = primary consolidation settlement (m or mm); C_c = compression index (dimensionless); e_0 = initial void ratio; H = layer thickness (m); σ'_0 = initial effective vertical stress at mid-layer (kPa); Δσ = stress increase at mid-layer (kPa)
Application
Normally consolidated clay (OCR = 1). This is the most frequently tested form on the board exam.
Example
OC clay: σ'_c = 200 kPa, σ'_0 = 120 kPa, Δσ = 60 kPa → σ'_f = 180 kPa < 200 kPa → use Cr only.
Formula
S_c = [C_r / (1 + e_0)] × H × log10[(σ'_0 + Δσ) / σ'_0]
Variables
C_r = recompression index; all other variables same as NC formula. Valid only when σ'_0 + Δσ ≤ σ'_c.
Application
Overconsolidated clay where the entire stress range stays below the preconsolidation pressure.
Example
OC clay: H = 3 m, Cr = 0.05, Cc = 0.30, e_0 = 0.75, σ'_0 = 80 kPa, σ'_c = 150 kPa, σ'_f = 220 kPa. Part 1 (OC range): [0.05/1.75] × 3000 × log(150/80) = 0.02857 × 3000 × 0.2730 = 23.4 mm. Part 2 (NC range): [0.30/1.75] × 3000 × log(220/150) = 0.1714 × 3000 × 0.1664 = 85.6 mm. Total S_c = 109 mm.
Formula
S_c = [C_r/(1+e_0)] × H × log10(σ'_c/σ'_0) + [C_c/(1+e_0)] × H × log10(σ'_f/σ'_c)
Variables
σ'_c = preconsolidation pressure (kPa); σ'_f = σ'_0 + Δσ = final effective stress (kPa); both Cr and Cc used in respective stress ranges.
Application
OC clay where loading crosses the preconsolidation pressure (σ'_f > σ'_c). Two-part computation required.
Exam Tips
- Always check OCR first: if OCR = 1, use Cc. If OCR > 1, compare σ'_f with σ'_c before choosing formula.
- Quick estimate for Cc from Atterberg limits (Terzaghi & Peck): Cc ≈ 0.009 × (LL – 10) for remolded clays. Not used in settlement calculation directly but useful for checking given values.
- On the board exam, settlement is typically asked in millimetres — convert H from metres to mm by multiplying by 1000 inside the formula.
- The formula can also be written as S_c = [Cc/(1+e_0)] × H × log10(σ'_f/σ'_0) — note σ'_f = σ'_0 + Δσ. Both forms are equivalent.
Key Points
- For a normally consolidated (NC) clay: S_c = [Cc / (1 + e_0)] × H × log10[(σ'_0 + Δσ) / σ'_0]
- For an overconsolidated (OC) clay where the final stress σ'_f = σ'_0 + Δσ still does not exceed σ'_c: S_c = [Cr / (1 + e_0)] × H × log10[(σ'_0 + Δσ) / σ'_0]
- For an OC clay where loading crosses the preconsolidation pressure (σ'_f > σ'_c): the settlement must be split into two parts — use Cr from σ'_0 to σ'_c, then Cc from σ'_c to σ'_f.
- The two-part OC formula: S_c = [Cr/(1+e_0)] × H × log10(σ'_c / σ'_0) + [Cc/(1+e_0)] × H × log10(σ'_f / σ'_c)
- The stress increase Δσ at mid-depth of the clay layer is typically estimated using Boussinesq stress distribution or the 2:1 method for preliminary calculations.
- H is the initial (uncompressed) thickness of the clay layer; e_0 is the initial void ratio at the start of loading.
Section Title
2. Primary Consolidation Settlement Equations
Common Mistakes
- Using the full layer thickness H instead of mid-layer depth when computing σ'_0 — σ'_0 must be evaluated at the midpoint of the clay layer.
- Applying Cc for the full stress range in an OC clay that crosses σ'_c — this overestimates settlement in the recompression range.
- Forgetting to add the two settlement components in the two-part OC calculation — only computing one part.
- Using the wrong stress ratio inside the log — the argument must be (final stress) / (initial stress), always > 1 so the log is positive.
- Mixing natural log (ln) with log base 10 — the e–log σ' plot uses log base 10 (common logarithm).
Formulas
Example
c_v = 3 m²/yr, H_dr = 2 m (double drainage from 4 m clay), t = 0.5 yr. Tv = 3 × 0.5 / 4 = 0.375
Formula
T_v = c_v × t / H_dr²
Variables
T_v = time factor (dimensionless); c_v = coefficient of consolidation (m²/yr or m²/s); t = elapsed time (yr or s); H_dr = drainage path length (m): equals H/2 for double drainage, H for single drainage
Application
Fundamental equation relating time to degree of consolidation. Used to find Tv, then look up or compute U. Can be rearranged to solve for t or c_v.
Example
U = 50% = 0.50. Tv = (π/4)(0.50)² = 0.7854 × 0.25 = 0.1963 ≈ 0.197 ✓
Formula
T_v = (π/4) × U² [U ≤ 0.60]
Variables
U = degree of consolidation expressed as a decimal (0 to 0.60). Valid for the parabolic early-time approximation.
Application
Solving for Tv when U is known and U ≤ 60%, or solving for U when Tv ≤ 0.2827.
Example
U = 90%. Tv = 1.781 – 0.933 × log10(10) = 1.781 – 0.933 × 1 = 0.848 ✓
Formula
T_v = 1.781 – 0.933 × log10(100 – U%) [U > 60%]
Variables
U = degree of consolidation in percent (60% to ~100%). log10 = common logarithm base 10.
Application
Solving for Tv when U is known and U > 60%, or inversely solving for U when Tv > 0.2827.
Example
Tv = 0.848, H_dr = 2 m (double drainage, H = 4 m), c_v = 3 m²/yr. t = 0.848 × 4 / 3 = 1.13 yr
Formula
t = T_v × H_dr² / c_v
Variables
t = time to reach degree of consolidation U (same units as c_v denominator — yr if c_v in m²/yr).
Application
Most common board form: find the time to reach U = 50%, 90%, etc.
Exam Tips
- Memorize the two benchmark pairs: Tv = 0.197 at U = 50% and Tv = 0.848 at U = 90%. These appear directly or as intermediate steps in most time-rate problems.
- If a problem says 'find the time for 50% consolidation', substitute Tv = 0.197 directly — no need to use the formula.
- Double drainage vs single drainage comparison: for the same Tv, doubling H_dr multiplies time by 4. This is a classic board-exam trick question.
- When solving for U given Tv: if Tv ≤ 0.2827, use U = sqrt(4Tv/π) × 100%. If Tv > 0.2827, use log formula: 100 – U = 10^[(1.781 – Tv)/0.933].
Key Points
- Terzaghi's one-dimensional consolidation theory assumes: homogeneous, saturated clay; Darcy's law governs flow; one-dimensional compression and drainage; constant c_v throughout consolidation.
- The time factor Tv is the dimensionless variable that links time, drainage path, and c_v: Tv = c_v × t / H_dr²
- For U ≤ 60%: Tv = (π/4) × U² (where U is in decimal form, e.g., U = 0.50 for 50%)
- For U > 60%: Tv = 1.781 – 0.933 × log10(100 – U%) where U is in percent
- Critical Tv–U pairs to memorize: U = 50% → Tv = 0.197; U = 90% → Tv = 0.848
- The boundary at U = 60%: Tv = (π/4)(0.60)² = 0.2827 — use this as the switching point between the two formulas.
- Rearranging for time: t = Tv × H_dr² / c_v
- Rearranging for c_v: c_v = Tv × H_dr² / t (used when interpreting lab consolidation test results)
Section Title
3. Time Rate of Consolidation — Terzaghi's Theory
Common Mistakes
- Using H instead of H/2 for H_dr in double drainage problems — this error makes t four times larger (quadratic effect).
- Applying the U ≤ 60% formula when Tv > 0.2827 — always check which branch to use.
- Mixing units of c_v (m²/s vs m²/yr) with time in years — always make units consistent before substituting.
- Forgetting that U must be in decimal form (not percent) in the Tv = π/4 × U² formula but in percent in the log formula.
- Using H_dr = H for double drainage in a 'find t given U = 90%' problem — this gives a time 4× too large.
Exam Tips
- PROBLEM 1 (NC Settlement): A 4 m NC clay layer: Cc = 0.30, e_0 = 0.80, σ'_0 = 100 kPa, Δσ = 80 kPa. S_c = [0.30/1.80] × 4000 × log(180/100) = 0.1667 × 4000 × 0.2553 = 170.2 mm. ANSWER: ≈ 170 mm.
- PROBLEM 2 (Time for 90% U): Same clay, c_v = 3 m²/yr, double drainage (H_dr = 2 m). t = Tv × H_dr² / c_v = 0.848 × (2)² / 3 = 0.848 × 4 / 3 = 1.13 yr. ANSWER: ≈ 1.13 years.
- PROBLEM 3 (Find U at t = 0.5 yr): c_v = 3 m²/yr, H_dr = 2 m. Tv = 3 × 0.5 / 4 = 0.375. Since Tv = 0.375 > 0.2827, use log formula: 100 – U = 10^[(1.781 – 0.375)/0.933] = 10^[1.406/0.933] = 10^1.507 = 32.1. U = 100 – 32.1 = 67.9% ≈ 68%. ANSWER: U ≈ 68%.
- PROBLEM 4 (OC Clay, two-part): H = 3 m, Cr = 0.05, Cc = 0.30, e_0 = 0.75, σ'_0 = 80 kPa, σ'_c = 150 kPa, Δσ = 140 kPa → σ'_f = 220 kPa > σ'_c. Part 1 (Cr): [0.05/1.75] × 3000 × log(150/80) = 23.4 mm. Part 2 (Cc): [0.30/1.75] × 3000 × log(220/150) = 85.6 mm. Total = 109 mm. ANSWER: ≈ 109 mm.
- PROBLEM 5 (Single vs Double Drainage comparison): Same clay, H = 6 m, c_v = 2 m²/yr. Double drainage: H_dr = 3 m, t_90 = 0.848 × 9/2 = 3.82 yr. Single drainage: H_dr = 6 m, t_90 = 0.848 × 36/2 = 15.26 yr. Ratio = 4:1. Classic board trap!
Key Points
- Board problems typically involve: (a) computing primary settlement of NC clay, (b) computing primary settlement of OC clay (one-part or two-part), (c) finding time to reach a given U%, or (d) finding U% at a given time.
- Always organize your solution: (1) identify soil type NC/OC, (2) identify drainage condition, (3) select correct formula, (4) substitute and solve.
- Stress increase Δσ is computed at the mid-depth of the clay layer unless otherwise stated.
Section Title
4. Board-Exam Worked Problems
Formulas
Example
C_α = 0.01, H = 3 m, t_1 = 5 yr, t_2 = 25 yr. S_s = 0.01 × 3000 × log(25/5) = 30 × 0.699 = 20.97 mm ≈ 21 mm
Formula
S_s = C_α × H × log10(t_2 / t_1)
Variables
S_s = secondary consolidation settlement; C_α = secondary compression index (dimensionless slope of strain–log time curve after primary consolidation); H = clay layer thickness (m); t_1 = time at end of primary consolidation; t_2 = time of interest (same units as t_1)
Application
Computing long-term creep settlement after primary consolidation is complete. Most relevant for organic clays and peats.
Exam Tips
- If the board question asks for 'total settlement' and gives C_α, remember to add S_s to S_c.
- The ratio C_α/Cc is approximately constant for a given soil (0.025 to 0.10); sometimes used to estimate C_α from Cc.
Key Points
- Secondary consolidation (creep) occurs after excess pore pressure has fully dissipated (U = 100%). It is due to plastic readjustment of clay particles at constant effective stress.
- Secondary settlement formula: S_s = C_α × H × log10(t_2 / t_1) where C_α is the secondary compression index, t_1 is the end of primary consolidation, and t_2 is the time of interest.
- C_α typically ranges from 0.0005 to 0.02 for inorganic clays; higher for organic soils and peats.
- For preliminary board-level problems, secondary consolidation is usually negligible compared to primary unless the problem specifically involves organic clay or peat.
- The 2:1 stress distribution method estimates Δσ at depth z below a loaded area (B × L) as: Δσ = Q / [(B + z)(L + z)], where Q = total load.
- Immediate (elastic) settlement occurs in all soils under load and is computed using elastic theory. For clays, it is typically small compared to primary consolidation and is added as S_i = q × B × (1 – ν²) / E_s × I_s.
- Total settlement = S_i (immediate) + S_c (primary consolidation) + S_s (secondary consolidation).
Definitions
Term
Secondary Compression Index (C_α)
Definition
Slope of the void ratio vs log(time) curve after primary consolidation is complete. Represents creep or plastic adjustment of clay fabric.
Importance
Critical for organic soils. For inorganic clays at board level, usually a secondary concern unless specifically asked.
Section Title
5. Secondary Consolidation and Additional Considerations
Common Mistakes
- Treating secondary consolidation as part of primary — they are distinct phases; secondary starts only after excess pore pressure is fully dissipated.
- Adding secondary settlement for inorganic clays when the problem only asks for primary — read the question carefully.
Connections
- Effective Stress Principle (Terzaghi): Consolidation is fundamentally a transfer of stress from pore water to soil skeleton (effective stress increases as pore pressure dissipates). Understanding σ' = σ – u is prerequisite to understanding consolidation.
- Permeability (Darcy's Law): c_v = k/(m_v × γ_w) — permeability k directly controls consolidation rate. Clays with higher k (silty clays) consolidate faster than pure clays. This connects the permeability chapter to consolidation.
- Soil Compressibility: The e–log σ' curve from the oedometer test yields both Cc and Cr directly. Understanding the oedometer/consolidation test procedure is the foundation for interpreting consolidation parameters.
- Foundation Engineering: Settlement predictions from consolidation analysis are directly used to check serviceability limit states of shallow and deep foundations. NSCP 2015 Section 304 specifies allowable settlements for structures.
- Stress Distribution (Boussinesq): The stress increase Δσ used in the settlement equation comes from Boussinesq theory or the 2:1 method — connecting the stress distribution chapter to settlement computation.
- Shear Strength of Clay: The preconsolidation pressure σ'_c is also the boundary between NC and OC behavior in the Cam Clay model, directly linking consolidation history to undrained shear strength (S_u ∝ σ'_c for NC clays).
- RA 1378 (Philippine Geotechnical Engineering Practice): Geotechnical investigations including consolidation testing are required for projects on soft ground, connecting the theoretical consolidation analysis to professional practice and legal requirements.
- Vertical Drains (PVD): Prefabricated vertical drains reduce the drainage path from H to the radial spacing, dramatically accelerating consolidation — an application of the same Tv formula with radial drainage corrections (Barron's theory).
Exam Strategy
For the PRC Civil Engineer Licensure Examination on Consolidation and Settlement, follow this proven attack strategy: (1) READ the problem fully — identify NC or OC clay, identify drainage condition (single or double), identify what is being asked (settlement magnitude or time rate). (2) For SETTLEMENT problems: Check OCR first. If OCR = 1 → use Cc formula directly. If OCR > 1 → compare σ'_f with σ'_c. If σ'_f ≤ σ'_c → use Cr formula. If σ'_f > σ'_c → use two-part formula. Always compute σ'_0 at mid-layer depth. (3) For TIME-RATE problems: Determine H_dr (H/2 for double, H for single). Compute Tv = c_v × t / H_dr². If Tv ≤ 0.2827, use parabolic formula for U. If Tv > 0.2827, use log formula for U. Alternatively, if U is given, compute Tv using the appropriate formula, then solve for t = Tv × H_dr² / c_v. (4) MEMORIZE benchmarks: Tv = 0.197 at U = 50%; Tv = 0.848 at U = 90%; switching point at Tv = 0.2827 (U = 60%). (5) Watch for UNIT TRAPS: c_v in m²/yr paired with t in years; H in metres but S_c often reported in mm (multiply H by 1000 or convert at the end). (6) DOUBLE-CHECK by estimating: For a 4 m NC clay with Cc ≈ 0.3 and a 50–100% stress increase, settlement should be in the 100–250 mm range. If your answer is 2 mm or 2 m, recheck. (7) Allocate time: A consolidation problem is typically worth 3–5 points in the Geotechnical Engineering section. Spend 3–5 minutes per problem. Nail the formula, substitute cleanly, and move on.
Quick Review Questions
A 6 m normally consolidated clay layer has Cc = 0.25, e_0 = 0.90, σ'_0 = 120 kPa, and Δσ = 100 kPa. Compute the primary consolidation settlement.
Step 1: Identify NC clay → use Cc formula. Step 2: Compute 1+e_0 = 1+0.90 = 1.90. Step 3: Compute stress ratio = (120+100)/120 = 220/120 = 1.833. Step 4: log10(1.833) = 0.2632. Step 5: S_c = (0.25/1.90) × 6000 × 0.2632 = 208 mm.
A clay layer has c_v = 2 m²/yr and H = 5 m with single drainage. Find the time (in years) for 50% consolidation.
Single drainage → H_dr = H = 5 m. For U = 50%, Tv = 0.197 (memorized benchmark). t = 0.197 × 25 / 2 = 2.46 years. Note: if this were double drainage (H_dr = 2.5 m), t = 0.197 × 6.25 / 2 = 0.616 yr — four times shorter.
For c_v = 4 m²/yr, double drainage, H = 8 m, find the degree of consolidation U (%) after 2 years.
Double drainage: H_dr = 8/2 = 4 m. Tv = c_v × t / H_dr² = 4 × 2 / 16 = 0.50. Tv > 0.2827 → log formula branch. 100 – U = 10^[(1.781 – 0.50)/0.933] = 10^1.3729 ≈ 23.6. U ≈ 76%.
An overconsolidated clay has σ'_c = 200 kPa. The current effective stress is σ'_0 = 120 kPa and a load will increase it to σ'_f = 180 kPa. Which index governs settlement and why?
Check OCR = 200/120 = 1.67 > 1 → OC clay. Check if σ'_f ≤ σ'_c: 180 ≤ 200 → YES, entirely in OC range. Therefore use Cr (not Cc). If σ'_f exceeded 200 kPa, a two-part calculation using both Cr and Cc would be required.
A clay layer is 10 m thick with double drainage. What is H_dr? How does changing to single drainage affect the time to reach any given degree of consolidation?
t = Tv × H_dr² / c_v. Tv and c_v are unchanged; only H_dr changes. Ratio of times = (H_dr,single)² / (H_dr,double)² = 10² / 5² = 100/25 = 4. Single drainage takes 4 times longer — a classic board exam trap.
Compute Tv for U = 75% using both the parabolic and log formulas. Which formula should you use?
Since U = 75% > 60%, the log formula must be used. If you mistakenly used the parabolic: Tv = (π/4)(0.75)² = 0.7854 × 0.5625 = 0.442 — this is incorrect for U > 60%. Always check the 60% boundary first.
What is the effect on primary settlement if c_v is doubled while all other parameters remain constant?
S_c = [Cc/(1+e_0)] × H × log(σ'_f/σ'_0) — this formula contains no c_v term. Settlement magnitude depends on Cc, e_0, H, and stresses only. c_v appears only in Tv = c_v × t / H_dr², which governs rate, not magnitude. A common conceptual trap on the board.
An OC clay has H = 4 m, Cr = 0.04, Cc = 0.28, e_0 = 0.70, σ'_0 = 100 kPa, σ'_c = 160 kPa, Δσ = 120 kPa. Compute S_c.
Step 1: σ'_f = 100 + 120 = 220 kPa. Step 2: Compare with σ'_c = 160 kPa → 220 > 160 → crossing preconsolidation. Step 3: Part 1 uses Cr from 100 to 160 kPa. Step 4: Part 2 uses Cc from 160 to 220 kPa. Step 5: Sum both parts for total settlement.
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