CELE Geotechnical Engineering — Consolidation and SettlementDetailed Explanation
Detailed explanations for CELE Geotechnical Engineering — Consolidation and Settlement. This page treats you like a serious reviewer: we unpack the concepts thoroughly, show worked examples of how Professional Regulation Commission (PRC) — Board of Civil Engineering frames Consolidation and Settlement questions, and explain the underlying reasoning that gets you to the right answer every time.
Exam context
Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Geotechnical Engineering section sits under a "Core" weighting, and Consolidation and Settlement is the 6th chapter in the 11-chapter CELE Geotechnical Engineering rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Geotechnical Engineering.
Consolidation and Settlement - Detailed Explanation
Consolidation and settlement is one of the most heavily tested topics in the PRC Civil Engineer Licensure Examination under Geotechnical Engineering. When a load is applied to a saturated clay layer, excess pore water pressure builds up and water gradually drains out, causing the clay skeleton to compress over time. This time-dependent volume change is called **primary consolidation settlement**. Unlike sand, which settles almost instantaneously, clay can take months to years to reach its final settlement — a critical concern in Philippine infrastructure projects built on soft marine clays (e.g., reclamation areas in Manila Bay, Pasig River flood plains, and Visayas coastal zones). This chapter covers: (1) computing the magnitude of primary consolidation settlement using the compression index for normally consolidated and overconsolidated clays, and (2) computing the time rate of consolidation using Terzaghi's 1-D consolidation theory. Mastery of these two aspects — magnitude and rate — is essential for passing the board exam and for professional practice under RA 544 (Civil Engineering Law).
Concepts
The Consolidation Process — Physical Mechanism
Consolidation is the gradual compression of a saturated clay layer as excess pore water pressure (EPWP) dissipates following an applied load. The physical analogy: imagine a wet sponge placed under a heavy book. Initially, the water in the sponge carries the load (high pore pressure). Over time, water squeezes out through the sponge pores, and the sponge skeleton gradually takes the load. The load transfer from pore water to the soil skeleton is the essence of consolidation. **Key stages:** 1. **Initial condition:** Load ΔP applied. Immediately, EPWP = ΔP, effective stress Δσ' = 0. 2. **During consolidation:** EPWP dissipates, effective stress increases. Partial settlement occurs. 3. **End of primary consolidation:** EPWP = 0, Δσ' = ΔP. Full primary settlement achieved. 4. **Secondary consolidation (creep):** Slow rearrangement of clay particles at constant effective stress. Governed by C_α (secondary compression index) — beyond the scope of most board items but good to know. **Drainage boundaries determine speed:** A clay layer draining from both top and bottom (double drainage) consolidates 4 times faster than one draining from one side only (single drainage), because the drainage path H_dr is halved and time is proportional to H_dr².
Examples
The drainage path is the maximum distance a water molecule must travel to escape the clay. With double drainage, the water only needs to travel to the nearest boundary — half the layer thickness.
Scenario
A 4 m saturated clay layer is sandwiched between sand layers above and below. Explain the drainage condition and state H_dr.
Solution
Because sand is highly permeable, both the top and bottom boundaries of the clay are drainage boundaries. This is **double drainage**. Therefore, H_dr = H/2 = 4/2 = 2 m.
The rock at the bottom acts as an impermeable boundary; no water can exit downward. Water must travel the full 5 m to the top sand layer.
Scenario
A 5 m clay layer sits on impermeable rock and has a permeable sand layer on top only. State H_dr.
Solution
Only the top boundary is a drainage boundary — **single drainage**. H_dr = H = 5 m.
Applications
- Determining whether a building foundation on soft clay will experience differential settlement.
- Design of vertical sand drains or prefabricated vertical drains (PVDs) to accelerate consolidation in reclamation projects.
- Predicting settlement timelines for embankments on soft ground (e.g., road embankments over Candaba Swamp, Pampanga).
- Back-calculation of c_v from field settlement monitoring data.
Misconceptions
- MISCONCEPTION: Settlement of clay starts immediately after loading. TRUTH: The soil skeleton begins to compress only as EPWP dissipates and effective stress increases.
- MISCONCEPTION: Double drainage means faster settlement by only 2×. TRUTH: It is 4× faster because time ∝ H_dr², and halving H_dr reduces t by a factor of 4.
- MISCONCEPTION: Consolidation and compaction are the same. TRUTH: Compaction is a mechanical process (dynamic, for unsaturated soils); consolidation is a hydraulic-flow-driven, time-dependent process in saturated soils.
Related Concepts
- Effective stress principle (Terzaghi)
- Permeability and Darcy's Law
- Soil compressibility and void ratio
- Preconsolidation pressure and OCR
- Terzaghi's 1-D consolidation theory
Common Exam Questions
Example
A clay layer is between two sand layers. What is H_dr if H = 6 m? Answer: H_dr = 3 m (double drainage).
Approach
Identify the drainage condition from the given soil profile, then state H_dr.
Question Type
Conceptual identification
Example
At t = 0, immediately after a load is applied to a saturated clay, the increase in effective stress is: (A) equal to Δσ, (B) zero, (C) equal to Δu, (D) cannot be determined. Answer: B — zero, because EPWP = Δσ at t = 0.
Approach
Remember that immediately after loading, all applied stress is carried by pore water, not the soil skeleton.
Question Type
True/False or multiple choice on mechanism
Key Points To Remember
- Consolidation is time-dependent; it is NOT immediate — this distinguishes clay from sand.
- EPWP = Δσ at t = 0; EPWP → 0 as t → ∞ (end of primary consolidation).
- Effective stress principle: σ' = σ − u (Terzaghi). Settlement is caused by increase in σ', not total stress.
- Double drainage: H_dr = H/2. Single drainage: H_dr = H. Time ∝ H_dr², so doubling H_dr quadruples consolidation time.
- Primary consolidation is the main component computed in board exams.
- Secondary consolidation (creep) occurs after EPWP = 0 and is governed by C_α.
Primary Consolidation Settlement — Magnitude
The magnitude of primary consolidation settlement depends on how much the void ratio changes when the effective stress increases. This is captured by the **compression index C_c** (for normally consolidated clays) or the **recompression index C_r** (for overconsolidated clays). **Void ratio vs. log σ' curve (e–log σ' diagram):** In the laboratory oedometer test, a clay sample is loaded in increments and the void ratio at each stress level is measured. Plotting e vs. log₁₀σ' gives a curve with two distinct segments: - **Recompression curve (OC range):** Gentle slope = C_r (typically C_r ≈ C_c/5 to C_c/10) - **Virgin compression line (NC range):** Steeper slope = C_c The breakpoint between the two is the **preconsolidation pressure σ'_c**. **Classification:** - **Normally Consolidated (NC):** Current effective stress σ'_0 = σ'_c (clay has never been more heavily loaded). Use C_c. - **Overconsolidated (OC):** σ'_0 < σ'_c (clay was once more heavily loaded, e.g., by a glacier or eroded overburden). Use C_r for the OC range, then C_c for the NC range. - **OCR = σ'_c / σ'_0** (Overconsolidation Ratio) **Settlement Equations:** For **NC clay** (σ'_0 + Δσ ≤ σ'_c is NOT satisfied — entire loading is in NC range): $$S_c = \frac{C_c}{1 + e_0} \cdot H \cdot \log_{10}\left(\frac{\sigma'_0 + \Delta\sigma}{\sigma'_0}\right)$$ For **OC clay, Case 1** — final stress stays within OC range (σ'_0 + Δσ ≤ σ'_c): $$S_c = \frac{C_r}{1 + e_0} \cdot H \cdot \log_{10}\left(\frac{\sigma'_0 + \Delta\sigma}{\sigma'_0}\right)$$ For **OC clay, Case 2** — final stress exceeds preconsolidation pressure (σ'_0 + Δσ > σ'_c): $$S_c = \frac{C_r}{1 + e_0} \cdot H \cdot \log_{10}\left(\frac{\sigma'_c}{\sigma'_0}\right) + \frac{C_c}{1 + e_0} \cdot H \cdot \log_{10}\left(\frac{\sigma'_0 + \Delta\sigma}{\sigma'_c}\right)$$ Note: The factor 1/(1 + e_0) converts the void ratio change to a strain, and H converts strain to settlement. **Empirical estimates of C_c (for board exam quick checks):** - Skempton (1944): C_c ≈ 0.009(LL − 10) for remolded clays (LL in %) - C_r ≈ C_c/5 to C_c/10 is a typical field approximation
Examples
Step 1: Compute the factor C_c/(1+e_0) = 0.30/1.80 = 0.1667. Step 2: Compute the stress ratio (100+80)/100 = 1.80. Step 3: Take log₁₀(1.80) = 0.2553. Step 4: Multiply: 0.1667 × 4000 × 0.2553 = 170.2 mm. Note: H was converted to mm (4 m = 4000 mm) to get S_c in mm directly.
Scenario
Board-Type Problem 1 (NC Clay): A 4 m NC clay layer has C_c = 0.30, e_0 = 0.80, initial effective overburden stress σ'_0 = 100 kPa, and a stress increase Δσ = 80 kPa. Find the primary consolidation settlement S_c.
Solution
S_c = [C_c / (1 + e_0)] × H × log₁₀[(σ'_0 + Δσ) / σ'_0] = [0.30 / (1 + 0.80)] × 4000 mm × log₁₀[(100 + 80) / 100] = [0.30 / 1.80] × 4000 × log₁₀(1.80) = 0.16667 × 4000 × 0.25527 = 170.2 mm ≈ 170 mm
The key step is checking whether the final stress (200 kPa) exceeds the preconsolidation pressure (160 kPa). Since it does, split the calculation at σ'_c. The first segment (100→160 kPa) uses C_r; the second segment (160→200 kPa) uses C_c. Sum both parts for total S_c.
Scenario
Board-Type Problem 2 (OC Clay, Case 2): A 3 m OC clay has C_c = 0.40, C_r = 0.08, e_0 = 0.75, σ'_0 = 100 kPa, σ'_c = 160 kPa, Δσ = 100 kPa. Compute S_c.
Solution
Check: σ'_0 + Δσ = 100 + 100 = 200 kPa > σ'_c = 160 kPa → Case 2 (two-part calculation). Part 1 (OC range: 100 to 160 kPa, use C_r): S₁ = [0.08 / (1 + 0.75)] × 3000 × log₁₀(160/100) = [0.08/1.75] × 3000 × log₁₀(1.60) = 0.04571 × 3000 × 0.20412 = 28.0 mm Part 2 (NC range: 160 to 200 kPa, use C_c): S₂ = [0.40 / (1 + 0.75)] × 3000 × log₁₀(200/160) = [0.40/1.75] × 3000 × log₁₀(1.25) = 0.22857 × 3000 × 0.09691 = 66.5 mm Total: S_c = S₁ + S₂ = 28.0 + 66.5 = 94.5 mm
Because the final stress (140 kPa) does not reach the preconsolidation pressure (160 kPa), the clay remains in its overconsolidated state throughout loading. Only C_r is used — the settlement is much smaller than Case 2.
Scenario
Board-Type Problem 3 (OC Clay, Case 1): Same clay as above but Δσ = 40 kPa only. Compute S_c.
Solution
Check: σ'_0 + Δσ = 100 + 40 = 140 kPa < σ'_c = 160 kPa → Case 1 (entire loading in OC range, use C_r only). S_c = [0.08 / 1.75] × 3000 × log₁₀(140/100) = 0.04571 × 3000 × log₁₀(1.40) = 0.04571 × 3000 × 0.14613 = 20.0 mm
Applications
- Estimating long-term settlement of buildings founded on soft clay deposits.
- Checking whether total settlement exceeds allowable limits per NSCP 2015 Table 305.1 (total settlement ≤ 25 mm for isolated footings on clay, differential ≤ 19 mm).
- Evaluating risk of damage to structures on reclaimed land (Manila Bay area).
- Designing preloading programs — preload raises σ'_c, making subsequent structural loads fall in the OC range, dramatically reducing settlement.
Misconceptions
- MISCONCEPTION: Use natural log (ln) in the settlement equation. TRUTH: Always log BASE 10 (log₁₀). Using ln overestimates by a factor of 2.303.
- MISCONCEPTION: For OC clay, always use C_r regardless of the final stress. TRUTH: If σ'_f > σ'_c, you MUST use C_c for the portion beyond σ'_c — ignoring this significantly underestimates settlement.
- MISCONCEPTION: C_c is the slope of the e vs σ' plot (arithmetic scale). TRUTH: C_c is the slope of the e vs. log₁₀σ' plot. On a linear scale, compressibility decreases with stress.
- MISCONCEPTION: H in the formula can be left in meters and S_c will come out in meters automatically. TRUTH: S_c comes out in the same units as H. If you want S_c in mm, use H in mm.
Related Concepts
- Oedometer (consolidometer) test procedure
- Preconsolidation pressure determination (Casagrande graphical method)
- Overconsolidation Ratio (OCR)
- Compression index empirical correlations (Skempton, Terzaghi)
- Allowable settlement limits (NSCP 2015 Section 305)
Common Exam Questions
Example
Given C_c = 0.35, e_0 = 0.90, H = 5 m, σ'_0 = 80 kPa, Δσ = 60 kPa → S_c = [0.35/1.90] × 5000 × log(140/80) = 0.1842 × 5000 × 0.2430 = 223.8 mm
Approach
Apply S_c = [C_c/(1+e_0)] × H × log(σ'_f/σ'_0) directly. Watch units of H.
Question Type
Direct computation — NC clay settlement
Example
If σ'_0 = 50, σ'_c = 120, σ'_f = 200 kPa: Part 1 uses C_r from 50→120; Part 2 uses C_c from 120→200.
Approach
Always check σ'_f vs σ'_c first. If σ'_f > σ'_c, split at σ'_c and compute two settlements.
Question Type
Two-part OC clay computation
Example
Clay with OCR = 2.5 and σ'_0 = 80 kPa → σ'_c = 2.5 × 80 = 200 kPa. If Δσ = 90 kPa, σ'_f = 170 kPa < 200 kPa → entire loading in OC range, use C_r.
Approach
Compute OCR = σ'_c / σ'_0. OCR = 1 → NC. OCR > 1 → OC. The question may give OCR directly.
Question Type
Identifying NC vs OC condition
Key Points To Remember
- Settlement formula uses log BASE 10 (log₁₀), NOT natural log (ln).
- For NC clay: use C_c throughout.
- For OC clay: use C_r while σ'_f ≤ σ'_c; switch to C_c once σ'_f > σ'_c (two-part calculation).
- OCR = σ'_c / σ'_0. If OCR = 1, the clay is NC. If OCR > 1, it is OC.
- C_c > C_r always. NC clays settle more than OC clays under the same load increment.
- The '1 + e_0' factor in the denominator normalizes the void ratio change relative to initial solids.
- H must be in consistent units with S_c (if H is in mm, S_c is in mm).
Time Rate of Consolidation — Terzaghi's Theory
Knowing how much settlement will occur (magnitude) is only half the problem. Engineers also need to know **when** the settlement will occur. Terzaghi's 1-D consolidation theory relates time, soil properties, drainage conditions, and the degree of consolidation achieved. **Time Factor T_v (dimensionless):** $$T_v = \frac{c_v \cdot t}{H_{dr}^2}$$ where: - c_v = coefficient of consolidation [m²/yr or m²/s] - t = elapsed time - H_dr = drainage path length (H/2 for double drainage; H for single drainage) **Degree of Consolidation U (%):** U represents the fraction of total primary settlement that has occurred at time t. $$U = \frac{S_t}{S_c} \times 100\%$$ where S_t = settlement at time t, S_c = ultimate primary settlement. **T_v – U Relationships:** The exact solution involves an infinite Fourier series, but two practical closed-form approximations are used: For **U ≤ 60%:** $$T_v = \frac{\pi}{4} U^2 \quad (U \text{ expressed as a decimal, e.g., 0.50 for 50\%})$$ For **U > 60%:** $$T_v = 1.781 - 0.933 \log_{10}(100 - U\%)$$ (Here U% is the percentage value, e.g., 70 for 70%) **Key T_v–U pairs to MEMORIZE for board exam:** | U (%) | T_v | |-------|------| | 10 | 0.008 | | 20 | 0.031 | | 30 | 0.071 | | 40 | 0.126 | | 50 | 0.197 | | 60 | 0.287 | | 70 | 0.403 | | 80 | 0.567 | | 90 | 0.848 | | 95 | 1.163 | | 99 | 1.781 | **Derivation of key values:** - At U = 50%: T_v = (π/4)(0.50)² = (π/4)(0.25) = 0.1963 ≈ **0.197** ✓ - At U = 90%: T_v = 1.781 − 0.933 × log₁₀(100 − 90) = 1.781 − 0.933 × log₁₀(10) = 1.781 − 0.933 × 1.0 = **0.848** ✓
Examples
Always identify H_dr first based on drainage condition. Look up (or compute) T_v for U = 90% = 0.848. Substitute into t = T_v H_dr²/c_v. Units: m²/yr gives t in years.
Scenario
Board-Type Problem 4 — Time for 90% consolidation: A 4 m clay layer drains from both top and bottom. c_v = 3 m²/yr. How long to reach 90% consolidation?
Solution
Double drainage → H_dr = 4/2 = 2 m At U = 90%: T_v = 0.848 t = T_v × H_dr² / c_v = 0.848 × (2)² / 3 = 0.848 × 4 / 3 = 3.392 / 3 = 1.13 years
Step 1: Compute T_v. Step 2: Check whether T_v corresponds to U above or below 60% by comparing with T_v at U=60% (= 0.287). Step 3: Since T_v = 0.375 > 0.287, use the logarithmic formula for U > 60%. Step 4: Solve algebraically for U.
Scenario
Board-Type Problem 5 — Degree of consolidation at a given time: Same clay (H = 4 m, double drainage, c_v = 3 m²/yr). Find U after t = 0.5 yr.
Solution
H_dr = 2 m T_v = c_v × t / H_dr² = 3 × 0.5 / 4 = 0.375 Since T_v = 0.375 > 0.287 (which corresponds to U = 60%), use the U > 60% formula: 0.375 = 1.781 − 0.933 × log₁₀(100 − U) log₁₀(100 − U) = (1.781 − 0.375) / 0.933 = 1.406 / 0.933 = 1.5070 100 − U = 10^1.5070 = 32.14 U = 100 − 32.14 = 67.9% ≈ **68%**
Once U is known, the settlement at time t is simply U × S_c (total primary settlement). This combines the magnitude and rate calculations.
Scenario
Board-Type Problem 6 — Settlement at a given time: Using the NC clay from Example 1 (S_c = 170 mm), how much settlement has occurred after 0.5 yr? Use the same c_v and drainage.
Solution
From Problem 5: U = 68% at t = 0.5 yr S_t = U × S_c = 0.68 × 170 = 115.6 mm ≈ **116 mm**
This problem emphasizes why drainage condition matters so much. The ratio of times equals the ratio of H_dr² values: (4/2)² = 4. Single drainage takes 4× longer than double drainage.
Scenario
Board-Type Problem 7 — Effect of changing drainage condition: If the 4 m clay in Problem 4 had only single drainage (resting on impermeable rock), how long for 90% consolidation?
Solution
Single drainage → H_dr = 4 m (full thickness) T_v = 0.848 (same U = 90%) t = 0.848 × (4)² / 3 = 0.848 × 16 / 3 = 13.568 / 3 = **4.52 years** Ratio: 4.52 / 1.13 = 4.0 (exactly 4× longer — because H_dr doubled from 2→4, and 2² = 4)
Applications
- Determining the construction schedule for embankments — when can the next lift be placed?
- Design of vertical drains (PVDs) to shorten consolidation time by reducing drainage path.
- Predicting when a building will reach its final settlement (important for connecting utilities and floor finishing).
- Monitoring field consolidation using settlement gauges and comparing with theoretical U–t curves.
- Stage construction planning for soft ground in Philippine expressway projects (NLEX, SLEX extensions).
Misconceptions
- MISCONCEPTION: Single and double drainage give the same result if the layer thickness is the same. TRUTH: Changing from single to double drainage (by adding a drainage layer at the bottom) cuts consolidation time to 1/4 of the original.
- MISCONCEPTION: U = T_v always. TRUTH: The T_v–U relationship is nonlinear. T_v = U only as a rough approximation at very low U values.
- MISCONCEPTION: Use the U ≤ 60% parabolic formula for any U. TRUTH: The parabolic formula T_v = (π/4)U² is only valid for U ≤ 60%. For U > 60%, the logarithmic expression must be used.
- MISCONCEPTION: c_v is dimensionless. TRUTH: c_v has units of area per time (m²/yr, cm²/s, etc.). Always check unit consistency.
Related Concepts
- Terzaghi's 1-D consolidation differential equation
- Coefficient of consolidation c_v (determined from oedometer test — log t or sqrt(t) methods)
- Degree of consolidation U and its relation to EPWP dissipation
- Isochrones — EPWP distribution with depth at various times
- Radial drainage (for vertical drains — extends to 2-D consolidation)
Common Exam Questions
Example
H = 6 m, double drainage, c_v = 1.5 m²/yr, U = 50%. → H_dr = 3 m, T_v = 0.197, t = 0.197×9/1.5 = 1.18 yr
Approach
1. Identify H_dr. 2. Find T_v for the given U (memorize 0.197 for 50%, 0.848 for 90%). 3. Solve t = T_v × H_dr²/c_v.
Question Type
Find time for a given U
Example
T_v = 0.20: Use U = √(4T_v/π) = √(4×0.20/π) = √(0.2546) = 0.505 → U = 50.5%
Approach
1. Compute T_v = c_v t/H_dr². 2. Check if T_v ≤ 0.287 (U ≤ 60%) or T_v > 0.287. 3. Apply correct formula.
Question Type
Find U at a given time
Example
S_c = 200 mm, T_v = 0.197 → U = 50% → S_t = 0.50 × 200 = 100 mm
Approach
1. Compute S_c (magnitude). 2. Compute T_v. 3. Find U. 4. S_t = U × S_c.
Question Type
Find settlement at time t (combined problem)
Key Points To Remember
- T_v = c_v × t / H_dr². Rearrange to find t = T_v × H_dr² / c_v.
- H_dr = H/2 for double drainage; H_dr = H for single drainage.
- Use T_v = (π/4)U² for U ≤ 60% (U as decimal). Use T_v = 1.781 − 0.933 log(100−U%) for U > 60%.
- Memorize T_v = 0.197 at U = 50% and T_v = 0.848 at U = 90% — these appear most often in board exams.
- U = S_t/S_c — if you know U and S_c, you can find settlement at any time t.
- c_v has units of [length²/time]. Keep units consistent throughout.
- Changing from double to single drainage quadruples the time for the same U (because H_dr doubles, H_dr² quadruples).
Overconsolidation Ratio (OCR) and Preconsolidation Pressure
The **preconsolidation pressure σ'_c** is the maximum past effective stress the clay has ever experienced. It is a memory of the soil's stress history. The **Overconsolidation Ratio (OCR)** quantifies this: $$OCR = \frac{\sigma'_c}{\sigma'_0}$$ where σ'_0 is the current effective overburden stress. **Physical causes of overconsolidation:** - Erosion of overlying soil (very common in Philippine hills and upland areas) - Past glaciation (not applicable in the Philippines, but common in textbooks) - Past desiccation / drying (surface clays exposed to sun during dry season) - Previous structures that were demolished - Groundwater lowering (increases effective stress temporarily) **Engineering significance:** - OC clay is STIFFER (smaller C_r vs. C_c) and stronger than NC clay at the same stress level. - For foundation design, knowing OCR tells you which compression index governs. - If a new structure's load keeps σ'_f < σ'_c (stays OC), settlement is much smaller. - If σ'_f > σ'_c (crosses into NC range), much larger settlement occurs — a critical design threshold. **Casagrande's Graphical Method (for determining σ'_c from oedometer test):** 1. On the e–log σ' curve, find the point of maximum curvature (Point A). 2. Draw a horizontal line and a tangent line at A. 3. Bisect the angle between the two lines. 4. Extend the virgin compression line (straight portion) upward. 5. The intersection of the bisector and the virgin line = σ'_c. (Board exams rarely test the graphical procedure directly, but OCR computations are common.)
Examples
The clay was once subjected to an effective stress of 225 kPa (perhaps from a thick layer of soil that was later eroded). Now at 75 kPa, it is much stiffer than NC clay.
Scenario
A clay sample from a depth where σ'_0 = 75 kPa shows σ'_c = 225 kPa from oedometer test. Compute OCR and classify.
Solution
OCR = σ'_c / σ'_0 = 225 / 75 = 3.0 Classification: Heavily overconsolidated clay (OCR > 2 is generally considered heavily OC).
Since the final stress never exceeds the preconsolidation pressure, the clay remains overconsolidated throughout loading. Only the recompression index C_r is needed.
Scenario
Board exam: An OC clay has σ'_c = 200 kPa, loaded from σ'_0 = 120 kPa to σ'_f = 180 kPa. Which index governs?
Solution
Check: σ'_f = 180 kPa < σ'_c = 200 kPa The entire stress increase (120→180 kPa) remains within the OC range. → Use C_r ONLY (Case 1).
Applications
- Evaluating whether preloading has been effective (measure new σ'_c after preload removal).
- Estimating past erosion depth from measured OCR in Philippine highlands.
- Foundation design decision: if load keeps clay in OC range, use lighter slab foundation; if it goes NC, deep foundation may be needed.
- Settlement prediction accuracy: using C_c when C_r is appropriate overestimates settlement by a factor of 5–10.
Misconceptions
- MISCONCEPTION: All Philippine clays are normally consolidated. TRUTH: Many Philippine upland and hillside clays are overconsolidated due to erosion and seasonal desiccation.
- MISCONCEPTION: OC and NC clays have the same stiffness. TRUTH: OC clays are significantly stiffer (lower compressibility) in the OC range — this is why preloading is used as a ground improvement technique.
Related Concepts
- Stress history of clay deposits
- Casagrande's graphical construction for σ'_c
- Preloading as a ground improvement method
- Undrained shear strength vs. OCR relationship
- Compression index C_c and recompression index C_r
Common Exam Questions
Example
σ'_0 = 80 kPa, OCR = 2.5, Δσ = 60 kPa → σ'_c = 200 kPa, σ'_f = 140 kPa < 200 kPa → use C_r.
Approach
Compute σ'_c from OCR and σ'_0. Compare σ'_f = σ'_0 + Δσ with σ'_c. Use C_r if σ'_f ≤ σ'_c; use two-part if σ'_f > σ'_c.
Question Type
Compute OCR and determine governing index
Key Points To Remember
- OCR = σ'_c / σ'_0. OCR = 1 → NC. OCR > 1 → OC. OCR < 1 is theoretically impossible (underconsolidated clays are a special case).
- For OC clay with final stress < σ'_c: use C_r only (small settlement).
- For OC clay with final stress > σ'_c: use C_r up to σ'_c, then C_c beyond (two-part).
- C_r is typically 1/5 to 1/10 of C_c — settlements in OC range are much smaller.
- σ'_c is determined from the oedometer test using Casagrande's graphical method.
- Preloading increases σ'_c artificially, converting future loads to the OC range and drastically reducing future settlement.
Practice Problems
The log₁₀(1.8333) is computed using a calculator: log₁₀(1.8333) = log₁₀(1.8333). Using log₁₀(1.8) = 0.2553 and log₁₀(1.85) ≈ 0.2672 → interpolate: 0.2553 + (0.33/0.05)×(0.2672−0.2553) ≈ 0.2631. Final answer: 208 mm. In board exam, carry 4 significant figures to minimize rounding error.
Problem
Exercise 1 (NC Clay Settlement): A 6 m normally consolidated clay layer has C_c = 0.25, e_0 = 0.90, σ'_0 = 120 kPa. A structural load causes a stress increase Δσ = 100 kPa at mid-layer. Compute the primary consolidation settlement S_c in mm.
Solution
Given: H = 6 m = 6000 mm, C_c = 0.25, e_0 = 0.90, σ'_0 = 120 kPa, Δσ = 100 kPa NC clay → use S_c = [C_c / (1 + e_0)] × H × log₁₀(σ'_f / σ'_0) σ'_f = 120 + 100 = 220 kPa S_c = [0.25 / (1 + 0.90)] × 6000 × log₁₀(220/120) = [0.25 / 1.90] × 6000 × log₁₀(1.8333) = 0.13158 × 6000 × 0.26316 = 207.9 mm ≈ **208 mm**
Use T_v = (π/4)U² because U = 50% ≤ 60%. Single drainage → H_dr = 5 m (full layer thickness). If it were double drainage (H_dr = 2.5 m), the time would be: 0.1963 × 6.25/2 = 0.613 yr — about 4× less.
Problem
Exercise 2 (Time for 50% consolidation — single drainage): A 5 m clay layer drains from the top surface only (impermeable base). c_v = 2 m²/yr. Find the time (in years) for 50% consolidation.
Solution
Given: H = 5 m, single drainage → H_dr = H = 5 m, c_v = 2 m²/yr, U = 50% At U = 50%: T_v = (π/4)(0.50)² = (π/4)(0.25) = 0.1963 t = T_v × H_dr² / c_v = 0.1963 × (5)² / 2 = 0.1963 × 25 / 2 = 4.908 / 2 = **2.45 years**
Key steps: (1) Always compute T_v first. (2) Check which formula branch by comparing T_v with 0.2827 (U=60% threshold). (3) For T_v > 0.2827, solve 1.781 − 0.933 log(100−U) = T_v algebraically. (4) Multiply U × S_c for settlement at that time.
Problem
Exercise 3 (Degree of consolidation after 2 years): A 8 m NC clay layer drains from both top and bottom. c_v = 4 m²/yr. Find U after t = 2 years. Then find settlement at t = 2 yr if S_c = 300 mm.
Solution
Given: H = 8 m, double drainage → H_dr = 4 m, c_v = 4 m²/yr, t = 2 yr Step 1: Compute T_v T_v = c_v × t / H_dr² = 4 × 2 / 4² = 8 / 16 = 0.50 Step 2: Determine formula branch Compare with T_v at U = 60%: T_v(60%) = (π/4)(0.60)² = (π/4)(0.36) = 0.2827 Since T_v = 0.50 > 0.2827, use U > 60% formula: Step 3: Solve for U 0.50 = 1.781 − 0.933 × log₁₀(100 − U) log₁₀(100 − U) = (1.781 − 0.50) / 0.933 = 1.281 / 0.933 = 1.3731 100 − U = 10^1.3731 = 23.61 U = 100 − 23.61 = **76.4%** Step 4: Settlement at t = 2 yr S_t = U × S_c = 0.764 × 300 = **229.2 mm**
Since σ'_f = 170 kPa does not exceed σ'_c = 200 kPa, the clay never enters the NC (virgin compression) range. Only C_r is used. Compare this with what S_c would be if we wrongly used C_c: S_c = [0.32/1.72] × 3500 × 0.327 = 212 mm — more than 5× overestimation. Correct identification of NC vs. OC is crucial.
Problem
Exercise 4 (OC Clay, Case 1): An overconsolidated clay layer is 3.5 m thick with e_0 = 0.72, C_r = 0.06, C_c = 0.32, σ'_0 = 80 kPa, σ'_c = 200 kPa. A new building applies Δσ = 90 kPa. Compute S_c.
Solution
Step 1: Check condition σ'_f = σ'_0 + Δσ = 80 + 90 = 170 kPa Compare with σ'_c = 200 kPa: 170 kPa < 200 kPa → **Case 1: entire loading in OC range** Step 2: Apply formula with C_r only S_c = [C_r / (1 + e_0)] × H × log₁₀(σ'_f / σ'_0) = [0.06 / (1 + 0.72)] × 3500 × log₁₀(170/80) = [0.06 / 1.72] × 3500 × log₁₀(2.125) = 0.034884 × 3500 × 0.32735 = **39.9 mm ≈ 40 mm**
This is the highest-level combined problem type. (a) Two-part OC clay: split at σ'_c = 100 kPa. (b) Use T_v = 0.848 (memorized) and t = T_v H_dr²/c_v. (c) Multiply. Always double check: σ'_f vs σ'_c comparison in (a) is the most error-prone step.
Problem
Exercise 5 (Combined — OC Clay Case 2 + Time Rate): A 4 m clay has C_c = 0.42, C_r = 0.07, e_0 = 0.85, σ'_0 = 60 kPa, σ'_c = 100 kPa, Δσ = 80 kPa. Double drainage. c_v = 2.5 m²/yr. (a) Compute total primary settlement S_c. (b) How long for U = 90%? (c) What is settlement at that time?
Solution
Part (a) — Magnitude: σ'_f = 60 + 80 = 140 kPa > σ'_c = 100 kPa → **Case 2 (two-part)** 1 + e_0 = 1 + 0.85 = 1.85 H = 4000 mm Part 1 (OC range, 60→100 kPa, use C_r): S₁ = [0.07/1.85] × 4000 × log₁₀(100/60) = 0.037838 × 4000 × log₁₀(1.6667) = 0.037838 × 4000 × 0.22185 = 33.6 mm Part 2 (NC range, 100→140 kPa, use C_c): S₂ = [0.42/1.85] × 4000 × log₁₀(140/100) = 0.22703 × 4000 × log₁₀(1.40) = 0.22703 × 4000 × 0.14613 = 132.7 mm S_c = S₁ + S₂ = 33.6 + 132.7 = **166.3 mm** Part (b) — Time for U = 90%: Double drainage → H_dr = 4/2 = 2 m T_v at U = 90% = 0.848 t = T_v × H_dr²/c_v = 0.848 × (2)²/2.5 = 0.848 × 4/2.5 = 3.392/2.5 = **1.357 years ≈ 1.36 yr** Part (c) — Settlement at U = 90%: S_t = U × S_c = 0.90 × 166.3 = **149.7 mm ≈ 150 mm**
Exam Preparation Tips
- MEMORIZE these two critical T_v–U pairs: T_v = 0.197 at U = 50% and T_v = 0.848 at U = 90%. These appear in approximately 70% of board exam problems on time rate.
- ALWAYS state H_dr before computing T_v. The most common error is forgetting that double drainage gives H_dr = H/2, not H.
- ALWAYS check σ'_f vs. σ'_c FIRST in any settlement problem. This determines whether you use C_r only, C_c only, or both in two parts. Writing this check explicitly in your solution shows systematic thinking.
- Use log BASE 10 exclusively in the settlement and U>60% formulas. Write 'log₁₀' or 'log' clearly, and if using a calculator, make sure it is set to common log, not natural log.
- In the U ≤ 60% formula T_v = (π/4)U², U must be a decimal (e.g., 0.50 for 50%). The U > 60% formula uses percentage (e.g., 90 for 90%). Mixing these up gives wrong answers.
- Practice the 3-step combined problem: (1) find S_c magnitude, (2) compute T_v and get U, (3) multiply U × S_c for S_t. Board exam frequently asks for 'settlement after X years.'
- Remember the drainage path–time relationship: if H_dr doubles (single vs. double drainage), consolidation time quadruples (because t ∝ H_dr²). Conversely, halving H_dr via PVDs reduces time to 1/4.
- For OC clay problems, if only C_c and OCR are given (not C_r), use C_r ≈ C_c/5 as an approximation unless told otherwise.
- Units discipline: if c_v is in m²/yr, H_dr in m, and H in m (converted to mm for S_c), then t comes out in years. Don't mix cm²/s with m²/yr without converting.
- The T_v formula T_v = 1.781 − 0.933 log(100−U) at U = 99%: T_v = 1.781 − 0.933(0) = 1.781. This is consistent because log₁₀(1) = 0. Note this at U → 100%, T_v → ∞ (complete consolidation takes infinite time).
- Philippine board exam context: Soft clay problems involving reclamation areas (Manila Bay) or coastal plains are common. Always identify if the clay is below groundwater table (saturated) before applying consolidation equations.
- Secondary consolidation (creep) uses S_s = [C_α/(1+e_p)] × H × log(t₂/t₁). While rarely the main question, it may appear as a sub-part. C_α is usually given as a fraction of C_c (typically 0.04C_c to 0.06C_c for inorganic clays).
In summary
Consolidation and settlement is a cornerstone topic in Geotechnical Engineering for the PRC Civil Engineer Licensure Examination. The subject tests two distinct but interconnected skills: computing **how much** a clay layer will settle (using C_c and C_r in the settlement equation) and computing **how fast** settlement occurs (using Terzaghi's T_v–U relationship with c_v and H_dr). **Core equations to commit to memory:** 1. S_c = [C_c/(1+e_0)] × H × log₁₀(σ'_f/σ'_0) — for NC clay 2. Two-part equation for OC clay when σ'_f > σ'_c (use C_r then C_c) 3. T_v = c_v × t / H_dr² — linking time to consolidation 4. T_v = (π/4)U² for U ≤ 60% and T_v = 1.781 − 0.933 log(100−U%) for U > 60% 5. **T_v = 0.197 at U = 50%; T_v = 0.848 at U = 90%** — board exam favorites **Critical decision points in any board exam problem:** - Is the clay NC or OC? (Check OCR or compare σ'_f with σ'_c) - Is drainage single or double? (Determines H_dr) - Which T_v–U formula branch? (Check if U ≤ 60% or U > 60%) - What units are consistent throughout? Philippine practice relevance under RA 544: Civil engineers must ensure that predicted settlements do not exceed the allowable limits in NSCP 2015 Section 305 (total settlement ≤ 25 mm for isolated footings on clay; differential ≤ 19 mm). Projects in soft ground areas — reclaimed land in Manila Bay, coastal Visayas, Mindanao river deltas — demand accurate consolidation analysis. Mastery of these concepts is not only a board exam requirement but a professional responsibility. Approach board exam problems systematically: identify parameters, classify the clay (NC/OC), determine drainage, apply the correct equation, and check units. With disciplined practice of the five solved examples and exercises in this chapter, you will confidently handle any consolidation problem on examination day.
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