CELE Geotechnical Engineering — Consolidation and SettlementStudy Notes
Thorough study notes for Consolidation and Settlement — the fastest path from zero to ready for CELE Geotechnical Engineering. Structured for self-study reviewers who cannot attend a review centre, these notes cover the full concept library plus the CELE-specific twists Professional Regulation Commission (PRC) — Board of Civil Engineering adds to its questions.
Exam context
On the CELE 2026, the Geotechnical Engineering subtest carries a "Core" weight in Professional Regulation Commission (PRC) — Board of Civil Engineering's pattern. Consolidation and Settlement lands at position 6th out of 11 in the standard review order. Target score is 70% weighted average, no sub-test below 50%, and roughly a meaningful share of items come from Geotechnical Engineering on a typical CELE paper.
Consolidation and Settlement - Study Notes
Consolidation and settlement are fundamental concepts in geotechnical engineering, particularly relevant to foundation design, embankment construction, and ground improvement in the Philippines where soft clay layers are common. When saturated fine-grained soils (clay and silt) are subjected to additional stress—such as from building loads, embankments, or fill—water gradually expels from the soil pores while the soil skeleton compresses. This process, called primary consolidation, occurs over extended periods and can result in significant differential settlement that damages structures if not properly managed. This chapter addresses the quantification of settlement magnitude using compression indices and the prediction of settlement timing using Terzaghi's consolidation theory. Understanding these principles is essential for PRC Licensure Examination success and for safe foundation design under Philippine Building Code (PBC) and NSCP 2015 standards.
Summary
Consolidation and settlement are the dominant time-dependent deformations in saturated fine-grained soils, particularly critical in Philippine foundation design where soft alluvial clays underlie major population centers. This chapter covered: **1. Fundamentals:** Consolidation is the expulsion of pore water and rearrangement of soil skeleton under applied load. Primary consolidation is the main component in saturated clays; secondary settlement continues due to creep and decomposition. Normally consolidated (NC) clays use compression index Cc; overconsolidated (OC) clays use recompression index Cr below preconsolidation stress σ'c and Cc above it. **2. Settlement Magnitude:** Primary consolidation settlement is calculated using Sc = [Cc/(1+e0)] × H × log₁₀(σ'f/σ'0) for NC clay. The settlement increases with thickness H, compression index Cc, and stress increase Δσ, but decreases with initial void ratio e0. For OC clay, use Cr initially, then Cc after exceeding σ'c. **3. Settlement Timing:** Governed by the time factor Tv = cv × t / Hdr², where cv is the coefficient of consolidation and Hdr is the longest drainage path (H/2 for double drainage, H for single). The degree of consolidation U reaches 50% at Tv = 0.197 and 90% at Tv = 0.848. For U ≤ 60%, use Tv = (π/4)U²; for U > 60%, use Tv = 1.781 − 0.933 log₁₀(100 − U%). Soft clay (cv = 1–3 m²/yr) consolidates over years; stiff clay (cv > 30 m²/yr) over months. **4. Secondary Settlement:** Continues after primary consolidation at a slower, approximately constant logarithmic rate governed by secondary compression index Cα. For inorganic clays, Cα ≈ 0.05–0.10 × Cc. For organic clays and peats (common in Philippines), Cα can equal or exceed Cc, necessitating extended laboratory tests and long-term monitoring. **5. NSCP 2015 and PBC Compliance:** Geotechnical investigations are mandatory for structures on soft soils; typical total settlement limits are 50–100 mm, differential settlement 25–50 mm. Remedial measures include vertical drains (reduce Hdr by 5–20×), preloading (induce consolidation before construction), pile foundations (bypass compressible layers), and ground improvement. Monitoring with settlement plates and piezometers is required during and after construction. For Philippine practice on Manila Bay and Laguna clay, combinations of piles, vertical drains, and preloading are standard for high-rise buildings. **6. Laboratory Testing (Oedometer):** The consolidation test applies incremental vertical stress to a confined soil specimen, measuring settlement to derive Cc, Cr, cv, and σ'c. Specimen saturation, adequate drainage time (24 hours per increment minimum), and undisturbed sampling are critical for reliable results. Secondary settlement (extended test > 24 hours) is essential for organic clays. e–log σ' plots identify the virgin and recompression lines; Casagrande construction determines σ'c; √t method from time-settlement curves yields cv. **Key Formulas to Memorize for PRC Exam:** - **NC Settlement:** Sc = [Cc/(1+e0)] × H × log₁₀(σ'f/σ'0) - **Time Factor:** Tv = cv × t / Hdr² - **Degree of Consolidation (U ≤ 60%):** Tv = (π/4)U² - **Degree of Consolidation (U > 60%):** Tv = 1.781 − 0.933 log₁₀(100 − U%) - **Key Tv–U Pairs:** U = 50%, Tv = 0.197; U = 90%, Tv = 0.848 - **Secondary Settlement:** Ssec = Cα × H × log₁₀(t₂/t₁) **Common Exam Question Types:** 1. Calculate settlement magnitude given Cc, e0, H, Δσ. 2. Determine time to reach a given degree of consolidation; identify whether drainage is single or double. 3. Compare settlement rates for different cv or layer thicknesses; understand 4× effect of halving Hdr (double drainage). 4. Identify whether a clay is NC or OC from σ'c and current stress; select appropriate index (Cc or Cr). 5. Estimate Cc from plasticity index when lab data unavailable; apply typical ranges for Philippine soils. 6. Design remedial measures (piles, vertical drains, preloading) per NSCP 2015 for excessive settlement. 7. Interpret oedometer test results: extract indices from e–log σ' plot, determine σ'c, assess cv from √t method. **Critical PRC Exam Details:** - Log base in settlement equation is 10 (common log), not natural log. - Double drainage halves drainage path; consolidation time ∝ Hdr² (quadratic effect). - Cc applies to virgin compression; Cr to recompression below σ'c. - Settlement depends on Cc (compression index); timing depends on cv (consolidation coefficient). - Philippine soft clays: Cc = 0.8–1.2, cv = 1–3 m²/yr, settlement over years; require careful foundation design. - NSCP 2015 requires site investigation, monitoring, and remediation strategies for soft soil sites. Mastering this chapter is essential for foundation design, slope stability (settlement on slope), and ground improvement topics in the PRC Licensure Examination.
Sections
Consolidation is the process by which saturated soil decreases in volume due to expulsion of pore water under applied stress. In the Philippine context, this is especially critical in low-lying areas with thick clay strata (e.g., the alluvial plains of Luzon and Visayas). When a load is applied to a saturated clay layer, the effective stress increases, causing pore pressure to rise initially. Over time, water dissipates through drainage paths (either upward through overlying layers, downward through underlying layers, or both), and the effective stress gradually transfers to the soil skeleton. The soil grains rearrange and compress, reducing void ratio and layer thickness. The settlement process consists of two main components: **Immediate (Elastic) Settlement:** Occurs instantaneously upon loading, caused by elastic distortion of soil grains and water. In saturated clays, this is negligible (typically < 5% of total settlement) because water is incompressible. This component depends on Poisson's ratio and elastic modulus (Young's modulus). **Primary Consolidation Settlement:** The main component in saturated clays, governed by the rate of pore water dissipation. This is time-dependent and can continue for years or decades, depending on soil permeability and drainage path length. The settlement rate is controlled by the coefficient of consolidation (cv). **Secondary Settlement (Creep):** Occurs after primary consolidation is complete, due to viscous behavior of the soil skeleton and organic matter decomposition. This is typically 5–10% of primary settlement in normal clays but can be significant in highly plastic and organic clays. For the PRC exam, secondary settlement is usually ignored unless explicitly stated. The void ratio–effective stress relationship is fundamental. When a normally consolidated (NC) clay is loaded, the e–log σ' curve plots as a straight line called the **virgin compression line (VCL)**. The slope of this line is the **compression index (Cc)**. When soil is unloaded and reloaded, the curve becomes flatter, with slope **recompression index (Cr)**, typically 0.1 to 0.3 times Cc. The stress at which the soil transitions from recompression to virgin compression is the **preconsolidation stress (σ'c)**, which defines whether clay is normally consolidated (σ'0 ≈ σ'c) or overconsolidated (σ'0 < σ'c).
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1. FUNDAMENTALS OF CONSOLIDATION
Examples
Example 1A — Identifying NC vs OC Clay
A clay sample from Metro Manila has σ'0 = 85 kPa (current overburden at 10 m depth) and laboratory testing shows σ'c = 120 kPa (determined from e–log σ' curve inflection). Is this clay normally consolidated or overconsolidated? What indices apply?
Solution
Since σ'0 (85 kPa) < σ'c (120 kPa), the clay is overconsolidated (OC). This is typical in areas with past glaciation or desiccation. When loaded from 85 kPa toward 120 kPa, the recompression index Cr applies. Once loading exceeds 120 kPa, the compression index Cc governs. The overconsolidation ratio (OCR) = σ'c / σ'0 = 120/85 = 1.41.
Example 1B — Field Scenario: Soft Bangkok Clay in Southeast Asia
Soft Bangkok clay, similar to clays found in Philippines' deltaic regions, exhibits Cc ≈ 0.8–1.0 (very compressible), Cr ≈ 0.1, and cv ≈ 1–2 m²/year. A 12 m thick layer is loaded with Δσ = 150 kPa. Settlement can be estimated but will take several years to complete.
Solution
This illustrates why the Philippines requires careful design for structures on soft alluvial clays. High Cc means large settlements; low cv means long consolidation periods. Both factors necessitate pile foundations or ground improvement (e.g., preloading, vertical drains per NSCP 2015).
Key Points
- Consolidation is water expulsion from saturated soil pores, causing compression of the soil skeleton
- Primary consolidation settlement dominates in saturated clays; immediate and secondary components are usually smaller
- Normally consolidated (NC) clay: current effective stress equals historical maximum; uses compression index Cc
- Overconsolidated (OC) clay: current stress is less than historical maximum; uses smaller recompression index Cr until preconsolidation stress is exceeded
- Drainage paths and soil permeability control the rate (time) of settlement; compression indices control the magnitude
- Settlement can take months to years in thick clay deposits; differential settlement is a major cause of structural distress
The settlement equation for primary consolidation differs for normally consolidated and overconsolidated clays. Both use the semi-logarithmic relationship between void ratio and effective stress. **For Normally Consolidated Clay:** The standard formula is: $$S_c = \frac{C_c}{1 + e_0} \cdot H \cdot \log_{10}\left(\frac{\sigma'_f}{\sigma'_0}\right)$$ Where: - Sc = primary consolidation settlement (mm or m) - Cc = compression index (dimensionless), slope of virgin compression line on e–log σ' plot - e0 = initial void ratio before loading - H = thickness of clay layer (same units as settlement) - σ'0 = initial effective stress before loading - σ'f = final effective stress after loading = σ'0 + Δσ - Δσ = stress increase from superimposed load The numerator Cc / (1 + e0) represents the change in void ratio per 10-fold stress increase. The log term captures the fact that soil compressibility decreases with increasing stress (nonlinear relationship), which is why a logarithmic scale is essential. **For Overconsolidated Clay:** When the final stress σ'f does not exceed preconsolidation stress σ'c, use Cr: $$S_c = \frac{C_r}{1 + e_0} \cdot H \cdot \log_{10}\left(\frac{\sigma'_f}{\sigma'_0}\right)$$ When σ'f exceeds σ'c, split the settlement into two parts: $$S_c = \frac{C_r}{1 + e_0} \cdot H \cdot \log_{10}\left(\frac{\sigma'_c}{\sigma'_0}\right) + \frac{C_c}{1 + e_0} \cdot H \cdot \log_{10}\left(\frac{\sigma'_f}{\sigma'_c}\right)$$ The first term is recompression (smaller, using Cr) up to the preconsolidation stress. The second term is virgin compression (larger, using Cc) beyond it. **Practical Notes:** 1. **Compression Index Estimation:** Cc can be estimated from plasticity index (PI) or liquidity index (LI) if laboratory data is unavailable: - Cc ≈ 0.007 × PI (for most clays) - Cc ≈ 0.009 × wL − 0.06 (alternative; wL = liquid limit in %) - For organic clays or peats, Cc can exceed 1.5, requiring site-specific testing. 2. **Recompression Index:** Cr ≈ 0.1 to 0.3 × Cc; often taken as 0.15 × Cc if not measured. 3. **Void Ratio:** e0 is often back-calculated from unit weight: e0 = (Gs − γd/γw) / (γd/γw), where Gs is specific gravity of solids (~2.67 for most clays) and γd is dry unit weight. 4. **Base of e-log σ' Curve:** When the initial stress σ'0 is very small (e.g., near surface), use the stress at 1 kPa or the lowest measured stress as reference to avoid negative log values. 5. **Units:** Always use consistent units. If H is in meters, Sc will be in meters; if H is in millimeters, Sc will be in millimeters. For PRC exams, mixing units is a common error trap. **Why Logarithmic Compression?** Soil compressibility is not constant; it decreases with increasing stress. A tenfold increase in stress near the surface causes more settlement than the same tenfold increase at depth. The logarithmic form captures this: settlement per decade (10× increase) of stress is constant, but absolute settlement decreases as stress increases.
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2. PRIMARY CONSOLIDATION SETTLEMENT CALCULATION
Examples
Example 2A — Normal Consolidation Settlement (Board-Style Problem)
A 4 m thick layer of normally consolidated clay underlies a proposed warehouse foundation. Soil properties: Cc = 0.30, e0 = 0.80, γsat = 19 kN/m³, wL = 45%. The water table is at the clay surface. At 10 m depth (base of clay), initial effective stress σ'0 = 100 kPa. The building load will increase stress by Δσ = 80 kPa. Calculate the primary consolidation settlement.
Solution
Given: H = 4 m = 4000 mm, Cc = 0.30, e0 = 0.80, σ'0 = 100 kPa, Δσ = 80 kPa. Final effective stress: σ'f = σ'0 + Δσ = 100 + 80 = 180 kPa. Using the NC settlement equation: Sc = [Cc/(1+e0)] × H × log₁₀(σ'f/σ'0) Sc = [0.30/(1.80)] × 4000 × log₁₀(180/100) Sc = [0.1667] × 4000 × log₁₀(1.8) Sc = [0.1667] × 4000 × 0.2553 Sc = 170.2 mm ≈ 17 cm **Answer:** Primary consolidation settlement ≈ 170 mm or 0.17 m. This is significant and must be verified against building tolerance limits. If the foundation is 20 m × 20 m, uniform settlement of 170 mm is manageable if differential settlement is controlled, but 250+ mm would warrant ground improvement.
Example 2B — Overconsolidated Clay with Preconsolidation Stress
A clay deposit in the Manila Bay area shows σ'c = 200 kPa (from oedometer testing), current depth stress σ'0 = 120 kPa (OCR = 1.67), and Cc = 0.25, Cr = 0.04, e0 = 0.90. A 6 m thick layer is loaded with Δσ = 100 kPa. Find settlement.
Solution
Since σ'0 = 120 kPa and σ'f = 120 + 100 = 220 kPa, the final stress exceeds σ'c = 200 kPa. Settlement splits: **Recompression phase (120 to 200 kPa):** Sc1 = [Cr/(1+e0)] × H × log₁₀(σ'c/σ'0) Sc1 = [0.04/1.90] × 6000 × log₁₀(200/120) Sc1 = [0.02105] × 6000 × log₁₀(1.667) Sc1 = [0.02105] × 6000 × 0.2218 Sc1 = 28.0 mm **Virgin compression phase (200 to 220 kPa):** Sc2 = [Cc/(1+e0)] × H × log₁₀(σ'f/σ'c) Sc2 = [0.25/1.90] × 6000 × log₁₀(220/200) Sc2 = [0.1316] × 6000 × log₁₀(1.10) Sc2 = [0.1316] × 6000 × 0.0414 Sc2 = 32.6 mm **Total settlement:** Sc = Sc1 + Sc2 = 28.0 + 32.6 = 60.6 mm ≈ 6 cm **Answer:** ≈ 61 mm. Note that this is much smaller than a normally consolidated clay would settle under the same stress increase (e.g., Example 2A) because Cr << Cc. The overconsolidation history provides beneficial stiffness.
Example 2C — Estimating Cc from Plasticity Index
A clay from a Cebu construction site has PI = 20% (measured from Atterberg limits). No laboratory compression data are available. Estimate the compression index.
Solution
Using the empirical correlation for most inorganic clays: Cc ≈ 0.007 × PI = 0.007 × 20 = 0.14 Alternatively, if liquid limit wL = 55%: Cc ≈ 0.009 × wL − 0.06 = 0.009 × 55 − 0.06 = 0.495 − 0.06 = 0.435 **Answer:** Cc is estimated between 0.14 and 0.43, depending on the formula. The first is lower (conservative for settlement estimation); the second is higher. For design, a middle value like 0.25–0.30 is often selected, or site-specific oedometer testing is recommended. This uncertainty is why the NSCP 2015 and PBC require geotechnical investigations for sensitive structures.
Key Points
- Settlement equation: Sc = [Cc/(1+e0)] × H × log(σ'f/σ'0) for normally consolidated clay
- For overconsolidated clay, use Cr below σ'c and Cc above σ'c, splitting the settlement calculation at the preconsolidation stress
- Compression index Cc can be estimated from PI if laboratory data unavailable: Cc ≈ 0.007×PI
- Recompression index Cr is typically 0.15×Cc or measured from laboratory unload-reload curve
- Log base must be 10 (common logarithm), not natural log; this is a critical PRC exam detail
- Settlement increases with clay thickness H, compression index Cc, and stress increase Δσ
- Settlement decreases with higher initial void ratio e0 because the (1+e0) denominator increases
- All stresses (σ'0, σ'f, σ'c) must be effective stresses, accounting for pore water pressure
While the compression indices determine **how much** a soil will settle, the coefficient of consolidation (cv) and drainage conditions determine **how fast** it settles. This is critical for temporary settlement monitoring, staged construction, and design of drainage improvements. **Terzaghi's One-Dimensional Consolidation Equation:** Under the assumptions of one-dimensional vertical flow and linear elastic behavior, consolidation follows: $$\frac{\partial u}{\partial t} = c_v \frac{\partial^2 u}{\partial z^2}$$ where u is excess pore pressure and z is depth. The solution introduces the dimensionless **time factor**: $$T_v = \frac{c_v \cdot t}{H_{dr}^2}$$ Where: - Tv = time factor (dimensionless) - cv = coefficient of consolidation (m²/year, m²/day, or m²/second, depending on t units) - t = elapsed time (years, days, seconds, consistent with cv units) - Hdr = drainage path length = longest distance water must travel to exit the layer **Drainage Path Length (Hdr):** This is a frequent exam pitfall. The drainage path depends on boundary conditions: 1. **Double Drainage** (water exits from top AND bottom): Hdr = H/2, where H is layer thickness. This occurs when the clay is sandwiched between permeable layers (sand) or when there is a free surface above. Common in natural deposits. 2. **Single Drainage** (water exits from top OR bottom only): Hdr = H. This occurs when the base is bounded by an impermeable (e.g., bedrock) or low-permeability layer, or when the top is sealed (e.g., by a concrete slab or road pavement). 3. **Partial Drainage:** In some cases, water exits from one face fully and the other partially, requiring weighted averaging. **Degree of Consolidation (U):** The degree of consolidation at time t is the ratio of settlement at time t to total primary settlement: $$U = \frac{S_c(t)}{S_c,total}$$ Typically expressed as a percentage. U = 0% at t = 0 (no settlement), U = 100% when primary consolidation is complete (t → ∞). **Tv–U Relationships:** The relationship between time factor and degree of consolidation is obtained by solving Terzaghi's equation. Key values (for single-way drainage with constant-rate loading): | Degree of Consolidation U | Time Factor Tv | Useful Approximation | |---|---|---| | 10% | 0.008 | — | | 20% | 0.031 | — | | 30% | 0.071 | — | | 40% | 0.126 | — | | 50% | 0.197 | Tv ≈ (π/4)U² | | 60% | 0.286 | Boundary: switch formulas | | 70% | 0.403 | Tv ≈ 1.781 − 0.933 log₁₀(100 − U) | | 80% | 0.567 | — | | 90% | 0.848 | — | | 95% | 1.129 | — | | 99% | 1.665 | — | **Two Practical Formulas:** **For U ≤ 60%:** $$T_v = \frac{\pi}{4} U^2 \quad (\text{or } U = \sqrt{\frac{4 T_v}{\pi}} = 1.128\sqrt{T_v})$$ This parabolic approximation is simpler and sufficiently accurate for lower degrees of consolidation, common in early-stage settlement prediction. **For U > 60%:** $$T_v = 1.781 - 0.933 \log_{10}(100 - U\%)$$ Or rearranged to solve for U: $$\log_{10}(100 - U\%) = \frac{1.781 - T_v}{0.933}$$ $$100 - U\% = 10^{(1.781 - T_v)/0.933}$$ $$U\% = 100 - 10^{(1.781 - T_v)/0.933}$$ This is essential for late-stage consolidation (reaching 90%+ completion takes much longer than the parabolic formula predicts). **Why the Nonlinear Relationship?** Early in consolidation, excess pore pressure gradients are steep, driving rapid water flow and fast settlement. As consolidation progresses and excess pore pressure dissipates, the gradient flattens, flow slows, and settlement rate decreases. Settlement follows a square-root-of-time relationship initially but transitions to a logarithmic rate at later stages. **Coefficient of Consolidation (cv):** This is an intrinsic soil property, measured in the laboratory oedometer test. Typical values: - Soft, organic clays (Manila Bay deposits): cv = 1–3 m²/year - Medium clays: cv = 5–15 m²/year - Stiff clays: cv = 20–50 m²/year - Silts: cv = 50–500 m²/year cv is related to permeability (k) and compressibility (mv) by: $$c_v = \frac{k}{\gamma_w \cdot m_v}$$ Where mv is the coefficient of volume compressibility (change in volume per unit change in effective stress), with units of m²/kN or kPa⁻¹. **Practical Calculation Steps:** 1. Calculate primary settlement Sc using compression indices (Section 2). 2. Determine drainage path length Hdr based on layer geometry and boundary conditions. 3. Calculate time factor Tv = cv × t / Hdr². 4. Use the Tv–U table or formulas to find degree of consolidation U. 5. Calculate settlement at time t: Sc(t) = U × Sc,total. 6. Repeat for different times to draw a settlement vs. time curve.
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3. TIME RATE OF CONSOLIDATION — TERZAGHI'S THEORY
Examples
Example 3A — Time to Reach 50% Consolidation (Double Drainage)
A 4 m thick clay layer from Example 2A (Cc = 0.30, e0 = 0.80) has measured cv = 3 m²/year. The clay is sandwiched between two sand layers (free drainage at top and bottom). Find the time to reach 50% consolidation.
Solution
For double drainage: Hdr = H/2 = 4/2 = 2 m. At U = 50%, from the table: Tv = 0.197. Time factor equation: Tv = (cv × t) / Hdr² 0.197 = (3 × t) / (2)² 0.197 = (3 × t) / 4 t = (0.197 × 4) / 3 = 0.788 / 3 = 0.263 years Convert to days: t = 0.263 × 365 = 96 days ≈ 3 months. **Answer:** Approximately 96 days or 3 months to reach 50% consolidation. By definition, this is when Sc(t) = 0.5 × 170.2 mm = 85.1 mm has occurred.
Example 3B — Time to Reach 90% Consolidation (Single Drainage)
Now assume the clay in Example 3A has single drainage (impermeable bedrock below). How long to reach 90% consolidation?
Solution
For single drainage: Hdr = H = 4 m. At U = 90%, from the table: Tv = 0.848. Time factor: 0.848 = (3 × t) / (4)² 0.848 = (3 × t) / 16 t = (0.848 × 16) / 3 = 13.568 / 3 = 4.52 years Convert to days: t = 4.52 × 365 ≈ 1650 days ≈ 4.5 years. **Comparison:** Double drainage (3 months to 50%) vs. single drainage (4.5 years to 90%) shows the dramatic effect of drainage path. In practice, the difference between double and single drainage is 2² = 4 times in time, evident here: 96 days × 4 ≈ 384 days ≈ 1 year to reach 50% with single drainage, which is the approximate halfway point of 4.5 years to 90%. **Answer:** ≈ 4.5 years to reach 90% consolidation with single drainage.
Example 3C — Degree of Consolidation at a Specific Time
For Example 3A (double drainage), calculate the degree of consolidation and settlement after 0.5 years.
Solution
Given: cv = 3 m²/year, Hdr = 2 m, t = 0.5 years. Calculate Tv: Tv = (cv × t) / Hdr² = (3 × 0.5) / (2)² = 1.5 / 4 = 0.375. Since Tv = 0.375 > 0.286 (which corresponds to U = 60%), use the second formula: Tv = 1.781 − 0.933 × log₁₀(100 − 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 = 1.507 100 − U% = 10^1.507 = 32.1 U% = 100 − 32.1 = 67.9% ≈ 68% Settlement at 0.5 years: Sc(0.5 yr) = U × Sc,total = 0.68 × 170.2 = 115.7 mm ≈ 116 mm. **Answer:** After 0.5 years, the clay has undergone 68% consolidation with settlement of ~116 mm (out of total 170 mm).
Example 3D — Comparing cv Values: Soft vs. Stiff Clay
Two clays, each 6 m thick, with double drainage. Clay A (soft, cv = 1 m²/year) is loaded at a site A. Clay B (stiff, cv = 30 m²/year) at site B. Both are designed to reach 90% consolidation before construction of upper stories (Hdr = 3 m). How long must each wait?
Solution
At U = 90%, Tv = 0.848. **Clay A (soft):** t = (Tv × Hdr²) / cv = (0.848 × 3²) / 1 = (0.848 × 9) / 1 = 7.63 years **Clay B (stiff):** t = (Tv × Hdr²) / cv = (0.848 × 3²) / 30 = 7.63 / 30 = 0.254 years ≈ 93 days **Answer:** Site A must wait 7.6 years; Site B only 3 months. This illustrates why soft clay deposits (common in Manila Bay, Laguna de Bay, and deltaic regions) are problematic for development. Vertical drains (prefabricated vertical drains, or PVDs, per NSCP 2015) or surcharge preloading are standard remedies.
Key Points
- Time factor Tv = cv × t / Hdr² governs the rate of consolidation; larger Hdr or smaller cv means slower consolidation
- Drainage path Hdr = H/2 for double drainage (both top and bottom), Hdr = H for single drainage (one face only)
- Double drainage halves Hdr, reducing time to reach a given degree of consolidation by a factor of 4 (since time ∝ Hdr²)
- For U ≤ 60%, use Tv = (π/4)U²; for U > 60%, use Tv = 1.781 − 0.933 log(100−U)
- At U = 50%, Tv = 0.197; at U = 90%, Tv = 0.848 — key values to memorize for PRC exams
- Consolidation time increases with layer thickness squared; a 10 m thick layer takes 4× longer than a 5 m layer (for same cv)
- Soft clays (like Manila Bay deposits) have low cv (1–3 m²/year), resulting in very long consolidation times (years to decades)
- Settlement is approximately proportional to √t at early stages (U < 50%) but logarithmic at later stages (U > 80%)
- Excess pore pressure dissipation is not uniform; it is fastest near drainage faces and slowest in the center
After primary consolidation is complete (U → 100%), the soil does not remain static. Secondary settlement continues over years and decades, driven by creep of the soil skeleton, reorientation of clay particles, and (in organic clays) decomposition of organic matter. While secondary settlement is typically 5–10% of primary settlement in inorganic clays, in organic clays (peats, silts with mica, volcanic ashes common in Philippine soils) it can equal or exceed primary settlement. **Secondary Settlement Equation:** A common model is: $$S_{sec} = C_{\alpha} \cdot H \cdot \log_{10}\left(\frac{t_2}{t_1}\right)$$ Where: - Ssec = secondary settlement (mm or m) - Cα = secondary compression index (ratio of secondary void ratio change per log-cycle of time) - H = layer thickness - t₁ = time at which primary consolidation is considered complete (often 100 days, or when U ≈ 95%) - t₂ = time of interest (months or years later) The secondary compression index Cα is measured from the oedometer test after primary consolidation ends, typically ranging from 0.01 to 0.10 for inorganic clays, and 0.05 to 0.20 for organic clays or peats. **Relationship to Cc and Cr:** Empirical correlations (for inorganic clays): - Cα ≈ 0.05 to 0.10 × Cc for normal clays - Cα ≈ 0.01 to 0.03 × Cc for stiff overconsolidated clays - For organic soils and peats, Cα must be measured; Cα / Cc ratios can be 0.5 or higher **PRC Exam Scope:** Secondary settlement calculations are less frequently tested than primary settlement and time-rate problems, but they may appear in long-term differential settlement scenarios, especially for structures on highly compressible soils or in regions with significant organic content. The NSCP 2015 and PBC typically require that secondary settlement be evaluated for sensitive structures and over the design life. **Key Differences Between Primary and Secondary Consolidation:** | Aspect | Primary Consolidation | Secondary Settlement | |---|---|---| | **Mechanism** | Pore water dissipation; effective stress increase | Soil skeleton creep; particle rearrangement; decomposition | | **Rate** | Follows diffusion equation; slows over time | Approximately constant rate on semi-log time plot | | **Equation** | Exponential/square-root of time (early), logarithmic (late) | Linear on log-time scale | | **Index** | Cc (compression), Cr (recompression) | Cα (secondary compression) | | **Typical Magnitude** | 90% of total in inorganic clays | 5–10% of primary in inorganic; 50–100% in organic | | **Practical Duration** | Weeks to years | Years to decades | | **When Measured** | Oedometer test, 0–24 hours (end-of-day roots) | Extended oedometer test, 24 hours to weeks | **Why It Matters for Philippine Practice:** In areas with highly compressible soils (Manila Bay, portions of Laguna, Mindanao deltaic regions), structures may experience continuing settlement for decades. Examples include: 1. **Long-span Bridges:** The San Juanico Bridge and other structures founded on soft clay have experienced secondary settlement monitoring. 2. **High-Rise Buildings:** Differential secondary settlement can cause cracks, door/window misalignment, and HVAC duct damage. 3. **Infrastructure:** Roads and railways on soft clay exhibit rutting and differential heave/settlement requiring periodic resurfacing (NSCP 2015 pavement design accounts for settlement). **Monitoring and Remediation:** To control secondary settlement: - Use pile foundations to bypass compressible layers (skin friction on clays should account for long-term shear strength reduction from pore pressure redistribution). - Preload the site with fill, allowing primary and initial secondary consolidation before construction (staged construction approach). - Install vertical drains (PVDs) to accelerate primary consolidation, reducing the overall time window in which secondary settlement occurs (NSCP 2015 foundation design standards address this). - Select building systems tolerant of differential settlement (articulated structures, flexible utilities).
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4. SECONDARY SETTLEMENT AND LONG-TERM BEHAVIOR
Examples
Example 4A — Secondary Settlement Over 20 Years
A warehouse on soft clay (Cc = 0.8, e0 = 1.0, H = 8 m) with primary settlement Sc = 320 mm. Laboratory testing on undisturbed samples shows Cα = 0.06 (Cα/Cc ≈ 0.075, typical for this clay). Find secondary settlement from 100 days to 20 years after loading.
Solution
Secondary settlement formula: Ssec = Cα × H × log₁₀(t₂/t₁) t₁ = 100 days (end of primary consolidation, approximately U ≈ 95% for this deposit) t₂ = 20 years = 20 × 365 = 7300 days Ssec = 0.06 × 8000 mm × log₁₀(7300/100) Ssec = 480 × log₁₀(73) Ssec = 480 × 1.863 Ssec = 894 mm ≈ 90 cm Compare to primary: Ssec ≈ 90 cm vs. Sc ≈ 32 cm. Secondary settlement is nearly 3× primary settlement over 20 years! **Answer:** ≈ 90 cm of secondary settlement. This demonstrates why organic and highly compressible clays require careful analysis. If differential secondary settlement occurs (e.g., one part of the building settles 90 cm, another 70 cm), cracking and structural distress result. **Remediation:** Preloading with fill for 1–2 years before building construction would trigger primary consolidation and reduce the long-term secondary settlement rate.
Key Points
- Secondary settlement continues long after primary consolidation ends, governed by soil creep and viscous behavior
- Secondary compression index Cα is typically 0.05–0.10 × Cc for inorganic clays, much higher for organic soils
- On a semi-log time plot, secondary settlement appears as a linear continuation after primary consolidation curve flattens
- Secondary settlement can dominate in organic clays and peats; laboratory testing must extend beyond 24 hours to quantify
- For sensitive structures (hospitals, data centers) and long design lives (50+ years), secondary settlement evaluation is required per NSCP 2015
- Preloading and vertical drains accelerate primary consolidation, reducing the impact of secondary settlement by compacting soils earlier in the site's history
- Philippine soft clay deposits (alluvial plains) commonly exhibit significant secondary settlement; pile foundations or ground improvement are standard remedies
The National Structural Code of the Philippines 2015 (NSCP 2015) and the Philippine Building Code (PBC) provide mandatory guidance for foundation design on compressible soils. Key provisions include: **Geotechnical Investigation Requirements:** Per NSCP 2015 Section 1802 (Geotechnical Site Characterization), structures subject to significant settlement must be supported by site-specific boreholes, laboratory testing (including oedometer tests for fine-grained soils), and geotechnical reports. The level of investigation depends on soil risk category and structure type: 1. **Standard Structures** (low to moderate risk): At least one boring per 2500 m² of site area; minimum 2 borings. 2. **Sensitive Structures** (hospitals, power plants, data centers): One boring per 1500 m² or closer; extended laboratory testing. 3. **High-Rise Buildings on Soft Soil:** Detailed 3D geotechnical model, pile load tests, and settlement monitoring provisions. **Allowable Settlement Limits:** NSCP 2015 provides guidance (from international standards): - **Total Settlement:** Typically limited to 50 mm for rigid structures on cohesive soils, up to 100 mm for flexible structures. - **Differential Settlement:** Typically 0.5–1.0 of the height-to-width ratio of the structure, or 25–50 mm, whichever is smaller. For high-rise buildings, often limited to 10–25 mm. - **Angular Distortion:** The ratio of differential settlement to distance between columns; typical limit is 1/500 to 1/300. **Remedial Measures per NSCP 2015:** When settlement exceeds allowable limits, the code permits: 1. **Pile Foundations:** Transfer load to deeper, less compressible layers. Skin friction on clay piles must account for long-term strength reduction (typically use effective stress methods with φ' and c'). Design per NSCP 2015 Section 1810. 2. **Vertical Drains (PVDs):** Prefabricated vertical drains installed in a grid pattern accelerate drainage and consolidation. The NSCP 2015 references ASCE and DIN standards for PVD design. Typical spacing is 1–2 m in a square or triangular pattern. A 10 m thick clay layer with Hdr = 5 m (original double drainage) can be reduced to an equivalent Hdr of 0.5–1.0 m with PVDs, reducing consolidation time by 25–100×. 3. **Preloading (Surcharge):** Temporary fill is placed to induce consolidation before building construction. The surcharge must equal or exceed the building load and be maintained for a period (typically 1–2 years for soft clay) until primary consolidation and early secondary consolidation occur. Preload is then removed to a level matching the final building load. 4. **Ground Improvement (Soil Replacement, Grouting):** Removal and replacement of highly compressible soils, or injection of cement/chemical grout to reduce compressibility. Expensive but effective for limited depths and areas. 5. **Lightweight Fill:** Using expanded shale, recycled plastic, or foam fills reduces the magnitude of stress increase, thereby reducing settlement (Δσ is smaller). Trade-off: larger foundation footprint or pile cap volume. **Monitoring and Acceptance Criteria:** During and after construction, settlement monitoring must be performed per NSCP 2015 Section 1807 for sensitive sites: - Installation of settlement plates (level surveying), piezometers (pore pressure), and inclinometers (lateral movement). - Baseline survey before construction and regular monitoring (weekly to monthly) during construction and for at least 1–2 years after. - Stop work criteria if observed settlements exceed 80% of allowable limits. - Extension of monitoring period for long-term secondary settlement in organic clays. **Differential Settlement and Structural Damage:** NSCP 2015 requires structural design to accommodate differential settlement. Strategies include: - **Articulated (Floating) Structures:** Allow sub-structures to settle independently; interconnections are flexible or slotted. - **Flexible Utilities:** Piping, ductwork, and electrical conduits designed with expansion loops and flex connections. - **Crack Control Reinforcement:** Per ACI 318 (adopted by NSCP 2015 for concrete), provide minimum reinforcement in slabs and walls subject to differential settlement. **Typical Philippine Scenarios:** 1. **Manila Bay Reclamation Areas:** Highly compressible bay mud (Cc = 0.8–1.2, cv = 1–2 m²/yr). Preloading with 1–2 m of fill for 2–3 years is standard before structure construction. Many high-rise buildings in Bonifacio Global City and Makati use piles to 30–50 m depth to bypass clay. 2. **Laguna de Bay Surroundings:** Peaty clays with Cα/Cc ≈ 0.4–0.6 and high secondary settlement. Ground improvement (PVDs + preload) often combined with piles for critical structures. 3. **Mindanao Deltaic Regions (Davao, Cagayan de Oro):** Similar to Manila Bay; large infrastructure projects employ detailed consolidation analysis and staged construction. 4. **Upland Clay Deposits (Luzon, Visayas):** Often overconsolidated due to past desiccation. Settlement is less problematic, but differential settlement due to clay lenses and variable thickness is common. **Common Board-Exam Pitfalls Related to NSCP:** - Failure to specify double vs. single drainage when calculating consolidation time. - Omitting secondary settlement in long-term analysis. - Using undrained shear strength (cu) for effective-stress pile design (NSCP 2015 requires effective stress methods for long-term stability). - Confusing settlement magnitude (controlled by Cc) with settlement timing (controlled by cv and Hdr). - Forgetting that preloading reduces future settlement but does not change total settlement (i.e., if preload equals building load, the structure settles mainly due to secondary consolidation afterward).
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5. PRACTICAL CONSIDERATIONS AND NSCP 2015 REQUIREMENTS
Examples
Example 5A — Evaluating Remedial Strategy for High-Rise Building (NSCP Framework)
A proposed 30-story office tower (40 m × 50 m footprint, estimated load 150 kPa on 4 m deep clay layer) is to be built on Manila Bay soft clay. Preliminary geotechnical data: H = 8 m, Cc = 0.9, e0 = 1.1, σ'0 = 40 kPa, cv = 2 m²/yr, double drainage initially. Preliminary estimate: Sc ≈ 420 mm over 5 years. NSCP 2015 limit for differential settlement: 30 mm over building height. Which remedial approach is preferred?
Solution
Step 1: Assess preliminary settlement without remediation. Δσ = 150 kPa; σ'f = 40 + 150 = 190 kPa; Sc ≈ [0.9/(1+1.1)] × 8000 × log₁₀(190/40) ≈ 0.43 × 8000 × 0.678 ≈ 2330 mm ≈ 2.33 m (not 420 mm as stated; likely the 420 mm was for partial load or different layer). For U = 95%: Tv = ~1.129; t = (1.129 × 4²) / 2 = 9 years (Hdr = 4 m for double drainage). Consolidation takes ~9 years. Differential settlement across 50 m width, if uniform consolidation, is low. But unequal loading (heavier on one side due to mechanical systems, shear walls) or variable soil thickness can cause differential settlement > 30 mm. Also, 2.33 m total settlement is excessive for a rigid structure (NSCP limit ~50 mm total for rigid structures). Step 2: Evaluate remedial options per NSCP 2015. **Option A: Vertical Drains + Preload** - Install PVDs in 1.5 m triangular grid, reducing effective Hdr from 4 m to ~0.5 m. - New consolidation time to U = 95%: t = (1.129 × 0.5²) / 2 = 0.14 years ≈ 52 days. - Preload with 2 m of fill (equal to 150 kPa) for 100 days; remove fill gradually before construction. - Primary consolidation complete; secondary settlement (if Cα = 0.07): Ssec over next 10 years ≈ 0.07 × 8000 × log₁₀(10 × 365 / 100) ≈ 560 × 1.46 ≈ 817 mm. Still significant but manageable by piles. **Option B: Piles to 30 m Depth** - Design bored/driven piles to 30 m (bypassing clay, founding in sand layer typical of Manila Bay geology). - Only upper 8 m of clay participates in skin friction; design accounts for long-term shear strength reduction. - Settlement from skin friction along 22 m of pile shaft: assume avg. shear stress τ = 15 kPa (long-term, φ' ≈ 25°, σ'v ≈ 100 kPa, τ ≈ 0.15σ'v), total skin friction ≈ 15 × π × 0.4 × 22 ≈ 400 kN per 0.4 m diameter pile. For 150 kPa load over 40×50 = 2000 m² footprint, total load ≈ 300 MN. Pile group (say, 30 piles) carries 10 MN each; skin friction settles elastically and through clay consolidation. Still requires settlement analysis. - Differential settlement controlled by uniform pile settlement behavior and structural stiffness (raft or pile cap distributes load). **Option C: Combination (Recommended per NSCP 2015)** - Install PVDs + preload (2 years, cost: ~2 million PHP for 2000 m² site) to compact clay and accelerate consolidation. - After preload removal, place 3–5 rows of piles (reduced pile count, lower cost) to 25 m depth, designed as semi-floating piles (skin friction on clay + end-bearing on sand). - With preconsolidated clay (higher cv equivalent due to PVDs and removed excess pore pressure), further consolidation is minimal. - Total settlement from pile group: elastic (< 10 mm) + residual consolidation (< 20 mm). Differential settlement, controlled by pile cap stiffness and load distribution, can be kept < 30 mm. **Conclusion:** Option C (PVDs + preload + piles) is typical for Philippine high-rise projects on soft clay. Cost and schedule (2-year preload period) are offset by reduced pile requirements and controlled settlement. NSCP 2015 Section 1810 (piles) and Section 1807 (monitoring) would apply. **Answer:** Combination of vertical drains, preloading, and piles is the practical choice per NSCP 2015 practice.
Key Points
- NSCP 2015 requires site-specific geotechnical investigation with boreholes and lab testing (including oedometer) for structures on soft soils
- Typical total settlement limits: 50 mm for rigid, 100 mm for flexible structures; differential settlement: 25–50 mm or 1/500 of distance between supports
- Vertical drains reduce effective drainage path by 5–20×, accelerating consolidation from years to months; spacing typically 1–2 m
- Preloading induces consolidation before construction; surcharge duration is 1–2 years for typical clay; preload = building load or higher
- Pile foundations bypass compressible soils; design accounts for long-term pore pressure redistribution and shear strength reduction
- Monitoring (settlement plates, piezometers, inclinometers) is mandatory during and after construction on soft soils; extend 1–2 years post-construction
- Differential settlement is the controlling design criterion; allowable differential settlement drives foundation system choice (floating vs. rigid piles)
- Philippine practice on soft clay: combine piles, vertical drains, and preloading for critical structures; purely shallow foundations are inadequate for loads > 50 kPa on clay
- ACI 318 reinforcement requirements apply for concrete structures; crack control based on allowable differential settlement and structural stiffness
The oedometer (also called odometer or consolidation cell) is the laboratory apparatus used to measure consolidation settlement and the consolidation indices (Cc, Cr, cv). Understanding how the test works and how to interpret data is essential for the PRC exam and for supervising laboratory work. **Oedometer Apparatus:** The oedometer consists of: 1. **Consolidation Ring:** A metal cylinder (typically 63.5 mm diameter, 20 mm height for standard tests) holding the soil specimen. 2. **Porous Stones:** Ceramic or sintered glass discs at top and bottom allow water drainage while preventing soil loss. 3. **Load Frame:** A lever or hydraulic system applies vertical load (stress) to the specimen. Typical test applies loads in increments: 10, 20, 40, 80, 160, 320, 640, 1280 kPa. Each load step is maintained for 24 hours (though extended tests may run to 1 week or longer for secondary settlement). 4. **Dial Gauge (or LVDT):** Measures vertical displacement (settlement) of the top platen, accurate to 0.01 mm. 5. **Water Supply:** Water bath maintains saturation; specimen remains submerged. **Test Procedure (Standard ASTM D2435 / ISO 17892-5):** 1. Trim undisturbed soil sample to ring height; measure initial thickness H0, mass, water content, and void ratio e0. 2. Place specimen in ring with porous stones above and below; saturate by flooding from bottom or allowing capillary absorption. 3. Apply seating load (~10 kPa); allow equilibration for 24 hours ("primary loading"). 4. Apply first stress increment (e.g., 10 kPa above seating); record dial readings at 0.25, 1, 2.25, 4, 9, 16, 25, 36, 49, 64, 100, 144, ... minutes (following √t method for time intervals) and at 24 hours. 5. Maintain each stress increment for 24 hours; increase stress to next level. 6. Continue through all stress levels (typically to 1280 kPa or failure), each 24 hours. 7. Unload in decrements (e.g., 640, 320, 160 kPa, etc.), holding each 24 hours to measure recompression and estimate OCR. 8. Optional: Re-load to determine virgin compression in an OC specimen. **Void Ratio Calculation During Test:** At each load step after 24 hours: $$e = e_0 - \frac{\Delta H \times (1 + e_0)}{H_0}$$ Where: - ΔH = cumulative vertical displacement from start of test - H0 = initial specimen height - e0 = initial void ratio **Compression Index (Cc) Determination:** 1. Plot e vs. log₁₀(σ') on semi-log paper (void ratio on linear y-axis, stress on log x-axis). 2. Identify the virgin compression line (VCL) — the straight portion on the plot at higher stresses after yielding of overconsolidated structure. 3. For a normally consolidated specimen, the VCL is the entire loading curve after seating. For an overconsolidated specimen, the VCL begins after the preconsolidation stress σ'c is exceeded. 4. Calculate Cc as the slope of the VCL: $$C_c = \frac{\Delta e}{\log_{10}(\sigma'_2 / \sigma'_1)}$$ Where σ'₁ and σ'₂ are two well-defined points on the VCL (e.g., stress at 50 kPa and 200 kPa, or any two points spanning at least one decade of stress). **Recompression Index (Cr) Determination:** The recompression line (RCL) is the curve or line during initial loading (for NC clay, very flat; for OC clay, evident) and unload-reload. Calculate Cr as the slope of the linear portion during unload-reload: $$C_r = \frac{\Delta e}{\log_{10}(\sigma'_{max} / \sigma'_{min})}$$ Where σ'max and σ'min are stresses spanning the recompression phase (e.g., unload from 640 kPa to 160 kPa, then reload to 640 kPa). **Preconsolidation Stress (σ'c) Determination:** For overconsolidated clay, σ'c is identified using the **Casagrande construction method**: 1. On the e–log σ' plot, find the point of maximum curvature (often by visual inspection or by identifying where the curve transitions from steep to less steep). 2. Draw a horizontal line through the point of maximum curvature. 3. Draw a tangent line to the curve at that point. 4. Bisect the angle between the horizontal and tangent; extend the bisector to intersect the extrapolated VCL. 5. Drop a vertical line from the intersection; this vertical line intersects the log(σ') axis at σ'c. Alternatively, some labs use the **log-log** or **second derivative** method. The exact method should be noted in the lab report. **Coefficient of Consolidation (cv) Determination:** From the dial readings during each load increment, plot displacement vs. √t (square-root of time) on a linear-linear plot. 1. **√t Method (Square-Root-of-Time):** Identifies time for 50% consolidation (t50) from the curve. - Plot dial reading vs. √t for the first 10–20 minutes of the load step. - The initial portion is approximately linear (zero-dimensional consolidation). - Identify the time t0 at which dial reading is zero (intersection of best-fit line with time axis). - Measure the initial slope of the √t curve. - Identify t90: the time corresponding to 90% of primary consolidation (approximately where the curve becomes curvilinear); t90 ≈ 1.27 × t50. - Calculate cv = (Tv × H²dr) / t50, where Tv = 0.197 at U = 50%. - Thus, cv = 0.197 × H²dr / t50, with H_dr = H/2 for double drainage in oedometer (water exits top and bottom). 2. **Log-Time Method (Logarithmic-Fitting):** Plots dial reading vs. log(t). - Identify the inflection point (transition between linear and flattening portions). - Measure t100 (time for 100% consolidation of this load step) from the curve. - cv = (Tv × H²dr) / t100. - Less commonly used for oedometer; more useful for field consolidation (piezometer data). **Typical Laboratory Report Data:** A standard oedometer test report includes: - **Initial Conditions:** Natural water content wn, dry unit weight γd, void ratio e0, degree of saturation Sr, specific gravity Gs. - **Consolidation Table:** Load, dial reading at each time interval, calculated void ratio, settlement. - **e–log σ' Plot:** With VCL, RCL, σ'c marked. - **Cc and Cr:** Values with standard deviation if multiple tests are available. - **cv Values:** From √t method at each load level, averaged or selected (often higher stress levels are more reliable). - **Discussion:** Assessment of NC vs. OC, OCR, sensitivity to disturbance, secondary settlement rate if extended test, and recommendations for field settlement prediction. **Common Laboratory Errors and Pitfalls:** 1. **Insufficient Saturation:** Entrapped air bubbles in pores reduce drainage and underestimate cv. Saturation checks include degree of saturation Sr (target > 95%), or back-calculation from test data (if cv seems unusually low). 2. **Specimen Disturbance:** Sampling and trimming of clay specimens causes structure damage. Undisturbed samples (U-100 or Osterberg tube) are essential; disturbed samples overestimate Cc and underestimate preconsolidation stress. PRC exam often asks whether test results apply to field behavior. 3. **Load Increment Too Large:** Stresses should increase gradually (not step from 10 kPa to 640 kPa in one jump) to avoid shock and ensure drainage equilibrium. 4. **Insufficient Drainage Time:** Some labs shorten the 24-hour hold period; inadequate time results in underestimated settlement and overestimated cv. ASTM D2435 requires minimum 24 hours; Philippine practice per NSCP 2015 typically follows this. 5. **Height Measurement Errors:** Oedometer ring height should be measured carefully before and after test. If not accounted for, calculated void ratio is in error, affecting all indices. 6. **Secondary Settlement Ignored:** For organic or peaty clays (common in Philippines), secondary settlement can dominate but is missed if test is not extended beyond 24 hours per load. Extended tests (to 3–7 days per increment) are recommended for sensitive structures. **Typical Values for Philippine Soils:** | Soil Type | Cc | Cr | cv (m²/yr) | e0 | Typical Location | |---|---|---|---|---|---| | Soft Manila Bay clay | 0.8–1.2 | 0.08–0.15 | 1–3 | 1.0–1.2 | Manila, Laguna | | Bangkok-type soft clay | 0.8–1.0 | 0.10–0.20 | 1–2 | 0.9–1.1 | Coastal areas | | Medium clay (Quaternary) | 0.2–0.5 | 0.03–0.10 | 10–20 | 0.6–0.9 | Interior lowlands | | Stiff clay (residual) | 0.10–0.25 | 0.02–0.05 | 30–100 | 0.4–0.7 | Upland regions | | Peat and highly organic | 1.5–3.0 | 0.30–0.80 | 0.5–2 | 2.0–5.0 | Laguna, swamps | **Quality Assurance for Board-Exam Preparation:** When reviewing lab data for a consolidation problem: 1. **Verify Test Conditions:** Check that soil is saturated, test duration is adequate (24 hours minimum), and specimen is undisturbed. 2. **Check Index Ranges:** If Cc < 0.05 or Cr > Cc, suspect lab error or misidentification of NC vs. OC. 3. **Compare cv to Literature:** If cv is 10× higher or lower than expected for the soil type, question whether pore pressure was truly zero at the end of each load increment. 4. **Estimate Settlement:** Use reported Cc and initial conditions to estimate settlement; if field settlement greatly exceeds this, secondary settlement or stress-dependent permeability may explain the difference. 5. **Assess Long-Term Behavior:** For clays, secondary settlement estimates (using Cα/Cc) should be checked; if secondary settlement over 20 years exceeds primary by > 5×, the lab result or field conditions need clarification.
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6. CONSOLIDATION TEST (OEDOMETER) AND LABORATORY DETERMINATION OF INDICES
Examples
Example 6A — Interpreting Oedometer e–log σ' Plot to Identify OC Clay and Extract Indices
Lab report for Manila Bay clay sample shows the following e–log σ' data (selected points):\nσ' (kPa): 10, 20, 40, 80, 160, 320, 640\ne: 1.05, 1.03, 0.98, 0.90, 0.78, 0.66, 0.54\n\nAfter unload from 640 to 80 kPa and reload to 320 kPa, recompression curve is parallel to initial loading below σ'c. Interpretation?
Solution
Step 1: Plot e vs. log₁₀(σ') and observe behavior. - From 10 to ~80 kPa: e changes from 1.05 to 0.90, a moderate slope (likely recompression line for OC clay). - From 80 to 640 kPa: e changes from 0.90 to 0.54, steeper slope (virgin compression line). - Inflection around 80–100 kPa suggests σ'c ≈ 100 kPa. Step 2: Calculate indices. **VCL (using 160 and 640 kPa points on virgin line):** Cc = (0.78 − 0.54) / log₁₀(640/160) = 0.24 / 0.602 = 0.399 ≈ 0.40 **RCL (if unload-reload data available):** Assume unload from 640 to 80 kPa shows parallel slope to initial loading; Cr ≈ 0.08 (visual estimate from curve steepness ratio Cr/Cc ≈ 0.2). **Preconsolidation Stress (Casagrande method):** Maximum curvature is at ~80–100 kPa; σ'c ≈ 100 kPa. Step 3: Verify OC status. If original depth of sampling was 8 m with γsat ≈ 19 kN/m³ and WT at surface: σ'0 ≈ 19 × 8 = 152 kPa. But σ'c ≈ 100 kPa < 152 kPa would mean normally consolidated, not OC. **Contradiction:** Either σ'c is higher (re-examine curve), or sample was from a shallower depth, or past geology (desiccation, erosion) explains OCR > 1. For an exam question, note: σ'0 > σ'c suggests either mis-identification of σ'c or special past loading. **Answer:** Cc ≈ 0.40, Cr ≈ 0.08, σ'c ≈ 100 kPa. The sample shows signs of overconsolidation, but the σ'c vs. current stress relationship should be clarified from geological context.
Key Points
- Oedometer test applies incremental vertical stress to confined soil specimen; measures settlement and void ratio change to derive Cc, Cr, cv, σ'c
- Void ratio is calculated from specimen height change: e = e0 − ΔH(1+e0)/H0, where ΔH is cumulative displacement
- Compression index Cc is slope of virgin compression line (VCL) on e–log σ' plot; typical range 0.2–1.2 for clays
- Recompression index Cr is slope of unload-reload curve; typically 0.15–0.3 × Cc
- Preconsolidation stress σ'c is determined by Casagrande construction method; identifies transition from recompression to virgin compression
- Coefficient of consolidation cv is derived from √t method (50% consolidation time) or log-time method; units must be consistent with time
- Standard 24-hour hold per load increment is essential; shorter periods underestimate settlement and overestimate cv
- Secondary compression index Cα is measured from extended tests (> 24 hours); essential for organic clays and long-term settlement prediction
- Specimen saturation, disturbance, and height measurement errors are common sources of unreliable results; quality control is critical
- Philippine practice: expect soft clays (Manila Bay, Laguna) with Cc = 0.8–1.2, cv = 1–3 m²/yr; upland clays with Cc < 0.5, cv > 30 m²/yr
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