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CELE Geotechnical EngineeringSoil ClassificationStudy Notes

Detailed study notes for CELE Geotechnical Engineering — Soil Classification. These are the kind of notes you would take if you were reviewing with someone who has already scored well on the CELE: organised by what Professional Regulation Commission (PRC) — Board of Civil Engineering tests first, followed by the nice-to-knows, and ending with the traps to avoid.

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. Soil Classification lands at position 2nd 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.

Soil Classification - Study Notes

Soil classification is a systematic method of grouping soils based on their physical properties, particularly grain size distribution and plasticity characteristics. For civil engineers, classification serves as the foundation for predicting soil behavior in construction, earthworks, and foundation design. The PRC Civil Engineer Licensure Examination emphasizes two classification systems: the Unified Soil Classification System (USCS) used in geotechnical engineering design, and the AASHTO system used in pavement and highway construction. This chapter develops the quantitative skills needed to determine gradation parameters (Cu and Cc), interpret Atterberg limits, and assign proper soil classifications for both systems. Mastery of these concepts is essential for solving board problems involving soil mechanics, foundation design, and earthwork specifications.

Summary

Soil classification is the foundational step in geotechnical engineering that allows engineers to predict soil behavior and select appropriate design and construction methods. The chapter has presented two primary systems: the **Unified Soil Classification System (USCS)**, which is the standard for general geotechnical work and design, and the **AASHTO system**, which is specific to highway and pavement applications. Key concepts covered include: (1) **Grain-size distribution parameters** (D₁₀, D₃₀, D₆₀, Cu, Cc) that characterize coarse-grained soils and determine whether they are well-graded or poorly graded; (2) **Atterberg limits** (LL, PL, SL) and the **plasticity index** that define the plastic behavior of fine-grained soils and allow classification on the plasticity chart; (3) the **plasticity chart** with the A-line (PI = 0.73(LL − 20)) that separates clays from silts and the LL = 50 line that separates low from high plasticity soils; (4) **USCS classification** with its two-letter codes (GW, GP, GM, GC, SW, SP, SM, SC, ML, MH, CL, CH, OL, OH, Pt) that provide detailed engineering description; and (5) **AASHTO classification** with its Group Index formula that rates soils from A-1 (best) to A-8 (peat) for subgrade use. For the PRC Civil Engineer Licensure Examination, the board will test your ability to: • Read grain-size curves accurately and calculate Cu and Cc • Determine well-graded vs. poorly graded classification (both criteria required) • Perform Atterberg limit tests and interpret LL and PI values • Plot fine-grained soils on the plasticity chart using the A-line relationship • Assign proper USCS two-letter classifications • Calculate AASHTO Group Index, observing all formula caps and rounding rules • Recognize and avoid common mistakes (forgetting Cc, wrong A-line slope, GI calculation errors) • Apply classification results to engineering decisions (foundation depth, pavement design, embankment material selection) Mastery of soil classification, combined with understanding of soil properties (shear strength, permeability, compressibility), forms the technical foundation for safe and economical civil engineering design in the Philippines, where tropical soils and challenging geologies (Manila Bay soft clay, residual soils, laterites) require careful site-specific investigation and design consideration.

Sections

Grain-size distribution describes the percentage (by mass) of soil particles passing through standard sieves of various sizes. The gradation curve is plotted on a semilogarithmic graph with sieve opening (diameter) on the logarithmic x-axis and percentage passing (finer than) on the arithmetic y-axis. From this curve, three characteristic diameters are identified: • D₁₀ (effective size): diameter at which 10% of soil particles are finer • D₃₀: diameter at which 30% of soil particles are finer • D₆₀: diameter at which 60% of soil particles are finer These values are used to compute two important shape parameters: **Uniformity Coefficient:** Cu = D₆₀ / D₁₀ The uniformity coefficient indicates the range of particle sizes. A high Cu indicates a wide size range (nonuniform), while a low Cu indicates most particles are similar in size (uniform). **Coefficient of Curvature (Gradation Coefficient):** Cc = D₃₀² / (D₁₀ × D₆₀) The coefficient of curvature describes the shape of the gradation curve between D₁₀ and D₆₀. It accounts for whether intermediate-size particles are present in appropriate proportions. **Well-Graded vs. Poorly Graded:** A soil is classified as **well-graded** if it satisfies BOTH conditions simultaneously: • For gravels: Cu ≥ 4 AND 1 ≤ Cc ≤ 3 • For sands: Cu ≥ 6 AND 1 ≤ Cc ≤ 3 If either condition fails, the soil is **poorly graded**. Well-graded soils have good particle size continuity, which typically results in better compaction, higher shear strength, and lower permeability. Poorly graded soils may be uniformly graded (most particles similar size) or gap-graded (missing intermediate sizes). **Physical Significance:** Well-graded soils compact more efficiently because smaller particles can fill voids between larger particles. This makes them ideal for embankments, road bases, and earth dams. Poorly graded soils, particularly uniformly graded sands, are more prone to liquefaction and settlement under seismic loading.

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1. Grain-Size Distribution and Gradation Parameters

Examples

Problem

EXAMPLE 1.1 — Uniformity and Curvature Calculation A sand sample has the following sieve analysis results: D₁₀ = 0.15 mm, D₃₀ = 0.45 mm, D₆₀ = 1.5 mm. Calculate the uniformity coefficient and coefficient of curvature. Determine if the sand is well-graded or poorly graded.

Solution

Step 1: Calculate the uniformity coefficient. Cu = D₆₀ / D₁₀ = 1.5 / 0.15 = 10 Step 2: Calculate the coefficient of curvature. Cc = D₃₀² / (D₁₀ × D₆₀) = (0.45)² / (0.15 × 1.5) = 0.2025 / 0.225 = 0.9 Step 3: Check gradation criteria for sand. For sand to be well-graded: Cu ≥ 6 AND 1 ≤ Cc ≤ 3 • Cu = 10 ≥ 6 ✓ (passes) • Cc = 0.9 < 1 ✗ (fails) Step 4: Classify. Since Cc = 0.9 falls outside the range [1, 3], the sand is **POORLY GRADED** despite having a high uniformity coefficient. The gradation curve is concave, indicating a deficiency in intermediate-size particles. Note: This is a common board-exam pitfall. Students often check only Cu and forget the Cc requirement. Both conditions must be satisfied simultaneously for a well-graded classification.

Problem

EXAMPLE 1.2 — Well-Graded Gravel A gravel has D₁₀ = 2 mm, D₃₀ = 8 mm, D₆₀ = 16 mm. Classify the gradation.

Solution

Step 1: Calculate uniformity coefficient. Cu = 16 / 2 = 8 Step 2: Calculate coefficient of curvature. Cc = (8)² / (2 × 16) = 64 / 32 = 2.0 Step 3: Check gravel criteria. For gravel: Cu ≥ 4 AND 1 ≤ Cc ≤ 3 • Cu = 8 ≥ 4 ✓ • Cc = 2.0, where 1 ≤ 2.0 ≤ 3 ✓ Step 4: Classify. The gravel is **WELL-GRADED (GW)** because both conditions are satisfied. The gradation curve is convex (S-shaped), indicating good representation of particle sizes throughout the range. This soil would be suitable for filter layers, embankments, or road bases.

Problem

EXAMPLE 1.3 — Poorly Graded Gap-Graded Soil A sample has D₁₀ = 0.5 mm, D₃₀ = 0.8 mm, D₆₀ = 2.0 mm. Evaluate the gradation.

Solution

Step 1: Calculate parameters. Cu = 2.0 / 0.5 = 4.0 Cc = (0.8)² / (0.5 × 2.0) = 0.64 / 1.0 = 0.64 Step 2: Check criteria (assuming sand: Cu ≥ 6). • Cu = 4.0 < 6 ✗ (fails uniformity criterion) • Cc = 0.64 < 1 ✗ (fails curvature criterion) Step 3: Classify. The soil is **POORLY GRADED (SP)**. The very low Cc (0.64) indicates a gap-graded distribution with most particles clustered at small sizes and a jump to larger sizes. Intermediate particles are scarce. This type of gradation is common in materials that have been sorted by wind or water (e.g., beach sand or dune sand).

Key Points

  • D₁₀, D₃₀, D₆₀ are read directly from the gradation curve at 10%, 30%, and 60% finer
  • Cu = D₆₀/D₁₀ measures the range of particle sizes (uniformity)
  • Cc = D₃₀²/(D₁₀·D₆₀) measures the shape of the size distribution (curvature)
  • Well-graded requires both Cu criterion AND 1 ≤ Cc ≤ 3 to be satisfied simultaneously
  • For gravels: Cu ≥ 4; for sands: Cu ≥ 6; if either criterion fails, soil is poorly graded
  • Well-graded soils have continuous particle size distribution and compact better
  • Poorly graded soils may be uniformly graded (narrow size range) or gap-graded (missing intermediate sizes)

The Atterberg limits are moisture content values that define transitions between the solid, plastic, and liquid states of fine-grained soils (silts and clays). These limits are determined by standardized laboratory tests specified in ASTM D4318 and are critical for understanding soil behavior during construction and over time. **Shrinkage Limit (SL):** The moisture content at which a soil, when dried, reaches its minimum volume. Below this moisture content, further drying does not cause additional volume change. The shrinkage limit is important for predicting cracking in clay embankments and foundations. **Plastic Limit (PL):** The moisture content at which a soil transitions from semi-solid behavior to plastic behavior. At this point, the soil can be rolled into a thread of 3 mm diameter on a glass plate without crumbling. Below the plastic limit, the soil behaves as a brittle material; above it, the soil can deform without cracking. **Liquid Limit (LL):** The moisture content at which a soil transitions from plastic to liquid behavior. At this limit, a pat of soil placed in a standard Casagrande cup requires 25 blows of a falling cam to close a standard groove. At the liquid limit, the soil begins to flow under its own weight. **Plasticity Index (PI):** PI = LL − PL The plasticity index represents the range of moisture content over which the soil exhibits plastic behavior. A high PI indicates a wide plastic range and generally signifies a highly plastic clay. A low PI indicates that the soil transitions quickly from plastic to liquid state. Soils with PI < 4 are classified as non-plastic. **Physical Interpretation:** The plasticity index correlates with clay mineralogy and has strong implications for engineering behavior: • PI < 4: Non-plastic or slightly plastic soils (silts, sandy silts) • 4 ≤ PI < 15: Low plasticity soils (lean clays, silty clays) • 15 ≤ PI < 30: Medium plasticity soils (typical clays) • PI ≥ 30: High plasticity soils (fat clays, montmorillonite-rich soils) Soils with high PI are more compressible, have lower shear strength when wet, and are more prone to shrinkage cracking during drying. The liquid limit LL = 50% is a threshold in both USCS and AASHTO systems, separating low-plasticity (L) from high-plasticity (H) soils.

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2. Atterberg Limits and Plasticity Index

Examples

Problem

EXAMPLE 2.1 — Plasticity Index Calculation A clay soil has a liquid limit of 52% and a plastic limit of 26%. Calculate the plasticity index and classify the soil as low- or high-plasticity.

Solution

Step 1: Apply the formula. PI = LL − PL = 52 − 26 = 26 Step 2: Classify by LL threshold. Since LL = 52 > 50, the soil is classified as HIGH-PLASTICITY (H). Step 3: Interpret. PI = 26 indicates a clay with a moderately wide plastic range. The soil is sensitive to moisture changes and will exhibit significant volume change with wetting and drying cycles. This clay is likely suitable for clay liners in landfills or retention ponds but problematic for foundation support without stabilization. Note: The combination of LL > 50 and PI > 20 typically indicates montmorillonite clay (smectite group), which is highly expansive. Special geotechnical investigation and design measures are required for such soils in the Philippines, where tropical weathering produces high LL clays.

Problem

EXAMPLE 2.2 — Non-Plastic Silt Determination A silt sample has LL = 38% and PL = 34%. Classify the soil's plasticity.

Solution

Step 1: Calculate PI. PI = 38 − 34 = 4 Step 2: Classify by plasticity range. Since PI = 4, the soil falls at the borderline between non-plastic and slightly plastic. Many standards classify PI ≤ 4 as non-plastic or very slightly plastic. Step 3: Interpret behavior. With such a narrow plastic range, this silt transitions rapidly from plastic to liquid state with small moisture increases. It is less compressible than clays but still retains some cohesion. The soil is moderately sensitive to pore pressure changes and may exhibit instability on steep slopes if saturated. Note: The liquid limit of 38% indicates this is a low-plasticity silt (ML classification). The narrow plastic range is characteristic of silts lacking significant clay minerals.

Problem

EXAMPLE 2.3 — High-Plasticity Clay A clay has LL = 68%, PL = 32%. Find PI and discuss engineering implications.

Solution

Step 1: Calculate PI. PI = 68 − 32 = 36 Step 2: Classify. LL = 68 > 50 (high-plasticity threshold) PI = 36 > 30 (high plasticity range) Classification: **CH (high-plasticity clay)** Step 3: Engineering implications. PI = 36 indicates a clay with very high clay content, likely montmorillonite or illite minerals. Engineering considerations: • High compressibility: significant settlement expected under loads • Low permeability: drainage is slow; pore pressures dissipate over months/years • High shrinkage potential: drying causes significant volume loss and cracking • Problematic foundation material: requires deep foundations or soil stabilization • Excellent liner material: used in retention pond design due to low permeability In Philippine geotechnical practice, such soils are common in low-lying areas of Metro Manila and are typically stabilized with lime or cement for construction purposes. The Maynila Clay formation exemplifies such high-PI soils.

Key Points

  • Atterberg limits (LL, PL, SL) define transition points between solid, plastic, and liquid states
  • Plasticity Index (PI) = LL − PL measures the range of moisture content for plastic behavior
  • Liquid limit (LL) is determined by the Casagrande cup test (25 blows to close groove)
  • Plastic limit (PL) is determined by rolling soil into 3 mm thread on glass plate
  • Shrinkage limit (SL) is the moisture content below which no additional volume change occurs with drying
  • LL = 50% is the threshold separating low-plasticity (L) from high-plasticity (H) soils
  • Higher PI indicates greater clay content and more compressible soil behavior
  • Non-plastic soils have PI < 4 (primarily silts and sands)

The plasticity chart is a plotting tool used to classify fine-grained soils based on their liquid limit and plasticity index. The chart's two key features are the **A-line** and the **LL = 50% vertical line**, which together create regions for each soil classification. **The A-Line Equation:** PI = 0.73(LL − 20) This empirical relationship, established through extensive laboratory testing, separates **clays** (soils plotting above the A-line) from **silts** (soils plotting below the A-line). The A-line has a slope of 0.73 and intersects the LL axis at LL = 20. **Physical Meaning:** Clays have higher plasticity relative to their liquid limit than do silts. For a given LL value, clay soils consistently exhibit PI values that plot above the A-line, while silt soils plot below. This distinction reflects differences in mineral composition and surface chemistry. Clays (primarily phyllosilicate minerals) have larger specific surface areas and greater water-holding capacity than silts (primarily quartz and feldspar). **Classification Regions:** The plasticity chart is divided into regions by the A-line and the LL = 50 vertical line: 1. **CL (low-plasticity clay):** Above A-line, LL < 50 2. **CH (high-plasticity clay):** Above A-line, LL ≥ 50 3. **ML (low-plasticity silt):** Below A-line, LL < 50 4. **MH (high-plasticity silt):** Below A-line, LL ≥ 50 Additional regions include: • **OL and OH:** Organic silts and clays (LL after oven-drying is less than original LL) • **CL-ML:** Borderline clays and silts (plotting near the A-line) **Using the Plasticity Chart in Practice:** To classify a soil using the plasticity chart: 1. Determine LL and PI from laboratory tests 2. Locate the point (LL, PI) on the chart 3. Determine which side of the A-line the point falls 4. Note whether LL < 50 or LL ≥ 50 5. Assign the appropriate two-letter code based on region **Practical Note:** In the Philippines, high-PI clays (CH) from tropical weathering are common. These soils often exhibit unusual behavior, including potential for quick-clay formation and significant volume change. The Maynila Clay and Manila Bay soft clay deposits are well-known examples in geotechnical practice.

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3. The Plasticity Chart and A-Line Relationship

Examples

Problem

EXAMPLE 3.1 — Classifying on the Plasticity Chart A fine-grained soil has LL = 45% and PI = 23%. Using the plasticity chart, classify the soil.

Solution

Step 1: Determine position relative to LL = 50 line. LL = 45 < 50, so the soil is on the LOW-PLASTICITY side of the vertical line. Step 2: Determine position relative to the A-line. A-line PI at LL = 45: PI_A = 0.73(45 − 20) = 0.73(25) = 18.25 Soil PI = 23 > 18.25, so the soil plots ABOVE the A-line. Step 3: Classify. Above A-line AND LL < 50 → **CL (low-plasticity clay)** Step 4: Engineering behavior. This clay has moderate plasticity and is suitable for general construction purposes. The PI of 23 indicates significant clay content. It will have moderate compressibility and volume change potential. Common in embankments and subgrades when properly compacted. Note: The soil is sometimes called a "lean clay" because PI < 30, indicating not all particles are clay-sized but there is a sufficient fine fraction with plastic properties.

Problem

EXAMPLE 3.2 — Borderline Clay-Silt Classification A soil is tested: LL = 32%, PI = 10%. Classify on the plasticity chart.

Solution

Step 1: Check LL threshold. LL = 32 < 50 (low-plasticity region). Step 2: Calculate A-line PI. PI_A = 0.73(32 − 20) = 0.73(12) = 8.76 Soil PI = 10 > 8.76 (just above A-line). Step 3: Classify. The point plots slightly above the A-line in the CL region, but very close to the borderline. The soil exhibits mixed characteristics of clay and silt. It is classified as **CL** but with recognition that it has borderline clay-silt properties. Alternative notation: **CL-ML** might be used if soil properties closely straddle the A-line. Step 4: Interpretation. This soil has weak plastic properties, likely a sandy clay or silty clay with low clay content. Shear strength will be influenced by both friction angle and cohesion. Settlement will be moderate. Permeability is intermediate between sand and clay.

Problem

EXAMPLE 3.3 — High-Plasticity Clay Classification A sample from a cut slope in Metro Manila: LL = 72%, PI = 38%. Classify and discuss implications.

Solution

Step 1: Position relative to LL = 50. LL = 72 > 50 (high-plasticity region). Step 2: Position relative to A-line. PI_A = 0.73(72 − 20) = 0.73(52) = 37.96 Soil PI = 38 > 37.96 (just above A-line). Step 3: Classify. Above A-line AND LL ≥ 50 → **CH (high-plasticity clay)** Step 4: Engineering implications for Philippine context. This is a typical Maynila Clay with very high plasticity. Geotechnical concerns: • Highly compressible: significant consolidation settlement expected • Very low permeability: slow drainage, high pore pressure dissipation times • Extreme shrinkage potential: dangerous for slope stability during dry season • Low undrained shear strength: may require preloading or soil improvement • High cost of foundation design: must use piles or deep foundations For construction in the Philippines, such soils are typically: - Stabilized with lime/cement for embankments - Replaced entirely for shallow foundation projects - Penetrated with deep piles for building foundations This exemplifies the challenge of tropical soil mechanics in the Philippines.

Key Points

  • The A-line equation is PI = 0.73(LL − 20); soils above this line are clays, below are silts
  • The LL = 50% line divides low-plasticity (L) from high-plasticity (H) soils
  • CL: above A-line, LL < 50 (lean clay, low plasticity)
  • CH: above A-line, LL ≥ 50 (fat clay, high plasticity)
  • ML: below A-line, LL < 50 (silt, low plasticity)
  • MH: below A-line, LL ≥ 50 (silt, high plasticity)
  • A-line slope of 0.73 reflects the higher plasticity of clays relative to their LL
  • The position on the plasticity chart directly indicates clay mineralogy and engineering behavior

The Unified Soil Classification System is the primary classification scheme used by geotechnical engineers in design and construction. It was developed jointly by the U.S. Bureau of Reclamation and the U.S. Army Corps of Engineers and has been adopted internationally, including in the Philippines. The USCS classifies soils into two major groups based on the percentage of particles finer than 0.075 mm (the #200 sieve). **Major Division — Coarse-Grained vs. Fine-Grained:** **COARSE-GRAINED SOILS (more than 50% retained on #200 sieve):** Further subdivided by the gravel/sand boundary (4.75 mm sieve): • **Gravel (G):** More than 50% of coarse fraction is larger than 4.75 mm (retained on #4 sieve) • **Sand (S):** 50% or more of coarse fraction is smaller than 4.75 mm (passing #4 sieve) Coarse-grained soils are subdivided into clean (lacking fines) or dirty (containing significant fines) based on the percent passing #200: • **Clean:** Less than 5% passing #200 - GW: well-graded gravel - GP: poorly graded gravel - SW: well-graded sand - SP: poorly graded sand • **Dirty (with fines):** 5% to 12% passing #200 (borderline) or more than 12% passing #200 - GM: gravel with silt - GC: gravel with clay - SM: sand with silt - SC: sand with clay For soils with 5–12% fines, both designations are sometimes used (e.g., SW-SM: well-graded sand with some silt). **FINE-GRAINED SOILS (50% or more passing #200 sieve):** Classified by type and plasticity using the plasticity chart: • **M (Silt):** Below the A-line on the plasticity chart - ML: low-plasticity silt (LL < 50) - MH: high-plasticity silt (LL ≥ 50) • **C (Clay):** Above the A-line on the plasticity chart - CL: low-plasticity clay (LL < 50) - CH: high-plasticity clay (LL ≥ 50) • **O (Organic):** Contains organic material; LL after drying < 4/5 of original LL - OL: low-plasticity organic clay or silt - OH: high-plasticity organic clay or silt • **Pt (Peat):** Primarily organic material; easily identified by dark color, fibrous structure, and distinctive odor **Classification Decision Tree:** The USCS classification requires the following information: 1. Percentage passing #200 sieve (determines coarse vs. fine) 2. Percentage passing #4 sieve (determines gravel vs. sand for coarse soils) 3. D₁₀, D₃₀, D₆₀ (determines Cu and Cc for coarse soils) 4. LL and PI (determines position on plasticity chart for fine soils) 5. Visual inspection (determines if organic or peat) **Physical Significance of Each Class:** **GW (Well-Graded Gravel):** Dense, excellent drainage, high shear strength, minimal settlement. Ideal for filter layers, embankment cores, road base courses. **GP (Poorly Graded Gravel):** Potentially uniform or gap-graded; still has good drainage but may not compact as well as GW. Used in fill and drainage applications. **GM (Gravel with Silt):** Silt fills voids between gravel particles; reduced drainage, moderate compressibility. Used in embankments with controlled permeability. **GC (Gravel with Clay):** Clay binds gravel particles; low drainage, higher cohesion, more stable on slopes. Used in embankment cores and dam construction. **SW (Well-Graded Sand):** Good drainage and compaction properties; often mixed with gravel (SW-GW) for road bases and filters. **SP (Poorly Graded Sand):** Uniform sizing; still drains well but potential for liquefaction in earthquake zones. Requires analysis for seismic hazard. **SM (Sand with Silt):** Silt reduces drainage; variable compressibility. May be problematic if silt content is high (>20%). **SC (Sand with Clay):** Clay provides cohesion; low permeability; suitable for embankment cores and dikes. **ML (Low-Plasticity Silt):** Moderate compressibility, low shear strength when saturated, potential for piping erosion in embankments. Requires stabilization for bearing layers. **MH (High-Plasticity Silt):** Similar behavior to ML but more compressible and less permeable. Often problematic for foundations. **CL (Low-Plasticity Clay):** Moderate to good bearing capacity, moderate compressibility, suitable for general foundation and embankment use. **CH (High-Plasticity Clay):** High compressibility, low permeability, very low strength when saturated, significant volume change. Requires special design consideration in the Philippines due to tropical weathering. **OL & OH (Organic Soils):** Highly compressible, low strength, very slow drainage. Generally unsuitable for foundations; must be removed or treated before construction. **Pt (Peat):** Extremely compressible, very high water content, essentially unsuitable for any structural support. Must be removed from building sites.

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4. Unified Soil Classification System (USCS)

Examples

Problem

EXAMPLE 4.1 — Coarse-Grained Soil Classification (Clean Sand) A sand sample analysis shows: 92% passing #4 sieve, 8% passing #200 sieve, D₁₀ = 0.2 mm, D₃₀ = 0.6 mm, D₆₀ = 2.0 mm. Classify using USCS.

Solution

Step 1: Determine if coarse or fine-grained. Percent passing #200 = 8% > 5%, so the soil has some fines and is not "clean." Percent retained on #200 = 92% > 50%, so it is COARSE-GRAINED. Step 2: Determine if gravel or sand. Percent passing #4 = 92% means 92% of the soil is finer than the #4 sieve. This is a high percentage, indicating SAND (>50% of coarse fraction is <4.75 mm). Step 3: Determine if clean or with fines. Percent passing #200 = 8% falls in the range 5–12%, so the soil is borderline or has fines. The soil has some silt or clay. Step 4: Calculate gradation parameters (for sand criterion: Cu ≥ 6, Cc between 1 and 3). Cu = D₆₀/D₁₀ = 2.0/0.2 = 10 Cc = D₃₀²/(D₁₀ × D₆₀) = (0.6)²/(0.2 × 2.0) = 0.36/0.4 = 0.9 Cu = 10 ≥ 6 ✓ Cc = 0.9 < 1 ✗ Since Cc does not satisfy the range [1, 3], the sand is POORLY GRADED → SP designation for the coarse fraction. Step 5: Determine fines type (requires plasticity test or visual inspection). Assuming silt fines (typical for clean beach sand): Final Classification: **SP-SM (poorly graded sand with silt)** Step 6: Engineering implications. This is a fine, poorly graded sand with some silt, typical of beach or dune environments. It has moderate permeability and potential for liquefaction under earthquake loading due to uniform grading. It is not suitable as-is for structural fill; blending with coarser material (gravel) is recommended to improve gradation and reduce liquefaction potential.

Problem

EXAMPLE 4.2 — Fine-Grained Soil Classification Using USCS Fine soil sample: 95% passing #200 sieve, LL = 48%, PL = 28%. Classify with USCS.

Solution

Step 1: Determine if coarse or fine-grained. Percent passing #200 = 95% ≥ 50%, so the soil is FINE-GRAINED. Step 2: Calculate plasticity index. PI = LL − PL = 48 − 28 = 20 Step 3: Determine position relative to LL = 50 line. LL = 48 < 50, so the soil is on the LOW-PLASTICITY (L) side. Step 4: Determine position relative to A-line. PI_A = 0.73(LL − 20) = 0.73(48 − 20) = 0.73(28) = 20.44 Soil PI = 20 < 20.44, so the soil plots BELOW the A-line. Step 5: Classify. Below A-line AND LL < 50 → **ML (low-plasticity silt)** Step 6: Engineering properties. This is a silt with low cohesion and moderate permeability. Shear strength is primarily frictional. Settlement is moderate; permeability is lower than sand but higher than clay. The soil would require stabilization (compaction, cement/lime treatment) if used as a bearing layer for shallow foundations. Suitable for embankment shells when properly compacted.

Problem

EXAMPLE 4.3 — Coarse-Grained Soil with Clay Fines Gravel sample analysis: 45% passing #4 sieve, 18% passing #200 sieve, D₁₀ = 5 mm, D₃₀ = 12 mm, D₆₀ = 25 mm, LL = 32%, PI = 18% (clay fines). Classify using USCS.

Solution

Step 1: Determine if coarse or fine-grained. Percent passing #200 = 18% < 50%, so this is COARSE-GRAINED. Percent retained on #200 = 82% > 50%, confirming coarse-grained classification. Step 2: Determine if gravel or sand. Percent passing #4 = 45% < 50%, meaning more than 50% of the coarse fraction is retained on #4 (larger than 4.75 mm). This is GRAVEL. Step 3: Evaluate fines content. Percent passing #200 = 18% > 12%, so the soil is "dirty" (has significant fines). The gravel contains clay fines. Step 4: Determine gradation for gravel (Cu ≥ 4, Cc between 1 and 3). Cu = 25/5 = 5 Cc = (12)²/(5 × 25) = 144/125 = 1.152 Cu = 5 ≥ 4 ✓ 1 ≤ Cc = 1.152 ≤ 3 ✓ Both criteria satisfied → WELL-GRADED GRAVEL → GW designation. Step 5: Classify the fines using plasticity chart. A-line PI = 0.73(32 − 20) = 0.73(12) = 8.76 Soil PI = 18 > 8.76, and LL = 32 < 50. This plots above A-line, low-plasticity → clay fines. Step 6: Final Classification. Well-graded gravel with clay fines → **GC (or GW-GC if borderline)** Step 7: Engineering properties. This is a strong, stable material suitable for embankment cores in earth dams. The clay fines provide cohesion and reduce permeability relative to clean GW, while the well-graded gravel framework provides high shear strength and controlled drainage. This is ideal for dam construction in the Philippines where impermeability is desired in the core and good compaction is achievable.

Problem

EXAMPLE 4.4 — Complete Classification of Tropical Soil (High-PI Clay) Soil from manila site: 88% passing #200 sieve, LL = 65%, PL = 28%, no visible organic material. Classify using USCS and discuss engineering implications.

Solution

Step 1: Determine if coarse or fine-grained. Percent passing #200 = 88% > 50% → FINE-GRAINED. Step 2: Calculate plasticity index. PI = 65 − 28 = 37 Step 3: Position relative to LL = 50 line. LL = 65 > 50 → HIGH-PLASTICITY (H) side of chart. Step 4: Position relative to A-line. PI_A = 0.73(65 − 20) = 0.73(45) = 32.85 Soil PI = 37 > 32.85 → above A-line. Step 5: Classify. Above A-line AND LL ≥ 50 → **CH (high-plasticity clay)** Step 6: Engineering implications for Philippine practice. This is typical Maynila Clay from Metro Manila and other low-lying areas. Geotechnical properties: • Unconfined compression strength: 30–60 kPa (very soft to soft) • Permeability: 10⁻⁷ to 10⁻⁹ cm/s (nearly impermeable) • Compression index (Cc): 0.8–1.2 (highly compressible) • Preconsolidation pressure: Often less than overburden pressure (normally consolidated) Design considerations for structures: • **Shallow foundations:** Not recommended; bearing capacity is low (100–150 kPa maximum) • **Recommended solution:** Deep piles extending to underlying sand/rock layers (typical pile depth 20–40 m in Manila) • **Embankments:** May be used but requires preloading and long consolidation periods (settlement occurs over 1–5 years) • **Slope stability:** Critical concern during wet season; undrained strength is low (cu ≈ 20–40 kPa) • **Soil improvement:** Lime stabilization is economical; 2–4% lime by weight increases strength by 100–200% This example illustrates why Manila Bay soft clay deposits pose special design challenges in Philippine geotechnical engineering and require investigation in every project.

Key Points

  • USCS divides soils into coarse-grained (>50% retained on #200) and fine-grained (≥50% passing #200)
  • Coarse soils are subdivided by gravel (>50% larger than 4.75 mm) or sand (≤50% larger than 4.75 mm)
  • Coarse soils are further classified as W/P (well/poorly graded) and M/C (with silt/clay fines)
  • Fine soils are classified by position on plasticity chart: M (below A-line) or C (above A-line)
  • Fine soils are subdivided by LL = 50 threshold: L (low-plasticity) or H (high-plasticity)
  • Two-letter code = grain size category + second descriptor (gradation or plasticity/fines type)
  • Clean coarse soils have <5% passing #200; dirty soils have >12% (or >5-12% for borderline)
  • USCS requires sieve analysis data (D₁₀, D₃₀, D₆₀) and Atterberg limits for proper classification

The AASHTO (American Association of State Highway and Transportation Officials) classification system is specifically designed to evaluate soils for use in highway pavement construction and embankments. While USCS is the primary system for general geotechnical engineering, AASHTO is the standard for road and highway projects, including projects in the Philippines designed to Philippine road standards. The system rates soils from A-1 (best subgrade material) to A-8 (peat and organic soils, unsuitable). **AASHTO Classification Groups:** **A-1 (Excellent to Good):** - A-1-a: Well-graded gravel and sand with little or no fines - A-1-b: Well-graded sand with little or no fines **A-2 (Good to Fair):** Sandy soils with moderate fines - A-2-4: Sandy materials with silt - A-2-5: Sandy materials with high-plasticity silt - A-2-6: Sandy materials with clay - A-2-7: Sandy materials with high-plasticity clay **A-3 (Fair):** Poorly graded sand with little fines **A-4 (Fair to Poor):** Silt soils **A-5 (Poor):** High-plasticity silt **A-6 (Poor to Very Poor):** Clay soils **A-7 (Very Poor):** High-plasticity clay - A-7-5: PI ≤ LL − 30 - A-7-6: PI > LL − 30 **A-8 (Unsuitable):** Peat and highly organic soils **The Group Index (GI):** Within each AASHTO classification, the **Group Index** provides a numerical refinement. A higher GI indicates worse subgrade performance. The Group Index is calculated using the formula: **GI = (F − 35)[0.2 + 0.005(LL − 40)] + 0.01(F − 15)(PI − 10)** Where: • F = percentage of soil passing the #200 sieve (expressed as a whole number, 0–100) • LL = liquid limit • PI = plasticity index **Calculation Rules:** 1. All terms in brackets are independent; calculate each separately 2. If a calculated term is negative, use zero 3. Each bracketed term in the formula has a maximum value: - (F − 35) ≤ 40, so use 40 if F > 75 - (LL − 40) ≤ 20, so use 20 if LL > 60 - (F − 15) ≤ 40, so use 40 if F > 55 - (PI − 10) ≤ 20, so use 20 if PI > 30 4. Round the final GI to the nearest whole number (0.5 rounds up) 5. The minimum value is 0 (negative results are reported as 0) **Interpretation:** The Group Index modifies the AASHTO rating: - GI = 0: Best subgrade material in that AASHTO class - GI = 20: Worst subgrade material in that AASHTO class For example: A-4-3 means AASHTO A-4 (silt) with Group Index 3; A-6(12) means A-6 (clay) with Group Index 12. The higher the GI within a class, the more problematic the material for subgrade use. **Use in Pavement Design:** In pavement thickness design (per AASHTO pavement design methods), the Group Index directly affects the required thickness of asphalt and base courses. A soil with A-1-a(0) requires minimal pavement thickness, while A-7-6(18) requires substantially thicker pavements due to poor load-bearing capacity and high volume change potential. **Relationship to USCS:** While AASHTO and USCS measure different aspects, there is general correspondence: - AASHTO A-1: Corresponds to USCS GW, SW - AASHTO A-2: Corresponds to USCS GM, SM, GC, SC - AASHTO A-3: Corresponds to USCS SP - AASHTO A-4: Corresponds to USCS ML - AASHTO A-5: Corresponds to USCS MH - AASHTO A-6: Corresponds to USCS CL - AASHTO A-7: Corresponds to USCS CH A soil can have one USCS classification and one AASHTO classification; they address different engineering questions.

Heading

5. AASHTO Soil Classification System and Group Index

Examples

Problem

EXAMPLE 5.1 — Group Index Calculation (Silt Soil) A silt soil has the following properties: 68% passing #200 sieve, LL = 42%, PI = 15%. Calculate the Group Index and classify using AASHTO.

Solution

Step 1: Identify soil as AASHTO A-4 or A-5 (silt). With PI = 15, the soil has low plasticity for a silt, suggesting A-4. Step 2: Prepare values for GI calculation. F = 68, LL = 42, PI = 15 Step 3: Calculate each term of the GI formula. GI = (F−35)[0.2+0.005(LL−40)] + 0.01(F−15)(PI−10) First term: (F−35)[0.2+0.005(LL−40)] = (68−35)[0.2+0.005(42−40)] = (33)[0.2+0.005(2)] = (33)[0.2+0.01] = (33)[0.21] = 6.93 Second term: 0.01(F−15)(PI−10) = 0.01(68−15)(15−10) = 0.01(53)(5) = 0.01(265) = 2.65 Step 4: Sum and round. GI = 6.93 + 2.65 = 9.58 ≈ 10 (rounded to nearest whole number) Step 5: Final classification. **AASHTO A-4(10)** — silt with Group Index 10 Step 6: Engineering interpretation. GI = 10 indicates a moderately poor subgrade material. For pavement design, this silt would require: - Compaction control during construction (typically ≥95% Standard Proctor density) - Thicker asphalt and base courses compared to A-1 or A-2 soils - Possible subgrade stabilization with 3–5% cement or lime - Moisture control to prevent swelling/rutting in service This is typical of silts found in Philippine river valleys and floodplains.

Problem

EXAMPLE 5.2 — Group Index for High-Plasticity Clay A clay sample: 82% passing #200, LL = 62%, PI = 28%. Calculate GI and classify.

Solution

Step 1: Classify as AASHTO A-7 (high-plasticity clay). With LL = 62 > 50 and high PI, this is clearly A-7. Step 2: Determine if A-7-5 or A-7-6. Check if PI ≤ LL − 30: PI = 28, LL − 30 = 62 − 30 = 32 Since 28 ≤ 32, this is **A-7-5**. Step 3: Prepare values. F = 82, LL = 62, PI = 28 Step 4: Calculate GI terms (applying caps as needed). First term: (F−35)[0.2+0.005(LL−40)] Check caps: (F−35) = 82−35 = 47, but cap is 40, so use 40 (LL−40) = 62−40 = 22, but cap is 20, so use 20 = (40)[0.2+0.005(20)] = (40)[0.2+0.1] = (40)[0.3] = 12.0 Second term: 0.01(F−15)(PI−10) Check caps: (F−15) = 82−15 = 67, but cap is 40, so use 40 (PI−10) = 28−10 = 18, which is ≤ 20 ✓ = 0.01(40)(18) = 0.01(720) = 7.2 Step 5: Sum and round. GI = 12.0 + 7.2 = 19.2 ≈ 19 (rounded) Step 6: Final classification. **AASHTO A-7-5(19)** — high-plasticity clay, Group Index 19 Step 7: Engineering implications. GI = 19 is very poor (near the worst possible for this class). This soil is unsuitable as subgrade without major improvement: • Very high volume change potential (swell/shrink) • Very low bearing capacity • Problematic moisture sensitivity For highway construction, this material would: - Require complete removal or replacement, OR - Require stabilization with 6–10% cement or 4–6% lime - Require thick pavement structure (12–18 inches of asphalt + 12–18 inches of base) - Mandate moisture control measures and surface drainage This exemplifies the Maynila Clay challenge in Metro Manila road projects.

Problem

EXAMPLE 5.3 — Group Index for Well-Graded Gravel A gravel material: 12% passing #200, LL = 28%, PI = 8%. Determine AASHTO classification and GI.

Solution

Step 1: Classify soil type. With 12% passing #200 and low PI, this is likely A-1 or A-2 (good coarse material). QL = 0.73(28−20) = 0.73(8) = 5.84 PI = 8 > 5.84 (above A-line, slight clay tendency) → Could be A-2-6 (sandy materials with clay) However, with only 12% fines and LL = 28, this material likely classifies as **A-1-b** (well-graded sand) or **A-2-4** (sandy silt). For this example, assume the coarse fraction analysis confirms GW (well-graded gravel), making it **A-1-a**. Step 2: Calculate GI for A-1-a material. F = 12, LL = 28, PI = 8 Step 3: Calculate GI terms. First term: (F−35)[0.2+0.005(LL−40)] = (12−35)[0.2+0.005(28−40)] = (−23)[0.2+0.005(−12)] Since (F−35) < 0, this term = 0 (per rule) Second term: 0.01(F−15)(PI−10) = 0.01(12−15)(8−10) Since both (F−15) and (PI−10) are negative, this term = 0 Step 4: Calculate. GI = 0 + 0 = 0 Step 5: Final classification. **AASHTO A-1-a(0)** — well-graded gravel, Group Index 0 Step 6: Engineering significance. GI = 0 represents the best possible subgrade material. This gravel would: - Require minimal pavement thickness - Be suitable as-is (or minimally processed) for road base - Provide excellent bearing capacity (>200 kPa) - Require no stabilization - Be specified for premium road and airfield applications Note: The low GI reflects the favorable gradation (well-graded), low fines content, and low plasticity fines. This is ideal material in the Philippines for highway upgrades where high-quality aggregate is available.

Problem

EXAMPLE 5.4 — Borderline A-2 Classification with GI A sandy soil: 22% passing #200, LL = 34%, PI = 9%. Classify and calculate GI.

Solution

Step 1: Determine AASHTO group. With 22% passing #200 (between 15 and 35% range) and moderate PI, this is in the A-2 range. Check plasticity: A-line PI = 0.73(34−20) = 10.22 Soil PI = 9 < 10.22 (below A-line → silt) With 15 ≤ F ≤ 35 and PI < 10: **A-2-4 (sandy silt)** Step 2: Calculate GI. F = 22, LL = 34, PI = 9 First term: (F−35)[0.2+0.005(LL−40)] = (22−35)[0.2+0.005(34−40)] = (−13)[0.2+0.005(−6)] = (−13)[0.2−0.03] Since (F−35) < 0, this term = 0 Second term: 0.01(F−15)(PI−10) = 0.01(22−15)(9−10) = 0.01(7)(−1) Since (PI−10) < 0, this term = 0 GI = 0 + 0 = 0 Step 3: Final classification. **AASHTO A-2-4(0)** — sandy silt, Group Index 0 Step 4: Engineering implications. Despite being A-2-4 (not as good as A-1), the GI = 0 indicates good subgrade properties: - Low plasticity (PI = 9) provides stability - Moderate fines content (22%) aids compaction and moisture retention - Good bearing capacity (150–200 kPa) - Suitable for base course with minimal additional stone - No stabilization required for normal traffic This material is common in Philippine alluvial plains and is frequently used in road upgrades. The GI = 0 indicates it is acceptable for highway use without treatment, though comparison with A-1 and A-2-6 materials might favor those options if available.

Key Points

  • AASHTO system rates soils A-1 (best) to A-8 (peat, unsuitable) for highway subgrade use
  • Group Index (GI) provides numerical refinement within each AASHTO class; higher GI = worse performance
  • GI formula: GI = (F−35)[0.2+0.005(LL−40)] + 0.01(F−15)(PI−10); each bracket is independent
  • F = percent passing #200 sieve (0–100); LL and PI come from Atterberg limit tests
  • Calculate each bracketed term separately; if negative result, use zero; apply maximum caps
  • Round final GI to nearest whole number; minimum value is 0 (report GI ≥ 0)
  • GI caps: (F−35)≤40, (LL−40)≤20, (F−15)≤40, (PI−10)≤20
  • AASHTO classification is required for highway projects; directly influences pavement thickness design

Effective soil classification requires a systematic approach to laboratory testing and data interpretation. This section synthesizes the concepts from previous sections and provides a workflow that mirrors board examination questions. **Step 1: Obtain Basic Soil Information** Begin with: • Representative soil sample (minimum 200–500 g for sieve analysis) • Visual description (color, smell, texture, visible particles) • Site location and depth • Previous boring data if available **Step 2: Perform Preliminary Visual Classification** Before detailed testing, make preliminary observations: • **Grain size:** Rub between fingers; sand feels gritty, silt feels like flour, clay sticks to fingers • **Mica content:** Glints in sunlight indicate silt or fine sand • **Organic material:** Distinctive smell, dark color, fibrous texture suggest OL, OH, or Pt • **Coarse fraction:** Estimate gravel vs. sand visually **Step 3: Perform Sieve Analysis (Coarse-Grained Soils)** If preliminary assessment suggests coarse-grained soil (>50% retained on #200): 1. Dry the sample (oven at 105°C until constant mass) 2. Pass through standard sieves (#4, #10, #40, #200) 3. Record percent retained on each sieve 4. Calculate percent passing #200 (if >50%, soil is actually fine-grained) 5. Determine grain-size curve and read D₁₀, D₃₀, D₆₀ 6. Calculate Cu and Cc 7. Determine if gravel (>50% coarse > 4.75 mm) or sand 8. Classify as well-graded (W) or poorly graded (P) based on Cu and Cc **Step 4: Perform Atterberg Limit Tests (Fine-Grained Soils)** If soil is fine-grained (≥50% passing #200): 1. **Liquid Limit (Casagrande cup method, ASTM D4318):** - Mix soil with distilled water to paste consistency - Place in cup and create groove with standard tool - Drop cup repeatedly; count blows to close groove at 1/2 inch - Typically 25 blows = LL; make multiple determinations - Plot moisture content vs. log(number of blows); LL is at 25 blows 2. **Plastic Limit:** - Roll small soil ball on glass plate into 3 mm diameter thread - Thread crumbles when moisture approaches PL - Record moisture content at crumbling; PL is average of multiple trials 3. **Calculate PI:** - PI = LL − PL - If PL cannot be determined (non-plastic), report PI = 0 **Step 5: Classify Fine-Grained Soil Using Plasticity Chart** With LL and PI values: 1. Locate point (LL, PI) on chart 2. Determine if above or below A-line: PI = 0.73(LL − 20) 3. Determine if LL < 50 or LL ≥ 50 4. Assign USCS classification: - Above A-line, LL < 50 → CL - Above A-line, LL ≥ 50 → CH - Below A-line, LL < 50 → ML - Below A-line, LL ≥ 50 → MH 5. Check for organic material (visual inspection, LL reduction test) **Step 6: Determine Fines Type in Coarse-Grained Soils (If ≥5% Fines)** For coarse soils with 5–12% or >12% passing #200: 1. Perform Atterberg limits on the fines fraction (soaked for 10 minutes, allowed to settle) 2. Plot on plasticity chart to determine if silt (M) or clay (C) 3. Add second letter to coarse classification: - GM: gravel with silt - GC: gravel with clay - SM: sand with silt - SC: sand with clay **Step 7: Final USCS Classification** Assemble the two-letter code: - First letter: G (gravel), S (sand), M (silt), C (clay), O (organic), Pt (peat) - Second letter: W/P (well/poorly graded), M/C (silt/clay fines), L/H (low/high plasticity) **Step 8: Determine AASHTO Classification and Group Index** For highway projects: 1. Refer to AASHTO classification table (A-1 to A-8) based on grain size and plasticity 2. Calculate Group Index if LL and PI are available: - GI = (F−35)[0.2+0.005(LL−40)] + 0.01(F−15)(PI−10) - Apply all caps and floor at zero 3. Report as AASHTO A-X(GI) **Board Exam Strategy:** **For USCS Problems:** - Always check BOTH Cu and Cc criteria for well-graded classification - Plot on plasticity chart for fine soils; use A-line equation precisely - Watch for borderline cases (CL-ML); be prepared to justify - Distinguish organic soils by oven-drying LL test **For AASHTO Problems:** - Apply the GI formula mechanically; watch bracket caps - F is always a percentage (0–100), not a decimal - Round GI to nearest whole number; never report negative - Understand the engineering meaning: higher GI = worse subgrade **For Combined Problems:** - A soil can be GW-GC (USCS) AND A-1-a(2) (AASHTO) simultaneously - USCS addresses engineering design; AASHTO addresses highways - Show both classifications if both systems are mentioned **Common Board Exam Mistakes to Avoid:** 1. **Forgetting Cc in well-graded determination:** Cu ≥ 4/6 alone is insufficient 2. **Misreading D₁₀, D₃₀, D₆₀ from curve:** Use semilog graph correctly 3. **A-line slope error:** Always use 0.73(LL − 20), not approximations 4. **Ignoring fines in coarse soils:** 5% passing #200 changes the classification letter 5. **GI formula errors:** Calculate each term independently; apply caps 6. **Rounding GI incorrectly:** Round to nearest whole number, minimum 0 7. **Confusing LL and PI thresholds:** LL = 50 is the dividing line; PI = 30 is not a threshold in USCS 8. **Forgetting to classify organic material:** OL/OH classification changes with LL reduction test **Quick Reference: USCS Classification Decision Tree** • If >50% passing #200 → Fine-grained (M/C/O/Pt) - If below A-line → M (silt) - If above A-line → C (clay) - If LL < 50 → add L (low-plasticity) - If LL ≥ 50 → add H (high-plasticity) • If <50% passing #200 → Coarse-grained (G/S + W/P + M/C if fines) - If >50% coarse > 4.75 mm → G (gravel) - If ≤50% coarse > 4.75 mm → S (sand) - If Cu and Cc both satisfy criteria → W (well-graded) - Otherwise → P (poorly graded) - If >12% passing #200 → add fines type (M or C) **Practice Approach:** For each board-style problem: 1. Extract given data (sieve percentages, D₁₀, LL, PL, etc.) 2. Identify what system is being asked (USCS, AASHTO, or both) 3. Determine grain-size category first (coarse vs. fine) 4. Work through the decision criteria step-by-step 5. Write the classification clearly (e.g., "GW-GC" or "A-6(11)") 6. Provide brief engineering context if requested (e.g., "suitable for embankment core") 7. Double-check calculations; especially GI formula and curve reading

Heading

6. Practical Classification Workflow and Board Exam Strategy

Examples

Problem

EXAMPLE 6.1 — Complete Classification of Board-Exam Problem A soil sample from a borrow pit for road construction has the following properties: • 22% passing #200 sieve • D₁₀ = 0.3 mm, D₃₀ = 0.9 mm, D₆₀ = 3.0 mm • LL = 30%, PL = 20% • 8% passing #4 sieve Classify the soil using USCS and determine the AASHTO classification with Group Index. Discuss suitability for highway base course.

Solution

PART A: USCS CLASSIFICATION Step 1: Determine coarse vs. fine-grained. Percent passing #200 = 22% < 50% → COARSE-GRAINED Percent retained on #200 = 78% > 50% ✓ Step 2: Determine gravel vs. sand. Percent passing #4 = 8% means 8% of soil is finer than 4.75 mm Since this is <50% of total soil and we have a coarse-grained soil, more than 50% of the coarse fraction is >4.75 mm → GRAVEL (G) Step 3: Calculate gradation parameters. Cu = D₆₀/D₁₀ = 3.0/0.3 = 10 Cc = D₃₀²/(D₁₀ × D₆₀) = (0.9)²/(0.3 × 3.0) = 0.81/0.9 = 0.9 Step 4: Classify gradation. For gravel: Cu ≥ 4 AND 1 ≤ Cc ≤ 3 • Cu = 10 ≥ 4 ✓ • Cc = 0.9 < 1 ✗ Gradation is POORLY GRADED (P) → GP designation for coarse fraction. Step 5: Determine fines type (22% passing #200 > 12%, so significant fines). Calculate PI = LL − PL = 30 − 20 = 10 A-line PI at LL = 30: PI_A = 0.73(30 − 20) = 7.3 Soil PI = 10 > 7.3 → above A-line (clay, not silt) Step 6: Final USCS classification. Coarsely graded (GP) with clay (C) fines → **GP-GC (poorly graded gravel with clay)** PART B: AASHTO CLASSIFICATION Step 7: Assign to AASHTO group. With 22% passing #200 and clay fines (not silt), this is in the A-2 range. PI = 10, LL = 30 (low-plasticity clay) Since 15 ≤ F ≤ 35 and fines are clay: **A-2-6 (sandy materials with clay)** Step 8: Calculate Group Index. F = 22, LL = 30, PI = 10 First term: (F−35)[0.2+0.005(LL−40)] = (22−35)[0.2+0.005(30−40)] = (−13)[0.2+0.005(−10)] = (−13)[0.2−0.05] Since (F−35) < 0, this term = 0 Second term: 0.01(F−15)(PI−10) = 0.01(22−15)(10−10) = 0.01(7)(0) = 0 GI = 0 + 0 = 0 Step 9: Final AASHTO classification. **A-2-6(0) — Sandy materials with clay, Group Index 0** PART C: ENGINEERING DISCUSSION Suitability for Highway Base Course: USCS Classification GP-GC Implications: • Poorly graded gravel with clay indicates: variable particle sizing, less ideal compaction • Clay fines provide some cohesion but reduce drainage • Not ideal for drainage layer but acceptable for base course AASHTO Classification A-2-6(0) Implications: • A-2-6 is classified as "Good" subgrade material • Group Index 0 is the best in that class (no plasticity/fines penalty) • Suitable for direct use as highway base without treatment • Moderate bearing capacity; meets standard pavement design requirements Conclusion: • **Acceptable for highway base course** with standard compaction (≥98% Standard Proctor) • No soil stabilization (lime/cement) required • Ensure drainage with adequately thick base and subbase • Recommended application: intermediate base course layer (4–6 inches under asphaltic concrete) • Not recommended for drainage layers due to clay fines Note: This material represents a typical Philippine alluvial gravel with clay coating, common in Bulacan and Pangasinan road projects.

Problem

EXAMPLE 6.2 — Fine-Grained Soil Complete Classification A soil boring from Metro Manila shows: 92% passing #200 sieve, LL = 54%, PL = 24%, color dark gray, no visible organic fibers. Classify using USCS and AASHTO. What are the main design concerns for a 20-story building foundation in this soil?

Solution

PART A: USCS CLASSIFICATION Step 1: Grain-size classification. Percent passing #200 = 92% ≥ 50% → FINE-GRAINED Step 2: Calculate plasticity parameters. PI = LL − PL = 54 − 24 = 30 Step 3: Determine position on plasticity chart. LL = 54 > 50 (high-plasticity region) A-line PI at LL = 54: PI_A = 0.73(54 − 20) = 0.73(34) = 24.82 Soil PI = 30 > 24.82 (above A-line) Step 4: Check for organic content. No visible organic fibers or distinctive smell. Assuming inorganic material (confirm with oven-drying LL test if needed). Step 5: Classify. Above A-line AND LL ≥ 50 → **CH (high-plasticity clay)** PART B: AASHTO CLASSIFICATION Step 6: Assign AASHTO group. With 92% passing #200, LL = 54 (>50), and PI = 30 (high): Check A-7-5 vs. A-7-6: A-7-5 if PI ≤ LL − 30: PI = 30, LL − 30 = 54 − 30 = 24 Since 30 > 24, this is **A-7-6 (high-plasticity clay)** Step 7: Calculate Group Index. F = 92, LL = 54, PI = 30 Apply caps before calculation: (F−35) = 92−35 = 57, capped at 40 → use 40 (LL−40) = 54−40 = 14, which is ≤ 20 → use 14 (F−15) = 92−15 = 77, capped at 40 → use 40 (PI−10) = 30−10 = 20, which is ≤ 20 → use 20 First term: (40)[0.2+0.005(14)] = (40)[0.2+0.07] = (40)[0.27] = 10.8 Second term: 0.01(40)(20) = 0.01(800) = 8.0 GI = 10.8 + 8.0 = 18.8 ≈ 19 (rounded) Step 8: Final AASHTO classification. **A-7-6(19) — High-plasticity clay, Group Index 19** PART C: FOUNDATION DESIGN IMPLICATIONS Geotechnical Concerns for 20-Story Building: 1. **Bearing Capacity Issues:** • Undrained shear strength (cu): ~20–40 kPa (very soft clay) • Allowable bearing pressure for shallow foundation: 75–100 kPa maximum • A 20-story building exerts pressures of 800–1500 kPa → SHALLOW FOUNDATIONS NOT VIABLE 2. **Settlement Concerns:** • Compression index (Cc): 0.8–1.2 (highly compressible) • Expected settlement: 30–60 cm for shallow foundation (unacceptable) • Differential settlement between building corners: significant, causing damage 3. **Consolidation Time:** • Coefficient of consolidation (cv): 1–5 × 10⁻⁸ m²/s (very slow) • Time for 90% settlement: 10–50 years • Building would experience long-term settlement during occupancy 4. **Permeability and Drainage:** • Permeability: 10⁻⁷ to 10⁻⁹ cm/s (nearly impermeable) • Slow pore pressure dissipation • High pore pressures during construction (undrained conditions) • Stability concerns during excavation **Recommended Solution:** • **DEEP PILING (REQUIRED)** - Extend piles to underlying sand layer or firm clay layer at depth - Typical pile depth in Metro Manila: 30–50 m (reached at N > 30 SPT blows) - Pile type: Cast-in-place bored piles (drilled caisson) preferred for large loads - Pile design: Skin friction from clay (cu × Nc) + tip bearing in firm stratum • **GEOTECHNICAL INVESTIGATION:** - Boring to minimum 50 m depth to confirm underlying strata - Laboratory testing: UU and CU triaxial, consolidation, permeability - Field testing: CPT and SPT every 3–5 m to identify firm layer - Piezometer installation to monitor groundwater and pore pressures • **DESIGN CONSIDERATIONS:** - Estimate total pile settlement (elastic + consolidation): typically 5–10 cm - Design building to tolerate differential settlement - Monitor during construction using settlement plates - Consider preloading of fill before building construction to reduce later settlement This example illustrates why the Maynila Clay deposits present special challenges in Manila and why all major buildings in Metro Manila use deep pile foundations rather than shallow foundations.

Key Points

  • Systematic workflow: visual assessment → sieve analysis (if coarse) → Atterberg limits (if fine) → chart classification → AASHTO if needed
  • Sieve analysis provides percent passing #200 (determines coarse vs. fine), D₁₀/D₃₀/D₆₀, Cu, Cc, and gradation (W vs. P)
  • Atterberg limits (LL and PL) determine PI and position on plasticity chart for fine-grained soils
  • USCS two-letter code: grain size category + second descriptor (gradation or fines/plasticity type)
  • Plasticity chart: A-line (PI = 0.73(LL−20)) separates clay (above) from silt (below); LL = 50 separates L (low) from H (high) plasticity
  • AASHTO GI formula must account for caps: (F−35)≤40, (LL−40)≤20, (F−15)≤40, (PI−10)≤20; round to nearest whole number
  • Common exam errors: ignoring Cc in well-graded test, misreading curve diameters, wrong A-line calculation, forgetting fines classification, GI rounding mistakes
  • A soil can have both USCS and AASHTO classifications; they answer different engineering questions
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