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CELE Geotechnical Engineering — Soil ClassificationDetailed Explanation

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Detailed explanation of Soil Classification for the CELE 2026. Full depth, full reasoning — exactly what you need when Professional Regulation Commission (PRC) — Board of Civil Engineering tests this chapter with applied or scenario-based questions in the CELE Geotechnical Engineering subtest.

Exam context

Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Geotechnical Engineering section sits under a "Core" weighting, and Soil Classification is the 2nd chapter in the 11-chapter CELE Geotechnical Engineering rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Geotechnical Engineering.

Soil Classification - Detailed Explanation

Soil classification is one of the most consistently tested topics in the PRC Civil Engineer Licensure Examination under Geotechnical Engineering. Its purpose is straightforward: by grouping soils according to grain size and plasticity, engineers can quickly anticipate engineering behavior — drainage, compressibility, shear strength, and suitability as a subgrade or fill material — without running expensive laboratory tests every time. Two systems dominate both practice and the board exam: the Unified Soil Classification System (USCS), which is the international standard used in structural and geotechnical design, and the AASHTO Classification System, which is specifically tailored for highway subgrade evaluation. Mastery of this chapter requires fluency in three areas: (1) reading and interpreting grain-size distribution curves to compute gradation parameters, (2) applying Atterberg limits and the plasticity chart to fine-grained soils, and (3) assigning the correct USCS symbol and AASHTO group (with group index) to a soil given its laboratory data. This review chapter covers all three areas with board-style worked problems and exam-focused guidance.

Concepts

Grain-Size Distribution and Gradation Parameters

The grain-size distribution curve (also called the gradation curve or particle-size distribution curve) is a semi-log plot where the x-axis is particle diameter in millimetres (log scale) and the y-axis is the percentage finer by mass (linear scale, 0–100%). Sieve analysis is used for coarse-grained particles (> 0.075 mm, i.e., retained on the No. 200 sieve), while hydrometer analysis (based on Stokes' Law settling velocities) is used for fine-grained particles (< 0.075 mm). From the gradation curve, three characteristic diameters are read directly: D₁₀ (diameter at which 10% of the soil by mass is finer), D₃₀ (30% finer), and D₆₀ (60% finer). D₁₀ is called the effective size because it controls the permeability of a soil — Hazen's formula k ≈ C(D₁₀)² is built on it. From these three diameters, two dimensionless shape parameters are computed. The Coefficient of Uniformity (Cᵤ) measures the spread of particle sizes: Cᵤ = D₆₀ / D₁₀. A high Cᵤ means a wide range of particle sizes is present (well-graded); a low Cᵤ (close to 1.0) means nearly all particles are the same size (uniformly graded or gap-graded). The Coefficient of Gradation or Coefficient of Curvature (Cᶜ) checks whether the gradation curve has a smooth, concave shape: Cᶜ = (D₃₀)² / (D₁₀ × D₆₀). A well-graded soil satisfies BOTH of the following criteria simultaneously: for GRAVEL, Cᵤ ≥ 4 AND 1 ≤ Cᶜ ≤ 3; for SAND, Cᵤ ≥ 6 AND 1 ≤ Cᶜ ≤ 3. If either criterion fails, the soil is poorly graded (uniform or gap-graded). A gap-graded soil has a flat horizontal section on the gradation curve, indicating a missing range of particle sizes; it fails the Cᶜ criterion even if Cᵤ is large. Board examiners frequently test this by giving a soil with a high Cᵤ but a Cᶜ outside [1, 3] — the correct answer is still poorly graded.

Examples

This is a classic board-exam trap: the large Cᵤ of 10 suggests wide grading, but the Cᶜ of 0.90 (less than 1.0) reveals the gradation curve is skewed — there is a deficiency of mid-range particles. Since both criteria must be met simultaneously, the sand fails and is classified SP. The physical interpretation is a gap or skew in the particle-size distribution.

Scenario

A sand sample from a borrow pit has the following sieve analysis results: D₁₀ = 0.15 mm, D₃₀ = 0.45 mm, D₆₀ = 1.50 mm. Compute Cᵤ and Cᶜ, and determine whether the sand is well-graded or poorly graded.

Solution

Step 1 — Compute Cᵤ: Cᵤ = D₆₀ / D₁₀ = 1.50 / 0.15 = 10.0. Step 2 — Compute Cᶜ: Cᶜ = (D₃₀)² / (D₁₀ × D₆₀) = (0.45)² / (0.15 × 1.50) = 0.2025 / 0.225 = 0.90. Step 3 — Check well-graded criteria for sand: Cᵤ = 10.0 ≥ 6 ✓; Cᶜ = 0.90 — NOT in [1, 3] ✗. Conclusion: The sand is POORLY GRADED (SP in USCS) because the Cᶜ criterion is not satisfied, despite a large Cᵤ.

Both the uniformity and curvature criteria are met for gravel (threshold Cᵤ ≥ 4). The Cᶜ of 2.67 is within the acceptable range, confirming a smooth, well-distributed gradation curve. In the field, GW gravels are excellent subgrade and base course materials.

Scenario

A gravel sample has D₁₀ = 0.50 mm, D₃₀ = 4.0 mm, D₆₀ = 12.0 mm. Classify the gradation.

Solution

Cᵤ = D₆₀ / D₁₀ = 12.0 / 0.50 = 24.0. Cᶜ = (4.0)² / (0.50 × 12.0) = 16.0 / 6.0 = 2.67. Check for gravel: Cᵤ = 24.0 ≥ 4 ✓; Cᶜ = 2.67, which is in [1, 3] ✓. Both criteria satisfied → WELL-GRADED GRAVEL (GW in USCS).

Applications

Selecting filter materials for drainage blankets and retaining wall drains — gradation must satisfy filter criteria based on D₁₅ and D₈₅.
Classifying base course and subbase aggregates for road pavement design per DPWH standard specifications.
Estimating hydraulic conductivity of sands using Hazen's formula: k (cm/s) ≈ C(D₁₀)², where D₁₀ is in cm.
Evaluating liquefaction potential — uniformly graded, clean, fine to medium sands (low Cᵤ) are most susceptible.
Specifying compaction fill — well-graded soils achieve higher maximum dry density at optimum moisture content.

Misconceptions

MISCONCEPTION: A high Cᵤ alone means the soil is well-graded. CORRECTION: Both Cᵤ (≥ 4 for gravel, ≥ 6 for sand) AND 1 ≤ Cᶜ ≤ 3 must be satisfied simultaneously.
MISCONCEPTION: D₅₀ (median diameter) is the most important characteristic diameter. CORRECTION: D₁₀ (effective size) is the most important for permeability and classification purposes.
MISCONCEPTION: Uniformly graded and poorly graded mean the same thing. CORRECTION: Uniformly graded (all particles similar size, low Cᵤ) is a sub-type of poorly graded; gap-graded (missing intermediate sizes, abnormal Cᶜ) is another sub-type.
MISCONCEPTION: The Cᵤ threshold is the same for gravel and sand. CORRECTION: Gravel threshold is Cᵤ ≥ 4; sand threshold is Cᵤ ≥ 6.

Related Concepts

Atterberg Limits and Plasticity Index
USCS Classification (GW, GP, GW-GM, SW, SP, SM, SC)
Hazen's permeability formula
Sieve analysis and hydrometer test procedures
AASHTO Classification A-1 to A-3

Common Exam Questions

Example

Given D₁₀ = 0.20 mm, D₃₀ = 0.60 mm, D₆₀ = 2.00 mm, find Cᵤ and Cᶜ and classify the gradation. Answer: Cᵤ = 10, Cᶜ = 0.90 → Poorly graded sand (SP).

Approach

Read D₁₀, D₃₀, D₆₀ from the problem, apply Cᵤ = D₆₀/D₁₀ and Cᶜ = (D₃₀)²/(D₁₀×D₆₀), then check both thresholds for either gravel (Cᵤ ≥ 4) or sand (Cᵤ ≥ 6) AND Cᶜ ∈ [1, 3].

Question Type

Compute-and-classify (numerical)

Example

Which of the following best describes D₁₀ of a soil? (a) Median grain size (b) Effective size (c) Mean particle diameter (d) Maximum particle size — Answer: (b) Effective size.

Approach

Identify what each parameter measures: D₁₀ = effective size, Cᵤ = uniformity, Cᶜ = curvature/gradation. Know which parameter controls permeability.

Question Type

Concept identification (multiple choice)

Example

A gravel has Cᵤ = 20 and Cᶜ = 3.5. Is it well-graded? Answer: No — Cᶜ = 3.5 > 3, so it is poorly graded (GP).

Approach

Be alert when Cᵤ is large but Cᶜ is not given between 1 and 3, or vice versa. Both criteria must be met.

Question Type

Well-graded vs. poorly graded trap question

Key Points To Remember

Cᵤ = D₆₀ / D₁₀ — measures the spread of particle sizes.
Cᶜ = (D₃₀)² / (D₁₀ × D₆₀) — checks smoothness/curvature of the gradation curve.
Well-graded GRAVEL: Cᵤ ≥ 4 AND 1 ≤ Cᶜ ≤ 3 (GW in USCS).
Well-graded SAND: Cᵤ ≥ 6 AND 1 ≤ Cᶜ ≤ 3 (SW in USCS).
BOTH criteria must be satisfied for well-graded classification — Cᵤ alone is insufficient.
D₁₀ is the effective size; used in Hazen's permeability formula.
A gap-graded soil has a flat (horizontal) segment on the gradation curve.
Gradation curve x-axis is log scale; y-axis is percent finer (linear).

Atterberg Limits and the Plasticity Chart

Atterberg limits define the moisture contents at which a fine-grained soil transitions between different states of consistency. The Liquid Limit (LL) is the moisture content (%) at which the soil transitions from plastic to liquid behavior — determined by the Casagrande percussion cup test (25 blows to close a groove of standard dimensions) or the fall cone test. The Plastic Limit (PL) is the moisture content at which the soil transitions from plastic to semi-solid behavior — determined by rolling the soil into 3.2 mm (1/8-inch) diameter threads. The Shrinkage Limit (SL) is the moisture content below which further drying causes no more volume change. The Plasticity Index (PI) = LL − PL is the range of moisture contents over which the soil behaves plastically. A high PI indicates highly plastic (cohesive) behavior; PI = 0 means the soil is non-plastic (NP). The Liquidity Index (LI) = (w − PL) / PI describes the current state of a natural deposit relative to its limits; LI > 1 indicates the soil is near or above its liquid limit (sensitive, possibly quick clay). The Toughness Index (TI) = PI / Flow Index relates to the toughness at the plastic limit. The Activity (A) = PI / (% clay fraction < 0.002 mm) characterizes clay mineralogy: A < 0.75 inactive clay (kaolinite), 0.75 ≤ A ≤ 1.25 normal, A > 1.25 active clay (montmorillonite/smectite — highly expansive, as found in some Philippine black cotton soils). The CASAGRANDE PLASTICITY CHART plots PI (y-axis) vs. LL (x-axis). Two lines are critical: (1) The A-line: PI = 0.73(LL − 20) — separates clays (ABOVE the A-line, symbol C) from silts and organic soils (BELOW the A-line, symbol M or O). (2) The vertical line at LL = 50 — separates low plasticity (L, LL < 50) from high plasticity (H, LL ≥ 50). A hatched zone near the A-line (PI between 4 and 7 with LL between 25 and 35) is the borderline CL-ML zone. The U-line: PI = 0.9(LL − 8) is the upper boundary of all naturally occurring soils — no soil should plot above the U-line.

Examples

The point (LL = 45, PI = 23) lies above the A-line, confirming clay mineralogy dominates. Since LL < 50, it is low plasticity. CL clays are common in Philippine engineering practice as foundation soils; they have moderate compressibility and moderate shear strength.

Scenario

A fine-grained soil has LL = 45% and PL = 22%. Compute PI, plot the point on the plasticity chart, and determine the USCS classification symbol.

Solution

Step 1 — Compute PI: PI = LL − PL = 45 − 22 = 23%. Step 2 — Check A-line at LL = 45: PI_A = 0.73(45 − 20) = 0.73 × 25 = 18.25. Actual PI = 23 > 18.25 → ABOVE the A-line → clay (C). Step 3 — Check LL vs. 50: LL = 45 < 50 → low plasticity (L). Conclusion: CL (low-plasticity clay).

CH (fat clay) is a highly compressible, highly plastic clay. In Philippine practice, CH soils are problematic as subgrade — they swell when wet and shrink when dry. They are found in low-lying areas and some volcanic soil formations.

Scenario

A soil has LL = 65% and PL = 30%. Classify using the plasticity chart.

Solution

PI = 65 − 30 = 35%. A-line at LL = 65: PI_A = 0.73(65 − 20) = 0.73 × 45 = 32.85. Actual PI = 35 > 32.85 → above A-line → clay (C). LL = 65 > 50 → high plasticity (H). Classification: CH (high-plasticity clay / fat clay).

The point (LL = 40, PI = 10) falls below the A-line, indicating silt or organic characteristics. Without organic content information, it defaults to ML. To distinguish ML from OL, the LL is measured before and after oven-drying: if LL after oven-drying < 0.75 × LL before oven-drying, the soil is organic (OL or OH).

Scenario

A soil has LL = 40% and PL = 30%. Determine PI and classify.

Solution

PI = 40 − 30 = 10%. A-line at LL = 40: PI_A = 0.73(40 − 20) = 0.73 × 20 = 14.6. Actual PI = 10 < 14.6 → BELOW A-line → silt or organic (M or O). LL = 40 < 50 → low plasticity (L). Classification: ML (low-plasticity silt) — assuming non-organic based on oven-dry LL test.

Applications

Identifying expansive (swelling) soils for foundation design — high PI and high activity (A > 1.25) indicate potential swell problems common in Philippine black cotton soils.
Specifying compaction characteristics — PI affects optimum moisture content and maximum dry density in Proctor compaction test.
Determining suitability for use as fill, liner material (CH clays make excellent landfill liners due to low permeability).
Evaluating sensitivity of marine clays — LI near or above 1.0 in undisturbed samples indicates sensitive or quick-clay conditions.
Road pavement design — AASHTO classifies fine-grained subgrade soils based on LL and PI; high PI soils require stabilization.

Misconceptions

MISCONCEPTION: The A-line equation is PI = 0.73 × LL. CORRECTION: The correct equation is PI = 0.73(LL − 20). Forgetting the '−20' shift is the most common calculation error.
MISCONCEPTION: All soils below the A-line are silts. CORRECTION: Organic clays (OL, OH) also plot below the A-line; distinguishing requires the oven-drying test.
MISCONCEPTION: LL = 50 means high plasticity starts exactly at PI = 50. CORRECTION: LL = 50 is the LL threshold; PI can be any value — what matters is where the point (LL, PI) plots relative to the A-line AND the LL = 50 vertical line.
MISCONCEPTION: A non-plastic (NP) soil can still be classified as CL or ML. CORRECTION: NP soils (PI = 0 or not measurable) that are fine-grained go directly to ML or MH by default — they cannot be classified as clays.
MISCONCEPTION: The U-line is the A-line for classifying clays. CORRECTION: The A-line classifies C vs. M/O. The U-line is only an upper bound check — no naturally occurring soil should plot above it.

Related Concepts

USCS fine-grained classification (ML, MH, CL, CH, OL, OH, Pt)
Liquidity Index and sensitivity of clays
Activity and expansive soils (swell potential)
AASHTO A-4 to A-7 classification based on LL and PI
Compaction characteristics (optimum moisture content vs. PI correlation)

Common Exam Questions

Example

LL = 55, PL = 28. PI = 27. A-line at LL = 55: 0.73(35) = 25.55. PI = 27 > 25.55 → above A-line → clay. LL = 55 > 50 → high plasticity. Answer: CH.

Approach

Compute PI = LL − PL. Compute A-line PI at the given LL using 0.73(LL − 20). Compare actual PI to A-line PI. Check LL against 50.

Question Type

Compute PI and locate on plasticity chart

Example

A soil has LL_natural = 60%, LL_oven-dried = 40%. Check: 40 < 0.75 × 60 = 45 → organic. It also plots above the A-line with PI = 25 → OH (organic high-plasticity).

Approach

If LL (oven-dried) < 0.75 × LL (natural/air-dried), classify as organic (O) rather than inorganic (M or C).

Question Type

Organic vs. inorganic identification

Example

PI = 30, clay fraction = 20%. A = 30/20 = 1.5 → active clay, likely montmorillonite — high swell potential.

Approach

Compute A = PI / clay fraction (%). Interpret: A < 0.75 = inactive (kaolinite); 0.75–1.25 = normal (illite); A > 1.25 = active (montmorillonite).

Question Type

Activity calculation and clay mineral identification

Key Points To Remember

PI = LL − PL (range of plastic behavior).
LI = (w − PL) / PI (current moisture state of a natural soil).
A-line equation: PI = 0.73(LL − 20) — ABOVE = clay (C); BELOW = silt or organic (M or O).
LL = 50 divides low plasticity (L) from high plasticity (H).
U-line: PI = 0.9(LL − 8) — upper limit for all natural soils.
Activity A = PI / (% clay fraction) — identifies clay mineralogy and swell potential.
A-line slope is 0.73, not 0.9 (that is the U-line) — do not confuse the two.
PI < 4 plots below the A-line regardless of LL → silt-like behavior (ML or MH).
CL-ML borderline: PI between 4 and 7, LL between 25 and 35.

Unified Soil Classification System (USCS)

The USCS (originally developed by Casagrande in 1948, standardized as ASTM D2487) classifies a soil with a two-letter symbol. The FIRST letter identifies the dominant particle type or behavior; the SECOND letter qualifies the gradation or plasticity. Classification starts with a key question: Is more than 50% by mass retained on the No. 200 sieve (0.075 mm)? YES → COARSE-GRAINED. NO (i.e., ≥ 50% passing No. 200) → FINE-GRAINED. COARSE-GRAINED SOILS: Further divided by whether the coarse fraction is predominantly gravel or sand — checked at the No. 4 sieve (4.75 mm). If more than 50% of the coarse fraction is retained on No. 4 → GRAVEL (G); otherwise → SAND (S). Then check fines content (% passing No. 200): If fines < 5% → use gradation criteria (Cᵤ and Cᶜ): well-graded = GW or SW; poorly graded = GP or SP. If fines > 12% → classify fines using plasticity chart: if fines plot below A-line or PI < 4 → GM or SM (silty gravel/sand); if fines plot above A-line and PI > 7 → GC or SC (clayey gravel/sand). If fines are between 5% and 12% → borderline: dual symbol (e.g., GW-GM, SP-SC). FINE-GRAINED SOILS: Go directly to the plasticity chart: ML (low-plasticity silt), MH (high-plasticity silt), CL (low-plasticity clay), CH (high-plasticity clay), OL (organic low-plasticity), OH (organic high-plasticity), Pt (peat). SPECIAL CASES: A soil with ≥ 50% passing No. 200 and LL > 110% or strongly organic in appearance → Pt (peat). Borderline symbols are used when a soil is near a boundary; this is common in the field where soil types transition gradually. The USCS system is used in structural design, geotechnical reports, dam and embankment design, and is the system referenced in most Philippine geotechnical practice.

Examples

With > 12% fines that are clayey, the gravel is classified GC. GC gravels have reduced permeability compared to clean GW gravels and are susceptible to frost heave. They are still usable as subgrade but may require drainage provisions.

Scenario

A soil has 65% retained on No. 200 sieve. Of the coarse fraction, 60% is retained on No. 4 sieve. Fines (passing No. 200) = 35%. The fines have LL = 38, PI = 15. Classify using USCS.

Solution

Step 1 — Coarse or fine? 65% retained on No. 200 → COARSE. Step 2 — Gravel or sand? 60% of coarse retained on No. 4 → GRAVEL → first letter G. Step 3 — Fines = 35% > 12% → use plasticity chart. A-line at LL = 38: PI_A = 0.73(38−20) = 13.14. Actual PI = 15 > 13.14 → above A-line → clayey → second letter C. Classification: GC (Clayey Gravel).

CH is the most commonly tested fine-grained USCS class in Philippine board exams. It represents highly compressible, highly plastic soils that shrink and swell significantly with moisture change. They are the most problematic soils for highway subgrade and shallow foundations.

Scenario

A soil has 72% passing No. 200 sieve, LL = 55, PL = 28. Classify using USCS.

Solution

Step 1 — 72% passing No. 200 → FINE-GRAINED. Step 2 — Compute PI: PI = 55 − 28 = 27. Step 3 — A-line at LL = 55: PI_A = 0.73(55−20) = 25.55. PI = 27 > 25.55 → above A-line → clay (C). Step 4 — LL = 55 > 50 → high plasticity (H). Classification: CH (High-Plasticity Clay / Fat Clay).

Despite a satisfactory Cᵤ of 9.0, the Cᶜ of 0.694 fails the curvature criterion. SP sands are common in beach and river deposits in the Philippines. They are susceptible to liquefaction under earthquake loading — a major concern in highly seismic areas like Metro Manila and Mindanao.

Scenario

A sand has D₁₀ = 0.20 mm, D₃₀ = 0.50 mm, D₆₀ = 1.80 mm. Only 3% passes the No. 200 sieve. Classify using USCS.

Solution

Step 1 — 3% passing No. 200 → COARSE (97% retained). Step 2 — Assume majority is sand (given as 'sand'): first letter S. Step 3 — Fines < 5% → use gradation. Cᵤ = 1.80/0.20 = 9.0 ≥ 6 ✓. Cᶜ = (0.50)²/(0.20 × 1.80) = 0.25/0.36 = 0.694 — NOT in [1, 3] ✗. Both criteria not met → POORLY GRADED → second letter P. Classification: SP (Poorly Graded Sand).

Applications

Foundation design: USCS class guides selection of bearing capacity equations and settlement analyses.
Embankment and dam fill specification — USCS symbol specifies acceptable fill materials in DPWH and NIA specifications.
Retaining wall drainage design — GM and SM soils may clog gravel drains; GW and SW are preferred as drain backfill.
Liquefaction susceptibility assessment under NSCP 2015 Section 208 (Earthquake Design) — SP and SM sands below the water table are most vulnerable.
Geotechnical investigation reports per Philippine standard practice (ASTM D2487 adopted locally).

Misconceptions

MISCONCEPTION: The No. 200 sieve divides coarse from fine at exactly 50% retained. CORRECTION: The split is at > 50% retained on No. 200 → coarse; ≥ 50% passing → fine. The equals sign matters — exactly 50% passing goes to the fine-grained side.
MISCONCEPTION: In USCS, 'M' stands for 'mud.' CORRECTION: M stands for Mo (Swedish for silt) — it refers to silt-like plasticity behavior, not muddy appearance.
MISCONCEPTION: If a sand has < 5% fines, it is always classified SW. CORRECTION: < 5% fines means use gradation criteria (Cᵤ and Cᶜ). It is SW only if both well-graded criteria are satisfied; otherwise SP.
MISCONCEPTION: OL and OH are classified above the A-line. CORRECTION: Organic soils (OL and OH) typically plot BELOW the A-line — they are silt-like in plasticity. This is distinct from organic content causing them to be below the A-line versus inorganic soils of similar LL.
MISCONCEPTION: All clays are fine-grained in USCS. CORRECTION: A clayey gravel (GC) or clayey sand (SC) is classified as coarse-grained in USCS, even though the fines are clay-type — the major constituent is still coarse-grained.

Related Concepts

Grain-size distribution and gradation parameters (Cᵤ, Cᶜ)
Plasticity chart and Atterberg limits
AASHTO Classification System
Engineering properties by USCS class (permeability, compressibility, shear strength)
Seismic liquefaction potential (NSCP 2015 Section 208)

Common Exam Questions

Example

Soil: 55% passing No. 200, LL = 32, PL = 26. Fine-grained. PI = 6. A-line at LL = 32: 0.73(12) = 8.76. PI = 6 < 8.76 → below A-line → M. LL < 50 → L. Also PI = 6 is between 4 and 7 at LL = 32 → borderline CL-ML. Answer: CL-ML.

Approach

Always follow the flowchart: (1) No. 200 test → coarse or fine? (2) If coarse → No. 4 test → G or S? Fines content → W/P vs. M/C. (3) If fine → plasticity chart → M or C, L or H.

Question Type

Full USCS classification from given data

Example

Which USCS soil class poses the greatest risk of liquefaction under seismic loading? Answer: SP (loosely deposited, poorly graded clean sand below water table).

Approach

Know the engineering behavior: CH = high compressibility + swell; MH = dilatant, high swell; SP = liquefaction; Pt = very high compressibility, not usable.

Question Type

Identify the engineering problem associated with a USCS class

Example

A gravel with 8% fines (below A-line) and Cᵤ = 15, Cᶜ = 2.1 → USCS: GW-GM.

Approach

When fines content is between 5% and 12%, assign a dual symbol combining the gradation-based and fines-based classifications, e.g., GW-GM.

Question Type

Dual symbol assignment

Key Points To Remember

No. 200 sieve (0.075 mm): splits coarse (> 50% retained) from fine (≥ 50% passing).
No. 4 sieve (4.75 mm): within coarse-grained, splits gravel from sand.
< 5% fines → use Cᵤ and Cᶜ for W or P designation.
> 12% fines → use plasticity chart for M or C designation.
5–12% fines → dual symbol (e.g., SW-SM, GW-GC).
Fine-grained: use plasticity chart + oven-drying test (for organic).
Second letters: W = well-graded, P = poorly graded, M = silty, C = clayey, L = low plasticity, H = high plasticity.
Pt = peat (highly organic, very high LL, not suitable for any structural use).
GW and SW are the best engineering soils; CH, MH, and Pt are the most problematic.

AASHTO Classification System and Group Index

The AASHTO (American Association of State Highway and Transportation Officials) system classifies soils specifically for their suitability as highway subgrade material. It divides soils into eight groups (A-1 to A-8) with A-1 being the best subgrade and A-7 (or A-8 for peat) being the worst. Group A-1 and A-3 are granular materials (≤ 35% passing No. 200); Groups A-4 through A-7 are silty-clay materials (> 35% passing No. 200). GROUP DESCRIPTIONS: A-1-a: well-graded gravel or gravel-sand mix, ≤ 50% passing No. 40, ≤ 15% passing No. 200, NP or PI ≤ 6. A-1-b: gravel, coarse sand, PI ≤ 6. A-2: borderline granular-silty-clay (25–35% passing No. 200), subdivided into A-2-4, A-2-5, A-2-6, A-2-7 based on LL and PI. A-3: fine sand, ≤ 10% passing No. 200, NP. A-4 to A-7: predominantly silt-clay soils. A-4 and A-5: primarily silty soils. A-6 and A-7: primarily clayey soils. A-7 is split into A-7-5 (PI ≤ LL − 30) and A-7-6 (PI > LL − 30; more plastic, more swell). A-8: peat, highly organic, not used as subgrade. CLASSIFICATION PROCEDURE: Use the AASHTO Classification Table in order from A-1 to A-7, assigning the first group that satisfies all criteria (from left to right in the table). THE GROUP INDEX (GI): Within A-4 to A-7 (and A-2-6, A-2-7), the group index refines the subgrade rating. GI = (F − 35)[0.2 + 0.005(LL − 40)] + 0.01(F − 15)(PI − 10). Where: F = percent of soil passing No. 200 sieve (expressed as a whole number, e.g., 60 for 60%), LL = liquid limit, PI = plasticity index. CAPS on partial terms to prevent negative contributions: (F − 35) is capped at 40 max; (LL − 40) is capped at 20 max and floored at 0; (F − 15) is capped at 40 max; (PI − 10) is capped at 20 max and floored at 0. RESULT: GI is rounded to the nearest whole number and floored at 0 (never negative). A GI of 0 is excellent subgrade; GI of 20 is very poor. The AASHTO group is written as, for example, A-6(11) where 11 is the GI.

Examples

With F = 60 > 35%, LL = 45%, and PI = 23%, this soil is in the A-4 to A-7 range with a GI of 11 — a moderately poor subgrade. The GI of 11 means it requires significant thickness of base and surface courses to perform adequately as a highway subgrade. In DPWH road design practice, subgrades with high GI require stabilization (lime, cement) before pavement construction.

Scenario

A soil has the following data: % passing No. 200 (F) = 60%, LL = 45%, PI = 23%. Compute the Group Index.

Solution

Step 1 — Identify variables: F = 60, LL = 45, PI = 23. Step 2 — First partial product: (F − 35) = 60 − 35 = 25 (< 40, no cap). [0.2 + 0.005(LL − 40)] = [0.2 + 0.005(45 − 40)] = [0.2 + 0.005 × 5] = [0.2 + 0.025] = 0.225. First term = 25 × 0.225 = 5.625. Step 3 — Second partial product: 0.01(F − 15)(PI − 10) = 0.01 × (60 − 15) × (23 − 10) = 0.01 × 45 × 13 = 5.85. Step 4 — GI = 5.625 + 5.85 = 11.475 ≈ 11. Answer: GI = 11.

This exercise illustrates the cap on (F − 35) and (F − 15). Without capping, the result would be 0.01 × 67 × 18 = 12.06 for the second term — significantly different. A GI of 17 indicates a very poor subgrade — typical of soft, highly plastic clays found in coastal and deltaic areas of the Philippines (e.g., Manila Bay reclamation areas).

Scenario

A soil has F = 82%, LL = 50%, PI = 28%. Compute the Group Index.

Solution

Step 1 — F = 82, LL = 50, PI = 28. Step 2 — First term: (F − 35) = 82 − 35 = 47; capped at 40. (LL − 40) = 50 − 40 = 10; within [0, 20], no cap. 40 × [0.2 + 0.005 × 10] = 40 × [0.2 + 0.05] = 40 × 0.25 = 10.0. Step 3 — Second term: (F − 15) = 82 − 15 = 67; capped at 40. (PI − 10) = 28 − 10 = 18; within [0, 20], no cap. 0.01 × 40 × 18 = 7.2. Step 4 — GI = 10.0 + 7.2 = 17.2 ≈ 17. Answer: GI = 17.

A-2-4 represents borderline granular soils with low plasticity. GI is defined as 0 for A-1, A-2-4, A-2-5, and A-3. For A-2-6 and A-2-7, a partial GI formula is used (only the second term applies). This soil, while not ideal, is usable as subgrade with adequate compaction.

Scenario

A soil has F = 25%, LL = 28%, PI = 8%. What is the AASHTO group?

Solution

Step 1 — F = 25% ≤ 35% → possibly granular group (A-1, A-3, or A-2). Step 2 — Check A-1-a: % passing No. 40 and No. 200 needed; with PI = 8 > 6, A-1 is ruled out. Check A-3: PI should be NP → not A-3. Check A-2 series: F = 25% (between 25–35%), LL = 28 < 40, PI = 8 < 10 → A-2-4. Step 3 — GI for A-2-4 = 0 (by definition). Answer: A-2-4(0).

Applications

DPWH highway and road design — subgrade classification determines required pavement thickness per the AASHTO pavement design guide.
Evaluation of borrow materials for road embankments — A-1 and A-2 soils preferred; A-6 and A-7 require evaluation of swell potential.
Specification of subgrade improvement — soils with high GI (> 10) typically require lime or cement stabilization.
Bridge approach fill design — AASHTO classification helps specify acceptable fill to minimize settlement and approach slab problems.
Environmental impact assessment for road projects — identifying A-7 and A-8 soils to avoid or remediate.

Misconceptions

MISCONCEPTION: F in the GI formula should be expressed as a decimal (e.g., 0.60). CORRECTION: F is expressed as a percentage number (e.g., 60 for 60%). Using 0.60 gives a completely wrong (near-zero) GI.
MISCONCEPTION: GI can be negative when terms are small. CORRECTION: GI has a floor of zero — if the formula gives a negative result, report GI = 0.
MISCONCEPTION: The AASHTO classification goes from best to worst in alphabetical order: A-1 through A-7. CORRECTION: The numerical order A-1 to A-7 is indeed from best to worst subgrade, but A-3 (fine sand) is actually a better subgrade than A-2-6 or A-2-7, despite its lower number.
MISCONCEPTION: AASHTO and USCS always give consistent ratings — a GW in USCS is always A-1 in AASHTO. CORRECTION: The two systems use different criteria and can give different ratings. A soil can be GW in USCS but still A-2-4 in AASHTO depending on fines content and plasticity.
MISCONCEPTION: The GI cap values apply to the entire formula, not to individual terms. CORRECTION: The caps apply to each partial term separately: (F−35) is individually capped at 40, (F−15) is individually capped at 40, (LL−40) has a range of [0, 20], and (PI−10) has a range of [0, 20].

Related Concepts

USCS Classification (for comparison and correlation)
Highway pavement design (AASHTO design method, CBR correlation)
Subgrade stabilization with lime and cement
California Bearing Ratio (CBR) — correlation with AASHTO groups
DPWH standard specifications for road and bridge construction

Common Exam Questions

Example

F = 70%, LL = 55%, PI = 30%. (70−35)=35 no cap; [0.2+0.005×15]=0.275; 35×0.275=9.625. 0.01×(70−15)×(30−10)=0.01×55×20=11.0 (PI−10=20, capped). GI=9.625+11.0=20.625≈21. But GI is capped at 20 by convention for A-7 → GI=20.

Approach

Plug F, LL, PI into GI = (F−35)[0.2+0.005(LL−40)] + 0.01(F−15)(PI−10). Apply caps carefully: (F−35) max 40, (LL−40) in [0,20], (F−15) max 40, (PI−10) in [0,20]. Round to nearest integer, floor at 0.

Question Type

Compute Group Index (most common numerical question)

Example

Soil A: GI=5; Soil B: GI=12. Soil B is the poorer subgrade and requires greater pavement thickness.

Approach

Compute GI for both soils. Higher GI = poorer subgrade. Identify which soil requires more pavement thickness or more improvement.

Question Type

Compare AASHTO group index for two soils

Example

LL = 50, PI = 25. Check: LL − 30 = 20. PI = 25 > 20 → A-7-6 (more plastic, more expansive).

Approach

Both are A-7 soils (> 35% fines, LL > 40 for A-7-5 or LL > 40 for A-7-6, PI > 10). Differentiator: A-7-5 has PI ≤ LL − 30; A-7-6 has PI > LL − 30.

Question Type

A-7-5 vs. A-7-6 differentiation

Key Points To Remember

AASHTO rates subgrade quality: A-1 (best) → A-7 (worst); A-8 = peat (unusable).
GI formula: (F − 35)[0.2 + 0.005(LL − 40)] + 0.01(F − 15)(PI − 10).
F = percent passing No. 200 (as a whole number, not a decimal).
Partial terms are capped: (F−35) max 40; (LL−40) range [0, 20]; (F−15) max 40; (PI−10) range [0, 20].
GI is always ≥ 0 (floor at zero); round to nearest integer.
A-7-5 vs. A-7-6: A-7-6 has PI > LL − 30 (more expansive/problematic).
A-3 is fine sand, non-plastic — classified BEFORE A-2 in the classification table despite being a 'worse' subgrade.
GI = 0 for A-1, A-2-4, A-2-5, and A-3 groups.
AASHTO classification always proceeds left-to-right in the standard table — assign the FIRST group whose criteria are all met.

Practice Problems

Both gradation criteria fail — Cᵤ is slightly below the threshold of 6 for sand, and Cᶜ is below 1.0, indicating the gradation curve is steep (most particles are nearly the same size). In the field, SP sands are loose, drain freely, but have low bearing capacity in loose state and are susceptible to liquefaction.

Problem

PROBLEM 1 (Gradation Parameters): A sieve analysis of a sand yields the following: D₁₀ = 0.18 mm, D₃₀ = 0.38 mm, D₆₀ = 0.95 mm. (a) Compute Cᵤ and Cᶜ. (b) Is the sand well-graded or poorly graded? (c) What is the USCS symbol, assuming fines content < 5%?

Solution

(a) Cᵤ = D₆₀ / D₁₀ = 0.95 / 0.18 = 5.28. Cᶜ = (D₃₀)² / (D₁₀ × D₆₀) = (0.38)² / (0.18 × 0.95) = 0.1444 / 0.171 = 0.845. (b) For sand: need Cᵤ ≥ 6 AND 1 ≤ Cᶜ ≤ 3. Cᵤ = 5.28 < 6 ✗ (fails). Cᶜ = 0.845 < 1 ✗ (also fails). Both criteria fail → POORLY GRADED. (c) USCS Symbol: SP (Poorly Graded Sand).

This soil represents the typical highly plastic marine clay found in Manila Bay reclamation and other coastal areas of the Philippines. CH clays typically have very low CBR values (< 3%), high compressibility indices (Cc), and require preloading or deep foundations to manage settlements. Under NSCP 2015, Section 304, site investigation must characterize such soils before foundation design.

Problem

PROBLEM 2 (Plasticity Chart & USCS): A fine-grained soil from a Manila Bay reclamation area has LL = 72% and PL = 31%. (a) Compute PI. (b) Locate the point on the plasticity chart and determine the USCS symbol. (c) What engineering problem is most associated with this classification?

Solution

(a) PI = LL − PL = 72 − 31 = 41%. (b) A-line at LL = 72: PI_A = 0.73(72 − 20) = 0.73 × 52 = 37.96. Actual PI = 41 > 37.96 → above A-line → clay (C). LL = 72 > 50 → high plasticity (H). USCS: CH (High-Plasticity Clay / Fat Clay). (c) CH soils are highly compressible and exhibit significant volume change (swell and shrinkage) with moisture variation. They are problematic for foundations, pavements, and retaining structures due to high settlement and differential movement.

The high fines content (75%), moderate LL (48%), and significant PI (26%) combine to give a GI of 16 — placing this soil in the very poor subgrade category. In DPWH practice, such soils typically require at least 150–200 mm of lime-treated subgrade before the base course. The AASHTO classification would likely be A-6 or A-7-6.

Problem

PROBLEM 3 (AASHTO Group Index): A highway subgrade soil has the following properties: % passing No. 200 sieve = 75%, Liquid Limit = 48%, Plasticity Index = 26%. (a) Compute the Group Index. (b) Rate the subgrade quality.

Solution

(a) F = 75, LL = 48, PI = 26. Term 1: (F−35) = 75−35 = 40 (exactly at cap, use 40). (LL−40) = 48−40 = 8; within [0,20]. First partial = 40 × [0.2 + 0.005 × 8] = 40 × [0.2 + 0.04] = 40 × 0.24 = 9.60. Term 2: (F−15) = 75−15 = 60; capped at 40. (PI−10) = 26−10 = 16; within [0,20]. Second partial = 0.01 × 40 × 16 = 6.40. GI = 9.60 + 6.40 = 16.0 ≈ 16. (b) GI = 16 → Very Poor subgrade. This soil requires significant pavement thickness or subgrade stabilization (lime treatment) before it can function as highway subgrade.

With 8% fines (between 5% and 12%), a dual symbol is required by USCS. The gradation indicates well-graded (SW criteria met), while the fines plasticity indicates clayey (SC criteria met). The dual symbol SW-SC is written with the gradation-based symbol first. Soils with dual symbols behave intermediate between the two end members.

Problem

PROBLEM 4 (Complete USCS Classification): A soil sample is analyzed with the following results: % passing No. 200 sieve = 8%, % of coarse fraction passing No. 4 sieve = 65% (i.e., most of the coarse fraction is sand-sized), D₁₀ = 0.10 mm, D₃₀ = 0.35 mm, D₆₀ = 1.20 mm. The fines have LL = 30 and PI = 12. Determine the USCS symbol.

Solution

Step 1 — 8% passing No. 200 → COARSE-GRAINED. Step 2 — 65% of coarse fraction passes No. 4 (sand-sized) → more than 50% of coarse fraction is sand → first letter S. Step 3 — Fines content = 8%, which is between 5% and 12% → BORDERLINE, use dual symbol. Step 4 — Gradation check: Cᵤ = 1.20/0.10 = 12.0 ≥ 6 ✓; Cᶜ = (0.35)²/(0.10×1.20) = 0.1225/0.12 = 1.021, in [1,3] ✓ → Well-graded → SW. Step 5 — Plasticity check: A-line at LL = 30: PI_A = 0.73(30−20) = 7.3. PI = 12 > 7.3 → above A-line → clayey → SC. Step 6 — Dual symbol: SW-SC.

A-7-6 is the most expansive and most problematic AASHTO soil class. The GI of 18 (out of a practical maximum of 20) indicates a very poor subgrade that will undergo significant volume change, rutting, and pumping under traffic loading. Philippine highway projects encountering A-7-6 soils typically require complete subgrade replacement or deep lime stabilization.

Problem

PROBLEM 5 (A-7-5 vs. A-7-6 + GI): A soil has F = 90%, LL = 55%, PL = 28%. (a) Compute PI. (b) Determine if this is A-7-5 or A-7-6. (c) Compute the Group Index.

Solution

(a) PI = LL − PL = 55 − 28 = 27%. (b) F = 90% > 35%; LL = 55 > 40%; PI = 27 > 10% → A-7 group. Check A-7-5 vs. A-7-6: LL − 30 = 55 − 30 = 25. PI = 27 > 25 → A-7-6. (c) GI computation: (F−35) = 90−35 = 55; capped at 40. (LL−40) = 55−40 = 15; in [0,20]. First term = 40×[0.2+0.005×15] = 40×0.275 = 11.0. (F−15) = 90−15 = 75; capped at 40. (PI−10) = 27−10 = 17; in [0,20]. Second term = 0.01×40×17 = 6.8. GI = 11.0 + 6.8 = 17.8 ≈ 18. Full designation: A-7-6(18).

Exam Preparation Tips

MEMORIZE THE A-LINE EQUATION EXACTLY: PI = 0.73(LL − 20). The most common error is using PI = 0.73 × LL. The '−20' shift is non-negotiable — without it, all plasticity chart placements are wrong.
NEVER FORGET THE CᶜCRITERION: In every 'well-graded' question, compute BOTH Cᵤ and Cᶜ. The board exam frequently presents a soil with a large Cᵤ but Cᶜ outside [1, 3] — the correct answer is still poorly graded (SP or GP).
FOR THE GI FORMULA, EXPRESS F AS A PERCENTAGE (e.g., 60, not 0.60). Using the decimal form will give a near-zero GI, which is clearly wrong for a fine-grained soil.
APPLY CAPS ON GI PARTIAL TERMS BEFORE MULTIPLYING: (F−35) cap 40; (LL−40) range [0,20]; (F−15) cap 40; (PI−10) range [0,20]. Practice with F > 75% to build habit of checking caps.
MASTER THE USCS FLOWCHART: Practice the decision sequence — No. 200 test first → if coarse, No. 4 test → fines percentage → gradation OR plasticity check. Draw this flowchart on scratch paper at the start of the exam.
KNOW THE SIEVE NUMBERS: No. 200 = 0.075 mm, No. 4 = 4.75 mm, No. 40 = 0.425 mm. These are frequently used without conversion in board problems.
PLASTICITY CHART COORDINATES ARE (LL, PI) — not (PI, LL). LL is always on the x-axis. Do not reverse these when checking the A-line.
FOR AASHTO A-7-5 vs. A-7-6: Compute LL − 30. If PI > LL − 30 → A-7-6 (more expansive, worse). If PI ≤ LL − 30 → A-7-5.
DUAL SYMBOLS IN USCS: When fines content is between 5% and 12%, assign a dual symbol (e.g., SW-SM). Write the gradation-based symbol first (SW, SP) and the fines-based symbol second (SM, SC).
ORGANIC SOIL IDENTIFICATION: If LL after oven-drying < 0.75 × LL before oven-drying → organic soil (OL if LL < 50, OH if LL ≥ 50). This is the definitive test for Pt, OL, OH classification.
GI IS ALWAYS ≥ 0: If calculation gives a negative number (possible when F < 35 and LL < 40 and PI < 10), report GI = 0.
REVIEW THE STANDARD USCS TABLE AND AASHTO TABLE: The board exam may provide these or may require recall. Familiarity with both tables shortens classification time significantly.
CORRELATE THE TWO SYSTEMS FOR CROSS-CHECKING: CH (USCS) typically corresponds to A-7-6 (AASHTO). CL corresponds to A-6. ML corresponds to A-4 or A-5. This correlation can help verify answers.
USE SIGNIFICANT FIGURES APPROPRIATELY: GI is reported as a whole number; Cᵤ and Cᶜ to two or three significant figures; PI to the nearest whole percent.

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In summary

Soil classification is a foundational skill that underpins almost every area of geotechnical engineering practice in the Philippines. From evaluating the bearing capacity of a building foundation in Manila's soft clay to designing the subgrade of a DPWH highway through Mindanao's lateritic soils, the ability to classify soils quickly and accurately from grain-size and plasticity data is non-negotiable. In the PRC board examination, classification questions consistently appear in the Geotechnical Engineering component, often combining numerical computation (gradation parameters, PI, GI) with conceptual interpretation (well-graded vs. poorly graded, clay vs. silt, A-7-5 vs. A-7-6). The most critical habits to build are: (1) always checking BOTH Cᵤ and Cᶜ for well-graded classification — neither criterion alone is sufficient; (2) using the exact A-line equation PI = 0.73(LL − 20) with the '−20' shift; (3) expressing F as a percentage number (not a decimal) in the GI formula and applying all four caps correctly; (4) following the USCS classification flowchart systematically — starting with the No. 200 sieve test and proceeding logically. With consistent practice on the five standard problem types (gradation parameters, plasticity chart classification, USCS full classification, AASHTO group index, and A-7-5/A-7-6 differentiation), a reviewee can confidently answer any classification question the board presents. Review the worked examples in this chapter until the procedure is automatic, and use the visual flowcharts as quick references during timed practice. Good luck on your board exam — the Philippine engineering profession needs well-prepared, analytically sharp civil engineers.

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