CELE Geotechnical Engineering — Soil ClassificationExam Answer Templates

Marks

1

Topic

Grain-Size Distribution Parameters

Difficulty

easy

Template Id

T1

Examiner Tip

One-mark questions require the formula in full — a vague verbal description without the equation will not earn the mark.

Model Answer

The coefficient of uniformity (Cu) is a dimensionless parameter that measures the spread of particle sizes in a soil. It is defined as: Cu = D60 / D10 where D60 is the diameter at which 60% of the soil (by mass) is finer, and D10 is the effective size (diameter at which 10% is finer).

Question Type

very_short_answer

Answer Structure

• Line 1: State the name and what it measures (1 mark — definition + formula)

Scoring Breakdown

Common Mark Deductions

• Writing Cu = D10/D60 (inverted ratio) — zero marks
• Stating the formula without identifying what D10 and D60 represent
• Confusing Cu with Cc

Key Phrases To Include

• Cu = D60 / D10
• D60
• D10
• effective size
• coefficient of uniformity

Marks

1

Topic

USCS Classification — Coarse-Grained Soils

Difficulty

easy

Template Id

T2

Examiner Tip

The board frequently tests whether examinees know the different Cu thresholds for gravel (≥4) versus sand (≥6). Always specify which soil type you are referring to.

Model Answer

For a sand to be classified as well-graded (SW) under the USCS, it must simultaneously satisfy both of the following: 1. Cu ≥ 6 2. 1 ≤ Cc ≤ 3 If either criterion is not met, the sand is classified as poorly graded (SP).

Question Type

very_short_answer

Answer Structure

• Line 1: State Cu criterion for sand (Cu ≥ 6)
• Line 2: State Cc criterion (1 ≤ Cc ≤ 3)
• Line 3 (optional but good): Consequence if not met → SP

Scoring Breakdown

Common Mark Deductions

• Stating Cu ≥ 4 (that is the gravel criterion, not sand)
• Forgetting the Cc criterion entirely
• Not specifying that BOTH must be satisfied simultaneously

Key Phrases To Include

• Cu ≥ 6
• 1 ≤ Cc ≤ 3
• well-graded
• SW
• USCS
• both criteria

Marks

2

Topic

Atterberg Limits and Plasticity

Difficulty

easy

Template Id

T3

Examiner Tip

For differentiation questions, present definitions side-by-side or in clear labeled sections. Examiners reward structured contrast over a paragraph that mixes both concepts.

Model Answer

Liquid Limit (LL): The moisture content (%) at which a soil transitions from the plastic state to the liquid (flow) state. It marks the upper boundary of the plastic range and is determined by the Casagrande cup test. Plasticity Index (PI): The range of moisture content over which a soil behaves plastically. It is calculated as: PI = LL − PL where PL is the plastic limit. PI quantifies the plasticity of a soil; a higher PI indicates a more plastic (clay-like) soil.

Question Type

short_answer

Answer Structure

• Line 1–2: Define LL with its physical meaning and test method [1 mark]
• Line 3–4: Define PI with formula PI = LL − PL and its engineering significance [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining PI without the formula PI = LL − PL
• Confusing LL with PL in the definition
• Not mentioning what a high PI indicates (more plastic/clay-like)

Key Phrases To Include

• PI = LL − PL
• plastic limit
• liquid limit
• plastic state
• liquid state
• Casagrande cup
• plasticity range

Marks

2

Topic

Atterberg Limits — USCS Plasticity Chart

Difficulty

medium

Template Id

T4

Examiner Tip

Always compute the A-line PI explicitly and compare numerically. Board examiners want to see the decision logic, not just the final symbol.

Model Answer

Given: LL = 38, PL = 20 Step 1 — Compute PI: PI = LL − PL = 38 − 20 = 18 Step 2 — Evaluate the A-line at LL = 38: PI_A = 0.73(LL − 20) = 0.73(38 − 20) = 0.73 × 18 = 13.14 Step 3 — Compare: PI = 18 > PI_A = 13.14 → soil plots ABOVE the A-line LL = 38 < 50 → LOW plasticity (suffix L) Above A-line, LL < 50 → CLAY of Low Plasticity USCS Classification: CL

Question Type

numerical

Answer Structure

• Step 1: Compute PI = LL − PL [0.5 mark]
• Step 2: Compute A-line PI at given LL using PI_A = 0.73(LL − 20) [0.5 mark]
• Step 3: Compare PI vs PI_A and state position (above/below A-line) [0.5 mark]
• Step 4: Apply LL < 50 rule and state final USCS symbol CL [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Not showing the A-line computation — just guessing CL earns zero for Step 2
• Using LL = 50 boundary incorrectly (LL = 38 < 50, so it is L, not H)
• Writing the USCS symbol without the full name

Key Phrases To Include

• PI = LL − PL
• PI = 18
• A-line: PI = 0.73(LL − 20)
• PI_A = 13.14
• above A-line
• LL < 50
• CL
• low-plasticity clay

Marks

2

Topic

USCS Classification — General Framework

Difficulty

easy

Template Id

T5

Examiner Tip

The #4 sieve (4.75 mm) separates gravel from sand within the coarse-grained branch. Mentioning this distinction shows depth of knowledge and can earn bonus regard from the examiner.

Model Answer

In the USCS (Unified Soil Classification System), the #200 sieve (opening ≈ 0.075 mm) serves as the boundary between coarse-grained and fine-grained soils: • If MORE than 50% of the soil (by dry mass) is RETAINED on the #200 sieve, the soil is COARSE-GRAINED (gravel G or sand S). • If 50% or MORE passes the #200 sieve, the soil is FINE-GRAINED (silt M, clay C, or organic O). The percent finer than #200 thus determines which classification branch — and consequently which criteria (gradation parameters for coarse soils, or Atterberg limits and the plasticity chart for fine soils) — are applied.

Question Type

short_answer

Answer Structure

• Line 1: State the sieve size/opening (0.075 mm) [0.5 mark]
• Line 2: State the coarse-grained rule (>50% retained) [0.5 mark]
• Line 3: State the fine-grained rule (≥50% passing) [0.5 mark]
• Line 4: Explain what classification branch follows each case [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Stating the threshold as >50% for fine-grained (should be ≥50%)
• Not mentioning what classification tools (Cu/Cc or Atterberg) follow from each branch
• Confusing #200 with #4 (the gravel-sand boundary)

Key Phrases To Include

• #200 sieve
• 0.075 mm
• 50% passing
• coarse-grained
• fine-grained
• USCS
• Atterberg limits
• gradation parameters

Marks

3

Topic

Grain-Size Distribution Parameters — USCS Coarse Classification

Difficulty

medium

Template Id

T6

Examiner Tip

Board problems frequently set up a soil where Cu is high but Cc is outside [1, 3] — the trap is concluding SW. Always check both criteria and explicitly state which one(s) fail.

Model Answer

Given: D10 = 0.15 mm, D30 = 0.45 mm, D60 = 1.5 mm Soil type: Sand (< 50% passing #200; < 50% retained on #4) Step 1 — Coefficient of Uniformity: Cu = D60 / D10 = 1.5 / 0.15 = 10 Step 2 — Coefficient of Gradation (Curvature): Cc = (D30)² / (D10 × D60) = (0.45)² / (0.15 × 1.5) = 0.2025 / 0.225 = 0.90 Step 3 — Well-Graded Criteria for SAND (USCS): Criterion 1: Cu ≥ 6 → Cu = 10 ≥ 6 ✓ (SATISFIED) Criterion 2: 1 ≤ Cc ≤ 3 → Cc = 0.90 < 1 ✗ (NOT SATISFIED) Conclusion: Both criteria must be satisfied simultaneously. Since Cc = 0.90 fails the curvature criterion, the soil is POORLY GRADED. USCS Classification: SP (Poorly Graded Sand)

Question Type

numerical

Answer Structure

• Step 1: Correct formula and computation of Cu = 10 [1 mark]
• Step 2: Correct formula and computation of Cc = 0.90 [1 mark]
• Step 3: Check both criteria explicitly; identify which fails; state SP [1 mark]

Scoring Breakdown

Common Mark Deductions

• Concluding well-graded based on Cu alone without checking Cc
• Arithmetic error in Cc: forgetting to square D30
• Using gravel criterion Cu ≥ 4 instead of sand criterion Cu ≥ 6
• Not identifying the soil as sand before applying the criteria

Key Phrases To Include

• Cu = D60/D10
• Cc = D30² / (D10 × D60)
• Cu = 10
• Cc = 0.90
• Cu ≥ 6
• 1 ≤ Cc ≤ 3
• Cc fails
• SP
• poorly graded sand

Marks

3

Topic

AASHTO Classification — Group Index

Difficulty

medium

Template Id

T7

Examiner Tip

Memorize the GI formula exactly — it appears in almost every board exam set. Also note that GI = 0 does NOT mean the soil is excellent; it means A-1 or A-3 class where the index is defined as zero.

Model Answer

AASHTO Group Index Formula: GI = (F − 35)[0.2 + 0.005(LL − 40)] + 0.01(F − 15)(PI − 10) where: F = percent of soil passing the #200 sieve (expressed as an integer, e.g., 60 not 0.60) LL = liquid limit of the soil (%) PI = plasticity index of the soil (%) Capping Rules: • (F − 35) shall not exceed 40; negative values → use 0 • (F − 15) shall not exceed 40; negative values → use 0 • (LL − 40) shall not exceed 20; negative values → use 0 • (PI − 10) shall not exceed 20; negative values → use 0 • Final GI: round to nearest integer; if negative → report as 0 Engineering Significance: A higher GI indicates a POORER subgrade material. GI = 0 corresponds to the best subgrade soils (A-1 and A-3 groups). A GI of 20 or more represents very poor subgrade suitable only with significant improvement or thick pavement design.

Question Type

short_answer

Answer Structure

• Line 1: Write the complete GI formula [1 mark]
• Lines 2–4: Define F, LL, PI with units and the integer convention for F [1 mark]
• Lines 5–7: State the capping rules and explain that higher GI = poorer subgrade [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using F as a decimal (e.g., 0.60) instead of an integer (60) — gives wrong GI by factor of 100
• Omitting the floor (GI cannot be negative)
• Omitting one of the two terms in the GI equation

Key Phrases To Include

• GI = (F − 35)[0.2 + 0.005(LL − 40)] + 0.01(F − 15)(PI − 10)
• percent passing #200
• F expressed as integer
• GI ≥ 0
• round to nearest integer
• higher GI = poorer subgrade

Marks

3

Topic

AASHTO Classification — Group Index Computation

Difficulty

medium

Template Id

T8

Examiner Tip

Always write the cap checks as a separate preliminary step. This earns partial credit even if the final arithmetic is wrong, and it shows the examiner you know the procedure.

Model Answer

Given: F = 60, LL = 45, PI = 23 Check caps before substituting: (F − 35) = 60 − 35 = 25 ≤ 40 ✓ (F − 15) = 60 − 15 = 45 > 40 → use 40 (LL − 40) = 45 − 40 = 5 ≤ 20 ✓ (PI − 10) = 23 − 10 = 13 ≤ 20 ✓ Substitute into GI formula: GI = (F − 35)[0.2 + 0.005(LL − 40)] + 0.01(F − 15)(PI − 10) = (25)[0.2 + 0.005(5)] + 0.01(40)(13) = 25[0.2 + 0.025] + 0.01 × 520 = 25(0.225) + 5.20 = 5.625 + 5.20 = 10.825 Round to nearest integer: GI ≈ 11 Note: GI is positive, so no flooring needed. Final Answer: GI = 11

Question Type

numerical

Answer Structure

• Step 1: Identify F, LL, PI and check all four cap conditions [1 mark]
• Step 2: Substitute correctly into the GI formula, applying the (F−15) cap of 40 [1 mark]
• Step 3: Arithmetic to get 10.825 and round to GI = 11 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not applying the cap on (F − 15) — using 45 instead of 40 gives wrong answer
• Rounding 10.825 to 10 (should round to 11 by standard rounding rules)
• Using F = 0.60 instead of F = 60

Key Phrases To Include

• F = 60
• (F − 15) capped at 40
• GI formula
• GI = 10.825
• rounded to 11
• GI = 11

Marks

3

Topic

USCS Classification — Coarse-Grained (Gravel)

Difficulty

medium

Template Id

T9

Examiner Tip

The gravel criterion is Cu ≥ 4, not 6. This distinction is a classic board trap — always identify whether the soil is gravel or sand before applying the well-graded thresholds.

Model Answer

Given: D10 = 0.5 mm, D30 = 4 mm, D60 = 12 mm Soil: Gravel (assumed > 50% retained on #4 sieve based on given data) Step 1 — Cu: Cu = D60 / D10 = 12 / 0.5 = 24 Step 2 — Cc: Cc = (D30)² / (D10 × D60) = (4)² / (0.5 × 12) = 16 / 6 = 2.67 Step 3 — Well-Graded Criteria for GRAVEL (USCS): Criterion 1: Cu ≥ 4 → Cu = 24 ≥ 4 ✓ SATISFIED Criterion 2: 1 ≤ Cc ≤ 3 → Cc = 2.67 ✓ SATISFIED Both criteria satisfied → WELL-GRADED GRAVEL USCS Classification: GW (Well-Graded Gravel)

Question Type

numerical

Answer Structure

• Step 1: Cu = D60/D10 = 24 [0.5 mark]
• Step 2: Cc = D30²/(D10×D60) = 2.67 [1 mark]
• Step 3: Check GRAVEL criteria Cu ≥ 4 and 1 ≤ Cc ≤ 3; both pass; GW [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Using Cu ≥ 6 (sand criterion) instead of Cu ≥ 4 (gravel criterion)
• Arithmetic error: not squaring D30 in Cc
• Writing GW without 'Well-Graded Gravel'

Key Phrases To Include

• Cu = 24
• Cc = 2.67
• Cu ≥ 4 for gravel
• 1 ≤ Cc ≤ 3
• both satisfied
• GW
• well-graded gravel

Marks

3

Topic

USCS Classification — Fine-Grained Soils

Difficulty

medium

Template Id

T10

Examiner Tip

For LL > 50 soils, the distinction between CH and MH depends entirely on the A-line. Compute PI_A every time — never skip this step.

Model Answer

Given: LL = 60, PL = 25; inorganic soil Step 1 — Compute PI: PI = LL − PL = 60 − 25 = 35 Step 2 — Evaluate A-line at LL = 60: PI_A = 0.73(LL − 20) = 0.73(60 − 20) = 0.73 × 40 = 29.2 Step 3 — Compare PI to A-line: PI = 35 > PI_A = 29.2 → soil plots ABOVE the A-line → CLAY (C) Step 4 — Determine plasticity suffix: LL = 60 > 50 → HIGH plasticity (suffix H) Step 5 — Also check U-line (optional but thorough): U-line: PI = 0.9(LL − 8) = 0.9(60 − 8) = 46.8 PI = 35 < 46.8 → below U-line ✓ (valid plot) USCS Classification: CH (High-Plasticity Clay / Fat Clay)

Question Type

numerical

Answer Structure

• Step 1: PI = LL − PL = 35 [0.5 mark]
• Step 2: A-line PI_A = 0.73(LL − 20) = 29.2 [1 mark]
• Step 3: PI > PI_A → above A-line → clay (C) [0.5 mark]
• Step 4: LL = 60 > 50 → high plasticity (H); Final: CH [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting to compute the A-line value and simply assuming CH
• Using LL = 50 incorrectly — LL = 60 > 50 so it is H
• Writing MH instead of CH — MH is below the A-line

Key Phrases To Include

• PI = 35
• PI_A = 0.73(LL − 20) = 29.2
• above A-line
• LL > 50
• high plasticity
• CH
• fat clay

Marks

5

Topic

AASHTO Classification — Group Index (Advanced)

Difficulty

hard

Template Id

T11

Examiner Tip

In a 5-mark GI problem, two marks are typically allocated to the cap checks and substitution alone. Show every cap check clearly with '→ cap at 40' notation. This visual makes partial scoring unambiguous.

Model Answer

Given: F = 82%, LL = 50, PI = 28 ─── Step 1: Apply Caps on Individual Terms ─────────────────────────── (F − 35) = 82 − 35 = 47 → cap at 40 → use 40 (F − 15) = 82 − 15 = 67 → cap at 40 → use 40 (LL − 40) = 50 − 40 = 10 ≤ 20 → use 10 (PI − 10) = 28 − 10 = 18 ≤ 20 → use 18 ─── Step 2: Substitute into GI Formula ───────────────────────────── GI = (F − 35)[0.2 + 0.005(LL − 40)] + 0.01(F − 15)(PI − 10) = (40)[0.2 + 0.005(10)] + 0.01(40)(18) = 40[0.2 + 0.05] + 0.01 × 720 = 40(0.25) + 7.20 = 10.00 + 7.20 = 17.20 ─── Step 3: Apply Floor and Round ───────────────────────────────── GI = 17.20 → round to nearest integer → GI = 17 GI > 0, so no flooring needed. ─── Step 4: Engineering Interpretation ───────────────────────────── GI = 17 indicates a POOR subgrade material (high GI = poor subgrade). With F = 82% and LL = 50, this soil likely classifies as AASHTO A-7 (subgroup A-7-5 since PI ≤ LL − 30 → 28 ≤ 50 − 30 = 20? No: 28 > 20 → A-7-6). Subgrade requires significant stabilization or thick pavement overlay. ─── Final Answers ────────────────────────────────────────────────── GI = 17 AASHTO Subgrade Quality: Poor (A-7-6)

Question Type

numerical

Answer Structure

• Step 1: Check and apply all four cap conditions — show each calculation [1 mark]
• Step 2: Write GI formula and substitute with capped values [1 mark]
• Step 3: Compute partial products and sum to get GI = 17.20 [1.5 marks]
• Step 4: Round to integer GI = 17 and confirm no negative flooring needed [0.5 mark]
• Step 5: Engineering interpretation — poor subgrade, A-7-6 classification [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not capping (F−35) = 47 at 40 — gives GI = 18.50 instead of 17.20
• Not capping (F−15) = 67 at 40 — leads to GI overestimate
• Using F = 0.82 (decimal) instead of F = 82 (integer)
• Rounding 17.20 to 17 incorrectly as 18 (it rounds to 17, not 18)
• Not providing engineering interpretation in a 5-mark question

Key Phrases To Include

• F = 82
• cap at 40
• (F−35) capped
• (F−15) capped
• GI formula
• 40 × 0.25 = 10.00
• 7.20
• GI = 17.20
• GI = 17
• poor subgrade
• A-7-6

Marks

5

Topic

USCS Classification — Sand Subgroups

Difficulty

hard

Template Id

T12

Examiner Tip

A neat table comparing all four symbols side-by-side shows mastery and earns full marks faster than four separate paragraphs. Add the mini-example for the SC/SM distinction — examiners reward applied demonstration.

Model Answer

All four symbols apply to soils classified as SAND (S) under USCS — i.e., more than 50% of the coarse fraction passes the #4 sieve, and more than 50% of the total sample is retained on the #200 sieve. ─── Classification Summary ───────────────────────────────────────── Symbol | Full Name | Second-Letter Criteria -------|----------------------|----------------------------------------------- SW | Well-Graded Sand | Cu ≥ 6 AND 1 ≤ Cc ≤ 3 (BOTH must be satisfied) SP | Poorly Graded Sand | Fails Cu ≥ 6 OR 1 ≤ Cc ≤ 3 (or both) SM | Silty Sand | Fines ≥ 12% passing #200 AND fines are SILTY | | (plot below A-line on plasticity chart, PI < 4) SC | Clayey Sand | Fines ≥ 12% passing #200 AND fines are CLAYEY | | (plot above A-line on plasticity chart, PI ≥ 7) ─── Decision Logic ───────────────────────────────────────────────── • If fines < 5% passing #200 → apply Cu and Cc criteria → SW or SP • If fines > 12% passing #200 → apply plasticity chart → SM or SC • If fines = 5–12% (borderline) → dual symbols possible (e.g., SW-SM) ─── Key Boundary: A-Line ────────────────────────────────────────── For SM vs SC, plot the fines on the plasticity chart: PI = 0.73(LL − 20) → A-line Fines below A-line (or PI < 4) → silty → SM Fines above A-line (and PI ≥ 7) → clayey → SC Example: A sand with 15% fines, LL = 30, PI = 10 A-line PI at LL=30: 0.73(30−20) = 7.3 PI = 10 > 7.3 → above A-line → CLAYEY fines → SC

Question Type

short_answer

Answer Structure

• Section 1: State that all four are sand (S) subgroups and define what makes a soil 'sand' under USCS [0.5 mark]
• Section 2: Define SW — Cu ≥ 6 AND 1 ≤ Cc ≤ 3 — both must be met [1 mark]
• Section 3: Define SP — fails one or both gradation criteria [0.5 mark]
• Section 4: Define SM — silty fines (below A-line), fines > 12% [1 mark]
• Section 5: Define SC — clayey fines (above A-line), fines > 12% [1 mark]
• Section 6: Decision rule for 5–12% fines zone (dual symbols) and worked mini-example [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing SM and SC — silty is BELOW the A-line, clayey is ABOVE
• Not specifying fines content thresholds (5% and 12%)
• Applying Cu ≥ 4 (gravel) instead of Cu ≥ 6 (sand) for SW
• No example or diagram to support the answer in a 5-mark question

Key Phrases To Include

• Cu ≥ 6
• 1 ≤ Cc ≤ 3
• SW
• SP
• SM
• SC
• A-line
• PI = 0.73(LL − 20)
• below A-line → silty
• above A-line → clayey
• fines > 12%
• dual symbol

Marks

2

Topic

Atterberg Limits — Plasticity Chart

Difficulty

easy

Template Id

T13

Examiner Tip

Sketch a small plasticity chart in the margin with the A-line drawn and two labeled regions. A quick diagram can replace a paragraph and earns just as many marks.

Model Answer

The A-line is an empirical boundary line on the Casagrande plasticity chart that separates inorganic clays from inorganic silts and organic soils. Equation of the A-line: PI = 0.73 (LL − 20) Physical Significance: • Soils plotting ABOVE the A-line: PI > 0.73(LL − 20) → predominantly CLAY-LIKE behavior (inorganic clays) → USCS prefix C (e.g., CL, CH). • Soils plotting BELOW the A-line: PI < 0.73(LL − 20) → SILT-LIKE or ORGANIC behavior → USCS prefix M or O (e.g., ML, MH, OL, OH). • The vertical line at LL = 50 divides low-plasticity (L) from high-plasticity (H) soils. Note: The A-line is valid only for fine-grained soils (≥ 50% passing #200 sieve).

Question Type

short_answer

Answer Structure

• Line 1: Define A-line as the boundary between clays and silts/organics [0.5 mark]
• Line 2: Write PI = 0.73(LL − 20) [0.5 mark]
• Lines 3–4: Explain above → C-group and below → M/O-group [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing the slope as 0.73 but starting from LL = 0 instead of LL = 20
• Saying below A-line → clay (reversed)
• Not mentioning that the A-line applies only to fine-grained soils

Key Phrases To Include

• PI = 0.73(LL − 20)
• A-line
• above A-line → clay
• below A-line → silt
• LL = 50
• CL/CH
• ML/MH
• fine-grained soils

Marks

5

Topic

USCS Classification — Full Worked Classification

Difficulty

hard

Template Id

T14

Examiner Tip

Case-study 5-mark questions reward a structured, step-numbered solution. Begin each step with a heading (Step 1, Step 2, etc.) and highlight the final classification in a box. This makes it impossible for the examiner to miss your conclusions.

Model Answer

Given: D10 = 0.08 mm, D30 = 0.30 mm, D60 = 2.4 mm % passing #200 = 18% (<50% → coarse-grained) LL = 28, PL = 18 ─── Step 1: Identify Coarse vs Fine ──────────────────────────────── 18% < 50% passing #200 → COARSE-GRAINED soil Assume % retained on #4 sieve data indicates more sand fraction → SAND (S) subgroup (assume > 50% of coarse fraction passes #4) ─── Step 2: Assess Fines Content ────────────────────────────────── Fines = 18% → between 12% and 50% → Fines content > 12% → must evaluate plasticity of fines ─── Step 3: Evaluate Plasticity of Fines ────────────────────────── PI = LL − PL = 28 − 18 = 10 A-line at LL = 28: PI_A = 0.73(28 − 20) = 0.73 × 8 = 5.84 PI = 10 > PI_A = 5.84 → fines plot ABOVE A-line → CLAYEY fines ─── Step 4: Compute Gradation Parameters (for reference) ────────── Cu = D60/D10 = 2.4/0.08 = 30 Cc = (D30)²/(D10 × D60) = (0.30)²/(0.08 × 2.4) = 0.09/0.192 = 0.469 Cc = 0.469 — outside [1, 3]; however, with fines > 12%, gradation criteria are NOT used; plasticity governs the second letter. ─── Step 5: Assign USCS Symbol ──────────────────────────────────── First letter: S (Sand) Second letter: C (Clayey — fines above A-line, fines > 12%) ─── Final Classification ─────────────────────────────────────────── USCS: SC — Clayey Sand ─── Engineering Implication ──────────────────────────────────────── Clayey sand is susceptible to moisture-induced strength loss and frost heave. As a road sub-base, compaction control and drainage are critical design considerations.

Question Type

case_study

Answer Structure

• Step 1: Confirm coarse-grained (18% < 50% passing #200); identify as sand [0.5 mark]
• Step 2: Note fines = 18% > 12% — plasticity chart governs [0.5 mark]
• Step 3: Compute PI = 10, A-line PI_A = 5.84, compare — above A-line → clayey [1.5 marks]
• Step 4: Compute Cu and Cc, note they are not used when fines > 12% [1 mark]
• Step 5: Assign SC with full name [0.5 mark]
• Step 6: Brief engineering implication relevant to road sub-base context [1 mark]

Scoring Breakdown

Common Mark Deductions

• Applying Cu/Cc criteria when fines > 12% — gradation parameters do not determine the second letter in this range
• Omitting the A-line check and guessing SM or SC without computation
• Not providing an engineering implication in a 5-mark case-study question
• Confusing 'more than 50% passing #200' with 'more than 50% retained' — misidentifying as fine-grained

Key Phrases To Include

• coarse-grained
• fines = 18%
• fines > 12%
• PI = 10
• A-line PI_A = 5.84
• above A-line
• clayey fines
• SC
• clayey sand
• plasticity governs

Marks

2

Topic

Atterberg Limits — Shrinkage Limit

Difficulty

medium

Template Id

T15

Examiner Tip

Referencing NSCP 2015 Section 304 (expansive soils) in a geotechnical answer signals code-awareness — boards reward this even when it is not explicitly required by the question.

Model Answer

Shrinkage Limit (SL): The shrinkage limit is the moisture content (expressed as a percentage of dry mass) at which a soil ceases to decrease in volume upon further drying. Below the SL, additional water loss does not cause further shrinkage; instead, air enters the voids. It is the lowest of the three Atterberg limits. Ordering of Atterberg Limits: SL < PL < LL Significance: 1. Soils with a low SL have a wide range over which they shrink upon drying — problematic for foundations on expansive soils. 2. The difference (PL − SL) is called the shrinkage index and quantifies the range of plastic-to-semi-solid transition. 3. Expansive clays (e.g., montmorillonite-rich soils common in some Philippine volcanic terrains) have high plasticity and low SL, leading to significant volume change with seasonal moisture variation — a critical consideration under NSCP 2015 Section 304 for foundation design on expansive soils.

Question Type

short_answer

Answer Structure

• Line 1–2: Define SL as the moisture content where volume change ceases upon drying [1 mark]
• Line 3–4: State SL < PL < LL ordering and at least one engineering significance [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining SL as the moisture content where soil becomes solid — not precise enough without mentioning volume change cessation
• Omitting the ordering SL < PL < LL
• Not connecting SL to any engineering application or soil behavior consequence

Key Phrases To Include

• shrinkage limit
• volume ceases to decrease
• lowest Atterberg limit
• SL < PL < LL
• expansive soils
• NSCP 2015 Section 304