CELE Geotechnical Engineering — Soil Properties and Phase RelationshipsExam Answer Templates

Marks

1

Topic

Basic Phase Ratios — Void Ratio

Difficulty

easy

Template Id

T1

Examiner Tip

The three-part formula (symbol, equation, unit/range note) in one sentence is the gold standard for 1-mark definition answers. Do not write a paragraph.

Model Answer

Void ratio (e) is the ratio of the volume of voids to the volume of solids in a soil mass: e = Vv / Vs. It is dimensionless and can exceed 1.0 for loose or highly plastic soils.

Question Type

very_short_answer

Answer Structure

• One sentence: symbol + formula + physical interpretation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing e = Vv/V (total volume) instead of Vv/Vs — this defines porosity, not void ratio
• Omitting the formula and giving only a verbal description
• Stating that e must be less than 1 — it can exceed 1 for soft clays

Key Phrases To Include

• volume of voids
• volume of solids
• e = Vv / Vs
• dimensionless

Marks

2

Topic

Basic Phase Ratios — Void Ratio and Porosity

Difficulty

easy

Template Id

T2

Examiner Tip

Board exams frequently test whether students can derive one from the other quickly. Memorise both forms so you can rearrange within 10 seconds.

Model Answer

Void ratio: e = Vv / Vs (ratio of void volume to solid volume; reference base is solids). Porosity: n = Vv / V (ratio of void volume to total volume; reference base is total soil volume). Relationship: e = n / (1 − n) and n = e / (1 + e).

Question Type

short_answer

Answer Structure

• Line 1: Define e with formula, noting that the reference base is the volume of solids [0.5 mark]
• Line 2: Define n with formula, noting that the reference base is total volume [0.5 mark]
• Line 3: Write both conversion formulas e = n/(1−n) and n = e/(1+e) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing only one conversion formula instead of both
• Confusing the denominators — using Vv as the denominator for e
• Stating n can exceed 1 (it is bounded 0 < n < 1, unlike e)

Key Phrases To Include

• e = Vv/Vs
• n = Vv/V
• e = n/(1−n)
• n = e/(1+e)
• reference base

Marks

2

Topic

Master Identity — Se = wGs

Difficulty

easy

Template Id

T3

Examiner Tip

This identity appears in almost every phase-relationship problem. Writing it at the top of your solution page signals strong subject mastery to the examiner.

Model Answer

Master identity: S · e = w · Gs Where S = degree of saturation (decimal), e = void ratio, w = gravimetric water content (decimal), Gs = specific gravity of solids. Physical limits: S = 0 → Dry soil: all voids are filled with air; Vw = 0. S = 1 → Fully saturated soil: all voids are filled with water; Va = 0.

Question Type

short_answer

Answer Structure

• Line 1: Write Se = wGs and define all four symbols [1 mark]
• Line 2: State S = 0 condition (dry) and S = 1 condition (saturated) with physical meaning [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing w as a percentage instead of a decimal in the identity
• Failing to define all four variables
• Stating S can be negative or greater than 1

Key Phrases To Include

• Se = wGs
• degree of saturation
• S = 0 dry
• S = 1 fully saturated
• all voids filled with water

Marks

3

Topic

Void Ratio, Porosity — Saturated Soil

Difficulty

easy

Template Id

T4

Examiner Tip

For soft Philippine marine clays (e.g., Manila Bay clay), void ratios above 1.0 are normal. Mentioning this shows engineering awareness and earns bonus impression marks.

Model Answer

Given: w = 38% = 0.38, Gs = 2.72, S = 1.0 (saturated) Step 1 — Void ratio using master identity: Se = wGs (1.0)e = (0.38)(2.72) e = 1.034 Step 2 — Porosity: n = e / (1 + e) = 1.034 / (1 + 1.034) = 1.034 / 2.034 n = 0.508 (50.8%) Answer: e = 1.034, n = 50.8% Note: e > 1 is physically valid for soft marine or volcanic clays common in the Philippines.

Question Type

numerical

Answer Structure

• Line 1: List given data and state S = 1.0 for saturated condition [0.5 mark]
• Line 2: Write Se = wGs and solve for e [1 mark]
• Line 3: Apply n = e/(1+e) and compute n [1 mark]
• Line 4: Box final answers with units/percentage [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Substituting w = 38 (not 0.38) in the formula — gives e = 103.4, a physically impossible result
• Using n = Vv/Vs instead of Vv/V for porosity
• Not stating S = 1.0 explicitly; examiner expects this to be written

Key Phrases To Include

• S = 1.0 (saturated)
• Se = wGs
• n = e/(1+e)
• e = 1.034
• n = 0.508

Marks

3

Topic

Dry, Moist Unit Weight and Degree of Saturation

Difficulty

easy

Template Id

T5

Examiner Tip

Parts (a), (b), (c) each earn one mark; tackle each independently. Even if part (a) is wrong, correct formulas in (b) and (c) using your (a) value still earn marks — this is called 'error carried forward' and Philippine board examiners apply it.

Model Answer

Given: Gs = 2.65, e = 0.60, w = 12% = 0.12, γw = 9.81 kN/m³ (a) Dry unit weight: γdry = Gs·γw / (1 + e) = (2.65)(9.81) / (1 + 0.60) γdry = 26.0 / 1.60 = 16.25 kN/m³ (b) Moist unit weight: γ = γdry(1 + w) = 16.25(1 + 0.12) = 16.25(1.12) γ = 18.2 kN/m³ (c) Degree of saturation: S = wGs / e = (0.12)(2.65) / 0.60 = 0.318 / 0.60 S = 0.53 (53%) Answers: γdry = 16.25 kN/m³, γ = 18.2 kN/m³, S = 53%

Question Type

numerical

Answer Structure

• Header: List all given values including γw = 9.81 kN/m³ [implicit mark]
• Part (a): Write formula γdry = Gsγw/(1+e), substitute, compute [1 mark]
• Part (b): Write formula γ = γdry(1+w), substitute, compute [1 mark]
• Part (c): Write S = wGs/e, substitute, compute [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting to convert w to decimal for the γ = γdry(1+w) formula
• Not stating γw = 9.81 kN/m³ — examiner may deduct for an unexplained constant
• Using γw = 10 kN/m³ without acknowledgement — use 9.81 unless told otherwise in the problem

Key Phrases To Include

• γdry = Gsγw/(1+e)
• γ = γdry(1+w)
• S = wGs/e
• γw = 9.81 kN/m³
• 16.25 kN/m³
• 18.2 kN/m³
• 53%

Marks

3

Topic

Unit Weights — Saturated Unit Weight Derivation

Difficulty

medium

Template Id

T6

Examiner Tip

Derivation questions in licensure exams reward systematic, step-by-step logic. Each intermediate line shown is a potential partial-credit point. Never skip steps.

Model Answer

Definition: The saturated unit weight (γsat) is the bulk unit weight of a soil when all voids are completely filled with water (S = 1.0). Derivation from phase diagram: Let total volume V = 1 + e (take Vs = 1 as unit volume). Volume of solids, Vs = 1 Volume of voids, Vv = e → all water since S = 1 Weight of solids, Ws = Gs·γw·Vs = Gs·γw Weight of water, Ww = γw·Vw = γw·e Total weight, W = Ws + Ww = (Gs + e)γw Therefore: γsat = W/V = (Gs + e)γw / (1 + e) This is the standard formula used in geotechnical engineering.

Question Type

short_answer

Answer Structure

• Line 1: State definition with S = 1 condition [1 mark]
• Lines 2–5: Set up unit phase diagram with Vs = 1 and derive weights [1 mark]
• Final line: Write boxed formula γsat = (Gs + e)γw/(1 + e) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Skipping the derivation and writing only the formula — loses the 1 mark for derivation
• Using Vv = n instead of Vv = e in the phase diagram
• Writing γsat = (Gs + Se)γw/(1+e) — correct for moist, but the question asks for saturated (S=1)

Key Phrases To Include

• S = 1.0
• all voids filled with water
• Vs = 1 (unit volume)
• γsat = (Gs + e)γw/(1 + e)

Marks

2

Topic

Submerged Unit Weight and Effective Stress

Difficulty

medium

Template Id

T7

Examiner Tip

Examiners reward candidates who connect γ' explicitly to effective stress theory (σ' = σ − u). This shows you understand WHY the formula matters, not just what it is.

Model Answer

The submerged unit weight (γ' or γsub) is the effective unit weight of a saturated soil mass submerged below the groundwater table: γ' = γsat − γw It represents the net downward unit weight after buoyancy (upward water pressure) is subtracted from the saturated unit weight. In effective stress calculations (σ' = σ − u), the pore water pressure u equals γw·h for a static water table. Using γ' directly accounts for this buoyancy and gives the effective overburden stress in a single step: σ' = γ'·z (below water table) without separately computing total stress and pore pressure.

Question Type

short_answer

Answer Structure

• Line 1: Define γ' with formula γ' = γsat − γw [1 mark]
• Line 2: Explain its role in effective stress — buoyancy effect [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing γ' = γmoist − γw instead of γsat − γw
• Saying 'used in saturated soil' without explaining the buoyancy or effective stress link
• Confusing buoyant unit weight with submerged density

Key Phrases To Include

• γ' = γsat − γw
• buoyancy
• effective stress
• below groundwater table
• σ' = γ'·z

Marks

3

Topic

Back-Calculation of Dry Unit Weight and Void Ratio

Difficulty

medium

Template Id

T8

Examiner Tip

When a problem gives γmoist and w, always go to γdry = γ/(1+w) first. This is the gateway to finding e, n, and S. Examiners expect this as the first formula written.

Model Answer

Given: γ = 19.0 kN/m³, w = 15% = 0.15, Gs = 2.70, γw = 9.81 kN/m³ (a) Dry unit weight: γdry = γ / (1 + w) = 19.0 / (1 + 0.15) = 19.0 / 1.15 γdry = 16.52 kN/m³ (b) Void ratio: From γdry = Gs·γw / (1 + e): 1 + e = Gs·γw / γdry = (2.70)(9.81) / 16.52 = 26.49 / 16.52 = 1.604 e = 1.604 − 1 = 0.604 Answers: γdry = 16.52 kN/m³, e = 0.604

Question Type

numerical

Answer Structure

• Header: List given data [implicit]
• Part (a): Write γdry = γ/(1+w), substitute, compute [1 mark]
• Part (b): Rearrange γdry = Gsγw/(1+e) to solve for e [1 mark]
• Box final answers [1 mark for correct values and units]

Scoring Breakdown

Common Mark Deductions

• Using w = 15 (not 0.15) in the denominator — gives γdry = 1.19 kN/m³, an absurd value
• Attempting to use Se = wGs without knowing S — introduces an extra unknown
• Arithmetic error in rearranging (1 + e) = Gsγw/γdry

Key Phrases To Include

• γdry = γ/(1+w)
• γdry = Gsγw/(1+e)
• rearranging for e
• 16.52 kN/m³
• e = 0.604

Marks

3

Topic

Saturated and Submerged Unit Weights

Difficulty

easy

Template Id

T9

Examiner Tip

The typical range for γsat of common soils is 18–22 kN/m³ and for γ' is 8–12 kN/m³. If your answer falls outside these ranges, recheck your arithmetic before finalising.

Model Answer

Given: e = 0.55, Gs = 2.66, γw = 9.81 kN/m³ Step 1 — Saturated unit weight: γsat = (Gs + e)γw / (1 + e) γsat = (2.66 + 0.55)(9.81) / (1 + 0.55) γsat = (3.21)(9.81) / 1.55 γsat = 31.49 / 1.55 γsat = 20.32 kN/m³ Step 2 — Submerged unit weight: γ' = γsat − γw = 20.32 − 9.81 γ' = 10.51 kN/m³ Answers: γsat = 20.32 kN/m³, γ' = 10.51 kN/m³

Question Type

numerical

Answer Structure

• Header: List e, Gs, γw [implicit]
• Step 1: Write γsat = (Gs+e)γw/(1+e), substitute, compute [1.5 marks]
• Step 2: Write γ' = γsat − γw, substitute, compute [1 mark]
• Box final answers with kN/m³ units [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Computing (Gs + e)/(1+e) first and then multiplying by γw — this is correct but risks arithmetic errors; multiply Gs+e by γw first
• Writing γ' = γsat − γw/2 (a confusion with average pressure) — always subtract one γw
• Omitting units kN/m³ from both final answers

Key Phrases To Include

• γsat = (Gs + e)γw/(1+e)
• γ' = γsat − γw
• 20.32 kN/m³
• 10.51 kN/m³

Marks

5

Topic

Comprehensive Phase Relationships — n, S, Gs given

Difficulty

medium

Template Id

T10

Examiner Tip

For a 5-mark problem, work systematically: solve each unknown in the order they are needed as inputs for the next step. Never jump steps. Examiners trace your logic line by line.

Model Answer

Given: n = 0.40, Gs = 2.68, S = 0.80, γw = 9.81 kN/m³ (a) Void ratio: e = n / (1 − n) = 0.40 / (1 − 0.40) = 0.40 / 0.60 e = 0.667 (b) Water content (using master identity Se = wGs): w = Se / Gs = (0.80)(0.667) / 2.68 = 0.5336 / 2.68 w = 0.199 = 19.9% ≈ 20.0% (c) Moist unit weight: Method 1 — Using dry unit weight first: γdry = Gs·γw / (1 + e) = (2.68)(9.81) / (1 + 0.667) = 26.29 / 1.667 γdry = 15.77 kN/m³ γ = γdry(1 + w) = 15.77(1 + 0.199) = 15.77(1.199) γ = 18.91 kN/m³ [Alternatively: γ = (Gs + Se)γw/(1+e) = (2.68 + 0.80×0.667)(9.81)/1.667 = (3.214)(9.81)/1.667 = 18.91 kN/m³ ✓] Answers: e = 0.667, w = 20.0%, γ = 18.91 kN/m³

Question Type

numerical

Answer Structure

• Header: List all given values and state γw = 9.81 kN/m³ [0.5 mark for completeness]
• Part (a): Write e = n/(1−n), substitute n = 0.40, compute e = 0.667 [1 mark]
• Part (b): Write Se = wGs, rearrange to w = Se/Gs, substitute, compute w = 0.199 [1.5 marks]
• Part (c): Compute γdry = Gsγw/(1+e) then γ = γdry(1+w), or use combined formula [1.5 marks]
• Box and label all three final answers with correct units/format [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using e = n directly (a very common and serious error) — e ≠ n
• Forgetting to convert w back to percentage for the final reported answer
• Using w = 0.199 in percent form (19.9%) inside the (1+w) formula — must use decimal 0.199

Key Phrases To Include

• e = n/(1−n)
• Se = wGs
• w = Se/Gs
• γdry = Gsγw/(1+e)
• γ = γdry(1+w)
• e = 0.667
• w = 20.0%
• γ = 18.91 kN/m³

Marks

5

Topic

Phase Relationships from Laboratory Measurements

Difficulty

hard

Template Id

T11

Examiner Tip

Laboratory-based problems (given mass and volume of sample) require computing Vs from Gs. This step is the key that unlocks void ratio, and many students lose marks by skipping it. Always compute Vs = Ms/(Gs·ρw).

Model Answer

Given: V = 100 cm³ = 100 × 10⁻⁶ m³, M = 190 g, Ms = 165 g (dry mass), Gs = 2.70, γw = 9.81 kN/m³ (ρw = 1.00 g/cm³) (a) Water content: Mw = M − Ms = 190 − 165 = 25 g w = Mw / Ms = 25 / 165 = 0.1515 = 15.15% (b) Void ratio: Volume of solids: Vs = Ms / (Gs·ρw) = 165 / (2.70 × 1.00) = 61.11 cm³ Volume of voids: Vv = V − Vs = 100 − 61.11 = 38.89 cm³ e = Vv / Vs = 38.89 / 61.11 = 0.636 (c) Degree of saturation: Volume of water: Vw = Mw / ρw = 25 / 1.00 = 25.0 cm³ S = Vw / Vv = 25.0 / 38.89 = 0.643 (64.3%) Verification: Se = (0.643)(0.636) = 0.409; wGs = (0.1515)(2.70) = 0.409 ✓ (d) Since S = 64.3% < 100%, the soil is partially saturated (three-phase system: solids + water + air). Answers: w = 15.15%, e = 0.636, S = 64.3%, partially saturated

Question Type

numerical

Answer Structure

• Header: Extract all data; note ρw = 1.00 g/cm³ [0.5 mark]
• Part (a): Compute Mw = M − Ms, then w = Mw/Ms [1 mark]
• Part (b): Compute Vs = Ms/(Gs·ρw), Vv = V − Vs, e = Vv/Vs [1.5 marks]
• Part (c): Compute Vw = Mw/ρw, S = Vw/Vv, include Se = wGs verification [1.5 marks]
• Part (d): One-sentence conclusion comparing S to 1.0 [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Not computing Vs from Gs — directly assuming Vs equals a fraction of V without calculation
• Using volume units inconsistently — mixing cm³ and m³ in the same equation
• Forgetting the verification step Se = wGs — this is a free half-mark that many students skip

Key Phrases To Include

• Mw = M − Ms
• w = Mw/Ms
• Vs = Ms/(Gs·ρw)
• Vv = V − Vs
• e = Vv/Vs
• S = Vw/Vv
• Se = wGs verification
• partially saturated

Marks

3

Topic

Physical Interpretation of Phase Changes — Clay Shrinkage

Difficulty

medium

Template Id

T12

Examiner Tip

Theory questions in licensure exams reward formula-referenced answers. For every change you state ('e decreases'), also cite the formula that shows it ('e = Vv/Vs, Vs constant, Vv ↓ ∴ e ↓'). This demonstrates engineering reasoning, not just memorisation.

Model Answer

During drying, clay particles draw together as pore water evaporates, reducing the total volume V and the volume of voids Vv while Vs (volume of solids) remains constant. Effect on void ratio: e = Vv/Vs. Since Vv decreases and Vs is constant, e DECREASES. The clay becomes denser. Effect on porosity: n = Vv/V. Both Vv and V decrease, but the ratio also decreases because V shrinks proportionally less than Vv in the early stages of shrinkage. n DECREASES. Effect on dry unit weight: γdry = Gs·γw/(1+e). As e decreases, (1+e) decreases, so γdry INCREASES — the soil becomes heavier per unit volume. Shrinkage limit: Below the shrinkage limit water content, volume ceases to change on further drying (air replaces expelled water instead of the fabric collapsing).

Question Type

short_answer

Answer Structure

• Opening sentence: Physical mechanism — pore water leaves, particles pack closer [0.5 mark]
• Three separate bullet/paragraph statements for e, n, and γdry, each with formula reference and direction of change [2 marks total, ~0.67 each]
• Note about shrinkage limit for complete understanding [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Stating that e and n increase during drying — this is incorrect for clays above the shrinkage limit
• Not anchoring the explanation to the phase-diagram formulas — answer appears vague
• Treating γdry as constant during shrinkage — it changes because V changes

Key Phrases To Include

• Vs constant
• Vv decreases
• e decreases
• n decreases
• γdry increases
• γdry = Gsγw/(1+e)
• shrinkage limit

Marks

1

Topic

Specific Gravity of Solids

Difficulty

easy

Template Id

T13

Examiner Tip

Even a 1-mark definition must have three elements: symbol, formula, and context. Three elements in three seconds is the target for very-short-answer items.

Model Answer

Gs is the ratio of the unit weight of soil solids to the unit weight of water: Gs = γs/γw. Typical range: 2.65–2.70 for most soils (2.60 for organic, up to 2.80 for heavy minerals). It is used in the master identity Se = wGs and in all unit-weight formulas to relate the weight of solids to their volume.

Question Type

very_short_answer

Answer Structure

• One compact answer: formula + typical range + usage [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating Gs is the ratio of soil mass to water mass — incomplete; must reference unit weights or densities
• Giving a range outside 2.60–2.80 without qualification
• Confusing Gs with bulk specific gravity or apparent specific gravity

Key Phrases To Include

• Gs = γs/γw
• 2.65–2.70
• specific gravity of solids
• Se = wGs

Marks

5

Topic

Application of Submerged Unit Weight — Effective Stress

Difficulty

hard

Template Id

T14

Examiner Tip

In board exam problems involving marine or riverside sites (common in Philippine geotechnical practice), always check whether the water table is at or above the surface. If yes, use γ' for effective stress and γsat for total stress. Manila Bay soft clay problems appear regularly in the licensure exam.

Model Answer

Given: γsat = 16.5 kN/m³, γw = 9.81 kN/m³, z = 4 m, water table at seabed (z = 0) (a) Submerged unit weight: γ' = γsat − γw = 16.5 − 9.81 = 6.69 kN/m³ (b) Effective vertical stress at z = 4 m: Since the water table is at the surface, the entire 4 m column of soil is submerged. σ'v = γ' × z = 6.69 × 4 = 26.76 kN/m² (c) Total vertical stress at z = 4 m: σv = γsat × z = 16.5 × 4 = 66.0 kN/m² Verification using σ' = σ − u: u = γw × z = 9.81 × 4 = 39.24 kN/m² σ' = σv − u = 66.0 − 39.24 = 26.76 kN/m² ✓ Answers: γ' = 6.69 kN/m³, σ'v = 26.76 kN/m², σv = 66.0 kN/m²

Question Type

numerical

Answer Structure

• Header: List given data and state water table condition [0.5 mark]
• Part (a): Write and compute γ' = γsat − γw [1 mark]
• Part (b): Compute σ'v = γ'·z with reasoning about fully submerged condition [1.5 marks]
• Part (c): Compute σv = γsat·z and verify via σ'v = σv − u [1.5 marks]
• Box all three answers with correct units [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using γ' = γsat − γw/2 (incorrect; always subtract one full γw)
• Applying γdry instead of γsat for total stress in submerged conditions
• Not performing the verification — loses the final 1 mark

Key Phrases To Include

• γ' = γsat − γw
• water table at surface
• σ'v = γ'·z
• σv = γsat·z
• u = γw·z
• σ' = σv − u
• verification

Marks

2

Topic

Void Ratio Back-Calculation and Saturated Unit Weight

Difficulty

medium

Template Id

T15

Examiner Tip

The transition from dry to saturated at constant void ratio is a classic two-part problem. The key insight — that e does not change with saturation — must be implicit in your solution by using the same e in both formulas.

Model Answer

Given: γdry = 16.0 kN/m³, Gs = 2.68, γw = 9.81 kN/m³ Step 1 — Void ratio: γdry = Gs·γw / (1 + e) 1 + e = Gs·γw / γdry = (2.68)(9.81) / 16.0 = 26.29 / 16.0 = 1.643 e = 0.643 Step 2 — Saturated unit weight: γsat = (Gs + e)γw / (1 + e) = (2.68 + 0.643)(9.81) / 1.643 = (3.323)(9.81) / 1.643 γsat = 32.60 / 1.643 = 19.84 kN/m³ Answers: e = 0.643, γsat = 19.84 kN/m³

Question Type

numerical

Answer Structure

• Step 1: Rearrange γdry formula to solve for e [1 mark]
• Step 2: Use e in γsat formula [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using a different e in Step 2 than computed in Step 1 — shows lack of linkage between steps
• Arithmetic error in (2.68 + 0.643) = 3.323 — careless addition
• Forgetting that e is unchanged when the soil becomes saturated (saturation changes S, not e)

Key Phrases To Include

• γdry = Gsγw/(1+e)
• rearranging for e
• γsat = (Gs+e)γw/(1+e)
• e = 0.643
• 19.84 kN/m³