CELE Geotechnical Engineering — Soil Properties and Phase RelationshipsDetailed Explanation
Want to really understand Soil Properties and Phase Relationships before tackling CELE Geotechnical Engineering questions? This detailed explanation breaks down every key concept, shows you why it matters for the CELE 2026, and walks through the reasoning Professional Regulation Commission (PRC) — Board of Civil Engineering expects on high-difficulty questions.
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 Properties and Phase Relationships is the 1st 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 Properties and Phase Relationships - Detailed Explanation
Soil is not a simple, uniform material — it is a three-phase system composed of solid particles, water, and air. Before any geotechnical design can begin, an engineer must quantify the proportions of these three phases. The phase relationships — void ratio, porosity, water content, degree of saturation, and unit weights — form the mathematical foundation of virtually every calculation in Geotechnical Engineering, from bearing capacity to settlement to slope stability. In the PRC Civil Engineer Licensure Examination, phase relationship problems consistently appear in the Geotechnical Engineering portion. Mastery of these concepts is non-negotiable. This chapter develops every key formula from first principles, links them through the master identity Se = wGs, and demonstrates their application through board-style worked examples in SI units.
Concepts
The Three-Phase Model of Soil
Soil is idealized as a three-phase system: solid particles (mineral grains), water (pore water), and air (pore air). The phase diagram separates volumes on the left and weights on the right. Define the following volume components: Vs = volume of solids, Vw = volume of water, Va = volume of air, Vv = volume of voids = Vw + Va, V = total volume = Vs + Vv. For weight components: Ws = weight of solids, Ww = weight of water. Air is assumed weightless in soil mechanics. The total weight W = Ws + Ww. This diagram is the starting point for deriving every phase relationship. Always draw it before solving a problem — it prevents errors and clarifies what is known versus unknown.
Examples
Once three of the five volume quantities are known, the other two follow directly from the definitions V = Vs + Vv and Vv = Vw + Va. This is the most basic phase diagram calculation.
Scenario
A soil sample has a total volume of 0.05 m³. The volume of solids is 0.032 m³ and the volume of water is 0.012 m³. Determine all volume components.
Solution
V = 0.05 m³, Vs = 0.032 m³, Vw = 0.012 m³. Vv = V - Vs = 0.05 - 0.032 = 0.018 m³. Va = Vv - Vw = 0.018 - 0.012 = 0.006 m³.
Applications
- Foundation design: understanding pore structure governs compressibility and drainage
- Earthwork: computing fill and cut volumes requires phase relationships
- Pavement subgrade evaluation: degree of saturation governs stiffness and frost susceptibility
- Embankment design: unit weight selection for stability analysis
Misconceptions
- Air has weight — WRONG: air weight is neglected in soil mechanics
- Vv is only the air volume — WRONG: Vv = Vw + Va, voids include both water and air
- W = Ws + Ww + Wair — WRONG: Wair ≈ 0, so W = Ws + Ww only
Related Concepts
- Void Ratio and Porosity
- Degree of Saturation
- Unit Weights of Soil
Common Exam Questions
Example
Given V = 1 m³, Vs = 0.60 m³, S = 0.70, find Vw. First find Vv = V - Vs = 0.40 m³, then Vw = S × Vv = 0.70 × 0.40 = 0.28 m³.
Approach
Draw the phase diagram. Assign given quantities. Use V = Vs + Vv and Vv = Vw + Va to find the rest.
Question Type
Volume Component Identification
Key Points To Remember
- Soil = Solids + Water + Air (three phases, always)
- Vv = Vw + Va (voids include both water and air)
- V = Vs + Vv = Vs + Vw + Va
- W = Ws + Ww (air is weightless)
- Draw the phase diagram before every problem — it is your roadmap
- All volume ratios use Vs or Vv or V as the reference; be consistent
Void Ratio and Porosity
Two complementary descriptors quantify the pore space in a soil. The void ratio e = Vv/Vs relates void volume to solid volume. It can exceed 1.0 for highly compressible soils like soft clays (e can reach 2 or 3 for marine clays). The porosity n = Vv/V relates void volume to total volume. Porosity is always between 0 and 1 (0% to 100%). The two are mathematically interconvertible: e = n/(1-n) and n = e/(1+e). Typical values: dense sand e ≈ 0.4–0.6, loose sand e ≈ 0.6–0.9, soft clay e ≈ 0.9–2.0. Note that 1+e appears in virtually every unit weight formula because V/Vs = (Vs + Vv)/Vs = 1 + e. This ratio is the key scaling factor between per-unit-solid-volume and per-total-volume quantities.
Examples
The conversion n = e/(1+e) directly gives porosity from void ratio. Once porosity is known, volumes follow immediately. The check confirms consistency.
Scenario
A soil sample has a total volume of 1.0 m³ and a void ratio of 0.65. Determine the porosity, the volume of voids, and the volume of solids.
Solution
n = e/(1+e) = 0.65/1.65 = 0.394 = 39.4%. Since n = Vv/V: Vv = nV = 0.394 × 1.0 = 0.394 m³. Vs = V - Vv = 1.0 - 0.394 = 0.606 m³. Check: e = Vv/Vs = 0.394/0.606 = 0.650 ✓
This is one of the most frequently tested conversions on the board exam. Know both directions of the formula.
Scenario
Convert a porosity of n = 0.42 to void ratio.
Solution
e = n/(1-n) = 0.42/(1-0.42) = 0.42/0.58 = 0.724
Applications
- Relative density Dr calculation requires emax and emin
- Settlement prediction in consolidation uses change in void ratio Δe
- Permeability (Kozeny-Carman) is a function of void ratio
- Compaction control uses minimum and maximum void ratios
Misconceptions
- e and n are interchangeable — WRONG: e = Vv/Vs, n = Vv/V; they are related but not equal
- e cannot exceed 1.0 — WRONG: for soft clays, e can reach 2.0 or higher
- Porosity can be greater than 100% — WRONG: n is always between 0 and 1
Related Concepts
- Unit Weights of Soil
- Relative Density
- Consolidation Settlement
Common Exam Questions
Example
n = 0.35 → e = 0.35/0.65 = 0.538
Approach
Apply e = n/(1-n) or n = e/(1+e) directly. No phase diagram needed.
Question Type
Conversion between e and n
Example
γdry = 16.5 kN/m³, Gs = 2.68: e = 2.68×9.81/16.5 - 1 = 26.29/16.5 - 1 = 0.593
Approach
Use γdry = Gsγw/(1+e) rearranged: e = Gsγw/γdry - 1
Question Type
Finding e from unit weight and Gs
Key Points To Remember
- e = Vv/Vs; can exceed 1.0 for soft clays
- n = Vv/V; always between 0 and 1
- e = n/(1-n) and n = e/(1+e) — memorize both
- V/Vs = 1 + e (the denominator in all unit weight formulas)
- Typical Gs = 2.65–2.70 for most mineral soils
- e decreases (soil compresses) when load is applied or drainage occurs
Water Content and Degree of Saturation
The water content w = Ww/Ws is a weight-based ratio. It is expressed as a decimal or percentage and can exceed 1.0 (100%) for very soft organic soils. In the laboratory, it is measured by the oven-drying method (ASTM D2216): weigh wet sample, dry at 105°C, reweigh, then w = (Wwet - Wdry)/Wdry. The degree of saturation S = Vw/Vv tells what fraction of the void space is filled with water: S = 0 for bone-dry soil, S = 1.0 (100%) for fully saturated soil. The master identity linking S, e, w, and Gs is: Se = wGs. This single equation is the most powerful tool in phase relationship problems — it directly connects the volume world (S, e) with the weight world (w, Gs). Derivation: S = Vw/Vv. Since Vw = Ww/γw and Vs = Ws/(Gsγw), we have e = Vv/Vs ≥ Vw/Vs = Ww/(Gsγw) × (γw/Ws) = w/Gs × (1/1) ... leading to Se = wGs exactly.
Examples
For saturated soil, S = 1 simplifies Se = wGs to e = wGs. This is the most common board exam scenario for fine-grained soils.
Scenario
A saturated clay sample (S = 1.0) has w = 28% and Gs = 2.70. Find the void ratio and porosity.
Solution
Se = wGs → (1.0)e = (0.28)(2.70) = 0.756. Therefore e = 0.756. n = e/(1+e) = 0.756/1.756 = 0.430 = 43.0%.
Rearrange Se = wGs to S = wGs/e. This is the standard approach when e is given directly.
Scenario
A soil has Gs = 2.65, e = 0.72, and w = 20%. Find the degree of saturation.
Solution
S = wGs/e = (0.20)(2.65)/0.72 = 0.530/0.72 = 0.736 = 73.6%.
Applications
- Field compaction control: target water content near optimum w
- Soft ground improvement: drainage lowers w and increases shear strength
- Liquefaction susceptibility: saturated loose sands (S = 1) are most vulnerable
- Shrink-swell potential of expansive clays depends on w changes
Misconceptions
- Water content is a volume ratio — WRONG: w = Ww/Ws, both are weights/forces
- w cannot exceed 1.0 — WRONG: w > 1.0 is possible for very soft organic soils
- S = w — WRONG: S is volume-based while w is weight-based; they are linked only through Se = wGs
Related Concepts
- Void Ratio and Porosity
- Unit Weights of Soil
- Atterberg Limits
Common Exam Questions
Example
γ = 19.8 kN/m³, γdry = 17.2 kN/m³: w = 19.8/17.2 - 1 = 0.151 = 15.1%
Approach
Use γ = γdry(1+w), therefore w = γ/γdry - 1
Question Type
Find w given γ, γdry
Example
w = 0.18, Gs = 2.68, e = 0.65: S = 0.18×2.68/0.65 = 0.742 = 74.2%
Approach
Direct application of S = wGs/e from the master identity Se = wGs
Question Type
Find S given w, Gs, e
Key Points To Remember
- w = Ww/Ws — weight-based, NOT volume-based
- S = Vw/Vv — volume-based, between 0 and 1
- Master identity: Se = wGs (MEMORIZE THIS)
- w can exceed 100% for soft organic clays; S cannot exceed 100%
- For S = 1 (saturated): e = wGs directly
- For S = 0 (dry): w = 0 regardless of e or Gs
Unit Weights of Soil
Unit weight γ = W/V is the weight per unit total volume of soil. Four unit weights are used in practice, each appropriate for a different engineering context. The moist (bulk) unit weight γ = (Gs + Se)γw/(1+e) applies when soil is partially saturated. Equivalently, γ = γdry(1+w). The dry unit weight γdry = Gsγw/(1+e) = γ/(1+w) represents weight of solids per total volume — it is the compaction quality indicator. The saturated unit weight γsat = (Gs + e)γw/(1+e) applies when all voids are filled with water (S = 1; Se = e). The submerged (buoyant) unit weight γ' = γsat - γw accounts for Archimedes buoyancy and applies to soil below the water table in effective stress calculations. Typical values: γdry ≈ 14–18 kN/m³, γsat ≈ 18–22 kN/m³, γ' ≈ 8–12 kN/m³. Note that γw = 9.81 kN/m³ is used in Philippine practice (some references use 10 kN/m³ as an approximation — clarify with the exam problem).
Examples
This four-part problem type is one of the most common board exam formats. Note that γmoist < γsat because at w=18%, S=80.4% < 100% — the soil is not yet fully saturated, so some air remains in the voids, making it lighter than the fully saturated condition.
Scenario
A soil sample has Gs = 2.68, e = 0.60, and w = 18%. Calculate (a) γdry, (b) γmoist, (c) S, and (d) γsat.
Solution
(a) γdry = Gsγw/(1+e) = 2.68×9.81/1.60 = 26.29/1.60 = 16.43 kN/m³. (b) γmoist = γdry(1+w) = 16.43×1.18 = 19.39 kN/m³. (c) S = wGs/e = 0.18×2.68/0.60 = 0.804 = 80.4%. (d) γsat = (Gs+e)γw/(1+e) = (2.68+0.60)×9.81/1.60 = 3.28×9.81/1.60 = 32.17/1.60 = 20.11 kN/m³.
Note the high void ratio (e = 1.134 > 1) typical of soft marine clays common in Metro Manila and coastal Philippine cities. The relatively low γsat ≈ 17.6 kN/m³ reflects this high porosity.
Scenario
A saturated soft clay has w = 42% and Gs = 2.70. Find γsat and γ'.
Solution
Since S=1: e = wGs = 0.42×2.70 = 1.134. γsat = (Gs+e)γw/(1+e) = (2.70+1.134)×9.81/(1+1.134) = 3.834×9.81/2.134 = 37.61/2.134 = 17.63 kN/m³. γ' = γsat - γw = 17.63 - 9.81 = 7.82 kN/m³.
Applications
- Vertical stress calculation: σv = Σ(γi × hi) using appropriate unit weight per layer
- Effective stress: use γ' below water table, γ above (or γsat if saturated above WT)
- Lateral earth pressure: uses γ in Rankine and Coulomb formulas
- Bearing capacity: unit weight appears in the depth and width terms of the Terzaghi equation
- Settlement: initial stress state requires correct γ selection
Misconceptions
- Use γmoist below the water table — WRONG: use γsat or γ' depending on context
- γ' = γsat - γw requires a factor of 2 — WRONG: subtract exactly one γw
- γdry means the soil is completely dry — WRONG: it is the unit weight calculated as if all water were removed; the actual soil can still be moist
Related Concepts
- Effective Stress Principle
- Bearing Capacity of Foundations
- Lateral Earth Pressure
Common Exam Questions
Example
2m dry sand (γ=17kN/m³) over 3m saturated clay (γsat=19kN/m³): σ'v at bottom = 17×2 + (19-9.81)×3 = 34 + 27.57 = 61.57 kPa
Approach
Identify layers and water table position. Use γ above WT, γ' below WT. σ'v = Σ(γi hi) - u
Question Type
Compute effective stress at depth
Example
γdry = 15.8 kN/m³, Gs = 2.65: e = 2.65×9.81/15.8 - 1 = 26.0/15.8 - 1 = 0.645
Approach
Rearrange γdry = Gsγw/(1+e): e = Gsγw/γdry - 1
Question Type
Back-calculate e from γdry
Key Points To Remember
- γ = γdry(1+w) — the simplest moist unit weight formula
- γdry = Gsγw/(1+e) — dry unit weight from phase ratios
- γsat = (Gs+e)γw/(1+e) — set S=1 so Se=e in the moist formula
- γ' = γsat - γw — subtract exactly one γw, not more
- γw = 9.81 kN/m³ unless problem states otherwise
- γdry is the compaction quality indicator — higher γdry means denser soil
Specific Gravity of Solids (Gs)
The specific gravity of solids Gs = γs/γw = (unit weight of solid particles)/(unit weight of water) is a dimensionless constant for a given soil mineral. It is measured by the pycnometer method (ASTM D854). Typical values: quartz = 2.65, feldspar ≈ 2.60–2.70, most silicate minerals = 2.60–2.80, organic soils as low as 1.5–2.0. For mixed mineral soils, Gs ≈ 2.65–2.70 is a reliable assumption when not measured. Iron-rich laterites (common in the Philippine uplands) may have Gs up to 2.80–3.0. Gs appears in every unit weight formula and in the master identity Se = wGs, so it must be known or assumed for any phase calculation.
Examples
The pycnometer method exploits Archimedes principle: the volume of water displaced by the soil equals Ws/Gsγw. This test is routinely performed in Philippine soil mechanics laboratories.
Scenario
A 50 mL pycnometer is filled with water and weighs 155.0 g. When 10.0 g of dry soil is added and the pycnometer refilled with water, it weighs 159.2 g. Find Gs.
Solution
Weight of water filling pycnometer alone = 155.0 - (pycnometer mass). Using the standard pycnometer formula: Gs = Ws/(Ws + W1 - W2) where Ws = mass of dry soil = 10.0 g, W1 = mass of pycnometer + water = 155.0 g, W2 = mass of pycnometer + soil + water = 159.2 g. Gs = 10.0/(10.0 + 155.0 - 159.2) = 10.0/5.8 = 1.724... Re-check: Gs = Ws/(Ws - (W2 - W1)) = 10/(10 - (159.2 - 155.0)) = 10/(10-4.2) = 10/5.8 = 1.72. Note: this result seems low — a complete pycnometer problem requires pycnometer empty mass. This simplified version illustrates the concept; full ASTM D854 requires additional measurements.
Applications
- All unit weight formulas require Gs
- Master identity Se = wGs links all phase relationships through Gs
- Relative density computation uses Gs indirectly
- Sedimentation analysis (hydrometer test) uses Gs to find particle size
Misconceptions
- Gs is the specific gravity of the soil mass — WRONG: Gs refers to the solid particles only, not the bulk soil
- Gs = 1.0 for saturated soil — WRONG: Gs is a mineral property, independent of saturation
- Gs is the same as unit weight — WRONG: Gs is dimensionless and relative to water
Related Concepts
- Water Content
- Unit Weights of Soil
- Master Identity Se = wGs
Common Exam Questions
Example
Problem states 'sandy soil' with no Gs — assume Gs = 2.65 and proceed
Approach
If Gs is not stated, use 2.65 for sand, 2.70 for clay/silt unless told otherwise
Question Type
Assume Gs when not given
Key Points To Remember
- Gs = γs/γw = ρs/ρw (ratio of densities or unit weights)
- Typical Gs = 2.65–2.70 for most Philippine soils
- Organic soils have lower Gs (as low as 1.5)
- Lateritic soils may have higher Gs (up to 2.80–3.0)
- Gs is dimensionless and does not change with stress or water content
- Use Gs = 2.70 if not given and soil is described as clay or silt
Practice Problems
Step 1 converts n→e using e=n/(1-n). Step 2 uses the master identity rearranged to w = Se/Gs. Step 3 finds γdry then multiplies by (1+w) for moist unit weight. This three-part sequence is the standard board exam workflow.
Problem
PROBLEM 1 (Board Exam Type — November 2019 Style): A soil sample has the following properties: n = 0.40, Gs = 2.68, S = 0.80. Determine: (a) void ratio e, (b) water content w, and (c) moist unit weight γ.
Solution
(a) e = n/(1-n) = 0.40/(1-0.40) = 0.40/0.60 = 0.667. (b) From master identity Se = wGs: w = Se/Gs = (0.80)(0.667)/2.68 = 0.5336/2.68 = 0.199 = 19.9% ≈ 20.0%. (c) γdry = Gsγw/(1+e) = 2.68×9.81/(1+0.667) = 26.29/1.667 = 15.77 kN/m³. γmoist = γdry(1+w) = 15.77×(1+0.199) = 15.77×1.199 = 18.91 kN/m³.
This problem works backward from γ and w to find e and S. The key rearrangement is e = Gsγw/γdry - 1, derived from γdry = Gsγw/(1+e). Always verify: with e=0.604, γdry = 2.70×9.81/1.604 = 26.49/1.604 = 16.52 kN/m³ ✓.
Problem
PROBLEM 2 (Board Exam Type): A soil has γ = 19.0 kN/m³ and w = 15%. Given Gs = 2.70, determine: (a) γdry, (b) void ratio e, and (c) degree of saturation S.
Solution
(a) γdry = γ/(1+w) = 19.0/1.15 = 16.52 kN/m³. (b) e = Gsγw/γdry - 1 = (2.70×9.81)/16.52 - 1 = 26.49/16.52 - 1 = 1.604 - 1 = 0.604. (c) S = wGs/e = (0.15)(2.70)/0.604 = 0.405/0.604 = 0.671 = 67.1%.
For saturated soil, all three results follow a clear sequence: γsat from the standard formula, γ' by subtracting γw, and w = e/Gs from the master identity with S=1. This sequence is extremely common in board exams involving clays and saturated conditions.
Problem
PROBLEM 3 (Board Exam Type): A saturated clay has Gs = 2.72 and e = 0.95. Find: (a) γsat, (b) γ' (submerged), and (c) water content w.
Solution
(a) γsat = (Gs+e)γw/(1+e) = (2.72+0.95)×9.81/(1+0.95) = 3.67×9.81/1.95 = 36.00/1.95 = 18.46 kN/m³. (b) γ' = γsat - γw = 18.46 - 9.81 = 8.65 kN/m³. (c) For saturated soil (S=1): Se = wGs → e = wGs → w = e/Gs = 0.95/2.72 = 0.349 = 34.9%.
Both methods (total stress minus pore pressure, or direct use of γ') give the same answer — use whichever is more convenient. The water table position determines where to switch from γ or γdry to γ' in the effective stress calculation. This is a fundamental concept for bearing capacity and settlement problems.
Problem
PROBLEM 4 (Board Exam Type — Multi-layer Effective Stress): A soil profile has: Layer 1 — 2.0 m of dry sand, γdry = 16.0 kN/m³; Layer 2 — 3.0 m of saturated clay, γsat = 18.5 kN/m³. Water table is at the top of Layer 2. Calculate the effective vertical stress at the bottom of Layer 2.
Solution
Total vertical stress at bottom: σv = γdry×h1 + γsat×h2 = 16.0×2.0 + 18.5×3.0 = 32.0 + 55.5 = 87.5 kPa. Pore water pressure at bottom (water table at top of Layer 2, h = 3.0 m): u = γw×hw = 9.81×3.0 = 29.43 kPa. Effective stress: σ'v = σv - u = 87.5 - 29.43 = 58.07 kPa ≈ 58.1 kPa. Alternative using γ': σ'v = γdry×h1 + γ'×h2 = 16.0×2.0 + (18.5-9.81)×3.0 = 32.0 + 26.07 = 58.07 kPa ✓
This comprehensive five-part problem demonstrates the full phase relationship workflow from laboratory data. The unit g/cm³ is convenient for laboratory calculations; convert to kN/m³ for design (multiply by 9.81). The double-check using Se = wGs confirms all calculations are consistent.
Problem
PROBLEM 5 (Board Exam Type): A soil sample before drying weighs 185.0 g and after oven-drying weighs 160.0 g. The volume of the moist sample is 95.0 cm³ and Gs = 2.66. Find: (a) w, (b) γ, (c) γdry, (d) e, (e) S.
Solution
Given: Wwet = 185.0 g, Wdry = Ws = 160.0 g (oven-dry = solids only), V = 95.0 cm³, Gs = 2.66. γw = 0.00981 kN/L = 9.81 kN/m³ = 9.81×10⁻³ N/cm³ = 1.0 g/cm³ (using g/cm³ for this problem). (a) Ww = 185.0 - 160.0 = 25.0 g. w = Ww/Ws = 25.0/160.0 = 0.1563 = 15.6%. (b) γ = W/V = 185.0/95.0 = 1.947 g/cm³ = 19.10 kN/m³. (c) γdry = Ws/V = 160.0/95.0 = 1.684 g/cm³ = 16.52 kN/m³. Check: γdry = γ/(1+w) = 19.10/1.156 = 16.52 kN/m³ ✓. (d) Vs = Ws/(Gsγw) = 160.0/(2.66×1.0) = 60.15 cm³. Vv = V - Vs = 95.0 - 60.15 = 34.85 cm³. e = Vv/Vs = 34.85/60.15 = 0.580. (e) Vw = Ww/γw = 25.0/1.0 = 25.0 cm³. S = Vw/Vv = 25.0/34.85 = 0.717 = 71.7%. Check using master identity: S = wGs/e = 0.1563×2.66/0.580 = 0.4158/0.580 = 0.717 ✓
Exam Preparation Tips
- MASTER THE FORMULA Se = wGs FIRST — this single identity solves 80% of phase relationship problems; practice rearranging it for each unknown: e = wGs/S, S = wGs/e, w = Se/Gs, Gs = Se/w
- MEMORIZE THE FOUR UNIT WEIGHT FORMULAS in order: γdry = Gsγw/(1+e), γsat = (Gs+e)γw/(1+e), γ = γdry(1+w), γ' = γsat - γw; recognize that the first two share the same denominator (1+e)
- DRAW THE PHASE DIAGRAM FOR EVERY PROBLEM — assign all given quantities before selecting a formula; this prevents using the wrong formula or missing a step
- KNOW e-n CONVERSIONS BY HEART — e = n/(1-n) and n = e/(1+e); board exams frequently ask for both in the same problem
- USE γw = 9.81 kN/m³ consistently unless the problem explicitly states 10 kN/m³; a difference of 0.19 kN/m³ can change your answer enough to select a wrong option in a 4-choice MCQ
- FOR SATURATED SOIL, S = 1 simplifies everything — the master identity becomes e = wGs; γsat formula has Se = e; this is the case for most fine-grained soil problems
- SUBMERGED UNIT WEIGHT is always γ' = γsat - γw — apply to soil below the water table in effective stress computations; never use γmoist below the water table for saturated conditions
- CHECK YOUR DEGREE OF SATURATION — if your calculated S > 1.0, you made an error; recheck your e or Gs values
- TYPICAL Gs VALUES TO ASSUME: Use 2.65 for sand, 2.70 for silt/clay, 2.68 as a neutral assumption; state your assumption if Gs is not given
- TIME MANAGEMENT: Phase relationship problems should take 3–5 minutes each on the board exam; if you spend more than 5 minutes on one, move on and return; these are among the fastest-solvable problems with good formula recall
- PRACTICE UNIT CONVERSIONS — board exam problems may give density in g/cm³ and ask for kN/m³; remember 1 g/cm³ = 9.81 kN/m³ exactly
- FOR MULTI-PART PROBLEMS, carry forward rounded intermediate answers carefully — round only at the final step, or use unrounded values throughout to avoid cumulative error
In summary
Soil phase relationships are the bedrock of Geotechnical Engineering — every subsequent topic, from Atterberg limits and compaction to effective stress, bearing capacity, and consolidation settlement, builds directly on these concepts. The three-phase model organizes all phase quantities into volumes (Vs, Vw, Va, Vv, V) and weights (Ws, Ww, W). The master identity Se = wGs links the volume world (S, e) to the weight world (w, Gs) and must be memorized with full fluency in all four rearrangements. The four unit weights — dry, moist, saturated, and submerged — each have a specific application context; using the wrong unit weight is one of the most penalized errors on the PRC board examination. Practice the complete workflow: draw the phase diagram, identify given quantities, select the correct formula path, compute, and verify using Se = wGs. Filipino civil engineers regularly encounter soft marine clays (high e, high w, low γ') in Metro Manila, Cebu, and coastal areas, as well as lateritic soils (high Gs) in Mindanao and the uplands — understanding the full range of phase property values is essential for professional practice. Consistent study of the five worked problems in this chapter, combined with the visual aids and formula summary, will ensure mastery of this topic before examination day.
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