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CELE Geotechnical EngineeringSoil Properties and Phase RelationshipsRevision Notes

Revision notes for CELE Geotechnical Engineering Soil Properties and Phase Relationships — designed for time-pressed reviewers. These notes skip the basics and focus on what Professional Regulation Commission (PRC) — Board of Civil Engineering consistently tests, so you spend your revision hours on the content most likely to appear on exam day.

Exam context

The Civil Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Civil Engineering and is scheduled for May and November 2026. The Geotechnical Engineering subtest is marked as "Core" in the official pattern, and Soil Properties and Phase Relationships appears in position 1st of 11 in the CELE Geotechnical Engineering review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent CELE 2026 papers have drawn roughly a meaningful share of questions from this subject.

Soil Properties and Phase Relationships - Revision Notes

Soil is a three-phase material consisting of solid particles, water (pore fluid), and air. Before any geotechnical analysis — bearing capacity, settlement, slope stability, or lateral earth pressure — an engineer must quantify how much of each phase is present. This chapter establishes the fundamental phase relationships (void ratio, porosity, water content, degree of saturation, specific gravity) and the unit-weight expressions derived from them. These relationships appear in virtually every topic of the PRC Civil Engineer Licensure Examination in Geotechnical Engineering. Master the phase diagram, the master identity Se = wGs, and the four unit-weight formulas, and you will solve any board problem in this area with confidence.

Sections

Formulas

Example

A soil specimen has Vs = 0.60 m³ and Vv = 0.40 m³. Then e = 0.40 / 0.60 = 0.667.

Formula

e = Vv / Vs

Variables

e = void ratio (dimensionless); Vv = volume of voids (m³); Vs = volume of solids (m³)

Application

Fundamental soil structure descriptor used in all unit-weight and settlement calculations.

Example

If e = 0.667, then n = 0.667 / 1.667 = 0.40 = 40%.

Formula

n = Vv / V = e / (1 + e)

Variables

n = porosity (dimensionless); V = total volume (m³); e = void ratio

Application

Used in permeability and compaction analyses. Converts between e and n.

Example

If n = 0.40, then e = 0.40 / 0.60 = 0.667.

Formula

e = n / (1 - n)

Variables

e = void ratio; n = porosity (as a decimal)

Application

Inverse conversion from porosity to void ratio.

Example

A 150 g moist sample, after oven drying, weighs 130 g. Then Ww = 20 g, w = 20/130 = 0.154 = 15.4%.

Formula

w = Ww / Ws

Variables

w = water content (decimal or %); Ww = weight of water (kN or g); Ws = weight of solids (kN or g)

Application

Determined by oven-drying test (ASTM D2216). Used in all unit weight and Se = wGs calculations.

Example

If Vw = 0.30 m³ and Vv = 0.40 m³, then S = 0.30/0.40 = 0.75 = 75%.

Formula

S = Vw / Vv

Variables

S = degree of saturation (decimal or %); Vw = volume of water (m³); Vv = volume of voids (m³)

Application

Indicates how filled voids are. Used in moist unit-weight formula.

Example

Given w = 0.18, Gs = 2.70, S = 1.0 (saturated): e = wGs/S = 0.18×2.70/1.0 = 0.486.

Formula

S·e = w·Gs

Variables

S = degree of saturation (decimal); e = void ratio; w = water content (decimal); Gs = specific gravity of solids

Application

The master identity. Use to find any one of the four parameters when the other three are known.

Example

For a mineral with γs = 26.5 kN/m³: Gs = 26.5/9.81 = 2.70.

Formula

Gs = γs / γw

Variables

Gs = specific gravity of solids; γs = unit weight of solid particles (kN/m³); γw = unit weight of water = 9.81 kN/m³

Application

Used in all unit-weight derivations. Determines density of solid mineral grains.

Exam Tips

  • Draw the phase diagram for every problem — label Vs, Vw, Va, Ws, Ww. This prevents errors in identifying which volumes and weights to use.
  • The master identity Se = wGs is the single most-tested equation — memorise it forward and backward.
  • When a soil is described as 'saturated,' immediately set S = 1.0, simplifying Se = wGs to e = wGs.
  • When a soil is described as 'dry,' set S = 0 and w = 0, meaning Vw = 0 and γ = γdry.
  • Always check: S ≤ 1.0 after computing S = wGs/e. If S > 1.0, recheck your inputs — a physical impossibility signals an arithmetic error.

Key Points

  • Soil is composed of three phases: solid particles (S), water (W), and air (A). The phase diagram separates these three components by volume and weight.
  • Total volume: V = Vs + Vv, where Vv = Vw + Va (void volume = water volume + air volume).
  • Total weight: W = Ws + Ww (air has negligible weight in engineering practice).
  • Void ratio e = Vv / Vs: ratio of void volume to solid volume. Range: loose sands ~0.8; dense sands ~0.4; soft clays can exceed 1.5.
  • Porosity n = Vv / V: ratio of void volume to total volume. Always between 0 and 1 (or 0% to 100%).
  • Water content w = Ww / Ws: weight of water divided by weight of solids — NOT volumes. Expressed as a decimal or percentage. Can exceed 100% in very soft clays.
  • Degree of saturation S = Vw / Vv: fraction of voids filled with water. S = 0 for completely dry soil; S = 1.0 (100%) for fully saturated soil.
  • Specific gravity of solids Gs = γs / γw = ρs / ρw. Typical range: 2.60–2.80; most mineral soils use 2.65–2.70.
  • The MASTER IDENTITY: S·e = w·Gs. This single equation links the four most-tested parameters.

Definitions

Term

Void Ratio (e)

Definition

The ratio of the volume of voids to the volume of solid particles in a soil mass: e = Vv/Vs.

Importance

Primary index property. Used directly in all unit-weight, settlement, and permeability formulas. Unlike porosity, e can exceed 1.0.

Term

Porosity (n)

Definition

The ratio of the volume of voids to the total volume of the soil mass: n = Vv/V.

Importance

Always between 0 and 1. Commonly used in permeability (Kozeny-Carman) and compaction contexts.

Term

Water Content (w)

Definition

The ratio of the weight of water to the weight of dry solids: w = Ww/Ws. Measured by oven-drying at 105°C.

Importance

Most commonly measured soil property in the laboratory. Directly affects shear strength, compressibility, and unit weight.

Term

Degree of Saturation (S)

Definition

The ratio of the volume of water to the volume of voids: S = Vw/Vv. Ranges from 0 (dry) to 1.0 (fully saturated).

Importance

Determines which unit weight formula to apply. S = 1.0 for saturated soils below the water table.

Term

Specific Gravity of Solids (Gs)

Definition

The ratio of the unit weight of solid particles to the unit weight of water: Gs = γs/γw.

Importance

A mineral constant, typically 2.65–2.70, used in Se = wGs and all unit-weight formulas. Measured by pycnometer test.

Section Title

The Phase Diagram and Basic Volume–Weight Ratios

Common Mistakes

  • Confusing void ratio e (can exceed 1.0) with porosity n (always less than 1.0) — never substitute one for the other directly.
  • Expressing water content as Vw/Vs or Vw/Vv — water content is ALWAYS weight-based: w = Ww/Ws.
  • Forgetting to convert w and S to decimals before applying Se = wGs — using percentages gives wrong answers.
  • Assuming Gs = 1.0 (the value for water) instead of 2.65–2.70 for solid mineral grains.
  • Using Gs = 2.65 when the problem explicitly states a different value — always use the given Gs.

Formulas

Example

Gs = 2.65, e = 0.60: γdry = 2.65×9.81/1.60 = 25.997/1.60 = 16.25 kN/m³.

Formula

γdry = Gs·γw / (1 + e)

Variables

γdry = dry unit weight (kN/m³); Gs = specific gravity of solids; γw = 9.81 kN/m³; e = void ratio

Application

Compaction control (compare to Proctor maximum dry unit weight). Back-calculate e from field γdry.

Example

Gs = 2.65, e = 0.60, S = 0.53: γmoist = (2.65+0.53×0.60)×9.81/1.60 = (2.65+0.318)×9.81/1.60 = 2.968×9.81/1.60 = 18.2 kN/m³.

Formula

γmoist = (Gs + S·e)·γw / (1 + e)

Variables

γmoist = moist unit weight (kN/m³); S = degree of saturation (decimal); all other symbols as before

Application

General formula applicable to any saturation condition.

Example

γdry = 16.25 kN/m³, w = 0.12: γmoist = 16.25×1.12 = 18.20 kN/m³.

Formula

γmoist = γdry·(1 + w)

Variables

w = water content as a decimal

Application

Quick conversion between dry and moist unit weight. Widely used when γdry and w are given.

Example

Gs = 2.70, e = 0.486: γsat = (2.70+0.486)×9.81/1.486 = 3.186×9.81/1.486 = 21.03 kN/m³.

Formula

γsat = (Gs + e)·γw / (1 + e)

Variables

γsat = saturated unit weight (kN/m³); e = void ratio; Gs = specific gravity

Application

Used for soils below the groundwater table when computing total stress.

Example

γsat = 21.03 kN/m³: γ' = 21.03 − 9.81 = 11.22 kN/m³.

Formula

γ' = γsat − γw

Variables

γ' = submerged (buoyant) unit weight (kN/m³); γw = 9.81 kN/m³

Application

Effective stress calculations below the water table. Represents net downward unit weight after buoyancy is deducted.

Example

γdry = 16.25 kN/m³, Gs = 2.65: e = (2.65×9.81/16.25)−1 = (26.00/16.25)−1 = 1.60−1 = 0.60.

Formula

e = (Gs·γw / γdry) − 1

Variables

Rearranged from γdry formula to back-calculate void ratio from a known dry unit weight.

Application

Field compaction verification — determine in-situ void ratio from density test results.

Exam Tips

  • Memorise the three principal formulas: γdry = Gs·γw/(1+e), γsat = (Gs+e)·γw/(1+e), γ' = γsat − γw. These cover 90% of board questions on unit weight.
  • Notice that γsat − γdry = e·γw/(1+e) = n·γw — this is a useful cross-check in multi-part problems.
  • If a problem gives γ and w but not e, find γdry = γ/(1+w) first, then back-calculate e = Gs·γw/γdry − 1.
  • For a quick sanity check: γ' ≈ γsat − 9.81 should give a value between 8 and 13 kN/m³ for most natural soils.
  • Board problems often give three of the four quantities {S, e, w, Gs} and ask for the fourth — apply Se = wGs directly.

Key Points

  • Four unit weights are used in geotechnical engineering: moist (bulk), dry, saturated, and submerged (buoyant/effective).
  • Dry unit weight γdry = Gs·γw / (1+e): derived by setting Ww = 0 in the phase diagram. Indicator of compaction state.
  • Moist unit weight γ = (Gs + Se)·γw / (1+e), equivalently γ = γdry·(1+w). Use when partial saturation exists.
  • Saturated unit weight γsat = (Gs + e)·γw / (1+e): obtained by setting S = 1.0 in the moist formula.
  • Submerged (buoyant) unit weight γ' = γsat − γw: represents the effective weight of soil submerged under water. Used in effective stress calculations below the water table.
  • Typical values (kN/m³): γdry = 14–18; γmoist = 17–21; γsat = 18–23; γ' = 8–13.
  • Unit weight of water γw = 9.81 kN/m³ (use this exact value unless the problem specifies 10 kN/m³).
  • The relationship between dry and moist unit weight: γdry = γ / (1+w). This is a board-exam shortcut for back-calculating void ratio from field measurements.

Definitions

Term

Dry Unit Weight (γdry)

Definition

The weight of solid particles per unit total volume: γdry = Ws/V = Gs·γw/(1+e).

Importance

Key indicator of compaction quality. Compared against Proctor maximum dry density in field control tests.

Term

Moist (Bulk) Unit Weight (γ)

Definition

Total weight (solids + water) per total volume: γ = W/V. The unit weight measured in most standard density tests.

Importance

Used to compute total stress in the unsaturated zone above the water table.

Term

Saturated Unit Weight (γsat)

Definition

Unit weight when all voids are filled with water (S = 1.0): γsat = (Gs + e)·γw/(1+e).

Importance

Used for soils below the groundwater table. Always greater than γdry and γmoist.

Term

Submerged (Buoyant) Unit Weight (γ')

Definition

Effective unit weight of soil submerged in water: γ' = γsat − γw. Accounts for Archimedes' buoyant force.

Importance

Critical for effective stress calculations in saturated profiles. Typical range: 8–13 kN/m³.

Section Title

Unit Weights of Soil

Common Mistakes

  • Subtracting γw from γmoist instead of γsat when computing buoyant unit weight — γ' is ALWAYS γsat − γw.
  • Forgetting to add 1 to e in the denominator of the unit weight formulas — (1+e) is in the denominator, not e alone.
  • Using γw = 10 kN/m³ when the problem does not specify this simplification — default is 9.81 kN/m³.
  • Mixing up γdry and γmoist in effective stress calculations — use γmoist above the water table and γsat below.
  • Computing γ' by subtracting γw from γmoist — this is physically incorrect; buoyancy applies only to saturated submerged soil.

Formulas

Example

n = e/(1+e) = 0.486/1.486 = 0.327 = 32.7%; γsat = (2.70+0.486)×9.81/1.486 = 21.0 kN/m³; γ' = 21.0−9.81 = 11.2 kN/m³.

Formula

Step-by-step for Problem 1: e = wGs/S = 0.18×2.70/1.0 = 0.486

Variables

S = 1.0 for saturated; w = 0.18; Gs = 2.70

Application

Finding void ratio of saturated soil from water content and specific gravity.

Example

γmoist = γdry(1+w) = 16.25×1.12 = 18.2 kN/m³; S = wGs/e = 0.12×2.65/0.60 = 0.53 = 53%.

Formula

Step-by-step for Problem 2: γdry = Gs·γw/(1+e) = 2.65×9.81/1.60 = 16.25 kN/m³

Variables

Gs = 2.65; e = 0.60; γw = 9.81 kN/m³

Application

Computing dry and moist unit weights from given Gs and e.

Example

e = Gs·γw/γdry − 1 = 2.70×9.81/16.52 − 1 = 26.49/16.52 − 1 = 0.603; S = wGs/e = 0.15×2.70/0.603 = 0.672 = 67.2%.

Formula

Step-by-step for Problem 3: γdry = γ/(1+w) = 19/1.15 = 16.52 kN/m³

Variables

γ = 19 kN/m³; w = 0.15

Application

Back-calculating dry unit weight and void ratio from field density and water content.

Example

w = Se/Gs = 0.80×0.667/2.68 = 0.199 = 19.9%; γmoist = (Gs+Se)·γw/(1+e) = (2.68+0.80×0.667)×9.81/1.667 = (2.68+0.534)×9.81/1.667 = 3.214×9.81/1.667 = 18.90 kN/m³.

Formula

Step-by-step for Problem 4: e = n/(1−n) = 0.40/0.60 = 0.667

Variables

n = 0.40; then apply Se = wGs

Application

Starting from porosity to find void ratio, then water content, then unit weight.

Exam Tips

  • In a 5-minute board problem, spend 30 seconds drawing the phase diagram — it saves time in the long run by preventing sign and assignment errors.
  • The sequence for most problems: Step 1 — Se = wGs (find missing parameter); Step 2 — γdry = Gs·γw/(1+e); Step 3 — apply γ = γdry(1+w) or γsat = (Gs+e)·γw/(1+e).
  • Unit weight problems often have three parts in sequence: the answer from Part 1 feeds into Part 2. Carry full decimal precision until the final answer.
  • Typical answer ranges: e = 0.4–1.5; n = 0.28–0.60; w = 5–80%; S = 0–100%; γsat = 17–23 kN/m³. Results outside these ranges usually indicate an error.
  • When the problem says 'determine the degree of saturation,' it is almost always asking you to use S = wGs/e after finding e from another given.

Key Points

  • Problem 1 — Saturated soil: Given w = 18%, Gs = 2.70, S = 100% (saturated). Find e, n, γsat, γ'.
  • Problem 2 — Partially saturated soil: Given Gs = 2.65, e = 0.60, w = 12%. Find γdry, γmoist, S.
  • Problem 3 — Back-calculation from field data: Given γ = 19 kN/m³, w = 15%, Gs = 2.70. Find γdry, e, S.
  • Problem 4 — From porosity: Given n = 0.40, Gs = 2.68, S = 0.80. Find e, w, γmoist.
  • Systematic approach: (1) Draw phase diagram; (2) Apply Se = wGs; (3) Compute unit weight formula; (4) Check S ≤ 1.0.
  • Always write out given data, identify the unknown, select the governing equation, substitute, and box the answer with units.

Definitions

Term

Phase Diagram Method

Definition

A schematic drawing that separates the soil into its three constituent phases (solid, water, air) by volume on one side and by weight on the other, allowing systematic computation of all phase relationships.

Importance

Ensures no volume or weight is double-counted. Highly recommended for board exam solutions to earn full partial credit.

Term

Back-Calculation

Definition

The process of starting from measured field values (γ, w) to determine index properties (e, S, n) rather than computing γ from index properties.

Importance

Commonly tested in board problems that describe a field density test result and ask for the void ratio or degree of saturation.

Section Title

Worked Board-Style Problems

Common Mistakes

  • Skipping the phase diagram and making volume/weight assignment errors that propagate through the entire solution.
  • Not converting percentage values of w and S to decimal form before substituting into formulas.
  • In Problem 3 type questions, computing e using γmoist instead of γdry in the formula e = Gs·γw/γdry − 1.
  • Mixing up n and e when given porosity — always convert n to e first using e = n/(1−n).
  • Reporting degrees of saturation greater than 100% without recognising the arithmetic error.

Formulas

Example

Given S = 0.70, e = 0.55, Gs = 2.68: w = Se/Gs = 0.70×0.55/2.68 = 0.1437 = 14.4%.

Formula

Summary: Se = wGs → e = wGs/S → S = wGs/e → w = Se/Gs

Variables

All forms of the master identity for direct problem-solving

Application

Use the form that isolates the unknown parameter from the three given parameters.

Example

γ = 19.5 kN/m³, w = 20%: γdry = 19.5/1.20 = 16.25 kN/m³.

Formula

γdry = γ / (1 + w)

Variables

γ = moist unit weight; w = water content (decimal)

Application

Fastest way to get γdry when γ and w are both given — avoids needing e as an intermediate step.

Exam Tips

  • Create a personal formula sheet with the five core equations: Se = wGs; e = n/(1−n); γdry = Gs·γw/(1+e); γsat = (Gs+e)·γw/(1+e); γ' = γsat − γw.
  • Practice converting between all four variables in Se = wGs in all four directions until it becomes automatic.
  • For multi-step problems, label every computed quantity and carry 4 significant figures throughout; round only at the final answer.
  • When given γsat and asked for γ', the answer is always γsat − 9.81 — this is a one-second computation, do not overthink it.
  • Review the solved examples in NSCP 2015 Volume I Appendix discussions and Punzalan/Gillesania review books for additional Filipino board-style practice.

Key Points

  • The four most-tested quantities are e, w, S, and Gs — tied by Se = wGs. Mastering this one equation handles half the phase-relationship problems.
  • Porosity n and void ratio e are different — they are linked by e = n/(1−n) and n = e/(1+e). Never write e when you mean n.
  • Water content is WEIGHT-based (Ww/Ws), not volume-based. This is a deliberate distractor in many board problems.
  • Submerged unit weight ALWAYS uses γsat, not γmoist: γ' = γsat − γw.
  • Gs for most mineral soils is 2.65–2.70; if not given, assume Gs = 2.70 for conservative calculations in saturated clay problems.
  • γw = 9.81 kN/m³ is the standard value. Some problems use γw = 10 kN/m³ for simplicity — use whichever is given.
  • A higher void ratio means a looser, more compressible, more permeable soil — qualitative reasoning questions also appear on the board.
  • When a clay shrinks (desiccation or consolidation), e decreases, n decreases, and γdry increases — Gs remains constant.

Definitions

Term

Master Identity

Definition

The equation Se = wGs that mathematically links degree of saturation, void ratio, water content, and specific gravity of solids.

Importance

The single most important equation in phase relationships. Appears directly or indirectly in nearly every board problem on this topic.

Term

Relative Density (Dr)

Definition

Dr = (emax − e) / (emax − emin). Describes how dense a granular soil is relative to its loosest and densest possible states.

Importance

While not derived from the three-phase diagram, Dr uses void ratio and appears in conjunction with phase relationships in board questions on granular soil characterisation.

Section Title

Common Board-Exam Pitfalls and Key Takeaways

Common Mistakes

  • Applying γ' = γsat − γw to soil above the water table — buoyancy only acts on fully submerged soil.
  • Using the saturated unit weight formula when the soil is only partially saturated — use the moist formula with the given S value.
  • Forgetting that e can exceed 1.0 for soft clays — this is physically valid and not an error.
  • Reporting final answers without units — board examiners expect kN/m³ for unit weight, % or decimal for w, S, and n.
  • Rounding intermediate results to 2 decimal places too early, causing propagated errors in multi-step problems.

Connections

  • Effective Stress Principle (σ' = σ − u): Relies directly on γdry, γmoist, γsat, and γ' for computing total and effective vertical stresses in soil profiles with or without a water table.
  • Consolidation and Settlement: The change in void ratio (Δe) drives primary consolidation settlement. Phase relationships define the initial and final e values used in settlement calculations.
  • Compaction: Proctor tests produce a curve of γdry vs. w. The optimum compaction condition is found using phase relationship formulas, particularly γdry = Gs·γw/(1+e) and Se = wGs.
  • Permeability (Darcy's Law): Hydraulic conductivity k is correlated to e and n (e.g., Kozeny-Carman equation). Phase relationships define the void ratio input.
  • Shear Strength (Mohr-Coulomb): Undrained shear strength of saturated clays depends on water content and void ratio at failure. Phase relationships define the initial state before shearing.
  • Slope Stability and Lateral Earth Pressure: Unit weights (γ, γsat, γ') from phase relationships are direct inputs to slope stability analyses and Rankine/Coulomb earth pressure calculations.
  • Soil Classification (USCS/AASHTO): Atterberg limits and grain-size analysis define soil classification, while phase relationships provide the engineering properties (γdry, e, w) that quantify the engineering behavior within each class.

Exam Strategy

For the PRC Civil Engineer Licensure Examination (Geotechnical Engineering), phase relationships typically account for 3–6 questions per board exam cycle, often embedded in multi-part problems. The optimal strategy is: (1) Memorise five core equations cold — Se = wGs, e = n/(1−n), γdry = Gs·γw/(1+e), γsat = (Gs+e)·γw/(1+e), γ' = γsat − γw. (2) Always draw a phase diagram for any problem involving more than one unknown — it takes 20 seconds and prevents 80% of errors. (3) Know the four conversion directions of Se = wGs: solve for e, S, w, or Gs depending on which three are given. (4) Watch for problems that give porosity n instead of void ratio e — convert with e = n/(1−n) before substituting. (5) Identify the saturation condition from keywords: 'saturated' → S=1, 'dry' → S=0 and w=0, 'submerged below water table' → use γsat and γ' for effective stress, not γmoist. (6) In multi-part problems, carry full precision (4 decimal places) throughout and round only the boxed final answer. (7) Cross-check computed S values — S > 1.0 is physically impossible and signals an arithmetic error to correct before time runs out.

Quick Review Questions

A saturated soil has water content w = 35% and specific gravity Gs = 2.70. What is the void ratio?

For saturated soil, S = 1.0. Apply the master identity: e = wGs/S = 0.35 × 2.70 / 1.0 = 0.945.

If a soil has void ratio e = 0.72, what is its porosity n?

n = e/(1+e) = 0.72/1.72 = 0.4186 ≈ 41.9%. Note: n and e are related but not equal.

A soil has Gs = 2.68, e = 0.55, and S = 0.65. What is the water content?

From Se = wGs: w = Se/Gs = (0.65 × 0.55)/2.68 = 0.3575/2.68 = 0.1334 = 13.34%.

Compute the dry unit weight of a soil with Gs = 2.65 and e = 0.75.

γdry = Gs·γw/(1+e) = 2.65 × 9.81 / (1+0.75) = 26.0/1.75 = 14.86 kN/m³.

A soil has γdry = 16.50 kN/m³ and water content w = 18%. What is the moist unit weight?

γmoist = γdry × (1+w) = 16.50 × 1.18 = 19.47 kN/m³.

A saturated clay has γsat = 20.5 kN/m³. What is the submerged unit weight?

γ' = γsat − γw = 20.5 − 9.81 = 10.69 kN/m³. Always subtract γw from γsat, not from γmoist.

A soil sample has γ = 18.8 kN/m³ and w = 14%. Find γdry.

γdry = γ/(1+w) = 18.8/1.14 = 16.49 kN/m³. This is the field back-calculation approach.

A soil has n = 0.35 and Gs = 2.66. If fully saturated, find γsat.

First, e = n/(1−n) = 0.35/0.65 = 0.5385. Then γsat = (Gs+e)·γw/(1+e) = (2.66+0.5385)×9.81/1.5385 = 3.1985×9.81/1.5385 = 20.38 kN/m³. [Exact: 20.38 kN/m³]

What does Se = wGs become for a completely dry soil?

For a dry soil, S = 0. The master identity gives 0·e = w·Gs, which means w = 0. This confirms that a completely dry soil has zero water content.

A soil has Gs = 2.70, e = 0.486, and S = 1.0. Compute γsat and γ'.

γsat = (2.70+0.486)×9.81/1.486 = 3.186×9.81/1.486 = 21.03 kN/m³. γ' = 21.03−9.81 = 11.22 kN/m³.

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