CELE Geotechnical Engineering Reviewer 2026

Answer: A. Solids, water, and air

Step 1 – Recall the definition: Soil is a particulate material composed of three distinct phases. Step 2 – Identify each phase: (a) Solid phase – the mineral skeleton or soil grains; (b) Liquid phase – pore water; (c) Gas phase – pore air (or other gases). Step 3 – Eliminate wrong options: 'Clay, silt, and sand' are particle-size classifications, not phases. 'Gravel, water, and organic matter' mixes classification with phase. 'Plasma' does not exist in soil. Step 4 – Confirm: Solids + Water + Air is the universally accepted three-phase model used in all geotechnical analyses (Das, Principles of Geotechnical Engineering; NSCP context).

Answer: B. e = 0.667

Step 1 – Identify given values: V = 1.0 m³, Vs = 0.60 m³. Step 2 – Find volume of voids: Vv = V − Vs = 1.0 − 0.60 = 0.40 m³. Step 3 – Apply the void ratio formula: e = Vv / Vs = 0.40 / 0.60 = 0.667. Step 4 – Check wrong options: e = 0.40 is the porosity n (Vv/V), not the void ratio. e = 1.50 would result from inverting the ratio (Vs/Vv). e = 0.60 equals Vs, not a ratio. Step 5 – Key distinction: Void ratio uses volume of SOLIDS in the denominator, while porosity uses TOTAL volume.

Answer: B. n = 0.444

Step 1 – Recall the conversion formula: n = e / (1 + e). Step 2 – Substitute: n = 0.80 / (1 + 0.80) = 0.80 / 1.80 = 0.444. Step 3 – Check: n must always be between 0 and 1 (it is a ratio of voids to total volume). ✓ Step 4 – Eliminate wrong options: n = 0.80 incorrectly equates n with e (they are different). n = 0.556 = 1 − n, which is the solid fraction. n = 1.25 > 1, which is impossible for porosity. Step 5 – Inverse check: e = n/(1−n) = 0.444/0.556 = 0.80 ✓.

Answer: B. w = 20.0%

Step 1 – Find mass of water: Mw = Mwet − Mdry = 180 − 150 = 30 g. Step 2 – Apply water content formula: w = Mw / Ms = 30 / 150 = 0.20 = 20%. Step 3 – Note: Water content is always expressed as a ratio of the WATER mass to the DRY SOIL (solids) mass, not to total mass. Step 4 – Check wrong options: w = 16.7% uses total mass in denominator (30/180) — a common mistake. w = 83.3% is Ms/Mtotal. w = 30% has no valid basis. Step 5 – Key reminder: w can exceed 100% in very soft clays and organic soils — it is not a volume fraction.

Answer: B. S · e = w · Gs

Step 1 – Derive from volume definitions: Volume of water Vw = Ww / γw = (w · Ws) / γw. Step 2 – Volume of solids: Vs = Ws / (Gs · γw). Step 3 – Degree of saturation: S = Vw / Vv = Vw / (e · Vs). Step 4 – Substituting: S = [w · Ws / γw] / [e · Ws / (Gs · γw)] = w · Gs / e, which rearranges to S · e = w · Gs. Step 5 – This identity is fundamental — it appears in nearly every phase-relationship problem on the board exam. Memorize it. The other options have incorrect variable positions.

Answer: A. 15.29 kN/m³

Step 1 – Formula: γdry = Gs · γw / (1 + e). Step 2 – Substitute: γdry = 2.65 × 9.81 / (1 + 0.70) = 25.997 / 1.70. Step 3 – Compute: γdry = 15.29 kN/m³. Step 4 – Check wrong options: 18.40 kN/m³ is a typical moist unit weight — it includes water contribution. 12.65 kN/m³ results from dividing Gs·γw by (1 + e)² (incorrect). 16.25 kN/m³ corresponds to e = 0.60 (Gs = 2.65), not 0.70. Step 5 – Physical sense check: γdry must be less than γsat and less than Gs·γw = 26.0 kN/m³. ✓

Answer: C. 21.0 kN/m³

Step 1 – Formula: γsat = (Gs + e) · γw / (1 + e). Step 2 – Substitute: γsat = (2.70 + 0.486) × 9.81 / (1 + 0.486) = 3.186 × 9.81 / 1.486. Step 3 – Numerator: 3.186 × 9.81 = 31.25 kN/m³. Step 4 – Divide: 31.25 / 1.486 ≈ 21.03 ≈ 21.0 kN/m³. Step 5 – This matches Example 1 from the reference notes. Note that for full saturation S = 1, so Se = e = wGs, and the saturated formula uses (Gs + e) in the numerator because all voids are filled with water contributing γw·e/(1+e) to the unit weight.

Answer: C. 18.20 kN/m³

Step 1 – Formula: γ = γdry × (1 + w). Step 2 – Convert w to decimal: w = 12% = 0.12. Step 3 – Substitute: γ = 16.25 × (1 + 0.12) = 16.25 × 1.12. Step 4 – Compute: γ = 18.20 kN/m³. Step 5 – Physical check: Moist unit weight must be greater than dry unit weight (water adds weight) and less than saturated unit weight. ✓ Wrong options: 16.25 is γdry (forgot to add water), 17.50 used w = 7.7% incorrectly, 19.00 overshoots the correct multiplication.

Answer: B. 11.19 kN/m³

Step 1 – Formula: γ' = γsat − γw. Step 2 – Substitute: γ' = 21.0 − 9.81 = 11.19 kN/m³. Step 3 – Physical meaning: When soil is submerged, buoyancy reduces the effective weight by γw per unit volume. Step 4 – Check wrong options: 21.0 kN/m³ is γsat (no buoyancy correction applied). 9.81 kN/m³ is just γw — this would imply the solids contribute nothing, which is wrong. 30.81 kN/m³ adds instead of subtracts γw. Step 5 – Important: Always subtract γw from γsat — never from γdry or γmoist — for the submerged condition.

Answer: B. S = 0.53 (53%)

Step 1 – Use the master identity rearranged: S = wGs / e. Step 2 – Substitute: S = (0.12 × 2.65) / 0.60. Step 3 – Numerator: 0.12 × 2.65 = 0.318. Step 4 – Divide: S = 0.318 / 0.60 = 0.53 = 53%. Step 5 – Check: S must be between 0 (dry) and 1 (saturated). ✓ Wrong options: S = 40% has no calculation basis here. S = 72% would result from using n instead of e in the denominator. S = 100% means fully saturated — but the moist void ratio and water content given do not satisfy Se = wGs at S = 1.

Answer: B. D₆₀ / D₁₀

Step 1 – Recall the formula: Cᵤ = D₆₀ / D₁₀, where D₆₀ is the diameter at which 60% of the soil mass is finer and D₁₀ (the effective size) is the diameter at which 10% is finer. Step 2 – Cᵤ measures how spread-out the grain sizes are. A large Cᵤ means a wide range of grain sizes (well-graded potential). Step 3 – Option A (D₃₀/D₁₀) is not a standard parameter. Option C is the formula for the coefficient of gradation (Cc), not Cᵤ. Option D is the inverse and has no standard classification meaning. Therefore, Option B is correct.

Answer: A. 0.90

Step 1 – Recall the formula: Cc = D₃₀² / (D₁₀ × D₆₀). Step 2 – Substitute the given values: Cc = (0.60)² / (0.20 × 2.0) = 0.36 / 0.40 = 0.90. Step 3 – Option B (10.0) is the value of Cᵤ = D₆₀/D₁₀ = 2.0/0.20 — a common mix-up. Option C (1.80) results from using D₃₀ directly instead of D₃₀². Option D (0.36) is just the numerator D₃₀² without dividing. Step 4 – Since Cc = 0.90 < 1, the sand fails the Cc criterion for well-graded classification.