CELE 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: A. 78.48 kPa

Step 1 — Identify the formula for gauge pressure: p = γh. Step 2 — Substitute values: p = 9.81 kN/m³ × 8 m = 78.48 kPa. Step 3 — Why not 88.29 kPa? That corresponds to h = 9 m (9.81 × 9). Why not 98.10 kPa? That is p at h = 10 m. Why not 68.67 kPa? That is p at h = 7 m. The key point: gauge pressure ignores atmospheric pressure and increases linearly with depth.

Answer: A. x = 3 and x = 4

Step 1 – Identify coefficients: a = 1, b = −7, c = 12. Step 2 – Apply the quadratic formula: x = [−(−7) ± √(49 − 48)] / 2 = (7 ± 1) / 2. Step 3 – Compute both roots: x = (7 + 1)/2 = 4 and x = (7 − 1)/2 = 3. Step 4 – Verify using Vieta's formulas: sum = 3 + 4 = 7 = −b/a ✓; product = 3 × 4 = 12 = c/a ✓. Common mistake: Students sometimes choose −3 and −4 by forgetting that for positive roots the middle term must be negative (−7x), meaning the factors subtract from x, giving (x − 3)(x − 4) = 0.

Answer: A. 75 m

Step 1: Identify given values — stadia intercept s = 0.75 m, stadia interval factor K = 100, additive constant C = 0, horizontal sight (α = 0°). Step 2: Apply the basic stadia formula for a horizontal sight: D = Ks + C. Step 3: Substitute — D = 100(0.75) + 0 = 75 m. Step 4: Since the sight is horizontal, no cosine correction is needed. Final Answer: D = 75 m. Common mistake — multiplying 0.75 × 10 instead of × 100, giving 7.5 m.

Answer: A. σ = My / I

Step 1 — Recall the derivation: plane sections remain plane under bending, so strain (and stress for elastic material) varies linearly with distance y from the neutral axis. Step 2 — Equilibrium of the cross-section gives σ = My/I, where M is the bending moment and I is the second moment of area about the neutral axis. Step 3 — Option B (VI/Q) is not a real formula. Option C (VQ/Ib) is the shear stress formula τ, not bending stress. Option D (M/A) has no distance term and gives the wrong units (N·mm/mm² = MPa only accidentally — but it ignores depth variation). Step 4 — The correct formula is σ = My/I.

Answer: B. 140 N

Step 1 – Identify the friction law: F_max = μ_s × N (Coulomb's law of static friction). Step 2 – Substitute values: F_max = 0.35 × 400 N = 140 N. Step 3 – Interpret the result: 140 N is the threshold; any applied horizontal force ≤ 140 N keeps the block stationary. A force exceeding 140 N causes sliding. Why the other options are wrong: 100 N corresponds to μ = 0.25 (not 0.35); 160 N would require μ = 0.40; 200 N would require μ = 0.50 — none match the given coefficient.

Answer: B. Ultimate Strength Design (USD / LRFD)

Step 1 — Identify the code: NSCP 2015 (National Structural Code of the Philippines) governs RC design in the Philippines. Step 2 — Recall the default method: NSCP 2015 adopts USD (also called LRFD — Load and Resistance Factor Design) as its primary method for concrete structures, following the framework of ACI 318. Step 3 — Why not WSD? WSD is the older 'alternate' or 'allowable stress' method. While still valid and examinable, it is no longer the primary method. Step 4 — Why not the other options? LSD per Eurocode and ASD per AISC 360 are design methods for structural steel, not the primary RC framework in the Philippines. Conclusion: USD/LRFD is the correct answer — loads are factored UP and strength is factored DOWN by φ.

Answer: B. Joint displacements and rotations

Step 1 — Distinguish force vs. displacement methods: Force methods (e.g., three-moment equation, virtual work) take redundant forces/moments as unknowns. Displacement methods take joint displacements (translations, Δ) and rotations (θ) as unknowns. Step 2 — Once θ and Δ are found, member-end moments are calculated by back-substitution into the slope-deflection equation. Step 3 — Why the other options are wrong: 'Member forces and reactions' are the final results, not the primary unknowns. 'Fixed-end moments' are inputs (loading terms), not unknowns. 'Redundant reactions' are the unknowns in force methods, not displacement methods.

Answer: A. Pn = Fy × Ag

Step 1 – Identify the two limit states: (1) tensile yielding on gross area, (2) tensile rupture on effective net area. Step 2 – For yielding, NSCP 2015 Section 504.2 (AISC 360 D2-1) states Pn = Fy·Ag, where Ag is the full, unreduced gross cross-sectional area. Step 3 – The resistance factor for yielding is φt = 0.90, so the design strength is φtPn = 0.90·Fy·Ag. Step 4 – Option B (Fu·Ae) is the rupture formula, not the yield formula. Option C incorrectly mixes φ into the nominal strength. Option D uses the net area An, which applies to rupture, not yielding. Yielding uses the gross area because, at early loading, the full section resists the force before any fracture occurs.

Answer: B. 30.0 MPa

Step 1: Identify given values. Diameter d = 150 mm, Failure load P = 530 kN = 530,000 N. Step 2: Compute the cross-sectional area of the cylinder. A = (π/4)(150)² = (π/4)(22,500) = 17,671 mm². Step 3: Apply the compressive strength formula. f'c = P/A = 530,000 / 17,671 = 30.0 MPa. Step 4: Verify units — N divided by mm² gives MPa. The answer is 30.0 MPa. Common mistake: forgetting to convert kN to N before dividing.