CELE Construction Management & Methods Reviewer 2026

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.

Answer: C. 0.50

Step 1: Recall that w/c = weight of water ÷ weight of cement, both in the same units. Step 2: Substitute values. w/c = 180 kg / 360 kg = 0.50. Step 3: The ratio is dimensionless. A w/c of 0.50 is a common mid-range value for structural concrete. Common mistake: Some students invert the ratio (cement/water), giving 2.00 — always put water in the numerator.

Answer: B. 180 kg

Step 1: Given w/c = 0.45 and cement = 400 kg. Need to find water content. Step 2: Rearrange the formula. Water = w/c × cement = 0.45 × 400 kg. Step 3: Water = 180 kg. Step 4: Check: 180/400 = 0.45 ✓. Common mistake: Some students divide cement by w/c (400/0.45 ≈ 889 kg), which is incorrect.

Answer: B. 32.7 MPa

Step 1: For f'c ≤ 35 MPa, apply both ACI formulas. Step 2: Formula 1: f'cr = f'c + 1.34s = 28 + 1.34(3.5) = 28 + 4.69 = 32.69 MPa. Step 3: Formula 2: f'cr = f'c + 2.33s − 3.5 = 28 + 2.33(3.5) − 3.5 = 28 + 8.155 − 3.5 = 32.66 MPa. Step 4: f'cr = max(32.69, 32.66) = 32.7 MPa (Formula 1 governs). Always evaluate BOTH formulas and take the larger value.

Answer: B. 35.1 MPa

Step 1: Given P = 620 kN = 620,000 N, d = 150 mm. Step 2: Compute area. A = (π/4)(150)² = 17,671 mm². Step 3: f'c = P/A = 620,000 / 17,671 = 35.08 ≈ 35.1 MPa. Step 4: Verify: 35.1 × 17,671 ≈ 620,052 N ✓. Common mistake: using d = 150 m instead of mm — always check units consistently.

Answer: C. Slump = 100 mm

Step 1: The slump test measures the consistency or workability of fresh concrete. Step 2: A higher slump value means the concrete is more fluid and easier to place — higher workability. Step 3: Among the choices, 100 mm is the largest slump value, indicating the highest workability. Step 4: Note: high workability does not mean high strength — adding water increases slump but REDUCES strength by increasing w/c. These two properties trade off against each other.

Answer: C. Increases strength and decreases workability

Step 1: The w/c ratio is the single most important factor controlling concrete strength. Step 2: Lower w/c means less free water relative to cement, producing a denser hydration product with fewer voids — resulting in HIGHER compressive strength. Step 3: However, less water also makes the mix stiffer and harder to place — LOWER workability (lower slump). Step 4: This is the fundamental strength-workability trade-off in concrete mix design. Chemical admixtures (plasticizers/superplasticizers) can be used to improve workability without increasing w/c.

Answer: B. Gradation (particle size distribution) of fine aggregate

Step 1: The Fineness Modulus (FM) is a single index number summarizing the particle size distribution of fine aggregates. Step 2: It is calculated as the cumulative percentage retained on standard sieves divided by 100. Step 3: A higher FM indicates coarser aggregate; a lower FM indicates finer aggregate. Typical FM for fine aggregate is 2.3 to 3.1. Step 4: This is important in mix design because finer aggregates require more paste (cement + water), affecting strength and economy of the mix.

Answer: B. Yield strength and ultimate tensile strength

Step 1: Steel reinforcement (deformed bars) must conform to ASTM A615 / PNS 49 grade requirements. Step 2: A tension test (pull test) on a sample bar directly measures the yield strength (fy) — the stress at which plastic deformation begins — and the ultimate tensile strength (fu). Step 3: These values, combined with percent elongation (ductility), confirm the bar grade (e.g., Grade 60: fy = 414 MPa). Step 4: Bend tests also supplement tension tests to verify ductility. Mill certificates must be backed by sample tension tests per ACI 318 and NSCP 2015 requirements.

Answer: B. 40.7 MPa

Step 1: f'c = 35 MPa, s = 4 MPa. Apply both ACI formulas. Step 2: Formula 1: f'cr = 35 + 1.34(4) = 35 + 5.36 = 40.36 MPa. Step 3: Formula 2: f'cr = 35 + 2.33(4) − 3.5 = 35 + 9.32 − 3.5 = 40.82 MPa. Step 4: f'cr = max(40.36, 40.82) = 40.82 ≈ 40.8 MPa. Formula 2 governs here. Note: when s is larger, Formula 2 more often governs. Always check both.

Answer: B. 150 m³/hr

Step 1: Identify givens — bucket capacity C = 1.5 m³, cycle time t = 30 s, efficiency = 50/60 = 0.833. Step 2: Compute cycles per hour = 3600 / 30 = 120 cycles/hr. Step 3: Apply the output formula: Output = C × (cycles/hr) × efficiency = 1.5 × 120 × 0.833 = 150 m³/hr. Step 4: Verify — 1.5 × 120 = 180 gross; 180 × 0.833 = 150 m³/hr. ✓ Common mistake: Using 60-minute efficiency (eff = 1.0) gives 180 m³/hr — always apply the working-minute factor.

Answer: C. 6 trucks

Step 1: Identify givens — truck cycle time = 24 min, load time = 4 min. Step 2: Apply the fleet-matching formula: N = truck cycle time / load time = 24 / 4 = 6 trucks. Step 3: Interpretation — each truck takes 4 min to load; in the 24 min one truck is away completing its cycle, 6 trucks rotating ensures the loader is never idle. Step 4: Round up if the result is not a whole number (not needed here). Common mistake: Using haul-only time instead of the full cycle time.