CELE Reinforced & Prestressed Concrete Reviewer 2026

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. 192 kN·m

Step 1 — Identify the governing load combination: For gravity loads (dead + live), NSCP 2015 Section 409 gives: M_u = 1.2M_D + 1.6M_L. Step 2 — Substitute values: M_u = 1.2(80) + 1.6(60). Step 3 — Compute: M_u = 96 + 96 = 192 kN·m. Step 4 — Check distractor 140: This is simply 80 + 60 — no load factors applied (a common mistake in WSD thinking). Step 5 — Check distractor 204: This comes from using 1.4D + 1.6L = 1.4(80) + 1.6(60) = 112 + 96 = 208 — wrong combination mixing two different load factors. Conclusion: M_u = 192 kN·m. Always apply 1.2D + 1.6L for dead + live gravity loading.

Answer: D. 0.90

Step 1 — Recall the NSCP 2015 φ table: Different actions have different φ values reflecting their relative reliability. Step 2 — For tension-controlled sections (where the net tensile strain ε_t ≥ 0.005), the section has ample ductility warning before failure, so φ = 0.90. Step 3 — Why not 0.75? φ = 0.75 applies to shear/torsion AND to compression-controlled spiral columns — not flexure. Step 4 — Why not 0.65? φ = 0.65 is for compression-controlled tied columns and bearing on concrete. Step 5 — Why not 0.60? φ = 0.60 is for plain (unreinforced) concrete. Conclusion: For a well-designed flexural beam (tension-controlled), φ = 0.90. This is one of the most frequently tested values in the board exam.

Answer: C. 0.65

Step 1 — Distinguish column types: NSCP 2015 gives different φ values for tied vs. spiral columns because spiral-confined columns have greater ductility. Step 2 — Tied columns: φ = 0.65 (lower φ reflects less ductility; failure is more sudden). Step 3 — Spiral columns: φ = 0.75 (higher φ because spiral confinement gives more ductile post-peak behavior). Step 4 — Why not 0.90? That is for tension-controlled flexure — completely different failure mode. Step 5 — Why not 0.70? 0.70 is not a standard φ value in NSCP 2015 for columns. Conclusion: Tied column → φ = 0.65; Spiral column → φ = 0.75. Memorize this pair — it is a classic board-exam trap.

Answer: D. 0.85

Step 1 — Recall the β₁ rule: β₁ = 0.85 for f'_c ≤ 28 MPa (NSCP 2015, following ACI 318). Step 2 — Check: f'_c = 21 MPa. Is 21 ≤ 28? YES. Step 3 — Therefore β₁ = 0.85. The reduction formula does NOT apply here because the threshold is 28 MPa. Step 4 — Common mistake: Some students apply the reduction formula starting at f'_c = 21 MPa — this is WRONG. The formula only kicks in when f'_c > 28 MPa. Step 5 — Boundary check: β₁ starts at 0.85, drops 0.05 per every 7 MPa above 28, but never goes below 0.65. Conclusion: β₁ = 0.85 for any f'_c ≤ 28 MPa, including 21 MPa.

Answer: B. 0.80

Step 1 — Identify the applicable formula: Since 28 < 35 ≤ 55 MPa, use: β₁ = 0.85 − 0.05 × (f'_c − 28) / 7. Step 2 — Substitute: β₁ = 0.85 − 0.05 × (35 − 28) / 7. Step 3 — Simplify: (35 − 28) / 7 = 7/7 = 1.0. Step 4 — Compute: β₁ = 0.85 − 0.05(1.0) = 0.85 − 0.05 = 0.80. Step 5 — Verify bounds: 0.65 ≤ 0.80 ≤ 0.85 ✓. Conclusion: β₁ = 0.80 for f'_c = 35 MPa. Notice that 35 MPa is exactly 7 MPa above the 28 MPa threshold, dropping β₁ by exactly one step of 0.05.

Answer: B. 24 870 MPa

Step 1 — Recall the formula: E_c = 4700√f'_c (MPa) for normal-weight concrete per NSCP 2015. Step 2 — Substitute: E_c = 4700 × √28. Step 3 — Compute √28: √28 = √(4 × 7) = 2√7 ≈ 5.292. Step 4 — Multiply: E_c = 4700 × 5.292 ≈ 24 872 MPa ≈ 24 870 MPa. Step 5 — Cross-check distractors: 21 500 MPa is too low (possible confusion with E_s/√f'_c approximation); 28 000 MPa confuses E_c with f'_c numerically; 30 100 MPa overestimates. Conclusion: E_c ≈ 24 870 MPa. This formula is NSCP 2015 Section 419.2.2 (ACI 318 equivalent).

Answer: B. n = 8

Step 1 — Compute E_c: E_c = 4700√28 ≈ 24 872 MPa (from Example 2 in the chapter). Step 2 — Apply the modular ratio formula: n = E_s / E_c = 200 000 / 24 872. Step 3 — Calculate: n = 8.04. Step 4 — Round: In practice, n is rounded to the nearest whole number → n ≈ 8. Step 5 — Physical meaning: n = 8 means steel is 8 times as stiff as concrete; in WSD transformed-section analysis, steel area is replaced by an equivalent concrete area of (n × A_s). Conclusion: n = 8 for f'_c = 28 MPa. This is a standard result you should memorize for WSD problems.

Answer: C. 0.85 f'_c

Step 1 — Understand the Whitney stress block: The real parabolic-trapezoidal concrete stress distribution is replaced by a simpler rectangular block for ease of calculation. Step 2 — The block has: (a) Depth a = β₁c (where c is the neutral-axis depth), and (b) Uniform stress intensity = 0.85f'_c. Step 3 — Why 0.85 and not 1.0? The factor 0.85 accounts for the difference between the strength of in-situ cast concrete and that of the laboratory cylinder (sustained loading, casting direction, etc.). Step 4 — Common mistake: Confusing β₁ (which adjusts the DEPTH a) with the 0.85 factor (which defines the STRESS intensity). These are two separate but related things. Step 5 — β₁ affects GEOMETRY; 0.85 affects STRESS MAGNITUDE. Conclusion: Stress intensity = 0.85f'_c. Always. The depth is a = β₁c. Never mix these up.

Answer: C. Tension-controlled

Step 1 — Find the yield strain: ε_ty = f_y / E_s = 415 / 200 000 = 0.002075. Step 2 — Recall NSCP 2015 strain limits: • Compression-controlled: ε_t ≤ ε_ty (≈ 0.002075 for Grade 415) • Transition zone: ε_ty < ε_t < 0.005 • Tension-controlled: ε_t ≥ 0.005 Step 3 — Compare: ε_t = 0.006 ≥ 0.005 → TENSION-CONTROLLED. Step 4 — Consequence: φ = 0.90 applies (full flexure reduction factor). Step 5 — Why not transition? Transition is 0.002075 < ε_t < 0.005. Our value 0.006 > 0.005, so we are fully tension-controlled. Conclusion: ε_t = 0.006 > 0.005 → Tension-controlled → φ = 0.90.

Answer: A. 95.6 kN

Step 1: Write the NSCP 2015 simplified formula: Vc = 0.17λ√f'c · bw · d. Step 2: Substitute λ = 1.0 (normal-weight), f'c = 25 MPa, bw = 250 mm, d = 450 mm. Step 3: √25 = 5.0 MPa^0.5. Step 4: Vc = 0.17 × 1.0 × 5.0 × 250 × 450 = 0.17 × 5.0 × 112,500 = 95,625 N ≈ 95.6 kN. The other options result from using wrong multipliers (e.g., 0.17 replaced by 0.22 or omitting one dimension). Always verify units: N, MPa, mm give Newtons directly.

Answer: C. 0.75

Step 1: Recall that NSCP 2015 (aligned with ACI 318) assigns different φ values for different failure modes. Step 2: For flexure (tension-controlled), φ = 0.90. Step 3: For compression-controlled sections, φ = 0.65. Step 4: For shear and torsion, φ = 0.75. This lower φ for shear reflects the brittle and sudden nature of shear failures compared to the more ductile flexural failures. Always use φ = 0.75 when computing φVn or φVc for shear checks.