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CELE Reinforced & Prestressed Concrete — Reinforced Concrete Beams: FlexureDetailed Explanation

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This is the "office hours" version of Reinforced Concrete Beams: Flexure for the CELE 2026. No shortcuts, no hand-waving — just a full unpacking of why Professional Regulation Commission (PRC) — Board of Civil Engineering cares about each concept and how the Reinforced & Prestressed Concrete section items tend to play out on exam day. Read this once, then hit the practice questions with real understanding.

Exam context

Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Reinforced & Prestressed Concrete section sits under a "Core" weighting, and Reinforced Concrete Beams: Flexure is the 2nd chapter in the 7-chapter CELE Reinforced & Prestressed Concrete rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Reinforced & Prestressed Concrete.

Reinforced Concrete Beams: Flexure - Detailed Explanation

Flexural design of reinforced concrete beams is the cornerstone of structural engineering practice in the Philippines and one of the most heavily tested topics in the PRC Civil Engineer Licensure Examination under the Structural Engineering and Construction (SEC) subject. This chapter develops the theory and design procedures for singly reinforced rectangular beams, doubly reinforced beams, and T-beams using the Ultimate Strength Design (USD) method as prescribed by the National Structural Code of the Philippines (NSCP 2015, Volume I) — which is essentially aligned with ACI 318-14. Mastery of the equivalent rectangular stress block, steel ratio limits, and the tension-controlled criterion is non-negotiable for board exam success. Every derivation here is grounded in equilibrium of internal forces; every number you compute has a physical meaning. Work through each solved example step by step before attempting the exercises.

Concepts

Equivalent Rectangular Stress Block (Whitney Stress Block)

When a reinforced concrete beam is loaded to its nominal flexural capacity, the concrete in the compression zone reaches a highly nonlinear stress distribution. Whitney (1942) — later codified in ACI 318 and adopted verbatim by NSCP 2015 Section 422.2 — replaced this curved stress distribution with an equivalent rectangular block of uniform intensity 0.85f'c and depth 'a', where a = β₁c. Here, c is the depth to the neutral axis from the extreme compression fiber, and β₁ is a factor that depends on f'c: • β₁ = 0.85 for f'c ≤ 28 MPa • β₁ = 0.85 − 0.05(f'c − 28)/7 for 28 MPa < f'c ≤ 56 MPa (but not less than 0.65) • β₁ = 0.65 for f'c > 56 MPa The rectangular block is equivalent to the actual parabolic-trapezoidal stress block in two ways: (1) it produces the same total compression force C = 0.85f'c·a·b, and (2) its centroid is at a/2 from the compression face, which is approximately the centroid of the actual stress block. Physical Meaning: The beam cross-section at ultimate has two internal forces in equilibrium — compression C acting at a/2 from the top, and tension T = Asfy acting at the centroid of the steel (effective depth d). The moment arm is (d − a/2), giving Mn = Asfy(d − a/2). Force Equilibrium (ΣFH = 0): C = T 0.85f'c·a·b = Asfy Solving for a: a = Asfy / (0.85f'c·b) This is the master equation. Everything else follows from it.

Examples

The computation is straightforward: first find a from force equilibrium, then compute Mn using the moment arm (d − a/2). Note that As = 3 × π/4 × 25² = 3 × 490.87 = 1472.6 ≈ 1473 mm². The moment arm here is 457.2 mm, significantly less than d = 500 mm — ignoring a/2 would overestimate Mn by about 9%.

Scenario

A singly reinforced rectangular beam has b = 300 mm, d = 500 mm, As = 1473 mm² (3-25 mm bars), f'c = 28 MPa, fy = 415 MPa. Compute the depth of the equivalent stress block 'a' and the nominal moment Mn.

Solution

Step 1: Compute β₁ Since f'c = 28 MPa ≤ 28 MPa, β₁ = 0.85 Step 2: Compute a a = Asfy / (0.85f'c·b) a = 1473(415) / [0.85(28)(300)] a = 611,295 / 7,140 a = 85.6 mm Step 3: Compute Mn Mn = Asfy(d − a/2) Mn = 1473(415)(500 − 85.6/2) Mn = 611,295 × 457.2 Mn = 279.4 × 10⁶ N·mm Mn = 279.4 kN·m

Applications

Computing the design moment capacity φMn of existing beams for load rating.
Checking whether a given beam section can resist factored design moments Mu.
Establishing the depth of the neutral axis for deflection calculations (cracked section analysis).
Foundation beam design where beam depth is often governed by soil bearing.

Misconceptions

Using f'c instead of 0.85f'c for the compression stress intensity — always 0.85f'c.
Confusing a (stress block depth) with c (neutral axis depth) — they differ by β₁.
Using total beam depth h instead of effective depth d in moment calculations.
Forgetting to convert kN·m to N·mm when substituting into formulas (multiply by 10⁶).

Related Concepts

Neutral axis depth and strain compatibility
Net tensile strain εt and tension-controlled criterion
Strength reduction factor φ
Steel area As from bar sizes

Common Exam Questions

Example

Find φMn of a 250×450 mm beam (d = 450 mm) with As = 1200 mm², f'c = 21 MPa, fy = 275 MPa. [Answer: a = 71.4 mm, Mn = 131.3 kN·m, εt = 0.0159 > 0.005 → φ = 0.90, φMn = 118.1 kN·m]

Approach

Given b, d, As, f'c, fy — compute a, then Mn, then check εt for φ, then compute φMn.

Question Type

Direct Computation of φMn

Example

A beam has As = 2000 mm², b = 400 mm, f'c = 35 MPa, fy = 415 MPa. Find c. [a = 70.2 mm; β₁ = 0.85 − 0.05(35−28)/7 = 0.80; c = 70.2/0.80 = 87.8 mm]

Approach

Use a = Asfy/(0.85f'cb) directly; then c = a/β₁.

Question Type

Find a, c, or the stress block depth

Key Points To Remember

The stress block intensity is always 0.85f'c, NOT f'c — a very common board exam error.
β₁ decreases as f'c increases above 28 MPa; minimum value is 0.65.
The stress block depth a = β₁c; the neutral axis depth c = a/β₁.
Effective depth d is measured from the extreme compression fiber to the CENTROID of tension steel — not to the bottom of the beam.
The lever arm for moment is (d − a/2), not d alone.
NSCP 2015 Section 422.2.2 specifies these stress block parameters.

Net Tensile Strain, Tension-Controlled Sections, and φ Factor

NSCP 2015 Section 421.2 (aligned with ACI 318-14 Section 21.2) classifies beam sections by the net tensile strain εt in the extreme tension steel at nominal strength: • Tension-controlled: εt ≥ 0.005 → φ = 0.90 • Transition zone: 0.004 ≤ εt < 0.005 → φ interpolated (0.65 to 0.90 for spiral; 0.65 to 0.90 for ties — see NSCP Table 421.2.2) • Compression-controlled: εt ≤ εy (≈ 0.002 for Grade 415) → φ = 0.65 (tied) or 0.75 (spirally reinforced) Computing εt uses similar triangles on the strain diagram, where concrete crushes at εcu = 0.003 at the extreme compression fiber: εt = 0.003 × (d − c) / c For a beam to be considered tension-controlled (and thus use φ = 0.90 for bending — the standard assumption for well-designed beams), the steel must yield well before the concrete crushes, giving large tensile strains. Why This Matters: Over-reinforced beams fail in a brittle, sudden manner — concrete crushes with little warning. Under-reinforced (tension-controlled) beams give ample deflection and cracking before failure — the preferred ductile failure mode. NSCP 2015 enforces ductility by requiring εt ≥ 0.005 (not just εt ≥ εy) for the full φ = 0.90. Practical Check: For most standard beams with ρ ≤ ρmax, εt will exceed 0.005. Always verify this when the problem gives you a section with high reinforcement.

Examples

The strain εt = 0.0119 is nearly 6 times the yield strain of 0.00208, confirming a highly ductile, under-reinforced section. The φ = 0.90 applies. Had we found εt between 0.004 and 0.005, we would need to interpolate φ using NSCP Table 421.2.2.

Scenario

For the beam in the previous example (b = 300, d = 500, As = 1473 mm², f'c = 28, fy = 415 MPa), verify the section is tension-controlled.

Solution

From earlier: a = 85.6 mm β₁ = 0.85 (since f'c = 28 MPa) c = a/β₁ = 85.6/0.85 = 100.7 mm εt = 0.003 × (d − c)/c εt = 0.003 × (500 − 100.7)/100.7 εt = 0.003 × 399.3/100.7 εt = 0.003 × 3.965 εt = 0.01190 Since εt = 0.0119 >> 0.005 → Tension-controlled ✓ φ = 0.90 φMn = 0.90 × 279.4 = 251.5 kN·m

Applications

Determining the correct φ factor before computing φMn — never assume φ = 0.90 without checking.
Evaluating existing beams that may be over-reinforced due to field changes.
Seismic design where ductility (large εt) is especially critical per NSCP 2015 Section 418.
Transition-zone φ interpolation for lightly reinforced compression members with bending.

Misconceptions

Assuming φ = 0.90 without checking εt — this is a common board exam error that leads to wrong answers.
Confusing 'balanced' (εt = εy) with 'tension-controlled' (εt = 0.005) — they are different conditions.
Using εcu = 0.003 at the tension steel side — it is 0.003 at the COMPRESSION face, and εt is at the tension steel.
Thinking that a section with As < As_balanced is automatically tension-controlled — you still need to verify εt ≥ 0.005 (not just εt ≥ εy).

Related Concepts

Steel ratio limits (ρb, ρmax, ρmin)
Strength reduction factor φ (NSCP 2015 Section 421.2)
Ductility requirements for seismic design
Strain compatibility in flexure

Common Exam Questions

Example

Is a beam with c/d = 0.40 tension-controlled? εt = 0.003(1−0.40)/0.40 = 0.003(0.60/0.40) = 0.0045. Since 0.004 ≤ 0.0045 < 0.005, it is in the TRANSITION zone — φ < 0.90.

Approach

Compute a → c → εt. Compare to 0.005 and 0.004.

Question Type

Classify the section (tension-controlled vs. compression-controlled)

Example

For b=300, d=500, f'c=28, fy=415: a_max = 0.85(0.375×500) = 159.4 mm; As_max = 0.85(28)(159.4)(300)/415 = 2745 mm². Verify: ρmax = 2745/(300×500) = 0.0183.

Approach

Set εt = 0.005, giving c/d = 0.375. Then a = β₁(0.375d), and As = 0.85f'c·a·b/fy.

Question Type

Find maximum steel for tension-controlled section

Key Points To Remember

εt = 0.003(d − c)/c — derived from linear strain diagram with εcu = 0.003 at top.
Tension-controlled: εt ≥ 0.005 → φ = 0.90 for flexure.
For Grade 415 steel, yield strain εy = fy/Es = 415/200,000 = 0.00208.
NSCP 2015 requires εt ≥ 0.004 as an absolute minimum for beams (Section 409.3.3.1).
An over-reinforced section (εt < εy) is NOT permitted for new beam design under NSCP 2015.
The tension-controlled limit εt = 0.005 corresponds to c/d = 0.375 (since 0.003/(0.003+0.005) = 0.375).

Steel Ratio Limits: ρb, ρmax, and ρmin

The steel ratio ρ = As/(bd) is a dimensionless measure of the amount of tension steel relative to the beam cross-section. NSCP 2015 (following ACI 318) sets three critical limits: **1. Balanced Steel Ratio ρb** The balanced condition occurs when concrete reaches its crushing strain (εcu = 0.003) exactly as the steel reaches its yield strain (εy = fy/Es). Using similar triangles on the strain diagram: cb/d = 0.003/(0.003 + εy) = 0.003/(0.003 + fy/200,000) = 600/(600 + fy) The balanced steel area: Ab = 0.85f'c·ab·b/fy where ab = β₁·cb Therefore: ρb = 0.85β₁(f'c/fy)[600/(600 + fy)] **2. Maximum Steel Ratio ρmax (Tension-Controlled Limit)** NSCP 2015 requires εt ≥ 0.005 for beams, giving c/d ≤ 0.375: ρmax = 0.85β₁(f'c/fy)(0.375) Note: This is NOT 0.75ρb (that was the old ASD/older USD code). The current NSCP/ACI 318 uses the εt = 0.005 criterion. For Grade 415 steel with f'c = 28 MPa: ρmax ≈ 0.0183. Important: 0.75ρb = 0.85β₁(f'c/fy)(0.375) only if εy = 0.002075; the two expressions are numerically close but conceptually different. **3. Minimum Steel Ratio ρmin** Prevents sudden (brittle) failure at first cracking, when the cracked section's moment capacity must exceed the cracking moment. NSCP 2015 Section 409.6.1.2: ρmin = max(1.4/fy, √f'c/(4fy)) In MPa units. For f'c = 28 MPa, fy = 415 MPa: • 1.4/415 = 0.00337 • √28/(4×415) = 5.292/1660 = 0.00319 • ρmin = 0.00337 **Design Requirement:** ρmin ≤ ρ ≤ ρmax ensures ductile, tension-controlled behavior. **Note on the Two ρmax Expressions:** Some older Philippine reviewers still use ρmax = 0.75ρb. For Grade 415 and f'c = 28 MPa: • 0.75ρb = 0.75 × 0.0288 = 0.0216 • ρmax from εt = 0.005 = 0.0183 The current NSCP 2015/ACI 318-14 value of 0.0183 is MORE CONSERVATIVE. Use the εt criterion for current board exams.

Examples

Note that ρmax = 0.0183 < 0.75ρb = 0.0216. The current NSCP 2015 code is more conservative. The actual steel ratio of 0.00982 is about 54% of ρmax, confirming a well-designed, lightly reinforced section with large tensile strains.

Scenario

For a beam with f'c = 28 MPa and fy = 415 MPa (β₁ = 0.85), compute ρb, ρmax, and ρmin.

Solution

Step 1: ρb ρb = 0.85β₁(f'c/fy)[600/(600+fy)] ρb = 0.85(0.85)(28/415)[600/(600+415)] ρb = 0.7225 × 0.06747 × 0.5911 ρb = 0.7225 × 0.03989 ρb = 0.02882 ≈ 0.0288 Step 2: ρmax (NSCP 2015, εt = 0.005 criterion) ρmax = 0.85β₁(f'c/fy)(0.375) ρmax = 0.7225 × 0.06747 × 0.375 ρmax = 0.7225 × 0.02530 ρmax = 0.01828 ≈ 0.0183 Step 3: ρmin 1.4/fy = 1.4/415 = 0.003373 √f'c/(4fy) = √28/(4×415) = 5.292/1660 = 0.003188 ρmin = max(0.003373, 0.003188) = 0.003373 ≈ 0.00337 Verification check (from Example 1): ρactual = 1473/(300×500) = 0.009820 Since 0.00337 ≤ 0.00982 ≤ 0.0183 → Section is ductile and tension-controlled ✓

Applications

Preliminary sizing of beam cross-sections — target ρ ≈ 0.5ρmax for economy and ductility.
Checking adequacy of existing beams reinforced before NSCP 2015 adoption.
Doubly reinforced beam design trigger: when ρ would exceed ρmax for a singly reinforced section.
Code compliance verification per NSCP 2015 Section 409.6 for all new designs.

Misconceptions

Using ρmax = 0.75ρb from old codes — current NSCP 2015 uses the εt = 0.005 criterion, giving a different (usually lower) ρmax.
Applying only one expression for ρmin instead of taking the MAX of two — always use the larger value.
Thinking ρb is the maximum allowed — it is NOT; sections at balanced failure are brittle.
Forgetting that ρmin applies even to heavily loaded beams where analysis might show ρ < ρmin — if so, use As = ρmin·b·d.

Related Concepts

Equivalent rectangular stress block
Net tensile strain εt and tension-controlled criterion
Doubly reinforced beams (when ρ exceeds ρmax)
NSCP 2015 Section 409.6 (minimum reinforcement)

Common Exam Questions

Example

For f'c = 21 MPa, fy = 275 MPa, β₁ = 0.85: ρb = 0.85(0.85)(21/275)(600/875) = 0.7225(0.07636)(0.6857) = 0.0379; ρmax = 0.7225(0.07636)(0.375) = 0.0207; ρmin = max(1.4/275, √21/[4×275]) = max(0.00509, 0.00416) = 0.00509.

Approach

Plug in formula directly. Remember 600/(600+fy) for ρb and 0.375 for ρmax.

Question Type

Compute ρb, ρmax, ρmin for given materials

Example

A beam with ρ = 0.025 when ρmax = 0.0183 is OVER the tension-controlled limit. εt < 0.005; φ < 0.90. Such a section is NOT permitted for new design under NSCP 2015.

Approach

Compute ρactual, compare to ρmin and ρmax. Then compute εt to confirm classification.

Question Type

Determine if a beam is under-reinforced or over-reinforced

Key Points To Remember

ρb = 0.85β₁(f'c/fy)[600/(600+fy)] — balanced ratio, concrete crushes as steel yields simultaneously.
ρmax = 0.85β₁(f'c/fy)(0.375) — maximum for tension-controlled (εt ≥ 0.005) under NSCP 2015.
ρmin = max(1.4/fy, √f'c/[4fy]) — minimum to prevent sudden failure at cracking.
Current ρmax from εt criterion is more conservative than 0.75ρb for typical steel grades.
For Grade 415 (fy = 415 MPa): 600/(600+415) = 600/1015 = 0.5911.
Always check both steel ratio limits AND εt in analysis problems.

Design of Singly Reinforced Rectangular Beams

The design problem is the inverse of analysis: given the factored moment Mu, beam width b, effective depth d, and material strengths f'c and fy, find the required steel area As. The design procedure uses the Coefficient of Resistance Rn and the steel ratio formula: **Step 1: Compute Rn** Assume φ = 0.90 (tension-controlled — verify later): Rn = Mu / (φ·b·d²) **Step 2: Compute ρ** From the quadratic relationship between Rn and ρ: ρ = (0.85f'c/fy)[1 − √(1 − 2Rn/(0.85f'c))] **Derivation of this formula:** From Mn = ρ·fy·b·d²(1 − ρfy/[1.7f'c]): Let Rn = Mn/(bd²). Solving for ρ gives the above expression. **Step 3: Check steel ratio limits** • If ρ < ρmin: use ρmin (As = ρmin·b·d) • If ρmin ≤ ρ ≤ ρmax: use computed ρ • If ρ > ρmax: the beam size is inadequate — increase d or b, OR use doubly reinforced design **Step 4: Compute As** As = ρ·b·d **Step 5: Select bars** Choose bar sizes such that the provided As ≥ required As. Verify bar spacing satisfies cover and spacing requirements (NSCP 2015 Section 425). **Step 6: Verify φMn ≥ Mu** With the selected As, recompute a, Mn, εt, and confirm φMn ≥ Mu. **Practical Notes:** • For preliminary design, choose d based on span/depth ratios (NSCP Table 409.3.1: d ≈ L/12 to L/16 for simply supported beams with fy = 415 MPa). • A typical target is ρ ≈ 0.4 to 0.6 times ρmax for a good balance of economy and ductility.

Examples

The required As = 1299 mm² is satisfied by 3-25mm bars (1473 mm²). The excess steel (13.4% more than required) increases φMn from 200 kN·m to 224 kN·m, providing an additional margin. This is normal when selecting discrete bar sizes. The εt = 0.0104 confirms tension-controlled behavior and validates φ = 0.90.

Scenario

Design the tension steel for a singly reinforced rectangular beam with Mu = 200 kN·m, b = 300 mm, d = 450 mm, f'c = 28 MPa, fy = 415 MPa.

Solution

Step 1: Compute Rn Mu = 200 kN·m = 200 × 10⁶ N·mm Rn = Mu/(φ·b·d²) = 200×10⁶/[0.90(300)(450)²] Rn = 200×10⁶/[0.90 × 300 × 202,500] Rn = 200×10⁶/54,675,000 Rn = 3.659 MPa Step 2: Compute ρ ρ = (0.85f'c/fy)[1 − √(1 − 2Rn/(0.85f'c))] 0.85f'c = 0.85 × 28 = 23.80 MPa 2Rn/(0.85f'c) = 2(3.659)/23.80 = 7.318/23.80 = 0.30748 √(1 − 0.30748) = √0.69252 = 0.83218 ρ = (23.80/415)(1 − 0.83218) ρ = 0.057349 × 0.16782 ρ = 0.009622 Step 3: Check steel ratio limits ρmin = 0.00337; ρmax = 0.0183 Since 0.00337 < 0.009622 < 0.0183 ✓ Step 4: Compute As As = ρ·b·d = 0.009622 × 300 × 450 = 1299 mm² Step 5: Select bars Use 3 – 25 mm bars: As_provided = 3 × 490.9 = 1473 mm² > 1299 mm² ✓ (Alternative: 3 – 28 mm bars = 3 × 615.8 = 1847 mm², also acceptable) Step 6: Verify a = 1473(415)/[0.85(28)(300)] = 611,295/7,140 = 85.6 mm c = 85.6/0.85 = 100.7 mm εt = 0.003(450 − 100.7)/100.7 = 0.003(3.468) = 0.01040 ≥ 0.005 ✓ Mn = 1473(415)(450 − 42.8) = 611,295 × 407.2 = 248.9 kN·m φMn = 0.90 × 248.9 = 224.0 kN·m Wait — φMn = 224.0 kN·m < Mu = 200 kN·m? Let us recheck: d = 450 mm. φMn = 224 kN·m > 200 kN·m ✓ (224 > 200, so the design is adequate.)

Applications

Design of floor beams in residential and commercial buildings under NSCP 2015.
Bridge girder design (with appropriate live load factors per AASHTO, but same flexure mechanics).
Retaining wall design where stem acts as a cantilever beam.
Foundation beam design for combined footings and mat foundations.

Misconceptions

Using Mu in kN·m directly in Rn formula without converting to N·mm — always multiply by 10⁶.
Forgetting to check ρ against ρmin — a low Mu might require more steel than the formula gives.
Not verifying φMn ≥ Mu after bar selection — the discrete bar area may differ from As_required.
Using the full beam width b for As when beam is part of a T-beam system — use the web width bw for As in doubly reinforced and T-beam designs.

Related Concepts

Factored loads and Mu from NSCP 2015 Section 405 load combinations
Bar spacing and concrete cover requirements (NSCP 2015 Section 425)
Doubly reinforced beam design
Deflection control and minimum depth (NSCP Table 409.3.1)

Common Exam Questions

Example

Mu = 150 kN·m, b = 250 mm, d = 400 mm, f'c = 21 MPa, fy = 415 MPa. Rn = 150×10⁶/[0.9(250)(400²)] = 4.167 MPa; ρ = (17.85/415)(1−√(1−2×4.167/17.85)) = 0.04301(1−√0.5329) = 0.04301(0.2700) = 0.01161; As = 0.01161×250×400 = 1161 mm². Use 4-20mm bars (As=1257mm²).

Approach

Compute Rn, then ρ, then As = ρbd. Check ρmin ≤ ρ ≤ ρmax.

Question Type

Find required As given Mu, b, d, f'c, fy

Example

As_required = 1200 mm². Options: 4-20mm (1257mm²) ✓; 3-25mm (1473mm²) ✓; 2-28mm (1232mm²) ✓. Select the most economical option and verify φMn.

Approach

After computing As_required, select bars with standard sizes; verify φMn ≥ Mu.

Question Type

Select bars and verify adequacy

Key Points To Remember

Rn = Mu/(φ·b·d²) — Rn has units of MPa; Mu must be in N·mm.
The ρ formula uses 0.85f'c in two places — both the coefficient and inside the square root.
If ρ > ρmax, do NOT just use ρmax for As — the beam needs redesign (larger size or doubly reinforced).
Always select bars to provide As_provided ≥ As_required; round up to next whole number of bars.
Verify the design: recompute φMn with actual As_provided ≥ Mu.
Standard bar diameters in Philippines: 10, 12, 16, 20, 25, 28, 32, 36 mm (deformed bars per ASTM A615).

Doubly Reinforced Rectangular Beams

A doubly reinforced beam has steel in both the tension zone (As) and the compression zone (A's at depth d' from compression face). Doubly reinforced beams are used when: 1. The design moment Mu exceeds the maximum moment capacity of a singly reinforced tension-controlled section (φMn,max = φ·0.85f'c·amax·b·(d − amax/2)). 2. Beam dimensions are architecturally fixed and cannot be increased. 3. Long-term deflection control is important (compression steel reduces creep deflection). **Analysis/Design Approach — Superposition:** The doubly reinforced section is split into two sub-couples: **Couple 1 (Singly Reinforced Equivalent):** • Compression concrete + Tension steel As1 • As1 provides the same capacity as a balanced singly reinforced section at ρmax • Mn1 = As1·fy·(d − amax/2) **Couple 2 (Steel-Steel Couple):** • Compression steel A's + Additional tension steel As2 • As2 = A's (if compression steel yields) • Mn2 = A's·f's·(d − d') **Compression Steel Yield Check:** Compression steel yields if: εs' = 0.003(c − d')/c ≥ εy = fy/200,000 Equivalently, compression steel yields if: c ≥ 600d'/(600 − fy) ... but for fy = 415 MPa: c ≥ 600d'/(600 − 415) = 600d'/185 = 3.243d' If compression steel does NOT yield, use actual stress f's = Es·εs' = 200,000 × εs' in lieu of fy. **Key Equations:** amax = β₁(0.375d) As1 = 0.85f'c·amax·b/fy Mn1 = As1·fy(d − amax/2) Mn2 = Mu/φ − Mn1 A's = Mn2/[f's(d − d')] As2 = A's·f's/fy As,total = As1 + As2 **Note on Compression Steel in Stress Block:** When computing the net compression force, deduct the concrete displaced by compression steel: Net force from compression steel = A's(f's − 0.85f'c) This is often 0.85f'c·A's subtracted from f's·A's. For simplicity, many textbooks use A's·f's when f's >> 0.85f'c (small error for Grade 415).

Examples

Note that in this example, compression steel is just below yielding (f's = 392 MPa vs. fy = 415 MPa — only 1.1% difference). The distinction matters for precision. When compression steel does yield (ε's ≥ εy), simply use f's = fy and the calculation is simpler. This example illustrates that the compression steel yield check is critical and cannot be skipped.

Scenario

A beam is limited to b = 300 mm, d = 500 mm, d' = 65 mm, f'c = 28 MPa, fy = 415 MPa. The required Mu = 380 kN·m. Design the reinforcement.

Solution

Step 1: Compute maximum singly reinforced capacity amax = β₁(0.375d) = 0.85(0.375×500) = 0.85(187.5) = 159.4 mm As1 = 0.85f'c·amax·b/fy = 0.85(28)(159.4)(300)/415 = 1,142,964/415 = 2,754 mm² Mn1 = As1·fy(d − amax/2) = 2754(415)(500 − 79.7) = 2754(415)(420.3) = 480.4 × 10⁶ N·mm × [Wait, recompute] = 2754 × 415 × 420.3 = 480,402,690 N·mm = 480.4 kN·m φMn1 = 0.90 × 480.4 = 432.4 kN·m Since Mu = 380 kN·m < φMn1 = 432.4 kN·m → Singly reinforced is sufficient! [Revised scenario for doubly reinforced: Let Mu = 500 kN·m] Since Mu = 500 kN·m > φMn1 = 432.4 kN·m → Doubly reinforced required. Step 2: Compute Mn2 needed Mn2 = Mu/φ − Mn1 = 500/0.90 − 480.4 = 555.6 − 480.4 = 75.2 kN·m Step 3: Check compression steel yields c = amax/β₁ = 159.4/0.85 = 187.5 mm (this is the tension-controlled limit c) ε's = 0.003(c − d')/c = 0.003(187.5 − 65)/187.5 = 0.003(122.5/187.5) = 0.001960 εy = 415/200,000 = 0.002075 Since ε's = 0.00196 < εy = 0.00208 → compression steel does NOT yield f's = 200,000 × 0.001960 = 392 MPa Step 4: Compute A's and As2 A's = Mn2/[f's(d − d')] = 75.2×10⁶/[392(500 − 65)] = 75.2×10⁶/[392×435] = 75,200,000/170,520 = 441 mm² As2 = A's·f's/fy = 441(392)/415 = 416 mm² Step 5: Total tension steel As,total = As1 + As2 = 2754 + 416 = 3170 mm² Use: 4-32mm bars (3217 mm²) for tension, 2-16mm bars (402 mm²) for compression Or: tension 6-28mm (6×616=3695mm²), compression 2-16mm (402mm²).

Applications

Transfer girders in high-rise buildings where depth is architecturally constrained.
Beams in tight ceiling-to-floor clearances where increasing depth is not possible.
Long-span beams where deflection control requires compression steel.
Moment resisting frames in seismic zones where symmetric reinforcement may be specified.

Misconceptions

Assuming compression steel always yields — always verify ε's ≥ εy = fy/Es.
Forgetting to subtract 0.85f'c from compression steel stress to account for displaced concrete.
Using d instead of (d − d') as the moment arm for the steel-steel couple.
Adding A's directly to As without accounting for yield condition of compression steel.

Related Concepts

Long-term deflection and compression steel (ACI 318-14 Section 24.2.4)
Singly reinforced beam capacity limits
Strain compatibility in doubly reinforced sections
Seismic detailing for doubly reinforced beams (NSCP 2015 Section 418)

Common Exam Questions

Example

b=300, d=500, f'c=21, fy=415: amax=0.85(0.375×500)=159.4mm; As1=0.85(21)(159.4)(300)/415=2065mm²; Mn1=2065(415)(420.3)=359.9kN·m; φMn1=324kN·m. If Mu=400kN·m>324kN·m → doubly reinforced needed.

Approach

Compute φMn,max for singly reinforced. If Mu > φMn,max, doubly reinforced required.

Question Type

Determine if doubly reinforced beam is needed

Example

c=187.5mm, d'=70mm: ε's=0.003(187.5-70)/187.5=0.001880 < εy=0.00208 → f's=200,000(0.00188)=376 MPa (does NOT yield).

Approach

Compute c at ρmax condition; find ε's = 0.003(c-d')/c; compare to εy.

Question Type

Check if compression steel yields

Key Points To Remember

Use doubly reinforced design only when singly reinforced section cannot carry Mu within ρmax.
Always check whether compression steel yields: ε's ≥ εy = fy/200,000.
If compression steel does not yield, use f's = 200,000·ε's (not fy) for A's calculations.
The compression steel is typically placed at d' = 60–75 mm from the compression face.
Total tension steel: As = As1 + As2; the compression steel area is A's.
Compression steel in long beams significantly reduces long-term deflection (ACI 318-14 Section 24.2.4).

T-Beam Flexural Design

In monolithic slab-beam construction, a portion of the slab acts as the compression flange of the beam, creating a T-shaped cross-section. This greatly increases the compression area and moment capacity compared to a rectangular beam of the same web width. **Effective Flange Width (NSCP 2015 Section 406.3):** For interior T-beams, the effective flange width bf is the smallest of: • Span/4 • bw + 8hf on each side (total: bw + 16hf) • Center-to-center spacing of adjacent beams For isolated T-beams, bf ≤ 4bw. **Two Cases in Flexural Analysis:** **Case 1: Stress Block Within the Flange (a ≤ hf)** If the equivalent stress block depth a is less than or equal to the flange thickness hf (slab thickness), the beam behaves as a rectangular beam of width bf: • a = Asfy/(0.85f'c·bf) • Mn = Asfy(d − a/2) • Check: if a ≤ hf → Case 1 confirmed **Case 2: Stress Block Extends into the Web (a > hf)** Split the compression zone into flange portion and web portion: • Flange force: Cf = 0.85f'c·bf·hf (but the web also carries some) • More precisely, use the T-section equilibrium: Asfy = 0.85f'c[bf·a − (bf − bw)·hf] ... wait, restate properly: C_flange = 0.85f'c(bf − bw)hf C_web = 0.85f'c·bw·a T = Asfy Force equilibrium: Asfy = 0.85f'c(bf − bw)hf + 0.85f'c·bw·a Solve for a: a = [Asfy − 0.85f'c(bf − bw)hf]/(0.85f'c·bw) Mn = C_flange(d − hf/2) + C_web(d − a/2) **Design Strategy:** First, assume the stress block is in the flange. Check a vs. hf. If a ≤ hf, done. If a > hf, use T-section equations. **Key Insight:** T-beams are usually in Case 1 for typical proportions because the large bf means a small a. Most NSCP/board problems for positive-moment T-beams fall in Case 1.

Examples

The large effective flange width (800 mm vs. only 300 mm web width) results in a very shallow stress block (54.5 mm), well within the 100 mm slab. T-beams are typically very efficient in positive bending because concrete compression area is plentiful. The εt = 0.0204 shows highly ductile behavior.

Scenario

A T-beam has bf = 800 mm, bw = 300 mm, hf = 100 mm (slab thickness), d = 500 mm. f'c = 28 MPa, fy = 415 MPa. As = 2500 mm². Check whether the stress block is within the flange and compute φMn.

Solution

Step 1: Assume stress block in flange a = Asfy/(0.85f'c·bf) a = 2500(415)/[0.85(28)(800)] a = 1,037,500/19,040 a = 54.5 mm Step 2: Check vs. hf Since a = 54.5 mm < hf = 100 mm → stress block is within the flange ✓ (Case 1 confirmed — treat as rectangular beam with b = bf = 800 mm) Step 3: Compute Mn Mn = Asfy(d − a/2) Mn = 2500(415)(500 − 27.25) Mn = 1,037,500 × 472.75 Mn = 490.5 × 10⁶ N·mm = 490.5 kN·m Step 4: Check εt c = a/β₁ = 54.5/0.85 = 64.1 mm εt = 0.003(500 − 64.1)/64.1 = 0.003 × 6.796 = 0.02039 >> 0.005 ✓ φ = 0.90 φMn = 0.90 × 490.5 = 441.5 kN·m

Applications

Floor beam design in reinforced concrete buildings (monolithic slab-beam systems).
Bridge deck girder analysis where the deck slab contributes to flexural capacity.
Inverted T-beams in precast construction.
Design of spandrel beams at the perimeter of floor systems.

Misconceptions

Using bw instead of bf for T-beam positive moment analysis — always check and use bf when a ≤ hf.
Applying T-beam effective width for negative moment sections — use bw only for negative moments.
Forgetting to check the Case 1 condition: a ≤ hf must be verified, not assumed.
Using full slab span as effective flange width instead of applying the three NSCP criteria.

Related Concepts

Effective slab width and flange participation
Rectangular beam analysis (Case 1 T-beam reduces to this)
Negative moment regions in continuous T-beams
Shear in T-beams (web carries shear, flange does not)

Common Exam Questions

Example

T-beam with span = 8 m, bw = 300 mm, hf = 120 mm, beam spacing = 2.5 m. bf = min(8000/4, 300+16×120, 2500) = min(2000, 2220, 2500) = 2000 mm.

Approach

Apply the three criteria from NSCP 2015 Section 406.3 and take the minimum.

Question Type

Determine effective flange width

Example

As=3000mm², bf=1500mm, bw=350mm, hf=120mm, d=550mm, f'c=28, fy=415: a=3000(415)/[0.85(28)(1500)]=34.9mm < 120mm → Case 1. Mn=3000(415)(550-17.5)=662.6kN·m. φMn=596.3kN·m.

Approach

Compute a using bf. If a ≤ hf, proceed as rectangular beam. If a > hf, use T-section equations.

Question Type

Check if a ≤ hf and compute φMn

Key Points To Remember

Effective flange width bf is the smaller of span/4, bw+16hf, or beam spacing.
Always check Case 1 first: compute a = Asfy/(0.85f'c·bf). If a ≤ hf, treat as rectangular beam with width bf.
For negative moment (hogging) in T-beams, the flange is in tension — use rectangular beam of width bw only.
The slab reinforcement in the effective overhanging flange must be checked for transverse bending.
hf = slab thickness; bw = web width; bf = effective flange width.
NSCP 2015 Section 406.3.2 — memorize the effective width rules for board exam.

Practice Problems

Key note: β₁ = 0.80 for f'c = 35 MPa (not 0.85). This is a frequent board exam trap — the problem specifies f'c > 28 MPa specifically to test whether you adjust β₁. With β₁ = 0.80, c is larger than it would be with β₁ = 0.85, but εt = 0.0117 is still well above 0.005, confirming tension-controlled. The high εt (nearly 6 times yield strain) indicates a lightly reinforced, very ductile section.

Problem

PROBLEM 1 (Analysis — Singly Reinforced Beam) A rectangular beam has b = 350 mm, d = 600 mm, reinforced with 4-28 mm bars. Material strengths: f'c = 35 MPa, fy = 415 MPa. Find: (a) depth of stress block a; (b) neutral axis depth c; (c) net tensile strain εt; (d) strength reduction factor φ; (e) design moment capacity φMn.

Solution

Given: b = 350 mm, d = 600 mm 4-28 mm bars: As = 4 × π/4 × 28² = 4 × 615.75 = 2463 mm² f'c = 35 MPa, fy = 415 MPa Step 1: Compute β₁ (f'c = 35 MPa > 28 MPa) β₁ = 0.85 − 0.05(f'c − 28)/7 = 0.85 − 0.05(35 − 28)/7 β₁ = 0.85 − 0.05(7/7) = 0.85 − 0.05 = 0.80 Step 2: (a) Compute a a = Asfy/(0.85f'c·b) a = 2463(415)/[0.85(35)(350)] a = 1,022,145/10,412.5 a = 98.2 mm Step 3: (b) Compute c c = a/β₁ = 98.2/0.80 = 122.7 mm Step 4: (c) Compute εt εt = 0.003(d − c)/c = 0.003(600 − 122.7)/122.7 εt = 0.003(477.3/122.7) εt = 0.003 × 3.890 εt = 0.01167 Step 5: (d) Determine φ εt = 0.01167 > 0.005 → Tension-controlled → φ = 0.90 Step 6: (e) Compute φMn Mn = Asfy(d − a/2) Mn = 2463(415)(600 − 49.1) Mn = 2463 × 415 × 550.9 Mn = 2463 × 228,623.5 Mn = 563.1 × 10⁶ N·mm = 563.1 kN·m φMn = 0.90 × 563.1 = 506.8 kN·m **Answer: a = 98.2 mm, c = 122.7 mm, εt = 0.0117, φ = 0.90, φMn = 506.8 kN·m**

The required As = 1686 mm² is well within ρmin and ρmax limits. The selected 4-25mm bars provide 16.4% more steel than required, resulting in φMn = 367.6 kN·m vs. Mu = 320 kN·m — a comfortable margin. In practice, the engineer may reduce bar size or count if cost is a concern, provided As ≥ 1686 mm².

Problem

PROBLEM 2 (Design — Required Steel Area) Determine the required tension steel area As for a singly reinforced rectangular beam subjected to a factored moment Mu = 320 kN·m. Given: b = 350 mm, d = 550 mm, f'c = 28 MPa, fy = 415 MPa. Use NSCP 2015 provisions.

Solution

Step 1: Compute Rn Mu = 320 kN·m = 320 × 10⁶ N·mm Rn = Mu/(φ·b·d²) [assume φ = 0.90 initially] Rn = 320×10⁶/[0.90 × 350 × (550)²] Rn = 320×10⁶/[0.90 × 350 × 302,500] Rn = 320×10⁶/95,287,500 Rn = 3.358 MPa Step 2: Compute ρ 0.85f'c = 0.85(28) = 23.8 MPa 2Rn/(0.85f'c) = 2(3.358)/23.8 = 6.716/23.8 = 0.28218 √(1 − 0.28218) = √0.71782 = 0.84725 ρ = (23.8/415)(1 − 0.84725) ρ = 0.057349 × 0.15275 ρ = 0.008760 Step 3: Check limits β₁ = 0.85 (f'c = 28 MPa) ρmin = max(1.4/415, √28/[4×415]) = max(0.003373, 0.003188) = 0.003373 ρmax = 0.85(0.85)(28/415)(0.375) = 0.7225 × 0.067470 × 0.375 = 0.01828 Since 0.003373 < 0.008760 < 0.01828 ✓ Step 4: Compute As As = ρ·b·d = 0.008760 × 350 × 550 = 1686 mm² Step 5: Select bars Option A: 4-25mm bars → As = 4 × 490.9 = 1963 mm² ✓ Option B: 3-28mm bars → As = 3 × 615.8 = 1847 mm² ✓ Option C: 5-22mm bars → As = 5 × 380.1 = 1901 mm² ✓ Select 4-25mm bars (1963 mm²) for standard bar use. Step 6: Verify with As = 1963 mm² a = 1963(415)/[0.85(28)(350)] = 814,645/8,330 = 97.8 mm c = 97.8/0.85 = 115.1 mm εt = 0.003(550 − 115.1)/115.1 = 0.003(3.779) = 0.01134 > 0.005 ✓ Mn = 1963(415)(550 − 48.9) = 1963 × 415 × 501.1 = 408.4 kN·m φMn = 0.90 × 408.4 = 367.6 kN·m > Mu = 320 kN·m ✓ **Answer: As_required = 1686 mm². Use 4-25mm bars (As = 1963 mm²); φMn = 367.6 kN·m ≥ 320 kN·m ✓**

This problem illustrates all three steel ratio limits and their practical implications. The proposed As = 3200 mm² gives ρ = 0.0222 > ρmax = 0.0137, making the section over-reinforced. The εt check (0.00194 < 0.005) confirms this — the section is not even in the transition zone, approaching compression-controlled. NSCP 2015 does not allow new beam designs with εt < 0.004. The minimum steel of 486 mm² (2-20mm bars) ensures the beam can carry loads beyond first cracking without sudden failure.

Problem

PROBLEM 3 (Steel Ratio Analysis) A beam section has b = 300 mm, d = 480 mm, f'c = 21 MPa, fy = 415 MPa. (a) Compute ρb, ρmax (NSCP 2015), and ρmin. (b) A designer proposes As = 3200 mm². Is this acceptable? (c) What is the minimum required As?

Solution

Step 1: β₁ for f'c = 21 MPa ≤ 28 MPa → β₁ = 0.85 (a) Steel ratio limits: ρb = 0.85β₁(f'c/fy)[600/(600+fy)] ρb = 0.85(0.85)(21/415)[600/(600+415)] ρb = 0.7225 × 0.050602 × 0.59113 ρb = 0.7225 × 0.029918 ρb = 0.02162 ρmax = 0.85β₁(f'c/fy)(0.375) ρmax = 0.7225 × 0.050602 × 0.375 ρmax = 0.7225 × 0.018976 ρmax = 0.01371 ρmin = max(1.4/415, √21/[4×415]) = max(1.4/415, 4.5826/1660) = max(0.003373, 0.002761) ρmin = 0.003373 (b) Check proposed As = 3200 mm² ρproposed = As/(bd) = 3200/(300×480) = 3200/144,000 = 0.02222 Since ρproposed = 0.02222 > ρmax = 0.01371 → NOT ACCEPTABLE under NSCP 2015. The section would be compression-controlled (εt < 0.005). Verify εt: amax = β₁(0.375d) = 0.85(0.375×480) = 0.85(180) = 153.0 mm c = 0.375d = 0.375(480) = 180 mm (at tension-controlled limit) For proposed: a = 3200(415)/[0.85(21)(300)] = 1,328,000/5,355 = 248.0 mm c = 248.0/0.85 = 291.8 mm εt = 0.003(480 − 291.8)/291.8 = 0.003(0.6451) = 0.001935 < 0.005 This confirms: over-reinforced, NOT PERMITTED for new design. (c) Minimum required As: As,min = ρmin·b·d = 0.003373 × 300 × 480 = 485.7 mm² Use 2-20mm bars: As = 2 × 314.2 = 628.4 mm² > 485.7 mm² ✓ **Answer: ρb = 0.0216, ρmax = 0.0137, ρmin = 0.00337. As = 3200mm² is NOT acceptable (over-reinforced). As,min = 486 mm².**

Note that all three criteria for bf gave similar values (1800 and 1900 mm) — the minimum is 1800 mm. The very shallow stress block (34.9 mm vs. 100 mm slab) and extremely high εt (0.0349) demonstrate why T-beams in positive bending are very efficient. The same As = 3600 mm² in a rectangular beam of bw = 300 mm would give a = 3600(415)/[0.85(28)(300)] = 209.3 mm — much deeper, well into Case 2 territory. The T-beam harnesses the slab's full width for compression.

Problem

PROBLEM 4 (T-Beam Analysis) An interior T-beam is part of a floor system with span = 7.2 m. Web dimensions: bw = 300 mm, d = 520 mm. Slab thickness hf = 100 mm. Beam spacing = 1.8 m center to center. As = 3600 mm². f'c = 28 MPa, fy = 415 MPa. Find φMn.

Solution

Step 1: Determine effective flange width bf (NSCP 2015 Section 406.3.2) Criteria for interior T-beam: (i) Span/4 = 7200/4 = 1800 mm (ii) bw + 16hf = 300 + 16(100) = 300 + 1600 = 1900 mm (iii) Center-to-center beam spacing = 1800 mm bf = minimum of (1800, 1900, 1800) = 1800 mm Step 2: Check Case 1 (assume a ≤ hf) a = Asfy/(0.85f'c·bf) a = 3600(415)/[0.85(28)(1800)] a = 1,494,000/42,840 a = 34.9 mm Step 3: Check vs. hf Since a = 34.9 mm < hf = 100 mm → Stress block is within the flange ✓ Treat as rectangular beam with b = bf = 1800 mm. Step 4: Compute Mn Mn = Asfy(d − a/2) Mn = 3600(415)(520 − 34.9/2) Mn = 3600 × 415 × (520 − 17.45) Mn = 3600 × 415 × 502.55 Mn = 750,308,100 N·mm Mn = 750.3 kN·m Step 5: Check εt c = a/β₁ = 34.9/0.85 = 41.1 mm εt = 0.003(520 − 41.1)/41.1 = 0.003(11.65) = 0.03494 >> 0.005 ✓ φ = 0.90 Step 6: φMn φMn = 0.90 × 750.3 = 675.3 kN·m **Answer: bf = 1800 mm (governed by span/4 and beam spacing); a = 34.9 mm (Case 1: a < hf); φMn = 675.3 kN·m**

This problem demonstrates the multiple-choice format common in PRC board exams. The Rn-ρ formula gives As = 2272 mm², closest to choice (B). The small discrepancy from 2180 mm² (which differs by 4%) could arise from rounding in the answer choices — always select the nearest value. The verification confirms εt > 0.005 and φ = 0.90 was correctly assumed.

Problem

PROBLEM 5 (Design with Rn — Multiple Choice Format) A simply supported beam carries a factored moment Mu = 450 kN·m. The beam cross-section is b = 400 mm, d = 580 mm. f'c = 28 MPa, fy = 415 MPa. Which of the following is closest to the required As? (A) 1850 mm²; (B) 2180 mm²; (C) 2420 mm²; (D) 2750 mm²

Solution

Step 1: Compute Rn Mu = 450 × 10⁶ N·mm Rn = Mu/(φbd²) = 450×10⁶/[0.90(400)(580²)] Rn = 450×10⁶/[0.90(400)(336,400)] Rn = 450×10⁶/121,104,000 Rn = 3.716 MPa Step 2: Compute ρ 0.85f'c = 23.80 MPa 2Rn/(0.85f'c) = 2(3.716)/23.80 = 7.432/23.80 = 0.31227 √(1 − 0.31227) = √0.68773 = 0.82930 ρ = (23.80/415)(1 − 0.82930) ρ = 0.057349 × 0.17070 ρ = 0.009793 Step 3: Check limits ρmin = 0.00337, ρmax = 0.01828 0.00337 < 0.009793 < 0.01828 ✓ Step 4: Compute As As = ρbd = 0.009793 × 400 × 580 As = 0.009793 × 232,000 As = 2272 mm² Closest answer: (B) 2180 mm² — but let us check (C) 2420 mm² |2272 − 2180| = 92 |2272 − 2420| = 148 Answer: **(B) 2180 mm²** is closest. **Answer: (B) 2180 mm² [Computed: 2272 mm²]** Verification with As = 2272 mm²: a = 2272(415)/[0.85(28)(400)] = 943,040/9,520 = 99.1 mm c = 99.1/0.85 = 116.6 mm εt = 0.003(580−116.6)/116.6 = 0.003(3.976) = 0.01193 > 0.005 ✓

Exam Preparation Tips

MEMORIZE THE FIVE KEY FORMULAS: a = Asfy/(0.85f'cb); Mn = Asfy(d−a/2); φMn = 0.90Mn (tension-controlled); Rn = Mu/(φbd²); ρ = (0.85f'c/fy)(1−√(1−2Rn/0.85f'c)). These five equations cover 80% of flexure problems.
ALWAYS CHECK β₁ FIRST: β₁ = 0.85 for f'c ≤ 28 MPa; β₁ = 0.85 − 0.05(f'c−28)/7 for 28 < f'c ≤ 56 MPa; minimum β₁ = 0.65. Many boards use f'c = 30, 32, 35 MPa — adjust β₁ accordingly.
NEVER SKIP THE εt CHECK: Always compute c = a/β₁ and εt = 0.003(d−c)/c. If εt ≥ 0.005 → φ = 0.90. If εt < 0.005, φ must be reduced. This step is tested directly in the boards.
UNIT CONSISTENCY: Mu in boards is given in kN·m. Convert to N·mm by multiplying by 10⁶ before substituting into Rn or Mn formulas. This is the #1 arithmetic error source.
MEMORIZE STEEL RATIO FORMULAS: ρb = 0.85β₁(f'c/fy)(600/600+fy); ρmax = 0.85β₁(f'c/fy)(0.375); ρmin = max(1.4/fy, √f'c/4fy). For fy = 415 MPa with f'c = 28 MPa: ρb = 0.0288, ρmax = 0.0183, ρmin = 0.00337.
T-BEAM SHORTCUT: Almost all positive-moment T-beam problems in the boards have a ≤ hf. Check this first by computing a = Asfy/(0.85f'c·bf). If it's less than the slab thickness, proceed as a rectangular beam with width bf — saves significant computation time.
MEMORIZE STANDARD BAR AREAS: 10mm=78.5mm²; 12mm=113mm²; 16mm=201mm²; 20mm=314mm²; 25mm=491mm²; 28mm=616mm²; 32mm=804mm²; 36mm=1018mm². These are often needed in the last step of design problems.
FOR MULTIPLE-CHOICE: In analysis problems, compute φMn first, then look for the nearest answer. In design problems, compute As_required — the boards usually ask for the nearest bar selection or the nearest As value from the given choices.
DOUBLY REINFORCED TRIGGER: If Mu/φ > Mn,max (where Mn,max = As1·fy·(d − amax/2) and amax = β₁(0.375d)), then doubly reinforced beam is needed. Know this threshold for common material combinations.
BOARD EXAM STRATEGY: Flexure problems in the PRC CE Board (Structural Engineering and Construction) typically appear 3–5 times per examination. They range from pure analysis (find φMn) to combined design (find As or select bars). Practice each type to build speed. Time yourself: a standard flexure problem should take 4–6 minutes maximum.
NSCP 2015 SECTION REFERENCES: Know Section 422.2 (stress block), Section 421.2 (φ factors), Section 409.6 (minimum reinforcement), Section 406.3 (T-beam effective width). Examiners may ask 'per NSCP 2015, what is the minimum steel ratio?' — know the formula and its code reference.
COMMON NUMERICAL VALUES TO MEMORIZE: For f'c=28MPa, fy=415MPa: β₁=0.85; ρb=0.0288; ρmax=0.0183; ρmin=0.00337; εy=0.00208. For f'c=21MPa, fy=415MPa: β₁=0.85; ρb=0.0216; ρmax=0.0137; ρmin=0.00337. These material combinations appear most frequently in Philippine board exams.

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In summary

Reinforced concrete beam flexure is built on three pillars that every CE board examinee must master: (1) the equivalent rectangular stress block with force equilibrium giving a = Asfy/(0.85f'cb) and Mn = Asfy(d − a/2); (2) the tension-controlled criterion εt ≥ 0.005 ensuring φ = 0.90 and ductile behavior per NSCP 2015; and (3) the steel ratio limits ρmin ≤ ρ ≤ ρmax preventing both sudden cracking failure and brittle concrete crushing. The design pathway — compute Rn, then ρ, then select bars, then verify φMn — is a structured algorithm that, with sufficient practice, can be executed accurately within the time pressure of the board examination. Doubly reinforced beams and T-beams extend these principles: the former by superposing two force couples when single reinforcement is insufficient, the latter by harvesting the slab's compression capacity through an effective flange width. The most important practical rule for T-beams — check whether a ≤ hf before assuming T-behavior — saves enormous computational effort in nearly all positive-moment cases. Throughout this chapter, the emphasis has been on understanding the physical basis of each equation rather than rote memorization, because board exam problems are designed to test conceptual understanding through numerical variations. Know why 0.85f'c is used (not f'c), why d is the effective depth (not the full depth h), why we need both ρmin and ρmax, and why the εt criterion replaced the old 0.75ρb rule. With these concepts firmly grounded, supplemented by systematic practice on the worked examples and exercises provided, you will approach flexure problems in the PRC Civil Engineer Licensure Examination with confidence and precision.

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