CELE Reinforced & Prestressed Concrete — Reinforced Concrete Beams: FlexureRevision Notes

Formulas

Exam Tips

• Memorize the stress block depth formula a = Asfy/(0.85f'cb) — it appears in virtually every beam problem.
• Always verify tension-controlled status: compute c, then εt = 0.003(d−c)/c, check ≥ 0.005.
• For f'c = 28 MPa and f'c = 21 MPa (common board values), β₁ = 0.85. For f'c = 35 MPa, β₁ = 0.85 − 0.05(35−28)/7 = 0.80.
• The lever arm (d − a/2) is typically 85–95% of d for lightly reinforced beams — use this as a quick sanity check.

Key Points

• At ultimate (flexural failure), concrete compressive stress is non-linear; the Whitney equivalent rectangular stress block simplifies this to a uniform stress of 0.85f'c over depth 'a', giving the same resultant force and location as the actual parabolic stress distribution.
• The stress block depth a = β₁·c, where c is the neutral-axis depth measured from the extreme compression fiber.
• β₁ depends on f'c (NSCP 2015, Section 422.2.2.4.3): β₁ = 0.85 for f'c ≤ 28 MPa; for f'c > 28 MPa, β₁ = 0.85 − 0.05(f'c − 28)/7 ≥ 0.65.
• The concrete strain at extreme compression fiber is assumed to be εcu = 0.003 at ultimate (NSCP 2015).
• Tension steel yields (fs = fy) in an under-reinforced (tension-controlled) section — this is the DESIRED failure mode.
• Force equilibrium: Compression force C = 0.85f'c·b·a must equal Tension force T = As·fy.
• The net tensile strain εt is measured at the centroid of the extreme tension steel layer, NOT the compression steel.

Definitions

Section Title

1. Fundamental Concepts: The Equivalent Stress Block (Whitney Block)

Common Mistakes

• Using h (total depth) instead of d (effective depth) — this ALWAYS overestimates capacity.
• Using f'c instead of 0.85f'c in the stress block formula — the 0.85 factor is mandatory.
• Assuming φ = 0.90 without checking εt — an over-reinforced beam has lower φ (as low as 0.65).
• Forgetting to reduce β₁ below 0.85 for f'c > 28 MPa.
• Computing Mn in N·mm and forgetting to convert to kN·m (divide by 10⁶).

Formulas

Exam Tips

• For fy = 415 MPa and f'c = 28 MPa (the most common board combination), memorize: ρb ≈ 0.0288, ρmax ≈ 0.0183, ρmin ≈ 0.00337.
• A quick check: if ρ < ρmax and a < β₁d/2, the section is almost certainly tension-controlled — but always compute εt explicitly.
• The 600/(600+fy) factor in ρb comes from similar triangles on the strain diagram with εcu = 0.003 — knowing the derivation helps you reconstruct the formula under exam pressure.
• When f'c = 35 MPa: √35 = 5.916; √35/(4×415) = 0.00356 vs 1.4/415 = 0.00337 → ρmin = 0.00356 (second expression governs).

Key Points

• The steel ratio ρ = As/(b·d) is a dimensionless index of how heavily a beam is reinforced.
• ρmin prevents sudden brittle failure when the concrete cracks: the beam must carry at least the moment that caused cracking.
• ρmax (tension-controlled limit) ensures ductile behavior by keeping the net tensile strain ≥ 0.005 at ultimate.
• ρb is the BALANCED steel ratio — theoretical value where concrete crushing and steel yielding occur simultaneously (εt = εy at the same load). Under NSCP 2015, ρb itself is NOT the design limit — ρmax < ρb.
• The NSCP 2015 / ACI 318-14 approach replaces the old '0.75ρb' maximum with the εt ≥ 0.005 requirement, resulting in ρmax = 0.85β₁(f'c/fy)(0.375).
• Always check BOTH ρmin and ρmax: a beam with ρ < ρmin fails brittlely at cracking; ρ > ρmax has reduced ductility and φ < 0.90.
• For Grade 60 steel (fy = 415 MPa) and f'c = 28 MPa: ρb ≈ 0.0288, ρmax ≈ 0.0183, ρmin ≈ 0.00337.

Definitions

Section Title

2. Steel Ratio Limits — NSCP 2015

Common Mistakes

• Using 0.75ρb as ρmax instead of the NSCP 2015 / ACI 318-14 εt-based limit (0.375 factor). These give slightly different answers — the board exam uses the current code.
• Taking only 1.4/fy for ρmin without checking √f'c/(4fy) — for higher f'c, the second expression governs.
• Confusing ρ (steel ratio) with the percentage of steel (ρ% = 100×ρ) — always work with the decimal form in formulas.
• Not rechecking ρmax after selecting a bar arrangement with more steel than required.

Formulas

Exam Tips

• In a timed board exam, if the problem gives you a standard section with ρ clearly below ρmax, you can state 'section is tension-controlled, φ = 0.90' after a quick check — but show the εt calculation for full credit.
• Check your answer order of magnitude: φMn for a 300×500 beam with moderate steel is typically 150–300 kN·m. If you get 1500 kN·m, you forgot to divide by 10⁶.

Key Points

• Analysis means: the beam cross-section is fully defined (b, d, As, f'c, fy are all known) and you must find the design moment capacity φMn.
• Step 1: Compute the stress block depth a. Step 2: Compute the neutral axis depth c. Step 3: Check εt and determine φ. Step 4: Compute Mn. Step 5: Apply φ.
• If εt ≥ 0.005: section is tension-controlled, φ = 0.90, proceed directly.
• If εt < 0.005: section is in the transition or compression-controlled zone; φ < 0.90 and the beam may be over-reinforced (not allowed for new designs).
• Always verify ρmin ≤ ρ ≤ ρmax as part of a complete analysis.
• Report the answer as φMn in kN·m (not N·mm or N·m — the board exam uses kN·m for beam moments).

Definitions

Section Title

3. Analysis Procedure (Given Section → Find φMn)

Common Mistakes

• Stopping at Mn and not applying φ, or vice versa — know which one is being asked.
• Computing a in meters instead of mm, leading to errors in subsequent steps.
• Rounding a to fewer than 2 decimal places too early — carry intermediate results to 4 significant figures.

Formulas

Exam Tips

• The term inside the square root [1 − 2Rn/(0.85f'c)] must be positive and ≤ 1.0 — if negative, the beam dimensions are insufficient (Rn too high).
• For a quick design estimate in the exam, the lever arm is approximately 0.9d for lightly reinforced beams: As ≈ Mu/(φ·fy·0.9d) — useful for checking order of magnitude.
• After selecting bars, always re-compute a and φMn to confirm φMn ≥ Mu with the actual As — some board problems check if the designed beam is adequate.

Key Points

• Design means: Mu (factored moment) is known, and you must determine the required As (and select bar sizes).
• The design is performed assuming tension-controlled (φ = 0.90) — verify after computing As.
• The Coefficient of Resistance Rn = Mu/(φ·b·d²) consolidates the problem into one dimensionless parameter.
• The quadratic solution for ρ uses the steel ratio formula derived from setting Mn = As·fy·(d−a/2) equal to Mu/φ.
• After computing As, always verify: (1) ρmin ≤ ρ ≤ ρmax, (2) εt ≥ 0.005 confirms φ = 0.90 assumption.
• Select actual bars (from NSCP/PNS standard sizes: 10, 12, 16, 20, 25, 28, 32, 36 mm diameter) with total area ≥ As_required.
• Standard bar areas (mm²): 10→78.5, 12→113.1, 16→201.1, 20→314.2, 25→490.9, 28→615.8, 32→804.2, 36→1017.9

Definitions

Section Title

4. Design Procedure (Given Mu → Find As)

Common Mistakes

• Forgetting to convert Mu from kN·m to N·mm (multiply by 10⁶) before computing Rn.
• Using d² instead of d² in the Rn formula — make sure to square the effective depth, not just multiply.
• Not checking ρmin after computing ρ — for lightly loaded beams, ρ required may be less than ρmin, requiring As = ρmin·b·d.
• Selecting bar arrangement with As_provided < As_required due to rounding errors — always verify As_provided ≥ As_required.
• Not rechecking the tension-controlled assumption after getting the final As.

Formulas

Exam Tips

• In many board problems, compression steel is assumed to yield (a simplifying assumption given in the problem) — use this shortcut only when explicitly stated or clearly supported by the strain check.
• The 'displaced concrete' correction — replacing f's with (f's − 0.85f'c) in the compression steel force — is tested in detailed analysis problems.

Key Points

• A doubly reinforced beam has BOTH tension steel (As) and compression steel (A's) at effective depths d and d' respectively.
• Used when: (1) Mu exceeds the maximum singly reinforced capacity (φMn,max) of the given cross-section, (2) sustained loads cause excessive creep deflection and compression steel controls long-term deflection, or (3) reversal of moment occurs (seismic frames).
• The analysis uses superposition: Pair 1 = singly reinforced couple (As1, b, d, a) + Pair 2 = compression steel couple (As2 = A's, d, d').
• CRITICAL CHECK: Verify whether compression steel yields (f's = fy) or not. Use strain compatibility: ε's = 0.003(c−d')/c. If ε's ≥ εy = fy/Es (e.g., 415/200,000 = 0.00208 for fy=415), compression steel yields.
• If compression steel does NOT yield: f's = Es·ε's = 200,000·[0.003(c−d')/c] — use this stress in equilibrium equations, leading to a quadratic in c.
• NSCP 2015 Section 422.2.2.4.1 requires compression steel to be enclosed by ties (lateral reinforcement) to prevent buckling.

Definitions

Section Title

5. Doubly Reinforced Beams

Common Mistakes

• Assuming compression steel always yields — must check ε's ≥ εy explicitly.
• Forgetting to account for the concrete displaced by compression steel: effective compression steel force = (f's − 0.85f'c)·A's, not simply f's·A's.
• Using d' as the concrete cover (e.g., 40 mm) instead of the distance from the compression face to the centroid of A's (e.g., 40+6+14 = 60 mm for a 28mm bar with 40mm cover and 12mm stirrup).

Formulas

Exam Tips

• The board exam T-beam check question: 'Is a ≤ tf?' — compute a assuming the full flange width. If yes, rectangular beam solution applies and the problem becomes straightforward.
• When a < tf, the T-beam moment capacity is significantly higher than a rectangular beam of width bw — verify your answer is reasonable by comparing to bw analysis.
• Effective flange width computation is a stand-alone board exam question type — know all three criteria.

Key Points

• In cast-in-place construction, floor slabs and beams are cast monolithically. The slab acts as a COMPRESSION FLANGE for positive moment, making the beam behave as a T-section.
• Effective flange width bf is LIMITED by NSCP 2015 Section 406.3 to prevent over-estimation of flange contribution: bf ≤ min(span/4; bw + 16tf; bw + clear spacing to adjacent beam).
• Key distinction: if the equivalent stress block depth a ≤ tf (flange thickness), the entire compression zone is within the flange — treat as a RECTANGULAR beam of width bf.
• If a > tf, the compression zone extends into the web — must use the T-beam formulas, splitting the compression zone into flange overhangs and web portions.
• T-beams are most relevant for POSITIVE moment regions (slab in compression). For NEGATIVE moment (slab in tension), only bw (web width) acts in compression — treat as rectangular beam of width bw.
• The flange area significantly increases the compression capacity, allowing thinner, more economical webs.

Definitions

Section Title

6. T-Beams

Common Mistakes

• Applying T-beam formulas to negative moment regions where the slab is in tension (the flange is cracked — only bw resists compression).
• Using the full slab width as bf without computing the NSCP 2015 effective width limits.
• Not checking a vs tf first — jumping directly to T-beam formulas when a < tf leads to unnecessary complexity.