CELE Reinforced & Prestressed Concrete — Reinforced Concrete Beams: Shear and TorsionRevision Notes

Formulas

Exam Tips

• Memorize the two primary gravity combinations: U = 1.4D and U = 1.2D + 1.6L — one of these governs most beam problems.
• In WSD problems, always set up the transformed section first (transform steel area to equivalent concrete area = nAs) before applying flexure formula.
• When the problem says 'service loads,' you are likely in WSD territory; when it says 'factored loads' or gives Mu/Vu/Pu directly, you are in USD territory.
• Board problems frequently ask: 'Find the required Mu' — this is purely a load-combination calculation, no section design needed.

Key Points

• WSD (Working Stress Design / Alternate Design Method): Service loads are applied directly to the structure. Computed stresses must remain below prescribed allowable stresses — typically fractions of material strength (e.g., 0.45f'c for concrete in flexure, 0.40fy for steel). The method assumes linear-elastic behavior throughout.
• USD (Ultimate Strength Design / LRFD): The NSCP 2015 primary design method. Service loads are amplified by load factors (e.g., 1.2D + 1.6L) to obtain factored loads (Mu, Vu, Pu). The nominal strength (Mn, Vn, Pn) is then reduced by a strength-reduction factor φ. The design criterion is φRn ≥ Ru.
• WSD uses the modular ratio n = Es/Ec to transform a composite RC section into an equivalent all-concrete transformed section for elastic analysis.
• USD explicitly accounts for the nonlinear, inelastic behavior of concrete at ultimate load, giving a more realistic and economical design.
• NSCP 2015 Section 409 governs the strength design method; WSD is retained as the Alternate Design Method in Appendix A of ACI 318 and is still tested in Philippine board exams.
• Key WSD allowable stresses: concrete in flexure fc = 0.45f'c; reinforcing steel fs = 0.40fy (or 0.50fy for some conditions per ACI App. A).
• Key USD load combinations (NSCP 2015 Section 409.4): U = 1.4D; U = 1.2D + 1.6L + 0.5(Lr or S or R); U = 1.2D + 1.6(Lr or S or R) + (L or 0.5W); U = 1.2D + 1.0W + L + 0.5(Lr or S or R); U = 0.9D + 1.0W; U = 1.2D + 1.0E + L; U = 0.9D + 1.0E.

Definitions

Section Title

The Two Design Philosophies: WSD vs. USD

Common Mistakes

• Mixing WSD allowable-stress limits with USD factored loads — these belong to entirely separate design worlds and must never be combined.
• Forgetting to check ALL governing NSCP load combinations; the board exam often provides D, L, W, and E, expecting you to identify the critical combination.
• Using 1.4D as the only load combination when L is present — 1.2D + 1.6L typically governs for typical beam problems.
• Applying WSD flexural allowable (0.45f'c) to an USD analysis problem.

Formulas

Exam Tips

• Memorize: Ec = 4700√f'c for normal-weight concrete. For f'c = 21: Ec ≈ 21,538 MPa; f'c = 28: Ec ≈ 24,870 MPa; f'c = 35: Ec ≈ 27,806 MPa.
• Commonly tested modular ratios: n = 9 for f'c = 21 MPa; n = 8 for f'c = 28 MPa; n = 7 for f'c = 35 MPa.
• For fr problems: remember the 0.62 coefficient (not 0.6 or 0.7).
• When a problem states 'normal-weight concrete' without giving wc, always use Ec = 4700√f'c directly.

Key Points

• Concrete compressive strength f'c is measured on a standard 150 mm × 300 mm cylinder cured for 28 days. Common Philippine values: 21, 24, 27.5, 28, 30, 35 MPa.
• Concrete is strong in compression (f'c) but has negligible tensile strength — typically 10–15% of f'c. For design, tensile strength of concrete is ignored in USD flexural design (cracked section assumption).
• Modulus of elasticity of normal-weight concrete (unit weight wc ≈ 23.5 kN/m³ or 2300 kg/m³): Ec = 4700√f'c (MPa). This is the NSCP 2015 / ACI 318 formula.
• For lightweight concrete: Ec = wc^1.5 × 0.043√f'c (MPa), where wc is in kg/m³.
• Reinforcing steel: Grade 275 (fy = 275 MPa) and Grade 415 (fy = 415 MPa) are most common in the Philippines. Grade 60 (fy = 415 MPa) in older US-reference problems.
• Modulus of elasticity of steel: Es = 200,000 MPa (constant, regardless of grade).
• Modular ratio n = Es/Ec; it is always rounded to the nearest whole number in WSD transformed-section problems.
• Rupture modulus (modulus of rupture) of concrete: fr = 0.62λ√f'c (MPa), where λ = 1.0 for normal-weight concrete. Used for cracking moment Mcr calculations.
• Yield strain of steel: εy = fy/Es (e.g., for Grade 415: εy = 415/200,000 = 0.00208).
• Ultimate concrete strain: εcu = 0.003 (assumed constant at extreme compression fiber at ultimate — a fundamental USD assumption per NSCP 2015 / ACI 318).

Definitions

Section Title

Material Properties: Concrete and Steel

Common Mistakes

• Using Ec = 4700√f'c for lightweight concrete — the correct formula uses wc^1.5.
• Not rounding n to the nearest integer in WSD problems — n = 8.04 becomes n = 8.
• Confusing f'c (cylinder strength) with fck (European characteristic cube strength) — Philippine practice uses cylinder strength.
• Using εcu = 0.005 instead of 0.003 — the 0.005 value defines tension-controlled limit for strain, not the ultimate concrete strain.

Formulas

Exam Tips

• The φ table is the single most tested concept in this chapter. Memorize: Flexure (TC) = 0.90; Shear = 0.75; Tied column = 0.65; Spiral column = 0.75.
• A common board trap: a beam with high reinforcement ratio may be in the transition zone — compute εt from the strain diagram using similar triangles.
• εt = (dt − c)/c × 0.003, where dt = distance from extreme compression fiber to extreme tension steel, and c = neutral-axis depth at nominal strength.
• If a board problem gives you a fully designed beam and asks 'Is the section tension-controlled?', compute c from a = As·fy/(0.85·f'c·b), then c = a/β1, then εt.

Key Points

• φ factors account for: variability in material strength, workmanship, dimensional tolerances, and the consequence of failure of a particular type of member.
• Tension-controlled sections (net tensile strain εt ≥ 0.005): φ = 0.90. These are under-reinforced beams where steel yields well before concrete crushes — the preferred, ductile failure mode.
• Compression-controlled sections (εt ≤ εty = fy/Es): φ = 0.65 for tied columns, φ = 0.75 for spiral columns. Compression failures are sudden and brittle — hence lower φ.
• Transition zone (εty < εt < 0.005): φ is linearly interpolated between the compression-controlled and tension-controlled φ values.
• Shear and torsion: φ = 0.75.
• Bearing on concrete surfaces: φ = 0.65.
• Plain concrete (no reinforcement): φ = 0.60.
• Post-tensioned anchorage zones: φ = 0.85.
• Spiral columns receive a higher φ (0.75 vs 0.65 for tied) because spiral confinement provides warning before failure and significantly increases concrete ductility.
• The net tensile strain εt is measured at the extreme tension steel layer, NOT at the centroid of the tension steel group, when the beam has multiple layers.

Definitions

Section Title

Strength-Reduction Factors φ (NSCP 2015)

Common Mistakes

• Using φ = 0.90 for columns — columns are compression-controlled (0.65 or 0.75), not tension-controlled.
• Using φ = 0.85 for shear — φ for shear and torsion is 0.75 per NSCP 2015 (some older references used 0.85; NSCP 2015 corrected this).
• Forgetting that spiral columns get φ = 0.75 while tied columns get φ = 0.65 at the compression-controlled limit.
• Applying φ = 0.90 to a section that is not verified as tension-controlled — always check that εt ≥ 0.005.

Formulas

Exam Tips

• A key board shortcut: the β1 reduction is exactly 0.05 per every 7 MPa above 28 MPa. Tabulate: f'c = 28→β1=0.85; 35→0.80; 42→0.75; 49→0.70; 55→0.65.
• In a singly-reinforced beam problem, find a from equilibrium (T = C): a = As·fy/(0.85·f'c·b), then c = a/β1. You rarely need to find c first.
• When the problem gives c directly (from a strain-compatibility problem), get a = β1·c and proceed to Mn = T·(d − a/2).
• Remember: 0.85f'c is the stress block INTENSITY; β1 adjusts the DEPTH — do not confuse the two 0.85 values.

Key Points

• The actual parabolic-triangular concrete stress distribution at ultimate is replaced by an equivalent rectangular (Whitney) stress block of uniform intensity 0.85f'c and depth a = β1·c.
• This substitution produces the same compressive force C = 0.85f'c·a·b and the same location of that force (at depth a/2 from the extreme compression fiber) as the actual distribution — hence 'equivalent.'
• β1 is a dimensionless factor that relates the equivalent stress block depth a to the true neutral-axis depth c: a = β1·c.
• NSCP 2015 / ACI 318-19 values of β1: β1 = 0.85 for f'c ≤ 28 MPa; for 28 MPa < f'c ≤ 55 MPa: β1 = 0.85 − 0.05(f'c − 28)/7; β1 = 0.65 (minimum) for f'c ≥ 55 MPa.
• The equivalent stress block simplifies moment calculation: nominal moment Mn = As·fy·(d − a/2) for a singly-reinforced rectangular beam.
• The 0.85 factor in 0.85f'c accounts for the difference between cylinder strength and in-place concrete strength (size, curing, load-rate effects).
• β1 decreases for higher-strength concrete because the actual stress distribution becomes more rectangular (less parabolic) and the equivalent block must be adjusted to maintain equivalence.
• The stress block applies to the WIDTH b of the compression zone (rectangular or effective flange width for T-beams).

Definitions

Section Title

The Equivalent Rectangular Stress Block and β₁

Common Mistakes

• Applying the β1 reduction formula for f'c BELOW 28 MPa — the formula only applies above 28 MPa; below 28 MPa, β1 = 0.85 always.
• Computing β1 below 0.65 for very high f'c — 0.65 is the minimum floor value; never use a value below 0.65.
• Confusing a (stress block depth) with c (neutral-axis depth) — they are related by a = β1·c but are NOT equal unless β1 = 1.0.
• Using the full f'c (not 0.85f'c) as the stress block intensity — the stress block intensity is always 0.85f'c.
• For f'c = 28 MPa exactly: β1 = 0.85 (not the start of the reduction — reduction starts ABOVE 28 MPa).

Formulas

Exam Tips

• In WSD board problems, the neutral-axis quadratic always has one positive and one negative root — take the positive root.
• If a problem asks for Icr directly, it almost always needs either the deflection or the WSD stress check — identify which before computing.
• Singly-reinforced WSD beam: the 3 key outputs are kd (NA depth), Icr, then fc and fs. Solve in that order.

Key Points

• In WSD, the RC beam cross-section is transformed into an equivalent all-concrete section by replacing steel area As with (n·As) of concrete, where n = Es/Ec.
• For the transformed section, the neutral axis (kd) is found by equating first moments of area about the neutral axis: (b)(kd)(kd/2) = nAs(d − kd).
• This gives: 0.5b(kd)² + nAs(kd) − nAs·d = 0 — a quadratic in kd solved by the quadratic formula.
• The moment of inertia of the transformed cracked section: Icr = b(kd)³/3 + nAs(d − kd)².
• WSD stress check: fc = M·kd/Icr ≤ 0.45f'c; fs = n·M·(d − kd)/Icr ≤ 0.40fy.
• USD and WSD give similar (but not identical) results for the same beam. USD typically uses material more efficiently — a key reason for its adoption.
• The cracked transformed section moment of inertia Icr is also used in both WSD and USD for service-load deflection calculations (Branson's equation).

Definitions

Section Title

Connecting WSD and USD: Transformed Section Analysis

Common Mistakes

• Forgetting to use the CRACKED transformed section (not the gross section) for WSD stress calculations — using Ig gives unconservative (too low) stresses.
• Setting up the neutral-axis equation incorrectly — the concrete in tension below NA is ignored; only (b·kd) in compression and nAs in tension are included.
• Using (n−1)As instead of nAs — (n−1)As is used for doubly-reinforced beams for the compression steel to avoid double-counting; for tension steel, use nAs.