5.5 Structural Design (Steel & Reinforced Concrete)

Key Takeaways

  • LRFD checks factored load demand against φRn (reduced strength); ASD checks service load against Rn/Ω (Ω = safety factor).
  • Tension-controlled reinforced-concrete flexure uses φ = 0.90; shear uses φ = 0.75 per ACI 318.
  • Concrete flexural design strength φMn must equal or exceed the factored moment Mu; ρ = As/(bd) is the steel ratio.
  • Steel members are designed for tension yield/rupture, compression buckling, and flexure per the AISC Steel Construction Manual.
Last updated: June 2026

Two Design Philosophies: ASD and LRFD

Structural design ensures capacity ≥ demand with a margin against failure. Two formats coexist:

  • Load and Resistance Factor Design (LRFD) — the modern probabilistic format. Amplify loads with load factors, reduce nominal strength with a resistance factor φ: φ·Rn ≥ Σ(γᵢ·Qᵢ) (e.g., φMn ≥ Mu).
  • Allowable Strength Design (ASD) — the traditional format. Compare service (unfactored) loads against a reduced allowable strength: Rn/Ω ≥ Σ Qᵢ, where Ω is the safety factor.

The two are calibrated to give similar designs; φ and Ω are inversely related (roughly φ·Ω ≈ 1.5 for steel: φ = 0.90 ↔ Ω = 1.67). LRFD applies different factors to dead vs. live load to reflect their different uncertainty, so it is usually more economical for high live-to-dead ratios.

Steel Member Design (AISC)

Steel design follows the AISC Steel Construction Manual (American Institute of Steel Construction). Three basic limit states:

  • Tension members — governed by the lesser of yielding on the gross area (Pn = Fy·Ag, φ = 0.90) and rupture on the net effective area (Pn = Fu·Ae, φ = 0.75).
  • Compression members (columns) — strength Pn = Fcr·Ag, where the critical stress Fcr depends on slenderness KL/r and the elastic buckling stress Fe = π²E/(KL/r)². Long slender columns fail by buckling, not yielding; φc = 0.90.
  • Flexural members (beams) — nominal moment Mn = Fy·Z (plastic) for compact, laterally braced sections (Z = plastic section modulus); φb = 0.90. Long unbraced compression flanges trigger lateral-torsional buckling, reducing Mn.

A recurring trap: using elastic section modulus S where plastic Z is required for a compact section's plastic moment capacity.

Reinforced-Concrete Flexure & φ Factors

Reinforced concrete is designed by ACI 318. A beam's nominal flexural strength for a singly reinforced rectangular section:

Mn = As·fy·(d − a/2), with the stress-block depth a = As·fy/(0.85·f'c·b)

  • As = tension-steel area, fy = steel yield strength, f'c = concrete compressive strength, b = width, d = effective depth.
  • The steel ratio ρ = As/(b·d) is kept between ρ_min and ρ_max so the section is tension-controlled (steel yields first → ductile warning before failure).

Design requires φMn ≥ Mu. ACI strength-reduction factors:

Actionφ
Tension-controlled flexure0.90
Shear and torsion0.75
Spiral-reinforced compression0.75
Tied-column compression0.65

Use φ = 0.90 only when the section is verified tension-controlled (net tensile strain ε_t ≥ 0.005).

Load Combinations & a Worked Concrete Example

Design uses factored load combinations from ASCE 7 (American Society of Civil Engineers Standard 7). The two most-tested LRFD combinations are 1.4D and 1.2D + 1.6L — the latter usually governs for gravity-loaded members (full detail in Section 5.6).

Worked example: A reinforced-concrete beam has b = 12 in, d = 20 in, As = 3.0 in², f'c = 4000 psi, fy = 60,000 psi. Find φMn.

  • a = As·fy/(0.85·f'c·b) = (3.0 × 60,000)/(0.85 × 4000 × 12) = 180,000/40,800 = 4.41 in.
  • Mn = As·fy·(d − a/2) = 3.0 × 60,000 × (20 − 2.21) = 180,000 × 17.79 = 3.20×10⁶ in·lb = 266.9 kip·ft.
  • Tension-controlled, so φ = 0.90: φMn = 0.90 × 266.9 = 240 kip·ft.

The design is adequate if the factored moment Mu ≤ 240 kip·ft. Always confirm the section is tension-controlled before applying φ = 0.90.

Shear, Detailing & Material Limits

Flexure is only half of beam design; shear and detailing complete it. In reinforced concrete the nominal shear strength combines concrete and stirrups:

Vn = Vc + Vs, with Vc ≈ 2·λ·√f'c·b·d (psi units) and Vs = A_v·fy·d/s

where A_v = stirrup area, s = stirrup spacing, λ = lightweight-concrete factor. Design requires φVn ≥ Vu with φ = 0.75. Minimum steel ratio ρ_min and maximum bar spacing guard against brittle cracking; development length L_d ensures bars are anchored to develop fy.

For steel, common grades are A992 (Fy = 50 ksi) for wide-flange shapes and A36 (Fy = 36 ksi) for plates. The modulus of elasticity is E = 29,000 ksi for all structural steel.

Material propertyTypical value
Steel modulus E29,000 ksi
A992 yield Fy50 ksi
Concrete f'c (normal)3,000–6,000 psi
Concrete modulus Ec57,000√f'c (psi)
Rebar yield fy60,000 psi (Grade 60)

Knowing these reference values lets you sanity-check Handbook lookups and spot mis-keyed inputs quickly.

** Strength is necessary but not sufficient — designs must also satisfy serviceability: deflection limits (commonly live-load span/360, total span/240), crack-width control in concrete, and drift limits under lateral load. These use service (unfactored) loads, never the factored LRFD combinations. Connections transfer the member forces and are designed for the same limit states as members: bolt shear and bearing, and weld strength, must develop the connected member's demand. In concrete, development length and lap splices play the analogous anchoring role.

The unifying design statement across steel and concrete, ASD and LRFD, is simply capacity ≥ demand with an appropriate safety margin — apply load factors to the demand side and resistance factors (or a safety factor Ω) to the capacity side, consistently within one method.

Test Your Knowledge

What is the modulus of elasticity E commonly used for structural steel in AISC design?

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Test Your Knowledge

Per ACI 318, what strength-reduction factor φ applies to a tension-controlled reinforced-concrete section in flexure?

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Test Your Knowledge

A singly reinforced beam has b = 10 in, As = 2.0 in², f'c = 4000 psi, fy = 60,000 psi. What is the depth of the equivalent (Whitney) stress block, a?

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Test Your Knowledge

In Allowable Strength Design (ASD), how is adequacy verified?

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D