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AISC LRFD Steel Design

Load and Resistance Factor Design (LRFD) for steel members per AISC 360-22. Design tensile strength (yielding + rupture, Chapter D), compression with the column curve (Chapter E), and flexural yielding (Chapter F) — each reported as the design strength φRn to compare against factored demand Pu / Mu.

ksi (A992: 65; A36: 58)
ksi (steel: 29000)
in²
in² (after holes)
in³
in³
— (compression members)
kips
kips
kips
kips
kip·in

Defaults: A992 wide-flange (σ_y = 50 ksi), A_g = 10 in², KL/r = 80. Output is design strength φR_n — your factored demand (P_u, M_u) must not exceed it. Compactness, LTB, and shear are separately governed by AISC F2–F11 and G.

Tension yielding (AISC D2-1): φt = 0.90
$$ P_n = \sigma_y \, A_g, \qquad \phi_t P_n = 0.90 \, \sigma_y \, A_g $$
Tension rupture (AISC D2-2): φt = 0.75
$$ P_n = \sigma_u \, A_e, \qquad \phi_t P_n = 0.75 \, \sigma_u \, A_e $$
Compression (AISC E3): elastic σ_e = π²E/(KL/r)²
$$ \sigma_{cr} = \begin{cases} (0.658^{\sigma_y/\sigma_e}) \sigma_y & \sigma_e \ge 0.44\sigma_y \\ 0.877\, \sigma_e & \text{otherwise} \end{cases}, \qquad \phi_c P_n = 0.90 \, \sigma_{cr} A_g $$
Flexural yielding (compact, AISC F2): M_n = M_p = σ_y Z_x; φ_b = 0.90
$$ \phi_b M_n = 0.90 \, \sigma_y \, Z_x $$
φ resistance factor in LRFD (0.90 tension yield / compression / flexure; 0.75 rupture) · P_n, M_n nominal strength · φP_n, φM_n design strength · A_g gross area · A_e effective net area = U × A_n where U is the shear-lag reduction factor.

LRFD vs ASD — they're equivalent now

AISC 360 unified ASD and LRFD in 2005. Both methods use the same nominal strength equations; they differ only in how safety is applied:

The relationship: φ × Ω ≈ 1.5. For tension yielding, φ = 0.90 and Ω = 1.67 → φΩ = 1.50. LRFD matches reliability theory more directly because the load and resistance uncertainties are factored separately; ASD bundles them into a single Ω.

Compression — the column equation

For KL/r < 4.71 √(E/σy) ≈ 113 for A36, the section yields before Euler buckling fully manifests — use the inelastic curve (0.658 raised to the σye ratio). For KL/r above that, use Euler with a 0.877 reduction for residual stress and initial out-of-straightness. In LRFD, multiply the resulting φcPn = 0.90 σcr Ag.

Flexure — beyond yielding

For compact, laterally-braced beams, Mn = Mp = σy Z. For non-compact beams, plastic capacity reduces (F2.2). For long unbraced lengths, lateral-torsional buckling (LTB, F2.3) reduces capacity further. The full Chapter F has 11 sub-sections covering different failure modes. This tool returns the compact, fully-braced φbMn = 0.90 σy Zx.

What this calculator doesn't check

Combined-force interaction (H1.1)

For members with axial + flexure: Pr/Pc + (8/9)(Mrx/Mcx + Mry/Mcy) ≤ 1.0 when Pr/Pc ≥ 0.2; else Pr/(2Pc) + Mrx/Mcx + Mry/Mcy ≤ 1.0. In LRFD, Pr = Pu (factored) and Pc = φcPn. Second-order (P-Δ / P-δ) effects require a direct analysis or B1/B2 amplification per AISC C2.

Reference: AISC 360-22 (2022). Specification for Structural Steel Buildings. AISC Steel Construction Manual, 16th ed. ASCE 7-22, load combinations §2.3. Salmon, C.G., Johnson, J.E. (2009). Steel Structures: Design and Behavior, 5th ed., Pearson.

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