Beam Design Guide — Engineering Reference

AISC 360 Chapter F beam design step-by-step: compact sections, Lp/Lr unbraced length, LTB, Cb factor, shear design, and deflection serviceability checks.

Overview

Steel beam design per AISC 360 Chapter F requires checking flexural yielding, lateral-torsional buckling (LTB), flange local buckling (FLB), and web local buckling (WLB). The designer selects a W-shape or built-up section, classifies it as compact, noncompact, or slender per Table B4.1b, then calculates the available flexural strength considering unbraced length and moment gradient.

Shear design per Chapter G and deflection serviceability checks complete the beam design workflow. For composite beams with concrete slabs, Chapter I governs the shear stud and effective width requirements.

Flexural strength and LTB

The nominal moment capacity depends on unbraced length Lb relative to two transition points:

The moment gradient factor Cb accounts for non-uniform bending moment distributions. For uniform moment Cb = 1.0; for typical gravity loading on simply supported beams Cb ranges from 1.14 to 1.67 per Eq. F1-1.

Worked example — W18x50 simply supported beam

Given: W18x50, A992 (Fy = 50 ksi), span L = 30 ft, uniform load w_u = 2.5 kip/ft (factored), lateral bracing at midspan only (L_b = 15 ft). Properties: Z_x = 101 in^3, S_x = 88.9 in^3, r_y = 1.65 in., r_ts = 1.98 in., J = 1.24 in^4, h_o = 17.4 in.

  1. Check compactness: b_f/(2t_f) = 7.50/(2 x 0.57) = 6.58 < 9.15 (compact). h/t_w = 16.86/0.355 = 47.5 < 90.6 (compact). Section is compact.
  2. Plastic moment: M_p = F_y x Z_x = 50 x 101 = 5050 kip-in = 420.8 kip-ft.
  3. L_p: L_p = 1.76 x r_y x sqrt(E/F_y) = 1.76 x 1.65 x sqrt(29000/50) = 1.76 x 1.65 x 24.08 = 69.9 in. = 5.83 ft.
  4. L_r (from AISC Eq. F2-6): L_r ≈ 16.6 ft (calculated from r_ts, J, S_x, h_o, c).
  5. Check LTB: L_b = 15 ft. Since L_p (5.83) < L_b (15.0) < L_r (16.6), inelastic LTB governs.
  6. C_b factor: For uniform load with midspan brace, C_b ≈ 1.30 (quarter-point moment method).
  7. Nominal moment: M_n = C_b x [M_p - (M_p - 0.7 x F_y x S_x) x (L_b - L_p)/(L_r - L_p)] = 1.30 x [5050 - (5050 - 0.7 x 50 x 88.9) x (15.0 - 5.83)/(16.6 - 5.83)] = 1.30 x [5050 - 1939 x 0.852] = 1.30 x 3398 = 4418 kip-in. But M_n cannot exceed M_p = 5050, so M_n = 4418 kip-in = 368.2 kip-ft.
  8. Design strength: phi x M_n = 0.90 x 368.2 = 331.3 kip-ft.
  9. Required moment: M_u = w_u x L^2 / 8 = 2.5 x 30^2 / 8 = 281.3 kip-ft. Since 281.3 < 331.3, OK.

Shear design

Web shear capacity per AISC 360 Chapter G: phi_v x V_n = 1.00 x 0.6 x F_y x A_w x C_v1, where A_w = d x t_w. For most rolled W-shapes with h/t_w <= 2.24 x sqrt(E/F_y) = 53.9 (A992), the web shear coefficient C_v1 = 1.0 and no transverse stiffeners are needed.

Continuing the example: V_u = w_u x L / 2 = 2.5 x 30 / 2 = 37.5 kip. phi x V_n = 1.00 x 0.6 x 50 x (18.0 x 0.355) x 1.0 = 191.7 kip >> 37.5 kip. Shear is OK by inspection.

Code comparison — beam flexural design

Parameter AISC 360-22 (F2) AS 4100 (Sec. 5) EN 1993-1-1 (6.3.2) CSA S16 (13.6)
Plastic moment M_p = F_y x Z_x M_sx = f_y x Z_x M_pl = f_y x W_pl M_p = F_y x Z_x
LTB reduction Linear interpolation L_p to L_r alpha_s slenderness factor chi_LT buckling curves Linear interpolation similar to AISC
phi / gamma phi_b = 0.90 phi = 0.90 gamma_M1 = 1.00 phi = 0.90
Moment gradient C_b (quarter-point) alpha_m moment modification C_1 (end moment ratio) omega_2 (equivalent to C_b)
Shear check Chapter G, C_v1 Section 5.11, V_v Clause 6.2.6, V_pl Clause 13.4

Deflection limits

Beam deflection is a serviceability check using unfactored service loads, not factored LRFD loads:

Condition Live Load Limit Total Load Limit Source
Floor beams L/360 L/240 IBC Table 1604.3
Roof beams (no plaster) L/240 L/180 IBC Table 1604.3
Supporting brittle finishes L/480 L/360 Common practice
Cantilevers L/180 L/120 Engineering judgment
Steel joist (SJI) L/360 (default) SJI specification

For the worked example above: service live load w_L = 1.2 kip/ft (assumed), delta = 5 x w_L x L^4 / (384 x E x I_x) = 5 x 0.1 x 360^4 / (384 x 29000 x 800) = 0.95 in. Limit = L/360 = 360/360 = 1.0 in. Since 0.95 < 1.0, deflection OK.

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Disclaimer

This page is for educational and reference use only. It does not constitute professional engineering advice. All design values must be verified against the applicable standard and project specification before use. The site operator disclaims liability for any loss arising from the use of this information.