Steel Beam Design Guide 2026 — Complete AISC 360-22 Walkthrough

A comprehensive guide to steel beam design per AISC 360-22, covering the full design workflow from load definition through final strength and serviceability checks. Includes a complete worked example with numeric calculations and code clause references.

Overview

Steel beam design is the most common task in structural engineering practice. Whether you are designing floor beams for a commercial building, roof purlins for an industrial shed, or crane runway girders, the fundamental workflow is the same: define the loading, analyse for actions, select a trial section, and verify that section against all applicable limit states.

This guide walks through the complete beam design process using AISC 360-22 (Specification for Structural Steel Buildings) with the LRFD methodology. Every step references the specific code clause, shows the governing equation, and is illustrated with a fully worked numeric example. The guide assumes familiarity with basic structural mechanics but explains code-specific nuances that trip up even experienced engineers.

The worked example uses a W18x35 floor beam — a section commonly encountered in US practice — and demonstrates that beam design is rarely governed by a single check. Flexure, lateral-torsional buckling, shear, deflection, and web local buckling each impose constraints that must be satisfied simultaneously.

1. Load Definition and Analysis (ASCE 7-22, AISC 360 §B)

Before selecting a section, you must establish the factored design actions. The distinction between service (unfactored) and factored loads is critical — mixing them is the single most common beam design error.

Governing load combination (LRFD, ASCE 7-22 §2.3.1):

wu = 1.2 D + 1.6 L + 0.5 (Lr or S or R)

For a typical floor beam carrying dead load D and live load L only:

wu = 1.2 D + 1.6 L

Analysis for a simply supported beam under uniform load:

Worked Example — Loads:

A simply supported floor beam spans 30 ft with a tributary width of 10 ft. Superimposed dead load = 60 psf (flooring, ceiling, MEP). Live load = 50 psf (office occupancy per ASCE 7 Table 4.3-1).

wD = 60 psf x 10 ft = 600 plf = 0.60 klf
wL = 50 psf x 10 ft = 500 plf = 0.50 klf
wu = 1.2(0.60) + 1.6(0.50) = 0.72 + 0.80 = 1.52 klf

Mu = 1.52 x 30^2 / 8 = 171.0 kip-ft = 2,052 kip-in
Vu = 1.52 x 30 / 2 = 22.8 kips

2. Section Classification (AISC 360-22 §B4)

Section classification determines which capacity equations apply. Per Table B4.1b, W-shapes are classified by comparing width-to-thickness ratios against limiting values lambda_p (compact) and lambda_r (non-compact/slender).

Our W18x35 has compact flanges and a compact web, so we can use the full plastic moment capacity Mp for flexure (AISC §F2.1) and the full plastic shear capacity Vp (AISC §G2.1).

For non-compact flanges (lambda_p < bf/2tf <= lambda_r), you must use Mn = Mp - (Mp - 0.7 Fy Sx) x (lambda - lambda_p) / (lambda_r - lambda_p) per §F3.2. For slender webs, tension field action in shear may be available (§G2.2), but the flexural capacity is capped at My = Fy Sx.

3. Flexural Capacity (AISC 360-22 Chapter F)

3.1 Yielding Limit State (AISC §F2.1)

For compact sections with full lateral bracing (Lb = 0, top flange continuously braced by the concrete slab):

Mn = Mp = Fy Zx
phi Mn = 0.90 Fy Zx

Worked Example — W18x35, fully braced (Lb = 0):

Zx = 66.5 in^3 (from AISC Table 1-1)
phi Mn = 0.90 x 50 ksi x 66.5 in^3 = 2,992.5 kip-in = 249.4 kip-ft
Utilization = Mu / (phi Mn) = 171.0 / 249.4 = 0.686  OK

At Lb = 0, flexure is at only 69% utilization — the beam has reserve capacity.

3.2 Lateral-Torsional Buckling (AISC §F2.2)

When the compression flange is NOT continuously braced, LTB must be checked. The capacity depends on the unbraced length Lb relative to two limiting lengths:

Lp = 1.76 ry sqrt(E / Fy)           — full plastic moment limit
Lr = 1.95 rts (E / 0.7 Fy) x ...    — inelastic/elastic LTB transition

Where rts is the effective radius of gyration (given in AISC Table 1-1 or computed per §F2.2).

For W18x35: Lp = 4.31 ft, Lr = 12.0 ft, rts = 1.58 in.

Three LTB regimes:

Worked Example — same W18x35, now with Lb = 8 ft (unbraced between floor joists at 8 ft spacing):

Since Lp (4.31 ft) < Lb (8 ft) <= Lr (12.0 ft), we are in the inelastic LTB regime.

First, compute Cb. For a simply supported beam under uniform load, the maximum moment is at midspan. The quarter-point moments are at x = 7.5 ft:

At x = L/4 = 7.5 ft:  M(x) = w x (L - x) / 2 = 1.52 x 7.5 x (30 - 7.5) / 2 = 128.25 kip-ft
   MA = 128.25 kip-ft
At x = L/2 = 15 ft:   Mmax = 171.0 kip-ft = MB
At x = 3L/4 = 22.5 ft: MC = 128.25 kip-ft

Cb = 12.5 x 171.0 / (2.5 x 171.0 + 3 x 128.25 + 4 x 171.0 + 3 x 128.25)
   = 2,137.5 / (427.5 + 384.75 + 684.0 + 384.75)
   = 2,137.5 / 1,881.0 = 1.136

Now compute the reduced LTB moment:

Mp = Fy Zx = 50 x 66.5 = 3,325 kip-in
0.7 Fy Sx = 0.7 x 50 x 57.6 = 2,016 kip-in

Mn = Cb [Mp - (Mp - 0.7FySx) x (Lb - Lp) / (Lr - Lp)]
   = 1.136 [3,325 - (3,325 - 2,016) x (8.0 - 4.31) / (12.0 - 4.31)]
   = 1.136 [3,325 - 1,309 x 3.69 / 7.69]
   = 1.136 [3,325 - 628.2]
   = 1.136 x 2,696.8 = 3,064 kip-in = 255.3 kip-ft
   Capped at Mp = 277.1 kip-ft (3,325/12)

phi Mn = 0.90 x 255.3 = 229.8 kip-ft
Utilization = 171.0 / 229.8 = 0.744  OK (was 0.686 at Lb=0)

The LTB reduction is modest (utilization goes from 0.69 to 0.74) because Lb = 8 ft is only moderately beyond Lp = 4.31 ft. For Lb = 20 ft (common in crane girders), the capacity would drop to roughly 35% of Mp — a 65% reduction.

4. Shear Capacity (AISC 360-22 Chapter G)

For rolled I-shapes, shear is resisted by the web. Per §G2.1:

Vn = 0.6 Fy Aw Cv1

where Aw = d tw (gross area of the web) and Cv1 = 1.0 when h/tw <= 1.10 sqrt(kv E / Fy).

For W18x35: h/tw = 46.2, kv = 5.0 (unstiffened web).

1.10 sqrt(5.0 x 29,000 / 50) = 1.10 x 53.85 = 59.2
h/tw = 46.2 < 59.2, therefore Cv1 = 1.0

Aw = d tw = 17.7 x 0.300 = 5.31 in^2
Vn = 0.6 x 50 x 5.31 x 1.0 = 159.3 kips
phi Vn = 1.00 x 159.3 = 159.3 kips (phi = 1.00 per §G1)

Utilization = Vu / (phi Vn) = 22.8 / 159.3 = 0.143  OK

Shear rarely governs for standard rolled W-shapes under uniform load. It can govern for short, heavily loaded spans (L/d < 8), coped beams, or concentrated loads near supports.

Tension field action (§G2.2): For webs with h/tw > 1.10 sqrt(kv E/Fy), post-buckling tension field action can be mobilised in interior panels with transverse stiffeners. This is unusual for rolled sections but common in plate girders.

5. Deflection Check (AISC 360 Serviceability, IBC §1604.3)

Deflection is a serviceability check, not a strength check. Use unfactored (service) loads, not LRFD factored loads.

For a simply supported beam under uniform load:

delta = 5 w L^4 / (384 E I)

Worked Example — live load deflection:

w_serv = 0.50 klf (live load only)
L = 30 ft = 360 in
E = 29,000 ksi
Ix = 510 in^4 (W18x35)

delta_LL = 5 x (0.50/12) x 360^4 / (384 x 29,000 x 510)
         = 5 x 0.04167 x 1.6796e10 / (384 x 29,000 x 510)
         = 3.499e9 / 5.681e9
         = 0.616 in

Limit = L / 360 = 360 / 360 = 1.00 in
Utilization = 0.616 / 1.00 = 0.616  OK

Total load deflection (dead + live):

w_total = 0.60 + 0.50 = 1.10 klf
delta_TL = delta_LL x (1.10 / 0.50) = 0.616 x 2.2 = 1.355 in
Limit = L / 240 = 360 / 240 = 1.50 in
Utilization = 1.355 / 1.50 = 0.903  OK

Total load deflection is at 90% of the limit — this is the governing check if no camber is specified. Camber of 3/4 in would reduce the apparent total deflection to (1.355 - 0.75) = 0.605 in, well within limits.

Deflection limit reference table (IBC 2024 Table 1604.3):

6. Combined Flexure and Shear (AISC 360-22 §G4, §H2)

For most beams, flexure and shear are evaluated independently. However, when both utilization ratios exceed 0.50, AISC §G4 requires an interaction check:

When Vu / (phi Vn) > 0.50:  reduce Mn for the effect of shear on the moment capacity.

For our example, Vu/(phi Vn) = 0.14 — far below the threshold. No interaction check needed.

7. Web Local Yielding and Crippling (AISC 360-22 §J10)

Concentrated forces (reactions, point loads) applied to the flange must be checked for:

• Web local yielding (§J10.2): Rn = Fyw tw (5k + lb) for interior loads, or Rn = Fyw tw (2.5k + lb) for end reactions
• Web crippling (§J10.3): Rn based on web slenderness and bearing length
• Sidesway web buckling (§J10.4): Check when compression flange is not laterally restrained at the load point

For the W18x35 with end reactions of 22.8 kips on a 4-inch bearing plate:

Rn (yielding) = 1.00 x 50 x 0.300 x (2.5 x 0.827 + 4.0) = 15 x 6.068 = 91.0 kips
phi Rn = 0.90 x 91.0 = 81.9 kips
Utilization = 22.8 / 81.9 = 0.278  OK

8. Summary — Complete Design Check for W18x35

Result: W18x35 passes all limit states. The governing check is total load deflection at 0.90 utilization.

If a shallower section is required for architectural coordination, W16x31 achieves 0.84 flexure utilization and 0.89 total deflection — still acceptable. The lighter W14x26 would fail deflection at 1.15 utilization.

9. Multi-Code Beam Design Comparison

Related Calculators

• Simple Beam Calculator — Interactive beam analysis with shear, moment, and deflection diagrams for any span and loading
• Steel Beam Capacity Calculator — Flexure, shear, and LTB checks per AISC 360, AS 4100, EN 1993, and CSA S16
• Beam Deflection Calculator — Service-level deflection checks against IBC, AS 1170, EN 1990 limits
• Continuous Beam Calculator — Multi-span beam analysis with moment redistribution
• Moment of Inertia Calculator — Section properties for any built-up or standard steel section

FAQ

Q: When does lateral-torsional buckling govern over yielding?

A: LTB governs when the unbraced length Lb exceeds Lp. For W18x35, Lp = 4.31 ft. At Lb = 0, the beam uses its full plastic moment (0.69 utilization). At Lb = 8 ft, capacity drops to 0.74 utilization. At Lb = 20 ft, the beam would enter elastic LTB per AISC 360-22 §F2.2 and capacity would be approximately 35% of Mp — a 65% reduction. Always verify unbraced length assumptions before finalising a beam size.

Q: Should I use ASD or LRFD for beam design?

A: LRFD (AISC 360 Chapter B) is the default for most US building design. It applies separate load factors (1.2D + 1.6L per ASCE 7) and resistance factors (phi = 0.90 for flexure). ASD uses Omega = 1.67 for flexure. For live-to-dead ratios near 3:1 (typical office floors), the two methods produce virtually identical designs. Use whichever your office standard specifies — but never mix ASD loads with LRFD resistance factors or vice versa.

Q: What Cb value should I use?

A: Always compute the actual Cb per AISC 360 §F1 rather than conservatively using Cb = 1.0. For uniform load on a simply supported beam, Cb = 1.14. For beams with reverse curvature bending (end moments in opposite directions), Cb can be as high as 2.3. Using Cb = 1.0 when the actual value is 1.3 means leaving 30% of LTB capacity on the table — overdesigning the beam unnecessarily.

Q: Can I rely on the concrete slab to brace the compression flange?

A: Yes, if the slab is positively connected to the steel beam via shear studs or a reliable mechanical connection, the top flange can be considered continuously braced (Lb = 0) for gravity loads. However, during construction (before the slab is placed and cured), the bare steel beam may have Lb equal to the full span. AISC 360 §F1 and AISC DG3 provide guidance on bracing requirements. For negative moment regions (e.g., at continuous beam supports), the bottom flange is in compression and requires separate bracing unless attachment to the column/connection provides restraint.