Steel Beam Design Example -- AISC 360-22 LRFD Worked Solution

Steel beam design requires checking flexural strength, shear strength, and serviceability deflection. This worked example walks through each check per AISC 360-22, starting with section classification through to the final pass/fail verdict.

Design parameters

Simply supported steel beam spanning 30 ft, uniformly loaded. W18x35 section, ASTM A992 steel (Fy = 50 ksi, Fu = 65 ksi). Dead load: 0.8 kip/ft (includes self-weight), live load: 1.2 kip/ft. Unbraced length Lb = 6.0 ft (braced at 5 points by joists).

Section properties (W18x35)

A = 10.3 in^2, d = 17.7 in, tw = 0.300 in, bf = 6.00 in, tf = 0.425 in, Ix = 510 in^4, Sx = 57.6 in^3, Zx = 66.5 in^3, ry = 1.33 in, h/tw = 44.0.

Step 1: Section classification

Flange: bf / (2*tf) = 6.00 / (2 * 0.425) = 7.06. Compact limit per AISC Table B4.1b: 0.38 * sqrt(E/Fy) = 9.15. Since 7.06 < 9.15, the flange is compact.

Web: h/tw = 44.0. Compact limit: 3.76 * sqrt(E/Fy) = 90.6. Since 44.0 < 90.6, the web is compact. The section is compact; the full plastic moment applies.

Step 2: Factored loads (LRFD)

wu = 1.2 _ 0.8 + 1.6 _ 1.2 = 0.96 + 1.92 = 2.88 kip/ft. Maximum moment at midspan: Mu = wu _ L^2 / 8 = 2.88 _ 30^2 / 8 = 324 kip-ft = 3888 kip-in. Maximum shear at support: Vu = wu _ L / 2 = 2.88 _ 30 / 2 = 43.2 kips.

Step 3: Flexural strength (AISC Chapter F)

For a compact section with Lb = 6.0 ft, first check Lp and Lr. Lp = 1.76 _ ry _ sqrt(E/Fy) = 1.76 _ 1.33 _ sqrt(29000/50) = 5.60 ft. Lb = 6.0 ft > Lp = 5.60 ft, so inelastic LTB applies.

Lr = 1.95 _ ry _ (E / (0.7Fy)) * sqrt( (JCw) / (Sxry) _ sqrt(1 + sqrt(1 + 6.76 _ (0.7Fy/E * Sx/(J*Cw))^2) ) ) = 17.8 ft.

Since Lp < Lb <= Lr, the nominal moment strength is linear interpolation: Mn = Cb * [Mp - (Mp - 0.7FySx) * (Lb-Lp)/(Lr-Lp)] <= Mp.

Cb = 1.14 (uniform load, simply supported, braced at 5 points). Mn = 1.14 * [3325 - (3325 - 0.75057.6) * (6.0-5.60)/(17.8-5.60)] = 1.14 _ [3325 - 2309 _ 0.0328] = 1.14 _ 3249 = 3704 kip-in. But Mp = Zx _ Fy = 66.5 * 50 = 3325 kip-in. Since Mn > Mp, Mn = Mp = 3325 kip-in (capped at Mp).

Design flexural strength phi*b * Mn = 0.90 _ 3325 = 2993 kip-in = 249 kip-ft. Mu = 324 kip-ft. Ratio = 324/249 = 1.30 — FAILS flexure.

Step 4: Shear strength (AISC Chapter G)

h/tw = 48.7 (using clear web depth). kv = 5.34 (stiffener spacing a/h > 3). 1.10 * sqrt(kv*E/Fy) = 61.3. Since 48.7 < 61.3, Cv = 1.0.

Nominal shear Vn = 0.6FyAwCv = 0.650*(17.7*0.300)_1.0 = 159.3 kips. phi_v _ Vn = 0.90 * 159.3 = 143.4 kips. Vu = 43.2 kips. Ratio = 43.2/143.4 = 0.30 — shear check passes.

Step 5: Deflection check (serviceability)

Live load deflection: Delta_LL = 5wL^4 / (384EI) = 5*(1.2/12)360^4 / (38429000*510) = 2.39 in. Live load limit: L/360 = 360/360 = 1.0 in. Since 2.39 > 1.0: FAILS. Ratio = 2.39.

Total load deflection: Delta_TL = 5*(2.0/12)360^4 / (38429000*510) = 3.98 in. Total load limit: L/240 = 360/240 = 1.5 in. Since 3.98 > 1.5: FAILS. Ratio = 2.65.

Summary

The W18x35 fails for this 30 ft span. A deeper section such as W24x62 would satisfy all checks. Design is governed by flexural strength and deflection, not shear.

Key takeaways

Unbraced length (Lb) is critical — 6 ft bracing is inadequate for a W18x35 spanning 30 ft. Deflection is typically the governing serviceability limit for long-span beams. Section compactness must be verified first, but compactness is rarely the limiting factor for A992 W-shapes.

Educational reference only. All beam designs must be independently verified by a licensed Professional Engineer. Results are PRELIMINARY — NOT FOR CONSTRUCTION.

Try It Yourself

Ready to try this yourself? Use our free Beam Capacity Calculator. Complete flexure, shear, LTB, and deflection checks per AISC 360-22, AS 4100, EN 1993, and CSA S16.

Need section properties? Browse the Section Properties Database for W, HSS, C, L, and WT sections with dimensions, Ix, Sx, Zx, and classification limits.