Beam Capacity Worked Example — W18x35 per AISC 360-22 LRFD

Problem: Determine the design flexural strength, shear strength, and deflection of a W18x35 (A992 steel, Fy=50 ksi) spanning 25 ft, simply supported, with a uniform dead load of 0.5 kip/ft and live load of 1.0 kip/ft. The compression flange is braced at 6ft intervals.

Step 1: Section Properties (W18x35)

From AISC Manual Table 1-1:

• A = 10.3 in²
• d = 17.70 in
• tw = 0.300 in
• bf = 6.00 in
• tf = 0.425 in
• Ix = 510 in⁴
• Zx = 66.5 in³
• Sx = 57.6 in³
• Lp = 4.71 ft (from Manual Table 3-2)
• Lr = 13.5 ft
• h/tw = 53.5

Step 2: Required Strength (LRFD Load Combos)

Combination 1: 1.4D (AISC 360 Basic) wu = 1.4 × 0.5 = 0.70 kip/ft Mu = wu×L²/8 = 0.70 × 25² / 8 = 54.7 kip·ft Vu = wu×L/2 = 0.70 × 25 / 2 = 8.75 kips

Combination 2: 1.2D + 1.6L (governs) wu = 1.2 × 0.5 + 1.6 × 1.0 = 2.20 kip/ft Mu = 2.20 × 25² / 8 = 171.9 kip·ft Vu = 2.20 × 25 / 2 = 27.5 kips ← governs

Step 3: Flexural Capacity (AISC 360 F2)

Lb = 6.0 ft (brace spacing) Lp = 4.71 ft < Lb = 6.0 ft < Lr = 13.5 ft → Inelastic LTB

Mp = Fy × Zx = 50 × 66.5 / 12 = 277.1 kip·ft

Cb = 1.0 (conservative for uniformly loaded simple span)

Mn = Cb × [Mp - (Mp - 0.7Fy×Sx) × (Lb - Lp)/(Lr - Lp)] = 1.0 × [277.1 - (277.1 - 0.7×50×57.6/12) × (6.0 - 4.71)/(13.5 - 4.71)] = 1.0 × [277.1 - (277.1 - 168.0) × 0.142] = 1.0 × [277.1 - 15.5] = 261.6 kip·ft

ϕbMn = 0.90 × 261.6 = 235.4 kip·ft

Mu / ϕbMn = 171.9 / 235.4 = 0.73 → OK (73% utilized)

Step 4: Shear Capacity (AISC 360 G2)

h/tw = 53.5

For h/tw ≤ 2.24√(E/Fy) = 2.24√(29000/50) = 53.9 → 53.5 < 53.9 → Cv1 = 1.0

Vn = 0.6 × Fy × Aw × Cv1 = 0.6 × 50 × (17.70 × 0.300) × 1.0 = 159.3 kips

ϕvVn = 1.00 × 159.3 = 159.3 kips

Vu / ϕvVn = 27.5 / 159.3 = 0.17 → OK (17% utilized)

Step 5: Deflection Check (Serviceability)

Service load (D+L) = 0.5 + 1.0 = 1.5 kip/ft

Δmax = 5×w×L⁴/(384×E×I) = 5 × 1.5 × (25×12)⁴ / (384 × 29000 × 510) = 1.07 inches

Allowable Δ = L/360 = 25×12/360 = 0.83 inches

Δactual = 1.07 in > 0.83 in → Deflection FAILS

Step 6: Redesign

Try W21x44 (Ix = 843 in⁴): Δ = 1.07 × 510/843 = 0.65 inches Δ/L = 0.65/(25×12) = L/462 < L/360 → OK

Check flexure: W21x44, Zx = 95.4 in³ Lp = 6.47 ft > Lb = 6.0 ft → Compact, full plastic Mp = 50 × 95.4 / 12 = 397.5 kip·ft ϕbMn = 0.90 × 397.5 = 357.8 kip·ft → 48% utilized

Use W21x44 (22% heavier section, but passes all checks).

Step 7: ASD Comparison

For reference, the ASD equivalent for this beam:

Service loads: D = 0.5 kip/ft, L = 1.0 kip/ft

Required strength (ASD): Ma = (0.5 + 1.0) × 25² / 8 = 117.2 kip·ft Va = (0.5 + 1.0) × 25 / 2 = 18.75 kips

Allowable moment (W18x35): Mn/Ωb = 261.6 / 1.67 = 156.6 kip·ft Ma / (Mn/Ωb) = 117.2 / 156.6 = 0.75 → OK

Allowable shear (W18x35): Vn/Ωv = 159.3 / 1.50 = 106.2 kips Va / (Vn/Ωv) = 18.75 / 106.2 = 0.18 → OK

The ASD utilization ratios (75% flexure, 18% shear) are consistent with the LRFD ratios (73% flexure, 17% shear), confirming the load factor ratio of approximately 1.5 between LRFD and ASD for this load combination.

Step 8: Effect of Unbraced Length on Capacity

The bracing arrangement significantly affects flexural capacity:

Adding a brace at midspan (reducing Lb from 25 ft to 12.5 ft) would increase capacity by approximately 85%, potentially allowing the W18x35 to work at spans where it otherwise would not.

Step 9: Non-Uniform Moment (Cb Factor)

The example above used Cb = 1.0 (conservative). For a uniformly loaded simple span, the actual Cb = 1.14 produces:

Mn_actual = 1.14 × 252.6 = 288.0 kip·ft ϕbMn_actual = 0.90 × 288.0 = 259.2 kip·ft Mu / ϕbMn = 171.9 / 259.2 = 0.66 → Even more conservative at 66% utilization

For other moment diagrams:

• A beam with concentrated loads at third-points: Cb ≈ 1.01
• A beam with equal end moments (uniform M): Cb = 1.0
• A beam with one end moment (triangular): Cb ≈ 1.67

Using the correct Cb is particularly important near the Lp/Lr boundary where a small increase in capacity can change the economics of the design.

Step 10: Shear Deflection (Shear Deformability)

For short, deep beams (span/depth < 10), shear deflection becomes significant and must be added to the bending deflection. The shear deflection for a uniformly loaded simple span is:

Δ_v = (w × L²) / (8 × G × A_w)

For the W18x35 at 25 ft: G = 11,200 ksi, A_w = d × t_w = 5.31 in² Δ_v = 1.5 × (300)² / (8 × 11,200 × 5.31) = 0.028 inches

Total deflection Δ_total = Δ_bending + Δ_v = 1.07 + 0.03 = 1.10 inches

Shear deflection adds only 3% for this case but can reach 10-15% for beams with span/depth ratios below 8 (deep transfer girders, crane runway beams). Per AISC Design Guide 11, shear deflection is negligible for span/depth > 10.

Step 11: ASCE 7 Load Combination Reference

The beam is checked using the ASCE 7-22 basic load combinations. For this floor beam with dead and live loads only, the governing combinations are:

1. 1.4D — controls when live loads are less than approximately half the dead load
2. 1.2D + 1.6L — controls for typical floor loads with live load exceeding dead load
3. 0.9D — used only when wind or seismic loads produce uplift or reversal

For beams supporting roofs, additional combinations including wind (1.0D + 1.0W, 1.2D + 1.0L + 0.5W), snow (1.2D + 1.0L + 1.0S), and seismic (1.2D + 1.0E + 0.5L) per ASCE 7-22 must be checked. The Steel Calculator beam capacity tool includes all applicable load combinations automatically when the load type is specified.

Step 12: Compactness Check

Verify that the W18x35 is compact per AISC Table B4.1b:

• Flange: bf/2tf = 6.00/(2×0.425) = 7.06. λp = 0.38√(E/Fy) = 9.15. 7.06 < 9.15 → Compact
• Web: h/tw = 53.5. λp = 3.76√(E/Fy) = 90.6. 53.5 < 90.6 → Compact

Since the section is compact with Lb < Lr, the beam reaches the full plastic moment modified for LTB. No local buckling reduction applies.

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Use the Beam Capacity Calculator to check your own beam section, span, and loading with AISC, AS 4100, EN 1993, or CSA S16.

Frequently Asked Questions

Does the beam capacity calculator check deflection automatically? Yes. The calculator reports both strength (bending + shear) and serviceability (deflection at L/360, L/240, or user-defined limit). The example above shows why deflection often governs: the W18x35 had 73% flexural capacity but failed deflection at L/462 vs L/360.

What code standards does the calculator support? AISC 360-22 LRFD, AS 4100-2020, EN 1993-1-1 (with UK NA defaults), and CSA S16-19. The example above uses AISC 360 F2 for flexure and G2 for shear.

How do I get more than 50 free calculations per day? Register for a free account. Registered users get 50 calculations/day. For higher limits, check the pricing page.

How does the Cb factor affect beam capacity? The Cb factor (lateral-torsional buckling modification factor) accounts for non-uniform moment diagrams. For a uniformly loaded simple span, Cb = 1.14, which increases nominal capacity by 14% compared to the conservative Cb = 1.0 assumption. Using the correct Cb is most impactful when Lb is near Lp — it can make the difference between an inelastic LTB check and a full plastic moment, saving 5-15% on beam weight.

See Also

• Beam Capacity Calculator
• Beam Deflection Calculator
• Steel Beam Sizes Reference
• Beam Design Guide
• Beam Span Reference