UK Beam Design Worked Example — 533UB Section in S355, 6.0 m Span, UDL

Full worked example for a UK Universal Beam (533UB) in S355 steel spanning 6.0 m under a uniformly distributed load, designed to BS EN 1993-1-1:2005 with UK National Annex. Cross-section classification per Clause 5.5, bending moment resistance Mc,Rd per Clause 6.2.5, shear resistance Vc,Rd per Clause 6.2.6, lateral-torsional buckling resistance Mb,Rd per Clause 6.3.2 with intermediate restraints, and serviceability deflection check.

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Design Problem

Problem: Verify a simply supported 533UB in S355 steel spanning L = 6.0 m between supports. The beam supports a composite floor slab that provides full lateral restraint to the top (compression) flange.

Loading:

Design to BS EN 1993-1-1 with UK NA:

Section Properties — 533UB

From SCI P363 (Blue Book) for UK sections:

Property Symbol Value Units
Depth h 529 mm
Flange width b 211 mm
Web thickness tw 10.2 mm
Flange thickness tf 15.6 mm
Root radius r 12.7 mm
Area A 131 cm²
Iy Iy 65,750 cm⁴
Iz Iz 2,450 cm⁴
Wpl,y Wpl,y 3,040 cm³
Wel,y Wel,y 2,486 cm³
Iw Iw 1.24 dm⁶
It It 126 cm⁴
Mass m 103 kg/m

fy = 345 MPa for flange (tf = 15.6 ≤ 16 mm — actual tf < 16, so fy = 355 MPa). Wait: tf = 15.6 mm ≤ 16 mm, so fy = 355 MPa.

Design Loads and Internal Forces

ULS Combination (UK NA Eq. 6.10b)

wEd = 1.35 × 12.0 + 1.5 × 20.0 = 16.2 + 30.0 = 46.2 kN/m

Maximum bending moment: My,Ed = wEd × L² / 8 = 46.2 × 6.0² / 8 = 207.9 kN·m

Maximum shear force: VEd = wEd × L / 2 = 46.2 × 6.0 / 2 = 138.6 kN

Cross-Section Classification (Clause 5.5)

ε = √(235/fy) = √(235/355) = 0.814

Flange (outstand in compression)

c = (b − tw − 2r) / 2 = (211 − 10.2 − 25.4) / 2 = 175.4 / 2 = 87.7 mm

c/tf = 87.7 / 15.6 = 5.62

Class 1 limit: c/tf ≤ 9ε = 9 × 0.814 = 7.33

5.62 ≤ 7.33 → Flange is Class 1

Web (internal in bending)

cw = h − 2tf − 2r = 529 − 31.2 − 25.4 = 472.4 mm

cw/tw = 472.4 / 10.2 = 46.3

Class 1 limit (pure bending): cw/tw ≤ 72ε = 72 × 0.814 = 58.6

46.3 ≤ 58.6 → Web is Class 1

Result: 533UB in S355 is Class 1 — plastic design is permitted.

Bending Moment Resistance (Clause 6.2.5)

Mc,Rd = Wpl,y × fy / γM0 = 3,040 × 10³ × 355 / 1.00 = 1,079 × 10⁶ N·mm = 1,079 kN·m

Utilisation: My,Ed / Mc,Rd = 207.9 / 1,079 = 0.193 — OK (19 % utilised)

The 533UB in S355 has substantial reserve capacity for this span and loading, as expected for a deep UB section at 6.0 m. This suggests a lighter section could be considered.

Shear Resistance (Clause 6.2.6)

Av = A − 2btf + (tw + 2r)tf = 13,100 − 2×211×15.6 + (10.2 + 25.4)×15.6

Av = 13,100 − 6,583 + 35.6×15.6 = 6,517 + 555 = 7,072 mm²

Min Av = η hw tw = 1.0 × (529 − 31.2) × 10.2 = 1.0 × 497.8 × 10.2 = 5,078 mm²

Av = 7,072 ≥ 5,078 — OK

Vpl,Rd = Av × fy / (√3 × γM0) = 7,072 × 355 / (1.732 × 1.00) = 2,510,560 / 1.732 = 1,449 kN

Utilisation: VEd / Vpl,Rd = 138.6 / 1,449 = 0.096 — OK (10 % utilised)

VEd / Vpl,Rd < 0.50 → No bending-shear interaction required per Clause 6.2.8(2).

Lateral-Torsional Buckling (Clause 6.3.2)

Condition 1 — In-service (composite slab provides continuous compression flange restraint)

No LTB check required. Mc,Rd = 1,079 kN·m governs.

Condition 2 — Construction stage (beam alone, lateral restraint at supports only, span 6.0 m)

Mcr for simply supported beam under UDL (C1 = 1.13 for UDL, kz = kw = 1.0):

Mcr = C1 × π²EIz / L² × √(Iw/Iz + L²GIt / (π²EIz))

Mcr = 1.13 × π² × 210,000 × 2,450×10⁴ / 6,000² × √(1.24×10¹²/2,450×10⁴ + 6,000²×81,000×126×10⁴/(π²×210,000×2,450×10⁴))

= 1.13 × 2.069×10⁶ × 2.450×10⁷ / 3.6×10⁷ × √(50,612 + 1.296×10¹⁴ × 8.1×10⁴ × 1.26×10⁶ / (2.069×10⁶ × 2.45×10⁷))

= 1.13 × 1.408×10⁶ × √(50,612 + 1.323×10²⁴ / 5.069×10¹³)

= 1.590×10⁶ × √(50,612 + 26,099)

= 1.590×10⁶ × √(76,711)

= 1.590×10⁶ × 277 = 440×10⁶ N·mm = 440 kN·m

λLT,bar = √(Wpl,y fy / Mcr) = √(1,079×10⁶ / 440×10⁶) = √2.452 = 1.566

For h/b = 529/211 = 2.51 > 2.0 → buckling curve c (αLT = 0.49)

ΦLT = 0.5 × [1 + 0.49 × (1.566 − 0.2) + 1.566²] = 0.5 × [1 + 0.669 + 2.452] = 0.5 × 4.121 = 2.061

χLT = 1 / [2.061 + √(2.061² − 1.566²)] = 1 / [2.061 + 1.339] = 1 / 3.400 = 0.294

Mb,Rd = χLT × Wpl,y × fy / γM1 = 0.294 × 1,079 = 317 kN·m

Construction stage moment (beam + wet concrete ≈ 12.0 kN/m): MEd,c = 1.35 × 12.0 × 6.0² / 8 = 73.0 kN·m

Utilisation: 73.0 / 317 = 0.230 — OK (23 %)

The 533UB is adequate for both in-service and construction conditions.

Serviceability Deflection Check

SLS imposed load: wSLS = qk = 20.0 kN/m

δ = 5wL⁴ / (384EIy) = 5 × 20.0 × 6,000⁴ / (384 × 210,000 × 65,750×10⁴)

= 5 × 20.0 × 1.296×10¹⁵ / (384 × 210,000 × 6.575×10¹¹)

= 1.296×10¹⁷ / (5.302×10¹⁹) = 6.1 mm

L/360 = 6,000/360 = 16.7 mm → 6.1 < 16.7 — OK (37 % utilised)

Total deflection (gk + qk = 32.0 kN/m): δtot = 6.1 × 32.0/20.0 = 9.8 mm

L/200 = 30.0 mm → 9.8 < 30.0 — OK

Summary

Check Resistance Demand Utilisation Status
Bending Mc,Rd 1,079 kN·m 207.9 kN·m 0.19 OK
Shear Vpl,Rd 1,449 kN 138.6 kN 0.10 OK
LTB construction 317 kN·m 73.0 kN·m 0.23 OK
Deflection L/360 16.7 mm 6.1 mm 0.37 OK

Conclusion: 533UB in S355 is adequate for the 6.0 m span with significant reserve capacity. A lighter section (e.g., 457UB or 406UB) could be considered for optimisation.

Design Resources

Frequently Asked Questions

What is the maximum span for a 533UB in S355?

The maximum span depends on loading, restraint conditions, and deflection limits. For a typical office floor loading (5.0 kN/m² imposed, 3.0 m tributary), a 533UB in S355 can span approximately 10-12 m before LTB governs at the construction stage or deflection exceeds L/200. At 6.0 m, the 533UB is lightly utilised at 19 % in bending. At 9.0 m, utilisation increases to approximately 40-50 % in bending and 60-70 % in construction-stage LTB, with deflection likely governing at approximately 20 mm for imposed load.

What lateral restraint is required for UK UB beams during construction?

The UK Building Regulations and BS EN 1090-2 require that steel beams during construction have adequate lateral restraint. For the 533UB with h/b = 2.51, LTB check at the construction stage shows Mb,Rd = 317 kN·m, which is 3.3× the construction moment. For beams with h/b > 2.0 at longer spans, intermediate restraint at third points or temporary propping is typically required. Permanent lateral restraint from the composite slab or secondary steelwork is the preferred solution.

How does the UK NA affect beam design compared to the recommended values?

The UK NA to BS EN 1993-1-1 adopts γM0 = 1.00 (same as recommended). The main difference from some other EU National Annexes is the use of Eq. 6.10b for ULS combinations rather than the more onerous Eq. 6.10. The UK NA also adopts UK-specific section sizes (UB/UC series per BS 4-1) rather than European IPE/HEA sections. The design methodology is otherwise identical.

What is the utilisation of a 533UB at 6.0 m span?

At 6.0 m span with typical office loading (12 kN/m dead + 20 kN/m imposed), the 533UB in S355 is lightly utilised: 19 % in bending, 10 % in shear, 23 % in LTB, and 37 % in deflection. This suggests a 457UB or 406UB could be more economical. The deep 533UB is typically used for spans of 8-12 m in UK practice.

Related Pages

Educational reference only. All design values are per BS EN 1993-1-1:2005 + UK National Annex and BS EN 10025-2:2019. Verify all values against the current editions of the standards and the applicable National Annex for your project jurisdiction. Designs must be independently verified by a Chartered Structural Engineer registered with the Institution of Structural Engineers (IStructE) or the Institution of Civil Engineers (ICE). Results are PRELIMINARY — NOT FOR CONSTRUCTION without independent professional verification.