FAQPage: "@type": "FAQPage" mainEntity: - "@type": "Question" "name": "What are the governing limit states for UK steel beam design per EN 1993-1-1?" "acceptedAnswer": "@type": "Answer" "text": "The governing limit states for UK steel beam design per EN 1993-1-1 are: (1) Cross-section resistance — bending moment resistance Mc,Rd (Clause 6.2.5), shear resistance Vpl,Rd (Clause 6.2.6), and combined shear+bending interaction (Clause 6.2.8). (2) Lateral-torsional buckling Mb,Rd (Clause 6.3.2) — reduces the bending capacity when the compression flange is not fully restrained. (3) Shear buckling Vb,Rd (Clause 6.2.6(6)) — for webs with hw/tw > 72ε/η. (4) Web bearing and buckling under concentrated loads (Clause 6.2.6 and Section 6 EN 1993-1-5). (5) Serviceability — deflection limits per UK NA (L/360 for brittle finishes, L/200 otherwise). For typical UK floor beams with deck-supported compression flanges, cross-section resistance and deflection usually govern." - "@type": "Question" "name": "When does lateral-torsional buckling govern UK beam design?" "acceptedAnswer": "@type": "Answer" "text": "Lateral-torsional buckling (LTB) governs when the compression flange is not fully restrained against lateral movement. This is most common in: (1) beams supporting precast concrete planks without positive connections to the top flange, (2) beams during construction before the slab is cast, (3) crane runway beams, (4) beams with top-flange loading where the load is not stabilising. For UK composite beams with profiled steel decking shear-connected to the top flange, LTB is typically not critical during the normal condition (composite stage). However, the construction condition (bare steel, unpropped) must always be checked for LTB. The non-dimensional slenderness λ_LT is calculated per Clause 6.3.2.2 and the reduction factor χ_LT per Clause 6.3.2.3." - "@type": "Question" "name": "How do I check shear buckling in UK beam webs?" "acceptedAnswer": "@type": "Answer" "text": "Shear buckling need not be considered when hw/tw ≤ 72ε/η (EN 1993-1-1 Clause 6.2.6(6)). For S355 (ε = 0.814, η = 1.0): hw/tw ≤ 72×0.814 = 58.6. For typical UKB sections: 457x191x67 has hw/tw = 37.3 — well below the limit. Even 914x305x289 has hw/tw = 53.8 — below the limit. Therefore, shear buckling is not a concern for standard UB sections in S355. The limit may be exceeded in plate girders with slender webs (hw/tw > 58.6 for S355), where shear buckling resistance per EN 1993-1-5 Clause 5 must be calculated using the rotated stress field method or the simple post-critical method." - "@type": "Question" "name": "What partial factors do I use for UK beam design?" "acceptedAnswer": "@type": "Answer" "text": "Per the UK National Annex to EN 1993-1-1: γM0 = 1.00 for cross-section resistance (bending, shear, axial). γM1 = 1.00 for member buckling resistance (lateral-torsional buckling, flexural buckling). γM2 = 1.10 for tension fracture at net section (UK-specific, lower than base EN 1993 recommendation of 1.25). For beam design, γM0 = 1.00 applies to the bending moment resistance Mc,Rd = Wpl × fy / γM0, and γM1 = 1.00 applies to the LTB resistance Mb,Rd = χLT × Wpl,y × fy / γM1. The γM2 factor is relevant only for tension members and connection design, not for beam flexural checks."

UK Steel Beam Design — BS EN 1993-1-1 Worked Example

Complete step-by-step guide to designing a simply supported steel beam under the UK steel design framework: BS EN 1993-1-1 with the UK National Annex. This guide walks through section classification, bending resistance, shear verification, lateral-torsional buckling, web bearing, and deflection — using a 254x146x31 UKB in S355 steel as a worked example.

Quick access: UK Design Guide → | UB/UC Sections → | UK Steel Grades → | Beam Calculator →

Worked Example — 254x146x31 UKB Floor Beam

Design Brief

A simply supported floor beam in a multi-storey office building in Manchester:

Step 1: Loading and Design Actions

Characteristic Line Loads on Beam

Tributary width = 3.0 m (half-span to each adjacent beam)

• Dead: gk = 4.5 × 3.0 = 13.5 kN/m
• Imposed: qk = 3.0 × 3.0 = 9.0 kN/m
• Self-weight: 0.31 kN/m (254x146x31 UKB mass per metre × 9.81/1,000)

Total characteristic: gk = 13.5 + 0.31 = 13.81 kN/m, qk = 9.0 kN/m

Design Value per EN 1990 UK NA

Use Expression 6.10 (STR — persistent/transient design situation):

wEd = γG × gk + γQ × qk = 1.35 × 13.81 + 1.50 × 9.0 = 18.64 + 13.50 = 32.14 kN/m

Where γG = 1.35 and γQ = 1.50 per UK NA Table NA.A1.2(B) (Expression 6.10).

Design Forces

MEd = wEd × L²/8 = 32.14 × 6.0²/8 = 144.6 kN·m

VEd = wEd × L/2 = 32.14 × 6.0/2 = 96.4 kN

Step 2: Section Properties — 254x146x31 UKB

From SCI P363 (Blue Book):

Material: S355J2 → fy = 355 MPa (tf = 8.6 mm ≤ 16 mm, no thickness reduction).

Step 3: Section Classification (EN 1993-1-1 Cl 5.5)

ε = √(235/fy) = √(235/355) = 0.814

Flange (Outstand in Compression)

c = (b − tw − 2r)/2 = (146.1 − 6.1 − 2×7.6)/2 = 62.4 mm

c/tf = 62.4/8.6 = 7.26

Class 1 limit: 9ε = 9 × 0.814 = 7.33 → c/tf = 7.26 ≤ 7.33

Flange is Class 1.

Web (Bending, Neutral Axis at Mid-Depth)

cw = h − 2tf − 2r = 251.4 − 2×8.6 − 2×7.6 = 219.0 mm

cw/tw = 219.0/6.1 = 35.9

Class 1 limit: 72ε = 72 × 0.814 = 58.6 → cw/tw = 35.9 ≤ 58.6

Web is Class 1 in bending.

Overall Classification

Section is Class 1 — full plastic moment resistance can be used. Plastic design permitted.

Step 4: Cross-Section Resistance

Bending Resistance (Cl 6.2.5)

Mc,Rd = Wpl,y × fy / γM0 = 346 × 10³ × 355 / 1.00

Mc,Rd = 122.8 × 10⁶ N·mm = 122.8 kN·m

Utility ratio: MEd / Mc,Rd = 144.6 / 122.8 = 1.178 — FAILS!

The 254x146x31 UKB is inadequate in bending. Utility ratio > 1.0. We need a larger section.

Try 305x165x40 UKB

Section properties: h = 303.8 mm, b = 165.1 mm, tw = 6.1 mm, tf = 10.2 mm, Wpl,y = 623 cm³

Classification: ε = 0.814

Flange: c = (165.1 − 6.1 − 2×8.9)/2 = 70.6 mm, c/tf = 70.6/10.2 = 6.92 ≤ 7.33 → Class 1

Web: cw = 303.8 − 2×10.2 − 2×8.9 = 265.6 mm, cw/tw = 265.6/6.1 = 43.5 ≤ 58.6 → Class 1

Section is Class 1.

Mc,Rd = 623 × 10³ × 355 / 1.00 = 221.2 kN·m

Utility ratio: 144.6 / 221.2 = 0.654 — OK.

Shear Resistance (Cl 6.2.6)

Shear area (rolled I-section, load parallel to web): Av = A − 2×b×tf + (tw + 2r)×tf

Av = 5,130 − 2×165.1×10.2 + (6.1 + 2×8.9)×10.2 = 5,130 − 3,368 + 244 = 2,006 mm²

but Av ≥ η × hw × tw = 1.0 × (303.8 − 2×10.2) × 6.1 = 1,729 mm² → OK, use Av = 2,006 mm².

Vpl,Rd = Av × (fy/√3) / γM0 = 2,006 × (355/√3) / 1.00 = 411 kN

Utility: VEd / Vpl,Rd = 96.4 / 411 = 0.235 ≤ 0.50

Since VEd ≤ 0.5Vpl,Rd, no reduction in bending resistance is required for shear interaction (Cl 6.2.8(2)).

Combined Shear and Bending (Cl 6.2.8)

Not required. VEd/Vpl,Rd = 0.235 < 0.5. Bending resistance is not reduced.

Step 5: Lateral-Torsional Buckling (Cl 6.3.2)

Although the top flange is restrained by the composite slab during normal conditions, we check LTB for the construction condition (bare steel, unpropped) as required by UK practice:

During construction: dead load only (composite slab not yet effective)

wEd,construction = 1.35 × (13.5 + 0.40) ≈ 18.77 kN/m (self-weight of 305x165x40 UKB = 0.40 kN/m)

MEd,construction = 18.77 × 6.0²/8 = 84.5 kN·m

Elastic Critical Moment (NCCI SN003a-EN-EU, SCI P362)

Assume top flange is laterally restrained by decking (conservatively: no rotational restraint from deck).

For a simply supported beam with uniform moment (ψ = 1.0 where ψ is end moment ratio), the elastic critical moment is:

Mcr = C1 × π²EIz/L² × √(Iw/Iz + L²GIT/π²EIz)

C1 = 1.00 for uniform moment (conservative for simply supported beam with UDL — actual equivalent uniform moment factor is approximately 0.88-0.94 depending on restraint conditions).

Using C1 = 0.94 (UDL with top flange laterally restrained at supports):

Iz = 515 cm⁴, Iw = 0.0259 dm⁶ = 25,900 cm⁶, IT = 14.5 cm⁴

E = 210,000 N/mm², G = 81,000 N/mm², L = 6,000 mm

Term 1 = π²EIz/L² = π²×210,000×515×10⁴/6,000² = 296.4 kN

Term 2 = √(25,900×10⁶/515×10⁴ + 6,000²×81,000×14.5×10⁴/(π²×210,000×515×10⁴))

= √(5,030 + 2,060) = √7,090 = 84.2 mm

Mcr = 0.94 × 296.4 × 0.0842 = 23.5 kN·m...

Actually, let me recalculate properly:

Mcr = C1 × π²EIz/L² × √(Iw/Iz + L²GIT/(π²EIz))

Iz = 515 × 10⁻⁸ m⁴ = 5.15 × 10⁻⁶ m⁴ Iw = 0.0259 dm⁶ = 2.59 × 10⁻⁸ m⁶ IT = 14.5 × 10⁻⁸ m⁴ = 1.45 × 10⁻⁷ m⁴ L = 6.0 m E = 210 × 10⁹ Pa, G = 81 × 10⁹ Pa

Mcr = 0.94 × (π² × 210×10⁹ × 5.15×10⁻⁶ / 6.0²) × √(2.59×10⁻⁸/5.15×10⁻⁶ + 6.0²×81×10⁹×1.45×10⁻⁷/(π²×210×10⁹×5.15×10⁻⁶))

Term A = π²×210×10⁹×5.15×10⁻⁶/36 = 297 × 10³ N

Term B = √(0.00503 + 6,263/10,690) = √(0.00503 + 0.5859) = √0.5909 = 0.7687 m

Mcr = 0.94 × 297×10³ × 0.7687 = 214.6 kN·m

Non-Dimensional Slenderness (Cl 6.3.2.2)

λ̄LT = √(Wpl,y × fy / Mcr) = √(623×10³ × 355 / 214.6×10⁶) = √(1.031) = 1.015

Reduction Factor χLT (Cl 6.3.2.3 — Buckling Curve b for rolled sections with h/b > 2)

For rolled I-sections, buckling curve b applies when h/b > 2.

h/b = 303.8/165.1 = 1.84 ≤ 2 → use buckling curve a (UK NA, rolled sections h/b ≤ 2).

Curve a: αLT = 0.21, λ̄LT,0 = 0.40, β = 0.75

ΦLT = 0.5[1 + αLT(λ̄LT − λ̄LT,0) + β×λ̄LT²] = 0.5[1 + 0.21(1.015 − 0.40) + 0.75×1.015²] = 0.5[1 + 0.129 + 0.773] = 0.951

χLT = 1/(ΦLT + √(ΦLT² − β×λ̄LT²)) = 1/(0.951 + √(0.951² − 0.75×1.015²)) = 1/(0.951 + √(0.904 − 0.773)) = 1/(0.951 + √0.131) = 1/(0.951 + 0.362) = 0.762

Buckling Resistance (Cl 6.3.2.1)

Mb,Rd = χLT × Wpl,y × fy / γM1 = 0.762 × 623×10³ × 355 / 1.00 = 168.6 kN·m

Utility (construction): MEd/Mb,Rd = 84.5/168.6 = 0.501 — OK.

Since the LTB check passes for the construction condition and the top flange is fully restrained during normal service, the 305x165x40 UKB is adequate. The governing limit state is cross-section bending (utility 0.654).

Step 6: Web Bearing and Buckling (EN 1993-1-5, Section 6)

Check web under concentrated support reaction (Rsd = VEd = 96.4 kN).

Assume 100 mm stiff bearing length (ss) at support.

Web Bearing Resistance (BS EN 1993-1-5 Cl 6.2)

FRd = fyw × tw × Leff / γM1

For an end support (Type c): Leff = (ss + n × k × tf) × (1/λ̄F)

Assuming the web is not susceptible to buckling (λ̄F ≤ 0.5):

Leff = ss + 2×tf(1 + √(m1 + m2))

m1 = fyf × bf / (fyw × tw) = 355 × 165.1 / (355 × 6.1) = 27.1

m2 = 0.02 × (hw/tf)² = 0.02 × (283.4/10.2)² = 0.02 × 772 = 15.4

For λ̄F ≤ 0.5 and end support: Leff = χF × ly / γM1 → perform simplified check:

FRd = 2 × (ss + tf) × tw × fyw / γM1 (simplified approach)

FRd = 2 × (100 + 10.2) × 6.1 × 355 / 1.00 = 477 kN

Utility: 96.4 / 477 = 0.202 — OK. Web bearing and buckling are not critical.

Step 7: Serviceability — Deflection

Deflection Under Variable Actions (Imposed Load Only)

qk = 9.0 kN/m (imposed load component only)

δq = 5 × qk × L⁴ / (384 × E × Iy)

= 5 × 9.0 × 6,000⁴ / (384 × 210,000 × 8,510 × 10⁴)

= 5 × 9.0 × 1.296×10¹⁵ / (384 × 210,000 × 85,100,000)

= 5.83×10¹⁶ / 6.87×10¹⁵ = 8.5 mm

Limit per UK NA: L/360 (brittle finishes — office with suspended ceiling and raised floor = brittle finishes assumed)

δlim = 6,000/360 = 16.7 mm

Utility: 8.5/16.7 = 0.51 — OK.

Total Deflection (Dead + Imposed, Long-Term)

For a simple office beam, total deflection is checked for visual acceptability.

δtotal = 5 × (gk + qk) × L⁴ / (384 × E × Iy)

= 5 × (13.81 + 9.0) × 1.296×10¹⁵ / 6.87×10¹⁵ = 21.5 mm

δlim,total = L/200 = 30 mm

Utility: 21.5/30 = 0.72 — OK.

Step 8: Summary of Checks — 305x165x40 UKB, S355

Final design: 305x165x40 UKB in S355J2 is adequate for all limit states. The governing limit state is cross-section bending at 65% utilisation.

EN 1993-1-1 Beam Design Flowchart

1. Define Actions (EN 1991)
   ↓
2. Apply Load Combinations (EN 1990 UK NA)
   ↓
3. Determine Design Forces (MEd, VEd, NEd)
   ↓
4. Select Trial Section (SCI Blue Book or span/depth ≈ 20-24)
   ↓
5. Classify Cross-Section (Cl 5.5, Table 5.2)
   ↓
6. Check Cross-Section Resistance (Cl 6.2)
   ├── Bending Mc,Rd (Cl 6.2.5)
   ├── Shear Vpl,Rd (Cl 6.2.6)
   ├── Shear + Bending Interaction (Cl 6.2.8)
   └── Axial (if applicable)
   ↓
7. Check Member Buckling (Cl 6.3)
   ├── LTB Mb,Rd (Cl 6.3.2)
   └── Flexural Buckling (Cl 6.3.1, if axial)
   ↓
8. Check Local Effects
   ├── Web Bearing/Buckling (EN 1993-1-5 §6)
   └── Flange Induced Buckling (EN 1993-1-5 §8)
   ↓
9. Check Serviceability
   ├── Deflection (UK NA limits)
   └── Vibration (if applicable)
   ↓
10. Finalise Section

Common UK Beam Pitfalls

Code References

• BS EN 1990:2002+A1:2005 — Basis of Structural Design (+ UK NA Table NA.A1.2(B))
• BS EN 1993-1-1:2005+A1:2014 — Design of Steel Structures: General Rules (Cl 5.5, 6.2, 6.3)
• BS EN 1993-1-5:2006+A2:2019 — Plated Structural Elements (§6 Web Bearing/Buckling)
• SCI P363 — Steel Building Design: Design Data (Blue Book) — section properties
• SCI P362 — Steel Building Design: Concise Eurocodes — worked examples and NCCI guidance
• NCCI SN003a-EN-EU — Elastic Critical Moment for LTB (Mcr formula and C factors)

FAQ

Why did the 254x146x31 UKB fail in bending?

The 254x146x31 UKB has plastic modulus Wpl,y = 346 cm³, giving Mc,Rd = 122.8 kN·m against a design moment of MEd = 144.6 kN·m (utility 1.18). The section is simply too small for a 6 m span with 3 m tributary width. The design was intentionally shown failing to demonstrate the iteration process. Real UK office beams in this span range would typically use 305x165x40 UKB (utility 0.65) or 356x171x51 UKB (utility ~0.44) depending on architectural depth constraints. A quick preliminary check using span/20 ≈ 300 mm depth suggests the 305 mm range is appropriate.

When do I need to consider the construction condition for LTB?

Always. UK structural engineers are required to check the bare-steel condition before the composite slab gains strength. During construction, the steel beam alone carries the wet concrete weight plus construction live load (typically 0.75-1.5 kN/m² per BS EN 1991-1-6). The compression flange may not be restrained until the deck is positively connected (shear studs welded through-deck). If LTB governs the construction condition, options include: (1) providing temporary lateral bracing at mid-span or third-points, (2) increasing the section size, (3) propping the beam during concreting, (4) sequencing the concrete pour to reduce the unsupported length in stages.

How do I handle beam-to-beam connections (secondary to primary)?

For simple connections (UK standard practice), secondary beams are designed as simply supported. The end reaction from the secondary beam is applied to the primary beam as a point load. The SCI P358 'Green Book' provides pre-designed partial-depth and full-depth end plate connections with tabulated capacities for standard UB sections. For the worked example, a 305x165x40 UKB with 96.4 kN reaction could use: (1) 200x10 partial-depth end plate with M20 Grade 8.8 bolts (SCI P358, connection capacity ~150 kN), or (2) full-depth end plate if rotational capacity is a concern. The primary beam must be checked for web bearing and buckling under the concentrated load from the secondary beam — this often governs primary beam design in UK frame construction.

What deflection limit applies to UK office beams?

Per the UK National Annex to EN 1993-1-1, the limit is L/360 for beams supporting brittle finishes and L/200 for other beams. For offices: (1) If the floor has a raised access floor and suspended ceiling (common in UK commercial offices), both are considered brittle finishes → use L/360 for imposed load deflection. (2) If the floor is a simple slab with no brittle finishes, L/200 may apply. (3) Many UK clients and architects specify L/360 regardless for all occupied floors, driven by tenant expectations of floor stiffness. (4) For total deflection (dead + imposed), L/200 is a common absolute limit to prevent visual sag. Dynamic performance (natural frequency > 3 Hz for offices per SCI P354) is a separate serviceability check for long-span and lightweight floors.