EN 1993-1-1 Beam Design — Worked Example

This worked example demonstrates the full verification of a simply supported steel beam under uniformly distributed loading, following EN 1993-1-1:2022 with UK National Annex parameters. Educational use only — always verify by a qualified engineer.

Problem Statement

Beam: 457×191×67 UKB, Grade S355JR (EN 10025-2) Span: 6.0 m, simply supported (pinned at both ends) Loading: w_Ed = 45 kN/m (factored ULS load, including self-weight)

Section properties (from SCI P363 Blue Book):

Material: S355JR steel to EN 10025-2

Step 1 — Section Classification (Table 5.2)

Flange Classification

The flange outstand is the compression part:

c = (b - tw - 2r) / 2 = (189.0 - 8.5 - 20.4) / 2 = 80.1 mm
c / tf = 80.1 / 12.7 = 6.31

Limiting c/tf ratios per Table 5.2 (internal compression part in bending):

Web Classification

The web depth between root radii:

cw = h - 2tf - 2r = 453.6 - 25.4 - 20.4 = 407.8 mm
cw / tw = 407.8 / 8.5 = 48.0

Limiting cw/tw ratios per Table 5.2 (web in pure bending):

Result: Section is Class 1 — plastic moment resistance may be used.

Step 2 — Moment Resistance Mc,Rd (Cl. 6.2.5)

For a Class 1 or 2 cross-section:

Mc,Rd = Wpl,y × fy / γ_M0
     = 1,470 × 10³ × 355 / 1.00
     = 521.9 × 10⁶ Nmm
     = 521.9 kNm

Utilization: M_Ed / Mc,Rd = 202.5 / 521.9 = 0.39 ✓ (39%)

The cross-section moment capacity is adequate by a wide margin. However, since the beam is laterally unrestrained, LTB will govern (Step 3).

Step 3 — Lateral-Torsional Buckling Mb,Rd (Cl. 6.3.2)

3a — Elastic Critical Moment Mcr

Using the SCI SN003 expression for doubly-symmetric I-sections:

Mcr = C₁ × (π² × E × Iz / L_c²) × √(Iw/Iz + L_c² × G × It / (π² × E × Iz))

From SCI guidance, C₁ = 1.13 for a simply supported beam with UDL and end restraints preventing warping (standard case).

π² × E × Iz / L_c² = 9.87 × 210,000 × 1,430×10⁴ / 6,000²
                    = 9.87 × 210,000 × 1,430×10⁴ / 36.0×10⁶
                    = 2.96×10¹³ / 36.0×10⁶
                    = 8.23 × 10⁵ N

Iw = 0.596 × 10¹² mm⁶
Iw / Iz = 0.596 × 10¹² / 1,430×10⁴ = 4.17 × 10⁴ = 41,700 mm²

G × It = 80,770 × 38.2×10⁴ = 3.09 × 10¹⁰ N·mm²

L_c² × G × It / (π² × E × Iz) = 36.0×10⁶ × 3.09×10¹⁰ / 2.96×10¹³
                              = 1.112×10¹⁸ / 2.96×10¹³
                              = 3.76 × 10⁴ mm²

√(Iw/Iz + correction) = √(41,700 + 37,600) = √(79,300) = 282 mm

Mcr = 1.13 × 8.23×10⁵ × 282 = 2.62 × 10⁸ Nmm = 262 kNm

3b — Non-Dimensional Slenderness

λ_LT_bar = √(Wpl,y × fy / Mcr)
         = √(1,470×10³ × 355 / 262×10⁶)
         = √(521.9×10⁶ / 262×10⁶)
         = √(1.99) = 1.41

3c — Reduction Factor χ_LT

Per UK National Annex to EN 1993-1-1 (NA 6.3.2.3), for rolled I-sections:

• α_LT = 0.21 (buckling curve a)

Φ_LT = 0.5 × [1 + α_LT × (λ_LT_bar - 0.2) + λ_LT_bar²]
     = 0.5 × [1 + 0.21 × (1.41 - 0.2) + 1.41²]
     = 0.5 × [1 + 0.21 × 1.21 + 1.99]
     = 0.5 × [1 + 0.25 + 1.99]
     = 0.5 × 3.24 = 1.62

χ_LT = 1 / [Φ_LT + √(Φ_LT² - λ_LT_bar²)]
     = 1 / [1.62 + √(1.62² - 1.41²)]
     = 1 / [1.62 + √(2.62 - 1.99)]
     = 1 / [1.62 + 0.79]
     = 1 / 2.41 = 0.415

3d — Buckling Resistance Moment

Mb,Rd = χ_LT × Wpl,y × fy / γ_M1
      = 0.415 × 1,470×10³ × 355 / 1.00
      = 216.6 × 10⁶ Nmm
      = 216.6 kNm

Utilization: M_Ed / Mb,Rd = 202.5 / 216.6 = 0.93 ✓ (93%)

The beam is adequate for LTB but the 93% utilization leaves only 7% margin. Consider increasing the section to 457×191×74 or 457×191×82 UB if future load increases are anticipated.

Step 4 — Shear Resistance Vc,Rd (Cl. 6.2.6)

4a — Shear Area

For a rolled I-section loaded in the major axis (Cl. 6.2.6(3)):

Av = A - 2 × b × tf + (tw + 2r) × tf
   = 8,550 - 2 × 189 × 12.7 + (8.5 + 20.4) × 12.7
   = 8,550 - 4,801 + 367
   = 4,116 mm²

Alternatively by the simplified formula Av = 1.04 × hw × tw (conservative): hw = h - 2tf = 453.6 - 25.4 = 428.2 mm Av ≈ 1.04 × 428.2 × 8.5 = 3,785 mm²

Use Av = 4,116 mm² (Cl. 6.2.6(3) formula, less conservative).

4b — Shear Buckling Check

hw / tw = 428.2 / 8.5 = 50.4

Limiting value per Cl. 6.2.6(6): 72 × ε / η = 72 × 0.814 / 1.0 = 58.6 (η = 1.0 per UK NA)

50.4 < 58.6 → No shear buckling check required ✓

4c — Shear Resistance

Vc,Rd = Av × (fy / √3) / γ_M0
      = 4,116 × (355 / 1.732) / 1.00
      = 4,116 × 204.9
      = 843.7 × 10³ N
      = 843.7 kN

Utilization: V_Ed / Vc,Rd = 135.0 / 843.7 = 0.16 ✓ (16%)

Shear is not critical for this beam.

Step 5 — Bending and Shear Interaction (Cl. 6.2.8)

Bending and shear interaction must be checked when V_Ed > 50% of Vc,Rd:

0.5 × Vc,Rd = 0.5 × 843.7 = 421.9 kN
V_Ed = 135.0 kN < 421.9 kN

Interaction not required ✓ — the reduced moment resistance does not need to be calculated.

Step 6 — Serviceability Deflection (Informative)

For reference, the serviceability deflection under characteristic load:

Assuming characteristic load w_ser ≈ 33.0 kN/m (estimated as w_Ed / 1.35 for UDL):

δ = 5 × w × L⁴ / (384 × E × Iy)
  = 5 × 33.0 × 6,000⁴ / (384 × 210,000 × 29,400×10⁴)
  = 5 × 33.0 × 1.296×10¹⁵ / (384 × 210,000 × 2.94×10¹¹)
  = 2.14×10¹⁷ / 2.37×10¹⁹
  = 9.0 mm

Span/360 = 6,000 / 360 = 16.7 mm δ = 9.0 mm < 16.7 mm → OK for L/360

Summary

Conclusion: A 457×191×67 UKB in S355 steel is adequate for the design loads governed by LTB at 93% utilisation. The bending resistance is the critical design check.

This worked example is for educational purposes. All designs must be verified by a qualified engineer. Use the EN 1993 beam design calculator to check other sections or load cases.

See Also

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