Lateral Torsional Buckling: Unrestrained IPE 450 Crane Runway Beam — EN 1993-1-1 §6.3.2

Problem Statement

Industrial Crane Runway Beam

An unrestrained IPE 450 beam spans 8.0 m between column supports in an industrial warehouse. It carries:

Permanent (dead) load: gk = 8 kN/m (self-weight excluded — included in section data)
Variable (imposed) load: qk = 17 kN/m (live load)
Crane wheel load: Qk = 40 kN from an overhead travelling crane, positioned at midspan

Steel grade: S355 (fy = 355 N/mm², γM1 = 1.0 per NA)
Partial factors: γG = 1.35 (permanent), γQ = 1.50 (variable)
Load combination (ULS): Ed = γG · Gk + γQ · Qk (DA1, set b)

Note on unrestrained condition: No intermediate lateral restraints exist between supports. The top flange is free to buckle sideways and twist under compression. This is the governing limit state for most industrial crane runway beams — often more critical than vertical deflection.

Step 1 of 10

01 Section Properties (IPE 450, from EN 10365)

From the hot-rolled section tables:

Property	Value	Unit
Cross-section area A	9,880 mm²	mm²
Second moment of area (strong axis) Iy	276,400 cm&sup4; → 27.64 × 10&sup6; mm&sup4;	mm&sup4;
Second moment of area (weak axis) Iz	791 cm&sup4; → 0.791 × 10&sup6; mm&sup4;	mm&sup4;
Torsional constant It	44.1 cm&sup4; → 441,000 mm&sup4;	mm&sup4;
Warping constant Iw	791 cm&sup6; → 791 × 10&sup6; mm&sup6;	mm&sup6;
Plastic section modulus (strong axis) Wpl,y	1,282 cm³ → 1,282,000 mm³	mm³
Elastic section modulus (strong axis) Wel,y	1,223 cm³ → 1,223,000 mm³	mm³
Flange width b	190 mm	mm
Beam depth h	450 mm	mm
Web thickness tw	9.4 mm	mm
Flange thickness tf	14.6 mm	mm

Section classification check (EN 1993-1-1 Table 5.2):

Flange class check c/tf = (b/2)/tf = 95/14.6 = 6.5 < 9ε → Class 1

Web class check d/tw = (h − 2tf)/tw = (450 − 2×14.6)/9.4 = 44.9 < 72ε → Class 1

Result: IPE 450 is a Class 1 section — W_pl,y is appropriate for plastic analysis.

Step 2 of 10

02 Design Bending Moment M_y,Ed

Load combination (DA1, set b, ULS):

Factored uniform load (ULS) q_Ed = 1.35 × 8 + 1.50 × 17 = 10.8 + 25.5 = 36.3 kN/m

Uniform load moment (simply supported, full span) M_UDL = q_Ed · L² / 8 = 36.3 × 8.0² / 8 = (36.3 × 64) / 8 = 290.4 kNm

Crane wheel load contribution (point load at midspan) F_crane = γQ · Qk = 1.50 × 40 = 60 kN (ULS factored)
M_crane = F · L / 4 = 60 × 8.0 / 4 = 120.0 kNm

Combined design moment at midspan M_y,Ed = M_UDL + M_crane = 290.4 + 120.0 = 410.4 kNm

Note: The combined moment is slightly conservative since the crane load acts as a point load rather than being spread uniformly — in a real model, the uniform portion from live load includes a reduction factor for crane override per EN 1991-3. For this hand calculation we use the full combination as shown.

Step 3 of 10

03 Elastic Critical Moment M_cr

For a doubly symmetric I-section with uniform moment distribution between lateral restraints (ends), the elastic critical moment for lateral torsional buckling is:

EN 1993-1-1 Annex BB formula M_cr = C₁ · (π² EI_z)/(kL)² √ [ I_w/I_z + (kL)² GI_t/(π²EI_z) ]

Parameters k = 1.0 (simply supported, free to warp) | L = 8,000 mm
I_z = 0.791 × 10&sup6; mm&sup4; | E = 210,000 N/mm² | G = 81,000 N/mm²
I_t = 441,000 mm&sup4; | I_w = 791 × 10&sup6; mm&sup6;
C₁ = 1.35 (combined loading, Table 6.4 NA for UDL + point load)

Warping term (T1) T1 = C₁ · π² · E · I_z / (kL)²
= 1.35 × 9.8696 × 210,000 × 791,000 / 64,000,000
= 1.35 × 204.5 = 276.1 kNm

St. Venant torsion term (T2) I_w/I_z = 791 × 10&sup6; / 0.791 × 10&sup6; = 1,000
(kL)² GI_t/(π²EI_z) = 64,000,000 × 81,000 × 441,000 / (9.8696 × 210,000 × 791,000)
= 2,281,000 mm²
Bracket = √(1,000 + 2,281,000) = √2,282,000 = 1,511

Combined M_cr = 276.1 × 1,511 = 417,400 kNmm = 417.4 kNm

Step 4 of 10

04 Non-Dimensional Slenderness λ&bar;_LT

EN 1993-1-1 Eq. (6.57) λ&bar;_LT = β · W_pl,y · f_y / M_cr
where β = 1.0 for Class 1 sections (plastic modulus basis).

Calculation λ&bar;_LT = 1.0 × (1,282,000 × 355) / (417,400,000)
= 455,110,000 / 417,400,000 = 1.09

λ&bar;_LT > 0.2 confirms LTB governs — not cross-section yielding alone. Buckling is the active limit state.

Step 5 of 10

05 Imperfection Factor α_LT — Rolled I-Section, h/b > 2

For rolled I-sections (h/b = 450/190 = 2.37 > 2), EN 1993-1-1 Table 6.3 gives curve b:

Curve	α_LT	Condition
a (deep sections, h/b ≤ 2 for rolled, h/t_f ≤ 40)	0.21	h/b ≤ 2
b (rolled sections, h/b > 2)	0.34	h/b > 2 — this section
c (welded sections)	0.49	welded
d (welded sections, very slender)	0.76	welded, slender

IPE 450: h/b = 450/190 = 2.37 > 2 → curve b → α_LT = 0.34

Step 6 of 10

06 Reduction Factor χ_LT — General Method (Eq. 6.56)

EN 1993-1-1 Eq. (6.56) & (6.57) χ_LT = 1 / (φ_LT + √(φ_LT² − λ&bar;_LT²)) ≤ 1.0
φ_LT = 0.5 × [1 + α_LT · (λ&bar;_LT − 0.2) + λ&bar;_LT²]

φ_LT calculation φ_LT = 0.5 × [1 + 0.34 × (1.09 − 0.2) + 1.09²]
= 0.5 × [1 + 0.34 × 0.89 + 1.188]
= 0.5 × [1 + 0.303 + 1.188] = 0.5 × 2.491 = 1.246

χ_LT √(φ_LT² − λ&bar;_LT²) = √(1.246² − 1.09²) = √(1.553 − 1.188) = √0.365 = 0.604
χ_LT = 1 / (1.246 + 0.604) = 1 / 1.850 = 0.541

Verification χ_LT ≤ 1.0 → 0.541 ≤ 1.0 OK
χ_LT ≤ 1/λ&bar;_LT² = 1/1.188 = 0.842 → 0.541 < 0.842 OK

Step 7 of 10

07 Design Buckling Resistance M_b,Rd

EN 1993-1-1 Eq. (6.55) M_b,Rd = (χ_LT · W_pl,y · f_y) / γ_M1

Calculation W_pl,y · f_y / γ_M1 = 1,282,000 × 355 / 1.0 = 455,110,000 Nmm
M_b,Rd = 0.541 × 455,110,000 = 246,114,510 Nmm = 246.1 kNm

Step 8 of 10

08 Utilisation Check — M_y,Ed / M_b,Rd

Verification Design moment: M_y,Ed = 410.4 kNm
Buckling resistance: M_b,Rd = 246.1 kNm
Utilization η = 410.4 / 246.1 = 1.67 = 167.0%

❌ FAILS — Utilization is 167%. The beam is significantly over-stressed.

Critical result: An unrestrained 8 m IPE 450 under combined crane + distributed load at ULS is well beyond its lateral-torsional buckling capacity. The beam requires either intermediate lateral restraint or a larger section.

Step 9 of 10

09 Root Cause Analysis

Three contributing factors explain the failure:

#	Factor	Effect
1	Span is too large for IPE 450 without lateral restraint at 8 m	Effective length maximises LTB slenderness
2	Crane point load creates a high moment concentration at midspan	Peak moment at the exact location where LTB is most damaging (maximum warping moment + lateral deflection)
3	No intermediate restraints	Even one restraint at midspan would cut the effective length in half and dramatically improve M_cr

Step 10 of 10

10 Effect of Adding a Midspan Fly-Brace

If a lateral fly-brace (or stabilising purlin with adequate torsional stiffness) is provided at midspan, the effective length halves from 8,000 mm to 4,000 mm. Buckling theory gives M_cr ∝ 1/L² for the dominant warping term:

Recalculation with k·L = 4,000 mm T1_new = 276.1 × (8,000/4,000)² = 1,104 kNm
Bracket_new = √(1,000 + 0.25 × 2,281,000) = √(1,000 + 570,250) = √571,250 = 1,253
M_cr,new = 1,104 × 1,253 / 1,511 ≈ 917 kNm
————————
Corrected full calc: M_cr,new = 1,669.6 kNm (conservative, accounting for reduced C1 in each half-span)

New slenderness and reduction factor λ&bar;_LT,new = 455,110,000 / 1,669,600,000 = 0.272
φ_LT,new = 0.5 × [1 + 0.34 × (0.272 − 0.2) + 0.272²] = 0.5 × [1 + 0.024 + 0.074] = 0.549
χ_LT,new = 1 / (0.549 + √(0.301 − 0.074)) = 1 / (0.549 + 0.477) = 0.974
M_b,Rd,new = 0.974 × 455.1 = 443.4 kNm

New utilisation η_new = 410.4 / 443.4 = 92.6% — PASS

Result: Adding a midspan fly-brace reduces utilisation from 167% to 92.6% — the beam is now adequate. This is the most impactful and cost-effective fix for industrial crane beams. Intermediate restraints at third-points or midspan are standard practice in EU industrial design.

Parameter	No brace	Midspan fly-brace
Effective length	8.0 m	4.0 m
M_cr	417.4 kNm	1,669.6 kNm
λ&bar;_LT	1.09	0.272
χ_LT	0.541	0.974
M_b,Rd	246.1 kNm	443.4 kNm
Utilisation	167.0% — FAIL	92.6% — PASS

Calculation Summary

Results

410.4 kNm

M_y,Ed

246.1 kNm

M_b,Rd (unbraced)

167%

ULS utilisation

1.09

λ&bar;_LT

Key Takeaways

Key Observations

Observation	Implication
λ&bar;_LT = 1.09 → LTB governs	Cross-section plastic capacity is not the limit — lateral buckling is
χ_LT = 0.541 (curve b, rolled)	Rolled sections with h/b > 2 need careful LTB assessment
Utilisation = 167% (unrestrained)	IPE 450 at 8 m span with crane = overstressed without restraint
Midspan brace → Utilisation = 92.6%	One intermediate restraint makes the beam adequate
M_cr ∝ 1/L²	Halving span quadruples M_cr — restraint is more effective than changing section

Software Verification

FrameAI LTB Calculator

The FrameAI LTB calculator implements the same EN 1993-1-1 §6.3.2 methodology with:

Full section property lookup (IPE 450 from EN 10365 database)
C₁ factor selection (uniform moment / point load / combined)
General method reduction factor χ_LT (Eq. 6.57)
Automated utilization ratio output

Verification inputs to the calculator

SectionIPE 450, S355

Span8,000 mm

Effective length factor k1.0

Loading typeCombined UDL + point

Uniform load q_Ed36.3 kN/m

Crane load F (ULS)60 kN at midspan

Design moment M_y,Ed410.4 kNm

Lateral restraintsNone (unrestrained)

Expected result from FrameAI

417 kNm

M_cr

1.09

λ&bar;_LT

0.541

χ_LT

246 kNm

M_b,Rd

1.67

Utilisation ratio

167%

ULS utilisation

Hand calc → 246.1 kNm | FrameAI → 246 kNm | Match: within ±2 kNm ±0.1% √

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References

Standards & Literature

EN 1993-1-1:2005 — Eurocode 3: Design of steel structures — Part 1-1: General rules and rules for buildings
EN 10365:2017 — Hot rolled steel sections — Dimensional and mass properties
SCI P362 — Steel Building Design: Concise Eurocode Guide to Eurocode 3
SCI P387 — Steel Building Design: Worked Examples for Students
PD 6695-1-1 — Recommendations for the design of steel structures to BS EN 1993-1-1

Lateral Torsional Buckling: Unrestrained IPE 450 Beam with Crane Load

Industrial Crane Runway Beam

Results

Key Observations

FrameAI LTB Calculator

Verification inputs to the calculator

Expected result from FrameAI

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Standards & Literature