Lateral-Torsional Buckling of Steel Beams

When a laterally unrestrained beam is subjected to bending about the major axis, there is a need to check for lateral-torsional buckling. Lateral-torsional buckling is a type of buckling that involves a combination of lateral deflection of beams and twisting, and typically occurs in open cross-sections.

The phenomenon occurs on the compression flange of the member and depends on factors such as the loading conditions, lateral restraint conditions, and geometry of the compression flange. Steel beams with sufficient lateral restraint to the compression flange may not need to be checked for lateral-torsional buckling. Cross-sections such as circular hollow sections or square box sections are also not susceptible to lateral-torsional buckling.

Lateral restraint to a steel beam in a building may be provided by;

• Concrete floor slab on beams (inclusive of composite metal deck)
• Sheeting or metal decking on roofs (spanning perpendicular to the beam)
• Secondary beams to primary beams
• Purlins on rafters
• Bracings, etc

In general, the bracing system assumed to provide effective lateral restraint must be capable of resisting an equivalent stabilising force q_d (defined in clause 5.3.3(2) of EC3), the value of which depends on the flexibility of the bracing system.

Design for Lateral-Torsional Buckling

The design bending moment is denoted by M_Ed (bending moment design effect), and the lateral-torsional buckling resistance by M_b,Rd (design buckling resistance moment). The design requirement is that M_Ed must be shown to be less than M_b,Rd, and checks should be carried out on all unrestrained segments of beams (between the points where lateral restraint exists).

The design buckling resistance of a laterally unrestrained beam (or segment of beam) should be taken as;

M_b,Rd = χ_LTW_yf_y/γ_m0

where W_y is the section modulus appropriate for the classification of the cross-section, as given below. In determining W_y, no account need to be taken for fastener holes at the beam ends.

W_y = W_pl,y for Class 1 or 2 cross-sections
W_y = W_el,y for Class 3 cross-sections
W_y = W_eff,y for Class 4 cross-sections
χ_LT is the reduction factor for lateral torsional buckling.

Solved Example

A simply supported primary beam is required to span 7m and to support two secondary beams as shown in the figure below. The secondary beams are connected through fin plates to the web of the primary beam, and full lateral restraint may be assumed at these points. Check the suitability of UKB UB 533 x 210 x 92 for the primary beam assuming grade S275 steel.

Let ∑M_B = 0;
7V_A – (350 × 5.7) – (375 × 1.3) = 0
V_A = 354.64 kN

Let ∑M_A = 0;
7V_B – (350 × 1.3) – (375 × 5.7) = 0
V_B = 370.36 kN

M_B = 354.64 × 1.3 = 461.032 kNm
M_C = (354.64 × 5.7) – (350 × 4.4) = 481.619 kNm

UB 533 x 210 x 92

E = 210000 N/mm²
G = 81000 N/mm²

Properties of UB 533 x 210 x 92
h = 533.1 mm
b = 209.3 mm
t_w = 10.1 mm
t_f = 15.6 mm
r = 12.7 mm
A = 11700 mm²
I_y = 55200 cm⁴
I_z = 2390 cm⁴
I_T = 7.57 x 10⁶ mm⁴
I_W = 1.6 x 10¹² mm⁶
W_el,y = 2070 cm³
W_el,z = 228 cm³
W_pl,y = 2360 cm³
W_pl,z = 356 cm³

Section Classification

ε = √235/f_y = √235/275 = 0.92

Web
c_w = d = h – 2t_f – 2r = 476.5 mm
c_w/t_w = 47.18
The limit for class 1 is 72ε = 66.24
c_w/t_w = 47.18 < 66.24
Therefore the web is class 1 Plastic

Flange
c = [(b – t_w – 2r)]/2 = [209.3 – 10.1 – (2 × 12.7)]/2 = 86.9 mm
c_f/t_f = 5.57
The limit for class 1 is 9ε = 9 × 0.92 = 8.28
5.57 < 8.28
Therefore the flange is Class 1 (plastic)
Therefore the beam section is class 1

Bending Resistance (Clause 6.2.5 BS EN 1993-1-1)
M_pl,y,Rd = (W_plf_y)/γ_m0 = (2360 × 10³ × 275)/1.0 × 10^-6 = 649 kNm

Maximum moment on the beam M_y,Ed = 481.619 kNm
481.619 < 649 kNm Ok

Shear Resistance (Clause 6.2.6 EN 1993-1-1)
Shear area A_v = A – 2bt_f + (t_w + 2r)t_f but not less than ηh_wt_w

A_v = 11700 – (2 × 209.3 × 15.6) + (10.1 + 2 × 12.7) × 15.6 = 5723.64 mm²
ηh_wt_w = 1.0 × 501.9 × 10.1 = 5069.19 mm²

Therefore take A_v = 5723.64 mm²
V_pl,Rd = [Av(f_y⁄√3)]/γ_m0 = [5723.64 (275⁄√3)]/1.0 × 10^-3 = 908.749 kN
V_Ed = 370.36 kN < 908.749 kN Ok

Bending and Shear Interaction (clause 6.2.8 BS EN 1993-1-1)
When shear force and bending moment act simultaneously on a cross-section, the effect of the shear force can be ignored if it is smaller than 50% of the plastic shear resistance.

0.5V_pl,Rd = 0.5 × 908.749 = 454.374 kN
370.36 kN < 454.374 kN, therefore the effect of shear on the moment resistance can be ignored.

Lateral torsional buckling (segment B – C)
L_cr,T = 4.4 m
h/b = 533.1/209.3 = 2.54 > 2.0

Therefore select buckling curve : c = 0.49 (Table 6.5 EC3)

Moment diagram of the point between restraints

Ratio of end moments ψ = 461.032/481.619 = 0.957

C₁ = 1.88 – 1.40ψ + 0.52ψ² = 1.01 < 2.7 Okay

M_cr = C₁ × (π²EI_z)/(kL² ) × [I_w/I_z + (kL²GI_T)/(π²EI_z )]^0.5

M_cr = 1.01 x [(π² × 210000 × 2390 × 10⁴)/4400²] × [(1.6 × 10¹²)/(2390 × 10⁴) + (4400² × 81000 × 75.7 × 10⁴)/(π² × 210000 × 2390 × 10⁴)]^0.5 x 10^-6 = 779.182 kNm

Non-dimensional lateral torsional slenderness λ_LT

λ_LT = √[(W_pl,yf_y)/M_cr ] = √[(2360000 × 275)/(779.182 × 10⁶] = 0.912

λ_LT,0 = 0.4, and β = 0.75
ϕ_LT = 0.5[1+ α_LT (λ_LT – λ_LT,0) + βλ_LT²]
ϕ_LT = 0.5[1 + 0.49(0.912 – 0.4) + 0.75 × 0.912²] = 0.937

χ_LT = 1/[ϕ_LT + √(ϕ_LT² – βλ_LT²)] but χ ≤ 1.0

χ_LT = 1/([0.937 + √(0.937² – 0.75 × 0.912²)] = 0.6938

M_b,Rd = χ_LTW_yf_y/γ_m0 = (0.6938 × 2360 × 10³ × 275)/1.0 × 10^-6 = 450.33 kNm

M_Ed/M_b,Rd = 481.619/450.33 = 1.06 > 1.0

Therefore the section is not okay to resist lateral torsional buckling on the primary beam.

References
Gardner L. (2011): Stability of Steel Beams and Columns (In Accordance with the Eurocodes and UK National Annex). SCI – Steel Construction Institute, Berkshire UK.