Free Tool · EN 1993-1-5 §6

Web Bearing & Buckling

Check the resistance of a steel web to transverse (concentrated) forces per EN 1993-1-5 §6. Covers load types a (between stiffeners), b (one flange, near end stiff bearing), and c (one flange, near unstiffened end). Computes kF, Fcr, ℓy, λ̄F, χF, FRd and the optional interaction check with concurrent bending moment (§7.2). IPE, HEA, HEB, UB. S235–S460.

h_w s_s F_Ed L_eff stiffener a (panel width)
Section Parameters
Length of stiff bearing: width of load distribution over flange. For a wheel: 0. For a stiffener: equal to stiffener width. Min 0.
Distance between transverse stiffeners. Used for k_F. For beams without intermediate stiffeners, use span length.
For load types b/c: distance from load centroid to the nearest end of the member. Leave 0 for type a.
Concurrent bending moment for §7.2 interaction check. Leave 0 to skip interaction.
EN: γM0 = γM1 = 1.0 (recommended values)
Web Bearing Results
23.9%
F_Ed / F_Rd
PASS
F_Rd (kN)626.4
F_Ed (kN)150
k_F6.031
F_cr (kN)1943.7
m₁20.93
m₂15.268
ℓ_y (mm)264.4
λ̄_F0.644
χ_F0.776
L_eff (mm)205.2
h_w (mm)373
t_w (mm)8.6
f_y (N/mm²)355
Interaction check (§7.2): η₁ + 0.8·η₂ ≤ 1.4
η₁ = M_Ed/M_Rd
η₂ = F_Ed/F_Rd
η₂ + 0.8·η₁
M_Rd (kNm)
Result
▶ Show step-by-step derivation
k_F = 6.031 (load type a) F_cr = 0.9 · kF · E · tw³ / hw = 1943.7 kN m₁ = bf/tw = 20.93 m₂ = 0.02·(hw/tf)² = 15.268 ℓ_y = ss + 2tf·(1+√(m₁+m₂)) = 264.4 mm λ̄_F = √(ℓy·tw·fy/Fcr) = 0.644 χ_F = 0.5/λ̄_F (≤1.0) = 0.776 L_eff = χ_F · ℓ_y = 205.2 mm F_Rd = fy · L_eff · tw / γM1 = 626.4 kN
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FAQ

What is web bearing and buckling under transverse forces?

When a concentrated load or reaction force is applied through the flange, it is distributed into the web over an effective length L_eff. The web must resist this without crippling (yield at the load point — bearing) or buckling (elastic instability of the web under the concentrated load). EN 1993-1-5 §6 gives a unified formula F_Rd = f_yw · L_eff · t_w / γ_M1 that covers both failure modes via the slenderness factor χ_F.

What are load types a, b, and c?

Type a (Fig. 6.1a): the load is applied between two transverse stiffeners (or end plates), supported through both flanges — most common in mid-span or interior panels. Type b (Fig. 6.1b): the load is on one flange near a stiffener or end with a stiff bearing — e.g. a crane wheel near an end plate. Type c (Fig. 6.1c): the load is on one flange at or near an unstiffened end — the most onerous case, giving k_F = 2.0.

How is the effective loaded length ℓ_y computed?

ℓ_y is the length of the web over which the transverse force is assumed to spread. Per Eq. 6.10–6.13: ℓ_y = s_s + 2·t_f·(1 + √(m₁ + m₂)) when m₂ > 0 (slender web), or ℓ_y = s_s + 2·t_f·(1 + √m₁) when m₂ = 0 (stocky). Here m₁ = f_yf·b_f / (f_yw·t_w) accounts for flange-to-web yield strength spread, and m₂ = 0.02·(h_w/t_f)² penalises slender webs.

What is the buckling coefficient k_F?

k_F is derived from Figure 6.1 and depends on load type and the panel aspect ratio h_w/a. For Type a: k_F = 6 + 2·(h_w/a)². For Type b: k_F = 3.5 + 2·(h_w/a)². For Type c: k_F = 2.0 (conservative constant for unstiffened ends). Higher k_F means more resistance — a stocky panel between close stiffeners benefits substantially.

What is the §7.2 interaction check?

When a beam is simultaneously loaded by a transverse force F_Ed and a bending moment M_Ed (and/or axial N_Ed), EN 1993-1-5 §7.2 requires: η₂ + 0.8·η₁ ≤ 1.4, where η₂ = F_Ed/F_Rd is the patch loading utilisation and η₁ = M_Ed/M_Rd is the bending utilisation (M_Rd = W_pl·f_y/γ_M0 for a Class 1/2 section). This interaction is most critical for crane runway girders and plate girders under wheel loads.

Is γ_M1 = 1.0 for all national annexes?

Yes for the EN, NL (NEN-EN 1993-1-5/NB), DE (DIN EN 1993-1-5/NA), and BE annexes. The UK NA also retained 1.0. Use the National Annex selector to apply the correct value if your project differs.