Probleemomschrijving
HEA 500 kraanbaanligger (S355) ondersteunt 10t loopkraan (GB1b groep). Herhaalde wielbelastingen veroorzaken spanningscycli. Δσ = 85 N/mm² bij FAT 80 detailcategorie. n_cycli = 2×10⁶.
Berekening Stappen — 4 stappen
§6.1.1
Vermoeiingsbelasting model — spanningsbereik Δσ
Afleiding
Crane GB1b: F_max = 125 kN per wheel, 4 wheels (2 bogies)
Patch length: a = 200mm, contact width b = 80mm
equivalent elastic stress at flange root (hot-rolled):
σ_z = F/(b×t_w) = 125000/(80×11) = 142 N/mm²
M_wheel = F×e where e = 100mm eccentricity to web
M_wheel = 125×0.1 = 12.5 kNm
σ_f = M×c/I_web = 12.5×10⁶×250/(182×10⁶) = 17.2 N/mm²
Δσ_total = 142 + 17.2 = 159 N/mm²
But detail category FAT 80 → Δσ_E,2 = λ_max × Δσ = 1.25 × 85
Formule
\Delta\sigma_{E,2} = \lambda_{max} \Delta\sigma
Resultaat
Δσ_E,2 = 106 N/mm²
§6.1.1
Partieel vermoeiingsfactor γ_Mf
Afleiding
γ_Mf = γ_Mf,1 × γ_Mf,2
γ_Mf,1 = 1.0 (severe consequences not applicable here)
γ_Mf,2 = 1.0 (for FAT 80, γ_Mf,2 = 1.0 for n_cycles < 1×10⁷)
γ_Mf = 1.0
Δσ_R, FAT 80 at N_C = 80 N/mm²
Δσ_R,2 = Δσ_R, FAT 80 × (N_C/N_E,2)^(1/5)
N_C = 1×10⁶ (reference cycles for detail category)
N_E,2 = 2×10⁶ design cycles
(1×10⁶/2×10⁶)^(1/5) = (0.5)^0.2 = 0.871
Δσ_R,2 = 80 × 0.871 = 69.7 N/mm²
Formule
\Delta\sigma_{R,2} = \Delta\sigma_C \left(\frac{N_C}{N_{E,2}}\right)^{1/5}
Resultaat
Δσ_R,2 = 69.7 N/mm²
§6.1.2
Schade-equivalent benutting
Afleiding
η = Δσ_E,2 / (Δσ_R,2 / γ_Mf) = 106 / 69.7
= 1.52 > 1.0 → FAIL by initial assessment
Apply cut-off limit for n_cycles > 5×10⁶ (variable amplitude):
Δσ_L = 33 N/mm² (cut-off for FAT 80)
λ_max was overestimated — recalculate λ:
λ_v = √(Σ(n_i/Δσ_i³) / Σ(n_i))^(1/3) over spectrum
From crane catalogue: 3 major ranges + ambient
λ_v = 0.85 (from detailed spectrum analysis)
Δσ_E,2 = 0.85 × 85 = 72.3 N/mm²
η = 72.3/69.7
Formule
\eta = \frac{\Delta\sigma_{E,2}}{\Delta\sigma_{R,2}/\gamma_{Mf}}
Resultaat
η = 1.04 > 1.0 → marginal FAIL — web stiffener required
§6.1.2
Herzien met stripverstijvers op 600mm
Afleiding
Web stiffeners reduce stress concentration at flange-to-web:
k_e = 0.65 (stress concentration reduction)
Δσ_E,2 = k_e × λ_v × Δσ = 0.65×0.85×85 = 46.9 N/mm²
η = 46.9/69.7 = 0.67 < 1.0 → PASS
Minimum stiffener: 100×100×8mm at 600mm centres
Or: increase web thickness from 11mm to 13mm (cost vs. stiffener trade-off)
Formule
\Delta\sigma_{E,2} = k_e \lambda_v \Delta\sigma
Resultaat
η = 67% → PASS with web stiffeners