Aufgabenstellung
HEA 300 (S355) Stütze, Länge 5.0m, gelenkig gelagert. Druck N_Ed = 850 kN, Biegemoment M_Ed = 120 kNm. Knickkurve a (warmgewalzt).
Berechnungsschritte — 4 Schritte
§6.2.4.1
Querschnittsklassifizierung
Herleitung
HEA 300: b=300mm, h=300mm, t_f=19mm, t_w=11mm
ε = √(235/f_y) = √(235/355) = 0.814
Flange: c/b = (300-11)/2/300 = 0.482
Class 1 limit = 9ε = 7.33 → 0.482 < 7.33 → Class 1
Web: c/t_w = (300-2×19)/11 = 23.8
Class 1 limit = 72ε = 58.6 → 23.8 < 58.6 → Class 1
Section is Class 1 plastic → use W_pl
Formel
\varepsilon = \sqrt{235/f_y}
Ergebnis
Class 1 plastic
§6.3.1.1
Schlankheit λ̄
Herleitung
A = 12550 mm², I_y = 182.6×10⁶ mm⁴
L_cr = L = 5000 mm (pinned-pinned)
i_y = √(I_y/A) = √(182.6×10⁶/12550) = 120.7 mm
λ_y = L_cr/i_y = 5000/120.7 = 41.4
λ_1 = π√(E/f_y) = π√(210000/355) = 76.4
λ_bar = λ_y/λ_1 = 41.4/76.4 = 0.542
Formel
\bar\lambda = \frac{L_{cr}}{i \cdot \lambda_1}
Ergebnis
λ̄ = 0.542
§6.3.1.2
Knickreduktionsfaktor χ
Herleitung
α = 0.21 (buckling curve a, hot-rolled I/H sections)
Φ = 0.5(1 + α(λ̄ - 0.2) + λ̄²)
= 0.5(1 + 0.21(0.542-0.2) + 0.542²)
= 0.5(1 + 0.072 + 0.294)
= 0.683
χ = 1/(Φ + √(Φ² - λ̄²)) = 1/(0.683 + √(0.467-0.294))
= 1/(0.683 + 0.416)
= 0.909
Formel
\chi = \frac{1}{\Phi + \sqrt{\Phi^2 - \bar\lambda^2}}
Ergebnis
χ = 0.909
§6.3.3
Kombinierte Interaktion
Herleitung
N_b,Rd = χ·A·f_y/γ_M1 = 0.909×12550×355/1.0/1000
= 4054 kN
M_b,Rd = χ_lt·W_pl·f_y/γ_M1
χ_lt = 1.0 (no LTB for pinned column)
W_pl,y = 951×10³ mm³
M_b,Rd = 1.0×951×10³×355/1.0/10⁶ = 337.6 kNm
Interaction formula (Annex A):
N_Ed/N_b,Rd + M_Ed/M_b,Rd = 850/4054 + 120/337.6
= 0.21 + 0.356
Formel
\frac{N_{Ed}}{\chi_y N_{Rk}/\gamma_{M1}} + \frac{M_{y,Ed}}{M_{y,Rk}/\gamma_{M0}} \leq 1
Ergebnis
56.6% utilisation → PASS