Aufgabenstellung
GL 24h Brettschichtholzträger, Querschnitt 100×320mm, Stützweite 5.2m. Last: g_k=2.0 kN/m, q_k=3.0 kN/m. Nutzungsklasse 1. Randträger → k_h=1.10.
Berechnungsschritte — 3 Schritte
§6.1.6
Biegespannung σ_m,y,d
Herleitung
f_m,k = 24 N/mm², γ_M = 1.3, k_mod = 0.9 (permanent+medium term)
f_m,d = k_mod × f_m,k / γ_M = 0.9×24/1.3 = 16.62 N/mm²
M_Ed = (1.35×2.0 + 1.5×3.0) × 5.2² / 8 = (2.7+4.5)×26.9/8
= 7.2 × 3.36 = 24.2 kNm
W_y = b×h²/6 = 100×320²/6 = 1710×10³ mm³
σ_m,y,d = M_Ed/W_y = 24.2×10⁶/1710×10³ = 14.15 N/mm²
Formel
\sigma_{m,y,d} = \frac{6 M_{Ed}}{b h^2}
Ergebnis
σ_m,y,d = 14.15 N/mm²
§6.1.6
k_h Modifikation für Randträger
Herleitung
h = 320 mm > 300 mm
k_h = min((600/h)^0.1, 1.1) = min((600/320)^0.1, 1.1)
= min((1.875)^0.1, 1.1) = min(1.068, 1.1)
= 1.068
Edge beam: also consider k_m,alpha = 1.0 for simply supported
No significant lateral torsional buckling: k_crit = 1.0
f_m,d,adjusted = k_h × f_m,d = 1.068 × 16.62 = 17.75 N/mm²
Formel
k_h = \min\left(\left(\frac{600}{h}\right)^{0.1}, 1.1\right)
Ergebnis
k_h = 1.068
§6.1.6
Ausnutzungsgrad
Herleitung
σ_m,y,d = 14.15 N/mm²
f_m,d,adjusted = 17.75 N/mm²
Ratio = 14.15/17.75 = 0.797
Also check deflection (SLS):
δ_inst = 5×q×L⁴/(384×E_0,g·I) = (5×7.2×5.2⁴)/(384×10800×1093×10⁴)
= 5×7.2×731/(415×1179×10⁴)
= 26316/4901×10⁴ mm
= 5.37 mm
L/250 = 20.8 mm → δ = 5.37mm < 20.8mm
Formel
U_{m} = \frac{\sigma_{m,y,d}}{k_h f_{m,k}/\gamma_M}
Ergebnis
77.2% bending, 26% SLS deflection → PASS