Aufgabenstellung
IPE 400 (S355) Einfeldträger, Stützweite L = 8.5m. Verbundplatte: ComFlor 60 130mm Aufbeton C30/37. Wirksame Plattenbreite b_eff = 1.2m. Kopfbolzen 16mm im Abstand 150mm.
Berechnungsschritte — 4 Schritte
§6.2.1.2
Wirksame Plattbreite b_eff
Herleitung
L_0 = 0.7 × L = 5.95m (continuous or simply supported)
b_ei = L_0/8 = 5.95/8 = 0.744m → capped at actual geometry
b_eff = Σb_ei = 2 × 0.6 = 1.2m (actual geometry governs)
b_0 = 0 (no transverse ribs)
For simply supported: b_eff = L/4 = 8.5/4 = 2.125m → 1.2m governs (actual)
Formel
b_{eff} = \sum b_{ei}
Ergebnis
b_eff = 1200 mm
§6.2.2.1
Plastische Nulllage
Herleitung
A_s = 8450 mm², f_y = 355 N/mm², f_cd = 20 N/mm²
C_f = A_s × f_y / γ_M0 = 8450×355/1.0 = 2999 kN
A_c = b_eff × h_c = 1200 × 130 = 156000 mm²
C_c,max = A_c × f_cd = 156000×20/1000 = 3120 kN
C_c = C_f = 2999 kN (full interaction)
a_pna = C_f/(b_eff×f_cd) = 2999×1000/(1200×20) = 125 mm < h_c=130mm
Plastic N.A. lies in concrete slab → Full interaction confirmed
Formel
x_{pna} = \frac{A_s f_y}{b_{eff} f_cd}
Ergebnis
125 mm from slab soffit — in concrete
§6.2.2.2
Biegetragfähigkeit M_pl,Rd
Herleitung
z = h_s + h_c - a_pna/2 = 400+130-62.5 = 467.5 mm
M_pl,Rd = C_f × z = 2999 × 0.4675
= 1402 kNm
Applied moment from analysis:
M_Ed = 1.35×(2.5×8.5²/8) + 1.35×(1.5×8.5²/8) + 1.5×(5.0×8.5²/8)
= 1.35×22.7 + 1.35×13.6 + 1.5×45.2
= 30.6 + 18.4 + 67.8 = 116.8 kNm
Utilization = 116.8/1402
Formel
M_{pl,Rd} = A_s f_y z
Ergebnis
8.3% utilisation → PASS
§6.6.3.1
Kopfbolzen Schubtragfähigkeit P_Rd
Herleitung
h_sc/d = 100/16 = 6.25 > 4 → k = 1
f_u(stud) = 450 N/mm² (typical headed stud)
P_Rd = 0.29·α·d²·√(f_ck·E_cm) / γ_V
α = 1.0 (h_sc/d > 3)
d = 16 mm
P_Rd = 0.29×1.0×16²×√(30×31900)/1.25/1000
= 0.29×256×√957000/1.25/1000
= 74.2 × 978 / 1250
= 58.1 kN per stud
Number required: n = C_f/P_Rd = 2999/58.1 = 51.6 studs
Installed: L/spacing = 8500/150 = 56.7 → 57 studs ≥ 52 → PASS
Formel
P_{Rd} = \frac{0.29 \alpha d^2 \sqrt{f_{ck} E_{cm}}}{\gamma_V}
Ergebnis
57 studs installed ≥ 52 required → PASS