Balkparameters
Voor benuttingsgraad η = MEd / Mb,Rd
Kipresultaten
73.07
Mb,Rd (kNm)
Mcr93.95 kNm
λ̄LT1.5404
φLT1.9144
χLT0.3278
Knikklasse
B (α = 0.34)
Mc,Rd (no LTB)222.94 kNm
Mb,Rd73.07 kNm
Wy (Class 1)628,000 mm³
1. Section: IPE 300 | S355 (fy = 355 N/mm²)
Iz = 6,038,000 mm⁴ | It = 201,000 mm⁴ | Iw = 6.26e+10 mm⁶
h/b = 2 → Buckling curve B (αLT = 0.34)
2. M_cr (EN 1993-1-1 Annex F, Eq F.2)
C1 = 1.13 | C2 = 0.45 | k = 1 | zg = 0 mm
Mcr = 93.95 kNm
3. λ̄LT (Eq 6.56)
λ̄LT = √(Wy · fy / Mcr) = √(628,000 × 355 / 93950000) = 1.5404
λ̄LT,0 = 0.2
4. χLT (§6.3.2.2 (general))
φLT = 0.5 × [1 + αLT(λ̄LT − λ̄LT,0) + β·λ̄LT²] = 1.9144
χLT = 1 / (φLT + √(φLT² − β·λ̄LT²)) = 0.3278
5. Mb,Rd (Eq 6.55)
Mb,Rd = χLT · Wy · fy / γM1 = 0.3278 × 628,000 × 355 / 1 = 73.07 kNm
Iz = 6,038,000 mm⁴ | It = 201,000 mm⁴ | Iw = 6.26e+10 mm⁶
h/b = 2 → Buckling curve B (αLT = 0.34)
2. M_cr (EN 1993-1-1 Annex F, Eq F.2)
C1 = 1.13 | C2 = 0.45 | k = 1 | zg = 0 mm
Mcr = 93.95 kNm
3. λ̄LT (Eq 6.56)
λ̄LT = √(Wy · fy / Mcr) = √(628,000 × 355 / 93950000) = 1.5404
λ̄LT,0 = 0.2
4. χLT (§6.3.2.2 (general))
φLT = 0.5 × [1 + αLT(λ̄LT − λ̄LT,0) + β·λ̄LT²] = 1.9144
χLT = 1 / (φLT + √(φLT² − β·λ̄LT²)) = 0.3278
5. Mb,Rd (Eq 6.55)
Mb,Rd = χLT · Wy · fy / γM1 = 0.3278 × 628,000 × 355 / 1 = 73.07 kNm
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