Balkparameters
Voor benuttingsgraad η
Kipresultaten
193.32
kNm
Mcr
161.93
kNm
Mb,Rd
1.3651
λ̄LT
0.4495
χLT
Curve c (αLT = 0.49)
§6.3.2.3 (rolled — modified, β=0.75, λLT,0=0.4)
Mc,Rd (no LTB)
360.25 kNm
C1 (load shape)
1.13
k (rotation) / kw
1 / 1
zg (load height)
200 mm
φLT
1.4353
f-factor (§6.3.2.3 Eq 6.58)
0.9892
Section class
Class 1 → Wy = 1310.0 cm³
h/b ratio
2.22
1. M_cr (Eq F.2)
C1 = 1.13, C2 = 0.45, k = 1, zg = 200 mm
M_cr = C1 · (π²EIz/(kL)²) · [√(Iw/Iz + (kL)²·G·It/(π²·E·Iz) + (C2·zg)²) − C2·zg]
M_cr = 193.32 kNm
2. λ̄_LT (§6.3.2, Eq 6.56)
λ̄_LT = √(Wy · fy / M_cr) = √(1310000 · 275 / (193.32 × 10⁶)) = 1.3651
3. Buckling curve (Table 6.4)
h/b = 2.22 → Curve c, α_LT = 0.49
4. Reduction factor χ_LT (§6.3.2.3 (rolled — modified, β=0.75, λLT,0=0.4))
φ_LT = 0.5·[1 + α_LT·(λ̄_LT − λ̄_LT,0) + β·λ̄_LT²] = 1.4353
χ_LT = 1/(φ + √(φ² − β·λ̄²)) = 0.4495
5. M_b,Rd (§6.3.2, Eq 6.55)
M_b,Rd = χ_LT · Wy · fy / γ_M1 = 0.4495 · 1310000 · 275 / (1 × 10⁶) = 161.93 kNm
C1 = 1.13, C2 = 0.45, k = 1, zg = 200 mm
M_cr = C1 · (π²EIz/(kL)²) · [√(Iw/Iz + (kL)²·G·It/(π²·E·Iz) + (C2·zg)²) − C2·zg]
M_cr = 193.32 kNm
2. λ̄_LT (§6.3.2, Eq 6.56)
λ̄_LT = √(Wy · fy / M_cr) = √(1310000 · 275 / (193.32 × 10⁶)) = 1.3651
3. Buckling curve (Table 6.4)
h/b = 2.22 → Curve c, α_LT = 0.49
4. Reduction factor χ_LT (§6.3.2.3 (rolled — modified, β=0.75, λLT,0=0.4))
φ_LT = 0.5·[1 + α_LT·(λ̄_LT − λ̄_LT,0) + β·λ̄_LT²] = 1.4353
χ_LT = 1/(φ + √(φ² − β·λ̄²)) = 0.4495
5. M_b,Rd (§6.3.2, Eq 6.55)
M_b,Rd = χ_LT · Wy · fy / γ_M1 = 0.4495 · 1310000 · 275 / (1 × 10⁶) = 161.93 kNm
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