Probleemomschrijving
5 verdiepingen stalen moment frame, zone 2 NL (a_g,R = 1.6 m/s²). Grondtype B (S = 1.2). Belangrijkheidsklasse II (γ_I = 1.0). DCM, q = 4.0. Totaal seismisch gewicht W = 8500 kN.
Berekening Stappen — 5 stappen
§3.2.1
Ontwerpgrondversnelling a_g
Afleiding
a_g,R = 1.6 m/s² (Zone 2 NL, NEN-NA EN 1998-1 NA Table 3.1)
γ_I = 1.0 (class II)
a_g = γ_I × a_g,R = 1.0 × 1.6 = 1.6 m/s²
S = 1.2 (ground type B)
SDS = a_g × S × (2.5/3) = 1.6×1.2×0.833 = 1.60 m/s²
But EN 1998-1 uses: S × a_g for elastic spectrum
S = 1.2 → a_gS = 1.92 m/s²
Formule
a_g = \gamma_I a_{g,R}
Resultaat
a_g = 1.6 m/s², S = 1.2, a_gS = 1.92 m/s²
§4.3.3.2
Laterale-kracht methode — basisreactie F_b
Afleiding
For T₁ = 0.65s < TC = 0.5s (estimate, check later):
S_d(T₁) = a_g × S × (1 + T/T_B×(2.5/q) - 1)/3)
No — EN 1998-1 elastic → design spectrum:
S_d(T) = a_g S [1 + T/T_B (3/q - 1)] for 0 < T < TB
For 0.65s > TC = 0.5s: S_d(T) = a_g S × 2.5/q / T² × TC² for T > TC
Actually: S_d(T) = a_g × S × 2.5/q for TB < T < TC
S_d(T₁) = 1.6×1.2×2.5/4.0 = 1.20 m/s²
F_b = S_d(T₁) × W × λ
λ = 0.85 (for W > 4800 kN, n > 3 floors)
F_b = 1.20 × 8500 × 0.85 / 1000 = 10.26 MN
Formule
F_b = S_d(T_1) \cdot W \cdot \lambda
Resultaat
F_b = 10.26 MN
§4.3.3.2
Verticale verdeling — F_i per verdieping
Afleiding
F_i = F_b × (s_i × h_i) / Σ(s_i × h_i)
s_i = 1.0 (uniform mass distribution)
Height at floor i: h_i = 3.6, 7.2, 10.8, 14.4, 18.0 m
Σ(s_i × h_i) = 3.6+7.2+10.8+14.4+18.0 = 54.0 m
F1 = 10.26 × 3.6/54.0 = 0.684 MN
F2 = 10.26 × 7.2/54.0 = 1.369 MN
F3 = 10.26 × 10.8/54.0 = 2.053 MN
F4 = 10.26 × 14.4/54.0 = 2.738 MN
F5 = 10.26 × 18.0/54.0 = 3.421 MN
Formule
F_i = F_b \frac{s_i h_i}{\sum s_i h_i}
Resultaat
Top floor governs: F5 = 3.42 MN (roof)
§4.4.3.2
Verdiepingsverplaatsing — P-Δ effecten
Afleiding
V_Story_5 = 3.42 MN
Δe = drift of top floor (from analysis)
θ = P-Δ sensitivity coefficient
θ = P×Δe/(V×H) where P = W_story, V = V_story, H = storey height
P_roof = 1700 kN (roof weight)
Δe (from pushover) = 38mm at roof
V_roof = 3.42 MN = 3420 kN
H_5 = 3.6m
θ = 1700×0.038/(3420×3.6) = 64.6/12312 = 0.0053 < 0.1
θ < 0.1 → P-Δ effects negligible
θ < 0.2 → second-order effects included in analysis
θ < 0.3 → multiply displacements by 1/(1-θ)
θ = 0.0053 → multiplier ≈ 1.005 → negligible
Formule
\theta = \frac{P \cdot \Delta_e}{V \cdot H}
Resultaat
θ = 0.005 → P-Δ negligible, design without amplification
§5.2.2.2
SCWB verificatie
Afleiding
SCWB ratio = ΣM_Rc / ΣM_Rb
Column: HEA 340, S355, Class 1 → M_Rc = W_pl,y × f_y/γ_M0 = 951×10³×355/10⁶ = 337.6 kNm
Beam: HEA 300, S355, Class 1 → M_Rb = W_pl,y × f_y/γ_M0 = 782×355/10⁶ = 277.6 kNm
Column overstrength factor Ω = 1.25 (S355, NL NA §5.2.2.2)
ΣM_Rc = 337.6 × 2 columns × 1.25 = 844 kNm
ΣM_Rb = 277.6 × 1 beam = 277.6 kNm
SCWB = 844 / 277.6 = 3.04 > 1.3 → PASS
Design is column-weak → redesign or add continuity plates
Formule
\frac{\sum M_{Rc}}{\sum M_{Rb}} \geq 1.3
Resultaat
SCWB = 3.04 > 1.3 → PASS