1. Geometrie & Nuttige Hoogte
$$ h = h_f + h_w = 25 + 90 = 115 \text{ mm} $$
$$ d = h - c - \frac{\phi}{2} = 115 - 20 - \frac{10}{2} = 90 \text{ mm} $$
2. Materiaaleigenschappen & Capaciteit Rib (ULS)
$$ f_{cd} = 0.85 \cdot \frac{25}{1.5} = 14.17 \text{ MPa}, \quad f_{yd} = \frac{500}{1.15} = 434.78 \text{ MPa} $$
$$ M_{Rd} = 78.54 \cdot 434.78 \cdot \left( 90 - \frac{10.04}{2} \right) \cdot 10^{-6} = \mathbf{2.9 \text{ kNm}} $$
$$ V_{Rd,c} = \mathbf{4.76 \text{ kN}} \quad \text{(volgens EC2 art. 6.2.2 ilmaar beugels)} $$
3. Eigen Gewicht & Resterende Capaciteit per Rib
$$ M_{Ed,g} = \frac{1.35 \cdot 0.308 \cdot 3^2}{8} = 0.47 \text{ kNm}, \quad V_{Ed,g} = \frac{1.35 \cdot 0.308 \cdot 3}{2} = 0.62 \text{ kN} $$
$$ M_{rest} = 2.9 - 0.47 = \mathbf{2.43 \text{ kNm}} $$
$$ V_{rest} = 4.76 - 0.62 = \mathbf{4.14 \text{ kN}} $$
4. Maximaal Toelaatbare Puntlast op 1 Rib (Maatgevend UGT)
$$ P_{ULS,max,M} = \frac{4 \cdot M_{rest}}{L} = \frac{4 \cdot 2.43}{3} = 3.25 \text{ kN (op buiging)} $$
$$ P_{ULS,max,V} = 2 \cdot V_{rest} = 2 \cdot 4.14 = 8.27 \text{ kN (op dwarskracht)} $$
$$ P_{ULS,max} = \min(P_{ULS,max,M}, P_{ULS,max,V}) = \mathbf{3.25 \text{ kN}} $$
5. Dimensionering Lastverdeelplaat (S235) voor $P_{app} = 10 \text{ kN}$
$$ P_{ULS,app} = 1.5 \cdot 10 = 15 \text{ kN} $$
$$ \text{Benodigde ribben} (N) = \lceil \frac{15}{3.25} \rceil = \mathbf{5 \text{ ribben}} $$
$$ \text{Afmeting Plaat } (L \times B) = \mathbf{1100 \times 1100 \text{ mm}}, \quad \text{Dikte } t = \mathbf{10 \text{ mm}} $$