Formula 5

\begin{array}{l} V_{c, Rd} = V_{pl, Rd} = \frac{A_{v} \cdot \frac{f_{y}}{\sqrt{3}}}{γ_{M 0}} = \frac{60.56 \cdot \frac{23.5}{\sqrt{3}}}{1.0} = 821.66 kN \\ \frac{\max V_{z, Ed}}{V_{pl, Rd}} = \frac{48.75}{821.66} = 0.06 \leq 1 \to Design is fulfilled . \\ \frac{h_{w}}{t_{w}} = \frac{315}{12.5} = 25.2 \leq 72 \cdot \frac{ε}{η} = 72 \cdot \frac{1.0}{1.2} = 60.00 \to No shear buckling design required . \end{array}

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