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2024-04-19

Formula 5: Shear Forces of Partial Cross-Section

\begin{array}{l} V_{y, 1} = \frac{100 \cdot (1.084 \cdot 12.268)}{12.268 \cdot 1.093} = 99.18 kN \\ V_{y, 2} = \frac{100 \cdot (0.009 \cdot 12.268)}{12.268 \cdot 1.093} = 0.823 kN \end{array}

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