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000711

16. Januar 2024

Formel 14

M_{cr} = 1, 127 \cdot \frac{π^{2} \cdot 21.000 \frac{kN}{{cm}^{2}} \cdot 10.140 {cm}^{4}}{{(1 \cdot 650 cm)}^{2}} \cdot (\sqrt{(\frac{1}{1}) \cdot \frac{2.883.000 {cm}^{6}}{10.140 {cm}^{4}} \frac{{(1, 0 \cdot 650 cm)}^{2} \cdot 8.076, 92 \frac{kN}{{cm}^{2}} \cdot 292, 5 {cm}^{4}}{π^{2} \cdot 21.000 \frac{kN}{{cm}^{2}} \cdot 10.140 {cm}^{4}} {(0, 454 \cdot 18 cm)}^{2}} - 0, 454 \cdot 18 cm)

M_{cr} = 115.310 kNcm = 1.153, 10 kNm

{\bar{λ}}_{LT} = \sqrt{\frac{W_{pl, y} \cdot f_{y}}{M_{cr}}} = \sqrt{\frac{2.683 {cm}^{3} \cdot 23, 5 \frac{kN}{{cm}^{2}}}{115.310 kNcm}} = 0, 739

ϕ_{LT} = 0, 5 \cdot [1 0, 34 \cdot (0, 739 - 0, 4) 0, 75 \cdot 0, 739^{2}] = 0, 762

χ_{LT} = \frac{1}{0, 762 \sqrt{0, 762^{2} - 0, 75 \cdot 0, 739^{2}}} = 0, 85 < 1

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