Wzór 6

\begin{array}{l} k_{A} = 0.1 (fixed column base) \\ k_{B} = \frac{\sum (\frac{E \cdot I_{col}}{l_{col}})}{\sum (\frac{α \cdot E \cdot I_{R}}{l_{R}})} \\ k_{B} = \frac{\frac{3, 300 kN / {cm}^{2} \cdot 160, 000 {cm}^{4}}{3 m} \frac{3, 300 kN / {cm}^{2} \cdot 160, 000 {cm}^{4}}{1.5 m}}{\frac{3 \cdot 3, 300 kN / {cm}^{2} \cdot 160, 000 {cm}^{4}}{3 m}} \\ k_{B} = 1.0 \\ l_{0} = 3 m \cdot \max \{\begin{cases} \sqrt{1 10 \cdot \frac{0.1 \cdot 1.0}{0.1 1.0}} = 1.38 \\ (1 \frac{0.1}{1 0.1}) \cdot (1 \frac{1.0}{1 1.0}) = 1.64 \end{cases}) \\ l_{0} = 4.91 m \end{array}

Czy mają Państwo jakieś pytania?