Dyskretyzacja:

Zgodnie z tym przemieszczenia:

\begin{aligned} d=\left[\begin{array}{c} 0 \\ 0 \\ d_3 \\ d_4 \\ 0 \\ 0 \\ 0 \\ d_8 \\ 0 \end{array}\right] \end{aligned}

Agregacja:

Jak łatwo zauważyć po wykreśleniu elementów macierz sztywności:

\begin{aligned} \begin{bmatrix} K_{33} & K_{34} & K_{38} \\ K_{43} & K_{44} & K_{48} \\ K_{83} & K_{84} & K_{88} \end{bmatrix} = \begin{bmatrix} k_{33.1} & k_{34.1} & 0 \\ k_{43.1} & k_{44.1} + k_{11.2} & k_{15.2} \\ 0 & k_{51.2} & k_{55.2} \end{bmatrix} \end{aligned} \begin{aligned} k_{33.1} &= \frac{E \cdot I}{l_1^3} \cdot 4 \cdot l_1^2 = 4.8 \cdot 10^4 \end{aligned} \begin{aligned} k_{34.1} &= \frac{E \cdot I}{l_1^3} = 0 \end{aligned} \begin{aligned} k_{43.1} &= \frac{E \cdot I}{l_1^3} = 0 \end{aligned} \begin{aligned} k_{44.1} &= \frac{E \cdot I}{l_1^3} \cdot A \cdot \frac{l_1^2}{I} = 3 \cdot 10^5 \end{aligned} \begin{aligned} k_{11.2} &= \frac{E \cdot I}{l_2^3} \cdot A \cdot \frac{l_2^2}{I} = 2 \cdot 10^5 \end{aligned} \begin{aligned} k_{15.2} &= \frac{E \cdot I}{l_2^3} = 0 \end{aligned} \begin{aligned} k_{51.2} &= \frac{E \cdot I}{l_2^3} = 0 \end{aligned} \begin{aligned} k_{55.2} &= \frac{E \cdot I}{l_2^3} \cdot 12 = 1.067 \cdot 10^4 \end{aligned}

Analogicznie macierz mas:

\begin{aligned} \begin{bmatrix} M_{33} & M_{34} & M_{38} \\ M_{43} & M_{44} & M_{48} \\ M_{83} & M_{84} & M_{88} \end{bmatrix} = \begin{bmatrix} m_{33.1} & m_{34.1} & 0 \\ m_{43.1} & m_{44.1} + m_{11.2} & m_{15.2} \\ 0 & m_{51.2} & m_{55.2} \end{bmatrix} \end{aligned} \begin{aligned} m_{33.1} &= \frac{\mu \cdot l_1}{420} \cdot 4 \cdot l_1^2 = 3.2 \end{aligned} \begin{aligned} m_{34.1} &= \frac{\mu \cdot l_1}{420} = 0 \end{aligned} \begin{aligned} m_{43.1} &= \frac{\mu \cdot l_1}{420} = 0 \end{aligned} \begin{aligned} m_{44.1} &= \frac{\mu \cdot l_1}{420} \cdot 140 = 28 \end{aligned} \begin{aligned} m_{11.2} &= \frac{\mu \cdot l_2}{420} \cdot 140 = 42 \end{aligned} \begin{aligned} m_{15.2} &= \frac{\mu \cdot l_2}{420} = 0 \end{aligned} \begin{aligned} m_{51.2} &= \frac{\mu \cdot l_2}{420} = 0 \end{aligned} \begin{aligned} m_{55.2} &= \frac{\mu \cdot l_2}{420} \cdot 156 = 46.8 \end{aligned}

Równanie dynamiki:

\begin{aligned} \begin{bmatrix} k_{33.1} & k_{34.1} & 0 \\ k_{43.1} & k_{44.1} + k_{11.2} & k_{15.2} \\ 0 & k_{51.2} & k_{55.2} \end{bmatrix} - \omega^2 \cdot \begin{bmatrix} m_{33.1} & m_{34.1} & 0 \\ m_{43.1} & m_{44.1} + m_{11.2} & m_{15.2} \\ 0 & m_{51.2} & m_{55.2} \end{bmatrix} &= \begin{bmatrix} -3.2 \cdot \omega^2 + 48000.0 & 0.0 & 0.0 \\ 0.0 & -70.0 \cdot \omega^2 + 500000.0 & 0.0 \\ 0.0 & 0.0 & -46.8 \cdot \omega^2 + 10666.67 \end{bmatrix} \end{aligned} \begin{aligned} \det \left( \begin{bmatrix} -3.2 \cdot \omega^2 + 48000.0 & 0.0 & 0.0 \\ 0.0 & -70.0 \cdot \omega^2 + 500000.0 & 0.0 \\ 0.0 & 0.0 & -46.8 \cdot \omega^2 + 10666.67 \end{bmatrix} \right) &= 0 \end{aligned} \begin{aligned} -10483.2 \cdot \omega^6 + 234517334.08 \cdot \omega^4 - 1176106683200.0 \cdot \omega^2 + 256000080000000.0 &= 0 \end{aligned}

Rozwiązania:

\begin{aligned} \omega_1 &= 122.474 \\ \omega_2 &= 84.515 \\ \omega_3 &= 15.097 \end{aligned}

W tym przypadku ze względu na kształt macierzy nie trzeba liczyć wyznacznika który byłby 6 stopnia, wystarczy przyrównać każdy element z osoba do zera

MES

Metody numeryczne - MES - Przykład 9

Przykład 9