Oznaczamy węzły:

Dla obu elemenetów zapisujemy K i T. Dla elementu 1 dodatkowo macierz obciążeń lokalną i globalną:

element 1, 1->3

\begin{aligned} L &= 4 \\ x_k &= 0 \\ x_p &= 0 \\ y_k &= 4 \\ y_p &= 0 \\ p &= 30 \end{aligned} \begin{aligned} c &= \frac{x_k - x_p}{L} = 0 \\ s &= \frac{y_k - y_p}{L} = 1 \\ \alpha &= \arcsin(s) = 90^\circ \end{aligned} \begin{aligned} K_1 &= \frac{EA}{L} \begin{bmatrix} c^2 & c \cdot s & -c^2 & -c \cdot s \\ c \cdot s & s^2 & -c \cdot s & -s^2 \\ -c^2 & -c \cdot s & c^2 & c \cdot s \\ -c \cdot s & -s^2 & c \cdot s & s^2 \end{bmatrix} = 10^2 \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 25 & 0 & -25 \\ 0 & 0 & 0 & 0 \\ 0 & -25 & 0 & 25 \end{bmatrix} \end{aligned} \begin{aligned} T_1 &= \begin{bmatrix} c & s & 0 & 0 \\ 0 & 0 & c & s \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \end{aligned} \begin{aligned} z &= \begin{bmatrix} \frac{p \cdot L}{2} \\ \frac{p \cdot L}{2} \end{bmatrix} = \begin{bmatrix} 60 \\ 60 \end{bmatrix} \\ z_1 &= T_1^T \cdot z = \begin{bmatrix} 0 \\ 60 \\ 0 \\ 60 \end{bmatrix} \end{aligned}

element 2, 2->3

element 2, 1->3 \begin{aligned} L &= 5 \\ x_k &= 0 \\ x_p &= 3 \\ y_k &= 4 \\ y_p &= 0 \end{aligned} \begin{aligned} c &= \frac{x_k - x_p}{L} = -0.6 \\ s &= \frac{y_k - y_p}{L} = 0.8 \\ \alpha &= \arcsin(s) = 53.13^\circ \end{aligned} \begin{aligned} K_2 &= \frac{EA}{L} \begin{bmatrix} c^2 & c \cdot s & -c^2 & -c \cdot s \\ c \cdot s & s^2 & -c \cdot s & -s^2 \\ -c^2 & -c \cdot s & c^2 & c \cdot s \\ -c \cdot s & -s^2 & c \cdot s & s^2 \end{bmatrix} = 10^2 \begin{bmatrix} 7.2 & -9.6 & -7.2 & 9.6 \\ -9.6 & 12.8 & 9.6 & -12.8 \\ -7.2 & 9.6 & 7.2 & -9.6 \\ 9.6 & -12.8 & -9.6 & 12.8 \end{bmatrix} \end{aligned} \begin{aligned} T_2 &= \begin{bmatrix} c & s & 0 & 0 \\ 0 & 0 & c & s \end{bmatrix} = \begin{bmatrix} -0.6 & 0.8 & 0 & 0 \\ 0 & 0 & -0.6 & 0.8 \end{bmatrix} \end{aligned}

Agregacja

\begin{aligned} K &= 10^2 \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 25 & 0 & 0 & 0 & -25 \\ 0 & 0 & 7.2 & -9.6 & -7.2 & 9.6 \\ 0 & 0 & -9.6 & 12.8 & 9.6 & -12.8 \\ 0 & 0 & -7.2 & 9.6 & 7.2 & -9.6 \\ 0 & -25 & 9.6 & -12.8 & -9.6 & 37.8 \end{bmatrix} \end{aligned}

Zapisujemy równanie MES

\begin{aligned} K \cdot d &= z + r \\ \left[ \begin{array}{cccccc} 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 25 & 0 & 0 & 0 & -25 \\ 0 & 0 & 7.2 & -9.6 & -7.2 & 9.6 \\ 0 & 0 & -9.6 & 12.8 & 9.6 & -12.8 \\ 0 & 0 & -7.2 & 9.6 & 7.2 & -9.6 \\ 0 & -25 & 9.6 & -12.8 & -9.6 & 37.8 \end{array} \right] &\cdot 10^2 \cdot \left[ \begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \\ -5 \cdot 10^{-2} \\ d_6 \end{array} \right] &= \left[ \begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 60 \end{array} \right] + \left[ \begin{array}{c} r_1 \\ r_2 \\ r_3 \\ r_4 \\ r_5 \end{array} \right] \end{aligned}

Po wykreśleniu zerowych wierszy i kolumn:

\begin{aligned} \left[ \begin{array}{cc} 7.2 & -9.6 \\ -9.6 & 37.8 \end{array} \right] &\cdot 10^2 \cdot \left[ \begin{array}{c} -5 \cdot 10^{-2} \\ d_6 \end{array} \right] = \left[ \begin{array}{c} 0 \\ 60 \end{array} \right] + \left[ \begin{array}{c} r_5 \\ 0 \end{array} \right] \end{aligned}

Co prowadzi do rozwiązania:

\begin{aligned} 3780.0 \cdot d_6 + 48.0 &= 60 \\ d_6 &= \frac{60 - 48}{3780} = 3.175 \cdot 10^{-3} \end{aligned}

Więc wektor przemieszczeń:

\begin{aligned} d &= \left[ \begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \\ -5 \cdot 10^{-2} \\ d_6 \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \\ -5 \cdot 10^{-2} \\ 3.175 \cdot 10^{-3} \end{array} \right] \end{aligned}

Siły węzłowe:

\begin{aligned} f_1 &= T_1 \cdot K_1 \cdot d_1 - T_1 \cdot z_1 \\ f_1 &= \left[ \begin{array}{ccc} 0 & 1 & 0 & 0 \\ 0 & 25 & 0 & -25 \\ 0 & 0 & 0 & 0 \\ 0 & -25 & 0 & 25 \end{array} \right] \cdot 10^2 \cdot \left[ \begin{array}{c} 0 \\ 0 \\ -5 \cdot 10^{-2} \\ 3.175 \cdot 10^{-3} \end{array} \right] - \left[ \begin{array}{ccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right] \left[ \begin{array}{c} 0 \\ 60 \\ 0 \\ 60 \end{array} \right] = \left[ \begin{array}{c} -67.94 \\ -52.06 \end{array} \right] \end{aligned} \begin{aligned} f_2 &= T_2 \cdot K_2 \cdot d_2 \\ f_2 &= \left[ \begin{array}{cc} -0.6 & 0.8 \\ 0 & -0.6 \cdot 0.8 \end{array} \right] \cdot \left[ \begin{array}{cccc} 7.2 & -9.6 & -7.2 & 9.6 \\ -9.6 & 12.8 & 9.6 & -12.8 \\ -7.2 & 9.6 & 7.2 & -9.6 \\ 9.6 & -12.8 & -9.6 & 12.8 \end{array} \right] \cdot 10^2 \left[ \begin{array}{c} 0 \\ 0 \\ -5 \cdot 10^{-2} \\ 3.175 \cdot 10^{-3} \end{array} \right] = \left[ \begin{array}{c} -65.08 \\ 65.08 \end{array} \right] \end{aligned}

Zilustrujmy na wykresie:

Reakcje

Sprawdzenie globalne

\begin{aligned} \Sigma X &= 0 \quad 39.05 - 39.05 = 0 \\ \Sigma Y &= 0 \quad 30 \cdot 4 - 67.94 - 52.06 = 0 \\ \Sigma M_2 &= 0 \quad 30 \cdot 4 \cdot 3 - 67.94 \cdot 3 - 39.05 \cdot 4 = 0 \end{aligned}

Równowaga węzła

\begin{aligned} \sin \alpha &= \frac{4}{5} \quad &\cos \alpha &= \frac{3}{5} \\ \Sigma X &= 0 \quad -39.05 + 65.08 \cdot \cos \alpha = 0 \\ \Sigma Y &= 0 \quad 52.06 - 65.08 \cdot \sin \alpha = 0 \end{aligned}

MES

Metody numeryczne - MES - Przykład 7

Przykład 7