Centralne wzory różnicowe dla zagadnienia jednowymiarowego

\begin{aligned} & f^I=\frac{1}{2 h}\left(-v_{i-1}+v_{i+1}\right) \\ & f^{I I}=\frac{1}{h^2}\left(v_{i-1}-2 v_i+v_{i+1}\right) \\ & f^{I I I}=\frac{1}{2 h^3}\left(-v_{i-2}+2 v_{i-1}-2 v_{i+1}+v_{i+2}\right) \\ & f^{I V}=\frac{1}{h^4}\left(v_{i-2}-4 v_{i-1}+6 v_i-4 v_{i+1}+v_{i+2}\right) \end{aligned}

Centralne wzory różnicowe dla belki zginanej

\begin{aligned} & \frac{d^4 v(x)}{d x^4}=\frac{q(x)}{E J} \Rightarrow \frac{v_{i-2}-4 v_{i-1}+6 v_i-4 v_{i+1}+v_{i+2}}{h^4}=\frac{q}{E J} \\ & M(x)=-E J \frac{d^2 v(x)}{d x^2} \quad \Rightarrow \quad \mathrm{M}=-\mathrm{EI} \cdot \frac{v_{i-1}-2 v_i+v_{i+1}}{h^2} \\ & T(x)=-E J \frac{d^3 v(x)}{d x^3} \quad \Rightarrow \quad T=-\mathrm{EI} \cdot \frac{-v_{i-2}+2 v_{i-1}-2 v_{i+1}+v_{i+2}}{2 h^3} \end{aligned}

Wielomiany Hermite’a

\begin{aligned} & H_1=1-3 \cdot \xi^2+2 \cdot \xi^3 \\ & H_2=L \cdot\left(\xi-2 \cdot \xi^2+\xi^3\right) \\ & H_3=3 \cdot \xi^2-2 \cdot \xi^3 \\ & H_4=L \cdot\left(\xi^3-\xi^2\right) \end{aligned}

Warunki brzegowe dla belki zginanej

	\begin{aligned} & v=0 \\ & v_i=0 \end{aligned}	\begin{aligned} & \frac{d v}{d x}=0 \\ & \frac{-v_{i-1}+v_{i+1}}{2 h}=0 \end{aligned}
	\begin{aligned} & v=0 \\ & M=-E J \frac{d^2 v}{d x^2}=0 \end{aligned}	\begin{aligned} & v_i=0 \\ & -E J \frac{v_{i-1}-2 v_i+v_{i+1}}{h^2}=0 \end{aligned}
	\begin{aligned} & \frac{d v}{d x}=0 \\ & T=-E J \frac{d^3 v}{d x^3}=0 \end{aligned}	\begin{aligned} & \frac{-v_{i-1}+v_{i+1}}{2 h}=0 \\ & -E J \frac{-v_{i-2}+2 v_{i-1}-2 v_{i+1}+v_{i+2}}{2 h^3}=0 \end{aligned}
	\begin{gathered} M=-E J \frac{d^2 v}{d x^2}=0 \\ T=-E J \frac{d^3 v}{d x^3}=0 \end{gathered}	\begin{aligned} & -E J \frac{-v_{i-1}-2 v_i+v_{i+1}}{h^2}=0 \\ & -E J \frac{-v_{i-2}+2 v_{i-1}-2 v_{i+1}+v_{i+2}}{2 h^3}=0 \end{aligned}

Dynamika belka

Macierz sztywnosci

\[ \mathbf{K} = \frac{EI}{l} \begin{bmatrix} 12 & 6l & -12 & 6l \\ 6l & 4l^2 & -6l & 2l^2 \\ -12 & -6l & 12 & -6l \\ 6l & 2l^2 & -6l & 4l^2 \end{bmatrix} \]

Macierz bezwładności

\begin{aligned} \mu & = \rho A \quad -\text{ masa na jedn. długości }[\mathrm{kg} / \mathrm{m}] \\ \mathbf{M} & = \frac{\mu l}{420} \begin{bmatrix} 156 & 22l & 54 & -13l \\ 22l & 4l^2 & 13l & -3l^2 \\ 54 & 13l & 156 & -22l \\ -13l & -3l^2 & -22l & 4l^2 \end{bmatrix} \end{aligned}

Dynamika rama

Macierz sztywnosci

\begin{aligned} \mathbf{k}=\frac{E I}{l^3}\left[\begin{array}{cccccc} \frac{A l^2}{I} & 0 & 0 & -\frac{A l^2}{I} & 0 & 0 \\ 0 & 12 & 6 l & 0 & -12 & 6 l \\ 0 & 6 l & 4 l^2 & 0 & -6 l & 2 l^2 \\ -\frac{A l^2}{I} & 0 & 0 & \frac{A l^2}{I} & 0 & 0 \\ 0 & -12 & -6 l & 0 & 12 & -6 l \\ 0 & 6 l & 2 l^2 & 0 & -6 l & 4 l^2 \end{array}\right]=\left[\begin{array}{cccccc} \frac{E A}{l} & 0 & 0 & -\frac{E A}{l} & 0 & 0 \\ 0 & \frac{12 E I}{l^3} & \frac{6 E I}{l^2} & 0 & -\frac{12 E I}{l^3} & \frac{6 E I}{l^2} \\ 0 & \frac{6 E I}{l^2} & \frac{4 E I}{l} & 0 & -\frac{6 E I}{l^2} & \frac{2 E I}{l} \\ -\frac{E A}{l} & 0 & 0 & \frac{E A}{l} & 0 & 0 \\ 0 & -\frac{12 E I}{l^3} & -\frac{6 E I}{l^2} & 0 & \frac{12 E I}{l^3} & -\frac{6 E I}{l^2} \\ 0 & \frac{6 E I}{l^2} & \frac{2 E I}{l} & 0 & -\frac{6 E I}{l^2} & \frac{4 E I}{l} \end{array}\right] \end{aligned}

Macierz bezwładności

\begin{aligned} \mathbf{m}=\frac{\mu l}{420}\left[\begin{array}{cccccc} 140 & 0 & 0 & 70 & 0 & 0 \\ 0 & 156 & 22 l & 0 & 54 & -13 l \\ 0 & 22 l & 4 l^2 & 0 & 13 l & -3 l^2 \\ 70 & 0 & 0 & 140 & 0 & 0 \\ 0 & 54 & 13 l & 0 & 156 & -22 l \\ 0 & -13 l & -3 l^2 & 0 & -22 l & 4 l^2 \end{array}\right]=\left[\begin{array}{cccccc} \frac{\mu l}{3} & 0 & 0 & \frac{\mu l}{6} & 0 & 0 \\ 0 & \frac{13 \mu l}{35} & \frac{11 \mu l^2}{210} & 0 & \frac{9 \mu l}{70} & -\frac{13 \mu l^2}{420} \\ 0 & \frac{11 \mu l^2}{210} & \frac{\mu l^3}{105} & 0 & \frac{13 \mu l^2}{420} & -\frac{\mu l^3}{140} \\ \frac{\mu l}{6} & 0 & 0 & \frac{\mu l}{3} & 0 & 0 \\ 0 & \frac{9 \mu l}{70} & \frac{13 \mu l^2}{420} & 0 & \frac{13 \mu l}{35} & -\frac{11 \mu l^2}{210} \\ 0 & -\frac{13 \mu l^2}{420} & -\frac{\mu l^3}{140} & 0 & -\frac{11 \mu l^2}{210} & \frac{\mu l^3}{105} \end{array}\right] \end{aligned}

Funkcje kształtu liniowy element trójkątny

\begin{gathered} N_i^e=\frac{1}{2 A^e}\left(\alpha_i^e+\beta_i^e x+\gamma_i^e y\right) \\ \alpha_i^e=x_j y_k-x_k y_j \quad \beta_i^e=y_j-y_k \quad \gamma_i^e=x_k-x_j \\ i \neq j \neq k \quad i, j, k=1,2,3 \end{gathered}

A - powierzchnia trójkąta
Indeksy we współczynnikach zmieniają się wg permutacji podstawowej czyli:

\[ \begin{aligned} \alpha_1 & = x_2 \cdot y_3 - x_3 \cdot y_2 \\ \beta_1 & = y_2 - y_3 \\ \gamma_1 & = x_3 - x_2 \\ \end{aligned} \] \[ N_1(x, y) = \frac{1}{2A} \cdot (\alpha_1 + \beta_1 \cdot x + \gamma_1 \cdot y) \] \[ \begin{aligned} \alpha_2 & = x_3 \cdot y_1 - x_1 \cdot y_3 \\ \beta_2 & = y_3 - y_1 \\ \gamma_2 & = x_1 - x_3 \\ \end{aligned} \] \[ N_2(x, y) = \frac{1}{2A} \cdot (\alpha_2 + \beta_2 \cdot x + \gamma_2 \cdot y) \] \[ \begin{aligned} \alpha_3 & = x_1 \cdot y_2 - x_2 \cdot y_1 \\ \beta_3 & = y_1 - y_2 \\ \gamma_3 & = x_2 - x_1 \\ \end{aligned} \] \[ N_3(x, y) = \frac{1}{2A} \cdot (\alpha_3 + \beta_3 \cdot x + \gamma_3 \cdot y) \]

warunek kompletności

\[ N_1(x, y) + N_2(x, y) + N_3(x, y) = 1 \]