Punkty są ponumerowane zgodnie z ruchem wskazówek zegara, zapisujemy ich współrzędne oraz pole trójkąta

\begin{array}{ll} x_1 = 0 & y_1 = 0 \\ x_2 = 2 & y_2 = 2 \\ x_3 = 0 & y_3 = 2 \\ A = \frac{1}{2} \cdot 2 \cdot 2 = 2 \end{array}

Wyznaczamy funkcje kształtu:

\begin{aligned} & \alpha_1 = x_2 \cdot y_3 - x_3 \cdot y_2 = 4 \\ & \beta_1 = y_2 - y_3 = 0 \\ & \gamma_1 = x_3 - x_2 = -2 \\ & N_1(x, y) = \frac{1}{2A} \left(\alpha_1 + \beta_1 \cdot x + \gamma_1 \cdot y\right) = \frac{-y + 2}{2} \\ \\ & \alpha_2 = x_3 \cdot y_1 - x_1 \cdot y_3 = 0 \\ & \beta_2 = y_3 - y_1 = 2 \\ & \gamma_2 = x_1 - x_3 = 0 \\ & N_2(x, y) = \frac{1}{2A} \left(\alpha_2 + \beta_2 \cdot x + \gamma_2 \cdot y\right) = \frac{x}{2} \\ \\ & \alpha_3 = x_1 \cdot y_2 - x_2 \cdot y_1 = 0 \\ & \beta_3 = y_1 - y_2 = -2 \\ & \gamma_3 = x_2 - x_1 = 2 \\ & N_3(x, y) = \frac{1}{2A} \left(\alpha_3 + \beta_3 \cdot x + \gamma_3 \cdot y\right) = \frac{y - x}{2} \end{aligned}

Zapisujemy wektor przemieszczeń:

\begin{aligned} d=\left[\begin{array}{c} 0 \\ 0 \\ 2 \cdot 10^{-3} \\ -4\cdot 10^{-3} \\ 0 \\ 0 \end{array}\right] \end{aligned}

Obliczamy D i B:

\begin{aligned} D & = \frac{E}{1 - \nu^2} \cdot \begin{bmatrix} 1 & \nu & 0 \\ \nu & 1 & 0 \\ 0 & 0 & \frac{1 - \nu}{2} \end{bmatrix} = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix} \cdot 10^{10} \\ \\ B & = \begin{bmatrix} \frac{\partial N_1(x, y)}{\partial x} & 0 & \frac{\partial N_2(x, y)}{\partial x} & 0 & \frac{\partial N_3(x, y)}{\partial x} & 0 \\ 0 & \frac{\partial N_1(x, y)}{\partial y} & 0 & \frac{\partial N_2(x, y)}{\partial y} & 0 & \frac{\partial N_3(x, y)}{\partial y} \\ \frac{\partial N_1(x, y)}{\partial y} & \frac{\partial N_1(x, y)}{\partial x} & \frac{\partial N_2(x, y)}{\partial y} & \frac{\partial N_2(x, y)}{\partial x} & \frac{\partial N_3(x, y)}{\partial y} & \frac{\partial N_3(x, y)}{\partial x} \end{bmatrix} \rightarrow \begin{bmatrix} 0 & 0 & \frac{1}{2} & 0 & -\frac{1}{2} & 0 \\ 0 & -\frac{1}{2} & 0 & 0 & 0 & \frac{1}{2} \\ \frac{1}{2} & 0 & \frac{1}{2} & 0 & -\frac{1}{2} & -\frac{1}{2} \end{bmatrix} \end{aligned}

Wyliczamy naprężenia:

\begin{aligned} \sigma=D \cdot B \cdot d=\left[\begin{array}{r} 20000 \\ 0 \\ -20000 \end{array}\right] 10^3 \end{aligned}

MES

Metody numeryczne - MES - Przykład 2

Przykład 2