We use the FEM equation in the form:

\begin{aligned} K \cdot d=w+z+r \end{aligned}

Stiffness matrix of a single element:

\begin{aligned} K_e=\frac{2 E I}{L^3}\left[\begin{array}{cccc} 6 & 3 L & -6 & 3 L \\ 3 L & 2 L^2 & -3 L & L^2 \\ -6 & -3 L & 6 & -3 L \\ 3 L & L^2 & -3 L & 2 L^2 \end{array}\right]=\left[\begin{array}{rrrr} 200 & 600 & -200 & 600 \\ 600 & 2400 & -600 & 1200 \\ -200 & -600 & 200 & -600 \\ 600 & 1200 & -600 & 2400 \end{array}\right] \end{aligned}

Global stiffness matrix

\begin{aligned} K=\left[\begin{array}{cccccc} 200 & 600 & -200 & 600 & 0 & 0 \\ 600 & 2400 & -600 & 1200 & 0 & 0 \\ -200 & -600 & 400 & 0 & -200 & 600 \\ 600 & 1200 & 0 & 4800 & -600 & 1200 \\ 0 & 0 & -200 & -600 & 200 & -600 \\ 0 & 0 & 600 & 1200 & -600 & 2400 \end{array}\right] \end{aligned}

Load, reaction, displacement vectors

\begin{aligned} z=\left[\begin{array}{c} \frac{10 \cdot 6}{2} \\ \frac{10 \cdot 6^2}{12} \\ \frac{10 \cdot 6}{2} \\ -\frac{10 \cdot 6^2}{12} \\ 0 \\ 0 \end{array}\right]=\left[\begin{array}{r} 30 \\ 30 \\ 30 \\ -30 \\ 0 \\ 0 \end{array}\right] \quad d=\left[\begin{array}{c} 0 \\ 0 \\ 0 \\ \varphi_2 \\ \Delta \\ 0 \end{array}\right] \quad r=\quad\left[\begin{array}{c} r_1 \\ r_2 \\ r_3 \\ 0 \\ r_5 \\ r_6 \end{array}\right] \end{aligned}

Substituting into the equation:

\begin{aligned} \left[\begin{array}{cccccc} 200 & 600 & -200 & 600 & 0 & 0 \\ 600 & 2400 & -600 & 1200 & 0 & 0 \\ -200 & -600 & 400 & 0 & -200 & 600 \\ 600 & 1200 & 0 & 4800 & -600 & 1200 \\ 0 & 0 & -200 & -600 & 200 & -600 \\ 0 & 0 & 600 & 1200 & -600 & 2400 \end{array}\right] \cdot\left[\begin{array}{c} 0 \\ 0 \\ 0 \\ \varphi_2 \\ \Delta \\ 0 \end{array}\right]=\left[\begin{array}{c} 30 \\ 30 \\ 30 \\ -30 \\ 0 \\ 0 \end{array}\right]+\left[\begin{array}{c} r_1 \\ r_2 \\ r_3 \\ 0 \\ r_5 \\ r_6 \end{array}\right] \end{aligned}

As can be seen by eliminating zero elements, we obtain a simple system of equations, we calculate and sketch

\begin{aligned} & 4800 \cdot \varphi_2-600 \Delta=-30 \\ & -600 \cdot \varphi_2+200 \Delta=r_5 \\ & \varphi_2=\frac{1}{4800}(-30+600 \cdot 0.02)=-3.75 \cdot 10^{-3} \quad \mathrm{rad} \end{aligned}

MES

Numerical methods - MES - Example 1

Example 1