We denote the nodal displacements:

Stiffness matrices for individual bars:

\begin{aligned} L &= 4 \\ K_1 &= \frac{2 \cdot 2 EI}{L^3} \begin{bmatrix} 6 & 3L & -6 & 3L \\ 3L & 2L^2 & -3L & L^2 \\ -6 & -3L & 6 & -3L \\ 3L & L^2 & -3L & 2L^2 \end{bmatrix} = \begin{bmatrix} 6.75 & 13.5 & -6.75 & 13.5 \\ 13.5 & 36 & -13.5 & 18 \\ -6.75 & -13.5 & 6.75 & -13.5 \\ 13.5 & 18 & -13.5 & 36 \end{bmatrix} \cdot 10^3 \\ L &= 6 \\ K_2 &= \frac{2 \cdot EI}{L^3} \begin{bmatrix} 6 & 3L & -6 & 3L \\ 3L & 2L^2 & -3L & L^2 \\ -6 & -3L & 6 & -3L \\ 3L & L^2 & -3L & 2L^2 \end{bmatrix} = \begin{bmatrix} 1 & 3 & -1 & 3 \\ 3 & 12 & -3 & 6 \\ -1 & -3 & 1 & -3 \\ 3 & 6 & -3 & 12 \end{bmatrix} \cdot 10^3 \end{aligned}

Global stiffness matrix:

\begin{aligned} K = \begin{bmatrix} 6.75 & 13.5 & -6.75 & 13.5 & 0 & 0 \\ 13.5 & 36 & -13.5 & 18 & 0 & 0 \\ -6.75 & -13.5 & 7.75 & -10.5 & -1 & 3 \\ 13.5 & 18 & -10.5 & 48 & -3 & 6 \\ 0 & 0 & -1 & -3 & 1 & -3 \\ 0 & 0 & 3 & 6 & -3 & 12 \end{bmatrix} \cdot 10^3 \end{aligned}

Displacement vector and right-hand side:

\begin{aligned} d &= \begin{bmatrix} 0 \\ 0 \\ 0 \\ d_4 \\ d_5 \\ 0 \end{bmatrix} \quad F = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 13 \\ 0 \end{bmatrix} \quad r = \begin{bmatrix} r_1 \\ r_2 \\ r_3 \\ 0 \\ 0 \\ r_6 \end{bmatrix} \end{aligned}

FEM equation:

\begin{aligned} K \cdot d &= F + r \\ \begin{bmatrix} 6.75 & 13.5 & -6.75 & 13.5 & 0 & 0 \\ 13.5 & 36 & -13.5 & 18 & 0 & 0 \\ -6.75 & -13.5 & 7.75 & -10.5 & -1 & 3 \\ 13.5 & 18 & -10.5 & 48 & -3 & 6 \\ 0 & 0 & -1 & -3 & 1 & -3 \\ 0 & 0 & 3 & 6 & -3 & 12 \end{bmatrix} \cdot \begin{bmatrix} 0 \\ 0 \\ 0 \\ d_4 \\ d_5 \\ 0 \end{bmatrix} \cdot 10^3 &= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 13 \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} r_1 \\ r_2 \\ r_3 \\ 0 \\ 0 \\ r_6 \end{bmatrix} \end{aligned}

After plotting the zero elements, we solve the system of equations

\begin{aligned} \begin{bmatrix} 48 & -3 \\ -3 & 1 \end{bmatrix} \cdot 10^3 \cdot \begin{bmatrix} d_4 \\ d_5 \end{bmatrix} &= \begin{bmatrix} 0 \\ 13 \end{bmatrix} \\ \begin{bmatrix} -3000 \cdot d_5 + 48000 \cdot d_4 \\ 1000 \cdot d_5 - 3000 \cdot d_4 \end{bmatrix} &= \begin{bmatrix} 0 \\ 13 \end{bmatrix} \\ d_4 &= 1 \cdot 10^{-3} \\ d_5 &= 1.6 \cdot 10^{-2} \end{aligned}

We calculate the reactions:

\begin{aligned} r &= \begin{bmatrix} 6.75 & 13.5 & -6.75 & 13.5 & 0 & 0 \\ 13.5 & 36 & -13.5 & 18 & 0 & 0 \\ -6.75 & -13.5 & 7.75 & -10.5 & -1 & 3 \\ 13.5 & 18 & -10.5 & 48 & -3 & 6 \\ 0 & 0 & -1 & -3 & 1 & -3 \\ 0 & 0 & 3 & 6 & -3 & 12 \end{bmatrix} \cdot 10^3 \cdot \begin{bmatrix} 0 \\ 0 \\ 0 \\ d_4 \\ d_5 \\ 0 \end{bmatrix} - \begin{bmatrix} 13.5 \\ 18 \\ -26.5 \\ 0 \\ 0 \\ -42 \end{bmatrix} \end{aligned}

We calculate the nodal forces in the elements:

\begin{aligned} f_1 &= \begin{bmatrix} 6.75 & 13.5 & -6.75 & 13.5 \\ 13.5 & 36 & -13.5 & 18 \\ -6.75 & -13.5 & 6.75 & -13.5 \\ 13.5 & 18 & -13.5 & 36 \end{bmatrix} \cdot 10^3 \cdot \begin{bmatrix} 0 \\ 0 \\ 0 \\ d_4 \end{bmatrix} = \begin{bmatrix} 13.5 \\ 18 \\ -13.5 \\ 36 \end{bmatrix} \\ f_2 &= \begin{bmatrix} 1 & 3 & -1 & 3 \\ 3 & 12 & -3 & 6 \\ -1 & -3 & 1 & -3 \\ 3 & 6 & -3 & 12 \end{bmatrix} \cdot 10^3 \cdot \begin{bmatrix} 0 \\ d_4 \\ d_5 \\ 0 \end{bmatrix} = \begin{bmatrix} -13 \\ -36 \\ 13 \\ -42 \end{bmatrix} \end{aligned}

We graphically present the solution:

FEM

Numerical methods - FEM - Example 5

Example 5