Take $EI = 2 \times 10^{-6} \space Nm^2 $

(i) Number of nodes and elements

(ii) Element matrix equation

$$ \frac{2EI}{h_e^3} \begin{bmatrix} 6 & -3h_e & -6 & -3h_e \\ -3h_e & 2h_e^2 & -3h_e & h_e^2 \\ -6 & 3h_e & 6 & 3h_e \\ -3h_e & h_e^2 & 3h_e & 2h_e^2 \end{bmatrix} \begin{Bmatrix} \omega_1 \\ \omega_2 \\ \omega_3 \\ \omega_4 \end{Bmatrix} = \frac{fh_e}{12} \begin{Bmatrix} 6 \\ -h_e \\ 6 \\ h_e \end{Bmatrix} + \begin{Bmatrix} Q_1 \\ Q_2 \\ Q_3 \\ Q_4 \end{Bmatrix} $$

For element 1

h$_e$ = 4m, EI = 2 x 10$^6$ Nm$^2$, f = -40 x 10$^3$ N/m

$ \frac{2EI}{h_e^3} = \frac{2 \times 2 \times 10^6}{4^3} = \frac{1}{16} \times 10^6 \\ \frac{fh_e}{12} = \frac{-40 \times 10^3 \times 4}{12} = -13.33 \times 10^3 $

$$ \frac{1}{16} \times 10^6 \begin{bmatrix} 6 & -12 & -6 & -12 \\ -12 & 32 & 12 & 16 \\ -6 & 12 & 6 & 12 \\ -12 & 16 & 12 & 32 \end{bmatrix} \begin{Bmatrix} \omega_1 \\ \omega_2 \\ \omega_3 \\ \omega_4 \end{Bmatrix} = -13.33 \times 10^3 \begin{Bmatrix} 6 \\ -4 \\ 6 \\ 4 \\ \end{Bmatrix} + 10^3 \begin{Bmatrix} Q_1 \\ Q_2 \\ Q_3 \\ Q_4 \end{Bmatrix} $$

$$ 10^3 \begin{bmatrix} 0.375 & -0.75 & -0.375 & -0.75 \\ -0.75 & 2 & 0.75 & 1 \\ -0.375 & 0.75 & 0.375 & 0.75 \\ -0.75 & 1 & 0.75 & 2 \end{bmatrix} \begin{Bmatrix} \omega_1 \\ \omega_2 \\ \omega_3 \\ \omega_4 \end{Bmatrix} = \begin{Bmatrix} -80 \\ 53.32 \\ -80 \\ -53.32 \end{Bmatrix} + \begin{Bmatrix} Q_1 \\ Q_2 \\ Q_3 \\ Q_4 \end{Bmatrix} $$

For element 2

h$_e$ = 2m, f = 0

$ \frac{2EI}{h_e^3} = \frac{2 \times 2 \times 10^6}{2^3} = \frac{1}{2} \times 10^6 $

$$ \frac{1}{2} \times 10^6 \begin{bmatrix} 6 & -6 & -6 & -6 \\ -6 & 8 & 6 & 4 \\ -6 & 6 & 6 & 6 \\ -6 & 4 & 6 & 8 \end{bmatrix} \begin{Bmatrix} \omega_3 \\ \omega_4 \\ \omega_5 \\ \omega_6 \end{Bmatrix} = 10^3 \begin{Bmatrix} Q_3 \\ Q_4 \\ Q_5 \\ Q_6 \end{Bmatrix} $$

$$ 10^3 \begin{bmatrix} 3 & -3 & -3 & -3 \\ -3 & 4 & 3 & 2 \\ -3 & 3 & 3 & 3 \\ -3 & 2 & 3 & 4 \end{bmatrix} \begin{Bmatrix} \omega_3 \\ \omega_4 \\ \omega_5 \\ \omega_6 \end{Bmatrix} = \begin{Bmatrix} Q_3 \\ Q_4 \\ Q_5 \\ Q_6 \end{Bmatrix} $$

(iii) Global matrix equation

$$ 10^3 \begin{bmatrix} 0.375 & -0.75 & -0.375 & 0.75 & 0 & 0 \\ -0.75 & 2 & 0.75 & 1 & 0 & 0 \\ -0.375 & 0.75 & 3.375 & -2.25 & -3 & -3 \\ -0.75 & 1 & -2.25 & 6 & 3 & 2 \\ 0 & 0 & -3 & 3 & 3 & 3 \\ 0 & 0 & -3 & 2 & 3 & 4 \end{bmatrix} \begin{Bmatrix} \omega_1 \\ \omega_2 \\ \omega_3 \\ \omega_4 \\ \omega_5 \\ \omega_6 \end{Bmatrix} = \begin{bmatrix} -80 \\ 53.32 \\ -80 \\ -53.32 \\ 0 \\ 0 \end{bmatrix} \begin{Bmatrix} Q_1 \\ Q_2 \\ Q_3 \\ Q_4 \\ Q_5 \\ Q_6 \end{Bmatrix} $$

(iv) Imposing boundary condition

$ \omega_1 = 0, \hspace{0.25cm} \omega_3 = 0, \hspace{0.25cm} \omega_5 = 0 \\ Q_2 = 0, \hspace{0.25cm} Q_4 = 0, \hspace{0.25cm} Q_5 = -P, \hspace{0.25cm} Q_6 = 0 \,\, (For \,\, balancing) $

By method of elimination

$$ 10^3 \begin{bmatrix} 2 & 1 & 0 \\ 1 & 6 & 2 \\ 0 & 2 & 4 \end{bmatrix} \begin{Bmatrix} \omega_2 \\ \omega_4 \\ \omega_6 \end{Bmatrix} = \begin{Bmatrix} 53.32 \\ -53.32 \\ 0 \end{Bmatrix} $$

By solving the equations, we get,

$\omega_2$ = 0.0355 rad

$\omega_4$ = -0.01778 rad

$\omega_6$ = 0.00889 rad

Now, $ 10^3 (3 \omega_4 + 3 \omega_6) = Q_5 = -P $

P = 26.67 N