Take $f = 1.5 kN/m; \space \space EI = 10^7 Nm^2; \space \space k = 3 \times 10^5 N/m$

(i) Number of nodes and elements

(ii) Element matrix equation $$ \frac{2EI}{h_e^3} \begin{bmatrix} 6 & -3h_e & -6 & -3h_e \\ -3h_e & 2h_e^2 & -3h_e & h_e^2 \\ -6 & 3h_e & 6 & 3h_e \\ -3h_e & h_e^2 & 3h_e & 2h_e^2 \end{bmatrix} \begin{Bmatrix} \omega_1 \\ \omega_2 \\ \omega_3 \\ \omega_4 \end{Bmatrix} = \frac{fh_e}{60} \begin{Bmatrix} 9 \\ -2h_e \\ 21 \\ 3h_e \end{Bmatrix} + \begin{Bmatrix} Q_1 \\ Q_2 \\ Q_3 \\ Q_4 \end{Bmatrix} $$

For element 1

h$_e$ = 1.2m, EI = 10$^7$ Nm$^2$, f = 0

$ \frac{2EI}{h_e^3} = \frac{2 \times 10^7}{1.2^3} = 1.157 \times 10^7 $

$$ 1.157 \times 10^7 \begin{bmatrix} 6 & -3.6 & -6 & -3.6 \\ -3.6 & 2.88 & 3.6 & 1.44 \\ -6 & 3.6 & 6 & 3.6 \\ -3.6 & 1.44 & 3.6 & 2.88 \end{bmatrix} \begin{Bmatrix} \omega_1 \\ \omega_2 \\ \omega_3 \\ \omega_4 \end{Bmatrix} = \begin{Bmatrix} Q_1 \\ Q_2 \\ Q_3 \\ Q_4 \end{Bmatrix} $$

$$ 10^6 \begin{bmatrix} 6.94 & -4.17 & -6.94 & -4.17 \\ -4.17 & 3.33 & 4.17 & 1.67 \\ -6.94 & 4.17 & 6.94 & 4.17 \\ -4.17 & 1.67 & 4.17 & 3.33 \end{bmatrix} \begin{Bmatrix} \omega_1 \\ \omega_2 \\ \omega_3 \\ \omega_4 \end{Bmatrix} = \begin{Bmatrix} Q_1 \\ Q_2 \\ Q_3 \\ Q_4 \end{Bmatrix} $$

For element 2

h$_e$ = 1.8m, EI = 10$^7$ Nm$^2$, f = -1.5 x 10$^3$ N/m

$ \frac{2EI}{h_e^3} = \frac{2 \times 10^7}{1.8^3} = 0.343 \times 10^7 \\ \frac{fh_e}{60} = \frac{-1.5 \times 10^3 \times 1.8}{60} = -45 $

$$ 0.343 \times 10^7 \begin{bmatrix} 6 & -5.4 & -6 & -5.4 \\ -5.4 & 6.48 & 5.4 & 3.24 \\ -6 & 5.4 & 6 & 5.4 \\ -5.4 & 3.24 & 5.4 & 6.48 \end{bmatrix} \begin{Bmatrix} \omega_3 \\ \omega_4 \\ \omega_5 \\ \omega_6 \end{Bmatrix} = -45 \begin{Bmatrix} 9 \\ -3.6 \\ 21 \\ 5.4 \end{Bmatrix} + \begin{Bmatrix} Q_3 \\ Q_4 \\ Q_5 \\ Q_6 \end{Bmatrix} $$

$$ 10^6 \begin{bmatrix} 2.06 & -1.85 & -2.06 & -1.85 \\ -1.85 & 2.22 & 1.85 & 1.11 \\ -2.06 & 1.85 & 2.06 & 1.85 \\ -4.17 & 1.11 & 1.85 & 2.22 \end{bmatrix} \begin{Bmatrix} \omega_3 \\ \omega_4 \\ \omega_5 \\ \omega_6 \end{Bmatrix} = \begin{bmatrix} -405 \\ 162 \\ -945 \\ -243 \end{bmatrix} + \begin{Bmatrix} Q_3 \\ Q_4 \\ Q_5 \\ Q_6 \end{Bmatrix} $$

(iii) Global matrix equation

$$ 10^4 \begin{bmatrix} 694 & -417 & -694 & -417 & 0 & 0 \\ -417 & 333 & 417 & 167 & 0 & 0 \\ -694 & -417 & 900 & 232 & -206 & -185 \\ 417 & 167 & 232 & 555 & 185 & 111 \\ 0 & 0 & -206 & 185 & 206 & 185 \\ 0 & 0 & -185 & 111 & 185 & 222 \end{bmatrix} \begin{Bmatrix} \omega_1 \\ \omega_2 \\ \omega_3 \\ \omega_4 \\ \omega_5 \\ \omega_6 \end{Bmatrix} = 10^2 \begin{bmatrix} 0 \\ 0 \\ -4.05 \\ 1.62 \\ -9.45 \\ -2.43 \end{bmatrix} \begin{Bmatrix} Q_1 \\ Q_2 \\ Q_3 \\ Q_4 \\ Q_5 \\ Q_6 \end{Bmatrix} $$

(iv) Imposing boundary condition

$ \omega_1 = \omega_2 = 0 \\ Q_3 = Q_4 = Q_6 = 0 $

Now, at free end cantilever beam is supported by spring, the shear force is equal to k x deflection

$ \therefore Q_5 = -k \omega_5 = -3 \times 10^5 \omega_5 = -30 \times 10^4 \omega_5 $

Now, by elimination method,

$$ 10^4 \begin{bmatrix} 900 & 232 & -206 & -185 \\ 232 & 555 & 185 & 111 \\ -206 & 185 & 206 & 185 \\ -185 & 111 & 185 & 222 \end{bmatrix} \begin{Bmatrix} \omega_3 \\ \omega_4 \\ \omega_5 \\ \omega_6 \end{Bmatrix} = 10^2 \begin{bmatrix} -4.05 \\ 1.62 \\ -9.45 \\ -2.43 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ -30 \times 10^4 \omega_5 \\ 0 \end{bmatrix} $$

By solving the equations, we get,

$\omega_5$ = -2.32 mm (Deflection at free end)