$Ay_2 = \frac{Ao+AL}{2} = \frac{80+20}{2} = 50 mm^2$
$A_1 = \frac{80+50}{2} = 65 mm^2$
$A_2 = \frac{50+20}{2} = 35 mm^2$
$\hspace{1cm}L_1=L_2 = 30 mm$
Elemental stiffness matrix,
$\hspace{0.6cm}k_1=\frac{AE}{L}$
$\begin{bmatrix}
\ 1 & -1 \\
\ -1 & 1 \\
\end{bmatrix} = 10^3$
$\begin{bmatrix}
\ 455 & -455 \\
\ -455 & 455\\
\end{bmatrix}$
$\hspace{0.6cm}k_2=\frac{AE}{L}$
$\begin{bmatrix}
\ 1 & -1 \\
\ -1 & 1 \\
\end{bmatrix} = 10^3$
$\begin{bmatrix}
\ 245 & -245 \\
\ -245 &245\\
\end{bmatrix}$
Global stiffness matrix,
$[k]= 10^3\begin{bmatrix}
\ 455 & -455 & 0 \\
\ -455 & 700 & -245 \\
\ 0 & -245 & 245 \\
\end{bmatrix}$
$\hspace{3cm}$[K][u]=[F]
$10^3$
$\begin{bmatrix}
\ 455 & -455 & 0 \\
\ -455 & 700 & -245 \\
\ 0 & -245 & 245 \\
\end{bmatrix}$
$\begin{Bmatrix}
\ u_1 \\
\ u_2 \\
\ u_3 \\
\end{Bmatrix}$=
$\begin{Bmatrix}
\ f_1 \\
\ f_2 \\
\ f_3 \\
\end{Bmatrix}$
Apply Boundary conditions ,
$\hspace{1cm}u_1 = 0 , f_3 = 0.5 \times10^3N$
$10^3$
$\begin{bmatrix}
\ 455 & -455 & 0 \\
\ -455 & 700 & -245 \\
\ 0 & -245 & 245 \\
\end{bmatrix}$
$\begin{Bmatrix}
\ u_1 \\
\ u_2 \\
\ u_3 \\
\end{Bmatrix}$=
$\begin{Bmatrix}
\ f_1 \\
\ 0 \\
\ 0.5 \\
\end{Bmatrix}\times 10^3$
$\hspace{2cm} \therefore 700u_2-245u_3=0$
$\hspace{2.2cm}-245u_2+245u_3 = 0.5$
$\hspace{2.6cm}\therefore u_2 = 0.0011mm$
$\hspace{2.9cm}u_3=0.00314mm$
Reaction , f_1 = 455u_1-455u_2
$\hspace{2.3cm}= 0- 455 \times (0.0011)$
$\hspace{2.3cm} = -0.5 KN$