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Given:

• Lengths are given in figure.

The equation for deflection at point B due to point A can be given by,

$\begin{array}{rcl}{t}_{B/A }& =& \frac{1}{2}\left(\frac{-PL}{2EI}\right)\left(\frac{L}{2}\right)\left(\frac{L}{6}\right)\\ & =& \frac{-P{L}^3}{48EI}\end{array}$

The equation for deflection at point C due to point A can be given by,

$\begin{array}{rcl}{t}_{C/A }& =& \frac{1}{2}\left(\frac{-PL}{2EI}\right)\left(\frac{L}{2}\right)\left(L\right)\\ & =& \frac{-P{L}^3}{8EI}\end{array}$

The equation for net deflection at point C can be given by,

${\Delta }_C=\left|t_{C/A}\right|-\left(\frac{L}{{\displaystyle \frac{L}{2}}}\right)\left|t_{B/A}\right|$

Substitute  the required values in the above equation to obtain the net deflection at C.

$$\begin{gathered} {\Delta }_C=\left|\begin{array}{rcl}& & \frac{-P{L}^3}{8EI}\end{array}\right|-\left(2\right)\left|\begin{array}{rcl}& & \frac{-P{L}^3}{48EI}\end{array}\right| \\ {\Delta }_C=\begin{array}{rcl}& & \frac{P{L}^3}{8EI}\end{array}-\begin{array}{rcl}& & \frac{P{L}^3}{24EI}\end{array} \\ {\Delta }_C=\begin{array}{rcl}& & \frac{P{L}^3}{12EI}\end{array} \end{gathered}$$

Therefore, the deflection at point C is $\begin{array}{rcl}& & \frac{P{L}^3}{12EI}\end{array}$.