Now it is known that the post has been shortened by $1.08 \mathrm{mm}$.

How far downward did the rigid plate move at $\mathrm{D}$ to cause this amount of contraction in the post?

$$\begin{gathered} ∆\mathrm{L}=(-1,800 \mathrm{\mu \epsilon }) \frac{1 \mathrm{mm}/\mathrm{mm}}{1\times {10}^{6 }\mathrm{\mu \epsilon }}(600 \mathrm{mm}) \\ =-1.08 \mathrm{mm} \end{gathered}$$