How calculate this 25 mm diameter bar when subjected to a force of 40 kN has an extension of 0.08 mm

on a gauge length of 200 mm. If the diametrical reduction is 0.003 mm, find the values of

E, G, K, Poisson's ratio.

Poission Ration, $v=\frac{\text { Lateral strain }}{\text { Long. strain }}$ $$ =\frac{0.003 / 25}{0.08 / 200}=\frac{0.003}{25} \times \frac{200}{0.08} $$ As per hooke's Law $$ \begin{aligned} \sigma &=\varepsilon \times E \\ \Rightarrow \quad E &=\frac{(40 \times 103) / \frac{\pi}{4}(25)^{2}}{0.08 / 200} \\ E &=203.7184 \mathrm{~GPa} \end{aligned} $$ $$ \begin{aligned} E &=3 K(1-2v) \\ \Rightarrow K &=\frac{203.718}{3 \times(1-0.6)}=169.765\mathrm{GPa} \\ \end{aligned} $$ Using Relation $$ \Rightarrow \quad \begin{aligned} E &=2G(1+v) \\ G &=\frac{203.718}{2 \times 1.3}=78.354\mathrm{GPa}\\ G &=78.35GPa \end{aligned} $$