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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.

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}