$$ \begin{array}{l}{\text { Data: } \mathrm{d}=20 \mathrm{mm}, \mathrm{L}=200 \mathrm{mm}, \delta \mathrm{L}=0.05 \mathrm{mm}, \delta \mathrm{b}=0.0025 \mathrm{mm}} \ {\text { Find: } \mu, \mathrm{G}} \ {\text { Calculate } \mu :} \ {\mu=\frac{\text {Lateral Strain }}{\text { Linear Strain }}} \ {\mu=\frac{(\delta d / d)}{\left(\delta_{L} / L\right)}} \ {\mu=\frac{(0.0025 / 20)}{(0.05 / 200)}} \ {\mu=0.5}\end{array} $$ $$ \begin{array}{l}{\text { Calculate E: }} \ {\qquad \begin{array}{rl}{\delta l=} & {\frac{P L}{A E}} \ {E=} & {\frac{P L}{A \times \delta L}=\frac{45 \times 10^{3}}{\frac{\pi}{4} \times 20^{2} \times 0.05}=572.957 \times 10^{3} \mathrm{N} / \mathrm{mm}^{2}} \ {\mathrm{E}=572.957 \times 10^{3}} & {\mathrm{N} / \mathrm{mm}^{2}}\end{array}}\end{array} $$ $$ \begin{array}{l}{\text { Calculate G: }} \ {\begin{array}{l}{E=2 G(1+\mu)} \ {572.957 \times 10^{3}=2 G(1+0.5)} \ {G=190.985 \times 10^{3} \mathrm{N} / \mathrm{mm}^{2}}\end{array}}\end{array} $$