The decrease in length is given as 0.075 cm and increase in width is 0.003 cm.

Given - L = 300 mm , b = 40mm, d = 40mm, $P = 400 \times 10^{3}N$, $\delta L = 0.075 mm$ , $\delta b = 0.03mm $

Solution:

Stress = $\delta$ = $\frac{P}{A}$ = $ \frac{400 \times 10^{3}}{40 \times 40}$ = $250 N/mm^{2}$

Strain = $e$ = $\frac{\delta L}{L}$ = $ \frac{0.75}{300}$ = $2.5 \times 10^{-3}$ Lateral Strain = $e_{lat}$ = $\frac{\delta d}{d}$ = $ \frac{0.03}{40}$ = $7.5 \times 10^{-4}$ Young's modulus = $E$ = $\frac{\delta }{e}$ = $ \frac{250}{2.5 \times 10^{-3}}$ = $1 \times 10^{5}$ Poisson's ratio = $\mu$ = $\frac{e_{lat}}{e_{lin}}$ = $ \frac{7.5 \times 10^{-4}}{2.5 \times 10^{-3}}$ = $0.3$

Determine the young’s modulus and Possion’s ratio of a metallic bar of length 25cm breadth 3cm depth 2cm when the beam is subjected to an axial compressive load 240KN. The decrease in length is given by 0.05cm and increase in breadth 0.002

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