A metal bar of 50 mm × 50 mm section is subjected to a tensile load of 400 kN. The elongation of a 200 mm gauge length is found to be 0.2 mm, and the decrease in thickness is 0.012 mm. The value of Poisson's ratio is:
Calculation of Poisson's Ratio Poisson's ratio is determined by the ratio of lateral strain to axial strain. Step 1: Calculate the axial strain The axial strain (\(\epsilon _{\text{axial}}\)) is calculated by dividing the elongation by the original gauge length. \(\epsilon _{\text{axial}}=\frac{\text{elongation}}{\text{original\ gauge\ length}}=\frac{0.2\text{\ mm}}{200\text{\ mm}}=0.001\) Step 2: Calculate the lateral strain The lateral strain (\(\epsilon _{\text{lateral}}\)) is calculated by dividing the decrease in thickness by the original thickness. \(\epsilon _{\text{lateral}}=\frac{\text{decrease\ in\ thickness}}{\text{original\ thickness}}=\frac{0.012\text{\ mm}}{50\text{\ mm}}=0.00024\) Step 3: Calculate Poisson's ratio Poisson's ratio (\(\nu \)) is calculated as the negative ratio of lateral strain to axial strain. \(\nu =-\frac{\epsilon _{\text{lateral}}}{\epsilon _{\text{axial}}}=-\frac{-0.00024}{0.001}=0.24\) Final Answer The value of Poisson's ratio is \(0.24\).