SIGMA = \frac{P}{A}; \epsilon = \frac{{{\rm{\DELTA }}l}}{l}\) Young’s modulus is given by\(E = \frac{\sigma }{\epsilon} = \frac{{PL}}{{A{\rm{\Delta }}l}}\) Calculation:Given l = 2M, d = 50 mm, Δl = 5 mm, P = 400 kN;\(A = \frac{\pi }{4}{d^2} = \frac{\pi }{4}{50^2} = 1963.495\;m{m^2}\) \(E = \frac{{400\; \times \;{{10}^3} \;\times \;2000}}{{1963.495\; \times\; 5}} = 81.487\;GPa\) Nearest option is 82 GPa

"> SIGMA = \frac{P}{A}; \epsilon = \frac{{{\rm{\DELTA }}l}}{l}\) Young’s modulus is given by\(E = \frac{\sigma }{\epsilon} = \frac{{PL}}{{A{\rm{\Delta }}l}}\) Calculation:Given l = 2M, d = 50 mm, Δl = 5 mm, P = 400 kN;\(A = \frac{\pi }{4}{d^2} = \frac{\pi }{4}{50^2} = 1963.495\;m{m^2}\) \(E = \frac{{400\; \times \;{{10}^3} \;\times \;2000}}{{1963.495\; \times\; 5}} = 81.487\;GPa\) Nearest option is 82 GPa

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A rod of length 2 m and diameter 50 mm is elongated by 5 mm when an axial force of 400 kN is applied. The modulus of elasticity of the material of the rod will be nearly

Engineering Physics Elastic Limit in Engineering Physics 9 months ago

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Concept:The stress and strain under uniaxial load are given by\(\SIGMA = \frac{P}{A}; \epsilon = \frac{{{\rm{\DELTA }}l}}{l}\) Young’s modulus is given by\(E = \frac{\sigma }{\epsilon} = \frac{{PL}}{{A{\rm{\Delta }}l}}\) Calculation:Given l = 2M, d = 50 mm, Δl = 5 mm, P = 400 kN;\(A = \frac{\pi }{4}{d^2} = \frac{\pi }{4}{50^2} = 1963.495\;m{m^2}\) \(E = \frac{{400\; \times \;{{10}^3} \;\times \;2000}}{{1963.495\; \times\; 5}} = 81.487\;GPa\) Nearest option is 82 GPa

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Posted on 23 Oct 2024, this text provides information on Engineering Physics related to Elastic Limit in Engineering Physics. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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