2G (1 + μ),E = 3K (1 – 2μ)Where E = Young’s modulus, G = Modulus of RIGIDITY, μ = Poisson ratio, K = Bulk modulus of elasticity.Calculation:Given:E = 1.25 × 105 N/mm2, μ = 0.25 E = 3K (1 – 2μ)\({\rm{K}} = \frac{{\rm{E}}}{{3\left( {1 - 2{\rm{\mu }}} \RIGHT)}} = \frac{{1.25\times 10^5}}{{3\left( {1 - 2 \times 0.25} \right)}} = 0.83\times 10^5\;{{N}}/{{{mm}}^2}\)The bulk modulus will be 0.83 × 105 N/mm2.

"> 2G (1 + μ),E = 3K (1 – 2μ)Where E = Young’s modulus, G = Modulus of RIGIDITY, μ = Poisson ratio, K = Bulk modulus of elasticity.Calculation:Given:E = 1.25 × 105 N/mm2, μ = 0.25 E = 3K (1 – 2μ)\({\rm{K}} = \frac{{\rm{E}}}{{3\left( {1 - 2{\rm{\mu }}} \RIGHT)}} = \frac{{1.25\times 10^5}}{{3\left( {1 - 2 \times 0.25} \right)}} = 0.83\times 10^5\;{{N}}/{{{mm}}^2}\)The bulk modulus will be 0.83 × 105 N/mm2.

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A material has a young’s modulus of 1.25 × 105 N/mm2 and a Poisson’s ratio of 0.25. The bulk modulus of the material will be

Engineering Physics Elastic Limit in Engineering Physics 8 months ago

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Concept:The relation between E, G, μ, and k will be,E = 2G (1 + μ),E = 3K (1 – 2μ)Where E = Young’s modulus, G = Modulus of RIGIDITY, μ = Poisson ratio, K = Bulk modulus of elasticity.Calculation:Given:E = 1.25 × 105 N/mm2, μ = 0.25 E = 3K (1 – 2μ)\({\rm{K}} = \frac{{\rm{E}}}{{3\left( {1 - 2{\rm{\mu }}} \RIGHT)}} = \frac{{1.25\times 10^5}}{{3\left( {1 - 2 \times 0.25} \right)}} = 0.83\times 10^5\;{{N}}/{{{mm}}^2}\)The bulk modulus will be 0.83 × 105 N/mm2.

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Posted on 11 Nov 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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