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10 June, 23:11

Consider a cylindrical nickel wire 2.3 mm in diameter and 2.6 * 104 mm long. Calculate its elongation when a load of 260 N is applied. Assume that the deformation is totally elastic and that the elastic modulus for nickel is 207 GPa (or 207 * 109 N/m2)

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  1. 11 June, 02:21
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    7.255 * 10 ⁻³ m

    Explanation:

    Using the formula

    elastic modulus = stress / strain

    strain = stress / elastic modulus

    stress = F / A

    A = πr² = 3.142 * (2.3 mm / 2 * (1 m / 1000 mm)) ² = 0.0000041553 m²

    Stress = 260 N / 0.0000041553 m² = 62570768.14 N/m²

    strain = 62570768.14 N/m² / 207 * 10 ⁹ N/m² = 3.023 * 10 ⁻⁴

    strain = ΔL / L

    where L = 2.6 * 10 ⁴ mm = 2.6 * 10 ⁴ mm * 1 m / 1000 mm = 26 m

    3.023 * 10 ⁻⁴ = ΔL / 26 m

    ΔL = 3.023 * 10 ⁻⁴ * 26 m = 7.86 * 10 ⁻³ m
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