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3 August, 03:18

Consider a cylindrical titanium wire 3.0 mm (0.12 in.) in diameter and 2.5 Ã 104 mm (1000 in.) long. calculate its elongation when a load of 500 n (112 lbf) is applied. assume that the deformation is totally elastic.

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  1. 3 August, 03:34
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    Given:

    Diameter, d of the titanium wire = 3.0 mm = 0.003 m

    Length, L = 104 mm = 0.0104 m

    Force applied F = 500 N

    To determine:

    The deformation, i. e. change in length ΔL

    Explanation:

    Stress on the wire = Force/Area

    F = 500 N

    A = π (d/2) ² = π (0.003/2) ² = 7.069*10⁻⁶ m²

    Stress = 500 N/7.069*10⁻⁶ m² = 70.731*10⁶ N/m2 = 0.0707 GPa

    Now, the Young's modulus for Ti = 107 GPa

    Young's modulus = Stress/Strain

    Strain = 0.0707/107 = 0.000660

    Now,

    Strain = ΔL/L

    ΔL = 0.000660*0.0104 = 6.87 * 10⁻⁶ m

    Ans: The elongation in the Ti wire would be 6.87 * 10⁻⁶ m
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