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15 November, 15:21

A cylindrical bar of steel 10.1 mm (0.3976 in.) in diameter is to be deformed elastically by application of a force along the bar axis. Determine the force that will produce an elastic reduction of 2.8 * 10-3 mm (1.102 * 10-4 in.) in the diameter. For steel, values for the elastic modulus (E) and Poisson's ratio (ν) are, respectively, 207 GPa and 0.30.

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  1. 15 November, 16:58
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    given d = 10.2x10-3m

    change in diameter d' = 3.4x10-6 m

    elastic modulus=207x109pa

    1/m=0.30

    we know that stress/strain=E

    but poissons ratio 1/m = lateral starin/longitudinal starin

    0.3 = (d'/d) / (L'/L)

    L'/L = 3.4x10-6 / (0.3x10.2x10-3)

    = 1.11*10-3

    E = (f/A) / (L'/L)

    force f = E*A * (L'/L)

    f = (1.11*10-3) * 207*109 * (3.1415 * (10.2*10-3) 2) / 4

    = 18775.2N = 18.775KN

    Explanation:

    The force that produce an elastic reduction is 18.775KN
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