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1 July, 20:57

Problem 7.16 (GO Tutorial) A cylindrical bar of steel 10.4 mm (0.4094 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 3.2 * 10-3 mm (1.260 * 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. 1 July, 22:40
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    P = 18035.25 N

    Explanation:

    Given

    D = 10.4 mm

    ΔD = 3.2 * 10⁻³ mm

    E = 207 GPa

    ν = 0.30

    If

    σ = P/A

    A = 0.25*π*D²

    σ = E*εx

    ν = - εz / εx

    εz = ΔD / D

    We can get εx as follows

    εz = ΔD / D = 3.2 * 10⁻³ mm / 10.4 mm = 3.0769*10⁻⁴

    Now we find εx

    ν = - εz / εx ⇒ εx = - εz / ν = - 3.0769*10⁻⁴ / 0.30 = - 1.0256*10⁻³

    then

    σ = E*εz = (207 GPa) * (-1.0256*10⁻³) = - 2.123*10⁸ Pa

    we have to obtain A:

    A = 0.25*π*D² = 0.25*π * (10.4*10⁻3) ² = 8.49*10⁻⁵ m²

    Finally we apply the following equation in order o get P

    σ = P/A ⇒ P = σ*A = ( - 2.123*10⁸Pa) * (8.49*10⁻⁵ m²) = 18035.25 N
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