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27 November, 02:08

A cylindrical bar of metal having a diameter of 19.7 mm and a length of 206 mm is deformed elastically in tension with a force of 47500 N. Given that the elastic modulus and Poisson's ratio of the metal are 65.1 GPa and 0.35, respectively, determine the following: (a) The amount by which this specimen will elongate in the direction of the applied stress. (b) The change in diameter of the specimen. Indicate an increase in diameter with a positive number and a decrease with a negative number.

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  1. 27 November, 05:02
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    a) ΔL = 4.93*10⁻⁴ m = 0.4931 mm

    b) Δd = 0.0165 mm

    Explanation:

    d = 19.7 mm = 0.0197 m

    L = 206 mm = 0.206 m

    P = 47500 N

    E = 65.1 GPa = 65.1*10⁹ Pa

    ν = 0.35

    a) ΔL = ?

    b) Δd = ?

    Solution:

    We get A as follows

    A = 0.25*π*d² = 0.25*π * (0.0197 m) ² = 3.048*10⁻⁴ m²

    Then we can apply the following equation (Hooke's Law):

    ΔL = P*L / (A*E)

    ΔL = (47500 N) * (0.206 m) / (3.048*10⁻⁴ m²*65.1*10⁹ Pa)

    ΔL = 4.93*10⁻⁴ m = 0.4931 mm

    εa = ΔL / L = 0.4931 mm / 206 mm = 2.39*10⁻³

    Now, we can use the formula

    ν = εl / εa ⇒ εl = ν*εa = 0.35*2.39*10⁻³

    ⇒ εl = 8.378*10⁻⁴

    If we know that

    εl = Δd / d ⇒ Δd = εl*d

    ⇒ Δd = 8.378*10⁻⁴*0.0197 m = 1.65*10⁻⁵ m = 0.0165 mm
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