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1 April, 20:40

A solid circular rod that is 600 mm long and 20 mm in diameter is subjected to an axial force of P = 50 kN The elongation of the rod is Ii = l. 40 mm. and its diameter becomes d' = 19.9837 mm. Determine the modulus of elasticity and the modulus of rigidity of the material. Assume that the material does not yield.

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  1. 1 April, 20:54
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    a) V = 0.354

    b) G = 25.34 GPA

    Explanation:

    Solution:

    We first determine Modulus of Elasticity and Modulus of rigidity

    Elongation of rod ΔL = 1.4 mm

    Normal stress, δ = P/A

    Where P = Force acting on the cross-section

    A = Area of the cross-section

    Using Area, A = π/4 · d²

    = π/4 · (0.0020) ² = 3.14 * 10⁻⁴m²

    δ = 50/3.14 * 10⁻⁴ = 159.155 MPA

    E (long) = Δl/l = 1.4/600 = 2.33 * 10⁻³mm/mm

    Modulus of Elasticity Е = δ/ε

    = 159.155 * 10⁶/2.33 * 10⁻³ = 68.306 GPA

    Also final diameter d (f) = 19.9837 mm

    Initial diameter d (i) = 20 mm

    Poisson said that V = Е (elasticity) / Е (long)

    = - (19.9837 - 20 / 20)

    2.33 * 10⁻³

    = 0.354,

    ∴ v = 0.354

    Also G = Е/2. (1+V)

    = 68.306 * 10⁹ / 2. (1 + 0.354)

    = 25.34 GPA

    ⇒ G = 25.34 GPA
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