A rod has a radius of 10 mm is subjected to an axial load of 15 N such that the axial strain in the rod is ????௫ = 2.75*10-6, determine the

Question

A rod has a radius of 10 mm is subjected to an axial load of 15 N such that the axial strain in the rod is ????௫ = 2.75*10-6, determine the modulus of elasticity E and the change in the rod’s diameter. Let Poisson’s ratio ν = 0.23.

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MichaelMet 3 years 2021-08-27T16:28:31+00:00 1 Answers 40 views 0

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  1. Euphemia
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    2021-08-27T16:30:02+00:00
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    Answer:

    Knowing we only have one load applied in just one direction we have to use the Hooke’s law for one dimension

    ex = бx/E

    бx = Fx/A = Fx/πr^{2}

    Using both equation and solving for the modulus of elasticity E

    E = бx/ex = Fx / πr^{2}ex

    E = \frac{15}{pi (10 * 10^{-3})^{2} * 2.75 * 10^{-6} } = 17.368 * 10^{9} Pa = 17.4 GPa

    Apply the Hooke’s law for either y or z direction (circle will change in every direction) we can find the change in radius

    ey = \frac{1}{E} (бy – v (бx + бz)) = -\frac{v}{E}бx

    = \frac{vFx}{Epir^{2} } = \frac{0.23 * 15}{pi (10 * 10^{-3)^{2} } * 17.362 * 10^{9} } = -0.63 *10^{-6}

    Finally

    ey = Δr / r

    Δr = ey * r = 10 * -0.63* 10^{-6} mm = -6.3 * 10^{-6} mm

    Δd = 2Δr = -12.6 * 10^{-6} mm

    Explanation:

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