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20 August, 07:46

A bar having a length of 5 in. and cross-sectional area of 0.7 i n. 2 is subjected to an axial force of 8000 lb. If the bar stretches 0.002 in., determine the modulus of elasticity of the material. The material has linear elastic behavior.

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  1. 20 August, 09:18
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    E=1.969 * 10¹¹ Pa

    Explanation:

    The formula to apply is;

    E=F*L/A*ΔL

    where

    E=Young modulus of elasticity

    F=Force in newtons

    L=Original length in meters, m

    A=area in square meters m²

    ΔL = Change in length in meters, m

    Given

    F = 8000 lb = 8000*4.448 = 35584 N

    L = 5 in = 0.127 m

    A = 0.7 in² = 0.0004516 m²

    ΔL = 0.002 in = 5.08e-5 m

    Applying the formula

    E = (35584 * 0.127) / (0.0004516*5.08e-5)

    E=1.969 * 10¹¹ Pa
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