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30 June, 07:49

A bar having a length of 5 in. and cross-sectional area of 0.7 in.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. 30 June, 10:11
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    Answer: 2,857,142.5lb/in²

    Explanation:

    The modulus of elasticity is defined as the ratio of tensile stress to tensile strain of a material.

    Mathematically,

    Modulus of elasticity = Tensile stress/tensile strain

    Tensile stress is the ratio of the force acting on a body to its unit cross sectional area.

    Stress = Force (F) / Area (A)

    Given force = 8000lb

    Area = 0.7in²

    Stress = 8000/0.7

    Stress = 11,428.57lb/in²

    Tensile strain is the ratio of the extension of a material to its original length.

    Strain = extension (e) / original length (Lo)

    Given extension = 0.002in

    Original length of the bar = 5in

    Strain = 0.002in/5in

    Strain = 0.004 (it has no unit)

    Therefore the modulus of elasticity = 11428.57/0.004

    = 2,857,142.5lb/in²
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