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1 December, 09:04

A bar has a length of 8 in. and cross-sectional area of 12 in2. Part A Determine the modulus of elasticity of the material if it is subjected to an axial tensile load of 10 kip and stretches 0.006 in. The material has linear-elastic behavior.

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  1. 1 December, 09:34
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    Answer: modulus elasticity = 4.9 * 10^6 N/in²

    Explanation: modulus elasticity = tensile stress / tensile strain.

    Tensile stress = force / area.

    Force = 10kip = 4.44 * 10^4N (note that 1kip = 4448.22N), area = 12 in²

    Tensile stress = 4.44 * 10^4 / 12 = 3.7 * 10³ N/in²

    Tensile strain = extension / length

    extension = 0.006in length = 8 in

    Tensile strain = 0.006 / 8 = 7.5 * 10^-4.

    Modulus = stress / strain = 3.7 * 10³ / 7.5 * 10^-4

    Modulus = 0.49 * 10^7

    modulus = 4.9 * 10^6 N/in²
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