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4 August, 21:02

n aluminium bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 30mm in diameter and is 100mm long. If modulus of elasticity for the aluminium is 85GN/m2, calculate the total contraction on the bar due to a compressive load of 180kN

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  1. 5 August, 00:52
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    It is to be noted here that the aluminium bar having some hole will not have any effect on the creation of strain in it due to stress.

    length of bar L = 600 mm

    =.6 m

    cross sectional area = 3.14 x. 02²

    =.001256 m²

    Compressive stress created = 180 x 10³ /.001256

    = 143312.1 x 10³ N / m²

    Modulus of elasticity = stress / strain

    85 x 10⁹ = 143312.1 x 10³ / l / L

    l / L = 143312.1 x 10³ / 85 X 10⁹

    = 1686.02 x 10⁻⁶

    l = L x 1686.02 x 10⁻⁶

    = 600 x 10⁻³ x 1686.02 x 10⁻⁶

    =.6 x 1686.02 x 10⁻⁶

    = 1011.61 x 10⁻⁶

    1.011 x 10⁻³ m

    1.011 mm.
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