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.

Question

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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Xavia 4 years 2021-08-28T00:12:28+00:00 1 Answers 9 views 0

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  1. hongcuc2
    0
    2021-08-28T00:13:35+00:00
    Reply

    Answer:

    Explanation:

    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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