An aluminum bar 600mm long with diameter 40mm, has a hole drilled in the center of the bsr.The hole is 30mm in diameter and is 100mm long. I

Question

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

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bonexptip 4 years 2021-08-20T19:35:37+00:00 1 Answers 6 views 0

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  1. Orla
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    2021-08-20T19:37:17+00:00
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    Answer:

    1.228 x 10^{-6} mm

    Explanation:

    diameter of aluminium bar D = 40 mm  

    radius of aluminium bar R = D/2 = 40/2 = 20 mm

    diameter of hole d = 30 mm

    radius of the hole r = d/2 = 30/2 = 15 mm

    compressive Load F = 180 kN = 180 x 10^{3} N

    modulus of elasticity E = 85 GN/m^2  = 85 x 10^{9} Pa

    length of bar L = 600 mm

    length of hole = 100 mm

    true length of bar = 600 – 100 = 500 mm

    area of the bar A = \pi r^{2} = 3.142 x 20^{2} = 1256.8 mm^2

    area of hole a = \pi (R^{2} - r^{2}) = 3.142* (20^{2} - 15^{2}) = 549.85 mm^2

    Total contraction of the bar = \frac{F*L}{AE} + \frac{Fl}{aE}

    total contraction = \frac{F}{E} * (\frac{L}{A} +\frac{l}{a})

    ==> \frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85}) = 1.228 x 10^{-6} mm

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