An 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 the modulus of elasticity for the aluminium is 85GN/m2, calculate the total contraction on the bar due to a compressive load of 180kN.​

Answer:

[tex]\delta = 0.385\,m[/tex] (Compression)

Explanation:

The aluminium bar is experimenting a compression due to an axial force, that is, a force exerted on the bar in its axial direction. (See attachment for further details) Under the assumption of small strain, the deformation experimented by the bar is equal to:

[tex]\delta = \frac{P\cdot L}{A \cdot E}[/tex]

Where:

[tex]P[/tex] - Load experimented by the bar, measured in newtons.

[tex]L[/tex] - Length of the bar, measured in meters.

[tex]A[/tex] - Cross section area of the bar, measured in square meters.

[tex]E[/tex] - Elasticity module, also known as Young's Module, measured in pascals, that is, newtons per square meter.

The cross section area of the bar is now computed: ([tex]D_{o} = 0.04\,m[/tex], [tex]D_{i} = 0.03\,m[/tex])

[tex]A = \frac{\pi}{4}\cdot (D_{o}^{2}-D_{i}^{2})[/tex]

Where:

[tex]D_{o}[/tex] - Outer diameter, measured in meters.

[tex]D_{i}[/tex] - Inner diameter, measured in meters.

[tex]A = \frac{\pi}{4}\cdot [(0.04\,m)^{2}-(0.03\,m)^{2}][/tex]

[tex]A = 5.498 \times 10^{-4}\,m^{2}[/tex]

The total contraction of the bar due to compresive load is: ([tex]P = -180\times 10^{3}\,N[/tex], [tex]L = 0.1\,m[/tex], [tex]E = 85\times 10^{9}\,Pa[/tex], [tex]A = 5.498 \times 10^{-4}\,m^{2}[/tex]) (Note: The negative sign in the load input means the existence of compressive load)

[tex]\delta = \frac{(-180\times 10^{3}\,N)\cdot (0.1\,m)}{(5.498\times 10^{-4}\,m^{2})\cdot (85\times 10^{9}\,Pa)}[/tex]

[tex]\delta = -3.852\times 10^{-4}\,m[/tex]

[tex]\delta = -0.385\,mm[/tex]

[tex]\delta = 0.385\,m[/tex] (Compression)

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