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.

(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:

Where:

- Load experimented by the bar, measured in newtons.

- Length of the bar, measured in meters.

- Cross section area of the bar, measured in square meters.

- 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: (, )

Where:

- Outer diameter, measured in meters.

- Inner diameter, measured in meters.

The total contraction of the bar due to compresive load is: (, , , ) (Note: The negative sign in the load input means the existence of compressive load)

(Compression)

a.   mass number from atomic number