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.

The total contraction on the bar is 1.2277 mm

Note: Kindly find an attached copy of part of the solution given below

Explanation:

Solution

Recall that:

The giving data for aluminum bar is sated below:

The total length = 600 mm

The diameter = 40 mm

The hole = 30 mm

The length of the hole = 100 mm

The elasticity (E) = 85 * 10^9 GPa

= 85 * 10^3 N /mm²

Now,

To determine the contraction we apply he following formula given below:

δL = Load *L/A * E

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explanation:

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