Question

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 1

$$\frac{\Delta l_1}{L-l}=\frac{4F}{E\pi D^2}\\\frac{\Delta l_1}{600-100}=\frac{4(180\cdot10^3)}{(85\cdot10^9)\pi (0.04)^2}\\\Delta l_1=0.843\ mm$$

$$\frac{\Delta l_2}{l}=\frac{4F}{E\pi (D^2-d^2)}\\\frac{\Delta l_2}{100}=\frac{4(180\cdot10^3)}{(85\cdot10^9)\pi ((0.04)^2-(0.03)^2)}\\\Delta l_2=0.385\ mm$$

$$\Delta l=\Delta l_1+\Delta l_2=0.843+0.385=1.23\ mm$$