Question

An aluminium bar 600mm long with diameter 40mm has a hole drilled in the centre of the bar. The hole is 30mm in diameter and is 100mm long. If the modulus of elasticity for the aluminium is 85GN/m^2
Calculate the total contraction on the bar due to a compressive load of 180kn

Answer 1

Answer:

The total contraction of the bar is 1.228 mm

Explanation:

The parameters given are;

Length of aluminium bar = 600 mm

Diameter of aluminium bar = 40 mm

Diameter of hole in aluminium bar = 30 mm

Length of hole = 100 mm

Modulus of elasticity of aluminium = 85 GN/m²

Applied compressive load = 180 kN

$\delta = \dfrac{P \times L}{A \times E}$

Where:

δ = Total contraction of the bar

A₁ for the hole section = $\frac{1}{4 }\times (40^{2}-30^{2}) \times \pi = 549.78 \, mm^2$

A₂ for the solid section = $\frac{1}{4 }\times 40^{2} \times \pi = 1256.64 \, mm^2$

L₁ = 100 mm

L₂ = 600 - 100 = 500 mm

$\delta = \dfrac{P}{E} \times\left ( \dfrac{L_1}{A _1} + \dfrac{L_2}{A _2} \right ) = \dfrac{180000}{85000} \times\left ( \dfrac{100}{549.78} + \dfrac{500}{1256.64} \right ) = 1.228 \ mm$

The total contraction of the bar = 1.228 mm.