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

The total contraction of the bar is 1.228 mm

Step-by-step 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 = (P * L)/(A * E)`

Where:

δ = Total contraction of the bar

A₁ for the hole section =
`(1)/(4 )* (40^(2)-30^(2)) * \pi = 549.78 \, mm^2`

A₂ for the solid section =
`(1)/(4 )* 40^(2) * \pi = 1256.64 \, mm^2`

L₁ = 100 mm

L₂ = 600 - 100 = 500 mm

`\delta = (P)/(E) *\left ( (L_1)/(A _1) + (L_2)/(A _2) \right ) = (180000)/(85000) *\left ( (100)/(549.78) + (500)/(1256.64) \right ) = 1.228 \ mm`

The total contraction of the bar = 1.228 mm.