Question

An aluminium bar 600mm long with diameter 40mm has a hole drilled in the centre of the bar.The hole is 3omm 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 compresive load of 180kn

Answer 1

Given that,

Length of bar = 600 mm

Diameter of bar = 40 mm

Diameter of hole = 30 mm

Length of hole = 100 mm

Modulus of elasticity = 85 GN/m²

Load = 180 kN

We need to calculate the area of cross section without hole

Using formula of area

$A=\dfrac{\pi\times d^2}{4}$

Put the value into the formula

$A=\dfrac{\pi\times40^2}{4}$

$A=1256.6\ mm^2$

We need to calculate the area of cross section with hole

Using formula of area

$A=\pi\times\dfrac{(d_{b}^2-d_{h}^{2})}{4}$

Put the value into the formula

$A=\pi\times\dfrac{(40^2-30^2)}{4}$

$A=549.77\ mm^2$

We need to calculate the total contraction on the bar

Using formula of total contraction

Total contraction = contraction in bar without hole part + contraction in bar with hole part

$Total\ contraction = \dfrac{F\times L_{1}}{A_{1}\times E}+\dfrac{F\times L_{2}}{A_{2}\times E}$

Where, F = load

L = length

A = area of cross section

E = modulus of elasticity

Put the value into the formula

$Total\ contraction=\dfrac{180\times10^3}{85\times10^{3}}(\dfrac{500}{1256.6}+\dfrac{100}{549.77})$

$Total\ contraction = 1.227\ mm^2$

Hence, The total contraction on the bar is 1.227 mm²