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=(\pi* d^2)/(4)`

Put the value into the formula

`A=(\pi*40^2)/(4)`

`A=1256.6\ mm^2`

We need to calculate the area of cross section with hole

Using formula of area

`A=\pi*((d_(b)^2-d_(h)^(2)))/(4)`

Put the value into the formula

`A=\pi*((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 = (F* L_(1))/(A_(1)* E)+(F* L_(2))/(A_(2)* E)`

Where, F = load

L = length

A = area of cross section

E = modulus of elasticity

Put the value into the formula

`Total\ contraction=(180*10^3)/(85*10^(3))((500)/(1256.6)+(100)/(549.77))`

`Total\ contraction = 1.227\ mm^2`

Hence, The total contraction on the bar is 1.227 mm²