Question

An aluminum rod if 20 mm diameter iselongated 3.5 mm along its

longitudinal direction by a load of 25KN. If the modulus of
elasticity of aluminum is E = 70 GPa,determine the original length
of the bar.

Answer 1

3.0772 m

Step-by-step explanation:

Given:

Diameter of the aluminium rod, d = 20 mm = 0.02 m

Length of elongation, δL = 3.5 mm = 0.0035 m

Applied load, P = 25 KN = 25000 N

Modulus of elasticity, E = 70 GPa = 70 × 10⁹ N/m²

Now,

we have the relation

`\delta L=\frac{\textup{PL}}{\textup{AE}}`

Now,

Where, A is the area of cross-section

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

or

A =
`(\pi)/(4)*0.02^2`

or

A = 0.000314 m²

L is the length of the member

on substituting the respective values, we get

`0.0035=(25000* L)/(0.000314*70*10^9)`

or

L = 3.0772 m