Answer:
3.0772 m
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
[tex]\delta L=\frac{\textup{PL}}{\textup{AE}}[/tex]
Now,
Where, A is the area of cross-section
A = [tex]\frac{\pi}{4}d^2[/tex]
or
A = [tex]\frac{\pi}{4}\times0.02^2[/tex]
or
A = 0.000314 m²
L is the length of the member
on substituting the respective values, we get
[tex]0.0035=\frac{25000\times L}{0.000314\times70\times10^9}[/tex]
or
L = 3.0772 m
