Respuesta :
Answer:
27009.56 mm
Explanation:
Given:
Diameter of the aluminium alloy bar, d = 12.5 mm
Length of the bar, L = 27 m = 27 × 10³ mm
Tensile force, P = 3 KN = 3 × 10³ N
Elastic modulus of the bar, E = 69 GPa = 69 × 10³ N/mm²
Now,
for the uniaxial loading, the elongation or the change in length (δ) due to the applied load is given as:
[tex]\delta=\frac{PL}{AE}[/tex]
where, A is the area of the cross-section
[tex]A=\frac{\pi d^2}{4}[/tex]
or
[tex]A=\frac{\pi\times12.5^2}{4}[/tex]
or
A = 122.718 mm²
on substituting the respective values in the formula, we get
[tex]\delta=\frac{3\times10^3\times27\times10^3}{122.718\times69\times10^3}[/tex]
or
δ = 9.56 mm
Hence, the length after the force is applied = L + δ = 27000 + 9.56
= 27009.56 mm
