Answer:
[tex]\Delta L = 0.1883 inch[/tex]
Explanation:
Given data:
d = 0.5 inch
[tex]L_i = 8 ft[/tex]
[tex]\sigma_{allow} = 20 kpsi[/tex]
we know that change in length is given as
[tex]\Delta L = \frac{PL}{AE}[/tex]
[tex]= \frac{P}{A}\times \frac{L}{E}[/tex]
[tex]= \sigma_{allow} \times \frac{L}{E}[/tex]
modulus of elasticity E for aluminium alloy is [tex]10.2 \times 10^6 psi = 10.2 \times 10^3 kpsi[/tex]
[tex]\Delta L = \frac{20 \times 8}{10.2\times 10^3}[/tex]
[tex]\Delta L = 0.01569 ft[/tex]
[tex]\Delta L = 0.1883 inch[/tex]
