Answer:
The stretched length is 0.18816 in.
Explanation:
Given that,
Distance = 0.5 in
length = 8 ft
Tensile stress = 20 kpsi
We need to calculate the stretched length
Using formula of length
[tex]\Delta L=\dfrac{PL}{AE}[/tex]
[tex]\Delta L=\sigma\times\dfrac{L}{E}[/tex]
Where, [tex]\sigma[/tex] = Tensile stress
L = length
E = young's modulus of aluminum
Put the value into the formula
[tex]\Delta L= 20\times10^3\dfrac{8}{10.2\times10^{6}}[/tex]
[tex]\Delta L= 0.01568\ ft[/tex]
[tex]\Delta L=0.01568\times12[/tex]
[tex]\Delta L=0.18816\ in[/tex]
Hence, The stretched length is 0.18816 in.
