Answer:
[tex]L_i = 130.44 mm[/tex]
Explanation:
Given data:
original diameter = 17.847mm
final diameter = 17.869 mm
final length = 75mm
elastic modulus = 109 GPa
shear modulus = 40 GPa
we know
[tex]E = 2G( 1 + \nu)[/tex]
solving for Poisson's ratio
[tex]\nu = \frac{E}[2G} - 1[/tex]
[tex]\nu = \frac{109}{2\times 40} -1 = 0.362[/tex]
we know that Poisson ration is given as
[tex]\nu = \rac{\frac{d_f -d_i}[d_i}}{\frac{L_f -L_i}[L_i}}[/tex]
[tex]0.362 = - \frac{\frac{17.869 - 17.847}{17.847}}{\frac{75 - L_i}{L-i}}[/tex]
solving initial length
[tex]L_i = 130.44 mm[/tex]
