Answer:
the maximum length of the specimen before deformation is 0.4366 m
Explanation:
Given the data in the question;
Elastic modulus E = 124 GPa = 124 × 10⁹ Nm⁻²
cross-sectional diameter D = 4.2 mm = 4.2 × 10⁻³ m
tensile load F = 1810 N
maximum allowable elongation Δl = 0.46 mm = 0.46 × 10⁻³ m
Now to calculate the maximum length [tex]l[/tex] for the deformation, we use the following relation;
[tex]l[/tex] = [ Δl × E × π × D² ] / 4F
so we substitute our values into the formula
[tex]l[/tex] = [ (0.46 × 10⁻³) × (124 × 10⁹) × π × (4.2 × 10⁻³)² ] / ( 4 × 1810 )
[tex]l[/tex] = 3161.025289 / 7240
[tex]l[/tex] = 0.4366 m
Therefore, the maximum length of the specimen before deformation is 0.4366 m
