A cylindrical specimen of some metal alloy having an elastic modulus of 124 GPa and an original cross-sectional diameter of 4.2 mm will experience only elastic deformation when a tensile load of 1810 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.46 mm.

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 for the deformation, we use the following relation;

= [ Δl × E × π × D² ] / 4F

so we substitute our values into the formula

= [ (0.46 × 10⁻³) × (124 × 10⁹) × π × (4.2 × 10⁻³)² ] / ( 4 × 1810 )

= 3161.025289 / 7240

= 0.4366 m

Therefore, the maximum length of the specimen before deformation is 0.4366 m

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