Question

A cylindrical specimen of copper having an original diameter of 15.2mm is pulled in tension with 9,500 N force, producing only elastic deformation.what is the resulting engineering stress in the specimen?​

Answer 1

The stress in the specimen will be 52.4 x 10⁶

Explanation :

Given diameter of the cylinder = 15.2 mm

=> Radius of the cylinder = 15.2/2 = 7.6 mm = 7.6 x 10⁻³ m

Area of the cylinder = πr² = 3.14 x (7.6 x 10⁻³)²

=> A= 181.3 x 10⁻⁶ m²

Given force applied, F = 9500 N

we know that stress is given as the ratio of force and area,

S = F/A

= 9500/(181.3 x 10⁻⁶)

= 52.4 x 10⁶ N/m²

Hence the stress in the specimen will be 52.4 x 10⁶

I hope this answer helps you, please feel free to ask any doubt regarding this in the comment section.