Question

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

Answer 1

Stress is difined as

Stress = External restoring force acting on per unit area, N/mm²

Given force F = 9500 N

Specimen is copper, with diameter d = 15.2 mm

As the specimen is in tension, the area onwhich the stress can be calculated is its cross-sectional area.

∴ Cross-sectional area of cylinder A = (π/4) × d²

∴ A = (3.14÷4) × 15.2²

∴ A = 181.3664 mm²  

∴ Stress = Restoring force / Area

as every action have equal and opposite reaction, the force applied will produce stress with equal restoring force.

∴ Stress = 9500 / 181.3664

∴ Stress = 52.38 N/mm²

Resulting Engineering stress on the specimen = 52.38 N/mm²