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?
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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
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