Question

A load of 500 kg hanging from a steel wire of length 3 meter and cross sectional area 0.2 centimeter square was found to stretch the wire by 4 millimeter. Calculate the stress strain and Y​

Answer 1

force applied by the load = 500 × 10 = 5000 N

area of cross - section of steel = 0.2 cm² = 0.00002 m²

length of steel wire = 3 m

∴ stress = force / area

σ  = 5000 / 0.00002 = 250000000 N/m²

∴ strain = change in dimension / original length

δ = 0.0004 / 3 = 0.00013333333 m

∵ γ  = stress / strain

γ = σ / δ = 250000000 / 0.00013333333 = 1.87

Answer 2

Stress is 250000000 N/m²

Strain is 0.00013333333 m

γ is 1.87

force applied by the load = 500 × 10 = 5000 N

area of cross - section of steel = 0.2 cm² = 0.00002 m²

length of steel wire = 3 m

Formula: stress = force / area

σ  = 5000 / 0.00002 = 250000000 N/m²

Formula: strain = change in dimension / original length

δ = 0.0004 / 3 = 0.00013333333 m

Formula: γ  = stress / strain

γ = σ / δ = 250000000 / 0.00013333333 = 1.87

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