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Stress, strain, elongation of steel rod

Short answer

Stress σ = P/A = 1500/4 = 375 kg/cm² (≈ 36.8 N/mm²). Strain ε = σ/E = 375/(2.1×10⁶) = 1.786×10⁻⁴. Elongation δ = ε·L = 1.786×10⁻⁴ × 300 cm = 0.0536 cm (0.536 mm).

A mild steel rod 3 m long having a cross-sectional area of 4 cm2^2 is subjected to an axial pull of 1500 kg. If EE for steel is 2.1×1062.1\times10^{6} kg/cm2^2, find stress, strain and elongation of the rod.

Given: L=3L = 3 m =300= 300 cm, A=4 cm2A = 4\ \text{cm}^2, P=1500P = 1500 kg, E=2.1×106 kg/cm2E = 2.1\times10^6\ \text{kg/cm}^2.

Stress σ=PA=15004=375 kg/cm2Strain ε=σE=3752.1×106=1.786×10−4Elongation δ=εL=1.786×10−4×300=0.0536 cm\begin{aligned} \text{Stress } \sigma &= \frac{P}{A} = \frac{1500}{4} = 375\ \text{kg/cm}^2\\ \text{Strain } \varepsilon &= \frac{\sigma}{E} = \frac{375}{2.1\times10^{6}} = 1.786\times10^{-4}\\ \text{Elongation } \delta &= \varepsilon L = 1.786\times10^{-4}\times300 = 0.0536\ \text{cm} \end{aligned}

Answer: Stress = 375 kg/cm² (about 36.8 N/mm²); strain = 1.786 × 10⁻⁴; elongation = 0.0536 cm = 0.536 mm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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