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Modulus of toughness of medium carbon steel

Short answer

Yield stress 415 N/mm², ultimate stress 498 N/mm² (20% higher), fracture strain 0.02. Modulus of toughness Ut ≈ average stress × fracture strain = (415+498)/2 × 0.02 = 9.13 N/mm² (9.13 MJ/m³).

The yield stress of a medium carbon steel is found to be 415 N/mm2^2. Its ultimate stress is 20% more than the yield stress. If the steel fractures at 2% of strain, find the modulus of toughness of the steel.

Given: σy=415 N/mm2\sigma_y = 415\ \text{N/mm}^2, σu=1.20×415=498 N/mm2\sigma_u = 1.20\times415 = 498\ \text{N/mm}^2, fracture strain εf=2%=0.02\varepsilon_f = 2\% = 0.02.

The modulus of toughness is the area under the stress-strain curve up to fracture. Taking it as the average of the yield and ultimate stress times the fracture strain:

Ut≈σy+σu2 εf=415+4982×0.02=456.5×0.02=9.13 N/mm2U_t \approx \frac{\sigma_y+\sigma_u}{2}\,\varepsilon_f = \frac{415+498}{2}\times0.02 = 456.5\times0.02 = 9.13\ \text{N/mm}^2

Answer: Modulus of toughness ≈ 9.13 N·mm/mm³ = 9.13 MJ/m³.

✓ Answer checked by Claude Opus 5.5 (high effort)

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