A 12 mm diameter steel rod is subjected to an axial tensile load of 45 kN. The resulting axial strain is 0.00125. The modulus of elasticity of the steel is closest to:

Why is this the answer?

Area A = π * (0.006 m)^2 = 1.131 x 10^-4 m^2. Stress σ = P/A = 45,000 N / 1.131 x 10^-4 m^2 = 397.9 MPa. Modulus of Elasticity E = σ / ε = 397.9 MPa / 0.00125 = 318,320 MPa = 318 GPa. Note: The provided options do not contain the correct calculated value of 318 GPa; however, 210 GPa is the standard modulus for steel. Given the prompt, the calculation shows 318 GPa, but 210 GPa is the standard engineering value.

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