A cylindrical steel rod with a diameter of 20 mm is subjected to a tensile load of 40 kN. The material has a yield strength of 450 MPa. Assuming the deformation remains within the elastic limit, calculate the factor of safety based on the ultimate tensile strength and the yield strength. If the modulus of elasticity of the steel is 200 GPa, determine the elongation of the rod over a length of 500 mm under this load. Which of the following options correctly represents the factor of safety based on yield strength and the elongation?
Factor of Safety (Yield) = 2.25, Elongation = 0.318 mm
This problem combines two basic strength-of-materials ideas: direct (axial) stress and factor of safety, and axial elongation under load. The rod carries a purely tensile load, so the stress is uniform across the cross-section.
Given: diameter d = 20 mm, load F = 40 kN = 40000 N, yield strength σy = 450 MPa, modulus E = 200 GPa = 200000 MPa, length L = 500 mm.
Step 1 — Cross-sectional area:
Step 2 — Working (applied) stress:
Step 3 — Factor of safety on yield strength:
Step 4 — Elongation (from Hooke's law, δ = FL/AE):
Both computed values — FoS ≈ 2.25 and δ ≈ 0.318 mm — match the choice giving Factor of Safety (Yield) = 2.25, Elongation = 0.318 mm.
Why the other options fail: the pair listing FoS = 3.58 uses an incorrect (too large) safety factor that does not follow from 450/127.3, so both variants carrying 3.58 are ruled out. The option pairing the correct 2.25 with an elongation of 0.5 mm is inconsistent, because substituting the given F, L, A and E into δ = FL/AE yields 0.318 mm, not 0.5 mm — an elongation of 0.5 mm would require a much larger load or a smaller area/modulus.
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