Axial stress and elongation of a steel rod

δ = PL/(AE). Keep units consistent: N and mm give MPa.

Worked answer.

Step 1: Cross-sectional area.
A = (π/4) d² = (π/4) × 20² = 314.16 mm².

Step 2: Stress.
σ = P / A = 50,000 N / 314.16 mm² ≈ 159.2 MPa (1 N/mm² = 1 MPa).

Step 3: Strain.
ε = σ / E = 159.2 / 200,000 ≈ 7.96 × 10⁻⁴ (dimensionless).

Step 4: Elongation.
δ = ε × L = 7.96 × 10⁻⁴ × 2000 mm ≈ 1.59 mm.

Using the direct formula: δ = PL/(AE) = (50,000 × 2000) / (314.16 × 200,000) ≈ 1.59 mm. The two methods agree.

Step 5: Safety check. Allowable stress = yield strength / FoS = 250 / 1.5 ≈ 166.7 MPa. The working stress of 159.2 MPa is below this, so the rod is safe, though with little margin. The actual factor of safety is 250 / 159.2 ≈ 1.57.

Check. The stress is well below yield, so the material is in the elastic range and using Hooke's law (σ = Eε) is valid.

Units tip. Working in N and mm keeps stress in MPa and E as 200,000 MPa, avoiding powers of ten. Write E = 200 GPa = 200,000 N/mm².

Common mistake. Using the radius instead of the diameter in (π/4)d², or mixing metres and millimetres, which gives an elongation off by a factor of 1000.

Likely follow-up. "What diameter would give exactly the allowable stress?" A = 50,000 / 166.7 ≈ 300 mm², so d = √(4 × 300/π) ≈ 19.5 mm. In practice you would pick the next standard size, so 20 mm is reasonable.