Axial deformation of a steel rod — elongation under tensile load A steel rod of diameter 2 cm and length 5 m is subjected to an axial tensile load of 3000 kg (kgf). If Young's modulus E = 2.1 × 10^6 kg/cm^2 (consistent units), compute the elongation of the rod.

Introduction / Context:Axial deformation calculations are fundamental for serviceability checks. Using consistent units with the material modulus avoids large numerical errors.

Given Data / Assumptions:

• Diameter d = 2 cm ⇒ radius r = 1 cm.
• Length L = 5 m = 500 cm.
• Axial load P = 3000 kgf.
• E = 2.1 × 10^6 kg/cm^2.
• Linear-elastic behavior; uniform stress; small strains.

Concept / Approach:Axial elongation for a prismatic bar:delta = (P * L) / (A * E)with area for a circle:A = π * r^2

Step-by-Step Solution:

Verification / Alternative check:Compute directly: delta = (P * L) / (A * E) = (3000 * 500) / (3.1416 * 2.1 × 10^6) cm ≈ 0.2275 cm.

Why Other Options Are Wrong:

• 0.2275 mm and 0.02275 mm: too small by factors of 10 and 100.
• 2.02275 mm: arithmetic inconsistency.
• None of these: unnecessary since 2.275 mm is correct.

Common Pitfalls:Mixing SI (N, MPa, mm) and gravitational (kgf, cm) units; forgetting to convert metres to centimetres with E in kg/cm^2.

Final Answer:2.275 mm