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.
Mechanical Engineering
Strength of Materials
Difficulty: Medium
Choose an option
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A2.275 mm
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B0.2275 mm
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C0.02275 mm
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D2.02275 mm
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ENone of these
Answer
Correct Answer: 2.275 mm
Explanation
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:
A = π * (1 cm)^2 = 3.1416 cm^2.Stress sigma = P / A = 3000 / 3.1416 ≈ 955 kg/cm^2.Strain epsilon = sigma / E = 955 / (2.1 × 10^6) ≈ 4.548 × 10^-4.Elongation delta = epsilon * L = 4.548 × 10^-4 * 500 cm ≈ 0.2274 cm = 2.274 mm ≈ 2.275 mm.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