Axial stress in a round steel rod under tensile load A steel rod of diameter 20 mm and length 5 m is subjected to an axial tensile force of 3000 kgf. If the measured elongation is 2.275 mm (not needed for stress), what is the average normal stress developed in the rod (in kg/cm²)?

Introduction / Context:
Direct (normal) stress under axial loading is one of the first checks for tension members. It is simply the applied force divided by the loaded cross-sectional area. Unit consistency is key to obtaining a correct value.

Given Data / Assumptions:

• Diameter d = 20 mm = 2 cm.
• Axial tensile force P = 3000 kgf.
• Uniform circular cross-section, no stress concentrations considered.
• Average normal stress σ = P / A.

Concept / Approach:

Compute area in cm² because the requested stress unit is kg/cm². For a circular section A = π d² / 4. Insert values and divide the load by the area.

Step-by-Step Solution:

Verification / Alternative check:

If you convert to N/mm² (MPa) for intuition: 1 kgf/cm² ≈ 0.09807 MPa, so 955.41 kgf/cm² ≈ 93.7 MPa, a reasonable stress level for mild steel in service.

Why Other Options Are Wrong:

• 9.5541 and 95.541 kg/cm² come from misplacing decimal points (area units confused).
• 9554.1 kg/cm² is off by a factor of 10 (using mm² but keeping kg/cm² units).
• 190.0 kg/cm² corresponds to using area ≈ 15.79 cm² (incorrect for 20 mm diameter).

Common Pitfalls:

• Mixing mm² and cm² when the requested stress unit is kg/cm².
• Using diameter in mm directly in a cm² area expression without conversion.

Final Answer:

955.41 kg/cm².