A vertical load of 1960 N is lifted at the free end of a straight steel wire. What is the minimum wire diameter (in mm) required so that the tensile stress does not exceed 100 N/mm^2?

Introduction / Context:
When a straight steel wire is subjected to a tensile load, the induced normal stress equals the load divided by the cross-sectional area. Designing the minimum diameter against an allowable stress is a fundamental sizing task in strength of materials and structural design.

Given Data / Assumptions:

• Tensile load, P = 1960 N.
• Allowable tensile stress, sigma_allow = 100 N/mm^2.
• Wire is prismatic, circular, and behaves elastically under the given load.
• Self-weight and stress concentrations are neglected; static loading only.

Concept / Approach:
Tensile stress sigma = P / A must not exceed sigma_allow. For a circular section, A = (pi * d^2) / 4. Solve for the minimum diameter d such that P / A <= 100 N/mm^2. Rearranging gives d = sqrt(4 * P / (pi * sigma_allow)). Choose the next available size that is equal to or greater than the computed minimum to ensure stress does not exceed the limit.

Step-by-Step Solution:
Step 1: Compute required area A_req = P / sigma_allow = 1960 / 100 = 19.6 mm^2.Step 2: Relate area to diameter: A = (pi * d^2) / 4 ⇒ d^2 = 4 * A / pi.Step 3: Substitute A_req: d = sqrt(4 * 19.6 / pi).Step 4: Numerically: 4 * 19.6 = 78.4; 78.4 / pi ≈ 24.96; sqrt(24.96) ≈ 4.996 mm.Step 5: Select a listed diameter ≥ 4.996 mm → 5.0 mm.

Verification / Alternative check:
Check stress with d = 5.0 mm: A = (pi * 25) / 4 ≈ 19.635 mm^2; sigma = 1960 / 19.635 ≈ 99.8 N/mm^2, which is within the 100 N/mm^2 limit.

Why Other Options Are Wrong:

• 4.0 mm, 4.5 mm: Areas too small, producing sigma > 100 N/mm^2.
• 5.5 mm, 6.0 mm: Safe but not the minimum required by the stress criterion.

Common Pitfalls:
Using area in m^2 instead of mm^2; forgetting to check the next higher standard size; confusing tensile stress limits with yield or ultimate strengths without safety factors.

Final Answer:
5.0 mm