Axial deformation of a steel rod – compute the required tensile force A steel rod 1000 mm long and 12 mm in diameter must elongate by 0.045 mm. If E = 2 × 10^6 kg/cm^2, what tensile force is required?

Introduction / Context:
Axial deformation under load for a prismatic bar is a fundamental strength-of-materials calculation. It is frequently used to size rods, bolts, and tendons for specified elongations or to check stresses from measured strains.

Given Data / Assumptions:

• Length L = 1000 mm = 100 cm.
• Diameter d = 12 mm = 1.2 cm.
• Required elongation Δ = 0.045 mm = 0.0045 cm.
• Modulus of elasticity E = 2 × 10^6 kg/cm^2.
• Linear elastic behavior; uniform cross-section and uniaxial loading.

Concept / Approach:
For an axially loaded bar, the relation between load P and elongation Δ is
Δ = (P * L) / (A * E)
where A is the cross-sectional area. Solve for P after converting all quantities into consistent units (centimeters here).

Step-by-Step Solution:
Compute area A = (π/4) * d^2 = (π/4) * (1.2)^2 cm^2.A = π * 0.36 ≈ 1.131 cm^2.Rearrange formula: P = (Δ * A * E) / L.Insert values: P = (0.0045 cm * 1.131 cm^2 * 2 × 10^6 kg/cm^2) / 100 cm.Evaluate numerator: 1.131 * 2 × 10^6 = 2.2619 × 10^6; multiply by 0.0045 → ≈ 10,178.8.Divide by 100 → P ≈ 101.8 kg (round to 102 kg).

Verification / Alternative check:
Back-calculate Δ using P = 102 kg: Δ ≈ (102 * 100) / (1.131 * 2 × 10^6) ≈ 0.0045 cm = 0.045 mm, confirming consistency.

Why Other Options Are Wrong:

• 166 kg and 204 kg overpredict the load (would produce larger elongations).
• 74 kg underpredicts the load.
• 128 kg is also inconsistent with the computed requirement.

Common Pitfalls:
Mixing units (mm vs cm) or using area in mm^2 with E in kg/cm^2; always convert to a consistent set before substitution.

Final Answer:
102 kg