Question

A piece of metal having a square cross section of area 400 mm\(^2\) is pulled with 40 kN force, producing only elastic deformation. If the Young's modulus of the metal is \(40 \times 10^9~\text{N/m}^2\), then the strain is:

• A. \(1 \times 10^{-3}\)
• B. \(1.5 \times 10^{-3}\)
• C. \(2.5 \times 10^{-3}\)
• D. \(4.0 \times 10^{-3}\)

Hint:

Strain in elastic materials can be calculated using \(\epsilon = \frac{\sigma}{Y}\), where \(\sigma = \frac{F}{A}\) and Y is Young's modulus.

Answer 1

The Correct Option is C

Step 1: Given quantities.
Area A = 400 mm² = 400 × 10⁻⁶ m², force F = 40 kN = 4 × 10⁴ N, Young's modulus Y = 40 × 10⁹ N/m².

Step 2: Stress-strain relation.
Y = Stress/Strain → Strain ε = Stress/Y, with Stress σ = F/A.

Step 3: Calculating stress.
σ = (4 × 10⁴) / (400 × 10⁻⁶) = (4 × 10⁴) / (4 × 10⁻⁴) = 1 × 10⁸ N/m².

Step 4: Calculating strain.
ε = (1 × 10⁸) / (40 × 10⁹) = 0.0025 = 2.5 × 10⁻³.

Step 5: Conclusion.
The strain is 2.5 × 10⁻³.