Question

What is the stress developed when a steel rod of radius \(10 \, \text{mm}\) is stretched by a \(100 \, \text{kN}\) force?

• A. \(159 \, \text{MPa}\)
• B. \(318 \, \text{MPa}\)
• C. \(100 \, \text{MPa}\)
• D. \(200 \, \text{MPa}\)

Hint:

\(\sigma = \frac{F}{A}\), and for circular section: \(A = \pi r^2\).
Always convert mm to m before calculation.

Answer 1

The Correct Option is B

Concept: Stress is defined as force per unit area: \[ \sigma = \frac{F}{A} \]
Step 1: Convert given values into SI units.
Force: \[ F = 100 \, \text{kN} = 100 \times 10^3 \, \text{N} \] Radius: \[ r = 10 \, \text{mm} = 10 \times 10^{-3} \, \text{m} \]
Step 2: Calculate cross-sectional area.
\[ A = \pi r^2 = \pi (10 \times 10^{-3})^2 = \pi \times 10^{-4} \, \text{m}^2 \]
Step 3: Compute stress.
\[ \sigma = \frac{100 \times 10^3}{\pi \times 10^{-4}} = \frac{10^5}{\pi \times 10^{-4}} = \frac{10^9}{\pi} \approx 318 \times 10^6 \, \text{Pa} \] \[ \sigma \approx 318 \, \text{MPa} \]
Conclusion:
Thus, the stress developed is \( \boxed{318 \, \text{MPa}} \).