A prismatic bar has a cross section of 25 mm × 50 mm and a length of 2 m. If the bar is subjected to an axial tensile force of 90 kN, the bar is elongated by 1.5 mm. What is the tensile stress?
- A3.6 N/mm²
- B7.2 N/mm²
- C36 N/mm²
- D72 N/mm²Correct
Solution
Concept
Normal (tensile) stress = axial load / cross-sectional area. Elongation and length are not needed.
Formula
σ = P / A
Step-by-step solution
1. Area A = 25 × 50 = 1250 mm². 2. Load P = 90 kN = 90000 N. 3. σ = P / A = 90000 / 1250. 4. σ = 72 N/mm².
Explanation
Tensile stress is the load divided by the area. Area = 1250 mm² and load = 90000 N, so stress = 72 N/mm². The length and elongation are extra data (useful only for strain or E).
Why the other options are wrong
A (3.6) and B (7.2): wrong unit conversion (decimal place error). C (36): area taken as double (2500 mm²) or load taken as half.
Shortcut
90000 / 1250 = 72. Divide by 1000 to get 90/1.25 = 72.
Common mistake
Using the extra data (2 m length, 1.5 mm elongation), or forgetting to convert kN to N.
Correct answer: D) 72 N/mm²