A hollow steel column carrying an axial load of 2·1 MN has an ultimate stress of 510 N/mm². The internal diameter of the column is 150 mm. Consider the factor of safety as 4. What is the external diameter of the column ?
- A201·49 mm
- B208·49 mmCorrect
- C214·49 mm
- D218·49 mm
Solution
Concept
Allowable stress and area of a hollow circular section
Formula
σallow = σult / FS; A = P / σallow = π (D² − d²) / 4
Step-by-step solution
1. σallow = 510 / 4 = 127.5 N/mm². 2. A = 2.1 × 10⁶ / 127.5 = 16 470.6 mm². 3. D² = 4A/π + 150² = 20 971.6 + 22 500 = 43 471.6. 4. D = 208.49 mm.
Explanation
The area needed to carry the axial load at the allowable stress is found first, then the outer diameter follows from the area of a ring.
Why the other options are wrong
Only D = 208.49 mm gives a ring area of 16 470 mm²; the other diameters give smaller or larger areas and therefore different safe loads.
Shortcut
D = √(4P/(π σallow) + d²).
Common mistake
Using the ultimate stress without the factor of safety.
Correct answer: B) 208·49 mm