A short column of external diameter D and internal diameter d is subjected to a compressive load P acting with an eccentricity 'e'. If the stresses at one of the extreme fibre is zero then the eccentricity has to be
- A. (D² + d²) / 8\pi D
- B. (D² + d²) / 8D
- C. (D² - d²) / D
- D. (D³ - d³) / 8D^2
Answer: B. (D² + d²) / 8D
Explanation: Condition for zero stress: Direct Stress = Bending Stress. P/A = Pe(D/2)/I. e = 2I/AD. I = \pi(D^4-d^4)/64. A = \pi(D²-d²)/4. e = 2 * [\pi(D²+d²)(D²-d²)/64] / [\pi(D²-d²)/4 * D] = (D²+d²)/8D.