For a column of length L is fixed at both the ends, corresponding Euler's critical load is
- A. $\pi² EI / L²$
- B. $2\pi² EI / L²$
- C. $3\pi² EI / L²$
- D. $4\pi² EI / L²$
Answer: D. $4\pi² EI / L²$
Explanation: For fixed-fixed column, effective length $l_e = L/2$. $P = \pi² EI / (L/2)^2 = 4\pi² EI / L²$.