According to Euler's column theory, the crippling load for a column length $$l$$ hinged at both ends, is
A. $$\frac{{{\pi ^2}{\text{E}}I}}{{{l^2}}}$$
B. $$\frac{{{\pi ^2}{\text{E}}I}}{{4{l^2}}}$$
C. $$\frac{{4{\pi ^2}{\text{E}}I}}{{{l^2}}}$$
D. $$\frac{{2{\pi ^2}{\text{E}}I}}{{{l^2}}}$$
Answer: Option A
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Both end fixed L = l/2 and Load = pi^2 * E * I/((l/2)^2),
Both end Hinged L = l and Load = pi^2 * E * I/(l^2),
one Fixed another Hinged L = l/Root 2,
one end fixed another free L = 2 * l.