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For a column of length L is fixed at both ends, corresponding Euler’s critical load is
$$\frac {\pi^2 EI}{L^2}$$
$$\frac {2\pi^2 EI}{L^2}$$
$$\frac {3\pi^2 EI}{L^2}$$
$$\frac {4\pi^2 EI}{L^2}$$
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