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When acolumn 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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