A hollow column has outside dia 200mm, it is 4.5m long and is fixed at both ends. Determine the safe load the column can carry. Use Rankine formula.

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written 6.4 years ago by

teamques10 ★ 69k

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$\frac{1}{\alpha} = 1600\\ \alpha = \frac{1}{1600}$

$E = 94\hspace{0.05cm}GPa = 94\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}MPa\hspace{0.05cm}(or)\hspace{0.05cm}N/mm^2\\ FOS = 4 L = 4.5\hspace{0.05cm}m = 4500\hspace{0.05cm}mm\\ A = \frac{\pi}{4}(D^2 - d^2) = 11.30\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}mm^2$

$I_{xx} = I_{yy} = \frac{\pi}{64}(D^2 - d^2)\\ \hspace{0.5cm} = 46.36\hspace{0.05cm}\times\hspace{0.05cm}10^6\hspace{0.05cm}mm^4$

$I = AK^2\\ K = \sqrt{\frac{I}{A}} = \sqrt{\frac{46.36\hspace{0.05cm}\times\hspace{0.05cm}10^6}{11.30\hspace{0.05cm}\times\hspace{0.05cm}10^3}}\\ K = 64.05\hspace{0.05cm}mm$

$P_{Rankine} = \frac{\sigma_{cr}\hspace{0.05cm}\times\hspace{0.05cm}A}{1 + \alpha (\frac{L_c}{K})^2}\\ \hspace{0.25cm} = \frac{550\hspace{0.05cm}\times\hspace{0.05cm}11.30\hspace{0.05cm}\times\hspace{0.05cm}10^3}{1 + \frac{1}{1600}(\frac{l}{2K})^2}\\ \hspace{0.25cm}\frac{550\hspace{0.05cm}\times\hspace{0.05cm}11.30\hspace{0.05cm}\times\hspace{0.05cm}10^3}{1 + \frac{1}{1600}(\frac{4.5\hspace{0.05cm}\times\hspace{0.05cm}10^3}{2\hspace{0.05cm}\times64.05})^2}\\ \hspace{0.25cm} = 3.5\hspace{0.05cm}\times\hspace{0.05cm}10^6\hspace{0.05cm}N = 3.5\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}KN$

$FOS = \frac{P_{Rankine}}{P_{Safe}}\\ 4 = \frac{3.5\hspace{0.05cm}10^3}{P_{Safe}}\\ P_{safe} = 875\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}KN$

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