A tapering chimney of hollow section is 30m height. Its internal diameter of base is 2.4m and the top 1.6m. It is subjected to wind pressure 2.2 KN/m2 of projected area. If weight of base is 400KN.

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written 6.1 years ago by

teamques10 ★ 68k

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$\textit {Wind pressure} = 2.2\hspace{0.05cm}KN/m^2\\ W = 400\hspace{0.05cm}KN$

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$y = \frac{\Sigma Ay_i}{\Sigma A} = \frac{840}{60} = 14\hspace{0.05cm}m$

$I_{yy} = \frac{\pi}{64} (D^4 -d^4)\\ \hspace{0.25cm} = \frac{\pi}{64} (2.4^4 - 0.8^4) \hspace{0.25cm} = 1.60\hspace{0.05cm}m^4$

Direct Stress $\sigma_d = \frac{W}{A} = \frac{400}{\frac{\pi}{4}(D^2 - d^2)}\\ \hspace{0.25cm} = 99.47\hspace{0.05cm}KN/m^2 = 99470\hspace{0.05cm}10^3\hspace{0.05cm}N/m^2$

$\textit{Wind force}\hspace{0.05cm}(P) = \textit{Wind pressure X Projected Area}\\ \hspace{02cm} = 2.2\hspace{0.05cm}\times\hspace{0.05cm}60\\ \hspace{02cm} = 132\hspace{0.05cm}KN $

$M = P . y\\ \hspace{0.25cm} = 132\hspace{0.05cm}\times\hspace{0.05cm}14\\ \hspace{0.25cm} = 1848\hspace{0.05cm}KN/m^2$

$\sigma_b = \frac{M.y}{I}\\ \hspace{0.25cm} = \frac{1848\hspace{0.05cm}\times\hspace{0.05cm}1.2}{1.6}\\ \hspace{0.25cm} = 1386\hspace{0.05cm}KN/m^2$

$\sigma_{max} = +1386\hspace{0.05cm}KN/m^2$

$\sigma_{min} = -1386\hspace{0.05cm}KN/m^2$

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