The wind intensity is $1.3 KN/m^2$ on projected area and specific weight of the measure is $22 KN/m^2$. Calculate the maximum and minimum stresses of the base of the chimney.

$\textit{Wind pressure} = 1.3\hspace{0.05cm}KN/m^2\\ W = 22\hspace{0.05cm}KN/m^2\\ y = \frac{2.5}{2} = 1250\hspace{0.05cm}mm$

(i) $\textit{Projected Area} = 24\hspace{0.05cm}\times\hspace{0.05cm}2.5\\ \hspace{02cm} = 60\hspace{0.05cm}m^2$

(ii) $\textit{Wind force} (P) = \textit{Wind pressure X projected area}\\ \hspace{02cm} = 1.3\hspace{0.05cm}\times\hspace{0.05cm}60\\ \hspace{02cm} = 78\hspace{0.05cm}KN = 78\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}N$

(iii)$\textit{self weight} = W\hspace{0.05cm}\times\hspace{0.05cm}Volume\\ \hspace{02cm} = W\hspace{0.05cm}\times\hspace{0.05cm}A\hspace{0.05cm}\times\hspace{0.05cm}L\\ \hspace{02cm} = 22\hspace{0.05cm}\times\hspace{0.05cm}[(1.25\hspace{0.05cm}\times\hspace{0.05cm}1.25) - (2.5\hspace{0.05cm}\times\hspace{0.05cm}2.5)]\hspace{0.05cm}\times\hspace{0.05cm}24\\ \hspace{02cm} = 2475\hspace{0.05cm}KN = 2475\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}N$

Direct stress $\sigma_d =\frac{W}{A} = \frac{2475\hspace{0.05cm}\times\hspace{0.05cm}10^3}{4.69\hspace{0.05cm}\times\hspace{0.05cm}10^6} = 527\hspace{0.05cm}\times\hspace{0.05cm}10^{-3}\hspace{0.05cm}N/mm^2$

$M = P\hspace{0.05cm}\times\hspace{0.05cm}y\\ \hspace{0.25cm} = 78\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}\times\hspace{0.05cm}12\hspace{0.05cm}\times\hspace{0.05cm}10^3\\ \hspace{0.25cm} = 936\hspace{0.05cm}\times\hspace{0.05cm}10^5\hspace{0.05cm}Nmm$

$I_{yy} = I_{yy(outer)} - I_{yy(inner)}\\ \hspace{0.25cm} = \frac{2.5\hspace{0.05cm}\times\hspace{0.05cm}2.5^3}{12} - \frac{1.25\hspace{0.05cm}\times\hspace{0.05cm}1.25^3}{12}\\ \hspace{0.25cm} = 3.05\hspace{0.05cm}m^4 = 3.05\hspace{0.05cm}\times\hspace{0.05cm}10^{12}\hspace{0.05cm}mm^4$

$\sigma_b = \frac{M .y}{I} = \frac{936\hspace{0.05cm}\times\hspace{0.05cm}10^6\hspace{0.05cm}\times\hspace{0.05cm}1250}{3.05\hspace{0.05cm}\times\hspace{0.05cm}10^{12}}\\ \hspace{0.25cm} = 0.383\hspace{0.05cm}N/mm^2$

$\sigma_{max} = \sigma_d + \sigma_b\\ \hspace{0.5cm} = 527\hspace{0.05cm}\times\hspace{0.05cm}10^{-3} + 0.383\\ \hspace{0.5cm} = 0.91\hspace{0.05cm}N/mm^2$

$\sigma_{min} = \sigma_d - \sigma_b\\ \hspace{0.5cm} = 0.144\hspace{0.05cm}N/mm^2$