Height of the chimney is 25m and subjected to wind pressure $1.5 KN/m^2$ , the density of measure is $21 KN/m^3$.

$\textit {Wind pressure} = 1.5\hspace{0.05cm}KN/m^2\\ \textit{density} \hspace{0.05cm}(\rho) = 21\hspace{0.05cm}KN/m^3\\ y_1 = \frac{25}{2} = 12.5\\ y_2 = \frac{5}{2} = 2.5$

(i) $\textit{Projected Area} = 25\hspace{0.05cm}\times\hspace{0.05cm}5\\ \hspace{02cm} = 125\hspace{0.05cm}m^2$

(ii)$\textit{Wind force}\hspace{0.05cm}(P) = \textit{Wind pressure X Projected Area}\\ \hspace{02cm} = 1.5\hspace{0.05cm}\times\hspace{0.05cm}125\\ \hspace{02cm} = 187.5\hspace{0.05cm}KN = 1875\hspace{0.05cm}\times\hspace{0.05cm}10^2\hspace{0.05cm}N$\

(iii) $\textit{Self Weight} = \rho\hspace{0.05cm}\times\hspace{0.05cm}Volume\\ \hspace{02cm} = \rho\hspace{0.05cm}\times\hspace{0.05cm}A\hspace{0.05cm}\times\hspace{0.05cm}L\\ \hspace{02cm} = 21\hspace{0.05cm}\times\hspace{0.05cm}\frac{\pi}{4}(D^2 - d^2)\hspace{0.05cm}\times\hspace{0.05cm}25\\ \hspace{02cm} = 6594\hspace{0.05cm}N$

Direct Stress $\sigma_d = \frac{W}{A} = \frac{6594}{12.56}\\ \hspace{0.25cm} = 525\hspace{0.05cm}KN/m^2$

$M = P . y_1\\ \hspace{0.25cm} = 187.5\hspace{0.05cm}\times\hspace{0.05cm}12.5\\ \hspace{0.25cm} = 2343.75\hspace{0.05cm}KN/m^2$

$I_{yy} = I_{yy(outer)} - I_{yy(inner)}\\ \hspace{0.25cm} = \frac{\pi}{64} (D^4 -d^4)\\ \hspace{0.25cm} = 26.70\hspace{0.05cm}m64$

$\sigma_b = \frac{M.y}{I}\\ \hspace{0.25cm} = \frac{2343.75\hspace{0.05cm}\times\hspace{0.05cm}2.5}{26.70}\\ \hspace{0.25cm} = 219.45\hspace{0.05cm}KN/m^2$

$\sigma_{max} = \sigma_d + \sigma_b\\ \hspace{0.25cm} = 525 + 219.45\\ \hspace{0.25cm} = 744.45\hspace{0.05cm}KN/m^2$

$\sigma_{min} = \sigma_d - \sigma_b\\ \hspace{0.25cm} = 525 - 219.45\\ \hspace{0.25cm} = 305.55\hspace{0.05cm}KN/m^2$