A masonry chimney having the shape of a frustum of a cone is 25m high. The external diameter at the top and the internal diameter at the bottom is 2 m. The chimney is 0.5 m thick at the base. If the weight of the chimney is 1800 kN find the uniform horizontal wind pressure that may act per unit projected area of the chimney in order tension at the base may be just avoided.

moment of Inertia of the base section about a diameter is
$ I = {\pi\ times(D^4-d^4)\over 64 }={\pi\times(3^4-2^4) \over 64 }= {3.191 m^4 }$

area of the base is given by ${\pi\times (D^2-d^2)\over2}={\pi\times (3^2-2^2)\over 4}={3.927 m^2}$

section of the base $Z = {2I\over D }= {2\times3.191\over 3}={2.127}$

direct stress due to the weight of the chimney ${1800000\over 3.927}={458365.16}$

AREA OF THE CHIMNEY IS GIVEN BY THE FOLLOWING FORMULAE

${h\times (D+d)\over 2}= {62.5}$

If p is the wind pressure of the shear force P then it acts at the centroid of the structure then ${P }= {62.5\times p}$ is the total wind pressure centroid of the trapezium is given by the following formula $\bar{y}= {(D+ 2\times d )\over(D+d)}{\times h \over 2}= {(3+ 2\times2)\times h\over 3+2 }= 11.65$ moment due to the pressure
$M =P\times \bar{y}$

${P{b}} ={\pm\ M\over Z}= {62.5 \times p\times11.67\over2.127}$ Pb =Po $ {62.5 \times p\times11.67\over2.127}$ =48365
so p =1336.87 $N\over m^2 $

Note: For Trapezium section centroid is