I am trying to solve the following problem:
A masonry chimney having the shape of a conical frustrum is 25 m high. The external diameter at the top and internal diameter at the bottom is 2 m. The chimney is 0.5 m thick at its base. If the weight of the chimney is 1800 kN, find the uniform horizontal wind pressure that may act on the chimney per unit projected area of the chimney in order for tension at the base to be just avoided.
I am using the equation:
$$\mathrm{Centroid} (\bar{y}) = \dfrac {A_{1}y_{1} + A_2y_{2}} {A_{1} + A_{2}}$$
where $A1$ and $A2$ are the shaded areas in the figure below. Observe that the object is a chimney, so it is hollow in the centre portion; we are not adding up the area of that.
The centroid becomes:
$$ \bar{y} = \dfrac{ (\dfrac{1}2 *0.5 *25) * \dfrac{25}3 + (\dfrac{1}2 *0.5 *25) * \dfrac{25}3}{(\dfrac{1}2 *0.5 *25) +(\dfrac{1}2 *0.5 *25) }$$
$$\bar{y} = 8.333$$
However, I created a 3D model of the solid chimney in CAD and found the centroid to be 8.035, which does not match my calculated centroid. It is close to 8.333 but not exactly the same as my $\bar{y}$.
When I calculate the total wind pressure, which centroid should I use? The one I calculated by hand from the 2D cross-section, or the one I generated in CAD from the 3D model?
The CAD generated centroid is for a 3D solid and $\bar{y}$ calculated is for that central section.