Question 3.8: A thin-walled conical tank is supported on a horizontal base...

From the general governing equation of a thin-walled pressure vessel, we have

\frac{\sigma_1}{r_1}+\frac{\sigma_2}{r_2}=\frac{p}{t}

Here, from Figure 3.14(b), it is noted that r_1= BC \text { and } r_2=\infty , Now,

\sigma_1=\frac{p r_1}{t}

where p is pressure at level of X−X’. Obviously, we note that

p=\gamma_{ L } h  \bullet \bullet \bullet \bullet \bullet BC = B \tan \alpha

From the given geometry,

\frac{ AD }{ AB }=\cos \alpha

or          \frac{h}{ AB }=\cos \alpha

or         AB = h secα

So, we can rewrite BC = h sec α tan α. Therefore,

\sigma_1=\frac{\gamma_{ L } h}{t} \times h \sec \alpha \tan \alpha=\frac{\gamma_{ L } h^2}{t} \times \frac{\tan \alpha}{\cos \alpha}

For finding out \sigma_2 , let us now consider the vertical equilibrium of the upper section of the tank cut along the line X−X’. The net force in upward direction must balance the force arising out of \sigma_2 in downward direction. Referring to Figure 3.15, upward thrust is nothing but the weight of the liquid contained in the shaded volume. If we assume BD = r’, the shaded volume is

\pi r^{\prime 2} h-\frac{1}{3} \pi r^{\prime 2} h=\frac{2}{3} \pi r^{\prime 2} h

Now, from Figure 3.14(b), BD r’ = h tan α. Therefore,

\begin{aligned} \text { Net upward thrust } & =\frac{2}{3} \pi(h \tan \alpha)^2 h \gamma_{ L } \\ & =\frac{2}{3} \pi h^3 \gamma_{ L } \tan \alpha \end{aligned}

Equating this with downward force (arising out of \sigma_2 ), we get

\frac{2}{3} \pi h^3 \gamma_{ L } \tan \alpha=\left(2 \pi r^{\prime} t \sigma_2\right) \cos \alpha^5           (since area of the circular base is 2πr’t )

But,

\frac{r^{\prime}}{h}=\tan \alpha \Rightarrow r^{\prime}=h \tan \alpha

Thus, we can rewrite the above equation as:

\frac{2}{3} \pi h^3 \gamma_{ L } \tan \alpha=2 \pi h t \tan \alpha \cos \alpha \sigma_2

or            \sigma_2=\frac{\gamma_{ L } h^2}{3 t \cos \alpha}

Thus, the stresses are

\sigma_1=\frac{\gamma_{ L } h^2}{t \cos \alpha} \quad \text { and } \quad \sigma_2=\frac{\gamma_{ L } h^2}{3 t \cos \alpha}

^5 We note that \sigma_2 stress lines are tangential to cone surface and they make angle α with the vertical. Obviously, their vertical components, defined by cos α, balance the upward thrust. It is easy to note that the horizontal components of forces balance themselves due to symmetry of the geometry.