Question 4.11: Derive an expression for the maximum shear stress τmax in th...

Let us take on tank wall an arbitrary biaxial stress element as shown in the figure. Considering the pressure vessel where biaxial stress-situation prevails, we can find out the general equations for circumferential stress \left(\sigma_1\right) and longitudinal stress \left(\sigma_2\right) . If w denotes weight per unit volume of water, t as thickness of the thin-walled conical tank, (h − y) the depth of the element concerned and α the apex angle of revolution, we can write

\text { Circumferential stress, } \sigma_1=\frac{w(h-y)}{t} \times \frac{y \tan \alpha}{\cos \alpha}            (1)

(Detailed derivation is left as an exercise for the students and they can take help from the previous chapter.)
Longitudinal stress is

\sigma_2=\frac{w}{2 t}\left\lgroup h y-\frac{2}{3} y^2 \right\rgroup \times \frac{\tan \alpha}{\cos \alpha}

One can find this by considering vertical equilibrium of the cone along with the weight of water column as shown shaded in the figure. Rearranging, we get

\sigma_2=\frac{w}{t} \frac{\tan \alpha}{\cos \alpha} \times\left\lgroup \frac{h y}{2}-\frac{y^2}{3} \right\rgroup                 (2)

The general expression for the magnitude of shear stress is

\tau=\frac{1}{2}\left(\sigma_1-\sigma_2\right)

For finding the maximum shear stress, obviously, we must have

\frac{ d \tau}{ d y}=0

or        \frac{h}{2}=\frac{4}{3} y

or        y=\frac{3 h}{8} \quad\left[\text { check that }\left( d ^2 \tau / d y^2\right)\right. \text { is negative] }

Substituting this value of y in Eq. (2), we get

\begin{aligned} \tau_{\max } & =\frac{w \tan \alpha}{2 t \cos \alpha}\left[\frac{3 h^2}{16}-\frac{2}{3} \times \frac{9 h^2}{64}\right] \\ & =\frac{3 w h^2}{64 t} \times \frac{\tan \alpha}{\cos \alpha} \end{aligned}

Thus, the maximum shear stress is

\tau_{\max }=\frac{3 w h^2}{64 t} \times \frac{\tan \alpha}{\cos \alpha}