Derive an expression for the maximum shear stress τmax in the plane of the wall of the conical tank filled with water which is supported along its upper edge as shown in Figure 4.22.

Question

Derive an expression for the maximum shear stress [latex] au_{\max } [/latex] in the plane of the wall of the conical tank filled with water which is supported along its upper edge as shown in Figure 4.22.

Accepted Answer

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 [latex] \left(\sigma_1 ight) [/latex] and longitudinal stress [latex] \left(\sigma_2 ight) [/latex]. 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

[latex] ext { Circumferential stress, } \sigma_1=\frac{w(h-y)}{t} imes \frac{y an \alpha}{\cos \alpha} [/latex] (1)

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

[latex] \sigma_2=\frac{w}{2 t}\left\lgroup h y-\frac{2}{3} y^2 ight group imes \frac{ an \alpha}{\cos \alpha} [/latex]

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

[latex] \sigma_2=\frac{w}{t} \frac{ an \alpha}{\cos \alpha} imes\left\lgroup \frac{h y}{2}-\frac{y^2}{3} ight group [/latex] (2)

The general expression for the magnitude of shear stress is

[latex] au=\frac{1}{2}\left(\sigma_1-\sigma_2 ight) [/latex]

[latex] =\frac{w an \alpha}{2 t \cos \alpha}\left[\frac{h y}{2}-\frac{2}{3} y^2 ight] \quad[ ext { from Eqs. (1) and (2)] } [/latex]

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

[latex] \frac{ d au}{ d y}=0 [/latex]

or [latex] \frac{h}{2}=\frac{4}{3} y [/latex]

or [latex] y=\frac{3 h}{8} \quad\left[ ext { check that }\left( d ^2 au / d y^2 ight) ight. ext { is negative] } [/latex]

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

[latex] \begin{aligned} au_{\max } & =\frac{w an \alpha}{2 t \cos \alpha}\left[\frac{3 h^2}{16}-\frac{2}{3} imes \frac{9 h^2}{64} ight] \ & =\frac{3 w h^2}{64 t} imes \frac{ an \alpha}{\cos \alpha} \end{aligned} [/latex]

Thus, the maximum shear stress is

[latex] au_{\max }=\frac{3 w h^2}{64 t} imes \frac{ an \alpha}{\cos \alpha} [/latex]

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

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