Investigate the stress conditions at any position in a thin-walled pressure vessel which is a volume of revolution of the curve y = x². The vessel is supported from the top and is subjected to an internal gas pressure p. Assume wall thickness to be t. Also, assume that one end of the vessel is

Question

Investigate the stress conditions at any position in a thin-walled pressure vessel which is a volume of revolution of the curve y = x². The vessel is supported from the top and is subjected to an internal gas pressure p. Assume wall thickness to be t. Also, assume that one end of the vessel is closed. Refer to Figure 3.16(a).

Accepted Answer

Let us take a small element ABCD as shown surrounding a point from Figure 3.16(b) and draw a tangent and normal to that point on the curve. Obviously

[latex] r_1=\frac{x}{\sin \psi}[ ext { refer Eq }(3.7)] [/latex]

[latex] r_1=\frac{x}{\sin \alpha} [/latex] (3.7)

where tan ψ = dy/dx = 2x because y = x². Therefore,

[latex] r_1=\frac{x}{2 x / \sqrt{1+4 x^2}}=\frac{\sqrt{1+4 x^2}}{2} [/latex]

and [latex] r_2=\frac{\left[1+( d y / d x)^2 ight]^{3 / 2}}{\left| d ^2 y / d x^2 ight|}=\frac{\left[1+(2 x)^2 ight]^{3 / 2}}{2}=\frac{\left(1+4 x^2 ight)^{3 / 2}}{2} [/latex]

Using the general equation

[latex] \frac{\sigma_1}{r_1}+\frac{\sigma_2}{r_2}=\frac{p}{t} [/latex]

and substituting [latex] r_1 ext { and } r_2 [/latex], we get

[latex] \frac{2 \sigma_1}{\sqrt{1+4 x^2}}+\frac{2 \sigma_2}{\left(1+4 x^2 ight)^{3 / 2}}=\frac{p}{t} [/latex]

or [latex] \frac{\sigma_1}{\sqrt{1+4 x^2}}+\frac{\sigma_2}{\left(1+4 x^2 ight)^{3 / 2}}=\frac{p}{2 t} [/latex] (1)

To find [latex] \sigma_2 [/latex], we can make a horizontal cut along X' - X' and consider the vertical equilibrium of the lower portion. It gives us the following equation:

[latex] \sigma_2(2 \pi x t) \sin \psi=p \pi x^2 [/latex]

This is because [latex] \sigma_2 [/latex] stress lines are parallel to the tangents to the vessel at the location where X'-X' cuts the vessel geometry. Therefore,

[latex] \begin{aligned} \sigma_2 & =\frac{p \pi x^2}{2 \pi x t \sin \psi} \ & =\frac{p x}{2 t} imes \frac{1}{\sin \psi} \ & =\frac{p x}{2 t} imes \frac{\sqrt{1+4 x^2}}{2 x} \ & =\frac{p}{4 t} \sqrt{1+4 x^2} \end{aligned} [/latex]

As y = x², we have

[latex] \sigma_2=\frac{p}{4 t} \sqrt{1+4 y} [/latex]

This is the desired longitudinal stress. Putting this value in Eq. (1), we obtain the value of [latex] \sigma_1 [/latex]

[latex] \frac{\sigma_1}{\sqrt{1+4 x^2}}=\frac{p}{2 t}-\frac{p}{4 t} imes \frac{\sqrt{1+4 x^2}}{\left(1+4 x^2 ight)^{3 / 2}} [/latex]

or [latex] \frac{\sigma_1}{\sqrt{1+4 x^2}}=\frac{p}{2 t}\left[1-\frac{1}{2\left(1+4 x^2 ight)} ight] [/latex]

or [latex] \sigma_1=\frac{p}{2 t}\left[\sqrt{1+4 x^2}-\frac{1}{2 \sqrt{1+4 x^2}} ight] [/latex]

[latex] =\frac{p}{2 t}\left[\sqrt{1+4 y}-\frac{1}{2 \sqrt{1+4 y}} ight] [/latex]

This is the required circumferential stress.

Question 3.9: Investigate the stress conditions at any position in a thin-...

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