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

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

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

r_1=\frac{x}{\sin \alpha}             (3.7)

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

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

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

Using the general equation

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

and substituting r_1 \text { and } r_2 , we get

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

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

To find \sigma_2 , we can make a horizontal cut along X’ – X’ and consider the vertical equilibrium of the lower portion. It gives us the following equation:

\sigma_2(2 \pi x t) \sin \psi=p \pi x^2

This is because \sigma_2 stress lines are parallel to the tangents to the vessel at the location where X’-X’ cuts the vessel geometry. Therefore,

\begin{aligned} \sigma_2 & =\frac{p \pi x^2}{2 \pi x t \sin \psi} \\ & =\frac{p x}{2 t} \times \frac{1}{\sin \psi} \\ & =\frac{p x}{2 t} \times \frac{\sqrt{1+4 x^2}}{2 x} \\ & =\frac{p}{4 t} \sqrt{1+4 x^2} \end{aligned}

As y = x², we have

\sigma_2=\frac{p}{4 t} \sqrt{1+4 y}

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

\frac{\sigma_1}{\sqrt{1+4 x^2}}=\frac{p}{2 t}-\frac{p}{4 t} \times \frac{\sqrt{1+4 x^2}}{\left(1+4 x^2\right)^{3 / 2}}

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

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

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

This is the required circumferential stress.