# Volumetric Strain In a Thin Spherical Pressure Vessel

Consider a thin spherical pressure vessel with a fluid inside at a gauge pressure of P.

The normal stress developed in the pressure vessel is given by $$\sigma = \frac{Pd}{4t}$$

where t = thickness , d = diameter

I was interested in determining the volumetric strain for the vessel so I took an element in the pressure vessel and tried finding the normal strains along x, y and z.

$$\epsilon_x = \frac{\sigma(1-\mu)}{E}$$ $$\epsilon_y = \frac{\sigma(1-\mu)}{E}$$ $$\epsilon_z = \frac{-2\sigma \mu }{E}$$

The volumetric strain of this element will be

$$\epsilon_v = \epsilon_x + \epsilon_y + \epsilon_z$$

Substituting

$$\epsilon_v = \frac{2\sigma (1-2 \mu )}{E}$$

However when I use this formula for finding the volumetric strain in practice problems, I don't get the correct answer. In the solution they use the formula for volumetric strain as

$$\epsilon_v = \frac{3\sigma (1-\mu)}{E}$$

How $$\epsilon_v = \frac{3\sigma (1-\mu)}{E}$$ is the correct formula for the volumetric strain in a thin spherical pressure vessel?

As you rightly said $$\sigma=\frac{Pd}{4t}$$
Strain would be: $$\epsilon=\frac{\sigma}{E}(1-\mu)=\frac{\delta d}{d}$$
$$V=\frac{\pi d^3}{6} \implies \boxed{\epsilon_v=\frac{\delta V}{V}=3\frac{\delta d}{d}=3\frac{\sigma}{E}(1-\mu)}$$
• Thanks for the derivation Abhiram. Can you help me identifying the flaw in my way of deriving volumetric strain? I mean, if I take an element anywhere in the spherical vessel won't the Volumetric strain for the element be $\epsilon_v = \epsilon_x + \epsilon_y+ \epsilon_z$? Commented Apr 29, 2022 at 11:42