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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.

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

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?

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As you rightly said $$\sigma=\frac{Pd}{4t}$$

Strain would be: $$\epsilon=\frac{\sigma}{E}(1-\mu)=\frac{\delta d}{d}$$

We also know volume of a sphere is:

$$V=\frac{\pi d^3}{6} \implies \boxed{\epsilon_v=\frac{\delta V}{V}=3\frac{\delta d}{d}=3\frac{\sigma}{E}(1-\mu)}$$

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  • $\begingroup$ 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 $? $\endgroup$ Commented Apr 29, 2022 at 11:42
  • $\begingroup$ Oh wait am I calculating the volumetric strain for the volume occupied by the vessel but not the volume in which the fluid is present. The volumetric strain that we are interested in is based on the change in storage volume of the vessel and not the change in occupied volume of the vessel. $\endgroup$ Commented Apr 29, 2022 at 11:59

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