The context

I’ve taken an interest in the design of pressure vessels for use in submarine applications. From a cursory survey of what is used in industry, it seems the de facto standard housing is a thick-walled cylinder (rather than a thin-walled cylinder which would be prone to buckling when loaded in hydrostatic compression). While the equations for stress and deformation in a thick-walled cylinder subject to internal and external pressures are well established (Young, Norton), I’ve also read that prestressed compound cylinders can result in a housing with a greater strength to weight ratio (Kharagpur, Sun).

Stress distributions in compound pressure cylinders

As a result, I’ve been trying to understand the derivation of the thick-walled cylinder equations to better understand how they can be applied and optimized.

The question

With the help of multiple online sources (primarily Kadapa and Katna), I’m confident that I understand the derivation of Lamé’s equations

$$\sigma_r = C_1 + \frac{C_2}{r^2}$$ $$\sigma_\theta = C_1 - \frac{C_2}{r^2}$$

This is where things get interesting. In both sources mentioned above, and indeed everywhere else I can find it, the equation for the change in radius of a differential shell is calculated from the hoop strain, rather than the radial strain as I would expect.

$$\varepsilon_\theta = \frac{\sigma_\theta - \nu \left( \sigma_r + \sigma_z \right)}{E}$$ $$u = \varepsilon_\theta r$$

While I’m satisfied this is correct, I don’t understand why. Why doesn’t the radial deformation result from the radial strain as I expect it should, according to

$$\varepsilon_r = \frac{\sigma_r - \nu \left( \sigma_\theta + \sigma_z \right)}{E}$$ $$u = \varepsilon_r r$$

What I've tried so far

I’ve tried evaluating both equations for $u$ for the simple case of a single cylinder subject to internal and external pressure (the boundary conditions given in every textbook).

Differential cylinder element showing stresses and deformation

While I haven’t compared the values to a more reliable source, the equation with hoop strain always produces deformations in the direction of the pressure differential (an external pressure causes the cylinder to contract). The equation using radial strain has so far always produced deformations of differing magnitude and in the direction opposite the pressure differential (an external pressure causes the cylinder to expand).

In both Kadapa and Katna, it is offhandedly mentioned that and $\varepsilon_r$ and $\varepsilon_\theta$ are equal on the bases that

$$\varepsilon_r = \frac{u}{r} = \frac{2 \pi u}{2 \pi r} = \frac{\delta\ circumference}{circumference} = \varepsilon_\theta$$

While I can’t find fault with this statement, I think I have shown that it is only true for thin-walled cylinders where the hoop stress is assumed to be constant throughout the wall as below:

Based on the methods presented in Kadapa, Katna, and others,

$$\varepsilon_r = \frac{du}{dr}$$ $$\varepsilon_\theta = \frac{u}{r}$$

Assume that

$$\varepsilon_r = \varepsilon_\theta$$

If so,

$$\frac{du}{dr} = \frac{u}{r}$$

This first order linear ODE has the solution

$$u = kr$$

As a result,

$$\varepsilon_\theta = \frac{u}{r} = k$$

  • 1
    $\begingroup$ Check what Timoshenko says... widely considered the defacto reference... $\endgroup$
    – Solar Mike
    Mar 22, 2019 at 22:53
  • $\begingroup$ This is not an answer or even a hint, but since you said you are interested in design of submarines ..., take a look here engineering.stackexchange.com/questions/23257/…, in the comments there is link, i guess you'll find it very interesting. $\endgroup$
    – user14407
    Mar 22, 2019 at 23:53
  • $\begingroup$ I'm not sure what you are asking, but you seem to have got the whole derivation backwards. The deformation is purely radial, so $u$ in polar coordinates is just a function of $r$. This deformation gives both an radial and hoop strains. The derivations are starting from the deformation, then finding the corresponding strains, and then finding the stresses, not the other way round. Of course you can "sort of" do the derivation the other way round for a thin cylinder by ignoring the radial stress, since it is small compared with the hoop stress. $\endgroup$
    – alephzero
    Mar 23, 2019 at 11:31

1 Answer 1


I think I've identified my misconception. As alephzero commented on my question, I was conflating the local strain with the overall radial deformation. The strain is defined as the normalized change in length of an infinitesimal line segment, and not as the change in length of a macroscopic dimension such as radius. As such, the results

$$\varepsilon_r = \frac{du}{dr}$$ $$\varepsilon_\theta = \frac{u}{r}$$

give me immediate access to the deformations as

$$u = \int \varepsilon_r dr = \varepsilon_\theta r$$

Thanks for pointing me in the right direction.


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