Imagine a spherical solid body with initially uniform density, to which we apply a uniform external pressure on its surface. How would the stress distribution look inside that sphere? Would the center of the sphere be more compressed, than its outer part? Or could it also be that the surface of the sphere which is in direct contact with whatever is applying pressure would experience more deformation, but the overall stress distribution inside the body would be uniform?

I would appreciate any intuition or references.

  • 1
    $\begingroup$ There is an answer here:physics.stackexchange.com/questions/70637/… $\endgroup$
    – mike stone
    Commented Mar 8, 2023 at 14:10
  • $\begingroup$ @mikestone That question is about a hollow pressure vessel; this question is about a solid sphere. $\endgroup$ Commented Mar 8, 2023 at 16:05
  • $\begingroup$ For a solid sphere, set $R_i = 0$ in the first answer to the linked question. $\endgroup$
    – mmesser314
    Commented Mar 8, 2023 at 16:08
  • $\begingroup$ The stress distribution would follow the local density changes in your sphere as a result of the pressure. If the pressure changes the density of sphere uniformly, then the stress distribution will also be uniform. $\endgroup$ Commented Mar 8, 2023 at 16:16
  • $\begingroup$ @mike stone thanks! That was still very usefull $\endgroup$ Commented Mar 8, 2023 at 18:54

2 Answers 2


If a solid uniform body (spherical or not) is subjected to a uniform pressure P, then the stress state inside is simply an equitrixial compressive normal stress of magnitude P.

As you note, the relative deformation will be greater with increasing distance from the center of mass.

  • $\begingroup$ Thanks! I was a bit too lazy to get into the literature, but looking at yours and the other answer, sounds like there is a standard procedure of computing stress and deformation inside the body that experiences pressure. Hope to find more in Landau's VII book $\endgroup$ Commented Mar 8, 2023 at 18:59

For a small volume $\Delta V = (\Delta r)(r\Delta\theta)(rsin(\theta))(\Delta\phi) = r^2sin(\theta)\Delta r\Delta\theta \Delta\phi$, the side forces are equal by symmetry. For the radial forces are in equilibrium:$$P_{r+\Delta r}((r+\Delta r)^2)sin(\theta)\Delta\theta \Delta\phi - P_rr^2sin(\theta)\Delta\theta \Delta\phi = 0$$ When $\Delta$'s go to zero, the pressures must be equal. So, for this symmetric situation, the pressure is constant inside the sphere.


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