So I have a sphere of radius $R$, and density $\rho$, elastic modulus $E$ and Poisson ratio $\nu$.

With no gravity it is a sphere, but when gravity comes with a direction $(Oz)$ it becomes an ellipsoid :

enter image description here

How can one find the values of $a$ and $b$ please ? Is there a possibility to find a dynamic relaxation equation : $\partial_t a =... \; ; \; \partial_t b=...$ ?

Another way to ask the question is how to find the energy that costs the deformation of the sphere ?

Cubic case :

If we look at a case of a cube of side $R_0$ that would deform into a cuboid shape (parallelepiped rectangle) of sides $R_0(1-\epsilon)$, $R_0(1+\nu\epsilon)$ (the base is a square), and if we use linear elasticity :

we have as the energy of deformation for a uniaxial stress:

$$W=\frac{1}{2}V\sigma \epsilon=\frac{1}{2}R_0^3(1-\epsilon)(1+\nu\epsilon)^2\sigma \epsilon=R_0^3(1-\epsilon)(1+\nu\epsilon)^2\frac{1}{2}\frac{(1-\nu)E}{(1+\nu)(1-2\nu)}\epsilon^2$$

where $\sigma$ is the stress and $\epsilon$ the deformation. Here I'm not sure of wich volume I should take the initial $R_0^3$ or $R_0^3(1-\epsilon)(1+\nu\epsilon)$; it actually doesn't change much since we are interested only by the second order $\epsilon^2$ ?

Now the gravity energy is varying from $G=R_0^2\int\rho g z dz=R_0^4g\rho/2 $ to $G'=R_0^4(1-\epsilon)^2(1+\nu\epsilon)^2\rho g /2$.

Then equalizing both energies $$G-G'=W$$ gives $\epsilon$.

$$\frac{(1-\nu)E}{(1+\nu)(1-2\nu)}\epsilon=2R_0\Big((1 - \nu)+(-1 + 4 \nu - \nu^2)\epsilon\Big)\rho g $$ $$\epsilon=\frac{2R_0(1 - \nu)\rho g}{(1 - 4 \nu + \nu^2)+\frac{(1-\nu)E}{(1+\nu)(1-2\nu)}}$$

From which we can conclude that the bigger the object is the more deformed it becomes.

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    $\begingroup$ If the sphere is resting on a flat, rigid surface, I don't think that it will deform into an ellipsoid under gravity even as a first approximation. The bottom of the sphere resting on the rigid surface will undergo much more elastic deformation than the top of the sphere. Since the deformations of the top and bottom halves of sphere will be much different in character, I don't think that assuming an ellipsoidal shape is appropriate since that would mean assuming that the top and bottom halves have the same shape. $\endgroup$ – user93237 Jun 19 '19 at 16:51
  • $\begingroup$ @SamuelWeir I was indeed thinking about a first order simplification. Would you have a quantitative argument to strengthen your intuition ? $\endgroup$ – J.A Jun 19 '19 at 16:54
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    $\begingroup$ I don't have a quantitative argument readily available, but I think that I cast enough doubt on the use of the ellipsoid approximation idea here that the justification burden lies on anyone wishing to use the ellipsoid approximation for this problem. $\endgroup$ – user93237 Jun 19 '19 at 18:26
  • $\begingroup$ Also, even for a cubic shape, the deformation will not be homogeneous. The strains will be greater near the bottom than near the top. $\endgroup$ – Chet Miller Jun 22 '19 at 0:19

I don't think that your reasoning is correct. I will add some approximations to compare the results for a cube and a sphere.


Let's do a first approximation assuming uniaxial stress and polynomial solutions. That means that the boundary conditions won't be exactly satisfied, but we can live with that.

For the assumptions mentioned above and fixing the coordinate system to the bottom of a cube with side $a$, the displacement is given by

$$u = \frac{\rho g}{E}\left(\nu x \left(a - z\right), \nu y \left(a - z\right), \frac{\nu \left(x^{2} + y^{2}\right)}{2} - \frac{z \left(2 a - z\right)}{2}\right)\, ,$$

the strain is

$$\epsilon = \frac{\rho g}{E}\left[\begin{matrix}\nu \left(a - z\right) & 0 & 0\\0 & \nu \left(a - z\right) & 0\\0 & 0 & - (a - z)\end{matrix}\right]\, ,$$

and the stress is

$$\sigma = \rho g\left[\begin{matrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & -(a - z)\end{matrix}\right]\, .$$

The original and (exagerated) deformed configuration are depicted below with the stress represented by the intensity of blue.

enter image description here

Now, the density of energy is

$$u_e = \frac{g^{2} \rho^{2} \left(a^{2} - 2 a z + z^{2}\right)}{2E}\, ,$$

and integrating

$$U_e = \int\limits_{-a/2}^{a/2}\int\limits_{-a/2}^{a/2}\int\limits_{0}^{a} u_e \mathrm{d}z\mathrm{d}y\mathrm{d}x = \frac{a^{5} g^{2} \rho^{2}}{6 E} \, .$$


Regarding the deformation of the sphere, this is a problem that is really difficult to solve because it involves contact that is a nonlinear phenomenon. The larger the weigth, the larger the contact zone.

Let's use a Hertz contact model to estimate the energy. In this case, we are assuming that most of the deformation is occuring near the contact zone while the zone far apart are undeformed. We are also going to assume that the table is flat (infinite radius of curvature) and rigid (infinite Young 's modulus).

For this considerations, the radius of the contact zone is

$$r_\text{contact} = F^{1/3} \left(\frac{D R}{2}\right)^{1/3}\, ,$$

with $D = \frac{3}{4} \frac{(1 - \nu^2)}{E}$. And, the penetration is given by

$$h = F^{2/3}\left(\frac{D^2}{R}\right)^{1/3}\, .$$

Thus, the energy is given by

$$U_e = h^{5/2}\frac{2}{5D} \sqrt{\frac{R}{2}}\, .$$

And the force is

$$F = \rho g V = \frac{4}{3}\pi \rho g R^3\, ,$$

leading to

$$U_e = \frac{4\sqrt{2}\pi^{5/3}}{15} R^{14/3}g^{5/3} \left[\frac{\rho^{5/3}(1-\nu^2)^{2/3}}{E^{2/3}}\right]\, ,$$

or using the diameter $a$:

$$U_e = \frac{ 2^{5/6}\pi^{5/3}}{120} a^{14/3}g^{5/3} \left[\frac{\rho^{5/3}(1-\nu^2)^{2/3}}{E^{2/3}}\right]\, .$$

The following image shows the y stress for the problem solved using the finite element method. For this case I obtained a stored energy that is 1.5 times the analytic version, but the mesh was not too fine since the problem is nonlinear.

enter image description here


So, we have different trends. Neglecting the factor that involves the Poisson's ratio, we have

\begin{align} &U_e^\text{cube} \propto \frac{a^5 g^2 \rho^2}{E} \, ,\\ &U_e^\text{sphere} \propto \frac{a^{14/3} g^{5/3} \rho^{5/3}}{E^{2/3}} \approx \frac{a^{4.67} g^{1.67} \rho^{1.67}}{E^{0.67}}\, . \end{align}

  • $\begingroup$ Thanks that's great ! Would you have an idea how to deal with larger deformations (meaning with a flatness $<1$ significantly) also plz ? $\endgroup$ – J.A Jul 3 '19 at 18:36
  • $\begingroup$ @J.A, for common materials I doubt that you reach large deformations. The factor $\rho g/E$ is a really small number, and I don't think that you reach large deformations unless you have huge pieces. $\endgroup$ – nicoguaro Jul 3 '19 at 19:20

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