# Pressure on a compressible solid, dynamics of the relaxation

I have a sphere made of a solid of bulk compressibility B, and of volumic mass $$\rho_0$$.

The pressure around that sphere is $$P_0$$.

At time $$t=0$$, we apply a pressure on the sphere $$P=P_0+\Delta P$$. The way we apply it is with some solid stuff (a vise).

I would like to know how the sphere is going to get deformed, if we assume a spherical symmetry.

I assume we have to write (at least) the following equations :

$$\left\{ \begin{split} & \partial_t(\rho v)+v\partial_r(\rho v)=-\partial_{r}p \\ & \partial_t \rho+\partial_r(\rho v)=0 \\ \end{split} \right.$$

but I don't know how to write $$\partial_r p$$, meaning how the stress depends on the radius knowing that we know it outside.

• You said the sphere is solid, but you seem to be writing equations for the motion of a fluid. For a solid, you would usually write equations in terms of the displacement (not the velocity) from the initial configuration, and the change in density doesn't appear - the mass of an infinitesimal element of the solid doesn't change when it moves or when it is deformed. – alephzero Nov 5 '18 at 12:31
• Are you saying that you are interested in the transient response if the external pressure is suddenly increased? – Chet Miller Nov 5 '18 at 13:17
• @alephzero Well I should have said that I have a visco-elastic material. Is it ok in that case ? – J.A Nov 5 '18 at 13:21
• @ChesterMiller exactly – J.A Nov 5 '18 at 13:21
• See my answer below. If you'd like me to complete the analysis, I will. – Chet Miller Nov 5 '18 at 21:13

The equations you have written so far are incorrect. In spherical coordinates (for a spherically symmetric deformation), the continuity equation should read: $$\frac{\partial \rho}{\partial t}+\frac{1}{r^2}\frac{\partial}{\partial r}\left(\rho r^2v\right)=0\tag{1}$$ And the equations of motion should read: $$\rho\left(\frac{\partial v}{\partial t}+v\frac{\partial v}{\partial r}\right)=\left[\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\sigma_{rr})-\frac{(\sigma_{\theta \theta}+\sigma_{\phi \phi})}{r}\right]$$and $$\sigma_{\theta \theta}=\sigma_{\phi \phi}\tag{2}$$where the $$\sigma$$'s are the components of the stress tensor. These are related to the displacement in the radial direction by the 3D version of Hooke's law.

(@Chemomechanics answer gives the steady state solution to these equations, but not the transient solution. The transient solution is anisotropic in stress and strain.)

COMPLETION OF THE FORMULATION

The principal directions of stress and strain in this symmetric problem are in the radial, latitudinal, and longitudinal directions. The principal strains are as follows: $$\epsilon _{rr}=\frac{\partial u}{\partial r}$$ $$\epsilon_{\theta \theta}=\epsilon_{\phi \phi}=\frac{u}{r}$$where u is the displacement in the radial direction.

From Hooke's 3D law for the stress-strain behavior of elastic solids, it then follows that the principal stresses are $$\sigma_{rr}=\frac{E}{(1+\nu)(1-2\nu)}[(1-\nu)\epsilon_{rr}+\nu(\epsilon_{\theta \theta}+\epsilon_{\phi \phi})]=\frac{E}{(1+\nu)(1-2\nu)}\left[(1-\nu)\frac{\partial u}{\partial r}+2\nu\frac{u}{r}\right]$$ $$\sigma_{\theta \theta}=\sigma_{\phi \phi}=\frac{E}{(1+\nu)(1-2\nu)}[(1-\nu)\epsilon_{\theta \theta}+\nu(\epsilon_{rr}+\epsilon_{\phi \phi})]=\frac{E}{(1+\nu)(1-2\nu)}\left[\nu\frac{\partial u}{\partial r}+\frac{u}{r}\right]$$where E is Young's modulus and $$\nu$$ is Poisson's ratio. If we substitute these relationships between the stresses and the displacements into Eqn. 2, the differential momentum balance equation, we obtain: $$\rho_0\frac{\partial ^2 u}{\partial t^2}=\frac{E(1-\nu)}{(1+\nu)(1-2\nu)}\left[\frac{\partial^2 u}{\partial r^2}+\frac{2}{r}\frac{\partial u}{\partial r}-2\frac{u}{r^2}\right]$$where quadratic terms in the displacement have been neglected.

Boundary and initial conditions on this problem are:

$$\sigma_{rr}=\frac{E}{(1+\nu)(1-2\nu)}\left[(1-\nu)\frac{\partial u}{\partial r}+2\nu\frac{u}{r}\right]=-(P-P_0)$$ at r = R, t > 0

and

$$u=0$$ at t=0, all r

• Great ! Actually the equation you wrote is analytically solvable. If we make it non dimensional and redefine $U=u~r^2$ , we obtain : $U_{tt}=U_{rr}-4U/r^2$. Then we use the separation of variable : $U(r,t)=f(r)g(t)$. And eventually we get two equation $g_{tt}=\beta g$ which implies $g(t)=G ~ e^{-\sqrt \beta t}$ and $f_{rr}-4 f /r^2 = \beta$ and we get something of the form $f(r)=a r^{2}+b r^{\alpha_1}+c r^{\alpha_2}$. – J.A Nov 6 '18 at 15:08
• Wow. Nice. I would have solved it numerically. The equation is basically a wave equation (as expected). – Chet Miller Nov 6 '18 at 15:15
• Just to make sure : u linearized the equation and dropped the cubic (and above) terms in the displacement ? And also, what with $\rho$. Do u assume we're rather incompressible, so $1/r^2 \partial_r (r^2 v)=0$ ? Or that we're close to $\rho_0$ and we can also make the approximation : $\rho=\rho_0 + \alpha_i (\sigma_{ii}+P_0)$ – J.A Nov 6 '18 at 17:52
• I retain only linear terms (drop quadratic terms and above) in displacement. Regarding the density in the inertia term, the deviation from $\rho_0$ is proportional to u, and, since the acceleration is already linear in u, the deviation times the acceleration is quadratic in u, and is thus negligible. – Chet Miller Nov 6 '18 at 18:07
• I should mention that the compressibility effects are fully taken into account in the Hooke's law stress equilibrium terms on the right hand side of the equation. For example, for an isotropic compression, the bulk modulus would be given by $B=3(1-2\nu)/E$. Of course, for the present transient compression, the deformation is not isotropic. – Chet Miller Nov 6 '18 at 18:25

Your system of differential equations also needs an equation for the pressure. To find this, you can use the definition for bulk compressibility

$$\frac{1}{\rho} \frac{\partial \rho}{ \partial p} = B$$.

For solids it is typical to assume that the density varies only slightly with pressure. $$\rho_0$$ is the density of the solid when the pressure is $$p=P_0$$. Thus, we can assume

$$\rho- \rho_0 =B\rho_0(p - P_0)$$

since $$\frac{\partial \rho}{ \partial p}$$ is a small quantity in comparison with $$\rho_0$$. In general, if the pressure difference $$\Delta P$$ is very high, the following holds (constant compressibility assumed):

$$\rho = \rho_0 \exp(B(p-P_0))$$.

• This is the steady state solution, and the OP is asking for the transient solution. – Chet Miller Nov 5 '18 at 20:37
• By solving above identity by $p$ and substituting into the Momentum equation, one obtains a closed set of equations – kryomaxim Nov 6 '18 at 7:45
• The above equation only applies if the deformation is isotropic (equal strains in the 3 principal directions). The equation is not valid for anisotropic deformations (unequal strains in the 3 principal directions). It is a relatively straightforward matter to show that, for the transient problem, the deformation is anisotropic. Therefore, for this transient problem, your suggested procedure is not valid. – Chet Miller Nov 6 '18 at 11:40