First and foremost, I might say horribly wrong things. Feel free to correct any inconsistencies in the following post.

Let's assume an incompressible, viscous, newtonian fluid that is not moving and enclosed by a solid surface.

The total stress tensor of an incompressible, viscous, newtonian fluid is defined by :

$\sigma^f(u,p) = -p.Id + \mu (\nabla u + \nabla u ^T)$

As the fluid is static, we have $\nabla u = 0$, hence the total stress tensor is reduced to its hydrostatic part $\sigma^f(u,p) = -p.Id$

In Fluid/Structure interaction settings, we define a fluid problem ($\mathcal{F}$), a solid problem ($\mathcal{S}$), and we define a system of coupling condition ($\mathcal{C}$) which are usually defined by a kynematic condition ($\mathcal{C}_1$) and a dynamic condition ($\mathcal{C}_2$), with $\mathcal{C}_1$ and $\mathcal{C}_2$ defined by :

  • $(\mathcal{C}_1) : u_f = u_s$

  • $(\mathcal{C}_2) : \sigma^f(u,p).n^f = \sigma^s(d).n^s$

Up until here, I do think that I understand what is happening.

In partitioned solvers, I read numerous times that the fluid solver is transfering forces to the solid solver, and that the solid solver is transfering displacement to the fluid solver. I am rather unsatisfied by this because of $\mathcal{C_2}$. Going back to the case of my incompressible enclosed fluid stated above (again, stop me if my interpretation is wrong), if one of the sides tries to compress the fluid (like a piston), the fluid will not move nor will the side of the solid. However, the intracavity pressure will rise, as will the stress inside the solid.

My question are the following :

  • Was I wrongly understanding how and what data had to be transfered during iterations of a partitioned FSI solver ?

  • Assuming a fluid problem where you would not like to compute the solid behavior. What type of boundary condition would be used to impose both the velocity of the fluid at the solid boundaries AND the stress ?

Thanks in advance for anything that could be useful to my understanding

  • $\begingroup$ "... if one of the sides tries to compress the fluid... the fluid will not move nor will the side of the solid..." Can you explain what you are trying to say here? If a piston pushes on the fluid, the solid would definitely be moving because it's a piston. And the fluid would be moving out of the way of the new position of the solid. This would send compression waves through the fluid towards the side opposite the chamber. This all serves to increase the pressure in the chamber. $\endgroup$ – tpg2114 Jun 1 '16 at 15:29
  • $\begingroup$ In the original setting, I assume the fluid to be incompressible. Under this assumption, the fluid cannot be compressed by the piston. I would assume that the piston force is transmitted through the interface and, following Pascal's law, the pressure increases everywhere, right ? $\endgroup$ – Al_th Jun 1 '16 at 15:46

Question 1: The fact that some solvers use that approach does not mean that it is robust to any condition. Indeed, the fluid subproblem as you state it has non-admissible BC. There are definitely solvers which are monolithic and solve both problems simultaneously, if in addition the pressure is imposed via a Lagrange multiplier approach then it should handle the case you consider.

Question 2: You would not prescribe both. You would prescribe either and obtain the other as a result of the computation, which is exactly what you do with your separate subproblems. But note that still, if you impose a Dirichlet BC on the whole of the boundary of an incompressible material, you have to check that the corresponding deformation preserves volume.

EDIT: If all BC are Dirichlet, then in an incompressible fluid the pressure is set up to an additive constant. This is a mathematical result: there is simply no information on the average pressure in the domain. This means that the pressure of a motionless fluid is undetermined in this case.

That said, depending on how you discretise the problem, the outcome will be different. Many fluid solvers impose Dirichlet BC weakly by means of a Lagrange multiplier, in the case you describe the result would be a volume-preserving effective displacement of the boundary -- but then if the solid solver does not take into account the final boundary displacement, you will have a mismatch.

  • $\begingroup$ 1. Yeah, I was indeed aware of monolithic approaches. In the monolithic approaches, I don't have any comprehension problem as all constraints are part of the system that is solved. 2. Thanks for your explanation but I'm still lacking comprehension. You say : "You prescribe one and you get the other as a result". If my piston cannot compress the fluid (aka u = 0 everywhere), I don't see how I could "get" a pressure that is linked to the force that I applied on one of the sides if I do not prescribe the force. $\endgroup$ – Al_th Jun 2 '16 at 6:15
  • $\begingroup$ OK, see my edit. $\endgroup$ – Joce Jun 2 '16 at 7:46

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