[MUSIC] Well we have simplified the general problem of fluid solid interactions by considering the limit case where the velocity of the fluid is very small so that we can neglect its influence. When considering the motion of a solid in still fluid, there is also a very interesting simple case which is that of small amplitudes. For instance, what is the influence of coupling with the external fluid on the vibrations of our flexible dam? Or even simpler, maybe coupling with the external fluid can modify the stability of an equilibrium position of a solid. We know it does, for instance, when a boat capsizes. You remember that we have a dimensionless parameter that quantifies the magnitude of the displacement of the solid. It is the displacement number, D =Ksi0 over L. The ratio of the typical displacement with the size of the solid. D, much smaller than 1, means that there is little change of shape in the solid and certainly this will be a simpler case to solve our coupled equations. Let us therefore explore the problem of a very small displacement number say D much smaller than 1. By the way, to make equations easier to read from now on I shall omit the bar over the variables. But remember they are dimensionless. [MUSIC] Because we have a small parameter D, we can write expansions of the quantities as zero order in D, first order, and so on. For instance, in the fluid, let us write the pressure, P, as P = P0+ D p and so on. P0 is the zero order term. And small p is the first order term in my expansion. The velocity may also be developed. I can already simplify this because as the other zero, U is equal to zero. Why? Because under my approximation, if the solid does not move, D =0 then, than the fluid does not either. So U equals simply 0 + D times u. In the solid now, the dimension of displacement Ksi, has already been written as D q phi, when we used a single mode approximation. I'll write this again as Ksi equals 0 + D q Phi. By putting all these variables in the equation, we should be able now to write them at the order 0 in D, at the order one in D, and so on. And hopefully, this will be much easier to solve. [MUSIC] Let us do the order zero first. That means what happens at D equals zero. In other terms, once I have inserted the extensions in my equations, I look only at the terms where D does not appear. Here are the equations in the fluid, I just substitute my expansions of U and P. And here is what I have at the order zero. The mass balance is satisfied by div of zero equals zero. The moment just relates now the pressure field P0 with the gravity. This shows that although there is no velocity in the fluid at that order there is a pressure which depends only on the gravity force. This is your well known Hydrostatic pressure. [MUSIC] The equations at the order one are much more interesting. I take the same equations. I make the same substitution. But I gather now only the terms that are proportional to D. I obtain then the equations on this velocity small u. and pressure small p, which are my variations of the pressure and velocity in the fluid when the solid moves. The mass balance is divergence of u equals 0. The momentum balance is much simplified. First the convective term in the acceleration disappears because it will give a D square term second the gravity is not here anymore because it is a constant, therefore, a zero order term not order one. These equations are satisfied in the fluid domain. Now, let us look at what happens at the interface. [MUSIC] At the interface, the kinematic condition is simple. The fluid velocity should be equal to the solid velocity. And so, because we have D coefficient on both sides, we get u = dq over dt times Phi. For the dynamic condition we have to be a bit careful. Let us see why. You remember that the dynamic interface condition that we have used allows to derive the fluid loading projected to the mode of interest. This was done by summing over the interface, the product of the local fluid loadings times the model shape phi. Note that this integration is of course done on the instantaneous position of the interface. But where is the interface? It is deformed, and that is the purpose fluid solid interactions. The point X has moved at the position X plus Dq Phi, and so we should also expand the geometry at the first order in D. We have to be careful there in considering this change of geometry. There exists a general way to do this using continuum mechanics. But we are going to do the simplest case. Let us imagine that the displacement of the solid is a pure translation. I mean, Phi is a vector, independent of the point of interest. In that case, I can take out the Phi from the integral, and focus on all of the terms. Moreover the normals, n, are unchanged. What I want to do is to take advantage of the smallest placement of interface to use quantities that are defined on the original a undeformed position of the interface. For instance, because D is small the pressure at the new position maybe developed as P equals P0 at at the new position, x plus Dq phi plus Dp at the new position x plus Dq phi, and so on. Hopefully things simplify, because the pressure of P0 at the new position may be estimated by the pressure at the original position plus its gradient times the displacement. And the first terms of the velocity remainsthe same as before. Now, all the quantities involved are defined as at the reference undeformed X. We now just have to put this P, and this U in the integral that defines the fluid loading. Let us look just at all the terms proportional to D the first order the terms. We have a lot of terms in the results. But what is important for us is that the integral is now on the original position of interface because all variables are taken there. Here is the full result. We have two very different terms in there and I can see now what they mean physically. The first term, what we have is a projected force on the solid, resulting from pressure and viscous stress in the flulid. Remember that this pressure and velocity small p and small u, are those directly caused by the motion of solid. We expected this. It is just the feedback from the fluid to the solid motion. The second term is quite different, and a bit surprising, I must say. It does not depend on p and u, but only on the order zero-pressure P0. It means that it is not related to the motion of the fluid induced by that of the solid. But to the motion of the solid in a pre-stressed fluid. This is very different. [MUSIC] This is a very important result. Imagine a fish under water. A resting fish that moves just a bit. He is experiencing two kinds of new forces on him. Some are due to the effect on him of the flow he's creating himself around him. And some forces are just due to the fact that he enters a zone where the pressure is different. We have computed this term, in the simplest case of a translation in the fluid, by expanding only the first stress, in terms of the position of the interface. Actually we can do the general case, but this is much more involved. The local normal may change, and the local area where stress applies may change, as an inflated balloon. And phi is generally not uniform. Here, I just give you the term corresponding to the gradient of P0 in the case of a marginal deformation of the body. Can we summarize all our first-order equations? Here they are. They will be much simpler to solve because they are linear. Why are they linear? Because we linearized the original equations when taking advantage of the small motion. What you see here are linear equations that couple the fluid and solid dynamics. They can be solved now. You might be a bit impatient to start solving real problems. I understand. But in this course, we want to solve cases, not just one by one, but as much as possible, all in the same time, if we can. Next, we are now going to see some very important results that can be derived on all fluid solid interaction problems at low reduced velocities in small motion, regardless of the shape or type of motion of the solid. For a fish a boat, or a dam. It was worth going through the equations. [MUSIC]