[MUSIC] Hello, we are presently considering small motions of a solid in a fluid, in the particular case where the main fluid velocity can be neglected. I have illustrated that by the oscillation of a boat on a lake or by the vibrations of an inflatable dam. These two conditions as you know, correspond to a specific region of the space of our dimensionless parameters. Small motions mean that the displacement number is much smaller than one. And negligible fluid velocity means that the reduced velocity is much smaller than D. Moreover, we have considered description of the motion of solid by a single mode. In this framework, the equations are much simplified. Here they are. On the fluid side, the solid side, and at the interface. They just relate the motion of the solid governed by q of t, and the corresponding motion in the fluid where we have the velocity small u and the pressure small p. Note as I'm now using the simple notation dot for the time derivation when a variable only depends on time. So q dot means dq over dt. We have shown that the fluid force acting on the solid consists of two terms. The second one is actually a simple stiffness force. Which depends only on the static pressure gradient in the fluid. And we computed this stiffness in the case of an ice cube or an iceberg. That was easy. Let us now explore the first term of our fluid force. What is it made of? It is a sum over the interface of the force resulting from the pressure p and the velocity u in the fluid. You can recognize the pressure loading and the viscous shear loading. So schematically, this term corresponds to the effect on the solid of the motion induced in the fluid, by the motion of the solid itself. It is a feedback term which depends on how the fluid reacts to the motion of the interface. To compute this force, of course we need now to compute the motion of the fluid. This was not the case for the stiffness terms we analyzed before. I recall here the equations that we have to solve in the fluid domain. We have the mass balance, and the momentum balance. At the interface we have the kinematic condition on the velocity. And the dynamic condition which is going to give me the force. For the sake of clarity, I have removed the stiffness term because I know how to compute it. These equations in the fluid domains are simple because I've considered only small motions, and everything is linearized. But there is a dimensions parameter which appears explicitly here and there, the Stokes number. What is it? You remember that we introduced the Stokes number in our new series of dimensionless parameters. It scaled the viscous effects. Actually, it is equivalent to the Reynolds number but not based on the mean velocity, which is neglected here, but based on the velocity of the solid. This is why it appears in the equations, to bring the effect of viscosity in the momentum equation and in the fluid force. What is the value of the Stokes number? [MUSIC] The Stokes number depends on the fluid viscosity, mu, the fluid density, rho, the length scale of the problem, L, and the velocity of elastic waves, c. Or equivalently remember a velocity defined by L over T solid. Imagine the case of oscillations of the small boat that I used before. We have L equals about one meter. Mu/Rho for water = 10 to the 6 meters squared per second. And the order of magnitude of the period of the oscillations is say a second. This would give me a Stokes number of 10 to the 6. This is a very large dimensionless number. And of course in such a case, the coefficient 1 / Stokes is going to be very small. And corresponding term in the equation can be neglected. In practice there are many cases where 1 / Stokes is going to be very small. So let us now make the approximation of high Stokes number in our equations. The equations simplify in several ways. First, in the fluid domain, the momentum balance equation simplifies to du / dt =-gradient of p. The viscous term being neglected. Second, in the dynamic condition at the interface, the effect of viscous shear is neglected. And we only have the sum over interface of the pressure loading times the boat shape. Finally, the kinematic condition at the interface must be changed. Why? Because in the absence of viscous effects, it is just impossible to enforce a condition on the tangential components of the velocity. The only condition that remains is that the normal velocities must be equal, meaning that there is no mass transfer or separation at interface. These equations have solutions in a very simple form. Note first that at interface the kinematic condition implies that the normal fluid velocity takes the form of a function of time, q dot, times a function of space, Phi n. This comes from the single mode description of the motion of the solid. We may then look for a fluid velocity that would have the same form, not only on the interface but everywhere in the fluid. Of course, the time dependence would be q dot. The space dependence is unknown and noted Phi u. The same idea: because pressure is related to accelerations, let us look for pressure field p(x,t) as q double dot times a function of x say Phi p of x. Then, the mass balance implies that div of Phi u = 0. In the momentum balance, the q double dot cancel out. And we have Phi u = minus grad of Phi p. At interface, the kinematic condition is simply Phi u n = Phi n. And in the fluid induced force, because I can take the q double dot out of integral space, we obtain a very simple form which I'm going to command now. [MUSIC] What we have here is a major result of fluid-structure interaction theory. As you can see, the term in brackets is actually a constant because we have the mass number times a sum over the interface of quantities that depend only on space. So in this approximation, the fluid force resulting from the motion of the solid is an inertia force and the coefficient of inertia is called the added mass. What is also remarkable is that the response of the fluid to the motion of the solid is instantaneous. I mean that the force at a given time only depends on the acceleration at that same moment. So, schematically, we can say that the problem on the left, where we have a solid moving in interaction with a fluid reduced to a very simple problem on the right of the same solid with just an augmented mass. This is much, much simpler. All we have to do is to compute a single quantity, this added mass, mA, once and for all. Then, for any motion along this mode, I know that the coupling with the fluid reduces to adding a mass effect on the dynamics of the mode. Before going further in this direction, Let us look back at the steps of simplification that we have done. Originally in the very general case, we had a set of dimensionless numbers that were the displacement number, the Froude number, the Reynolds number, the Cauchy number, and the reduced velocity. We then choose an alternative set. That would be more suited to the case of Rubio's velocity, and that was the displacement number, the dynamic Froude number, the Stokes number, the mass number, and the reduced velocity. At this stage, no assumption was made. We then define the domain of low reduced velocity where the proper velocity of the fluid could be neglected. Then Ur did not appear in the equations anymore. We then look at the case of motions of small amplitudes. D much smaller than 1. And D cancelled out everywhere in the equations. And now by assuming low viscous effects, the Strokes number disappears. In this framework of approximations, we have a fluid force that only depends on the dynamic Froude number and on the mass number. And that force is made of a stiffness term and then an added mass term. So to summarize, we know that coupling with a still fluid for small vibrations, and without viscous effect results in just an increased stiffness and an increased mass of the vibrating system. We know how to compute the added stiffness. Let us see next how we can compute in practice the added mass. [MUSIC]
