LCS 4b - Mathematical modeling of translational mechanical systems

In the first part of this lecture we talked about mathematical modeling of mechanical systems with translational displacements. We discussed a simple mechanical system and obtained its mathematical model. We introduced the concept of mechanical impedance. In this part of the lecture we shall talk about mathematical modeling of a more complex system that is described by this diagram. We have object m1 and another object with mass m2. This object with mass m1 is connected to a spring with spring parameter k1. One end of this spring is fixed and cannot move. The other end of the spring is connected to this mass m1. There is another spring with the spring constant k2. One end of the spring is connected with mass m1. The other end of the spring is connected with mass m2. There is a viscous damper. One end of this damper is connected with mass m1 and other end of this mass is connected with mass m2. We have a third spring. One end is connected with mass m2 and the other end of this spring is fixed. These two objects with mass m1 and m2 are moving on this fixed surface. Therefore, there is also friction between this moving object m1 and this fixed structure. Likewise, there is also friction between this object and fixed structure. The friction coefficients are Fv1 and Fv2. We want to obtain a mathematical model for this mechanical system. The first step in obtaining mathematical model in mechanical systems is to identify the linearly independent displacements. So, we can see that this end of the spring is fixed and cannot move and this end is moving when some force F of t is applied on this object. So, let's call this displacement as x1 of t. As you remember, we were considering rigid masses. That is, this end of the mass and this end, both of them have the same displacement. Therefore, this displacement is x1. Also, this end of spring will have the same displacement but the other end can have different displacement. Thus, we name the displacement of this end as x2. Likewise, for the damper, this end has displacement x1. The other end can have different displacement, let's call it x2, the same as this mass. Both the ends of this mass, this side and this side, they have the same displacement and that is therefore x2. For this spring, this end has displacement x2 and the other end is fixed. You remember that for obtaining mathematical model, we first draw a free body diagram but over here, situation is a little bit more complex. In the previous example, one end of the spring was fixed and the other end was moving but over here, both the ends of the spring are moving. Likewise, both the ends of the damper are moving. So, how to analyze such situations? To analyze such situations, what we do is, we apply superposition. That is, in obtaining the free body diagram for this displacement, what we do is, first we keep this object fixed, that is, this object is not moving and determine all the forces which are acting on this object. In the second step, what we do is, we keep this object fixed and move this object in this direction and determine all the forces which are acting on this object. This is demonstrated over here. We have object m1 and if we keep this m2 fixed and move m1 in this direction, what are the forces acting on this object? Firstly, there is a force applied by this spring. If this object is moving towards right, this spring will apply a force towards left and what will be the magnitude of that force? The magnitude of this force will be proportional to this displacement x1. That is, the force exerted by this spring on this object is towards left and its magnitude is k1 into x1 of s. We are directly writing everything in s domain. Similarly, the force exerted by this friction that is proportional to velocity and that is given by Fv1 s x1 of s. There is external force applied to this object which is F of t. Now, the force due to inertia that is proportional to acceleration in the object and it always opposes the direction of displacement. Therefore, its direction is towards left and its magnitude is proportional to acceleration in Laplace domain s square multiplied by x1 of s. Now, to determine the force due to this damper, we have first kept this object fixed and moved this towards right. If this object is moving towards right, this damper will apply a force towards left and the magnitude will be proportional to the derivative of this displacement r. In Laplace domain, that force is given by Fv3 into s multiplied by x1 of s. Similarly, the force exerted by this spring, we have kept this fixed and we are moving this towards right. So, if this object moves towards right, this spring will exert a force on this object towards left and its magnitude will be proportional to the displacement x1 of s. This diagram, free body diagram shows the forces which are acting on this object when this object was kept fixed and this was moved in the direction of x1. Now, if this is moved and this is kept fixed, then this diagram shows the forces acting on this object. If we move this object towards right, this spring will tend to pull this object, that is, it will exert a force towards right and the magnitude of that force is given by k2 x2 of s. Likewise, this viscous damper will also exert a force towards right and the force exerted by this damper will be proportional to the derivative of the displacement r. In Laplace domain, the force exerted by this damper is Fv3 s into x2 of s. So, this is free body diagram for this object when this was kept stationary and this was moved towards left and this is free body diagram for the same object when this object was kept stationary and this object was moved in the direction of x2. The overall free body diagram is obtained by combining these two free body diagrams which is done on the next slide. Here, these are the two free body diagrams which were shown on the previous slide. By combining the two, we have this free body diagram, that is, k1 and k2, both the forces are towards left and therefore are lumped together in this term. Likewise, Fv1 and Fv3, both are acting towards left and therefore are lumped over here. This external force F of s and this is the force due to inertia, kx2 is this force and this term is over here. That is, the overall free body diagram for the first displacement x1 is obtained by combining these two free body diagrams. Now, once we have this free body diagram, writing the corresponding force balance equation is not a difficult task. Force balance equation can be obtained by writing the sum of forces acting towards left that will be equated to sum of the forces acting towards right. That is, the sum of the forces acting towards left and sum of the forces acting towards right. This force is acting towards right but it is brought to the other side of equality, therefore a negative sign over here. Up to this point, we have only modeled, we have obtained the differential equation or the equation corresponding to displacement x2. The same procedure will be repeated for the other. That is, we had obtained free body diagram for this object and now we obtain free body diagram for this displacement. Again, the procedure is same, we apply superposition, first we keep this object stationary and move it in the direction of x2 and determine what forces are acting on this object and then we keep this object stationary and move it in this direction to determine the forces acting on this object. So, those forces are shown over here. This free body diagram corresponds to the situation when this object was kept fixed and this was moved in the direction of x2. If we move it in the direction of x2 and keep it stationary, what will be the direction of force that will be exerted by this spring on this object. If this object is moving towards right, this spring will apply a force towards left and the magnitude of force will be proportional to this displacement. That is written over here, k2 x2 of s. Likewise, for this friction, if this object moves towards right, the friction force will oppose the direction of displacement and the magnitude will be proportional to the velocity of this object or in Fs domain, we have Fv2 s into x2 of s. For this damper, if this is kept fixed and this moves towards right, this damper will apply a force towards left and magnitude of the force will be proportional to the derivative of x2. That is, this force is Fv3 s x2. The force due to inertia, that is given by m2 s square x2 of s. And if this object moves towards right, this spring will apply a force towards left, the magnitude of that force will be proportional to this displacement x2. Therefore, the force exerted by this spring is k3 x2 of s. This is free body diagram for this displacement, when this object was kept stationary and this was moving towards right. Now, we repeat the procedure. We keep this object fixed and move this object in this direction and determine what forces are acting on this object. So, if we move this object towards right, this spring will tend to push the object, that is, this spring will apply a force on this object towards right and magnitude of the force will be proportional to the displacement or the compression in the spring, that is, k2 x1 of s. Similarly, for this damper, if this object is stationary and this object moves towards right, this damper will tend to push this object, that is, it will tend to apply a force towards right, the magnitude of that force will be proportional to derivative of this displacement or in s domain, the force is Fv3 s x1 of s. So, these are two free body diagrams obtained by first keeping this object stationary and this moving and then this diagram obtained by keeping this object stationary and this object moving. The overall free body diagram for this displacement is obtained by adding these two free body diagrams and that is shown on the screen. We have the forces, all the forces shown over here obtained from the two free body diagrams on the previous slide. Thus, the forces acting towards, sum of the forces acting towards left is equal to sum of the forces acting towards right and when these terms, these terms brought to the left hand side, so we have a negative sign over here. Since there is no external force acting on it, so here we have this term equal to zero. So, overall mathematical model for the mechanical system is described by these two equations. These equations were explained on previous slides and these two equations comprise of the total overall mathematical model for that mechanical system. Now, if you are interested in obtaining a transfer function between x2 of s and F of s, we can proceed in different ways. We can either eliminate x1 of s from both the expressions and finally obtain a relation between x2 and F of s. Otherwise, we can apply Cramer's rule and by Cramer's rule the transfer function is given by this expression where delta is this determinant. You need to do this at your home. This is your homework. Expand, apply the Cramer's rule to obtain this transfer.