 Okay, good afternoon. So, myself I am Professor Shobik Banerjee. We are both here right now and for the next hour or so, what we are planning to do is we have covered so many different topics. So, last week we covered a huge number of topics in great details on statics. We started with basics of vector mechanics, equivalent force movement systems, then we did centroid, distributed force systems, 2D equilibrium, machines and frames, then 3D equilibrium, trusses, friction, complex friction, principle of virtual work and principle of minimum potential energy. So, these were the topics which we discussed in great details. Then we also took various quizzes, solved a lot of tutorial problems and then starting this week we discussed a lot of dynamics, particle kinetics, particle kinematics, impulse momentum theorem, work energy theorem, kinematics of rigid bodies, kinetics of rigid bodies. And vibration dynamics and vibrations, we did that today. So, now there are so many different topics, essentially if you think about it then it is almost close to two semesters worth of syllabus that we had essentially covered in 10 days. So, what we will now try to do is if you have any particular queries, any questions, any doubts which were not clear either during the lectures or which some questions were asked on Moodle, some answers were posted and if you want to ask them again. So, our suggestion is that we will follow the best optimal mode because the time is limited. Why do not you send those queries on the chat? We will have a look at them and we will start answering those questions one by one. So, I request all the coordinators to collate the questions from their respective centers, send them our way via chat and we will start to answer them. Yes, so one of the questions that has been posted, what is the effect of intrinsic damping on natural frequency? So, first of all let us understand what is intrinsic damping? It is driven by the material properties, nothing else. So, mostly it is the material damping that is referred to intrinsic damping. So, for example, let us say you take a continuous system. Now, in civil structure remember we talked about that water tank problem. In the water tank problem if you have that tower and tank on the top, then essentially we are trying to do that as a single degree of freedom system. The question is where does damping come into play? Damping will come into play through the material. So, therefore, ultimately we do have still the equation of motion that is mx double dot plus kx plus cx dot equals to 0. So, now that c that is really coming from the material damping now. So, in the sense nothing happens you know that omega d equals to omega n square root of 1 minus xi square. So, that whatever we have derived that damped frequency equals to omega n square root of 1 minus xi square. So, ultimately your xi that for civil structure which is coming through the material damping can be up to 5 percent that is the damping ratio. So, ultimately xi would be at max to be 0.05. So, as such it is going to affect the damped frequency of vibration. So, there are two questions one is from 1010. Please explain inertial and non-inertial frame of rotation. See I think you know we discussed that in terms of particles actually know how D'Alembert principle is coming into play. And we said that that if you look at fixed reference frame let us say x y and the particle is accelerating with respect to that fixed reference frame x y and z. Then remember there is a resultant force and the acceleration direction is along the resultant direction. So, particle actually see an acceleration which is its absolute acceleration. So, we can use sum of force equals to mass times acceleration. Now, in a non-inertial frame what is happening that let us say observer itself is in the moving body. So, that means what we now said we have another system that is small x and small y. Now, if that frame is actually moving with the particle then what happens that it is not just sum of force that is acting on the particle, but we do need to take into account that there is a force which is a fixed reference frame inertia force that we call the inertia force that should enter as negative mass times acceleration. So, as such then we said that that means as an observer you are actually at rest that means you are actually at equilibrium. So, this will be equilibrium and this will be motion. So, that is the difference you know in simple sense between the inertial frame and non-inertial frame. So, these two is going with the delambert and this is going with along with the Newton's second law. Now, other question that is being posed is let us say 1148 1148 I think 1148 the question is what is the basis of number of members of truss that is the number of basis of 2 n minus 3. So, remember that m equals to 2 n minus 3. Now, this one we have really adopted to construct a simple truss. What was a simple truss? Remember to form a truss first of all I need minimum of 3 members and 3 pins. So, therefore if you count now this is satisfied why because number of member is equals to 3 ok number of joint is equals to 3. Therefore, we see that this relationship is satisfied. So, this is called basic truss it is a simple truss. Now, main point is that if I really draw if I extrude 2 members and connect by pin then remember this is again a simple truss. So, what was the basic philosophy behind simple truss? A simple truss is internally rigid internally rigid means under the action of the load if I do apply any load the internal member should not change angle that means the truss will not collapse. So, that is internally rigid. Now remember the number 3 appears therefore what is happening if I want to make this problem statically determinate then what do I have to do? If I want to make it statically determinate and properly constrained then I do need 3 reactions. So, do need at least 3 reactions has to be there, but these 3 reactions has to be in such a way that the entire truss will behave like a rigid body and it should be properly constrained. So, what we said therefore most appropriate would be putting a hinge and let us say putting a roller at this end. So, in this way this truss will be properly constrained as well as it is a statically determinate now why it is determinate? Now m plus 3 equals to 2 n. So, the number 3 that is the additional unknown that is coming from the reactions 3 reactions right. Now remember then what we have discussed that it is not always necessary that we have to always have 3 reactions that is absolutely not true. So, what can also happen most general class could be m plus r equals to 2 n where m is the number of forces r is the number of reactions and 2 n is the n is the number of joints. So, therefore, what can happen is the total number of unknowns should be total number of equilibrium equation. So, that was the main purpose remember when this is coming into play we can think of problem even just think of is this a truss very simple example remember if I make it a hinge here and if I make it a hinge here again I think that is not coming. So, I am just drawing it here just take a look suppose I make it a hinge here and if I make it a hinge here just see whether it is a statically determinate problem or not. So, in this way I cannot really say that m equals to 2 n minus 3 and all that what I need to look at I have to make sure that m plus r equals to 2 n remember there is one hinge here one hinge here 4 reactions are coming into play, but even with the 4 reactions I do have a statically determinate system. So, it can be solved for a loading p all the member forces and all the reactions can be solved clear. So, m equals to 2 n minus 3 that is just taken to define a simple truss and the number 3 is when you have exactly 3 unknowns that is 3 reactions supporting the truss to make it properly constrained and statically determinate, but it is not always necessary that we have to always go through m equals to 2 you know m plus 3 equals to 2 n we can go to more general m plus r equals to 2 n to solve this problem this is also statically determinate properly constrained. So, from that we have actually went and looked into also partially constrained improperly constrained truss and then we looked at when m plus r is basically you know less than 2 n right like that we have studied also that is partially constrained ok. Is that clear now? Can you give some examples for 1, 2, 3, 4 and 5 degrees of freedom systems? Can we give now a small problem to solve 2 degree of freedom system if you would like? So, can you derive in this case what are the equations of motion is it possible to derive using the D'Alembert principle concept? What are the equations of motion for this system? How do you define the degree of freedom for a dynamic system? So, I think if you can simply you know give us the equations of motion for this system that will be a good exercise for all of us to understand how the you know whatever we have discussed how that can be taken up for further you know studies for 2 degree of freedom or even 3 degree of freedom systems ok. So, can we try to can we expect some answers for this? Now, we gave a little bit of task if any of the centers please help your colleagues also those who are asking that will be appreciated ok. It is not that we also you know try to answer all the questions any remote center please go ahead and try to pose the solution of this problem which is now being shown ok. So, can you please one of the centers at least help your other colleagues just the equation of motions for this ok just writing down the answer, but you can give it a try thing one of the participants some of you have given the correct answers I am looking at center number 1100 1100 yes the answer is correct. So, m 1 x 1 double dot plus k 1 plus k 2 x 1 negative k 2 x 2 equals to 0 then another equation of motion will be m 2 x 2 double dot negative k 2 x 1 plus yeah let me go over a couple of doubts ok. So, one doubt has I was meaning to answer it yesterday somehow could could not get around to do that. So, the one doubt is that that yesterday we discussed about instantaneous center of rotation ok. So, suppose we have a rigid body like this what we discussed yesterday is suppose there is a point a which has a velocity a in this direction and a point b which has a velocity in this direction then how do we find out if these two velocities have different directions then we saw that we just draw a perpendicular from this line and draw a perpendicular from this vector and this particular point where those two perpendicular lines intersect we said that this was c or the instantaneous center of rotation. But now a new question but another question is that what if this v a and v b have the same direction. Now, in that case let us take this point a which has the same velocity which has the same direction for the velocity as at point b this is v b and this is v a. In that case the instantaneous center of rotation has to be along this line perpendicular, but a location cannot be immediately obtained. So, wait one way to obtain that location is to join these two outer extremities and this point c will be the instantaneous center of rotation. And now what is immediately clear is that by proportionality theorem that a c by b c should be equal to magnitude of velocity of a divided by magnitude of velocity of b because these are similar triangles. And if you know what is the magnitude of velocity at a magnitude of velocity at b then straight away you can find out what is the instantaneous center of rotation that it has to be perpendicular to this direction and the length should be how much such that this ratio will be satisfied. So, that is the answer. Now, what happens if v a is equal to v b? If v a is equal to v b then this becomes like this and you realize that instantaneous center of rotation is now at infinity or in other words this is pure translational motion and there is no rotational component that is involved here. So, I hope that this point is clear. There is a question that is asked by center number 1298 and the question is in virtual work principle we have the assumption of small deflections. So, I take it that this is actually a question. So, it is very important that the deformation that are provided or the not a deformations, but the virtual displacements that are provided are small. There is a very detailed and a complex answer for this even though for example, if you look at this simple example for example, if you take a beam like this with a roller and a pin support and we apply some force then if you want to apply principle of virtual work to find out what is the reaction at this support b. Then if you recall what we had done that we release this support and this became our active force diagram or a f d and this was replaced by reaction at b. This was p and this was q and what we did is that that we did we provided this a small rotation of delta theta about point a and then what we saw is that that because this rotation was small. And also because of from whatever discussions we had done yesterday about the kinematics of rigid body we said that if this length is l by 2 and this is l by 2 then this vertical displacement at point b is simply equal to l delta theta and the vertical displacement at the midpoint where the load p is applied is simply l by 2 into delta theta. So, the rotation was small now you may ask me that what if I give a large in virtual if I give a large rotation that a rotation is this. You will see that if the rotation is large then this assumption about the displacement of this point perpendicular to this line and the magnitude means l delta theta will no longer be true. If this and exerts as I leave it to you you give it a very large rotation of theta then what will happen is that all these points even the point where the horizontal force is applied will get a reasonable amount of displacement in the horizontal direction. And now if you try to apply the principle of virtual world you will see first of all that when the rotation is large this will be arc of a circle this was point p. Now this point p actually moved here. So, rather than moving just upwards it also had a simultaneously a horizontal direction. So, what will happen is that when we use principle of virtual world this load which is applied here q will also do virtual world. Similarly, the virtual displacement of this point will no longer be equal to l times delta theta it will be equal to l sin theta. And now you try to solve this problem you will realize is that you will get complete garbage answers. And there is a very deep reason for that it is that infinite symbol rotation is a vector and you can then express everything in terms of that vector. Whereas, a finite rotation is a tensor and then everything will go wire go hay wire if you use that. So, in the principle of virtual world all the displacements and particularly rotations should be infinite symbol or very small and small means small compared to the overall dimensions of the structure. That is another question why principle of transmissibility cannot be applied on a truss problem. What we do I think can any of the remote centers could you please answer to this question why principle of transmissibility cannot be applied on a truss problem. Any of the remote centers can you physically explain then raise your hand please you can also try to respond to your colleagues. If you raise your hand raise means just say yes on the chat then I am going to go to that center is that center is ok. Let us see also you would like to hear from you as well what is your thought. Yes go principle of transmissibility is applicable only for mechanics not for some not for internal analysis purpose. If you take a rod you take the forces external forces on both ends that rod is under suppose compression. So, if you use principle of transmissibility you use those forces and drag them on the other end what you see as a tension sorry compression will become tension or if it is under compression if you drag them on the other end that will be external vice versa it will change. So, principle of transmissibility is only applicable for external analysis purpose and not for internal deformation or internal forces. Yes so that was a so the main point is that so physically what it is if you treat the entire truss as a rigid body ok. Then principle of transmissibility can be used to get the reactions let us say I want to solve for the reactions by treating the entire truss as a rigid body, but when it comes to the internal members I cannot put that force external force internal analysis when I am performing then I cannot take it to the individual member of the truss and try to solve for the unknown forces. So, everyone listen to the answer so we also physically explain that therefore principle of transmissibility cannot be used ok for truss analysis when we are looking at the internal members right trying to solve for the forces. There are again questions coming in the form of in case of a truss ok. Why do we consider pin joint in truss while in real structure we provide weld or bolt joint which is fixed. Now remember we have explained this and the point is that there are two simultaneous things happening one is that all the members are connected at a joint and they are concurrent ok that is the first thing. Second thing load is also applied at a joint ok. The other issues are of course that we are not applying any lateral load on the members which are very slender in nature. Now again it is a very much it is a assumption that we assume it pin the reason being we want to say that since the load is applied at joint and we have concurrent force system coming into play therefore what happens basically we do not really have any movement in the joint although in real sense there could be some movement in the joint but joints are you know always have a gusset plate right and they are able to take this movement. So it is again a very you know rough assumption why we are assuming the joint to be pin. Remember the majority will be carried by the axial force the major force that we see is the axial force I am not arguing that there is no bending you know part to it there is a bending ok but if you think of the energy point of view most of the energy will be taken by the axial energy and bending energy will be very small. So but concept really is that load is applied at the joint and we have concurrent force system and therefore at the joint if you look at the equilibrium point of view there is no movement ok although you may have bolted or fixed connection over there. Remember there is no lateral load that is being applied at the member ok. Last but not the least since we have covered the structural dynamics and those who want to take this forward let us say I have not discussed this but you know for both civil engineering applications and let us say for mechanical engineering application and so on I will just give two different problems ok and that can be you know you can take this forward what I would urge all of you for ground excitation ok. So I am just keeping two problems ok. Suppose very basic problem in civil engineering SDOF system I have a lumped mass here two columns, columns can be taken as springs so that means you have stiffness for this column ok. Suppose I excite it UGT that is the ground displacement that is equals to let us say it has some form that is you know ground acceleration is coming in the form of let us say U0 omega square sin omega bar t that is one problem. So in other words you do have a ground acceleration in this case ok. The other problem let us say suppose I have a vehicle ok and that vehicle this wheel it is again you have the stiffness KK ok and you have mass alright. Now suppose it moves at a velocity V and suppose this ground undulation that can be described so therefore you have a periodicity let us say of the undulation is L that length is L and also let us say that ground undulation can be described by this D right and L this vehicle is moving at a constant speed can you not attempt to solve the equations of motion ok. So this is something that will go naturally in the next phase of structural dynamics when we talk about forced vibration. In both cases I do have a forced vibration problem how this force is coming very simple example is earthquake is happening in this case let us say so I have ground acceleration in this case right which has this form again a periodic form. In this case the vehicle is moving at a constant velocity again a undulated surface ok. So therefore I do have ground displacement can be derived right so in some sign omega bar t. So question is what would be the equation of motion for this mass or this system that will be a natural part that we should take forward ok. Now I am not going to really look at how the problem will be solved but remember in this case what is important to realize I again repeat what is important to realize the net displacement of the mass in both cases that has to be decomposed into total you know the two parts the net displacement should be the ground displacement ok plus the relative displacement of the mass with respect to the ground that means one component will be just directly from the ground motion right whatever ground displacement form it has so mass is simply going to be moved by that amount plus you have the relative displacement how that is coming that is relative motion that is due to spring ok. So the spring force that enters that will enter through this ok and then you have the rigid motion that rigid motion of the mass will be determined by the ground motion ok. So I think we have now you know plenty of discussions but there are long list of questions and I do not think we can go through all of this but just to say that we can always take these things forward all the structural dynamics part and so on so forth but ultimately these two problems will also give a very very good understanding of how to go about the equations of motion for these class of problems ok and I think now we can have a tea break.