 Good afternoon everyone. So, what we are going to do in this session particularly, first of all I am going to you know kind of spend maybe half an hour discussing the properly constrained, improperly constrained and partially constrained truss, because we have done already truss yesterday and only thing that we have kept in mind that truss is statically determinate and properly constrained therefore, we are able to solve all the forces ok. Now, what we do here also just to give the students a bit flavor that when truss can go actually unstable. So, that is what we are now going to discuss. So, remember that for the truss to be properly constrained first of all it should be able to stay in equilibrium for any combination of moving. And not only that both the global equilibrium as well as internal equilibrium will be satisfied ok. So, what we have said yesterday that we have a simple truss let us say and simple truss is always internally rigid. So, it cannot deform or undergo rigid body motion inside of it right. Now, when it is resting on support we have to make sure that support is appropriately given that means at least we need more than 3 supports for a simple truss we need 3 supports to balance it, but those supports has to be properly given it has to be properly constrained such that no motion is allowed. So, that is in terms of global equilibrium ok. Now, there could be trusses where also it can you know have internal instability and that is also possible. So, what it means is that let us say I have m plus r equals to 2n, but it is still possible that the truss is not stable ok. However, if m plus r is less than 2n then we can always say that the truss is unstable. So, we call in that time that truss is partially constrained. So, m plus r less than 2n that means it is never going to satisfy the equilibrium condition and therefore, it will be partially constrained ok. And when m plus r is equals to or greater than 2n there is a also a possibility that truss is unstable in that case we will just say it is improperly constrained. Now, remember one thing that both partially constrained and improperly constrained cannot be classified as statically determinate or indeterminate because in both cases truss is unstable. We cannot say now it is a statically determinate or it is a statically indeterminate. When we say it is a partially constrained or it is a improperly constrained. So, as I said that m plus r equals to 2n is just a thumb rule to make sure ok can I start thinking about that it is a statically determinate truss, but it is not necessary or it is not a sufficient condition that the truss has to be a stable truss ok. So, we will now just you know kind of continue various problems here. So, first let us look at these examples. So, I have two examples here and the main intention of these two problems are just to bring out you know the essence that is simple truss can also be unstable externally unstable ok. So, what has been done here in the first problem as we see it is a simple truss, but we have put it in two rollers. That means what happens I need to have at least three supports that is absolutely missing here. So, we have two supports that means sum of force along x equals to 0 is not balanced. Therefore, it is a unstable truss and not only that as we can see that m plus r equals to 2n if I go by that logic then in this case m plus r is actually less than 2n ok. So, it is a partially constrained truss although it is a simple truss, but it is not supported properly that means there is a inadequacy in the external constraints there is no problem internally, but still the truss can collapse. So, this will be called partially constrained, but that partial constraint is coming from the external side not the internal side of the truss ok. In the second problem now this is also a simple truss and simple truss I only have three reactions. So, one would think that this is also a stable truss because in this case m plus r is indeed equals to 2n, but remember what will happen again what is missing in this case? Once I apply the load you know like this in an inclined manner then I am unable to balance this truss in the x direction right. So, m plus r is still equals to 2n, but it is a improperly constrained truss. So, the point is m plus r 2n that is just a thumb rule if m plus r is less than 2n it is definitely a partially constrained it will always collapse. If m plus r equals to 2n it is not necessary that it is a stable truss it can still collapse, but we will call it improperly constrained ok. So, just let us do some exercise right now very simple exercises we will be doing. So, all we have to answer that whether this truss is partially constrained fully constrained or improperly constrained. So, let us go with the you know some counting from the very beginning. So, how about the first one? It is fully constrained right properly constrained why because I have a simple truss here at the very first case and simple truss is supported by three proper reactions because this reaction now A x reaction can absorb this type of force also right. So, in the x direction it can also be balanced in the x direction. So, the first answer is that it is a properly constrained truss and also we can solve all of these member forces therefore, it is a statically determinate as well. So, how about the second one? Second one is partially constrained ok why because I have taken out this right member here now what happens to the truss that is important. Now, again we have to kind of you know refresh our memory on method of sections and method of joints very quickly to analyze this kind of truss do I actually have internal equilibrium or not ok. So, in this case what happens we can if we just quickly go by the method of joints what happens this member is under compression. Now, I just use a method of section instantly. So, I will cut the truss here if this member is in compression then how the force will act on this part of the body it will be no it is towards the joint if it is in compression right. Now, this part is a simple truss no issue this part is a simple truss, but I am going to apply a compressive force here that is coming through the member and once I do the movement balance about this point at the left joint then nothing is going to happen right I mean what is going to happen there is no movement balance about that point because what is happening now it is a contradictory in nature because I am telling this member is a compression right. Now, if I apply this compression it is now telling that moment about a should give me this force equals to 0, but that does not happen that means if that force actually presents right that means it can actually turn the truss about the left hinge left in connection therefore, the truss will deform see thinking from the point of view of equilibrium or forces is very easy then that of actually kinematic deformation how the kinematic deformation is taking place that is always complex and you have observed that in the friction problem also, but if you can just remember from the force concept it is always easy I have always number of equilibrium equations must be satisfied you know number of unknown should be number of equilibrium equation ok. So, now that is not happening in this case so, this is my partially constrained truss now how about this one this is not partially just be careful here I have added one this member so that means here I had taken it out again I put it back here. So, it is actually has to be said that m plus r still equals to 2 n, but it is not stable truss same logic applies as that of before ok. So, it will be called improperly constrained just the difference in name partially and improperly ok. Now, how about this just understand first that do we have a simple truss hidden here which portion of the truss is simple truss from here up to here it is a simple truss right. So, if I apply a method of section right here that simple truss is properly supported by a pin support and by a roller support that means I have actually three reactions properly placed that means no matter what happens no matter how I apply the load on that portion of the truss it is always going to sustain the load. So, what I have to now worry about ok what happens to right here this member as I apply the method of section this is also rigid body just a straight line vertical member it is supported by a roller and it is supported by you can say that two forces are actually coming from this, but remember in this case no matter how you apply the loading this truss will be this part will also be stable ok no matter how you apply the loading. So, this will be completely constrained that means properly constrained truss ok. So, we understand the concept basically from the simple truss wherever I have simple truss I am just going to keep that ok and if that is properly constrained then I can say that ok that part is properly constrained then I look at the other part any question yes. Because the section anything then from right hand side again if you take the moment about that point then there is no cost to buy in that moment which one right side right support. This one if I use method of section and you are looking at this side now for this force what is happening it is just going to the support right. Top member is 0 top member is a 0 force member as per the method of joint for this loading now you can incline the loading that means you can just put a horizontal load. So, what happened this member will go to 0 now this member will go to 0 if we put a horizontal loading this member will be 0 if this we take out let us say we put a horizontal force then this member will be 0 right and ultimately force will be transferred here, but this part is stable ok. So, any combination of loading therefore we will hold good for this truss. Now, if you put a inclined load then one force will be taken by this member and the other force will be taken by this member there is no way you will be able to say that there is a non equilibrium of that part. So, sum of force along x 0 sum of force along y 0 sum of moment about a point has to be balanced ok. Any where you take even for that part now some part you know some forces may go to 0 some member forces that is fine it is coming from equilibrium though ok yes it is statically determined that we have first said that count the number of members and number of unknown right. So, I have number of member forces are 16 right and number of reactions are 4. So, total 20 and look at the number of joints we have that is 10. So, that that is that is fine, basic truss is triangle. No it is not, but it was not always true it was not always true what we have said also yesterday that we keep doing that extrusion right we keep extruding 2 members from the basic truss and we make the simple truss. Now, here what is the realization that up to which point I have the simple truss. So, if I can take that out that simple truss I take out and make sure that that is properly balanced then I go to the other part of it and try to analyze whether that is also properly balanced no matter what kind of combination of loading I give. But the complete truss is a truss no. Complete truss is a truss truss has no problem. So, that way then. But I cannot call it the complete truss is not simple truss complete truss itself is not a simple truss. But the equation m plus r equal to n is not satisfied no. Why count no it is not satisfied it is minus 1 why m r 16 there are 4 reactions and number of forces are member is number of members are 16 ok that is ok that is fine see what we are analyzing here in this case that it could be externally indeterminate that is fine. But see what is happening that I have more than 3 supports, but question is whether that is solvable or not it is solvable because I can use the equilibrium for joints I can establish 20 you know equations and 20 unknowns is that clear ok. Sir, one second this truss is externally indeterminate. This truss is both way determinate externally no externally determinate yes we cannot solve the all the reactions by taking the global equilibrium, but then you can go to the joint you know method of joints you can still find the forces. So, it is not you know it is not a good idea to look into even that manner external what is externally I am trying to analyze that whether the entire truss can be solvable or not that is my objective yes right. Now, there could be more than 3 supports or there could be you know 4 supports 5 supports I do not care what I care ultimately if I can find all the forces right and that is fine found here. Now, let us look into this one very simple exercises quickly because we should not take more time for this remember that there is a simple truss right here simple truss right here again you can think this of a compound truss it is a good compound truss example, but what is happening here for that compound truss I need 6 right unknowns that is not happening because one side is a roller one side needed to be hinged in order for to be a compound truss valid compound truss clear. So, as soon as I apply the force in this case and you can see clearly that it is going to actually go to this direction roller is going to move. So, as a thumb rule you can start doing this M plus R you know whether it is 2 N or not it is actually M plus R less than 2 N, but logically when we are applying we can also try to look into we have learned the compound truss am I really getting a stable compound truss or not it is not ok. So, that means it is a unstable and partially constrained. So, the answer is correct now this one I get some hint here how about M plus R that concept do I have the number of you know unknowns equals to number of equilibrium equations now question is ok M plus R equals to 2 N in this case M plus R equals to 2 N. So, can I now see it is a statically determinate and how do I prove if it is if it is how do I show that that it is a statically determinate or properly constrained and if it is not then how do I show it. So, the basis is let us say I do very quickly we can you can try to look at ok we have already learned the method of joints. So, let us say I look into this member how about the force in this member this is a tensile right. So, if this is tensile if this force is tensile then what is this force that is also tensile right. So, therefore, B C is a tensile force question is which part is actually a you know simple truss logic if you look at this one this one this truss right internally it is stable this portion of it no issue internally it is stable. In fact, there is a redundancy here there is a extra member right. So, up to this portion of the truss is properly you know internally stable now you apply that tensile force. So, if I use method of section there is a tensile force here acting which is outward from joint B right then if I take a moment about again the left joint what happens it can rotate about that left joint right. That means equilibrium is not satisfied once I take a moment about that left joint clear. Therefore, it is a improperly constrained truss at the end. See m plus r is still equals to 2 n, but truss is not stable because it can it is violating the equilibrium about you know that left joint. Excuse me sir. Yes sir suppose the P is upward. Suppose P is upward does not matter it has to be valid for any combination of loading. Then the P on the left of B is suppose upward. Fine in one if I want to make a truss stable it has to hold good for any combination of loading. So, if it fails in one combination of loading I can say it is a unstable truss. So, there is no way we can even discuss it further ok right because I have given you a loading for which it is failing therefore, no other questions comes into play is that clear because that is already failed. If it is then that case it is ok, but it is violating the rule for the other combinations right. It has to be stable for all combinations clear. Yeah, what was the issue? How it is supported that cantilever truss? But it has always to four reactions at the one end right. Excuse me sir. So, that truss we are saying as a cantilever truss just like a beam we are supporting that truss along the vertical wall. But vertical wall should have a if you are defining that as a cantilever truss fine. Yeah. You should have a pin here and I should have a roller one end. Yeah. You cant have anything more you know other than that. Yeah, definitely. If you just make a pin then that is unstable, but you do have a roller know that will balance the force. Yeah. Is that clear? Yeah. Ok. Excuse me sir. Yes. The moment we will be shifting this right support to below point B it will be stable. That you can which one? If the roller is shifted to below B that is just left of that. Yeah. And the truss will be stable, no? Yeah. Yeah. If we move it. In case the situation will be like this in that condition. Yes. It is a possibility. It is a possibility. Yes. Because in that case it will be able to take the load. Yeah. Thank you. Ok. That was. Yeah. To change the direction of load in vertical direction. Hmm. And cut the suction at whatever it cut the B C. In that case it will satisfy the. That may. That may. Yeah. That may. But see what I am. Whatever is the force irrespective of force it should be stable. Stable. Now you are putting the force downward. Downward. If it was initially upward then we have to check the by taking the forces in both direction. You can check because when you design a truss. Yeah. I have to make sure the truss is properly stable under all combinations of loading. Right. Yeah. Yeah. Now. It is a by chance here force is downward P. Yes. If it was the. Upward P. Upward direction. Then we have said it is ok. It is ok. It is ok under that particular loading. That particular loading. That particular loading. Yes. That means for total to check the assembly we have to change the direction of. You have to. If it is not given in the problem you have to also make sure checking all the you know integrity we have to make sure that it is a properly stable truss. Ok. It is not a simple analysis but what I am giving here just given type of loading is given let us say if for that we can at least claim that it is not stable then fine ok. So, at least we are checking for one load combination that is not valid. Any other any other questions ok. So, this one quickly. So, it is a very similar problem we have done actually. First just found M plus R equals to 2 and if that is valid or not first of all. That is ok. Now can we identify the simple truss here please. So, A D F G is a A D G F is a simple truss. If you really take a method of section if we apply through D E and G H right. So, that side is a properly constraint truss because it has a in connection and it has a roller and looks like all the force can be balanced there ok. Now we are again concerned about the right hand side of it G H. So, will it be balanced under any combination of loading just do not look at the combination I gave because in this case at least we have sorted out that one part of it is a simple truss and it is going to you know be stable under any kind of loading. So, I am now just concerned about this particular member E H and the connections it is getting from the surroundings right. So, is that going to be always stable or not? It is going to be always stable you can try to figure out if one member on the top goes under tension the other member when the bottom will be under compression. So, that will balance the moment all the time balance the horizontal reaction all the time. It does not matter how I apply the role right clear ok. So, that member E H will always be you know will have a couple acting on it that two member will act as a couple right, but that will be balanced against the forces that is being acting on that particular member ok. Where are the please problem also towards the end this one is very interesting you are not doing please now, but let us say how about this one. So, I said yeah both both are pin connections here both has pins here just you know again you can try to think of you know method of section we cannot apply. How many are improperly constant? Raise hands may be. Is it really difficult I mean can we can we do a method of section here quickly please can you tell me can you tell me which portion is a simple truss which portion is a simple truss internally rigid is not that this portion right. So, I will definitely cut this truss this is also a basic triangular truss right this is a basic truss. So, what can happen if I really cut this right here I am just supposed to get the force here what is this force going to act? Sir. Sure 0 tensile compressive I am getting all sorts of answer that cannot happen just do join this is going to be a. It is going to be compressive. So, what happens now you can clearly see now it is going to be compressive right. So, therefore, I will now put this force simple right here like this towards the joint right and I will take a moment about again left hand what happens unstable it can rotate about that point right therefore, it is a improperly constant ok.