 What I will do? I have looked at some queries which I thought could be useful to discuss again ok. So, one query was and I tried to you know I did explain it is that when you actually have a let us say when we are constructing the truss as I said that joints are usually bolted or welded ok. However, we are assuming that it is pinned. Now, what is the purpose of that? Now, why it is so? First of all just try to think of it. So, I will just try to explain it again in a brief detail there are two important things. See as I was saying that joints that are assumed to be pinned ok. So, that was the assumption again. Now, why it is so? Although it is bolted or welded together the first thing to understand that when we are looking at a truss we already said that the load must be carried by the joint. So, that is the first thing load can only be applied at the joint it cannot be applied anywhere ok. Because what was the reason of to do so? First was members are slender ok that is one thing. If I apply a lateral load let us say I have a member and I apply a lateral load it can break because they are very thin in the outer plane direction therefore, there will be tendency it cannot take the excessive bending. Now, remember from the mechanics point of view what was important is that when I have a concurrent force system. So, if I really take this joint what is happening that I have a set of forces in these joints that are actually concurrent in nature. Now, what does concurrent means? That means they are all collinear in nature right passing through the same point therefore, the net moment is 0. So, although they are bolted or welded that means actually if you see that this is a truss member this is a let us say truss member ok. And then there is another truss member you put a gusset plate here and put some bolt or maybe weld it ultimately remember the force that is passing through the centroid has to be concurrent. That means if it is so therefore, there is no moment about this point. Although remember due to construction error there may be some eccentricity. Now what can happen due to that eccentricity? Yes fine there may be some moment coming into play here, but this joint will be designed to take this moment. Remember what is predominant? Predominant action will be always axial in the member that is very very important ok. Now you may have some out of plane moment, but that is very mild and that can come remember you may have a wind load it is actually a distributed load in the member, but we still take that to the joint. So, that distributed load on the member that this member will be able to sustain. So, ideal thing is from the mechanics point of view we are always going to have a concurrent force system. The major issue is that load has to be applied at joint ok. So, again in engineering mechanics we are only making certain assumption as I have said in the very first you know discussion that we are really looking at a bare bone model with a assumption ok. Now how good that is that we all know because so many traces are designed and we have been designing it based on only pin joint assumption because we know the logic is that if the force system remains concurrent then I do not have the moment here, but if I let us say you know if the force system is not concurrent I am applying some load in this member and I have a weld or you know fixed then definitely it will start deform like this. So, it can get an excessive deformation right then problem demands that I have some moment coming into play and I have to design both the member as well as the joint for that moment because now you have a proper fixity here and that fixity will take this load ok, but in this case that is for the truss we are actually applying the load right here ok. So, that was one of the queries I had. So, I thought again it will be better to you know redo this remember you could you know when we look at the structural analysis I can give you a truss like this, but I will say these are properly fixed ok. Now if it is properly fixed then remember even if I apply the load like this. So, it will try to deform that means if I apply like this it can take actually this kind of loads and we will try to deform like this, but this will still be 90 degree. So, this is concept delivered in frame that this angle the change in angle right although it is changing its angle, but ultimately whatever angle it was before right if this is 90 degree original angle has to be retained. So, 90 degree becomes 90 degree, but now since you are applying a lateral load right then you will have the deformation. Now that problem will be more you know complicated because now you are trying to go to the actual deformation of the body and in order to solve for the moments you have to link this deformation right you have to link this deformation in order to get this moment, but here make sure that we are talking about statically determinate problem it is perfectly you know concurrent system of forces and it is not taking any moment at the joint. So, before we proceed to the tutorial what I will do I will try to you know kind of take it to more advanced level as I was discussing before and during the tutorial session we can again have some round of discussion. So, I will try to finish it very quickly just to take home you know we can give at independent thinking also towards this line. So, now as I was saying that we have already demonstrated the concept of simple truss simple truss is very very you know important aspect the reason being a simple truss is always internally rigid ok that means we treat this as a perfect rigid body it can only undergo pure translation or pure rotation, but we do not we will not have based on the rigid body concept the truss should not collapse that means there should not be excessive internal deformation ok. Remember in rigid body mechanics we always have an internal deformation, but if that deformation is small we just ignore that there is a small deformation and we treat this as a rigid body ok. So, therefore now in compound truss as you can see what is compound truss two or more simple trusses are added together. So, now how this trusses are jointed together can we identify the simple trusses we can clearly see a b c is a simple truss d e f is another simple truss. So, that means this is a perfect rigid body this is also a rigid body right both are in triangular in shape now they are jointed by several members. So, we treat this problem we treat this problem as if that there are two simple trusses and some members are simple simply connecting this trusses ok. Now, how do I separate this trusses? So, think of like this I have two reactions here I have one reaction here right that is three reactions I have. So, externally it is perfectly supported internally I have simple trusses no problem. Now look at here I have to connected by three members why this three members are used if I treat this as a rigid body and if I treat this as a rigid body how many equilibrium equations I have two force and one moment that means per rigid body I have three equilibrium equations. Therefore, I have total six equilibrium equations that means I can have total of six unknowns to solve this problem to make it statically determinate and completely constrained. Therefore, I can actually add minimum of three additional members in such a way it becomes completely constrained. So, what is happening I have added this three members and you can clearly see now what happens if I just detach this rigid body because it is a simple truss if I detach this rigid body I am going to have perfectly six unknowns 1 2 3 right what else 1 2 3. So, total six unknowns I have is that concept clear now think of this that is another way of joining again I have a simple truss here A D C again D E F these are two simple trusses now these two simple trusses that means these two rigid bodies as if taken and being pinned here and I have one member only required to make it completely constrained and statically determinate. Now, again I have to prove that why it is statically determinate right. So, therefore, what I need since I have two rigid bodies I have to come up with six unknowns and that six unknowns should be put in such a manner that it is completely constrained it cannot collapse. So, we can clearly see now how these unknowns are coming into play I have one unknown two unknown three unknown four five six remember here Newton's third law is applied Newton's third law is applied when I am detaching it ok. Next can we talk about this now just one more simplification what has been done now I have just taken out this if I take out this remember now I have five unknowns here to bring in six unknown and make it properly stable truss I will make this roller to be a hinge and you can think of it if I apply any kind of load this cannot now go away remember if I would have had a roller here if I just have a roller instead of a hinge if we apply a load since this is spin this rigid body can rotate about this spin it will try to translate in this direction to prevent that what I have done I am bringing in another horizontal reaction in the form of a hinge. So, ultimately again you can show that if I treat this as a rigid body if I treat this as a rigid body I exactly have six unknowns one two three four five six. So, my idea would be given a loading if I give a external load first I would should try to solve for these unknown forces what that will do eventually remember neither I am comfortable using method of sections neither I am using comfortable using method of joints I can immediately get some forces right here if I solve for these two forces ok. So, with this concept I will just demonstrate one simple example now look at this truss can we identify what are the simple trusses in this truss within this truss it is a compound truss I am supposed to find out the member forces one two three. Now, can you think of that how complex it is in terms of method of joints or method of sections ok. So, think of this now therefore, the concept of compound truss can easily solve for these unknown forces how because I can clearly see this is a simple truss A D E that is a simple truss similarly F B C this is also a simple truss. So, therefore, what I will do I will simply try to detach these two remember reactions can always be found from the global equilibria I have no issue to find out the reactions at least for this problem remember loading pattern is also not symmetric this is also P to P P to P ok. So, as I said the simple trusses are A D E and B C F. So, we have to detach to expose these three unknowns ok and draw the free body diagram of individual trusses showing the unknowns forces once you detach it and take the equilibrium of each of this truss that means, you treat them as a rigid body because they are simple truss and just see whether you know we can solve for the forces. So, ultimately what is happening if I detach this truss let us say I take this A D E. So, I take this A D E therefore, once I detach it what are my unknown forces this one this member force this member force and this member force all others are known because this can be solved this is already given. So, therefore, just looking at the equilibrium of this I should be able to just solve for this three unknowns because it is a rigid body which has got three unknowns. So, either I choose this one or either you I choose this one it would not matter. However, you can clearly see there would be some trick to solve this problem. Now, suppose I am interested in finding out what is my F 3 what do I do. So, member force 3 can be easily found by taking the moment about A you see in one go I can find out what is the force in F 3. So, take the moment about A therefore, I can get the solution for F 3. So, now if I ask you then how do I solve for F 1 I will simply consider this truss take the moment about B. So, in the next what is done if I want to solve F 1 I will just take the this truss as a rigid body and I take moment about B because F 2 and F 3 passes through B. So, therefore, I will get F 1 now what you can do to solve for F 2 remember F 1 and F 3 both are solved. So, now left is F 2 again you can just use some of force along x direction to make it 0. So, that is how we analyze the compound truss and just remember the concept is very easy we identify the simple truss and we simply detach those simple truss by saying now this simple truss as a rigid body it is a proper rigid body and we just show the free body diagram of all the forces that are acting on those simple trusses. Another important concept would be as I was saying that m plus r equals to 2 n is that a necessary and sufficient condition for a truss to be properly constrained or not I know m plus r equals to n can I say it is a statically determinate as well as properly constrained the answer is no ok. Now why remember for a truss to be properly constrained it should be able to stay in equilibrium for any combination of loading that is very important it should be able to stay in equilibrium for any combination of loading equilibrium implies both global equilibrium as well as internal equilibrium. Now remember if it is a simple truss then internal equilibrium is always going to be maintained ok then only thing I have to worry about whether it is supported properly that means if the global equilibrium is you know the reactions are given properly so that global equilibrium is maintained or not. Remember if m plus r is less than 2 n that means number of unknown forces these are total number of unknown force is less than number of equilibrium equation 2 times n then the truss is most definitely partially constrained that means truss has to be an unstable one in certain loadings ok. But let us say m plus r greater than equals to 2 n is no guarantee that the truss is stable and may be termed improperly constrained that means the point is although I may have m plus r equals to 2 n or m plus r greater than equal to 2 n but it does not guarantee the truss is stable. So it is simply termed that improperly constrained not partially but improperly ok. And remember m plus r less than 2 n it cannot be statically determinate in any case because I cannot use equilibrium equations to solve for all unknown forces. But in this case if m plus r equals to 2 n it may or may not be statically determinate ok. So m plus r equals to 2 n it is a necessary condition for the truss to be statically determinate but not a sufficient to be it sufficient it has to be properly constrained also it can be improperly constrained ok. So now think of this very simple simple examples we can take here this truss if you look at it internally rigid it is a simple truss. It is supported by two reactions if you apply a load like this what happens it will simply start rolling in this direction because I have not adequately constrained it I had to put a hinge right here or right here that means I had to give a support reaction horizontally that means now what is happening as you see for simple truss m equals to 2 n minus 3 I need 3 unknowns because m plus 3 equals to 2 n. So 3 has to be unknown as support reaction now I have only 2 unknowns as support reaction. So the truss will collapse collapse in the sense it will collapse means it will start walking in this direction it will start rolling in this direction but internally nothing will happen internally it is still stable only problem it got is external constrained problem. So the truss will be defined as partially constrained partially constrained based on external inadequacies similarly if you look at this I am still supporting it into three reactions but these three reactions are not adequate to make it a stable truss. So again we will term that now in this case remember m plus r equals to 2 n because m plus 3 equals to 2 n okay you have three reactions it is a simple truss you needed three reactions but remember it is still not stable because it can still walk if I put a load like this that means if I apply horizontal force it will work. So I will term that now improperly constrained. So this was partially because it has less number of unknowns it has equal number of unknowns than equations but it is improperly constrained. So both of this truss are unstable. So let us try to check ourselves that have we understood this concept remember this partially constrained improperly constrained internally we also have to satisfy. So what I have demonstrate so far there is some inadequacy in the external constraints. Now there could be inadequacy in the internal constraints also that means internally it may not be rigid that means internally somewhere the equilibrium conditions can be violated. So let us look at the first problem right here the first problem as you can see do I have a simple truss the answer is yes I do have a simple truss m equals to 2 n minus 3 is completely satisfied also make sure I took a basic triangle and I am erecting two members joining by pin with that mechanism I am constructing this truss it is absolutely fine simple truss it can sustain any load in any direction without deforming internally that means there should not be internal collapse. So therefore what we need to look at whether it is constrained properly from the external side is it externally constrained properly the answer is yes why not because I have one reaction this way another one this way and another one is roller. So it is a statically determinate because m plus r equals to 2 n as well as I have maintained the support conditions properly it is a simple truss supported by three forces correctly it is a properly constrained. Now let us look at this I have simply taken out one member what is the problem now remember that means you have reduced the number of unknown by one so it is definitely partially constrained if you count m plus r that should be less than 2 n that is for sure it is a partially constrained. Now how it is unstable use the knowledge of method of joints if you cannot understand let us see if we cannot understand based on how it is how this truss can actually collapse we can do a very simple mechanism first identify which portion is a simple truss the simple truss is really this portion right here that is my simple truss. Now question is if this is simple truss is it adequately constrained the answer is no why it is not remember what will happen if you apply you know this force. So what is the force in this member it will be compressive there will be a compressive force see if I take out this simple truss and draw the free body diagram there will be a compressive force here take a moment about this point nothing to balance that means this portion of the truss can actually rotate about this hinge that means truss is highly unstable and that had to happen because we already made it partially constrained now this is non-rigid truss that means internally it is not rigid you might have given proper adequate support here and here but internally it has got problem I needed one more member here similarly let us look at this what I have done I have just added one more member just one member here one member here but still it will do the same thing that it was doing here there is a force P take the moment about this equilibrium is violated moment about this point is not 0 for this rigid body remember this force is passing through this point therefore in this case m plus r is still equals to 2n but it is violating the condition. So next thing would be we can take a look at quick problems here let us just try to see this one can you tell whether it is partially constrained improperly constrained or properly constrained as I said always look at the simple truss where is my simple truss what is what portion is the simple truss I will circle it is not that a simple truss remember this is the portion is a simple truss it is properly supported you have one hinge one reaction that means in this truss I can apply any amount of load any direction it will not it will sustain that load. Now what about this one we have to see if I detach this member and try to interrogate the forces it is perfectly fine under the any action of load you can have a tension here compression here right if I apply let us say horizontal load does not matter. So ultimately what will happen this load will be simply transmitted to this member. So no matter what kind of load you apply in this body at the joint it is going to survive ok. So therefore the entire truss will be completely constrained. So the logic is that first I understand if I the simple truss portion find out if that is properly constrained if it is properly constrained it is fine no problem then I look at the other part try to study for various combination of loading only one combination is given for this load we can this load will just take you know go this way. So it is properly fine you apply a load like this load will simply transmitted here this is 0 force now. So this load which is horizontal this simple truss is going to take that load. So no issue so it is a completely constrained and statically determined. How about this one can you identify the problem very simple think of it this part is a simple truss this part is a simple truss as if you can treat this as a compound truss where is the problem I needed a member here I do not have that member. So it is a partially constrained truss you can count M plus R should be less than 2 N. So it is a partially constrained you can see clearly if I apply this load it will simply try to walk because this entire portion this right portion can rotate about this pin very easily and this part can rotate about this pin. So if we apply a load let us say it will simply start rolling in this direction. So it is a partially constrained now how about this one I have given it a hint this portion is a simple truss but with a higher you know unknown fine. So it is a redundant truss it has an additional member no issue. So this part remains a very properly rigid similarly this part also is very rigid this is a basic triangular truss. Now what is holding these two trusses at these two members can you try to count the number of members and number of joints and try to see if M plus R equals to 2 N you will be surprised that M plus R in this case is exactly equals to 2 N no problem M plus R is still 2 N. Now what is the problem with this truss then use the knowledge of method of section remember what will happen if I adopt method of section and try to find out the force in BC what this force should be just take this portion of the truss adopt method of section find out this force you can see by method of section if I take a moment about this point this couple P times this distance will balance this force. So that force nature has to be tensile this force has to be tensile is not that if it is tensile. So in this portion then how it is going to act it is going to act outward so it will be this way now look at this portion of the truss. So I have a force BC which is tensile so if I take a moment about this point nothing to balance take moment about this point the truss will simply rotate. So that means equilibrium is not violated as I take the moment about this. So what you can do you justify M plus R equals to 2 N but you start solving it by equilibrium approach you will be able you will be surprised the somewhere equilibrium is violated that means the truss is improperly constrained I have not properly constrained the truss. So with this I will conclude the lecture part of it and now I can take couple of questions before we go to simple let us say tutorial problems is there any discussions on the compound truss partially improperly properly as I said it requires little bit of independent thinking but idea is very simple the idea is very simple even if M plus R equals to 2 N it does not guarantee that the equilibrium conditions will be satisfied the truss can be still not properly constrained it can be improperly constrained and when M plus R less than 2 N we said it is anyway partially constrained it is a unstable truss ok or for compound truss we know how to solve it now I have shown couple of problems yes 1039 so is it possible to apply method of section in all cases no it depends on the type of problem you are solving your question is can I apply method of section in all cases ok see method of section is usually applied if I want to interrogate if I want to find out force in certain members for example think of a long truss right think of a very long truss I am interested in finding out some member force in the middle of this truss so instead of going joint wise starting from the left end to the middle I can simply use the concept of method of section ok so that what happens in the process you are treating a big chunk of the truss as a rigid body so whatever forces are inside I am not concerned about that because I now have a left free body I now have a right free body that means we have a rigid body entirely and we do not care what are the forces inside because I am never cutting that out I am never exposing those forces so through method of section I can only expose a few forces and try to solve it ok. Level 173 sir normally to solve the problems and the trusses so you have two conditions one is 2n but 2n is less than m plus r another one is 2n is equal to m plus 3 you have to check both conditions before solving the problem remember m equals to 2n minus 3 that is m plus 3 equals to 2n your question is there are two condition one is m plus r equals to 2n the other one is m plus 3 equals to 2n right which one should I choose my suggestion would be m plus r equals to 2n is the most general case ok so r is always number of unknowns remember m plus 3 equals to 2n that is valid only for if you have a simple truss and then you have three unknown reactions ok so m plus r equals to 2n can be applied for any truss now remember m plus r could be 2n but the truss can still be not properly constrained and that is what we have discussed it may not be statically determinate ok. Hello at times I find difficulty in solving the cantilever truss by applying the method of sections ok there was a question like this when I see what is a cantilever truss now cantilever truss we do not really you know teach it that way in IIT but I will just explain what it is what I understood probably so truss looks like a cantilever truss looks ok and I do not see what is the problem of solving this kind of problem in fact ok we have discussed this in the method of in joints when we did this kind of a problem so it can be typically you have basically a hinge here and we have a reaction here so it more or less same ok so we have two reactions and we have three reactions these reactions can be solved does not matter if I do this ok ultimately you see I can always start from this joint remember this entire thing is a rigid body and I can start from this joint or even I can start from this joint ok so it can always be solved so there is no unique about the cantilever truss ok. .