 So, we are going to cover truss which is often you know applied mostly applied in civil infrastructures and as you see both of us are you know from civil engineering department. So, now we are going to keep this session very short. So, you have basically morning session I am going to cover the truss and truss will include method of joints, method of sections and also some other type of truss problems will also be solved. In addition to that we are also going to look at stability and unstability of a truss that means, when the truss is properly constrained and not properly constrained. So, with that we will have the one and half hour let us say lecture and then we will continue to do the tutorial problems. So, now see as you see truss you know almost everywhere in roads, in bridges, you know in foot over bridge and truss has been used. Now, why do we use a truss? Now, to summarize I would say that to design a truss first of all is very simple and also notice that all the truss members that we have they are very slender in nature. So, in that way what we can do we can actually make it more economical. So, there are two aspects to it one is the simple analysis that is what we are going to learn as well as also we can make it very economical compared to frame structures. Frame structures will have lot of complicated analysis and also we have to worry more about the moment being taken. So, sometimes sections are designed very heavily. So, let us just try to go through the important features of a truss. So, remember that truss consist of long and slender straight members connected at joints. So, we call A, B, C, D as you see here these are all joints and all the members are connected to the joints. Now, we can simply say that truss is nothing more than assembly of triangle. So, if you keep on adding few triangles so that makes a truss. But in actual practice what we see that how the connections are done here you can see here that there will be a gusset plate. So, we call it gusset plate and all the members are going to come and then we are going to make it either bolted or riveted. So, therefore, a fixity is induced there is some kind of fixity in this joint. However, during the analysis we say the members are pinned together and the reason we do this because it much easier to constructs this kind of joints than that of actual pin joints. However, pin joints are also seen you can see how the three members are connected by pin joints. That means they will be able to rotate that means small rotation is allowed. However, no bending is allowed no bending flexural deformation of the member will be allowed at the joint. So, they can actually try to rotate about this joints. Now, remember when we say that these members are pin jointed that means AC is a pin joint you know AD they are jointed by the pins at the two ends and I do not apply any load lateral load on the members then we can clearly say that all these members will be two force members. So, what is two force member as we have learnt that two force members are going to carry equal and opposite load right and their line of action has to be same as if there is only one load going through it. But in order to make the equilibrium possible we are saying that equal and opposite force with same line of action is there ok. So, if this is happening then the if the member is in tension then we are going to show like this if the member is under compression then we are going to show like this. So, remember as I said before also the members of a truss are slender and not capable of supporting large lateral loads. That means in a truss structure I will never try to put any load on the member laterally they will simply bend and they can break only thing it can do it can take a large amount of axial load. So, the members are designed for only axial loads and the loads must be applied at the joints again as we can see the members are always going to be light in weight and therefore very economical. Now remember that when we look at bridges we will see that there are 2 truss here you can see the 2 trusses are actually joined together how they are joined. So, this is kind of bridge deck that bridge deck now where the probably if you think of a railway bridge the train will pass through it. So, you can see or even it is a foot bridge then human being is crossing through this bridge. So, what is happening now this bridge deck can take the lateral load. So, if you look at the arrangements we are going to have some you know longitudinal members. So, these are we call you know take the bending these are called stringers and these are floor beams. Now here these members can actually take the lateral load or rather you know out of plane load that is coming from the weight of the body or weight of the vehicle and these are going to be transferred to the joints of the truss ok. Therefore, as we clearly demonstrated that loads must be applied at the joints in a truss. Now there are a variety of truss that is that are being used in civil engineering applications as you can see here that all of this truss are actually possible. However, those are given different types of names now depending on the applications depending on the you know person who has discovered this kind of truss first or even based on you know their shapes. So, all of this you can see clearly their assemblage of triangles, but you know it is not necessarily that all the time that a truss has to be a assembly of triangles ok. Now we are going to discuss about how do I define this truss when we are trying to do the analysis ok. So, in order to make this kind of truss table what would be the necessary condition for us. So, if we look at it carefully we can clearly see that how I am going to construct a truss. Let us think of this that I have a triangle and remember in this triangle I have three members and all members are pin connected that means I have a pin connection at A, pin connection at B and pin connection at C. Therefore, all these members are two force member. Now that is my basic triangular truss. So, if I consider the basic truss and now what I am going to do I am simply going to add two members and one connection that means two members are extruded and they are connected by a pin at D ok. Likewise I can again extend this that means I can extrude two members let us say D E and C E and again connected by a pin. So, therefore, I am going to form this kind of truss as I have demonstrated earlier ok. So, you can see now if that happens if I can construct a truss using this philosophy then I can say that I am following a thumb rule and that thumb rule is m equals to 2n minus 3 where m is the total number of members and n is the number of joints. So, as you can see here I have in this case five members how many joints I have I have four joints right. So, 2 times 4 minus 3 that is going to give me 5 and I will call this now this truss will be called a simple truss as long as this logic is maintained and I can construct this truss. However, remember as I said before that it is not necessary that to maintain this logic I have to always come up with a triangle assembly. For example, in this case you can see that B E and D is a basic triangular truss ok and from that I am going to extrude two members right. So, this part becomes a simple truss you can see here right and here you see that it is now still I can do m equals to 2n minus 3 in this case and I can come up with this kind of structure this kind of truss. So, the idea is I have to still call this is a simple truss, but it is not necessarily made up only of triangles. However, in practice we tend to always maintain the triangular shape. Now, what does a simple truss would do? What do I mean by simple truss? Remember what happens if I have a simple truss for example, a basic truss like this or this or that if I support it properly using three reactions. So, proper three reactions are induced in this truss it will be perfectly stable it can sustain any amount of load and also observe that if we construct a truss like this right. Then if we apply any kind of loading which can come from the external load or support reactions the internal geometry will not change that means a simple truss is always internally rigid. So, what is most important that if it is a simple truss it is always internally rigid. So, the internal rigidity will always be maintained that means no angle here would change if you apply the load the truss internally will not collapse. However, if you look at this truss now if I really apply a load what can happen? We can clearly see that there will be tendency that this joint can basically come to here and this will be shifted out. So, therefore, this truss can collapse. So, a simple truss will be always internally rigid it will not internally collapse, but if you try to form any other type of truss without this mechanism we have to be extremely careful and therefore, it can internally collapse. Now, how do I analyze a truss? Remember just going back to it suppose this is a truss where I have 4 joints and 5 members. So, you can see this is simple truss where I have m equals to 2n minus 3 that is number of members must be equals to 2 times joints minus 3. So, before we analyze this truss we have to make sure that what are the supports I have what kind of supports I have. So, here you have a pin connection. So, there is a hinge here and we have a roller or a rocker here. So, that means first I identify the support reactions. So, as we can see I have A x and A y and here we have B y. So, these are the 3 support reactions. Now, to analyze the truss by method of joints we will always try to consider the equilibrium of joints. So, what will happen if I keep on dismembering the I want to isolate the members. So, therefore, if I isolate what we are going to have we are either going to have a compressive force on it or a tensile forcing. Now, as per the Newton's third law suppose this member is in compression then on the joint if I look at the joint A it is simply going to be transferred back to the joint. So, there will be an opposite force that is going to the joint. So, in the joint it is going to be inward. So, a compressive force in the member will be inward in the joint whereas, if you look at it carefully we see that a tensile member that means let us say if it is under tension the tensile member the tension force will go as a force in the joint which is outward. So, a compressive force on the joint will be always inward a tensile force in the joint will be always outward and that will tell me that then if I know the nature of the member forces then I should be able to transmit these forces into the joints. So, likewise what is happening here if I keep on dismembering all the members if I isolate the members and transfer these forces back onto the joints and I look at the joint equilibrium. Now, how many joint equilibrium I have? So, per joint remember I have 2 equilibrium equations because at the joint remember at the joint I have concurrent system of forces. You can see here therefore there is no moment equation only thing that I have sum of force along x equals to 0 and sum of force along y equals to 0. So, in that way in this problem how many joints I have I have 4 joints therefore I have total of 8 equilibrium equations. So, I can create 8 equilibrium equations in this case. So, on the basis of that if I have 8 equilibrium equations then I should be able to solve for maximum of 8 unknowns. The question is do I have 8 unknowns in this problem? The answer is yes because we can see clearly that I have 5 member forces as unknown. So, 5 unknown forces from the member and I have 3 reactions from the support as unknown. So, total unknowns are actually 8 and I can have 8 equilibrium equations. So, we have to adopt to this logic. So, what it means essentially for a simple truss we said that m equals to 2 n minus 3. Now, if that simple truss is supported properly by 3 reactions that means I will call this as r reaction equals to r therefore m plus r equals to 2 n. So, as long as m plus r equals to 2 n where m is the number of members r is the number of reactions and n is the number of joints as long as this is maintained I can say that the truss is statically determinate. And it is in this problem completely constrained because the way the reactions are being acting here that will make the truss absolutely stable. It cannot move in any direction. So, therefore this truss will be statically determinate and completely constrained. However, remember that this statement right here m plus r equals to 2 n does not guarantee that the truss will be properly constrained. That means it can still be an unstable truss and therefore in that case if it is so I cannot say that it is statically determinate as well because in that case what will happen basically all the equilibrium equations will not be satisfied. So, the idea is that when we start solving the problem I have to ensure to make it at least statically determinate I have to say m plus r equals to 2 n. Now, it may be unstable truss if it is unstable then equilibrium equations may be violated and we will see that towards the end of this session. So, for the time being let us assume that the problem that we are going to target can be solved by the static equilibrium equations and they are properly constrained. That means I should have m plus r equals to 2 n as well as the reactions should be acting in such a way that the truss will not be able to make any movement in any directions. Now to start the analysis remember one of the way to solve would be I have to show each joint I can choose and I can start solving the problem by saying sum of force equals to x equals to 0 sum of force along y equals to 0 and it will be always advisable that we start from a joint where I have only two unknowns. For example, in this problem by taking the global equilibrium I can solve for the reactions first. So, reaction will let us say be solved then I will be interested in finding out what are the member forces and if it is so then I will just go to this joint right here let us say then I have only two unknowns here the member forces and I can solve for this member forces by saying by taking the equilibrium along the x direction and along the y direction at this joint. Now remember if you keep doing this what will ultimately happen that for each joint I will be able to construct a force polygon and the force polygon can be drawn. So, this will give you a instant visual feedback that how my joint forces should look like. For example, in this case so it will always give you a closed force in a closed force polygon. So, the closed force polygon means there is no actually resultant because resultant vanishes resultant equals to 0 and that is the equilibrium. So, what will happen you can see here that if I look at this joint particularly which is being depicted here we see here that reaction is first solved and reaction has to be upward. Now if the reaction remains upward and I want to construct a force polygon in this case you know closed polygon then this member has to be under compression such that on the joint it is pointing inward ok. So, this will be F that is the compressive force on the member and this has to be tensile force on the member and thereby I have a closed polygon whose resultant is equals to 0. So, once we start from this joint let us say and I can get an instant visual feedback that yes this member has to be under compression and this member has to be under tension. We could also do that using the simple sum of force along x equals to 0 and sum of force along y equals to 0, but this is going to be unique that you have to always construct with these directions. The direction will remain same I am not saying that you cannot do it other way around it is possible, but the direction cannot be altered. These directions cannot be altered it cannot be you know taken to the other direction. It has to be always a closed circuit and to maintain that closed circuit direction of the forces has to be like this. Similarly, once I solve this joint then I can immediately move to the next joint and first I know the AF. So, I draw the AF now to make a force polygon I have to say EF this way and BF this way such that the circuit is closed the polygon is closed. So, in this way instantly I will know that what are the members are in tension and what are the members are in compression. We could go to more and more complex joints you can clearly see that ultimately in all cases if the joint equilibrium is satisfied we should be able to come up with a force polygon which will look like either a triangle or a polygon. Now what we are going to study we are going to study the joints under special loading conditions. Now to do so remember in my mind I am always trying to take the joint equilibrium and try to see what relationship exist between various members that are connecting at a joint. So, I could either do sum of force along x equals to 0 or sum of force along y equals to 0. Remember x and y are always two perpendicular axis and they can be oriented in any manner. It is not always necessary that x will be horizontal and y will be vertical. Just keep that in mind we can either look at sum of force along x equals to 0 or y equals to 0 or we can take the methodology of the force polygon whichever way we like it we can do it. So, what happens force in opposite members intersecting in two straight lines. So, see A E is a member A C is a member A D and A B they are intersecting in two straight lines. What will happen we can clearly see that A B is equals to A D and A C is equals to A E. So, that means whenever we have opposite members intersecting at a joint and the joint does not have any additional load that means external load then we can always say that forces in the opposite members are equal. Similarly, how about this joint now you have two members which is collinear and then one member is like this. In this member you can see the force is acting which is collinear to this member A C. Now how do I determine what relationship exist between the member forces or with the load. Now remember if I take a horizontal equilibrium then it immediately tells me that A D equals to D B and if A D is equals to D B then from the vertical equilibrium I will have A C must be equals to P that means this member has to be under compressor. Now how about this one now here no external loads are applied then immediately what will happen this member will be equals to 0. So, the force in the third member will be 0 if there is no force acting along that member otherwise that member will simply be equals to the load that is acting along that member. Now if you come to the next one forces in two members connected at a joint are equal if the members are aligned. Now if the members are aligned I can clearly say that the forces are going to be equal in this member otherwise they will be 0 that means in this case if this is the you know condition of a joint I have only two members connecting like this then both members will go to 0. Now we are going to take a very simple exercise to see that if I understand the spacing special loading conditions in the joint then analysis of truss could be very simple that means what just inspecting various joints I can immediately get some of the member forces. So, some member forces can be immediately obtained as long as any of these conditions are met. So, suppose this is a truss it is a simple truss as you can see here load is applied it has a hinge joint here it has a roller support here. So, two reactions now you can see clearly the members that are marked in green color. So, for example if I just look at this joint looking at this joint I can immediately tell that C B is equals to 0. So, the member force in this member is 0 force in this member is 0 similarly if I look at this joint I can say the member force is going to be 0. Now what happens at K if I look at this joint K j is 0 then immediately I can jump to joint j if I go to the joint j now joint j is under which condition now since j k is already 0 joint j comes in under this category you see now when it is under this category then if you see if I take the equilibrium perpendicular to this axis. So, that means if I take equilibrium perpendicular to B D right then AC has to be 0. So, therefore in this case j i also should be equals to 0. Now I cannot predict any other joints in this case or in other words no other joints are actually falling under this category. However, there is a joint that is going to this one. So, since I know j i let us say I know j i already can I come to i if I come to i joint i then what is happening then I see this is under this category I have two members like this and third member is aligned with the load. So, third member is aligned with the load and as we know therefore, I H must be equals to 20 kilo Newton. So, firstly what we have learnt is that it will be simple to understand the 0 force member. So, 0 force members are BC, IJ and JK and we can clearly see the 0 force members are quickly obtained if this condition is there or this condition is there where by AC is also equals to 0. So, going by this I can always say that what are my 0 force member in the truss. So, we will take a quick example of 0 force member let us try to identify what are the 0 force member in this truss for the given loading. Can we identify quickly the answer is yes how about if I just first try to understand that we have to make sure the we know how the reactions are going to come into play. So, one reaction here will be vertical reaction here and vertical reaction here there is no horizontal reaction. So, therefore, now we go to the special conditions. So, I can clearly see to go to the 0 force member if I really go to this member right here I can clearly say that IE equals to 0. Similarly, once IE is known let us say then again I can go to join T now join T is under that special condition where if you take a equilibrium perpendicular to HC I am going to get EB equals to 0. So, EB is 0, IE is 0 what else is 0 in this case. How about IJ is it a 0 force member the answer is yes now you have a vertical reaction here right and if you look at the joint equilibrium here there is nothing to balance. So, IJ has to be also equals to 0 similarly if IJ is equals to 0 then IH is also equals to 0 from joint I then you go to let us say on the other side similarly the logic that we have say there is a if you look at the joint equilibrium here of this joint I have a reaction here that reaction will be taken by this load right here. However, there is nothing to balance this force. So, therefore, this will also be equals to 0 anything else could be 0 yes if this is equals to 0 if any other FG equals to 0 then GH is also equals to 0. So, we can take one more examples of this. So, try to identify quickly what are the 0 force member very immediately we can say LG is 0, Ni is 0, CAH is 0, EJ is 0, DE is 0. So, all of this has to be looked at very quickly very thoroughly and very quickly. So, you can clearly see that all of these joints are falling under those special category and therefore, we can very easily identify the 0 force members on a truss. Now, before I get into the method of joints is there any points to discuss here before we go to the detail analysis of a truss that try to find out the member forces can we have any kind of discussions here. 1139 go ahead for your question. You have shown such a simple truss sir and there you are saying that two members are required to form a two members and one joint, but there you have not defined the truss definition. No, you are directly given m is equal to 2 n minus 3 m is equal to 2 n minus 3 yes. So, the logic is that now see before even get into the truss. So, your question is why did I call it a simple truss. So, it is a truss it truss yes. So, the idea is see truss is what as I said see you can when we are explaining we can say the truss is just a formation of triangles. If we have triangles on the system right I can say it is a some kind of truss remember I am assuming that all the joints are pin also that has to be a condition. Now, simple truss when I am illustrating for the simple truss I am saying that when you think of a basic triangular truss if I have a just a triangle you apply any kind of load it should not be internally deformed first I make sure that the body that I have chosen is properly rigid. If I cannot say the body is rigid then I cannot say that that is a good truss that is a stable truss because internally if the it is collapsing then there is no point of discussion. So, to define a proper internally rigid truss we came to the definition of a simple truss. Now, simple truss that is a thumb rule that m is equals to 2 n minus 3 because you are always extruding from the 2 members from the joints and connecting it by a another pin connection. So, m equals to 2 n minus 3 that logic is always being maintained. However, it was not necessary that it will always form know the triangles I have also shown an example but by m equals to 2 n minus 3 it is not going to be proper triangle but remember that if I follow that logic it is always going to be internally rigid that means no matter how I apply the load the internal angle will not change that is the rigid body concept and then we can solve for the forces external forces. So, internal rigidity is being maintained in a simple truss. Now, to make the entire truss to be stable then we have to see that what kind of support reactions are necessary. Now, if you support a simple truss by 3 reactions then it is fine. I want to add one sentence to format truss 3 basic bars are required and 3 joints are required. That is the basic triangular truss. You are absolutely right. I need minimum I need 3 members and 3 joints. Sir, the problem that we just discussed in that we were finding the one we were finding the 0 force member sir. If you look into the half part of it yesterday we had a problem that was solved by Mandar Inamda sir. In that case he had particularly considered a particular beam to be 0 force member whereas in this case we are not doing the same. So, the last problem that we just saw where we were considering the 0 force members. Sir, the link DE that we have, in that case it was yesterday he had the same configuration but it was horizontal. This is a vertical configuration. He had a horizontal configuration back then. So, I would just like to know how do we decide upon a particular member being a 0 force member and not being a 0 force member in the loading conditions, similar loading conditions. So, let me understand your problem question clearly. So, because the last problem means you are talking about the one now in the slide. The problem one this one. Sir, the other one. Yes, because your voice is breaking just now go ahead. The question was why it is a 0 force member i.e. Sir, the problem is that why if DE is a 0 force member why not DD is a 0 force member in this case. So, now your question was why DE is a 0 force member but not BD. Remember all of the time I am looking at a joint equilibrium. So, what I am saying here that if I consider a joint and these member forces are transmitting the force into the joint that is the first part. Now, I am simply considering this joint has to be in equilibrium. Now, I have already proved in this from this side that i e is equals to 0. Now, if i e is equals to 0 then if I come to this joint just think of this joint now has this force this force and this force. So, E h, E c and E b. Now, E h and E c they are equal to each other just think of taking an equilibrium along the E c axis. Now, if this is so then E c must be equals to E h. But if I take a equilibrium perpendicular to this line of E h and E c then what happens then there is nothing to balance no force there is to balance for E b. So, that means that is equals to 0 come with the same logic this side. What happens now I have to start from joint g first of all is it clear why E b is equals to 0 did you follow me. Now, let us come to this side I am starting from joint g can you please tell what is the force in g d. It will be equal to p. Yes absolutely great. So, now if this force is equals to p remember in this case it was 0 now this is p. Now, you think of same way that means what happens if I take a vertical equilibrium about this line h. If I consider h and I try to take a perpendicular equilibrium what that would give me that would give me now a relationship between b d and p. Therefore, b d cannot be equals to 0 is that clear now. So, it is all about equilibrium about any two axis that is what we are doing because remember now joint is simply a particle and on that particle I am applying concurrent system of forces because all the member forces are coming and meeting to this joint it becomes a concurrent force system and the resultant has to be 0 that means what sum of force along x equals to 0 sum of force along y equals to 0. All I am saying that x y can be any two perpendicular axis need not to be horizontal and vertical. Therefore, since you have said that d g is equals to p if I take a ideally consider an equilibrium perpendicular to a h then you can see that d b of some angle must be equals to p of some angle. So, therefore, d b is not non-zero is that clear now. Yes sir yes sir 1 1 3 1 in problem 1 the full bottom core member appears to be in 0 force in practical how will it be possible that the full member f j will not have any force neither tension or compression. So, your question is that there are so many members let us say there are so many members that are going to 0. So, let us say I have you know in the bottom cord of that truss bottom members all are 0 how is that possible. So, we do not think like this that is a special condition of a loading for a given loading the member forces are going to 0, but in a practical when we are designing a we have to design it based on all possible combinations of loading. So, all possible combinations of loading will not give to be lead to 0 force. So, it depends on I may get some forces to be 0 for a given condition of loading given set of external loading, but it may not be 0 all the time is that clear now. So, for a particular loading it is just 0, but it is not necessary that there will be always 0 depends on the loading condition. Still the force p acts at g downwards it will pull the joint d downwards the joint d will pull b and a downwards and hence a a b c will come in compression and hence f h i j will go in tension that is what I feel. No fine no whatever it goes that is a different issue which member will be in which compression or tension it is fine, but see remember your truss is a simple truss it is properly supported supported by the reactions properly it is a properly constraint and stable truss. That means no matter whatever tension compression coming into play it is fine, but the truss will not deform truss cannot deform it is being a rigid body internally rigid. Now, there are actual deformation that will take of course place, but what I am trying to say actual deformation is what we study in the you know when we go to the next level of course on structure analysis there will study the actual deformation, but in this case engineering mechanism rigid body concept is intact that means that truss can as a whole move that means as a whole it can translate or as a whole it can rotate that is the rigid body, but I cannot say that individual parts are actually you know doing the tension and compression that means individual parts are actually deforming in such a way that entire truss will collapse I cannot say that. So, that is the whole point of same and we will come to that in this session itself towards the end when it is not going to be stable and when it is going to be stable internally even. Just a small one more doubt in the initial slides of internally rigid trusses. Which one? Simple trusses one deep. Yes. I think third force light it might be. Yeah go ahead. Yes sir. Yes. That figure in that figure we have shown one force acting at B dash B is the force actually applied on the structure this is the force being applied now consequence is that see that brings in the concept of why a simple truss is rigid and why other truss is not rigid. This is not a simple truss first of all. This is not a simple truss because it does not start from a basic truss triangular truss look at the concept of simple truss here. First I must identify a basic triangular truss that means that has to be formed by three members and three pins. Then I start doing this operation. Now here that is not maintained first of all and the consequence is very very simple. If you apply a load every joint is a pin what will happen? This joint can rotate this joint can rotate. There is no member that this joint will transfer the force to member to make it simple truss I needed a member BD. I needed a member BD I repeat to make it a simple truss I need a member BD. If I do not have BD consequence is severe the truss will collapse because there is no load to support this externally applied load. You can see now just from the intuition this joint will try to move this way because there is a pin here it will try to adjust themselves. So, this will simply undergo rigid body rotation this is also coming down it can deform into different shape depending on to what kind of load you are applying. But in this truss if you apply a load it is not going to deform in any manner like the way it is shown. So, internally this is rigid you are see there is a small deformation that deformation is not what we are talking about that deformation is always present that comes in strength of materials. When we solve for the forces remember that deformation does not do anything unless otherwise it is a very very large deformation. Because engineering mechanics we started with a rigid body concept based on the fact that the body can only translate and rotate. For example, if you have a beam right a beam you apply a load of course there is a deformation that is a bending deformation to the beam. But when we are solving the forces where that is where is that bending we are not taking into account right. So, here problem is very different first we proved that this entire truss itself is a internally rigid truss it cannot change any angle means it will not collapse under the action of load it will not collapse the word collapse is used in this case it will collapse it cannot sustain how can it sustain this load. So, simple truss is always internally rigid and this is not internally rigid because you have pins they are able to rotate itself ok. So, there is a actual deformation that actual deformation is the next study. But there is a real deformation the real deformation if it takes place real means that is a rigid body deformation that means we are really going to share the collapse stage ok yes. Sir, we preferably go for the perfect frames always means it is satisfying the condition m is equal to 2 n minus 3. Why? Sometime we may need to go for redundant frames also. So, how to ensure how to think? So, now I got your questions you are saying now first of all this is a truss ok I am not talking about frame here your question is what is a redundant frame redundant is simply statically indeterminacy even you can think in terms of truss perspective suppose due to severe load one member fails right then the truss become unstable if I make it perfectly statically determinate then if one member fails then the truss cannot sustain the load it will collapse. So, what we do we add few extra members that will make it indeterminate. So, that will be the redundant members. So, we prefer to always make truss redundant you are absolutely correct we always prefer to make a truss with lot of indeterminacy because if one member fails at least the truss will not collapse ok yeah, but we are not going to solve indeterminate problem and that will bring in the essence of what your colleague was telling before when you want to solve indeterminate problem we have to look at the kinematics actual deformation of the truss ok then actual deformation will come into play and that will couple to solve for the extra unknown forces ok that is in structural analysis sir in practice whether we shall prefer the redundant frames then it is always desired in case of a truss it is desired we tend to do it, but it is all depends on the designer in practice you will always try to decide on a redundant frame you are and in fact there are lot of research how to make a building redundant you know more redundancy, but the you have to minimize the redundancy also it is not that you want to make it extremely redundant because your economy comes into play also ok 1, 2, 7, 9 go ahead for your question. Yes sir what is the ease of zero force member in a truss why we why we incorporate zero force members in the truss? Your question is why do we incorporate zero force member on truss my answer is we do not incorporate zero force member it is a consequence of the external loading as a designer I will always try to make different kind of loading conditions for that loading conditions I design the truss now I have given a problem in such a way that some members are becoming zero due to that loading it is not the incorporation of zero force member, but if I apply some other loading on to the truss they will not be zero no they will not be zero ok Is there any criteria on the number of zero zero force members no, no, no there is nothing like zero force member incorporation it is a process depending on how you apply the load in the structure ok.