 So let us start with the important features of a class, it usually a class consists of a norm and a standard state member connected to join, no member is continuous to it and we can simply say that formation of a class is nothing but simply assemblage of triangle which is connecting many triangles to a central format. The basic assumption that is made is although class joins they are bolted or welded connections, they are actually assumed to be a in connection. So that is the basic assumption we made and the basis for that assumption is that firstly we can see that these members are long and slender in nature and also the loads that we apply in a truss that must pass through the joint therefore what happens at the joint we are going to get a concurrent system of forces. So each and every member since they are connected by pins at the two ends we are going to call it a two force member as we know and therefore at the end that is at the joint there is only one force that is acting that is single force with no couple. So usually we go through also the notation of tension and compression. So in the member if we look at it if the members are actually pulled then we say it is a tension if it is pushed then we say it is a compression. So as we said that we will never see a truss where the members are actually subject to any kind of lateral load. Later what happens eventually if we want to you know make a space framework for example as you see this is a kind of bridge example several trusses are jointed together. So what it is shown here that we have two trusses they are jointed together and how they are jointed is actually with the help of stringers and also we have sorry floor beams and also we have the stringers that is running around. Now what happens as we can see that all the movement is happening actually in this area. So that area can be subjected to lateral loads. So that means these members can actually be in bending and so on. But eventually those members specially these floor beams they are attached to the joint of the truss and therefore all the loads that is acting here they should be transferred to the joint of the truss. So now there are various types of trusses and specially as we can see here clearly that these are again nothing more than you know assemblage of triangles. These are special roof trusses those are used and named after you know various people who first discovered this kind of truss. And this is actually bridge trusses there are different kind again they are you know named as based on either shapes or maybe when they are first used based on whoever first made those kind of trusses. So the first thing to understand here that what is a simple truss a simple truss is simply a triangle connected by pins that is the first thing that is the you know sorry that is the basic truss. So basic truss is always going to be a triangle. Now from the basic truss we can actually construct the main trusses which will call the simple truss. So how the simple truss is formed for example that if we successively add two members and one connection we put. So B D and C D has been erected from the basic truss basic triangle and they are connected by D ok. So like this we can actually now think of it that we can also erect another two members maybe let us say D E and C E and connect by a pin. And if we keep doing this operation we will eventually form a simple truss ok. Now what is most important to understand is that a simple truss is always internally rigid that is a very important you know concept that we use. So that means a simple truss cannot be deformed under the action of any load. For example now this is not a simple truss. So in this case we can clearly see if I really apply a load because this end and this end is pin. So it is actually you know rotations are allowed about the joint and it will deform. So therefore in case of a simple truss if it is properly supported by at least 3 reactions then we can say that it is properly rigid truss. It is internally rigid as well as externally rigid and it will be properly constrained. So under the action of any kind of load it cannot deform. So the basic thumb rule for a simple truss is that m equals to 2 n minus 3 where m is the number of members and n is the number of joints and you can always show this that this is the you know simple thumb rule that we can use to verify that it is a simple truss that has to follow. So now there are you know how to analyze a truss. So now we have found ok this is simply is going to be assemblage of triangles and it is going to be supported by hinge connections or let us say roller connections. So what are the methods I am going to use? There are two methods first one is the method of joints and the second one would be method of sections. So the method of joints what we need to do basically this is the applied load and these are the reactions. Remember since it is a simple truss it is internally rigid that means I can assume that entire truss is a rigid body. I need not to bother about what are the members internally connected as such. Therefore what I can do? I can calculate the reactions just by considering that this is a entire rigid body and I should be able to solve for the reactions. That is the first thing. Second thing is that when we really want to solve it we try to create free body diagram for each and every member and the pin ok. So as we can see here that any member I see that AC let us say this AC member we have used this you know directions to represent it is in compression. Now if this is in compression so we will represent it like this way and therefore as per Newton's third law at the pin or at the joint it will simply be reversed. So right here it is shown in a reverse way here also at C it is shown in a reverse way. Now how do I solve it? So first of all we count the number of members present. So number of members is equals to 5 in this problem. Number of reactions equals to 3 ok. Now what is the then total number of equilibrium equations that are available? Because remember each pin is under equilibrium. Each of these joints will be under equilibrium and therefore per joint since it is a set of concurrent forces we can say that there are two equations per joint and they will be simply sum over force along x equals to 0 and sum over force along y equals to 0. So per joint basically I have two equilibrium conditions to use. Therefore total number of equilibrium conditions in this case should be 8. So therefore the total number of unknowns including the member forces and the reactions is equals to total number of equilibrium equations and therefore we can say the truss is statically determinate and why it is completely constrained that is basically we have already said that simple truss is internal rigid. So no internal deformation is allowed first of all and also here we can clearly see that it is supported by three reactions ok and these reactions will never allow the truss to move anywhere ok. So it is externally rigid as well as internally rigid and therefore it is completely constrained. Now here what is being presented is basically not very commonly used but we can apply actually you know equilibrium conditions in a different way because as we know that these are actually concurrent system of force set at each and every joint. We always have concurrent system of process and if it is under equilibrium then we say that resultant force must be equals to 0. That means what will happen we can actually create a force triangle let us say we talk about joint A right here. In joint A we have actually let say the reaction is known right. So we put the reaction and we use the triangle you know we construct this force triangle in this way such that these two member forces can be known. So either we can see that it is a close circuit all the time we are going to get a close circuit therefore the resultant force is actually equals to 0 and it helps us to visualize also that ok what is the nature of the force in the members. For example here we can see that nature of the forces at A B that has to be tensile whereas this has to be compressive. So creating this force triangle is very simple although it may not be unique in the way why it is may not be unique. See I can actually in this case see I can set this you can choose any direction. So you can actually go you know any downward we can choose. So therefore it need not to be necessarily unique. But what it helps us to do is that quickly inspect that what are the you know different forces I am getting in the member and we can always if let us say I have two unknown forces then that can always be solved. Knowing the one force I can always solve the other two forces simply by using the triangle rule. Now if the number of forces actually you know are more let us say more than 2 then I may not want to use it because in that way I will create the polygons I will keep creating the closed polygons. So as such you know it is not very mandatory but this is just a graphical way to solve the problem also ok. But in general we tend to solve it using method of you know in the using the equilibrium conditions that is sum of force along x 0 and sum of force along y equals to 0. Now since we have learnt the equilibrium conditions sum of force along x 0 and sum of force along y 0 we can actually keep that in mind and try to see look at joints under special conditions. So in this case actually four members are connected and also observe that forces in the opposite members intersecting in two straight lines there are four members but they are intersecting in two straight lines. So we can clearly say that force in A D must be equals to force in A B and force in A E must be equals to force in A C that is true based on the graphical approach that is true based on the equilibrium conditions. Now this is very important because when you see actually a truss we should be able to use this special conditions to immediately get a feedback that what kind of relationship exists or maybe some forces can also be solved directly. So now let us look at the other type of condition where we have two opposite members the forces in the two opposite members are equal when a load is aligned with a third member. So what it means that in this case actually a load is applied at the joint but that load is parallel to A C. So therefore, I know for sure that force in A C must be equals to P and it is going to be compressive in and also if I do sum of force along x 0 then A B must be equals to A D. This is very important if actually no load is applied at the joint then we can always say the third member that is third member A C is equal is actually force is 0 so it is a 0 force member. Similarly if two members are connected but they are aligned then forces will be equal in this case if they are not aligned then forces in this both of these members should be equals to 0 if and only if there is no external load applied at the joint. So with that we can actually see that how do I find typically the 0 force members in a truss. So we can do one small exercise here so again this is a simple truss let us say connected by a hinge here and connected by roller here. So we basically have two reactions here and one reaction here and we can clearly see what are my 0 force members they are actually marked here in green. So for example now this joint comes under which condition here actually this joint C will simply come under this condition therefore we can say the force in B C is equals to 0. Similarly we have the same situation at K that joint will also tell K J equals to 0. Now as soon as K J equals to 0 then J joint again comes under this and we can say that J i will be equals to 0 because if I take a perpendicular equilibrium right with respect to let us say this member. So if I take a perpendicular equilibrium then immediately we can say that J i equals to 0 ok. So with that we will just take up a very small problem to identify the 0 force member in the truss. So start with let us say joint I I is equals to 0 if I E equals to 0 you can immediately go to joint E therefore B E equals to 0 if you take a perpendicular equilibrium B E has to be equals to 0. Is there anything else I can do on that side I J is equals to 0 why I J is equals to 0 because there is only roller reaction that will balance this one and therefore I J will also be equals to 0 if I J is 0 then I H equals to 0 then F G is equals to 0 why F G is equals to 0 because there is only vertical force in this truss there is no horizontal force so that is going to be equals to 0 and therefore G H is equals to 0. So it is just the visual inspection that we are doing and we are applying the special conditions that we have left. Number B to 0 is it not necessary that angle made by B with the C H no no no not necessary at all see what we are looking at I am inspecting from these joints. So first we have said that I E equals to 0 yes right and then once I go to joint E now I have three forces these two are equal but this one if we take a equilibrium perpendicular to C E right there is no force to balance that force B E. Yes suppose some force is acting on the point C. No if that is acting then no still does not matter if we are concerned about the B E it does not matter if I apply a force here sir it is always happening at the joints. No if some force is acting at some point A or C the member B has to be equal to B has to be perpendicular to the C H to have the force in the B to be 0. It is perpendicular what is it has to be no no no no I am asking I am making my clear doubt clear is B has to be perpendicular to the C H. No no not absolutely not answer is no why should it be. It should be sir. No because you just look at the joint equilibrium over there see only way B can be non-zero I cannot comment. Similarly I also have the doubt if the forces are just see here if the forces are like this cross which one yeah. In this case also this member should be at 90 degree no. No no no no not at all why because it is all there are four forces here they are all concurrent forces why it is necessary to be 90 degree you just take the sum of force along x 0 then you will equate eventually F A must be equals to F A D. Suppose if angle B A E is less than 90 degree yeah will that condition applicable? It is applicable applicable it is applicable there is no issue as sir because we are always looking at the equilibrium and there are four forces at that particular joint and we are all all we are saying that sum of force along x equals to 0 sum of force along y equals to 0. So, what relationship exists between these forces right you can also visual inspect you are actually going to create a polygon that is a closed polygon there is no R. So, R equals to 0 and that polygon tells me immediately that F A D must be equals to F A B and F A C must be equals to F A E. I have a answer of that question sir can I explain? Yeah you please go ahead. Now analysis of a joint when you consider the joint that will act as a concurrent force system okay now at a joint I there are three forces now one is H I second one is I J and third one is I now when three forces acting at a joint out of those three two are collinear then third incline force is 0 to maintain the equilibrium of that joint whatsoever is that inclination right. So, at I three forces are there therefore I is 0 as soon as I is 0 then we are looking at the joint E at a joint E three forces are there again one is H E second one is C E and third one is B E out of those three two are collinear then whatsoever is the inclination of a third force to maintain the equilibrium of that joint it has to be 0 okay thank you so that is what no no it is immaterial whatsoever is a force acting at a point C we are just analyzing the joint E only we are going to analyze the joint E only we are not considering other forces which are acting at a different joints yes yes no force matlab only how to concentrate over the three forces are there or not more than three by observation we can't say it by calculation it may be 0 but by observation we can't say about it okay thank you so again we consider this trust now so same logic we will apply if we can start with joint A let us say if we come to this joint then we can clearly see that A F equals to 0 right similarly we go to this joint joint E remember both have to be equals to 0 under the special condition okay C H is 0 n i is 0 lg is 0 all are under special conditions yes sir before we go into the details of these problems could you please tell what is the advantage why we really use the trust see the trust is basically used or we can think of that way that one is the economical design aspect right so that is the first thing and difference you know in the sense that since trust members are slender right and loads are applied at the joint then you can actually use very simple analysis than that of the frame sir my question is what is the difference between trust and frame sir that is the exactly what we should try to think of it that is what I am saying that frame will carry the moment see frame takes the load in the member itself in a frame I can apply the load in the member but in a trust I can't do that it is it is it is not two force members see the reason is that only difference is that because of this lateral load that is in the frame right and joints are assumed to be rigid so there is a moment at the joint excuse me sir yes sir why we need to provide this zero force member even though absolutely good question so the answer is why do I need this zero force member remember I am just here trying to say that there is a this load is being applied but remember a trust can be subjected to any kind of load right so if I change the load direction or the late load position then I need those members to held the equilibrium of the trust in see then then then you are actually trying to say if there is a small deformation then what happens can the member be held see as long as my force equilibrium is concerned nothing is going to be problematic under this force under this load even if there is a zero force member but remember if I really try to take out this member and construct this trust is it a visible first of all it is not feasible because this member will automatically collapse so to construct that trust I need these members what will happen if I take this member out this can rotate about this point all of these are hinge pin right so as long as it is a pin therefore this member will automatically collapse so trust is formed to make sure that all combinations of the load should held the trust in equilibrium sir yes sir what to do when the forces are in literal direction like wind force or earthquake force does not matter no so it will always be earthquake force is a different issue that comes from the ground but when when wind force are coming forces are acting then we just concentrate the forces at the joints all the forces that we analyze here are said that they are going to be acting at the joints ok so all the considerations that we will make load is going to apply at the joints ok so let us move on to the method of joints we will just solve some simple problems so we consider this trust again there are three forces here and if we look at the geometry of this trust we can see the based on the loading that we have this trust can be said to be symmetric in other words the force in this side of the member should be equals to force in this side of the member so it is a symmetric due to the loading that is being applied and the type of supports that is being put so as such you see what we can quickly do when we want to do the method of joints either we can go joint by joint that is one way of doing it so we go from joint to joint but from where do I start I should start from a joint where two members at least we have two unknown forces that is a conventional right approach that we follow so that means I can start from joint a so if I consider joint a I can automatically say that the force in a b equals to 0 right so just go to this one so we start from this joint now how to start the problem this I have used already the special conditions right I said if a b equals to 0 because you know ideally we are doing some of force along x 0 at that joint but remember I have to first make sure the tensile and compressive that notation has to be correct so either tensile force or compressive force how it is going to be presented that has to be absolutely clear to us so tensile force I could assume that I will start by saying that forces are actually tensile in nature if it is tensile in nature in the member then from at the joint it will be shown outward so I will assume the forces are in tension to start with and then take the equilibrium whatever sign I will get that will if it if I get a positive sign then it is tensile if I get a negative sign it is going to be compressive remember I can make a tensile as compressive and you know compression as a tensile as let us say negative and compressive as a positive I can start with that no problem but I have to I have to be consistent whatever I use throughout see in this case we will try to always assume that forces are you know member forces are in tensile in nature and therefore I start the problem from there so in this case what has been found is that f ad equals to negative 5 okay now once this is found then I will go to also joint e why I am going to joint t remember joint e has more than two forces but these force can be immediately known so if I go to joint t then I can always say that fbe must be equals to 12 and fbe is equals to it is it is actually tensile in nature right and this must be equals to this one so f ed must be equals to f e f so that is the joint equilibrium I have at joint e so we keep solving like this then we can see that clearly I already know this one and that one I have also found out this one so f ab is known fbc is known by the symmetry f e fbe is known therefore I am left with bd and bf so I can solve for fbd and fbf at this joint b okay so here again every time I am doing this task in this particular problem making use the symmetry nature of the class so last joint would be let us say joint d so what is left so far is the d e see what is left here is d e right I have already found ad I have found ab I have found bd and I have found d what is left is really d e so I can simply jump on to this joint d and once I go to the joint d right I take sum of force along x equals to 0 so since I already know fbd then I can find fd I can further check my answer whether I got it correctly or not so check will be basically we can do sum of force along y equals to 0 that will give me the reaction right that reaction is now equals to 21 now that reaction can also be found at the very beginning by considering the total truss to be rigid right so that we can do at the very beginning here we can see clearly due to this force it is going to be reaction is going to be equally distributed so that should be equals to 21 by considering joint a we got f ad phi compressive yes then ad joint d also it has to be compressive at joint d it has to be compressive no joint d it has to be compressive you are correct so you can show it like this yes if you already show it like this then we don't have to add that negative sign yes okay but already we have confirmed it's a nature over there what was that already we have confirmed nature of f ad for that earlier free body diagram there there are processes I mean you can carry over either the sign in each and every step we can carry through that or we can change the direction and then you forget about the sign then we just say okay it is not negative five it is now positive five initially we initially we are assuming all the forces are tensile right initially we are assuming all the forces if your answer is positive then our assumption is correct that is tensile yes if it is negative then compressive it is so already we have confirmed about force f ad that is compressive then again why to assume it as a tensile in the next joint hello we have already assumed the signs we have already assumed the signs it's like a mathematical problem I am just simply using sum up there is minus we have assumed tensile force and the tensile force is negative five this is f ad we have initially found minus five yes if in the last equation if we don't consider f ad to be tensile what is shown in here then we cannot replace f ad as minus five in the last equation in that case we have to replace f ad as plus five but f ad once we have got minus five we have to put if it equal to minus five that is why I have to show in that way no you are you are sticking with the sign or a nature of the force that is important that is both both are interrelated both are interrelated as such when we do this one yes see there is nothing called positive and negative and that is what we precisely tell to students yes but what happens that if you are carrying forward the nature of the force then you have to keep changing that the keep changing that you know direction every time you solve one step and carry forward it to next step but if someone wants to have the sign from the beginning he or she is absolutely welcome to do so yes no no no no issue at all negative means something is wrong over there and we are writing compressive over there right now so we have to replace that negative sign with compressive force so as such as I said that sign does not make any any kind of you know issues here as such because that is just irrelevant in this case we can simply also look at by the nature of force okay so we can in principle reverse the you know direction if we found a compression in the previous step we can simply reverse the direction see what it means I can also start with APD this way no so I will get a positive sign how does it matter it does not matter okay so just be consistent with whatever we do but I think you know as long as we keep on exercising this we should be able to have a grasp what is going on on the subsequent steps see this is another problem sir for the previous problem is there any typing mistake that is what I am trying to understand FDB we are getting 55.25 but you are given is minus 34 where where can you please identify FDB you consider the whole trust and calculate the reaction at D reaction at D D off of the load because load is symmetry yeah so that is point a reaction is 21 at D and also F is there any typing mistake here FAD is negative 5 but when you analyze the giant D we are getting the answer is FDB is 25.25 when we are analyzing giant D joint D okay okay after getting the reaction of 21 already is equal 21 we are getting because of symmetrical loading here right here so if I substitute APD with sign and APD with sign am I getting this answer here what is the problem I am trying to find out what is this force here right to do the equilibrium that is just checking I mean whether I am getting back the reaction correctly or not it is just a check as such we have found out all the forces in the members right I am just checking okay if I analyze the joint D now then am I going to get the reaction back or not just a check sir sir one question is there so what changes we have to made in this trust if suppose a dynamic force comes into the picture if if dynamic forces comes into the picture you just take the inertia forces effect no lump mass assumption or this assumption so why we are all the time focusing about the because we have we have already said that we are going to use the static at the beginning so dynamics will only come later on okay but is the dynamics coming I have not talked about anything dynamic loading here then I have to take into account inertia forces right but in actual practice the things may be different now everything is going to be different yeah everything is going to be different k minus omega square m will be the net thing that I will be looking at okay so you have to generate the stiffness matrix you have to generate the mass matrix and we have to solve that okay that we will talk on Friday do not worry but not probably trust but let us try to do in a simplified manner okay so just another problem of method of joints so as such we are going to follow the same logic so we will start from this joint and once that these two forces are found we can keep moving and try to find the all member forces now I first step you know we can solve also for the reactions if someone wants to let us start the problem from this end then we can solve the reactions and we can also since two forces are known in this particular joint C as reactions then we can solve for the other two unknown forces that is always another approach so we can that is an alternative we can also proceed from the other end but see what we look here in the method of joints that we always try to start from you know joint where two member forces are unknown that we always tend to do but remember as such for a simple trust like this with which is supported by three reactions it is always possible to find out the reactions at the beginning also okay so this is just quickly done here how do I find the reactions at the supports let us say we have C x and C y and then here we have the rollers so that is E y and we can again use the entire truss as a rigid body and we take the moment about C we take sum of force along x equals to 0 and sum of force along y equals to 0 so three conditions of equilibrium will give us the reactions three reactions three unknown reactions that is Re C x and C y. Sir sorry to interrupt you sir I have no doubt sir suppose I calculate first the reaction forces at C and another joint is and their support is E. E support roller there was a roller at that support right. So I am finding the support reactions and then I proceed to C there are two unknowns followed by E again there are two unknowns then I move to D again there will be two unknowns and then I move to that center point center point B. So in the next when I start from joint A build the answer match absolutely it has to match you go anywhere around because you are talking about a equilibrium right see ultimately what it is we have how many members are there let us look at this way how many members are there 1 2 3 4 5 6 7 I have seven members how many reactions I have three right so total number of unknowns are 10 10 okay okay and what we are doing is really we are taking the joint equilibrium so now we are saying per joint I can establish two equilibrium conditions so how many joints do I have I have five joints I can get exactly 10 equations 2 times 5 so 10 equations should be able to solve this 10 unknowns you can do it in a computer way or whatever so basically we can even adopt a let us say express it excel sheet that will also solve you 10 equations with 10 unknowns but what we are doing here that this truss is not complex so we can start from anywhere we like as long as there are two unknown forces okay clear so just quickly here the you know just to show that first part of this problem let us say we have joint A here right in joint A I have two unknowns and immediately I can create this force triangle so there is a 2000 Newton here and to create the force triangle I will just simply say F AB and F AD right that closes the triangle so resultant force is 0 and from this we can use the laws of science to get the values of also F AD and F AD so this is alternative approach right this is just we are using the laws of science to get the value of F AB and F AD nothing more than again it is actually a joint equilibrium sum of force along x equals to 0 sum of force along y equals to 0 in any case now since we are all in the same page so once this is found then we can actually go to join D and we can solve for DB and DE then we can you know look at joint B from joint B we can try to find out the other two unknown forces that it F B and F BC and then we can come to join T we can find out F EC now as it is shown here that this was the earlier discussions the nature of the forces are actually reversed at the joint and if we look at the nature of if we reverse the nature of force then we should consider this in this case all the time when we are establishing the equation we have to consider it to be positive ok so here nature of force is already being taken into account because this was compressive so it was taken inward to the joint so this is just an analysis method the second method is the method of section now why do I use the method of section first of all suppose where the forces in a very few members need to be calculated then we can adopt this strategy of method of section remember in a method of section what happens that again first we run this section and we split the truss into 2 halves so therefore these forces are exposed now and we will have a left free body diagram or right free body diagram remember this part right here is a rigid body this also is a rigid body because these are my basic triangular trusses so I do have two rigid bodies now being disconnected and I expose the you know these two parts and therefore we can draw the free body diagram of this side on the left free body diagram should be shown like this where I have only three forces as unknown forces these two are known loading so now we can actually solve for these three forces the difference between the method of joints and method of section is that a portion of the truss that is being cut out is already a rigid body and therefore since we treat this as a rigid body it is no longer a particle like the pin was in the method of joint it was just a particle there and we have the concurrent force systems now this became a rigid body so therefore we are again going back to the 2D equilibrium so we can actually use sum of force along x 0 sum of force along y 0 and moment about any point O is equals to 0 so these three equations will again be used to solve for the unknown forces so simple way to look at it suppose I want to solve for the force in FBD let us say I want to find the force here then what do I have to do take moment about E I want to find FCE then I should take a moment about D equate it to 0 so like that we can you know do this so we will now try to take a more complex problems to realize how powerful this method is when it comes to calculation of just one or two forces within the truss anywhere limiting criteria for passing a section it can't cut the members twice twice yes single member can't be cut twice that's the only way then you are not actually doing the method of section so just try to see for example if I just pass this around I take this way then it does not make any sense you want to always create two halves out of it and one single member should not be cut twice then you are actually having the same equilibrium no no that's more than three unknown should not be there maximum unknown set that section should be no no no that is not true actually legally that is not true you can have number of forces more than three also when you do the method of that should be calculated but idea is that you should cut a section in such a way that you can solve for at least three unknowns okay we'll come to that point just keep in mind that we'll actually try to interrogate a truss like this sir the section is always a straight line or it can be a curved line also it does not matter what we are doing is see if I do like this it does not matter all we are saying that it is being cut and there is a force that force is exposed so how it really does not matter if I go like this or if I go like that we are simply exposing the forces on those members and a force has an infinite line of action a force has an infinite I don't think that is we don't have to okay so we take an example of a thing truss now this is a simple roof truss that is being shown what is asked here that we have to find out C D C E and B D these are the three member forces we want to find out C D C E and B D these are the three member forces I have to find out okay so remember to do the to do the method of section firstly we must solve for the reactions because again in method of sections we want to detach the two parts of it right so therefore these reactions have to be known to us okay so when we run the section I will just show it here so we have these three forces are actually exposed right so we have once we pass a section through B D C E C D and C E then basically we are now looking at the left free body diagram of it now why don't we consider the right free body diagram there is a right side part also because that is going to bring in more loads into the picture so it is always appropriate to consider the part of the section which has lesser number of loads okay so here we can clearly see that to solve these forces these three forces we can now make use of the equilibrium conditions for example to solve the FC I should take a moment about D okay so moment about D was made to be 0 so the number came out to be 22.5 kilo Newton that is tensile why it is tensile because I have again assumed all the member forces to begin with is in tensile nature so they are all pointing outward from the joint okay so then we take sum of moment about A equals to 0 that will give me F C D so if I do a moment about A equals to 0 just watch here then what is happening basically these two forces are actually passing through the joint A F B D and F C they are passing through joint A so they will not contribute to the moment the contribution that I get from the Y component of F C D and the 6 kilo Newton load okay so that is how I get the force C D then we do sum of force along Y equals to 0 that will be solved for the F B D okay so remember the approach that we are really following that we are simply treating this particular portion as a rigid body and we are using making use of the equilibrium conditions. Let us say I have this kind of truss so it is called an inverted k truss so inverted k truss you can see those members okay so those are actually represented right here and all of these are pinned here so H E and B are actually pin connections okay so H E and B these are pin connections here and all everywhere we have the pin connections this is really we have a hinge here and we have a roller here. Now I want to find out what is the force in members G J and I K using method of section that is what the point comes into play how do I choose my section here in this case so that I can solve for G J and I K in a very fast manner how do I run the section again please repeat G J and I K yes so the section will be run so it will be really carved section here we can see it is going to be carved section that will cut through G J G H H I and I K yes we will expose four unknowns remember there is no member that is cut twice all the members were only cut once so that is okay no problem with it so it answers actually two questions are first one is can I have more than three unknowns yes you can have more than three unknown second ones that can I cut at a section can I cut a particular member twice absolutely no you cannot okay so the section will eventually look like this okay now once this once I cut it out I have now two choices either to consider the upper free body diagram or to consider the lower free body diagram now that is always left with the you know person who is solving it whichever way he or she fills it okay so in this case we are actually going to choose the lower part of it so to do so we are going to solve first first the problem is solved to get the reactions at the support so we are going to get this J X J Y and K Y okay so these three forces we have first got by using the global equilibrium okay the way we do it and then using the method of section we can clearly see that if I take a moment about G so let us take the moment about G then what happens these three forces that is G J this force then you have H G this force and H I all are passing through G so they will not contribute to the moment so these unknown forces get cancelled right and we can simply get the value of K I directly okay similarly I can take the moment about I and once we do the moment about I then we can get the value of G J now this is a very you know interesting problem as such to answer again two different queries that were raised now we will talk about the different types of truss now it is a compound truss what is a compound truss compound truss is composed of two or more simple trusses that means two or more simple trusses they are actually jointed together to form the compound truss so for example what are my simple trusses here A B C is a simple truss and D E F is the another one so these two trusses are jointed how they are jointed by three members right here and also remember it is placed on a support that is a hinge support and also we have a roller support here now to solve this using let us say equilibrium method of joint or method of section it is going to be very difficult for example if I just ask you what is the force can you tell me the force in this particular member I want to calculate this force it will be a bit difficult to solve so what this problem the approach that we are going to adopt here that we are going to again separate these two trusses remember that we have now two rigid bodies one rigid body is A B C and the other rigid body is D E F these are two simple trusses right and if we have two rigid bodies and if we use the equilibrium conditions then how many unknowns I can solve no for two rigid bodies six unknowns now in compound truss this is quite a natural logic that is going to be followed okay that means once I separate these two trusses I have I am separating really two rigid bodies therefore if I have six unknowns exposed I should be able to solve those a priori using the equilibrium condition and once I solve these six unknowns then only I can go and try to use the method of section or method of joint for individual trusses so therefore once I dismember this how it is going to look like so remember if I take out the A B C and D E F then what are the forces going to be exposed F B D right F B E F C along with that I also have three reactions so therefore how many unknowns I have six and these six unknowns can be solved very easily by considering the equilibrium of each of these trusses see once I solve this once I solve this I can easily use the method of joint to get these four right so as such using you know method of joints and method of sections it is very difficult to analyze this compound truss so a priori we try to get the connection forces that is being established in the compound truss similarly we can go to a next problem look at this is also a compound truss so this truss A D C and D E F how I am going to you know take these two truss out so how I am going to do this what happens at joint D in joint D right here I should have two forces right okay so if I disjoint these two trusses then you can see that this is going to be all the unknown forces that I am going to have so that means here it is going to look like this let us say DX and DY that will be simply be altered as per the Newton first law so we are going to get a DY here and DX in this way right similarly here we are going to get this unknown force and we have another three so again what we are going to see here we are actually having the six unknown forces yes so assumptions of trusses we cannot cut the joint also I am not cutting the joints here I am simply detaching it see think of I have two rigid bodies I have said simple trusses are rigid bodies right okay so I say let us say this is a body this is a body and I have simply made a pin here I want to take this out so what happens this will not a stable truss what so this will not remain a stable truss I should not be relieved because a simple truss supported by three reactions properly is always a stable truss no issue but according to assumptions joints cannot be cut we are just saying these are two simple rigid bodies and we are simply trying to take it out there is a pin right so that pin is actually allowing so actually we are passing the sections that is no that is not I am not passing any section whatsoever then you can take one of the load right in one one one of the trusses that is always done right so but ultimate idea is now I am telling I am not telling that answer what it needs to be done if we have a really force applied at this joint and the reason I am not telling you will see around five o'clock okay so we will take a quiz and there will be actually some kind of load here okay so similarly now here again take on both the parts of the no not necessary because you are automatically these forces will be balanced these force will take care of you know that distribution that is coming into play okay so only one side of the truss you can take the truss yes will be having the load vertical load yes the other part will go yes why we are doing this thing no not not every time it is possible not every time it is possible see for example if I want to find out let's say you know force in these member yes it is not that easy or let's say force in these members it is not that easy at all sir we can take this section in some cases you may but in some problem you may not see that's the whole issue I mean see it is we are trying to say that these classes can be first separated like what you have done in the case of inverted k truss what you have shown the example in the previous yes yes in the same scenario you can also make it cut passing depending on yes that is that is possible I am not saying I am not claiming to get some of the forces we may be able to do that I am not saying that it is absolutely impossible to do that but let me tell you honestly in this truss yeah in this truss specifically in this truss can we actually use method of section remember first of all what is the problem if I apply load anywhere this truss has how many reactions four reactions can you solve that a priority I have four unknowns with one single rigid body it is not possible to solve be the point so in order to solve the reactions I have to exercise this I have to separate it and I have to make sure there are six unknowns with six forces such that I can solve for the reaction is that answer your question see it is not that for all compound trusses it is required but this truss specially when you have two hinges I would not be able to solve for all the reactions a priority therefore it will be appropriate to get the reaction I have to exercise this I have to separate the truss okay and I have to expose the six unknown and solve from the two rigid bodies clear okay I will just take a quick problem here so in this case what is being asked is that to find out the forces in one two and three these are the three members we have to find the forces and one comment here all of these are continuous member they are not cutting each other at these intersections now how do I solve for this one two and three these three forces easiest way to solve I am here again see we are trying to understand what would be the easiest approach to solve this problem does the method of section work answer is no you try to solve it using method of section now you won't be able to do it can I use a method of joint that's also not possible because there is no joint that gives me two unknown forces see the logic okay so therefore what we have to do here now I have to understand that this is a compound truss what are my two simple trusses here what are the two simple trusses so great so as long as this is understood then we can go forward so ultimately we have two simple trusses ADE and BCF these are the two simple trusses which I have to detach ADE and BCF so next just 8 so if you have confusion let me just clarify some of you are claiming that ACF is a simple truss that is not possible okay there are joints here intermediate it's written here it can only be ADE and BCF AC that is what I want to find out so if I choose this one is a simple truss what happens just this is the simple truss BC and that is a total a truss right okay and also I have ADE that's a total truss can I repeat sir remember they are not intersecting here these are now intersection here these are continuous member we understood you are saying that ADE is one separate truss is it ADE is a simple truss then how about the member BD and AC this number and EF this is just I am connecting to connect the trusses I am using these members can you show the separated trusses of those two I will of course so that is the whole point otherwise you won't be able to solve it remember this is my simple truss one simple truss this is the another simple truss and I am simply joining these two trusses using three different members and also we have the supports okay AC is a member then ADE also there should be a force from D to AC it is passing here F1 and F1 look at this okay and then you have F2 and F2 here you have F3 and F3 here these are just forces because all I am saying that I am trying to separate these two trusses no sir they are connected by the members you are you are showing the member ADE let us say it is represented with a force at A as F1 what you are showing in the picture this is the force what is the simple truss first of all what is the logic for simple truss m equals to 2n minus 3 so let us not confuse there okay so first identify the simple truss then you can make any number of connections you want first identify the simple trusses properly if you cannot identify the simple trusses this problem is not going anywhere so actually I have exposed here there are three forces these are the three members taken into account to join these two trusses and also I have the reactions as usual so there will be two reactions here and one reaction right here okay so total six unknowns are again in place I have two simple trusses that means two rigid bodies and I can solve for these six unknowns I think because of the loading pattern Ax is not represented there no because Ax will be automatically 0 why sir you take the entire truss say sum of force along x equals to 0 does not matter you apply no how the things will change it is a joint you are applying everything in the joint right so if I really apply a force here nothing changes that will be just taken into the simple truss when I dismembered but in this problem there is no horizontal load so it has to be x equals to 0 there is horizontal there will be a horizontal force you solve the problem in a similar manner if you have the horizontal reaction what a big deal I am trying to find out the forces in F1 F2 and F3 right and that can always be done by taking the equilibrium of this individual trusses see ultimately I do have a Ax here but that Ax is said to be equals to 0 absolutely 0 because that is clear from the equilibrium of the total truss right so from the equilibrium of the total truss I can always say that Ax equals to 0 I should also be able to find out what is Ay and I should also be able to find out what is dy so these are these can be solved a priority also for this particular problem just by considering the total truss as a rigid board ok and then once you want to find out these three forces then we dismember the two trusses simple trusses and simply find out by the equilibrium of this truss total unknown at 6 when we separate see remember when we separate it what is happening I have a Ax Ay F1 F2 F3 5 and only thing that is remaining now is dy so total 6 unknowns are there no issue total 6 unknowns are there ok it has to be otherwise I would not be able to solve this this is a statically indeterminate if it is not there for example if I make a hinge at B if I make a hinge at B and let us say I also apply a horizontal load then problem becomes statically indeterminate so that m plus r equals to 2n that logic is always going to apply for any truss problem but m plus r equals to 2n it is a just you know you cannot say that truss is stable or not based on that logic itself but is that picture clear for the compound truss so simply think of simple truss just extract the simple truss and simple truss remember is always a internally rigid if it is internally rigid all we have to take care is the fact that I am going to stabilize that truss based on the external reactions that is being put in ok.