 Let us continue with whatever topic we had done yesterday on equilibrium of rigid bodies but specifically we will discuss about frames and machines. So what I want to emphasize is that in yesterday's class we briefly discussed about what is meant by rigid body and what is meant by equilibrium of a rigid body we saw that for any given free body some of all the forces should be equal to 0 for it to be in equilibrium and some of all the movement should be equal to 0. So when both of these criteria are satisfied we say that a rigid body is in equilibrium. Then yesterday we briefly discussed about what is the role of supports in equilibrium we saw that there are various type of supports and there can be various types of linkages which can be used to keep structures or rigid bodies in equilibrium under the action of various forces. Then we saw that there are constraints and there are reactions and as a result there is an intricate relationship between the kinematics which really is deals with motion and the corresponding reactions. So what we saw is that if a particular constraint constraints a degree of freedom which is a kinematic degree of freedom then it gives rise to the corresponding reaction and if a support moves freely in a particular degree of freedom without by allowing full motion of that particular degree of freedom then there is no reaction from the support in that direction. So then we discussed about free body diagrams we discussed free body diagrams of very simple structures. Now we solved a few problems and this is where we stopped yesterday this is the material which I wanted to cover yesterday but I could not get the time. So let us begin briefly with what we mean as a two force member this is an extremely important concept that we emphasize to the students that recognizing that particular components of a structure are two force members can greatly simplify a problem. So what there are various examples of two force members say what is the definition of a two force members really is that that we have a linkage no matter what that linkage what kind of shape it has and it is connected at two ends with typically a pin joint what does that mean that there can only be a force that can acting here there can only be a force that can act here and if because these forces are much larger compared to the self weight of this member then in that case we can assume this as a two force member that is the line of action of this reaction and this reaction they pass along the line joining these two ends. We will discuss briefly in some details this for example is a door stopper this is another example of a two force member so this is a part of a bridge they are connected to each other so this is a linkage these are two pins so this is another example of two force member this is one particular some particular form of a machine which is used to some variant of a player in which what you have is that this link is connected by two screws here and this is another example of a two force member and the most popular example of two force member and the most and one of the most useful examples of two force members that we will see is this hydraulic cylinder and what that is will briefly looked in some details. Now what is a two force member so two force member if you draw the free body diagram of the two force member what we had seen earlier now this is a pin connection this is a pin connection and what we say is that that the force that are transmitted to the pin connection now we saw yesterday that at a pin connection the force because the pin connection is restricting the degree of freedom in the horizontal direction as well as degree of freedom in the vertical direction we can replace that particular kinematic constraint with two forces one in the horizontal direction one in the vertical direction or as many of you pointed out yesterday we can replace that with a force acting in an unknown direction similarly at this point since this is a two force member since this is a pin connection you will see that there will be forces possible in vertical direction or in horizontal direction or in other words there can be a reaction acting or there can be a force acting on it in a particular direction. Now what is the criteria for this member to be in equilibrium now clearly for this to be in equilibrium some of all the forces should be equal to 0 now let us say that if this force is vertical then this force also has to be vertical now think about it if this is vertical this is vertical clearly equilibrium condition is obeyed in the horizontal direction equilibrium condition is obeyed for force balance in the vertical direction but it is not obeyed for torque balance because this positive f and negative f they create a torque or a moment or a couple whatever way you want to see about it so plus f minus f so f into the distance will be the anticlockwise couple that will be created and as a result because of this anticlockwise couple this body would not be in equilibrium and we can convince ourselves that only possible way in which this two force member can be in equilibrium is if the line of action of this force is along the line connecting these two joints and they should be equal and opposite now if that is satisfied note that we can take our x axis along this direction there is no force in the y direction so equilibrium in the y direction is automatically satisfied and equilibrium in x direction is also satisfied because this is plus f this is minus f some is 0 and because these two forces are acting along the same line they do not cause any couple and as a result the moment balance is also satisfied so this is one way in which the forces can act the other way they can act is that instead of acting outwards they can act inwards now one point that for example we emphasize to the students a lot is that two force members is independent of what is the shape of this member we do not worry that if this is L shape it is a C shape it is a straight line only thing we should be worried about that we emphasize to the students is what is the line which is connecting the two joints and when we find out that line the only thing that the two force member the only thing that we another thing that we observe is that are there any other forces acting on this member other than these two end forces if not then it is a two force member and irrespective of the shape of the member we have the line of action of the forces along the line which is joining the two ends where the force act it is a very very important concept very useful concept and we will discuss this concept we will you solve a lot of problems on two force members. Now next very important example is the example of a hydraulic cylinder okay so what does a hydraulic cylinder do so if you have seen many construction sites so this for example caterpillar is one big company which produces many of these cranes now what did they do is that there are various functions affiliated with that they have a hydraulic cylinder like this which is operated by a pneumatic device okay some fluid device is in is present inside which raises or lowers what is called as a boom and because of that you can change the configuration of the structures make this go up make this go down make this turn twist and do all manner of things that are required for construction purposes. So I have a mathematical demo here okay unfortunately my computer is having some problems so let me fix it by the break during the break and I can show you a nice animated demo of how this hydraulic cylinder typically works and how does it change configurations okay now next concept that we come to is equilibrium of a three force body now this is a very useful concept but it can be a little bit problematic so in principle it is very nice but in practice okay one has to be very careful when you are using it so before we go on to the details let us see what is a three force body the three force body essentially is so like a two force member where the forces acted only at two points so one force acted at one point other force acted at another point we join the line connecting the two points where the force acted and the line and what we saw is that for the body to be in equilibrium the force two forces should pass along the line joining the two points of action of those two forces or in other words the line of action of the two forces should be exactly with the same and the forces should be equal and opposite now in the three force member what we have is that that the force acts at one point a another force acts at another point B and third force acts at a third point C now we know from our geometry that when two force acts on it when two lines they can always meet at a point so we extend this line so this is the line of action of force one so this is line this is line of action of force two so these two meet at point D okay and they will always meet at point D now the third force for this body to be in equilibrium better also pass through point D because if it does not pass through point D then what will happen is that we can take force balance about point D and these two forces do not cause any moment point D whereas this force if it does not pass through point D will cause a moment about point D and will cause the body to rotate just nullifying our assumption that the body is in equilibrium so as a result if three forces act on a body at three different points which are not in the straight line then they should meet at a point now this idea can be used to solve problems in an elegant manner although it is not pretty intuitive at least I feel it is not always useful we will we will see that for example we will solve a few problems later on where you can very nicely use this concept but we will see that it need not be always used in a in a simple manner and we will discuss that with an example okay so let us do one simple example here so what we have this is sample problem from BJ 10 so we have a rod AB okay the rod has some weight okay which effective weight which passes through the center of mass of the rod through the center of AB in the vertical direction now this rod we want to hoist it up by hinging around point A and how are we going to hoist it we are applying a tension through a rope which is connected at point B and we are pulling on it now what we want to know that when there this rod in is in this configuration okay the weight of the rod is 10 kg and when the rod is in this particular configuration all the angles are given this is 45 degree the angle the rope makes with the rod is 25 degree we are asked to find out what is the tension in this rope and what is the effective reaction acting at point A effective reaction is the total reaction A square plus A y square the square root of that and the direction of that reaction is what we are asked to find out now this problem is a really straight forward problem you can use force balance movement balance three equations of equilibrium why because there are three unknowns here A x A y there is another unknown here which is the tension so three equations we can write for one free body diagram in 2D and three equations three unknowns this problem is very easily solved but now what we do is that let us use the concept of three force body所 and find out what is the reaction that acts at point A and what is the tension that acts at what is the what is the tension that act in this work visible to one thing said in this problem we know what is the direction of this rope we also know what is the direction and what is the point through which the weight act which is your so we know the line of action of two forces and we need to find out what are what is it this force and what line of action of this reaction. Now, we use simple geometry. So, weight passes through G center, A G is equal to G B, this angle is 45 degrees. So, this is the line of action of the weight, this is the tension, this is the line of action of the tension. We know that these two will meet at point C, by definition two straight lines will always meet, if they are not parallel they will always meet at a point. Now, this third reaction should pass through this point, because if it does not, then this will cause a moment about point C and the body will no longer be in equilibrium. Now, all it remains for us is to understand, what is this unknown angle alpha that is reaction makes with the horizontal and to do that we just use some geometry. How do we use that? It is pretty simple, but kind of complicated that it is prone to some errors, because the geometry is not straight forward, all the numbers involved 25 degrees and all are not perfectly good numbers like 30 degrees, 60 degrees or 45 degrees. There may be some issue, but overall conceptually the process is straight forward. So, B F is this distance, what is this B F is nothing but A B cos 45 straight forward, we get this value of B F. What we want to find out is, what is this distance C D? Now, what is this distance C D? This distance C D is nothing but equal to A E, because this is the center line, this passes to the center. So, by simple geometry C D will be equal to A E, which equal to half of A F, A F is given to us why? Because that is what we found out that B F is equal to A F, A F is equal to A B cos 45. So, we know what is C D. Now, what we do is that, we find out what is this tiny distance B D, which is nothing but C D cos 25 plus 45. So, it is C D cos 25 plus 45. So, we get this answer equal to 0.515 and ultimately we find out what is this C E, which is nothing but equal to D F, which is B F minus B D. So, after doing all this somewhat involved trigonometry, what we finally get is what is the value of C E. Now, I know what is the value of C E, we have already found out what is the value of A E. And so, we can use simple trigonometry to say that tan of this unknown alpha will be equal to 2.313 divided by 1.414, which is C divided by A E. We get what is tan alpha and from that we can find out alpha, which in this problem turns out to be 58.6. It is just use of geometry, but only thing is that you have to go through multiple steps. Now, what we know? We know that the line of action of this reaction is 58.6 degree. We also know from the parallelogram law or Lame's theorem that all these three should form a nice triangle in this form. So, this is reaction R whose angle is now known, weight acted vertically. We also know that this tension T acted at an angle of 25 degrees with respect to this rod, but because we now know all these inclinations, we can also immediately find out what is the overall inclination of this tension with respect to this reaction. So, we know this triangle with all these angles and we also know what is the weight of the rod, which is nothing but W times G, which is 98.1 Newton. And so, using sin law or Lame's theorem, whatever you want to call it, we immediately write it. So, T divided by 31.4, sin of 31.4 is equal to R divided by sin of 110 is equal to 98.1 divided by 38.6 and then we can immediately find out what is the tension. We immediately know what is the reaction, total reaction and we also know the angle. So, this is one way in which you can solve problems where force acts exactly at 3 points. So, these are typically 3 force members or 3 force bodies and we can use this conceptually simple ideas to solve this problem, but note one thing, that if we had used principle of force balance and movement balance, then we could have got an answer in just 3 lines. Just take torque about this point, you can immediately find out what is the tension, take equilibrium in the y direction, you get reaction in the y direction, equilibrium in the x direction, immediately you get reaction in the x direction and you are done. Whereas, here you have to use somewhat involved geometry. So, it is a matter of taste, if you want to use it or not use it, but we will see when we solve certain problems that when the geometry is nicer, then this method becomes extremely powerful. But in this case, I believe that this method does not carry significant advantage over doing simple force balance and movement balance. Now, let us come back. So, after discussing all these things, let us come to the main topic of today's workshop, of today's session. In yesterday's lecture, we discussed only about single rigid bodies which are supported by various kind of constraints. But the more interesting problems are the problems in which there are multiple rigid bodies that are connected to each other. And we will see that there are many challenging problems that arise because of that. And we will see that what are the various methods, what are the intuitive techniques that we use in order to understand how to find out the reactions from the supports and the reaction between linkages when we have multiple rigid bodies that are connected to each other. So, this is one simple demo. So, this is a simple demo of an exercise machine. So, this is a simple exercise machine. What we have here is that this particular component, we use, there is a weight that is tied here. We apply forces here. This is a fulcrum. We apply force such that this is free to rotate. And then in particular configuration, typically exercise is done in a quasi-static manner so as to get the maximum out of it. So, this entire rod is in equilibrium around this line for in essentially each and every configuration. Now, what we do in typical engineering mechanics problem is that that we have to emphasize this to the students that we have a real life problem. We have a real world problem. What we do is that we create a caricature out of it. Now, this real exercise machine, we create a cartoon of this. And we see is that that this cartoon what do we have that this hinge is like this. And this essentially is just a demo problem we show through the students so that they can visualize it. And they can see that for example, what are happening here. So, this is a screw over which the rotation can happen. See this is the screw. This is the exercise component. This is the hole which is slotted through the screw. And these are the various connections we can think about them. If you look from the side, what is happening here is that. So, these are the various angles from which we see the machine. And ultimately what we do final piece of the modeling that these are the these are the loads that we exert from our side. This is the weight. These are the reactions that exert that this portion exerts at this joint. And this is one reaction. This is other reaction. And if they are symmetric then we can deal with this exercise machine only from one side. And this becomes our 2D problem. This is a pin connection. These are the reactions that are exerted by one part on the other part. So, there is this bench. There is the fulcrum. So, bench exerts load on the fulcrum. How through this connection? And these are the reactions at the connection because it is a pin joint. It prevents translation in the horizontal direction as well as in the vertical direction. And so, it will give rise to these two reactions. So, this is AX, this is AY. This is the weight. This is the force. And now this is how we have taken this real life problem and converted that into a 2D caricature problem, which we will now analyze using various methods. Now, let us come to what we discussed as structures. So, the topic of today's lecture is structures and machines. We will now come to simple structures. Now, for equilibrium of structures made of several connected parts, we have to consider the internal forces as well as the external forces. If you recall all the problems we solved yesterday, the forces typically were all external except the problem about the cylinder and a man walking into the cylinder, where we also had one internal force. Otherwise, all the forces were external. Now, the internal forces for the equilibrium structure made of several connecting parts, the internal forces as well as external forces are considered. Now, what we do is that in the interaction between the connected parts, Newton's third law states. So, if we draw the free body diagram of this complete structure, we cut the cable here. So, what we have? This cable is replaced with a tension T. We take this structure, isolate it. This is a hinge joint. Now, in a hinge joint, what happens is that the translation in the X direction is prevented. The translation in the Y direction is prevented that will give rise to two reactions AX and AY. So, we draw here. If we isolate this member CEF, note one thing that C is connected to DA with a pin connection. So, note that the rotation is free, but the two translations are constrained. So, as a result on this member CEF, we will have two forces acting in the vertical horizontal direction. Now, Newton's third law dictates that this member will exert an equal and opposite force on the member to which is connected. So, if this is horizontal, this is horizontal, but in the opposite direction, this is downwards, this is upward equal and opposite. So, this is the free body diagram. Note one thing that BE is now a two force member Y because the force acts only through these joints. We are neglecting the weight of this member. So, BE, we say that these are these. So, this should be the line of action of the two forces along BE plus minus like this. Now, this is assumed to be in compression. If the force at joint B acts in this direction, Newton's third law dictates that the force that this exerts on this vertical member should be equal and opposite and at joint E, the force direction on member BE is this. So, by Newton's third law, it should be equal and opposite at this point. So, these are the free body diagrams of all the individual members and then we analyze all these free body diagrams depending on what is asked of us to find out what are the corresponding forces and what are the corresponding reactions. So, under structures, there are three typical categories that we encounter. One is trusses. So, trusses are typically formed of two force members which professor Banerjee will teach or will cover in tomorrow's class. What we are going to cover in today's class are frames and machines. So, frames are the portion which contains at least one multi force member. So, what is a multi force member? That the forces do not act only at the ends in this manner. The moment there is some other force acting in between, this becomes a multi force member that we cannot immediately say that what is the direction of the forces acting at this joint, what is the direction of forces acting at this joint whereas, here only the magnitude is unknown but the direction is completely known. And then some of the most interesting problems come under this category machines and we will discuss some of these problems in today's class. We come to analysis of frames. So, not a big deal. The theory nothing, just take this frame whatever free body diagram which draw a few moments ago, essentially the philosophy remains the same. In principle you are given a frame, you are given loads on the frame, the simplest and the most naive thing to do is just break open everything. Just go for the bro, open each and every joint. See what possible members is a two force member. This is a two force member. So, the forces should act along the line generally the two force member. This B is connected to member BE, it is also connected to DA. So, what should happen? If the force on BE acts in this direction equal and opposite force should act at point B which is now common also to joint to member DA. So, just these simple things, what happens? Joint C is common both to the vertical member DCBA, it is also common to CEF. So, horizontal vertical equal and opposite on this member just as simple as that. But the trick lies not in just mindlessly disassembling that is what we have to emphasize and that is what we all emphasize our students. But there are certain tricks in order to understand that what are the appropriate free body diagrams to choose. Now, think about it. If you want to analyze this particular frame then you can just disassemble everything and see what are the unknowns. 2 unknowns here, 3, 4, this 5, 6. So, you say there are 6 unknowns. How many free body diagrams you can draw? 1 free body diagram, 2 free body diagram. Note that this is not a free body diagram because you cannot do anything about it plus minus it is a self-equibrating member. You cannot draw, you cannot write down extra equations of equilibrium for this. But we have 6 unknowns, 6 equations great, we can solve. But the point is that if you mindlessly solve just like draw free body diagrams like this, write 6 equations, 6 unknowns you will see that most of the equations will be completely coupled to each other and if you have a computer in front of you, if you have a software like Mathematica at your disposal, you are great. You just type in all the equations, solve them using Mathematica, gone. 6 by 6 matrix it will solve in 2 seconds. But if you want to solve it by hand, this is a procedure which is fraught with errors. So, because of that the trick, because the concept is very simple, but the trick is in really identifying what is it that is asked of us and what should be the appropriate free body diagrams that we should draw for this particular trust member. So, we will discuss that in a few moments. Now, there is this one extremely interesting concept which comes only in multiply connected members. This concept never arises when you have a single rigid body. Now, there are these structures which are frames, which cease to be rigid when detached from their supports. We will demonstrate this with a simple example. This example is taken from B and J, but there are infinite of such examples that you can have a look at. Just see. So, we have a member AC, we have a member CB. Now, one of the things that we have noted that students have a quick tendency. Oh, this is a member, this is a pin member, pin member, this is a 2 force member, but this is not a 2 force member. Why? Because there is a vertical load that is passing through the center. So, the force is not acting at two points, it is acting at three points. So, this is not a 2 force member, this is a regular non 2 force member. Same goes for member CB. It is not a 2 force member. Now, what happens is that just use simple geometry. This is a triangle, this is a triangle, this length is fixed, this is fixed, this is fixed. Now, if you want to change the configuration of this triangle, think about it. If you want to move this position C somewhere else, because A and B are completely fixed, then what you have to do? You cannot do that without extending or compressing any of these members. So, because of that, this configuration is said to be a rigid configuration. Why? The meaning simply being that this configuration cannot be changed without deforming any individual member just by simple combination of rigid motions. So, this is a rigid frame, but think about it. If you replace this point B with a roller, then what happens? You can easily move this, you can rotate this, you can slide this and you can see that this triangle, you can change its configuration without changing the length of any of the members. Now, in that case, it is scissors to be rigid. It is called as a non-rigid member or if I remove both of the supports, you will see that any support you release, this will become a non-rigid. What does that mean? That the configuration can be changed without changing the length of any of the members. Now, what happens? In this kind of configurations, you note that you cannot find even the reactions at support A and support B. You cannot take one full free body diagrams and can get all these four reactions because it is a pin joint, two reactions, pin joint, two reactions. We cannot get four reactions because if we take the full free body diagram, then we have three equations, four unknowns. We are done, but this is not a statically indeterminate problem. Why? Because we can draw two separate free body diagrams, this and this, two unknowns here, two unknowns here, C y, C x equal and opposite because these are connected to each other equal and opposite acting here. 1, 2, 3, 4, 5, 6. So, we have six equations, sorry six unknowns, three equations for this, three equations for this, six equations, six unknowns. This is a statically determinate structure, but the interesting thing about this which you never saw, which you never saw in the previous problems is that that unless we do an internal analysis, unless we isolate a member which is inside, we can never solve this problem. We can never get all the reactions and all the forces. So, that is means that these are the frames which cease to be rigid where detached from their supports. And when we do principle of virtual work, you will see that how can we actually visualize that if we release any support, what is the resulting kinematics of this particular frame. Now, these non-rigid structures are also referred to in mechanical engineering as mechanisms. So, sometimes I will use the word mechanisms, sometimes I will use the word non-rigid structures, but they are interchangeable and they mean the same. Now, there are more of these examples. Look at this structure. This structure is perfectly rigid. It is a perfectly stable structure when you have a proper connection here, a pin connection at point A, a pin connection at point C, but think about it. The moment you make this as a roller or this as a roller, anything or you release this, you completely release this, then what you can convince yourself, think about it, is that this is a quadrilateral and in a quadrilateral you can always make infinitesimal changes and change its shape without actually deforming any of the members. This is you convince yourself. You can draw simple diagrams for yourself and you can convince that if I isolate this structure from joint A and joint C, this resulting structure we can deform, we can change its configuration. We can bring it in, we can bring it out without changing the length of any individual members and as a result upon removing it from the supports, this becomes a non-rigid structure and then it becomes clear that if I want to find out all the internal forces, forget about internal forces. If I want to find out even the reactions 1, 2, 3, 4, we cannot find out without separating or without drawing the free body of any internal member and then drawing the appropriate free body diagram and the corresponding equations of equilibrium. Now note how many unknowns we have here, 1, 2, 3, 4. This is a 2 force member, 5, 6, 7, 8, 9. So we have 9 unknowns. We have 9 unknowns and how many free body diagrams we have, 1, 2, 3. So 3, 4 free body diagrams, 3 into 3 unknowns, 3 equations, 9 equations, 9 unknowns. This is a statically determinate structure and there is another subtle point which I want to emphasize but this is in too much details. This is not a very straightforward point that even though the number of equations is equal to number of unknowns, let us see if we get the time we will discuss some of those problems. Even then there is no guarantee that this structure is a stable structure or it is not a mechanism. The number of equations and number of unknowns being equal is no guarantee that the structure is stable but by and large the problems that we will encounter in engineering mechanics if this criteria is satisfied then you are good to go. And note another thing that many of us had a question yesterday about what happens that for example if some number of equations is more than is number of equations is less than the number of unknowns. If that is true then we can clearly say that there is no unique solution to this problem or there are infinite of solutions to that problem and the unique solution now is found out only by using techniques that we typically deal with in structural mechanics or solid mechanics. Now note one thing that there is another case we just saw that the number of equations is equal to number of unknowns. Now what happens if the number of equations is actually more than the number of unknowns. The number of equations becoming more than the number of unknowns will clearly be seen if you replace this C which is a pin joint with a roller. And you can mathematically demonstrate using principles of linear algebra that this will become a mechanism and there will be some forces that can act on the structure and the structure can no longer will equilibrium and the equations of equilibrium hence will have no solution. So these are the points to keep in mind they will become more clear as we proceed further. This is another example of the structure which is completely rigid which means that you cannot change the configuration of the structure without making changes in the length of any of the members. But if you release it from the supports it ceases to be a rigid structure and in this structure even to get these simple things as the reactions at point D and reactions at point A you have to go internally and analyze the forces in any of the members. So with this preamble let us solve a couple of problems or what we do is that we can have a 5 minute brief question and answer session and then we will go to problem solving. 1145, internal hinge where does an internal hinge arise? Okay so it is not a question of internal hinge arising or not it is a part of the problem that if you are designing your structure in such a way. So just look here this configuration it is not a question of when does an internal hinge arise maybe it is a matter of terminology. We have made this structure in such a way that these 2 joints are connected with the pin connection. Now pin connection like we discussed for the support reactions. What did we discuss here? This is also a pin connection. This is also a pin connection. So there is strictly speaking no difference but the main difference here is that that here this joint C we are completely restricting the motion of these joints in the horizontal and in the vertical direction whereas at internal points D what we are ensuring is that that there is no separation. Now see this point D which is a hinge is common to BC and it is common to ED. So there are 2 members for which this hinge is common. Now the presence of an internal hinge what does it do? That it prevents relative separation between BC and ED in the horizontal direction as well as in the vertical direction and as a result it gives rise to when we draw the appropriate free body diagram to forces in the horizontal direction forces in the vertical direction when you isolate them. But note one thing that because it is a hinge it is free to rotate. This angle is free to change and as a result this connection is incapable of providing any internal movements. So when we draw the free body diagram we just replace this internal hinge with 2 forces in x and y direction. So when you say it arises I take it to mean that what happens when you put an internal hinge here as far as the free body diagram of that structure is concerned. I hope that answers the question if not you can ask further. One doubt is there all pin joints should be there on top in this problem or shall we fix only any fixed in between that A and F? It does not matter it can be anywhere see because for example we look at all our structures around us right. For example just look our simple plials which we used to carry utensils. The pin joint is in between somewhere okay you can look at so many different structures okay bookshelf so many other things where the pin joint can be in between. Only see what we want is this we want to design our structure in such a way that ultimately it does not become a mechanism it is rigid. For example if we build a structure like this okay this is a pin joint okay this is another pin joint it is like this and then I put 2 hinges here okay I can build a structure like this okay so I can put a pin joint here I can put a pin joint here also like this. This is I can say that this is a side bracing okay but note one thing that this structure is no good structure why because this can just deform like this okay deformation will be less that. It is not deformation deformation I use the wrong word you can change the configuration without changing the length of any members the length of all members will remain the same but the effective configuration changes so this is a mechanism or a non rigid structure. So what we want to do is that we want to put whatever things that are required appropriately so that the final structure remains in equilibrium and if you put more constraints for example instead of putting a pin we put a weld then the structure becomes statically indeterminate okay so instead of pin you can put a weld there nothing wrong but only thing is that now the structure becomes statically indeterminate it is fine statically indeterminate if you put a weld instead of a pin in this case okay we take one last question and then we proceed further we will have another extensive discussion afterwards we will take one more question yes yes what are the moments in the joints of the frame sir so that is a good question so this is what I emphasized say for example that is the connection okay so look at this thing in this structure for example this is one part of a structure if these two members let me say that this is member a b okay and this is member cd I am only isolating one part of the structure the remainder of the structure can be here can be here and so on I just zoom in onto this joint now note here that if this is a pin joint we again ask ourselves the question what is this pin joint doing now this pin joint o is common to dc it is common to ab what is this pin joint doing because of this pin joint we cannot separate these two members horizontally there cannot be any gap that can be created there cannot be a vertical gap that can be created because of the presence of this joint and what we had seen that if we constrain any degree of freedom that will give rise to a corresponding force now because we are constraining relative motion between these two in the vertical direction as well as in the horizontal direction if we write down the free body diagram of ab you can in principle I am not saying that you will always get horizontal and vertical reaction sometimes it can be horizontal sometime it can be vertical it all depends on what is the applied load on the system but in principle this joint is capable of providing two reactions like this ax and ay now your question is that why is there not an internal moment and that is same thing that if this joint is well lubricated which we assume most of the time that the screw for example we are used to put it or a rivet we are using is not rusted for example it is well lubricated then the rotation is free so that degree of freedom is not constrained and as a result we say that there is no torque because why because the relative rotation between these two members is free and so we do not put an external torque or an external moment at this point this we do not put it whereas instead of having a pin joint if we put a weld then we are restricting even the relative rotation and in that case when we draw the free body diagram I have to put this external moment at joint is it clear so I think what we do now is that so there are few problems that I want to solve we will solve them for say 15-20 minutes and then we will what we will do today is we will not do a tutorial wholesale tutorial we will like take problems in batches okay so let us move back to our lecture we will have an extended extensive discussion after that so this is one simple problem note one thing that this frame problem is not like this problem which is rigid externally which is rigid when supported and when it is removed it is non-rigid you can convince yourself okay do some geometrical thinking and that geometrical thinking I am sure most of us are used to but what I am saying is that this we emphasize to the students that you do some geometrical thinking that if I remove this from the supports what happens can I change the configuration of this without deforming or without changing the length of any member in this one the students can be convinced okay if you tell them that okay just think about it what way the members can move this is a rigid member even when you are releasing it from the support and then we are putting and then we know that if you have a rigid body from all our exercises we did yesterday that if you have a rigid body we put a support such that all the reactions are not parallel to each other neither do they intersect at one point then the rigid body is appropriately support in this case two reactions horizontal vertical only horizontal they are not intersecting at a point they are not all parallel to each other so we are done this internally it is a rigid member okay it is not a mechanism and externally it is properly supported so it is a proper structure which is statically determinate let us quickly do a count on what are the equations what are the unknowns one unknown two three unknowns internally is a hinge four five okay two force members six so we have six unknowns how many free body diagrams can be drawn max one two so two free bodies six equations so six equations so per free body we have three equations so three into two six number of unknowns is six number of equation six statically determinate structure okay so we are good now we want to find out in this case okay that for this member what are the components of force exerted at C on member BCD so what are the reactions at C which are exerted at BCD so essentially the the problem demands that we need to draw the free body diagram for BC and D now let us ask ourselves a question if we draw the free body diagram only for BCD just BCD then how many unknowns do we have we have one reaction which is the unknown we have two unknowns here three and we have this force unknown four so we have four unknowns and one free body diagram so what does that mean and that what we emphasize to the student that even though we are asked to find out the reactions acting at point C on this member just drawing one free body diagram will not be enough why because this free body diagram has four unknowns and we have only three equations so we need extra free body diagrams and now how what extra free diagram just ask ourselves if we know this horizontal reaction at B then clearly for this free body diagram the only unknowns left are this one two and this three so three equations three unknowns we are done now the most naive thing to do would be for example I take this free body diagram expose all the unknowns one two three four five six sorry one two three four five take this free body diagram one two three four but then what will happen is that if you write down the equations of equilibrium separately for this and for this you will get a bunch of simultaneous equations there is nothing wrong with that in principle it is perfectly fine but in practice this procedure is prone to errors so we choose our free body diagram judiciously we say okay let us find out the reaction at member B now we ask ourselves a simple question what if this roller were not there okay what if this roller were not there we can immediately convince ourselves is that this force will tend to create a moment which is an anticlockwise direction about point O and as a result this entire member will tend to tilt like this and this will try to press inside now what we see is that what is this member be doing it is preventing the motion of this thing in the horizontal direction so we expect that because it is trying to move inwards without the presence of the roller the reaction will act outwards in principle there is nothing wrong with taking the reaction acting in the other direction also but our intuition tells us that this will without the presence of roller this will try to move inside and as a result the reaction our intuition tells us will act in the outward direction so what we do is that typically is we try to do here is we try to emphasize to the students that use your intuition okay don't rely fully on it but use it as much as possible to get some idea about what will be the directions whatever directions you obtain do they make sense or not make sense so our intuition now tells you tells us that the direction is in this way now what we do we assume our direction in this way take torque about point A when we torque take torque of this entire free body diagram about point A you can easily do this and we can find out what is the we don't need to find out A, Y and A, X okay we don't need this for this problem but say we got it taking moment about point A we immediately obtain what is the reaction B our assumed direction is this the value we obtain is 300 which is positive so our intuition about the reaction being in the inward direction was right okay so that's how we confirm our intuition now what do we want we draw the free body diagram for this member now this particular slide from Bear and Johnston is extremely good why because as far as teaching students is concerned it is telling for example that what are the possible errors that the students do one of the most common errors that we have seen students do is that is failing to recognize that this is a two-force member so they would just say that it's a member it's a pin connection so these are some of the mistakes which typically students make and you would also be aware this is just that I am reminding ourselves that what kind of mistakes are possible so one very common mistake is that students don't realize that is a two-force member is a pin joint degree of freedom restricted in horizontal vertical direction replace it by 2 okay so they do that that is clearly an error second thing for example that this they may forget for example the vertical reaction that is another error another thing what they can do is that again fdx fdy so even add this this error and this error put together you can do this error so typically as we all know that a free body diagram for this that 300 Newton reaction which we had obtained just a few moments ago because joint C is a pin joint it is preventing relative separation between this member and this member in both the horizontal and vertical direction we have to replace that when we draw a free body diagram the kinematic constraints are replaced with corresponding reactions C x C y and now for this member what do we have we have 1 2 3 unknowns the direction of this force is known straight forward now what do we do for this free body diagram take moment balance about point F sorry about point C we can immediately find out what is fdE when we find out what is fdE we are done now we can take force balance in the horizontal direction we can find out what is C x we can take force balance about a vertical direction we can find out what is C y and for extra confirmation what we can do is find out this C x and C y for this free body diagram we take moment balance about point D and that will come out to be 0 so that is our confirmation so very simple problem straight forward problem the only reason we took it is that that many a times even simple problem student make those errors because they do not recognize that there is a 2 force member that is lurking there okay so we can get this and we are done so what another check is that that we and a better check again okay actually this is a really good check is that that so far the 2 free body diagrams that we have used is the complete free body diagram and this diagram now a check better check will be that we have never used this free body diagram but we have used all the information so using 2 free body diagrams we are not going to get any additional information so what we can do is that we can now take just this member and use all the values that we have obtained and just see that if this moment balance force balance stays or not and if our entire procedure is clear then this balance will happen and this is a consistency check for us just to be sure that there are no errors in whatever we have done okay this is not going to give us any new information this is only giving to going to give us a check that if this is equal to 0 everything balances then our approach our procedure is perfectly right now the last topic okay this is just the last topic there are plenty of problems that we are going to solve but the last topic that we are going to deal with is machines what are machines typically a simple example of this machine okay is this is a plier for example what is this what does this plier do that we apply some force P and that force P is effectively transmitted to this structure a small structure this can be a wire this can be a bearing anything it can be ball bearing anything so this force from here is transmitted to this point now what is the purpose of a machine the purpose of a machine is that that for example if I want to cut something or want to crush something here for example this can even be a bitter nut cracker so if I want to crush something here I need to apply huge amount of force and that particular amount of force okay for example most of us humans okay unless somebody is a super human okay unless you are a crish for example you cannot do this thing but now what you do is that you use something even better we use a machine and when we use this machine what happens is that we keep the lever of this much larger compared to the lever of this we can draw now the free body diagram of this particular component we see that this is the pin connection which is separate a preventing the separation between the top and the bottom portion so that will lead to two reactions now if you take for this free world diagram torque about point O what you will see is that at P times A is equal to Q times B or Q or the force that is generated at this point will be equal to A by B times B now if we keep this A reasonably large what you will see is that A by B is equal to P and if this A by B ratio is large then that means that a small force can be greatly magnified okay as this one as the corresponding jaw force so with this much of the preamble we will now solve some problems okay