 We will now go into the tutorial mode. So this problem is not done that given in the tutorial that is uploaded on the course website. It is one very interesting problem which tells you the concept of frames which are rigid when connected and which are and which cease to be rigid when the supports are removed. What we will do is that we will look at this problem and try to attempt and we will spend around 10 minutes on this. You try and solve this at least goes to the procedure and then we will have a discussion at the end of this problem. So we will do that so there are two problems that I want to discuss with. So this problem and this problem. So these two problems I want to discuss so let us do that before I solve the problem. So many of you are saying that give this in SI units. So that is my limitation. The figure is available to me in imperial units. So I have to give it because if I try to convert everything into SI units it is a huge pain but on the same time just note that if we consistently choose all the units that all the dimensions either in inches or in feet all the forces in pounds and the corresponding moments in pounds feet or pound inches then there is absolutely no difference between using these or SI units. Only thing is that we have to be consistent. If we use some set of units we have to stick with those units. We will prefer SI units of course but there is absolutely no problem using them. You have to just be consistent that we use force as pound then always as pounds unit as inches then always as inches then everything will be consistent. Now with this much what we are asked in this problem that we want to know what is the reaction at D and a force in member BF. So note one thing that you will see how many unknowns are present here 1, 2, 3, 4, 5, 6. So we have 3 unknowns and we have 2 free body diagrams that are possible A, B, C, D, E, F. 2 means not just these but definitely 2 free body diagrams are available to us. So 2 free body diagrams is equivalent to 6 equations. So 6 equations and 6 unknowns and we are good. But now think about it that if we take this complete thing as a free body diagram then there are 4 equations and 4 unknowns. So there is no way that we can obtain what are the horizontal and vertical both reactions at A and D. So what we do? We take the full structure, we draw the free body diagram for that, we draw the free body diagram for that structure and then what do we do? We can find out by taking moment about point A what is the horizontal reaction at point A. We get that. Now what the question we ask ourselves is that we get that reaction. Now let us say what is happening in here? What if this were a roller and not a hinge? In that case what happens that this thing in that case if this were a roller as opposed to a hinge then the equilibrium in the vertical direction or moment balance about any other point would not be possible. Why? Because these are 2 forces their effective component will act in the vertical direction if there is nothing to balance that then this will just slide down. So what is that particular hinge doing is that it is preventing the translational motion of this particular member in the horizontal direction and preventing it for slipping down and that is done. So we get dx which comes out to be 9.59 pounds then we draw the free body diagram of this and there is various way of solving it. One simple way of solving is what we do? We are also asked to find out what is force in member bf. This is fbf. So we write down equation of equilibrium in the horizontal direction. We already know what is dx then we will get one simultaneous equation in fce fbf. Then the second equation we write is moment balance about this point why? Because a dy does not come into picture. We only get equation in terms of fce and fbf. From these two equations we can solve and obtain that fbf is equal to 357 pounds and once we get fbf what we can do? We just do equilibrium force balance in the y direction or take moment about e to be equal to 0 whatever both way you can do and what you will get is that the dy comes out to be 315 pounds and so the total reaction at d will be square root of dy square plus y square which is 1009 pounds approximately and angle it will be tan inverse of dy by dx and the corresponding force in member bf is approximately 357 point reasonably state forward problem but just proof of principle that we had discussed one particular method of solving this problem and this is a problem that we actually went ahead and solved. Now a bigger challenge okay this I leave it to you is to visualize that if I remove this structure from support a and support d how can that structure deform without changing the length of any of the member so that is a challenge that if you can figure that out then for example we can generate a huge amount of insight into overall behavior of the structure that we remove it from the supports and then ask ourselves how does overall structure deform or how what kind of mechanism does the structure go into okay. Now with this let us take a few questions okay and then we will move on to the next problem. 1047. Sir what is the difference between pin joint and hinge joint? Pin joint and hinge joint again are interchangeably used okay so pin joint only thing one quick thing that I can say is many times but it will be it also becomes very clear from the context okay typically they are the same but when you have a fixed support to which we connect this typically this people call as a hinge joint but you can say it as pinned also it does not matter and typically when this happens two members are connected you say that as a pin okay but these are equivalent okay. One more question in a two force member the forces act along the line of actions what if the forces are acting line of actions the line of action is the forces they act always along line of action by definition only thing in a two force member it means is that that if you have a member like this where the forces can act only at this point and at this point then equilibrium demands that if we join the line connecting these two points then the four should act either like this that this should be the line of the action of the forces or they should act like this that is what two force member demands that you know the two points at which the forces act we join those two points and the line of action of the forces is along that line and the forces are equal and opposite at one point and at the other point that is what it means okay fine we will move to the next problem so this problem now is a part of tutorial so this is another frames problem what we are asked is that that these are the dimensions that are given that dimension of each pulley each pulley as a radius of 250 millimeters and we are asked to find out determine the components of reaction at D and E just the components of reaction at D and E you have a string this pulley you assume them to be either frictionless or free to rotate both of them are totally fine so for this problem we are asked to find out what are the components of reaction at D and E okay. So let us take another 5 to 10 minutes max and then we will get back to this problem and after that we can start working on machines problem okay so welcome back this is a very simple problem not unlike what we had done previously the only complication with student face when they see a problem like this is if you do not do a judicious choice of free body diagram then you will see that there are too many components here there is a pulley there is another pulley here this member other member so if you completely disentangle this free body diagram without if you means the students just disentangle this free body diagram without paying attention appropriately to what are the reactions okay or what is asked here that we are asked only the reactions at D and E so if you mindlessly just remove these things everything that the number of free body diagrams you will get will be this 1, 2, 3, 4, 4 free body diagrams you will get and the number of unknowns will be just too much and another thing that happens is that at joint A you see that member 1, member 2 and pulley so there are 3 rigid bodies that are connected at point A so we cannot immediately say that remove one member and say that the reaction here is Rx and reaction is Ry and equal and opposite act on the other that works only when 2 members are pinned at a point but here there are 3 members pulley, ABD and AC so all 3 are pinned at that point so it may become a complete nightmare if you disentangle everything and then try to solve this problem on the other hand if you recognize that the only thing we want is reaction at D and E so first thing we ask ourselves as we did previously that if you take the full free body diagram we can take for that free body diagram moment balance about point D we can get horizontal reaction at point E then we can do force equilibrium in the horizontal direction you will see that dx will be plus or minus Ex sorry will be minus of Ex and then the question we ask ourselves is this what if this E where a roller instead of a pin joint now this if this E where a roller instead of a pin joint then what will happen is that just look here if you draw the free body diagram only for AC E okay only for AC E we draw the free body diagram then what we realize is that that first of all because we had neglected friction either at the joint or overall in the pulley okay we have neglected that friction as a result the tension is uniform all throughout so this is 4.8, 4.8, 4.8 so this tension is also equal to 4.8 when we make a cut here we remove the joint A what we see is that that there are 2 possible reactions here FAY FAX this is the tension and if point E or the connection E where a roller instead of a hinge connection then in that case what happens is that that there is nothing here to balance this vertical force and this structure would not be in equilibrium so what we do is that that we then understand that what is the role of this pin connection here is to provide us with a reaction here and then very easy we know this tension we take moment balance of all these forces for this free body diagram about point A and then we get sigma MA is equal to 0 and we immediately can find out what is FEY which will come out to be minus 2.7 kilo Newton or it will act downwards in this case okay so we had assumed a direction to be upwards value came minus so the resultant force is plus 2.7 acting downwards and then we have we already found out what is the reaction in the horizontal direction by taking for the field free body diagram moment balance about point D and we immediately get what is F DX by equilibrium in the horizontal direction and F DY by equilibrium for this free body diagram in the vertical direction okay so the entire idea is that once we recognize what is the approach what is the thing that is asked reaction at D and E then what is the appropriate free body diagram once we figure that out the problem becomes very simple but if we disentangle each and every free body diagram write the free body diagram for this pulley for the other pulley for this member for this member then the number of equations number of complications will be too much and this really simple problem can become a nightmare so this is the point that we want to emphasize to this problem so if there are any questions okay we can have a brief discussion. Sir my question is related to the session that we had yesterday I could not ask this question to you my question is what is the difference between twisting and pure torsion for example they are all both related to each other first of all this is strictly speaking not a topic of engineering mechanics but I will briefly mention this so if you take a rod like this okay for simplicity let us take it to be a circular rod and if you apply a moment like this what is this moment using this is equivalent to putting in 3D a torque like this okay or so this torque is called as a twisting torque okay and the corresponding effect what is what happens because of this torque is that this sections they rotate or they twist and the effect of this twisting torque is by definition is to create a twist as simple as this or for example you can take a screwdriver and if you want to use it okay to turn a screw then what we do is that we apply this twisting torque and what does that do is that it twist this screw okay and make it rotate like this is that okay. So what is torsion then pure torsion this is torsion what happens is that torsion there is a relative deformation okay so what happens is if there is a straight line like this if you draw a straight line imaginary straight line on this member okay and when you deform you will see that this straight line deforms in the form of a helix so that particular deformation is called as torsion okay this is a topic of solid mechanics strength of material advanced solid mechanics okay but this is typically the torsion that if you draw a line like this you will see that it twist when you apply this twisting torque and that particular thing is called as torsion that this member is undergoing torsion 1 to 9 to go ahead for your question sir a beam is supported at two ends one is a pin joint support in support yes pin support and and next is roller support with inclined at 30 degree with the horizontal okay roller support inclined with 30 degree with the horizontal inclined at 30 degree with the horizontal okay now beam is loaded now beam is loaded in such a manner that there is there are horizontal and vertical forces means load is inclined okay mean load load loads are inclined and vertical both both type of forces are now my question is that that that support reaction at pin point billet be horizontal and vertical both support reaction component of roller at second end so it just depends on what is the inclination of this okay so what is the final force will all depends on what is this angle theta and what is this angle beta okay so whatever these angles are what point it is acting a they will decide that whether there will be a horizontal reaction present or not present but only thing we can say clearly is that the free body diagram has to be drawn like this and the final value of the horizontal reaction if it is plus or minus that we cannot immediately say at least it is not immediately clear to me it depends on so many parameters this is f there is this angle theta this distance is a this distance is b it is not immediately clear by just looking at this figure that if this will be 0 or not it will depend on this theta it will depend on this angle beta and it will depend on what where the load is acting it is acting at a or at b we have to solve this problem is a straight forward problem which you can solve and then figure out if this reaction is 0 or not 0 okay but you have to solve the problem but the reaction the free body diagram will look like this right so we will have to consider a you have to consider horizontal reaction has to be considered you cannot throw that away it may come out to be 0 later but if you want to be safe you should never throw it away just like that 1136 okay yeah pk8 where is it okay okay this problem yes if q is equal to 0 there if you consider the free body diagram of joint f only then bf must be equal to 0 the force in the member bf must be equal to 0 why but if you consider the equilibrium of joint f but the point is that is not because see that idea okay I will tell you I can see your point okay so what you are trying to say is that what you are trying to say okay it is a settle point but it is a very important point okay good thanks for raising this so this is your member horizontal member it is connected by pin in this like this and this is q okay now what happens is that that we have this member we have a screw and we have this other member and on the screw we have applied load q now strictly speaking if you draw free body diagrams of this what will happen is that this will be a 2 force member so that is how it will let me redraw it again okay that this is horizontal member this is this and it is connected now the important thing is that that we draw the free body diagram for the screw itself now the screw is connected to this 2 force member say this is the force acting like this force acting like this okay now look at this member b now b member you note that the bottom member is not a 2 force member so there is no restriction that the force it transmits from here to here is only in the horizontal direction there is no restriction like that it can as well transmit a force like this such that these 2 balance each other the problem is that if this were a truss in which the bottom member is also 2 force then what happens is that then it becomes compulsory for the bottom member to have a force only along this direction and then what you have is that if you take the balance force balance on the screw then you will see that one force is acting like this other force is acting like this okay and for now the balance of the screw in the equivalent x and y direction both of them will become 0 this will become 0 but the main point is that that this bottom member is not a 2 force member so there is no compulsion that the force acting at this point from the bottom member should only be along this it can be in any direction okay so that is what will happen so the bottom member will exert a force like this on the screw equivalent opposite will act on that member and then you are done so this problem arises only if the bottom member is a 2 force member this is a 2 force member and then when q is 0 both forces are 0 but in this case it would not be because bottom member is very capable of exerting a force like this on the screw okay is that clear the main reason being that at the bottom member is not a 2 force member that is the idea that value is equal to 0 why did you give the value why we have to ask Bayer and Johnston they gave that 0 value so is nothing is stopping us from giving q equal to 0 right you can use it for a non-zero q also you can as well solve the problem yeah that is the point right yes that is the point that see this is a very important point okay what the point you are raising is a very important point let me go to the slide it is a very important point it is a subtle point but important point that when this structure okay this structure itself in itself is not rigid but when you connect it at point A and D with with hinge connections or pin connections then this complete structure becomes rigid and by definition a structure is rigid means that no matter what combination of loading you apply to this structure it will provide appropriate reactions to keep the structure in balance or in equilibrium so no matter what you apply you have to convince yourself that when the structure is like this with two two connections here pin connections then there is no way you can change this configuration as a mechanism or without changing the length of any of the members so this is a kinematic approach to this thing and you can also convince yourself that once that is true then no matter what load you apply you will see that appropriate reactions will come and to really convince yourself you have to use linear algebra what you have to do is assume some random loading write down all equations of equilibrium the right hand side in a matrix form in terms of all the reactions you will see that the matrix that comes will be a non-singular matrix with all non-zero eigenvalues okay so that is going into linear algebra and that is a rigorous way in which you can see that this particular structure can withstand any possible loading at least as far as getting simple reactions is concerned okay so it is a very important point okay that you can apply any load and you will see that appropriate reactions and unique reactions will come for this structure but if I make this pin and I make this into a roller then any apply application of q is will not do certain cases if I apply p horizontal load then this structure will be in equilibrium but if I apply load in this direction then you will see that the structure will not be in equilibrium so that is a very subtle but very important point that a structure not being a mechanism or a structure being rigid under these conditions it implies that no matter what loading you apply the supports and the linkages will provide appropriate reactions to keep the structure in equilibrium okay it is a subtle point but it is a very important point okay I hope it is clear that you can apply any load this structure will stay in equilibrium 1063 go ahead for your question can you please elaborate about two force and three force principle okay I will quickly elaborate that definitely okay it is very important so two force members the simple idea is as follows okay so let us take that in the context of a structure let us take a structure like this okay so this is the cable okay and suppose we have applied some uniformly distributed load here some point load here okay and these are pin connection so this is member a b c d e f g h okay I am just drawing it in details so that things become more clear now note here that how many members do we have I have one member g e b a okay then we have second member e d f and we have a third member b c d now what is so special about this member b c d if we draw the free body diagram of this member itself okay let us draw the free body diagram we do this okay we drawing the free body diagram now whatever what what does our what does our understanding tell us that this b is a pin connection okay this b is a pin connection so it is preventing relative separation between the the vertical member and the horizontal member on and this member b c d both horizontal and vertical reactions are separated so at point b you should have or you can have in principle two reactions but then immediately you also see that from point d because also a pin connection you can have two possible reactions okay now for this structure to be in equilibrium okay what should happen is that let me call this as f y f x f and let me call this as g y and g x okay this is the most general case okay you have two reactions coming here from the pin joint two reactions coming here from the pin joint now what we do is we play a trick like this let us join these two points and yesterday we discussed that our coordinate axis did not be vertical and horizontal let us say that my new coordinate axis this I choose purely for my convenience let us say that this is my coordinate axis let me call this as x prime and this coordinate axis let me call as y prime now I have this f x f y g x g y well I can as well have my f x prime f f y prime g x prime g y prime clearly because this is the coordinate axis and nobody stopping me from doing that because it is a free country I can choose my coordinate axis so we do this so we say that these two forces are acting like this okay so this is my f x prime and this is my f y prime similarly at this point also I will say that this is my g x prime and this is my g y prime now note what should happen for equilibrium okay forces okay sum of all the forces in the y prime direction should be equal to 0 for this member to be in equilibrium now as a result what will happen that g y prime plus f y prime should be equal to 0 so if I redraw this again what do I get I am doing this very slowly I know okay so if you are if you already have understood please bear because I think because you ask the question I should go in proper details so let me redraw it again what should happen is that that if this direction is like this equilibrium mandates that what this should be equal and opposite similarly equilibrium in the x direction also mandates that these two should be equal and opposite this is B C D so note that these two will balance each other and since their line of action is the same they do not cause any moment but note these two forces this force is f this force is f now they are equal and opposite to each other and they are not acting in the same line so if this distance is a then what are they going to do they are going to create a couple f a about any point and so this structure won't be in equilibrium if this f is positive or negative non-zero so the only way this structure can be in equilibrium is if this becomes zero if this becomes zero and the only forces that remain are these forces along the line joining the two points where the forces act so ultimately we say that we only worry about the line joining the two points at with the forces act and the forces can be either this or they can be that is the only way in which this structure can stay in equilibrium so that is the concept behind two force members is it clear okay and a three force member actually three force member is not indispensable three force member only thing it says is this that if you have a body subjected to forces acting at three points or three lines okay so this is the line of action of one force f 1 this is the line of action f 2 of force 2 by definition when two forces are not parallel what do they do they have to intersect at a point let us call that point as o now if another force acts on the member only one force not more than one for only another force third force act on this member then if the line of action of that force is like this suppose then what happens if I take the moment balance of all these forces about point o then these two forces by definition do not cause any moment but this force which is not passing now through this point will cause a moment and there is nothing to resist that moment so what it says is that if two forces are acting and a third force is also acting then that third force should have a line of action which should only be this so they should meet at one point so if the four if three forces are acting along three lines of action then three body just means the equilibrium dictates that all three meet at one point and now additionally if you know the direction of two forces like we knew in our problem then we can use geometrical considerations by ensuring that all these three forces meet at one point and get some geometrical criteria to find out what the unknown forces or unknown reactions are but this is not extremely mandatory okay but two force members okay is extremely useful concept and that simplifies our life a lot in every problem so I hope this helps okay are there any practical limitations for what practical limitations the two force member practical limitation only means that we neglect the self-weight of the members but typically if the forces involved at the joints are much more as compared to the weight of the members then we can neglect them three force members you have to do a lot of visualization and geometry so if you are good at geometry you are good at trigonometry fantastic okay but many a times three force member problems can be very easily solved using simple force balance and movement balance but when some geometry is very perfect like things are meeting at right angles then three force member concept becomes very very powerful okay even with simple visualization you can immediately see what should be the third force and so on okay we will solve such problems when we do friction okay but if you have good visualization if you have good geometry skills go for it but it is a very idiosyncratic thing so let us come back to what we define as to the machine problems so there are some extremely interesting problems here so one of the simpler problems that we are going to do now is a particular mechanism now what does this mechanism involve so you have a rigid stand here to which you have a device that is attached okay so when this device so this device is used to hoist the car up so initially this device is perfectly horizontal you raise the car up and then use this hydraulic cylinder okay now what does this hydraulic cylinder do okay to increase the length of this boom or decrease the length of this boom and buy that by change the inclination of this angle and let the car go up now what we are asked in this problem that to keep the car in equilibrium in this position okay all the dimensions are given to you okay we are asked to find out what will be the corresponding force in this hydraulic cylinder to keep this entire assembly in equilibrium okay the weight is given okay the 3300 kg which passes through this point now let me briefly explain what this hydraulic cylinder is and what it does so for that I will refer you to a simple animation which I have taken from there is this website called demonstration.wolf.com so from this company which makes this very powerful symbolic computation numerical computation software called Mathematica so on this website there are a number of demonstrations and a number of animations with which for example we can get some visualization about various mechanisms now this is a simple mechanism that this is the hydraulic cylinder this portion is called as the boom this is another hydraulic cylinder this portion is the boom of this hydraulic cylinder now look at it this is a tilt boom so this is a tilt boom what do we do we change the angle of this sorry this is a tilt boom what happens here is that we can decrease the length of this and lower this or we can increase the length of this portion and we can increase this we can also move the position of this and then we can either lower it down by reducing the size of this boom or we can extend it up by increasing the size of this boom so this is the particular way in which this hydraulic cylinder works that a hydraulic cylinder has a boom and that boom can be either decreased or increased to raise and lower other components of the structure okay so with this in mind now we ask ourselves that in this position if we increase this then this will go up if we decrease this this will go down and that is the way that this mechanism works and now what we are asked to find out is what is the force in this hydraulic cylinder okay for this particular assembly now before we solve this problem let us look at what this mechanism is let us ask ourselves various questions now what is this hydraulic cylinder doing okay if this hydraulic cylinder were not present what will happen so a quick question will be that hydraulic cylinder is not there then this will try to flip about point C but then we ask ourselves a question what happens if we don't have a rigid ground underneath point D if there is no rigid ground underneath point D then this entire assembly will tend to rotate about point O so that is the first observation the second observation is that what does the ground do the ground now prevents this point from going down how does it do that it applies a resistance in the vertical direction because of which the moment the rotation of this structure okay this structure about point O is prevented so then we immediately know that without ground this will tend to rotate in the clockwise direction about point O so this point will try to go down so clearly would expect the reaction at point D to be upwards and then the free body diagram becomes extremely apparent to us so first free body diagram we will draw okay is what is take this entire assembly this is a pin joint there is a horizontal force there is a vertical force this is a roller support only a vertical force for this entire assembly we can take moment balance about point O and then find out what is a vertical reaction at point D now the second question we ask ourselves is that what is this hydraulic cylinder doing what will happen if this hydraulic cylinder is not present now if the hydraulic cylinder is not present then just look at this member DCB what will happen there is a force that is acting here this force will tend to rotate this member DCB about point C so what hydraulic cylinder is doing is that it is providing you a counter force which then can exert an opposite torque or opposite moment about point C and prevent the rotation of this member so immediately we know to find out what is the force in this hydraulic cylinder what should we do this hydraulic cylinder can be approximated as a two force member why because it is pinned at point A it is pinned at point B and typically the forces that are exerted by the hydraulic cylinder are much more than compared to the weight of the hydraulic cylinder as a result we neglect the weight of the cylinder and so the force acts at point A force acts at point B so the effective direction of the force should be along AB so that is essentially a two force member so what we do is we draw a free body diagram for this take moment balance about point C and we are done okay then we can find out what is the force in this hydraulic cylinder so if you go here first portion is the complete free body diagram of this we take moment balance for this about point O we get what is the vertical reaction now what do we do the vertical reaction will try to rotate member DCB about joint C and this hydraulic cylinder which is a two force member okay it is a good idea it is a perfectly fine idea to approximate that as a two force member so two force member the force acts along the direction so we have assumed that the hydraulic cylinder is in compression so it is exerting a pushing force at point B and then for this member number of unknowns 1 2 3 we do not care about these reactions so for this free body diagram we take moment balance about point C and immediately find out what is the force in the hydraulic cylinder so this is a typical problem that involves the hydraulic cylinder okay so let us now okay solve some simple problems involving hydraulic cylinders okay so this is one problem involving hydraulic cylinder what I want you to do is we will go over to the statement of the problem and then we try to attempt that what is the best way or what are the appropriate free body diagrams that we should draw and what will be the corresponding forces that will be generated by the hydraulic cylinder so we have this assembly these are rollers this vertical is a rigid pillar and this assembly what is it used for this assembly is used to carry this weight of 1.5 mega grams what is mega grams 10 to the power 6 grams up and down now what happens is that that this assembly there are 1 2 3 4 4 rollers the weight of this assembly okay has to be carried by a mechanism what is this mechanism we have a two force member hf why is it a two force member that the forces are acting only at point h and only at point so it is a two force member the force acts along hf this two force member okay hoists this up and how does it transmit its force that this two force member is connected to another member edf at joint e and what do we do that this joint edf okay is free to rotate about point e and this then is propped up by a hydraulic cylinder which is pinned at point c and connected at joint d so this is a two force member and what we want to find out is for this entire assembly given all the geometry and given all the loads what is the force that acts on it okay or acts in the hydraulic cylinder is the question that we are asking okay so why don't we take so before we do that okay are there any questions okay about this problem okay so if you have a question only related to this problem please ask otherwise general question send it on chat okay so these are the typical problem involving hydraulic cylinders this is a reasonably involved problem okay so let me go over this problem a few moments from now but why don't we attempt this problem at least this problem is very similar to the problem which we discussed just now the solution everything will be uploaded all the solutions will be at the end of the lecture slides this is problem number 4 okay we discussed this briefly why don't you attempt that okay we will have around say 10 minutes for that and then if you have any questions in the meanwhile okay you please send through the chat okay I will try to answer questions right now and at the end of the problem let me briefly discuss the solution and then if you have questions only about this problem then you ask general question please send via chat so I can see from the chat that many of you have obtained the correct answer so the concept is pretty much clear to a lot of people I guess okay so it's a simple problem if we understand what is the mechanism doing otherwise it can get pretty messy now if you look at the top free body diagram for this portion then there are four rollers so in principle like some student may be misguided into thinking that there are four reactions okay and there is a this is a two force member so five so one structure has five reactions and then overall if you count the number of equation number of unknowns you may feel that this is actually a statically indeterminate problem and in fact it may very well be so instead of having four rollers I can have 16 rollers like a one another set of rollers like three more set of rollers any number of rollers you can have but what is the purpose of this hydraulic cylinder if let us ask ourselves a question what happens if this link FH is not present now if this link FH is not present then there is nothing to prop this assembly and keep it in equilibrium so this FH is essentially providing a vertical force to keep the system in equilibrium now let us ask ourselves that what is this hydraulic cylinder we are not doing now what this what is link is doing it is providing a vertical force to keep this in equilibrium so we immediately know that the free body diagram of interest is this top free body diagram and the corresponding equation of equilibrium is nothing but equilibrium in the vertical direction and when we do that what will we get is immediately the force in the link we will immediately get the force in the link because it is a two force member the direction is known we take the vertical component of that and balance that with the weight of this assembly and we are done but now the next question is that what is this hydraulic cylinder doing what happens if this hydraulic cylinder were not present now if this hydraulic cylinder were not present then this link would exert a force on this member now this link has a component along this line EF as well as perpendicular to this line EF now the component of this force along line EF will not cause any torque but this perpendicular component will tend to rotate this assembly EDF about point E so what is this hydraulic cylinder doing the hydraulic cylinder is precisely preventing that rotation about point E now with this logic we immediately know that to find out force in the hydraulic cylinder the free body of interest should be EDF and the corresponding equation of equilibrium should be for this free body diagram using the force that we obtained from the previous free body diagram and take torque or moment about point E and then we immediately find out what is the force in the hydraulic cylinder so that is the logic of this that if you go step by step and note that this may have for example more rollers like 3 more rollers here 3 more rollers here and it does not matter as far as a force in the hydraulic cylinder is concerned the number of wheels here is completely immaterial okay so even though we increase the number of wheels and make the problem statically indeterminate as far as finding the force in the hydraulic cylinder is concerned we are good okay it does not matter at all and once we use that logic then with this logic what becomes clear to us is what is the appropriate free body diagram and what is the appropriate equation of equilibrium we do that and we are done the solution is pretty much shown here okay this will be all uploaded and the answer comes up to be 59.7 kilo Newton in compression which many many colleges have gotten okay so I think that the point has been driven home properly. Now this is a reasonably interesting problem okay what we have this is a solved problem in beer and Johnstone but why I have taken this is it is a very very interesting problem you need to think reasonably well and decide what are the appropriate free body diagrams in order to get your answer now what we have in this case is we have a platform okay we have a platform and how this platform is propped up that this platform has assembly shown on this side and on the other side at the back. So two identical assemblies on two sides okay which are working and keeping this platform in equilibrium so as a result what we are doing is that we are exploiting the symmetry of the problem and say that half of the weight is transmitted to this assembly half of the weight is transmitted to the other assembly which is exactly the same on the other side and so we deal with a planar problem. Now what do we have just look at one side we have this platform which is hollowed out and this platform has two wheels which are connected by a link and these two links these two wheels are connected by one member like this another member like this such that e b c g is a parallel gram now this assembly has another member ad okay it has another member ad and a hydraulic cylinder which is attached to one point at h and to another point at d and what is this hydraulic cylinder doing if this hydraulic cylinder were not there then what will happen this parallel gram can keep deforming okay this becomes a mechanism and this this weight can no longer be carried it will keep going down and down this this angle theta will keep decreasing and the platform will finally come down so that is what this hydraulic cylinder that is what this hydraulic cylinder is doing this hydraulic cylinder is effectively is effectively making sure that this parallel gram does not deform. Now let us ask ourselves a question that what is this link ad doing if this link ad were not there just remove it then what happens is that if you apply any force in the horizontal direction there may be wind load there may be any other lateral perturbation then this platform cannot stay in equilibrium why because these are wheels okay these are rollers they can only exert reaction in the vertical direction and cannot prevent the motion of this thing in the horizontal direction so we we see that this two force member what does it do it prevents this assembly from moving sideways but in this platform you will see that if you draw the free body diagram for this platform then since there is no force in the horizontal direction this ad is a two force member and it will become zero so essentially what is this hydraulic cylinder doing this hydraulic cylinder is making sure that without the in the absence of the hydraulic cylinder this becomes a two force member this becomes a two force member and we saw that in a two force member the force acts along the direction of the two force members now what do we have here we have that the reaction that these members will exert at B will be parallel will be like this along along this direction the reaction exerted by this member GC at this point along will be along GC so in the absence of hydraulic cylinder and in the absence of any lateral load this becomes zero there is no hydraulic cylinder so these are two force members and these two force members will only exert load parallel along this direction but note that this load W also has a component which is perpendicular to that direction and because of that if these two are parallel force member this assembly okay will not be able to exert a reaction in a direction perpendicular to that which is actually being produced by this load W and the presence of this hydraulic cylinder is essentially making this member EB not to be a two force member if there were no DH then this becomes a two force member why because AD is also zero but in the presence of DH this is no longer a two force member and it and this join and this member now can exert a force both along BE and perpendicular to BE at joint B now with this logic okay with this thinking we have in mind what I want you to do is find out what is a force in the hydraulic cylinder you can obtain this answer in terms of this angle theta A and all the dimensions given you can assume that the angle this angle that the hydraulic cylinder DHE okay which it makes with the horizontal take that angle to be equal to alpha just for simplicity and get a symbolic answer that let us say that this angle DHE is alpha then can you get a symbolic answer that what is the force in the hydraulic cylinder in terms of this theta this alpha and the dimension A to A of this problem okay and a weight W if you have any questions okay we will work on this for around 10 15 minutes max 15 minutes and then we will get back and solve this problem fully if you have any questions send them via chat and at the end of 10 minutes we will again see if there are any difficulties and then try to address them okay so I can see from the chat that at least 10 colleges have been able to answer this question okay so it is a very nice question for the simple reason is that we had discussed a few moments ago that some free bodies need to be judiciously chosen now how do we choose the free bodies we take the free body diagram of this platform by removing everything from inside so what are the forces acting on it this is the force coming from the two force member here we are assuming that a two force member is actually pulling on the joint which means that it is in tension that is the assumption the two wheels will exert vertical forces and W it should be W by 2 okay acting downwards because there is no horizontal force what happens because there is no horizontal force this FAD is the only force providing a component in the horizontal direction that becomes 0 now come here okay then we draw the free body diagram for the set of wheels these this set of wheels connected by a linkage now what happens there you see here that GC is a two force member why because there is no force acting anywhere only acting at the joints which are pin connections so this is a two force member force acting along this but because of the presence of the hydraulic cylinder BE is no longer a two force member so at point B force can exert we can be exerted by this member both along this and perpendicular to this so this is how we choose our coordinate axis we don't choose X and Y horizontally but we choose X and Y in this inclined direction and our logic is that that in the absence of hydraulic cylinder what is happening is that both are becoming two force member so the two forces are parallel to each other so any force which has a component perpendicular to those forces will essentially sees the which we will cause the system to go away from equilibrium now what is this hydraulic cylinder doing it is making this force this member to be a non two force member so it can exert a force also in the perpendicular direction so our equation is very clear for this free body diagram write down equation of balance in this inclined direction and what you will immediately see is that a cost component of this force will be nothing but this reaction now third this force is present this force is present what is happening if the hydraulic cylinder were not there first of all this will become 0 but then this will not be in equilibrium but because of hydraulic cylinder this force comes into being and then this force tries to create a moment for this free body diagram about point E and tends to rotate that and a hydraulic cylinder precisely prevents that rotation so we immediately know what is the free body diagram now that this is the free body diagram and the corresponding equation of equilibrium is moment balance about point E so we just say that FB2 into 2A will be balanced by if this is alpha this is theta then this angle is theta plus alpha so what is the component of this force perpendicular to this FH into sin theta plus alpha we balance them together and what do we get that the force in the hydraulic cylinder should be equal to W by 2 cos theta that 2 I forgot so W by 2 cos theta sin theta plus alpha okay so please remember this 2 okay which I had forgotten here so it will be W cos theta by 2 sin theta plus alpha and what that is can be obtained okay so we can obtain DH we know A we know theta we know L we can obtain DH by using cos rule and once we obtain DH we can just use sin rule to find out what is sin of theta plus alpha okay that sin L divided by sin theta plus alpha will be equal to DH divided by sin theta and from the sin rule we can immediately obtain what this ratio is and then we are done and answer it comes out to 5.15 kilo Newton which many of you have obtained properly okay so the entire idea okay even before getting the final answer is to recognize that what is every component doing what happens so anytime we want to analyze the problem because we want to analyze the problem appropriately because there are so many different free body diagrams that are possible we can break down everything every pulley this 2 force member this pulley this platform this member this member so many ways if we can break down but then that will only increase the confusion so the best idea is to ask ourselves that what if we remove this hydraulic cylinder what happens and that immediately gives us a clue about what is happening and the presence of this is doing what so in this case we immediately realize that in the absence of this these two reactions are parallel for this free body diagram for the rollers and this is providing a component okay in the vertical direction so we immediately know what is the free body diagram and what is the appropriate equation of equilibrium so if you think logically like this then the free body diagrams that we need to draw and the corresponding coordinate system that we need to use and equations of equilibrium we need to write will become apparent if you think in terms of what happens if you do not have those particular component if you think from that point of view and see what is happening to the system that is tending to collapse it will be a two force member it cannot provide reaction in the direction that you want then the free body diagrams as well as the appropriate equations of equilibrium will become clear okay so this is the point I think we should emphasize to the students that we do not mindlessly draw the free body diagrams but we think properly that what is happening what is every component doing and then once we use that logic then the corresponding equations of equilibrium and the free body diagrams for which we are writing those equilibrium will become much clear and be much more logical okay so one or two quick questions about this problem okay general problems I have answered a lot of them still some problems are remaining I will answer them via chat okay one zero four eight go ahead for your question sir in the previous problem we had a doubt actually okay yes the point is in the point let me go to the previous problem basically you have shown us the pinch yeah please continue yeah previous problem I am going to that slide point point point is basically shown as a pin joint is a pin joint yeah you have not considered it anywhere but you have not considered anywhere for solving the okay okay okay you know what it is a good point I should have drawn here ex and y okay my mistake okay in a hurry okay I just disregarded it okay clearly you are totally right okay please note that at E because it is a pin joint there will be two possible reactions you are totally right totally right okay my mistake okay so all the participants please take note of that at point E there can be a horizontal reaction there can be a vertical reaction but this does not come into picture why because what we did is that for this free body diagram we took a torque or moment about point E as a result these reactions did not play a role okay so that is the reason why I forgot to mention them but right it is my mistake clearly there should be reaction here horizontal and vertical reaction yes fully agree with you sir we are having a doubt in previous problem that is problem number five okay let me there is a table if we remove this table problem number five okay yes point B in previous problem this problem we consider the point B and if you apply a concentrated load at this point B we apply a concentrated load at point that vertical load at point B and suppose we want to make the free body diagram of link P E and link BC then load P which in which link we will consider the external load P okay so okay so let me draw it okay let me draw it clearly okay sure sure sure so what is happening is this this is a wheel this is the one member connecting these two wheels okay and what do we have we have this member connected here and what we this is not a free body diagram yet and what we saw from the previous how do I go to the slide okay what we saw from yeah what we saw from this free body diagram is that these two wheels exert upward force on the platform and by Newton's third law they get downward force acting on them okay so now this where the forces acting on them so this is RA yes please this is RB now the question you are asking is this that when we draw the free body diagram for this assembly what do we do if you want to draw a full free body diagram note one thing that this is a little bit complicated problem why because at this joint you have one member okay so we have this one member I want to draw a free body diagram of link BC and link you want to draw the free body diagram of link BC link BC and link B is separately okay link BC and B separately okay okay okay so you can do that so what you can do is this so then essentially what you are doing is that you are essentially drawing this roller cuff free body diagram separately is cuff free body diagram separately is cuff free body diagram separately now what you will see is that at this point there are two possible forces which is this P and this this force and equal and opposite force will act on this like this this link here is a two force member so it will have some force acting on it in principle I have assumed it to be compression and then this is RA so this will be the set of free body diagrams okay that I will draw if I want to draw the roller cuff free body diagram and this link of free body diagram and this member cuff free body diagram separately so this is how it will look like and then we have to write down equation of equilibrium for this roller okay that is how this will look like and then this roller also is in equilibrium so there is an equation of equilibrium that you have to write for this roller yes but to prevent this confusion okay so in order to prevent because we do not care about this what we do is that we draw yes please if we do not have the roller we have only F pin joint at this end and we are applying just a vertical load P then what will the free body diagram of these two links separately means P in which member we will consider the P oh if you oh I see your point you are saying that if you I totally see you are saying that if I apply here I see your point I see your point okay so you are asking this now it is not exactly related to this problem okay that is what got confused me confused okay I get your point so your point is this that you have something like this yeah but you are you are asking me a special question that where if I do this if I apply a force here and now I want to draw the free body diagram for this or this what do I do so what you are asking is this that if I have something like this one member like this another member like this and a force exactly acting at the joint F so the easiest trick to do is you assume that this force is acting just a little bit like this on this that is the easiest trick to do or it is equivalent to say that this is acting just little bit like this on this member and both all these three are equivalent okay so these we are doing so this is the most convenient way to draw the free body diagram so when you draw the free body diagram in this case this force will be acting on this member and in this case the force will act on this only thing is that you cannot do both of this you cannot do this also and this also you either decide that this force belongs completely to the horizontal member or it belongs completely to this inclined member and be done so both are completely legit ways of doing it if you act make it act exactly at the joint then what will happen is that you will also have to take equilibrium of the screw also or the joint so in order to prevent that complication you may just decide that the force acts very close to the joint but on the horizontal member or very close to the joint but on the inclined member both are equivalent but you have to be consistent you do this and you always stay with this you cannot do this and this both okay means we have to consider only on one link only you cannot do it on both that is not right you either you make a choice that either you want to distribute here like this or here but you cannot like do this and this both that is double counting that is not allowed think about it and you will convince yourself that doing both of them is not the right thing to do but you do you attach it to this or this both are equally fine and you will see you try doing both ways you will see that answer will not change so one question okay that why only vertical reactions at point B and C just note that we have assumed that this is a roller connection that is an idealization okay and as a result because it is a roller connection there can be only force acting in a vertical direction so if you want to go into more details we can go into as many details as possible but to cut the long story short this wheels are free to rotate about this joint so as a result they can only exert reaction in the vertical direction because if there is a horizontal force also then the equilibrium for this wheel about the hinge point will not be will not be obeyed as a result the reactions can be exerted only in the vertical directions at point B and C then the force acting in link AD some colleges have asked what is the force acting in link AD clearly from this free body diagram for the platform okay if we take equilibrium in the horizontal direction because these rollers are exerting forces only in the vertical direction if now we take the equation of equilibrium in the horizontal direction what you will see is that that because for this problem there is no force acting on the platform in the horizontal direction the FAD horizontal component should be 0 which in turn implies that FAD should also be equal to 0 okay huh so one other problem is that the one question that is raised is that the problem says that there are two hydraulic cylinders so this is the problem that I had emphasized that this platform is a 3D problem so imagine this platform is actually 3D platform which is coming out of the plane of the paper on one side we have an assembly like this and on the other side we have an assembly like this so those are the two cylinders exact copy of this and what we are looking at is that we are saying we are exploiting the symmetry of the problem saying that only half of the weight is distributed on half side other half is distributed on the other side and so only one cylinder okay so I can see that all the questions are essentially answered now there is one problem so this problem is a very nice problem I leave it to you for solving when you go back okay this is almost a replica of the problem which just did a couple of moments ago okay I request you okay to when you go back just have a look at it to just clear your concept okay so let me briefly discuss this problem in the next five minutes okay and we will start our quiz after that so what we have here okay is a mechanism for what I will do is that okay it will be too much time to discuss this so why do not I take a couple of questions if more time permit we will discuss that tomorrow or in the buffer session okay so let us not worry about it let us ask let us go and have more discussion we do not have enough time VJ 10 could you explain me that how the it is only carries only only one reaction which which C sorry GC okay okay okay think about it okay so we are taking that GC is a two force member why now GC has one pin connection at G and another pin connection at C okay now if we look at if you draw the free body diagram we had just seen that what is a two force member that there are forces acting only at the end points and nowhere in between now this member is satisfying all that criteria because we are neglecting the self-made of these members and because it is satisfying all the criteria it is a two force member and in a two force member what we had seen is that what is the direction of the force the direction of the force is along the line joining the two points where the force is acting so the force at side G the force acts at C and the line joining them is along GC so the force has to act along GC okay that is a simple logic why GC is a two force member and why GC will exert if we remove this GC from the connection then you have to replace that with an equivalent force and an equivalent force will be in the direction of GC so that is a simple logic sir in the case of bolted joint what kind of support we have to assume so bolted joint okay if you have a bolted joint okay so this is one thing which many have many of us have been been asking so if you have let us say a simple example like this that we have a joint which for example is welded or bolted for example as opposed to or it is pinned okay so this is weld this is pinned now what is the difference the difference here is that that look at these members AB CD again look here AB CD this point C is common to AB as well as CD now note what is happening at this joint what is happening at this joint is that that there is no relative separation possible okay there is no relative separation possible between AC and BD both in the vertical direction and the horizontal direction not only that this angle 90 degrees can also not change because this is a fixed joint or a welded joint and as a result when we replace when we draw the free body diagram of this member what do we say that because all these three relative kinematic degrees of freedom are constrained we have to replace when we remove them by three unknown reactions why because all the three degrees of freedom are constrained and on ACB what do we do Newton's third law tells me that it acts equal and opposite like this but whereas if I come to this the relative separation in the x and y direction is prevented but this angle 90 degrees is free to change in principle as a result the only possible reactions are this no moment because why because the angle the relative angle is free to change and correspondingly according to Newton's third law equal and opposite so this is the difference between a hinge joint and a welded joint or a bolted joint here the rotation is fixed here the rotation is free okay fine some other queries okay you can post on Moodle and I will try to answer those queries too okay then.