 Let us begin with today's course okay which is principle of virtual work. Now before we define what is virtual work let us briefly try to understand okay or let us try to brief brush up our concepts on what is the concept of work of a force. So in the past many many lectures what we had discussed we had discussed about force we had discussed about moment two forces forming a couple okay all these things we had discussed in the past many many lectures. But now there is a completely complementary approach okay which was developed historically by people like Leibniz Hamilton and so on okay it is an approach which seems to be fundamentally very different than the approach which was taken by Newton and whose laws we have been using for the problems that we have been working on for the last few days. But there is an alternative and really really beautiful approach which is which instead of defining a force the fundamental quantity that is defined there is the work of a force. Now how you define work of a force now the work of a force okay acting on a small particle is defined as that if a force F acts and the particle undergoes some displacement dr a very small displacement under the action of that force then the work of that force is defined as a scalar product of the force and the vector displacement F dot dr. So now we all know that scalar product simply can be written as the magnitude of this force into ds which is the displacement or the magnitude of dr times the cos alpha which is the angle between the force and the displacement and some special cases why are these special cases very important we see that when alpha is equal to 0 then the force and the displacement are in the same direction and as a result a same force for a given infinitesimal or a small displacement produces a work which is maximum and just given as F times ds where ds is actually the magnitude of this vector displacement okay or it is just the ds or the distance is goes from point a to point a prime second case is when alpha equal to pi in that case force and the displacements are exactly opposite to each other so the work done is negative now this kind of thing clearly happens if you take a particle and raise it against gravity so gravity acts downwards but because we raise it upwards gravitational force does the work that is negative okay more about that later. So with this special cases we can also derive this okay I will not go into the all the details of this derivation but if you have a rigid body which is subjected to a couple how do we define a couple let us remind ourselves at point a we have a force minus F acting in this direction at another point B we have exactly opposite of this force F acting on this direction now the resultant sum of these two forces is 0 so they do not cause any translational effect on this body but now what do they do is they combine to form a couple given by F times r which is a perpendicular distance between these two now if you just go through the derivation to this system what we do is that we put a translation of dr1 such that a goes to a prime b goes to b prime and we form a parallel in addition to that if you also rotate this body now about point a prime then by an angle delta theta then this simple derivation shows you that the work done by a couple will be given by nothing but m which is F times r into d theta which is the small rotation. So these are the two important concepts that the work of a force which is F dot ds and the work of a couple okay and in this case let me emphasize that we are talking about a planar body and for a planar body d theta is clearly nicely defined and the work done for that couple when the couple has a magnitude m and we have a rotation d theta then the work done is equal to m times d theta okay so far so good. Now let us come to the principle of what we call as a principle of virtual work now what is the principle of virtual work and first of all before we define that principle what do we mean by virtual work so virtual work is a concept that we have developed for our own understanding. So what we do is that suppose we have a point particle which is subjected to various loads F1, F2, F3 and so on now to this point particle okay what we say is that we imagine that we apply some kind of displacement now when we say we imagine to apply that displacement the reason I will make it clear the reason for that imagining is that that these forces if they are not balanced with respect to each other they will end up moving this particle but the virtual displacement has nothing to do with the displacement that these forces can produce on this particle in fact this virtual displacement is completely dependent on us we say that we provide this particle A with a virtual displacement small delta r now this small delta is just used as opposed to this D to distinguish that this is actually a virtual work this is not real displacement particle subjected to various forces we just say that we apply this imaginary virtual displacement and the idea is that when we do that we ask ourselves that for this virtual displacement what is the corresponding work that is done and now because the displacement was virtual the work is now called as virtual work and the corresponding virtual work is given by F1 delta r F2 delta r plus F3 delta r and so on or in other words we can take this delta r out because of the commutative property of the dot product and this will become F1 plus F2 plus F3 delta r now this is the resultant force r so our virtual work is simply given by the resultant force of all the forces acting on the particle dot dr which is the virtual displacement we provide to this particle now we define what is virtual displacement we define what is virtual work now we define what is the principle of virtual work is a very innocuous looking statement very simple looking statement but it is extremely powerful so if now a particle is in equilibrium then what happens then sum of all the forces will be equal to 0 what does that imply that the total virtual work of forces acting on the particle is 0 for any delta r because r bar is 0 so for any delta bar delta r bar this virtual work is 0 that is one statement now you turn the argument around and say if a particle has various forces F1, F2, F3 acting on it now if I apply any virtual displacement and correspondingly we say that the virtual work for any arbitrary virtual displacement is 0 for a given set of forces then what will happen is that r dot delta r bar is equal to 0 for any arbitrary delta bar the immediate implication is that that r bar should also be equal to 0 so what it means is that that r bar equal to 0 means virtual work is 0 for any delta r bar and virtual work is 0 for any delta bar for given capital r bar means that capital r bar should be equal to 0 so what this simple statement means is that that Newton's laws of equilibrium and principle of virtual work both are equivalent so Newton's law implies principle of virtual work and principle of virtual work implies Newton's laws so that is why they are equivalent and there are various scenarios in which we can apply them appropriately and these methods are a far more powerful than the application of Newton's laws. This principle can be extended also to rigid bodies okay I am not going into the derivation of that that if a rigid body is in equilibrium the total virtual work done of the external forces acting on the body is 0 for any virtual displacement of the body just we have extended it now what is the additional difference between a particle and a rigid body the additional difference is that that for a particle there is no rotation so when we take for example what is the virtual work done on a particle some rotations of the rigid body may also be involved in addition to translations and we know that the work done for translation is simply f dot dr but we can also provide virtual rotations and a virtual work done corresponding to a virtual rotation of now instead of d theta delta theta will be nothing but m which is the moment or the couple acting on the body times delta theta okay as simple as that so a couple acting on a body times delta theta will give you what is the virtual work provided by the moment. Now we go one step ahead now instead of having just one rigid body if you have a many many rigid bodies which are connected to each other by various linkages so look at the point so what we had done previously is that we had done equilibrium of a particle previously when we did simple equilibrium then we did equilibrium of rigid bodies and then we went ahead and did equilibrium of multiply connected rigid bodies so here also principle of virtual work is defined for a particle it is defined for a rigid body okay and then one step we go ahead and say we define this principle even for a bunch of connected rigid bodies and the principle of virtual work works even well for that and we will do a bunch of problems and then whatever this statement is it will become extremely clear and you will see that there are various problems in which if you use Newton's law the problems will be very very very difficult to solve but those problems become extremely transparent and extremely easy when you apply principle of virtual work. Now when we have multiply connected systems okay we want to understand two things the first thing is that that a multiply connected system what does it have it has various supports it has linkages it has supports this system is subjected to external forces now there will be reactions that will be produced at the support now to various of the systems we can provide virtual displacements in different possible ways and the idea is that to make the best use of the principle of virtual work we need to decide that what is the appropriate virtual displacement that virtual displacement is fully in our control but we have to judiciously choose what virtual displacement we can provide on the structure such that whatever we require okay for example if I want to find out what is what is the horizontal reaction here or what is the vertical reaction here then we need to provide appropriate displacements and so on in order to obtain what is the corresponding unknown reaction. So in this case reactive and internal forces okay so long as we provide virtual displacements and virtual rotations appropriately okay reactive and internal forces do not do any work now the virtual displacements are to be given carefully so that the active forces okay what are the active forces only the forces that do the work are called as active forces so we have to provide virtual displacements in such a way that the active forces or the forces that do the virtual work are only the known forces which are given to us and the forces that we are interested in obtaining for example a reaction here. Now in this case in equilibrium using Newton's laws force balance moment balance what do we do we draw a free body diagram the free body diagram is the heart of 3D equilibrium 2D equilibrium and so on but when we come to principle of virtual work the corresponding equivalent is called as the active force diagram and what that is will become clear to us very soon okay but just one point that we have to provide virtual displacements to the system such that there are some forces which do work the forces which do work are called as active forces and the corresponding diagram that we draw is called as the active force diagram. Now let us take a very very very simple example okay this is virtual work for a rigid body the virtual work for a particle is trivial very straightforward whatever is involved in the virtual work for a simple particle we had discussed that okay in the previous slides that we had displayed but now let us come to one step ahead and discuss what is the virtual work for a rigid body. Let us take a simple example the simple example is we have this rigid triangle which is supported by a hinge support or a pin support at point O and it is supported at the other end by a roller support R now what we are asked we are asked a very simple question we want to find out that when this body is subjected to a horizontal force at the top P what are the reactions okay what is the vertical reaction R at this point what is the horizontal reaction at point O and what is the vertical reaction at point O. So these are the three questions okay that we want to answer so let us begin with the first question that what is the reaction at R and can we obtain that reaction using principle of virtual work now we all know how to do this problem using principle of equilibrium moment balance force balance and so on what do we do we just realize that there are two unknowns here we realize because a roller there is one unknown we remove this supports okay and replace this with two forces here with one force here and now what do we do because we are interested in finding out what is the reaction at this point we take for this free body diagram that results some of all the moments about point O and then we will immediately find out that the reaction R will be given by P A divided by B straight forward. Now what we do is that in the principle of virtual work we draw what is called as a active force diagram we are interested in finding out what is the reaction at this point R so what do we do we just release this support this roller we remove it and replace it by R but note that while drawing the AFD or the active force diagram we do not remove this support okay that is the clear difference between these kind of diagrams called the active force diagrams and the free body diagrams that we had discussed earlier so this is not a free body diagram we have released this support why because we want to find out what is the corresponding reaction here but we are not releasing this support the reason will become clear very soon so what we do is this we want to find out what is this reaction R we know what is this force P now there are many different virtual displacements virtual rotations that I can provide to this rigid body what I can do is that okay I release this I can just translate this body fully then what does that mean I have taken I have released this support I can take this body and move it in the horizontal direction okay which also means that I have released this support in the horizontal direction but what happens is that if we do then the corresponding reaction at point O because this point O also has a horizontal virtual displacement we will end up doing the work instead what we do is this we think cleverly and say what if I apply to this active force diagram a virtual rotation about point O okay this dotted line then shows us that what is the position of this thing after we provide a virtual rotation about point O for this free body diagram note now the active forces here is only R the active force here is only P so what is the displacement here okay this is O we give it a small rotation delta theta then the displacement here will be A times delta theta if this point is not clear right now okay hold on to that thought in the next few slides we will again come back to it that why when you rotate this entire object okay virtually by a rotation of delta theta the displacement of this point in what direction it will be in the left direction given by a delta theta now what is the displacement of this particle in the vertical direction it will be simply B times delta theta now what is the work done the displacement here because of this rotation is in the left direction force is in the right direction I have made a error here so the work done will be positive work okay displacement is exactly opposite to it so the work will be minus P A delta theta okay please note of this error here what is the virtual displacement in the vertical direction what is the direction of the force again vertical so R times B okay so R times B delta theta R is the force B delta theta is a displacement is the corresponding virtual work done by this force now what about the reactions here okay we have assumed that all the joints are frictionless because of this there is no internal moment so there is no internal work due to rotation there is a one horizontal force here which is possible but since point O has no virtual displacement this does not do any work similarly there is a vertical force here but since the we have provided to this body a virtual displacement in such a way that point A has 0 virtual displacement even in the vertical direction the vertical reaction at O also does not do any work so the only forces doing work is this P and this displacement R for the virtual rotation that we have provided and so these are the active forces we get this simple equation since this is true for any arbitrary delta theta we immediately see that minus P A plus R B equal to 0 or R will be equal to P A divided by B similarly now by virtual translations in the X and Y direction we can equivalently obtain what is the virtual reaction at O and what is the virtual reaction at X and giving those virtual translations will be equivalent to doing sigma of forces in the X or the horizontal direction and some of all the forces in the vertical directions okay. So we had done now two discussions very trivial discussion about principle of virtual work when we have a point particle trivial then principle of virtual work when we have an extended rigid body there principle of virtual work is somewhat non-trivial why because for a particle the only degrees of freedom that are present are translation in one direction and the translation in other direction so only two degrees of freedom but for an extended rigid body in two dimensions or in a plane there is another degree of freedom that is possible that is a rotation and that is one extra distinction that is there between a point particle and between an extended rigid body. Now we go one step ahead and say okay so work done by force okay when we have multiple connected bodies okay when we have multiple connected bodies then what happens okay first we note that what are the forces that do not do work so if the pin is frictionless for example then as we had seen here if this pin is frictionless which we idealize almost in this case okay or we say it is well lubricated so we neglect friction in that case the rotation there okay will not do any virtual work reaction at a frictionless surface due to motion of a body along that surface for example if this surface is frictionless and we have a virtual displacement that we provide to this particle or to this mass then the then the there is no horizontal force or no friction force the only force that is present is the reaction in the y direction and because the displacement is perpendicular to displacement what is virtual work F bar dot data r since force is perpendicular to this the work done by the normal by the normal force corresponding to the virtual displacement applied parallel to this plane will be equal to 0 okay reaction at a frictionless surface due to motion of body on surface 0 weight of a body with center of mass or center of gravity moving horizontally is also 0 why because if the body moves horizontally okay then the gravity acts perpendicular to the displacement and as a result it does not do any work and friction force of a wheel moving without slipping okay so this is somewhat controversial topic we will go into discussions of that a little bit later okay so it is not really controversial but it may be little bit complicated so we will hold on to that thought for the time being now think about it some of work done by several forces may be 0 if suppose we have 2 bodies which are connected by a frictionless spin now if you try to change this angle there is no internal work done at this joint why because internally this joint is frictionless so there is no internal moment that is being applied so this relative change of angle between these 2 links will not do any external works if bodies are connected by an inextensible chord okay then there is no extension of this chord so this T1 this T1 okay will not do any work because there is no stretching and as a result there is no effective displacement for this chord so when we provide some virtual displacement this internal forces will not do any work additionally internal forces which happen inside the body okay rigid body also do not do any work so we keep this in mind when we will apply principle of virtual work to various systems and try to find out okay support reactions or forces required for stability and so on okay so there are a variety of problems which we will solve in the next couple of hours. Now let us come to this concept of degrees of freedom degrees of freedom concept we had discussed briefly when we did Newton's laws force balance moment balance and so on but degrees of freedom is a concept which is extremely important when we when we use principle of virtual work so degree of freedom in this context is the total number of independent coordinates required to specify the complete location of every member of the structure or every point on the structure so okay I have just read out this definition and let me just discuss okay a simple one degree of freedom system. What do we have let us say that if we have two rods okay which are connected by pin here pin connection between these two rods and a roller connection here now clearly if we specify the displacement of this wheel okay that this wheel moves by some small amount delta v then we can think about it very clearly and what will happen is that that when this displacement is given then for this body to move without extending or without deforming any of these members or without opening up any of these joints okay there is only one particular way in which the body can be displaced okay so we specify one small displacement here all the other displacements are automatically specified same in this problem if we specify the displacement of this one particular particle so this is a pin support this is free to move in the vertical direction this joint this one part this one member is connected to another member by a pin joint here and ultimately there is a pin connection here now if you think about it okay just pay some attention okay and if you do that what will happen is that if you specify the vertical displacement of this point then all the displacements of this other points are automatically specified so such systems are called as one degree of freedom systems and there are these other systems which we are not going to deal with in this course but for example if you have two linkages which are connected like this then there are two independent angles theta 1 and theta 2 okay so these are two degree of freedom systems so do not worry about this we are not going to discuss this problems for the current course they are not relevant for us right now but one degree of freedom problems are extremely important and a simple idea is that by just specifying one displacement or one angle all the coordinates of the other points of the structure are also automatically specified that is what we mean when we say that the system is a one degree of freedom system so to summarize the virtual work done by external active forces or an ideal mechanical system in equilibrium is 0 for all virtual displacements consistent with the constraints now the consistent with the constraints only so that okay this point will again become clear okay so that the reactions at the supports do not do work okay or the reaction that we do not care about do not do any virtual work this point will become more clear when we actually solve some problems we more importantly assume that these are ideal systems all surfaces joints etc or frictionless okay and with this much of discussion okay we just briefly mention that why principle of virtual work okay we have spent enough time for discussing problems where we do force balance where we do moment balance and so on so why do we need to use principle of virtual what is the big deal the big deal is that for various complex mechanisms we will solve some of those problems we will see that drawing free body diagrams for these complex mechanisms will become very cumbersome because it will involve a huge number of free bodies and the solution procedure will become very difficult and highly prone to errors whereas many such problems we are going to solve a few such problems will be straight forwardly solve using principle of virtual work okay and many a times what happens is that in many problems we need to draw multiple free body diagrams in equilibrium in order to obtain some reactions or some joint support reactions link reactions and so on but here we will see that by choosing appropriate virtual displacements we can make sure that whatever forces we want to find out okay they can be obtained in one shot without like writing down various free body diagrams and multiple simultaneous equations okay. Another thing so very important point is that that this kind of analysis will be stepping stone okay in solid mechanics structural mechanics when or in approximate powerful methods like finite element method where principle of where Newton's law okay will be absolutely useless in order to take care of such problems okay with this okay let us discuss this very simple thing because we are going to use this idea again and again and again the topic is small rotations of rigid bodies now let us say we have a rigid rod AB okay the radius the length of this rod is R now this rod suppose we rotate by some angle theta about this point O so this is the arc that the that the rod covers this is the final position of the rod we have since just rotated the rod about point O this radius remain the same now we ask ourselves questions okay that these two angles alpha okay this angle is theta what are these two angles alpha because these two distances are the same these two angles are the base angles of isosceles triangle and they are the same we ask ourselves that when this point B now moves to B prime what is the horizontal separation between these two points clearly geometry tells us that this horizontal separation will be R times 1- cos theta we also ask ourselves a question that what is the vertical displacement between B and B prime clearly it is R sin theta is equal to the vertical displacement now note one very important thing that for small theta okay when this rotation is very small then this theta is small sum of all these angles is 180 what does that mean that alpha becomes approximately 90 degrees so this line will be approximately perpendicular to AB when which means that the displacement of B to B prime is approximately perpendicular to the original line okay when this angle of rotation is very small moreover we can write this 1- cos theta as 2 sin square theta by 2 and for small theta that can be written as R theta square by 2 similarly the vertical displacement R sin theta will be written as R times theta and when we write Y is equal to R theta okay why because sin theta approximately equal to theta now note one thing that when angle theta is very small then what happens that theta square will be smaller when compared to theta take a very small number less than 1 you take square of that the number becomes even smaller so with this approximation what we say is that that for a rigid body okay which is rotated about point O in this case rigid line AB then the resulting virtual the resulting displacement of point B from B to B prime is in the vertical direction because this angle alpha is almost 90 degrees and the corresponding displacement in the vertical direction is given by R times theta when this angle theta is small and the corresponding horizontal displacement of this point B okay to the first order is 0 okay to the first order means that this is not theta to the power 1 it is theta to the power 2 it is very small so we neglect that now where are we going with that so a generic observation is that if you have a rigid body okay at a particular angle then using the ideas that we discussed in this previous slide what we will see is that that if this rod AB is given a small rotation delta theta now delta theta because we are planning to apply virtual rotations to our problems okay our collection of rigid body problems we need to know that what are the displacements that arise because of the virtual rotations that we provide now see here if this angle is theta to begin with AB makes an angle theta with respect to the horizontal we provide a small virtual rotation delta theta to this AB what do we get that AB is equal to AB prime is equal to L which is the length of the rod and from the previous observation this displacement of B to B prime is perpendicular to rod AB okay this is perpendicular what is the magnitude of this displacement B B prime it is nothing but L delta theta just recall this okay that this displacement is nothing but L delta theta and now since this angle is theta this angle will also be equal to theta now we want to find out that at an arbitrary angle theta what is now the vertical component of this displacement and what is the horizontal component of this displacement and what we will see is that that the horizontal component delta X is nothing but B B prime okay which is delta into sin theta clearly which will be given as L sin theta delta theta similarly delta Y okay displacement of this B prime B to B prime in the vertical direction will be what will be B B prime which is equal to delta into cos theta so that will be delta cos theta will be this vertical displacement which now can also be written as L cos theta delta theta now note one thing there is a very powerful solution that comes out of it note here if I want to find out what is delta Y okay instead of writing it as L sin theta delta theta we write it as L times sin theta as one particular combine this L sin theta together into delta theta but what is L sin theta L sin theta is nothing but this vertical length of this line and what this thing is telling you is that if you have a rotation about point A of this body then the horizontal displacement will simply be given by this projection okay this projection multiplied by delta theta now what does that mean that if I have a rigid body okay instead of having just one line if you have one complete triangle then or a complete rigid body like this what it means is that that any point which lie along this line okay you take that line drop a perpendicular to this okay from this point a draw a perpendicular the distance of that perpendicular will be equal to L sin theta delta theta and corresponding displacement in the horizontal direction will be nothing but that distance which is L sin theta into delta theta look at the other way around what is delta Y okay think about this delta Y is L cos theta delta theta so we bring this L cos theta together what is L cos theta is nothing but this distance A B and what it is telling you is that that instead of having just A B if you have a full triangle A B P okay then this tells you immediately that any point along this line okay any point along this line will have the same displacement in the virtual direction in the vertical direction which is given by this A P which is L cos theta into delta theta so if we have a triangle like this A B P which is provided a small rotation of delta theta about point O then all these points along this line will have the same displacement in the vertical direction given by L cos theta into delta theta so this is this distance what is this distance on this line we drop a perpendicular from A to P and that distance into delta theta will be the corresponding displacement in the vertical direction for any point on this line if we have a rigid body given by A B P and the virtual rotation is provided about point O and the general principle is that a displacement in a general direction if I want to find out that at point B what is the displacement in some inclined direction that will be nothing but drop a perpendicular from this point O to that direction that distance multiplied by delta theta will be nothing but the displacement of this point in that particular direction if this point is not clear okay we will have a brief discussion session and this point we can make it clear by drawing a few more figures okay but just think about it that the displacement in a general direction is the projection in the perpendicular direction times delta theta and if you are not comfortable even with this then the best is join that point okay the displacement will be perpendicular and then find out what is the component in the horizontal direction vertical direction or component of the displacement along any arbitrary direction okay so these are some examples of various mechanisms which we will see can be solved very nicely using principle of virtual work now let us come to our favorite problem of ladder leaning against a wall okay and we also refer to this problem da as a ladder problem so what is the problem assuming friction lens contacts everywhere so one contact two contact determine the magnitude of P for equilibrium okay what are the forces acting on it W reaction in the horizontal direction which is the normal reaction here reaction in the vertical direction which is again the normal reaction here now what we are asked is that that if the contacts are friction less determine the magnitude of P okay because if you do not apply this force P we had seen multiple number of times this body cannot remain in equilibrium so we are asked to find out that at what P this body will remain in equilibrium now the question is that there are 3 unknown reactions 2 unknown reactions one is this reaction here one is another reaction here and what we want to find out the given force W okay is a known force but we want to find out that what P should be applied such that this body remains in equilibrium now to this rigid body we can provide all manner of virtual rotations we can provide this rigid body a virtual translation in the horizontal direction but why is that not good because this point will also have a displacement and that will also do work this point will also do work so that unnecessary point which reaction we are not interested in that will end up doing virtual work and needlessly make our equations difficult on the other hand if you take this particle and rotate sorry and displace in the vertical direction then what happens this point moves vertically upwards okay the reaction is only in this direction so no work is done okay on the other hand okay this unknown reaction okay which is acting from this point will do work and so as a result we do not get but P will not do any work why because the displacement of every point is in the in the vertical direction so no work is done by P so we are unnecessarily getting what is the reaction at this point whereas what we want is displacement is the particular force at this point so what we do is this we note here we think about it that what should be the appropriate virtual rotations like those are completely in my control that we should give to this ladder AB such that only this W does work and the force that is applied here does work okay that is the simple question we ask ourselves okay now you may suggest that okay there is another option we have we can take this ladder AB and provide it a small rotation about point B okay a small delta theta about point B good we are good now but then that is not good why because this point A has now moved to A1 and it has a vertical it has a displacement in this direction how much is the displacement it has we had seen earlier that the displacement it will have will be simply l cos theta into delta theta that will be the vertical displacement and a horizontal displacement delta x will be l sin theta into delta theta so this displacement will do work against the reaction acting here but whereas this poor P does not do any work so again we are this this kind of virtual displacement is again no good for us but then what we do is that we play a trick and say aha we do not want this gap to be there so what we do is that we give it one rotation about point A about point B and then provide a translation by how much by translation of delta x such that this gap closes and delta x is equal to l sin theta delta theta now when this gap closes what is the final displacement or the final virtual displacement of this green ladder AB the final virtual displacement is this point A prime B prime is this line A prime B prime what has happened here that this point B had moved to B prime how much distance same distance delta x because this is a complete translation which is equal to l cos theta delta theta that is the virtual displacement of this point B what is the virtual displacement of point A only in the y direction okay because now the gap is being closed and the virtual displacement that A goes from A to A prime in the y direction by a virtual displacement given by l cos theta delta straight forward okay this is this distance multiplied by delta theta which is l cos theta A to delta theta so what do we see here is that delta x by delta y okay delta x by delta y will be equal to tan theta now what is what are the forces that are doing virtual work the normal reaction here is it doing any virtual work no why because the reaction is upwards whereas the virtual displacement is in the perpendicular direction so no work what about the virtual work done by the normal reaction at A is it doing any work no why because the virtual reaction the normal reaction here is in this direction but the points displacement is only in the vertical direction and we had said that there is no friction so there is no work done by this normal reaction here is the work done by the center point yes most definitely why because when we provided a rotation this point moved up by a distance of delta y so w which is the center of this line will move up by delta y by 2 so this displacement of point o the virtual displacement was in the upward direction of magnitude delta y by 2 so the virtual work done will be downward force multiplied by the upward displacement because force and displacements are opposite work done is minus w into delta y by 2 then what do we see is the work with the force that is applied here p does it do any work clearly it will do work as p times delta x why because p is in this direction delta x is in this direction so p times dx is a virtual work done by this particular force so ultimately what is the equation that we have we will have w into delta y by 2 plus p times delta x is equal to 0 or in other words we will see that p will be equal to w by 2 delta y by delta x but what is delta y by delta x we have seen from all our geometrical considerations here that delta y by delta x is nothing but 1 by tan theta so we write here as this becomes w by 2 and theta and this is exactly the answer that we get if we do a simple equilibrium problem so this is how we have used okay in this simple problem principle of virtual work to find out what is the reaction at point p now what we will do okay we will do another simple problem okay what is that simple problem okay the simple problem is as follows okay this is a simple crank problem what is happening here is that the pressure p driving the piston of a diameter 100 nanometer is 1 Newton per millimeter square so we have been provided a diameter of this piston we have been provided what is the pressure so in other words we multiplied the area into the pressure we are provided what is the force that acts here now this is free to slide this particular portion okay this piston is free to slide in the horizontal direction because we neglect the friction or this joint is well lubricated so this is free to slide but now what we are doing is that that this piston okay when you apply some force here what it will do it will try to bring this inverse here but now what we do is that to prevent that motion from happening we put some weight w and what is asked of us is that given this angle is 10 degrees AB is pinned to BC okay and there is a hinge connection so this is 300 nanometer this is 300 nanometer what is the weight that the system will hold if the overall friction is neglected okay so very simple problem if you use principle of if you use Newton's laws force balance moment balance and so on you will clearly recognize that AB is a two force member BC is a two force member you can take equilibrium of joint A and can easily obtain what is the force that is required to keep this entire body in equilibrium okay now we ask ourselves that this simple problem how do we do this using principle of virtual work okay and answer is as follows so remember this problem how do we have we have AB and C okay so this is our A this is B and this is C now note one thing okay what we are doing here is that that this ABC is our original configuration this A is a pin support C okay we replace this P okay we say that this is a roller support to which we are providing this much force now what we say is that to this structure okay we draw this vertical line and say now let us say in our previous ladder problem what did we have we had one ladder leaning against a wall here now we say that instead of having one ladder leaning against a wall we have two ladders leaning against the wall and then we say that we provide the same virtual displacement that we had provided to the ladder that this point A moves to A1 this point C moves to C1 okay and then this point B moves upward along the imaginary mid-plane and goes to B1 and we are done but we are not yet done why because this point A okay if we give this entire structure a virtual displacement in this direction okay then this point A moves inwards and as a result the horizontal reaction at point A will do some work which we do not want because we are not interested in the work done by the forces at this joint A here so what we say is that okay so from our ladder problem we know that if this is delta x2 okay this is delta y2 is equal to delta y1 equal to y we know that delta x1 by delta y1 okay delta x1 by delta y1 is equal to tan theta 1 and delta x2 okay this is delta x2 by delta y2 is equal to tan theta 2 now what we say is that that this is not good enough for us because we do not want this point to move then what we do is that we just play simple trick we will give this entire assembly a translation by how much such that we connect this gap and this point A ultimately does not go anywhere it gets back to its original position so what we had done is that we had disassembled a system given it some virtual displacements and then ultimately given it the final translation such that the final virtual displacement is such that the work done at this point is 0 this point B moved up so the weight applied here could do some virtual work whereas this point C moved inwards and as a result the force which is coming from this piston will also end up doing work so for this active force diagram the only points where the work is done is because of the weight acting here and because of the force acting in the horizontal direction so we can get one very simple equation okay and immediately can find out what is the weight that we need to support now how much translation we need to give we need to give translation of delta x2 now note here that this relation we already know the total displacement that will happen at point C will be equal to this initial displacement because of the first virtual displacement we had given and after the translation it gets another component of displacement which is this delta x2 and so the total virtual displacement of point C will be given as delta x1 plus delta x2 we write these two equations and ultimately what do we see that the total displacement delta xc of this point will be nothing but tan theta 1 plus tan theta 2 into delta y and what is the vertical displacement of this point will be nothing but just simply delta y okay and ultimately when we come here what does the principle of virtual work tell you that a work acting was downwards minus w into delta y plus force coming from the piston multiplied by delta xc which is the horizontal displacement of the piston both are in the same direction and this is equal to 0 is what our principle of virtual work tell us or w will simply be equal to f piston divided into delta xc divided by delta y and we had seen from our slides from all these relations that delta xc is clear related with delta y like this so we immediately can find out what is the weight that the force in that piston can support okay so these two are very simple illustrative examples of principle of virtual work okay now let us do this one simple problem okay it is simple when we think in terms of virtual work but we had seen that when we had done this problem earlier using Newton's laws then a lot of thinking had to go into it but now we will come to this problem and solve it using principle of virtual work okay so what do we have here if I repeat the problem quickly in a couple of moments what do we see here that we have a platform there is a vertical slot which is frictionless and this platform is free to slide up and down okay but what happens is that on this platform we want to keep some weight 2w which is supported by an assembly on two sides if this assembly inside is not there then this platform cannot be supported on this slot but what do we have is that that in this platform inside this platform okay there is a hollow cavity in which we have frictionless wheels okay 1, 2 and 3 frictionless wheels okay second we need not okay so we have two wheels here everything is frictionless internally and now what we want to know is that to keep this entire assembly in equilibrium what should be the force that should be exerted by this hydraulic cylinder on this middle member such that this assembly of w weight w remains in equilibrium at an angle theta when the length of each member is given to be 2l and this point a is at the midpoint of this line so this distance is l. So we know how to do this problem using the equilibrium approach now let us see how to do this problem using the principle of virtual work okay so think about it okay to do this problem using principle of virtual work what do we realize is that that we need to find out what is the force in this hydraulic cylinder so let us remove this force or remove this hydraulic cylinder and expose this force let us say that this is the force that acts from the hydraulic cylinder okay we remove it but now this weight is a known quantity all the reactions here all the internal reactions they are all unknown reactions and we do not care about them the only thing that we want is the force in this hydraulic cylinder which we represent by f of h now what do we do we want to give this entire system a virtual displacement in such a way that only this work okay this weight does work and a force in the hydraulic cylinder does work we are not bothered about anything else other than this so how do we give a virtual displacement the way to give a virtual displacement we all know that these are parallelogram okay this distance okay all these distances remain same so if we give all these rods a small rotation in the clockwise direction of delta theta then what will happen is that you will see that all these initial rods which are represented like this okay we lower the platform down okay and form into another set of parallel lines like this all these distances remain same why because we cannot change the distances between these two points otherwise these internal forces will also do work so we need to keep provide a virtual displacement in such a way that no internal forces do any work only work is done only by the weight w so this virtual displacement is good for us and a virtual displacement along this direction is good for us because this is the unknown quantity which we also want to keep track of okay so now look here if look at all these points so if one of these rods move down by an angle delta theta then it is very clear that the platform will come down by an amount how much what did we saw that if you want to find out how much amount it comes down by what do we do we drop okay a perpendicular line okay from here okay to this direction okay so because we want to find out what is the vertical what is the displacement in the vertical direction we just draw a vertical line see what is the horizontal projection what is that it is nothing but l cos theta okay so l cos theta is the vertical projection into delta theta so 2l cos theta into delta theta is how much the platform will move down by now we want to ask ourselves and all of these center lines should also move by the same amount delta theta okay so so if this goes down by 2l delta theta this also goes down by the same amount because this is a parallelogram if this is theta all the angle should remain theta because otherwise you will make otherwise the parallel nature of these things will not be possible. Now you think about it okay look at this central hydraulic cylinder what do we have this force is acting at an angle alpha with respect to horizontal the displacement of this point O what we want to find out that what is the work done by this force now what is the work done by the force the work done by the force will be nothing but the force into the virtual displacement in that direction now how do we find out that virtual displacement okay so what we do is that from point A okay this is the direction in which we want to find out what is the virtual displacement the rotation of this rod happens about point A so what do we do from point A we drop a line which is perpendicular to this equivalently from this point we drop a line which is perpendicular to a parallel line both are equivalent now what is the distance of this perpendicular line from here to here it will be nothing but l sin theta plus alpha this angle becomes theta plus alpha and this distance becomes l sin theta plus alpha so this distance of the projection A to this line will become l times sin theta plus alpha that is now the displacement in the direction or along this direction of the hydraulic cylinder now what do we have the work done at the weight position was how much this came down by an amount 2l cos theta delta theta the weight applied was w the force and the displacement are in the same direction so the work done will be w 2l cos theta delta theta okay and what is the work done in the direction of the hydraulic cylinder the displacement was in what direction it was in the inward direction l sin theta plus alpha delta theta we have assumed that the force in the hydraulic cylinder is compressive so it is pushing so the force is in this direction the displacement that we have written is in this direction so the work done will be minus fh into l sin theta plus alpha and we see that this is true for any arbitrary delta theta so these two cancel out and we see that the total force in the hydraulic cylinder is nothing but w divided by w cos theta divided by sin theta plus alpha which is precisely the answer that we had obtained in the problem that we had done earlier using Newton's laws of equilibrium force and moment balance.