 So there are some questions that are posted, so now what I will do is that I will see what are the questions that are posted and try and answer those questions and after that we will have a brief discussion okay. So let us answer a few questions okay, so from center 1, 2, 7, 5 they have asked what is the physical significance of virtual world, there is no clear cut way in which you can say that there is a physical significance per se because virtual work is essentially a mathematical concept and we have devised this concept purely for our convenience or in other words what you can say is that you can take Newton's laws as fundamental and then derive virtual work principle from that or you can say that my virtual principle is a fundamental principle and then derive Newton's laws from that both are perfectly equivalent okay. So physical significance of virtual work, I do not think I can answer the proper answer for this question exist why because it is like asking what is the physical significance of Newton's law, physical significance means that Newton's laws has particular implications in the real world but these are axioms and those axioms now you see that there are some empirical problems in the real world which follow those laws. So the physical significance essentially is that that in any system in equilibrium you apply a virtual displacement but virtual displacement by definition is something that you do not really apply. So it means in this case that you try to apply virtual displacement in your mind okay and you will see that certain nice properties of all those forces will come into picture okay this answer is a bit vague I understand but so is the question. So if you have any more questions please send it on chat, second please explain solution of simple truss using principle of virtual work, yes we will do that there is one example problem and within a few moments we will solve that problem okay. In fact why do not I go to that problem okay. So let us look at this problem number 9 what is asked here is that determine force in member CD by using the method of virtual work okay. So we can definitely solve this problem using principle of joints, principle of sections and what not okay. So Professor Banerjee had taught you a bunch of methods it is a very state forward problem you can solve it using a variety of techniques but we want to solve it by principle of virtual work why because we just want to understand okay that what this principle of virtual work what can it do and what we are asked to find out is force in member CD by using method of virtual work. Now what we do is this we replace we want to find out force in this member CD so we remove this member CD and replace that with two unknowns okay why because when we remove this member assuming that CD is in tension what do we get we get a pulling force adjoint C and an equal and opposite pulling force adjoint D okay so FCD, FCD. Now we want to find out that given this W what is this force F now think about it okay think about it that once okay that this is a great beauty of principle of virtual work okay. Suppose if you have a truss like this incredibly beautiful this principle okay so let us see how many joints does this have 1, 2, 3, 4, 5 so J is equal to 5 how many members does it have 1, 2, 3, 4, 5, 6, 7 so M is equal to 7 how many unknown reactions do we have 2 here because it is a hinge support and 1 here because it is a roller so R is equal to 3 so the number of equations that we will have is 2 J is equal to 10 and M plus R is also equal to 10. So the number of equations number of unknowns are same so we would say that okay great this truss looks good it is not statically indeterminate but then the second question arises that is this mechanism stable or not well it is clearly stable why because this triangle okay is the basic triangle to which we add one joint one more triangle another joint one more triangle so we have created a system of triangles beginning with a triangle so this entire truss is actually a rigid structure now to this rigid structure we are providing support reactions which are not parallel to each other and which are not meeting at one point so this entire assembly is perfectly well constrained okay it is a properly constrained system now suppose this member CD let us remove this member okay so we get very happy say okay everything is good and in that happiness we say we remove this member CD now when we remove this member CD what happens number of members now becomes 7 minus 1 so the number of equations okay still remains 10 because the number of joints have not changed but the number of unknowns now become 9 so what do we have we have number of equations that is less with that is more than the number of joints and we clearly have a feeling that this assembly is not stable okay this assembly is not stable so not stable means what that it is a mechanism now what is the meaning of a mechanism we had discussed that a mechanism is something where we can change the configuration of the structure okay so we can change the configuration of the structure without deforming any individual member and if that can be done we say that this is a mechanism and clearly this is a mechanism why because if you draw the displacement diagram for this you can provide a virtual displacement like this but this virtual displacement okay this dotted lines can as well be a real displacement for that structure that we can take a trust like this where this member is removed and we can just deform it like this very easily without changing the length of any member why because for this statically determinate structure upon removing one member okay it suddenly become a mechanism that it is freely possible to deform it like this and principle of virtual work essentially makes use of that fact what it says is this that if you do not have any internal resisting forces then under the application of this force W what will happen I provide it a virtual displacement which is very similar to the kind of mechanism this structure will deform into what is the work done by this W if I provide a virtual rotation delta theta then this AE okay we will move exactly vertically to this and the displacement will be how much this distance AE which we say A into delta theta now if this internal force were not there then what will happen principle of virtual work tells me that the sum of all virtual works done by the active forces here this is not an active force because this virtual displacement is not does not produce any displacement here it does not produce any displacement in the vertical direction here so no work done here only work done is by W okay and how much is the work done by it is W times delta E which is the delta E is the displacement in the vertical direction which is nothing but A delta theta so if there were no resisting force coming from this truss member boom this system cannot be in equilibrium because for this system to be in equilibrium weight itself has to be 0 is W has to be equal to 0 but then what is this internal mechanism doing that when you provide put a virtual displacement on the system which is very similar to the kind of mechanism this structure will undergo in the absence of this joint okay this internal displacement will end up in creating virtual work from this internal forces and now they will go in now how much is the vertical displacement here what we saw is that take this line of action okay rotation is happening about point A so drop from A to this a perpendicular what is the distance the distance is nothing but A into sin 60 okay so A into sin 60 or A into theta is the displacement in the in this direction which direction will it will it be left or right if the rotation is clockwise direction the virtual displacement is to the right and then what is the work done fcd into A sin theta delta theta plus fcd into A sin theta delta theta so what is this internal force doing this internal force is also providing a counter virtual work to the virtual work done by the applied force and together this can make the system okay make the total virtual work done to be equal to 0 and we can find out that this is the fcd for this member so see what we have learnt from here what we have learnt from here is that for a statically determinate properly constrained truss we remove any member we remove any support that becomes a mechanism and as a result what is that reaction doing there that when the structure becomes a mechanism okay the structure can deform in that particular mechanism okay in that particular manner okay in this case when we remove this the structure can deform like this now what we do is that we put that itself as the virtual displacement and then we see that in order to prevent that mechanism from really happening okay this internal force okay for example are doing virtual work and providing a counter virtual work to the work done by this force and as a result maintain the system in equilibrium if there were no force here there will be no reaction here and you can never make this go to equal to 0 and the system is not in equilibrium okay so principle of virtual work the kinematic degrees of freedom of the structure and the stability of the structure or all this intricately linked with each other okay so this is the point okay so now third question is asked by center 1101 in the virtual work method we can find out only one unknown so what if more unknown is there what we can do now note one thing okay I keep on saying statically determinate structure that these are all statically determinate structures that we are doing it now in a statically determinate structure what it means is that if I remove even one constraint okay say for example if I remove this roller support boom this structure is now free to rotate about point A it becomes a mechanism so what what can I do is that if I want to find out vertical reaction at point B I remove this support replace that by reaction but then think to myself that if this vertical support were not there what possible motions can this structure have without deforming any member the only motion it can have is rotation about point A now what we do is we provide a small vertical virtual rotation about point A and then this reaction will do some work this will do some work and we can put all those works together and say that for equilibrium the reaction should be such that the total virtual work has to be 0 and we will get that so to cut the long story short if a structure is statically determinate if you want to find out some reaction okay if you want to find out one reaction you can always remove that one reaction make the structure a mechanism give the structure virtual displacement like a mechanism and find out what is the overall virtual work done so you can always do it one after the other okay in a statically determinate structure for example come to this problem if you think very clearly okay this is not a statically determinate structure why because there are three of them if you think very clearly about it if you count the number of equations number of unknowns and so on you will see that the number of unknowns okay I leave that as an exerces for you that the number of unknowns is more than the number of equations okay that the structure strictly is statically indeterminate but think about it if I add 10 more of this lines okay 1, 2, 3 there are 3 more 3 of this links if I add 10 more of them this degree of static indeterminacy of the structure keep on increasing but note one thing that if we do not provide this hydraulic cylinder then even though I have huge number of huge amount of indeterminacy in the system the structure still is a mechanism why because all of this parallel lines then can deform okay that angle can keep changing and the structure can come down as a mechanism so we exploit the property and say that in the absence of this hydraulic cylinder okay what will happen I can deform the resulting structure in such a way that all the lengths remain the same but the structure changes configuration because it is a mechanism and what is this hydraulic cylinder doing that hydraulic cylinder is essentially preventing that mechanism how is it preventing the mechanism that it is providing this force from the hydraulic cylinder and in the language of principle of virtual work okay that this is doing some virtual work and to counter that it is providing a reaction and making sure that this has a counter and the total virtual work is 0 such that there is enough reaction produced here and the system remains in equilibrium so to cut the long story short okay in the virtual work method if there are more than one unknowns we can always choose okay appropriate mechanisms okay we remove that particular constraint okay and choose appropriate unknowns for example in this case if I ask to find if you are asked to find out that what is the horizontal force here then what do we do we do not give virtual displacement here we remove this support and replace this by a roller and a horizontal force and now instead of making sure that this points remain here we make sure that in the virtual work principle point C remains stationary and point A now ultimately goes exactly it does the same thing that we had done previously to point C we now do that to point A and there are various examples that we will look into that depending on what reaction you want to for a statically determined structure you can always figure out what is the appropriate virtual displacement to give such that we get exactly that particular unknown in the triangle problem okay we will ask answer this one last question one it is by centre 1 to 2 4 in the triangle problem first problem if you are considering moment at O how to consider the direction of rotation okay let me briefly explain that see principle of virtual work Newton's laws they are mathematically all equivalent but as far as the implementation is concerned okay in some cases principle of virtual work can be quite complicated whereas in some cases use of Newton laws can be very complicated and principle of virtual work can be really straightforward okay so let us look at the first example problem that we had discussed okay something like this we just make it very short this was a hinge support this was a roller support we were given a force P here this is A this is B now what we did is this okay we want to find out what is the reaction at this support so what we do we just remove this replace that when unknown reaction are now there infinite number of ways in which I can provide virtual displacements to this structure but what I want is that I want the active forces or the forces that do work to be only this P and this R so we give it a virtual displacement in such a way that we rotate this assembly of course in a virtual manner in the clockwise direction we can also put anti-clockwise about this joint O and not what do we do now look here that we had discussed that this is O this is A this is B now this OA is perpendicular okay so this is the line point about this rotation happens this is directly I am joining them we want to find out what is the displacement in the perpendicular direction which will be nothing but simply delta theta into B now we want to find out what is the horizontal displacement of this point because this force is in the horizontal direction what do we do we just see that this line goes straight up okay so the vertical display the horizontal displacement of this point will be nothing but a delta theta in the horizontal direction now the only virtual work is done by this force and this reaction so what do we have P minus A delta theta is the virtual work done by force B plus R B delta theta should be equal to 0 or in other words PA divided by B is equal to R and this is precisely equivalent to take in moment balance about point O is the point that I wanted to make okay so with this we will have we will visit a 34 centers okay before taking a break in another 10 minutes if you have any more questions please send it via chat okay we are very glad to answer them 1 3 0 5 yes why force in member FCD you have taken in both the cases in calculation as positive because we just took that as a sign convention okay we assume that the member CD is in tension okay I have perfectly justified in doing that we assume that the member CD is in tension okay I can as well say that is in compression but it does not matter and then what we do is that when we replace this member here okay replace the CD we have to put the appropriate forces so since CD is in tension okay clearly the force that will act at joint C coming from CD should be a pulling force in the inward direction similarly the force acting at joint D coming from the from the stress should be in the inward direction so because I assumed this to be in tension okay it this is the direction but when we solve the problem it turns out that FCD is minus W by 2 sin theta which means that what I had assumed previously as tension was not right and actually this force was compression it is just a matter of taking a sign convention like taking a reaction upward or downwards or sideways or left side left side or right side if it is not appropriate the sign will come out to be negative just as simple as that okay thank you sir yeah 1 1 2 go ahead considering both the method which is more reliable sir both are reliable both are equivalent okay both Newton's laws and this both are equivalent but reliable when you say reliable what do you mean by that both are equally both are the same methods so when you say reliable okay so in what context are you using that word reliable dependable it is dependable okay so both are dependable but the problem for example we will solve a couple of other problems where you will see that there are certain complex mechanisms okay in which for example Newton's laws okay for example will be much more difficult to implement this method will be far more easier again for example this problem look at this problem this problem is clearly clearly much more transparent using principle of virtual work it is literally two line answer whereas to solve it is in Newton's laws we have to draw at least three free body diagrams and to make sure the coordinate axis is oriented properly so on if not you will get lot of equations and if you don't draw free body diagrams appropriately you may not be able to solve the problem whereas here once we have an idea about the how the virtual displacements can be given to the system this is just one line if you think about it one line two line three line you get the answer immediately so for complex mechanisms principle of virtual work okay is a great boon okay no question about that okay but there are some other problems okay in which Newton's laws may be better may be applied in much more easier way for example problem find the tutorial okay that is one example where principle of virtual work is actually not that great an idea so it all depends on the problem but one great thing about the principle of virtual work is that if the students they understand how to visualize the mechanisms okay that okay if I remove some structure then this is way in which can in which the structure can go into some mechanism then the physical intuition becomes perfect then by just looking at a problem okay he can immediately figure out he or she can immediately figure out oh this structure is stable this structure is not stable this structure is definitely statically indeterminate you can figure out that a structure is statically indeterminate but these particular forces can be obtained so if you can understand how the different mechanisms can happen in a structure upon removing one or one member okay then that is a great boon you can immediately figure out that what forces are obtainable and what forces are not obtainable if the structure is statically indeterminate and so on or if it is a mechanism so principle of virtual work combined with Newton's laws okay if you have a comprehensive understanding then nothing can beat that 1175 go ahead for your question sir virtual work is alternate method or it is specific for some type of problem no virtual virtual work method can always be used it can it is equivalent of using principle of is equivalent of using force balance and movement balance they are equivalent but only thing is that that for certain problems especially for problems without fiction and complicated mechanisms okay virtual work is highly dependable okay in the sense that dependable in the sense that it makes the problem solution very transparent very easy and you do not have to draw huge number of free body diagrams which may not be very clear if you want to apply Newton's laws force balance and movement balance but it is not that some problems you can solve with this others you can't in principle all problems can be solved using both methods in engineering mechanics but you will see that in structural mechanics solid mechanics or infinite element method okay for approximate methods you need to use principle of virtual work okay in order to use compatibility conditions but that is beyond the scope of this course but as far as you make is concerned both are equivalent but some problems can be much better solved with virtual work some problem much better solved with Newton's laws thanks 1139 go ahead for your question in that trust problem okay in the trust problem you have shown the d theta so d theta is same for both sides yes yes yes I had shown it same for both sides okay so I did not explain that so let me briefly explain you want the answer to that right why is it same on both sides yeah so the it is the reason is this okay sorry sorry I should have mentioned that think about this okay point E okay is common to this triangle also and to this triangle also now let us say that if we apply delta theta 1 here and delta theta 2 here then what will happen is that that point E when you look at this triangle will move down by delta theta 1 into a whereas when you look at point E from this triangle it will move down by delta theta a into delta theta 2 and if delta theta 1 is not equal to delta theta 2 what you will see that there is a split that will come that these two triangles will separate from each other and when they separate from each other the internal force that is acting here we will also do some virtual work and that has to be taken into account so in order to prevent that we make sure that both this delta 2 delta theta are the same and as a result is point E which is common to these both triangles moves together that is the reason so that we are not creating any openings in the structure is that okay one more question yeah theta is the original angle theta is the original angle theta is the original angle theta is the change the theta I should not say like this okay so let me put it this way okay so I think this is the I can see I see where you are going I see where I'm going so let me put it this way that this delta theta is not actually the change in this angle okay it is not change in that delta that these two lines okay one line and two line that angle is not changing on the other hand we are giving this entire triangle a rigid rotation about point A. So I think that may be a point of confusion I agree. It is not that delta theta is the change in theta of that angle I should have made this as some delta alpha. This angle theta it does not change okay. So please sorry for that confusion. This angle is theta but this delta theta is not the change in that angle on the other hand it is a complete rigid rotation given to this triangle for rotation about A. Because a triangle you cannot change its angle okay without deforming the members okay because triangle is rigid okay. Is that point clear? That is C A E is theta and C dash A E dash is also theta. No which one which one? C A E C A E is theta. Yes yes that is also theta yes yes yes and C dash this is C dash okay. So C dash A yes C dash E E prime is also theta yes this is C dash this okay because it is a triangle for change in the angle of a triangle C prime A E prime yes that is also theta because for a triangle triangle is rigid what does that mean that in order to change that relative angle we need to deform one or more of the members okay. Which you do not want to do here because then in the internal forces we will also do work. Okay so that both angles remain the same. Thank you sir. Yeah thanks. Okay so we take a break for 20 minutes.