 What we are going to do now I think we should try to look at some of the tutorial problems maybe two three problems and then we can see actually how everything goes on and we can try to explore the partially constrained improperly constrained and properly constrained. This is what looks to ask that that particular aspect may be dropped and we can just go through these simple exercises as a tutorial part and then we will see that whether that can be taken into account. Since we have covered the method of joints method of sections and compound truss. So, let us just you know try to go through these problems first ok. We go through that tutorial problems quickly just go through these few problems maybe you know we do not want to solve it completely. I just need some part answer let us say one member or two members if we can just solve those then we will know that for sure everyone is in same page. Hello sir. Please tell compressive or tensile ok. Mk tensile 100, Lm compressive 80, Lk tensile 120. Sir fmk. Lk is how much? Lk compressive 120. Yeah that is ok. Sorry sir compressive 120. Mk 100 tensile. Compression, compression sir. Mk should be compressive. Flm 80 tension. Yes ok so this is done let us say everyone got it I mean is it ok. So shall we move to the next one. Just simple exercises we want to go through just give me the force in member bd let us say in fact all of this can be solved. So what kind of you know we can use the method of section what kind of section is required. Fbd. Fbd please tell tensile or compression. Tensile or compressive. It is tensile. So answer should be should be tensile 180 and how it is done that section we have taken through bd, de and eg. If we take this section through bd, deb and eg then take the moment about e right. So we take the moment about e by taking this section bd, de and eg. So the answer should be Fbd is 180 kilo Newton. Tensile. Yes so this is the section we are looking at that is how the free body of this look like ok. So move to the next one. So again just one quick aspect of it that ad, ad, ae these members are continuous ok. There is no intersection ad, ae, bf and cf all are continuous members. If we can get the force Fab at least that is more than enough to solve this problem. Just get the force Ab that is just one of these forces. Get the reactions first ok get the reactions first because there is a roller and a hinge. So all the reactions can be solved in this problem. So statically determinate problem also m plus r equals to 2n. Just take the equilibrium of the entire truss to find out the reactions at the supports. What is Ay by the way? Ay. Ay is 0, Ax is 40 in this direction right. Ax you have to show the direction also ok. It should be leftward Ax is 40 kilo Newton leftward and Fy is equals to 70 kilo Newton. Is that solved by everyone? Ok so 70 kilo Newton upward in Fy. Ay should be equals to 0 and Ax should be equals to 40 kilo Newton leftward ok. Now we want to really find out the force in Ab. So can we identify you know the basic trusses here or rather simple truss. So can everyone identify that there are two trusses? Simple trusses or basic truss. So we have ade as a basic truss and Fbc as a basic truss. We need to take these two trusses out and we have to show this all the forces on those two trusses ok. So draw the free body diagram of these two trusses. So all the forces. Sir Fab 70s. Fab would be 0. Fe, Fe is 70 very good ok. So Ab is equals to 0. So just draw the free body diagram of this particular simple truss ade just draw the free body diagram. Take that truss out and draw the free body diagram. And remember in that truss what are the unknown forces I have. See if I take this truss out let us say I want to take this truss out I already know the reactions. So only unknowns are Fab is one unknown, Fdc is one unknown and Fef should be another unknown right. So there are three unknowns and you can actually find all of these if we detach that simple truss ade. I do not call it simple truss but it is a basic truss now ok it is just a triangle. Is that clear? Just detaching this truss we have already solved for Ax and Ey right that was already solved at the beginning. So now once we do sum of force along x0 we should get Fcd equals to 0 first of all. Then we take a moment about A to solve for Fef and then we can take force along y equals to 0 to solve for BA. So BA will be equals to 0. So there is a should be here this equation there is a equation here. So this one is really coming from sum of force along y equals to 0 after we solve this two. There is no section it is a compound truss problem we discussed. See now once these forces are solved you can actually do the joint equilibrium to get the others ok. So it can be combined also the basic you know idea here is that we need to detach the simple trusses get the unknowns exposed. And it is always true if you have two simple trusses in if the problem is statically determinant then we should have six unknowns. Six unknowns should be exposed overall ok. Is that clear compound truss ok? So shall we move on to one more but you know the problem this problem is going to be a little bit messy in the in the regard that it is going to be very complex geometry. But we do not necessarily have to solve all of these forces let us say but can we get the idea how to solve for the forces that are asked in the question ok. Because as such geometry would be bit complex and it will take long time to solve. So we will just try to understand the procedure how to get the forces. Remember there are two pins here. So ultimately if you want to solve for the reactions from the global equilibrium it is not possible ok. So you have to detach it again. So if I detach this truss let us say what is E y? Once I detach these two trusses and let us say I am asking that at the joint E right once I detach it I have E x and E y that will be transferred to the others simple truss also. What is E y? What is the value of E y? You detach it first. So what is happening I am detaching it. Since it was connected by the pin once I detach I show it E x E y right in one truss in the other truss it will be just reversed ok. So what is that value E y I am looking at? E y should be 0 y it is a symmetric problem right. So that is a very important part here because if we have to solve for the reactions we have to detach this problem. By the way can we get the force in C E? Can we get the force in C E? How do you solve it? Is it possible through method of section very quickly? Just a hint using method of section even I do not have to you know detach the truss as such. Can I solve for F C? That is a good method of section exercise actually. Hello sir yes sir F C will be same as E x. So if you take moment about A that will give you E x directly and that will be equal to F C. So why F C should be equals to E x? That line of action is same but there are two other members right. They are equally inclined. So you can do the joint equilibrium here. The other way around. Sir one can even take the section. What I am asking let us say we do not find the E x even can I find C E? Sir we can even take the section along C D, C E and K E cutting those members. That is what exact then. Yeah take moment about A you will get directly C E. See I really do not have to detach the truss if I want to solve for C E. See I can really run a section through C D, C E and K E. So just cut it and then what do we do? We take the moment about A because we can see that C D and K B these two forces are passing through A. They will not contribute to the moment. So see C E can be solved even you know going getting into compound truss and do the procedure we follow ok. So that is one clue even C E can now let us say we solve for E x from there also if we try to detach this truss and solve for E x we can get the equilibrium of this joint. But there will be three unknowns right here and that will actually not solve the problem for C E in that way right. That one is because that will be only equal and opposite. So you have to go on the top maybe you solve for the D and then you come back ok. So the idea is that if you use the method of section we can get the C E. But in order to get A K and K E then we have to do something else ok. So now you have to really look at what is your E x from E x once we detach this truss let us say I detach this truss then I can actually find what is A x right and A y then we can come and do a joint equilibrium here to find what is A K A B right. A B is in the question or not? A B was not in the question so K E ok. So then we have to actually go to these joints individually A K was there yes A K was there no. So quickly you know C E is found right just using the method of section let us say we find the C E first. So that is found. At E we can find out a horizontal reaction. Yes. Then draw the free body diagram of a entire left part. Yes. At A hinge support is there are two reactions are there. So we can find out those support reactions. That is what that is what we have to do exactly. Yes and then take the section C D C E and K E. No but see C D C E and K E that section it is irrespective of if I want to find out let us say C E right it is irrespective of you know whether I get the reactions here or not just understand that part. If I want to get the member for C E it is irrespective of whether I find the reactions here or not because what I am doing I will just cut this right and take the moment about A. Yes. But if I want to find out K E then you have to go on that way right. Yes. So we now have to find out what is E X. Yes. And once we find the E X then we can find what is A X. Yes and Y also. Use the Y also. And then take the section and get the remaining also by applying conditions of A. Yes. So that is the that is the approach is. Thank you. So the compound trust as I said that you know we have to detach the trust to in order for get this connection forces those are very important. Remember UI will be 0 ok just because of the symmetry. So cut this and we get the E X can be found actually. Sum of moment about A if you do E X can be found. E X will be 0.943 times L again it is a very complex geometry is there. And from E X I can find out horizontal it is actually should be A X ok. So A X can be found out right. And then you can do whatever way you like it I mean you can do the joint equilibrium from one or you can even do method of sections because ultimately basic structure is solved because reactions are solved that is very important ok. So that this problem was basically given the purpose was that ok we have a compound trust but maybe some members in a force can be solved even I do not want to get into the compound trust issues.