 I think we will go to the tutorial. So, let us just try to do ourselves a couple of problems. Tutorial one problem is displayed. So, again determine the forces in members f h, e h, e g, l m, m k and l k. So, you could either do by method of joints or you could do by method of sections as well. So, I just need the numbers at least for if we can just work out f h, e h and e g that will be more relevant. Just solve f h, e h and e g, take any method, but again the main point here these 3 members are actually at the middle of the truss. So, therefore, as we all know that method of section would be more appropriate to solve for these 3 forces. If you are done, you can just type your answer in the module or in the chat type your answer in the chat. For f h, e h and e g, I will be interested in the one of these forces at least. Please mention in the bracket tensile or compressive. Tensile or compression should be mentioned. Do not put plus minus sign. As I said plus minus sign is very relevant. Please always indicate your answer by t or c. So, I see some of you are giving the correct answers. So, f h will be 160 kilo Newton compressive. I am displaying the solution because we do not take much time here because as I said you can do it as a homework also, but glad to see that many of you are posting the correct answers. So, the purpose is to solve here. So, let us just look at the solution. It is being displayed already. So, just look at it how the method of sections are being f h, e h and e g can be solved by passing this section as you can see here. So, this has been solved. We can easily take let us say I want to solve for f h. So, I can take a moment about e. Once I take a moment about e, we can solve it. So, 160 kilo Newton compressive because the reason it is negative because I have chosen it already in tension. So, this is already considered positive, but it came out as negative. Therefore, it is a compressive force. Similarly, f e h is 200 kilo Newton and for f l k and f m k you can simply use joint m and joint l. So, that will lead to these two problems m k and l k can be solved by based on method of joints. So, solution will be posted on the module as well. So, we will move to the next one just one more quick exercise as such it will be simple. We understand there may be some problem viewing the solution plus we do not worry the solution is posted on the web. We have main purpose of the tutorial session as I said before also that we want to do some brainstorming together that is all. So, we want to solve a problem in a definite period of time. There may be some homework or exercises that can be done separately you know at home or as a homework, but here we have to get the main concept and we have to find an answer within a definite amount of time. Can we now do this? Let us say I want to find b d e and d e then what kind of section I am going to take. So, I am getting answers from different centers. Both b d e and d e will be tensile force and their answer is 180 kilo Newton. Both b d e and d e they are tensile in nature and magnitude is 180 kilo Newton. Someone just wrote 180 kilo Newton compression that is not correct, should be tensile force. So, the simplest way to solve would be you see the point here is you could get the reaction here, but if you avoid if you try to avoid even solving for the reactions you know you should take first a section through b d d e and e g. So, we should take a section through b d d e and e g and take the upper half of it. So, take the equilibrium of the upper half of the body. So, let us try to look at the solution here. So, solution is displayed here in your video. Try to take a section through b d d e and e g that will expose these forces b d e and e g will be exposed and you can take the upper half of it that means a b e c that portion we have taken out. So, now it is you can see clearly I have to find out f b d e and f d e. So, as I can see clearly just very fast solution I can obtain by simply taking the equilibrium along the x direction. So, if you look at the equilibrium along the x direction. So, that is 180 kilo Newton remember here I have already taken this as a tensile force because this is outward direction. So, member is in tension therefore, I have f d e solved and therefore, now f b d how do I solve f b d simply take moment about e. So, idea all the time is how can I solve the given unknown using one equilibrium condition. So, if you take moment about e then I can solve f b d directly that will be 180 kilo Newton is that clear. So, now next I am going to give a little bit challenging problem and just try to spend maybe few more minutes. So, now you will see that method of joints and method of sections will fall apart or it is difficult just one hint a b c b e f everywhere you have pins and c f b f they are continuous members. So, there is no pins at the intersection. So, b c you know c f b f a d a e they are continuous members there is no pin connection at the intersection. You can attempt to solve this problem using method of joints or method of sections and try to see what is the consequence or you can try to identify what are the simple trusses is that a compound truss can I identify two simple trusses that are jointed together by various members. So, that is the hint once you identify the two simple trusses you can detach them and try to solve this problem. First part you must solve for the reactions from the global equilibrium we can solve for the reactions from the global equilibrium. So, I try to adopt the concept of compound truss here a d e and b c f these are two simple trusses you see basic triangular truss a d e and b c f. Now, simply detach them and try to work on the equilibrium you will see you will see how simple it is to solve for the desired forces. So, far I have only received probably one correct answer and that answer is f a b equals to 0. The force in member a b equals to 0 I got now people are writing actually you need more time just take another two three minutes at least give me one or two forces based on the concept delivered. I think I will start showing the solution as I said try to get the concept some of you are giving correct answer a b is 0 f. So, only thing here I see f a b if you want to f b c will be 0 also. So, just try to see how we have done this problem first we get the reactions definitely. So, the first part here is to get the reactions get the reactions a y will be 0 vertical reaction at a is 0 you will have a x and then you will have a reaction here also. So, now what has been done just detach one of this truss I mean I have detached here a d e if you detach a d e now you will see there are three unknowns that is f c d f e f and f b a. So, therefore these three unknowns can be solved by taking the equilibrium of this. So, the free body that I am really looking for is shown here remember this is a d e which is a basic triangular truss that is the free body I am really looking at. Once you get the reactions from the past part just analyze this one we should be able to get f a b very quickly f c d just take the sum of force along x 0. So, that is the you know basic thing that we have discussed how to solve the compound truss. Now, you can look at the solution afterwards, but ultimately idea was can we solve a compound truss problem that is done. So, I will give one more problem, but that is a homework problem you can look at the next problem and we can discuss this even towards the end of this workshop. So, we can you can simply look at there is another problem I am sharing with you just the next problem is again a compound truss problem. So, this problem you can work it out as a homework problem just take it home we will push the solution, but remember what will happen I will just give a hint a d e is a simple truss and e f i is another simple truss they are connected here. As I said remember there is a hinge here hinge here I have two unknowns here I have two unknowns here problem just cannot be solved using method of section. So, easily or method of joints I cannot start because first I have to solve for the reactions. Now, remember there are two reactions here as we discussed this was in the lecture problem also during the lecture we discussed there are two reactions can you tell can you type in in the chat what is the value of E y what is the reaction in the vertical direction as you detach this two bodies the problem is symmetric. So, just think about if I detach this what is the vertical reaction at y e what is the vertical reaction at e what is the value of E y once that is understood lot of centers now responding the E y if you detach it and draw the free body the vertical reaction will be 0 and that will be the you know starting point to solve this problem then you can try to use method of sections and all that method of joints to get the member forces as such one more thing can you solve for C e without even going to the concept of compound trust can I solve for the C e directly C e can be solved by taking a section through C d C e and E k if I do that take a section through C d C e and E k use method of section here and take a moment about a take a moment about a you see C d and k e will pass through a therefore, you can solve C e even if you do not want to go to the compound trust. Now, once C e is solved now you see I have to solve from here actually to solve the other unknowns like k e and d or whatever other forces are asked. So, that is the approach so one thing is that still you are able to calculate the C e through method of section but to get the other forces now I have to get the horizontal reaction at e to solve for k e I could I should get the horizontal reaction at e.