 Okay, we can just complete the method of joints since we have you know gone through many questions now. So one of the important issue remember I again repeat the discussion on the internal rigidity that is very very important. The reason we are choosing a simple truss is because it is internally rigid. So what happens this truss internally rigid means it can rotate about point to overhaul let us say but it cannot no angle can change internally and the truss will actually collapse okay. So that is the basis of internal rigidity but internal rigidity does not guarantee that the truss is going to be stable unless otherwise you support it properly think of it. If I just give it a roller if I say let us say I have a roller at A and roller at C and I apply a horizontal force at B is that a stable truss answer is no the reason is I have a roller here let us say I have a roller here and I am applying a horizontal load. So there is no external reaction in the horizontal direction that will balance the truss. But remember that internally it is still rigid internally it is not going to change any configuration even if it is collapsing. So that is internal rigidity so it will collapse as a rigid body okay. So going back to the method of joints again we will just take up now simple problem. So we have studied that for a given force P if I just apply a force P there may be members which can go to 0 okay and we know how to decide on those 0 force members. Now let us try to study this problem so ultimately we have again a simple truss you can just look at the basis I can start from a basic triangular truss and I can form this truss with the concept we have developed just extrude 2 members and go on like that and also you can see that M equals to 2N minus 3 that is satisfied on this truss okay. So M equals to 2N minus 3 is a thumb rule for the truss to be a simple truss but remember it has to always made on the basis of 2 members extruded and then connected by a pin. So now let us look at how do I solve this problem so 5 kN, 20 kN, 5 and 12 these are the forces that is coming in the joint okay and now you have a hinge support here and we have a roller support here. Remember what is happening in this problem we have only vertical loads therefore reaction should also be vertical there is no horizontal reaction fine. So this is the truss is symmetric about this axis that means the member forces B A must be equals to member force B C, B D should be equals to B F and E D should be equals to E F. Now what importantly comes that when I am going to talk about method of joints there are 2 things to discuss here either I can assume that all the members are in tension to begin with to begin the process I can just say all the members are in tension that means when I am going to the joint equilibrium that means I am taking the equilibrium of joint then I will show all the forces outward from the joint is that clear. So that is the conventions you can now choose that let us say I choose tension to be positive however you may want to choose you know tension to be negative also but point that I am trying to say just let us say we strict ourselves to tension to be positive and we carry this forward. So for the time being we are bypassing the information on the nature of force that means if I can identify that the member is actually in compression I could show that inward into the joint but I am not going to do that at this point. So I will show all the forces member forces to be tensile in the nature and I will say that tensile force is positive okay and I am going to carry forward the sign as we move from one joint to the other joint. So firstly as we know that I can since it is a simple truss therefore the entire truss is a rigid body therefore I can calculate the reactions I have no problem by taking the global equilibrium okay so I will calculate the reactions 21 and 21 okay. So next thing is that remember I will always start the process from where I have two member forces unknown that means AB and AD are unknown here. So therefore I will go to this joint and remember these forces applied as it is but member forces are taken as tensile I do not know what the outcome would be I will know the answer when I do the joint equilibrium but remember I have used tension to be always positive and tension must be shown outward to the joint. So what will happen if I do the equilibrium of joint A I will get FAB equals to 0 sum of force along X equals to 0 will give FAB equals to 0 that was also my special condition okay. However if I do sum of force along Y equals to 0 then FAD is coming to be negative 5 since I have assumed tension to be positive then negative 5 has to be compression and what I will do I will not change the sign in the next step suppose I want to go to this joint I am not going to change the sign okay and we will show this later on. So next thing is that since this joint is done remember due to symmetry FAB is equals to FBC okay that need not to be done the next joint I have gone also joint E. So let us say joint E will solve for the problem of the member force FEB. So FEB all the forces remember I am always showing as a tensile force with a positive okay if after performing the equilibrium I get a negative answer then it is actually a compressive force okay. So ultimately you can see that FEB is going to be 12 similarly now what I can do since I have gone to a complex joint I have gone to joint B remember FBA is solved right FBA is 0 same as FBC FEF is already solved that is FBE right FBE is 12 that was in tension. So what I will do I will again show everything as a tensile here as I have shown here so all are positive as such and I am going to carry forward whatever sign I got from the previous step of these forces so I just set the joint equilibrium again. So remember if I do a horizontal joint equilibrium then FBD is equals to FBF and for the vertical equilibrium now pay attention to this vertical equilibrium suggest FBD and FBF are equal so 2 times FBD cosine of theta plus 20 plus FBE equals to 0 I will simply take the value of FBE from this substitute here with sign not the magnitude we also with sign if it would have been compressive I am going to take the negative sign from here and substitute back here my ultimate outcome would be FBD must be equals to FBA equals to negative 34 clear. So ultimately in the process we have actually solved for all the member forces okay now let us look at joint D I am yet to calculate I think FBD is not calculated FED was not calculated right so to calculate the FED just check what I have done here to calculate FED that was my still unknown I have taken help of joint D joint D remember has a support reaction that is coming the way it is this is external load but FAD FBD FD everything was shown as tension although I know this is in compression and I am trying to find out what is the answer for this this is in compression this is in compression but I showed it as a tension I then use the equation like this this is the horizontal equilibrium right okay so FDE plus FDB cosine of this angle should be equals to 0 remember interestingly FDB that sign I have taken with sign so negative 34 will be substituted and therefore FDE will become positive that means FDE is actually in tension clear so the one way of doing it that you do not change the nature of force as you go from the joint to joint just strict with your sign convention whatever you get with sign you bring it to the next step do not reverse the force on the joint do not reverse the direction of the force on the joint that means if you got a compressive force in the previous joint do not just say it is inward in the joint and do the equilibrium in that case remember if you do inward then actually you have already incorporated the sign so during the equilibrium sign is not coming into play because equilibrium does not know sign okay now check would be whether this joint is actually under equilibrium okay then did I actually satisfied that vertical equilibrium is 0 if I check that you can clearly see yes absolutely correct so if I do the vertical equilibrium and bring in all the signs here then you should get it equals to 0 clear so this is assuming all member forces are tensile at the beginning and we are carrying we are choosing tension to be positive from the very beginning and we are consistently adopting that towards the you know thoroughly in the analysis the other way of solving it as I said that you can first calculate one joint and you can reverse the you know show it in the direction it is actually be in the next joint okay so likewise let us look at the second problem so in the second problem what we have essentially this is the truss again remember it is a simple truss it is internally rigid and it is supported properly you can see clearly that you have one hinge here one roller here okay so there is no way that it will you can make this structure unstable no matter how you apply the force it is always going to sustain that load okay so now you see that how I start the process process is simple that first I try to calculate the reaction because entire truss is a rigid body I should be able to get the reactions okay then what do I do remember that I have only two unknown forces here right so I can start from this joint and then subsequently if this joint is solved I can go to join D solve for these two then I can move to join this one solve for these two come here solve for these two and then we can do again a cross check right here okay so in that way methodology will be systematic now here only thing I am saying that I am not going to adopt the sign convention that I have done in the beginning so here I am going to strict myself with a nature of force nature of force means is it a tensile or compressive and I am going to always show that in the joint accordingly so the to start with as I said calculate the reactions so I am skipping this because everyone knows this is a rigid body and how to get the equilibrium of this that has already been done into the equilibrium so those reactions can be obtained then you look at it the joint a if I start from joint a what is happening if I want to draw this force polygon that means here it is a triangle I can clearly use the triangle rule even naturally what will happen you can clearly see that this member will be in tension and this member will be in compression so AB will be a tensile and AD will be under compression okay so therefore you can also use the equilibrium of this to get the value of AB and AD okay so whenever it is shown like this this is AD that means on the joint it has to be inward to the joint and here FAB the outcome will be it is outward from the joint so I get this answer then I move to the next joint can I do that triangle again so I know already FAD right to make it close circuit I must have FDB like this the directions are already frozen you cannot otherwise close the circuit you could also do a you know sum of force along x0 sum of force along y0 okay so ultimately what is happening in this case using the triangle rule you can also get the answer the point that I am trying to make remember what has happened since I know FAD from this which was in under compression I have actually shown this like that in this joint so look at a joint D I have actually reversed right so it was nature of force that is being shown that is a compressive force on AD on the joint it is inward so let us now move on to joint so that is just using the force polygon or triangle rule if you want to do however you can always do sum of force along x sum of force along y so let us look at joint B now if I go to joint B what are my known known was already FAB that was known see I am showing it as a tensile because it was tension similarly FBD was found as a tensile tension so I have shown is as a tension now I have no choice I have to assume what these two are however you can use also force polygon does not matter this will give you the direction a priori for example in this case both will be actually in tension you can show that using the equilibrium so sum of force along x0 sum of force along y0 I can clearly see if you even use polygon concept this will be in tension this will be under compression so now you move on to this joint joint E so in the joint E again remember once I go to joint E I already know the solution this solution was known this solution was known right so it was shown so this was compressive force so it is showing acting in this joint in compressive manner that is inward in the joint similarly this was compressive inward in the joint I do not know the solution of this yet let us say then I just do sum of force along x equals to 0 I get this answer so it is again going to be a compressive if you do a polygon it you are going to find that FEC FEC has to be in under compression because to close the polygon okay so next what we can do we can simply check at C so we can do a check at C that whether the equilibrium is satisfied so these are the two checks I am doing to say that okay joint equilibrium is satisfied at C so the point that I am trying to make here or what we have studied that I have two choices the first choice is I can start with all joints you know all forces are to be under tension I will use positive notation for that and I will keep following that sign convention throughout the joint equilibrium the other way of doing it first I start from a joint where there are only two unknown forces I decide on their sign first that means using the polygon rule or even joint equilibrium once those two nature of the forces are frozen I can go to the next joint where again two member forces are unknown but on that joint I am reversing the you know sign you know I am taking I am showing the force depending on their nature that means on the joint if it is compressive it should be in what if it is tensile it should be out clear so any question regarding method of joints now 1178 1178 Sir if it is find out the internal hinge whether it is acting at any where how to create with that the internal hinge was solved in the 2D equilibrium or not usually these are typically sure question is internal hinge if there is internal hinge then how I solve the problem right see internal hinges are typically used you know kind of guard or beams okay the remember what internal hinge cannot take the moment the only issue is that internal hinge cannot take the moment so it is a almost like a pin joint okay so when you say let us say you have a beam and two beams are connected by a internal hinge and you apply a load then again you can treat them as two separate rigid bodies okay so you can simply treat that two separate rigid bodies but remember when you disjoint these two bodies when you disconnect the two bodies right then you have only the vertical reaction or horizontal reaction that is going from one body to the other body okay now think like this so I will just I just want to use just to deliver the concept of internal hinge see the for internal hinge typically so let us assume that this end is fixed okay so how do you separate this body let us say this now first of all remember for internal hinge if I have an internal hinge here okay so what happens you see that first of all make sure whether it is a statically determinate problem or not that is the first thing how do you ensure it is a statically determinate or not so fixed end here we will have three reactions and you detach the body here I have one reaction right so in internal hinge since there is no movement the force will be transferred like this so you have only two reactions horizontal reaction and vertical reactions remember as per Newton third's law what are my this is equals to this this is equals to this so ultimately what I have one two three four five six so total six unknowns and two rigid bodies okay so two rigid body that means three equilibrium equations each okay so I have really total six equations so this is a perfectly statically determinate problem and I can solve it now in truss we do not say it is an internal hinge but think of like this and we are going to come to this suppose I have a truss like this it has internal member but remember let us say you know this is a simple truss I have one truss like this then I have another truss like this okay now if I have something like this so there let us say this is a simple truss and this is a simple truss okay now there is no concept of internal hinge but suppose I take these two trusses and joint this two together here so therefore what will happen you can clearly see that see therefore we can have a truss I can have a truss like this and then I can have a truss like this okay now they are jointed together therefore you know something like this will come okay so now this is as you are saying it could be called as an hinge here this is internal hinge now to make it perfectly stable what I need I need appropriate support but concept remains same now you are treating as such two rigid bodies right as if I have two rigid bodies and therefore I require six unknowns this will already take two unknowns this will already take two unknowns then I can support it properly by saying I have hinge here and I have hinge here okay and here remember at this particular joint what is happening and we are actually going to study this next we are going to do that next also so for internal hinge analysis remember that we do not have any moment about there however we have only two reactions that will be transferred from the one body to the other body okay okay so we will we are going to study the internal hinge also when we talk about compound truss just hold on to that okay okay but for beam and for truss they will be different okay but mechanism remains same mechanism is going to be same okay now what we are going to do we are going to go to a next analysis technique that is method of section okay so we can again analyze a truss by method of section now why do we need it although we have method of joints which can solve for the truss if it is a statically determinate the method of sections however used to solve let us say a particular member force or a few member forces in a truss okay now systematically what it does if I want to get a solution of few member forces in a truss it may not be desired that the truss is too long and I am asking somewhere you know some forces in the middle it may not be desired that I start analyze the truss from one end and go and try to find out the member forces you know at the middle of it okay so that is the whole idea so method of joints in that case can be very lengthy process but method of joints can solve the problem so what we are going to do we are going to now take method of sections so what we do idea is that we will run a section and we will treat the bodies that are on the left side as a rigid body and the right side also as a rigid body so if I cut few members here I am going to expose the member forces okay so if I expose the member forces then I can again simple treat this one as a rigid body remember this was a simple truss I made a cut here right so this part is a rigid body this part is a rigid body because both of them still are simple truss rather they are basic triangular truss now so therefore now due to this cut due to this section that is running nn I am going to have fbd fb and fc okay so these are the three forces that are exposed I have my unknown forces exposed then suppose now I want to solve for what is the force in bd can I solve it what is the force in bd remember this rigid body now have three unknowns and you can always solve it by sum of force along x0 sum of force along y0 and moment about a point equals to 0 right that is the whole concept now how do I solve fbd in a fraction of second I can simply take the moment about point e so join these two forces at point e and take the moment about e and that will solve for fbd in terms of p1 and p2 so let us look at this concept carefully now for some bigger problem okay so remember you can take different sections depending on what you are going to solve however in a section if I have three unknowns if I have three unknowns I should be able to solve for all the three unknowns because the left free body diagram or the right free body diagram we will have only three unknowns and they can be solved using the standard equilibrium equations that is sum of force along x0 sum of force along y equals to 0 and moment about a point equals to 0 now in this problem it is the example of a pink truss how I am going to solve it the force in member cd ce and bd needs to be solved so cd ce and bd if I want to solve this firstly remember this is again a simple truss and you can say that it is a thorough rigid body here proper rigid body supported by you know the the reactions here is hinge and here you have a roller so I am going to get two reactions here okay and this can be first solved since it has a symmetrical loading pattern therefore you can say that total load divided by 2 will be the reaction here total load divided by 2 will be the reaction here okay there is no horizontal force so suddenly ax would be equals to 0 so now how do I cut the section in order to find out bd ce and ce and cd basically as we see we have to run a section like this so that these three member forces are exposed so once we expose these three members as we see here fbd fcd and fc is exposed right now we can treat this small part fbc as a rigid body so now this becomes a rigid body subject to these set of forces where I have three forces as unknown and therefore I should be able to solve it so all the three forces can now be solved okay so how they are solved now there are few tricks involved for example can I get the solution for fce can fce be solved directly the answer is yes how do I do that by taking moment about d if I take moment about d remember these two unknowns are passing through d therefore they will not contribute to the moment so we take the moment okay and then what we can see is that this is the force in that member remember the answer that is coming is actually positive that means it is on the direction that is chosen and it is shown it was taken earlier as tensile force so it remains as a tensile force clear similarly what we can do sum of moment about a equals to 0 now if I take sum of moment about a equals to 0 okay then fbd and fc is not going to contribute any moment about a therefore fcd can be solved the answer will be a plus 6.25 that means it is going to be tensile that means it is actually the direction that was chosen that direction stands so therefore it is a tensile force now if you simply do sum of force along y equals to 0 then we are going to solve for the fbd okay and fbd will come out to be a compressive there will be a negative sign in this case okay it is going to be negative here just and therefore that will be a compressive because we have taken fbd as tensile earlier right so therefore if I get a negative sign that will be compressive force so ideally it is very simple so ultimately I am just cutting a section exposing three unknowns and taking the equilibrium of the left free body or the right free body now in this case I have taken the left free body because that will make it simpler I could have also taken the right free body that would be more complicated to analyze because there are many external loads on the right segment now how about this now this is a k trace inverted k trace you can see here so what is shown here just let us say one side of it one frame of it and it is a k trace all of these are pin joint as it is now you can see that I am interested in finding out force in member g j and i k that means this force right here and i k right here now can I solve it by method of joints answer is yes to solve by method of joints remember I could have started from this joint right here and progressively I can solve for all the forces but that will be very tedious because I am interested in finding out g j ok so I have to solve all the forces in order to get into this g j because although I will know the reactions I cannot start from any of these joints even if I know the reactions because it is a rigid body simple truss and you can see even if I start from here I cannot start the method of joint because there are three unknown forces so method of section will be very appropriate but how do I take the section remember my objective is to find out g j and i h and it is not necessary that a section has to be always a straight line so I can take a section in a very different manner so that these two unknown forces can be solved so remember what we can do think of just detaching this part just simply try to detach this part so that means this portion I need to cut out this triangle g j and k because I know the reactions once I take this out then the type of section now I am talking about is kind of curved section that section will run through g j g h h i and i k now that will solve the problem and we will see interestingly what is happening so that is my section now the section is x prime x okay and once I do this section what is happening remember in the process what will happen I will have this truss intact this is my basic triangular truss I will expose this force that is g j h g h i and k i these are the forces going to be exposed now you think of this is a rigid body it will have its own reactions so reactions will be j x k y and j y so that we can solve from the equilibrium of the entire truss at the beginning okay then now interestingly what will happen if I just consider the free body diagram of this truss with the forces unknown forces and the reactions acting and I take a moment about point g so if I take a moment about point g this force and these two forces are not going to contribute okay so only thing that is coming up is I have one unknown force to solve that is k i take a component multiply by this distance right so if I take the moment about this point only k y is my unknown all others are going to be known and I can solve for k i okay similarly to solve for g j I can simply take moment about i if I take moment about i then you can clearly see that g j can be solved okay so the idea is that method of section when we are choosing the method of section depending on what is demanded in the problem the unknown forces that is demanded in the problem we have to choose the section appropriately in order to get those forces and it is not necessary that I have to always expose you know I have to always take a straight cut I can also make a card section to get the unknowns right that is desired and here also you have observed that in the process I am actually exposing four forces four unknown forces are exposed right although I can still solve for two unknown forces that is g j and i k so method of section is clear now