Trusses || Introduction & Basic Concept ||

hi everyone how's life this is one us and I'm back again with another video on engineering mechanics and today in this session we are going to be taking up crosses what they exactly are what is the math how the axial forces are worked out or in which members the axial forces work out a zero all of these things and much more coming up in today's session now first of all to just start off you first need to have a feel of the concept you need to have a feel of this structure which is known as a stress first of all let me just read out a definition for all of you so that you can you can get to start a relationship with truss so just listen to this definition so it's it's a system of connected members which have been arranged in a certain manner to serve two purposes those two purposes are number one number one to withstand the loads load withstanding let me just try this load withstand okay wait stand or you can also say in a certain way you have to support the loads and the second purpose that these huge system of different members the system of combination do is that they share the loads share and transfer the loads so these are basically the two purposes or the two objectives the two objectives of a truss let me just like the two objective of a truss now there are plenty of examples which I can give you but before before giving you those examples let me tell you something well so far in mechanic's whatever we learn or so far in mechanic's what are the what is the nature of the force that you have actually dealt with in case of a general case of forces on a plane in case of even in friction and even in some other problems even in when I taught you the lecture based on work there also in all the problems the one thing which was common between the forces that those forces all those forces were of external nature all those forces were of external nature let me just try to extend but in this case when you talk about trusses when you talk about system of members connected in a certain way okay in that case the forces well they are going to be internal the reason being very simple because because of this internal force we can actually work out the internal stresses and from these internal stresses we can actually work out the dimensions of the members dimension of members for designing a particular structure for designing a particular structure a structure or a truss you can say and there are some other assumptions also I will come to that in this article but for now let me just summarize what is trust Trust is nothing but a system of members connected in a particular way to to serve two objectives number one being is to support the loads whatever load is placed at the joint all the members should be such they they need to support the loads and secondly to share in the transfer the loads so these are the two objectives of any structure of any trust in particular let's see now let us talk about some of the beautiful examples of trust let me just wrap this so that we can move forward and let me just show you let me just show you I'm not going to be drawing them rather I will be using this software to show you some examples so here they are but there are going to be two examples and there is an example within an example first of all just take a look at this some of the roof trusses now this is something that you would find in Switzerland or even Kashmir to the roof of the houses okay the anatomy is something like this if you take a look at this this is nothing but a roof truss but a prat a roof truss okay then take a look at this one this is how a roof truss and finally we have this Fink roof truss beautiful is it now let us take a look at some of the bridge trusses our trusses are pretty much pretty much implementable in case of a bridge design and just take a look at this this is Pratt then we've got how then you have got a Ferren we've got pal Timor we've got a truss all of these are beautiful designs and in order to work out these designs carefully you need to understand the axial forces if I were to say axial forces that won't be right the correct word would be internal axial forces that is something which we need to work out in each and every problem of plane truss why am I telling plane truss because the truss are the problems of trust that we are going to be working out will be such that the members will be along a particular plane and preferably if this is X and this is y we are going to be working out all the forces acting along the members of a truss on the X Y plane let me make it very clear we are going to be dealing with all the problems based on a plane stress and if I get some time I will be making videos on space stress also that is a 3d version of a truss you can say that so these are basically all the examples of a roof truss as I told you kashmir switzerlan roof houses and not to me and then we've got preach trust in the form of water and baltimore k truss beautifully ok one more thing now let us let us get into the details let me just write something about the math of this so these two topics are over what stress you know system of connected members arranged in a certain particular way to serve two purposes one is to withstand the load and second is to transfer the load okay then let us talk about the math of dress because whenever you try to arrange members in its particular way they may collapse also they might not collapse they may collapse they might be rigid they might not be rigid a particular truss can be defined as a perfect dress and a particular truss cannot be defined as a particular truss or a perfect rest let's say so is there a way is there a math is a mathematical equation which in help us which can really help us in determining which truss has let's say which stress is perfectly rigid or let's say which stress is a perfect one which stress is collapsible that means on application of load it will collapse that is something which is absolutely undesirable okay as an engineer you need to make sure that the truss should be non collapsible it may be over rigid but it has to be non collapsible okay so now I'm gonna be writing this equation I just just take a look at this equation M plus 3 equals to 2 G ok so what Sam let me just write this first of all M is nothing but the number of members number of members secondly what about j j is nothing but the number of joints so you were to use this formula always whenever you come across any problem based on trust and whether if you need to identify whether it is a collapsible truss or a non collapsible tress whether it is a rigid truss or a non rigid truss or an even what do you call under rejecters you need to use this equation and let me tell you how this equation can essentially be used first of all M is the number of members J is the number of joints so if you've got a trust like this let's say you've got our support over here then you've got a roller support here okay and then you've got another one you need to work out whether this is a perfect truss or not so you simply need to use this equation if if if m plus 3 is equal to 2j it's the case of a perfect US perfect truss you can also call it a KA you can also call it a rigid truss okay so let us try to work out whether this over here is a perfect trust or not so how many members so there are 1 2 & 3 let me just write this the number of members are 3 so so so M is going to be equal to 3 how many joints 1 2 3 very simple n is equal to 3 just like to plug in all these values into this equation if you do so in that case LHS is equal to how much M that is 3 3 plus 3 is 6 LH is going to be equal to 6 and if you put that value of n over here not n this is M this is number of joints that is J so 3 times of 2 is 6 so LHS is equal to 6 and RHS is also equal to 6 so when LHS is equal to RHS then that is the case of perfect trust then that is the case of a rigid truss okay you don't have to worry as a design engineer you don't have to worry and by the way this topic is extremely important for all the civil engineers ok because civil engineers are very important even for mechanical engineers whenever in the erection of even a plant power plant let's say erection of an industry the startings the starting phase of an industry civil engineers are required might be for surveying and once surveying is done then it is followed by direction of the plant it's very important to set up the plant civil engineers are required and to run that plant continuously to keep the machines operating always you need mechanical engineers ok so that's it and now another one if M plus 3 is greater than 2 J in that case you would call that arrangement as an over rigid one over rigid crust remember all these things these are very fundamental you might come across objective problems where you will be given four options for figures and you need to work out which one of them is an overrated truss and that's how you need to work out ok you can you simply need to you simply need to compare the LHS with the RHS if the LHS is equal to the RHS it's a perfectly rigid truss or a perfect trust me you see if LHS is more than RHS it's an over rigid truss that means that means we'll just write a statement I wrote a beautiful statement has more members than required to be rigid has more members than required to be rigid more members let me just write this members then required to be rigid very beautiful statement let me explain that to you once again so if the value of LHS which works out as greater than RHS in that case it is suggesting that first of all it is an over rigid truss that which it has more members than required to be a perfect truss to be a rigid truss simple and final case M plus three less than two J if this is the case then this is known as an under rigid truss okay make sure that your trust does not fall into this category otherwise there is going to be a lot of devastation okay and you might lose your job so this has less members has less members less members than required for a rigid truss then required for a rigid truss as simple now I just just when I started this topic I told you that designing a stress is not enough okay connecting different members together and making a system of members and calling that a truss is not enough you need to know that on application of load whether that truss is going to collapse or not so let me just make a claim if this is the case when LHS is equal to RHS in that case the truss will be non collapsible let me just write this non collapsible when it's an over rigid case when L it is as greater than R edges again this is an on collapsible keys but when allegis is less than RHS when M plus 3 is less than 2 G in that case that truss is collapsible and when such a thing happens your courier may also collapse as a civil engineer so make sure this never happens ok so this was all about the mathematics of cross now let me talk about all the basic assumptions so so so here we go let me rub all of this and please note all of this down pause the video and silently take a screenshot ok keep on taking screenshots or keep on making notes also your wish happened ok now another important thing let us talk about basic assumptions here we go basic assumptions so I'll be just reading them one by one and they'll pop up in the screen over here so what are the basic assumptions ok so the first assumption is listen to this very carefully and better to write it down the number one assumption is joints are assumed to be in connections something like this something like this there is a member over here and how can I draw it it is another member right below and then there is a pin pin right here so so this pin over here all the joints have to be connected with the help of pins this is basically the assumption for our first assumption ok second one second assumption is that the loads will be applied at the joints only which essentially means load are to be applied at the joints which means it's going to be something like this yeah this is where you can apply the roots this is absolutely correct in case of a truss but when you do this this is not alone you cannot apply the load along the length of the member this is your second assumption okay for a truss okay once this is done let me talk about the third assumption this third assumption is on your screen the members are straight to force members so what do we mean by that to force members members are straight and do force members something like this watch this this is a member and this is also a member so by two Force member I mean to say a force it's a pair I this these forces are such that they are trying to come press the member and the forces could be like this okay like this okay and these are also collinear secondly the forces could be like this they will have try to pull the bar in that case we call it to be a tensile tensile force let's say okay again the members are straight to force members so that was it weight of the members these members their weights are extremely small okay so the weight of the members are not supposed to be taken into account the weights are going to be extremely small what is the next one truss is statically determinate have you ever heard about the sentence not sentence what a word statically determinate something which is statically calculate able statically statically determinate what the hell is this statically determinate what does this mean when you talk about statically you are basically referring to statics okay what is determinate I will tell you from statics we know that there are three equations of equilibrium summation of FX 0 and let me just talk about these two equations summation of FY 0 which essentially means with these two equations with these two equations of statics alone you will be able to work out the reactions as well as the axial forces in members again let me talk about this fifth assumption it says that this truss is statically determinate so what this essentially means is that all the values of axial forces in the members can be worked out simply by applying the equations of statics alone nothing else needs to be worked out you don't need to know how much how much deformation has taken place in any one particular or in any particular member you don't need to know that you just need to apply the equations of statics and you will be able to work out all the axial forces in different members of a given terms if you are able to do so if you are able to implement all these things then that truss is statically determinate we will say that okay so these were all the basic assumptions now tell you let me tell you some some cheat codes which are going to be awesome to help you to help you find find zero force members what is this one about well you just need to it let me wrap this so there are going to be two cheat codes and let me just list them as case one and case two so first of all let us talk about case one so when you talk about case well just take a look at this figure so we've got a member over here exactly and we also have a member here okay done and they've been connected with each other with the help of a simple pin we know that very well so if you've got if you've got a joint that means there is obviously one joint with two members okay and also without any external force without any external force or load let's say load or force whatever you may refer it to us then in such a case in such a case the force amongst these members let's say member 1 and member 2 then in that case f1 is equal to f2 and by the way both of them are going to be equal to 0 listen to this one second actually can trend inside your memory if you've got a joint and if there are two members at a particular joint and there is no force there is no force acting on this joint there is no force acting on this joint in that case in that case this member remember one and remember two they have zero Newton as their axial force is absolutely no actual force in those two members let me just explain you that with the help of an example let's say we put a truss something like this and here we go we've got something like this arrangement yeah done and we've got this let's say we apply a force over here and let's say this is member 1 and this is member 2 and we've got an arrangement like this isn't it in that case what will happen is just take a look f1 if you see this is the joint this is d joint and these are the two members member 1 and member 2 these two members are at the joint and there is no other force external force acting on this joint in that case f1 is going to be equal to f2 which is also going to be equal to zero so this is how you need to apply cheat code one two members at a particular joint with no external force acting at the joint in that case the force in both the members in both the members is going to be equal to so you know this is how I need to apply they remember pause the video take a screenshot okay and once the video is over you can obviously make your notes okay let's talk about case two very interesting and by the way if you don't believe all these things and you can obviously work them out by using the method of joints which I'll be taking up in the next video don't worry everything will follow in a particular sequence and by the way all the problems based on method of joins method of sections they have been covered up the theory portion was left based on trust and lot of the students actually made a comment dead cert to make a theory video on trust that's exactly what I'm doing okay okay case - well in case - it's very interesting just take a look at this so there are going to be three members now remember one member too and this is going to be member three let me just mark them one two and this is three so obviously there is a joint that we are going to be focusing on now this joint there are three members okay connected at this joint right now even here there is no external force on the joint let me state the conditions really properly so we've got three members at the joint okay and two members two members out of the three members two members are collinear collinear if you watch carefully here member one and member to our member one and member two are collinear when that is so and third member third member is well at a certain angle you don't have to worry about that third member will obviously be at a certain angle when that is the case when that is the case then when that is the case then the force in the third member is going to be working out as 0 let me summarize let me summarize there are a whole lot of things in this and this sheet code it goes like this so there are going to be three members acting at a joint fine with no external force acting on this joint fine with two members : ear that is member and member to collinear okay pick on all these points when that is so then the third member member three will have zero Newton as the internal axial force remember these two things these are going to be very important especially especially when you come across objective problems based on plane trusses so these were all the cheat codes okay to cheat codes you can apply them pretty easily that we just explained this would be help of a simple example then you can have a better idea and let me just set up this so make sure if you have not noted this down pause the video take a screenshot having let's do an example okay by taking examples we can actually bolster our concept so very will so you've gotta trust over here let me just draw this it's it's something like this yeah like this here yeah yeah here we go and what else like this like this toy toy done then there are members or members about forces so these were these supports so they've been replaced by these support reactions basically and then there is a one D one single force acting and this is force P now let's say let's say let's take a look at this portion this joint you can clearly see this member over here let's call this as one let's call this is to remember never I remember - they're collinear fine and actually adjoin g joint g if you if you focus carefully let this be in the member three okay so there are three members all right three members then one and two are collinear secondly there is no external force in that case the third member third member or the force in the third member is obviously comfortable toute is zero that means this is the zero member force so this is actually easy to work out this is more of a hit and trial approach and I'll be taking up an example okay and we'll be implementing both the cases and we'll be trying to work out as to which were are the members where the axial force becomes zero let us take up that example so guys here is the problem what you need to do is you need to find the zero force members you need to work out just by the hidden trial approach and just by the to cheat codes which I have given to you with the help of those cheat codes we need to work out the name of those members where the axial force is going to be equal to zero again let me summarize there were two cheat codes one based on two members at a joint while the other based on three members at our joint okay and the in the first case two members were at the joint with no external force on the joint and at the same time those two members were not collinear remember this they were not co-linear but in the third case in the third case there are three members acting at a particular joint with no external force and two memories were collinear and the third member will have zero action for serial Newton action force now let us take a look at this so just try to it is a randomly start from where are we gonna start just take a look at this take a look at this we just take a look at this there are two members a B and V D and there is another member of BC so if you watch carefully a B and BD are co-linear and this is the third member these two are collinear with no external force on B in that case this third member will have zero axial force so via what are funds first answer 0 axial force ok it's it's going to be that simple zero action for so first one is bc b c and where did we focused on the joint it was B so let me just underline second okay so we have worked out essentially one member where the axial force was zero another M it is random you take a look this has become zero okay anything else anything else yeah we've got one more we've got one more so if you watch carefully just try to focus on this member member G so if you watch EG&G eye now on member G there are three members acting that is f G a G and IG there is no external force also on this member on this joint G and at the same time these two members that is member eg and Gi they are collinear that means this third member is going to have zero axial force so the second option is if G and where did we focus we focused on G so that's done anything else let me just think let me just think let me just think every is also done okay just focus on this joint beautiful focus on this joint if you watch carefully HJ & GL these are collinear members and actually there are three members on this very joint chain two of which are collinear and there is even no external force that means this third member that is JK will have zero axial force let me point out J was taken as the joint where we focused okay fourth answer so if you watch carefully if you watch carefully just take a look on this joint team can you work something out okay any member in association with K that is K and K ie KJ is already done and KH one of these members is actually zero so if you watch carefully this has already been utilized JK is already zero okay JK is already zero so I've marked a zero over here so essentially if this is zero and there were initially four members KN kg aah and ki there were actually four members out of which one number is automatically 0 that means we are left with again three members so you can apply cheat code to so if you want to apply cheat code to you're going to have three members klk I and K H three members not and two of which are collinear with no external force acting on this member in that case this third member will have zero axial force this third member will have zero actual force and that is your member H K and where did we focus we focused on this joint okay done very easy another one hie hie even also V zero just take a look at this focus on H just trying to focus on H if you watch carefully H joint H has four members connected to it JH k h IH f h out of which one member is already 0 so essentially they are left with only three members that is JH IH and FH so if you watch carefully fh + JH they are collinear and there is no external force on it that means this third member will have zero axial force in that case this H ie is also be going to be equal to zero the force in each eye is also going to be equal to zero so this is also done very easy when did if we focus we focused on H ok so these are all the zero Force member there is also one more member left zero force members isn't this easy ok so we got one two three four five one more is left which one is left let me just check this over here this focus how many members one two three four out of which one is one is practically zero that means we are left with one two and three members so we got three members on a joint two members are collinear and there is no external force that means this third member will automatically have zero axial force so we've got the answer answer is f I we have focused on AI now some of you guys might be thinking why did not why I did not consider this jointly even if you focus on join T even if you focus on join T you let me see that there are four members and there is already a load acting so you cannot apply either of the cheat code even if you look focused on this joint c.join c has essentially there are three members 1 2 & 3 this one is already 0 so you cannot practically take into use so there are three members three members but there is already a force acting and whenever there is a force acting you cannot apply either of the cheat code this is the basic idea so that's the final call this basically hit internal approach you can keep on analyzing joined by John and then you can work out all the answers pretty easily so this was just a basic introduction as to what the concept of process remember let me just summarize when you talk about a truss you're talking about the system of connected members which has two prime objectives one it is to to support and withstand the load and second is to transfer the loads okay then then we saw some examples we saw some examples based on a roof truss and also bridge truss then we we spoke about the math of cross remember M plus 3 is equal to 2j perfect truss M plus 3 greater than 2j over rigid truss and M plus 3 less than 2 J under rigid truss the initial two options were non collapsible with the final option that is the under rigid truss that was collapsible isn't it then we learnt the basic assumptions basic assumptions beautiful pin connections are there between the members and then we spoke about that the loads have to be applied at the joint only if you don't apply the load Rd joint then we won't consider that member or we won't regard that member as a truss not remember but we won't regard that structure as a truss we spoke about that the members are straight to force members then the weight of the members are negligible and finally the D truss is going to be statically a what do you call determinate which is essentially means that on applying the equations of statics you can work out the reactions as well as the internal forces then you learn to cheat codes one was based on two members cheat code number one said if you've got two members this this at a joint with no external force on the joint and those two members are not collinear in that case both the members are going to have zero shear forces then there was another cheat code which had one joint and three members three members two members were collinear there was no external force in the third member is obviously going to have zero internal axial force and then we apply all of those speed codes to work out this very example so try to do this on a plain sheet of paper and that's it that's it for today I'm gonna see you again with a video on method of joints and that is going to be beautiful beautiful and so I'm gonna be meeting you guys tomorrow until then take care have a nice day keep learning keep watching thank you

Channel: Manas Patnaik

Views: 69,873

Keywords: GATE, ESE, trusses

Id: -1w4_4Sr2kg

Length: 36min 34sec (2194 seconds)

Published: Fri May 17 2019