Design Procedure of Cotter Joint , Socket and Spigot Joint – Design of Machine


Hello Friends here in this video we will see the design procedure of cotter joint for that purpose here I have a diagram of the cotter joint now this diagram it is also called as cotter joint or socket and spigot joint now as we see in this diagram socket and spigot joint it consists of three main parts the first is this socket which I have drawn it is red in color here this red portion shows the socket inside that socket we have another part called as spigot or spigot end which is inserted into the socket the third part is the quarter which we are using to assemble the socket and speaker so now when we have this kind of joint it is used in both the applications that is when we have tensile load which is a pull type of load and when we have compressive load which is a push type of load so in both the applications this joint is applicable so now if I look into this diagram here I have this small D this is the diameter of the rod which is connected I can see this black portion it is nothing but spigot so here I have one rod which is attached to the spigot and then another rod which is attached to the socket now here we can see this is load P applied on both the sides that is here it is applied to the spigot end and here it is applied to the socketed so this diagram shows socket and speaker joint under tension here we are applying pull type of force and as I have said before this joint is applicable in both the applications that is in tensile as well as in compressive applications so now when we have D as the diameter of the rod here I will explain all the notations now this D 1 it is the outer diameter of the socket as we can see here socket is a member or we can say it is a part which is hollow it means it will have two diameters one external other internal diameter so the external diameter of the socket is d1 as we can see here and internal diameter of socket is equal to d2 now this internal diameter of socket is equal to the external diameter of spigot which is d2 so d1 is the outer diameter of socket d2 is zero diameter of socket next here D 3 d3 is the diameter of this spigot collar this end which I am attaching here this is called as the spigot collar d4 it has a diameter of socket collar as we can see here d4 is the diameter of socket collar next B this is called as mean width of quarter now quarter is that joint which is having slope on one side and it is straight on the other side so we take the mean with of quarter as B next we have a which is the distance from the end of slot here the slot is ending to the end of rod this is a next we have see C is approximately taken as the thickness of socket collar II it is the thickness of socket next p1 is the thickness of spigot collar and then T will be the thickness of quarter which is thickness is not visible in this view now after I have explained this diagram next we can start with the designing procedure for this quarter joint now how to design the quarter joint first of all that is the question see whenever we have any component we have to check where the failure will take place it means failure is going to take place at that region where we have internal weakness or if we have small area because if the area is small at that location stress will be maximum and the material will be failing at that small area so how an object fails it will fail due to internal cracks it will fail if there is hole or we can say any kind of slot in that part an object can fail if it has less area so here before designing this soccer dance we got joint or also called a squatter joint if I look at this diagram D is the diameter which has the least value out of all the diameters this diameter of rod on the spigot and diameter of rod on the socket they have the smallest diameter so when we will pull this socket and speaker joint there are chances of this rod to break under tension so we will start this by designing the rod first so I will start with the step number one this is design procedure of quarter joint so now step number one as I have explained that rod is the area which has minimum diameter so the socket and speaker joint it would be failing along the rod so I will say that the first step is design of rod now when I am starting with the design of rod if I can explain it here that once we are pulling this rod there are chances of this rod to break along this diameter and when I see the area after breaking it will look like the cross-section area I will draw the diagram here I have diameter D and this is the failing area so this failing area of the rod is called as resisting area and this resisting area as I can see it is area of a circle so it will be PI by 4 into d square next if I want to design the rod then I need to know the strength of this rod that is how much load this rod can resist so I will say that there for strength of rod considering tensile failure is given by now when I am writing strength see strength means the load which any machine part can resist here in this case we are having a rod so we need to find out how much is the load carrying capacity of this rod for that purpose I can use the relation that since stress which is denoted by Sigma is equal to load upon area so therefore load becomes stress into area now this small concept I will be using here as well I will write down therefore since I have written strength strength means the load carrying capacity here it is denoted by P strength is equal to it is stress into area here area is PI by 4 d square multiplied by stress is since I am considering tensile failure so stress would be tensile stress so this completes my step number one and I will give this as equation number one I will say that therefore from equation number one D that is diameter of Rod can be calculated this completes our step number one now after completing step one that is we have designed the rod now after designing this rod now I will design this spigot end which is here it is shown in black I design this spigot so my step number two it is design of speed god and now see speaker 10 it is a possibility that spigot may fail under three conditions the first one is tension second is shear and third is crushing I will explain them one by one so starting with the first one that is considering failure of spigot under tension now this is the first type of failure for the spigot I will draw the diagram I will explain just the 3d diagram how a spigot looks like spigot is a member which is inserted into the socket and Here I am drawing this spigot collar and finally the rod which is attached now and this spigot will have a slot here like this now this is the diagram of the spigot it looks like this and when I am considering failure of spigot under tension in that way what happens as I have explained that failure occurs at that location where we are having holes slot any irregularity cracks or minimum dimensions now if I look at the spigot end here there is a slot in which the Cotter is inserted so there is a possibility that spigot may fail along this slot that is it can break when the load is applied that is when the spigot is pulled from this end there are chances that it can break into two halves at this slot and when I turn after it has filled it has broken into two parts when I turn it then the area which we are seeing it would be called as a resisting area now I’ll draw the diagram of that resisting area resisting area would be circular in section it would be looking something like this now this is the area which we will get after the spigot has broken into two parts here I can give the dimensions to these d2 is the diameter which is of the spigot and here this slot we can see this is the thickness of quarter which is inserted into this slot now this area is called as resisting area now this resisting area it can be written as capital a is equal to from the complete circle we will subtract this rectangular portion because this portion is hollow it is a slot so here area will be PI by 4 D 2 square that is from the area of circle I will be subtracting this area of rectangle which is d 2 into T so this is the resisting area for spigot which fails under tension now I’ll write down strength of speak out because I want to know how much load it can resist in tension so that is called a strength and therefore strength of spiggott under tension as I have explained strength is nothing but load and that load is nothing that is stress into area so therefore load P will be equal to here I have got the area I will write its value PI by 4 D 2 square minus D 2 into T this area will be multiplied by stress and here as we can see we are considering the failure under tension so the stress which we have that will be tensile stress so this is my second equation and I can say that from this equation from equation number 2 D 2 or T that is either the outer diameter of the spigot or thickness of quarter this thickness here thickness of quarter can be calculated so this was the first step in my step number 2 that is Here I am finding the possibilities in which the spigot can fail the first possibility is failure under tension now I will see considering failure of spigot end under shear that is the second part in this step so next I will say that again this is our step number 2 that is design of spigot and now in this the second kind of failure I am considering it is considering sheering failure of spigot now I will explain how the shear failure takes place so you in this diagram itself since here I have this 3d diagram now I am checking the possibilities of failure at first there was a possibility that spigot can split into two parts at this slot now the second possibility is when I am applying this load P there are chances of this portion of the spigot to go along with the load and then what happened since here I have quarter in between so this portion which is d2 into a this will slide in the opposite direction so that is called a shearing that is if I apply the load on to towards right then as there is the presence of quarter this much portion will go along with the load and d2 into a will slide in the opposite direction so this is called a shearing so here the area it will share in this way I can show it on this diagram itself this slot will be extended like this since this is a curved portion it can slide in this manner this is the shearing action here which I have drawn that is in this entire slot portion after this it will be removed so this slice comes out and that is called as shearing action now I will draw this resisting area the resisting area is it is in the form of a rectangle where I have this distance horizontal distance as a this vertical distance is the diameter of spigot which is d2 and here the area which I will get this is the shearing area and here this is a case of double shear because shearing takes place on both the side of this plate so this is the shearing area which is in double shear as I have explained this portion will slide out so as I am getting one A one surface here another surface is behind so it is a case of double shear I can say that therefore this is the resisting area so resisting area capital A it will be equal to 2 I am using 2 here multiplying by 2 because it is a case of double shear shearing on both the sides of the plate so that is d2 into a this is the shearing area so I can say that therefore shearing strength of spiggott that will be given by C load strength is nothing but the load carrying capacity and it is stress into area here since I am considering shearing failure so the stress will be shear stress denoted by tau multiplied by resisting area now after getting the shearing strength I can say that therefore P will be equal to area is 2 into D 2 into a multiplied by stress which is shear stress and this becomes my third equation now I can say that from Equation number 3 this distance a can be calculated so now in step two we have seen two kinds of failure first was tensile second was shearing now the third kind of failure of spiggott that is called as crushing so again we are into step number two that is the design of spigot the third kind of failure is considering crushing of spigot end now when I am considering crashing I will first draw the area and then explain this area which I am drawing it is the resisting area here this spots which I am showing here these are the crushing of spigot since d2 is the diameter of spigot here I have t as the thickness of this slot now I can explain crushing in this 3d diagram see here we have a slot and when the Cotter is inserted in this slot like I can superimpose the quarter here now when we have this quarter it will result in there are chances since quarter is also made up of metal this spigot is also made up of metal so when there is metal to metal rubbing that is called as crushing and because of that this spigot would be converted into fine powder that is at this slot some pieces of spigot will get disintegrated so here it looks like this when it is crushed so this is the crushing area and I will say that there for crushing area is the resisting area so resisting area capital a that is equal to it is area of a rectangle so d 2 into T now there for crushing strength of spigot or spigot that will be given by as I have told strength means the load carrying capacity so here I am finding the crushing load carrying capacity and that will be P is equal to it is stress into area here since it is crushing I am considering since we are taking crushing into account so load will be crushing stress into area so finally load will be equal to area is D 2 into T crashing stress is Sigma CR I will call this as my fourth equation and I can see that from Equation number 4 Sigma CR that is crushing stress can be checked so with this I complete step number 2 that is the design of spiggott now after I have designed speak out one thing is left that is called as the spigot collar now I will design this this is the spigot collar so after step number 2 my step number 3 is design of spigot collar now when I see this spigot collar there are chances when the load is applied spigot collar will remain at its position but this portion d2 it can go along with the load shearing the spigot collar here that is at the junction it is here where it will be shearing that is when the load is moving in this direction it will take this spigot along with it and then there will be shearing of this we got collar so now this is like shearing action so I will say that spigot collar is failing so considering shearing failure of spigot Koller now how it fails that diagram I will draw so that we can understand that how the spigot collar will look when it fails this is the failure diagram here I can explain it in a very clear manner that is this spigot end it goes ahead along with the load and here again there is a slot in this spigot portion so now it is very much clear by looking at this diagram that at first the spigot was looking like this when the load is trying to pull this spigot end it will shear this we got collar in this way the spigot will move ahead and this is the spigot collar so this is the shearing action we can say for the spigot collar now if I denote this outer diameter as d2 and actually it is d2 so now as I am considering shearing failure of spigot collar so this is the shearing area this is the shearing area which we have I will say that let p1 be the thickness of this spigot collar this shearing area is like this is an example like if we have some papers and we want to punch a hole in that paper that is also like a shearing action that punch is actually we can assume it like a spigot and thickness of papers is like spigot collar and we want to punch this so this example of the spigot collar shearing is like an example of number of sheets of paper in which we are punching a hole so in that case the area would be which is the shearing area that will be this circumference which is PI into D 2 multiplied by the thickness T 1 this is the shearing area so I will say that therefore shearing strength of spigot collar since I am finding shearing strength strength means load it is stress into area as we are interested in shearing so this will be shear stress tau into area so therefore P will be equal to area is PI D 2 into T 1 here PI D 2 into T 1 multiplied by shear stress so this becomes my fifth equation and I can say that from this equation number 5 the thickness of spigot collar can be found out so from Equation number 5 P 1 can be calculated so this completes my step number 3 now I am moving forward to step number 4 and in that we will be designing the socket design Oh socket this is my step number four now I will explain I will just give a quick recap of the diagram here was the socket in speaker joint this black portion is the spigot along with the rod and spigot collar we have already designed it now we will be designing this red portion that is the socket and again as I have taken three kinds of failure for spigot failure in tension failure in shear and failure in crushing the same three kinds of failure I will analyze for the socket as well so let us get started with it design of socket first considering failure of socket under tension now I will draw the resisting area of the socket when it fails under tension and then I’ll explain how that area has come here again since socket is also a circular portion this is the resisting area I can draw the section here this red portion it shows the material whereas whichever is hollow here we do not have any material so now your d1 is the external diameter of socket and d2 is its internal diameter which is also the outer diameter of spigot this is left as the thickness of slot or we can say thickness of quarter so now this is the resisting area which we have now how we are getting this area is that when we have this socket hence we got joint and load is applied at the socket socket moves towards left so as I have explained failure takes place where there is internal slot there are holes cracks or small area so if we can see socket also has a slot so it will be failing along this slot that is when the load is applied it can break along this slot into two halves and after breaking how the area looks like it will be like this so now I can say that this resisting area is I’ll denote it by capital a that is the area now it is we can say PI by 4 outer diameter square that is d1 square – internal diameter square that is d2 square from the hollow circle I would be subtracting these Oshin now these are nothing but from the hollow potion how to get this potion since this total height is d1 from d1 if I subtract d2 I will be getting these two regions and multiplying them by T so here it is minus d1 minus d2 will give me these heights and multiplying it by T so this is the resisting area which we are getting here so I will say that therefore since we are finding the failure under tension so tensile strength of socket strength is load which is stress into area here we are finding the tensile strength so it will be tensile stress into area so therefore P is equal to area is here I have already calculated it is PI by 4 D 1 square minus D 2 square minus D 1 minus D 2 into T this complete is the resisting area and multiplying it by tensile stress so this equation which I have got I will call this as equation number 6 so from equation number 6 I can find out the outer diameter of socket which is D 1 D 1 can be calculated now again in step number 4 now I will consider the second type of failure of socket that is failure of socket under shear it is step number four design of socket in that the second condition that is considering shearing failure of socket now when the socket is shearing I’ll explain it again on to the diagram since we are applying load on the socket towards left because of this load what can happen this portion of socket can go along with the load and here this portion will be shearing so now we are checking how this socket fails under shear when the load is applied since I have a quarter here this left portion of the socket can go along with the load but the portion which is here it will go into the opposite direction it will be a shearing action like I am pulling this socket and this portion it gets on to the right hand side so here d 4 minus D 2 into this C which is the thickness of this socket collar this is the shearing area I will explain it here I am drawing the diagram for the socket under shearing failure this is the failing area or we can say resisting area now this internal diameter is D – that is the internal diameter of socket and its external diameter is d 4 and here the thickness of socket this is C this area which I am getting it is a case of double shear now why it is double shear is that as socket spigot whatever we are using here they are all circular in shape so if it is shearing from one side as we can see the surface here another surface is there behind so that two surfaces are there and that is why it is the case of double shear so this is the resisting area I can say that there for resisting area capital a it will be equal to it is the area of these rectangles at the centre we do not have any material it is hollow so it is d 4 minus D 2 into T here I have C so I will write down C into C and I am multiplying it by 2 because this is a case of double shear shearing on both the sides so here this is the area which I am getting so after this I will say that as we are considering shearing so therefore shearing strength of socket since we are finding strength when that two shearing strength so therefore strength means load and that is equal to stress into area your stress will be shear stress into area the resisting area and now therefore this P will be equal to area is it is 2 into D 4 minus D 2 into C which is the ad I have written here now multiplying it by shear stress tau so this becomes my equation number 7 so I can say that from therefore from equation 7 either d4 or C whichever of them is known first or C can be calculated so this was the second step in step number 4 considering shearing failure of socket now as I have taken first figured three kinds of failure failure under tension shear and crushing same failures I am taking for socket failure of tension is over failure and shearing is over now failure under crushing so again step number four that is the design of socket considering failure under crushing now how crushing takes place since we have quarter which is made up of metal socket will also be made up of metal as we can see this land portion it is in contact with the socket so there are chances when the load is applied there will be metal to metal rubbing at this socket in and it will fail under crushing so I’ll draw the crushing failure area which will be the resisting area here this portion would be crushed as I as we know in case of spigot crushing means there will be small fine metal powders which get developed and then the material has failed so now this external diameter that is D for internal diameter is d 2 and here I have thickness of quarter so therefore resisting area capital a it will be equal to D 4 minus D 2 into T that is area of this these two rectangular portion that is the failing areum so after that I can say that there for crushing strength of socket strength P is equal to stress into area as we are taking crushing failure so stress will be crushing stress Sigma CR into area so therefore P becomes area is d 4 minus D 2 into T multiplied by Sigma CR now this is equation number 8 and I can say that therefore from Equation 8 d 4 can be calculated it means first we would be considering crushing failure we would be getting the value of D 4 next shearing pillar that is when we are designing the socket and speaker joint first we will get d 4 and then by considering shearing failure we will get the value of C as I have explained in equation number 7 now this completes the design of socket the last step which is step number 5 that is the design of quarter because as we know incidence we got joined we have three main parts so socket spigot and quarter so step number five will explain the design of quarter now I can explain here that since quarter is this middle portion which I have drawn in blue when I am applying the load at the spigot it is going towards right to the socket it is going towards left because of these two opposite forces when I am pulling this rod there are chances of this quarter to shear so as we can see here when I am pulling this spigot and socket in opposite directions one portion of quarter will go along with the spigot other two portion will go along with the socket so this is a kind of shearing failure it will break at the junction so how it breaks that diagram I will draw that is when we are pulling this quarter it will fail under shearing it will fail in this manner that is it will break into three parts and the area which I am getting these are the shearing area so I’ll see that for design of quarter considering shearing failure of quarter this is the shearing failure one portion this portion in the middle it will go along with the spigot spigot that is two spigot and these two portion they will go to socket socket is pulling the top and bottom portion spigot is pulling the middle portion so the failure has taken place and here as we can see we are getting one surface here one surface will be below so this is an example of double shear and so the resisting area if I convert this if I can explain it in a more simple way that is the area which I will be getting here it will be in the form of a rectangle this is the mean width of quarter because it is breaking from middle so this is the mean width here I have thickness of quarter so this is the resisting area since it is double shear so therefore a will be equal to B into T multiplied by two since it is a case of double sheer so therefore shearing strength of quarter strength is P is equal to stress into area as we are taking shearing of quarter so therefore stress will be shear stress multiplied by area so therefore P will be equal to area is 2 into B into T stress is tau I will call this as equation number 9 so therefore from equation number 9 B that is mean width of quarter can be calculated so here I complete step number 5 that is design of quarter we had started with the design of rod that was the design procedure we were designing socket and spigot end or it is also called as quarter joint in that the first thing was rod so in step number five we have in step number one we have started with designing of rod next in step number two it was the design of spigot end we have seen the failure of spigot in three ways that is in tension shear and crushing next are step number four it was design of socket and before that step 3 it was design of spigot collar I have explained each and every terms here how it fails under shearing next step number four it was the design of socket and finally the step number five which we have seen just now design of quarter so with these five steps we can easily design the socket and speaker joint also called as the quarter joint and with this we complete this design and this video I hope everything is understood

100 thoughts on “Design Procedure of Cotter Joint , Socket and Spigot Joint – Design of Machine

  1. Brilliant sir… I dont really praise people that much but you are too damn good i couldn't resist writing this comment

  2. Sir etna bhi kiya jrurat h advertisement ki ap to student ko pda nhi rhi kmane ka jriya bna rhe ho.. sir plz ek hi advertisement rhne do etni kiya jrurat h ..?. Bekar bat h

  3. Sir in step no.3 Design of Spigot collar aapne sirf Shearing btaya Crushing nhi crushing nhi hoga to d3 kaise find hoga

  4. thanks a lot sir,this is even better teaching style as compared with our college teacher,i covered a 1 chapeter.thanks a lot

  5. Sir your explanation is very very clean and good.. and I need sleeve and cotter design process as in this manner sir please……..

  6. 14:03 how can u say the length of the rectangle is exactly equal to the diameter of circle?? The length of rectangle will be little small than the dia of circle.

  7. 23:41 i think all the faces of cotter which is in contact with spigot and socket will be feelings crushing stress and will be converted in fine powders so why only rectangular cross-section is taken there?

  8. Dear sir,
    How to calculate d3 and what is the gap between the cotter add C.

    Note:
    I have one doubt cotter inserted means cotter and c portions will be coincide their is no gap here.

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  10. sir failure of spigot collar in crushing ??? not explained!
    crushing failure =. P / (π/4)×(d3^2-d2^2)

  11. Flattered…what a lecture…it didn't felt like it was 51 min long….I was totally into it…finally learned cotter joint and it's design

  12. Design of beding stress in cotter please explain that one also and sleeve and cotter joints also explain sir I follow all ur vedios Sir

  13. Sir me aapke sab videos dekhta hu. Bahot helpful hote hey. But agar aap sab write-up ka kam pahile hi ek sheet pe kar le to video ki length choti ho jayegi. Aur fir video jyada lengthy or boring nahi lagega. Bas drawing explain karte huye hi nikalna. Usee sab achhe se samaj me aayega.👍 (Aap drawing bahot achhe se nikalte ho)👌

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