Design Procedure of Knuckle Joint – Design of Machine


Hello friends here in this video we will see the design procedure for knuckle joint now for that purpose here I have a diagram of a knuckle joint this joint is used in case of tensile applications that is where we are applying pull type of load as it is clearly seen from the diagram knuckle joint consists of single eye end which I have written here it is hollow then we have double eye end now the other name of double eye end it is also called as fork end since it is like a fork now this single eye would be inserted into the double end and then we are having this knuckle pin which will be attached and here also there is a pin so this system becomes a locked system the assembly gets completed next on single iron here we are applying load P which is towards right on double lined here we are applying a load which is towards left so because of these two action of opposite forces in two directions the knuckle joint it gets pulled and this knuckle joint would be designed only for static load not for dynamic load now the major dimensions which I can see here is the rod which has the minimum diameter it is denoted by small D small T is the thickness of this single I on T 1 is the thickness of this double line or Fokin next year I will denote small year small D 1 has a diameter of single iron and fork and the diameter would be same knuckle pin diameter I will keep it equal to the diameter of this rod so now let us analyze the failure one by one since we can see in this diagram small D is the minimum diameter and failure will take place at minimum area so first step number one will be design of this rod where the area is minimum it means knuckle joint there is a possibility in it to fail at this rod first so step number one in design procedure of knuckle joint in this design procedure of knuckle joint are step number one is design of rod and once the rod fails it would be looking like this that is circular in section having its diameter as small D now this is the resisting area and it is a is equal to PI by 4 d square because when we are pulling this rod the failure area which we will see it will be in the form of a circle D is the diameter of this rod so here I have the resisting area after this I can see that therefore considering failure of Rodd under tension now this knuckle joint is used only I’m mentioning only it is used in case of tensile application tensile loads because if we apply compressive load to the knuckle joint there are chances of this joint to slip because it will provide angular motion also so it is applicable only for tensile applications now considering failure of Rod under tension therefore I will write down tensile strength of rod it will be given by now as I have explained strength is nothing but load carrying capacity P and that is equal to stress into area since here we are taking tensile failure under consideration so it will be tensile stress into area so therefore P is equal to area is PI by 4 d square multiplied by Sigma T so here I will call this as equation number 1 and therefore I can say that from Equation 1 B that is the diameter of this rod can be calculated now this completes step number 1 in step number 2 I will be designing this single I end so in step number 2 it is design of single I end now when we are considering single I end yella here I will draw the diagram of it in a 3d manner that how it looks like this diagram which I am drawing it is for a single I end it will look something like this and here we have a rod which is attached to it this single and it is hollow it means the knuckle pin can be inserted here so single line looks something like this now I’ll see that let this outer diameter of the single eye end be denoted by D 1 and this inner diameter is nothing but diameter of knuckle pin here I will say that this is the diameter of knuckle pin denoted by small D now once I have designed in step number 1 the diameter of Rod after that I will say that therefore let diameter of knuckle pin be equal to the diameter of rod so therefore the diameter of knuckle pin diameter of knuckle pin that will be equal to the diameter of rod and diameter of rod is D so here I have this hollow portion where the knuckle pin will be inserted internal diameter of single ion now I am see I will see in how many cases this single iron can fail the first chance is that when I am applying load at the single I end this portion it can split into half means the single I end would be splitting into half now we will see the failure of this single iron there are chances of the single iron to split into two halves like this from its center it would be splitting and when I draw the failure diagram it will look something like this this is the single I end after it has failed under tension and another portion would be in front of it so here the single ion would be splitting into two parts like this so this is a tensile failure and here we are getting this as the resisting area these portion they are the resisting area I will draw this resisting area here now here I’ll see that this is the diameter of knuckle pin which would be inserted that is the internal diameter of single ion this is the external diameter which is denoted by D 1 and here is the thickness which is denoted by T so this is the resisting area so I’ll say that sin considering the first kind of failure that is tensile failure for single I end the resisting area will be since the resisting area capital A that will be equal to D 1 minus D that is area of these potion which I have section so it will be D 1 minus D into T the same resisting area which we have here so therefore tensile strength of single I end strength as we are considering things I so it will be strength is a load carrying capacity that is stress tensile stress into area because tensile strength we are trying to get so therefore P will be equal to area is D 1 minus D into T we have calculated the resisting area multiplied by tensile stress so this is my second equation now I can see that from this equation we can get D 1 which is the outer diameter for this single I end I will say that therefore from Equation number 2 either I can get D or I can get T which is the thickness of single line D or T can be calculated because either I can fix any one dimension by using the relation which is called as empirical relation and then I can find any one of the values so this was regarding the 10 zile consideration after this again in step number 2 that is design of single I end next I will consider sheering failure so considering sharing failure now what is meant by shear is that when we are applying a load to this single I end there are chances of it to slide from here that is this section it will be sheared off so it will look something like this when I draw that so now this is the area and here I say I will say that this is the centre of the knuckle joint the centerline of the knuckle pin now this much is the radius which I’ll say that this is small our radius of knuckle pin and that is nothing but d by 2 and this is the outer radius which I will call it as R 1 that is equal to d 1 by 2 and here I have this as the thickness of single I n so this is the resisting area now after getting this resisting area since I will get 2 pairs of area one will be here another will be behind so I will say that therefore resisting area capital A it will be equal to here I have R 1 minus this R into T area of this rectangle which is shearing so R 1 is d 1 by 2 minus RS d by 2 x 2 since this is a case of double shear and next I will multiply it by T then I can simplify this area becomes this 2 I can take common denominator and it will get cancelled out so area is D 1 minus D into T this is the resisting area under shear so after this I can see that there for shearing strength of single I end shearing strength strength is load since we are finding shearing so that is shear stress into area which is the resisting area so therefore P is equal to area is we have found out D 1 minus D into T and stress is tau so this becomes my equation number 3 so I can see that from Equation number 3 either we can check D 1 D we can calculate the value of D 1 or T or we can check shear stress so we can say that equation number 3 is used for checking of shear stress tau or D 1 now this was the shearing failure of single line next we will see crushing failure crushing of single I end now single iron how it will crush is that from the diagram single ironed and knuckle pin both are made up of metal so here we are seeing the crushing of single I end as it is seen from this diagram single I end is in with contact with the knuckle pin both are made up of metals so when we are applying this tensile load there are chances of metal to metal rubbing between single-aisle and knuckle pin and that results in crushing crushing is a phenomena in which the metal they get converted into small fine powders and if it is so then the object is failing so here I will draw the crushing failure diagram the area which we would be getting it is the projected area which will get crushed like this here t is the thickness of single I end and this is the diameter of knuckle pin denoted by small D so this is crushing area also called as resisting area now I will say that after this this crushing area is area of a rectangle so therefore considering crushing of single I n therefore I will write it on to the next page crushing strength of single I end since via writing strength strength means the load carrying capacity load is denoted by P and we want to calculate crushing strength so therefore it is load is equal to crushing stress into area so therefore area it is d into T multiplied by Sigma C R so this is my equation number 4 and I can say that from this equation number four we can check the value of Sigma C R so from Equation number four Sigma C R that is crushing stress can be checked so here we complete the designing of single I n that is this was step number two now as we have seen for single iron there were three kinds of failure enzyme failure shear failure and crushing failure same three kinds of failure will be there for double I end as well so now in step three we will design the double I end in step number three we have design of double I end now for double I end also the first kind of failure which we will consider it is considering tensile failure of double I end now how it will fail in tension if I can explain it with a diagram double I end is in this way it will look like this if I draw the 3d diagram for it now you’re this double and also carries a rod and it is hollow in section so this is the double I end also called as the fork end now when we are checking for tensile failure it will break into two halves from its centre like this if I can explain that with the help of the original diagram here I have when we are pulling this double iron towards left there are chances that it can break into half from this center line break into two pieces like it was there for single I n so the shearing area will be yeah here the tensile area I can draw it this it will break into two halves like this the area will look like this upper I will break like this into two halves and similar kind of failure we will see at the bottom end also so there are two failing areas I will draw that area one area will be at the top other will be at the bottom it will look like this because here I have drawn this I end it will break into two halves this is the area at the top similarly one area will be there at the bottom so we are getting 2 pairs of area this is the center of the double I end or folk in now the thickness of the folk and this is t1 and this diameter is same as the diameter of single iron that is this is equal to d1 and this is the diameter of knuckle pin D so that I am getting two rectangles here okay it is here and the other rectangle here so I will say that there is two pair of area so since resisting area capital A it will be equal to it is d1 minus small D into t1 now I am multiplying by 2 because here I have two such areas so this is the resisting area now I can see that therefore tensile strength of double I end that will be equal to P is equal to since we want tensile strength it will be tensile stress into area so therefore tensile strength P is equal to area which I have written here I will write this again D 1 minus D into T 1 x 2 into tens is now this becomes my equation number equation number five deviously four equations are over now I will say that therefore from equation number five we can check the value of tensile stress so Sigma T can be checked so this was the first step in the design of double iron considering tensile failure now we will consider shear failure step number three design of double I end considering shear failure now when it is shearing the example is like this since we have the fork India when we are applying a load here there are chances of these two rectangles to shear and how it will look like it will be removed from this surface that is they will be shearing like this Here I am drawing the shearing area I am explaining how when it shears it will look like so this is the sharing area the plates they would be moving away that is when we are pulling the fork in these plates they would come out from the top and from the bottom end so the sharing area is like a rectangle these are two rectangles now if I consider this blue line as the center of the knuckle joint or we can say knuckle pin which I have denoted in the diagram here this is the center line I will measure the distances from the center so from the centre of knuckle pin till the outer radius this is the outer radius of double-lined which is equal to the outer radius of single I end so here I will mark the dimensions this is the outer radius which is R 1 that is equal to D 1 by 2 next this is the inner radius which is R is equal to D by 2 so I am getting the area in such a way that at the top also it is double shear and even at the bottom we are having double shear so now the resisting area will be therefore resisting area a it will be equal to this is outer radius minus inner radius that is d 1 by 2 minus D by 2 into this is the thickness of double I end that is t1 and here also this is this thickness is T 1 so this D 1 minus D 1 by 2 minus D by 2 into thickness now I will multiply by 2 for the first upper fork and again multiply by 2 because at both the cases we have double shear so this two will get common and which get cancelled out so the final area is 2 into D 1 minus D into T one of the two will get cancelled out so this is the resisting area so I will say that after this therefore shearing strength of double I end it will be equal to strength is the load carrying capacity as we know shearing therefore it is shear stress into area so this is the strength equation so finally I can say that P is equal to area we have got 2 into D 1 minus D into T multiplied by shear stress so this is equation number 6 and I can say that from equation number 6 tau that is shear stress can be calculated so after we have seen shearing failure of double iron next we will see crushing failure of double I end so again in step number 3 we have the design of double I end that is considering crushing failure I’ll draw the diagram for that when it will crush it will look something like this at the top at the bottom I have to crushing area and it would be crushed in such a way that is these dots which I am showing these are actually fine powders of metal which would be formed when the double iron gets crushed so here are the fine powders of metal which would be formed and when these fine powders are formed it means the double iron would be failing so here is the thickness which is T 1 for top and bottom folk next here after getting this T 1 next D is the diameter as we can see for this double iron when it is crushing it is in contact with the knuckle pin so here I have diameter of knuckle pin that is equal to small D so this D into T 1 which is here the thickness this is the crushing area because it is in contact with this knuckle pin so it will crush form fine powders here so after this I will say that this is the diameter of knuckle pin so therefore these are the failing area or I can say crushing area so these crushing area they are a is equal to D into T 1 and like this I have one more area so multiply it by 2 so this is the crushing area therefore crushing strength of double I end that will be equal to since strength is load is equal to here we are taking crushing into consideration so load is equal to crushing stress which is Sigma CR into area that is the resisting area so now therefore I will say that P is equal to area 2 into D into t1 into crushing stress so this is equation number 8 and I can say that therefore from Equation number 8 Sigma CR that is crushing stress can be checked so this completes our step number 3 that is design of double line is over now the last component which is left it is the knuckle pin so I will say that next is step number 4 design of knuckle pin knuckle pin it will be failing under shear so considering sheering failure of knuckle pin now here I will draw the diagram how the pin fails very quickly now knuckle pin would be sharing in this form that is it will be breaking into three parts as we can see one portion the middle portion of the knuckle pin is with the single I n so this portion will share with single ion and the top and bottom portion are with double ion so they will go towards left and it will be sharing it this way so the shearing area is a circle here I can draw the cross section area which is shearing D is the diameter for this knuckle pin so therefore resisting area a is equal to PI by 4 d square since it is a case of double shear because we have one sharing area here one is at the bottom so this is a case of double shear so here I have to multiply it by two and finally I can say that therefore shearing strength of knuckle pin that will be equal to load P is equal to stress since we are considering sheering so we have shear stress into area so therefore P is equal to area is here this is 2 into PI by 4 d square multiplied by tau which is the shear stress and this gives me equation number 9 so here I can say that therefore from equation number 9 either D that is the diameter of knuckle pin or tau can be calculated so with this we complete the four steps and in this video we have seen the designing of knuckle joint into four easy steps that is starting with the design of road design of single ion design of double iron and finally the design of knuckle plane and I hope whatever the explanation was there that is understood in a simplified manner

100 thoughts on “Design Procedure of Knuckle Joint – Design of Machine

  1. Yesterday was my exam on design of machine elements and I think I wrote my exam well, your multiple videos saved me, you saved me

    Thank you sir !!

  2. How is P called the tensile strength? I thought maximum stress value was called strength. Shouldn’t P be maximum load withstanding capacity?

  3. thanku very much sir i really enjoy your teaching technic …..i understand whole part of this topic as well as you teach.

  4. Thanks a lot sir dme was a. Mountain for me to cross Away but only because of you i am abl to crossing it…. U t the hero u r a (ropeway) for crossing the mountain 💐🔥

  5. Sir please upload videos on the topic Gears,brakes ,clutches etc..
    your videos are very easy to understand…
    Last semester ur videos saved me from backlog…

  6. Sir Jo double eye end me shear meaap area liye hai usme bhi to 't1' hi aayega na just like tension in double me ya sirf 't' aayega

  7. Sir you have expalind toatal 8 equation but you mentioned in your video 9 equation you have skip eqation 7

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  9. Sirr great explaination sir. I haven't got this in my college but after watching this video I find it very simple. Thanks for such a great video. 🙏🙏🙏

  10. DOM subject main jo drawing hai vo dimensions ke sath hogi ya free hand drawing hogi I am in 5th semester now

  11. what about bending failure of the pin????? (And Sir, i like the way you explain but please stop giving 10 20 ads in between a single video as it is really distracting and irritating.. Thank you)

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