noc20_me45_lec09

Hello and welcome to the 9th lecture of the Machining Science course . So, let me remind you that we have discussed the Merchant circle diagram and how to find out the stress strain rate ok and how to find out theangle which is the share plane angle and how to measure the shear plane angle and how to analytically find out the shear plane angle . I would like to make a small correction here that in the shear strain rate we said that this is delta S by delta Y into 1 by delta t . So, we can write this not like this, but delta S by delta t I am correcting that please make that correction delta t this is actually the V s right, this is the V s and this will be instead of delta t therefore, it will be divided by 1 by delta Y. So, please make this correction that this can be written as delta S by delta t which is V s that is how we can find out this is the V s and this has to be instead of delta t this has to be therefore, delta Y ok . So, what we can find out is the finally, the shear strain rate is equal to V s divided by delta Y . So, if we know the V s for any given delta Y you can find out what is the strain rate. So, now based on whatever we have discussed so far I would like todiscuss some of the numerical examples. So, that you could find out that how the marchants theory or whatever we have learned so far can be applied in practice. For example a mild steel workpiece is being machined at a cutting speed of 200 meter per minute with a tool which is given in this way . So, this is if you understand that this is in the ASA if you remember we have discussed it it is in the coordinate system . So, this is the back rake side rake and so on so forth. So, all these angles are given 6 angles and this is the tool nose radius in inch . The depth of cut and uncut thickness are 0.5 millimeter and 0.2 millimeter respectively, if the average value of coefficient of friction between the chip and the tool is given to be 0.5 and the shear stress of the work material is 400 Newton per millimeter square , then determine the shear plane angle and determine the cutting and the thrust component of the machining force. We know the Merchants Merchant's First Relationship . So, from here we can say that thethis also we can know that this is tan inverse of mu this is coefficient of friction which is given as 0.5 . So, you put the value of 0.5 and you find out what is the value of the lambda which is the friction angle by the way there it was beta, but this is the friction angle lambda here it is made . So, from here we can find out that the phi is equal to 90 minus this is given as the phi all right and phi we can find out from the from this value sorry this is given as 26.57 we found out is the lambda and this is the rake angle. So, the rake angle is given as the 8 degree. Now, why not to take that 0 degree, because this is the back rake angle, this is the side rake angle and as you understand that the side rake angle we said it influences the cutting force in the power not the back rake angle remember that . So, therefore, the alpha s is taken as the alpha value and this is put in here as 8 degree, lambda we found out as 26.57 we have put it in here from the Merchant's first relationship we can find out the phi. We have to find out the phi because then the F s will be according to the formula that we have seen earlier, w is d by sin gamma p we have seen that also in thecutting diameter, this is the plane like this ok shear plane and here we have said that this is the d this is the depth of cut or this is equivalent also we said the t 1 this is in the orthogonal case all right. This is the tool and here we said that the width would be the width of the material all right . So, therefore, w will be the depth of cut divided by sin of the gamma p, gamma p is the principal cutting edge angle and if this is the tool, then this is the principal cutting edge and this is the gamma p ok . So, therefore, from here we can find out what is the value of the w this is that is the width and knowing the width all other values are given, t 1 is given, tau s is given, phi we found out and we know that then F s will be this . If we know the F s then theresultant force are from the Merchants we can find out which is equal to F s divided by this, you remember that we said F s is equal to R and the cos of phi plus lambda minus alpha. This is the friction force they are sorry friction angle, there we said as beta by the way ok the same thing beta or lambda. So, this is the friction angle . Now, from here we can find out what is the resultant force, if we know the resultant force all right and from here we know the the formula of the F c and the F t again from the Merchants all right . Then the F c can be found out as R into cos of lambda minus alpha, R we found out lambda we found out 26.57 and alpha is given as 8. So, this is the F c . F t will be R of this and R we found out , lambda we found out, alpha we alpha is given so it can be found out. There is an alternate way you can see that this way also we can find out it will the result will be the same , that is you take the formula which we have derived during the Merchant circle diagram discussion and from here all these values will be known because F s we found out put the value of the F s and so on and we will find out that these values are the same . So, this is an alternate way of solving this problem. Next example problem 2, during orthogonal turning of a mild steel work piece of 20 millimeter diameter with the 150 rpm and with a 0 degree rake angle , the forces normal to the shear plane and thrust are found to be 85.6 and 35 , normal force is 85.6 Newton and thrust is 35 thrust force. If the chip thickness is twice the uncut thickness, estimate the power consumption in Watt . Now,first let us find out what is the cutting velocity, this is pi, D, N all these are given. So, it will be this value ok either it is in meter per minute or you can convert it into meter per second . Then F c which is from Merchants ok, this formula F because the F n is equal to F c sin phi plus F t cos phi we have derived them, from here the F c is equal to this, this is given that chip thickness ratio is 0.5 because the chip thickness is twice the uncut thickness. So, it is this therefore, for 0 degree rake angle it is given here all right we can find out the tan phi is equal to r which is 0.5 which is we found out from here. Therefore phi is equal to tan inverse of 0.5 is 26.56 degree and then we can find out the F c because the F n is given, F t is given, phi we found out ok and we can find out therefore, the value of the F c, V c, V c we found and F c is here. So, product of this and this will be the power. So, this is in Newton meter per second which is what ok and there is an alternate solution you can also see that that this way using the Merchant's diagram or the Merchant's formula we can also find out and the the you know result will be the same power . Third problem is for an orthogonal cutting the following data are given, cutting force is given, thrust force is given, rake angle is given , chip thickness and width of cut shear plane angle these are given, determine strength of the workpiece material friction angle without using Merchant's First Relationship ok . Now, the phi we know that we have derived this formula from here we can find out that what is the value of the r which is 0.419 here it is coming which ist 1 by t 2 this is the chip thickness ratio we are talking about r, now it is given the t 2 is 1.5 millimeter t 1 is this therefore, and because r we found out. So, chip thickness ratio into the into thet 1. So, this is t into t 2 sorry this is the t 1 . So, therefore, the area is this which we can also find out as 3.088 millimeter square and then the F s can be written in this way this is the formula taken from the Merchants all right and we have the values known F c F t phi is known . Therefore, the strength of material will be F s force upon area which is the strength and we know the F s we know the as we can find out the um strength of material F s we can find out from here this is given how to find out the F s because we found out here . F is equal to F c sin phi sin alpha plus F t cos alpha this is directly taken from marchants and all these values are now known we can find out what is the value of the F which is the friction force and the normal force ok . So, therefore, friction force divided by normal force will be the coefficient of friction and here the mu can be found out by F by N hence the friction angle which is the lambda tan inverse of mu which is this . So, whatever has been asked this we found out simply here knowing the uh you know the relationship that we have already derived earlier ok. Next to this we will be discussing another model which is based on another principal rather than the Merchants principal. So, once again let me remind you that the Merchants principal is based on the minimum power consumption ok and here this is thin zone model also, because they are assuming that the deformation is occurring in a thin zone and as I have said earlier that normally the plastic deformation occurs in a zone and not in in a plane ok . So, they are saying that this zone is very thin and this can be assumed to be a plane like in the Merchants . So, this solution is based on the following assumptions that the deformation takes place under the plane strain condition , deformation occurs in a thin zone as I said. So, this is another assumption this is the shear plane and not the shear zone the workpiece material is rigid perfectly plastic and it is behavior is independent of temperature and strain rate this is the third assumption . Next they are assuming that the inertia effects are negligible which is a steady state deformation , tooltip is sharp; that means, there is no nose radius and the continuous chip is found all right , stresses are uniform on the rake face as in case of the Merchants . So, as you can see that this is also a 2 D I mean this is not a 3 dimensional it is a 2 dimensional because they are saying that the tool is sharp mu is constant ok and the stresses are uniform and so on. Now, the cutting forces are transmitted, here what this is the principal is this cutting force are transmitted through the triangular plastic zone ABC ok , meaning that they are saying that if they are there is a resultant force acting on the chip from the workpiece through the shear plane , for this to be transmitted towards the rake face of the tool it has to go through a triangular plastic zone ABC where no deformation occurs, because they considered that there must be a stress field within the chip to transmit the cutting forces from the shear plane to the tool face. Once again I will explain it to you the the resultant force has to be transmitted from here to here ok . Now this transmission can take place through a plastic zone where no deformation occurs because of this force then only this can be transmitted if this force creates deformation it will be observed. So, therefore, they are saying that this is the ABC is the plastic zone where the material is plastically deformed to the maximum ok , in the ABC the entire material is in the plastic state stressed up to the yield point, yield point B in that diagram that we have discussed in the very beginning. Shear plane AB this is the shear plane is a slip line since maximum shear stress occurs here, as you can see thatAB is the shear plane and we said that the maximum shear deformation occurs along the shear plane . Other slip lines must be perpendicular to this line ok these lines are also the slip lines, because this is maximally stressed as we said ABC ok this is the plastic zone within the chip through which the forces are transmitted from the work to the tool ok. BC this one is the free surface since no force is transmitted to the chip after it has crossed the line BC , you understand what I was saying is that this is maximally stress and the force is transmitted that is it up to here and then there is no force transmitted there the C chip is moving ok because of the force transmitted up to this. So, BC is the free surface since no force is transmitted to the chip after it has crossed the line BC . Slip line must meet the surface at a 45 degree this is the slip line theory and in the theory you can find out that the free surface here BC makes an angle of 45 degree with the slip line that is the AB . AB is the slip line that makes an angle of 45 degree with the free surface which is BC, this is according to the slip line field theory all right. So, we are not discussing the slip line field theory in details . So, this is the application of the slip line field theory, which is called the Lee and Shaffer relationship. So, they havetaken a principal which is different than the principal taken by the Merchant and Ernst here they are saying that there is a slip line field theory and according to that there will be astress field through which the forces are transmitted from the shear plane to the rake face of the tool ok . Now, if we have this then the Mohr circle construction , Mohr circle diagram you all are familiar with is a convenient means of relating stresses on any plane to the principal stress . So, I have also shown it to you that the Mohr circle diagram is placed on the center is placed on the um normal stress or the or the direct stress and perpendicular to that is the shear stress all right . Now, since plane BC is a stress free , Mohr circle must pass through the origin b . So, as you can see that if it is a surface plane BC the stress can be given and as a point here on the periphery . This is the 2 2 D and I showed it to you earlier also that in the 2 D case it can be shown on the periphery of this Mohr circle . So, this BC stress on this is b which is located here because it is a stress free all right there is no stress 0 normal stress 0 shear stress . Point a, c, d, f are displaced from b by 90 degree which are point a, c, d and f these are here ah. So, this is the a that you can see, this a is here ok, d is this plane ok, this plane is f this plane is accordingly. So, these are the planes a, these are the lines either vertically here or normal to that this these are displaced from b by 90 degree ok this will be twice the angle of physical plane here it is 45 degree . So, therefore, in this the a, d, c and f they will be from b they will be at a 90 degree angle they will be inclined um displaced by a 90 degree angle . Now, e this is e ok this is inclined to face d this one sorry this is e and this is d d is inclined to e by an angle of phi minus alpha you can find out that this angle is phi minus alpha because this is the rake angle, this is the shear plane angle. So, this angle will be 5 minus alpha let us say this is nu nu is equal to 5 minus alpha . So, if it is nu therefore, in the stress plane the angle subtended by the arc ae this one this will be at the center which is the 2 nu ok. So, this angle this angle is the 2 nu angle, because the nu we are saying that this is the number ok 5 minus alpha this is with respect to the point b . So, with respect to center this angle is the 2 nu or the 2 phi minus alpha. Assuming uniform stress tau and normal stress sigma on the rake face the friction angle we can find out here. And this will be the tan inverse of lambda by sorry tau by it it is here from this you can find out that this is the value and this will be equal to the e s ok, e s divided by s into b. So, this divided by s into b this is related to the tan of this and tan of this is equal to the mu. So, you can find out the mu which is F by N ok and F by N will be here . Therefore eos this is the point, e o s e o s this angle plus this angle this angle is 2 mu or the 2 phi minus alpha this is equal to see here this is nothing eos plus this this is the angle we mean to say ok do not confuse that . This angle plus this angle is equal to 2 lambda you can see that all right this is 2 lambda at the center because here it is lambda . So, 2 lambda plus 2 nunu is equal to phi minus alpha. So, therefore, this is equal to pi by 2 and this makes 2 lambda plus 2 phi minus 2 alpha is equal to pi by 2 ok. So, this is the equation which is derived by the Lee and Shaffer and you can see that this is very close to the Merchant's relationship. Although the Merchant's relationship is based on the minimum power consumption during the machining process ok and here in this case the Lee and Shaffer says that there should be a stress field through which the um forces are transmitted from the shear plane to the rake face of the tool . So, I once again would like to repeat that that here what is happening is that this stress field is made. In this stress field these are all the slip lines where the maximum deformation occurs ok and each of these lines is given a name like for example, this is d e a b c f and so on . So, here there all maximally deformed and they aremaking an angle of 45 degree with the free surface . So, therefore, in theMohr circle it will be to twice that which is 90 degree. So, therefore, we said that a, d, c and f they arefrom the b ok they are at a 90 degree angle. Now then we what we said is this e this one is inclined to this by an angle of nu and here it will be the twice the nu 2 of this and then if you sum up this angle and this angle which is 90 degree from here you can find out this relationship . So, the relationship is not a very difficult to find out only thing is that you have to construct the Mohr circle diagram, from the stress field and the machining dimension machining diagram that the Lee and Shaffer has proposed . Now pleaseall the time you should also keep in mind that this is valid only when these assumptions are met, if any of these assumptions is not mating in that case the Lee and Shaffer relationship will not be valid ok. After that we will be discussingthe topic which is the friction in metal cutting ok, why we are discussing this becausein the normal case when we have the sliding friction there the F and the N, because of the normal force the friction force F. So, the F by N we always assumed to be constant; that means, the F is proportional to N this is what happens in the case of the sliding friction. We will see in thefriction in metal cutting through a model that how it differs in the case of the metal cutting and how it differs from the sliding friction particularly where the F by N remains constant let see this. See the nature of sliding friction if we have to uh consider that in here we have to mention that since the solid surfaces have asperities for example, this . So, when the 2 surfaces meet each other they will actually not meetalong the area which is the real area , but they will be mating at some point on the asperities as you understand this ok . The real area of contact differs from the apparent area apparent area is the geometrical mating area . In case so, therefore, what we are saying is that, the real area of contact this is different than the area of contact on which they are mating because they are mating on the asperities and not the full full on the full area ok . So, therefore, the real area is different from the geometrical mating area. In case when the load increases the asperity deformation becomes fully plastic and the real area of contact is then a direct function of the applied load independent of the apparent area or the geometrical area of the surfaces . So, what is said is that when the load is increasing then these these points on which the 2 surfaces are mating on the asperities they become initially elastically deform and then as the load is increasing further the elastic deformation becomes plastic deformation. In that case the real area of contact is then a direct function of the applied load N and this will be independent of the apparent area of or geometrical area of the surfaces . So, we can write down that A r ok which is the A r is the real area of contact all right that is what we said that the real area of contact is directly proportional to N which is equal to N by uh the yield stress of the softer material , out of these 2 whichever is the softer material the yield stress of that material has to be considered . So, therefore, the real area of contact will be equal to the N divided by yield stress of the softer material . Now, during sliding, shearing of the welded asperities occurs the mechanism described by the addition theory of friction meaning that once the normal load is applied as we said that asperities become so to say welded because they are the normal force is increasing and then when 2 bodies move on each with respect to each other these asperities welded asperities will be sheared off ok and this is known by the addition theory of friction . Now, the addition theory of friction will give you an equation of the uh F that is a friction force is equal to tau into A r , A r is the real area of contact and therefore, mu which is F by N this can be given as tau A r divided by N we found out as the A r into the yield stress of the softer material. So, which is nothing, but A r is getting cancelled so, the lambda sorry tau divided by yield stress of the software material sigma y . So, this equation shows that mu is independent of the apparent contact . So, that is the basic idea in the case of the sliding normal sliding friction when the 2 materials slide on each other and then this how it can be applied to them to the machining and how themu happens in the case of the machining that we will be discussing in the next class. Thank you for your attention .