 I welcome you all for module 3 and lecture 3. Now, in this lecture we will solve some numerical problems on fit and tolerance, so that we can understand the concept more clearly. Now, let us start the problem number 1. The limits shown on a drawing for the mating hole and shaft are given here. For the hole basic size is 50 and the limits are shown here. And for the shaft, again the basic size is 50 and these are the limits upper limit and lower limit. Now, we have to find the allowance and then we have to state what is the type of fit. We have to calculate what is the greatest possible amount of clearance or interference over to blackboard. In the problem, the basic size of the shaft are given. The basic size is 50 millimeter and limits for upper limit and lower limit for hole and shaft are also mentioned. So, the maximum metal limit on the hole is 50 millimeter. So, this is the maximum metal limit for the hole and the minimum metal limit for the hole is 50.046 millimeter. So, this corresponds to the maximum metal limit and this is the minimum metal limit. So, and then we have a shaft where in the basic size is 50. Again the limits are given in the problem. The minimum size of the shaft is 49.971 millimeter and the maximum size of the shaft is 49.99 millimeter. And this is the tolerance zone for shaft and this is the tolerance zone for the hole. Now, since the minimum size of the hole is greater than the maximum size of the shaft, we have a clearance fit. And now, the allowance is the difference between the maximum metal limits of hole and shaft. So, the maximum metal limit for the hole is 50 millimeter and maximum metal limit for the shaft is 49.99 millimeter. So, the difference algebraic difference gives the 11. So, that is 0.01 millimeter. Now, the amount of clearance is the difference between the maximum size of hole that is 50.046 millimeter and minimum size of shaft that is 49.971 millimeter. So, this difference gives the greatest amount of clearance that is 0.075 millimeter. Now, we will move to problem number 2, wherein the limits of hole and shaft combinations are given. The basic size of hole and shaft is 80 millimeter and then the limits for hole and shaft are also mentioned in the problem. We have to find the type of fit and we have to find the allowance. And also, we have to determine the greatest possible amount of clearance or variance. Now, from the given problem, we can understand that the minimum metal limit for shaft is 80.071 millimeter and minimum metal limit for hole is 80.035 millimeter. So, that is shown in the picture here. This is the hole. The minimum size of the hole is 80 millimeter and maximum size of the hole is 80.035 millimeter and this corresponds to the minimum metal limit for hole. And we have a shaft here and the minimum size of the shaft corresponding to minimum metal limit is 80.071 millimeter and then the maximum size of the shaft is 80.093 millimeter corresponding to the maximum metal limit for hole. Now, since the minimum size of the shaft is greater than the maximum size of the hole, we have interference fit. Now, maximum metal limit for shaft is equal to 80.093 millimeter and then maximum metal limit for hole is 80 millimeter corresponding to this and then the difference gives the allowance difference between the maximum metal limit for shaft and maximum metal limit for hole gives us the allowance that is 80.093 millimeter minus 80 millimeter is equal to 0.093 millimeter and then greatest interference is equal to allowance is equal to 0.093 millimeter. Now, we will move to problem number 3, wherein the conditions given are maximum metal limit for the shaft is 100.026 millimeter, maximum metal limit for the hole is 100.0 millimeter, minimum metal limit for the hole is 100.036 millimeter and minimum metal limit for the shaft is 100.003 millimeter. We are required to find the type of fit and we have to find the allowance also. Now, over to blackboard. Now, these are the given conditions, maximum metal limit for shaft is 100.026 millimeter, maximum metal limit for hole is 100 millimeter. So, if hole and shaft are made to maximum metal limits, then we have interference fit. Now, from this picture we can understand that the maximum metal limit for shaft is 100.026 and minimum metal limit for shaft is 100.003 millimeter and this is the hole, the maximum metal limit for hole is 100 millimeter and minimum metal limit for hole is 100.036 millimeter. This is the tolerance zone for hole and this is the tolerance zone for shaft. Now, if both the hole and shaft are made to maximum metal limits, then we will be having interference fit. That is, if the hole is made to maximum metal limit, it will be 100 millimeter and if the shaft is made to maximum metal limit, it will be 100.026 millimeter. That means, shaft will be greater than the hole size. So, we will be having interference fit and now, minimum metal limit for hole is 100.036 millimeter, minimum metal limit for shaft is 100.003 millimeter. If both hole and shaft are made to minimum metal limits, then we will be having a clearance fit. That means, minimum metal limit for shaft is 100.003 millimeter and minimum metal limit for hole is 100.036 millimeter. That means, hole size will be bigger than the shaft size. So, we will be having a clearance fit and then allowance is the difference between the maximum metal limits of hole and shaft that is equal to 0.026 millimeter. Now, we will move to problem number 4. A 15 millimeter shaft is made to rotate in a bush. The tolerances for both shaft and bush are 0.050 millimeter. Determine the dimension of the shaft and bush to give an allowance of 0.075 millimeter with a hole basis system over to board. Now, in the case of hole basis system, lower deviation of hole will be 0. So, the lower limit of hole is equal to 50 millimeter which is given in the problem. So, higher limit of hole is equal to lower limit of hole plus tolerance. That is, lower limit of hole is 50 millimeter. To get the higher limit, we have to add the tolerance, then we get 50.050 millimeter. Now, high limit of shaft that is this value, high limit of shaft is equal to low limit of hole minus allowance, lower limit of hole minus allowance. Lower limit of hole is 50 millimeter and allowance is given in the problem that is 0.075 millimeter. So, we get higher limit of shaft is equal to 49.925 millimeter. So, higher limit of shaft is 49.925 millimeter. Now, in order to get the lower limit of shaft, we have to subtract tolerance from the higher limit of shaft. That is, we have to subtract this tolerance of 0.05 millimeter from the high limit of shaft that is 49.925. Then, we get the lower limit of shaft is equal to 49.875 millimeter. The problem is the figure below shows the assembly of a shaft and its mating hole. This is the shaft which is having the basic size of 30 millimeter and the limits are given. The minimum size of the shaft is 29.89 millimeter and the maximum size of the shaft is 29.95 millimeter and this is the bush fitted in the housing and the size of the hole are given here. The basic size is 30 millimeter and the minimum size of the hole is 30 millimeter and maximum size of the hole is 30.06 millimeter. Now, we have to give the appropriate dimensions for the following that is hole tolerance we have to calculate, shaft tolerance we have to calculate and then we have to calculate minimum clearance, maximum clearance, maximum metal condition for hole and minimum metal condition for shaft. Now, we have the solution here. Hole tolerance is the difference between these two that is 30.06 millimeter minus 30 millimeter will give us the hole tolerance of 0.06 millimeter. Similarly, to find the shaft tolerance we have to find the difference between the maximum size of the shaft and minimum size of the shaft that is 29.95 minus 29.89 will give us the shaft tolerance of 0.06 millimeter and then we have to find the minimum clearance between the shaft and the hole. So, this is equal to minimum size of the hole that is equal to 30 millimeter and minus maximum size of the shaft that is equal to 29.95 millimeter the difference gives us 0.05 millimeter and then maximum clearance is equal to maximum size of the hole minus minimum size of the shaft. So, maximum size of the hole is 30.06 millimeter and minimum size of the shaft is 29.89 millimeter the difference is 0.170 millimeter is the maximum clearance we get and then maximum metal condition for hole is equal to lower limit of the hole that is 30 millimeter and then minimum metal condition for the shaft is equal to lower limit of the shaft that is equal to 29.89 millimeter. So, all these values are indicated in the picture given here. So, this is the hole, lower limit of hole is 30 millimeter, upper limit of the hole is 30.06 millimeter and we have the shaft with lower limit that is 29.89 millimeter and high limit of shaft is 29.95 millimeter. The difference between the minimum size of shaft and maximum size of hole gives us the maximum clearance and then the difference between the maximum size of the shaft and minimum size of the hole will give us the minimum clearance. So, here we can see some of the preferred hole basis fits as per American standard B 4.2. Now, you can see here we have the basic hole that is h 11, h 9, h 8 like this we have taken the basic hole wherein the deviation is 0. That means the lower limit of the hole will be equal to the basic size. Similarly, we have the preferred shafts here C 11 and then D 9, F 7, G 6, H 6. You can see here again we have the basic shaft here wherein the deviation is 0. Now, if you take h 11, C 11 combination you can see that there is a large clearance here. So, this is the minimum size of the shaft hole. So, this is the hole and this is the minimum size of the hole and this gives us the maximum size of the hole and similarly this gives us the minimum, I am sorry. So, this will give us the maximum size of the shaft and this gives the minimum size of the shaft. So, now, the difference between the minimum size of the hole and maximum size of the shaft will give us the minimum clearance and similarly the difference between the maximum hole and the minimum shaft will give us the maximum clearance here. Now for h 7, capital H 7 and lowercase h 6, we can see that the clearance is 0 and from C 11, D 9, F 7, these combinations, these combinations will give us the clearance fit. Similarly, these combinations of hole and shaft will give us, for example, h 7 and 6 will give us transition fit. You can see here there is some overlap here, the tolerance zones of hole and shaft they are overlapping. So, we get the transition fits and then we have the combination h 7, p 6, h 7, h 6, h 7, u 6, all these combinations will give us the interference fits. So, again here, this is the maximum size of the hole and this is the minimum size of the hole and this is the hole tolerance and similarly we have the shaft tolerance. So, this difference will give us minimum interference and this difference that is maximum size of the shaft and then minimum size of the hole. So, this distance will give us the maximum interference and here hole basis system is shown with letters. That means, the h hole is considered here, the basic hole is considered here where deviation is 0. That means, the basic size of the lower limit of the hole. So, this is the lower limit of the hole. So, it is coinciding with the basic size. So, that is why this is the basic hole with letter h and this is the tolerance for the hole and this line indicates the 0 line. Now, we can see that we have varying the shaft sizes. We have A shaft, B shaft, C shaft like this. So, up to h, we can have clearance fit. For example, one example is given here h 7 d 10. So, this is h 7 and then d 10, this will give us the clearance fit and then the other combinations like h 7 j k m n p up to this, we can have the transition fit. One example is given here h 7 m 6. So, this will provide us the transition fit. Similarly, these combinations R, S, T, U, V, X, Y, Z, Z, A, Z, B, Z, C, these shafts with h hole will give us interference fit. So, depending upon the application, we can select appropriate fit. Now, this is the graphical representation of the shaft basis system as per NCB 4.2. Now, you can see here the shafts are basic shafts. That means the deviation in this case is 0. That means upper size of the shaft is coinciding with the basic size. Hence, the deviation is 0. Now, we have various shafts h 11, h 9, different grades of tolerance that is the different IT grades or as the number increases, as the IT grade number increases, we can see that the tolerance value will increase. As the number reduces, the tolerance band is reducing. That means as they move, as we decrease the IT grades from h 11, 9, 7, 6 like this, we have to select the precision machine tools to control the sizes properly to provide the tight tolerances. So, when the tolerance grade increases like IT 8, 9, 10 like that, the tolerance gets widened so that we can select not so precision machine tools and the cost of production will decrease. Again the selection of fit and tolerance, it depends upon the application that is needed. Similarly, we have, we are varying the shaft sizes here C, D, F, G, H. Now, you can see here this is the basic hole H wherein the deviation is 0. That means the lower size of the hole is coinciding with the basic size. And this 11 indicates the IT grade, IT 11, IT 9, IT 8. As the number decreases, tolerance band reduces. As the number increases, tolerance band increases. So, for this particular IT 11, this is the hole tolerance and for this H 11, this is the shaft tolerance and you can see there is a wide gap between the C shaft and H shaft and C hole and this is the minimum clearance what we can get. And this corresponds to the maximum size of the hole and this corresponds to the minimum size of the shaft. So, this will give us the maximum clearance. And as we move towards right side, you can see here the hole sizes are lesser than the shaft size. So, we are getting interference. For example, we have this hole U7 and this is the maximum size of the hole and this is the minimum size of the hole and this is the tolerance, hole tolerance. And this is the H6 is the shaft wherein we have the shaft tolerance. This is the shaft tolerance and this is the maximum size of the shaft and this corresponds to the minimum size of the shaft. And now, the minimum size of the shaft is greater than the maximum size of the hole. Hence, we are getting the interference fit. Now, this is the minimum amount of interference and this gives us the maximum amount of interference. And again depending upon the what type of holes and shaft combination we get, we have we can get the clearance fit, transition fit and interference fit. Now, this is ANSI preferred shaft basis fits as per ANSI B 4.2. Now, we have taken the basic shaft and where the shaft size is fixed and this is the tolerance provided for the shaft. Now, this is the basic line zero line. It is corresponding to the basic size of the shaft. So, the deviation is zero. Hence, this is the basic shaft and then we have the shaft basis fits. Now, we are varying the holes, hole sizes we are varying. This is A hole, B hole with the tolerance zone. Now, from A to H, we can get the clearance fit. So, H6 D 10. So, this shaft is H shaft and 6 is the 80 grade. H6 D 10 will give us the clearance fit. And from J to P with H shaft will give us the transition fit and then from R, R hole to ZC with the H shaft we can get interference fit. Now, this is a description of preferred fits, a summary of preferred fits. Clearance fit, transition fit and interference fit and depending upon what type of fit required, whether we require loose running fit or free running fit or closed running fit, sliding fit, locational clearance fit. So, depending upon the type of fit, whether we want clearance fit or transition fit or interference fit, we can select the appropriate combinations and these combinations are the preferred fits. And these combinations H 11, C 11, they give more clearance and H 7, U 6, U 7, H 6, they give the more interference. Now, this table shows preferred hole basis metric clearance fits as per ANSI B 4.2. Now, we can understand, this is a readymade table depending upon the type of fit required, we can select the appropriate fits. Basic sizes are given here 1, 1.2, 1.6, 1.62 like this up to 500 millimeter. The basic sizes are given with the maximum value of the maximum value and minimum for both shaft and hole, they are mentioned here. And we have loose running fit, if we want loose running fit, what is the combination we should select. If we want free running fit, what combination we can select, all those things we can get here. And for example, if the basic size is 20 and if we select H 11 hole, then the hole minimum size will be 20, since it is H hole, the deviation is 0. So, the minimum size of the hole will be equal to the basic size. So, minimum size is 20 millimeter and maximum size will be equal to 20.130. So, the clearance, hole tolerance, hole tolerance will be 130 microns. Similarly, depending upon the type of fit required, we can select the locational clearance, sliding clearance etcetera, etcetera. And one more thing we can understand is as the number is reducing H 11, H 9, H 8, H 7, the amount of tolerance also reduces. For example, here for H 11 20 basic size, the tolerance for the hole is 130 microns. Whereas, for the same basic size, if we select H 9 hole that is IT grade, IT 9 tolerance grade, we can see the tolerance is reduced to 52 microns from 130 micrometers. And then if we go to H 8 hole, we can see the tolerance becomes 33 micrometers. And for H 7 hole, with the same basic size of 20 millimeter, we have the tolerance of 21 microns. So, as the IT grade number reduces, the tolerance becomes tight. Now, this table, it gives the basic size from 30 millimeter to 500 millimeters. Similarly, we can have ready made tables in the ANSI standard for hole basis metric transition at the interference phase and then the shaft basis metric clearance phase, shaft basis metric transition and interference phase. So, one can get the ready made tables from the standard and can use it. Now, this table 4, it gives the formula for fundamental deviation for shafts, shaft sizes up to 500 millimeter. So, depending upon the shaft destination, whether we are using A shaft, B shaft, C shaft, etcetera, etcetera. What is the upper deviation and what is the lower deviation? So, that we can find in micrometers. For example, if I select the shaft D, the upper deviation will be equal to minus 16 times D power of 0.44 micrometer, where D is in millimeter. So, like this, using this table, one can find the formula for the fundamental deviation. Now, this table shows the various grades of tolerances used as per IS 919 standard. So, these are the tolerance grades IT5, IT6 and IT2, IT16 and the second row indicates the values of the tolerance values in terms of tolerance unit, international tolerance unit I. Now, this table gives us 18 grades of tolerances as per IS 919. Now, also this table gives the field of use of these grades. For example, IT01, IT0, IT1 up to IT6, these grades are used for limit gauges and measuring instruments wherein very tight tolerances are specified and IT5 to IT12, they are used for precision and general engineering fits like the bush bearing and then for fitting the bearings in the housings, for fitting the pulleys on to the shaft, etc., etc. And IT11 to IT16, they are used for semi-finished products. Now, how do we designate the holes and shafts? And the combination of hole and shaft is shown here. So, this is the way in which the hole is designated, this 50H8. So, here this first number 50 indicates the basic size of the hole and then H indicates the fundamental, this letter indicates the fundamental deviation. For H whole, the fundamental deviation is 0 that we have already studied and this number 8 indicates the IT grade. So, this dictates as what is the tolerance that is allowed on the hole. Similarly, for the shaft, this 50 indicates the basic size in the millimeter and F indicates the fundamental deviation for F whole and 7 indicates the IT grade. So, IT grade 7, using this we can find what is the tolerance. And then for the combination of hole and shaft that is for fit, this is how we represent the fit. This number indicates the basic size of hole and shaft. This indicates what is the type of hole. This 8 is IT grade for hole and this small f, this is for shaft. So, 7 indicates the international tolerance grade. Now, so tolerance zone for H shaft. So, H shaft is considered here, wherein the deviation, fundamental deviation is 0. Now, for H shaft, various tolerance grades are shown here like 5, IT 5, IT 6, IT 7 up to 11, IT 5 to IT 11. Now, as the number increases, IT number increases, the tolerance value will also increase. You can see here for IT 5, this is the amount of tolerance. For IT 6, this is the amount of tolerance. IT 11, you can see we have a very large tolerance of 160 microns. And this tolerance value, it also depends upon what is the basic size. So, this particular picture is drawn for a basic size of 50 millimeter with H shaft and with IT grade, IT 5 to IT 11. So, these numbers indicate the tolerance. Now, we can see here for H 6, this is the amount of tolerance. So, from here to this, this is 20 micrometer tolerance. So, this will be, this is 10 micron and then this is 15 micron. So, this tolerance value will be approximately 16 or 17 microns. So, that we can check with the table 3. H 6, 50 millimeter basic size, 50 H 6, 50 is the basic size. So, basic size is 50 millimeter and H 6 is the hole. We have H 6 here. Now, since we are using basic shaft, the deviation is 0. That means the maximum size of the shaft is equal to the basic size, that is 50 millimeter. And we can see here, the minimum size of the shaft is 49.984. So, the difference between these two is 16 microns. So, this 16 microns is the tolerance provided on the shaft. So, that we can observe in this picture. 16 microns is the tolerance. So, we can use the readymade graphs like this or readymade tables for getting the tolerance value. Now, we have to calculate the international tolerance grades depending upon what is the grade that is provided, whether it is I T 5, I T 6 or I T 16 like that. Depending upon this, we have to find the standard tolerance unit. For that, we use the formulae that is given here. I, that is standard tolerance unit is equal to 0.45 times cube root of D plus 0.001 times D, where D is the geometric mean of the diameter step. So, what is the diameter of diameter step? We can understand from this. D is millimeter. It is the geometric mean of lower diameter and upper diameter. That means, you can see here, this table gives geometric mean of diameter steps, various steps that are allowed as per standard. So, 0 to 3 is one step, 6 to 10 is another diameter step. So, this is D 1. First number is D 1 and second number is D 2. So, if we have diameter of 25, it falls within this step of 18 to 30 millimeter. Then, D 1 becomes 18 millimeter and D 2 will be 30 millimeter. Then, using this relationship, we can find the geometric mean D. Now, this particular picture shows the various machining processes and associated I T grades. Now, we can see here, different processes are listed here. So, starting from very precise machining like lapping, honing, etcetera, super finishing, cylindrical grinding, diamond turning and then very not so precision operations like drilling, die casting, forging, etcetera, etcetera. Now, when we say I T grade 5, immediately we can select what is the process which is suitable for the, to produce that I T 5 grade. You can see here, if we say I T 5, then we have so many machining operations which can satisfy the tolerance values specified by I T 5. We can select honing, super finishing, reaming like this. So, the fine machining processes we have to select. If we say I T 10, then these are the machining processes we can select to get the tolerance value which is specified by I T 10. The processes like boring, sawing, milling, etcetera, we have to. So, that means when the designer says the particular I T grade, immediately we can use this table and we can find what is the appropriate machining operation. Now, we will conclude at this point. In this lecture, we studied about the various tables, readymade tables and charts, using which we can find the tolerance values to find the upper limit and lower limit of shaft and holes and then we studied about the different I T grades and what are the applicable machining processes suitable for various I T grades. In the next lecture, we will continue with some numerical problems. Thank you.