 I welcome you all for module 3 lecture 5. In this lecture, we will be discussing about the various aspects of positional tolerances and then we will see some numerical problems to understand the different kinds of fits and tolerances. Now, we will move to positional tolerance. Now, what happens is in the manufacturing of the work pieces, sometimes we specify some drilled holes and we also specify the location of that hole. In this case, this surface is taken as datum A and this surface is taken as B. So, the center of the hole that is to be drilled has some specification. This distance from this particular corner is 3 millimeter and from here it is 7.5 millimeter. Now, in the drilling of hole, exactly at that particular location as specified here may be difficult or it may not be possible because of various reasons like inaccuracy in the machine tool or because of the vibration or because of the thermal deformation, the tool, drill tool may not exactly coincide with the specified center. So, it may deviate from the specified center. So, we have to specify what is the deviation that is allowed. So, that is done by using the positional tolerance. Now, we can see here this is the hole that is to be drilled and its values are given here. The diameter of the hole is 2.5 millimeter and with some tolerance is also given. So, now, what is the deviation that is allowed for center? That is mentioned using this particular symbol. Now, this circle with the plus symbol, circle with plus is the symbol used to specify the positional tolerance and then in this case it is specified by the deviation that is allowed is 0.7 units, 0.7 millimeter. That means, the center of the hole, this is the 0.7 millimeter and the diameter within 0.7 millimeter diameter anywhere the center can lay. So, satisfying this maximum material condition, if we enlarge this portion, say this is the required center, desired center is this one which is specified by 7.5 units and 3 units with reference to A and B. Now, the positional tolerance is given as 0.7 units. That means, the drill can go here or the center can move anywhere within this particular circle. Now, because of this, what happens is, say we have some patterns like this. We have to draw, say, 3 holes. Now, because the center will be varying, so this is the center for A, hole A and this is the center for hole B. Now, we have the tolerance, positional tolerance for this hole A, positional tolerance for B. Similarly, positional tolerance for C. What happens is, the center can vary anywhere here. Say, the hole is drilled at this particular, taking this center within the positional tolerance, then we get the hole like this and in the second case, the center is taken somewhere here and the hole is drilled like this. Then, when we join these two centers, we get the line like this. There may be some angle between the desired location and the actual location. There is some angle. If this is within the allowed tolerance, then it can be accepted. Now, that is what I have shown here, the patterns of holes. So, we have the hole that is to be drilled and this is the positional tolerance for the center. Similarly, we have four holes and depending upon how the center of the hole varies, we may get the patterns like this. Then, sometimes what happens is, height is also important. Now, you can see here, this is the positional tolerance for the center of the hole in this particular plane. Now, when we consider the height, now we get a cylinder like this. So, within this volume, the center point can lay anywhere. That means, now when the second hole can see here, the center of the hole is at the extreme end. That means, instead of, this is the ideal condition where the desired center and the actual center, they are coinciding and here there is a deviation from the desired position. That means, center is of the hole that is drilled is here and in this case, there is angle, but again the center is within the allowed deviation. If it is within the allowed deviation, then we can accept the workpiece. This is the true position and this is the extreme attitude, axis of the actual hole that is drilled. We will solve some numerical problems on fit and tolerances. Now, this is the first problem. So, where in the basic size given is 25 millimeter and this diameter, it falls in the diameter step of 18 to 30 millimeter. So, we can find the mean diameter using this relationship, square root d 1 times d 2. So, the mean diameter is 23.238 millimeter and then we can find the tolerance unit using this relationship. So, we have to feed the value of d in millimeter, then we get the tolerance unit that is 1.307 micrometers. Now, for H 8 hole from table 5, we find that I t 8 tolerance grade is I t 8 and for this I t 8 tolerance value is 25 times I. So, we have this 25 times I is equal to 1.307. So, tolerance value for the hole H 8 hole is the 32.75 micrometer or 0.033 millimeter. Now, we know that for H hole that is hole basic shaft, basic hole, we have the fundamental deviation is 0. Now, for shaft d 9 from table 5, I t 9 is equal to 40 I. So, tolerance value for d 9 shaft is 40 times I, I is 1.307. So, this will give us 0.0523 millimeter. So, for d 9 shaft, fundamental deviation that is upper deviation from table 4 is equal to minus 16 times d power 0.44 micrometers. So, we have to feed the value of d that is 23.238, then we get the fundamental deviation of minus 63.86 micrometer or minus 0.064 millimeter. Now, we can find the limits for hole and shaft, the limits for hole that is upper limit, upper limit for, lower limit for hole is 25 millimeter because it is coinciding with the basic size. And upper limit is to the lower limit, we have to add the tolerance value, then we get the upper limit that is 25.033 millimeter. Similarly, we can find limits for shaft, lower limit is 25 millimeter that is the basic size. From this, we have to deduct the fundamental deviation as well as the tolerance value, then we get the lower limit for the shaft 24.884 millimeter. Similarly, we can find the upper limit for the shaft. So, for getting that, this is the 0 line and this is the basic size 25 millimeter. From this, we have to direct fundamental deviation that is 0.04, then we get the upper limit for the shaft that is 25 minus fundamental deviation of 0.064 will give us 24.936. So, this is the upper limit for the shaft. So, pictorially all the values are shown here, this is h 8 hole with the tolerance of 0.033 millimeter and this 0 line this distance indicates the basic size. And to the basic size, if we had this tolerance value, we get upper limit for hole that is 25.033 millimeter. Similarly, we can find the upper limit for shaft as well as lower limit of shaft. Now, we can calculate the minimum clearance and maximum clearance. Now, we can see, we get the minimum clearance by subtracting the upper limit of the shaft from the lower limit of the hole that is nothing but fundamental deviation which is equal to 0.064 is also equal to allowance value. So, minimum clearance is 0.064 millimeter and then to get the maximum clearance, what we have to do is, we have to add tolerance value for shaft, fundamental deviation value and tolerance value for the hole, then we get the maximum clearance of 0.149 millimeter. Now, we will move to the second problem 50 h 7 p 6. So, basic size is 50 millimeter and this falls in the diameter step of 30 to 50 millimeter. So, we can calculate the mean diameter which will be equal to 38.7 millimeter. And then tolerance unit, we can calculate using this relationship. We have to feed the value of D that is 38.7 millimeter in this equation, then we get tolerance unit is equal to 1.559 micrometer. And then for h 7 hole from table number 5, the tolerance grade is I t 7 for this tolerance value is 16 times I. So, where I is the tolerance unit with the value of 1.559 micrometer, then we get tolerance for hole that is equal to 0.025 millimeter. Now, for h hole fundamental deviation is 0. So, now, we can find the limits for hole, lower limit is equal to 50 millimeter and upper limit for the hole is lower limit of the hole plus tolerance. So, that will give us 50.025 millimeter. Now, coming to the shaft, we have shaft p 6 from table number 5 for tolerance grade 6 that is I t 6, we get a tolerance value of 10 times I, where I is equal to 1.559. So, we get 15.59 micrometer as the tolerance. So, tolerance for the shaft is 0.015 millimeter. Now, fundamental deviation for p shaft from table number 4 is equal to I t 7 plus 0 to 5. Now, I am taking only this I t 7 plus 0. So, 16 times I, I is equal to 1.559, this is equal to 24.94 micrometer, which is nothing but 0.025 mm. Now, from table number 5 for p 6 shaft that is the I t grade number is 6. So, I t 6 is equal to 10 times I. So, this will give us 0.015 mm. So, this is the tolerance value for the shaft. Now, limits for the shaft upper limit we can find by adding basic size and the fundamental deviation and the tolerance value that is basic value basic size is 50 millimeter and the fundamental deviation is 0.025 millimeter plus tolerance value is 0.15 millimeter. So, when I add these numbers we get 50.040 millimeter as upper limit for the shaft. Similarly, lower limit for the shaft is equal to basic size plus fundamental deviation. So, basic size is 50 millimeter and fundamental deviation is 0.025 millimeter. When I add these two values we get 50.025 millimeter and this is the lower limit for the shaft. Now, all the values are pictorially shown here. This is the zero line with reference to which all other dimensions are shown here. This is the basic size 50 millimeter and we have this H7 hole where the deviation is 0. So, the lower limit of the hole is equal to 50 millimeter. We have to add this tolerance value for hole to this basic size to get the upper limit. Now, similarly we have calculated the lower deviation for the shaft P6 shaft that is equal to 0.025 millimeter and the tolerance for the P6 shaft is 0.015 millimeter. Now, we can observe that here the maximum size of the hole is equal to the minimum size of the shaft. That means, the interference is equal to 0. It will just slide. There is no gap between a hole surface and the shaft surface by application of a little pressure. So, the shaft will move into the hole. Now, what is the maximum interference? So, that we get by we have to deduct the minimum size of the hole from the maximum size of the shaft. That is, maximum size of the shaft is 50.040 minus minimum size of the hole is 50. So, this difference gives 0.04 millimeter. So, this is the amount of maximum interference we get. Now, we will move to the next problem where the hole shaft combination is 25 h 8 m 9. So, the basic size is 25 millimeter and this falls in the diameter step of 18 to 30 millimeter. So, the mean diameter is 23.238 millimeter and then we can calculate the tolerance unit using this relationship and this will be equal to 1.31 micrometer. And for h 8 hole from table number 5, it 8 grade is equal to 25 I and this is equal to 32.75 micrometer. So, tolerance value for hole is 0.033 millimeter. And for h hole, we know that fundamental deviation is equal to 0. So, the lower limit for the hole is equal to the basic size that is 25 mm. And upper limit for hole is equal to basic size plus tolerance that is 25 millimeter plus tolerance value is 0.033 millimeter. Then, we get 25.033 millimeter. This is the upper limit for the hole. And coming to the shaft m 9 from table number 5 for I t grade number 9, the tolerance value is 40 I. This is equal to 52.28 micrometer which is equal to 0.052 millimeter. So, tolerance for shaft is 0.052 millimeter. And then the fundamental deviation for shaft m from table number 4 is equal to I t 7 minus I t 6. This is equal to 16 times I minus 10 times I. So, if we feed the value of I, we get fundamental deviation of 0.008 millimeter. Now, once we find this fundamental deviation and tolerance, we can find the limits for the shaft. That is lower limit for the shaft is equal to basic size plus lower deviation that is basic size is 25 millimeter plus lower deviation is 0.008 millimeter. This will give us 25.008 millimeter. Similarly, upper limit for shaft is equal to lower limit for the shaft is equal to lower limit plus tolerance value. That is lower limit for the shaft is 25.008 plus tolerance value is 0.052. Then we get 25.060 as the upper limit for the shaft. All these dimensions are indicated in this picture. So, this is the basic size 25 millimeter and 0 line. And this is h 8 whole with tolerance value of 0.033 millimeter. And this is the m 9 shaft with tolerance value of 0.052 and the lower deviation is 0.008 millimeter. And the difference between the maximum size of the shaft and minimum size of the hole will give us the maximum interference. So, that is equal to 0.060 millimeter. And the maximum clearance we get by subtracting the lower size of the shaft from maximum size of the hole. So, then we get maximum clearance of 0.025 millimeter. Now, from this picture we can understand that there is overlapping of tolerance zone of hole with the shaft. So, depending upon the actual size of the shaft and actual size of the hole, we may get interference fit or we may get clearance fit. So, this is the case of transition fit. Now, we will move to the next problem, wherein the basic size is 60 millimeter and this falls in the step of 50 to 80. And then the mean diameter is 63.25 mm. And we can find the tolerance unit using this relationship. And when we feed this 63.25 in the equation, we get tolerance unit of 0.002 millimeter. Now, for h 8 hole from table number 5, the I t grade number is 8. So, I t 8 is equal to 25 i. So, we have to feed the value of tolerance unit that is 1.853 micrometers. Then we get the tolerance value for the hole that is 0.046 mm. And we know that for h hole, the fundamental deviation is 0. So, upper limit for hole, we can get by adding tolerance value to the basic size. So, the basic size is 60 millimeter and the tolerance value for the hole is 0.046 millimeter. Then the upper limit for the hole is 60.046 millimeter. Similarly, for the lower limit for the hole is equal to the basic size that is 60 millimeter. Now, moving to the shaft M 9, the tolerance grade suggested for the M 9 shaft, for M 9 shaft, the tolerance grade that is suggested is I t 9. So, from table number 5, tolerance grade I t 9 is equal to 40 times I. So, tolerance value for M 9 shaft becomes 0.074 millimeter. Now, the fundamental deviation for M shaft for that is nothing but lower deviation from table number 4, we can calculate by using this relationship I t 7 minus I t 6, this is equal to 6 times I, this is equal to 0.011 millimeter. Now, after finding the fundamental deviation and the tolerance value for the shaft, now we can find the limits for the shaft. The upper limit for the shaft is equal to basic size basic size of the shaft plus lower deviation that is allowed and then we have to add the tolerance. So, then we get the upper limit for the shaft. That means the basic size is 60 millimeter and then the lower deviation is 0.01 millimeter and then tolerance value is 0.074. If we add all these three values, we get upper limit to be 60.085 millimeter. Similarly, we can find the lower limit for the shaft that is upper limit of the shaft minus tolerance. So, will give us the lower limit. So, that is equal to 60.011 millimeter. So, all the values are shown in this pictorial representation. So, now we can see there is overlapping of the zones. The tolerance zone for shaft is overlapping with the tolerance zone for hole. So, we have a transition fit. So, now we can find what is the maximum interference and what is the maximum clearance. So, maximum interference we get by adding the lower deviation and the this tolerance for the shaft that is equal to 0.085 millimeter. Similarly, the clearance we can find by subtracting the lower limit of the shaft from maximum size of the hole. So, that maximum clearance will be 0.035 millimeter. That means depending upon the actual size of the hole and actual size of the shaft, we may get interference fit or we may get clearance fit. Now, we will move to the next problem wherein the basic size is 25 millimeter and we are using a H hole with 80 grade of 8 and then we have we are using a S shaft with tolerance grade of 6. So, the 25 basic size it is in the step of 18 to 30. So, the mean diameter is 23.238 millimeter and then we can find the tolerance unit by feeding the value of D. We get 1.31 micrometer as the tolerance unit and then for H 8 hole from table number 5 we are using 80 grade 8. So, 80 8 is equal to 25 I. So, tolerance value for the hole is 0.033 millimeter and for H hole fundamental deviation is 0. So, the upper limit for the hole is equal to basic size plus tolerance value that is equal to 25 plus 0.033 millimeter which is nothing but 25.033 millimeter is the upper limit of the hole. And since the fundamental deviation is 0 for H hole, the lower limit for the hole is equal to basic size of the hole that is 25 millimeter. Now, moving to the shaft S 6, we from the table number 5 we have 86 is equal to 10 times I. So, this will be equal to 13.1 micrometer. So, tolerance for shaft is equal to 0.013 millimeter. Now, fundamental deviation for the shaft from table number 4 it is nothing but lower deviation this is equal to I T 8 plus 1. So, this relationship can be used if the diameter mean diameter is less than 50 millimeter. So, I T 8 is equal to 25 times I plus 1. So, this is equal to 0.034 millimeter. So, this is the tolerance that is allowed on the shaft fundamental deviation for the shaft. Now, once after finding the tolerance value and the lower deviation for the shaft, we can find the lower limit for the shaft and upper limit for the shaft. Lower limit of the shaft is equal to basic size plus lower deviation. So, this will give the lower limit for the shaft that is lower limit of the that is basic size is 25 millimeter plus lower deviation is 0.034. So, we get the lower limit of shaft is equal to 25.034 millimeter and similarly we can find the higher limit for the shaft. So, that we can find by adding tolerance value to the lower limit of the shaft that is lower limit is 25.034 millimeter plus tolerance value is 0.013 millimeter. So, higher limit of the shaft is equal to 25.047 millimeter. Now, we can understand that the lower deviation for the shaft is greater than the tolerance value of the hole. That means, the lower limit of the shaft is greater than the upper size of the hole. So, we are getting an interference fit here. So, the maximum interference we can get by adding lower deviation to the tolerance value that is 0.047 millimeter and minimum interference we can calculate by subtracting the maximum size of the hole by the minimum size of the shaft that is 25.034 minus 25.033 millimeter this is equal to 0.001 millimeter. That means, there is an interference of 1 micrometer. So, in order to fit the shaft s 6 shaft into the h 8 hole we have to apply some force and then we can get the interference fit. Now, we will move to the next problem wherein the basic size specified is 25 millimeter. So, in problem 6e we have calculated the i value that is 1.31 micrometer and upper limit for the hole h 8 hole 25 h 8 hole is 25.033 millimeter and lower limit for hole is equal to 25.000 millimeter. So, these values we have already calculated in problem number 6e for 25 h 8 combination and now moving to shaft u 7 the tolerance grade is 87 this is equal to 16 times high. So, we get the tolerance value of 0.021 millimeter and fundamental deviation for u shaft is equal to 87 and this is equal to again 0.021 millimeter. So, after finding the tolerance value for the shaft and fundamental deviation now we can find the lower limit for shaft this is equal to basic size of the shaft plus lower deviation that means the this is the shaft u 7 shaft with tolerance of 0.021 millimeter and the lower deviation is 0.021 millimeter. Now, we can find the lower limit for the shaft this is equal to basic size of shaft is 25 add to the basic size we have to add this fundamental lower deviation that is 0.021 mm then we get lower limit for the shaft that is 25.021 millimeter and then to get the upper limit of the shaft we have to add tolerance value to the lower limit of the shaft then we get higher limit of shaft to be 25.042 millimeter. Now, all values are indicated on this picture this is H 8 hole with tolerance value of 0.033 millimeter and the lower size of the hole is equal to basic size and we by adding the tolerance value to the basic size we get upper limit of the hole and then the since there is a overlapping of the tolerance zone for u shaft with the tolerance zone for H 8 hole we have a transition fit. Now, the maximum interference we can get by adding the lower deviation that is 0.021 mm with the tolerance value for the shaft. So, that is equal to 0.042 millimeter and by subtracting the lower limit of the shaft from maximum hole size we get the maximum amount of clearance that is equal to 0.012 millimeter we will mount to the next problem 25 H 7 u 6. So, the basic size is 25 millimeter and it is in the step of 18 to 30. So, mean diameter is 23.238 millimeter and the tolerance unit can be calculated and it is equal to 1.31 micrometer for H 7 hole the tolerance grade is 87 this is equal to 16 times I this is equal to 21 micrometer. So, tolerance value for hole is 0.021 millimeter and for H hole fundamental deviation is equal to 0. So, the lower limit for hole is equal to basic size that is 25 millimeter and higher limit for hole is equal to basic size plus tolerance that is 25.021 millimeter. Now, moving to the shaft u 6 from table number 5 the tolerance grade suggested is 86. So, 86 is equal to 10 times I. So, tolerance value for the shaft is 0.013 millimeter. Now, fundamental deviation for the shaft can u shaft can be calculated by using this relationship which is available in table number 4. So, the fundamental deviation varies from 87 to plus D. So, here I am taking only 87 only. So, this is nothing but 16 times I is equal to 0.021 millimeter. So, now we can find the limits for the shaft lower limit for the shaft is equal to basic size of the shaft that is 25 millimeter and then we have to add lower deviation that is 0.021 millimeter then we get lower limit of the shaft as 25.021 millimeter. Now, higher limit for the shaft is equal to lower limit plus tolerance value that is 25.021 millimeter plus 0.013 millimeter then we get higher limit of shaft is equal to 25.034 millimeter. Now, all these values are indicated here this is the basic size 25 millimeter 0 line and h 7 hole with the tolerance value of 0.021 millimeter and the lower limit lower deviation for the u shaft is 0.021 millimeter and tolerance value for u 6 shaft is 0.013 millimeter. Now, we can see here the lower limit of the shaft is equal to the upper limit of the hole. So, there is the minimum clear minimum interference is 0. There is no clearance and no interference and maximum difference we can calculate by adding lower deviation with the tolerance value for the shaft then we get maximum interference of 0.034 millimeter. Now, in all these problems we discussed about how to find the tolerance value then how to find the fundamental deviation and then how to calculate the tolerance value for both shaft and hole and then how to find the limits for shaft and hole and then how to calculate the maximum interference or minimum maximum clearance or whatever it is. Now, we have taken the examples of different basic sizes like 25 mm, 50 mm, 60 mm and then we have considered the basic hole of h with different I t grades like I t 7 and I t 8 and also we have taken different shafts m shaft, d shaft, p shaft, s shaft, u shaft with difference different tolerance grades of 7, 6, 9 etc. So, we have covered all the types of the clearance fits all the types of fits like clearance fit, transition fit and interference fit. In this lecture we discussed about the positional tolerances and what is their effect on the pattern of holes etc. and then we also saw some numerical problems to understand different kinds of fits in a better way. Thank you.