 Hello friends, I am Mr. Sanjeev B. Naik working in Walchand Institute of Technology, Solapor. In this video, I am going to solve the problem based on design and natural tolerances which are used in engineering practice. So at the end of session, learners will be able to solve the similar problems on design and natural tolerances which are commonly used in engineering practice. So this is an example where the shaft is produced on particular manufacturing setup. The diameters of the shaft that produce on that setup are going to vary in their basic dimensions and that is why they are normally assumed to be distributed with a mean value of 40.02 mm having a standard division of 0.02 mm. So it is assumed that the variation provided by manufacturing setup follows normal distribution curve having its mean value 40.02 mm and having standard division of 0.02 mm. Whereas the designer has specified the limit of the design tolerance for the shaft diameter as 40 plus 0.05 mm as upper dimension limit and 40 minus 0.03 mm as a lower dimension unit where 40 is basic size. So the designer has provided basic size 40 mm on which he has specified the upper and lower dimension variation as a design tolerance. Now as per the requirement of design tolerance if this is been manufactured on particular manufacturing process which has got its own and natural tolerance then what will happen about rejection that is whatever quality is expected by the designer that maintaining the dimension between this limit is possible for all the parts produced or not because the design tolerance is different and natural tolerance is different that we can analyze over here. So first we decide whether the there is any rejection if at all and if it is rejection how many shafts are likely to be rejected in terms of percentage. So percentage shafts likely to be rejected so this is what expected answer. So for that what we do we first calculate absolute tolerances just by establishing what is known as lower limit for natural tolerance. We know that as the natural tolerance is distributed by normal distribution curve which is expressed as lower limit and upper limit with the value mean minus 3 sigma is a lower limit of that curve and that is 39.96 mm whereas the upper limit of natural tolerance will be mean plus 3 sigma. So total range of dispersion is between these two limits lower limit and upper limit. So upper limit is mean plus 3 sigma and that results to be 40.08 mm and 39.96 mm. So for natural tolerance we establish two limits upper and lower limit of dimensions in which all the shafts will vary which are been manufactured. Similarly for design tolerances directly we have specified the limit as x1 how if it is lower limit we get 40 minus 0.03 mm is a minimum dimension which is been allowed by the designer to maintain the quality and that is why it is 39.97 whereas the upper limit is 40 plus 0.05 and that is why 40.050 mm will be the upper limit of design tolerance. So by knowing these two tolerance we can further work out with the requirements. So just understand and just recall what is the significance of upper limit and lower limit of dimensions in manufacturing considerations in product design or in acceptance of the product as a quality. So just we have seen that we have established these two limits natural tolerance and lower limit and upper limit design tolerance lower limit and upper limit. Now these tolerance that is the difference between these two is distributed as a mean curve. So this is a normal distribution curve between which we decide the lower and upper limit. So this is a curve which is referred for natural tolerance because it is a manufacturing process. Actually product is manufactured according to this variation which is distributed between minus 3 sigma to plus 3 sigma and that sigma is converted into normal deviated z value because the standard curve is converted into real application of practical application in which we convert standard deviation standard into required standard deviation that is value of z. So that is what a natural tolerance. So this natural tolerance is between minus 3 to plus 3 that is what a natural tolerance is being shown which exists with its lower limit value 39.96 and upper limit value 40.08 mm so which is shown over here. So this gives the natural tolerance dispersion. So whereas what is design tolerance? Design tolerance is 39.97 lower limit and 40.05 upper limit. So this difference between two values is actually design tolerance. So means what designer is expected variation between these two limits which are accepted for its functional requirement. If he is not lying in this region then it is not accepted so it is rejected. So this gives certain idea about considering lower limit upper limit of design and natural tolerance is that there can be certain rejection over here because manufacturing dimension is more than expected design dimension as a limit and that is why the shaft which will have dimensions more than 40.05 till 40.08 these are not accepted so it is rejected. So this way this analysis will help us. So we have to establish this in terms of z so that area under the curve which is provided by standard tables or standard charts considering standard normal curve by central limit theorem and that is what standard dimension areas are known. So to make it that what we calculate first what is the standard division of this dimension with respect to mean where z is 0 so that we calculate at x1 and that value of z1 is given as x1 minus mu upon sigma standard division like this 39.970 is lower limit and mu is 40.02 standard division is this so minus 2.5 so z value at this means what this level is lying at 2.5 minus value of z here this value this we can establish. Similarly we can calculate z according to this line as 40.0 upper limit that gives z2 plus 1.5 so this is a z value plus 1.5 and that is why the area under the curve between standard divisions minus 2.5 to plus 1.5 which is value of z is within design tolerance which is going to be accepted as a required quality of the shafts and anything beyond this like this it is a rejection. So here any dimension lies after 40 point something towards this it is not accepted so this is a rejected number of shafts as a percentage this is a rejected number of shafts as a percentage so area under the curve directly reflects the percentage of either rejection or acceptance and that is why if I want to know how much percentage of shaft is rejected I want to calculate area shaded area this area plus this area this is my need of problem to solve and as design tolerance is less than natural tolerance the first answer is that there is a rejection so as design tolerance is less than natural tolerance there is always rejection. So rejection is going to occur definitely and that is equal to the area under this shaded curve that we can calculate so percentage of rejection of shafts produced will correspond to the shaded area under the curve as shown. So we calculate from standard the shaded area so the first area we calculate between z is equal to 0 to 2.5 so z is equal to 0 to 2.5 and then from 0 to 1.5 so total area is between minus 2.5 to 1.5 that is been obtained between 0 and this value and between 0 and this value from standard table it is obtained as 0.4938 plus 0.432 that is area between this area that is what between the quality norms so it is accepted and that is why the area of rejection is obtained considering entire area under the curve which is 1 that is 100 minus this value which is accepted so it gives 1 minus 0.927 and that results to be 0.073 area so this area plus this area is aggregate to nearly 7.3% so percentage of rejection of shaft is 7.3% so this is the way we can first identify as the design tolerance is less than natural tolerance there is a rejection inherent rejection and that rejection total is about 7.3% so if thousand shafts are produced on this manufacturing system nearly 73 shafts will be rejected so that way the decision making analysis can be made by using this statistical technique in considering design and natural tolerances this is my reference thank you