Designing Capable and Reliable Products Episode 2 Part 9 ppsx

B. Later mechanical deformation Appendix VI 345 C. Adhesive bonding 346 Appendix VI D. Brazing and soldering Appendix VI 347 E. Resistance welding 348 Appendix VI F. Fusion welding Appendix VI 349 Appendix VII Blank conformability analysis tables A. Variability risks (q m and q a ) results table B. Conformability matrix Appendix VII 351 Appendix VIII Assembly problems with two tolerances The analysis of assembly problems with just two tolerances is common and often includes the requirement to ®nd the potential interference or clearance between two components. The following is an example of such a problem, involving the insertion of a deep drawn brass tube and a drilled alloy steel bore, the details of which are given in Figure 1 (Haugen, 1980). It is evident from the dimensions set on the tube and bore that the tolerances assigned are anticipated to be `worst case'. An analysis of the brass tube is performed to assess the likely capability using CA. Referring to the process capability map for deep drawing shown in Figure 2, and determining the material, m p , and geometry, g p , variability risks gives: Adjusted tolerance  Design tolerance t m p  g p  Æ0:05 1:7  1:2Â1:05 Æ0:023 mm A risk value of A H  1:3 is interpolated from the process capability map above and because there are no surface ®nish constraints, q m  1:3. An estimate for the shifted standard deviation was given in Chapter 3 as:  H t  t Á q 2 m 12  0:05  1:3 2 12  0:007 mm Similarly, for the alloy steel bore the process capability map for drilling is shown in Figure 3. Note that the tolerance in the case of a drilled dimension is given as a `' only: Adjusted tolerance  Design tolerance T m p Á g p  0:05 1  1:5Â1:05 0:032 mm The standard deviation for one half of the tolerance can be estimated by:  H b %  T 2  Á q 2 m 12  0:025  2:5 2 12  0:013 mm Using the algebra of random variables we can solve the probability of interference between the two tolerance distributions, assuming that the variables follow a Figure 1 Details of the tube and bore Figure 2 Process capability map for deep drawing Figure 3 Process capability map for drilling Appendix VIII 353 Normal distribution and that they are statistically independent. The problem is gov- erned, in fact, by the amount of negative clearance. This translates into a dierence between the mean of the two random variables,  t and  b (see Figure 4). The mean of this dierence distribution, , is given by:    b ÿ  t  11:525 ÿ 11:45  0:075 mm The standard deviation of the dierence distribution between the random variables is given by:   H2 t   H2 b  0:5 0:007 2  0:013 2  0:5  0:015 mm From the Standard Normal Distribution (SND) it is possible to determine the prob- ability of negative clearance, P. The area under the curve to the left of  b ÿ  t  0 relates to the probability of negative clearance. This area can be found from the SND table (Table 1, Appendix I) by determining the Standard Normal variate, z, where: z   x ÿ      0 ÿ 0:075 0:015  ÿ5:00 The area È SND z and hence the probability of negative clearance P % 0:0000003. Therefore, the probability that the parts will interfere on assembly is practically negligible. Also see Haugen (1980), Kolarik (1995) and Shigley (1986) for examples of this type. Figure 4 Statistical representation of the interference problem 354 Appendix VIII [...]... ‡ 1 !  % … ÿ xo† Á ÿ 0: 92 6 Maximum Extreme Value Type I distribution  %  ‡ 0:57 721 57  % 1 :28 25 498  Minimum Extreme Value Type I distribution  %  ÿ 0:57 721 57  % 1 :28 25 498  Note the parameters for the 3-parameter Weibull distribution, xo and , can be estimated given the mean, , and standard deviation, , for a Normal distribution (assuming ˆ 3:44) by: xo %  ÿ 3:1 394 473  %  ‡ 0:3530184... exp xÿ   ÿI . mean () and standard deviation () Lognormal distribution   exp     2 2   exp 2  2 2 ÿexp 2   2  0:5 2- parameter Weibull distribution  %  Á ÿ 1  1  !  %  Á  ÿ0: 92 6 See. x 1   2 p exp  ÿ x ÿ  2 2 2  ÿI< x < I where   mean and   standard deviation. Lognormal distribution f x 1 x  2 p exp  ÿ lnxÿ 2 2 2  0 < x <. 1:5Â1:05 0:0 32 mm The standard deviation for one half of the tolerance can be estimated by:  H b %  T 2  Á q 2 m 12  0: 025  2: 5 2 12  0:013 mm Using the algebra of random variables we