Designing Capable and Reliable Products Episode 2 Part 10 ppsx

The stress can be approximated by a Normal distribution with parameters: L $ N66:66; 6:72MPa Determining the second order terms in the variance equation will provide more accurate values for the mean and standard deviation of the function under certain circumstances. B. Finite difference method Substituting y  x 1 ; x 2 ; FFF; x n  into the variance equation for the output of the function, and expanding for n variables gives:  y %   n i 1  @y @x i  2 Á  2 x i  0:5   @y @x 1  2 Á  2 x 1   @y @x 2  2 Á  2 x 2 ÁÁÁ  @y @x n  2 Á  2 x n ! 0:5 The ®nite dierence method can be used to approximate each term in this equation by using the dierence equation for the ®rst partial derivative (see Figure 2). The values of the function at two points either side of the point of interest, k, are determined, y k 1 and y k ÿ1 , which are equally spaced by an increment Áx. The ®nite dierence equation approximates the value of the partial derivative by taking the dierence of these values and dividing by the increment range. The terms subscripted by i indicate that only x i is incremented by Áx i for calculating y k 1 and y k ÿ1 , holding the other independent variables constant at their k value points.  @y @x i  k % y k 1 ÿ y k ÿ1  i 2Áx i Figure 2 The ®nite difference principle Appendix XI 365 The expanded variance equation to determine the standard deviation of the stress in the rectangular bar problem above was:  L   @L @F  2 Á  2 F   @L @a  2 Á  2 a   @L @b  2 Á  2 b ! 0:5 The variables F, a and b were are all assumed to be random in nature following the Normal distribution with parameters: F $ N100; 10kN a $ N0:03; 0:0003m b $ N0:05; 0:0004m Consider the point x k to be the mean value of a variable and x k 1 and x k ÿ1 the extremes of the variable. The extremes can be determined for each variable by assuming they exist Æ4 away from the mean which covers 99.99% of situations. For each variable in the problem above, the extremes become: F k 1  100 000 410 000140 000 MPa F k ÿ1  100 000 ÿ410 00060 000 MPa ÁF  410 00040 000 MPa a k 1  0:03 40:00030:0312 m a k ÿ1  0:03 ÿ40:00030:0288 m Áa  40:00030:0012 m b k 1  0:05  40:00040:0516 m b k ÿ1  0:05  40:00040:0484 m Áb  40:00040:0016 m For the ®rst variable, F, in the variance equation:  @L @F  % y k 1 ÿ y k ÿ1  2ÁF Letting the variable F be its maximum value and the variables a and b kept at their mean values gives y k 1 when applied to the stress function: y k 1  140 000 0:03 Â0:05  93 333 333 and y k ÿ1  60 000 0:03 Â0:05  40 000 000 Therefore:  @L @F   93 333 333 ÿ40 000 000 2 Â40 000  666:66 For the second variable, a, in the variance equation:  @L @a  %  y k 1 ÿ y k ÿ1  2Áa y k 1  100 000 0:0312 Â0:05  64 102 564 366 Appendix XI y k ÿ1  100 000 0:0288 Â0:05  69 444 444  @L @a   64 102 564 ÿ69 444 444 2 Â0:0012 ÿ2:226 Â10 9 For the third variable, b, in the variance equation:  @L @b  % y k 1 ÿ y k ÿ1  2Áb y k 1  100 000 0:03 Â0:0516  64 599 483 y k ÿ1  100 000 0:03 Â0:0484  68 870 523  @L @b   64 599 483 ÿ68 870 523 2 Â0:0016 ÿ1:335 Â10 9 Substituting into the expanded variance equation gives:  L 666:66 2  10 000 2  ÿ2:226  10 9  2  0:0003 2   ÿ1:335  10 9 Â0:0004 2  0:5  L  6:72 MPa Again, the mean stress,  L can be approximated by substituting the mean values for each variable into the original stress function:   %  x 1 ; x 2 ; FFF; x n  Therefore:  L % F ab   F  a Á  b  100 000 0:03 Â0:05  L  66:66 MPa The stress can be approximated by a Normal distribution with parameters: L $ N66:66; 6:72MPa If second order partial derivatives are required for a more accurate solution of the variance equation, then these terms can be approximated by:  @ 2 y @x 2 i  k % y k ÿ1  y k 1 ÿ 2y k  i Áx i  2 Solving for the second order terms in the variance equation for both the mean and standard deviation is shown below for the above example:  @ 2 L @F 2   40 000 000 93 333 333 ÿ266 666 666 40 000 2  6:25  10 ÿ10  @ 2 L @a 2   69 444 444 64 102 564 ÿ266 666 666 0:0012 2  1:4839  10 11 Appendix XI 367  @ 2 L @b 2   68 870 523 64 599 483 ÿ266 666 666 0:0016 2  5:3388  10 10 The variance equation with second order terms is:   %   n i 1  @ @x i  2 Á  2 x i  1 2  n i 1  @ 2  @x 2 i  2 Á  4 x i  0:5 Therefore, substituting in values gives:  L  666:66 2  10 000 2  1 2 6:25 Â10 ÿ10  2  10 000 4  ÿ2:226 Â10 9  2  0:0003 2  1 2 1:4839 Â10 11  2  0:0003 4  ÿ1:335 Â10 9  2  0:0004 2  1 2 5:3388 Â10 10  2  0:0004 4  P T T R Q U U S 0:5  6:72 MPa Using the ®rst and second order terms in the variance equation gives exactly the same answer. For dierent conditions, say where one variable is not dominating the situation as above for the load, then the use of the variance equation with second order terms will be more eective. Continuing for the mean value:   %  x 1 ; x 2 ; FFF; x n  1 2  n i 1 @ 2  @x 2 i Á  2 x i  L  66:66  10 6  1 2 6:25 Â10 ÿ10 Â10 000 2 1:4839 Â10 11   0:0003 2 5:3388 Â10 10 Â0:0004 2   L  66:68 MPa Therefore, the stress can be approximated by a Normal distribution with parameters: L $ N66:68; 6:72MPa C. Monte Carlo simulation Monte Carlo simulation is a numerical experimentation technique to obtain the statistics of the output variables of a function, given the statistics of the input variables. In each experiment or trial, the values of the input random variables are sampled based on their distributions, and the output variables are calculated using the computational model. The generation of a set of random numbers is central to the technique, which can then be used to generate a random variable from a given distribution. The simulation can only be performed using computers due to the large number of trials required. Rather than solve the variance equation for a number of variables directly, this method allows us to simulate the output of the variance, for example the simulated dispersion of a stress variable given that the random variables in the problem can be characterized. 368 Appendix XI Typically, the random values, x, from a particular distribution are generated by inverting the closed form CDF for the distribution Fx representing the random variable, where: x  F ÿ1 u where u  random number generated from 0 to 1. Many distributions can be represented in closed form except for the Normal and Lognormal types. The CDF for these distributions can only be determined numeri- cally. For example, the 3-parameter Weibull distribution's CDF is in closed form, where: Fx1 ÿ exp  ÿ  x ÿxo  ÿxo    where xo  expected minimum value,   characteristic value, and   shape par- ameter. Therefore, for the 3-parameter Weibull distribution, the inverse CDF with respect to u is: x  xo  ÿ xoÿln1 ÿu 1= We can use Monte Carlo simulation to determine the mean and standard deviation of a function with knowledge of the mean and standard deviation of the input variables. Returning to the problem of the tensile stress distribution in the rectangular bar, the stress was given by: L  F ab where L  stress, F  load, and a and b  sectional dimensions of the bar. The variables L, a and b were all assumed to be random in nature following the Normal distribution with parameters: F $ N100; 10kN a $ N0:03; 0:0003m b $ N0:05; 0:0004m Because the Normal distribution is dicult to work with, we can use a 3-parameter Weibull distribution as an approximating model. Given the mean, , and standard deviation, , for a Normal distribution (assuming   3:44), the parameters xo and  can be determined from: xo %  ÿ3:1394473%   0:3530184 However, using these approximations, it is still assumed that the output variable will be a Normal distribution. For each variable F, a and b in the stress equation, therefore: xo F  100 000 ÿ 3:139447310 00068 605:5N  F  100 000  0:353018410 000103 530:2N  F  3:44 xo a  0:03 ÿ 3:13944730:00030:02906 m Appendix XI 369  a  0:03  0:35301840:00030:03011 m  a  3:44 xo b  0:05 ÿ 3:13944730:00040:04874 m  b  0:05  0:35301840:00040:05014 m  b  3:44 The inverse CDF is then used to generate the random numbers following the 3- parameter Weibull distribution, as given earlier. This is shown below in the subroutine for a Monte Carlo simulation using 10 000 trials. The subroutine is written in Visual Basic, but can be easily translated to other computer languages. It requires the declaration of two label objects to display the mean and standard deviation. Running the program gives the mean and standard deviation of the stress as: L $ N66:59; 6:74MPa (for one particular set of trials) Monte Carlo simulation is useful for complex equations; however, when possible it should be supported by the use of Finite Dierence Methods. Monte Carlo Simulation code (written in Visual Basic) Dim L() As Variant (declaration of variables) Dim SUM As Variant Dim VAR As Variant Dim MEAN As Variant Dim STD As Variant Dim F As Variant (declaration of variables Dim a As Variant for function) Dim b As Variant ReDim L(10000) (number of trials) For I%  1 To 10000 Let F  68605.5 + (103530.2 ÿ 68605.5) * ((ÿLog(1 ÿ Rnd)) ^ 0.2907) (inverse CDF's for Let a  0.02906 + (0.03011 ÿ 0.02906) * ((ÿLog(1 ÿ Rnd)) ^ 0.2907) each variable) Let b  0.04874 + (0.05014 ÿ 0.04874) * ((ÿLog(1 ÿ Rnd)) ^ 0.2907) Let L(I%)  F / (a * b) (function goes here) Next I% Let SUM  0 Let ST  0 Let VAR  0 For I%  1 To 10000 (equations to determine Let SUM  SUM + L(I%) the mean and Next I% standard deviation) Let MEAN  SUM / 10000 For I%  1 To 10000 Let VAR  VAR + (L(I%) ÿ MEAN) ^ 2 Next I% Let STD  Sqr(VAR / 10000) Label1.Caption  Format(MEAN, ``#0.000000'') (output mean) Label2.Caption  Format(STD, ``#0.000000'') (output standard deviation) 370 Appendix XI D. Sensitivity analysis Three dierent methods can be used to solve the variance equation as previously shown. The variance equation is also a valuable tool with which to draw sensitivity inferences, providing the contribution of each variable to the overall variability and so determining the key design variables. For example, suppose we are interested in knowing the variance contribution of each variable in the stress associated with the tensile loading on a rectangular bar. The governing stress was determined by: L  F ab where L  stress, F  load, and a and b  sectional dimensions of the bar. The variables F, a and b have Normal parameters: F $ N100; 10kN a $ N0:03; 0:0003m b $ N0:04; 0:0004m Using the variance equation, the standard deviation of the stress function can be found using the method of partial derivatives:  L   1  a  b  2 Á  2 F   ÿ  F  2 a  b  2 Á  2 a   ÿ  F  a  2 b  2 Á  2 b ! 0:5 Substituting the values in the variance equation gives:  L   1 0:03 Â0:05  2  10 000 2   ÿ 100 000 0:03 2  0:05  2  0:0003 2   ÿ 100 000 0:03 Â0:05 2  2  0:0004 2 ! 0:5  2 L  4:4444  10 13  4:4444  10 11  2:8444  10 11  2 L  4:5173  10 13 This value equals the variance of the stress, or the standard deviation squared. The variance contribution in percent of the load variable, F, to the variance of the stress then becomes: 4:4444 Â10 13 4:5173 Â10 13  100  98:39% The variance contribution in percent of the dimensional variable, a, to the variance of the stress then becomes: 4:4444 Â10 11 4:5173 Â10 13  100  0:98% The variance contribution in percent of the dimensional variable, b, to the variance of the stress then becomes: 2:8444 Â10 11 4:5173 Â10 13  100  0:63% Appendix XI 371 Plotting this data as a Pareto chart gives Figure 3. It shows that the load is the domi- nant variable in the problem and so the stress is very sensitive to changes in the load, but the dimensional variables have little impact on the problem. Under conditions where the standard deviation of the dimensional variables increased for whatever reason, their impact on the stress distribution would increase to the detriment of the contribution made by the load if its standard deviation remained the same. Design variable Variance contribution (%) 100 90 80 70 60 50 40 30 20 10 0 Figure 3 Pareto chart showing the variance contribution of each design variable in the tension bar problem 372 Appendix XI Appendix XII Simpson's Rule for numerical integration Typically in engineering we are required to ®nd the area under a curve where y  f x between limits as shown in Figure 1. Direct integration is sometimes dicult, and the use of numerical integration techniques helps in this respect. One commonly used technique which gives high accuracy is Simpson's Rule (more correctly called Simpson's 1 3 Rule). Here the area under the curve is divided into equal segments of width, h. For an even number of segments, m, we can divide the range of interest (MAX ÿ MIN) into ordinates x 0 to x m , where the number of ordinates is odd. At each ordinate the functional values are then determined, for example y 0  f x 0  as shown in Figure 2 using four segments (m  4). Simpson's method then uses second order polynomials to connect sets of three ordinates representing segment pairs to determine the composite area. The total area, A, under the curve y  f x between the limits is then approximated by the following equation: A   h 3  f x 0 4  m ÿ1 i 1;3;5;FFF f x i 2  m ÿ2 j 2;4;6;FFF f x j f x m  ! Figure 1 Area, A, under the curve y  fx between the limits MIN and MAX where: h  MAX ÿMIN m The higher the value for m, the more accurate the result will be; however, the time for computation will increase. The use of computers makes the task relatively simple though. The subroutine at the end of this appendix is written for Visual Basic, but can be easily translated to other computer languages. It requires the declaration of two textbox objects and one label object to display the area, A. Example 1 ± Numerical integration of the Stress±Strength Interference (SSI) equation In Section 4.4.1 we described a means of determining the reliability, R, when the loading stress, L, is represented by CDF in closed form (for all the distributions except for the Normal and Lognormal type) and the strength, S, is represented by its PDF. R   I 0 FLÁ f SdS If, for example, the loading stress, L, was given by a 3-parameter Weibull distribution, the CDF would be: FL1 ÿ exp  ÿ  x ÿxo L  L ÿ xo L   L  and the strength, S, was represented by a Normal distribution, the PDF would be: f S 1  S  2 p exp  ÿ x ÿ S  2 2 2 S  Figure 2 Graphical representation of Simpson's Rule using multiple segments 374 Appendix XII [...]... standard deviation of the stress function can be found using the method of partial derivatives:       ! 1 2 2 F 2 2 F 2 2 0:5 L ˆ Á F ‡ ÿ 2 Á a ‡ ÿ Á b a b a b a 2 b Substituting the values in the variance equation gives:   2 2 1 100 000 L ˆ  10 00 02 ‡ ÿ 0:03  0:05 0:0 32  0:05  2 !0:5 100 000 2 2  0:0003 ‡ ÿ  0:0004 0:03  0:0 52 2 ˆ 4:4444  101 3 ‡ 4:4444  101 1 ‡ 2: 8444... equations: xoL ˆ 350 ÿ 3:1394473…40† ˆ 22 4: 42 MPa L ˆ 350 ‡ 0:3530184…40† ˆ 364: 12 MPa L ˆ 3:44 The reliability equation then becomes:    ! 3:44 ! …I x ÿ 22 4: 42 1 …x ÿ 500 2 p exp ÿ Rˆ 1 ÿ exp ÿ Á dx 364: 12 ÿ 22 4: 42 224 : 42 2…50 2 50 2 which can be simpli®ed to:     ! ! …I x ÿ 22 4: 42 3:44 …x ÿ 500 2 1 ÿ exp ÿ Á 0:00798 exp ÿ dx Rˆ 139:7 5000 22 4: 42 Using Simpson's Rule outlined above,... MAX ÿ MIN ˆ 47:56 m 375 376 Appendix XII and x0 ˆ 22 4: 42 ˆ MIN x1 ˆ 22 4: 42 ‡ 47:56 ˆ 27 1:98 x2 ˆ 27 1:98 ‡ 47:56 ˆ 319:54 x3 F F F ; x10 ˆ 700 ˆ MAX From Simpson's Rule: !   mÿ1 m 2 ˆ ˆ h f …xi † ‡ 2 f …xj † ‡ f …xm † Aˆ f …x0 † ‡ 4 3 i ˆ 1;3;5;FFF j ˆ 2; 4;6;FFF  ˆ 47:56 3  0 ‡ 4…0 ‡ 0:000154 ‡ 0:005985 ‡ 0:004135 ‡ 0:000076†‡ ! 2 0:000003 ‡ 0:001756 ‡ 0:007 828 ‡ 0:000884† ‡ 0:000003 ˆ 0:988335 which... Part 2 ± Guide to Managing the Design of Manufactured Products London: BSI BS 7850 19 92: Part 1 ± Guide to Management Principles London: BSI BS 7850 1994: Part 2 ± Guidelines for Quality Improvement London: BSI BS EN 20 286 1993: Part 2 ± ISO System of Limits and Fits ± Tables of Standard Tolerance Grades and Limit Deviations for Holes and Shafts London: BSI BS EN ISO 9000 1994: Quality Management and. .. 22 4: 42 3:44 f …G† ˆ exp ÿ 139:7 Let's use 10 segments again, giving: hˆ MAX ÿ MIN 1 ÿ 0 ˆ ˆ 0:1 m 10 377 378 Appendix XII and x0 ˆ 0 ˆ MIN x1 ˆ 0:1 x2 ˆ 0 :2 x3 F F F ; x10 ˆ 1 ˆ MAX Applying Simpson's Rule to the function f …G† gives:   ! mÿ1 m 2 ˆ ˆ h Aˆ f …xi † ‡ 2 f …xj † ‡ f …xm † f …x0 † ‡ 4 3 i ˆ 1;3;5;FFF j ˆ 2; 4;6;FFF   ! 0:1 0 ‡ 4…0 ‡ 0:000001 ‡ 0:0000 32 ‡ 0:000747 ‡ 0:0179 62 ˆ ˆ 0: 021 598... Reliability, 329 ±336 Cagan, J and Kurfess, T R 19 92: Optimal Tolerance Allocation over Multiple Manufacturing Alternatives Advances in Design Automation, 2, ASME DE-Vol 44 -2, 165±1 72 Calantone, R J., Vickery, S K and Droge, C 1995: Business Performance and Strategic New Product Development Journal of Product Innovation Management, 12( 3), 21 4 22 3 Carter, A D S 1986: Mechanical Reliability, 2nd Edition... 174: 62 ÿ ln GŠ0 :29 1 The survival equation for the stress distribution is:         x ÿ xoL L x ÿ 22 4: 42 3:44 H ˆ exp ÿ or H ˆ exp ÿ 139:7 L ÿ xoL Substituting these expressions into equation 4.53 gives the reliability, R, as:     …1 f343:03 ‡ 174: 62 ‰ÿ ln GŠ0 :29 1 †g ÿ 22 4: 42 3:44 R ˆ 1 ÿ exp ÿ dG 139:7 0 where the actual function to be integrated is:     f343:03 ‡ 174: 62 ‰ÿ ln GŠ0 :29 1... International 1997a: Handbook No 1 Properties and Selection: Irons, Steels and High Performance Alloys, 10th Edition OH: ASM International ASM International 1997b: Handbook No 2: Properties and Selection: Non-ferrous Alloys and Special Purpose Materials, 10th Edition OH: ASM International Ayyub, B M and McCuen, R H 1997: Probability, Statistics and Reliability for Engineers Boca Raton: CRC Press 3 82 References... Harvard University Press Amster, S J and Hooper, J H 1986: Statistical Methods for Reliability Improvement AT&T Journal, 65 (2) , 69±76 Andersson, P 1993: Design for Quality ± As Perceived by Industry In: Proceedings ICED '93, The Hague, 1 123 ±1 126 Andersson, P 1994: Early Design Phases and their Role in Designing for Quality Journal of Engineering Design, 5(4), 28 3 29 8 Andersson, P A 1996: Process Approach... with and an appropriate value re¯ecting the problem should replace it For argument's sake, we will give it a value of 700 MPa Therefore, the reliability can be determined given that:     ! ! x ÿ 22 4: 42 3:44 …x ÿ 500 2 f …x† ˆ 1 ÿ exp ÿ Á 0:00798 exp ÿ 139:7 5000 and the limits are MIN ˆ 22 4: 42 and MAX ˆ 700 Now applying Simpson's Rule, let's start with a relatively low number of segments, m ˆ 10: . 1  @ @x i  2 Á  2 x i  1 2  n i 1  @ 2  @x 2 i  2 Á  4 x i  0:5 Therefore, substituting in values gives:  L  666:66 2  10 000 2  1 2 6 :25 10 10  2  10 000 4   2: 226 10 9  2 Â. Â0:05  2  10 000 2   ÿ 100 000 0:03 2  0:05  2  0:0003 2   ÿ 100 000 0:03 Â0:05 2  2  0:0004 2 ! 0:5  2 L  4:4444  10 13  4:4444  10 11  2: 8444  10 11  2 L  4:5173  10 13 This. value:   %  x 1 ; x 2 ; FFF; x n  1 2  n i 1 @ 2  @x 2 i Á  2 x i  L  66:66  10 6  1 2 6 :25 10 10  10 000 2 1:4839 10 11   0:0003 2 5:3388 10 10 Â0:0004 2   L  66:68 MPa Therefore,