THÔNG TIN TÀI LIỆU
Therefore, substituting these into the reliability equation gives: R I S xo L 1 ÿ exp ÿ x ÿ xo L L ÿ xo L L ! Á 1 S 2 p exp ÿ x ÿ S 2 2 2 S ! dx where the limits of integration are from the expected minimum value of the stress, xo L , to in®nity. Because the loading stress function as an output from the variance analysis is characterized by a Normal distribution, it is advantageous to estimate the parameters for a 3-parameter Weibull distribution, xo and , given the mean, , and standard deviation, , for a Normal distribution (assuming 3:44), where: xo % ÿ 3:1394473% 0:3530184 Consider the situation where the loading stress on a component is given as L $ N350; 40MPa relating to a Normal distribution with a mean of L 350 MPa and standard deviation L 40 MPa. The strength distribution of the component is S $ N500; 50MPa. It is required to ®nd the reliability under these conditions. From the above equations: xo L 350 ÿ 3:139447340224:42 MPa L 350 0:353018440364:12 MPa L 3:44 The reliability equation then becomes: R I 224:42 1 ÿ exp ÿ x ÿ 224:42 364:12 ÿ 224:42 3:44 ! Á 1 50 2 p exp ÿ x ÿ 500 2 250 2 ! dx which can be simpli®ed to: R I 224:42 1 ÿ exp ÿ x ÿ 224:42 139:7 3:44 ! Á 0:00798 exp ÿ x ÿ 500 2 5000 ! dx Using Simpson's Rule outlined above, the maximum limit, I, is dicult to work 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: f x 1 ÿ exp ÿ x ÿ 224:42 139:7 3:44 ! Á 0:00798 exp ÿ x ÿ 500 2 5000 ! and the limits are MIN 224:42 and MAX 700. Now applying Simpson's Rule, let's start with a relatively low number of segments, m 10: h MAX ÿ MIN m 47:56 Appendix XII 375 and x 0 224:42 MIN x 1 224:42 47:56 271:98 x 2 271:98 47:56 319:54 x 3 FFF; x 10 700 MAX From Simpson's Rule: 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 ! 47:56 3 0 40 0:000154 0:005985 0:004135 0:000076 20:000003 0:001756 0:007828 0:0008840:000003 ! 0:988335 which gives the reliability, R, as: R 0:988335 Increasing the number of segments, m, increases the accuracy and using the computer code for m 1000 gives the answer R 0:990274. Comparing answers with that derived from the coupling equation, the Standard Normal variate is: z ÿ 500 ÿ 350 50 2 40 2 p ÿ2:34 From the SND, Table 1 in Appendix I, the area under the curve, Èz, and hence the probability of failure, P, is found to be 0.009642. The reliability is then given by: R 1 ÿ P 1 ÿ 0:009642 0:990358 The bene®t using the approach over other methods is that it can be used to solve the general reliability equation for combination of distribution, including when there are multiple load applications. This is only possible when the loading stress is described in closed form. Example 2 ± Integral transform method We have a means of determining the interference of two distributions in a similar way as that given above but applying the integral transform method described in Section 4.4.1, where: R 1 ÿ 1 0 HdG where H represents the loading stress distribution and G the strength distribution. 376 Appendix XII Suppose again that both the stress and strength distributions of interest are of the Normal type, where the loading stress is given as L $ N350; 40MPa and the strength distribution is S $ N500; 50MPa. The Normal distribution cannot be used with the integral transform method, but can be approximated by the 3-parameter Weibull distribution where the CDF is in closed form. It was determined above that the loading stress parameters for the 3-parameter Weibull distribution were: xo L 224:42 MPa L 364:12 MPa L 3:44 Similarly, the strength parameters for the Weibull distribution can be approximated by the same method to: xo S 343:03 MPa S 517:65 MPa S 3:44 The CDF for the strength distribution is given by: Fx1 ÿ exp ÿ x ÿ xo S S ÿ xo S S which represents the probability of failure, P. The probability of surviving, or the reliability, is therefore 1 ÿ P, or: G exp ÿ x ÿ xo S S ÿ xo S S Solving for x to give the inverse of this function gives: x xo S S ÿ xo S ÿln G 1= S or x 343:03 174:62ÿln G 0:291 The survival equation for the stress distribution is: H exp ÿ x ÿ xo L L ÿ xo L L or H exp ÿ x ÿ 224:42 139:7 3:44 Substituting these expressions into equation 4.53 gives the reliability, R, as: R 1 ÿ 1 0 exp ÿ f343:03 174:62ÿln G 0:291 g ÿ 224:42 139:7 3:44 dG where the actual function to be integrated is: f Gexp ÿ f343:03 174:62ÿln G 0:291 g ÿ 224:42 139:7 3:44 Let's use 10 segments again, giving: h MAX ÿ MIN m 1 ÿ 0 10 0:1 Appendix XII 377 and x 0 0 MIN x 1 0:1 x 2 0:2 x 3 FFF; x 10 1 MAX Applying Simpson's Rule to the function f G gives: 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 ! 0:1 3 0 40 0:000001 0:000032 0:000747 0:017962 20 0:000005 0:000162 0:0034120:565802 ! 0:021598 R 1 ÿ A 0:978402 Increasing the number of segments, m, again increases the accuracy and using the computer code provided for m 1000 gives the answer R 0: 991182. This compares well with the answers given in the previous example, although we are using a 3- parameter Weibull distribution to model the stress and strength, when in fact they are of the Normal type. Area under a Function Calculated using Simpson's Rule (written in Visual Basic) Dim Ord As Integer (declaration of variables) Dim m As Integer Dim x( ) As Variant Dim y( ) As Variant Dim MAX As Variant Dim MIN As Variant Dim b( ) As Integer Dim c( ) As Variant Dim Sum As Variant Dim A As Variant Let MAX = Text1.Text (maximum limit of integration) Let MIN = Text2.Text (minimum limit of integration) Let Ord = 1001 (number of ordinates) Let m = Ord - 1 (number of segments) ReDim x(m + 1) For I% = 2 To m Let x(I%) = MIN + (I% - 1) * ((MAX - MIN) / m) Next I% Let x(m + 1) = MAX 378 Appendix XII Let x(1) = MIN ReDim y(m + 1) ForI%=1Tom+1 Let y(I%) = f (x(I%)) (function of x goes here) Next I% ReDim b(m + 1) For I% = 2 To m Step 2 Let b(I%) = 4 Next I% For I% = 3 To (m - 1) Step 2 Let b(I%) = 2 Next I% Let b(1) = 1 Let b(m + 1) = 1 ReDim c(m + 1) ForI%=1Tom+1 Let c(I%) = b(I%) * y(I%) Next I% Let Sum = 0 ForI%=1Tom+1 Let Sum = Sum + c(I%) Next I% Let A = (MAX - MIN) * Sum / (3 * m) (Simpson's Rule) Let Label1.Caption = Format(A, "#0.00000000") (output area under function) Appendix XII 379 References Abbot, H. 1993: The Cost of Getting it Wrong. 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Bohnenblust, H. and Slovic, P. 1998: Integrating Technical Analysis and
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