Lecture #26 (tape #26) Prof. John W. Sutherland. Oct. 24, 2001

Transcription

1 Lecture #26 (tape #26) Prof. John W. Sutherland Oct. 24, 2001

2 Process Capability The extent to which a process produces parts that meet design intent. Most often, how well our process meets the engineering specifications. Process capability -- when we quote a number for this we don t want it dependent on time. Rule: Never assess process capability until the process is "in-control"

3 Process Variability & Specifications Upper specification Lower specification Process variation is small relative to the width of the engineering specifications

4 Upper specification Lower specification Upper specification Lower specification

5 Cylinder Boring - Case Study

6 X-double-bar = R-bar = 6.61 σˆ X = R d 2 = 6.61 / = Histogram shows individuals normally distributed Specifications are 199 +/- 4 :

7 Z(lo) = ( )/ = prob =.0407 Z(hi) = ( )/ = prob =.8586 Capability = 4.07% % = 18.2% Process is not capable (want % > 99.73% as a min.) Would centering the process at the nominal value help?? Could calculate probability for this case as well. What action should we take??

8 Specifications & Control Limits Specification Limits Characteristic of the part in question Based on functional considerations Compare to individual part measurements Establish part s conformability to design intent Control Limits Characteristic of the process in question Based on process mean and variability Dependent on sample size, n, and α risk Establish presence/absence or special causes (local faults) in the process

9 Putting Specifications on Control Charts

10 Process Capability Indices ( USL LSL) C p 6σ X = Want 1 Capability Index C pk ( USL µ X ) ( LSL µ X ) Z USL = Z σ LSL = X σ X Z min = min[ Z USL, Z LSL ] Want 1 C pk = Z min 3

11 Example #1 Mean = 130, Sigma-X = 10, Nominal = 145, Specs:

12 Shift mean to 145 Example #

13 Example #3 A: Mean = 145, Sigma-X = 15 B: Mean = 130, Sigma-X =

14 Assembly Tolerances ? We might naively set the assembly tolerance by simply adding tolerances: Concerned that parts 1, 2, & 3 might be selected right at the tolerances - want assembly to be ok

15 Individuals & Assemblies Let s assume specs are 4σ from the mean/nominal Probability of a point at or below -4σ = Probability of simultaneously obtaining 3 such points: ( ) 3 = 2.7 E-14 (1 in 37 trillion!!)

16 Describing the Assembly µ 1 =1.5 σ 1 = µ 2 =1.0 σ 2 = µ 3 =1.25 σ 3 = X A = X 1 + X 2 + X 3 µ A = µ 1 + µ 2 + µ 3 = σ A = σ 1 + σ 2 + σ 3 = σ A =

17 Assembly Distribution µ A =3.75 σ A = Assembly If we again assume that the specs are 4σ A from the mean/nominal, then the tolerance is This differs significantly from that obtained by adding!!!

18 Random assembly Forces at Work Statistical (so far normal) distribution of part dimensions Additive Law of Variances In our example we also assumed that the process was centered at the nominal value. Tolerances at 4σ

19 Another Assembly Example ? 0.3? 0.3? How do we obtain the tolerances on the individual parts? Divide by 3 = 0.003?? Let s use the relations that we have developed to obtain the unknown tolerance. Assume 1=2=3

20 Remember that X A = X 1 + X 2 + X 3 Mean of individual distributions at 0.30 Assembly has tolerance of If tolerances are at 4σ A, then σ A =?? Since σ A = σ 1 + σ 2 + σ 3 = 3σ p, σ p =?? If we again put the specs for the individual parts at 4σ p, this turns out to be

21 ind. parts