424, TME 424, MSC 424)

Tolerance Design 1 ME222/424 MSC 424 TME 424 PD Funkenbusch (ME 222/424, TME 424, MSC 424) Tolerance Design 2  Where do tolerances come from?  ...
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Tolerance Design 1

ME222/424 MSC 424 TME 424

PD Funkenbusch (ME 222/424, TME 424, MSC 424)

Tolerance Design 2

 Where do tolerances come from?  

History Absolute vs. Statistical  tolerance as acceptable range

 Summing tolerances  



Worst-case Statistical More complex systems

 Tolerance Design methodology 

Tolerancing based on variance control

PD Funkenbusch (ME 222/424)

Interchangeable parts 3

 Inventor? Depends on your

interpretation. But maybe as early as 1040’s AD (Chinese moveable type).  Eli Whitney helped popularize

http://p2.img.cctvpic.com

the idea for the manufacture of guns.  “Took off” in the 1800’s as a

cornerstone of mass production.  Tolerances make sure parts will

fir together. http://3.bp.blogspot.com PD Funkenbusch (ME 222/424)

Tolerance as acceptable range 4

 Set maximum and minimum values for some

characteristic (e.g. part length).  

LSL = Lower Specification Limit USL = Upper Specification Limit

 For simplicity we will assume that the average value (m)

is midway between the LSL and USL.

 Tolerance = D  

LSL = m - D USL = m + D

 Range of values  m ± D PD Funkenbusch (ME 222/424)

m±D 5

 User (customer) friendly  Immediate sense of the values likely to be encountered.  Quick measure of “quality”  Small D values are immediately impressive  But what does the tolerance actually mean?  Two common methods to specify  Absolute  Statistical PD Funkenbusch (ME 222/424)

Absolute (method one) 6

 Inspect all components  Reject (discard) all those outside of the tolerance

(m ± D)

 All parts (that pass) will be within the tolerance  Absolute limit

PD Funkenbusch (ME 222/424)

Statistical (method two) 7

 Analyze a sample of components. Determine average (m) , standard

deviation (s), and distribution.

 Set tolerance so that only a small, specified fraction of components

will be outside of the range.

 For simplicity here, we will assume that all distributions are normal.  Examples:  D = 1 s  68.3 % in tolerance  average ± one standard deviation  D = 3 s  99.73 % in tolerance  D = 6 s  99.9999998% in tolerance 



D = 4.5s  99.99966% in tolerance

Note: D = 4.5s (evaluated over an extended time period) is sometimes used as the cut-off point for “Six Sigma Quality”  less than 3.4 DPMO (defects per million opportunities)  “rule of thumb” to adjust for drift in product mean

PD Funkenbusch (ME 222/424)

Summing tolerances 8

 Often concerned about how

to sum tolerances.

m2 ± D2

m1 ± D1

 Determine final tolerance

based on component tolerances

 Determine how to adjust

component tolerances to achieve a desired final tolerance 

?±? PD Funkenbusch (ME 222/424)

Add in information on costs to find the most cost effective approach

Example 9

m1 ± D1

m2 ± D2

m3 ± D3

m4± D4

 Fit three components

end to end into a slot machined into a fourth component.  Want to determine the

“gap” remaining. mg ± Dg

 

m2 ± D2

PD Funkenbusch (ME 222/424)

Average length of gap Tolerance on the gap

Gap 10

 Average 

mg ± Dg

 add & subtract based on geometry

 mg = m4 - m1 - m2 - m3  Tolerance 



 always sum Uncertainty increases with each component included

 Different ways of summing  

PD Funkenbusch (ME 222/424)

“Worst case” “Statistical”

“Worst-case” summation 11

 Tolerance  sum D’s  Dg = D1 + D2 + D 3 + D4

mg ± Dg

PD Funkenbusch (ME 222/424)

 With absolute tolerances,

this ensures that all the gaps will be within the tolerance (i.e. mg  Dg )  Can also use with

statistical tolerances. In this case it is possible to have product outside of tolerance  but the probability will be small.

“Statistical” summation 12

 Tolerance  add

D2

‘s

 Most meaningful with

statistical tolerances.

 Dg2 = D12 + D22 + D32 + D42  Assuming all distributions

mg ± Dg

PD Funkenbusch (ME 222/424)

are normal and that the component tolerances are set to the same number of s’s (e.g. D = 3s for all components), then the product tolerance should correspond to that of the components.

Where does this (summing squares) come from? 13

 From statistics, know that uncorrelated variances

can be added:  stotal2 = s12 + s22 + s32 + s42 …

 Multiply thru by a constant squared, n2:  n2stotal2 =n2s12 +n2s22 +n2s32 +n2s42 …  But D = ns is how we defined the statistical tolerance,

so:

 Dproduct2 = D12 + D22 + D32 + D42 … PD Funkenbusch (ME 222/424)

Numerical example 14

m1 ± D1

m2 ± D2

m3 ± D3

m4± D4

 Fit three components

end to end into a slot machined into a fourth component. Comp.

mg ± Dg

m2 ± D2

PD Funkenbusch (ME 222/424)

m (mm) D (mm)

1

10

0.1

2

30

0.3

3

20

0.3

4

61

0.2

Average gap 15

Comp.

m (mm) D (mm)

1

10

0.1

2

30

0.3

3

20

0.3

4

61

0.2 mg ± Dg

PD Funkenbusch (ME 222/424)

 Add & subtract based on

geometry  mg = m4 - m3 - m2 - m1  mg

= 61 – 20 – 30 – 10 = 1 mm

Tolerance on gap 16

Worst case

Statistical

 Dg = D1 + D2 + D3 + D 4

 Dg2 = D12 + D22 + D32 + D42

=0.1 + 0.3 + 0.3 + 0.2 = 0.9 mm  mg  Dg  0.1 to 1.9 mm

= (0.1)2 + (0.3)2 + (0.3)2 + (0.2)2 = 0.23  Dg = 0.5 mm  mg  Dg  0.5 to 1.5 mm

PD Funkenbusch (ME 222/424)

Tolerance type vs. summation method 17

Tolerance type Summation method

Absolute

Statistical

Worst-case (add D’s)

All product within tolerance

Some product out of tolerance (generally small %)

Statistical (add D2 ’s)

Difficult to estimate (“depends”)

Fraction in tolerance related to components’ fraction in tolerance

PD Funkenbusch (ME 222/424)

Complications 18

 Simple approach works well for a linear stacking of components as

shown in the example.

 May not apply to:  components with other “features” shape imperfections, roughness…  more complicated geometries  3-D, rotations…  properties other than length  variability caused by other sources (e.g. environmental conditions)  etc.

 As examples,  variability in the output voltage of an electrical circuit because of differences in component properties (resistances, capacitances, etc.) and geometry/assembly  variability in engine performance due to component wear, ambient temperature, fuel quality, etc.

PD Funkenbusch (ME 222/424)

Approaches to dealing with more complex systems 19

 Modeling   

Need a good mathematical model May be able to solve analytically, depending on complexity Alternatively use the model to “test” different combinations of component /environment values Monte Carlo random sampling based on frequency of occurrence. Generally requires large numbers of samples (1,000s or 10,000s)  must be practical with the model  Various systematic approaches, e.g. Tolerance design 

 Experimental 

Need to be able to identify and monitor or adjust sources of variability Analyze data collected “in the field”  is suitable data available?  Systematic testing, e.g. Tolerance design 

PD Funkenbusch (ME 222/424)

Random sampling (Monte Carlo) example 20 P trabecular

t

cortical

D

 Simple model for axial loading of a long bone. [This

is taken from a “Case Study” in Bartel, Davy, and Keaveny’s Orthopedic Biomechanics textbook. However, it is just an isostrain model for uniaxial compression.]

PD Funkenbusch (ME 222/424)

Random sampling (Monte Carlo) example 21

 The load supported on the cortical portion of the bone,

Pc, is given by:

 Where P is the total applied load, t is the thickness of the

cortical shell, D is the diameter of the trabecular centrum, and Ec and Et are the modulus of the cortical and trabecular bone, respectively.  Because of each of the terms in this equation will vary from person to person, the load on the cortical bone will also vary. PD Funkenbusch (ME 222/424)

22

 The table below gives some information on each of the

parameters. Parameter

Average value

Coefficient of variance

P (N)

1.5e3

20%

Et (Pa)

3.0e8

20%

Ec (Pa)

1.7e10

10%

t(m)

3.5e4

25%

D (m)

3.0e-2

25%

 Average values (except for P) are taken from Bartel et al.

and are nominally for vertebra. The rest of the values are “stand-ins” (more or less made up), just to illustrate. PD Funkenbusch (ME 222/424)

Results (different each time!) 23

Mean Square Error

Average 1.20E+03

1.40E+05 1.20E+05

1.15E+03

1.10E+03

MSE (N*N)

Load (N)

1.00E+05

1.05E+03

8.00E+04 6.00E+04 4.00E+04

1.00E+03

2.00E+04 9.50E+02

0.00E+00 1

10

PD Funkenbusch (ME 222/424)

100 N

1000

10000

1

10

100 N

1000

10000

Compare with a tolerance design (8 TC) 24

MSE  7.09 e4

Average  1.09 e3 1.20E+03

1.40E+05 1.20E+05

1.15E+03

1.10E+03

MSE (N*N)

Load (N)

1.00E+05

1.05E+03

8.00E+04 6.00E+04 4.00E+04

1.00E+03

2.00E+04 9.50E+02

0.00E+00 1

10

PD Funkenbusch (ME 222/424)

100 N

1000

10000

1

10

100 N

1000

10000

Tolerance Design 25 TOLERANCE AS VARIABILITY

PD Funkenbusch (ME 222/424)

Taguchi’s approach to quality 26

 Genichi Taguchi (1924 – 2012)   

quality “guru”, responsible for many innovations developed many tools and methods Tolerance design was Taguchi’s last resort method for improving quality

 Taguchi’s concept of quality 

 

Taguchi equated “quality” with reducing the variance (s2) in the final product Didn’t believe in using fixed “tolerances” (i.e. cutoff values) So Tolerance design focuses on reducing s2 , without considering % in/out of tolerance 

Can be applied to non-normal distributions, but need to be cautious about converting to a “D” and estimating % in tolerance

PD Funkenbusch (ME 222/424)

Tolerance Design  concept 27

 Assume that proportionality between variance in

components and final (product) variance still holds, but with a proportionality constant (sensitivity) added  stotal2 = h1s12 + h2s22 + h3 s32 + h4 s42 …  Experiment  



estimate variance for the product determine contribution of each component variance to the total  decide how to best improve tolerance (i.e. reduce variance) as needed h values show sensitivity of final product variance to tolerance (variance) of each component  think about the units…

PD Funkenbusch (ME 222/424)

Tolerance design experiment 28

 For each component, input

specific values

TC

measured A

B

C

...

response

 

1

-1

-1

-1

Y1

2

-1

-1

+1

Y2

3

-1

+1

-1

Y3

4

-1

+1

+1

Y4

...

+1

-1

-1



match variance of component (“levels” -1 ,+1)  m ± s

 Experiment tests different

combinations of component levels

 Measure the response of the

TC = Treatment condition, one “run” of the experiment A, B, C = different components -1, +1 = represent two different component values to be used in experimentation PD Funkenbusch (ME 222/424)

product 



 variation in these values provides estimate of the total product variance.  also determine contribution of each component to total

Matrix selection 29

 Design of matrix is important  Design Of Experiments (DOE)  Usually 2-level  Can include other (non-component) sources  Matrix size (# of TC)  Between ~ (n+1) and 2n  n = number of components  Much smaller than Monte Carlo style methods  Large matrix provides more/better data (rare)  But smaller sizes are still useful (common) PD Funkenbusch (ME 222/424)

Example (Throttle handle)  From “Designing experiments

for tolerancing assembled products”, Soren Bisgaard. Technometrics (1997), 142152

30

 Components  



 Friction in a throttle handle

of outboard motors  too much or too little  need to improve the tolerance

 Matrix size 



 Tracked friction by measuring

torque to turn the handle

 Three components in the

assembly

PD Funkenbusch (ME 222/424)

Knob, handle, and tube But multiple dimensions on the knob (three), and handle (three) Total of seven dimensions to tolerance



Minimum size  8 Maximum size  128 Chose to use 64 

Relatively conservative/expensive

Throttle handle (experimental details) 31

 Knobs (dimensions A, B, C)  m - s and m + s for each dimension  Eight possible combinations  Manufactured all eight combinations  Handles (dimensions D, E, F)  m - s and m + s for each dimension  Eight possible combinations  Manufactured four of the combinations

 Tube (dimension G)  Manufactured two tubes, one with m - s and one with m + s  Tested all combinations of these components  8 x 4 x 2 = 64 combinations  64 TC  Treatment condition  assembled one combination of components and measured torque PD Funkenbusch (ME 222/424)

Throttle handle (key results) Source

s

%

A

Variance (contrib. to stotal2) 0.0028 29.50

B

0.0023

56.91

8.40

C

0.0024

10.04

1.48

D

0.0037

107.33

15.83

E

0.0030

45.16

6.67

F

0.0043

86.27

12.74

G

0.0040

342.25

50.52

Total

--

677.46 100.00

4.36

32

 A, B, C  knob

dimensions  D, E, F  handle dimensions  G  tube dimension  G is main contributor to

variance (>50%)  

Part tolerances (length)

Variance of torque (force-length)2

PD Funkenbusch (ME 222/424)

Best bet to improve performance But also depends on relative costs

Throttle handle (predicting improvement) Source

s

A

Variance (contrib. to D2) 0.0028 29.50

B

0.0023

56.91

13.52

C

0.0024

10.04

2.38

D

0.0037

107.33

25.49

E

0.0030

45.16

10.73

F

0.0043

86.27

20.49

G

0.0040 0.0020 --

342.25 85.56 677.46 420.77

Total

%

7.01

20.32 99.95

33

 Consider the effect of

halving the tolerance (i.e. s) for G  



Variance of G (s2 ) will be reduced to ¼ Contribution from G to total with, therefore also be reduced to ¼ Total Variance for the throttle torque should be reduced to ~ 421 

PD Funkenbusch (ME 222/424)

677 – ¾ (342) = 421

Predictive equation Source

s

A

Variance (contrib. to D2) 0.0028 29.50

B

0.0023

56.91

13.52

C

0.0024

10.04

2.38

D

0.0037

107.33

25.49

E

0.0030

45.16

10.73

F

0.0043

86.27

20.49

G

0.0040 0.0020 --

342.25 85.56 677.46 420.77

Total

Part tolerances (length)

%

7.01

20.32 99.95

Variance of torque (force-length)2

PD Funkenbusch (ME 222/424)

34

stotal2 = hA sA2 + hBsB2 + hCsC2 + h D s D2 …  hG sG2

= contribution of G to total = SSG = 342.25

 hG

= 342.25/ sG2 = 342.25/(0.0040) 2 = 2.14 x 10 7

 Reduce sG to 0.0020

SSG= hG sG2 =2 .14 x 10 7 x (0.0020) 2 = 85.56

Summary 35

 Definitions of tolerance   

Based on % in/out of tolerance Absolute  all in tolerance Statistical  known % out of tolerance

 Summation of tolerances  

“worst-case”  summation of tolerances “statistical” summation of the squares

 Tolerance design  



Based on reducing the product variance Assumes product variance is proportional to component variances DOE to estimate total product variance, component contributions, and the effects of changing component tolerances

PD Funkenbusch (ME 222/424)