tolerance stack-up analysis ppt

Tolerance interpretation

Dr. Richard A. Wysk  ISE316 Fall 2010

Agenda • Introduction to tolerance interpretation • Tolerance stacks • Interpretation

Tolerance interpretation • Frequently a drawing has more than one datum  – How do you interpret features in secondary or  tertiary drawing planes?  – How do you produce these?  – Can a single set-up be used?

TOLERANCE STACKING Case #1 1

4

2

1.0±.05

3

1.5±.05

1.0±.05

?

What is the expected dimension and tolerances? D1-4= D1-2 + D2-3 + D3-4 =1.0 + 1.5 + 1.0 t1-4 = ± (.05+.05+.05) = ± 0.15

TOLERANCE STACKING Case #2 1

4

2

1.0±.05

3

1.5±.05 3.5±.05

What is the expected dimension and tolerances? D3-4= D1-4 - (D1-2 + D2-3 ) = 1.0 t3-4 =

(t1-4 + t1-2 + t2-3 )

t3-4 = ± (.05+.05+.05) = ± 0.15

TOLERANCE STACKING Case #3 1

4

2

1.0’±.05

3

?

1.00’±0.05

3.50±0.05

What is the expected dimension and tolerances? D2-3= D1-4 - (D1-2 + D3-4 ) = 1.5 t2-3 = t1-4 + t1-2 + t3-4 t2-3 = ± (.05+.05+.05) = ± 0.15

From a Manufacturing Point-of-View Case #1

1

4

2

1.0±.05

3

1.0±.05

?

1.0±.05

Let’s suppose we have a wooden part and we need to saw. Let’s further assume that we can achieve .05 accuracy per cut.

How will the part be produced?

Mfg. Process 3 2

Let’s try the following (in the same setup) -cut plane 2 -cut plane 3

Will they be of appropriate quality?

So far we’ve used Min/Max Planning • We have taken the worse or best case • Planning for the worse case can produce some bad results – cost

Expectation • What do we expect when we manufacture something? PROCESS DRILLING REAMING SEMI-FINISH BORING FINISH BORING

DIMENSIONAL  ACCURACY  + 0.008 - 0.001 + 0.003 + 0.005 + 0.001

POSITIONAL  ACCURACY  0.010 (AS PREVIOUS) 0.005 0.0005

COUNTER-BORING (SPOT-FACING)

+ 0.005

0.005

END MILLING

+ 0.005

0.007

Size, location and orientation are random variables • For symmetric distributions, the most likely size, location, etc. is the mean

2.45 2.5

2.55

What does the Process tolerance chart represent? •  Normally capabilities represent + 3 s • Is this a good planning metric?

An Example Let’s suggest that the cutting process produces 2) dimension where (this simplifies things)

( ,

=mean value, set by a location 2=process

variance

Let’s further assume that we set = D1-2 and that =.05/3 or 3 =.05 For plane 2, we would surmise the 3 of our parts would be good 99.73% of our dimensions are good.

We know that (as specified) D2-3 = 1.5

.05

If one uses a single set up, then (as produced)

and .95 1.0

D1-2

1.05 D1-2

2.45 2.5

2.55

D1-3

D2-3 = D1-3 -

D1-2

What is the probability that D2-3 is bad? P{X1-3- X1-2>1.55} + P{X1-3- X1-2<1.45} Sums of i.i.d. N( , ) are normal N(2.5, (.05/3)2) +[(-)N(1.0, (.05/3)2)]= N (1.5, (.10/3)2) So D2-3

1.4

1.5

1.6

The likelihood of a bad part is

P {X2-3 > 1.55}-1 P {X2-3 < 1.45} (1-.933) + (1-.933) = .137

As a homework, calculate the likelihood that

D1-4 will be “out of tolerance” given the same logic.

What about multiple features? • Mechanical components seldom have 1 feature -- ~ 10 – 100 • Electronic components may have 10,000,000 devices

Suppose we have a part with 5 holes • Let’s assume that we plan for + 3 s for each hole • If we assume that each hole is i.i.d., the P{bad part}

= [1.0 – P{bad feature}]5 = .99735 = .9865

Success versus number of features 1 feature = 0.9973 5 features = 0.986 50 features = 0.8735 100 features = 0.7631 1000 features = 0.0669

Should this strategy change?