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Tolerance interpretation PowerPoint Presentation

Tolerance interpretation

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Tolerance interpretation

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Dr. Richard A. Wysk

IE550

Fall 2008

- Introduction to tolerance interpretation
- Tolerance stacks
- 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?

Case #1

1

4

2

3

1.0±.05

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

Case #2

1

4

2

3

1.0±.05

1.5±.05

3.5±.05

TOLERANCE STACKING

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

Case #3

1

4

2

3

?

1.0’±.05

1.00’±0.05

3.50±0.05

TOLERANCE STACKING

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

Case #1

1

4

2

3

1.0±.05

1.0±.05

1.0±.05

?

From a Manufacturing Point-of-View

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?

Let’s try the following (in the same setup)

-cut plane 2

-cut plane 3

3

2

Mfg. Process

Will they be of appropriate quality?

- We have taken the worse or best case
- Planning for the worse case can produce some bad results – cost

PROCESS

DIMENSIONAL ACCURACY

POSITIONAL ACCURACY

DRILLING

+ 0.008

- 0.001

0.010

REAMING

+ 0.003

(AS PREVIOUS)

SEMI-FINISH BORING

+ 0.005

0.005

FINISH BORING

+ 0.001

0.0005

COUNTER-BORING (SPOT-FACING)

+ 0.005

0.005

END MILLING

+ 0.005

0.007

- What do we expect when we manufacture something?

2.45

2.5

2.55

- For symmetric distributions, the most likely size, location, etc. is the mean

- Normally capabilities represent + 3 s
- Is this a good planning metric?

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 3of our parts would be good 99.73% of our dimensions are good.

.95

1.0

1.05

2.45

2.5

2.55

D1-2

We know that (as specified)

D2-3 = 1.5 .05

If one uses a single set up, then

(as produced)

and

D1-2

D1-3

D2-3 = D1-3 - D1-2

1.4

1.5

1.6

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

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.

- Mechanical components seldom have 1 feature -- ~ 10 – 100
- Electronic components may have 10,000,000 devices

- 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

1 feature = 0.9973

5 features = 0.986

50 features = 0.8735

100 features = 0.7631

1000 features = 0.0669