Tolerance interpretation

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Tolerance interpretation

• Dr. Richard A. Wysk
• IE550
• Fall 2008

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Agenda

• Introduction to tolerance interpretation
• Tolerance stacks
• Interpretation

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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?

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TOLERANCE STACKING
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
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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
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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
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From a Manufacturing Point-of-View
Lets suppose we have a wooden part and we need
to saw. Lets further assume that we can achieve
? .05 accuracy per cut.
How will the part be produced?
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Mfg. Process
Will they be of appropriate quality?
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So far weve used Min/Max Planning

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

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Expectation

• What do we expect when we manufacture something?

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Size, location and orientation are random
variables

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

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What does the Process tolerance chart represent?

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

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An Example
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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
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What is the probability that D2-3 is bad?
PX1-3- X1-2gt1.55 PX1-3- X1-2lt1.45
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
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The likelihood of a bad part is P X2-3 gt
1.55-1 P X2-3 lt 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.
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What about multiple features?

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

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Suppose we have a part with 5 holes

• Lets assume that we plan for 3 s for each hole
• If we assume that each hole is i.i.d., the
• Pbad part 1.0 Pbad feature5
• .99735
• .9865

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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

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Should this strategy change?